Jacaranda Physics 1 VCE Units 1 & 2 4th ed. [4 ed.] 9780730373155

Table of contents :
HALF TITLE PAGE
TITLE PAGE
COPYRIGHT PAGE
CONTENTS
ABOUT THIS RESOURCE
ACKNOWLEDGEMENTS
UNIT 1 WHAT IDEAS EXPLAIN THE PHYSICAL WORLD?
1 Thermodynamic principles
1.1 Overview
1.1.1 Introduction
1.1.2 What you will learn
1.2 Explaining heat using the kinetic theory
1.2.1 What is heat?
1.2.2 Linking energy and heat: the kinetic theory of matter
1.3 Measuring and converting temperature
1.3.1 Measuring temperature
1.3.2 The Kelvin scale
1.4 Transferring heat
1.4.1 Conduction
1.4.2 Convection
1.4.3 Radiation
1.5 Thermal equilibrium and the laws of thermodynamics
1.5.1 Laws of thermodynamics
1.6 Specific heat capacity
1.6.1 Specific heat capacity
1.6.2 Latent heat
1.6.3 Evaporation
1.7 Review
1.7.1 Summary
1.7.2 Key terms
1.7.3 Practical work and investigations
2 Thermodynamics and climate science
2.1 Overview
2.1.1 Introduction
2.1.2 What you will learn
2.2 Earth's energy systems
2.2.1 Energy from our Sun
2.2.2 Heat from within
2.2.3 Energy in balance
2.2.4 How much energy does the Earth get from the Sun?
2.3 The enhanced greenhouse effect
2.3.1 The natural greenhouse effect
2.3.2 Fossil fuels’ fingerprints
2.3.3 The Earth’s energy budget
2.3.4 Feedback
2.4 Climate models
2.4.1 Scientific principles that underlie climate
2.4.2 How to know if a climate model is accurate?
2.4.3 Investigating a simple climate model
2.5 Investigating issues related to thermodynamics
2.5.1 Issues related to thermodynamics
2.5.2 Evaluating resources
2.6 Review
2.6.1 Summary
2.6.2 Key terms
2.6.3 Practical work and investigations
2.6 Exercises
2.6 Exercise 1: Multiple choice questions
2.6 Exercise 2: Short answer questions
2.6 Exercise 3: Exam practice questions
2.6 Exercise 4: studyON topic test
UNIT 1 | AREA OF STUDY 1 REVIEW
AREA OF STUDY 1
How can thermal effects be explained?
PRACTICE SCHOOL-ASSESSED COURSEWORK
3 Concepts used to model electricity
3.1 Overview
3.1.1 Introduction
3.1.2 What you will learn
3.2 Electric circuits
3.2.1 Fundamentals of electricity
3.2.2 Electric circuits
3.3 Current
3.3.1 Defining current
3.3.2 Describing current direction
3.3.3 Measuring electric current
3.3.4 Modelling an electric circuit
3.3.5 How rapidly do electrons travel through a conductor?
3.4 Voltage
3.4.1 The conventional point of view
3.4.2 Measuring potential difference or voltage drop
3.5 Energy and power in an electric circuit
3.5.1 Energy transformed by a circuit
3.5.2 Power delivered by a circuit
3.5.3 Providing energy for the circuit
3.6 Resistance
3.6.1 Resistors
3.6.2 Ohm's Law
3.6.3 Ohmic and non-ohmic devices
3.6.4 Heating effects of currents
3.6.5 Power and resistance
3.7 Review
3.7.1 Summary
3.7.2 Key terms
3.7.3 Practical work and investigations
4 Circuit analysis
4.1 Overview
4.1.1 Introduction
4.1.2 What you will learn
4.2 BACKGROUND KNOWLEDGE Electric circuit rules
4.2.1 Circuit diagrams
4.2.2 Circuit rules
4.2.3 Conservation of electrical energy
4.3 Series and parallel circuits
4.3.1 Resistors in series
4.3.2 Resistors in parallel
4.3.3 Short circuits
4.3.4 The voltage divider
4.4 Non-ohmic devices in series and parallel
4.4.1 Non-ohmic devices
4.4.2 Transducers and sensors
4.4.3 Potentiometers
4.5 Power in circuits
4.6 Review
4.6.1 Summary
4.6.2 Key terms
4.6.3 Practical work and investigations
4.6 Exercises
4.6 Exercise 1: Multiple choice questions
4.6 Exercise 2: Short answer questions
4.6 Exercise 3: Exam practice questions
4.6 Exercise 4: studyON topic test
UNIT 1 | AREA OF STUDY 2 REVIEW
AREA OF STUDY 2
How do electric circuits work?
PRACTICE SCHOOL-ASSESSED COURSEWORK
5 Using electricity and electrical safety
5.1 Overview
5.1.1 Introduction
5.1.2 What you will learn
5.2 Household electricity and usage
5.2.1 Household use of electricity
5.2.2 Power ratings
5.2.3 Paying for electricity
5.3 Electrical safety
5.3.1 A shocking experience
5.3.2 Resistance of the human body
5.3.3 The effect of current
5.3.4 The effect of current path
5.3.5 Time of exposure
5.3.6 In the event of a shock
5.3.7 Safety in household circuits
5.3.8 Earthing
5.3.9 Residual current device
5.3.10 Double insulation
5.4 Review
5.4.1 Summary
5.4.2 Key terms
5.4.3 Practical work and investigations
5.4 Exercise 1: Multiple choice questions
5.4 Exercise 2: Short answer questions
5.4 Exercise 3: Exam practice questions
6 Origins of atoms
6.1 Overview
6.1.1 Introduction
6.1.2 What you will learn
6.2 Early developments of the Big Bang Theory
6.2.1 Scientific theories
6.2.2 Discovering the universe of galaxies
6.2.3 The expansion of space
6.2.4 Measuring expansion
6.2.5 Einstein and general relativity
6.3 Further developments of the Big Bang Theory
6.3.1 An expanding, cooling universe
6.4 Measurements of the universe
6.4.1 The astronomical unit
6.4.2 The parsec
6.4.3 The light-year
6.4.4 Working in metres
6.4.5 Numbers and time
6.5 The formation of the first atoms
6.5.1 Looking back in time
6.5.2 The early universe
6.5.3 Nucleogenesis (the birth of atoms)
6.6 Review
6.6.1 Summary
6.6.2 Key terms
6.6.3 Practical work and investigations
6.6 Exercises
6.6 Exercise 1: Multiple choice questions
6.6 Exercise 2: Short answer questions
6.6 Exercise 3: Exam practice questions
6.6 Exercise 4: studyON topic test
7 Particles in the nucleus
7.1 Overview
7.1.1 Introduction
7.1.2 What you will learn
7.2 The discovery of subatomic particles
7.2.1 The Standard Model of particle physics
7.2.2 Gauge bosons
7.2.3 Fermions
7.2.4 Matter and antimatter
7.2.5 Bosons and the strong and weak nuclear forces
7.2.7 Prediction of new particles
7.3 Nuclear radiation
7.3.1 Atoms and isotopes
7.3.2 Half-life
7.4 Types of nuclear radiation
7.4.1 Alpha (?) decay
7.4.2 Beta (?) decay
7.4.3 Gamma (?) decay
7.4.4 Decay series
7.5 Review
7.5.1 Summary
7.5.2 Key terms glossaryAn
7.5.3 Practical work and investigations
7.5 Exercises
7.5 Exercise 1: Multiple choice questions
7.5 Exercise 2: Short answer questions
7.5 Exercise 3: Exam practice questions
7.5 Exercise 4: studyON topic test
8 Energy from the atom
8.1 Overview
8.1.1 Introduction
8.1.2 What you will learn
8.2 Energy from mass
8.3 Energy from the nucleus
8.3.1 Binding energy
8.3.2 Nuclear fission
8.3.3 Nuclear fusion
8.4 Energy from accelerating charges
8.4.1 Matter and light
8.4.2 Light from atoms
8.5 Review
8.5.1 Summary
8.5.2 Key terms
8.5.3 Practical work and investigations
8.5 Exercises
8.5 Exercise 1: Multiple choice questions
8.5 Exercise 2: Short answer questions
8.5 Exercise 3: Exam practice questions
8.5 Exercise 4: studyON topic test
UNIT 1 | AREA OF STUDY 3 REVIEW
AREA OF STUDY 3
What is matter and how is it formed?
PRACTICE SCHOOL-ASSESSED COURSEWORK
UNIT 2 WHAT DO EXPERIMENTS REVEAL ABOUT THE PHYSICAL WORLD?
9 Analysing motion
9.1 Overview
9.1.1 Introduction
9.1.2 What you will learn
9.2 Describing movement
9.2.1 Distance and displacement
9.2.2 Speed
9.2.3 Converting units of speed
9.2.4 Velocity
9.2.5 Instantaneous speed and velocity
9.2.6 Acceleration
9.3 Analysing motion graphically
9.3.1 Position-versus-time graphs
9.3.2 Velocity-versus-time graphs
9.3.3 Acceleration-versus-time graphs
9.3.4 Working with motion graphs
9.4 Equations for constant acceleration
9.4.1 Deriving the equations algebraically
9.4.2 Deriving the equations graphically
9.4.3 Problem-solving steps
9.5 Review
9.5.1 Summary
9.5.2 Key terms
9.5.3 Practical work and investigations
9.5 Exercises
9.5 Exercise 1: Multiple choice questions
9.5 Exercise 2: Short answer questions
9.5 Exercise 3: Exam practice questions
9.5 Exercise 4: studyON topic test
10 Forces in action
10.1 Overview
10.1.1 Introduction
10.1.2 What you will learn
10.2 Forces as vectors
10.2.1 Describing a force
10.2.2 Force due to gravity (Fg)
10.2.3 Friction (Ffr)
10.2.4 Forces from fluid motion
10.2.5 The normal force (FN)
10.2.6 Compression and tension materials
10.2.7 Free body diagrams
10.2.8 The net force (Fnet) or sum of forces
10.3 Newton's First Law of Motion
10.4 Newton's Second Law of Motion
10.4.1 Applying Newton's second law in real life
10.4.2 Falling down
10.5 Newton's Third Law of Motion
10.5.1 Forces in pairs
10.5.2 Moving forward
10.6 Forces in two dimensions
10.6.1 Vehicles on horizontal surfaces
10.6.2 Vehicles on inclined planes
10.6.3 Connected objects
10.7 Momentum and impulse
10.7.1 Momentum
10.7.2 Impulse
10.7.3 Impulse from a graph
10.7.4 Follow through
10.7.5 Protecting that frail human body
10.8 Torque
10.8.1 Torque or the turning effect of a force
10.9 Equilibrium
10.9.1 Equilibrium or keeping still
10.9.2 Strategy for solving problems involving torque
10.9.3 Types of structures: cantilevers
10.10 Review
10.10.1 Summary
10.10.2 Key terms
10.10.3 Practical work and investigations
10.10 Exercises
10.10 Exercise 1: Multiple choice questions
10.10 Exercise 2: Short answer questions
10.10 Exercise 3: Exam practice questions
10.10 Exercise 4: studyON topic test
11 Energy and motion
11.1 Overview
11.1.1 Introduction
11.1.2 What you will learn
11.2 Impulse and momentum
11.2.1 Impulse and momentum in collisions
11.2.2 Modelling real collisions
11.3 Work and energy
11.3.1 The concept of energy
11.3.2 Getting down to work
11.3.3 Force-versus-distance graphs
11.4 Energy transfers
11.4.1 Kinetic energy
11.4.2 Potential energy
11.4.3 Gravitational potential energy
11.4.4 Strain potential energy
11.4.5 Conservation of energy
11.5 Efficiency and power
11.5.1 Efficiency
11.5.2 Power
11.6 Review
11.6.1 Summary
11.6.2 Key terms
11.6.3 Practical work and investigations
11.6 Exercises
11.6 Exercise 1: Multiple choice questions
11.6 Exercise 2: Short answer questions
11.6 Exercise 3: Exam practice questions
11.6 Exercise 4: studyON topic test ONLINEonlyicon.pdf
UNIT 2 | AREA OF STUDY 1 REVIEW
AREA OF STUDY 1
How can motion be described and explained?
PRACTICE SCHOOL-ASSESSED COURSEWORK
UNIT 2 | AREA OF STUDY 2 OPTIONS
OBSERVATION OF THE PHYSICAL WORLD
12 What are stars?
12.1 Overview
12.1.1 Introduction
12.1.2 What you will learn
12.2 BACKGROUND KNOWLEDGE Changing views of the universe
12.2.1 Galileo's telescope
12.2.2 The universe
12.2.3 The discovery of galaxies
12.2.4 Where we live
12.2.5 Information from the stars
12.3 Using electromagnetic radiation to collect information
12.3.1 Collecting information about the universe
12.3.2 Electromagnetic radiation and the speed of light
12.3.3 Properties of electromagnetic waves
12.3.4 Frequency of light
12.3.5 Electromagnetic waves
12.4 Spectroscopy
12.4.1 Extracting information from the light
12.4.2 Helium
12.4.3 Composition of the Sun
12.4.4 What produces the spectra?
12.4.5 Absorption spectra
12.4.6 Emission spectra
12.5 Measuring distances to stars and galaxies
12.5.1 Methods used to measure distances
12.5.2 Parallax
12.5.3 The Gaia spacecraft
12.5.4 Star brightness and distance
12.5.5 The red shift method
12.6 The properties of stars
12.6.1 The Sun
12.6.2 The mass of stars
12.6.3 Diameters of stars
12.6.4 Classifying stars
12.7 Nuclear fusion
12.7.1 The energy source of a star
12.7.2 What happens next?
12.8 The Hertzsprung–Russell diagram
12.8.1 The temperature of stars
12.8.2 Spectral type
12.8.3 Hertzsprung–Russell diagrams
12.8.4 Schwartzchild radius
12.8.5 Black holes
12.9 Galaxies
12.9.1 Comparing the Milky Way galaxy to other galaxies
12.10 Review
12.10.1 Summary
12.10.2 Key terms
12.10.3 Practical work and investigations
12.10 Exercises
12.10 Exercise 1: Multiple choice questions
12.10 Exercise 2: Short answer questions
12.10 Exercise 3: Exam practice questions
12.10 Exercise 4: studyON topic test
13 Is there life beyond Earth's solar system?
13.1 Overview
13.1.1 Introduction
13.1.2 What you will learn
13.2 Spectroscopic analysis of light
13.2.1 Electromagnetic radiation
13.2.2 Production of light
13.2.3 Dispersion: producing colour from white light
13.2.4 Spectroscopy
13.2.5 Spectral type
13.3 Locating extrasolar planets (exoplanets)
13.3.1 The search for exoplanets
13.3.2 The astrometric method
13.3.3 The Doppler effect
13.3.4 Measuring radial velocity
13.3.5 The transit method
13.3.6 Microlensing
13.3.7 Trends in the data
13.3.8 Finding the right star
13.4 Conditions for life beyond Earth's solar system
13.4.1 Life based on carbon
13.4.2 The importance of water
13.4.3 The habitable zone
13.4.4 Planet size
13.5 The search for extraterrestrial intelligence
13.5.1 The Fermi paradox
13.5.2 The Drake equation
13.5.3 The search for extraterrestrial intelligence (SETI)
13.5.4 Radio telescopes
13.5.5 Deciding where and how to search
13.5.6 Dealing with the data
13.5.7 Shouting out
13.6 Review
13.6.1 Summary
13.6.2 Key terms
13.6.3 Practical work and investigations
13.6 Exercises
13.6 Exercise 1: Multiple choice questions
13.6 Exercise 2: Short answer questions
13.6 Exercise 3: Exam practice questions
13.6 Exercise 4: studyON topic test
14 How do forces act on the human body?
14.1 Overview
14.1.1 Introduction
14.1.2 What you will learn
14.2 Forces in the human body
14.2.1 Equilibrium — staying in place
14.2.2 Rotational equilibrium
14.2.3 Centre of mass
14.2.4 Centre of gravity
14.2.5 Lever models of human joints
14.2.6 Types of forces
14.3 Stress and strain
14.3.1 Stress
14.3.2 Strain
14.3.3 Stiffness
14.3.4 Young's modulus of elasticity
14.4 Elastic and plastic behaviour of materials
14.4.1 Strain energy
14.4.2 Toughness
14.4.3 Elastic behaviour
14.4.4 Plastic behaviour
14.5 Materials for prostheses
14.5.1 The behaviour of living tissues
14.5.2 Common artificial materials
14.5.3 Metal in prosthetic limbs
14.5.4 The effect of temperature on materials
14.5.5 Limitations of ductile materials
14.6 Performance of artificial limbs
14.6.1 Composite materials
14.6.2 Arm prostheses
14.6.3 Joint replacements
14.6.4 Materials in dentistry and stents
14.7 Review
14.7.1 Summary
14.7.2 Key terms
14.7 Exercises
14.7 Exercise 1: Multiple choice questions
14.7 Exercise 2: Short answer questions
14.7 Exercise 3: Exam practice questions
14.7 Exercise 4: studyON topic test
15 How can AC electricity charge a DC device?
15.1 Overview
15.1.1 Introduction
15.1.2 What you will learn
15.2 Voltage, current and resistance
15.2.1 AC voltage and current
15.2.2 Cathode ray oscilloscope
15.2.3 Using a multimeter
15.3 Converting AC to DC
15.3.1 Diodes
15.3.2 Capacitors
15.3.3 Half-wave rectification
15.3.4 Full-wave rectification
15.3.5 Ripple voltage
15.4 Voltage regulators
15.4.1 Internal resistance
15.4.2 Choosing a suitable voltage regulator
15.4.3 Voltage regulator using a Zener diode
15.4.4 Power dissipation
15.4.5 Controlling voltage in circuits
15.4.6 Basic voltage dividers
15.5 Transducers and diodes
15.5.1 Thermistors
15.5.2 Photonics
15.5.3 Light-emitting diodes (LEDs)
15.5.4 Light-dependent resistors (LDRs)
15.5.5 Photodiodes
15.5.6 Phototransistors
15.6 Data transfer
15.6.1 Communication carriers
15.6.2 Modulation
15.6.3 Decoding signals
15.7 Review
15.7.1 Summary
15.7.2 Key terms
15.7.3 Practical work and investigations
15.7 Exercises
15.7 Exercise 1: Multiple choice questions
15.7 Exercise 2: Short answer questions
15.7 Exercise 3: Exam practice questions
15.7 Exercise 4: studyON topic test
16 How do heavy things fly?
16.1 Overview
16.1.1 Introduction
16.1.2 What you will learn
16.2 Forces acting on an aircraft
16.2.1 Forces in flight
16.2.2 Newton's laws of motion
16.3 Aerodynamic lift
16.3.1 Movement of fluids
16.3.2 Fluid speed and pressure
16.3.3 The Bernoulli principle
16.3.4 Aerofoil characteristics
16.3.5 Newton's Third Law of Motion
16.3.6 Flying upside down
16.4 Thrust and drag
16.4.1 Producing thrust
16.4.2 Types of drag
16.4.3 Lift-to-drag ratio
16.4.4 Faster than sound
16.5 Manipulating flight
16.5.1 Lift and drag coefficients
16.5.2 Turning effect of a force
16.6 Controlling an aircraft
16.6.1 Flight controls
16.6.2 Stages of flight
16.7 Review
16.7.1 Summary
16.7.2 Key terms
16.7.3 Practical work and investigations
16.7 Exercises
16.7 Exercise 1: Multiple choice questions
16.7 Exercise 2: Short answer questions
16.7 Exercise 3: Exam practice questions
16.7 Exercise 4: studyON topic test
17 How do fusion and fission compare as viable nuclear energy power sources?
17.1 Overview
17.1.1 Introduction
17.1.2 What you will learn
17.2 Nuclear fusion
17.2.1 Proton–proton and deuterium–tritium fusion reactions
17.2.2 Binding energy
17.2.3 Fusion reactors
17.2.4 Availability of deuterium and tritium
17.2.5 Fusion reactor designs
17.3 Nuclear fission
17.3.1 Fission fragments
17.3.2 Fission chain reactions
17.3.3 Achieving a chain reaction
17.4 Nuclear bomb design
17.4.1 Fission bombs
17.4.2 The effects of nuclear weapons
17.4.3 Immediate effects
17.4.4 Long-term effects
17.4.5 Fusion bombs
17.5 Nuclear energy as a power source
17.5.1 Nuclear power plants
17.5.2 Generating electricity
17.5.3 Australia's nuclear reactor
17.5.4 Fast breeder reactors
17.5.5 Fukushima Daiichi nuclear disaster
17.5.6 Nuclear wastes
17.5.7 Energy released
17.5.8 Percentage of mass that is transformed into energy
17.5.9 Environmental impact of a fusion reactor
17.5.10 Conclusion
17.6 Review
17.6.1 Summary
17.6.2 Key terms
17.6.3 Practical work and investigations
17.6 Exercises
17.6 Exercise 1: Multiple choice questions
17.6 Exercise 2: Short answer questions
17.6 Exercise 3: Exam practice questions
17.6 Exercise 4: studyON topic test
18 How is radiation used to maintain human health?
18.1 Overview
18.1.1 Introduction
18.1.2 What you will learn
18.2 Radiation and the human body
18.2.1 Electromagnetic radiation and particle radiation
18.2.2 What are X-rays?
18.2.3 Production of X-rays
18.2.4 Hard and soft X-rays
18.2.5 Ionising radiation and living things
18.2.6 The effects of ?, ? and ? radiation on humans
18.3 The use of radiation in diagnosis and treatment
18.3.1 Radioactivity as a diagnostic tool
18.3.2 Choosing the right medical radioisotope
18.3.3 Labelling with isotopes
18.3.4 Targeting body organs
18.3.5 Some applications of radioisotopes
18.3.6 Production of medical radioisotopes
18.3.7 Safety issues
18.4 X-rays in medical diagnosis
18.4.1 Medical imaging using X-rays
18.4.2 Effect of X-radiation on the body
18.4.3 Using conventional X-rays as a diagnostic tool
18.4.4 Imaging parts of the body
18.5 Medical imaging with CT, PET, MRI and ultrasound
18.5.1 CT scanning
18.5.2 Ultrasound
18.5.3 Positron emission tomography (PET)
18.5.4 Magnetic resonance imaging (MRI)
18.5.5 Strengths and limitations of imaging techniques
18.6 Review
18.6.1 Summary
18.6.2 Key terms
18.6 Exercises
18.6 Exercise 1: Multiple choice questions
18.6 Exercise 2: Short answer questions
18.6 Exercise 3: Exam practice questions
18.9 Exercise 4: studyON topic test
19 How do particle accelerators work?
19.1 Overview
19.1.1 Introduction
19.1.2 What you will learn
19.2 Particle accelerators and the production of light
19.2.1 Synchrotron
19.2.2 Design features of the Australian Synchrotron
19.2.3 Technical features of the Australian Synchrotron
19.2.4 Synchrotron radiation and X-rays
19.3 Accelerator technology and the development of modern particle physics
19.3.1 The cyclotron
19.3.2 Early synchrotrons
19.3.3 The development of the Standard Model of particle physics
19.3.4 Features of the Large Hadron Collider
19.4 Current and future applications of accelerator technology for society
19.4.1 Data analysis
19.4.2 Applications of particle accelerators
19.4.3 Current and proposed future of collider technologies
19.5 Review
19.5.1 Summary
19.5.2 Key terms
19.5 Exercises
19.5 Exercise 1: Multiple choice questions
19.5 Exercise 2: Short answer questions
19.5 Exercise 3: Exam practice questions
19.5 Exercise 4: studyON topic test
20 How can human vision be enhanced?
20.1 Overview
20.1.1 Introduction
20.1.2 What you will learn
20.2 Behaviour of light
20.2.1 Light travels in straight lines
20.2.2 Objects and images
20.2.3 Looking in the mirror
20.3 Refraction
20.3.1 The behaviour of light when changing media
20.3.2 Bending of light
20.3.3 Snell's Law
20.3.4 The speed of light in glass
20.3.5 Total internal reflection and critical angle
20.4 Manipulating light for a purpose
20.4.1 Cameras
20.4.2 The thin lens formula and magnification
20.5 Extending vision
20.5.1 Telescopes
20.5.2 Microscopes
20.6 Light and the eye
20.6.1 Human and animal vision
20.6.2 Correcting eye defects
20.6.3 Restoring sight
20.7 Review
20.7.1 Summary
20.7.2 Key terms
20.7.3 Practical work and investigations
20.7 Exercises
20.7 Exercise 1: Multiple choice questions
20.7 Exercise 2: Short answer questions
20.7 Exercise 3: Exam practice questions
20.7 Exercise 4: studyON topic test
21 How do instruments make music?
21.1 Overview
21.1.1 Introduction
21.1.2 What you will learn
21.2 Sound as a wave
21.2.1 Longitude waves
21.2.2 Wave speed
21.2.3 The wave equation
21.2.4 Diffraction
21.2.5 Reflection of waves
21.2.6 Standing waves
21.3 Measuring sound intensity
21.3.1 Intensity
21.3.2 Inverse square law
21.3.3 Sound intensity levels
21.4 How does the human ear respond to sound?
21.4.1 Structure of the human ear
21.4.2 The effect of frequency on hearing
21.4.3 Loudness level curves
21.5 How instruments create sound
21.5.1 Stringed instruments
21.5.2 Wind and brass instruments
21.5.3 Standing waves in pipes open at both ends
21.5.4 Standing waves in pipes closed at one end
21.5.5 The human voice
21.6 Characteristics of sound
21.6.1 Loudness
21.6.2 Pitch
21.6.3 Timbre
21.6.4 Resonance
21.6.5 Beats
21.6.6 Consonance
21.6.7 Psychoacoustics
21.7 Review
21.7.1 Summary
21.7.2 Key terms
21.7.3 Practical work and investigations
21.7 Exercises
21.7 Exercise 1: Multiple choice questions
21.7 Exercise 2: Short answer questions
21.7 Exercise 3: Exam practice questions
21.7 Exercise 4: studyON topic test
22 How can performance in ball sports be improved?
22.1 Overview
22.1.1 Introduction
22.1.2 What you will learn
22.2 Collisions and conservation of momentum
22.2.1 Coefficient of restitution
22.2.2 Making the ball go faster
22.2.3 Billiard ball collisions
22.3 Friction and rolling balls
22.3.1 Friction
22.3.2 Sliding and rolling
22.3.3 Velocity and angular speed
22.3.4 Moment of inertia
22.3.5 Applying rotational kinematics to a billiard ball
22.3.6 Rolling friction
22.4 Maximising flight
22.4.1 Modelling a golf swing
22.4.2 Flight through the air
22.4.3 What goes up must come down
22.4.4 Air resistance and terminal velocity
22.4.5 The swerve of balls in flight
22.4.6 A spinning rough ball
22.5 Review
22.5.1 Summary
22.5.2 Key terms
22.5.3 Practical work and investigations
22.5 Exercises
22.5 Exercise 1: Multiple choice questions
22.5 Exercise 2: Short answer questions
22.5 Exercise 3: Exam practice questions
22.5 Exercise 4: studyON topic test
23 How does the human body use electricity?
23.1 Overview
23.1.1 Introduction
23.1.2 What you will learn
23.2 Electrical signals in the human body
23.2.1 Why do humans conduct electricity?
23.2.2 Charges, electrical potential and human cells
23.2.3 Electrical stimulus and cells
23.3 Electrical signalling and the nervous system
23.3.1 The nervous system
23.3.2 Electrical pulses in the body: the action potential
23.3.3 How electrical pulses move along nerve cells
23.3.4 Sensation and stimulus — getting an action potential started
23.4 The heart as an electrically powered pump
23.4.1 Stimulating the heart
23.4.2 Measuring heart function — the electrocardiogram
23.4.3 Heart beat and the ECG
23.4.4 Heart problems — fibrillation and defibrillation
23.4.5 The artificial pacemaker
23.5 The effects of electricity applied to the body
23.5.1 How does the human body resist current flow?
23.5.2 Current and the human body
23.5.3 Electrosurgery
23.6 Applications of electricity involving the human body
23.6.1 Monitoring electrical activity in the brain
23.6.2 Lie detectors and biotherapy feedback
23.6.3 Neural stimulation: The bionic eye project
23.6.4 Bypassing neural damage: Bionic control devices and activated muscles
23.6.5 Retraining neural networks
23.7 Review
23.7.1 Summary
23.7.2 Key terms
23.7.3 Practical work and investigations
23.7 Exercises
23.7.1 Exercise 1: Multiple choice questions
23.7 Exercise 2: Short answer questions
23.7 Exercise 3: Exam practice questions
23.7 Exercise 4: studyON topic test
24 Practical investigation
24.1 Overview
24.1.1 Introduction
24.1.2 What you will learn
24.2 Key science skills in Physics
24.2.1 The scientific method: why do we conduct investigations?
24.2.2 Using a logbook
24.2.3 Developing aims and questions
24.2.4 Formulating hypotheses and making predictions
24.2.5 Planning and undertaking experiments
24.3 BACKGROUND KNOWLEDGE Variables
24.4 Concepts specific to investigation, key terms and representations
24.4.1 Concepts specific to investigation
24.4.2 Key terms
24.4.3 Physics representations
24.5 Scientific research methodologies and techniques
24.5.1 Characteristics of scientific research methodologies
24.5.2 Techniques of primary qualitative and quantitative data collection
24.5.3 Choosing techniques relevant to an investigation
24.5.4 Precision, accuracy, reliability and validity
24.5.5 Identification of uncertainty
24.6 Health and safety guidelines
24.6.1 Health and safety guidelines
24.6.2 Ethics
24.7 Methods of organising, analysing and evaluating primary data
24.7.1 Organising primary data
24.7.2 Analysing primary data
24.7.3 Evaluating primary data
24.7.4 Sources of uncertainty and error
24.7.5 Limitations of data and methodology
24.8 Models and theories to understand observed phenomena
24.8.1 Models
24.8.2 Theories
24.9 Nature of evidence and key findings of investigations
24.9.1 Nature of evidence: supporting or refuting a hypothesis, model or theory
24.9.2 The key findings of investigations
24.10 Conventions of scientific report writing and scientific poster presentation
24.10.1 Conventions of report writing
24.10.2 Terminology and representations
24.10.3 Symbols
24.10.4 Equations and formulas
24.10.5 Units of measurements
24.10.6 Significant figures
24.10.7 Standard abbreviations
24.10.8 Acknowledgement of references
24.10.9 Presenting a scientific poster
24.10.10 Practical investigation checklist
24.11 Review
24.11.1 Summary
24.11.2 Key terms
24.11 Exercise 1: Multiple choice questions
24.11 Exercise 2: Short answer questions
24.11 Exercise 3: Exam practice questions
GLOSSARY
APPENDIX 1
Periodic table
APPENDIX 2
Astronomical data
Answers
1 Thermodynamic principles
2 Thermodynamics and climate science
3 Concepts used to model electricity
4 Circuit analysis
5 Using electricity and electrical safety
6 Origins of atoms
7 Particles in the nucleus
8 Energy from the atom
9 Analysing motion
10 Forces in action
11 Energy and motion
24 Practical investigations
INDEX

Citation preview

JACARANDA

PHYSICS

VCE UNITS 1 AND 2 | FOURTH EDITION

1

JACARANDA

PHYSICS

VCE UNITS 1 AND 2 | FOURTH EDITION

1

DAN O’KEEFFE GRAEME LOFTS JANE COYLE MICHAEL ROSENBROCK ROSS PHILLIPS PETER NELSON BARBARA MCKINNON PETER PENTLAND GARY BASS DANIELA NARDELLI PAM ROBERTSON JILL TACON JON PEARCE

C O NT R I B U T I N G AU T H O R S

MURRAY ANDERSON MOSES KHOR

This edition published 2020 by John Wiley & Sons Australia, Ltd 42 McDougall Street, Milton, Qld 4064 Typeset in 11/14 pt Times LT Std © Clynton Educational Services Pty Ltd (Graeme Lofts), Dan O’Keefe, Peter Pentland, Ross Phillips, Gary Bass, Daniella Nardelli, Pam Robertson, Jill Tacon and Jon Pearce 2020 The moral rights of the authors have been asserted. ISBN: 978-0-7303-7315-5 Previous editions of this title have been published as follows. Physics 1 VCE Units 1 and 2 and eBookPLUS © Clynton Educational Services Pty Ltd (Graeme Lofts), Dan O’Keefe, Peter Pentland, Ross Phillips, Gary Bass, Daniella Nardelli, Pam Robertson, Jill Tacon and Jon Pearce 2015 Jacaranda Physics 1 Third Edition © Clynton Educational Services Pty Ltd (Graeme Lofts), Dan O’Keefe, Peter Pentland, Ross Phillips, Gary Bass, Daniella Nardelli, Pam Robertson, Jill Tacon and Jon Pearce 2008 Jacaranda Physics 1 Second Edition © Clynton Educational Services Pty Ltd (Graeme Lofts), Dan O’Keefe, Peter Pentland, Ross Phillips, Gary Bass, Daniella Nardelli, Pam Robertson, Jill Tacon and Jon Pearce 2003 Jacaranda Physics 1 © Clynton Educational Services Pty Ltd (Graeme Lofts), Dan O’Keefe, Peter Pentland, Pam Robertson, Barry Hill and Jon Pearce 1996 Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that the educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL). Reproduction and communication for other purposes Except as permitted under the Act (for example, a fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All inquiries should be made to the publisher. Trademarks Jacaranda, the JacPLUS logo, the learnON, assessON and studyON logos, Wiley and the Wiley logo, and any related trade dress are trademarks or registered trademarks of John Wiley & Sons Inc. and/or its affiliates in the United States, Australia and in other countries, and may not be used without written permission. All other trademarks are the property of their respective owners. Front cover image: © Denis Belitsky/Shutterstock; © maxchu / Getty Images Illustrated by various artists, diacriTech and Wiley Composition Services Typeset in India by diacriTech Printed in Singapore by Markono Print Media Pte Ltd All activities have been written with the safety of both teacher and student in mind. Some, however, involve physical activity or the use of equipment or tools. All due care should be taken when performing such activities. Neither the publisher nor the authors can accept responsibility for any injury that may be sustained when completing activities described in this textbook.

10 9 8 7 6 5 4 3 2

CONTENTS About this resource ..................................................................................................................................................................................................

ix

Acknowledgements ..................................................................................................................................................................................................

xi

UNIT 1

WHAT IDEAS EXPLAIN THE PHYSICAL WORLD?

1

AREA OF STUDY 1 HOW CAN THERMAL EFFECTS BE EXPLAINED?

1 Thermodynamic principles 1.1 1.2 1.3 1.4 1.5 1.6 1.7

3

Overview.............................................................................................................................................................................................. Explaining heat using the kinetic theory .................................................................................................................................... Measuring and converting temperature ..................................................................................................................................... Transferring heat ............................................................................................................................................................................... Thermal equilibrium and the laws of thermodynamics .......................................................................................................... Specific heat capacity ..................................................................................................................................................................... Review .................................................................................................................................................................................................

2 Thermodynamics and climate science 2.1 2.2 2.3 2.4 2.5 2.6

3 4 9 13 18 22 30

35

Overview.............................................................................................................................................................................................. Earth’s energy systems ................................................................................................................................................................... The enhanced greenhouse effect ................................................................................................................................................ Climate models ................................................................................................................................................................................. Investigating issues related to thermodynamics ..................................................................................................................... Review .................................................................................................................................................................................................

35 37 45 54 56 58

AREA OF STUDY 1 REVIEW

Practice examination

...................................................................................................................................

64

Practice school-assessed coursework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 AREA OF STUDY 2 HOW DO ELECTRIC CIRCUITS WORK?

3 Concepts used to model electricity 3.1 3.2 3.3 3.4 3.5 3.6 3.7

Overview.............................................................................................................................................................................................. 70 Electric circuits .................................................................................................................................................................................. 71 Current ................................................................................................................................................................................................. 76 Voltage ................................................................................................................................................................................................. 83 Energy and power in an electric circuit ...................................................................................................................................... 86 Resistance .......................................................................................................................................................................................... 91 Review ................................................................................................................................................................................................. 103

4 Circuit analysis 4.1 4.2 4.3 4.4 4.5 4.6

70

112

Overview.............................................................................................................................................................................................. 112 BACKGROUND KNOWLEDGE Electric circuit rules .............................................................................................................. 113 Series and parallel circuits ............................................................................................................................................................. 119 Non-ohmic devices in series and parallel ................................................................................................................................. 130 Power in circuits ................................................................................................................................................................................ 135 Review ................................................................................................................................................................................................. 137

CONTENTS v

5 Using electricity and electrical safety 5.1 5.2 5.3 5.4

145

Overview.............................................................................................................................................................................................. 145 Household electricity and usage .................................................................................................................................................. 146 Electrical safety ................................................................................................................................................................................. 153 Review ................................................................................................................................................................................................. 161

AREA OF STUDY 2 REVIEW

Practice examination

...................................................................................................................................

167

Practice school-assessed coursework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 AREA OF STUDY 3 WHAT IS MATTER AND HOW IS IT FORMED?

6 Origins of atoms 6.1 6.2 6.3 6.4 6.5 6.6

Overview.............................................................................................................................................................................................. 176 Early developments of the Big Bang Theory ............................................................................................................................ 177 Further developments of the Big Bang Theory ........................................................................................................................ 186 Measurements of the universe ..................................................................................................................................................... 189 The formation of the first atoms ................................................................................................................................................... 195 Review ................................................................................................................................................................................................. 203

7 Particles in the nucleus 7.1 7.2 7.3 7.4 7.5

207

Overview.............................................................................................................................................................................................. 207 The discovery of subatomic particles ......................................................................................................................................... 208 Nuclear radiation ............................................................................................................................................................................... 221 Types of nuclear radiation .............................................................................................................................................................. 227 Review ................................................................................................................................................................................................. 235

8 Energy from the atom 8.1 8.2 8.3 8.4 8.5

176

242

Overview.............................................................................................................................................................................................. 242 Energy from mass ............................................................................................................................................................................. 243 Energy from the nucleus ................................................................................................................................................................. 247 Energy from accelerating charges ............................................................................................................................................... 256 Review ................................................................................................................................................................................................. 261

AREA OF STUDY 3 REVIEW

Practice examination

...................................................................................................................................

266

Practice school-assessed coursework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271

UNIT 2

WHAT DO EXPERIMENTS REVEAL ABOUT THE PHYSICAL WORLD?

273

AREA OF STUDY 1 HOW CAN MOTION BE DESCRIBED AND EXPLAINED?

9 Analysing motion 9.1 9.2 9.3 9.4 9.5

Overview.............................................................................................................................................................................................. 275 Describing movement ..................................................................................................................................................................... 276 Analysing motion graphically ........................................................................................................................................................ 287 Equations for constant acceleration ........................................................................................................................................... 299 Review ................................................................................................................................................................................................. 305

10 Forces in action 10.1 10.2 10.3 10.4

275

313

Overview.............................................................................................................................................................................................. 313 Forces as vectors ............................................................................................................................................................................. 314 Newton’s First Law of Motion ....................................................................................................................................................... 325 Newton’s Second Law of Motion ................................................................................................................................................. 327

vi CONTENTS

10.5 Newton’s Third Law of Motion ...................................................................................................................................................... 337 10.6 Forces in two dimensions .............................................................................................................................................................. 342 10.7 Momentum and impulse ................................................................................................................................................................. 351 10.8 Torque .................................................................................................................................................................................................. 360 10.9 Equilibrium .......................................................................................................................................................................................... 362 10.10 Review ................................................................................................................................................................................................. 368

11 Energy and motion 11.1 11.2 11.3 11.4 11.5 11.6

374

Overview.............................................................................................................................................................................................. 374 Impulse and momentum ................................................................................................................................................................. 375 Work and energy ............................................................................................................................................................................... 381 Energy transfers ................................................................................................................................................................................ 385 Efficiency and power ....................................................................................................................................................................... 397 Review ................................................................................................................................................................................................. 402

AREA OF STUDY 1 REVIEW

Practice examination

...................................................................................................................................

408

Practice school-assessed coursework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413 AREA OF STUDY 2 OPTIONS: OBSERVATION OF THE PHYSICAL WORLD

12 What are stars? 13 Is there life beyond Earth’s solar system? 14 How do forces act on the human body? 15 How can AC electricity charge a DC device? 16 How do heavy things fly? 17 How do fusion and fission compare as viable nuclear energy power sources?

18 How is radiation used to maintain human health? 19 How do particle accelerators work? 20 How can human vision be enhanced? 21 How do instruments make music? 22 How can performance in ball sports be improved? 23 How does the human body use electricity?

CONTENTS vii

AREA OF STUDY 3 PRACTICAL INVESTIGATION

24 Practical investigation

430

24.1 Overview.............................................................................................................................................................................................. 430 24.2 Key science skills in Physics ......................................................................................................................................................... 431 24.3 BACKGROUND KNOWLEDGE Variables .................................................................................................................................. 440 24.4 Concepts specific to investigation, key terms and representations .................................................................................. 442 24.5 Scientific research methodologies and techniques................................................................................................................ 446 24.6 Health and safety guidelines ......................................................................................................................................................... 454 24.7 Methods of organising, analysing and evaluating primary data ......................................................................................... 457 24.8 Models and theories to understand observed phenomena ................................................................................................. 469 24.9 Nature of evidence and key findings of investigations .......................................................................................................... 472 24.10 Conventions of scientific report writing and scientific poster presentation .................................................................... 474 24.11 Review ................................................................................................................................................................................................. 485

Appendix 1: Periodic table ..................................................................................................................................................................................... 496 Appendix 2: Astronomical data ............................................................................................................................................................................ 498 Glossary ....................................................................................................................................................................................................................... 499 Answers ........................................................................................................................................................................................................................ 505 Index .............................................................................................................................................................................................................................. 542

viii CONTENTS

ABOUT THIS RESOURCE Jacaranda Physics has been reimagined to provide students and teachers with the most comprehensive resource on the market. This engaging and purposeful suite of resources is fully aligned to the VCE Physics Study Design (2016–2021).

Formats Jacaranda Physics is now available in print and a range of digital formats, including: Print

learnON

eBookPLUS

Printed textbook with free digital access code inside

learnON is our immersive and

The eBookPLUS is an electronic version of the student text

learning platform

eGuidePLUS

PDF

Downloadable The eGuidePLUS includes PDFs available everything from the eBookPLUS with eBookPLUS with additional resources designed for teachers

Fully aligned to the VCE Physics Study Design tailored exercise sets at the end of each sub-topic additional background information easily distinguished from curriculum content • practice SACs clearly linked to each outcome • practice exams for each area of study. • •

FIGURE 3.7 (a) Symbols for circuit components (b) Diagram of the simple electric circuit (a)

(b) Connecting wire

Light globe

What you will learn

+ – Battery

apply concepts of charge (Q), electric current (I), potential difference (V), energy (E) and power (P), in electric circuits explore different analogies used to describe electric current and potential difference

I=

Q

FIGURE 3.4 kite and key electricity experiment

scientist, introduced the concepts of positive and negative electricity.

Switch

conducted the kite and key electricity experiment during a lightning

, V=

t

E

, P=

Q

E

= VI

t

when he was trying to electrocute a turkey with a condenser, a device

justify the use of selected meters (ammeter, voltmeter, multimeter) in circuits. it was acceptable to see how big an animal they could electrocute.)

Source:

a Practical investigation logbook and

3.2 EXERCISE

, are included in this topic to provide opportunities to

To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au.

Resources (doc-31856)

regions called orbits or shells.

(doc-32178)

Key concepts approach Students can easily understand which aspect

(doc-32179)

FIGURE 3.5 (a) The structure of an atom (electron shell) (b) An atom showing orbits and shells To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0028).

(a)

(b) Proton (positive) Electron cloud

3.2 Electric circuits Nucleus

Apply concepts of charge (Q), electric current (I), potential difference (V), energy (E) and power (P), in electric circuits. Explore different analogies used to describe electric current and potential difference.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au.

Proton Neutron

studyON: Practice exam questions

Electron Neutron (neutral)

Key Knowledge dot points highlighted at the

1. After a plastic pen is rubbed with a piece of wool it can be used to attract small pieces of paper. Describe what has happened in terms of electric charge. 2. After rubbing a balloon on your clean dry hair, the balloon should try to stick to your hair when you try to remove it. Explain why this occurs. 3. If you separately rub two balloons on your hair and then hold them near each other what will happen? Explain why this occurs. 4. After walking across a nylon carpet in woollen socks and then touching a metal doorknob it is possible to get an electric shock. Explain why this occurs. 5. she slowly moved the balloon away from the can it started to roll and follow the balloon. Describe why this happened. 6. Imagine you are an electron. Describe your journey around the closed circuit of a torch, beginning at the negative terminal of a cell. 7. A doorbell connected to a battery comprises a button at the door, the bell and wires. The bell only sounds after the button is pushed. Why doesn’t the bell always sound?

Electron (negative)

Fully worked solutions and sample responses are available in your digital formats.

Electric charge (in terms of the basic structure of matter) Electric charge and gas.

TOPIC 3 Concepts used to model electricity 75

electron. TOPIC 3 Concepts used to model electricity 73

Concepts in each sub-topic.

• • while for others it was a force of repulsion. TOPIC 3 Concepts used to model electricity

Engaging resources and rich media learning styles, including: • videos and interactivities embedded at the point of learning • practical investigations complete with printable logbook • a variety of worksheets and activities allow students to apply their knowledge of the content.

Teacher-led videos Videos of both sample problems and practical investigations led by experienced teachers allow students to better consolidate their learning.

ABOUT THIS RESOURCE ix

Inspiring students to become independent learners This resource, on the immersive digital platform learnON, encourages self-driven student learning through: • fully worked solutions for every question providing students with immediate feedback • progress tracked automatically in learnON allowing students to highlight areas of strength and weakness topic summary, key terms and practice questions at the end of every topic • sample problems with fully worked solutions set out in the THINK–WRITE format. •

Prepare for exams with every relevant past exam question since 2006! studyON is now included in learnON and as a printable exam revision booklet

A wealth of teacher resources Jacaranda Physics empowers teachers to teach their class their way with the extensive range of teacher resources including: • quarantined topic tests and SACs that are easily customisable • practical investigation Test Maker support with demonstrative Create custom tests videos, laboratory for your class from the information, expected results entire question pool — and risk assessments including all subtopic, • work programs topic review and past • curriculum grids. VCAA exam questions

Visibility of student results Detailed breakdown of student results allows you to identify strengths and weaknesses across various topics and sub-topics.

x ABOUT THIS RESOURCE

ACKNOWLEDGEMENTS The authors and publisher would like to thank the following copyright holders, organisations and individuals for their assistance and for permission to reproduce copyright material in this book. VCE Physics Study Design content is copyright Victorian Curriculum and Assessment Authority (VCAA), reproduced by permission. VCE® is a registered trademark of the VCAA. The VCAA does not endorse this product and makes no warranties regarding the correctness or accuracy of its content. To the extent permitted by law, the VCAA excludes all liability for any loss or damage suffered or incurred as a result of accessing, using or relying on the content. Current VCE Study Designs and related content can be accessed directly at www.vcaa.vic.edu.au. Teachers are advised to check the VCAA Bulletin for updates.

Images • Alamy Australia Pty Ltd: 73/© Alamy Stock Photo; 253/© blickwinkel; 73/© ClassicStock / Alamy Stock Photo; 184/© INTERFOTO / Alamy; 357/© Javier Larrea; 160/© sciencephotos; 243/© Sueddeutsche Zeitung Photo; 147/© Yevhen Rychko / Alamy Stock Photo; • ANSTO: 257/© ANSTO; • ATLAS Experiment: 425/ATLAS Experiment © 2014 CERN • Bhavesh Chauhan: 450 (top middle)/© Bhavesh Chauhan; • Creative Commons: 358/© Brady Holt; 151/© EECA; 57 (top), 151/© EECA / Commonwealth of Australia 2019; • Creatives Commons: 259/© CK-12 Foundation - Christopher Auyeung; • Dan OKeeffe: 430/© Dan OKeeffe; • European Space Agency Comm Dep: 196/© ESA and the Planck Collaboration; • Getty Images: 229/© Andrew Johnson; 1/© lovelyday12 • Getty Images Australia: 319 (left)/© LEMAIRE Stéphane / hemis.fr; 92/© Science Source; • gmstockstudio: 56/© gmstockstudio; • Image Addict: 79/© imageaddict.com.au; • IPCC: 55/© IPCC; • iStockphoto: 69/© Lee Chin Yong; 273/© Qiang Chen; 435/© Stanislovas Kairys; • Jim Thalassoudis: 260/© Jim Thalassoudis; • NASA: 419/ ALMA ESO/NAOJ/NRAO / NASA / ESA 218/© NASA; • NASA/ public domain: 188 (top), 197/© NASA • Pixabay: 15 (b)/ Ozgur Coskun; 421/stokpic/229 images; 466 (b)/Pixaline • Public Domain: 248, 249/© Atomic Heritage Foundation; 5/© Public Domain; • Science Photo Library: 180 (middle)/© Ann Ronan Picture Library; 179/© Harvard College Observatory; 450 (top left)/© MARTYN F. CHILLMAID; 11 (left)/© Science Photo Library; • Shutterstock: 15 (a)/© Juergen Faelchle; (b)/Ozgur Coskun; 81, 84, 93/© (a) Vladimir A Veljanovski (b) Peter Pentland; 159/ Caron Badkin; 256/ Adrio Communications Ltd; 320/© Alexandr Shevchenko; 420/ design36; 422/Everett Historical; 424/ Semnic; 426/ MJTH; 427/ Halfpoint; 428/Graham Hunt/ProSports; 429, 433/Sergey Nivens; 450 (bottom)/© ALEXEY FILATOV; 317/© Allexxandar; 471/© Amanda Carden; 279/© Andrey Armyagov; 418/© Ververidis Vasilis; 449 (bottom left)/© Andrey Suslov; 156 (right)/© Antonio Guillem; 37/© Aphelleon; 159, 207/© Newscom; 397/© cassiede alain; 222/© CHEN WS; 44/© cherezoff; 434 (bottom)/© chinasong; 374/© conrado; 466 (a)/© Hodyachaya Alla; 449 (top left)/© Constantine Pankin; 282/© CP DC Press; 448 (bottom left)/© Craig Walton; 25/© D. Pimborough; 271/© danpastia; 188 (bottom)/© David Fowler; 451 (bottom)/© desdemona72; 470 (top)/© Designua; 389/© Diego Barbieri; 43/© dikobraziy; 443/© Dr Project; 155/© Egor_Kulinich; 224/© endeavor; 70/© f11photo; 456/© Feng Yu; 489/© Friday24; 414/© FS Stock; 212 (left)/© general-fmv; 326 (bottom)/© Georgios Kollidas; 334/© Germanskydiver; 448 (bottom right)/© Guy J. Sagi; 275/© hfzimages; 492/© ilkercelik; 340/© Ivan Smuk; 145/© Joel_420; 11 (right)/© kaband; 457/© Katsiuba Volha; 449 (bottom right), 465 (bottom)/© KinoMasterskaya; 434 (top)/© Kletr; 466 (top)/© litchima; 259/© magnetix; 356 (top)/© manzrussali; 463/© marekuliasz; 361/© MARGRIT HIRSCH; 318 (top)/© mariolav; 356 (bottom)/© maxpro; 34/© Mihancea Petru; 153/© Mike Focus; 451 (top)/© milanzeremski; 355/© Mitch Gunn; 3/© mkrol0718; 469/© Molekuul_be; 178 (top)/© Molodec; 156 (left)/© MPIX; 14/© Nagy-Bagoly Arpad; 176/© NASA images; 470 (bottom)/© Nasky; 465 (top)/© NatalieIme; 57 (bottom)/© NavinTar; 393/© Neale Cousland; 35/© Nightman1965; 112/© Olga Kashubin; 449 (top right)/© oTTo-supertramp; 6/© photastic; 326 (top)/© Photohunter; 174/© Phovoir; 394/© picturepartners;

ACKNOWLEDGEMENTS xi

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xii ACKNOWLEDGEMENTS

UNIT 1 WHAT IDEAS EXPLAIN THE PHYSICAL WORLD? Ideas in physics are dynamic. As we continue to learn and apply experimental techniques that have been used over the centuries, our knowledge of the physical world improves and theories evolve. In this unit, you will be introduced to the fundamental physics concepts of thermodynamics, electricity and matter, and will explore how physicists use these fundamental ideas and models in an attempt to understand and explain the world we live in, even those things that are beyond what we can see. You will learn how physics examines and explores concepts and theories that dictate the behaviour of the smallest particles to the expanse of the universe, and uses this understanding to create a model of our universe. AREA OF STUDY

OUTCOME

TOPICS

1. How can thermal effects be explained?

Apply thermodynamic principles to analyse, interpret and explain changes in thermal energy in selected contexts, and describe the environmental impact of human activities with reference to thermal effects and climate science concepts.

1. Thermodynamics principles 2. Thermodynamics and climate science

2. How do electric circuits work?

Investigate and apply a basic DC circuit model to simple battery-operated devices and household electrical systems, apply mathematical models to analyse circuits, and describe the safe and effective use of electricity by individuals and the community.

3. Concepts used to model electricity 4. Circuit analysis 5. Using electricity and electrical safety

3. What is matter and how is it formed?

Explain the origins of atoms, the nature of subatomic particles and how energy can be produced by atoms.

6. Origins of atoms 7. Particles in the nucleus 8. Energy from the atom

Source: VCE Physics Study Design (2016–2021) extracts © VCAA; reproduced by permission.

AREA OF STUDY 1 HOW CAN THERMAL EFFECTS BE EXPLAINED?

1

Thermodynamic principles

1.1 Overview Numerous videos and interactivities are available just where you need them, at the point of learning, in your digital formats, learnON and eBookPLUS at www.jacplus.com.au.

1.1.1 Introduction In this topic you will revisit your understanding of energy, utilise your knowledge of the particle theory and apply practical skills of measuring temperature. By the end of this topic you should be able to identify and apply the zeroth and first laws of thermodynamics to describe how the energy within a system is shared and transferred. You will be able to use the particle theory to explain internal energy, temperature and heat transfer. You will describe and explain internal energy as the random motion of particles in a substance. You will test and manipulate the relationships that dictate temperature changes in a substance (specific heat capacity) and changes of state (latent heat) and consider how these concepts are engineered to our benefit. FIGURE 1.1 A cold day in two temperature scales. A balloon caught on a twig shrinks as the translational kinetic energy of the air inside is transferred to the surrounding air, losing the ability to stop the air pressure pushing it in.

TOPIC 1 Thermodynamic principles

3

1.1.2 What you will learn KEY KNOWLEDGE After completing this topic you will be able to: • convert temperature between degrees Celsius and kelvin • describe the Zeroth Law of Thermodynamics as two bodies in contact with each other coming to a thermal equilibrium • describe temperature with reference to the average kinetic energy of the atoms and molecules within a system • investigate and apply theoretically and practically the First Law of Thermodynamics to simple situations: Q = ∆U + W • explain internal energy as the energy associated with random disordered motion of molecules • distinguish between conduction, convection and radiation with reference to heat transfers within and between systems • investigate and analyse theoretically and practically the energy required to: • raise the temperature of a substance: Q = mc∆T • change the state of a substance: Q = mL • explain why cooling results from evaporation using a simple kinetic energy model. Source: VCE Physics Study Design (2016–2021) extracts © VCAA; reproduced by permission.

PRACTICAL WORK AND INVESTIGATIONS Practical work is a central component of learning and assessment. Experiments and investigations, supported by a Practical investigation logbook and Teacher-led videos, are included in this topic to provide opportunities to undertake investigations and communicate findings.

Resources Digital documents Key science skills–VCE Units 1–4 (doc-31856) Key terms glossary (doc-31857) Practical investigation logbook (doc-31853)

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0026).

1.2 Explaining heat using the kinetic theory KEY CONCEPTS • Explain internal energy as the energy associated with random disordered motion of molecules. • Describe temperature with reference to the average kinetic energy of the atoms and molecules within a system.

1.2.1 What is heat? It may surprise you to consider that heat was not considered a form of energy until quite late in human history. Fire was one of the four elements in ancient times and it was clear that when substances combusted, heat was released. Thus, for centuries heat was considered a substance that could be transferred between objects and this was the starting point for scientific thought. In 1667 Johann Joachim Becher proposed that a substance he called phlogiston was released during combustion. Later, in 1789, the French scientist Antoine Lavoisier published a treatise on chemistry in which he described heat as an invisible, tasteless, 4 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

odourless, weightless fluid that he called calorific fluid, which flowed from hot to cold objects. Although now defunct as an explanation of heat, the measure of the calorie is still used. In 1798 Benjamin Thompson, later to be called Count Rumford, conducted an experiment on the nature of heat. The barrel of a cannon is made by drilling a cylindrical hole in a solid piece of metal. Rumford observed the piece of metal and the drill became quite hot. He devised an experiment to investigate the source of the heat and how much heat is produced. Rumford put the drill and the end of the cannon in a wooden box filled with water. He measured the mass of water and the rate at which the temperature rose. He showed that the amount of heat produced was not related to the amount of metal that was drilled out. He concluded that the amount of heat produced depended only on the work done against friction. He said that heat was in fact a form of energy, not an invisible substance that is transferred from hot objects to cold objects. Instead a hot object had heat energy, in the same way as a moving object has kinetic energy or an object high off the ground has gravitational potential energy.

AS A MATTER OF FACT Count Rumford was born Benjamin Thompson in FIGURE 1.2 Count Rumford Massachusetts in 1753. By the age of 16 he was conducting experiments on heat. By 1775, when the American War of Independence began, he was already a wealthy man and of some standing in his community. He joined the British side of the war, becoming a senior advisor. While with the army, he also investigated and published a paper on the force of gunpowder. At the end of the war, he moved to England, where he was known as a research scientist. A few years later he moved to Bavaria, in what is now southern Germany, and spent 11 years there. He moved in royal circles and eventually became Bavaria’s Army Minister, tasked with reorganising the army. As part of those duties he investigated methods of cooking, heating and lighting. He developed a soup, now called Rumford’s soup, as a nutritious ration for soldiers. He also used the soup to establish soup kitchens for the poor throughout Bavaria. For his services he was made a Count of the Holy Roman Empire, taking the name ‘Rumford’ from his birth place. On return to England, his activities included: (i) redesigning an industrial furnace, which revolutionised the production of quicklime, a component of cement and is used for lighting (‘limelight’) (ii) edesigning the domestic fireplace to narrow the chimney at the hearth to increase the updraught, resulting in greater efficiency and no smoke coming back into the room (iii) inventing thermal underwear, a kitchen range and a drip coffee pot. With Joseph Banks and others, Rumford established the Royal Institution (RI) in London as a scientific research establishment with a strong emphasis on public education. Initial funding came from the ‘Society for Bettering the Conditions and Improving the Comforts of the Poor’, with which Count Rumford was centrally involved. Famous member scientists in its early years included Humphrey Davy and Michael Faraday. Fifteen Nobel Prize winners have worked at the RI and 10 chemical elements were discovered there.

Rumford’s ideas about heat were not adopted for a few decades. But in 1840 James Prescott Joule conducted a series of experiments to find a quantitative link between mechanical energy and heat. In other words, how much energy is required to increase the temperature of a mass by 1 °C? Joule used different methods and compared the results. • Using gravity. A falling mass spins a paddle wheel in an insulated barrel of water, raising the temperature of the water. • Using electricity. Mechanical work is done turning a dynamo to produce an electric current in a wire, which heats the water.

TOPIC 1 Thermodynamic principles 5

• • •

Compressing a gas. Mechanical work is used to compress a gas, which raises the gas’s temperature. Using a battery. Chemical reactions at the battery terminals produce a current, which heats the water. Using gravity. Measure the temperature of water at the top and bottom of a waterfall. Joule obtained approximately identical answers for all methods. This confirmed heat as a form of energy. To honour his achievement, the SI unit of energy is the joule (J). The unit joule is used to measure the: • kinetic energy of a runner • light energy in a beam • chemical energy stored in a battery • electrical energy in a circuit • potential energy in a lift on the top floor • heat energy when water boils. One joule is approximately the amount of energy needed to lift a 100 gram apple through a height of 1 metre. The usual metric prefixes make the use of the unit joule more convenient. For example: 1 kJ (kilojoule) = 103 J

1 MJ (megajoule) = 106 J

1 GJ (gigajoule) = 109 J

The chemical energy available from a bowl of breakfast cereal is usually hundreds of thousands of joules and is more likely to be listed on the packet in kilojoules. The amount of energy needed to boil an average kettle full of cold water is about 500 kJ. Examples of 1 joule include: • kinetic energy of a tennis ball moving at about 6 m s−1 • heat energy needed to raise the temperature of 1 gram of dry air by 1 °C • heat energy needed to raise the temperature of 1 gram of water by 0.24 °C • change in gravitational potential energy when an apple falls 1 metre to the ground • amount of sunlight hitting a square centimetre every 10 seconds when the Sun is directly above • amount of sound energy entering your eardrum at a loud concert over 3 hours • amount of electrical energy used by a plasma TV screen while on standby every 2.5 seconds • energy released by the combustion of 18 micrograms of methane.

1.2.2 Linking energy and heat: the kinetic theory of matter Joule showed that a measure of energy was directly related to a change in temperature. But what happens to this energy within a material? The kinetic theory of matter enables us to explain the effect of energy on a material. The kinetic theory of matter, which considers all objects as assemblies of particles in motion, is an old one, first described by Lucretius in 55 AD. The kinetic view of matter was developed over time by Hooke, Bernoulli, Boltzmann and Maxwell. The evidence for the existence of particles includes the following. • Gases and liquids diffuse; that is, a combination of two gases or two liquids quickly becomes a mixture, for example, a dye spreading in water. Even solids can diffuse; if a sheet of lead is clamped to a sheet of gold, over time the metals merge to a depth of a few millimetres. • The mixing of two liquids gives a final volume that is less than the sum of their original volumes. • A solid dissolves in a liquid.

6 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 1.3 Iodine crystals sublimate (turn directly into a gas) when heated.

This diagram shows a gas jar with iodine crystals.

As the crystals warm up, they produce a purple gas that diffuses throughout the jar.

After a long period of time, the crystals have completely sublimated.

The kinetic theory of matter assumes that: • all matter is made up of particles in constant, random and rapid motion • there is space between the particles. The energy associated with the motion of the particles in an object is called the internal energy of the object. The particles can move and interact in many ways, so there are a number of contributions to the internal energy.

Gases In a gas made up of single atoms, such as helium, the atoms move around, randomly colliding with each other and the walls of the container. So, each atom has some translational kinetic energy (see figures 1.4 and 1.5). However, if the gas is made up of molecules with two or more atoms, the molecules can also stretch, contract and spin, so these molecules also have other types of kinetic energy called vibrational kinetic energy and rotational kinetic energy (see figure 1.5).

FIGURE 1.4 Moving single atoms have translational kinetic energy.

Liquids Like a gas, molecules in a liquid are free to move, but within the confines of the surface of the liquid. There is some attraction between molecules, which means there is some energy stored as molecules approach each other. Stored energy is called potential energy. It is the energy that must be overcome for a liquid to evaporate or boil.

Solids In a solid, atoms jiggle rather than move around. They have kinetic energy, but they also have a lot of potential energy stored in the strong attractive force that holds the atoms together. This means that a lot of energy is required to melt a solid. This is seen in Figure 1.6.

TOPIC 1 Thermodynamic principles 7

FIGURE 1.5 The movements of a molecule Translational kinetic energy

Vibrational kinetic energy

Rotational kinetic energy

Spinning Bending

Moving Stretching

FIGURE 1.6 In a solid, atoms can jiggle around an essentially fixed location.

When the internal energy of a substance increases, this means the particle movement increases. Heating or cooling is reflected in a change in the particle movement. Heat energy will always transfer from a hotter substance to a colder substance as the collisions of the faster moving particles transfer translational kinetic energy to the slower moving particles. 8 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

TABLE 1.1 The internal energy of different objects Internal energy Movement that is not related to temperature

Movement that is related to temperature

Atoms in a gas

None

Moving and colliding

Molecules in a gas

Spinning, stretching, compressing and bending

Moving and colliding

Molecules in a liquid

Spinning, stretching, compressing and bending

Moving and colliding

Atoms in a solid

Pulling and pushing

Jiggling

Energy types

Other types of kinetic energy, potential energy

Translational kinetic energy

1.2 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. Use the particle theory to explain why you can smell what’s for dinner from the front door of your house. Explain the difference between translational kinetic energy and other types of internal energy. Use the particle theory to explain what happens when your cup of tea cools down over time. Cling film on a warm bowl of soup placed in the fridge gets sucked down when it cools. Use the particle theory to explain this. 5. James Joule showed that mechanical energy could be transformed into the internal energy of a substance or object. The temperature of a nail, for example, can be raised by hitting it with a hammer. List as many examples as you can of the use of mechanical energy to increase the temperature of a substance or object. 6. Explain in terms of the kinetic particle model why a red-hot pin dropped into a cup of water has less effect on the water’s temperature than a red-hot nail dropped into the same cup of water. 7. Explain why energy is transferred from your body into the cold sea while swimming even though you have less internal energy than the surrounding cold water. 1. 2. 3. 4.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

1.3 Measuring and converting temperature KEY CONCEPT • Convert temperature between degrees Celsius and kelvin.

We have learnt that heat is described as energy and that energy can be measured using the joule. We also know that heating a substance causes the particles within that substance to move faster. This movement is erratic and impossible to measure individually. Instead we consider the average increase in translational kinetic energy.

1.3.1 Measuring temperature Temperature is a measure of the average translational kinetic energy of particles. The other contributions to the internal energy do not affect the temperature. This becomes important when materials melt or boil because the added heat must go somewhere, but the temperature does not change.

TOPIC 1 Thermodynamic principles 9

Measuring temperature is a relatively simple process. Our bodies tell us when it is hot or cold. Our fingers warn us when we touch a hot object. However, for all that, our senses are not reliable. Try the following at home. Place three bowls of water in front of you. Put iced water in the bowl on the left, water hot enough for a bath in the bowl on the right and room temperature water in the one in the middle. Place each hand in the two outer bowls, leave them there for a few minutes, then place both hands in the middle bowl. As you would expect, your left hand tells you the water is warmer, while your right hand tells you it is colder. Thermometers were designed as a way to measure temperature accurately. A good thermometer needs a material that changes in a measurable way as its temperature changes. Many materials, including water, expand when heated, so the first thermometer, built in 1630, used water in a narrow tube with a filled bulb at the bottom. The water rose up the tube as the bulb was warmed. German physicist Daniel Fahrenheit replaced the water with mercury in 1724. Liquid thermometers now use alcohol with a dye added. Fahrenheit developed a scale to measure the temperature, using the lowest temperature he could reach, an ice and salt mixture, as zero degrees and the temperature of the human body as 100 degrees. Fahrenheit also showed that a particular liquid will always boil at the same temperature. Swedish astronomer Anders Celsius developed another temperature scale in 1742, which is the one we use today. Celsius used melting ice and steam from boiling water to define 0 °C and 100 °C for his scale. A third temperature scale was proposed in 1848 by William Thomson, later to be ennobled as Lord Kelvin. He proposed the scale based on the better understanding of heat and temperature that had developed by that time (see page 9). This scale uses the symbol ‘K’ to stand for ‘kelvin’. Other materials, including gases and metals, also expand with temperature and are used as thermometers. A bimetallic strip is two lengths of different metals, usually steel and copper, joined together. The two metals expand at different rates, so the strip bends one way as the temperature rises, or the other as it cools. A bimetallic strip can be used as a thermometer, a thermostat or as a compensating mechanism in clocks.

THE LOWEST TEMPERATURE In 2015, the temperature of a cloud of 100 000 rubidium atoms was reduced to 50 × 10−12 kelvin (above absolute zero). The average speed of the atoms was less than 70 micrometres per second.

Properties that change with temperature and can be employed in designing a thermometer are: • electrical resistance of metals, which increases with temperature • electrical voltage from a thermocouple, which is two lengths of different metals with their ends joined; if one end is heated, a voltage is produced (see figure 1.7) • colour change; liquid crystals change colour with temperature (see figure 1.8) • colour emitted by a hot object; in steel making, the temperature of hot steel is measured by its colour (see figure 1.9).

FIGURE 1.7 This thermocouple is connected to a voltmeter, which reads differing voltages as the thermocouple changes temperature. Iron wire Joint of two different metals

Copper wire

Heat source

10 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 1.8 This liquid crystal thermometer indicates body temperature when the liquid crystals change colour. The thermometer is registering 37 °C.

FIGURE 1.9 This steel is nearly 1000 °C and has recently been poured into a mould to shape it. The steel will continue to glow until it has cooled to about 400 °C.

1.3.2 The Kelvin scale The kinetic theory of matter is the origin of the Kelvin temperature scale. If temperature depends on the movement of particles, then the slower they move, the lower the temperature. When the particles stop moving, the temperature will be the lowest that is physically possible. This temperature was adopted as absolute zero. But how do we measure it and what is its value? In the early 1800s gases were a good material to work with to explore the nature of matter. An amount of gas in a glass vessel could be heated and the variables of temperature, volume and pressure to keep the volume fixed could be easily measured. Joseph Gay-Lussac and Jacques Charles independently investigated how the volume of gases changed with temperature if they were kept under a constant pressure. They found that all gases kept at constant pressure expand or contract by 1/273 of their volume at 0 °C for each Celsius degree rise or fall in temperature. From that result you can conclude that if you cooled the gas, and it stayed as a gas and did not liquefy, you could cool it to a low enough temperature that its volume reduced to zero. The temperature would be absolute zero. According to their experiments, absolute zero was −273 °C. Nowadays more accurate experiments put the value at −273.15 °C. In kelvin, absolute zero is 0 K. The increments in the Kelvin temperature scale are the same size as those in the Celsius scale, so if the temperature increased by 5 °C, it also increased by 5 K. The conversion formula between the two temperature scales is: kelvin = Celsius + 273 T (kelvin) = T (Celsius) + 273

TOPIC 1 Thermodynamic principles 11

Volume

FIGURE 1.10 By extrapolating his trend line back, Kelvin was able to establish absolute zero at –273 °C. With modern equipment absolute zero is now determined to be –273.15 °C.

ion

olat

rap

Ext

Celsius scale (°C) −273

−200

0

−100

100

200

300 Temperature

Kelvin scale (K)

0

73

173

273

373

473

573

TABLE 1.2 Some temperatures on the Kelvin and Celsius scales Event

Temperature K

°C

Absolute zero

0

−273

Helium gas liquefies

4

−269

Lead becomes a superconductor

7

−266

Nitrogen gas liquefies

63

−210

Lowest recorded air temperature on the Earth’s surface (Vostok, Antarctica)

184

−89

Mercury freezes

234

−39

Water freezes

273

0

Normal human body temperature

310

37

Highest recorded air temperature on the Earth’s surface (Death Valley, USA)

330

57

Mercury boils

630

357

Iron melts

1535

1262

Surface of the Sun

5778

5505

Note: In 1968, the international General Conference on Weight and Measures decided that kelvin temperatures do not use the ° symbol, unlike Celsius and Fahrenheit temperatures. SAMPLE PROBLEM 1

What is the kelvin temperature at which ice melts? Teacher-led video: SP1 (tlvd-0003) THINK

WRITE

Recall the relationship between the two temperature scales. 2. Ice melts at 0 °C. Substitute 0 °C into the conversion formula.

T(kelvin) = T(Celsius) + 273

1.

3.

State the solution.

12 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

T(kelvin) = T(Celsius) + 273 = 0 + 273 = 273 K Ice melts at 273 K.

PRACTICE PROBLEM 1 The temperature of the universe predicted by the cosmic microwave background is 3 K. What is this in degrees Celsius?

1.3 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Why is the Celsius scale of temperature commonly used rather than the Kelvin scale? 2. What is the main advantage of an absolute scale of temperature? 3. Estimate each of the following temperatures in kelvin: (a) the maximum temperature in Melbourne on a hot summer’s day (b) the minimum temperature in Melbourne on a cold, frosty winter’s morning (c) the current room temperature (d) the temperature of cold tap water (e) the boiling point of water. 4. The temperature of very cold water in a small test tube is measured with a large mercury-in-glass thermometer. The temperature measured is unexpectedly high. Suggest a reason for this. 5. Carbon dioxide sublimates, that is, goes directly from solid to gas, at −78.5 °C. What is this temperature in kelvin? 6. The temperature of the surface of Mars was measured by the Viking lander and ranged from 256 K to 166 K. What are the equivalent temperatures in degrees Celsius?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

1.4 Transferring heat KEY CONCEPTS • Distinguish between conduction, convection and radiation with reference to heat transfers within and between systems. • Identify regions of the electromagnetic spectrum as radio, microwave, infra-red, visible, ultraviolet, X-ray and gamma waves.

From our understanding of the particle theory and its relationship to heat we know that energy is always transferred from a region of higher temperature to a region of lower temperature. There are many situations in which it is necessary to control the rate at which the energy is transferred. • Warm-blooded animals, including humans, need to maintain their body temperature in hot and cold conditions. Cooling of the body must be reduced in cold conditions. In hot conditions, it is important that cooling takes place to avoid an increase in body temperature. • Keeping your home warm in winter and cool in summer can be costly, both in terms of energy resources and money. Applying knowledge of how heat is transferred from one place to another can help you find ways to reduce how much your house cools in winter and heats up in summer, thus reducing your energy bills.

TOPIC 1 Thermodynamic principles 13

•

The storage of many foods in cold temperatures is necessary to keep them from spoiling. In warm climates most beverages are enjoyed more if they are cold. The transfer of heat from the warmer surroundings needs to be kept to a minimum. FIGURE 1.11 Reducing heat transfer to and from buildings saves precious energy resources, and reduces gas and electricity bills.

There are three different processes through which energy can be transferred during heating and cooling: conduction, convection and radiation (see figure 1.12). FIGURE 1.12 Heat is transferred by conduction, convection and radiation. Convection

Convection

Condu

ction

Conduction

Radiation

1.4.1 Conduction Conduction is the transfer of heat through a substance as a result of collisions between neighbouring vibrating particles. The particles in the higher temperature region have more random kinetic energy than those in the lower temperature region. As shown in figure 1.13, the more energetic particles collide with the less energetic particles, giving up some of their kinetic energy. This transfer of kinetic energy from particle to particle continues until thermal equilibrium is reached. There is no net movement of particles during the process of conduction. Solids are better conductors of heat than liquids and gases. In solids, the particles are more tightly bound and closer together than in liquids and gases. Thus, kinetic energy can be transferred more quickly. Metals are the best conductors of heat because free electrons are able to transfer kinetic energy more readily to other electrons and atoms. 14 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 1.13 Conduction is the transfer of thermal energy (heat) due to collisions between neighbouring particles. Direction of heat transfer

High temperature

Low temperature

Materials that are poor conductors are called insulators. Materials such as polystyrene foam, wool and fibreglass batts are effective insulators because they contain pockets of still air. Air is a very poor conductor of heat. If air is free to move, however, heat can be transferred by a different method — convection. FIGURE 1.14 (a) Insulators like rock mineral wool have excellent thermal and acoustic properties. (b) Fibreglass insulation is made from recycled glass bottles, sand and other materials. (a)

(b)

1.4.2 Convection Convection is the transfer of heat through a substance as a result of the movement of particles between regions of different temperatures. Convection takes place in liquids and gases where particles are free to move around. In solids, the particles vibrate about a fixed position and convection does not generally occur, except under specific conditions (e.g. in the extreme temperature and pressure in the Earth’s mantle). The movement of particles during convection is called a convection current. Faster moving particles in hot regions rise while slower moving particles in cool regions fall. The particles in the warm water near the flame in figure 1.15 are moving faster and are further apart than those in the cooler water further from the flame. The cooler, denser water sinks, forcing the warm, less dense water upwards. This process continues as the warm water rises, gradually cools and eventually sinks again, replacing newly heated water. FIGURE 1.15 Purple particles from a crystal of potassium permanganate carefully placed at the bottom of the beaker are forced around the beaker by convection currents in the heated water. Beaker

Small potassium permanganate crystal

Convection current

Water Gauze mat Tripod Bunsen burner

Bunsen burner Heat-proof mat

TOPIC 1 Thermodynamic principles 15

Convection currents are apparent in ovens that do not have fans. As the air circulates, the whole oven becomes hot. However, the top part of the oven always contains the hottest, least dense air. As the air cools, it sinks and is replaced by less dense hot air for as long as the energy source at the bottom of the oven remains on. Fans can be used to push air around the oven, providing a more even temperature. Home-heating systems use convection to move warm air around. Ducted heating vents are, where possible, located in the floor. Without the aid of powerful fans, the warm air rises, circulates around the room until it cools and sinks, being replaced with more warm air. In homes built on concrete slabs, ducted heating vents are in the ceiling. Fans are necessary to push the warm air downwards so that it can circulate more efficiently. In summer, loose fitting clothing is more comfortable because it allows air to circulate. Thus, heat can be transferred from your body by convection as the warm air near your skin rises and escapes upwards.

FIGURE 1.16 Convection currents circulate warm air pushed out by home heating systems. The warm air rises, circulates around the room until it cools and sinks, being replaced with more warm air.

Heater

1.4.3 Radiation Heat can be transferred without the presence of particles by the process of radiation. All objects with a temperature above absolute zero (0 K) emit small amounts of electromagnetic radiation. Visible light, microwaves, infrared radiation, ultraviolet radiation and X-rays are all examples of electromagnetic radiation. All electromagnetic radiation is transmitted through empty space at a speed of 3.0 × 108 m s–1 , which is most commonly known as the speed of light. Electromagnetic radiation can be absorbed by, reflected from or transmitted through substances. Scientists have used a wave model to explain much of the behaviour of electromagnetic waves. These electromagnetic waves transfer energy, and reflect and refract in ways that are similar to waves on water. What distinguishes the different types of electromagnetic radiation from each other is: • their wavelength (the distance the wave takes to repeat itself) • their frequency (the number of wavelengths passing every second) • the amount of energy they transfer. 16 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 1.17 Radiant heat can be transmitted, absorbed or reflected Transmitted heat

Absorbed heat

Radiated heat

Reflected heat

Transmitted radiant heat Clear objects, like glass, allow light and radiant heat to pass through them. The temperature of these objects does not increase quickly when heat reaches them by radiation.

Absorbed radiant heat Dark-coloured objects tend to absorb light and radiant heat. Their temperatures increase quickly when heat reaches them by radiation.

Reflected radiant heat Shiny or light-coloured surfaces tend to reflect light and radiant heat away. The temperature of these objects does not change quickly when heat reaches them by radiation.

These properties in turn determine their ability to be transmitted through transparent or opaque objects, their heating effect and their effect on living tissue. Figure 1.18 shows the electromagnetic spectrum and demonstrates that higher energy radiation corresponds to low wavelength. FIGURE 1.18 The electromagnetic spectrum. All objects emit some electromagnetic radiation. Increasing frequency Increasing energy

TV Shortand wave CB FM radio radio radio

AM radio 104

102

100

Microwaves and radar 10−2

Visible light

Radio waves Infra-red radiation

10−4

10−6

Ultraviolet radiation

10−8

X-rays

10−10

Gamma rays

10−12

Wavelength (metres) Increasing wavelength

Decreasing wavelength

Why do hot objects emit electromagnetic radiation? All matter is made up of atoms. At any temperature above absolute zero, these atoms are moving and colliding with each other. The atoms contain positive and negative charges. The motion of the atoms and their collisions with other atoms affect the motion of the electrons. Because they are charged and moving around, the electrons produce electromagnetic radiation. Electrons moving in an antenna produce a radio signal, but in a hot object the motion is more random with a range of speeds. So, a hot object produces radiation across a broad range of wavelengths. If its temperature increases, the atoms move faster and have more frequent and more energetic collisions. These produce more intense radiation with higher frequencies and shorter wavelengths.

FIGURE 1.19 Electromagnetic radiation from a hot body

1.4 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. Explain with the aid of a well-labelled diagram how heat is transferred through a substance by conduction. Why are liquids and gases generally poorer conductors of heat than solids? Explain how convection occurs in a liquid that is being heated from below. Why is it not possible for heat to be transferred through solids by convection? At what speed does radiant energy move through space? What is significant about this speed? When you swim in a still body of water on a hot afternoon there is a noticeable temperature difference between the water at the surface and the deeper water. (a) Explain why this difference occurs. (b) If the water is rough, the difference is less noticeable. Why? 7. Standing near the concrete wall of a city building after a hot day you can instantly feel its warmth from a few metres away. (a) How is the energy transferred to you? (b) What caused the building to get hot during the day? 8. Why is it not practical to drink hot coffee in an aluminium picnic cup? 9. Why do ducts in the ceiling need more powerful fans than those in the floor? 1. 2. 3. 4. 5. 6.

TOPIC 1 Thermodynamic principles 17

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

1.5 Thermal equilibrium and the laws of thermodynamics KEY CONCEPTS • Describe the Zeroth Law of Thermodynamics as two bodies in contact with each other coming to a thermal equilibrium. • Investigate and apply theoretically and practically the First Law of Thermodynamics to simple situations: Q = ∆U + W. • Explain why cooling results from evaporation using a simple kinetic energy model.

Energy is always transferred from a region of high temperature to a region of lower temperature until both regions reach the same temperature. When the temperature is uniform, a state of thermal equilibrium is said to exist. So when a hot nail is dropped into a beaker of cold water, energy will be transferred from the hot nail into the water even though the hot nail has less total internal energy than the water. When thermal equilibrium is reached, the temperature of both the water and the nail is the same. The particles of water and the particles in the nail have the same amount of random translational kinetic energy. Figure 1.20 shows how the kinetic particle model can be used to explain the direction of energy transfer in the beaker. What is implicit in the above discussion on thermal equilibrium and internal energy, is the subtle, but important, point made by James Clerk Maxwell that ‘All heat is of the same kind’. FIGURE 1.20 The particles in the nail have more kinetic energy (on average) than those that make up the water. They collide with the particles of water, losing some of their kinetic energy and increasing the kinetic energy of individual particles of water. The temperature of the surrounding water increases.

Water Nail Some of the kinetic energy of the particles in the nail is transferred to the water as the particles in the nail collide with particles of water.

High kinetic energy, high temperature Low kinetic energy, low temperature

18 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Water

FIGURE 1.21 When you swim in a cold pool, energy is transferred from your body into the water. The water has much more total internal energy than your body because there is so much of it. However, the particles in your body have more random translational kinetic energy that can be transferred to the particles of cold water. Hopefully, you would not remain in the water long enough for thermal equilibrium to be reached.

Some energy is also transferred to the air.

Water temperature 15 °C

Swimmer’s body temperature 37 °C

Energy is transferred to the water.

Energy is transferred to the water.

1.5.1 Laws of thermodynamics Three laws of thermodynamics were progressively developed during the nineteenth century. In the twentieth century it became apparent that the principle of thermal equilibrium could be seen as the logical underpinning of these three laws. Consequently, the Zeroth Law of Thermodynamics became accepted.

Zeroth Law of Thermodynamics Consider three objects: A, B and C. It is the case that A is in thermal equilibrium with B, and C is also in thermal equilibrium with B. Since ‘all heat is of the same kind’, it follows that A is in thermal equilibrium with C. In practice this means that all three objects, A, B and C, are at the same temperature and the law enables the comparison of temperatures.

FIGURE 1.22 If objects are in contact, they will reach thermal equilibrium. The middle object is known to be in thermal equilibrium with each outer object; thus, each outer object must also be in thermal equilibrium.

B in equilibrium with C

A A in equilibrium with B

B

C

Therefore, A and C are in thermal equilibrium. If they were brought into contact, there would be no net heat transfer.

TOPIC 1 Thermodynamic principles 19

First Law of Thermodynamics The First Law of Thermodynamics states that energy is conserved and cannot be created or destroyed (see figure 1.23). If there is an energy change in a system, all the energy must be accounted for. The First Law of Thermodynamics says: Change in the internal energy Heat energy added = of the air to the system ∆U = Q ∆U = Q − W or Q = ∆U + W

− −

Work done by the system W

Where: Q is the heat energy added to the system, in joules W is the work done by the system, in joules ∆U is the change in internal energy, in joules. Note: The words in italics, ‘of ’, ‘to’ and ‘by’, and the minus sign are important in the equation as Q and W can be either positive or negative.

If heat is added to the system (+Q) or work is done on the system (–W), then the internal energy increases and ∆U is positive. If heat is removed from the system (–Q) or work is done by the system (+W), then the internal energy decreases and ∆U is negative. Consider a volume of air inside a balloon that is placed in direct sunlight. The air inside the balloon will get hotter and the balloon will expand slightly. The energy from the Sun heats the air inside the balloon, increasing the kinetic energy of the air molecules. The air molecules lose some of this energy as they repeatedly collide with the wall of the balloon, forcing it outwards. The First Law of Thermodynamics applies to many situations: cylinders in a car engine, hot air balloons, food consumption, pumping up a tyre and the weather (see table 1.3). Consequently, the word ‘system’ is often used as a generic name when discussing thermodynamics.

FIGURE 1.23 Energy is conserved. The heat energy added to the system is equal to the energy removed from the system.

Qin

∆U = 0 ⇒W = Qin – Qout

Qout

TABLE 1.3 First Law of Thermodynamics as it applies to different situations Circumstance

Example

Result

If a system absorbs heat

Energy from sunlight

Q>0

If a system releases heat

When you sweat

Q0

If the surroundings do work on the system

Pumping up a tyre

W 0, Q < 0 and W > 0 B. ∆U < 0, Q < 0 and W > 0 C. ∆U > 0, Q < 0 and W > 0 D. ∆U < 0, Q < 0 and W < 0

32 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

For two substances to reach thermal equilibrium, which of the following must they have? A. The same internal energy B. The same kinetic energy C. The same translational kinetic energy D. None of the above 9. What is the specific heat capacity a measure of? A. How much internal energy a substance can hold B. How much energy is required to raise the temperature of 1 kilogram of a substance by 1 °C C. How much energy is required to raise the temperature of 1 kilogram of a substance by 100 °C D. The temperature change for 1 kilogram of a substance when its internal energy has increased by 1 J 10. When the cooling curve of wax is examined a flat section on the graph is found. This section is due to which of the following? A. A pause in energy reaching the system causing no temperature change B. The wax changing from a solid to a liquid and consuming extra energy without a temperature change C. A fault in the measuring equipment as this type of flat section would never occur D. The wax changing state from liquid to solid releasing extra energy without a temperature change 8.

1.7 Exercise 2: Short answer questions 1. 2. 3. 4. 5. 6. 7.

Why can’t you put your hand on your own forehead to estimate your body temperature? If today’s maximum temperature was 14 °C and tomorrow’s maximum temperature is expected to be 28 °C, will tomorrow be twice as hot? Explain your answer. Explain with the aid of a well-labelled diagram how convection occurs in a liquid that is being heated from below. The daytime temperature of an area can decrease for several days after a major bushfire. Why does this happen? The microwave cooking instructions for frozen pies state that pies should be left to stand for two minutes after heating. What happens to the pie while it stands? Why do conventional ovens without fans have heating elements at the bottom? What is the advantage of having an oven with a fan? In the figure, two beakers are filled Blocks removed from plungers Poorly Perfectly with the same gas. A plunger is insulated insulated fitted so that no gas escapes; friction is negligible between the plunger and the beaker walls. The block Gas Gas is removed from each plunger, the A B plunger moves upward. a. In each case, does the gas do work or is work done on the gas? Explain your reasoning. b. Is there a larger transfer of thermal energy as heat between gas A and the surroundings or between gas B and the surroundings? Explain your reasoning in a few sentences. c. For the expansion of gas A, how do the work and heat involved in this process affect the internal energy of the gas? Explain your reasoning in a few sentences. d. For the expansion of gas B, how do the work and heat involved in this process affect the internal energy of the gas? Explain your reasoning.

TOPIC 1 Thermodynamic principles 33

Two insulated containers are connected by a valve. The valve is closed. One container is filled with gas, the other is a vacuum. The valve is opened. Is there any change in temperature? Is there any change in the internal energy? What are the values of Q and W in this situation? 9. Explain why vegetables cook faster by being steamed than boiled. 10. In hot weather, sweat evaporates from the skin. a. Where does the energy required to evaporate the sweat come from? b. Why do you feel cooler if there is a breeze? c. Why is the cooling effect greatly reduced if the weather is humid rather than dry? 8.

1.7 Exercise 3: Exam practice questions Question 1 (2 marks) Adam says ‘A thermometer measures the average temperature between itself and the object it is measuring’, while Katie says ‘A thermometer directly measures the temperature of the object’. Explain why each is wrong. Question 2 (4 marks) The image on the right is of a heat sink. These are used to cool electrical devices such as computers. They are made from aluminium, usually painted black and have thin fins as shown. a. Discuss the design of the device with respect to your understanding of heat transfer. 2 marks b. Discuss the material used to manufacture a heat sink by commenting on its thermodynamic properties. 2 marks Question 3 (3 marks) An 800 gram rubber hot-water bottle that has been stored at a room temperature of 15 °C is filled with 1.5 kilograms of water at a temperature of 80 °C. Before being placed in a cold bed, thermal equilibrium between the rubber and water is reached. What is the common temperature of the rubber and water at this time? (Assume that no energy is lost to the surroundings. The specific heat capacity of rubber is 1700 J kg−1 K−1 . The specific heat capacity of water is 4200 J kg−1 K−1 .) Question 4 (3 marks) How much energy does it take to completely convert 2.0 kilograms of ice at −5.0 °C into steam at 100 °C? Assume no energy loss to the surroundings. Question 5 (2 marks) Explain in terms of the kinetic particle model why you can put your hand safely in a 300 °C oven for a few seconds, while if you touch a metal tray in the same oven your hand will be burned.

1.7 Exercise 4: studyON topic test Fully worked solutions and sample responses are available in your digital formats.

Test maker Create unique tests and exams from our extensive range of questions, including practice exam questions. Access the Assignments section in learnON to begin creating and assigning assessments to students.

34 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

AREA OF STUDY 1 HOW CAN THERMAL EFFECTS BE EXPLAINED?

2

Thermodynamics and climate science 2.1 Overview Numerous videos and interactivities are available just where you need them, at the point of learning, in your digital formats, learnON and eBookPLUS at www.jacplus.com.au.

2.1.1 Introduction In this topic, you will apply many of the concepts you were introduced to in topic 1 to develop a more sophisticated understanding of the Earth’s energy systems. You will be introduced to two relationships that enable physicists to determine the power radiated from objects given their temperature (Stefan–Boltzmann law) and the peak wavelength re-radiated from an object given its temperature (Wien’s law). You will consider how these relationships are used to examine the Earth’s thermal energy. By the end of this topic you will be able to explain how atmospheric gases create the greenhouse effect. You will describe and apply thermal equilibrium to the Earth’s flow of thermal energy. You will analyse changes in the thermal energy of the surface of the Earth and of the Earth’s atmosphere. Finally, you will analyse the evidence for the influence of human contributions to the enhanced greenhouse effect, including surface materials and the balance of gases in the atmosphere. FIGURE 2.1 Incoming radiation is reflected off ice back into space, but is absorbed by the water. Consider how increasing air and ocean temperatures will change this scene.

TOPIC 2 Thermodynamics and climate science 35

2.1.2 What you will learn KEY KNOWLEDGE After completing this topic you will be able to: • describe electromagnetic radiation emitted from the Sun as mainly ultraviolet, visible and infrared • calculate the peak wavelength of the re-radiated electromagnetic radiation from Earth using Wien’s law: 𝜆max T = constant • compare the total energy across the electromagnetic spectrum emitted by objects at different temperatures such as the Sun • describe power radiated by a body as being dependent on the temperature of the body according to the Stefan–Boltzmann law, P ∝ T 4 • model the greenhouse effect as the flow and retention of thermal energy from the Sun, Earth’s surface and Earth’s atmosphere • explain how greenhouse gases in the atmosphere (including methane, water and carbon dioxide) absorb and re-emit infrared radiation • analyse changes in the thermal energy of the surface of Earth and of Earth’s atmosphere • analyse the evidence for the influence of human activity in creating an enhanced greenhouse effect, including affecting surface materials and the balance of gases in the atmosphere • apply thermodynamic principles to investigate at least one issue related to the environmental impacts of human activity with reference to the enhanced greenhouse effect • explain how concepts of reliability, validity and uncertainty relate to the collection, interpretation and communication of data related to thermodynamics and climate science. Source: VCE Physics Study Design (2016–2021) extracts © VCAA; reproduced by permission.

PRACTICAL WORK AND INVESTIGATIONS Practical work is a central component of learning and assessment. Experiments and investigations, supported by a Practical investigation logbook and Teacher-led videos, are included in this topic to provide opportunities to undertake investigations and communicate findings.

Resources Digital documents Key science skills — VCE Units 1–4 (doc-31856) Key terms glossary (doc-31861) Practical investigation logbook (doc-31858)

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0027).

36 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

2.2 Earth’s energy systems KEY CONCEPTS • Describe electromagnetic radiation emitted from the Sun as mainly ultraviolet, visible and infrared. • Calculate the peak wavelength of the re-radiated electromagnetic radiation from Earth using Wien’s law: 𝜆max T = constant. • Compare the total energy across the electromagnetic spectrum emitted by objects at different temperatures such as the Sun. • Describe power radiated by a body as being dependent on the temperature of the body according to the Stefan–Boltzmann law, P ∝ T 4 .

2.2.1 Energy from our Sun You will recall that hot objects produce electromagnetic radiation across a broad range of wavelengths. The hottest object in our environment is our sun. Nearly all of the energy available to Earth comes from the Sun, whose energy output is 3.86 × 1026 J s−1 (3.86 × 1026 W). This is known as its luminosity. A tiny portion of this energy hits the Earth, heating and lighting it. When we examine the light from the Sun, we see the characteristic spectrum of light produced by hot objects, known as blackbody radiation. FIGURE 2.2 Nearly all the energy available to Earth comes from the Sun.

Figure 2.3 shows the characteristic shape of a black body plotted against its electromagnetic (EM) emission. Note that the highest section of the peak is in the visible section of the EM spectrum; however, a significant amount of heat energy (infrared) is clearly illustrated, as well as a smaller but not insignificant amount of light in the ultraviolet section.

TOPIC 2 Thermodynamics and climate science 37

FIGURE 2.3 The characteristic blackbody shape of the solar spectrum. The electromagnetic radiation emitted from the Sun is mainly ultraviolet, visible and infrared. UV visible

Near-IR

Medium-IR

Solar irradiance (W/m2/nm)

2.0

1.5

1.0

0.5

0.0 200 400

700

1500 2000 Wavelength (nm)

1000

2500

3000

What is blackbody radiation? During the late nineteenth century, scientists conducted investigations into how much radiation was produced across the light spectrum and how this distribution changed with temperature. The results are displayed in figure 2.4. FIGURE 2.4 The variation in intensity versus wavelength produced for different hot objects. Note that not only does the intensity peak higher as temperature increases, but the spectrum moves to the left and becomes bluer (dotted line).

1.0

6000 K 0.5

0.5 5000 K

4000 K 3000 K 0 0

5000 Ultraviolet Visible

10 000 Infra-red Wavelength (× 10−10 m)

38 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

15 000

0 20 000

Intensity (relative scale)

Intensity (relative scale)

1.0

In figure 2.4, the graphs for the four different temperatures are generally the same shape. Starting from the right with long wavelengths, there is very little infrared radiation emitted. As the wavelength gets shorter, the radiation produced increases to a maximum; finally, as the wavelength shortens even further the amount of radiation drops quite quickly. The graphs for higher temperatures have a peak at a shorter wavelength and a much larger area under the graph, meaning a lot more energy is emitted (intensity). This shape is typical of blackbody radiation and all heated objects exhibit these characteristics to some degree. Early researchers such as Jozef Stefan were keen to find patterns and relationships in the data and to be able to explain their observations. In 1879, Stefan compared the area under the graph for different temperatures. This area is the total energy emitted every second across all wavelengths, in other words, the power. He found that the power was proportional to absolute temperature to the power of 4, that is, P ∝ T 4 . This means that if the absolute temperature of a hot object doubles from 1000 K to 2000 K, the amount of energy emitted every second increases by 24 (2 × 2 × 2 × 2 = 16 times). Using this relationship, Stefan was able to estimate the temperature of the surface of the Sun as 5430 °C or 5700 K, which is very close to the value known today of 5778 K. Ludwig Boltzmann later proved this from a theoretical standpoint, and so the P ∝ T 4 relationship is called the Stefan–Boltzmann law. Stefan–Boltzmann law P ∝ T4

This relationship applies to all objects, but the constant of proportionality depends on the size of the object and other factors. It is this property of any hot object that has enabled astronomers to determine the surface temperatures of stars.

SAMPLE PROBLEM 1

When iron reaches about 480 °C it begins to glow with a red colour. How much more energy is emitted by the iron at this temperature, compared to when it is at a room temperature of 20 °C? b. How much hotter than 20 °C would the iron need to be to emit 10 times as much energy? a.

Teacher-led video: SP1 (tlvd-0007)

THINK a. 1.

Convert the temperature to kelvin.

WRITE a.

Temperature of hot iron: T(kelvin) = T(Celsius) + 273 = 480 °C + 273 = 753 K Temperature of cold iron: T(kelvin) = T(Celsius) + 273 = 20 °C + 273 = 293 K

TOPIC 2 Thermodynamics and climate science 39

2.

Calculate the ratio.

3.

State the solution.

b. 1.

2.

Ratio of power (hot to cold) = ratio of temperatures to the power of 4 ( )4 Phot Thot = Pcold Tcold )4 ( 753 = 293 ≈ 44 The hot iron emits 44 times as much energy every second as it does when it is at room temperature. b. Temperature of cold iron: T(Kelvin) = T(Celsius) + 273 = 20 °C + 273 = 293 K Ratio of power (hot to cold) = ratio of temperatures to the power of 4 ( )4 Thot Phot = Pcold Tcold )4 ( Thot 10 = 293 1 Thot 10 4 = 293

Convert the temperature to kelvin.

Calculate the ratio.

1

Thot = 293 × 10 4 1

3.

Use your calculator to evaluate 10 4 , then use that value to calculate Thot .

4.

Convert the temperature to Celsius.

5.

State the solution.

1

10 4

≈ 1.778

Thot = 293 × 1.778 ≈ 521 K T(Celcius) = 521 K − 273 = 248 °C The iron would need to be at 248 °C to emit 10 times the energy every second it does at 20 °C.

PRACTICE PROBLEM 1 The Sun has a surface temperature of 5778 K and radiates energy at a rate of 3.846 × 1026 W. How much energy would a star of similar size radiate if its surface temperature was 8000 K?

In 1893, Wilhelm Wien (pronounced Veen) was able to show that as the temperature increased, the wavelength of maximum intensity of energy emitted decreased, and indeed the two quantities were inversely proportional. That is, the wavelength is proportional to the inverse of the temperature. This can be seen in figure 2.5b.

40 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 2.5 The wavelength is proportional to the inverse of the temperature. These graphs illustrate that as the temperature increased, the wavelength of maximum intensity of energy emitted decreased. (b)

λmax (m)

λmax (m)

(a)

0

0

1 Temperature

Temperature (Kelvin)

Wien’s law can be written as: 𝜆max T = constant The value of this constant is 2.90 × 10−3 mK (metre-kelvin). SAMPLE PROBLEM 2

At what wavelength is the peak intensity of the light coming from a star whose surface temperature is 11 000 K (about twice as hot as the Sun)? b. In what section of the spectrum is this wavelength? a.

Teacher-led video: SP2 (tlvd-0008)

THINK a. 1.

2.

State Wien’s law, and rearrange the equation to make 𝜆max the subject. Substitute the known values into the equation to find 𝜆max .

WRITE a.

𝜆max T = constant constant 𝜆max = T Constant = 2.90 × 10−3 mK, T = 11 000 K constant T 2.90 × 10−3 mK = 11 000 K = 2.64 × 10−7 m

𝜆max =

3.

b. 1.

State the solution.

Refer to figure 2.4.

The light coming from a star whose surface temperature is 11 000 K has a peak intensity at a wavelength of 2.64 × 10−7 m. b. 264 nm is beyond the violet end of the visible spectrum, so it is in the ultraviolet section of the electromagnetic spectrum.

TOPIC 2 Thermodynamics and climate science 41

PRACTICE PROBLEM 2 Determine the surface temperature of a star that emits light at a maximum intensity of 450 nm.

2.2.2 Heat from within Although the sun is our primary source of FIGURE 2.6 New crust is formed at a ridge and returns to energy on Earth, our secondary source of the mantle at a trench. energy comes from within the Earth. The Ridge Lithosphere heat energy within the Earth comes from Trench the radioactive decay of elements such Trench as uranium. This heat energy is not evenly Asth eno distributed and hot spots occur under the sp he re mantle. Mantle Energy transfer by convection is common in gases and liquids, but it can also occur in 700 km solids under the right conditions. The high temperatures, about 2000 °C, and pressures in Outer core the Earth’s mantle are enough to make solid rock move, only very slowly of course. The hot lighter rock at these points slowly rises, Inner while denser rock at colder spots slowly sinks. core This sets up a convection cell in the Earth’s mantle with the surface crust moving horizontally across the Earth. The speed of the rock movement is a few centimetres per year. The molten rock wells up at mid-ocean ridges and moves out. The rock eventually meets the edge of a continental plate and cools further, becoming denser, then sinks back towards the mantle in a deep ocean trench.

2.2.3 Energy in balance Every object attempts to achieve thermal equilibrium — that is, a balance between energy absorbed and energy emitted. Imagine a steel ball placed in direct sunlight as shown in figure 2.7. When light energy heats and warms an object, like a steel ball, its particles start to get more excited and its translational kinetic energy starts to increase. As a result, the steel ball emits thermal radiant energy (usually in the infrared). The absorbed energy increases the temperature of the ball; the increased temperature means that the ball will radiate more thermal energy. If the amount of energy radiated is less than the amount absorbed, the temperature of the ball will continue to increase, leading to more energy being emitted, until a temperature is reached where energy in and energy out balance.

42 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 2.7 Achieving thermal equilibrium is a balance between energy absorbed and energy emitted.

Energy out from thermal radiation Energy in from sunlight

Steel ball

Similarly, if a cloud moves in front of the Sun, the energy absorbed by the ball would suddenly decrease to less than the amount of energy being emitted. The temperature of the ball would drop and continue to fall until the amount of energy emitted matched the energy absorbed and a new equilibrium was reached. The Earth as viewed from space is like the steel ball. The energy falling on the Earth from the Sun is fairly constant. At the equator, an average of about 684 joules of energy from the Sun hits each square metre of the Earth’s surface every second; that is, 684 W m–2 , where 1 watt is a unit of power or the rate of energy delivery and equals 1 joule per second. This value varies from day to day by as much as 2 W m–2 , as well as having an approximate 11-year cycle of a similar magnitude. With the Stefan–Boltzmann relationship, P ∝ T 4 , it is possible to use the amount of energy the Earth radiates into space to calculate the temperature of the Earth as observed from space.

MILANKOVITCH CYCLES While a prisoner of war in World War I, Milutin Milankovic postulated several types of changes in the Earth’s movement around the Sun that could affect the amount of solar radiation the Earth receives and its distribution. These changes can affect climate on a time span of many thousands of years and possibly explain the occurrence of ice ages. Some of the types of changes include: • variation in the elliptical shape of the Earth’s orbit (eccentricity) with a cycle time of about 413 000 years • precession of the Earth’s axis of rotation; like any spinning top, the axis itself rotates, once every 26 000 years • the tilt of the axis (obliquity), which ranges from 22° to 24.5° every 41 000 years. FIGURE 2.8 The variation in solar radiation that the Earth is exposed to has changed the climate over tens and hundreds of thousands of years. These variations explain the ice ages the Earth has experienced in its past but do not explain the current global warming. Obliquity/tilt

Precession

22.1°– 24.5° Eccentricity

431 000 years

41 000 years

26 000 years

All these changes are due to gravitational interactions in the solar system between the Earth, the Sun and other planets. It is thought that these factors may explain the long-term cooling trend the Earth has been in over the last 6000 years. From the slowness with which these changes occur, none can explain the unprecedented global warming in recent decades.

2.2.4 How much energy does the Earth get from the Sun? The Sun is directly overhead the equator at midday on the equinox. At this place and time, the solar radiation is about 1368 W m–2 . But because the Earth turns, producing night and day, this value has to be halved to 884 W m–2 . Also, the Earth is curved with the North and South Poles receiving much less light than the equator over a full year. This requires the number to be halved again. So the average solar radiation across the Earth is 342 W m–2 as shown in figure 2.9. About 100 W m–2 of this radiation is reflected straight back into space by the white surfaces of clouds and ice sheets. This leaves 242 W m–2 to heat up the Earth. TOPIC 2 Thermodynamics and climate science 43

FIGURE 2.9 At the Earth’s equator at midday, the light intensity from the Sun is 1368 W m–2 . The average over night and day and from pole to pole is 342 W m–2 . Whole Earth average 342 W m−2 Earth North Pole

1368 W m−2 at equator at midday Equator

South Pole 684 W m−2 at equator average

2.2 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. A 100 W light globe has a tungsten filament with a temperature of 2775 K when switched on. (a) How much radiation does the filament emit at 20 °C? (b) The voltage on the light globe is reduced to increase the lifetime of the filament. The temperature of the filament is now 2000 K. What is the power saving? (c) The voltage is now increased so that the power output is 200 W. What is the new filament temperature in kelvin? 2. (a) A piece of iron has a yellow glow when it reaches 1150 °C. How much more energy is emitted every second at this temperature compared to when the iron glows red at 480 °C? (b) At what temperature in degrees Celsius would the iron give off 10 times as much energy as it does at 480 °C? 3. What is the wavelength of the light with the peak intensity from our solar system’s closest neighbouring star, Proxima Centauri, which has an average surface temperature of 3042 K? 4. Our Sun gives off most of its light in the ‘yellow’ portion of the electromagnetic spectrum. Its 𝜆max is 510 nm. Calculate the average surface temperature of the Sun. 5. The Earth’s surface has an average temperature of 288 K. What is the wavelength of maximum emission from the Earth’s surface? 6. The human body has a surface temperature of about 37 °C. (a) What is the wavelength at which the human body emits the most radiation? (b) In what part of the spectrum is this wavelength? 7. Suppose the surface temperature of the Sun was about 12 000 K, rather than about 6000 K. (a) How much more thermal radiation would the Sun emit? (b) What would happen to the Sun’s wavelength of peak emission?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

44 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

2.3 The enhanced greenhouse effect KEY CONCEPTS • Model the greenhouse effect as the flow and retention of thermal energy from the Sun, Earth’s surface and Earth’s atmosphere. • Explain how greenhouse gases in the atmosphere (including methane, water and carbon dioxide) absorb and re-emit infrared radiation. • Analyse changes in the thermal energy of the surface of Earth and of Earth’s atmosphere.

The Stefan–Boltzmann relationship, P ∝ T 4 , enables the absolute temperature of an object to be determined if you know how much energy it is radiating. For the 242 W m–2 that the Earth radiates, this gives a temperature of 255.6 K, that is, –18 °C. This seems incorrect; the Earth is not that cold! What causes the difference between this calculated temperature (which is the temperature of the Earth observed from space) and the temperature we observe at the surface? The explanation is that we ignored the greenhouse gases in the atmosphere.

2.3.1 The natural greenhouse effect The average global surface air temperature since the last ice age has been about +15 °C. This means that the greenhouse gases of water vapour (H2 O), carbon dioxide (CO2 ) and others have been a blanket that has provided an extra 33 °C of warmth. FIGURE 2.10 The necessary temperature (–18 °C) for the Earth to radiate 242 W m–2 actually occurs at an altitude of about 5 kilometres above the surface. So, the Earth’s surface is warmed by a 5-kilometre thick blanket! Upper atmosphere: –18 °C

Greenhouse blanket of H2O and CO2

Surface air temperature: 15 °C

Earth

How do water vapour and carbon dioxide act as a blanket to trap heat? If you shine a broad spectrum of light through a gas, some colours will be absorbed. The colours absorbed are specific to the substance, which means each substance produces its own unique pattern of absorption bands. These absorption bands are like fingerprints and can be used to identify molecules. The light in the absorption band has been absorbed by the molecule and then subsequently re-emitted, but in a random direction (see figure 2.13). This means the spectrum of light from the source will have gaps where absorption has occurred. The light from the other parts of the spectrum passes through the gas without a change in direction. Ultraviolet light has more energy than visible light, which has more energy than infrared, but none of them are energetic enough to break up molecules. These three types of radiation have only enough energy TOPIC 2 Thermodynamics and climate science 45

to stretch, twist and spin molecules. Molecules with two atoms such as nitrogen (N2 ) and oxygen (O2 ) have very strong bonds. When they absorb ultraviolet light, their bonds stretch. The energy of visible light and infrared radiation is too low to affect such molecules. This means the two gases are transparent to visible light and infra-red. H2 O and CO2 have three atoms in each molecule, so they are more flexible than N2 and O2 . These molecules can bend, whereas atoms with only two molecules cannot. When H2 O and CO2 absorb infrared light, their bonds stretch and bend. Other molecules with more than two atoms, such as methane (CH4 ), absorb infrared radiation in the same way. Figure 2.11 shows the different ways water vapour molecules bend and stretch and figure 2.12 shows the different ways carbon dioxide molecules bend and stretch. FIGURE 2.11 The water molecule has three different ways of stretching or bending, as well as oscillations about three axes. Stretch

Stretch

Bend

Oscillations

x

y

z

FIGURE 2.12 The carbon dioxide molecule has three different ways of stretching. Stretch

Stretch

Bend

The oxygen atoms at the ends of the bonds in carbon dioxide are heavier than the hydrogen atoms at the ends of the bonds in water, so the bonds in the two molecules stretch and bend differently. This means each molecule will absorb different parts of the infra-red spectrum. Consequently, water vapour and carbon dioxide contribute independently to the greenhouse effect. Once a gas molecule has absorbed infra-red radiation coming from the Earth’s surface, it re-emits the radiation but, importantly, in a random direction (see figure 2.13). So, for the gas as a whole, some radiation goes back down to the Earth to increase its temperature and some is directed towards the top of the atmosphere and out into space. However, other molecules further up in the atmosphere can absorb this radiation and re-emit more back to the Earth. The overall effect is that more than half the radiation emitted by the gas comes back to the Earth’s surface.

46 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 2.13 Infra-red radiation from the Earth’s surface is absorbed by greenhouse gases and re-emitted in all directions. Space

Reflected light

Solar radiation

O2 Atmosphere

O2

N2 N2

N2

O2

N2

N2

O2 N2

O2

N2 N2

O2

N2

CO2

O2 N2

N2

O2

N2

O2

N2

H2O

H2O

O2

O2

N2

CH4

O2 N2

N2

CO2

CH4

N2

ouse Greenh gases

O2 O2

N2 CH4

N2 O2

O2 N2

H2O

CO2

O2

Infra-red

N2 O2

Earth

Heated by absorbed solar radiation

Human activities, such as the burning of fossil fuels (coal, oil and natural gas), agriculture and land clearing, are increasing the concentrations of greenhouse gases in the atmosphere. This increase is sometimes called the enhanced greenhouse effect (see figure 2.14). FIGURE 2.14 Comparison of the greenhouse effect and the enhanced greenhouse effect Greenhouse effect

Enhanced greenhouse effect Sun Sun

Sun Sun

Earth

Earth

Increased carbon dioxide concentration in the atmosphere means that more of the wavelengths that carbon dioxide absorbs will be re-emitted back to Earth, increasing the temperature of the Earth.

TOPIC 2 Thermodynamics and climate science 47

Figure 2.15 is the same solar spectrum as figure 2.3 but also shows the: • spread of wavelengths radiated by the Sun at 6000 K (these are the wavelengths in sunlight) • spread of wavelengths received on the surface of the Earth, demonstrating the impact of the atmosphere on the light we receive from the Sun, and the ‘holes’ in the spectrum because of greenhouse gas absorption • absorption spectrum for water vapour • absorption spectrum for carbon dioxide. FIGURE 2.15 Characteristic blackbody shape showing the irradiance at sea level and thus the effect of the atmosphere. Note the ‘holes’ at particular frequencies due to absorption from atmospheric gases.

Near-IR

UV visible

Medium-IR

2.0

Solar irradiance (W/m2/nm)

Energy at the top of the atmosphere 1.5

Energy at sea level

Absorbed by H2O vapor

1.0

0.5

0.0 200 400

Absorbed by CO2

700

1000

1500 2000 Wavelength (nm)

2500

3000

It is worth clarifying the vertical scale on the absorption spectra. The spectra show that for some wavelengths 100% of the radiation is absorbed. This means that in a container holding only that gas, no radiation of that wavelength would pass through without interacting with a molecule. However, in the atmosphere with a mixture of gases, and with CO2 , H2 O and CH4 at low concentrations, much of the radiation with these wavelengths has a good chance of passing through without ever hitting one of these molecules. The gases CO2 , H2 O and CH4 make up a very small proportion of the atmosphere, so the infrared radiation emitted from the Earth’s surface has a good chance of reaching outer space without being absorbed. However, with increased emissions of CO2 and CH4 , interactions are more likely to occur.

2.3.2 Fossil fuels’ fingerprints Atoms with the same number of protons but different numbers of neutrons are called isotopes. Isotopes of an element have identical chemical behaviour because they have the same number of electrons. However, because different isotopes have different masses, their reactions may proceed at different speeds. Heavier isotopes generally react slower. For example, carbon has two main isotopes: 1. carbon-12 with six neutrons 2. carbon-13 with seven neutrons. The process of photosynthesis in plants turns carbon dioxide and water into carbohydrates. This reaction occurs more quickly with carbon-12 than it does with carbon-13. This means the proportion of carbon-13 in the plant is less than what was in the atmosphere at the time the plant was growing. 48 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Over millions of years, dead plants are turned into fossil fuels, such as coal, oil and natural gas. When these fossil fuels are burnt, the carbon, which has relatively little carbon-13, is released back into the atmosphere. Tree rings and ice cores can be used to determine the ratio of carbon-13 to carbon-12 of the Earth’s atmosphere going back thousands of years. These can then be compared to the current atmosphere. The data show the ratio of carbon-13 to carbon-12 is much lower now than in the past. The ratio started to decline about 1850, around the time fossil fuel use began to increase. The low amount of carbon-13 in our atmosphere is evidence that the increased CO2 is produced by humans using fossil fuels.

2.3.3 The Earth’s energy budget The Earth’s energy budget is an accounting of the incoming energy from the Sun, where it goes in the atmosphere and at the surface, and how the outgoing energy leaves the Earth. Note: This is called an energy budget, even though the numbers are measured in watts per square metre, which is actually the amount of energy passing through an area of one square metre every second. Figure 2.16 shows all the different paths energy takes from arriving at the Earth, mainly as visible light, to being emitted as longwave infrared radiation or reflected straight back into space. FIGURE 2.16 The Earth’s energy budget from the International Panel on Climate Change (IPCC) 107

Reflected solar radiation 107 W m−2

235

Incoming solar radiation 342 W m−2

342

Reflected by clouds, aerosol and atmospheric gases 77

Emitted by atmosphere

165

Emitted by clouds 67

Absorbed by atmosphere

24

Latent 78 heat

350

24 Thermals

40 Atmospheric window Greenhouse gases

Reflected by surface 30 168 Absorbed by surface

30

Outgoing longwave radiation 235 W m−2

78 Evapotranspiration

390 Surface radiation

40

324 Back radiation

324 Absorbed by surface

The numbers in figure 2.16 can be used to check the energy balance between energy coming in and energy going out — not only of the whole Earth, but also of just the atmosphere or just the Earth’s surface. In each case, the energy coming in should equal the energy going out (see table 2.1). TABLE 2.1 The energy balance between energy coming in and energy going out For the Earth as a whole

For the Earth’s surface

Incoming energy = 342 W m−2

Incoming energy = 168 + 324 W m−2

Outgoing energy = 77 + 30 + 165 + 30 + 40

= 492 W m−2 Outgoing energy = 24 + 78 + 350 + 40

= 342 W m−2 = incoming energy

= 492 W m−2 = incoming energy

TOPIC 2 Thermodynamics and climate science 49

Table 2.2 provides a description and the contribution to the energy budget of most of the terms in figure 2.16. TABLE 2.2 Energy budget of the Earth by energy category

Energy category

Amount (W m–2 )

Description

Incoming solar radiation

342

Visible light with some infra-red and ultraviolet from the Sun

Outgoing longwave radiation

235

Infra-red radiation from many sources heading out to space

Reflected solar radiation

107

Visible light from two sources heading out to space (30 + 77 = 107)

Reflected by clouds etc.

77

Visible light reflected by clouds, aerosol and atmospheric gases

Atmosphere window

40

Infra-red radiation passing through the atmosphere because it is in the part of the spectrum that is not absorbed by the main greenhouse gases

Emitted by clouds

30

Infra-red radiation from clouds heading out to space

Latent heat

78

Energy used or released when there is a change of state

Back radiation

324

Infra-red radiation emitted downwards by greenhouse gases in the atmosphere

Absorbed by surface

492

Visible light, infra-red and ultraviolet radiation from two sources absorbed by the Earth’s surface (168 + 324 = 492)

Thermals

24

Rising currents of hot air

Evapotranspiration

78

Moisture in plants released as water vapour

Surface radiation

390

Infra-red radiation emitted by the Earth’s surface into the atmosphere

SAMPLE PROBLEM 3

Using table 2.2, show that conservation of energy holds for the Earth system. Teacher-led video: SP3 (tlvd-0009) THINK

WRITE

Use table 2.2 and determine total energy in. 2. Use table 2.2 and determine total energy out.

Total incoming solar radiation = 342 W m−2

1.

Energy conservation holds if energy in = energy out. 4. State the solution. 3.

Total outgoing energy = Outgoing longwave radiation + Reflected solar radiation = 235 + 107 = 342 W m−2 Energy in = 342 W m−2 = energy out As the energy into the Earth system is equal to the energy out, conservation of energy holds for the Earth system.

50 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

PRACTICE PROBLEM 3 Using figure 2.16, find the missing amounts in the following table. Amount (W m–2 )

Energy category

Description

Reflected by surface

?

Visible light reflected by snow, land and sea ice

Absorbed by atmosphere

?

Incoming radiation that is absorbed by gases in the atmosphere

2.3.4 Feedback In any complicated system with many interacting components, such as the Earth’s climate, sometimes one part changes a second part, and a change in the second part can then change the first part. This is called feedback. A common audio example of feedback is when a person using a microphone walks in front of a loudspeaker. The microphone picks up the signal from the speaker, which is then amplified and fed back into the speaker, which is again picked up by the microphone, and so on, producing a loud high-pitched noise. This is called positive feedback. Negative feedback is also possible; for example, a ‘governor’ or control device is used in engines, where excess speed is used to reduce the input to keep the speed constant. The climate has examples of both positive and negative feedback, seen in table 2.3. TABLE 2.3 Positive and negative feedback in the climate Positive feedback Evaporation: Increasing sea temperatures leads to more evaporation of water, a greenhouse gas, which increases the temperature of the air and sea.

Negative feedback Thermal radiation: Increasing surface temperature emits more infra-red radiation, which cools the Earth.

The risk of positive feedback is that there is a TABLE 2.4 Albedo of different materials ‘tipping point’ beyond which the system can become Material Albedo unstoppable, as in the loudspeaker example above. Another example of positive feedback in the Water 0.08 Earth’s climate is the physical quantity ‘albedo’. Sea ice 0.5–0.7 Albedo is a measure of the proportion of incoming radiation that is reflected without being absorbed. It is Fresh snow 0.8–0.9 usually expressed as a number between 0 and 1, with Clouds 0.4–0.8 1 being a perfect reflector and 0 a perfect absorber. The significance of albedo for the climate is the Forest 0.1 role it plays in determining the amount of sea ice in Desert 0.3 the Arctic. Sea ice is very reflective but the water it is floating in is not. Increased air temperature Green grass 0.25 leads to more sea ice melting, which means that more of the incoming radiation hits water rather than ice and is absorbed. This increased energy absorption then heats up the atmosphere, which leads to more sea ice melting, and so the feedback loop continues.

TOPIC 2 Thermodynamics and climate science 51

FIGURE 2.17 The extent of Arctic sea ice in September each year. The solid line shows the average of climate model predictions, the shaded area is the spread of the predictions from the different models. Arctic September sea ice extent: observations and model runs 10.0

Average of climate models

Sea ice extent (106 Km2)

8.0

6.0 Observations 4.0

2.0

0.0 1900

1950

2000

2050

2100

Year

Resources Digital documents Investigation 2.1 Examining the sun’s spectrum (doc-31859) Investigation 2.2 The colour of temperature (doc-31860) Teacher-led video

Investigation 2.2 The colour of temperature (tlvd-0806)

Video eLesson

Albedo (eles-2515)

2.3 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Why doesn’t all of the radiation from the Sun that enters the Earth’s atmosphere reach the surface? 2. Why is the enhanced greenhouse effect a threat to life on Earth? 3. Is the majority of the heating of the Earth’s atmosphere due to the transfer of radiant energy from the Sun or from the Earth’s surface? 4. Identify the properties of water that cause the variation in climate over the Earth’s surface. 5. The following figure shows the radiance of the Sun (orange, left-hand scale) and the Earth (red, right-hand scale) and the absorption spectra for water vapour (H2 O) and carbon dioxide (CO2 ). Visible UV

Infra-red 0.4

0.7

52 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

(a) Determine the range of wavelengths emitted by the Earth that are absorbed by: i. carbon dioxide ii. water vapour. (b) Which wavelengths emitted by the Earth are absorbed by carbon dioxide and not by water vapour?

8 2000

4 1000

6000 K (Left-hand scale)

255 K (Right-hand scale)

Radiance

Radiance

6

2

0 Absorption (%)

100 80 60 40 20 0 100 Absorption (%)

Water vapour

80

Carbon dioxide

60 40 20 0 0.1

0.2

0.3 0.4 0.5

1.0

2 3 4 5 Wavelength ( μm)

10

20

30 40 50

100

6. (a) Which carbon dioxide molecule would move faster in the atmosphere: a lighter one with a carbon-12 atom or a heavier one with a carbon-13 atom? (b) Which molecule do you think is more likely to dissolve in the ocean? Give a reason. 7. Using figure 2.16 in section 2.3.3, calculate the total incoming energy and total outgoing energy for the atmosphere and show that each equals 519 W m–2 . 8. (a) Estimate the albedo of: i. the asphalt road surface ii. crops iii. the roof of your house. (b) Indicate whether each of the following is an example of positive or negative feedback. Explain your reasons. i. The release of methane from melting permafrost ii. Effect of clouds

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

TOPIC 2 Thermodynamics and climate science 53

2.4 Climate models KEY CONCEPT • Analyse the evidence for the influence of human activity in creating an enhanced greenhouse effect, including affecting surface materials and the balance of gases in the atmosphere.

2.4.1 Scientific principles that underlie climate The scientific principles that underlie climate have been understood since the nineteenth century. They include: • Newton’s laws of motion that explain the movement of gases and liquids, such as the existence of high- and low-pressure regions, the formation of cyclones, the flow of ocean currents • thermodynamics principles that explain heat transfer within the atmosphere, and between the atmosphere, the oceans and the land • gas laws and solubility in liquids from chemistry that explain the behaviour of gases within the atmosphere and their interactions with the oceans. These principles are known precisely with mathematical relationships, some of which you will come across in the VCE Physics and VCE Chemistry courses. The development of the computer meant these principles and their mathematical relationships could be applied to the biggest problem on Earth, the Earth itself. A climate model is an attempt to apply these relationships to the atmosphere of the whole Earth as well as the surface features of land, ice and sea. The model attempts to calculate aspects of the climate that are important to humans, such as rainfall and humidity, sea level rise, ocean acidity, wind strength and, of course, air temperature. Climate models are also able to calculate future trends in these aspects of climate and then investigate the effect on these trends of changes such as reducing greenhouse gas emissions or aerosol use.

2.4.2 How to know if a climate model is accurate? A climate model starts as hundreds of mathematical equations that describe all the physical and chemical interactions within the atmosphere, and between the atmosphere, the land, sea and ice. These equations are applied to a model of the Earth’s atmosphere as it was at an earlier time, say 1900, to calculate the expected climate conditions for the next 100 years. These calculations are then compared with the actual climate conditions to see how close the climate model comes to reality. If the model is out, then it is back to the drawing board to redesign the model. If the model is accurate, it can be used to calculate the future climate. Different countries and scientific organisations each produce their own climate models. The basic scientific principles are the same, but the models vary in their subtlety and complexity, and the available computing power. The solid black line in each graph in figure 2.18 is the global average surface air temperature from 1900 until 2000. The models were run twice. 1. The first run assumed the carbon dioxide concentration stayed constant at its value in 1900. This produced the blue graph in figure 2.18a. The solid blue line is the average of all the different models. 2. The second run included the actual carbon dioxide concentration over the century. This produced the yellow graph in figure 2.18b. The solid red line is the average of all the different models. The red line in figure 2.18b is a close fit to the actual observed temperature change. It is apparent from comparing the two graphs that increased carbon dioxide concentration has led to an increase in the global temperature.

54 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 2.18 Model comparisons against actual temperature from 1900 to 2000 (b)

Natural changes only 1.0

Observations

0.5

0.0

Models −0.5 Pinatubo Agung

Santa Maria −1.0 1900

1920

1940

1960

El Chichon

1980

2000

Temperature difference from 1900 (°C)

Temperature difference from 1900 (°C)

(a)

Natural changes and greenhouse gas emissions 1.0

Observations

0.5

0.0

Models −0.5 Pinatubo Agung

Santa Maria −1.0 1900

1920

1940

1960

El Chichon

1980

2000

Year

Year

Source: IPCC

The close match of figure 2.18b with the observations validates the design of the climate models and gives credence to the predictions for the decades and centuries ahead. The models can also be used to investigate how long the Earth’s atmosphere would take to respond to a significant reduction in carbon dioxide emissions. With this information, governments can act together to bring about change. The models not only produce a global average but also specific information for each region of the Earth, so local responses to climate change can be planned. Climate models can be used to investigate the long-term effects of international agreements on controlling greenhouse emissions on surface air temperature, sea level rise and rainfall.

2.4.3 Investigating a simple climate model The energy budget shown in table 2.2, in section 2.3.3, has been converted into a rudimentary spreadsheet model of the climate, which you can use to investigate the effect of future scenarios such as: • a decrease in sea ice, leading to a reduction in the amount of radiation reflected straight back into space • increased atmospheric carbon dioxide concentration, leading to a smaller proportion of infra-red radiation from the atmosphere being emitted into space and more being emitted back to Earth.

Resources Digital document A simple climate change model (doc-16168)

2.4 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. Use figure 2.18 in section 2.4.2 to answer questions 1–5. 1. Pinatubo, El Chichon, Agung and Santa Maria are all volcanic eruptions. (a) What was the effect of the eruptions on the average global surface air temperature? (b) Suggest a mechanism for this effect.

TOPIC 2 Thermodynamics and climate science 55

2. There are some stretches of a number of years where the models and the actual temperature data differ significantly. (a) Identify those stretches of years. (b) Thinking of the history of each of those years, is there a possibility that the collection of temperature data to obtain a global average may have been unreliable? 3. Looking at the two graphs, when do you think the increased carbon dioxide concentrations started to affect the global average surface air temperature? 4. The Earth’s surface is made up of solid rock, liquid oceans and a gaseous atmosphere. (a) Which do you think would be better to use to measure the temperature of the Earth’s surface? (b) Why do you think the air temperature is used? 5. Although the IPCC graphs show that long-term global warming is occurring, the temperature increase is not steady. What factors make the increase so erratic?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

2.5 Investigating issues related to thermodynamics KEY CONCEPTS • Apply thermodynamic principles to investigate at least one issue related to the environmental impacts of human activity with reference to the enhanced greenhouse effect. • Explain how concepts of reliability, validity and uncertainty relate to the collection, interpretation and communication of data related to thermodynamics and climate science.

2.5.1 Issues related to thermodynamics You have studied many aspects of thermodynamics in this and the previous topic that affect our lives in many ways, including: • storing food • designing our homes, as well as heating and cooling them keeping our bodies at an even temperature • • selecting energy sources for our economy • most importantly, understanding and addressing climate change. • It is therefore appropriate that you explore an issue in some depth, to identify the thermodynamic principles in use and to evaluate relevant public policy and social commentary for their scientific accuracy.

56 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 2.19 Applying thermodynamic principles is vital when designing a house.

ISSUES RELATED TO THERMODYNAMICS Some possible topics for you to research are listed here. • Compare domestic heating and cooling technologies. Are the brochures and advertising physically plausible? • Investigate the design of energy efficient houses. How do building materials differ? Are star ratings reliable? Should design features such as double glazing and orientation in relation to the Sun be mandatory? • Compare appliances by their technology, efficiency and emissions. Are the energy ratings useful? • Research solar thermal technology. Is generating electricity by solar thermal technology a feasible alternative for Australia? • Investigate geo-engineering solutions to tackle climate change. Are any of them feasible? • Examine the treatments for hypothermia (body core temperature < 35 °C) and hyperthermia (body core temperature > 38 °C). How have they changed over time and how do they work? • Compare different technologies for food preparation, for example, microwave oven versus convection oven, as well as different fuel options. These broad topics may need to be narrowed down to focus on a specific aspect if the task is to be manageable.

2.5.2 Evaluating resources When collecting information from the internet or elsewhere, it is important to determine how reliable and accurate the information is. Each item should be evaluated to see how useful it will be. Some aspects worth considering are as follows.

Reliability

• • • • • •

Relevance. Is the information central to your purpose? Up to date. When was the information produced? Has it been superseded by new information? Expertise. Is the information produced by someone with appropriate qualifications? Do they know what they are talking about? Source. Where did the author obtain their information? Audience. Who is the intended audience and is the resource suited to your purpose? Bias. Is there evidence of overstatement, selective quoting or other tricks designed to mislead?

TOPIC 2 Thermodynamics and climate science 57

Validity

• • •

Accuracy. Are physics concepts correctly used? Argument. Is the argument logically sound and strong? Are deceptive strategies used, for example, oversimplification or false selections? Analysis. Are physics relationships applied correctly?

Uncertainty

• •

Graphs. Are the graphs misleading? Data. Do the data lack precision? Are only selected data used?

2.5 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Apply thermodynamic principles to investigate at least one issue related to the environmental impacts of human activity with reference to the enhanced greenhouse effect.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

2.6 Review 2.6.1 Summary • • • • • • • • • • • • • •

The graph of the energy contribution of different wavelengths of emitted radiation has a characteristic shape. For a given temperature, there is a specific wavelength at which the most energy is emitted. Its symbol is 𝜆max . The graph of the energy contribution of different wavelengths for a higher temperature has a lower 𝜆max and a larger area under the graph. 𝜆max is inversely proportional to the temperature measured in kelvin (𝜆max T = constant). The amount of energy emitted per second is called power. The area under the graph of energy contribution against wavelength is a measure of power. The area under the graph is proportional to the kelvin temperature raised to the power of 4. This can be expressed as P ∝ T 4 . The Earth on average, over the day and across the globe, receives about 342 joules of energy every second in each square metre. About 100 J of the energy the Earth receives is reflected back into space by ice, snow and clouds. Without greenhouse gases in the atmosphere, the surface temperature of the Earth would be –18 °C. Most of the radiation emitted by the Earth is infrared radiation. The main greenhouse gases are water vapour (H2 O) and carbon dioxide (CO2 ), with methane (CH4 ) making a small contribution along with other molecules with more than two atoms. The relative flexibility of the greenhouse gas molecules allows them to absorb the infrared radiation emitted by the Earth’s surface. After absorbing the infrared radiation, the greenhouse gases re-emit it in all directions; some upwards out into space, but most back down to the Earth to heat the surface. Some of the wavelengths absorbed by the greenhouse gases are common to each, but each particular gas absorbs some wavelengths that are unique to that gas.

58 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

•

• • • • • •

The energy that comes to the Earth from the Sun is balanced by the energy the Earth radiates back into space. The energy from the Sun is absorbed by the atmosphere, the land surface and the oceans. Some of this energy returns to the Earth’s surface after being emitted by the greenhouse gases and increases the surface temperature. Feedback mechanisms are processes in complex systems where the output from the system can affect an input to the system. The feedback can be either positive or negative. Both types occur in the Earth’s atmosphere. The albedo of a surface is the proportion of incoming radiation that is reflected with no change in wavelength. Climate models are based on the scientific principles of motion, thermodynamics and the chemistry of gases and liquids. Climate model calculations are very consistent with the historical record. Climate model calculations show that the increase in global average surface air temperature in recent decades is explained by the increased atmospheric carbon dioxide concentration.

Resources

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0027).

2.6.2 Key terms Absorption bands are the range of wavelengths in the electromagnetic spectrum that are absorbed by atmospheric gases. Absolute temperature is the temperature of an object taken in the scale using absolute zero. Albedo is the proportion of solar radiation reflected by a surface. A black body is an object that absorbs all radiation that falls on it. Blackbody radiation is the characteristic radiation emitted by a black body when heated. Eccentricity is a measure of how elliptical an object’s orbit is. Enhanced greenhouse effect is the greenhouse effect due to human civilization which is greater than that caused by natural forces alone. Feedback is when a system’s input is fed by its previous output. Isotopes are atoms containing the same number of protons but different numbers of neutrons. Luminosity is the amount of radiated electromagnetic energy emitted by a light-emitting or luminous object. Negative feedback occurs when the response to the feedback is in the opposite direction to the input. Obliquity is a measure of the angle tilt of a planet against its plane of orbit. Photosynthesis is a chemical reaction that takes place in the chloroplasts of a plant cell consuming carbon dioxide and releasing oxygen. Positive feedback occurs when the response to the feedback is in the same direction to the input. Precession is a change in direction of the rotational axis of a spinning object. Radioactive decay is the process in which unstable isotopes spontaneous decomposes into a stable daughter isotope through the emission of radiation.

Resources Digital document Key terms glossary (doc-31861)

TOPIC 2 Thermodynamics and climate science 59

2.6.3 Practical work and investigations Investigation 2.1 Examining the sun’s spectrum Aim: To show the different parts of the Sun’s visible spectrum and to compare this to the other sections of the electromagnetic spectrum the Sun emits Digital document: doc-31859

Investigation 2.2 The colour of temperature Aim: To show that different colours have different thermal effects and to consider the implications of this for our planet Digital document: doc-31860 Teacher-led video: tlvd-0806

Resources Digital document Practical investigation logbook (doc-31858)

2.6 Exercises To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au.

2.6 Exercise 1: Multiple choice questions (

P1 P2

)

for two stars of the same diameter if star 1 had half the What would be the ratio of power surface temperature of star 2? 1 A. 2 1 B. 4 1 C. 8 1 D. 16 2. In relation to figure 2.4, which statement is true? A. The star at 3000 K would appear ‘bluer’ than the star at 6000 K because its peak is closer to the blue end of the visible spectrum. B. The star at 6000 K would appear ‘bluer’ than the star at 3000 K because its peak is closer to the blue end of the visible spectrum. C. The star at 6000 K would appear ‘redder’ than the star at 3000 K because its peak is closer to the red end of the visible spectrum. D. Star colour cannot be determined by temperature. 1.

60 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Which of the following is not a feature that influences climate change on Earth? A. The seasons B. The properties of water C. The tilt of Earth’s axis D. The geological features (land mass and oceans) 4. What is the major cause of the greenhouse effect? A. Gases in the ozone layer absorb heat from the Earth’s surface. B. Gases in the ozone layer absorb heat from the Sun. C. Gases in the atmosphere absorb heat from the Earth’s surface. D. Gases in the atmosphere absorb heat from the Sun. 5. The enhanced greenhouse effect is best described as which of the following? A. The change to the natural greenhouse effect brought about by variations in solar radiation B. The change to the natural greenhouse effect brought about by changing levels of gases in the atmosphere due to human activities C. The change to the natural greenhouse effect brought about by changing levels of gases in the atmosphere due to variation in the plant growth D. The change to the natural greenhouse effect brought about by changes in the Earth’s orbit and inclination 6. Which of the following statements is true? A. All bodies with temperatures higher than absolute zero emit radiation. B. The higher the temperature of the object, the higher the frequency of electromagnetic radiation it emits. C. The atmosphere and the surface of the Earth show a temperature range between 200 K and 300 K. D. All of the above 7. What would happen if the Earth did not emit infrared radiation? A. It would have no effect on the Earth’s climate. B. The Earth would be slightly warmer. C. The Earth would be too hot to support life. D. The Earth would emit visible light instead. 8. On a warm sunny day, the Sun’s radiation melts very little snow on the slopes of an alpine ski resort. Why? A. Snow, being a solid, is not affected by radiant heat. B. The snow particles are unable to transfer the translational kinetic energy and so cannot warm up. C. Snow is a type of water and due to its high latent heat capacity does not melt easily. D. The white snow reflects up to 90% of the radiant energy of the Sun. 9. Electromagnetic (EM) radiation from the Sun is primarily in the visible light range on the EM spectrum. EM radiation from the Earth is primarily in which portion of the EM spectrum? A. Infrared B. Visible C. Microwave D. Ultraviolet 10. Molecules such as water (H2 O) and carbon dioxide (CO2 ) absorb infrared radiation much better than oxygen (O2 ) or nitrogen (O2 ) molecules. Why is this? A. They are more flexible. B. They are less flexible. C. They are cooler. D. They are hotter.

3.

TOPIC 2 Thermodynamics and climate science 61

2.6 Exercise 2: Short answer questions The Sun gives out energy at a rate of 4.0 × 1026 W and has a surface temperature of 5778 K. Antares, a similar sized star to the Sun, has a surface temperature of 3700 K. At what rate does Antares release energy? b. What colour would you expect Antares to be in comparison to our sun? Soil temperature over a small area is found to be 40.0 °C during the day and is measured to lose energy at a rate of 479 W. At night the rate of energy loss is 330 W. What temperature is the soil at night? a. A violet star has a spectrum with a peak intensity at a wavelength of 4.00 × 10−7 m. Determine the temperature at the surface of this star. b. A red star has a spectrum with a peak intensity at a wavelength of 7.00 × 10−7 m. Determine the temperature at the surface of this star. Two stars have identical diameters. One has a temperature of 5800 K; the other has a temperature of 2900 K. What are the colours of these stars? Which is brighter and by how much? Why does water take much longer to increase in temperature due to exposure to sunlight than the same mass of soil in an identical container? Use the Earth’s energy budget illustrated in figure 2.16 to show that the total amount of energy entering the atmosphere is equal to the total amount of energy leaving it. It has been falsely argued that since the bulk of the greenhouse effect is caused by water, the other greenhouse gases are insignificant. Use the absorption spectra of CO2 to show that this is not the case. What are the similarities and differences between the energy from the Sun that is absorbed by the Earth and the energy the Earth radiates into space? Estimate the quantity of radiant energy that would fall on your body (with an approximate height of 1.5 m and a width of 0.50 m) if you were to lie in the afternoon sun on the beach for 30 minutes. Assume that each square metre is receiving radiant energy at the rate of 1200 W. Explain any additional assumptions that you have made in making your estimate.

1. a.

2. 3.

4. 5. 6. 7. 8. 9.

2.6 Exercise 3: Exam practice questions Question 1 (7 marks) The following graph shows how 𝜆max (the wavelength of the peak of the radiation spectrum) for a range of stars varies with their surface temperatures.

λmax (μm)

0.6

0.4

0.2

0 0

5000

10 000 15 000 Temperature (K)

20 000

Use values from the graph to confirm Wien’s law. b. Estimate the surface temperature of a star whose intensity peaks at a wavelength of: i. 0.4 𝜇m ii. 0.27 𝜇m. a.

62 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

3 marks 1 mark 1 mark

Estimate the peak wavelength for a star with a surface temperature of: i. 15 000 K 1 mark ii. 5550 K. 1 mark Question 2 (5 marks) The specific heat capacity of water is over 4 times that of sand. a. What effect does this have on the heating and cooling of water and sand? 1 mark b. Explain why, on a hot day, sand is too hot to stand on in bare feet while the water in the sea can be too cold for some people. 2 marks c. Why is the temperature of the sand of an inland desert almost always greater than that of the sand on a beach at the same latitude? 2 marks c.

Question 3 (3 marks) When the Sun is directly overhead, each square metre of the Earth’s surface receives 1368 J of radiant energy each second. On a typical winter’s weekday, Victoria consumes about 4500 MW of electrical energy. What area of solar collectors (of 15% efficiency) would be needed to provide energy at the same rate while the Sun is directly overhead? Question 4 (2 marks) In the tropical regions of the Earth, more radiant energy is received from space than is lost. At the poles, more radiant energy is lost than is received. This would suggest that average temperatures in the tropics should be continually increasing while the average temperatures at the poles should be continually decreasing. Explain why this doesn’t happen. Question 5 (6 marks) The following table compares the specific heat capacities as well as reflection and absorption properties of water and dry sand. The density of water is 1000 kg m−3 while the density of dry sand is 2400 kg m−3 . Property Specific heat capacity (J kg

−1

−1

K )

Reflectivity (%) Depth to which 50% of the radiant energy is absorbed

Water

Dry sand

4200

820

3

15

11 m

1 mm

When the Sun is directly overhead, each square metre of the Earth’s surface receives radiant energy at the rate of 1368 W. Assuming all of the energy that is not reflected is absorbed, what average temperature increase would be expected during a period of 1 hour in: a. water to a depth of 11 m 3 marks b. sand to a depth of 1 mm? 3 marks

2.6 Exercise 4: studyON topic test Fully worked solutions and sample responses are available in your digital formats.

Test maker Create unique tests and exams from our extensive range of questions, including practice exam questions. Access the assignments section in learnON to begin creating and assigning assessments to students.

TOPIC 2 Thermodynamics and climate science 63

UNIT 1 | AREA OF STUDY 1 REVIEW

AREA OF STUDY 1 How can thermal effects be explained? OUTCOME 1 Apply thermodynamic principles to analyse, interpret and explain changes in thermal energy in selected contexts and describe the environmental impact of human activities with reference to thermal effects and climate science concepts.

PRACTICE EXAMINATION STRUCTURE OF PRACTICE EXAMINATION Section

Number of questions

Number of marks

A

20

20

B

6

20 Total

40

Duration: 50 minutes Information:

• • •

This practice examination consists of two parts. You must answer all question sections. Pens, pencils, highlighters, erasers, rulers and a scientific calculator are permitted. You may use the VCAA Physics formula sheet for this task.

Resources Weblink VCAA Physics formula sheet

Additional practice examination information Specific heat capacities

Water Aluminium

Specific latent heat of fusion

Water

Specific latent heat of vaporisation

Water

4.2 × 103 J kg–1 K–1

0.92 × 103 J kg–1 K–1

3.3 × 105 J kg–1 2.3 × 106 J kg–1

SECTION A — Multiple choice questions All correct answers are worth 1 mark each; an incorrect answer is worth 0. 1. The average kinetic energy of atoms and molecules are given by which of the following? A. Their temperature B. The heat input into them C. Their work output D. The heat output from them

64 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

2. The freezing point of water could be represented by which of the following? A. 0 K B. 0 °F C. 273.15 °C D. 273.15 K 3. A system consisting of a rubber balloon and the air inside has experienced an increase in its internal energy. Which of the following could be assumed? A. Heat input into the system is less than the work done by the system. B. Heat output from the system is greater than the work done on the system. C. Heat input into the system is greater than the work done by the system. D. Heat output from the system is greater than the work done by the system. 4. Two bodies, X and Y, are in thermal equilibrium with each other. One of the bodies, X, is brought into contact with a third body, Z, and they are found to also be in thermal equilibrium. Which of the following statements is correct? A. X and Z must have the same mass as they are in thermal equilibrium with each other. B. Y and Z must have the same specific heat capacity because they would be in thermal equilibrium. C. Y and Z would have the same temperature because they would be in thermal equilibrium. D. There is insufficient information about specific heat capacities to make any correct conclusion. 5. Which of the following is correct when comparing particles in a gas with those in a liquid? A. They are closer together and more energetic in a gas. B. They are further apart and less energetic in a gas. C. They are closer together and less energetic in a liquid. D. They are further apart and more energetic in a liquid. 6. Water and aluminium, each with a mass of 1 kilogram, are placed in their own insulated containers. Initially, they are both at the same temperature and are each heated by 1 kJ. Assuming the heat is evenly distributed across each mass, which of the following statement is correct about their final temperatures? A. The final temperatures of both water and aluminium are the same. B. The final temperature of the water is lower than that of the aluminium. C. The final temperature of the water is higher than that of the aluminium. D. There is insufficient information to conclude about their final aluminium. 7. The latent heat of fusion, denoted by the symbol L, is the amount of energy required to transform the state of 1 kilogram of substance from solid to liquid, and vice versa. Which of the following could be used as a unit for L? A. J–1 kg B. J kg C. J m–3 D. J kg–1 8. Meena would like to raise the temperature of a 2.2 kilogram bar of aluminium from 22 °C to 44 °C. What is the amount of energy, in kJ, required to achieve this (assuming 100% efficiency)? A. 4.45 kJ B. 8.90 kJ C. 44.5 kJ D. 89.0 kJ 9. An electric kettle supplies 67 kJ of thermal energy to 750 mL of water at an initial temperature of 18 °C. Determine the final temperature of the water assuming that all the thermal energy is transferred into the water. A. 21.3 °C B. 34 °C C. 37.3 °C D. 39.3 °C

UNIT 1 Area of Study 1 Review 65

10. How much energy, in MJ, is liberated when 1.3 kilograms of steam condenses into water at 100 °C? A. 0.43 MJ B. 2.3 MJ C. 3 MJ D. 4.3 MJ 11. At the Mawson Antarctic Research station, an electric heater supplies 37 GJ of energy to melt ice to produce water for consumption by the scientists there. How much water, in kg, at 0 °C would be produced by this amount of energy? A. 1.12 × 104 kg B. 1.12 × 105 kg

C. 1.12 × 106 kg

D. 1.12 × 107 kg 12. An electric hotplate with a temperature of 750 K emits 1200 W of thermal energy. What is the expected power emitted if the filament temperature is doubled to 1500 K? A. 12 000 W B. 19 200 W C. 24 000 W D. 29 200 W 13. Thermite is a reaction between aluminium powder and iron oxide, which generates temperatures of about 2200 °C. What is the wavelength of the radiation emitted by thermite? A. 1.32 × 10−5 m B. 1.17 × 10−5 m

C. 1.32 × 10−6 m D. 1.17 × 10−6 m

14. A pyrometer is a device that senses temperature from the radiation emitted from a surface. One such device detected a peak radiation of 6.5 μm from an object. What is the surface temperature of the object? A. 173 °C B. 273 °C C. 346 °C D. 446 °C 15. Which grid energy supply does not contribute to the enhanced greenhouse effect? A. Coal powered electric generator B. Natural gas powered electric generator C. Nuclear powered electric generator D. Diesel powered electric generator 16. Which group of atmospheric gases contain only greenhouse gases? A. Carbon dioxide, water vapour, methane, argon B. Oxygen, nitrogen, methane, carbon dioxide C. Nitrous oxide, ozone, chlorofluorocarbon D. Methane, water vapour, carbon dioxide, neon 17. Albedo, which refers to the reflection of solar radiation, plays an important role in the Earth’s climatic system. Which of the following has the lowest albedo? A. Clouds B. Sea surface C. Polar ice caps D. Snow-capped mountains

66 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

18. Which of the following is an example of a positive feedback mechanism that enhances the effect of global warming on climate change? A. Rising ocean temperatures increase evaporation, which condenses into clouds to reflect the incoming solar radiation. B. Rising ocean temperatures increase evaporation, which increases the amount of water vapour, a greenhouse gas. C. Rising ocean temperatures disrupts the natural flow of ocean currents to transfer heat from the tropical regions to the polar regions, resulting in a new ice age. D. Rising ocean temperatures increase evaporation, which creates a humid greenhouse-like condition, which absorbs even more solar radiation. 19. What is the mechanism for heat transfer in convection? A. Vibrational transfer of kinetic energy B. Emission of electromagnetic radiation C. More energetic particles colliding with less energetic particles D. Movement of fluid due to differences in buoyancy 20. Greenhouse gases such as methane (CH4 ) and carbon dioxide (CO2 ) absorb infrared radiation emitted from the surface of the Earth, unlike gases such as oxygen (O2 ) or nitrogen (N2 ). What is this absorption of infrared radiation due to? A. These compounds contain a carbon atom, hence the term ‘carbon pollution’. B. Molecules with three or more atoms bend and stretch, and can better absorb infrared radiation. C. There is a higher number of atoms in those molecules. D. Compounds absorb infra-red radiation better than pure elements such as oxygen or nitrogen.

SECTION B — Short answer questions Question 1 (5 marks) A small piece of beef and a small piece of chicken are placed in an oven, each with a meat thermometer stuck into it to measure the temperature at its centre. After 30 minutes in the oven the thermometer in the chicken indicated the temperature to be 100 °C. After a further 30 minutes, the thermometer on the beef indicated the temperature at its centre to also be 100 °C. They remained in the oven to cook for a further 60 minutes. During this period, the thermometers steadily indicated 100 °C. a. The piece of beef and the piece of chicken could be said to be at thermal equilibrium after this time. By referring to the kinetic theory of matter, explain what is meant by ‘thermal equilibrium’. 2 marks b. Discuss why it may be possible to estimate the temperature of the air in the oven near these two masses of meat, even if the oven does not have a thermometer. Hence, give an estimate of the temperature of the air in the oven near these two masses of meat. 3 marks Question 2 (3 marks) Sue and Mel are preparing for a science fair at their school. They inflated a giant rubber balloon with air and attached it to a window where it was exposed to sunlight. After an hour, Sue noticed that the balloon was bigger. She said to Mel, ‘The internal energy of the air in the balloon is lower than when we attached it to a window. This is because the air inside has done work to expand the balloon’. a. Mel disagreed with Sue. What explanation might Mel give to show that Sue is incorrect? 2 marks b. What would be necessary for the internal energy of the air in the balloon to remain the same? 1 mark Question 3 (4 marks) Anton is carrying out an experiment to determine the quantity of aluminium rivets with an initial temperature of 23 °C required to cool 250 grams of boiling water at 100 °C inside an insulated container. Assuming the experiment was carried out with negligible heat loss, both the aluminium rivets and the water will reach thermal equilibrium at 90 °C.

UNIT 1 Area of Study 1 Review 67

a. Calculate the magnitude of heat lost by the water to the aluminium as it cools from 100 °C to the equilibrium temperature of 90 °C. 2 marks b. The amount of heat lost by the water is the heat gained by the aluminium. Determine the mass of aluminium rivets as it warmed up from the initial temperature of 23 °C to the equilibrium temperature of 90 °C. 2 marks Question 4 (2 marks) Kym and Shan are observing light emitted from different light-emitting diodes (LED). One of the LED emitted light at twice the electromagnetic radiation frequency of another LED. Kym said ‘According to Wien’s Law, if the electromagnetic radiation frequency of one LED is twice that of the other, it would also be at twice the absolute temperature of the other’. Shan disagrees. What explanation might Shan offer to show that Kym is incorrect. 2 marks Question 5 (2 marks) Habib’s electric barbeque radiates 2250 W of energy when the heating element is at a temperature of 850 °C. What is the expected power output when the heating element is at a temperature of 900 °C? 2 marks Question 6 (4 marks) The following diagram shows a part of the Earth’s energy budget, which is balanced. All numerical values are in W m–2 . Thermal radiation into space

Sun

Directly radiated from Earth’s surface

D

235 Solar radiation absorbed by Earth

C

Energy in the atmosphere

Non-greenhouse gas absorption 102

67 Greenhouse gas absorption 350

A

B

492

Earth’s land and ocean surface

a. Considering the total incoming solar radiation and the amount absorbed by the Earth’s atmosphere, how much is absorbed by the Earth’s land and ocean surface (box A)? 1 mark b. Considering that the energy budget is balanced, how much energy is re-radiated downward back to the Earth’s land and ocean surface from the atmosphere (box B)? 1 mark c. How much outgoing radiation from the Earth’s land and ocean surface is directly radiated into outer space (box C)? 1 mark d. How much thermal radiation from the Earth’s atmosphere goes into outer space (box D)?

68 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

1 mark

PRACTICE SCHOOL-ASSESSED COURSEWORK ASSESSMENT TASK — A REPORT ON A SELECTED SCIENTIFIC PHENOMENON In this task, you will be required to report on the phenomenon of the greenhouse effect, linking this to thermodynamic principles. • This practice SAC requires you to write a report; a structure set of questions is supplied to assist you to write your report. • You may use the VCAA Physics formula sheet and a scientific calculator for this task. • You may conduct research before commencing your write-up to assist you in this task. Total time: 50 minutes (5 minutes reading, 45 minutes writing) Total marks: 30 marks

A WORLDWIDE CRISIS — THE IMPACT OF THE GREENHOUSE EFFECT Humans have a dramatic impact on the environment and unfortunately it is often a detrimental one. From deforestation to pollution, the environment is being continually affected, creating huge scientific and ecological problems. Since the Industrial Revolution, human impact on increasing greenhouse gases has been particularly evident, and has lead to global warming and damage to ecosystems around the world. In Australia, the Great Barrier Reef has been significantly affected by coral bleaching caused by this global warming. Write a report with reference to the following concepts. You should use subheadings to clearly frame your response. 1. Identify and describe the types of electromagnetic radiation emitted from the Sun. 2. Explain the roles of conduction, convection and radiation in moving heat around in Earth’s mantle and surface and the Earth’s atmosphere. Identify which of these is most important in contributing to the greenhouse effect. 3. Provide background information on the greenhouse effect, with a clear link to global warming using thermodynamic principles. 4. Explain how greenhouse gases absorb and re-emit infrared radiation, and describe why different greenhouse gases are involved. 5. Clearly describe the evidence of the effect of human activity in creating an enhanced greenhouse effect. 6. Describe how you would collect evidence on the enhanced greenhouse effect, explaining how you can ensure reliability and validity while minimising uncertainty. 7. Apply thermodynamic principles to investigate one of the following issues (using evidence and data) related to the impacts of human activity on the enhanced greenhouse effect. In your response, you need to show a clear link to theory and describe how it may have an impact on the greenhouse effect. You should also suggest solutions that can help minimise the enhanced greenhouse effect for your given issue. • Proportion of national energy use due to heating and cooling of homes • Comparison of the operation and efficiencies of domestic heating and cooling systems: heat pumps, resistive heaters, reverse-cycle air conditioners, evaporative coolers, solar hot water systems and/or electrical resistive hot water systems • Possibility of homes being built that do not require any active heating or cooling at all • Use of thermal imaging and infra-red thermography in locating heating losses in buildings and/or system malfunctions; cost savings implications • Determination of the energy ratings of home appliances and fittings: insulation, double glazing, window size, light bulbs, and/or electrical gadgets, appliances or machines Cooking alternatives: appliance options (microwave, convection, induction), fuel options (gas, • electricity, solar, fossil fuel) • Automobile efficiencies: fuel options (diesel petrol, LPG and electric), air delivery options (naturally aspirated, supercharged and turbocharged) and fuel delivery options (common rail, direct injection and fuel injection)

Resources Digital document School-assessed coursework (doc-32273)

UNIT 1 Area of Study 1 Review 69

AREA OF STUDY 2 HOW DO ELECTRIC CIRCUITS WORK?

3

Concepts used to model electricity 3.1 Overview Numerous videos and interactivities are available just where you need them, at the point of learning, in your digital formats, learnON and eBookPLUS at www.jacplus.com.au.

3.1.1 Introduction What would your world be like without FIGURE 3.1 Electricity is an integral part of modern life. Consider electricity? Would your mobile phone how the processes of transfer and transformation of energy occur work or would you be able find your in electric circuits. way around your house on a dark night? The transfer of electrical energy into other forms of energy helps us in many ways. The electricity that is transferred to heat in order to warm us on cold nights or the electricity that is transferred to move an electric bicycle is current electricity that has been created by the movement of electric charge. Not all electric charge moves or flows. Static electricity can be created by the removal of electrons from a material. The discharge of static electricity, as occurs during a lightning storm, can be dramatic and sometimes dangerous. In an electric circuit, electric charges move in an organised way. A battery or other energy source separate electric charge and cause a current to flow. In a simple circuit connected to a battery, the overall movement of the electric charges is in one direction. Electric charges move in some materials better than others, particularly in metals that are used as conductors. You should already be able to recognise and connect simple circuits such as those containing batteries and globes. At the end of this topic you should be able to describe the concepts of electric charge and the effects of current and voltage in the transformation of electrical energy into other forms of energy.

70 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

3.1.2 What you will learn KEY KNOWLEDGE After completing this topic you will be able to: • apply concepts of charge (Q), electric current (I), potential difference (V), energy (E) and power (P), in electric circuits • explore different analogies used to describe electric current and potential difference • investigate and analyse theoretically and practically electric circuits using the relationships: I=

Q t

, V=

E Q

, P=

E t

= VI

• justify the use of selected meters (ammeter, voltmeter, multimeter) in circuits • model resistance in series and parallel circuits using – current versus potential difference (I–V) graphs – resistance as the potential difference to current ratio, including R = constant for ohmic devices. Source: VCE Physics Study Design (2016–2021) extracts © VCAA; reproduced by permission.

PRACTICAL WORK AND INVESTIGATIONS Practical work is a central component of learning and assessment. Experiments and investigations, supported by a Practical investigation logbook and Teacher-led videos, are included in this topic to provide opportunities to undertake investigations and communicate findings.

Resources Digital documents Key science skills — VCE Units 1–4 (doc-31856) Key terms glossary (doc-32178) Practical investigation logbook (doc-32179)

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0028).

3.2 Electric circuits KEY CONCEPTS • Apply concepts of charge (Q), electric current (I), potential difference (V), energy (E) and power (P), in electric circuits. • Explore different analogies used to describe electric current and potential difference.

3.2.1 Fundamentals of electricity Electric charge (in terms of the basic structure of matter) Electric charge is a basic property of matter. Matter is all the substance that surrounds us — solid, liquid and gas. You have probably experienced that small electric shock that happens when you touch a metal rail after walking across carpet. This type of phenomenon has been observed for thousands of years. Objects such as glass, gemstones, tree resin and amber can become ‘electrified’ by friction when they are rubbed with materials such as animal fur and fabrics to produce a spark. Indeed, the words ‘electric’ and ‘electricity’ are derived from the Greek word for amber, electron.

TOPIC 3 Concepts used to model electricity 71

Experiments in the early 1700s showed that: • both the object and the material became ‘electrified’ or ‘charged’ • when charged objects were brought near each other, for some objects there was a force of attraction, while for others it was a force of repulsion. FIGURE 3.2 (a) Two positively charged objects repel each other. (b) Oppositely charged objects attract each other. (a)

(b)

It was quickly observed that like-charged objects repelled each other while unlike-charged objects attracted each other. Charged objects exert a force on each other. The force between two stationary charged objects is called an electrostatic force. The direction of the electrostatic forces between electric charges act such that: • two positive charges repel one another • two negative charges repel one another • a positive charge and a negative charge attract one another. This is summarised as: like charges repel; unlike charges attract. Note: Neutral objects carry an equal amount of positive and negative charge and do not attract or repel other neutral objects. Electrostatic forces can be observed in the production of static electricity, in which electric charges become imbalanced. When a neutrally charged glass rod is rubbed with a neutrally charged silk cloth, electrons are transferred from the rod to the cloth. As shown in figure 3.3, the rod therefore becomes more positively charged (from losing electrons) and the cloth becomes more negatively charged. Therefore, they become attracted to one another as they have unlike charges. Experiments like these demonstrate that electric charge can be moved, while being neither created nor destroyed. Electric charge is conserved. FIGURE 3.3 Electric charge is conserved. Glass rod

+ +

− −

Positively charged glass rod

+ +

− +

− +

− +

− Silk cloth

72 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

− + −

Negatively charged silk cloth

Possible explanations for these observations abounded but further experiments could not determine the correct theory. In the mid-1700s Benjamin Franklin suggested that positively charged fluid was transferred from the silk to the glass leaving the silk negative and the glass positive. Although a negatively charged fluid was equally plausible, Franklin’s status as an eminent scientist ensured that the existence of a positive fluid was accepted. All developments in electrical engineering for the next 150 years were based on this convention. By 1897, when J.J. Thomson demonstrated that the negatively charged electron was responsible for electricity, it was too late to change the convention and all the associated labelling of meters.

THE RESEARCH OF BENJAMIN FRANKLIN Benjamin Franklin (1706–90), a US political leader, inventor and scientist, introduced the concepts of positive and negative electricity. His research on lightning, including discharges obtained with a kite, helped to establish its electrical nature. In 1752, Benjamin Franklin conducted the kite and key electricity experiment during a lightning storm (see figure 3.4). He advocated the use of lightning rods as protective devices for buildings. Franklin was almost killed one day when he was trying to electrocute a turkey with a condenser, a device used to store charge. (Members of genteel society at the time thought it was acceptable to see how big an animal they could electrocute.)

FIGURE 3.4 Benjamin Franklin’s kite and key electricity experiment

All matter is made up of atoms. Atoms in turn are made up of smaller particles called protons, neutrons and electrons. Protons and neutrons are found in the nucleus while the electrons move in well-defined regions called orbits or shells. FIGURE 3.5 (a) The structure of an atom (electron shell) (b) An atom showing orbits and shells (a)

(b) Proton (positive) Electron cloud

Nucleus

Proton Neutron Electron Neutron (neutral)

Electron (negative)

Protons and electrons possess a characteristic known as electric charge; because of their electric charge, these particles exert electric force on each other. Protons carry a positive charge and electrons carry a negative charge. The positive charge on a proton is equal in magnitude to the negative charge on an electron, meaning that the negative and positive charges neutralise each other. Neutrons have no electric charge; that is, they are uncharged or neutral. TOPIC 3 Concepts used to model electricity 73

Conductors and insulators Some materials are made up of atoms which are fixed together in such a way that all their electrons are bound tightly to the nucleus and are not free to travel through the material. Such substances are termed electric insulators. An insulator is a material that contains no charge carriers. If an insulator is given an electrostatic charge at a particular area on the insulator, the charge will remain at that area. Common examples of insulators include dry air, glass, plastics, rubber and ceramics. A conductor is a material that contains charge carriers; that is, charged particles can move and travel freely through the material. Examples include: • metals — materials whose outer electrons are so loosely bound to the nucleus that they are effectively free to move easily through the material. Usually this movement is random but when the free electrons are forced to flow in one direction, a current can be created • salt solutions, whose charged particles, such as ions, are free to move through the solution. Note: Pure water is not a conductor as it does not contain any ions. Another important group of materials are semiconductors, which allow electrons to move freely under certain conditions. Silicon is commonly used in the construction of semiconducting photovoltaic cells. When the Sun shines on these cells in solar panels they generate electricity.

Resources Digital document Investigation 3.1 The Van de Graaff generator (doc-31918)

3.2.2 Electric circuits An electric circuit is a closed loop of moving electric charge. For a circuit to be useful, it must contain a load. A load is a device where electrical energy is converted into other forms of energy to perform a task such as heating something, or providing light, sound or mechanical energy from a motor. This happens in devices such as toasters, speakers, lamps and motors. A load is anything that is doing a job. To convert electrical energy into other forms of energy, loads resist the flow of current through them. This transforms some or all the potential energy stored in the current. A simple circuit can be made up of: • a source of energy, such as a battery • conductors, such as wires • some kind of load, where energy is transformed from electrical energy into other forms such as heat, sound, light and movement by devices • a switch that stops or allows the flow of electricity in the circuit. In figure 3.6, note: • For the globe to light up there must be a complete FIGURE 3.6 A simple electric circuit conducting path between the terminals of the battery. The switch must be closed. • The battery is necessary for a current to flow around the circuit. The ability of the battery to cause a current to flow is often referred to as its voltage. • The battery has two terminals, marked positive and negative. • The light globe resists the flow of the current. As a result of this resistance, the current causes the light globe to heat up to such an extent that it gives off light. • The wires connecting the light globe to the battery do not heat up.

74 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

The symbols shown in figure 3.7a are used when representing electric circuits in diagrams. Therefore, the circuit shown in figure 3.6 can be represented more simply as shown in figure 3.7b. FIGURE 3.7 (a) Symbols for circuit components (b) Diagram of the simple electric circuit (a)

(b) Connecting wire

Light globe

+ – Battery

Switch

3.2 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. After a plastic pen is rubbed with a piece of wool it can be used to attract small pieces of paper. Describe what has happened in terms of electric charge. 2. After rubbing a balloon on your clean dry hair, the balloon should try to stick to your hair when you try to remove it. Explain why this occurs. 3. If you separately rub two balloons on your hair and then hold them near each other what will happen? Explain why this occurs. 4. After walking across a nylon carpet in woollen socks and then touching a metal doorknob it is possible to get an electric shock. Explain why this occurs. 5. After rubbing a balloon in her hair, Chris brought it very close to an aluminium can lying on a flat table. When she slowly moved the balloon away from the can it started to roll and follow the balloon. Describe why this happened. 6. Imagine you are an electron. Describe your journey around the closed circuit of a torch, beginning at the negative terminal of a cell. 7. A doorbell connected to a battery comprises a button at the door, the bell and wires. The bell only sounds after the button is pushed. Why doesn’t the bell always sound?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

TOPIC 3 Concepts used to model electricity 75

3.3 Current KEY CONCEPTS • Apply concepts of charge (Q), electric current (I), potential difference (V), energy (E) and power (P) in electric circuits. • Justify the use of selected meters (ammeter, voltmeter, multimeter) in circuits. • Explore different analogies used to describe electric current and potential difference.

3.3.1 Defining current Electric current is the movement of charged particles from one place to another, and is measured as the rate of flow of charge. The charged particles may be electrons in a metal conductor or ions in a salt solution. Charged particles that move in a conductor can also be referred to as charge carriers. There are many examples of electric currents. Lightning strikes are examples of large currents. Nerve impulses that control muscle movement are examples of small currents. Charge flows in household and automotive electrical devices such as light globes and heaters. Both positive and negative charges flow in cells, in batteries and in the ionised gases of fluorescent lights. The solar wind is an enormous flow of protons, electrons and ions being blasted away from the Sun. Not all moving charges constitute a current. There must be a net movement of charge in one direction for a current to exist. In a piece of metal conductor, electrons are constantly moving in random directions. Until there is a net movement of charge in one direction, as happens when a metal wire is connected in a closed circuit with a power source (such as a battery) and a load (such as a light bulb), there is no current. As water flows down a river there are millions of coulombs of charge moving with the water molecules, but there is no electrical current in this case because equal numbers of positive and negative charge are moving in the same direction. For there to be a current in a circuit there must be a complete conducting pathway around the circuit and a device to make the charged particles move. When the switch in the circuit is open, the pathway is broken and the current stops almost immediately. Electric current is a measure of the rate of flow of charge around a circuit. It can be expressed as: I=

Q t

Where: I is the current, in amperes Q is the quantity of charge flowing past a point in the circuit, in coulombs t is the time interval, in seconds. The unit of current is the ampere (A). It is named in honour of the French physicist André-Marie Ampère (1775–1836). One ampere is the current in a conductor when 1 coulomb of charge passes a point in the conductor every second. The unit for charge is the coulomb (C), named after the French physicist Charles-Augustin de Coulomb (1736–1806). One coulomb of charge is equal to the amount of charge carried by 6.24 × 1018 electrons. The charge carried by a single electron is equal to −1.602 × 10−19 C. The −1.602 × 10−19 charge possessed by an electron is the smallest free charge possible. All other charges are whole-number multiples of this value. This so-called elementary charge is equal in magnitude to the charge of a proton. The charge of an electron is negative, whereas the charge of a proton is positive. 76 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

SMALL CHARGES AND QUARKS Charges smaller than that carried by the electron are understood to exist, but they are not free to move as a current. Particles such as neutrons and protons are composed of quarks, with one-third of the charge of an electron, but these are never found alone.

SAMPLE PROBLEM 1

What is the current in a conductor if 10 coulombs of charge pass a point in 5.0 seconds? Teacher-led video: SP1 (tlvd-0010) THINK

WRITE

1.

Current is the rate at which charge, Q flows in the circuit. Give values for Q and t.

Q = 10 C, t = 5.0 s

2.

Substitute values for Q and t into the formula Q I= . t

I=

Current is measured in ampere (A), where: 1 A = 1 C s−1 . 4. State the solution. 3.

Q t 10 C = 5.0 s I = 2.0 C s−1 I = 2.0 A The current in the conductor is 2.0 A.

PRACTICE PROBLEM 1 What is the current passing through a conductor if 15 coulombs of charge pass a point in 3.0 seconds?

SAMPLE PROBLEM 2

How much charge passes through a load if a current of 3.0 A flows for 5 minutes and 20 seconds? Teacher-led video: SP2 (tlvd-0011) THINK

To find the charge, Q, passing through the circuit Q transpose the formula I = making Q the subject. t 2. Give values for I and t. Note: Be sure to convert the time to seconds. 3. Substitute values for I and t into the formula. 1.

Convert to scientific notation, to 2 significant figures. 5. State the solution.

4.

WRITE

Q t Q = It I = 3.0 A, t = 5 × 60 + 20 = 320 s I=

Q = It = 3.0 A × 320 s Q = 960 C Q = 9.6 × 102 C 9.6 × 102 C charge passes through the load. (Note: As the solution is quite small in magnitude, leaving the solution as 960 C instead of in scientific notation is acceptable).

TOPIC 3 Concepts used to model electricity 77

PRACTICE PROBLEM 2 For how long must a current of 2.5 amperes flow to make 7.5 coulombs of charge pass a point in a circuit? In real circuits, currents of the order of 10−3 A are common. To describe these currents, the milliampere (mA) is used. One milliampere is equal to 1 × 10−3 amperes. To convert from amperes to milliamperes, multiply by 1000 or by 103 . To convert from milliamperes to amperes, divide by 1000 or multiply by 10−3 . SAMPLE PROBLEM 3

Convert 450 mA to amperes. Teacher-led video: SP3 (tlvd-0012) THINK

WRITE

1.

To convert mA to A divide by 1000 (or multiply by 10–3 ).

450 mA = 0.450 A 1000

2.

State the solution.

450 mA is equal to 0.450 A.

PRACTICE PROBLEM 3 Convert 280 mA to amperes.

3.3.2 Describing current direction When the battery was invented by Alessandro Volta in 1800, it was accepted that electric current was the movement of positive charge. It was assumed that positive charges left the positive terminal of the battery and travelled through a conductor to the negative terminal. This is called conventional current. In reality, the charge carriers in a metal conductor are electrons moving from the negative terminal towards the positive terminal of the battery. The effect is essentially the same as positive charges moving in the opposite direction. When dealing with the mechanisms for the movement of electrons, the term ‘electron current’ is used. FIGURE 3.8 Conventional current direction

+

FIGURE 3.9 Electron current direction

+ −

78 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

−

Direct current (DC) refers to circuits where the net flow of charge is in one direction only. The current provided by a battery is direct current, which usually flows at a steady rate. Alternating current (AC) refers to circuits where the charge carriers change direction periodically, moving backwards and forwards. The electricity obtained from household power points is alternating current. FIGURE 3.10 Electron current direction: (a) direct current and (b) alternating current (a)

Direct current (DC) I

Battery

(b)

Resistor

Alternating current (AC) I

Resistor

Battery

I

I

3.3.3 Measuring electric current Electric current is measured with a device called an ammeter. This must be placed directly in the circuit so that all the charges being measured pass through it. This is known as placing the ammeter in series with the circuit. Ammeters are designed so they do not significantly affect the size of the current in the circuit, their resistance to the flow of current being negligible. FIGURE 3.12 A needle-deflection ammeter

FIGURE 3.11 The circuit diagram symbol for an ammeter

A

FIGURE 3.13 A ammeter in series Ammeter +

A

−

DC load

+ DC source −

Whereas some school laboratories might use needle-deflection meters, most now use digital multimeters. Digital meters can measure voltage drop and resistance as well as current. Each quantity has a few settings to allow measurement of a large range of values. Labels on multimeters may vary but those given below are most common. TOPIC 3 Concepts used to model electricity 79

FIGURE 3.14 A digital multimeter, which can measure current, voltage drop and resistance

FIGURE 3.15 Measuring current and voltage drop Ammeter A

Battery

Resistor

V

Voltmeter

If you are using digital multimeter, the following instructions generally apply. • The black or common socket, labelled ‘COM’, is connected to the part of the circuit that is closer to the black or negative terminal of the power supply. The red socket, labelled ‘VΩmA’, is used for measuring small currents and is connected to the part of • the circuit that is closer to the red or positive terminal of the power supply. • The red socket, labelled ‘10A MAX’ or similar, is used for measuring large currents — see warning below. • The dial has a few settings, first choose the setting for current, labelled ‘A’, with the largest value. If you want more accuracy in your measurement, turn the dial to a smaller setting. • If the display shows just the digit ‘1’, the current you are trying to measure exceeds the range of that setting and you need to go to a higher setting. WARNING: While for most quantities, multimeters are quite tolerant of values beyond a chosen setting, care must be taken when measuring current. Multimeters have a fuse that can blow if the current exceeds the rated value. For this reason, they have two red sockets. One socket is exclusively for use when measuring currents in the range 200 mA to 10 A. This is labelled ‘10A MAX’. (Some multimeters may be able to measure up to 20 A.) The other red socket is for currents less than 200 mA as well as the other quantities of voltage and resistance.

3.3.4 Modelling an electric circuit One way to understand something we can’t see is to use a model. A good scientific model uses objects and phenomena that we can see and understand or have experienced to explain things that we cannot see.

The hydraulic model of current Most circuits have metal conductors, which means that the charge carriers will be electrons. Metal conductors can be considered to be a three-dimensional arrangement of atoms that have one or more of their electrons loosely bound. These electrons are so loosely bound that they tend to drift easily among the atoms. Metals are good conductors of both heat and electricity because of the ease with which these electrons are able to move, transferring energy as they go. Diagrammatically, the atoms are represented as positive ions (atoms that have lost an electron and have a net positive charge) in a ‘sea’ of free electrons. When the ends of a conductor are connected to a battery, the free electrons drift towards the positive terminal. The electrons are attracted by the positive terminal and indeed accelerate, but constantly bump into atoms so on average they just drift along (see figure 3.16). The flow of electrons through a metallic conductor can also be modelled by the flow of water through a pipe (see figure 3.17).

80 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 3.16 The motion of free electrons through a metal. Note: Only two of the free electrons have been shown.

− − − − − −

+ + + + + + + + + − + + + + + + − + + +

+ + + + + +

+ + + + + +

+ + + + + +

+ + + + + +

FIGURE 3.17 The hydraulic model for current flow. One cupful of water in one end of the pipe means one cupful out the other end.

+ + + + + +

Electrons cannot be destroyed, nor, in a closed circuit, can they build up at a point. Therefore, if electrons are forced into one end of a conductor, an equal number will be forced out the other end. This is rather like pouring a cupful of water into one end of a full pipe. It forces a cupful of water to come out the other end. Note that when water is put in one end of a pipe it is not the same water that comes out the other end, because the pipe was already full of water. Other models are sometimes used. For example, the bicycle chain model. In this model the chain represents the circuit and the links in the chain represent electrons. When the pedals are turned the chain moves and energy is transferred to the rear of the bicycle to move the rear wheel. The moving chain represents the movement of electrons around the circuit. Note that the transformation of energy from the pedals to the rear wheel is virtually instantaneous. The energy transfer from the pedals does not depend on particular chain links travelling from the pedals to the wheel. Similarly, the energy transfer in an electric circuit does not depend on particular electrons travelling to the load. Overall the transfer of electrical energy is faster than the movement of electrons in the conducting wires.

Resources Interactivity

The hydraulic model of current (int-0053)

Video eLesson The hydraulic model of current (eles-0029) Weblink

Calculating an electron’s drift velocity

3.3.5 How rapidly do electrons travel through a conductor? The speed of electrons through the conductor depends on the cross-sectional area of the conductor, the number of electrons that are free to move, the electron charge and the size of the current. For example, if a current of 10 A passes through a copper wire of cross-sectional area 1 mm2 , the electron speed is 0.16 mm s−1 or 1.6 × 10−4 m s−1 . This speed is known as the drift velocity (since the electrons are drifting through the wire), and is quite small. SAMPLE PROBLEM 4

How long will it take an electron to travel from a car’s battery to a rear light globe if it has a drift velocity of 1.0 × 10−4 m s−1 and there is 2.5 metres of metal to pass through? (Electrons travel from the negative terminal of the battery through the car body towards the circuit elements.) Teacher-led video: SP4 (tlvd-0013)

TOPIC 3 Concepts used to model electricity 81

THINK

The drift velocity, v, equals the distance travelled, d, divided by the time taken to travel the distance, t. d 2. Transpose v = to make t the subject. t

1.

3.

Substitute the known values into the formula and solve for t.

4.

State the solution.

WRITE

v=

d t

vt = d d t= v v = 1.0 × 10−4 m s−1 , d = 2.5 m d t= v 2.5 m = 1.0 × 10−4 m s−1 = 2.5 × 104 s It would take 25 000 seconds, which is more than 7 hours.

PRACTICE PROBLEM 4 How long will it take an electron to travel to a headset from a console if it has a drift velocity of 7.4 × 10−5 m s−1 and there is 1.2 metres of copper wire to pass through?

3.3 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. State the difference between conventional current and electron current. 2. What is the difference between direct current and alternating current? 3. A steady direct current of 2.5 A flows in a wire connected to a battery for 15 seconds. How much charge enters or leaves the battery in this time? 4. Convert 45 mA to amperes. 5. Convert 2.3 × 10−4 A to milliamperes. 6. Convert 450 𝜇A to amperes (1𝜇A = 1 × 10−6 A). 7. Is current used up in a light globe? Explain your answer. 8. A car light globe has a current of 3.5 A flowing through it. How much charge passes through it in 20 minutes? 9. What is the current flowing through an extension cord if 15 C of charge passes through it in 50 seconds? 10. The drift velocity is directly proportional to the current in the conductor. If electrons have a drift velocity of 1.6 × 10−4 ms−1 for a current of 10 A in a certain conductor, what would be their velocity if the current was 5.0 A?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

82 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

3.4 Voltage KEY CONCEPTS • Apply concepts of charge (Q), electric current (I), potential difference (V), energy (E) and power (P), in electric circuits. • Justify the use of selected meters (ammeter, voltmeter, multimeter) in circuits.

A battery is a source of energy that enables electrons to move around a circuit. Inside the battery a chemical reaction separates electrons, leaving one terminal short of electrons and therefore with an excess of positive charge. The other battery terminal has an excess of electrons and so is the negative terminal. Batteries are rated by their voltage (V). This is a measure of the amount of energy the battery gives to the separated charges. Energy (E) is measured in joules; charge (Q) is measured in coulombs. So a 9-volt battery gives 9 joules of energy to each coulomb of charge. Voltage is the energy provided to each coloumb of charge: V=

E or E = VQ Q

SAMPLE PROBLEM 5 a. b.

How much energy does a 1.5 V battery give to 0.50 coulombs of charge? The charge on an electron is 1.6 × 10−19 coulombs. How much energy does each electron have as it leaves a 1.5 V battery? Teacher-led video: SP5 (tlvd-0014)

THINK

The energy is given by E = VQ. 2. Substitute the known values into the formula and solve for E.

a. 1.

3.

State the solution.

The energy is given by E = VQ. 2. Substitute the known values into the formula and solve for E.

b. 1.

3.

State the solution.

WRITE

E = VQ V = 1.5 V, Q = 0.50 C E = 1.5 × 0.50 = 0.75 J The battery would give 0.75 joules of energy to 0.50 coloumbs of charge. b. E = VQ V = 1.5 V, Q = 1.6 × 10−19 E = 1.5 × 1.6 × 10−19 a.

= 2.4 × 10−19 J Each electron would have 2.4 × 10−19 joules of energy.

PRACTICE PROBLEM 5 A mobile phone battery has a voltage of 3.7 V. During its lifetime, 4000 coulomb of charge leave the battery. How much energy did the battery originally hold?

TOPIC 3 Concepts used to model electricity 83

Note: In many technologies, such as X-ray machines and particle accelerators, the energy of electrons needs to be determined. The number 1.6 × 10−19 joules is inconvenient, so another energy unit is used. It is called the ‘electron volt’, abbreviated eV, where 1 eV = 1.6 × 10−19 joules. The electrons at the negative terminal of a battery are attracted to the positive terminal, but the chemical reaction keeps them apart. The only way for the electrons to get to the positive terminal is through a closed circuit. The energy the electrons gain from the chemical reaction is transferred in the closed circuit as the electrons go through devices such as light globes, toasters and motors. Once back at the positive terminal, the chemical reaction in the battery transfers the electrons across to the negative terminal, and then the electrons move around the circuit again.

3.4.1 The conventional point of view Looking from the perspective of conventional current — that is, FIGURE 3.18 The circuit symbol positive charge carriers — the current would go in the opposite for a battery showing direction of direction, as seen in figure 3.18. conventional current In the circuit in figure 3.19a, positive charges at A, the positive terminal, would leave with energy, travel anticlockwise and arrive + – at F with no energy. At the positive terminal, A, a coulomb of charge has 9 J of energy; its voltage is 9 V. The wire, AB, from the battery to the globe is a good conductor, so no energy is lost between A and B and the coulomb of charge still has 9 J of energy. In the globe, as the current goes from B to C, the coulomb of charge loses 3 J of energy and now has 6 J of energy left at C. The conducting wire from C to D has no effect, so the coulomb of charge arrives at the motor, DE, with 6 J of energy. This energy is used up in the motor so that at E the coulomb of charge has 0 J of energy. The charge then moves on to F, the negative terminal, where the battery re-energises the charge to go around again. The graph in figure 3.19b shows the energy held by 1 C of charge — that is, the voltage — as the charge moves around the circuit from A to F.

I

FIGURE 3.19 Voltage around a simple circuit 9V A

(b) F

Current

Globe B

Voltage (V) or energy per coulomb charge (J C –1)

(a)

D

6

3

0

Motor C

9

E

A

B

C

D

E

F

A

Voltage is also called the electric potential. Using the hydraulic model, at A the charge is like water in a high dam with gravitational potential energy that can be released when the dam opens. The charge at A has an electric potential of 9 V or 9 J for every coulomb, which can be released when the switch is closed.

3.4.2 Measuring potential difference or voltage drop The difference in voltage between any two points in the circuit can be measured with a voltmeter. This is called the potential difference, or voltage drop. The voltmeter must be connected across a part of the circuit in parallel. If the voltmeter was connected to points A and B in the circuit in figure 3.19, it would display zero, as there is no difference in the potential or voltage between those two points because charges do not use any energy moving between A and B. If instead it was connected across the globe at BC, it 84 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

would show a voltage drop of 3 V (9 − 6 = 3 V). This means that in the globe 3 joules of electrical energy are given up by each coulomb of charge and transformed by the globe into light and heat. Voltmeters are designed so that they do not significantly affect the FIGURE 3.20 The circuit size of the current passing through the circuit element. For this reason, diagram symbol for a voltmeter the resistance of the circuit elements involved. Resistance will be discussed later in this topic. The circuit diagram symbol for a voltmeter is shown in figure 3.20. As discussed in section 3.3.3, most school laboratories now use digital multimeters, which FIGURE 3.21 To measure the current through the can generally measure both AC and DC voltages resistor, an ammeter is connected in series. To measure (see figure 3.21). To measure DC voltages, use the voltage drop across the resistor, the voltmeter is one of the settings near the ‘V’ with a bar beside it. connected in parallel. • The black or common socket, labelled Ammeter ‘COM’, is connected to the part of the circuit that is closer to the black or negative A terminal of the power supply. • The red socket, labelled ‘VΩmA’, is used for measuring voltages and is connected Resistor Voltmeter V Battery to the part of the circuit closer to the red or positive terminal of the power supply. The other red socket, labelled ‘10A MAX’ is • for large currents only. • The dial has a range of settings; when first connecting the multimeter, choose the setting with the largest value. If you want more accuracy in your measurement, turn the dial to a smaller setting. • If the display shows only the digit ‘1’, the voltage you are trying to measure exceeds the range of that setting and you need to go to a higher setting.

V

3.4 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. What is the voltage supplied by a battery that gives 1.05 J of energy to 0.70 C of charge which passes through it? 2. Complete the following table by filling in the missing values. Potential difference

Energy

Charge

32 J

9.6 C

4.0 J

670 mC

9.0 V

3.5 C

12 V

85 mC

4.5 V

12 J

240 V

7.5 kJ

3. How much electrical potential energy will 5.7 𝜇C of charge transfer if it passes through a voltage drop of 6.0 V? 4. A 6.0 V source supplies 3.6 × 10−4 J of energy to a quantity of charge. Determine the quantity of charge in coulombs and microcoulombs.

TOPIC 3 Concepts used to model electricity 85

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

3.5 Energy and power in an electric circuit KEY CONCEPTS • Apply concepts of charge (Q), electric current (I), potential difference (V), energy (E) and power (P), in electric circuits. Q E • Investigate and analyse theoretically and practically electric circuits using the relationships: I = , V = , t Q E P = = VI. t

3.5.1 Energy transformed by a circuit Electric circuits transform electrical energy into other forms of energy. Since the potential difference is a measure of the loss in electrical potential energy by each coulomb of charge, the amount of energy (E) transformed by a charge (Q) passing through a load can be expressed as: E = QV E , where V is the potential difference across the load. Q The amount of charge passing through a load in a time interval t can be expressed as:

since V =

Q = It

E = Vlt Where: E is the energy transferred, in joules V is the potential difference, in volts I is the current, in amperes t is the time, in seconds.

86 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

SAMPLE PROBLEM 6

What is the potential difference across a heater element if 3.6 × 104 J of heat energy is produced when a current of 5.0 A flows for 30 seconds? Teacher-led video: SP6 (tlvd-0015) THINK 1.

WRITE

Recall the formula that calculates the amount of energy produced.

E = VIt

Transpose the formula to make the potential difference, V, the subject. 3. Substitute the known values into the formula and solve.

V=

2.

E It

E = 3.6 × 104 J, I = 5.0 A, t = 30 s 3.6 × 104 J 5.0 A × 30 s 36 000 = 150 = 240 V The potential difference is 240 V.

V=

4.

State the solution.

PRACTICE PROBLEM 6 What is the potential difference across a light globe if 1.44 × 103 J of heat is produced when a current of 2.0 A flows for 1 minute?

3.5.2 Power delivered by a circuit In practice, it is the rate at which energy is transformed in an electrical load that determines its effect. The brightness of an incandescent light globe is determined by the rate at which electrical potential energy is transformed into the internal energy of the filament. Power is the rate of doing work, or the rate at which energy is transformed from one form to another. Power is equal to the amount of energy transformed per second, or the amount of energy transformed divided by the time it took to do it. Power can be expressed as: P=

E t

Where: P is the power delivered, in watts (W) E is the amount of energy transformed, in joules t is the time interval, in seconds. 1 watt = 1 joule per second = 1 J s−1

TOPIC 3 Concepts used to model electricity 87

By substituting the formula for energy, E = VIt, into the formula for power, P = P=

E , we find that: t

VIt t

⟹ P = VI

This is a particularly useful formula because the potential difference V and electric current I can be easily measured in a circuit. In real circuits, large power measurements are common. It is sometimes useful to use the unit kilowatt, where 1 kilowatt = 1 × 103 watts. When converting from watts to kilowatts, divide by 1000. When converting from kilowatts to watts, multiply by 1000.

SAMPLE PROBLEM 7

What is the power rating of an electric heater if a current of 5.0 A flows through it when there is a voltage drop of 240 V across the heating element? Teacher-led video: SP7 (tlvd-0016) THINK

WRITE

Use the formula for power, P = VI. 2. Substitute the known values into the formula and solve for P.

P = VI V = 240 V, I = 5.0 A P = VI P = 240 V × 5.0 A = 1200 W P = 1.2 kW

1.

Remembering that 1 kW = 1000 watts, convert to kW by dividing the number of watts by 1000. 4. State the solution. 3.

The power rating is 1.2 kW.

PRACTICE PROBLEM 7 What is the power rating of a CD player if it draws a current of 100 mA and is powered by four 1.5 V cells in series?

88 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Transposing formulae You can use the following triangles to assist transposing formulae P = VI, Q = It and E = QV. Power formula triangle

Variants of the power formula triangle

P

Q

E

V × I

I × t

Q × V

For example, if you wish to transpose the formula P = VI to make I the subject, cover the pronumeral you want to be the subject, in this case I. What is visible in the triangle shows what that pronumeral equals. In this example: I=

P V

This method can also be used for any formula of the form x = yz.

SAMPLE PROBLEM 8

How much energy is supplied by a mobile phone battery rated 3.7 V and 1200 mAh? Note: ‘mAh’ stands for milliamp hours, which means that the battery would last for one hour supplying a current of 1200 mA or two hours at 600 mA. Teacher-led video: SP8 (tlvd-0017) THINK 1.

Recall the formula, E = VIt, and state the known values.

2.

Convert the current from mA to amperes by dividing by 1000. Convert the time to seconds.

3.

Substitute the known values and solve for E.

Convert the number of joules to kJ by dividing by 1000. 5. State the solution. 4.

WRITE

E = VIt V = 3.7 V, I = 1200 mA, t = 1 hour 1200 I = 1200 mA = A = 1.2 A 1000 t = 60 × 60 = 3600 s E = 3.7 V × 1.2 A × 3600 s = 16 000 J E = 16 kJ There are 16 kJ of energy supplied.

PRACTICE PROBLEM 8 A 3.7 V mobile phone battery has an energy capacity of 14 000 joules. In a talk mode test, the battery lasted for 6 hours. What was the average current?

ELECTRON VOLT In many technologies, such as X-ray machines and particle accelerators, the energy of electrons needs to be determined. The number 1.6 × 10−19 joules is inconvenient, so another energy unit is used. It is the called the ‘electron volt’, abbreviated eV, where 1 eV = 1.6 × 10−19 joules.

TOPIC 3 Concepts used to model electricity 89

Resources Digital document Investigation 3.2 Energy transferred by an electric current (doc-31862) Teacher-led video Investigation 3.2 Energy transferred by an electric current (tlvd-0807)

3.5.3 Providing energy for the circuit The purpose of ‘plug-in’ power supplies, batteries or cells is to provide energy for the circuit. Such devices are said to provide an electromotive force, or emf. The term electromotive force is misleading since it does not refer to a ‘force’, measured in newtons. It is a measure of the energy supplied to the circuit for each coulomb of charge passing through the power supply. The circuit symbol for emf is 𝜀. The unit of emf is the volt (V) because it is a measure of energy per coulomb. A power supply has an emf of X volts if it provides the circuit with X joules of energy for every coulomb of charge passing through the power supply. The rate at which an emf source supplies energy to a circuit is the product of the emf and current. The amount of energy (E) supplied to the charge passing through the power supply is equal to the amount of energy given to each coulomb, or emf (𝜀), multiplied by the amount of charge (Q) passing through the power supply: E = 𝜀Q This can be rewritten as: E = 𝜀It since Q = It and t is the time interval during which energy is transferred. The power delivered to the charge passing through the power supply can therefore be expressed as: E t ⇒ P = 𝜀I P=

This is the formula used to determine the rate at which a source of emf supplies energy to a circuit. SAMPLE PROBLEM 9

A 12 V car battery has a current of 2.5 A passing through it. At what rate is it supplying energy to the car’s circuits? Teacher-led video: SP9 (tlvd-0018) THINK 1.

Recall the formula, P = VI, and state the known values.

As the energy supplied to the circuit equals the emf of the battery, substitute the known values for V and I into the formula P = VI and solve. 3. State the solution. 2.

90 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

WRITE

P = VI V = 12 V, I = 2.5 A P = VI = 12 × 2.5 = 30 W The battery is supplying 30 W.

PRACTICE PROBLEM 9 At what rate is energy being supplied to a 3.0 V light when it is drawing a current of 4.0 A?

3.5 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Calculate the current drawn by: (a) a 60 W light globe connected to a 240 V source (b) a 40 W globe with a voltage drop of 12 V across it (c) a 6.0 V, 6.3 W globe when operating normally (d) a 1200 W, 240 V toaster when operating normally. 2. The element of a heater has a voltage drop of 240 V across it. (a) In terms of energy what does this mean? (b) How much energy is transformed into thermal energy in the element if 25 C of charge flow through it? 3. A rear window demister circuit draws 2.0 A of current from a 12 V battery for 30 minutes. (a) How much energy is transformed by the rear window? (b) What is the power rating of the car demister? 4. How long will it take a 600 W microwave oven to transform 5.4 × 104 J of energy? 5. What is the power rating of an electric radiator if it draws a current of 10 A when connected to a 240 V AC household circuit? 6. An electric jug is connected to a 240 V supply and draws a current of 3.3 A. How long would it take to transfer 3.2 × 104 J of energy to its contents? 7. What is the emf of a battery that provides 9.0 J of energy to 6.0 C of charge?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

3.6 Resistance KEY CONCEPT • Model resistance in series and parallel circuits using — current versus potential difference (I–V) graphs — resistance as the potential difference to current ratio, including R = constant for ohmic devices.

The resistance of a material or device is a measure of how difficult it is for a current to pass through it. The higher the value of resistance, the harder it is for the current to pass through the material or device. The resistance, R, of a substance is defined as the ratio of the voltage drop, V, across it to the current, I, flowing through it. R=

V I

TOPIC 3 Concepts used to model electricity 91

The SI unit of resistance is the ohm (symbol Ω). One ohm is the resistance of a conductor in which a current of 1 ampere results from the application of a constant voltage drop of 1 volt across its ends. 1 Ω = 1 V A−1 The ohm is named in honour of Georg Simon Ohm (1787–1854), a German physicist who investigated the effects of different materials in electric circuits. Resistance is a material property and it is temperature dependent. In general, the resistance of a metal conductor increases with temperature. Usually, the increases will not be significant over small temperature ranges and most problems in this text ignore any temperature and resistance changes that might occur. One example of the effect of a change in temperature on resistance can be seen in the tungsten filament of an ordinary light globe. When operating normally, the filament reaches a temperature of 2500 °C. The globe is filled with inert gases to prevent the filament from burning or oxidising. Tungsten is used because it has a high melting point. The filament is coiled to increase the length and it has a very small cross-sectional area so that the resistance of the filament is increased. As the temperature of the filament increases, its resistance increases due to an increase in tungsten’s resistivity. FIGURE 3.22 Georg Simon Ohm

FIGURE 3.23 A 240-volt, 60-watt globe Argon and nitrogen (at low pressure)

Glass bulb

Fuse Tungsten filament

Electrical contacts

THE LIE DETECTOR The lie detector, or polygraph, is a meter that measures the resistance of skin. The resistance of skin is greatly reduced by the presence of moisture. When people are under stress, as they may be when telling lies, they sweat more. The subsequent change in resistance is detected by the polygraph and is regarded as an indication that the person may be telling a lie.

Resources Digital documents Investigation 3.3 The current-versus-voltage characteristics of a light globe (doc-31863) Investigation 3.4 Dependence of resistance on length of resistance wire (doc-31864)

92 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

3.6.1 Resistors In many electrical devices, resistors are used to control the current flowing through, and the voltage drop across, parts of the circuits. Resistors have constant resistances ranging from less than 1 ohm to millions of ohms. There are three main types of resistors. ‘Composition’ resistors are usually made of the semiconductor carbon. Wire-wound resistors consist of a coil of fine wire made of a resistance alloy such as nichrome. The third type is the metal film resistor, which consists of a glass or pottery tube coated with a thin film of metal. A laser trims the resistor to its correct value. FIGURE 3.24 (a) Carbon or ‘composition’ resistors (b) Wire-wound resistor (c) Metal film resistor (a)

(b)

(c)

Laser-cut grooves (to adjust resistance) Glass or pottery tube Wire conductor

Coloured bands

Metal film

Cap

Some large resistors have their resistance printed on them. Others have a colour code to indicate their resistance, as shown in figure 3.25 and table 3.1. The resistor has four coloured bands on it. The first two bands represent the first two digits in the value of resistance. The third band represents the power of ten by which the two digits are multiplied. The fourth band is the manufacturing tolerance. FIGURE 3.25 A resistor, showing the coloured bands Tolerance

First digit Second digit

Conductor

Power of ten multiplier

TOPIC 3 Concepts used to model electricity 93

TABLE 3.1 The resistor colour code Colour

Digit

Multiplier

Black

0

100 or 1

Brown

1

101

Red

2

102

Orange

3

103

Yellow

4

104

Green

5

105

Blue

6

106

Violet

7

107

Grey

8

108

White

9

109 10−1

Gold Silver

10

−2

Tolerance

±2%

±5% ±10% ±20%

No colour

SAMPLE PROBLEM 10

What is the resistance of the following resistors if their coloured bands are: red, violet, orange and gold b. brown, black, red and silver? a.

Teacher-led video: SP10 (tlvd-0019) THINK a. 1.

Remember when holding a resistor to read its value, keep the gold or silver band on the right and read the colours from the left.

WRITE Tolerance

a.

First digit Second digit

2.

Using table 3.1, establish the first two digits.

3.

Using table 3.1, establish the multiplier.

4.

Using table 3.1, establish the tolerance.

5.

State the solution.

94 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Conductor

Power of ten multiplier

Red = 2 Violet = 7 Hence, the first two digits are 27. The third band is orange, which means multiply the first two digits by 103 . The resistance is: 27 × 103 = 27 000 Ω = 27 kΩ. The fourth band is gold, which means there is a tolerance of 5%. 5% × 27 000 Ω = 1350 Ω The true value is 27 × 103 kΩ ±1350 Ω.

b. 1.

Remember when holding a resistor to read its value, keep the gold or silver band on the right and read the colours from the left.

Tolerance

b.

First digit Second digit

2.

Using table 3.1, establish the first two digits.

3.

Using table 3.1, establish the multiplier.

4.

Using table 3.1, establish the tolerance.

5.

State the solution.

Conductor

Power of ten multiplier

Brown = 1 Black = 0 Hence, the first two digits are 10. The third band is red, which means multiply the first two digits by 102 . The resistance is: 10 × 102 = 1000 Ω = 1 kΩ The fourth band is silver, which means there is a tolerance of 10%. 10% × 1000 Ω = 100 Ω The true value is 1.0 × 103 Ω ±100 Ω.

PRACTICE PROBLEM 10 What are the resistances and tolerances of resistors with the colour codes: a. orange, white, black, gold b. green, blue, orange, silver c. violet, green, yellow, gold?

Resources eLesson

Resistance (eles-2516)

Interactivity

Picking the right resistor (int-6391)

3.6.2 Ohm’s Law Georg Ohm established experimentally that the current I in a metal wire is proportional to the voltage drop V applied to its ends. When he plotted his results on a graph of V versus I, he obtained a straight line.

FIGURE 3.26 Graphs of voltage versus current for two different metal wires V Metal A

I∝V Metal B

0

I

TOPIC 3 Concepts used to model electricity 95

The equation of the line shown in figure 3.26 is known as Ohm’s Law and can be written: V = IR where R is numerically equal to the constant gradient of the line. This is known as the resistance of the metal conductor to the flow of current through it. Remember that the SI unit of resistance is the ohm (Ω). You can use the triangle method for Ohm’s Law. V I × R

Convert the quantity/pronumeral you want to be the subject, for example, R. What is visible in the triangle shows what the pronumeral equals. The resistance, R, can also be expressed as: R=

V I

SAMPLE PROBLEM 11

A transistor radio uses a 6.0 V battery and draws a current of 300 mA. What is the resistance of the radio? Teacher-led video: SP11 (tlvd-0020)

THINK 1.

WRITE

From Ohm’s Law the resistance, R, can be found.

V = IR R=

V I

2.

State the known values and convert the current into amperes by dividing by 1000.

V = 6.0 V, I =

3.

Substitute the values for V and I and solve to find R.

R=

4.

State the solution.

300 mA = 0.300 A 1000

6.0 V 0.300 A = 20 Ω The resistance of the radio is 20 Ω.

PRACTICE PROBLEM 11 A 240 V kitchen appliance draws a current of 6.0 A. What is its resistance?

Resources Weblink Ohm’s Law app

96 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

3.6.3 Ohmic and non-ohmic devices An ohmic device is one for which, under constant physical conditions such as temperature, the resistance is constant for all currents that pass through it. A non-ohmic device is one for which the resistance is different for different currents passing through it. The graph in figure 3.26 has voltage on the y-axis and current on the x-axis. The graph is drawn this way so that the gradient of lines for the metals A and B gave the resistance of each. However, accepted convention graphs the independent variable on the x-axis and the dependent variable on the y-axis. So in 1 the graph in figure 3.27a the gradient equals . R FIGURE 3.27 The current-versus-voltage graphs for (a) an ohmic resistor and (b) a diode, which is a non-ohmic device (a)

(b)

I

I 1 I= R V

( )

0

V

V

Resources Digital document Investigation 3.5 Ohmic and non-ohmic devices (doc-31865) Teacher-led video Investigation 3.5 Ohmic and non-ohmic devices (tlvd-0810)

Non-ohmic devices Many non-ohmic devices are made from elements that are semiconductors. They FIGURE 3.28 are not insulators as they conduct electricity, though not as well as metals. Common Circuit symbol semiconductor elements are silicon and germanium, which are in Group 14 of the for a diode periodic table. Many new semiconductor devices are compounds of Group 13 and Group 15 elements such as gallium arsenide. A diode is formed by joining two differently doped materials together. A diode allows current to flow through it in only one direction. This effect can be seen in the current–voltage graph for a diode in figure 3.29b, where a small positive voltage produces a current, while a large negative or reverse voltage produces negligible current. Light-emitting diodes (LEDs) are diodes that give off light when they conduct. They are usually made from gallium arsenide. Gallium nitride is used in blue LEDs. Thermistors are made from a mixture of semiconductors so they can conduct electricity in both directions. They differ from metal conductors, whose resistance increases with temperature, as an increase in a thermistor’s temperature increases the number of electrons available to move and the resistance decreases.

TOPIC 3 Concepts used to model electricity 97

FIGURE 3.29 (a) Circuit symbol for a thermistor (b) Resistance-versus-temperature graph for a thermistor (a)

(b)

Resistance (kΩ)

2

1

0

20

40 Temperature (°C)

60

Light-dependent resistors (LDRs) are like thermistors, except they respond to light. The resistance of an LDR decreases as the intensity of light shining on it increases. The axes in the graph for an LDR in figure 3.30 have different scales to the other graphs. As you move from the origin, each number is 10 times the previous one. This enables more data to fit in a small space. FIGURE 3.30 (a) Circuit symbols for an LDR (b) Graph of resistance-versus-light intensity for an LDR, on a logarithmic scale (a)

(b)

or

LDR resistance (Ω)

10 000

1000

100

10 0.1

1

10

100

1000

Illumination (lux)

Note: A logarithmic scale is a non-linear scale and is commonly used as an effective way of displaying data that cover a large range of values on one graph. By plotting the log of resistance and the log of illumination, we are able to graphically plot two parameters.

3.6.4 Heating effects of currents Whenever a current passes through a conductor, thermal energy is produced. This is due to the fact that the mobile charged particles — for example, electrons — make repeated collisions with the atoms of the conductor, causing them to vibrate more and producing an increase in the temperature of the material. This temperature increase is not related to the direction of the current. A current in a conductor always generates thermal energy, regardless of which direction the current flows. Examples of devices that make use of this energy include radiators, electric kettles, toasters, stoves, incandescent lamps and fuses. 98 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

NICHROME Nichrome is a heat-resistant alloy used in electrical heating elements. Its composition is variable, but is usually around 62% nickel, 15% chromium and 23% iron.

3.6.5 Power and resistance Recall that the rate at which energy is dissipated by any part of an electric circuit can be expressed as: P = VI Where: P is the power I is the current V is the voltage drop. This relationship can be used, along with the definition of resistance, R = formulae describing the relationship between power and resistance: R=

V , to deduce two different I

V V , V = IR, I = I R

Substituting V = IR into the formula P = VI: P = VI = (IR) I Thus: P = I2 R [1] V Substituting I = into the formula P = VI: R P = VI( ) V P=V R Thus: V2 [2] P= R You now have three different ways of determining the rate at which energy is transferred as charge flows through a voltage drop in an electric circuit: P = VI

P = I2 R

P=

V2 R

In addition, the quantity of energy transferred, E, can be determined using: E = VIt = I2 Rt =

V2 t R

These formulae indicate that in conducting wires with low resistance, very little energy is dissipated. If the resistance, R, is small and the voltage drop, V, is small, the rate of energy transfer is also small. TOPIC 3 Concepts used to model electricity 99

SAMPLE PROBLEM 12

A portable radio has a total resistance of 18 Ω and uses a 6.0 V battery consisting of four 1.5 V cells in series. At what rate does the radio transform electrical energy? Teacher-led video: SP12 (tlvd-0021) THINK

WRITE

Recall that power is the rate of energy use and use the formula containing the variables P, V and R. 2. Substitute the known values into the formula and solve for P.

P=

1.

V2 R

V = 6.0 V, R = 18 Ω V2 R (6.0 V)2 P= 18 Ω = 2.0 W The radio transforms energy at 2.0 W.

P=

3.

State the solution.

PRACTICE PROBLEM 12 What is the power rating of an electric jug if it has a resistance of 48 Ω when hot and is connected to a 240 V supply?

SAMPLE PROBLEM 13

A pop-up toaster is labelled ‘240 V, 800 W’. a. What is the normal operating current of the toaster? b. What is the total resistance of the toaster while it is operating? Teacher-led video: SP13 (tlvd-0022) THINK a. 1.

The three variables P, V and I are given in the formula for power.

2.

Transpose the formula to make I the subject.

3.

Substitute the known values into the formula and solve for I.

4. b. 1.

2.

State the solution. The three variables P, V and R are given in the formula for power. Transpose the formula to make R the subject.

100 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

WRITE a.

P = VI P V V = 240 V, P = 800 W I=

800 W 240 V = 3.33 A The normal operating current 3.33 A. V2 b. P = R I=

R=

V2 P

3.

Substitute the known values into the formula and solve for R.

4.

State the solution.

V = 240 V, P = 800 W (240 V)2 R= 800 W = 72 Ω The resistance is 72 Ω.

PRACTICE PROBLEM 13 A microwave oven is labelled ‘240 V, 600 W’. a. What is the normal operating current of the microwave oven? b. What is the total resistance of the microwave oven when it is operating?

3.6 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. How much energy is provided by a 6 V battery if a current of 3 A passes through it for 1 minute? 2. Complete the following table by filling in the missing values. Potential difference

Current

Resistance

8.0 A

4.0 Ω

22 mA

2.2 kΩ

12 V

6.0 Ω

240 V

8.0 × 104 Ω

9.0 V

6.0 A

1.5 V

45 mA

3. What are the resistances and tolerances of resistors with the colour codes: (a) blue, brown, orange, gold (b) yellow, white, green, silver (c) brown, red, red, gold? 4. The following graph shows the current-versus-voltage characteristic for an electronic device.

Current (mA)

80 60 40 20

0

(a) (b) (c) (d)

0.2 0.4 0.6 Voltage drop across diode (V)

0.8

Is this device ohmic or non-ohmic? Justify your answer. What is the current through the device when the voltage drop across it is 0.5 V? What is the resistance of the device when the voltage drop across it is 0.5 V? Estimate the voltage drop across the device and its resistance, when it draws a current of 20 mA.

TOPIC 3 Concepts used to model electricity 101

5. At what rate is thermal energy being transferred to a wire if it has a resistance of 5 Ω and carries a current of 0.30 A? 6. Calculate the resistance of the following globes if their ratings are: (a) 240 V, 60 W (b) 6.0 V, 6.3 W (c) 12 V, 40 W. 7. What is the power rating of an electric jug if it has a resistance of 48 Ω when hot and is connected to a 240 V supply? 8. A thermistor has the characteristic curve shown in the following graph. 8

Resistance (kΩ)

7 6 5 4 3 2 1 0

10

20 30 40 50 60 70 80 90 Temperature (°C)

(a) What is the resistance of the thermistor at the following temperatures? i. 20 °C ii. 80 °C (b) What is the temperature when the thermistor has the following resistances? i. 4 kΩ ii. 1.5 kΩ 9. A temperature sensing system in an oven uses a thermistor with the characteristics shown in the following graph, on a logarithmic scale. 100 000

Resistance (Ω)

10 000

1000

100

10

0

50 100 150 200 250 300 Temperature (°C)

(a) What is the resistance of the thermistor when the temperature in the oven is 100 °C? (b) What is the temperature in the oven when the resistance of the thermistor is 400 Ω?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

102 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

3.7 Review 3.7.1 Summary • • • • • •

An electric circuit is a complete conducting path containing an energy supply and a load. Q Current, I, is the rate of flow of charge, where I = t Direct current always flows in one direction. Alternating current periodically reverses its direction in a circuit. Conventional current is defined as flowing from the positive to the negative terminal of a supply, even though the charge is usually carried by electrons travelling in the opposite direction. Electric current is measured with an ammeter. The potential difference across part of an electric circuit is a measure of the electrical potential energy lost by charge carriers. This can be expressed as: V=

• • • • • • •

E Q

Potential difference is measured using a voltmeter. The electromotive force (emf) of a power supply is a measure of the amount of energy supplied to the circuit for each coulomb of charge passing through that supply. Resistance is the opposition provided by a substance to the flow of current through it. Ohm’s Law states the current flowing in a metal wire varies directly with the voltage drop across the conductor and inversely with the resistance of the conductor. The graph of voltage drop versus current at a constant temperature is a straight line for an ohmic device. The amount of energy transformed in a device during a given time interval can be calculated using the equation E = VIt. Power is the rate at which work is done, or at which energy is transformed from one form into another. The power delivered to a device in an electric circuit can be calculated using the equation: P = VI = I2 R =

•

V2 R

Non-ohmic devices such as LDRs, LEDs, diodes and thermistors do not obey Ohm’s Law; their resistance is not constant.

Resources

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0028).

TOPIC 3 Concepts used to model electricity 103

3.7.2 Key terms Alternating current (AC) refers to circuits where the charge carriers move backwards and forwards periodically. An ammeter is a device used to measure current. The ampere is the unit of current. A charge carrier is a charged particle moving in a conductor. A conductor is a material that contains charge carriers; that is, charged particles can move and travel freely through the material. Conventional current is defined as the movement of positive charges from the positive terminal of a cell through the conductor to the negative terminal. The coulomb is the unit of electric charge. A diode is a device that allows current to pass through it in one direction only. Direct current (DC) refers to circuits where the net flow of charge is in one direction only. Electric charge is a basic property of matter. It occurs in two states: positive (+) charge and negative (−) charge. An electric circuit is a closed loop of moving electric charge. Electric current is the movement of charged particles from one place to another. In an electric insulator the electrons are bound tightly to the nucleus and are not free to travel through the material. Electromotive force (emf) is a measure of the energy supplied to a circuit for each coulomb of charge passing through the power supply. Electron current is the term used when dealing with the mechanisms for the movement of electrons. An electrostatic force is the force between two stationary charged objects. An ion is a charged particle. A light-dependent resistor (LDR) is a device that has a resistance which varies with the amount of light falling on it. A light-emitting diode (LED) is a small semiconductor diode that emits light when a current passes through it. A load is a device where electrical energy is converted into other forms to perform tasks such as heating or lighting. A model is a representation of ideas, phenomena or scientific processes; can be a physical model, mathematical model or conceptual model. A neutral object carries an equal amount of positive and negative charge. A non-ohmic device is one for which the resistance is different for different currents passing through it. An ohmic device is one for which, under constant physical conditions such as temperature, the resistance is constant for all currents that pass through it. The potential difference, or voltage drop, is the amount of electrical potential energy, in joules, lost by each coulomb of charge in a given part of a circuit. Power is the rate of doing work, or the rate at which energy is transformed from one form to another. The resistance, R, of a substance is defined as the ratio of voltage drop, V, across it to the current, I, flowing through it. A resistor is used to control the current flowing through, and the voltage drop across, parts of a circuit. A switch stops or allows the flow of electricity through a circuit. A thermistor is a device that has a resistance which changes with a change in temperature. A voltmeter is a device used to measure potential difference.

Resources Digital document Key terms glossary (doc-32178)

104 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

3.7.3 Practical work and investigations Investigation 3.1 The Van de Graaff generator Aim: To investigate electrostatic charge Digital document: doc-31918

Investigation 3.2 Energy transferred by an electric current Aim: To calculate how much energy is transformed from electrical potential energy into the internal energy of a load Digital document: doc-31862 Teacher-led video tlvd-0807

Investigation 3.3 The current-versus-voltage characteristics of a light globe Aim: To design and construct a circuit that will enable you to find the current through the globe for a suitable range of voltages Digital document: doc-31863

Investigation 3.4 Dependence of resistance on length of resistance wire Aim: To investigate how the resistance of a resistance wire varies with length Digital document: doc-31864

Investigation 3.5 Ohmic and non-ohmic devices Aim: To explore the resistance properties of a resistor and a light globe Digital document: doc-31865 Teacher-led video tlvd-0810

Resources Digital document Practical investigation logbook (doc-32179)

TOPIC 3 Concepts used to model electricity 105

3.7 Exercises To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au.

3.7 Exercise 1: Multiple choice questions Which best describes the transfer of electrical energy in a circuit? A. Electric charge is moved in the direction of the current. B. Free electrons pass all the way through the circuit to the load. C. Free electrons dissipate their energy. D. The battery pushes positive charges through the circuit. 2. Electron charge was removed from an object. Which of the following could represent the amount of charge removed? A. 1.5 × 10−19 C B. 3.0 × 10−19 C C. 4.8 × 10−19 C D. 5.4 × 10−19 C 3. For 25 seconds a battery supplies a constant current of 5.0 A to a circuit. What best represents the amount of charge leaving the battery? A. 0 C B. 0.20 C C. 5.0 C D. 125 C 4. In the circuit shown which best describes the current passing through the ammeter? 1.

24 W A

12 V

Zero, the current is used up in the 24 W load. The current is 2 A. C. Zero, the switch is open. D. The current is non-constant. 5. The current and voltage for an object in a circuit was collected and plotted on a graph. A.

B.

Voltage (V)

4 3 2 1 0

1

106 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

2

3 4 5 Current (A)

6

7

What is the resistance of this object? 0.60 Ω B. 0.67 Ω C. 1.50 Ω D. 1.70 Ω Each electron is given 1.44 × 10−18 joules of energy by a battery. What is the elecromotive force provided by the battery? A. 1.5 V B. 3.0 V C. 9.0 V D. 11 V A 240 V rice cooker draws a current of 1.25 A. How much energy is provided to the cooker in 1 minute? A. 19.2 J B. 300 J C. 3.6 kJ D. 18 kJ At what rate is a 7.2 V battery supplying energy to a tablet device when it is drawing a current of 1200 mA? A. 6.0 J B. 6.0 W C. 8.6 J D. 8.6 W When a 240 V microwave rated at 1200 W is operating what is the current? A. 5.0 mA B. 6.0 mA C. 5.0 A D. 48.0 A For the microwave described in question 9 what is the total resistance when operating? A. 5 Ω B. 6 Ω C. 20 Ω D. 48 Ω A.

6.

7.

8.

9.

10.

3.7 Exercise 2: Short answer questions During a storm a lightning bolt discharges 3 million kJ of energy to Earth in 0.75 ms. The discharge involves the movement of 15 C of charge. What is the potential difference between the lightning source and Earth? 2. Consider a 3.0 V battery. a. How much energy does it supply to: i. one electron ii. one coulomb of charge? b. The battery is used in a mobile telephone. A 1-minute conversation uses the energy transferred by 0.04 C of charge. At what rate is energy being transferred to the telephone? 3. An electric kettle operating off a 240 V power supply uses 2.7 kW when boiling water. a. What is the current in the kettle? b. When the kettle is on for 2.5 minutes how much energy does it use? 4. a. What is the voltage drop across a 44 Ω resistor carrying a current of 2.5 A? b. What would be the effect of connecting the same resistor to a larger power supply?

1.

TOPIC 3 Concepts used to model electricity 107

5.

A children’s toy comprises a 9 V battery, a switch and an electric motor in a simple circuit as shown in the following figure. The motor is labelled ‘9 V, 25 mA’. A

F

E D

Motor B

C

Voltage (V) or energy per coulomb charge

When the switch is closed: a. What is the maximum rate at which the battery can supply energy to the motor? b. By copying the axes shown, sketch a graph showing the energy held by a coulomb of charge as it moves around the circuit from A to F, then returning to A.

9 6 3 0 A

c.

B

C

D

E

F

A

By copying the axes shown, complete the graph for current around the circuit.

Current (A)

40 30 20 10 0 A

B

C

D

E

F

A

What is the resistance of the motor when operating at its maximum capacity? 6. A handheld fan is powered by two 1.5 V batteries. When tested at room temperature 15 C of charge flowed through it every minute. a. What was the current in the device? b. What is the resistance of the device? c. At what rate is the device using energy? 7. A circuit comprises a 1.2 kW heating element, a switch and a voltage source. a. Draw a circuit diagram demonstrating how to measure both the voltage across the heater and the current flowing through it? b. If the voltage supplied is 240 V, what is the current passing through the heater? d.

108 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

8.

The voltage supplied to a tungsten globe is varied. The current and the potential difference across the lamp are measured and the results plotted in the following graph.

Current (A)

0.4 0.3 0.2 0.1 1 0

2

4

6

8

10

Voltage (V)

Calculate the resistance of the globe when the current is 0.2 A. Would you describe the globe as ohmic or non-ohmic? Justify your answer. 9. The following graph shows the current versus voltage relationship for a non-ohmic device. 60 a. What is a non-ohmic device? 50 b. What is the current through the device when the voltage drop across it is 0.3 V? 40 c. What is the resistance of this device when the voltage 30 drop across it is 0.3 V? 20 10. a. What is the resistance of an 800 W toaster when a current of 3.3 A is flowing? 10 b. If it takes 40 seconds to brown the toast, how much energy is used? 0 Current (mA)

a. b.

0.1 0.2 0.3 0.4 0.5 Voltage (V)

3.7 Exercise 3: Exam practice questions Question 1 (2 marks) To model the energy supply, current and load in an electrical circuit a teacher and students stand in a circle loosely holding a loop of rope. The teacher models the energy supply by pulling the rope around while the students act as conductors by allowing the rope to readily move through their hands. One student, Luke, is asked to represent a globe in the circuit by making it more difficult for the rope to slide through his hands. An observer suggests that in an electric circuit the electrons must make their way to the globe before it lights up and that Luke must wait until the rope has travelled all the way from the teacher before he feels the pull on the rope. In terms of both the model and how electricity behaves in a circuit explain whether you agree or disagree with the observer.

TOPIC 3 Concepts used to model electricity 109

Question 2 (6 marks) Use the following circuit diagram to answer the questions. 24 V

72 W

a. b. c. d. e. f.

Draw an arrow to show the direction of the electron current. 1 mark Show where you would connect an ammeter to measure the current flowing through the globe. 1 mark Calculate the current flowing in the circuit. 1 mark When the switch is closed what is the voltage drop across it? 1 mark Show where you would connect a voltmeter to confirm your prediction of the voltage drop across the switch. 1 mark Explain what happens when the switch is opened. 1 mark

Question 3 (3 marks) An unknown electrical component is labelled ‘3.0 V, 2.4 W DC ONLY’.

Voltage (V)

4.0 3.0 2.0 1.0

0

0.5 Current (A)

1.0

Describe what DC ONLY means. 1 mark What is the maximum current that the component can safely tolerate? 1 mark c. A student places another component in a simple circuit to measure its voltage current characteristics, recording the results as shown in the following graph. Calculate the resistance when the current is 600 mA. 1 mark

a. b.

Question 4 (2 marks) An electric drill has a rechargeable 18 V battery that can store up to 6 MJ. When fully discharged it takes the battery 8 hours to fully recharge. a. Calculate the total charge needed to fully charge the battery. 1 mark b. Calculate the average current drawn when charging the battery. 1 mark

110 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Question 5 (4 marks) The following graph shows the current-versus-voltage characteristic for an electronic device.

Current (mA)

20

10

0

0.5

1.0

Voltage (V)

Is this device ohmic or non-ohmic? Justify your answer. What is the current through the device when the voltage drop across it is 0.5 V? c. What is the resistance of the device when the voltage drop across it is 0.5 V? d. Estimate the rate at which energy is being used when the voltage drop across it is 0.5 V. a.

b.

1 mark 1 mark 1 mark 1 mark

3.7 Exercise 4: studyON topic test Fully worked solutions and sample responses are available in your digital formats.

Test maker Create unique tests and exams from our extensive range of questions, including practice exam questions. Access the assignments section in learnON to begin creating and assigning assessments to students.

TOPIC 3 Concepts used to model electricity 111

AREA OF STUDY 2 HOW DO ELECTRIC CIRCUITS WORK?

4

Circuit analysis

4.1 Overview Numerous videos and interactivities are available just where you need them, at the point of learning, in your digital formats, learnON and eBookPLUS at www.jacplus.com.au.

4.1.1 Introduction In topic 3, an electric circuit was described as a number of electrical conductors connected to form a conducting path. A circuit can contain one or more sources of emf (electromotive force) to provide energy to the circuit. If the conductors form a continuous closed path through which a current can circulate, the circuit is said to be a closed circuit, as shown in figure 4.1. If there is a break in the path so that charge cannot flow, for example at a switch, the circuit is said to be an open circuit. In many simple electric circuits, electrical energy is used for heating one or more loads. The temperature increase in a load occurs because the charge carriers make repeated collisions with the atoms in the load. This increases the internal energy of the load and its temperature rises. The electrical energy transformed in the load originally came from some source of emf, for example a battery, laboratory power pack or power point in the home. Examples of this type of circuit include a torch (where the conductor in the filament gets so hot that it emits light), a toaster plugged into a power point and a car demister circuit. The current flowing in these circuits depends on the resistance of the loads. The torch globe, the heating element in the toaster and other loads make it difficult for a current to flow. In this topic you will look at what happens in different types of electric circuits and in the devices within these circuits. FIGURE 4.1 Decorative lights that are connected in series, such as those on the Melbourne Star Observation Wheel, can contain shunts which allow the circuit to remain closed if the filament breaks.

112 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

4.1.2 What you will learn KEY KNOWLEDGE After completing this topic you will be able to: • model resistance in series and parallel circuits using • resistance as the potential difference to current ratio, including R = constant for ohmic devices • equivalent effective resistance in arrangements in – series: RT = R1 + R2 + ... + Rn 1 1 1 1 – parallel: = + + ... + RT R1 R2 Rn calculate and analyse the effective resistance of circuits comprising parallel and series resistance and • voltage dividers • compare power transfers in series and parallel circuits. • investigate and apply theoretically and practically concepts of current, resistance, potential difference (voltage drop) and power to the operation of electronic circuits comprising resistors, light bulbs, diodes, thermistors, light dependent resistors (LDRs), light-emitting diodes (LEDs) and potentiometers (quantitative V analysis restricted to use of I = and P = VI) R • investigate practically the operation of simple circuits containing resistors, variable resistors, diodes and other non-ohmic devices • describe energy transfers and transformations with reference to transducers. Source: VCE Physics Study Design (2016–2021) extracts © VCAA; reproduced by permission.

PRACTICAL WORK AND INVESTIGATIONS Practical work is a central component of learning and assessment. Experiments and investigations, supported by a Practical investigation logbook and Teacher-led videos, are included in this topic to provide opportunities to undertake investigations and communicate findings.

Resources Digital documents Key science skills — VCE Units 1–4 (doc-31856) Key terms glossary (doc-32180) Practical investigation logbook (doc-32181)

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0029).

4.2 BACKGROUND KNOWLEDGE Electric circuit rules BACKGROUND KNOWLEDGE • The sum of the currents flowing into a junction is equal in magnitude to the sum of the currents flowing out of that junction: Iin = Iout • In any closed loop of an electrical circuit, the sum of the voltage drops must equal the sum of the emfs in that loop.

TOPIC 4 Circuit analysis 113

4.2.1 Circuit diagrams A circuit diagram shows schematically the devices used in constructing an electrical circuit. Table 4.1 shows the symbols commonly used in drawing circuits. Figure 4.2 is a diagram of a circuit comprising a battery, switch and resistor.

FIGURE 4.2 A circuit diagram

TABLE 4.1 Symbols used in circuit diagrams Circuit component

Symbol

Circuit component

Connection between conductors

Resistor with sliding contact to give a variable resistance

Terminal

Semiconductor diode*

Conductors not connected*

or

Conductors connected*

or

Single pole switch (open) Button switch (open)

or

Earth*

Voltmeter

or Battery

Ammeter

Variable power supply*

Symbol

Incandescent lamp*

or

Resistor

V

A

or

Light-dependent resistor (LDR)

or

Variable resistor*

or

Thermistor (heat-dependent resistor)

*The first of the two alternative symbols is used in this book.

4.2.2 Circuit rules Accounting for electrons Electric charge is conserved. At any point in a conductor, the amount of charge (usually electrons) flowing into that point must equal the amount of charge flowing out of that point. Electrons do not build up at a point in a conductor, nor will they magically disappear. You don’t get traffic jams in electric circuits. In figure 4.3, Ia + Ic = Ib + Id + Ie .

114 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

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FIGURE 4.3 Five wires soldered at a junction Ib Ia

Ie Ic Id

The sum of the currents flowing into a junction is equal in magnitude to the sum of the currents flowing out of that junction: Iin = Iout

SAMPLE PROBLEM 1

Calculate the magnitude and direction of the unknown current in the following figure, showing currents meeting at a junction.

1.0 A 2.5 A

I 1.3 A 4.0 A

Teacher-led video: SP1 (tlvd-0023) THINK

From the diagram, calculate the total currents flowing into the junction. 2. From the diagram, calculate the total currents flowing out of the junction. 3. Recall that Iin = Iout . The unknown values must be a flow out of the junction. Solve for the unknown current out of the junction. 1.

4.

State the solution.

1.0 A + 4.0 A = 5.0 A WRITE

2.5 A + 1.3 A = 3.8 A

Iin = Iout 5.0 A = 3.8 A + x x = 5.0 A − 3.8 A = 1.2 A The unknown current flowing out of the junction is 1.2 A.

TOPIC 4 Circuit analysis 115

PRACTICE PROBLEM 1 Find the unknown current at the junction in the following figure. State the direction of the current in each case.

3.0 A

1.5 A Ia

SAMPLE PROBLEM 2

Find the values of currents a, b, c, d, e and f as marked in the following figure. f E 6.5 mA

d

D

C c

e

2.1 mA b

B A a

7.9 mA

15.3 mA

Teacher-led video: SP2 (tlvd-0024) THINK

WRITE

Recall the formula Iin = Iout . 2. Determine the current at each junction.

Iin = Iout 15.3 mA = 7.9 mA + a a = 15.3 mA − 7.9 mA a = 7.4 mA 7.9 mA + 7.4 mA = b b = 15.3 mA 15.3 mA = c + 2.1 mA c = 15.3 mA − 2.1 mA c = 13.2 mA 13.2 mA = d + 6.5 mA d = 13.2 mA − 6.5 mA d = 6.7 mA 2.1 mA = e e = 2.1 mA d + 6.5 mA + e = f f = 6.7 mA + 6.5 mA + 2.1 mA f = 15.3 mA The values of the currents are: a = 7.4 mA, b = 15.3 mA, c = 13.2 mA, d = 6.7 mA, e = 2.1 mA, f = 15.3 mA.

1.

3.

State the solution.

116 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

PRACTICE PROBLEM 2 Find the values of currents a and b as marked in the following figure. 11.1 A

3.0 A

5.7 A

b B

a

4.2.3 Conservation of electrical energy Around a circuit electrical energy must be conserved. In any closed loop of a circuit, the sum of the voltage drops must equal the sum of the emfs in that loop.

SAMPLE PROBLEM 3

Calculate the unknown voltage drop Vbc in the following figure. a

Vab = 5.2 V V = 9.0 V b

Vbc

c

Teacher-led video: SP3 (tlvd-0025) THINK 1.

This circuit is a closed loop and the sum of the voltage drops within it must equal the voltage supplied by the battery.

WRITE

V = the sum of the voltage drops V = 9.0 V 9.0 V = Vab + Vbc = 5.2 V + Vbc

2.

State the solution.

Vbc = 3.8 V The unknown voltage drop is 3.8 V.

TOPIC 4 Circuit analysis 117

PRACTICE PROBLEM 3 Calculate the unknown voltage drop Vab in the following figure. 4.8 V V

12 V

Vab

4.2 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Explain what is meant by the following terms, as they relate to electric circuits. (a) Junction (b) Current (c) Voltage drop (d) Conductor 2. Find the missing currents in the following figures. State the direction of the current in each case. (a)

(b) 2.9 A 7.3 A

Ib

3.7 A

1.3 A 2.8 A

Ic

4.2 A

3. Calculate the voltages Vcd and Vef in the following figure. g f

24V

d

Vef

Vcd c

e b

Vab = 8V a

Fully worked solutions and sample responses are available in your digital formats.

118 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

4.3 Series and parallel circuits KEY CONCEPTS • Model resistance in series and parallel circuits using: • resistance as the potential difference to current ratio, including R = constant for ohmic devices • equivalent effective resistance in arrangements in – series: RT = R1 + R2 + ... + Rn 1 1 1 1 – parallel: = + + ... + RT R1 R2 Rn • calculate and analyse the effective resistance of circuits comprising parallel and series resistance and voltage dividers.

There are two ways in which circuit elements can be connected: in series and in parallel. When devices are connected in series, they are joined together one after the other. There is only one path for the current to take. When devices are connected in parallel, they are joined together so that there is more than one path for the current to flow through. Many devices can be connected in series and parallel. These include resistors and cells.

4.3.1 Resistors in series When a number of resistors are placed in series, some basic rules can be derived. There is only one path for the current to flow through. Therefore, in figure 4.4, the current in R1 equals the current in R2 and in R3 . I1 refers to the current in R1 ; I2 to the current in R2, and so on. Similarly, V1 is the voltage drop across R1 ; V2 is the voltage drop across R2 , and so on.

FIGURE 4.4 Resistors connected in series R1

R2

R3

I

I = I1 = I2 = I3 Since V = IR: V1 = IR1 V2 = IR2 V3 = IR3 The total voltage drop, VT , across resistors in series is equal to the sum of the voltage drops across each individual resistor. VT = V1 + V2 + V3 ⇒ VT = IR1 + IR2 + IR3 ⇒ VT = I(R1 + R2 + R3 ) Since VT = IRT (where RT is the effective resistance of all three resistors), the effective resistance offered by resistors in series is found by obtaining the sum of the individual resistances: RT = R1 + R2 + R3 This means that the effective resistance of a circuit is increased by adding an extra resistor in series with the others. The resistance of a series circuit is greater than that for any individual resistor.

TOPIC 4 Circuit analysis 119

SAMPLE PROBLEM 4

Find the effective resistance of a circuit comprising three resistors, having resistance values of 15 Ω, 25 Ω and 34 Ω, connected in series. Teacher-led video: SP4 (tlvd-0026) THINK

WRITE

When connected in series the effective resistance of the circuit is equal to the sum of the individual resistors. 2. Substitute the resistance values. 3. Calculate the effective resistance. 4. State the solution.

RT = ∑ Ri

i

1.

1

RT = 15 Ω + 25 Ω + 34 Ω RT = 74 Ω The effective resistance of circuit is 74 Ω.

PRACTICE PROBLEM 4 Find the effective resistance of a circuit comprising three resistors, having resistance values of 1.2 kΩ, 5.6 kΩ and 7.1 kΩ.

SAMPLE PROBLEM 5 a

In the series circuit in the figure on the right, the emf of the power supply is 100 V; the current at point a, Ia , equals 1 A; and the value of R2 is 60 Ω. Find the: a. current at point b b. voltage drop across R2 c. voltage drop across R1 d. value of R1 .

R1 b

R2

ε

c

Teacher-led video: SP5 (tlvd-0027) THINK

WRITE

The current is the same at all points along the series circuit. Therefore, the current at point b, Ib , is 1 A. b. Substitute known values in the equation for the voltage drop across the resistor, V2 = IR2 .

a. Ib

a.

c.

This is a series circuit so the voltage of the battery equals the sum of the voltage drops around the circuit. So 𝜀 = V1 + V2 .

d.

The resistance can be found using the relationship V1 = IR1 and making R the subject.

= 1A

V2 = IR2 = 1 A × 60 Ω = 60 V c. 𝜀 = V1 + V2 100 V = V1 + 60 V b.

d.

V1 = 40 V V1 = IR1 40 V = 1 A × R1 R1 = 40 Ω

120 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

PRACTICE PROBLEM 5 A 24 V battery supplies the energy to two objects connected in a series circuit as shown in the figure on the right. The current at point a, Ia , is 0.6 A and the value of the known resistor is 15 Ω. Find the: a. current at point b b. voltage drop across the known resistor c. voltage drop across the unknown resistor, R1 d. value of R1 .

0.6 A R1

a

b

24 V

c 15 Ω

4.3.2 Resistors in parallel In a parallel branch of a circuit, there is more than one path for the current to flow through. The total current flowing into the parallel section of a circuit equals the sum of the individual currents flowing through each resistor. IT = I1 + I2 + I3 As can be seen in figure 4.5, the left-hand sides of all the resistors are connected to point A, so they are all at the same voltage. This means that all charges on that side of the resistors have the same amount of electrical potential energy. Similarly, the right-hand sides of the resistors are connected to point B, therefore, they also are at the same voltage. This means that each resistor in a parallel section of a circuit has the same voltage drop across it. VT = V1 = V2 = V3 FIGURE 4.5 A parallel branch of a circuit I1

R1

A IT

B I2

R2

I3

R3

IT

In a parallel section of a circuit, the total current equals the sum of the individual currents and the voltage drops across each resistor are the same. It is possible to derive an expression for the effective resistance, RT , of a parallel section of a circuit. IT = I1 + I2 + I3 V V V V ⇒ = + + RT R1 R2 R3 (since I =

V for each resistor and the whole section of the circuit) R

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Dividing both sides of the expression for the effective resistance by V gives: ⇒

1 1 1 1 = + + RT R1 R2 R3

This means that the reciprocal of the effective resistance is equal to the sum of the reciprocals of the individual resistances. The effective resistance is less than the smallest individual resistance. The more resistors there are added in parallel, the more paths there are for the current to flow through, and the easier it is for the current to flow through the parallel section.

Modelling resistors in parallel One way to help understand this concept is to use the hydraulic model. Current is represented by water flowing in a pipe. Resistors are represented as thin pipes. The thinner the pipe, the greater the resistance; therefore, less water can flow in the circuit. A conductor is represented by a large pipe through which water flows easily. The source of emf is represented by a pump that supplies energy to the circuit. If there is only one thin pipe, as in figure 4.6a, it limits the flow of water. Adding another thin pipe beside the first, as in figure 4.6b, allows more water to flow. The total resistance offered by the two thin pipes in parallel is less than that offered by an individual thin pipe. FIGURE 4.6 The hydraulic model for resistors in parallel showing (a) a circuit with one ‘resistor’ and (b) a second ‘resistor’ added in parallel, which allows more current to flow and reduces the effective resistance (a)

(b)

Pump

Pump

Resources Digital document eModelling: Exploring resistors in parallel with a spreadsheet (doc-0047) Weblinks

DC circuit water analogy Simple electric circuits

SAMPLE PROBLEM 6

What is the effective resistance of three resistors connected in parallel if they have resistance values of 5.0 Ω, 10 Ω and 20 Ω? Teacher-led video: SP6 (tlvd-0028) THINK 1.

The three resistors are connected in parallel, so substitute the known values in: 1 1 1 1 = + + RT R1 R2 R3

122 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

1 1 1 1 = + + RT 5.0 Ω 10 Ω 20 Ω

WRITE

2.

Solve for RT .

3.

State the solution.

1 2 1 4 + + = RT 20 Ω 20 Ω 20 Ω 7 1 = ⇒ RT 20 Ω 20 Ω = 2.9 Ω ⇒ RT = 7 The effective resistance of the three resistors connected in parallel is 2.9 Ω. ⇒

Note: The effective resistance of a set of resistors connected in parallel is always less than the value of the smallest resistor used. Adding resistors in parallel increases the number of paths for current to flow through, so more current can flow and the resistance is reduced. R If there are n resistors of equal value, R, the effective resistance, RT , will be RT = . n

PRACTICE PROBLEM 6 Four resistors having values of 5 Ω, 5 Ω, 15 Ω and 20 Ω are connected in parallel. Calculate their effective resistance. SAMPLE PROBLEM 7

Consider the parallel circuit shown in the following figure. The emf of the power supply is 9.0 V, R2 has a resistance of 10 Ω and the current flowing through the power supply is 1.35 A. Find: a. the voltage drop across R1 and R2 b. I2 , the current flowing through R2 c. I1 , the current flowing through R1 d. the resistance of R1 e. the effective resistance of the circuit.

V=9V

IT = 1.35 A I1 I2

R1

R2 = 10 Ω

Teacher-led video: SP7 (tlvd-0029) THINK a. 1. 2. b. 1.

For a parallel circuit, V1 = V2. State the solution. Find the current in the resistor using the relationship V = IR and making I the subject.

State the solution. c. 1. Recall the formula for the current leaving the battery. 2. Make I1 the subject and substitute the known values. 2.

WRITE

V1 = V2 = 9 V The voltage drop across R1 and R2 is 9 V. V b. I2 = R2 9V ⇒ I2 = = 0.90 A 10 Ω I2 , the current flowing through R2 , is 0.90 A. c. IT = I1 + I2 a.

I1 = IT − I2 = 1.35 A − 0.90 A = 0.45 A

TOPIC 4 Circuit analysis 123

3. d. 1.

2. e. 1.

State the solution. Find the resistance by transposing the formula V = IR to make R the subject.

State the solution. The resistors are in parallel.

2.

Substitute the known values.

3.

RT is the reciprocal.

4.

State the solution.

I1 , the current flowing through R1 , is 0.45 A. V d. R1 = I1 9V ⇒ R1 = 0.45 A R1 = 20 Ω The resistance of R1 is 20 Ω. 1 1 1 e. = + RT R1 R2 1 1 1 ⇒ + = RT 10 Ω 20 Ω 1 3 ⇒ = RT 20 Ω 20 Ω ⇒ RT = 3 = 6.7 Ω The effective resistance of the circuit is 6.7 Ω.

PRACTICE PROBLEM 7 Consider the parallel circuit shown in the figure on the right. The emf of the power supply is 24 V and the heater R1 has a resistance of 40 Ω. When the switch is closed the current flowing is 0.8 A. Find: a. the voltage drop across R1 and R2 b. I1 , the current flowing through R1 c. I2 , the current flowing through R2 d. the resistance of R2 e. the effective resistance of the circuit.

24 V

0.8 A

I1

I2

R1 = 40 Ω

R2

SAMPLE PROBLEM 8

Find the effective resistance when a 10.0 Ω resistor is placed in parallel with a 10.0 kΩ resistor. Teacher-led video: SP8 (tlvd-0030) THINK 1.

The two resistors are connected in parallel, so substitute the known values in: 1 1 1 = + RT R1 R2

124 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

WRITE

1 1 1 = + RT R1 R2 1 1 = + 10.0 Ω 10 000 Ω

1 1000 + 10 000 Ω 10 000 Ω 1001 = 10 000 Ω 10 000 Ω RT = 1001 = 9.99 Ω The effective resistance is 9.99 Ω. =

2.

The reciprocal gives the value of the resistance.

3.

State the solution.

PRACTICE PROBLEM 8 Find the effective resistance when a 1.2 kΩ resistor is placed in parallel with a 4.8 kΩ resistor. Note: Adding a large resistance in parallel with a small resistance slightly reduces the effective resistance of that part of a circuit. Parallel circuits are used extensively. Australian households are wired in parallel with an AC voltage of 230 V. This is equivalent to a DC voltage of 230 V, and all the formulae that have been presented so far can be used for analysing AC circuits. The advantage of having parallel circuits is that all appliances have the same voltage across them and the appliances can be switched on independently. If appliances were connected in series, they would all be on or off at the same time; and they would share the voltage between them, so no appliance would receive the full voltage. This would present problems when designing the devices, as it would not be known what voltage to allow for. Car lights, front and rear, are wired in parallel for the same reason. If one lamp ‘blows’, the other lamps will continue functioning normally.

Resources Digital documents Investigation 4.1 Series circuits (doc-31866) Investigation 4.2 Parallel circuits (doc-31867)

4.3.3 Short circuits A short circuit occurs in a circuit when a conductor of negligible resistance is placed in parallel with a circuit element. This element may be a resistor or a globe. The result of a short circuit is that virtually all the current flows through the conductor and practically none flows through the circuit element. Because there is effectively no voltage drop across the wire, there is also no voltage drop across the circuit element and no current flows through it. Think of what would happen in the hydraulic model if a conducting pipe were placed beside a thin pipe. This situation is represented in both ways in figure 4.7. In this case, the current through the power supply passes through R1 , but then flows through the short circuit, effectively avoiding R2 and R3 .

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FIGURE 4.7 (a) Circuit diagram showing a short circuit (b) Hydraulic model of a short circuit (a)

(b)

R1

R1

R2

R2

Short circuit

Pump R3 R3

SAMPLE PROBLEM 9

The following figure shows a 10 kΩ resistor that has been short circuited with a conductor of 0 Ω resistance. Calculate the effective resistance of this arrangement.

R1 = 0

R2 = 10 kΩ

Teacher-led video: SP9 (tlvd-0031) THINK 1.

The two resistors are connected in parallel, so substitute the known values in: 1 1 1 = + RT R1 R2

2.

The reciprocal gives the value of the resistance. State the solution.

3.

1 1 1 = + RT R1 R2 1 1 = + 0 Ω 10 000 Ω =∞ ⇒ RT = 0 Ω The effective resistance is 0 Ω.

WRITE

PRACTICE PROBLEM 9 Find the effective resistance when a 1.2 kΩ resistor is placed in parallel with a 4.8 kΩ resistor and a 500 Ω resistor.

126 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

4.3.4 The voltage divider The voltage divider is an example of resistors in series. It is used to divide or reduce a voltage to a value needed for a part of the circuit. A voltage divider is used in many control circuits, for example, turning on the heating in a house when the temperature drops. The voltage divider has an input voltage, Vin , and an output voltage, Vout . A general voltage divider is shown in figure 4.8. FIGURE 4.8 A general voltage The current, I, flowing through R1 and R2 is the same since R1 and divider R2 are in series. Vin = I (R1 + R2 ) R1

⇒I=

Vin R1 + R2

Vin

Vout = IR2 R2

Vin ⇒ Vout = × R2 R1 + R2

Vout

This can be rewritten as: ] R2 Vin Vout = R1 + R2 [

More generally: V out =

[

] resistance across which V out is taken R V in = out V in sum of all resistances Rtotal

If R1 and R2 are equal in value, the voltage will be divided equally across both resistors. If R1 is much greater than R2 , then most of the voltage drop will be across R1 .

Resources Digital document Investigation 4.3 Determining emf and internal resistance (doc-31868) Teacher-led video Investigation 4.3 Determining emf and internal resistance (tlvd-0813) Interactivity

Voltage dividers (int-6392)

SAMPLE PROBLEM 10

Calculate the value of the unknown resistor in the voltage divider shown in the following figure, if the output voltage is required to be 4.0 V.

TOPIC 4 Circuit analysis 127

6.0 V

R1 = 2.2 kΩ

R2

Vout

0V

Teacher-led video: SP10 (tlvd-0032) THINK

WRITE

1.

Substitute known values in the formula, RV Vout = 2 in for a two resistor voltage divider. R1 + R2

2.

Solve by making R2 the subject.

3.

R2 Vin R1 + R2 6.0 V × R2 ⇒ 4.0 V = 2.2 kΩ + R2 8.8 kΩV + 4.0 VR2 = 6.0 VR2 2.0 V R2 = 8.8 kΩ V Vout =

R2 = 4.4 kΩ The value of the unknown resistor in the voltage divider is 4.4 kΩ.

State the solution.

PRACTICE PROBLEM 10 Calculate the value of the unknown resistor in the voltage divider in sample problem 10 if the output voltage is to be 1.5 V.

4.3 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Find the effective resistance of the following sets of resistors if they are connected in series. (a) 2.7 Ω, 9.8 Ω (b) 12 Ω, 20 Ω, 30 Ω (c) 1.2 kΩ, 3.2 kΩ, 11 kΩ 2. Find the voltage at the points a, b, c and d in the following figure, given that Vbc is 5.0 V. b a c V = 9.0 V

d

128 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

3. Find the unknown quantities in the series circuit shown in the following figure. Ia R1 = 4.0 Ω

V1

V = 6.0 V

R2

V2

1.0 A

4. Find the effective resistance when the following resistors are connected in parallel. (a) 30 Ω, 20 Ω (b) 5.0 Ω, 10 Ω, 30 Ω (c) 15 Ω, 60 Ω, 60 Ω 5. Two 10 Ω resistors are connected in parallel across the terminals of a 15 V battery. (a) What is the effective resistance of the circuit? (b) What current flows in the circuit? (c) What is the current through each resistor? 6. Three resistors of 60 Ω, 30 Ω and 20 Ω are connected in parallel across a 90 V power source as shown in the following figure.

60 Ω

ε = 90 V

30 Ω

20 Ω

(a) Calculate the effective resistance of the circuit. (b) Find the current flowing through the source. (c) What is the current flowing through each resistor? 7. Find the unknown quantities in the parallel circuit shown in the following figure. Ia I1

R1 = R 6.0 Ω 2

V1

V = 6.0 V

I2

V2

1.5 A

8. Three resistors, having resistance values of 6 Ω, 18 Ω and 9 Ω, are connected in parallel across a 36 V power supply. (a) What current flows through each resistor? (b) What total current flows in the circuit? (c) What is the effective resistance of the circuit? 9. The following figure shows an arrangement of switches and globes connected to a source of emf. G2

S3

S4

S2 S1

G3

G1

TOPIC 4 Circuit analysis 129

Which globes would light up if the following sets of switches are closed? (a) Switch S2 only (b) Switch S3 only (c) Switch S4 only 10. (a) Find the output voltage for the voltage divider shown in circuit (a). (b) What is the output voltage of circuit (b) if a load of resistance 4.4 kΩ is connected across the output terminals of the voltage divider? (b)

(a) R1 = 2.2 kΩ

R1 = 2.2 kΩ

Vin = 6.0 V

Vin = 9.0 V R2 = 2.2 kΩ

Vout

R2 = 4.4 kΩ

Vout

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

4.4 Non-ohmic devices in series and parallel KEY CONCEPTS • Calculate and analyse the effective resistance of circuits comprising parallel and series resistance and voltage dividers. • Investigate and apply theoretically and practically concepts of current, resistance, potential difference (voltage drop) and power to the operation of electronic circuits comprising resistors, light bulbs, diodes, thermistors, light dependent resistors (LDRs), light-emitting diodes (LEDs) and potentiometers (quantitative V analysis restricted to use of I = and P = VI). R • Investigate practically the operation of simple circuits containing resistors, variable resistors, diodes and other non-ohmic devices. • Describe energy transfers and transformations with reference to transducers.

4.4.1 Non-ohmic devices Non-ohmic devices do not obey Ohm’s Law. Their current-versus-voltage characteristics can be presented graphically. V The value of is not constant for non-ohmic devices. I The rules for series and parallel circuits still apply when analysing circuits containing non-ohmic devices, such as light bulbs, diodes, thermistors, light dependent resistors (LDRs) and light-emitting diodes (LEDs). Devices in series have the same current and share the voltage. Devices in parallel have the same voltage and the current is shared between them. The actual values of the voltage or current are obtained from the current-versus-voltage (V–I) graphs for the devices.

130 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 4.9 Current-versus-voltage characteristic for a silicon diode

80

Current (mA)

60

40

20

0

0.2

0.4

0.6

1.0

0.8

1.2

Voltage drop across diode

SAMPLE PROBLEM 11 5 Current (A)

The following figure shows the current-versus-voltage graph for two electrical devices. If X and Y are in parallel and the current through X is 2 A, calculate: a. the voltage across Y b. the current through Y.

X

4

Y

3 2 1 0 5

10 15 20 Voltage drop (V)

25

Teacher-led video: SP11 (tlvd-0033)

THINK

WRITE

As X and Y are in parallel, the voltage across X equals the voltage across Y. Use the graph to find the values. b. Again use the graph.

a.

When the current through X is 2 A, the voltage is 10 V, so the voltage across Y is also 10 V.

b.

When the voltage across Y is 10 V, the current through Y is seen to be 3 A.

a.

TOPIC 4 Circuit analysis 131

PRACTICE PROBLEM 11 The following figures show two electrical devices, A and B, connected in a simple circuit and the current-versus-voltage graph for A and B. A

A

B

I (mA)

100 80 60 40 20 0

B 2

4

6

8

10

V

When the current through A is 80 mA, calculate: a. the voltage across B b. the current through B.

4.4.2 Transducers and sensors Transducers are devices that convert energy from one form to another. They can be affected by, or can affect, the environment. The word transducer comes from the Latin for ‘to lead across’. Table 4.2 lists a range of transducers and their energy conversions.

TABLE 4.2 Examples of transducers Transducer

Energy conversion From

To

Solar cell

Light

Electrical

Loudspeaker

Electrical

Sound

Microphone

Sound

Electrical

LED

Electrical

Light

Antenna

Electromagnetic

Electrical

Thermocouple

Heat energy (temp. difference)

Electrical

Peltier cooler

Electrical

Heat energy (temp. difference)

Piezoelectric gas lighter*

Stored mechanical energy due to pressure

Electrical

*The piezoelectric effect is also used in strain gauges to measure the stress in construction materials and in the accelerometers found in video game controllers and guidance systems.

Sensors are a subset of transducers where the energy conversion is to electrical, that is, to a variation in voltage. Some sensors generate the voltage directly, for example, piezoelectric devices. Other sensors whose resistance changes, such as LDRs and thermistors, use a voltage divider circuit.

132 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

SAMPLE PROBLEM 12

The resistance of a thermistor changes with temperature as shown in the following graph. In the voltage divider circuit shown, the thermistor is one of the two resistors; the other is a resistor with a fixed resistance value.

Resistance (kΩ)

9.0 V 2 R 1

Voltage sensitive switch 0V

0 20

40

60

Temperature (°C)

As the temperature drops, the resistance of the thermistor increases. As the thermistor’s resistance increases, its share of the voltage from the power supply also increases, while that of the fixed value resistor will decrease. A voltage sensitive switch is placed across the thermistor. It is built to turn on a heater when the voltage across the thermistor is greater than 6 V. Your task as the circuit designer is to determine the resistance value required for the fixed-value resistor to turn on the heater at 19 °C. Teacher-led video: SP12 (tlvd-0034)

1.

Determine the resistance of the thermistor at 19 °C using the graph.

2.

Substitute the resistance into the voltage divider equation.

3.

Solve for R.

4.

State the solution.

WRITE

From the graph, at 19 °C the thermistor has a resistance [ of 1.5 ]kΩ (1500 Ω). R2 Vin Vout = R1 + R2 ] [ 1.5 kΩ 6V = × 9V R + 1.5 kΩ 6 × (R + 1500) = 1500 × 9 6R + 9000 = 13 500 6R = 4500 R = 750 Ω The resistance required for the fixed-value resistor to turn on the heater at 19 °C is 750 Ω.

PRACTICE PROBLEM 12 The resistance-versus-temperature characteristics of a thermistor are shown in the following graph. The thermistor is connected in parallel with a fixed resistor similar to that shown in the circuit in sample problem 12. The switch across the thermistor is designed to turn on a warning light when the voltage across the thermistor is 4 V or greater. Determine the resistance value required for the fixed value resistor to turn on the warning light at 20 °C.

3 R (kΩ)

THINK

2

1

0

10

20

30

40

50

T (°C)

TOPIC 4 Circuit analysis 133

SAMPLE PROBLEM 13

To reduce the heating bill from the heater in sample problem 12, it is decided that the heater should be turned on when the temperature is below 18 °C. Should the value of the fixed-value resistor be increased or decreased? Explain. Teacher-led video: SP13 (tlvd-0035) THINK

WRITE

The voltage to turn on the switch will still be 6 V, so the voltage across the two resistors will be unchanged. The ratio of their resistance values will therefore also be the same. From the graph in sample problem 12 it can be seen that at 18 °C the thermistor’s resistance will be greater than it was at 19 °C. So to keep the ratio the same, R must increase. This can also be explained using current. As the resistance of the thermistor is higher at the lower temperature, there will be less current through both resistors. As the voltage drop across R is to remain the same, its resistance will need to be greater (V = IR).

The resistance must be increased.

PRACTICE PROBLEM 13 The thermistor and voltage sensitive switch from sample problem 12 are to be used for a cooling system. The cooling system is to turn on when the temperature is greater than 24 °C. To which resistor, the thermistor or the fixed resistor, should the voltage sensitive switch be connected so that the voltage is greater than 6 V for temperatures greater than 24 °C? Explain. What should be the value of the fixed resistor?

4.4.3 Potentiometers A potentiometer, called a ‘pot’ for short, is a variable voltage divider. It consists of a fixed resistor, usually a length of wire, with a contact that can slide up and down, varying the amount of resistance in each arm of the voltage divider. Potentiometers are commonly used as controls in radio equipment, either as a slide control or in a rotary form. They are also the basis of joysticks in game controls.

134 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 4.10 (a) The symbol for a potentiometer (b) A potentiometer (a)

(b)

4.4 EXERCISE

1. The voltage-versus-current characteristic graph for a non-ohmic device is shown on the right. (a) What is the device’s current when the voltage drop across it is 100 V? (b) What is the voltage drop across the device when the current through it is 16 mA? (c) What is the resistance of the device when it carries a current of 16 mA? 2. The device described in question 1 is placed in series with a 5.0 kΩ resistor and a voltage drop is applied across the combination. This arrangement is shown in the following figure. The current in the resistor is measured to be 6.0 mA. (a) What is the voltage drop across the resistor? (b) What is the current in the device? (c) What is the voltage drop across the device? (d) What is the total voltage drop across the device and the resistor? 3. The device described in question 1 is now placed in parallel with a 5.0 kΩ resistor and a new voltage is applied across the combination. This arrangement is shown in the figure on the right. The current in the resistor is measured to be 20 mA. (a) Calculate the voltage drop across the resistor. (b) What is the voltage drop across the device? (c) What is the current in the device? (d) What is the total current in the circuit?

Voltage (V)

To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 160 140 120 100 80 60 40 20 2 4 6 8 10 12 14 16 Current (mA) Non-ohmic device R = 5.0 kΩ A I = 6.0 mA Non-ohmic device

IR = 20 mA

R = 5.0 kΩ

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

4.5 Power in circuits KEY CONCEPT • Compare power transfers in series and parallel circuits.

Recall that the power being used in a circuit element is the product of the voltage drop across it and the current through it: P = VI. The total power being provided to a circuit is the sum of the power being used in, or ‘dissipated by’, the individual elements in that circuit. It does not matter if the elements are connected in series or in parallel. PT = P1 + P2 + P3 = . . .

TOPIC 4 Circuit analysis 135

SAMPLE PROBLEM 14

A household electrical circuit is wired in parallel. Find the total current flowing in the circuit if the following appliances are being used: a 600 W microwave oven, a 450 W toaster and a 1000 W electric kettle. Household circuits provide a voltage drop of 230 V across each appliance. Teacher-led video: SP14 (tlvd-0036) THINK

WRITE

1.

The total power being used in the circuit is the sum of the power used in each component.

PT = 600 + 450 + 1000 = 2050 W

2.

Transpose the formula P = IV to make I the subject, then substitute the known values.

IT =

3.

State the solution.

PT V 2050 W = 230 V = 8.91 A The total current flowing in the circuit is 8.91 A.

PRACTICE PROBLEM 14 A household electrical circuit is wired in parallel. Find the total current flowing through a household circuit when the following devices are being used: a 400 W computer, a 200 W blender, a 500 W television and a 60 W lamp.

4.5 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Three resistors of value 25 Ω, 15 Ω and 10 Ω are connected in series to a 10 V power supply. (a) Calculate the current in the circuit. (b) What is the voltage drop across each resistor? (c) At what rate is energy being transformed in each resistor? (d) What is the total power rating of the circuit? 2. The three resistors in question 1 are connected in parallel. (a) Calculate the current in the circuit. (b) What is the voltage drop across each resistor? (c) At what rate is energy being transformed in each resistor? (d) What is the total power rating of the circuit? 3. Design a dimmer switch circuit for a light. Does the circuit consume more power when the light is bright or dull? Justify your answer.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

136 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

4.6 Review 4.6.1 Summary • • • • • • •

An electric circuit is a closed conducting path containing a source of emf. In a series circuit the components are placed one after the other. There is only one path for the charge carriers to flow through. In a series circuit the current is the same at all points; the sum of the potential drops is equal to the emf of the circuit. The effective resistance of resistors placed in series is equal to the sum of the individual resistances. In a parallel circuit, there is more than one path for the current to flow through. Resistors in parallel have the same potential difference across them; the current in each resistor adds up to the total current flowing into the parallel branch. In a parallel arrangement of resistors, the reciprocal of the effective resistance is equal to the sum of the reciprocals of the individual resistances: 1 1 1 = + RT R1 R2

• • • • • •

A short circuit occurs when a conductor of negligible resistance is placed in parallel with a circuit element and stops current from flowing through it. Circuits containing non-ohmic devices can be analysed using the rules for series and parallel circuits with their current-versus-voltage characteristic graphs. The total power used in a circuit equals the sum of the powers used in individual devices. A voltage divider is used to reduce an input voltage to some required value. A voltage divider consists of two or more resistors arranged in series to produce a smaller voltage at its output. The output of a voltage divider can be calculated using the equation: ] R2 Vin Vout = R1 + R2 [

•

A transducer is a device that can be affected by, or affects, the environment.

Resources

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0029).

4.6.2 Key terms A charge carrier is a charged particle moving in a conductor. A conductor is a material that contains charge carriers; that is, charged particles can move and travel freely through the material. Electric charge is a basic property of matter. It occurs in two states: positive (+) charge and negative (−) charge. (page 3) An electric circuit is a closed loop of moving electric charge. Electromotive force is a measure of the energy supplied to a circuit for each coulomb of charge passing through the power supply.

TOPIC 4 Circuit analysis 137

A load is a device where electrical energy is converted into other forms to perform tasks such as heating or lighting. A non-ohmic device is one for which the resistance is different for different currents passing through it. Devices connected in parallel are joined together so that one end of each device is joined at a common point and the other end of each device is joined at another common point. Power is the rate of doing work, or the rate at which energy is transformed from one form to another. The resistance, R, of a substance is defined as the ratio of voltage drop, V, across it to the current, I, flowing through it. A resistor is used to control the current flowing through, and the voltage drop across, parts of a circuit. Devices connected in series are joined together one after the other. A switch stops or allows the flow of electricity through a circuit. A thermistor is a device that has a resistance which changes with a change in temperature. A light-dependent resistor (LDR) is a device that has a resistance which varies with the amount of light falling on it. A potentiometer is a variable voltage divider, with a fixed resistor that can slide up and down. Transducers are devices that convert energy from one form to another form. The voltage divider is used to reduce, or divide, a voltage to a value needed for a part of the circuit.

Resources Digital document Key terms glossary (doc-32180)

4.6.3 Practical work and investigations Investigation 4.1 Series circuits Aim: To observe the voltage and current in a circuit set up in series Digital document: doc-31866

Investigation 4.2 Parallel circuits Aim: To observe the voltage and current in a circuit with resistors set up in parallel Digital document: doc-31867

Investigation 4.3 Determining emf and internal resistance Aim: To determine the emf and internal resistance of circuits with old and new dry cells with resistors of varying resistances Digital document: doc-31868 Teacher-led video: tlvd-0813

Resources Digital document Practical investigation logbook (doc-32181)

138 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

4.6 Exercises To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au.

4.6 Exercise 1: Multiple choice questions Use the following information to answer questions 1 to 3. Three resistors of 20 Ω, 30 Ω and 50 Ω are connected in series across a 9.0 V battery. 1. What is the effective resistance of the three resistors? A. 0.01 Ω B. 0.33 Ω C. 100 Ω D. 330 Ω 2. What current flows in the circuit? A. 90 mA B. 27 mA C. 9 A D. 27 A 3. What is the total voltage drop across the circuit? A. 0 V B. 0.09 V C. 9 V D. 900 V Use the following information to answer questions 4 to 6. Three resistors of 6 Ω, 18 Ω and 9 Ω are connected in parallel across a 36 V power supply. 4. What current flows through the 9 Ω resistor? A. 2 A B. 4 A C. 6 A D. 12 A 5. What is the effective resistance of the circuit? A. 3 Ω B. 4.5 Ω C. 12 Ω D. 33 Ω 6. What total current flows in the circuit? A. 2 A B. 4 A C. 6 A D. 12 A 7. Three resistors of value 25 Ω, 15 Ω and 10 Ω are connected in series to a 10 V power supply. What is the current in the circuit? A. 0.20 A B. 0.40 A R C. 0.70 A D. 1.0 A 20 V 8. In the circuit on the right what is the voltage drop between A and B? 4R A. 0 V B. 4 V C. 16 V D. 20 V 9. What is the value of the resistor shown in the circuit on the right? A. 3 kΩ B. 4 kΩ 3 kΩ 6V C. 6 kΩ D. 12 kΩ 10. Which behaviour is more likely when a thermistor is connected in R an electrical circuit? A. The resistance of the thermistor increases as the operating temperature increases. B. The ratio of the voltage across the thermistor to the current flowing through it is constant. C. When the potential difference across the thermistor is increased the current flowing through it is also likely to increase. D. The behaviour of the thermistor is not temperature dependent.

A

B

4V

4.6 Exercise 2: Short answer questions 1.

In the series circuit shown in the figure on the right, Vab = 20 V, R2 = 30 Ω, and Ia = 2.0 A. Find values for: a. Ib b. Vbc c. R1 d. R T , the effective resistance of the circuit e. V, the emf of the battery.

V d a R1 b R2 c

TOPIC 4 Circuit analysis 139

2.

3.

In the parallel circuit shown in the following figure, I1 = 0.20 A, R1 = 60 Ω and IT = 0.50 A. Find values for: a. Vab b. Vcd c. 𝜀 d. I2 e. R2 .

a I1

ε

R1

I2 c

d R2

IT

When the following circuit is powered by an 18 V power supply, what is the current through the 6 Ω resistor? 18 V

6Ω

3Ω

12 Ω

Two lamps labelled as 12 V 9 W and 12 V 24 W are connected in parallel with a 12 V battery. a. What is the effective resistance of the circuit? b. Will the lamps be equally bright? Explain your answer. 5. Consider the following circuit. 4.

20 Ω

24 V

60 Ω

What current passes through each resistor? How much power is dissipated in each resistor? The switch is now closed. c. What current passes through each resistor? d. How much power is dissipated in each resistor? 6. Some lighting circuits are controlled by two switches in different locations. Design a DC circuit to show how this could be achieved. a.

b.

140 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

b

7.

Find the voltage drop between A and B in each of the following voltage divider circuits. (a)

(b) 60 Ω

20 Ω

10 V

9V A

A

40 Ω

30 Ω B

B 40 Ω

8.

Find the value of the unknown resistor in the voltage dividers shown. (a)

(b) R

10 kΩ 10 V

9.0 V 5.0 kΩ

2.5 V

R

6.0 V

A student connects a variable resistor and a lamp in series in a circuit to a fixed power supply. a. What will happen in the lamp as the resistance of the variable resistor is increased? b. As the resistance of the variable resistor is increased will the power consumed in the circuit change? Justify your answer. The lamp and the variable resistor are now connected in parallel. c. What will happen in the lamp as the resistance of the variable resistor is increased? d. What will happen to the power consumed in the circuit? 10. A thermistor has the temperature-versus-resistance characteristic shown by the bottom curve in the following graph. It is placed in the voltage divider shown in the circuit diagram. 9.

+9.0 V 100 000

R1

Resistance (Ω)

10 000

R2 = thermistor

1000

Vout

0V 100

10

0

50

100

150

200

250

300

Temperature (°C)

What is the resistance of the thermistor when the temperature is 150 °C? b. What is the value of the variable resistor if the temperature is 200 °C and Vout is 6 V? a.

TOPIC 4 Circuit analysis 141

4.6 Exercise 3: Exam practice questions Question 1 (5 marks) Consider two people connected in series with a 25 V source across them. The first person, R1 , has a resistance of 1000 Ω and the second person, R2 , has a resistance of 1500 Ω. a. Determine the current through each person. 1 mark b. What is the voltage drop across each person? 1 mark The two people now connect themselves in parallel with the 25 V power source. c. Determine the current through each person. 2 marks d. What is the voltage drop across each person? 1 mark Question 2 (2 marks) Prove mathematically that the resistance of a short-circuited resistor branch is 0. Question 3 (3 marks) Consider the circuit on the right. a. Find the value of the unknown resistor. b. What is the power of the circuit?

1 mark 1 mark

R 9V b

5 kΩ

8 Vout

a

The resistor is now replaced by a variable resistor as shown. c. What happens to the voltage drop Vab , when the resistance of the variable resistor decreases and the other resistance is unchanged? 1 mark b

a

142 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Question 4 (3 marks) A thermistor has the temperature-versus-resistance characteristics shown by the top curve in the following graph. It is placed in the voltage divider shown in the following circuit diagram. +9.0 V 100 000 R1

10 000

Resistance (Ω)

R2 = Thermistor 1000

Vout

0V

100

10

0 50

a. b.

100 150 200 Temperature (°C)

250

300

What is the resistance of the thermistor when the temperature is 200 °C? 1 mark Calculate the value of the variable resistor in the voltage divider if the temperature is 100 °C and the output voltage is 4.5 V. 2 marks

Question 5 (3 marks) a. Find the value of R2 in the voltage divider in the following figure that would give an output voltage of 2.0 V.

2 marks

R1 = 3.0 kΩ Vin = 6.0 V R2

Vout = 2.0 V

TOPIC 4 Circuit analysis 143

The resistor R2 is now replaced by the thermistor with a temperature-versus-resistance curve as shown by the middle curve in the following graph.

100 000

Resistance (Ω)

10 000

1000

100

10

0 50

100

150

200

250

300

Temperature (°C)

b.

What is the value of Vout when the temperature of the thermistor is 100 °C?

4.6 Exercise 4: studyON topic test Fully worked solutions and sample responses are available in your digital formats.

Test maker Create unique tests and exams from our extensive range of questions, including practice exam questions. Access the assignments section in learnON to begin creating and assigning assessments to students.

144 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

1 mark

AREA OF STUDY 2 HOW DO ELECTRIC CIRCUITS WORK?

5

Using electricity and electrical safety 5.1 Overview Numerous videos and interactivities are available just where you need them, at the point of learning, in your digital formats, learnON and eBookPLUS at www.jacplus.com.au.

5.1.1 Introduction In this topic you will see how the basic rules for series and parallel circuits can be applied to household use of electricity. The safe use of electricity will also be discussed, as well as the effects of electric shocks on the human body. Parallel circuits are usually preferred over series circuits because each device in a parallel circuit can be turned on and off independently. Devices in parallel circuits also have the same voltage drop. In series circuits, on the other hand, the devices have the same current flowing through them and the voltage is shared across the circuit. If one device is switched off, all the devices go off. FIGURE 5.1 Light-duty all-electric and plug-in hybrid vehicles (combining electric drives with combustion engines) reduce reliance on petrol, increase efficiency and reduce greenhouse gas emissions.

TOPIC 5 Using electricity and electrical safety 145

5.1.2 What you will learn KEY KNOWLEDGE After completing this topic you will be able to: • model household (AC) electrical systems as simple direct current (DC) circuits • explain why the circuits in homes are mostly parallel circuits. • model household electricity connections as a simple circuit comprising fuses, switches, circuit breakers, loads and earth • compare the operation of safety devices including fuses, circuit breakers and residual current devices (RCDs) • describe the causes, effects and treatment of electric shock in homes and identify the approximate danger thresholds for current and duration • apply the kilowatt-hour (kW-h) as a unit of energy. Source: VCE Physics Study Design (2016–2021) extracts © VCAA; reproduced by permission.

PRACTICAL WORK AND INVESTIGATIONS Practical work is a central component of learning and assessment. Experiments and investigations, supported by a Practical investigation logbook and Teacher-led videos, are included in this topic to provide opportunities to undertake investigations and communicate findings.

Resources Digital documents Key science skills — VCE Units 1–4 (doc-31856) Key terms glossary (doc-32182) Practical investigation logbook (doc-32183)

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0030).

5.2 Household electricity and usage KEY CONCEPTS • Model household (AC) electrical systems as simple direct current (DC) circuits. • Explain why the circuits in homes are mostly parallel circuits. • Model household electricity connections as a simple circuit comprising fuses, switches, circuit breakers, loads and earth. • Apply the kilowatt-hour (kW-h) as a unit of energy.

5.2.1 Household use of electricity Houses connected to the main electrical grid are supplied with an AC voltage of 230 VRMS at a frequency of 50 Hz. The term ‘230 VRMS ’ means that the AC voltage produces the same heating effect when applied across a conductor as would a DC voltage of 230 V applied across the same conductor. The actual value of the voltage oscillates between +325 V and –325 V. ‘Root mean square’ (RMS) refers to the mathematical process by which the equivalent DC voltage is calculated. A‘frequency of 50 Hz’ means that the full cycle is completed 50 times each second. The voltage supplied is sinusoidal in nature.

146 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Frequency ( f ) is a measure of how many times per second an event happens. The unit for frequency is the hertz (Hz). One hertz means one cycle or event per second. Period (T) is the amount of time one cycle or event takes, measured in seconds. Period is the reciprocal of frequency: T=

1 1 or f = f T

A frequency of 50 Hz means that the period is 0.02 seconds, as shown in figure 5.2. FIGURE 5.2 The variation of voltage with time for a supply of 230 VRMS , 50 Hz

325

Voltage (V)

230 Amplitude (Vpeak) 0

−325

0.01

0.02

0.03 Time (s)

T

Electricity is fed into the home through FIGURE 5.3 Many houses have a meter and circuit underground cables or overhead lines. It breaker in their meter box. The circuit breaker is in the enters the house through a switchboard. This bottom left corner. contains a mains switch which can cut off the supply of electricity to the house. There is also a meter that measures the amount of electrical energy transformed in the house. From the meter, the electricity enters a fuse box or circuit breaker box where it is divided among a number of parallel circuits. The role of fuses and circuit breakers is dealt with later in this topic. The structure of household circuits differs from the structure of DC circuits studied in previous topics. Household circuits make use of the earth to complete the circuit. The active wire in a circuit is connected to the 230 VRMS supply at the switchboard. Its voltage oscillates periodically between +340 V and –340 V relative to a reference voltage called ‘earth’. The earth is defined as having a voltage of 0 V. The neutral wire is connected to the neutral link at the switchboard, which is connected to the earth through the supply wires and via a metal rod driven into the ground at the switchboard. The neutral wire is always at 0 V. The voltage drop between the active and the neutral wire oscillates between +340 V and –340 V.

TOPIC 5 Using electricity and electrical safety 147

When an appliance is connected between the active wire and the neutral wire, current flows backwards and forwards between the active and neutral wires through the device, supplying it with energy. Conventional current flows from a high voltage to a low voltage. When the active wire is positive, the current flows from the active to the neutral wire and so to the earth. When the active wire is negative, the neutral wire at 0 V will have the higher voltage and the current will flow from the neutral to the active wire. In lighting circuits, only the active and neutral wires are used if there are no metal fittings. The current oscillates through the filaments of the globes, transforming energy. If metal fittings are used, they must be earthed. A typical lighting circuit is shown in figure 5.4. In power circuits, a third wire is used. This is called the earth wire. It connects the case of the appliance being used to the earth as a safety device. Its function is discussed in section 5.3.8. Otherwise, a power circuit operates in exactly the same way as a lighting circuit, with the current oscillating between the active and neutral wires through the appliance. Figure 5.5 shows how three-point power sockets are connected in a typical power circuit. FIGURE 5.4 A typical household lighting circuit

FIGURE 5.5 A typical power circuit

Input from supply wires A N Fuse box

Neutral link

N

A

Neutral link

Fuse

Fuse (or circuit breaker)

Active

Three-point socket Active Earthing connection at the home

Neutral Earth Neutral

Figure 5.6 shows the connection of the earth wire to the case of an appliance. FIGURE 5.6 The connection of the earth wire to the case of an appliance Appliance case

A

N Neutral link

Fuse

Active Neutral Earth

Note that in both the lighting and power circuits the switch is in the wire connecting the device to the active wire. A switch in the neutral wire would also turn off the device, but the functional parts of the device, such as the element of a toaster, would still be directly attached to the active wire and ‘live’. If you were to touch anything in contact with the active wire while you were in contact with the ground, there would be a voltage drop across you and a potentially lethal current could flow through you.

148 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

For this reason, if ever it is necessary to tamper with the functional parts of an electrical device, it must be unplugged first. The three-point socket may have been wrongly wired, and it is not worth taking the risk. Different-coloured insulating plastic is used to distinguish the three wires from each other. In the modern system, the active wire is brown, the neutral wire is blue and the earth wire is striped green and yellow. In the old system, red was used for the active, black for the neutral and green for the earth wires. FIGURE 5.7 A three-point socket. The top left point is the active connection (brown wire), the top right point is neutral (blue wire) and the bottom point is the earth (green/yellow wire). The switch is connected in the active wire.

Active Neutral

Switch Earth

Neutral Active Earth

5.2.2 Power ratings The total power used in an electrical circuit, be it series or parallel, is the sum of the power used by each device in the circuit. Electrical appliances or devices are given a power rating, which is usually printed on them. SAMPLE PROBLEM 1

A toaster is rated at 1400 W, 230 V. What current does the toaster draw when operating normally? b. What is the resistance of the toaster element when hot? a.

Teacher-led video: SP1 (tlvd-0037) THINK a. 1.

Use the relationship between power, voltage and current to make I the subject.

WRITE a.

P = VI P I= V

TOPIC 5 Using electricity and electrical safety 149

2.

Substitute the known values and solve for I.

3.

State the solution.

b. 1.

Use the relationship between voltage, resistance and current to make R the subject.

2.

Substitute the known values and solve for R.

3.

State the solution.

P = 1400 W, V = 230 V P I= V 1400 W = 230 V = 6.09 A When operating normally the toaster draws 6.09 A. V = IR V R= I V = 230 V, I = 6.09 V R= I 230 V = 6.09 A = 37.8 Ω The resistance of the toaster element when hot is 37.8 Ω.

Alternatively: The resistance could also have been calculated using the V2 . relationship: P = R 2. Transpose to make R the subject. 1.

3.

Substitute in the values, V = 230 V and P = 1400 W, and solve.

P=

V2 R

V2 P 2302 R= 1400 = 37.8 Ω R=

PRACTICE PROBLEM 1 A compact fluorescent light globe is rated at 15 W, 230 V. a. Calculate the current through the globe when it is operating normally. b. Calculate the resistance of the globe when it is operating normally.

5.2.3 Paying for electricity The meter on a household switchboard is used to measure how much electrical energy has been consumed on the premises. The amount of electrical energy used in a household can be determined by multiplying the rate of power transformation by the time. Since power is equal to the voltage drop multiplied by current, and the voltage drop across a household is 230 V, the meter on the switchboard records the total current that has passed through the premises over a certain period of time. This amount is converted into the amount of energy ‘consumed’ or transformed. For domestic and commercial electrical consumption the joule is too small a measure. The unit used for measuring energy in these cases is the kilowatt-hour (kW-h). This is the amount of energy transformed by a 1000 W appliance when used for 1 hour. 150 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

SAMPLE PROBLEM 2

How many joules does 1 kilowatt-hour represent? Teacher-led video: SP2 (tlvd-0038) THINK 1.

2.

WRITE

Energy (kW-h) = power (kW) × time (hrs)

1 kW-h = 1 kW × 1 h = 1000 W × (60 × 60) seconds = 3.6 × 106 J = 3.6 MJ 1 kW-h represents 3.6 MJ.

State the solution.

PRACTICE PROBLEM 2 An oven used 8 kW-h of energy. In joules how much energy was used?

The cost to consumers of 1 kW-h of electrical energy can be found on electricity accounts. Find out what 1 kW-h of electrical energy costs in your household. To calculate the cost of running a particular appliance, calculate the energy in kW-h by multiplying the power rating of the appliance, in kilowatts, by the number of hours that it was used for.

FIGURE 5.8 Many appliances have a sticker outlining their energy rating and the average kW-h of energy consumed per year.

If the power rating of the appliance is unknown the energy consumed can be found using the formula: E = VIt Where: E = the energy consumed V = the voltage drop across the device I = the current flowing through the device t = the time in seconds. The amount of energy is then converted into kW-h by dividing by 3.6 × 106 .

To calculate the cost, use the formula: cost = energy × rate

TOPIC 5 Using electricity and electrical safety 151

SAMPLE PROBLEM 3

A television draws 0.37 A of current when connected to a 230 V supply. What is the power rating of the television? b. How much energy does the television consume if it is operated for 5 hours a day for 4 weeks? c. What is the cost of running the television for this period of time if the consumer is charged at a rate of 16.381 cents per kW-h? a.

Teacher-led video: SP3 (tlvd-0039)

Substitute in P = VI.

P = 230 V × 0.37 A P = 85 W The power rating of the television is 85 W. b. Power = 85 W Time = 5 hours per day for 28 days = 140 h Energy = 85 W × 140 h = 11 900 W-h = 11.9 kW-h The television consumes 12 kW-h if it is operated for 5 hours a day for 4 weeks. c. Cost = 11.9 kW-h × 16.381 cents = 195 cents = $1.95 The cost of running the television is $1.95.

THINK

WRITE

a. 1.

a.

State the solution. b. 1. energy = power × time 2.

2. c. 1.

2.

State the solution.

cost = energy × rate State the solution.

PRACTICE PROBLEM 3 A home sound system consumes 2.4 W of electric power when it is on standby and connected to a 230 V supply. a. Calculate the current that flows through the system when it is on standby. b. Calculate the energy consumed by the system if it is left on standby for one week. c. Calculate the cost of leaving the system on standby for one week if electricity is priced at 12 cents per kW-h.

Resources Weblink Operating costs of electrical appliances

5.2 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. 2. 3. 4. 5.

What is meant when it is said that a house is supplied with electricity at 230 VRMS , 50 Hz? Why is an overload in a household circuit potentially dangerous? What coloured insulation is used for the active, neutral and earth wires in modern houses? Sketch a power point and plug. Label the active, neutral and earth in each case. Why do many appliances need to be connected to both the neutral and earth wires?

152 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

6. When is the earth wire used in household lighting circuits? 7. What is the cost of running a 300 W refrigerator for a year (365 days) if the refrigerator operates on average for 12 hours a day, and electricity costs 31.28 cents per kW-h? 8. The following table gives the power consumption of various products when they are on standby. Product Laptop computer

Power (W) 14.5

Modem

3.4

Cordless phone equipment

3.7

DVD player

2.4

Television

6.2

(a) Calculate the energy used by each product if it is left on standby for one year. (b) Calculate the mass of greenhouse gases produced by these products if they are left on standby for one year, assuming that 1 kW-h of energy produces greenhouse gases that are equivalent to 1.444 kilograms of CO2 .

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

5.3 Electrical safety KEY CONCEPTS • Compare the operation of safety devices including fuses, circuit breakers and residual current devices (RCDs). • Describe the causes, effects and treatment of electric shock in homes and identify the approximate danger thresholds for current and duration.

5.3.1 A shocking experience An electric shock is a violent disturbance of the nervous system caused by an electrical discharge or current through the body. There are various factors that contribute to the severity of an electric shock. The first of these is the size of the voltage involved. Also, the human body is far more sensitive to alternating current than direct current. Voltages as low as 32 V AC and 115 V DC can be fatal. It is not the voltage alone that causes damage to the human body. When you slide across a car seat, you can generate a voltage of several thousand volts. When you get out of the car and touch the ground, you are discharged and experience a shock,

FIGURE 5.9 A victim of electric shock in a home accident possibly caused by overloading the circuit

TOPIC 5 Using electricity and electrical safety 153

but with no serious consequences. The voltage drop across a person is one factor in determining the seriousness of an electric shock, but clearly other factors are involved. The following information refers to shocks involving alternating currents with a frequency of 50 Hz.

5.3.2 Resistance of the human body One contributing factor to the severity of an electric shock is the resistance of the human body. The interior of the body is a good conductor of electricity. The tissues and fluids beneath the skin conduct electricity due to the presence of ions in the fluids. The skin provides the main resistance to the flow of electricity in the body. The resistance of the skin ranges in value from 106 Ω for dry skin, to 1500 Ω for a person with wet hands, and about 500 Ω for a person sitting in a bath. This is a good reason for keeping electrical appliances away from water in the bathroom; tap water is also a reasonably good conductor. Resistance to current flow is offered by the skin up to about 600 V, at which point the skin is punctured and offers little resistance to current flow. Breaks or cuts in the skin also reduce the resistance of the skin. One of the main reasons for the high resistance of skin is the poor contact that is made between the skin and the electrical source. Water improves the contact. In hospitals, a conducting gel is used when a good electrical contact is required — for example, when using an electrocardiogram.

5.3.3 The effect of current The second and most important factor to be considered in respect to the severity of an electric shock is the amount of current flowing through the body. This is important because impulses within the nervous system are themselves electrical in nature. Even very small currents passing along nerves make muscles contract. Skeletal muscles (muscles attached to the bones) work in pairs. To raise your forearm, for example, the biceps muscle contracts and the triceps muscle relaxes. To lower your forearm, the biceps muscle relaxes and the triceps muscle contracts. This arrangement of muscles is shown in figure 5.10. One effect of passing a small current through the body is to make muscles contract. Another effect stimulates the nerves that send pain signals to the brain, causing the painful sensations associated with shocks.

FIGURE 5.10 Diagram showing the biceps and triceps in the human arm Humerus Tendon

Triceps relaxing

Biceps contracting Radius

Ulna

AS A MATTER OF FACT You may have heard of someone who received an electric shock being ‘thrown across the room’. This is not due to any explosion, but to the violent contraction of the person’s muscles.

A current of 9 mA AC across the chest causes shock. A current of twice that amount causes difficulty in breathing. A current of 20 mA causes muscles to become paralysed: they contract and stay contracted. A person unfortunate enough to touch a live conductor with the palm of their hand will grip onto the conductor and not be able to let go. Some electricians, if unsure whether a wire is live, may bring the back of the hand towards the wire. Any shock they receive will contract the muscles so that the hand is pulled away from the wire. This procedure is definitely not recommended. A current as low as 25 mA through the trunk of the body can cause fibrillation. This is the disorganised rapid contraction of separate parts of the heart so that it pumps no blood, and death soon follows. Sometimes fibrillation subsides when the external voltage is removed.

154 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

HEART STARTER Defibrillation is a medical intervention technique carried out on victims of heart attack. If the cardiac monitor shows that fibrillation is occurring, a current of 20 A at 3000 V is passed through the heart for about 5 milliseconds. This produces a major contraction of all the muscles in the heart, which usually jolts them back into the proper rhythm. The shock is applied above and below the heart via two large electrodes called paddles. Conducting gel is used to make good contact with the body. It is important that the operator and other staff are well insulated from the patient.

FIGURE 5.11 Automatic external defibrillators are vital to use on victims on heart attacks

5.3.4 The effect of current path The third factor affecting the severity of an electric shock is the path of the current through the body. Respiratory arrest generally requires the current to pass through the back of the head. Ten milliamps of current through the forearm muscles make them contract sufficiently to hold the victim to any live conductor he or she is gripping. The most dangerous pathway for current is through the trunk of the body.

5.3.5 Time of exposure The final factor contributing to the severity of a shock is the time of exposure to the current. The longer the current flows through the body, the greater the damage to tissue will be. Table 5.1 shows what effects current size and time duration have on the heart. TABLE 5.1 Electric shock current-versus-time parameters Current (mA)

Time (ms)

Effect

50

10–200

Usually no dangerous effect

50

> 4000

Fibrillation possible

100

10–100

Usually no dangerous effect

100

> 600

Fibrillation possible

500

> 40

Fibrillation possible

5.3.6 In the event of a shock The first priority, when helping a victim of electric shock, is to make sure that the victim is not still connected to the electrical source. If the person is still connected and you grab them, you could be electrocuted too. (Electrocution is death brought about by an electrical shock.) Your muscles may contract and you will not be able to let go of the victim, becoming a victim yourself. Turn off the electric circuit or knock the victim away from the live conductor with an insulating material, for example, a wooden chair. Artificial respiration should be given to the victim if breathing has ceased. Respiratory failure is a common cause of death among shock victims. This is true of people who have been struck by lightning also. An ambulance should be called immediately. TOPIC 5 Using electricity and electrical safety 155

5.3.7 Safety in household circuits Every year many lives are lost and much property is damaged or destroyed because of electrical ‘accidents’ or through electrical faults in both industrial and domestic situations. Accidents occur because basic safety precautions are not followed in dealing with electricity. The effects of electricity on the human body have already been discussed. This section looks at some common electrical faults and the safety devices employed to reduce the danger to people. One common pitfall is overloading a circuit (see figure 5.12). In parallel circuits, the total current flowing in the circuit is the sum of the individual currents flowing through the devices in the circuit. Too many appliances operating on a single power circuit will produce a large current in the conducting wires. The wires will get hot and melt their insulation, potentially causing a fire in the walls or ceilings of the building. A short circuit can occur when frayed electrical cords (see figure 5.13) or faulty appliances allow the current to flow from one conductor to another with little or no resistance. This allows the current to increase rapidly, with the same results as an overload. Cheap extension cords are another source of overheating. They are not designed to carry more than 7 A safely and exceeding this amount may result in the insulation melting and allow arcing to occur. FIGURE 5.12 Overloading of an electric circuit

FIGURE 5.13 Frayed electrical cords

Fuses and circuit breakers In domestic applications, each circuit is protected by either a fuse or a circuit breaker. A fuse is a short length of conducting wire or strip of metal that melts when the current through it reaches a certain value. The most common type of fuse is the plug-in type illustrated in figure 5.14. This has a ceramic body with metal prongs projecting from each end. A short piece of special fuse wire connects the metal prongs. When the current through the fuse exceeds a predetermined value, the wire melts, or ‘blows’, breaking the circuit. (Figure 5.16 shows a plug that includes a cartridge fuse.) A circuit breaker carries out the same function as a fuse. It breaks the circuit when the current through the circuit exceeds a particular value. The advantage circuit breakers have over fuses is that they can be reset easily. There are two types of circuit breaker available: thermal and electromagnetic.

156 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 5.14 A plug-in fuse Ceramic or plastic body

Fuse wire

Metal contacts

When a current flows through a thermal circuit breaker, like the one illustrated in figure 15.15, the heating coil heats the bimetallic strip. The two metals in the bimetallic strip expand at different rates when heated, causing the strip to bend. When the current exceeds the predetermined amount, the bimetallic strip bends so much that it opens the catch and the circuit is broken. Because of the time it takes to heat up the strip, these circuit breakers will not trip if the current surge is of a short duration. This type of circuit breaker is not satisfactory if a short circuit occurs and offers little assistance in preventing electrocutions. FIGURE 5.15 A thermal circuit breaker (b)

(a) Trip bar

Trip bar

Latch

Latch Contacts open

Contacts closed Heating coil

Heating coil

Line

Line Bimetallic element

Bimetallic element

Metal heats and bends to open contacts on overload

The electromagnetic type uses the magnetic effects of electric currents: it uses an electromagnet to lift the catch and break the circuit. The bigger the current is in the coil, the stronger the electromagnetic force will be on the lever system. Again, these circuit breakers are designed to break the circuit at predetermined values of the current. To prevent this type of circuit breaker tripping when a short-duration current surge occurs, the switching mechanism is usually restrained in some way. The magnetic circuit breaker will trip almost instantly when a heavy overload occurs. It provides good protection against short circuits. Both fuses and circuit breakers are placed in the active wire at the meter box. Light circuits are generally designed to take a maximum safe current of 5A, whereas power circuits have a maximum safe current of 15 A. SAMPLE PROBLEM 4

A kitchen circuit has the following appliances operating in it: a 1000 W toaster, a 312 W refrigerator, a 1200 W kettle, a 600 W microwave oven and a 60 W juicer. The circuit is protected by a 15 A fuse and is connected to a 230 V, 50 Hz supply. a. What is the current flowing through the fuse when all the appliances are operating at the same time? Will the fuse ‘blow’? b. Will the fuse ‘blow’ if a 2400 W heater is used at the same time as the other appliances? Teacher-led video: SP4 (tlvd-0040)

THINK a. 1.

First, calculate the current through each appliance using their power ratings and the P formula P = VI or I = . V

WRITE a.

Toaster: 1000 W I= = 4.35 A 230 V Refrigerator: I=

312 W = 1.36 A 230 V

TOPIC 5 Using electricity and electrical safety 157

Kettle: I=

1200 W = 5.22 A 230 V

Microwave oven: I=

600 W = 2.61 A 230 V

Juicer: I=

4.35 + 1.36 + 5.22 + 2.61 + 0.26 = 13.8 A

The total current in the circuit is the sum of the individual currents of the appliances. Add the currents to determine the current through the fuse. If the current is less than the capacity of the fuse, the fuse will not melt. 3. State the solution. 2.

P . V 2. If the current is larger than 15 A it is at risk of ‘blowing’.

b. 1

3.

Calculate the additional current is using I =

State the solution.

60 W = 0.26 A 230 V

The circuit is protected by a 15 A fuse, so the fuse will not melt. b.

2400 W = 10.4 A 230 V The 2400 W heater will draw an additional 10.4 A. Total current in the circuit: 10.4 A + 13.8 A = 24.2 A 24.2 A is much greater than 15 A, so the fuse will blow. I=

PRACTICE PROBLEM 4 A bathroom fan, light and heater system consists of one 75 W light globe, four 150 W heat lamps and one 100 W fan. It is connected to a 230 V supply. a. Calculate the total current through the system when the fan and light globe are in use. b. Calculate the total current through the system when the fan and two heat lamps are in use. c. Calculate the total current through the system when all the devices are in use.

5.3.8 Earthing The earth wire is another safety measure used for power circuits (see figure 5.16). It connects the metal chassis of an appliance to the earth, which is at 0 V. This connection is made via a metal rod driven into the ground at the switchboard. An electrical fault could occur if the active wire were to come into contact with the metal case of an appliance. The case would then carry an AC voltage, and anyone touching the case would receive a shock. The earth wire provides a lower resistance conducting path to the earth than the appliance and the person. The low resistance involved produces a large current in the circuit and the fuse blows or the circuit breaker trips.

158 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

The earth wire does not provide the most reliable protection. As can be seen in figure 5.17, the amount of time it takes to blow a fuse depends on the size of the current. A quicker method of breaking the circuit is needed if lives are to be saved. If the current is flowing through a person to the earth, the following principles should be followed: the current should be as small as possible and the time of exposure to the current should be as short as possible. Fuses are not designed to meet these requirements. Their main function is to prevent fires in buildings due to the overheating of wires when they carry too great a current.

FIGURE 5.16 A three-pin plug with its back cover removed. The earth wire is covered in a green and yellow coating. Also note the cartridge fuse inside the plug.

FIGURE 5.17 Time before a typical 10 A fuse ‘blows’, as a function of (RMS) current 100 80 60 40

Time for fuse to ‘blow’ (s)

20

10 8 6 4 For 10 A fuse 2

1 0.8 0.6 0.4

0.2

0.1 6

8

10

20

40

60

80

100

Current (A)

5.3.9 Residual current device The residual current device is illustrated in figure 5.18. It operates by making use of the magnetic effects of a current and is similar to a transformer. The current in the active wire flows in the opposite direction to the current in the neutral wire. Both currents pass through the iron loop. When the current in the active wire is equal in magnitude to the current in the neutral wire, each wire produces a magnetic field. These fields are equal in magnitude, but opposite in direction, and have no overall effect.

TOPIC 5 Using electricity and electrical safety 159

FIGURE 5.18 A residual current device

Relay Coil

Appliance

Active

IA

Neutral

IN

Iron core

Earth

If IA = IN

Nothing happens.

If IA > IN

A magnetic field in the coil produces a current and the relay opens the switches.

However, if there is an electrical fault and a residual current flows to the earth via the earth wire or a person, the current in the active will be greater than in the neutral. The residual current is the difference between the active and neutral currents. The magnetic effects of the two currents will no longer cancel. A current is then produced in the relay circuit and both the active and neutral wires of the circuit are broken by a switch. A residual current device operates in about 40 milliseconds, limiting the current to 30 mA. At such values the shock will be perceptible, but not likely to have any harmful effects. The residual current device is useful only when the current flows to earth, not if the current flows through the person between the active and neutral wires.

5.3.10 Double insulation Hand-held electrical tools and appliances, such as electric drills and hair dryers, are protected by double insulation. These appliances have only a two-pin plug, using only the active and neutral wires. They should not be earthed. The symbol on the casing of an appliance means that it is double insulated. As the name implies, double insulation means that the accessible metal parts cannot become live unless two independent layers of insulation fail. The inner layer is called the functional insulation. This layer has both electrical insulation and heatresisting properties. The outer layer is called protective insulation and often forms part of the casing.

FIGURE 5.19 An electric hair dryer with the double insulation symbol seen on the casing

160 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Resources Digital documents Investigation 5.1 Examination of an electrical device (doc-31869) Investigation 5.2 Model circuits (doc-31870) Teacher-led video

Investigation 5.1 Examination of an electrical device (tlvd-0814)

Video eLesson

Electrical safety (eles-2517)

Weblink

Electrical safety

5.3 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. What is the difference between a shock and electrocution? 2. Describe factors that reduce the resistance of human skin. 3. Why is the amount of current flowing through the body important in determining the severity of an electric shock? 4. What is fibrillation? 5. What would happen if you touched a shock victim who was still conducting an electrical current? 6. How is the severity of a shock related to the time of exposure? 7. What is ‘double insulation’? 8. A worker touched an overhead power line and was electrocuted. A newspaper reported the incident: ‘He touched the cable and 50 000 V of electricity surged through his body.’ Evaluate this statement.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

5.4 Review 5.4.1 Summary • •

• • • • • • •

Household electricity is provided as alternating current with a frequency of 50 Hz. Household circuits include an active wire that oscillates between +325 and –325 volts relative to the earth, which is defined as having a voltage of zero. Household circuits also include a neutral wire, which is connected to the earth. Electric current flows backwards and forwards between the active wire and the neutral wire. In many circuits an earth wire is used as a safety device to connect the case of an appliance directly to the earth. The rules relating to series and parallel circuits can be applied to both AC and DC circuits. The kilowatt-hour is a unit widely used to measure the amount of electrical energy consumed. An electric shock is a violent disturbance of the nervous system caused by an electrical discharge or current through the body. The severity of an electric shock depends on a number of factors, including current, pathway through the body and time of exposure. Fuses and circuit breakers are safety devices that use different methods to break an electric circuit when a dangerous level of current flows through it. The residual current device opens switches in the active and neutral wires when the currents in these wires become unequal due to an electrical fault. It is designed to protect against electrocution and operates more quickly than a typical fuse. TOPIC 5 Using electricity and electrical safety 161

Resources

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0030).

5.4.2 Key terms The active wire in a circuit is connected to the 240 VRMS supply at the switchboard. A circuit breaker carries out the same function as a fuse by breaking the circuit when the current through it exceeds a certain value. The earth wire is used in power circuits as a safety device; it connects the case of the appliance being used to the earth. An electric shock is a violent disturbance of the nervous system caused by an electrical discharge or current through the body. Electrocution is death brought about by an electrical shock. Fibrillation is the disorganised, rapid contraction of separate parts of the heart so that it pumps no blood; death may follow. Frequency is a measure of how many times per second an event happens. A fuse is a short length of conducting wire or strip of metal that melts when the current through it reaches a certain value, breaking the circuit. A kilowatt-hour (kW-h) is the amount of energy transformed by a 1000 W appliance when used for one hour. The neutral wire in a circuit is connected to the neutral link at the switchboard, which is connected to the earth. Period is the amount of time one cycle or event takes, measured in seconds. A residual current device operates by making use of the magnetic effects of a current to break a circuit in the event of an electrical fault. A short circuit can occur when frayed electrical cords or faulty appliances allow the current to flow from one conductor to another with little or no resistance. The current increases rapidly, causing the wires to get hot and potentially cause a fire.

Resources Digital documents Key terms glossary (doc-32182)

5.4.3 Practical work and investigations Investigation 5.1 Examination of an electrical device Aim: To examine an electrical device and report on how it functions and what safety features it has Digital document: doc-31869 Teacher-led video: tlvd-0814

Investigation 5.2 Model circuits Aim: To apply the principles of series and/or parallel circuits to either household or car electrical systems through designing a circuit Digital document: doc-31870

Resources Digital documents Practical investigation logbook (doc-32181)

162 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

5.4 Exercises To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au.

5.4 Exercise 1: Multiple choice questions 1.

2.

3.

4.

5.

6.

7.

8.

Which colour wire carries the larger voltage in an electrical cable connected to a 230V AC supply? A. All carry the same voltage. B. Green and yellow stripes C. Brown D. Blue Which best describes the reason for building fuses into electrical devices? A. Fuses are not used in electrical devices. B. To protect the device from damage C. To protect the circuit from current overload D. To protect against electric shock Under what condition is a fuse in a circuit most likely to blow? A. When the current is too large B. When the voltage is too large C. When the current is too small D. When the voltage is too small Which statement best describes the function of a circuit breaker? A. A circuit breaker diverts excess current to other parts of the circuit. B. A circuit breaker is a resettable switch. C. A circuit breaker comprises many fuses that are resettable. D. A circuit breaker must be replaced after a circuit has blown. What current is drawn by a 1.84 kW sandwich maker when it is connected to a 230 V supply? A. 0.3 A B. 2.8 A C. 8.0 A D. 14.7 A When charged at 30 cents per kW-h, what will be the cost of leaving a 250 W computer switched on for two days? A. $1.80 B. $3.60 C. $5.40 D. $7.20 An electric handtool with a power rating of 2 kW has a built-in fuse. Which of the following fuses would be most suitable? A. 3 A B. 5 A C. 7 A D. 9 A Which of the following correctly describes the appearance and function of the earth wire? A. It is coated in blue plastic and connects the appliance to the earth. B. It is coated in brown plastic and provides a lower resistance conducting path to the earth than the appliance and the person. C. It is coated in green and yellow plastic and connects the case of the appliance being used to the earth as a safety device. D. It is coated in green plastic and is a safety measure in electric circuits.

TOPIC 5 Using electricity and electrical safety 163

Many hair dryers have a double insulation symbol ( ) found on them. What does it mean for a hair dryer to have double insulation? A. It has a two-pin plug with no neutral wire. B. There are two layers that separate the live wire and the external metal casings. C. There is a single insulating layer coating the wire, but it is double thickness. D. It has double the number of wires than usual (two neutral, two earth and two live). 10. What is the first priority when someone is a victim of electric shock? A. Calling an ambulance B. Making sure the victim is not still connected to the electric source C. Giving immediate CPR D. Making sure that a defibrillator is available 9.

5.4 Exercise 2: Short answer questions 1.

2.

3.

4.

5. 6. 7.

8. 9.

10.

A 3.6 kW domestic electric hot water system typically heats water for 3 hours per day. a. How much energy will this appliance consume in one year? b. If the cost of electricity is 28 cents per kW-h, how much will it cost to provide hot water for one year? Two 10 A speakers are connected to a 230 V sound system. a. Should these speakers be connected in series or parallel? Justify your answer. b. How much power will these speakers use? c. If the cost of electricity is 32 cents per kW-h, how much will it cost to operate the speakers at full volume for 6 hours? d. There is a fuse in each speaker. What size fuse should be used? Justify your answer. An oil heater is rated at 1000 W and runs off 230 V supply. a. What current does the heater draw? b. What is the effective resistance of the heater? c. If electricity costs 17 cents per kW-h, what does it cost to run the heater for 5 hours? Some household appliances such as electric kettles are encased with metal. When a person touches the metal casing they do not get a shock. a. What stops someone from getting a shock when they touch the metal casing? b. If someone did get a shock, why would this occur? a. What is meant by the terms electric shock and electrocution? b. Under what circumstances is an electric shock likely to be fatal? Describe the difference in operation and advantages of a fuse in a circuit and a circuit breaker. A bedroom circuit has the following appliances operating in it: a 150 W gaming console, a 140 W television, a 60 W lamp, a 55 W DVD player and a 200 W mini fridge. The circuit is protected by a 15 A fuse and it is connected to a 230 V, 50 Hz supply. a. What is the current flowing through the fuse when all the appliances are operating at the same time? b. Will the fuse ‘blow’ if a 2400 W air conditioner is used at the same time as the other appliances? Why is the length of time a current is flowing through the body important in determining the severity of an electric shock? An individual uses a hair dryer immediately after getting out of the shower to dry their hair, with their hands still wet. Explain why having wet hands leads to a greater risk of electric shock than having dry hands. Compare the operation of residual current devices and circuit breakers and how they improve safety of electrical circuits.

164 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

5.4 Exercise 3: Exam practice questions Question 1 (3 marks) A person is holding an electric appliance that is connected to a 230 V AC supply. If the insulation between the case and the rest of the appliance is short circuited, the person could receive an electric shock. Describe the factors that will determine the severity of the shock they could receive? Question 2 (5 marks) An appliance with a metal case is unsafely wired directly into a simple AC circuit as shown in the following figure. Casing

Appliance

A

N

On the diagram show what safety features should be provided to help avoid electric shocks and possible damage to the appliance. b. For the safety features you identify, explain how they improve the safety of the system. a.

3 marks 2 marks

Question 3 (3 marks) The metal cased appliance in question 2 is to be re-wired to a three-pin plug so that it can be plugged into a domestic 230 V AC socket. On the following diagram show where each wire should be connected to the appliance.

N

E A

Fuse

Question 4 (6 marks) The following table gives the power consumption of various products when they are on standby. Product Laptop computer

Power (W) 14.5

Microwave

4.2

Laser printer

8.5

Set top box Television

11.2 6.2

TOPIC 5 Using electricity and electrical safety 165

Calculate the energy used by each product if it is left on standby for one year. 3 marks Calculate the mass of greenhouse gases produced by these products if they are left on standby for one year, assuming that 1 kW-h of energy produces greenhouse gases that are equivalent to 1.444 kg of CO2 . 2 marks c. If you were charged 50 cents per kW-h, how much would you be required to pay in one year if the devices were left on standby? 1 mark

a. b.

Question 5 (5 marks) A person stands on a rubber mat that connects an electric appliance to a 230 V mains power outlet. However, the appliance, which has a metal casing, has short circuited. a. What is likely to happen if the person has a resistance of 600 kΩ? Justify your answer. 2 marks b. What is the smallest voltage on the case of the appliance that could cause fibrillation of the heart? 1 mark c. What would most likely happen if the person stands on a wet floor in bare feet where their resistance is effectively 3 kΩ? Justify your answer. 2 marks

5.4 Exercise 4: studyON topic test Fully worked solutions and sample responses are available in your digital formats.

Test maker Create unique tests and exams from our extensive range of questions, including practice exam questions. Access the assignments section in learnON to begin creating and assigning assessments to students.

166 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

UNIT 1 | AREA OF STUDY 2 REVIEW

AREA OF STUDY 2 How do electric circuits work? OUTCOME 2 Investigate and apply a basic DC circuit model to simple battery-operated devices and household electrical systems, apply mathematical models to analyse circuits, and describe the safe and effective use of electricity by individuals and the community.

PRACTICE EXAMINATION STRUCTURE OF PRACTICE EXAMINATION Section

Number of questions

Number of marks

A

20

20

B

3

20 Total

40

Duration: 50 minutes Information:

• • •

This practice examination consists of two parts. You must answer all question sections. Pens, pencils, highlighters, erasers, rulers and a scientific calculator are permitted. You may use the VCAA Physics formula sheet for this task.

Resources Weblink VCAA Physics formula sheet

SECTION A — Multiple choice questions All correct answers are worth 1 mark each; an incorrect answer is worth 0. 1. Students in a class are each provided with a battery. Each battery has an emf of 15 V. In an experiment, a student measured that a total 375 C of charge has passed through her battery. How much energy has been supplied by the battery to these charges? A. 1625 J B. 2650 J C. 5625 J D. 6250 J Use the following information to answer questions 2 and 3. A model electric motor is connected with a 12 V battery. When the motor is running normally, the current through the motor is measured to be 2.5 A. A. 2.0 Ω

2. What is the resistance of the motor? B. 4.8 Ω

C. 7.2 Ω D. 9.6 Ω UNIT 1 Area of Study 2 review 167

3. How much energy was consumed by the electric motor after 30 seconds? A. 75 J B. 360 J C. 720 J D. 900 J 4. An electrical pump operating normally has 9 A of current flowing through it. The electrical resistance of the pump is 8 Ω. Which of the following is the best estimate of its power? A. 72 W

B. 576 W C. 648 W D. 764 W 5. Anja and Cris are provided with a thermistor for a physics investigation. The characteristic curve of the thermistor is shown as follows.

Resistance (kΩ)

20

15

10

5

0

10

20

30

40

Temperature (oC)

They measured the resistance of the thermistor when the temperature is 25 °C. What is the expected value of the resistance? A. 5 kΩ B. 8 kΩ C. 12 kΩ 6. An LED is connected in series with a 290 Ω resistor and a 9.0 V battery. The diode has a switch-on voltage of 3.20 V and is forward-biased in this circuit. Which of the following is the best estimate of the current in the circuit? D. 15 kΩ

A. 2.0 mA B. 11 mA C. 20 mA 7. An ohmic device with a resistance of 7.5 Ω is connected to a circuit, and the current through the device and the voltage across it are measured. Which of the following combination of voltage and current could be approximate readings of this set up? D. 31 mA

A. V = 20.0 V and I = 3.2 A B. V = 24.0 V and I = 4.2 A

C. V = 28.8 V and I = 3.8 A D. V = 32.4 V and I = 4.8 A

168 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

8. The following diagram shows part of an electrical circuit with the values of the current in different parts of the circuit. 0.8 A

4.8 A

I

1.2 A

Which of the following is closest to the value of the current I? A. 3.6 A B. 4.4 A C. 5.6 A D. 6.8 A 9. The following diagram is a battery circuit. When the switch is closed, the voltages across the different resistors are shown on the diagram.

2.5 V

+ 9V

V

3.5 V

Which of the following is closest to the value of the voltage V? A. 2.5 V B. 3.0 V C. 3.5 V D. 4.0 V Use the following information to answer questions 10 and 11. Charlene and Deen found a box full of electrical resistors with resistances of either 7 Ω or 12 Ω. There are no other types of resistors. They decide to construct an electrical circuit with a laboratory power supply and connecting wires.

10. Charlene and Deen connected a 7.00 Ω resistor and a 12.0 Ω resistor in parallel and measured their total resistance. Which of the following is closest to the value of the reading on the resistance meter? A. 4.42 Ω B. 5.42 Ω

C. 6.42 Ω D. 7.42 Ω

UNIT 1 Area of Study 2 review 169

11. One electrical circuit requires a resistance of 20.0 Ω. Which of the following combination of 7.00 Ω and 12.0 Ω resistors could give them the required 20.0 Ω? A. Series connection of a 7.00 Ω and 12.0 Ω resistors

B. Parallel connection of two 7.00 Ω resistors, then in series with a 12.0 Ω resistor

C. Parallel connection of two 12.0 Ω resistors, then in series with a 7.00 Ω resistor

D. Parallel connection of two 12.0 Ω resistors, then in series with two 7.00 Ω resistors

12. The following circuit shows a 12 V battery connected with two parallel resistors, each with a resistance of 16 Ω.

+ 16 Ω

12 V

16 Ω

I

Which of the following is closest to the value of the current, I? A. 0.75 A B. 1.5 A C. 3.0 A 13. The following circuit shows a 12 V battery connected in series with a 3.0 Ω resistor and another resistor with an unknown resistance, R. The current, I, is 1.5 A. D. 4.5 A

3.0 Ω

+ 12 V

R 1.5 A

Which of the following is closest to the value of R? A. 4.0 Ω

B. 5.0 Ω

C. 6.0 Ω D. 8.0 Ω

14. The following circuit shows a 12 V battery connected with two parallel resistors, each of which has an unknown resistance, R. The current, I, is 2.4 A.

+ R

12 V

2.4 A

170 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

R

A. 7.2 Ω

Which of the following is closest to the value of R?

B. 8.4 Ω

C. 10 Ω

D. 14 Ω 15. The following circuit is a voltage divider circuit with a 9 V battery connected to a thermistor in series with a 4 kΩ resistor. The output voltage, Vout , is 6 V.

+

R

9V

4 kΩ

6V

Which of the following is closest to the value of the resistance of the thermistor, R? A. 1 kΩ B. 2 kΩ C. 3 kΩ D. 4 kΩ 16. The Australian Standards specify that the AC electrical supply to households has a root mean square voltage of 230 V. What does this mean in relation to the power supply? A. It has a peak voltage of 230 V. B. It has a peak-to-peak voltage of 460 V. C. It will provide the same heating effect as an DC supply of 325 V. D. It will provide the same heating effect as an DC supply of 230 V. 17. What colour is the active wire in household electrical wiring? A. Blue B. Brown C. Green and yellow D. Red 18. A dishwasher is rated at 1800 W and is operated for 2.5 hours. The cost of energy is 30 cents per kW-h. How much does it cost to operate the dishwasher? A. 75 cents B. 135 cents C. 75 000 cents D. 135 000 cents 19. What is the operating principle of a residual current device (RCD)? A. It measures the difference between the current of the active wire and the current of the neutral wire. B. It melts due to the dangerously high current. C. It provides an alternative path to current that flows through the casing of an electrical appliance. D. It activates an electromagnet to switch off the circuit. A. 500 Ω

20. Which of the following values is in the range of the electrical resistance of a dry hand? B. 2000 Ω

C. 500 kΩ D. 2 MΩ

UNIT 1 Area of Study 2 review 171

SECTION B — Short answer questions Question 1 (8 marks) Consider the following circuit.

Switch

Battery

Light globe

A battery is connected to a light globe and a switch. The emf of the battery is 4.2 V. The switch is currently open. a. What is the voltage across the switch?

1 mark

b. What is the current flowing out of the battery? c. How much energy does the battery supply to each coulomb of charge?

1 mark 1 mark

The switch is now closed. During each second, 1.4 C of charge pass through the battery when the switch closed. 1C = 6.24 × 1018 electrons.

d. State the potential difference across the switch when it is closed.

1 mark

e. Calculate the number of electrons that flow through the battery each second.

1 mark

f. What is the power that the battery is transferring to the circuit?

1 mark

g. How much electrical potential energy is transformed into light and heat by the globe in 3 minutes?

1 mark

h. Calculate the resistance of the light globe.

1 mark

Question 2 (8 marks) Consider the following circuit where a 15 V battery is connected to a switch S and there are three light globes with the labelled resistances. S

6Ω + 15 V

3Ω

172 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

2Ω

b. Calculate the current flowing through the 6 Ω light globe. a. Determine the total resistance of the circuit.

2 marks

c. What is the potential difference across the 2 Ω light globe?

2 marks

d. Find the current flowing through the 3 Ω light globe. e. Calculate the power supplied by the battery.

2 marks 1 mark 1 mark

Question 3 (4 marks) An electrician decided to break open an old toaster and examine the various safety features. a. Within the plug, he noted the presence of an earth wire. Describe the appearance of the earth wire, and explain how this works as a safety feature. 2 marks The electrician decides the toaster is still fine to use and plugs it back into his kitchen circuit. The circuit already has a 500 W fridge, a 800 W coffee machine and a 1000 W fan plugged into it. This is connected to a 230 V supply and protected by a 10.5 A fuse. b. If the toaster is 700 W, will the fuse blow when the toaster is added to the circuit? Justify your response using calculations. 2 marks

UNIT 1 Area of Study 2 review 173

PRACTICE SCHOOL-ASSESSED COURSEWORK ASSESSMENT TASK — PROPOSED SOLUTION TO A SCIENTIFIC OR TECHNOLOGICAL PROBLEM In this task, you will be required to investigate and apply a basic DC circuit model to household electrical systems, apply mathematical models to analyse circuits and describe the safe and effective use of electricity by individuals. • A scientific calculator is permitted. Total time: 50 minutes (5 minutes reading and 45 minutes writing) Total marks: 40 marks

SUPPLYING ELECTRICAL POWER TO A SHED A farmer, Alisa, is considering how to supply electrical power to a shed containing farming equipment. The farming equipment is designed to work safely from a power supply with a voltage of 30 V to 50 V DC. The main problem is that the shed is 1 kilometre from the power lines that service her farmhouse. Alisa has two choices: 1. to purchase a generator and position it next to the shed 2. to construct additional power lines that connect the shed to the existing power lines that service her farmhouse.

Part A: Using a generator A review of generators shows that they come with different attributes, which need to be analysed to assist Alisa in solving her problem. 1. One generator states that it is an AC generator while a second model states that it is a DC generator. In what way are the two generators different in terms of the electrical current they produce when connected to a load such as a piece of farming equipment? Alisa reviews two DC generators, X and Y, as a possible solution to her problem. 2. Alisa finds that DC generator X outputs a voltage of 32 V, while generator Y outputs a voltage of 48 V. Explain how the generators differ by clearly explaining in terms of energy transfer what the different voltages mean. 3. Each generator has a cut-off safety switch if the current drawn exceeds the maximum specified current. The maximum current of the 32 V generator is 36 A and is slightly lower in the 48 V generator, at 32 A. Which generator, X or Y, is capable of producing the greatest power output at maximum current? Use an appropriate calculation to support your choice. 4. What would be the electrical resistance of farming equipment if 36 A were drawn from the 32 V generator? 5. What would be the electrical resistance of farming equipment if 32 A were drawn from the 48 V generator? Alisa measures the electrical resistance of the farming equipment and finds it to be 1.20 Ω. 6. If Alisa connected the farming equipment to each of the generators in turn, what current would each generator supply to the equipment? 7. (a) How much power would the farming equipment consume when using the 32 V generator? (b) How much power would the farming equipment consume when using the 48 V generator? (c) The farming equipment is required to run for 8 hours per day. What is the total energy delivered to the farming equipment if generator X is used? Express your answer in joule and kW-h. 8. (a) Is there a problem using either of the generators with the existing farming equipment? (b) Of the two generators she has reviewed, which one would be more suitable? Explain your choice.

Part B: Using power lines The potential difference of the power lines to Alisa’s farmhouse is 240 V DC. It would normally be an AC voltage, but for this assessment we will simplify the task and make the voltage a 240 DC voltage. 9. If Alisa is to use power lines, how much cable (in km) will she need to supply electricity to the shed from her farmhouse?

174 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

10. She finds that she can purchase the cable, but it has a resistance of 1.60 Ω per km. What would be the total resistance of the power lines? 11. Draw a circuit of a 240 V DC power supply and power lines that connects to the farming equipment. In your diagram include a switch S that can isolate the farming equipment from the 240 V power supply. Alisa has previously measured the resistance of the farming equipment to be 1.20 Ω. 12. What would be the total resistance of the circuit containing the power lines and the farming equipment? 13. Determine the current that would be in the circuit if the switch S were closed and the farming equipment turned on. 14. (a) By determining the voltage across the farming equipment, if the switch were closed, assess whether the system would be safe. Remember that the farming equipment is designed to operate safely with a voltage of between 30 V and 50 V. (b) If you find the system would not be safe, give one possible modification that would change the situation and explain how it would enable this. If you find the system would be safe to operate, explain why.

Part C: Proposing a solution 15. Sum up your findings from Part A and B, identifying which of the two choices you would recommend to solve Alisa’s problem. Justify your response ensuring you compare both choices.

Resources Digital document School-assessed coursework (doc-32274)

UNIT 1 Area of Study 2 review 175

AREA OF STUDY 3 WHAT IS MATTER AND HOW IS IT FORMED?

6

Origins of atoms

6.1 Overview Numerous videos and interactivities are available just where you need them, at the point of learning, in your digital formats, learnON and eBookPLUS at www.jacplus.com.au.

6.1.1 Introduction Scientists find it useful to analyse matter in terms of the atoms that form it. This is the basis of chemistry, and our understanding of biology and geology. These atoms contain a dense nucleus of protons and neutrons surrounded by a cloud of electrons. But atoms did not always exist; they only exist under the right conditions. Through an amazing journey of exploration, physicists are now confident that the first atoms formed about 13.8 billion years ago, a mere 380 000 years after the universe itself came into being. This topic explores how physicists have come to this conclusion. To understand the origin of atoms, we need to understand how the universe began. The Big Bang is the name given to the theory that scientists use to explain why the universe is as it is. The Big Bang model of the universe is a triumph of decades of observation, measurement, theory and scientific exploration. However, there is still a lot that is not known, and there are alternative interpretations. The universe is a very active research field with new and surprising discoveries being made and new questions being asked. FIGURE 6.1 Spectacular panoramic view of the Carina nebula and the unique star Eta Carinae in the heart of the nebula

176 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

6.1.2 What you will learn KEY KNOWLEDGE After completing this topic you will be able to: • describe the Big Bang as a currently held theory that explains the origins of the universe • describe the origins of both time and space with reference to the Big Bang Theory • explain the changing universe over time due to expansion and cooling • apply scientific notation to quantify and compare the large ranges of magnitudes of time, distance, temperature and mass considered when investigating the universe • explain the change of matter in the stages of the development of the universe including inflation, elementary particle formation, annihilation of anti-matter and matter, commencement of nuclear fusion, cessation of fusion and the formation of atoms. Source: VCE Physics Study Design (2016–2021) extracts © VCAA; reproduced by permission.

PRACTICAL WORK AND INVESTIGATIONS Practical work is a central component of learning and assessment. Experiments and investigations, supported by a Practical investigation logbook and Teacher-led videos, are included in this topic to provide opportunities to undertake investigations and communicate findings.

Resources Digital documents Key science skills — VCE Units 1–4 (doc-31856) Key terms glossary (doc-32184) Practical investigation logbook (doc-32185)

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0031).

6.2 Early developments of the Big Bang Theory KEY CONCEPTS • Describe the Big Bang as a currently held theory that explains the origins of the universe. • Describe the origins of both time and space with reference to the Big Bang Theory. • Explain the changing universe over time due to expansion and cooling.

6.2.1 Scientific theories A theory in science is not a wild guess. It is the best attempt by scientists to explain the results of experiment and observation. In this subtopic we look at the major evidence about the nature of the universe and the particles of matter that form it. This evidence is used by scientists to form what is known as the Big Bang Theory. A good theory can be tested by observation and often helps scientists predict phenomena yet to be observed. The Big Bang Theory has allowed scientists to explain observations of the universe and make successful predictions of what would be observed in the future.

TOPIC 6 Origins of atoms 177

6.2.2 Discovering the universe of galaxies In the early twentieth century, significant progress was made in our understanding of atoms and subatomic particles with the discovery of the nuclear atom, made up of protons, neutrons and electrons. At the same time, our understanding of the universe was advancing at a tremendous pace. At the turn of the twentieth century, there was little comprehension of how large the universe was. Astronomers knew there were a very large number of stars and that these stars seemed to be clustered into a region of space known as the Milky Way. It was not yet clear whether the Milky Way was the extent of the universe, or just one population of stars among many.

FIGURE 6.3 The Milky Way

178 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 6.2 The Big Bang Theory of the universe is a triumph of decades of observation.

In 1912, Henrietta Leavitt (1868–1921) made FIGURE 6.4 Henrietta Leavitt investigated the a discovery that was about to shine light on this. relationship between period and luminosity in Cepheid variables. She investigated a type of star called a Cepheid variable. These stars vary in brightness overtime in a regular way. Leavitt studied the two clouds of stars called the Large Magellanic Cloud and the Small Magellanic Cloud. She reasoned that the stars in each of these ‘clouds’ would all be approximately the same distance from us, meaning that any variation in their brightness could be attributed to an actual difference in luminosity, rather than just being the result of varying distance. As with all light sources, stars become dimmer the further away they are. Working with the stars from the Magellanic Clouds proved to be a very powerful technique. Leavitt plotted a graph of the maximum luminosity of the stars versus the period of their variation in brightness. She discovered a clear relationship between the luminosity and period: the more luminous the Cepheid variable, the longer its period. The period is the time it takes for one occurrence of a repeating event. For example, the period of the Earth rotating on its axis is 24 hours; the period of the motion of the second hand on a clock is 60 seconds. FIGURE 6.5 The vertical axis measures the brightness of the star. As they are all in the same body of stars, they are similar distances from us meaning that the brightness varies in the same way as the luminosity. The horizontal axis shows increasing period plotted on a logarithmic scale; the scale is 10 to the power of the numbers on the axis.

13.0

m

M −4.35

14.0

−3.35

15.0

−2.35

16.0

−1.35

17.0 0.0

0.5

1.0 1.5 Logarithm of period

2.0

−0.35 2.5

By 1919, Harlow Shapley (1885–1972) had used this relationship to determine how far the Earth is from groups of stars called globular clusters and the Large Magellanic Cloud. All he needed to do was measure the period of the variation in brightness of the Cepheid variable stars in these clusters and use Leavitt’s relationship to determine the luminosity of the star. By comparing this luminosity to the brightness he could measure the distance to the stars, and hence the distance to the clusters they were in. Most of the stars in the Milky Way lie in a plane in the shape of a spiral. Shapley found that the globular clusters group around the centre of the Milky Way in a sphere.

TOPIC 6 Origins of atoms 179

Globular clusters are spherical clusters of thousands, and sometimes millions, of stars. Shapley used his mapping of globular clusters to determine the general shape and size of the Milky Way galaxy and found that the clusters formed a spherical shell whose centre lay in the direction of the constellation of Sagittarius. This suggested that the centre of the galaxy was in this direction and that our solar system was located towards the edge. Until Shapley’s measurements, the solar system was thought to be near the centre of the galaxy. Shapley’s estimate of the size of the galaxy included the Magellanic Clouds and assumed that the Andromeda nebula was within the Milky Way. He had made quite an error in his estimate. By this time thousands of nebulae in the shape of spirals and ellipses had been catalogued. Some of them seemed to be very small. How could they all be in our galaxy? From 1919 to 1926, Edwin Hubble (1889–1953) examined the Andromeda nebula in great detail using the large telescopes at Mount Wilson, California. He took long exposure photographs of the spiral arms and counted numerous novae (singular nova; bright stars that were not in previous photographs) and Cepheid variables. Using these, Hubble established the size and distance of the Andromeda nebula from Earth and found that it lay well beyond our galaxy and was comparable in size to the Milky Way. It seemed that our galaxy was only a tiny portion of the universe after all. One of the most significant results in the history of cosmology came from the work of Shapley and Hubble. In 1919, Shapley noticed that the velocities of nebulae, later found to be galaxies, indicated that they were nearly all moving away from us. Hubble explored further and, in 1929, showed that the more distant the galaxies were from us, the faster they were moving away. The speeds were enormous; one measurement suggested that a galaxy was moving away from us at 42 000 km s–1 , more than one-tenth of the speed of light!

FIGURE 6.6 Globular cluster

FIGURE 6.7 Edwin Hubble

FIGURE 6.8 The Andromeda galaxy contains most of its stars in a flat disk similar to the Milky Way. The globular clusters are not restricted to this disk.

180 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

SAMPLE PROBLEM 1

The discovery that stars form enormous structures called galaxies and that the rate galaxies recede from the Earth increases with their distance from us depended on a discovery by Henrietta Leavitt. Describe her discovery and how it enabled these great discoveries by Edwin Hubble and his collaborators. THINK 1.

Identify what Henrietta Leavitt discovered.

2.

Link this discovery to the use of it by Hubble and others to discover galaxies and their recession.

WRITE

Leavitt discovered that a type of star called a Cepheid variable varied in luminosity in such a way that the more luminous the star, the longer its period of variation. This provided a method to determine the luminosity of the star and therefore its distance could be determined by the difference between its luminosity and its brightness when viewed from Earth. Hubble used this discovery to show that clouds of stars were well beyond the limits of the Milky Way galaxy and therefore separate galaxies in their own right. He also found that the more distant the galaxies, the faster they were receding from us.

PRACTICE PROBLEM 1 Henrietta Leavitt discovered that Cepheid variables varied in luminosity over a period of time which increased the more luminous the stars were. Knowing this relationship, astronomers could determine the luminosity of the star by measuring its period. Why can’t astronomers simply measure the luminosity of a star by observing how bright it is?

6.2.3 The expansion of space Everywhere that astrophysicists looked in the sky, they found galaxies moving away from us. This information was astonishing and very unexpected. All galaxies are experiencing the same thing. If people in another galaxy looked at us, they would see us racing off in the opposite direction. In fact, it is not the galaxies themselves that are moving away; rather, space is expanding and taking the galaxies with it! Galaxies do also move through space due to gravitational interactions with other galaxies; we know that the Milky Way is currently colliding with the relatively tiny Sagittarius Dwarf Elliptical Galaxy. But these motions of neighbouring galaxies do not explain why distant galaxies are moving away from each other with a speed that increases with distance. Galaxies and tightly bound clusters of galaxies do not expand as they are held together by gravity. The Milky Way is part of a cluster of galaxies known as the Local Group. The Local Group includes over 30 galaxies, the two largest being Andromeda and the Milky Way. The Large and Small Magellanic Clouds are also members of the Local Group. The Andromeda galaxy is actually approaching us at enormous speed due to gravitational attraction, but as we look at more distant galaxies beyond the Local Group, the expansion of space dominates gravitational forces. Hubble jumped to the obvious conclusion. All of the galaxies are racing away from each other, so if we imagine running time backwards, we would see that they had all come from the same spot. It was as though TOPIC 6 Origins of atoms 181

the universe was born from some form of explosion that threw space and matter out in all directions. This was the first scientific evidence that the universe had a beginning. Prior to this, scientists assumed that the universe had existed forever. This was the beginning of time and space, which has expanded ever since.

6.2.4 Measuring expansion Hubble measured the expansion of space using a phenomenon called the red shift. Light emitted by a star or galaxy that is moving away from us is shifted towards the red end of the spectrum. The faster the galaxy moves away, the greater the red shift. Stars or galaxies moving towards us have their light shifted towards the blue end of the spectrum, which is called blue shift. FIGURE 6.9 Light from astronomical objects observed through a spectroscope reveals spectra showing red shift. The scale indicates the wavelength in nanometres. 400

500

600

700

400

500

600

700

Distant galaxy

Medium-distance galaxy

Close galaxy

Close star

The visible spectrum of light contains red, orange, yellow, green, blue, indigo and violet. The spectrum continues into invisible forms of radiation, including infra-red at lower frequencies than red and ultraviolet at higher frequencies than violet. Hubble observed that the red shift was greater for more distant galaxies. The red shift of all distant galaxies could be explained if space were expanding. As the light from a distant source travels through space en route to Earth, the space it passes through stretches, increasing the light’s wavelength. The longer the light travels through space (that is, the further away the galaxy), the greater the increase in wavelength due to expanding space, and so the greater the red shift. This effect can be quickly demonstrated using a rubber band to represent space. Mark a rubber band every 2 mm along a 2 cm length. The distance between neighbouring marks represents the wavelength of light. As you stretch the rubber band a little, you will see each mark move away from all of the others. As time goes on the universe keeps expanding, like stretching the rubber band further. As a consequence, the wavelength of light from distant galaxies gets longer over time.

182 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 6.10 (a) The rubber band before stretching (b) The stretched rubber band — each dot has moved away from all of the other dots (a)

(b)

SAMPLE PROBLEM 2

What major discovery about the nature of our universe did Hubble make when he considered the red shift effect in his analysis of spectra from galaxies? THINK 1.

Identify what Hubble saw when observing the red shift of galaxies.

2.

Connect his observation with the nature of the universe.

WRITE

When Hubble measured the red shift of distant galaxies he noticed that it increased the further the galaxy was from us. This was interpreted by physicists as meaning that the universe must be expanding. Prior to this discovery they thought it was static.

PRACTICE PROBLEM 2 Explain how the light from distant galaxies has come to be red shifted before it reaches us?

SAMPLE PROBLEM 3

Explain why expansion is observed on the large scale for distant galaxies, but not on the relatively small scale of nearby galaxies. Teacher-led video: SP3 (tlvd-0043) THINK 1.

State that expansion is seen when observing distant galaxies.

2.

Explain the difference in gravitational effects over different distances.

WRITE

When Hubble measured the motion of more distant galaxies, he found that they were moving away from us, but some nearer ones were moving towards us. Gravity is a force that weakens with distance. Gravity can cause galaxies near to each other to move toward each other; but for galaxies separated across a greater distance, the gravity becomes too weak and the expansion of space overrides its effect.

PRACTICE PROBLEM 3 What did Hubble observe about all of the galaxies that were not near to the Milky Way?

TOPIC 6 Origins of atoms 183

UNDERSTANDING WAVES Waves can be described in terms of their period, frequency, speed and wavelength. The wavelength is the distance between successive corresponding parts of a periodic wave — the length of one cycle. The period of a periodic wave is the time for one part of the wave to travel one wavelength. The frequency of a periodic wave is the number of cycles that occur every second. FIGURE 6.11 Transverse periodic waves in a piece of string

Wavelength

Crest

Direction of wave movement

Amplitude

Amplitude Position of undisturbed medium

Trough

Direction of particle motion

6.2.5 Einstein and general relativity In parallel to the experimental work of Shapley, Hubble and other FIGURE 6.12 Aleksandr Friedman astronomers, theoretical physics was making tremendous progress. In particular, Albert Einstein published his General Theory of Relativity in 1915, which provided a new way of understanding gravity. Until then, scientists had used Isaac Newton’s theory of gravity, which essentially stated that masses exert an attractive force on each other. Einstein’s theory was that mass curves space and other masses move in response to the curvature of space. However, there was a problem with both theories. Newton wondered why the universe was not collapsing, given that all masses attract each other, and decided that the universe must be infinite. The idea that the universe is infinite in size and has existed forever endured until the Big Bang Theory was developed. Einstein’s theory did not provide a satisfactory answer to the collapsing universe problem either, so he inserted what became known as a cosmological constant in his equations to counter the curvature of space. Beginning in 1917, scientists proposed other cosmological models based on the General Theory of Relativity. Aleksandr Friedman (1888–1925) found that there were a number of solutions to the equations of general relativity, each of which is equally valid in terms of the theory, but of course only one could represent our universe. In some solutions the universe expands forever, in others the universe would expand for a time and then collapse back on itself. This theoretical work led Willem de Sitter (1872–1934) to encourage Hubble to measure the red shift of the galaxies.

184 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Georges Lemaître (1894–1966) was both a Catholic priest and professor of physics. He was fascinated by what physics might say about the birth of the universe. In 1927, he used general relativity to produce an early version of what became known as the Big Bang Theory.

6.2 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. What is a galaxy? 2. Using Cepheid variables of equal distance from Earth, Henrietta Leavitt discovered a relationship between their luminosity and their period of variation. (a) Describe this relationship. (b) What does this relationship enable astronomers to measure? (c) Why did she need to use Cepheid variables that were similar distances from the Earth to determine the relationship? 3. What did Hubble notice about all the more distant galaxies? 4. Which of these spectra of three distant galaxies would be from the most distant galaxy? Give your reason.

5. Which great theory of Albert Einstein’s was used in the development of the Big Bang Theory? 6. It is easy to read about the red shift of the galaxies in all directions and jump to the conclusion that the Earth is at the centre of an expanding universe. Use the rubber band analogy, with each dot on the rubber band representing a galaxy, to explain why this is not so. 7. Why did proof of the expansion of space still come as such a surprise to scientists, even though those working with the General Theory of Relativity had already shown an expanding universe was possible? 8. Why did Einstein adjust his theory with a cosmological constant? 9. The light from the Andromeda galaxy is blue shifted. Explain why it is not red shifted like the light from most galaxies.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

TOPIC 6 Origins of atoms 185

Resources Digital document

Investigation 6.1 Expansion of the universe (doc-31871)

Teacher-led videos Investigation 6.1 Expansion of the universe (tlvd-0816) Video eLesson

The expansion of the universe (eles-0038)

Interactivity

Shifting stars (int-6395)

6.3 Further developments of the Big Bang Theory KEY CONCEPTS • Describe the Big Bang as a currently held theory that explains the origins of the universe. • Describe the origins of both time and space with reference to the Big Bang Theory. • Explain the changing universe over time due to expansion and cooling.

6.3.1 An expanding, cooling universe By 1929 there was both a theoretical basis for the expansion of space, through the work of Lemaître, de Sitter and Friedman, and the observational evidence of the red shift of galaxies to support it; however, it was by no means generally accepted. Hubble plotted the velocity (red shift) of galaxies versus their distance from Earth and ambitiously fitted a straight line to the data, well aware that the distance calculations had large uncertainties. If the galaxies really did fit this straight line rule, then it would be easy to judge the distance to other galaxies; simply measure their red shift and divide by the gradient of the line. This gradient became known as Hubble’s 1 constant (H) and the relationship between velocity and distance as Hubble’s Law ( gives physicists a H means of calculating the age of the universe). FIGURE 6.13 Hubble’s data. The solid dots are the results for galaxies treated individually and the solid line is the line of best fit for these data. The dashed line is fitted to the circles, which are the result of treating galaxies in clusters. One parsec (pc) is 3.09 × 1016 m or 3.26 light-years. Notice the group of blue-shifted galaxies at about 2.5 × 105 parsecs. This corresponds to the Andromeda galaxy and its satellites.

Velocity (km s−1)

1000

500

0

106 Distance (pc)

2 × 106

Source: Adapted from Edwin Hubble, ‘A relation between distance and radial velocity among extra-galactic nebula’, Proceedings of the National Academy of Science, vol. 15, no. 3, 15 March 1929, Mount Wilson Observatory, Carnegie Institution of Washington. Communicated 17 January 1929.

186 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

The value of Hubble’s constant has been measured with increasing accuracy since Hubble’s time and discoveries in recent years have established its value to a small margin of uncertainty. Using the simplest scenario — that the universe has expanded at a constant rate — Hubble’s early measurements put the age of the universe at 2 billion years, but the age of the Earth appeared to be greater than that. Improvements in measurement, including a better understanding of Cepheid variable stars and measuring the red shift of much more distant galaxies, have since put the age of the universe at about 13.8 billion years. SAMPLE PROBLEM 4

Use the solid line in the graph of Hubble’s data to estimate the age of the universe. Compare this with his estimate of 2 billion years. THINK 1.

WRITE –1

As the unit for speed here is km s , convert the distance to kilometres.

2.

Hubble’s constant is the gradient of the graph.

3.

The age of the universe is the reciprocal of Hubble’s constant.

4.

State the answer.

2 × 106 pc = 2 × 106 × 3.09 × 1013 = 6.2 × 1019 km 1100 H= 6.2 × 1019 = 1.78 × 10−17 s−1 1 t= H 1 = 1.78 × 10−17 = 5.62 × 1016 s = 1.8 × 109 years The universe is 1.8 × 109 years old. This is 2 billion years to one significant figure, so consistent with Hubble’s estimate.

PRACTICE PROBLEM 4 Given the currently accepted age of the universe is 13.8 billion years, is Hubble’s trendline with this early data too steep or not steep enough in gradient?

Resources Weblink The expansion of the universe and Brian Schmidt

Other evidence for the expansion model of the universe included surveys showing that galaxy density in distant space was greater than the density of galaxies closer to Earth. This is expected with the expansion model because when we observe distant galaxies, we see them as they were billions of years ago — due to the limit of the speed of light — when the universe had undergone much less expansion. The young universe must have been very hot and we will learn that a significant event called recombination resulted in photons (light) being able to travel freely for the first time without interacting with electrons. The wavelength of a photon depends on its energy. As the universe expanded, the wavelengths of the photons would have expanded, stretching out to form microwaves with much less energy than when the photons were initially released. These microwaves have become known as cosmic microwave background radiation (CMB) and correspond to a background temperature of about −270 °C or 2.7 K. Arno Penzias and Robert Wilson discovered this radiation in 1965 when they were trying to TOPIC 6 Origins of atoms 187

eliminate some noise coming from their radio telescope. FIGURE 6.14 Penzias and Wilson discovered The identification of this noise as the missing energy the cosmic microwave background radiation left over from the beginning of the universe was the (CMB) in 1965. turning point for the Big Bang Theory. The theory had predicted the existence of greatly red shifted energy left over from the early universe. With the discovery of CMB, this energy had been found — cooled from 3000 K at recombination to 2.7 K today by the expansion of the universe. The name ‘Big Bang’ came from an early opponent of the theory, Fred Hoyle, in 1950. It is not a particularly accurate description as it implies that the creation of the universe was like an enormous explosion pushing matter through space, which is not the case. Instead of pushing matter through space, space and time were created in the Big Bang and the expansion of the universe with time carries matter with space. In summary we have discovered that: • the universe is vast, filled with billions of galaxies each containing billions of stars • these galaxies are moving away from each other at a rate that increases the more distant the galaxies are • the more distant galaxies (the older ones) are more densely packed than those closer to us • the predicted CMB has been detected. Most scientists concluded that these discoveries meant the universe had a beginning — in a relatively small volume — and has expanded ever since. This idea is the Big Bang Theory.

STEPHEN HAWKING AND THE BIG BANG The Big Bang Theory raises many questions. What happened FIGURE 6.15 Stephen Hawking before the Big Bang? What is outside the universe? What is the universe expanding into? Famous physicist Stephen Hawking, in collaboration with others, posed answers to these questions. He reminded us that in the past people contemplated what would happen if you sailed off the edge of the world. That question turns out to not need an answer because we have a completely different view of the shape of the world; it is a globe, so we never reach an edge. The question of what happened off the edge of the world only arose because of our misunderstanding of the shape of the world, thinking that it was flat. In his PhD thesis in the mid-1960s, Stephen Hawking proved that Einstein’s Theory of General Relativity required the universe to have a beginning in what was called a ‘singularity’. A singularity is a point of infinite density that is achieved when we think of what happens if we run time backwards so that the expanding universe collapses back to its beginning. This was a stunning result for general relativity. If the universe was initially very dense, it must also have been incredibly hot and that energy should still be found throughout the universe. This energy was found soon after Hawking’s initial work on singularities in the form of the CMB. This radiation is central to the subject of this topic, the origin of atoms. Later Hawking showed that time becomes like another dimension of space under extreme conditions. It makes no more sense to ask what happened before the Big Bang than to ask what is south of the South Pole. Hawking asked us to imagine the passing of time as being like decreasing degrees of latitude away from the South Pole. The latitude is 90° at the South Pole and as we move north it becomes 89°, 88°, and so on. There is no 91° of latitude and there is no ‘before the Big Bang’. Time began with the Big Bang and space has no edge to fall off, or to step outside of, to discover what the universe is expanding into.

188 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

6.3 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. What is the connection between the observed cosmic microwave background and the Big Bang Theory? Explain how Hubble’s observation of receding galaxies led to the idea that the universe had a beginning. What does the Big Bang Theory say about time and space before the beginning of the universe? What does the Big Bang Theory have to say about space since the beginning of the universe? Explain how the energy left over from the Big Bang has cooled as a result of the universe’s expansion. How does the Big Bang Theory explain Hubble’s observation that the more distant the galaxy, the greater its red shift? 7. Why would the galaxies have been more densely packed in the universe in the past? 1. 2. 3. 4. 5. 6.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

6.4 Measurements of the universe KEY CONCEPT • Apply scientific notation to quantify and compare the large ranges of magnitudes of time, distance, temperature and mass considered when investigating the universe.

The universe is on a scale that is very difficult to comprehend with the usual numbers we are accustomed to. For example, when measuring distance on Earth, measurements in metres or kilometres are sufficient for most situations. For small objects millimetres or centimetres are often used. Using these units, even the vastness of the circumference of the Earth is not too unwieldy. The circumference of the Earth is around 40 000 km. It is a big number, but manageable. When looking beyond the Earth, physicists and astronomers find other units and notations easier to work with.

6.4.1 The astronomical unit One of the earliest measurements of the vastness of space was the distance from the Earth to the Sun. This was an important measurement because it could be used to work out the distance to the other planets. The main reason for Captain James Cook’s voyage, which eventually led him in 1770 to the east coast of what is now Australia, was to measure this distance by observing a transit of Venus at Tahiti. The average distance from the Earth to the Sun is known as an astronomical unit (AU), as shown in figure 6.16. It provides an alternative unit for measuring distance on the scale of the solar system. To see the effectiveness of the astronomical unit consider the following: 1 AU = 150 000 000 km or 150 000 000 000 m

TOPIC 6 Origins of atoms 189

FIGURE 6.16 The distance from the Earth to the Sun is 1 AU.

Venus Mercury

Sun

1

AU

Earth

The average distance from the Sun to Saturn is 1.4 billion km or 1 400 000 000 000 m. When performing calculations with these numbers it is easy to make a mistake, such as including an extra zero. The metre, or even the kilometre is not a very useful reference on the scale of the solar system. It can be easier to understand that Saturn is 9.6 AU, or 9.6 times as far as the Earth is from the Sun. However, the astronomical unit is not very helpful on the scale of the universe. The nearest star is 270 000 AU. The distance to the nearest large galaxy is 160 000 000 000 AU.

SAMPLE PROBLEM 5

The planet Mars has an elliptical orbit. It is 1.38 AU from the Sun at its closest (perihelion) and 1.67 AU from the Sun at its furthest (aphelion). Calculate these distances in metres. Teacher-led video: SP5 (tlvd-0045) THINK

WRITE

Recall the number of metres in 1 AU. 2. Multiply this value by the number of astronomical units. 3. State the solution in metres.

1 AU = 150 000 000 000 m At aphelion: 1.67 × 150 000 000 000 At perihelion: 1.38 × 150 000 000 000 At aphelion: 250 500 000 000 m At perihelion: 207 000 000 000 m

1.

PRACTICE PROBLEM 5 a. Venus has an average orbital radius of 0.72 AU. What is the average distance of Venus from the Sun in metres? b. The distance from the Earth to the Moon averages about 385 000 kilometres. What is this in astronomical units?

190 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

6.4.2 The parsec Measurement of the distance to nearby stars uses a technique called parallax. This involves observing the tiny change in position of the nearby star against the distant background stars as the Earth moves around the Sun. This change in position is so tiny that it is very difficult to measure, even with the largest telescopes. The lack of parallax observed is one of the reasons that it took a long time to accept that the Earth moved around the Sun, rather than the other way around as most ancients believed. The change in position is measured as an angle, but no changes as big as 1 degree were observed. Even one-sixtieth of that, one minute of arc, was too large a unit. Astronomers measure parallax in fractions of seconds of arc. The distance of an object that would cause one second of parallax as the Earth moved around the Sun is called a parallax second, or parsec (pc). The nearest star, Proxima Centauri is 1.3 parsec from Earth. The parsec is equivalent to 30 900 000 000 000 000 m. FIGURE 6.17 1 parsec is the distance of a nearby star (object) with a parallax angle of 1 arc second. Earth

Parallax angle = 1 arc second

1 AU

Nearby star Sun Distance = 1 parsec

SAMPLE PROBLEM 6

Alpha Centauri is 1.339 pc from Earth and Rigel is 265 pc from Earth. How many times further away is Rigel than Alpha Centauri? Teacher-led video: SP6 (tlvd-0046) THINK 1.

What is the distance from Earth to Rigel and Alpha Centauri in the same units?

Divide the distance to Rigel by the distance to Alpha Centauri. 3. State the solution.

2.

WRITE

265 pc and 1.339 pc 265 = 197.9 1.339 Rigel is about 198 times as far away as Alpha Centauri.

PRACTICE PROBLEM 6 The Andromeda galaxy is 778 000 pc from the Earth, while the Large Magellanic Cloud is 48 500 pc away. How many times further away is the Andromeda galaxy than the Large Magellanic Cloud?

TOPIC 6 Origins of atoms 191

6.4.3 The light-year A unit often used in the measurements of distances in space is the distance that light travels in one year. The light-year is 9 460 000 000 000 000 m. The light-year is informative because it describes how long it has taken the light to reach Earth. A galaxy that is observed as 4 billion light-years away is observed as it was 4 billion years ago. That galaxy will not currently be four billion light-years away as it has moved relative to the Earth in the meantime. The unit of measurement takes this complication into account. 1 light-year = 9 460 000 000 000 000 m = 9.46 × 1015 m SAMPLE PROBLEM 7

The Andromeda galaxy is measured to be 2.5 million light-years away. How long has it taken the light from it to reach our eyes and telescopes? Teacher-led video: SP7 (tlvd-0047) THINK

WRITE

Determine the distance in light-years. 2. Write that distance as the number of years the light has travelled.

2.5 million light-years The light has travelled for 2.5 million years to reach us.

1.

PRACTICE PROBLEM 7 How long has it taken light to reach the Earth from Betelgeuse, which is measured as being 643 lightyears away?

6.4.4 Working in metres Astronomical units, parsecs and light-years are excellent units for working on astronomical scales but ultimately scientists need to be able to relate distances to common units like metres. This involves dealing with very large numbers. Rather than write them out in full as has been done so far in this subtopic, scientists use scientific notation (also known as standard form). This is a number between one and ten multiplied by multiples of ten. For example, the light-year is the unwieldly 9 460 000 000 000 000 m. In scientific notation this is written as 9.46 × 1015 m. A number like 1 000 000 000 000 becomes 1012 . This is so much quicker to write and is much easier to read. Scientific notation makes it easy to compare very large numbers. It makes use of the powers of ten. 100 = 1 101 = 10 102 = 100 103 = 1000 ... and so on Try comparing the following numbers 9 450 000 000 000 and 3 600 000 000 000. It is much easier to compare 9.45 × 1012 and 3.6 × 1012 . Immediately it becomes clear that the first number is bigger than 192 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

the second by a factor of less than three. To find that factor exactly all that is needed is to perform the 9.45 calculation = 2.625. The rest of the number (1012 ) cancels out so can be ignored. If the numbers 3.6 did not have the same power of ten, they can be written so that they do. Consider the numbers 2.6 × 1016 and 1.2 × 1015 . To compare the numbers the first could be written as 26 × 1015 , then they can be compared on the same scale. Big numbers are everywhere in distance in the universe, from the diameters of stars to the circumference of a galaxy. Scientific notation is an effective way of managing these large numbers. SAMPLE PROBLEM 8

Add the numbers 2.5 × 108 and 1.2 × 109 . Teacher-led video: SP8 (tlvd-0048) THINK

WRITE

Write the two numbers so the powers of ten are the same. 2. Now that they have common powers of ten, add the numbers together. 3. Write with the multiple of ten. 4. Rewrite in scientific notation if necessary.

2.5 × 108 1.2 × 109 = 12 × 108 2.5 + 12 = 14.5

1.

14.5 × 108 1.45 × 109

PRACTICE PROBLEM 8 Add 2.3 × 107 and 4.3 × 106 .

6.4.5 Numbers and time The universe is not only grand on the scale of distances; the scale of time is also difficult to comprehend. The age of the universe is currently measured to be close to 13.8 billion years (1.38 × 1010 years). The next subtopic will address the earliest moments of the universe. The normal practice of dividing time into seconds is not very helpful. An event like the emergence of gravity occurred 0.000 000 000 000 000 000 000 000 000 000 000 000 000 000 1 seconds after the universe began. 0.000 000 000 000 000 000 000 000 000 000 000 000 000 000 1 is much easier to read as a power of ten. 1 = 100 0.1 = 10−1 0.01 = 10−2 0.001 = 10−3 ... and so on The time when the force of gravity is understood to have first been active as a separate force in the universe was when the universe was 10−43 seconds old. Temperatures and masses also require extremely large and small numbers to describe them. Temperatures range from 2.7 K in deep space to 1.47 × 1032 K, the temperature at which physics as we know it cannot explain what happens — the Planck temperature. Masses vary enormously too. An electron has a mass of 9.1 × 10−31 kg, while the observable universe has an estimated mass of 1.5 × 1053 kg.

TOPIC 6 Origins of atoms 193

SAMPLE PROBLEM 9

The mass of the Sun is 1.989 × 1030 kg. Write this number in full decimal notation. THINK

WRITE

Identify the number that needs to be converted. 2. Write the digits without the power of ten, then move the decimal point 30 places to the right, adding zeros where necessary.

1.989 × 1030 kg 1 989 000 000 000 000 000 000 000 000 000 kg

1.

PRACTICE PROBLEM 9 The Earth is understood to be 4.54 × 109 years old. Write this age in full decimal notation.

SAMPLE PROBLEM 10

The temperature at the centre of the Sun is approximately 15 000 000 K. Write this number in scientific notation to two significant figures. THINK 1. 2. 3. 4. 5.

Identify the number to be converted. Write down the first digits to the appropriate number of significant figures. Add a decimal point to make the number between 1 and 10. Count the number of places the decimal point would have to move to obtain the original number. Write the number as the product of part 3 and 10 to the power of part 4, including units.

WRITE

15 000 000 K 15 1.5 7 1.5 × 107 K

PRACTICE PROBLEM 10 The mass of Saturn is 568 300 000 000 000 000 000 000 000 kg. What is the mass of Saturn in scientific notation to four significant figures?

6.4 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. The average distance from the Sun to Neptune is 30.1 AU. Convert this distance to metres. 2. The distance from the Earth to Mercury averages 77 million kilometres. How far is that in astronomical units? 3. The solar system lies about 8 kpc from the centre of the galaxy. The radius of the galaxy is approximately 15 kpc. Describe our location within our galaxy. 4. Sirius, the brightest star in the night sky, is about 8.6 light-years from Earth. How long has it taken the light that we see on Earth to reach us from Sirius? 5. The distance from the Sun to Neptune is 4.495 billion kilometres. Write this in scientific notation in metres. 6. The light from the most distant galaxy observed has travelled 1.3339 × 1010 years to reach us. Write this number in decimal notation.

194 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

7. When the universe was a second old the Big Bang Theory estimates its temperature was about 10 000 000 000 K. Write this number as a power of ten. 8. The mass of Earth is about 5.98 × 1024 kg and the mass of the Moon about 7.35 × 1022 kg. Without using a calculator determine the total mass of the Moon and Earth to three significant figures. 9. The mass of Jupiter is 1.90 × 1027 kg and the mass of Earth is 5.98 × 1024 kg. Calculate the ratio of mass of Jupiter to mass of Earth to two significant figures.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

6.5 The formation of the first atoms KEY CONCEPT • Explain the change of matter in the stages of the development of the universe including inflation, elementary particle formation, annihilation of antimatter and matter, commencement of nuclear fusion, cessation of fusion and the formation of atoms.

6.5.1 Looking back in time To look into space is to look back in time. This is a consequence of the speed of light and the vast distances involved. Light travels faster than anything else in the universe, but it still only travels at just under 300 000 000 metres per second. The distance light travels in a year is called a light-year. The light from Earth’s nearest star, other than the Sun, takes more than four years to reach us, so we see it as it was more than four years ago. When we look out to the nearest large galaxies we see them as they were over 2.5 million years ago. This is still recent in the universe’s history. Looking at the most distant galaxies, we see them as they were over 13 billion years ago. These distant galaxies are incredibly faint and our current telescopes cannot see further. But even if we could, we don’t expect to see much more because physicists believe that the first stars would have only begun to shine at this time. A little over 13 billion years ago the matter in the universe was hydrogen and some helium in vast clouds, collapsing due to gravity on their way to producing the first stars and galaxies. This period is known as the Dark Ages (borrowed from a term used to describe the period in history following the fall of the Roman Empire). The Dark Ages marks the period from 380 000 years following the beginning of the universe until stars began to shine 150–800 million years later. To look further back in time, we don’t see objects, we see the remnants of an event. This is CMB, the energy in the form of photons left over from the early universe. Prior to the universe reaching 380 000 years of age, this energy was trapped in a state of constant scattering, a bit like light in a cloud. It was not water molecules that scattered this light but the charged particles that filled the universe. These protons and electrons were too energetic, too hot, to combine to form atoms until the expansion of the universe had cooled it enough for the bonds between electrons and protons and the other nuclei to form. This event is called recombination and marks the time when atoms first formed 380 000 years after the beginning of the universe. The CMB provides physicists with a picture of what the universe was like at 3.8 × 105 years of age because the light that formed it was free to travel through the universe from that time. The CMB has been red shifted uniformly in all directions by the expansion of the universe.

TOPIC 6 Origins of atoms 195

FIGURE 6.18 The CMB as measured by the Planck satellite. The differences in colour indicate temperature differences that correspond to differences in the density of the universe when it was 380 000 years old.

Source: ESA and the Planck Collaboration

SAMPLE PROBLEM 11

What event was the origin of the cosmic microwave background? Teacher-led video: SP11 (tlvd-0051) THINK 1.

Recall what the cosmic microwave background is.

2.

Recall what event resulted in the cosmic background radiation.

WRITE

The cosmic microwave background radiation is a uniform low temperature radiation throughout space in all directions. Energy from the Big Bang was constantly scattered by charged particles until the universe was cool enough for electrons to remain bonded to the hydrogen and helium nuclei that had formed. This event produced the first atoms and is called recombination. The photons of energy were able to propagate through space after this point and it is the greatly red shifted photons that are observed as the cosmic microwave background.

PRACTICE PROBLEM 11 Why is it not possible to view the universe using light as it was prior to recombination?

6.5.2 The early universe The CMB marks the earliest observation we have of the universe. To understand what happened prior to this event, physicists rely on particle physics and the Theory of General Relativity to help them make sense of their observations of the CMB and the universe that followed. Particle physics experiments, such as those 196 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

in the Large Hadron Collider that discovered the Higgs boson at CERN, create conditions that existed for particles in the early universe, so physicists do have experimental evidence for much of what happened prior to recombination. FIGURE 6.19 Diagram of the evolution of the universe (observable part) from the Big Bang (left) to the present

Source: NASA WMAP Science Team

Let’s piece the story together from the beginning until the formation of the first atoms, summarised in figure 6.20.

Planck era

The first 10−43 seconds This is known as the Planck era and current theories of physics cannot explain what happened in the conditions that were present. The universe was too small, too dense, too hot and existed for too short a time for current physics to say anything precise about it. Physicists refer to what existed prior to when the universe was 10−43 seconds old as a singularity. In this period space and time began. Gravity became a distinct force at the end of the Planck era. The temperature was 1032 degrees Celsius and the universe was 10−35 cm across.

Grand unified era

10−43 seconds to 10−36 seconds This tiny interval of time in the early universe is known as the grand unified era. During this period, physicists believe that the strong nuclear force, the weak nuclear force and the electromagnetic force did not yet exist as separate forces. The first matter began to form, but for each particle of matter, there was a particle of antimatter. As soon as a particle was formed, it would meet an antiparticle and be annihilated.

TOPIC 6 Origins of atoms 197

Inflation era

10−36 seconds to 10−32 seconds This next period of time is known as the inflation era. During this time, a period of exponential expansion has been proposed to explain a number of features of the universe. The universe is thought to have expanded in size by a factor of 1026 , so that the observable universe was about 10 cm across. This rapid expansion explains, among other things, why the CMB is so uniform, being close to 2.7 degrees above absolute zero in all directions. Alan Guth proposed this radical idea in 1980. It was a response to some of the limitations of the standard Big Bang model that worked well for most of the evidence, but fell short when it came to explaining the relative uniformity of the CMB in all directions and what is known as the ‘flatness’ of the universe. At first, inflation may look like a crazy idea invented just to explain away problems. However, in all of the time since, no-one has come up with a more successful explanation of why the visible universe is so uniform in temperature or why the universe appears so ‘flat’. In a flat universe, parallel lines remain an equal distance apart. In a curved universe, they could be more like north–south lines on the surface of the Earth, which meet at the poles. It is thought that rapid expansion during this era smoothed out deviations in the flatness of the universe that would otherwise have grown with time. A flat universe only just avoids eventual gravitational collapse. The small variations in the cosmic microwave background detected by the Planck orbiting observatory are consistent with a universe that underwent a period of inflation and resulted in the clumping of matter into clusters of galaxies.

FIGURE 6.20 Timeline of the early universe

10 10

10 5

10 0 Hadron era 10−6 seconds to 3 mins Quarks form protons and neutrons.

Quark era

10−12 seconds to 10−6 seconds The quark era was when particles began to appear in large numbers. These included quarks, electrons and neutrinos. Most particles still formed in pairs with an antiparticle, but a bias towards particles resulted in matter that was not annihilated through contact with antimatter. The source of this bias is not well understood. 198 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

10 –5

10

–10

10 –15

Quark era 10−12 seconds to 10−6 seconds Quarks, electrons and neutrinos emerge in large numbers.

10 –20 Electroweak era 10−32 seconds to 10−12 seconds The strong nuclear forces emerge. The Higgs boson provides particles with inertial mass.

10 –25

10 –30

Inflation era 10

Electroweak era

10−32 seconds to 10−12 seconds The electroweak era was the period when the strong nuclear force came into play. The Higgs boson formed, enabling particles to have mass.

Nucleogenesis 3 mins to 20 mins Protons and neutrons fuse to form helium-4.

–35

10−36 seconds to 10−32 seconds Period of exponential expansion

Grand unified era 10−43 seconds to 10−36 seconds Force of gravity emerges. Matter annihilated by antimatter.

10–40

10–45

10–50

Planck era 10−43 seconds Physics is unable to describe what happened in this time. The universe was a singularity.

Hadron era

10−6 seconds to 3 minutes The hadron era was the period when the temperature of the universe had dropped to the point where three quarks could form protons and neutrons (particles like neutrons and protons made from three quarks are called hadrons). Most are annihilated by contact with their antiparticles, and leptons such as electrons dominate.

Nucleogenesis 3 minutes to 20 minutes Nucleogenesis occurs. During this era, annihilation with antimatter became less significant and the critical phase of fusion occurred. During these few minutes most of the nuclei in the universe formed. The following section outlines this in detail. FIGURE 6.21 Temperature and size of the universe over time Time (years) 10−50

10−40

10−30

10−20

10−10

1010

1

Temperature of universe (K)

1030

Inflationary model 20

10

1010

Current interior of Sun Boiling water

1 1060 1050 Radius of observable universe

1040 30

Distance (m)

10

1020 1010 1 10−10 10−20

Radius of observable universe (inflationary model)

10−30 10

−40

B

C

10−50 10−60 10−45

10−35

10−25

10−15

10−5

105

Time (seconds) Inflation era =

10−36

s to 10

−32

s

Light nuclei form

1015 Present time Universe transparent to light

TOPIC 6 Origins of atoms 199

SAMPLE PROBLEM 12

What separates the quark era from the other eras of the early universe? Teacher-led video: SP12 (tlvd-0052) THINK 1.

Recall the details of the quark era.

2.

Identify the details that distinguish it from other eras.

WRITE

The quark era was when particles such as quarks, electrons and neutrinos began to appear in large numbers and matter began to dominate over antimatter. Prior to the quark era particles were annihilated by their antiparticles as soon as they formed. Quarks existed as individual particles. In subsequent eras, quarks have been only found combined with other quarks to form larger particles like protons and neutrons.

PRACTICE PROBLEM 12 What distinguishes the electroweak era from other eras of the early universe?

6.5.3 Nucleogenesis (the birth of atoms) In 1948, George Gamow (1904–1968) and Ralph Alpher (1921–2007) proposed a model that explained how over 99% of the atoms found in the universe could have formed. The protons and neutrons in the early period of the universe readily interacted with the abundance of electrons and neutrinos present, causing them to change from proton to neutron, and vice versa. This produced equal numbers of each, but as the universe cooled, fewer protons interacted with electrons with sufficient energy to form neutrons. This resulted in more protons being formed than neutrons in a ratio of about 7 to 1. This formed a universe of hydrogen. Under the right temperature and pressure protons and neutrons can fuse, one at a time, to form helium-4. These conditions were present when the early universe was about 400 seconds old, allowing nuclei up to a mass number of 4 to form. This produced: • hydrogen (1 proton) • deuterium (an isotope of hydrogen with 1 proton and 1 neutron) • tritium (an isotope of hydrogen with 1 proton and 2 neutrons) • helium-3 (2 protons and 1 neutron) • helium-4 (2 protons and 2 neutrons). No stable isotope with a mass number of 5 exists (this has been tested in the laboratory), so fusion of nuclei beyond helium-4 was very limited. Tiny quantities of lithium-7 and beryllium-7 were produced through fusion of helium nuclei, but the step to heavier elements involving the fusion of three helium nuclei to form carbon takes too long. The universe rapidly cooled through its expansion to the point where the conditions no longer supported fusion. This provided the young universe with its composition of about 75% hydrogen and 25% helium by mass. To understand the prediction of the proportion of hydrogen to helium, consider the ratio of protons to neutrons, which is 7:1. To form a helium-4 nucleus, 2 protons and 2 neutrons are required. According to the 7:1 ratio, for every 2 neutrons in the universe, there were 14 protons. The formation of helium-4 would take 2 neutrons and 2 protons, leaving 12 protons in the mix. That leaves 4 nucleons in helium and 12 in hydrogen, or 25% of the mass in helium and 75% in hydrogen. The fusion in these first moments of the

200 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

universe was so rapid and complete that virtually all of the available neutrons went into helium-4. Only a tiny proportion remained as deuterium or tritium (which decays to helium-3) so this simple calculation gives a very good prediction of the composition of the universe prior to star formation.

FIGURE 6.22 In the early universe protons outnumbered neutrons 7 to 1.

Fusion

For nearly 20 minutes, conditions in the universe were sufficiently hot and dense for protons to fuse with neutrons ensuring that virtually all neutrons were bound in helium-4 nuclei. One of the strongest pieces of measured evidence for the Big Bang model of the universe (along with the observed expansion and CMB) is this predicted proportion of hydrogen and helium. As astrophysicists measure the proportions of the elements in regions of the universe not greatly affected by fusion in stars, the elements are found in this predicted abundance. So we have the formation of nuclei in the early universe, but there are no atoms. That will take more time because although the universe has cooled sufficiently for nuclei to form, it is still way too hot for electrons to stay bound to those nuclei. Before atoms could form, 380 000 years would pass. Table 6.1 outlines significant events in the early universe. TABLE 6.1 Significant events in the early universe Time since beginning of universe (seconds)

Temperature (K)

10−36 to 10−32

Event Inflation occurs

10−12 to 10−6

1016

Elementary particles including quarks and leptons form

10−6 to 100

1012

Annihilation of antimatter and matter leave relatively small amount of matter

102

109

Commencement of nuclear fusion

103

Cessation of fusion

1013 (380 000 years)

3000

1013 to 1016 16

10

Dark Ages (stars yet to form)

(800 000 000 years)

First stars and galaxies form; most atoms re-ionised

1017 (9.3 billion years) 1018 (13.82 billion years)

Formation of atoms (recombination), CMB produced

Earth and solar system form 2.7

Today

It was not until 800 million years into the universe’s life that the story of the atom resumed, when the first stars formed and new elements began to form in nuclear fusion in their interiors. In the centres of these enormous stars, the temperature and pressure was sufficient for long enough for fusion to continue beyond the formation of helium-4, resulting ultimately in the genesis of all of the elements in nature. TOPIC 6 Origins of atoms 201

SAMPLE PROBLEM 13

A section of space contained 20 neutrons prior to nucleogenesis. How many helium-4 nuclei would have formed in this section of space during nucleogenesis? Teacher-led video: SP13 (tlvd-0053) THINK

WRITE

Recall how many neutrons and protons are required to make a helium-4. 2. Recall that there were seven protons for every neutron.

2 neutrons and 2 protons per helium-4

1.

3.

Count how many helium-4 nuclei could have formed from this raw material.

7 protons for every neutron were present, so in a space containing 20 neutrons we could expect 7 × 20 = 140 protons — plenty to make helium-4 from every neutron available. Each helium-4 requires 2 neutrons, so 20 neutrons would produce 10 helium-4 nuclei.

PRACTICE PROBLEM 13 A section of space formed 2 × 1030 helium-4 nuclei in the nucleogenesis era of the early universe. How many hydrogen-1 nuclei would have been produced in the same section of space?

Resources Weblinks NASA — the Big Bang Planck data The early universe

6.5 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. Outline how the quark era differed to the electroweak era. What first happened in the hadron era? When did the force of gravity first exist as a defined force? How long into the universe’s existence did protons and neutrons begin to appear in significant numbers? Why were particles in the earliest moments of the universe so short lived? Provide two problems that the inflation hypothesis solves. What prevented nuclear fusion occurring prior to 3 minutes and later than 20 minutes into the life of the universe? 8. Explain how the Big Bang Theory predicts a ratio of hydrogen to helium of 3:1 in terms of mass. 9. What prevented electrons from binding to nuclei to form atoms in the first 380 000 years of the universe’s existence? 1. 2. 3. 4. 5. 6. 7.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

202 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

6.6 Review 6.6.1 Summary • • • • • • • • • • • •

•

Our galaxy, the Milky Way, is only one of billions of galaxies. All distant galaxies are moving away from us, and the further away they are, the faster they are receding. The expansion of space provided the first evidence that the universe had a beginning, now determined to have occurred 13.8 billion years ago. In the earliest moments of the universe, there was a period called inflation, where the size of the universe grew exponentially within a fraction of a second. After about 20 minutes, the universe consisted of hydrogen and helium nuclei with 25% of the mass in helium and 75% in hydrogen. Atoms first appeared 380 000 years into the universe’s history when the universe was cool enough for electrons to stay bound to nuclei. Light was then free to travel through the universe, which is now visible as the cosmic microwave background radiation. Production of heavier elements did not occur until the first stars began fusing hydrogen and helium in their cores, 800 000 000 years after the beginning of the universe. The Big Bang is the name given to the theory that describes the universe beginning from a point of infinite density and expanding to create space and time as we see it today. The key evidence for the Big Bang includes: the expansion of the universe, the higher density of galaxies in the past, the proportion of elements in the universe and the cosmic microwave background radiation. Stephen Hawking showed that the universe may have no boundaries of space or time — time began with the universe, and space and time outside of the universe have no meaning. When measuring distance, temperature, mass and time in the universe, the numbers can be unmanageably large or small. To reduce this problem numbers can be expressed in scientific notation where a number between 1 and 10 is multiplied by a power of 10. For very small numbers, the power of ten is negative. Alternative units for distance are used in measuring space. The astronomical unit (AU) is the distance from Earth to the Sun. It is useful for measuring the solar system. The parsec (pc) is the distance that produces a parallax angle of one arc second as the Earth moves around the Sun. A light-year is the distance light travels in one year.

Resources

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0031).

6.6.2 Key terms The Big Bang is the name for the dramatic beginning of the universe from an infinitely dense, small point. A Cepheid variable is a type of star that has a relationship between its period of variation in luminosity and its maximum luminosity. They are useful for measuring distances to galaxies. The cosmic microwave background is radiation that can be found in all regions of empty space. It is left over energy from the Big Bang. Cosmology is the study of how the universe began, has evolved and will end.

TOPIC 6 Origins of atoms 203

A galaxy is a collection of hundreds of billions of stars. Stars are organised into galaxies, with very few stars existing between galaxies. A globular cluster is a very old, densely packed cluster of stars in the shape of a sphere. Hubble’s constant is the constant of proportionality relating the speed that galaxies are receding from Earth and their distance from Earth. Hubble’s Law states that the speed of recession of galaxies is proportional to their distance from Earth. Recombination is the event when electrons could remain bound to nuclei to form atoms. Prior to 380 000 years after the Big Bang, electrons had too much energy due to the temperature to remain bound to nuclei. Red shift is the increase in wavelength that results from a light source moving away from the observer.

Resources Digital document Key terms glossary (doc-32184)

6.6.3 Practical work and investigations Investigation 6.1 Expansion of the universe Aim: To model the expansion of the universe in two dimensions, using the surface of a balloon Digital document: doc-31871 Teacher-led video: tlvd-0816

Resources Digital document Practical investigation logbook (doc-32185)

6.6 Exercises 6.6 Exercise 1: Multiple choice questions Which of the following correctly describes scientific theories? A. Well thought out scientific explanations B. Guesses C. Beyond dispute D. Measured facts 2. What does the Big Bang Theory predict? A. The ratio of hydrogen to helium is 3:2. B. Galaxies are further apart as we look further into space. C. There is a relationship between distance and red shift of distant galaxies. D. The expansion of the universe will cause the universe to heat up. 3. Which of the following facts did Hubble use to measure distances to galaxies? A. Super novae have a known luminosity. B. Some stars have a luminosity related to their period of variation. C. Globular clusters contain old stars. D. The CMB is nearly uniform.

1.

204 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

4.

5.

6.

7.

8.

9.

10.

What did Hubble discover that was important evidence for the Big Bang Theory? A. Red shift of galaxies increases with distance. B. Red shift of galaxies decreases with distance. C. The CMB is nearly uniform. D. There is three times the amount of hydrogen by mass than there is helium. What is the explanation for the photons of the CMB having much less energy that 13 billion years ago? A. The photons have worn out with age. B. Photons are more dispersed through the expanded universe. C. They have lost energy through interaction with matter. D. Their wavelengths have been stretched as they travel through the expanding universe. Which of the following numbers is the smallest? A. 3.4 × 108 B. 2.34 × 109 C. 3.39 × 108 D. 3 × 108 Which of the following is not written in correct scientific notation? A. 25 × 103 B. 2.3 × 1025 C. 1.006 × 106 D. 7 × 10–7 Which of the following is correct? A. Matter formed before the quark era. B. The strong nuclear force emerged in the inflation era. C. The first atoms formed during nucleogenesis. D. Electrons existed in the quark era. When did the Dark Ages in cosmology occur? A. Before recombination B. After recombination and prior to the first stars C. During inflation era D. Before the quark era When did inflation occur most dramatically? A. Prior to the appearance of gravity B. After atoms formed C. After the strong nuclear force emerged D. After the grand unified era

6.6 Exercise 2: Short answer questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

What did Hubble discover in 1929 that led to the formulation of the Big Bang Theory? The light from the Andromeda galaxy is blue shifted. Explain why it is not red shifted like the light from most galaxies. How is the Big Bang different from normal explosions in terms of space and time? List the key evidence in favour of the Big Bang Theory. Why was the CMB discovery so important in establishing the Big Bang Theory? Why is scientific notation preferred when writing very large and very small numbers? The inflation era is the time when the universe was between 10−36 and 10−32 seconds old. Write these times in decimal notation. Why did nucleogenesis in the early universe only produce light nuclei like hydrogen and helium? List the following events in chronological order: particle–antiparticle annihilation, ignition of the first stars, nuclear fusion, inflation, the formation of atoms. What metaphor did Stephen Hawking use to help explain how time has a beginning? TOPIC 6 Origins of atoms 205

6.6 Exercise 3: Exam practice questions Question 1 (4 marks) a. Sketch a graph of red shift versus distance that summarises Hubble’s observations of galaxy red shifts. b. How did Hubble determine the speed of recession of galaxies? c. What type of star did Hubble use in his distance measurements for the galaxies? Question 2 (3 marks) a. Describe the source of the cosmic microwave background radiation. b. The variations in CMB measured by the Planck space observatory are about 0.000 57 K. Write this in scientific notation.

2 marks 1 mark 1 mark 2 marks 1 mark

Question 3 (1 mark) What tools do physicists use to improve their understanding of what happens to matter in the conditions in the early universe? Question 4 (1 mark) What needs to happen to energy in order to explore earlier periods? Question 5 (1 mark) As physicists explored the earliest moments of the universe they noticed that the four fundamental forces — gravity, weak nuclear force, strong nuclear force and electromagnetic force — were not always separate. In what order did the forces appear after the beginning of the universe? Question 6 (4 marks) For just a few minutes in the early universe, nuclear fusion could take place. a. What were the products of this fusion? b. Why were the products limited to these?

6.6 Exercise 4: studyON topic test Fully worked solutions and sample responses are available in your digital formats.

Test maker Create unique tests and exams from our extensive range of questions, including practice exam questions. Access the assignments section in learnON to begin creating and assigning assessments to students.

206 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

2 marks 2 marks

AREA OF STUDY 3 WHAT IS MATTER AND HOW IS IT FORMED?

7

Particles in the nucleus

Numerous videos and interactivities are available just where you need them, at the point of learning, in your digital formats, eBookPLUS and learnON at www.jacplus.com.au.

7.1 Overview 7.1.1 Introduction The nucleus was first described in 1911 by Ernest Rutherford. Earlier discoveries such as X-rays in 1895 by Wilhelm Röntgen, radioactivity by Henri Becquerel in 1896 and new radioactive elements in 1898 by Marie and Pierre Curie were both explained by and made possible by the nuclear model of the atom developed by Rutherford. By further investigating the nucleus, scientists have been able to identify a set of fundamental particles that are the building blocks of all matter. These particles, and the way they interact with each other, form the Standard Model of particle physics. Understanding the Standard Model is essential in identifying some of the major challenges we face, whether they be scientific, economic, technological, environmental or medical. That understanding is the focus of this topic. The next topic will focus on how the nucleus relates to energy production. FIGURE 7.1 The Super-Kamiokande neutrino observatory is situated 1000 metres underground to insulate it from other subatomic particles. Neutrinos are incredibly difficult to detect, with observatories such as Super-Kamiokande detecting only a handful each month.

TOPIC 7 Particles in the nucleus 207

7.1.2 What you will learn KEY KNOWLEDGE After completing this topic you will be able to: • explain nuclear stability with reference to the forces that operate over very small distances • describe the radioactive decay of unstable nuclei with reference to half-life • model radioactive decay as random decay with a particular half-life, including mathematical modelling with reference to whole half-lives − + • apply a simple particle model of the atomic nucleus to explain the origin of 𝛼, 𝛽 , 𝛽 and 𝛾 radiation, including changes to the number of nucleons − + • explain nuclear transformations using decay equations involving 𝛼, 𝛽 , 𝛽 and 𝛾 radiation • analyse decay series diagrams with reference to type of decay and stability of isotopes • relate predictions to the subsequent discoveries of the neutron, neutrino, positron and Higgs boson • describe quarks as components of subatomic particles • distinguish between the two types of forces holding the nucleus together: the strong nuclear force and the weak nuclear force • compare the nature of leptons, hadrons, mesons and baryons • explain that for every elementary matter particle there exists an antimatter particle of equal mass and opposite charge, and that if a particle and its antiparticle come into contact, they will annihilate each other to create radiation. Source: VCE Physics Study Design (2016–2021) extracts © VCAA; reproduced by permission.

PRACTICAL WORK AND INVESTIGATIONS Practical work is a central component of learning and assessment. Experiments and investigations, supported by a Practical investigation logbook and Teacher-led videos, are included in this topic to provide opportunities to undertake investigations and communicate findings.

Resources Digital documents Key science skills — VCE Units 1–4 (doc-31856) Key terms glossary (doc-32186) Practical investigation logbook (doc-32265)

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0032).

7.2 The discovery of subatomic particles KEY CONCEPTS • Relate predictions to the subsequent discoveries of the neutron, neutrino, positron and Higgs boson. • Describe quarks as components of subatomic particles. • Distinguish between the two types of forces holding the nucleus together: the strong nuclear force and the weak nuclear force. • Compare the nature of leptons, hadrons, mesons and baryons. • Explain that for every elementary matter particle there exists an antimatter particle of equal mass and opposite charge, and that if a particle and its antiparticle come into contact they will annihilate each other to create radiation.

208 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

7.2.1 The Standard Model of particle physics The Standard Model of particle physics describes all known elementary particles of matter and three of the four known fundamental forces that describe their behaviour. There are seventeen particles named in the Standard Model, all of which can be classified into two types of particles: fermions and bosons. Fermions describe the fundamental particles that make up all matter, while bosons are the mediators (cause) of interactions between other particles. A fundamental particle is one that is not made up of other smaller particles. FIGURE 7.2 The Standard Model of particle physics

7.2.2 Gauge bosons There are four fundamental forces of physics that describe the way particles interact with each other. They are, in order of increasing strength: • gravity. The force of attraction between any two objects with mass, it is the weakest of the four forces and acts over an infinite distance. • weak nuclear force (or weak interaction). This force causes radioactive decay. It is much stronger than gravity, but weaker than electromagnetism and the strong force. The weak nuclear force acts over very small distances of magnitude 10−18 metres, or distances less than the width of a proton. • electromagnetism. The force that governs electric fields and magnetic fields, it is the second strongest force after the strong nuclear force and acts over an infinite distance. • strong nuclear force (or strong interaction). This force is responsible for holding protons and neutrons together in the nucleus by acting between the fundamental particles that make up the proton and neutron. It acts over distances of magnitude 10−15 metres, which is about the size of the nucleus.

TOPIC 7 Particles in the nucleus 209

We know that a particle can experience a force due to another particle, even though the two particles are not in direct contact. In the Standard Model, force-carrying particles, called gauge bosons (or bosons), are thought to ‘carry’ forces from one particle to another. Three of the four fundamental forces are governed by the exchange of bosons. Electromagnetism, the strong nuclear force and the weak nuclear force each have their own boson, so gauge bosons come in three main forms: photons, which carry the electromagnetic force, gluons, which essentially hold quarks together and are therefore considered to carry the strong nuclear force, and the W and Z bosons, which are responsible for the weak nuclear force. As yet, no boson has been found that carries the gravitational force and so gravity is the one force that the Standard Model does not yet explain.

7.2.3 Fermions The twelve fundamental particles of matter are called fermions, which are defined by their quantum numbers (such as charge). Fermions can be classified into two groups: quarks and leptons. FIGURE 7.3 The families of subatomic particles and their relationship to matter and atoms

Fermions (matter)

Leptons

Quarks

Hadrons

Mesons

Baryons

Nuclei

Atoms

Leptons Leptons are the simplest and lightest of the subatomic particles. The different types of leptons are shown in table 7.1. Leptons interact using the weak nuclear force. They do not experience the strong nuclear force.

210 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Leptons are fundamental particles, that is, they have no internal structure, although muons and tau particles decay into electrons. The neutrinos accompany any interaction of their heavier partner. TABLE 7.1 Leptons Name

Mass

Charge

𝜇–

About 200 × mass of electron

Negative

1869 1936

Negative

1977

10−13 s

Electron neutrino

ve

About 277 × mass of electron

Negative

Negligible

Neutral

1956

Stable

Muon neutrino

vµ

Negligible

Neutral

1962

Stable

Negligible

Neutral

2000

Stable

Electron Muon Tau

Tau neutrino

Symbol e– 𝜏

vτ

First observed

Half-life Stable 10−6 s

The electron is found in atoms and determines the chemical properties of elements. The muon decays to an electron according to the equation: 𝜇 − → e− + ve + v𝜇 . The ve particle is an anti-electron neutrino. The bar above the symbol indicates that it is an antiparticle (discussed later). Note: The neutrino that is produced in beta decay — that is, when a neutron decays into a proton — is actually an anti-electron neutrino. Muons have a number of industrial uses. They are more penetrating than X-rays and gamma rays and are non-ionising, so they are safe for humans, plants and animals. Their better penetrating power means that, for example, they can be used to investigate cargo containers for shielded nuclear material. Muons have also been used to look for hidden chambers in the pyramids. Muon detectors were used at the Fukushima nuclear complex to determine the location and amount of nuclear fuel still inside the reactors that were damaged by the Japanese tsunami in 2011. The tau particle was discovered some time later than the muon. The unusual feature of this particle is that it decays into two pions, which are discussed later. The decay equation is 𝜏 − → 𝜋 − + 𝜋 0 + v𝜏 . The negative pion, 𝜋 − , then decays into an electron, while the neutral pion, 𝜋 0 , decays to two gamma rays. SAMPLE PROBLEM 1

How many different leptons are there in the Standard Model? Name them. THINK

WRITE

Look up leptons in table 7.1. 2. List the names from table 7.1.

There are six leptons in the Standard Model. The six leptons are: • electron • muon • tau • electron neutrino • muon neutrino • tau neutrino.

1.

PRACTICE PROBLEM 1 Compare the life times of the leptons.

TOPIC 7 Particles in the nucleus 211

Quarks We know that all matter is made up of protons, TABLE 7.2 Quarks neutrons and electrons, however, protons and Multiple of First neutrons can be broken down further into Quark Symbol Charge proton mass observed fundamental particles called quarks. Up u + 32 0.003 1968 The quark model has six different quarks, each with different masses and a fraction of the Down d − 31 0.006 1968 charge of the electron. Each quark has its own 2 Charm c +3 1.3 1974 antiparticle. Quarks have rather unusual names, which are shown in table 7.2 along with their Strange s − 31 0.1 1968 charges and mass. Top t + 32 184.0 1995 The top quark has the same mass as a 1 Bottom b −3 4.5 1977 gold atom! Quarks are the only particles that interact using the strong nuclear force. Leptons do not experience the strong nuclear force. Quarks have never been found on their own, they have only been found bound together with other quarks to form hadrons.

Hadrons Hadrons are distinctive because they are much heavier than the leptons, but much more importantly they all have an internal structure. Hadrons are made up of different combinations of quarks. Hadrons that are a combination of two quarks are called mesons. The other hadrons are combinations of three quarks and are called baryons.

Mesons There are over 60 different types of mesons. They play a role in nuclear interactions but have very short half-lives, so they are very difficult to detect. Each meson also has an antiparticle. Mesons are composed of one quark and one antiquark. A positive pion (𝜋 + ) is made of one up quark and one down antiquark to give a charge of +1. While its antiparticle, 𝜋 – , is made of one up antiquark and one down quark to give a charge of −1.

Baryons

Baryons include the proton and neutron as well as about 70 other different particles. Only the proton and neutron are stable, with all other baryons having extremely short half-lives. Each baryon also has its own antiparticle. Baryons have three quarks. A proton is made up of two up quarks and one down quark to give a charge of +1. A neutron consists of one up quark and two down quarks to give a charge of zero. FIGURE 7.4 The quark structure of the neutron and proton

FIGURE 7.5 The different types of fermions

Leptons (e.g. electrons)

Baryons (e.g. protons and neutrons)

Mesons

Hadrons Q (co uarks ) m bi ne to fo r m h a dro n s Fermions

A proton is made up of two up quarks and one down quark to give a charge of +1. A neutron consists of one up quark and two down quarks to give a charge of zero. 212 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

SAMPLE PROBLEM 2

Particles of matter can be classified as either mesons, baryons or leptons. Which of these is not part of an atom? THINK 1.

WRITE

Look at figure 7.3, a classification key for subatomic particles. Locate ‘Atoms’ and determine which is not part of an atom: a meson, baryon or lepton.

According to the key, atoms are made up of a combination of nuclei, which in turn are made up of baryons and leptons. Mesons are found on a different branch, therefore, a meson is not part of an atom.

PRACTICE PROBLEM 2 According to figure 7.3, nuclei are made of baryons. What, therefore, are two particles from the baryon family?

Why a quark model? In the late nineteenth century, when visible light was shone through a gas of atoms of a particular element, a spectra of black lines was observed. Each element produced a unique pattern of these lines, called an absorption pattern (see figure 7.6). FIGURE 7.6 A continuous spectrum and two different ways of producing an element’s fingerprint Continuous spectrum

Emission line spectrum

Hot gas

Cold gas

Absorption line spectrum

TOPIC 7 Particles in the nucleus 213

In topic 2, the greenhouse effect was explained by describing how H2 O and CO2 molecules respond to particular infra-red wavelengths of electromagnetic radiation. The lines in the atomic absorption patterns suggested that there was some complexity or structure inside the atom. This structure was discovered early in the twentieth century. Similarly, the absorption patterns for H2 O and CO2 molecules tell us something about how the molecules are put together. More information about the internal structure of the nucleus can be determined by the energy of alpha particles emitted through radioactive decay. This energy is specific to the nucleus undergoing decay. If a system is showing evidence that it can have only certain energy values, then it must have a structure, that is, be made up of smaller particles. During the 1960s it was discovered that when protons and neutrons were hit by a beam of particles, a type of spectra was evident, much like molecules, atoms and nuclei. This meant that protons and neutrons are made up of even smaller particles. This was the beginning of the quark model. In 1961, Murray Gell-Mann and Kazuhiko Nishijima developed a classification of all the known subatomic particles that predicted a new type of particle called a quark, which was found a few years later. Then Gell-Mann and George Zweig developed their idea further into the quark model.

SAMPLE PROBLEM 3

Like atoms and nuclei, protons and neutrons are not fundamental particles. They can be explained as a combination of smaller particles called quarks. How do physicists know that atoms, nuclei, protons and neutrons are not fundamental particles? THINK

WRITE

Identify the observations that physicists have about protons and neutrons. 2. How does this information imply that protons and neutrons are not fundamental particles?

In a similar way to atoms and nuclei, a spectrum of allowable energies has been detected for protons and neutrons. Atoms produce electromagnetic spectra containing particular allowable energies that are interpreted as energy released by electrons as they change state in the atom. If the atom had no structure, but was a fundamental particle, this would be impossible as there would be no electrons or energy states. Similarly with protons and neutrons, the existence of possible energy states within these particles reveals that they have internal structure. They are not fundamental. As protons and neutrons also have a spectrum of distinct energy states they are not fundamental, but are made up of smaller constituent parts known as quarks.

1.

PRACTICE PROBLEM 3 How many energy states could a fundamental particle be in?

214 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

7.2.4 Matter and antimatter In 1928, Paul Dirac developed a mathematical theory to explain the properties of the electron: its mass, charge, role in beta decay and its energy inside the atom. The calculated values for these properties exactly fitted their measured values, but Dirac’s equations suggested there may be two solutions — a negatively charged electron and a positively charged electron, with the same mass but opposite charge. Indeed, he raised the possibility of there being another form of matter, which he called ‘antimatter’. Every particle, whether lepton, hadron or boson has an antiparticle. The best known example is probably the antiparticle of the electron, known as the positron. These antiparticles have important practical uses and are a key component of medical scanning through positron emission tomography (PET) scans. Particles and their antiparticles have the same mass but opposite charge. To indicate an antiparticle of an uncharged particle, a bar is placed above the symbol of the matter particle; for example, a tau neutrino has the symbol v𝜏 and the antitau neutrino has the symbol v𝜏 . To indicate the antiparticle of a charged particle, the symbol is given the opposite sign; for example, the symbol for the electron is e− and the symbol for the positron is e+ . The electron is a lepton and the positron its antiparticle. Hadrons, such as protons, neutrons and mesons also have antiparticles. If the hadron is charged, its antiparticle has the same mass but opposite charge. The neutron has no charge so the antineutron also has no charge. Mesons have some interesting examples because they are made from a quark and an antiquark. If the quark and antiquark are of the same type, known as flavour (for example, an up quark and an anti-up quark), then the antiparticle is exactly the same particle! Bosons, apart from the W are chargeless. These bosons are their own antiparticles. The antiparticle of the + W is the W− , and vice versa. When a particle and its own antiparticle come together, they annihilate each other, producing photons of energy. SAMPLE PROBLEM 4

The antiparticle of a proton is an antiproton. It is made from two up antiquarks and a down antiquark. What is the charge of the antiproton? Show that this is consistent with the charges of its component quarks. Teacher-led video: SP4 (tlvd-0062) THINK 1.

Recall that a charged particle has the opposite charge of its antiparticle.

2.

Look up the charges of the up and down quarks and find the charges of the antiquarks.

3.

Add the charges of the three antiquarks that make an antiproton.

4.

Compare this total with the charge of the antiproton.

A proton has a charge of +1 so an antiproton has a charge of −1. 2 An up quark has a charge of + and a down 3 1 quark has a charge of − . Therefore, an up 3 2 antiquark has a charge of − and a down 3 1 antiquark has a charge of + . 3 An antiproton is made of two up antiquarks and one down antiquark. These have a total charge of 2 1 2 − + − + = −1. 3 3 3 The sum of the antiquark charges is the same as the antiproton charge as expected. WRITE

TOPIC 7 Particles in the nucleus 215

PRACTICE PROBLEM 4 Compare the properties of an antiproton with those of an electron.

7.2.5 Bosons and the strong and weak nuclear forces The strong force acts between quarks and holds protons and neutrons together to form the different nuclei. The way that this force acts between quarks is through the exchange of massless particles called gluons. The strong force acts over a tiny distance so quarks and hadrons made from them must be very close for gluon exchange to occur. The strong nuclear force is the strongest of the four fundamental forces. The weak nuclear force is much stronger than gravity but much weaker than electromagnetic force and the strong nuclear force. It acts over only one thousandth of the distance of the strong nuclear force. It is the force that results in beta decay, where protons become neutrons, or neutrons become protons. Other forces do not change particles from one type to another like this. To achieve this change a boson, either a W+ or a W− , is exchanged. The weak force acts on all particles involved in beta decay, the nucleon, the electron/positron and the neutrino/antineutrino. The strong nuclear force does not act on leptons such as electrons and neutrinos. A third boson can be exchanged in the weak force. The uncharged Z particle is exchanged when scattering of leptons, like neutrinos, changes their momentum. SAMPLE PROBLEM 5

How does the strong nuclear force differ from the weak nuclear force in terms of particles it acts on and the distance over which it acts? Teacher-led video: SP5 (tlvd-0063) THINK 1.

What particles does the strong nuclear force act on?

2.

What particles does the weak nuclear force act on?

3.

Over what distances do the strong and weak forces act?

WRITE

The strong nuclear force acts on quarks and particles made from quarks — hadrons, including protons and neutrons. The weak nuclear force acts on leptons such as electrons and neutrinos, as well as quarks. While the strong force holds quarks to each other, the weak force changes properties of the particle. (It can change an up quark to a down and it can change the momentum of a neutrino.) The strong force acts over a distance of the order of a proton diameter. The weak force acts over a distance only about one-thousandth of this distance.

PRACTICE PROBLEM 5 There is a particle known as a pion. It is a meson made from combinations of pairs of up and down quarks and antiquarks. The half-life of the pion made from an up quark and a down antiquark is 2.6 × 10−8 seconds. They decay into muon and muon neutrinos. What force acts to cause this decay?

7.2.6 Explaining the strong nuclear force The atomic nucleus is held together by the strong nuclear force due to the exchange of gluons between quarks. The hadrons (such as the proton and neutron) made up of these quarks also experience the strong force. 216 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

In the 1930s, Hideki Yukawa was seeking an explanation for the properties of the strong nuclear force that exists between protons and neutrons inside the nucleus. It was known that this force had a very short range, with each proton or neutron attracted only to its near neighbours, not the whole nucleus. Yukawa suggested that a previously unobserved particle acted as ‘glue’ between pairs of protons in the nucleus, as well as between other pairings. To fit the known features of the strong force, he determined the properties of this unobserved particle. He said it should: • be about 200 times the mass of the electron • have the same charge size as the electron • come in two types: positive and negative • have a very short half-life of about a millionth of a second • interact very strongly with nuclei. In 1936 Carl Anderson’s group found such a particle in cosmic ray showers. This particle was named the muon. However, while the muon satisfied the first four of Yukawa’s properties, it became apparent that its interaction with nuclei was very weak, so the muon was not a good candidate to explain the strong nuclear force. A few years later, Cecil Powell investigated cosmic ray showers at high altitude in the Pyrenees and the Andes. These observations were higher up in the cascade of collisions that cosmic rays set off when they hit the atmosphere. At this altitude, Powell found another particle that better fit the needs of the strong nuclear force. This particle is called the 𝜋-meson or pion. Shortly after Powell’s discovery, the pion was also detected in the laboratory when carbon nuclei were bombarded with high-energy alpha particles. After this time, most new particles were found in laboratories using particle accelerators.

7.2.7 Prediction of new particles The positron Natural radioactivity was first discovered when beta particles exposed photographic plates. One of the other technologies used to investigate radioactivity was the gold leaf electroscope. Electroscopes show the presence of electric charge. A charged electroscope slowly loses charge due to the ions in the air produced by radioactive elements in the Earth’s crust. It was thought that this effect would decrease with height above the ground.

Number of ions detected every second

FIGURE 7.7 The number of charged particles from cosmic rays varies by altitude above the Earth’s surface. Charged electroscope − − − − − − − − − − −

80

− − − − − − − −

60

40

− − − − − − − −

− − − − − − − − − −

20

0

2

4 6 Altitude (km)

Negative charges

Gold leaf is repelled

8

TOPIC 7 Particles in the nucleus 217

However, in 1909 it was found that the intensity of radiation was greater on top of the Eiffel Tower. Balloon flights then showed the intensity continued to increase with height, suggesting that the radiation may originate from space. So, the name cosmic rays was coined. Initially cosmic rays were called ‘rays’ because they were thought to be like light. However, even though they are now known to be particles, the name has stuck. Further investigation over the following decades showed that the particles entering the Earth’s atmosphere were mainly protons. The particles seemed to come from beyond the solar system from all points of the sky. Indeed they are now thought to originate in supernovae and the centre of galaxies. They are also extremely fast and energetic. The energy of these protons is 40 million times the energy of the protons in the Large Hadron Collider used to produce the Higgs boson. When these protons with their massive energy hit an atom in the upper atmosphere, they cause a cascade of successive collisions that produces a shower of charged particles and gamma rays at the Earth’s surface. On average, cosmic rays contribute about 16% of your exposure to ionising radiation from natural sources. This exposure increases the more you fly in a plane and the higher you fly. In 1933 Carl Anderson was investigating the charged particles in cosmic ray showers and observed a particle that had the same mass as the electron, but with a positive charge. He had discovered a new particle, the positron. The chamber Anderson used to detect this charged particle was placed in a strong magnetic field so that a positively charged particle would curve one way and a negatively charged particle would curve the other way. In this experimental set up, an incoming gamma ray collides with a nucleus; the energy of the gamma ray is converted into mass, using E = mc2 , but because charge needs to be conserved, two particles of opposite sign are produced. FIGURE 7.8 Showers of particles form as cosmic rays hit the atmosphere. p — proton n — neutron π+, π−, π0— pions μ+, μ−— muons e−— electron e+— positron v — neutrino γ — gamma ray

Top of the atmosphere p

π−

π+ π+ n

π0

γ v

γ

μ+

v

e−

e+ − e

μ−

e+

Source: NASA

The reverse process is also possible. An electron and a positron, or indeed, any particle and its antiparticle, can collide and annihilate each other, producing two gamma rays.

218 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 7.9 A gamma ray produces an electron and a positron. The spiralling is due to the slow loss of energy as the track is created.

FIGURE 7.10 A particle and antiparticle annihilate each other with their mass, producing two gamma rays.

A more energetic electron–positron pair

Invisible gamma ray Positron

Electron

Gamma ray

Electron −

+ Positron Before Energy = 2mec2

After gamma ray

Resources Interactivity Electrons and positrons (int-6394)

The neutron When Rutherford was developing the concept of the nucleus, it was known that an alpha particle had exactly twice the positive charge of a hydrogen ion, so presumably contained two protons, but was almost exactly four times as heavy. There were a number of explanations for this anomaly. One was that the extra mass was made up of proton–electron pairs, which would effectively have zero charge; another was that there was an as yet unknown neutral particle in the nucleus. It was only in the 1930s that this neutral particle, called the ‘neutron’, was discovered. In 1930, Walter Bothe and Herbert Becker fired alpha particles at beryllium and detected what they thought was gamma radiation. The husband-and-wife team Frédéric Joliot and Irène Joliot-Curie (daughter of Marie Curie) placed hydrogen-rich paraffin wax in front of this ‘gamma radiation’ and observed the ejection of protons. While they explored the possibility of very high-energy radiation, James Chadwick showed that this was virtually impossible; instead, he demonstrated that a single neutron was ejected when the alpha particle entered the beryllium nucleus, which in turn knocked on a proton in a simple billiard-ball-like collision. FIGURE 7.11 Discovery of the neutron Unknown particles

Protons

Ionisation chamber Source of α particles Be

Paraffin wax

TOPIC 7 Particles in the nucleus 219

The discovery of the neutron now provided an explanation for the existence of isotopes (atoms with the same atomic number but different atomic mass due to different numbers of neutrons).

The neutrino Alpha particles are emitted with particular energies that are unique to the host nucleus, whereas beta particles are emitted with any energy up to a maximum. When examining decay reactions, scientists found that not all of the energy was accounted for. The possible explanations were: • the Law of Conservation of Energy, one of the foundations of physics, did not apply in nuclear processes • a second particle, as yet undetected, was emitted. This idea was proposed by Wolfgang Pauli, who said the particle must have no charge, as all the charge was accounted for, and have negligible mass. Enrico Fermi named the particle ‘neutrino’, from the Italian for ‘little neutral one’. Fermi incorporated Pauli’s suggestion into a theory of 𝛽 decay that not only built on Dirac’s work but also derived a mathematical relationship between the half-life of a particular decay and the maximum energy of the emitted 𝛽 particle. This relationship matched the experimental data, which was convincing evidence for the existence of the neutrino, although it was not actually detected until 1956. The neutrino has the symbol v, which is the Greek letter nu. The antineutrino has the symbol v. The complete 𝛽 – decay process is: 1 n 0

→ 11p + −10e + v

Beta decay will be further explored in section 7.4.2.

The Higgs boson Since the positron was observed in 1932, confirming a prediction from theory, particle after particle that has been predicted has been observed. By the end of 2000, all predicted particles had been observed in experiment except for the Higgs boson. This particle had been predicted in 1964 to solve problems with existing theory. Scientists had developed a theory that made sense of what was known about the particles of matter that made up the known universe. However, it only worked if certain particles had zero mass. Some of those particles were known to have mass so something was missing. Peter Higgs, François Englert and three others proposed that there was a field through space, now known as the Higgs field, that particles with mass interacted with via a particle called the Higgs boson. If the Higgs boson was found, the theory of the Standard Model would match experimental results. To search for this elusive particle, the largest particle accelerator ever built was constructed at CERN on the boarder of Switzerland and France. More than 100 countries are involved in the Large Hadron Collider, which has cost more than US$13 billion. In July 2012 scientists announced that this enormous investment had paid off and the final piece in the Standard Model had been observed. Higgs and Englert were awarded the Nobel Prize for Physics in 2013, for the prediction they had made nearly fifty years earlier. SAMPLE PROBLEM 6

Name four particles whose existence was predicted by scientific theory before they were observed or detected. THINK 1

This topic highlights four particles that were predicted prior to discovery.

WRITE

The neutron, neutrino, positron and Higgs boson were all predicted by scientific theory prior to their detection. If they had been found not to exist, a new theory would have been required as the theory would not have been consistent with observation.

220 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

PRACTICE PROBLEM 6 Refer to topic 6. What is another example of a phenomenon that was predicted by theory prior to being observed?

7.2 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. The Large Hadron Collider used to discover the Higgs boson is an internationally run experiment that has cost billions of dollars. Comment on the importance of scientific theory in its construction. 2. Explain the difference between the terms baryon and hadron. 3. How many quarks does the Standard Model predict? Name them. 4. How many leptons does the Standard Model predict? Name them. 5. Compare the forces holding electrons in an atom with those holding protons and neutrons together in the nucleus. 6. Neutrinos are very difficult to detect and most that hit the Earth pass right through. Protons and electrons do not pass through. Explain the differences. 7. From your knowledge of the neutron, describe an antineutron. 8. What would happen if a neutron and an antineutron came together? 9. Antimatter was discussed in topic 6 when describing matter in the first moments of the universe. Explain why matter appeared for such brief moments before being destroyed in the early universe, and why most of the matter in the form of protons, neutrons and electrons has existed since the first minutes of the universe’s existence. 10. Hadrons and quarks were eras of the early universe discussed in topic 6. What were the defining features of those eras? 11. The lambda baryon (Λ) was discovered by researchers at the University of Melbourne in 1950. Its quark composition is uds. What is its charge?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

7.3 Nuclear radiation KEY CONCEPTS • Explain nuclear stability with reference to the forces that operate over very small distances. • Distinguish between the two types of forces holding the nucleus together: the strong nuclear force and the weak nuclear force. • Describe the radioactive decay of unstable nuclei with reference to half-life. • Model radioactive decay as random decay with a particular half-life, including mathematical modelling with reference to whole half-lives.

7.3.1 Atoms and isotopes Nuclear radiation, as the name suggests, is radiation emitted from the nucleus of an atom. In order to explain the mechanisms that release such radiation, it is important to understand a little about the structure of the atom.

TOPIC 7 Particles in the nucleus 221

Atoms All matter is made up of atoms. Each atom consists of a tightly FIGURE 7.12 The structure of a packed, positively charged centre, called the nucleus, which is typical atom surrounded by a ‘cloud’ of negatively charged electrons. The Proton particles in the nucleus are known collectively as nucleons, but (positive) there are two different types, protons and neutrons. Protons are positively charged, neutrons are chargeless, and both are about Electron cloud 2000 times heavier than the electrons that surround the nucleus. Scientists name atoms according to the number of protons in Nucleus the nucleus. For example, all atoms with six protons are called carbon, all atoms with 11 protons are called sodium, and all atoms with 92 protons are called uranium. A substance consisting of atoms that all have the same name is called an element. Each Electron element’s name has its own shorthand symbol that scientists use. (negative) Neutron Carbon has the symbol ‘C’, sodium ‘Na’ and uranium ‘U’. It is (neutral) very important that the upper or lower case of the letters used in the symbols is kept the same. The names of all the elements, and their symbols, can be found in the periodic table in figure 7.13. FIGURE 7.13 The periodic table Group 18

Group 1

Period 1

1 metals Hydrogen H 1.0

Group 2

Period 2

3 Lithium Li 6.9

4 Beryllium Be 9.0

11 Sodium Na 23.0

12 Magnesium Mg 24.3

Period 4

19 Potassium K 39.1

Period 5

Period 3

Period 1

1 Hydrogen H 1.0

2 Helium He 4.0

Group 3

Group 4

Group 5

Group 6

20 Calcium Ca 40.1

21 Scandium Sc 45.0

22 Titanium Ti 47.9

23 Vanadium V 50.9

24 Chromium Cr 52.0

25 Manganese Mn 54.9

37 Rubidium Rb 85.5

38 Strontium Sr 87.6

39 Yttrium Y 88.9

40 Zirconium Zr 91.2

41 Niobium Nb 92.9

42 Molybdenum Mo 96.0

Period 6

55 Caesium Cs 132.9

56 Barium Ba 137.3

57–71 Lanthanoids

72 Hafnium Hf 178.5

73 Tantalum Ta 180.9

Period 7

87 Francium Fr (223)

88 Radium Ra (226)

89–103 Actinoids

104 Rutherfordium Rf (261)

105 Dubnium Db (262)

Alkali metal

Group 7

Key Atomic number Name Symbol Relative atomic mass

Group 8

Group 15

2 Helium He 4.0

Group 13

Group 14

Group 16

Group 17

5 Boron B 10.8

6 Carbon C 12.0

7 Nitrogen N 14.0

8 Oxygen O 16.0

9 Fluorine F 19.0

10 Neon Ne 20.2

13 Aluminium Al 27.0

14 Silicon Si 28.1

15 Phosphorus P 31.0

16 Sulfur S 32.1

17 Chlorine Cl 35.5

18 Argon Ar 39.9

Group 9

Group 10

Group 11

26 Iron Fe 55.8

27 Cobalt Co 58.9

28 Nickel Ni 58.7

29 Copper Cu 63.5

30 Zinc Zn 65.4

31 Gallium Ga 69.7

32 Germanium Ge 72.6

33 Arsenic As 74.9

34 Selenium Se 79.0

35 Bromine Br 79.9

36 Krypton Kr 83.8

43 Technetium Tc (98)

44 Ruthenium Ru 101.1

45 Rhodium Rh 102.9

46 Palladium Pd 106.4

47 Silver Ag 107.9

48 Cadmium Cd 112.4

49 Indium In 114.8

50 Tin Sn 118.7

51 Antimony Sb 121.8

52 Tellurium Te 127.6

53 Iodine I 126.9

54 Xenon Xe 131.3

74 Tungsten W 183.8

75 Rhenium Re 186.2

76 Osmium Os 190.2

77 Iridium Ir 192.2

78 Platinum Pt 195.1

79 Gold Au 197.0

80 Mercury Hg 200.6

81 Thallium Tl 204.4

82 Lead Pb 207.2

83 Bismuth Bi 209.0

84 Polonium Po (210)

85 Astatine At (210)

86 Radon Rn (222)

106 Seaborgium Sg (266)

107 Bohrium Bh (264)

108 Hassium Hs (267)

109 Meitnerium Mt (268)

110 Darmstadtium Ds (271)

111 Roentgenium Rg (272)

112 Copernicium Cn (285)

113 Nihonium Nh (280)

114 Flerovium Fl (289)

115 Moscovium Mc (289)

116 Livermorium Lv (292)

117 Tennessine Ts (294)

118 Oganesson Og (294)

61 Promethium Pm (145)

62 Samarium Sm 150.4

63 Europium Eu 152.0

64 Gadolinium Gd 157.3

65 Terbium Tb 158.9

66 Dysprosium Dy 162.5

67 Holmium Ho 164.9

68 Erbium Er 167.3

69 Thulium Tm 168.9

70 Ytterbium Yb 173.1

71 Lutetium Lu 175.0

93 Neptunium Np (237)

94 Plutonium Pu (244)

95 Americium Am (243)

96 Curium Cm (247)

97 Berkelium Bk (247)

98 Californium Cf (251)

99 Einsteinium Es (252)

100 Fermium Fm (257)

101 Mendelevium Md (258)

102 Nobelium No (259)

103 Lawrencium Lr (262)

Group 12

Lanthanoids

Alkaline earth metal Transition metal Lanthanoids Actinoids

57 Lanthanum La 138.9

58 Cerium Ce 140.1

59 60 Praseodymium Neodymium Pr Nd 140.9 144.2

Unknown chemical properties Post-transition metal Metalloid Reactive non-metal Halide Noble gas

Actinoids 89 Actinium Ac (227)

90 Thorium Th 232.0

91 Protactinium Pa 231.0

92 Uranium U 238.0

Isotopes Not all atoms of the same name (and therefore the same number of protons) have the same number of neutrons. For instance, it is possible to find carbon atoms with six, seven and eight neutrons in the nucleus along with the six protons that make it carbon. These different forms of an element are called isotopes. To avoid confusion about which isotope is being referred to, scientists have a few standard ways of writing them. The number of nucleons, or mass number, of the particular isotope is used. The isotope of carbon with six protons and six neutrons is written as carbon-12 or 12 C, whereas the isotope of carbon with eight

222 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

neutrons is written as carbon-14 or 14 C. Sometimes the number of protons, or atomic number, is written directly underneath the mass number, although this is not necessary (for example, 146C ). SAMPLE PROBLEM 7

Write the name of the isotope of an atom with 90 protons and 144 neutrons. THINK 1.

WRITE

What element has 90 protons?

What is the mass number of this isotope? 3. Write the name of the isotope.

2.

Checking the periodic table, the element thorium (symbol Th) has 90 protons. 90 + 144 = 234 The isotope of an atom with 90 protons and 144 neutrons is thorium-234 or 234 Th.

PRACTICE PROBLEM 7 What is the name of the isotope of an atom with 26 protons and 30 neutrons?

What holds the nucleus together? The force that holds electrons around a nucleus is called an electrostatic force. Electrostatic forces increase as charges move closer together. Electrostatic attraction exists between unlike charges; electrostatic repulsion exists between like charges. So it seems strange that the positive charges inside a nucleus don’t repel each other so strongly that the nucleus splits apart. In fact, two protons do repel each other when they are brought together but in the nucleus they are so close to each other that the force of repulsion is overcome by an even stronger force — the strong nuclear force. While the strong nuclear force is a very strong force, it can act over incredibly small distances only. Inside a nucleus, the nucleons are sufficiently close that the pull of the strong nuclear force is much greater than the push of the protons repelling each other, and therefore the nucleus remains intact. Figure 7.14 shows how the strength and sense of the strong nuclear force changes with the separation of the nucleons. Its ability to hold two nucleons together is strongest at around 1 × 10−15 metres but is virtually zero at about 2.5 times this distance.

Repulsion

1

2

3

4

Separation of nucleons (10−15 m)

Attraction

Force between nucleons

FIGURE 7.14 How the strong nuclear force between two nucleons varies with the separation of the nucleons

TOPIC 7 Particles in the nucleus 223

Unstable nuclei

Number of neutrons

Some nuclei are unstable. Isotopes that contain FIGURE 7.15 This graph shows which nuclei are stable unstable nuclei are called radioisotopes. (yellow) and which are unstable (other). The reason that some nuclei are unstable can N be explained by considering the interaction Unstable isotopes can emit various types of radiations of the nucleons with the fundamental forces. Though the strong nuclear force acts over very short distances only, the electromagnetic force continues to act with decreasing strength 126 as the distance between charged particles increases. This is one reason that many nuclei are unstable. In a large nucleus, the combined Z=N effect of all the protons in the nucleus results in an electrostatic repulsion on some protons 82 greater than the strong force holding those Type of decay protons in the nucleus. This nucleus is unstable and cannot remain like this forever. This β+ explains why the number of neutrons in large 50 β− atoms is greater than the number of protons; α neutrons are chargeless, so do not contribute Fission to the electromagnetic repulsion. This trend is 28 Stable nuclide apparent in figure 7.15. 14 The weak nuclear force and the gravitational 6 force are not significant in determining the stability of nuclei as they are comparatively 6 14 28 50 82 weak over the distance of nucleon separation. Number of protons In order to become more stable, the nucleus in a radioisotope emits different types of nuclear radiation that increase the stability of the nucleus. The different types of radiation will be explained in detail in subtopic 7.4. TABLE 7.3 Types of forces, their relative strengths and ranges Force Strong nuclear Electric and magnetic Weak nuclear Gravity

Strength

Range

1

–15

1 137 1 1000 000 1 38

10

10

m (diameter of a nucleus)

Type Attractive

Infinite

Attractive or repulsive

10–18 m (fraction of proton diameter)

Attractive

Infinite

Attractive

SAMPLE PROBLEM 8

The existence of nuclei larger than a single proton depends on forces holding neutrons and protons to together. Describe the nature of this force. THINK 1.

A nucleus larger than a proton includes protons and neutrons that are somehow held together.

WRITE

There is an attractive force called the strong nuclear force that acts between protons and neutrons.

224 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Z

2.

Protons repel each other as they have like charge.

3.

Describe the nature of the force.

The strong nuclear force can only hold a nucleus together when it is stronger than the electromagnetic force pushing the protons apart. The force holding protons and neutrons together to form nuclei must be an attractive force that works between neutrons and protons and is stronger than the electromagnetic force at close range.

PRACTICE PROBLEM 8 A helium-3 nucleus has two protons and one neutron. Describe the forces acting within it.

7.3.2 Half-life It is not possible to predict exactly when a given unstable nucleus will decay. However, we can predict what proportion of a certain number of nuclei will decay in a given time. It is rather like tossing a coin. We can’t be sure that a given toss will result in a tail but we can predict that from 1000 tosses about 500 will result in tails. Scientists know that it will take 24 days for half of a group of unstable thorium-234 nuclei to decay to protactinium-234. The time taken for half a group of unstable nuclei to decay is called the half-life. Halflives vary according to the isotope that is decaying. They range from microseconds to thousands of millions of years. TABLE 7.4 Table of half-lives Element

Half-life

Indium-115

1.3 × 109 years

Potassium-40 Uranium-235 Actinium-227

6 × 1014 years

7.1 × 108 years 22 years

Thorium-227

18 days

Carbon-11

20 minutes

Thallium-207

4.8 minutes

Magnesium-23

11 seconds

Polonium-212

3 × 10−7 seconds

Mathematicians and scientists often use graphs with the same basic shape as the one in figure 7.16. It shows what is known as a decay curve. This type of curve often appears in science. It is called exponential decay.

TOPIC 7 Particles in the nucleus 225

Looking at figure 7.16, in the first few days the number of atoms decaying every day is quite high but towards the end the number is quite low. If there was a Geiger counter near the source at the beginning, it would be clicking quickly but near the end you would wait days for the next click. In fact, a graph of the count rate (the number of clicks per second) will have exactly the same shape and will show the same half-life. The number of decays per second of a radioactive source is a measure of its activity and is measured in becquerels (Bq).

FIGURE 7.16 Graph showing decay of 1000 thorium-234 nuclei. The time taken for half of the original nuclei (500 nuclei) to remain is called the half-life. It can be seen from this graph that the half-life is 24 days. After the fourth half-life (at time = 96 days) it can be predicted that one-sixteenth of the thorium-234 (about 62 or 63 nuclei) would remain undecayed. 1000

Item

Activity (Bq)

1 adult human

3000

1 kg coffee

1000

1 kg granite

1000

1 kg coal ash (used in cement)

2000

1 kg superphosphate fertiliser

5000

Number of 234Th nuclei

TABLE 7.5 Activities of some everyday items

500

0 24 t1

50

Time (days)

– 2

SAMPLE PROBLEM 9

Technetium-99 is often used for medical diagnosis. It has a half-life of 6 hours. A patient has a small amount of the isotope injected into the bloodstream. What fraction of the original amount will remain after: a. 12 hours b. 48 hours? Teacher-led video: SP9 (tlvd-0056) THINK a. 1.

How many half-lives have passed?

How much would be left after halving it that many times? 3. State the solution. 2.

b. 1.

How many half-lives have passed?

2.

How much would be left after halving it that many times?

3.

State the solution.

226 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

12 = 2 half-lives 6 ( )2 1 1 1 1 = × = 2 2 2 4

WRITE a.

One-quarter would be left after 12 hours. 48 b. = 8 half-lives (6 )8 1 1 = 2 256 1 would be left after 48 hours. 256

PRACTICE PROBLEM 9 Cobalt-60 is radioactive with a half-life of 5.27 years. It is produced from cobalt-59 by bombardment with neutrons at a nuclear reactor. It emits a low-energy beta particle followed by two high-energy gamma rays. It is used in the sterilisation of medical equipment, in radiotherapy and in industrial 1 of its initial applications. A cobalt-60 source will need to be replaced when its activity decreases to 16 value. How long will this take?

7.3 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Name the isotope that has an atomic number of 11 and contains 12 neutrons. 2. How many protons and neutrons are there in one atom of each of the following isotopes? (a) Hydrogen-2, also known as deuterium (b) Americium-241 (c) Europium-164 3. What force acts between a proton and a neutron and under what conditions does it act? 4. The half-life of caesium-134 is 2.06 years. What fraction of caesium is left after 10.3 years? 5. What is meant when an isotope is described to be stable? 6. If the strong nuclear force acts between all nucleons and is much stronger than the electromagnetic force at the distance of two neighbouring nucleons in a nucleus, explain how a nucleus might be unstable. 7. 100 grams of a radioactive isotope is delivered to a hospital. Three hours later there is only 6.25 grams left. What is the half-life of the isotope? 8. Sketch a graph of the decay of the isotope in question 7. 9. Why is half-life used for radioactive isotopes rather than just stating how long it takes to decay? 10. The activity (decays per second) of a radioactive isotope halves with each half-life. Explain why that would be the case. 11. You need 20 grams of a radioactive isotope with a half-life of three hours. How much would you need to buy if it takes 24 hours to deliver it?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

7.4 Types of nuclear radiation KEY CONCEPTS + • Apply a simple particle model of the atomic nucleus to explain the origin of 𝛼, 𝛽 – , 𝛽 and 𝛾 radiation, including changes to the number of nucleons. + • Explain nuclear transformations using decay equations involving 𝛼, 𝛽 – , 𝛽 and 𝛾 radiation. • Analyse decay series diagrams with reference to type of decay and stability of isotopes.

Unstable isotopes can emit various types of radiation while ‘striving’ to become more stable. There are three naturally occurring forms of nuclear radiation: 𝛼, 𝛽 and 𝛾 (pronounced alpha, beta and gamma). Each type of radiation was named with a different Greek letter because, when the different types of radiation were discovered, scientists did not know what they consisted of. The emissions are described as decay processes because the nucleus changes into a different nucleus and the change is irreversible.

TOPIC 7 Particles in the nucleus 227

7.4.1 Alpha (𝛼) decay

During 𝛼 decay an unstable nucleus ejects a relatively large particle known as an 𝛼 particle. This actually consists of two protons and two neutrons, and so may be called a helium nucleus. The remainder of the original nucleus, known as the daughter nucleus, is now more stable. FIGURE 7.17 Alpha decay: a nucleus ejects a helium nucleus. Parent nucleus

Daughter nucleus

Alpha particle

The number of protons in the nucleus determines the elemental name of the atom. The daughter nucleus is therefore of a different element. For example, uranium-238 decays by emitting an 𝛼 particle. The uranium-238 atom contains 92 protons and 146 neutrons (238 – 92 = 146). It emits an 𝛼 particle, with two protons and two neutrons. The original nucleus is left with four less nucleons: 90 protons (92 – 2 = 90) and 144 neutrons (146 – 2 = 144). As the daughter nucleus now has 90 protons, it is called thorium and has the symbol Th. This particular isotope of thorium has 234 nucleons (90 protons and 144 neutrons) and is more correctly written as thorium-234. The information in the previous paragraph can be written much more effectively in symbols. This is called the decay equation: 238 U 92

or

→

238 U 92

234 90Th

→

+ 42He + energy

234 90Th

+ 𝛼 + energy

The ejected 𝛼 particle is relatively slow and heavy compared to other forms of nuclear radiation. The particle travels at 5–7% of the speed of light: roughly 2 × 107 metres each second. Every object that moves has a form of energy known as kinetic energy, or energy of motion. Because the 𝛼 particle is moving, it has kinetic energy. That energy is written into the decay equation. In addition to having energy, the 𝛼 particle has an overall charge of +2 because it contains two protons. This charge means the particle can be deflected by electric or magnetic fields — properties that helped scientists determine what an 𝛼 particle consisted of. The mass and charge of the alpha particle ensure that it cannot travel more than ten centimetres through the air without interacting with other matter including atoms and molecules. A sheet of paper will stop most alpha particles. For the same reason, its ability to penetrate the skin of animals and humans is extremely limited. However, if an alpha emitting source decays inside the body, it can do significant damage. As a result, alpha emitters are rarely used in nuclear medicine although its lethal properties are being put to work in a targeted way to destroy cancer tissue.

Properties of alpha radiation

228 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Alpha particles are a strongly ionising form of radiation. As a charged particle it can remove electrons from other atoms it approaches, which can result in chemical changes in body tissue and genetic material.

RADIOACTIVITY IN SMOKE DETECTORS Some smoke detectors include a sample of the isotope americium-241, an alpha emitter. The alpha particles ionise the air, which sustains a small current between charged plates in the detector. When smoke blocks the access of the alpha particles to the air molecules between the charged plates, the current can no longer flow, which triggers an alarm to go off.

FIGURE 7.18 Smoke alarms are set off when smoke blocks the path of alpha particles

Discovering the nucleus Ernest Rutherford was one of the central players in the investigation of radioactivity. By 1908, Rutherford and his team had determined that alpha particles were doubly charged helium ions and that they were moving very fast — at about 5% of the speed of light. They quickly realised that these particles would be ideal ‘bullets’ to investigate the structure of the atom. Two of Rutherford’s younger colleagues, Hans Geiger and Ernest Marsden, fired alpha particles at a very thin foil of gold, about 400 atoms thick, and measured their angle of deflection. Nearly all of the particles either went straight through or suffered a very small deflection, but about 1 in every 8000 came back. FIGURE 7.19 Geiger and Marsden’s gold foil experiment (a)

Source of alpha particles

(b)

Beam of alpha particles

Gold foil

Alpha particles Nucleus

Atoms of gold foil

Lead shield Fluorescent screen

The positively charged 𝛼 particle was repelled and deflected by the electrostatic interaction with the positive charges in the atom. Rutherford calculated that for an 𝛼 particle to be turned around, these positive charges would need to be concentrated in a very small volume, which he called the ‘nucleus’. He calculated that the radius of such a nucleus would be about 10−14 metres, and the radius of an atom was about 10−10 metres. Rutherford’s nuclear model of the atom had nearly all the mass of the atom in the central nucleus and the much lighter electrons ‘orbiting’ around it. However, this model was incomplete because it did not fully explain the mass of an atom. The neutron was subsequently discovered. TOPIC 7 Particles in the nucleus 229

7.4.2 Beta (𝛽) decay

Two types of 𝛽 decay are possible. The 𝛽 − particle is a fast moving electron that is ejected from an unstable nucleus. The 𝛽 + particle is a positively charged particle with the same mass as an electron and is the electron’s antiparticle, the positron. Positrons are mostly produced in the atmosphere by cosmic radiation, but some nuclei do decay by 𝛽 + emission. FIGURE 7.20 𝛽 − decay: a nucleus ejects an electron. Parent nucleus

Daughter nucleus

+ energy

β− particle

In 𝛽 − decay, an electron is emitted from inside the nucleus. Since nuclei do not contain any electrons this might seem strange, but it is in fact true! There is no change whatsoever to the electrons in the shells surrounding the nucleus. Some very interesting changes take place inside a nucleus to enable it to emit an electron. One of the neutrons in the nucleus transforms into a proton and an electron. The proton remains in the nucleus and the electron is emitted and called a 𝛽 − particle: 1 0n

→ 11p + −10e

The resulting daughter nucleus has the same number of nucleons as the parent, but one less neutron and one more proton. An example of 𝛽 − decay is the decay of thorium-234. This isotope is the result of the 𝛼 decay of uranium-238. The nucleus is more stable than it was before the emission of the 𝛼 particle but could become more stable by emitting a 𝛽 − particle. During this second decay, the mass number of the nucleus is unchanged (234). The number of protons, however, increases by one when a neutron changes into a proton and an electron. There are now 91 (90 + 1) protons in the nucleus, so the atom must be called protactinium-234. In the next section we will learn about the neutrino. Beta minus decay is accompanied by the emission of another particle called an antineutrino. Beta positive decay is accompanied by the emission of a neutrino. Neutrinos are particles with no charge and nearly zero mass. The decay equation is written as: 234 234 − 90 Th → 91Pa + 𝛽 + 𝜈 + energy 234 0 or 234 90 Th → 91Pa + −1e + 𝜈 + energy

In 𝛽 + decay, the positron is also emitted from inside the nucleus. In this case, strange as it may seem, the proton changes into a neutron and a positron with the neutron staying in the nucleus: 1 p 1

→10 n + +10e + v + energy

230 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

The resulting nucleus has one less proton but the same number of nucleons. An example of 𝛽 + decay is sodium-22 decaying to neon-22: 22 Na 11

0 →22 10 Ne + +1e + v + energy

Properties of beta particles

Beta particles are very light when compared to alpha particles. They travel at a large range of speeds — from that of an alpha particle up to 99% of the speed of light. Just like 𝛼 particles, 𝛽 particles are deflected by electric and magnetic fields. The lower charge, mass and higher speed of the beta particle ensures that it can penetrate much further through air and body tissues than alpha particles. They can travel through centimetres of human skin and a metre or two of air. They can be stopped by a few centimetres of water or a few millimetres of aluminium. Beta radiation is useful for precisely measuring the thickness of paper in production and for treating some types of cancers. It is also a form of ionising radiation. SAMPLE PROBLEM 10

Write down the complete decay equation in each of the following.

230 a. 234 92 U → 90 Th + ? + energy

210 b. 210 82 Pb → 83 Bi + ? + energy 11 c. 11 6 C→ 5 B + ? + energy

THINK

What is the mass number of the missing particle? 2. What is the atomic number of the missing particle? 3. Determine the missing particles from this information. 4. Write the equation.

a. 1.

What is the mass number of the missing particle? 2. What is the atomic number of the missing particle? 3. Determine the missing particles from this information. 4. Write the equation.

b. 1.

Teacher-led video: SP10 (tlvd-0057)

234 − 230 = 4

WRITE a.

92 − 90 = 2 𝛼 particle

The number of particles in the nucleus has decreased by 4, while the number of protons has decreased by 2. This implies that an 𝛼 particle, or helium nucleus, has been released. The full 230 4 equation is 234 92U → 90T h + 2H e + energy b. 210 − 210 = 0 82 − 83 = −1 𝛽 – particle

This equation cannot show an 𝛼 emission, as the mass number remains constant. The atomic number has increased, indicating that a proton has been formed, and therefore 𝛽 – decay has occurred. The equation becomes: 210 210 0 82P b → 83B i + −1e + energy TOPIC 7 Particles in the nucleus 231

What is the mass number of the missing particle? 2. What is the atomic number of the missing particle? 3. Determine the missing particles from this information. 4. Write the equation.

c. 1.

c.

11 − 11 = 0

6−5=1

𝛽 + particle

The mass number stays the same, but there is one less proton, so it must be 𝛽 + decay. The equation becomes: 11 C → 115B + +10e + energy 6

PRACTICE PROBLEM 10 Write the equations for: a. the alpha decay of americium-241 b. the 𝛽 − decay of platinum-197 c. the 𝛽 + decay of magnesium-23.

7.4.3 Gamma (𝛾) decay

This form of radioactive decay is quite different from either 𝛼 or 𝛽 decay. During 𝛾 decay a small packet of electromagnetic energy called a 𝛾 ray, or photon, is emitted, rather than a particle. Gamma emission occurs after another form of nuclear decay has taken place. Following a decay, the arrangement of protons and neutrons in the nucleus may not be ideal and the nucleus may need to release some extra energy to become more stable. Before the release of this energy, the nucleus is known as ‘excited’. An excited nucleus is denoted by an asterisk (*) after the symbol for the element. The excess energy is emitted as a 𝛾 ray. FIGURE 7.21 Gamma decay: an excited nucleus ejects a photon. Parent nucleus

Daughter nucleus

Gamma ray

One example of 𝛾 decay occurs after lead-210 emits a 𝛽 − particle and becomes bismuth-210. The excited daughter nucleus goes on to emit a 𝛾 ray: 210 ∗ 83 Bi

232 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

→210 83 Bi + 𝛾

This 𝛾 ray is a packet of excess energy. It has no mass and no charge and is not deflected by electric or magnetic fields. Because it is a photon, or packet of electromagnetic energy, it travels at the speed of light. Gamma radiation is extremely penetrating. A shield of a few centimetres thickness absorbs about 90% of gamma rays. Gamma radiation is ionising like alpha and beta radiation. These properties of gamma radiation make if very useful in medicine. It can be used to trace physiological processes by injecting or swallowing chemical compounds with gamma emitters in the compound. The movement of the compound through the body can be traced with cameras outside the body detecting the gamma rays. A newer technique involves using positron emitters. When the positrons are emitted they meet electrons and both are annihilated, producing gamma rays that can be detected outside the body and an FIGURE 7.22 Radioactive series of uranium-238. The half-life image constructed by computer. This is given beside each decay. is called positron emission tomography, N or PET scans. Gamma rays are also a 238 U powerful means of killing α 145 4.5×109 y cancerous tissue. 234

Properties of gamma radiation

Th

7.4.4 Decay series

234

1.2 min α 230

140

Pa β 234

U

2.5×105 y

Th

α

226

8×104 y Ra α 1600 y

222Rn

Number of neutrons

In its ‘quest’ to become stable, an isotope may have to pass through many stages. As a radioactive isotope decays, the daughter nucleus is often radioactive itself. When this isotope decays, the resulting nucleus may also be radioactive. This sequence of radioisotopes is called a decay chain or decay series. Uranium-238 undergoes 14 radioactive decays before it finally becomes the stable isotope lead-206. Two other decay chains, one starting with uranium-235 and another with thorium-232, also end with a stable isotope of lead. Another decay chain once passed through uranium-233, but this chain is almost extinct in nature now due to its shorter half-lives. Not all naturally occurring radioisotopes are part of a decay series. There are about 10 with atomic numbers less than that of lead, for example, potassium-40. How can there be radioisotopes isolated in the periodic table? A look at the half-lives of potassium-40 and indium-115 reveals the answer (see table 7.4). Their half-lives are so long, greater than the age of the Earth, that they are still decaying from when they were formed in a supernova explosion billions of years ago.

β

24 d

α

135

3.8 d 218Po

β 3.0 m α 218 At α 214Pb β 27 m 214Bi 20 m β 214 130 Po α 210 Tl α 1.6×10−4 s β 210 Pb 1.3 m β 22 y 210Bi 5.0 d β 210 206 Po Tl α α 125 138 d β 206Pb

120 80

85

Key α β y m d s

90 95 Number of protons

alpha decay beta decay year(s) month(s) day(s) second(s)

100

Z

TOPIC 7 Particles in the nucleus 233

SAMPLE PROBLEM 11

Using the decay series, comment on whether the daughter nucleus is always more stable than its parent. THINK

WRITE

How can I tell how stable a nucleus is? 2. Do the daughter nuclei always have longer half-lives than the parent nucleus?

The longer the half-life, the more stable the nucleus. Often, in a decay chain, the daughter nucleus has a shorter half-life than the parent. For example, uranium-238 has a much longer half-life than its daughter thorium-234. In a decay chain, a nucleus may decay to a less stable nucleus. This, however, is a step towards increased stability as the final nucleus is always stable.

1.

3.

State the solution.

PRACTICE PROBLEM 11 a. Which is the most stable nucleus in the uranium-238 decay chain apart from lead-206, which is stable? b. Why do you think the decay chain starts with a relatively stable nucleus?

Nuclear transformations Alpha and beta decay are natural examples of nuclear transformations. The numbers of protons and neutrons in the nucleus change during these processes. Artificial nuclear transformations are also possible. These are done either to investigate the structure of the nucleus or to produce specific radioisotopes for use in medicine or industry. The first artificial transformation was made by Ernest Rutherford, who fired alpha particles at nitrogen atoms to produce an isotope of oxygen. 14 N 7

+ 42 He → 178 O + 11 H

This result raised the intriguing possibility of achieving the alchemist’s dream of changing lead into gold. Although prohibitively expensive, it appears to be theoretically possible. The building of particle accelerators in the early 1930s enabled charged particles such as protons and alpha particles to be fired at atoms as well as alpha particles, but with the advantage that their energy could be pre-set. The limitation of both these particles is that since they are positively charged, they have to be travelling at very high speed to overcome the repulsion of the positively charged nucleus. This problem was overcome with the discovery of the neutron in 1932. The neutron, which has no net charge, can enter the nucleus at any speed. Both protons and neutrons are used today to produce radioisotopes. Particle accelerators firing positive ions produce neutron-deficient radioisotopes such as thallium-201 ( t 1 = 73 days), which is used to show damaged

heart tissue, and zinc-65 (t 1 = 244 days), which is used as a tracer to monitor the flow of heavy metals in mining 2

effluent. Neutrons from nuclear reactors produce neutron-rich radioisotopes such as iridium-192 (t 1 = 74 days), 2

which is used to locate weaknesses in metal pipes, and iodine-131 (t 1 = 8 days), which is used in the diagnosis 2

2

and treatment of thyroid conditions. Particle accelerators are also used to produce new elements with atomic numbers greater than that of uranium. The hunt is on for a new stable element. In 2007, calcium-48 ions were fired at californium-249 atoms to produce the element with atomic number 118. Only three atoms were produced and, as the half-life of this isotope is 0.89 milliseconds, they don’t exist anymore.

234 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Resources Digital documents eModelling: Numerical model of a decay series (doc-0045) Investigation 7.1 Radioactive decay (doc-31872) Investigation 7.2 Background radiation (doc-31873) Teacher-led video Investigation 7.1: Radioactive decay (tlvd-0817) Video eLesson

Nuclear stability and radiation (eles-2518)

7.4 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Write the nuclear equation for the 𝛼 decay of radon-222. 2. Write the nuclear equation for the 𝛽 decay of lead-214. 3. Write the nuclear equation for the 𝛾 decay of cobalt-60. 4. (a) A particle is missing from the right-hand side of a nuclear equation. The atomic numbers on the left-hand side of a nuclear equation add to 20. The sum of the atomic numbers on the right-hand side of the equation add to 18. What is the atomic number of the missing particle? (b) The mass numbers of the equation in part (a) add to 50 on the left-hand side. The sum of the mass numbers on the right-hand side of the equation add to 46. What is the mass number of the missing particle? (c) What is the missing particle? 5. Identify two particles or emissions in nuclear equations that have a mass number and atomic number of zero. 6. Energy is released in nuclear decays other than in the form of gamma rays. What form does this energy take? 7. The atomic number counts the number of protons in the nucleus of an atom. In nuclear equations some particles have atomic numbers but no protons. Given an example and explain what it means.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

7.5 Review • • •

7.5.1 Summary • • • • •

There are two types of fundamental particles of matter: quarks and leptons. There are six leptons. They are the electron, muon and tau particle, and each has its own neutrino. There are six quarks. They all have an electric charge, which is a fraction of the charge size of the electron. Three have a charge of + 32 , and three have a charge of − 31 . Their masses vary significantly. The quarks combine to form particles called hadrons, of which there are two types: mesons and baryons. Mesons are composed of one quark and one antiquark. There are many mesons. They can be positively charged, neutral or negatively charged, and have short half-lives. Baryons are composed of three quarks. There are a large number of different types of baryons, with a range of masses and charges ranging from +2 to −2. Some of the subatomic particles were predicted well before they were detected. Nuclear radiation is emitted from the nucleus of unstable atoms (radioisotopes) that are striving to become more stable.

TOPIC 7 Particles in the nucleus 235

• • • • • • • • •

There are four types of nuclear decay: 𝛼, 𝛽 – , 𝛽 + and 𝛾 radiation. 𝛼 particles are released during 𝛼 decay. 𝛼 particles are relatively slow-moving particles that are equivalent to a helium nucleus and can be represented as 42 He. After 𝛼 decay, the mass number of the daughter nucleus is four less than that of the parent nucleus and the atomic number is two less. 𝛽 − particles are released in 𝛽 − decay. 𝛽 − particles are high-speed electrons released from the nucleus when a neutron transforms into a proton, an electron and an antineutron. After 𝛽 − decay, the mass number of the daughter nucleus is the same as that of the parent nucleus, but the atomic number is one more than that of the parent nucleus. 𝛽 + particles are released in 𝛽 + decay. 𝛽 + particles are high-speed positrons emitted with a neutrino from the nucleus when a proton transforms into a neutron. The atomic number of the daughter nucleus is one less than the parent nucleus; the mass number remains the same. 𝛾 radiation is electromagnetic radiation that is emitted when an excited nucleus becomes more stable. 𝛾 rays are emitted during 𝛼 and 𝛽 decay. In all nuclear transformations, atomic and mass numbers are conserved. Half-life is the time for half of a group of unstable nuclei to decay. It is different for every isotope. The shorter the half-life of an isotope, the greater the activity — that is, the greater the number of decays per second. Activity decreases over time as less and less of the isotope remains. Activity is measured in becquerels (Bq). Isotopes may pass through a sequence of decays in order to become stable. Such a sequence is called a decay chain, or decay series. The force that holds nucleons together in a nucleus of an atom is called the strong nuclear force. It acts over a very short distance and is strong enough to overcome the electrostatic force of repulsion that exists between the protons of a nucleus when nucleons are very close together.

Resources

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0032).

7.5.2 Key terms glossary An 𝛼 particle is a relatively slow-moving decay product consisting of two protons and two neutrons. It is equivalent to a helium nucleus and so can be written as 42 He. 𝛼 particles carry a positive charge. A 𝛾 ray is the packet of electromagnetic energy released when a nucleus remains unstable after 𝛼 or 𝛽 decay. 𝛾 rays travel at the speed of light and carry no charge. The atomic number of an atom is the number of protons in its nucleus. Baryons are hadrons with three quarks. Cosmic rays are very energetic charged particles that enter our atmosphere. They are mainly protons and originate from beyond the solar system. A daughter nucleus is the nucleus remaining after an atom undergoes radioactive decay. It is more stable than the original nucleus. A decay chain also known as a decay series, is the sequence of stages a radioisotope passes through to become more stable. At each stage, a more stable isotope forms. The chain ends when a stable isotope forms. A decay curve is a graph of the number of nuclei remaining in a substance versus the time elapsed. The half-life of a substance can be determined by looking at the time that corresponds to half of the substance remaining. A decay equation is a representation of a decay reaction. It shows the changes occurring in nuclei and lists the products of the decay reaction. An electron is a negatively charged particle found around the nucleus of an atom. An element is a substance that consists only of atoms of the same name.

236 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

An excited nucleus is one that does not have an ideal arrangement of protons and neutrons in its nucleus. An excited nucleus emits 𝛾 radiation to become more stable. Hadrons are composite particles made up of either two or three quarks. A half-life is the time taken for half of a group of unstable nuclei to decay. Isotopes are atoms containing the same number of protons but different numbers of neutrons. Isotopes of an element are atoms containing the same number of protons but different numbers of neutrons. Leptons are the simplest and lightest of the subatomic particles. They are fundamental particles with no internal structure. Mass number describes the total number of nucleons in an atom. Mesons are hadrons with two quarks. A neutron is a nucleon with no charge. The collective name for the particles found in the nucleus of an atom is nucleon. The nucleus is the solid centre of an atom. Most of the mass of an atom is concentrated in the nucleus. A positron is a positively charged particle with the same mass as an electron. A proton is a positively charged particle in the nucleus of an atom. Quarks are the fundamental particles that combine to form hadrons. A radioisotope is an unstable isotope. The strong nuclear force is the force that holds nucleons together in a nucleus of an atom. It acts over only very short distances. The weak nuclear force is the force that explains the transformation of neutrons into protons, and vice versa.

Resources Digital document Key terms glossary (doc-32186)

7.5.3 Practical work and investigations Investigation 7.1 Radioactive decay Aim: To analyse the radioactive decay of a source Digital document: doc-31872 Teacher-led video: tlvd-0817

Investigation 7.2 Background radiation Aim: To measure the background radiation in the classroom at school Digital document: doc-31873

Resources Digital document Practical investigation logbook (doc-32265)

TOPIC 7 Particles in the nucleus 237

7.5 Exercises To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au.

7.5 Exercise 1: Multiple choice questions 1.

2.

3.

4.

5.

6.

7.

8.

Which of the following shows the quark composition of a proton? A. Up, up, down B. Up, down, down C. Up, up, up D. Down, down, down Which of the following types of particles is not made of quarks? A. Mesons B. Baryons C. Hadrons D. Leptons There are four fundamental forces: i. gravity ii. electromagnetic force iii. weak nuclear force iv. strong nuclear force. What is the correct order, from strongest to weakest, at the distance between two nucleons in a nucleus? A. i, ii, iii, iv B. iv, iii, ii, i C. iv, ii, iii, i D. iv, ii, i, iii Which of the following particles was discovered experimentally before being predicted theoretically? A. Higgs boson B. Positron C. Electron D. Neutron Which of the following forces is not significant in the stability of nuclei? A. Gravity B. Electromagnetic force C. Strong nuclear force D. Weak nuclear force How is the half-life of an isotope best described? A. Half of the time it would take for the whole sample to decay B. The time it would take for half of the existing sample to decay C. The time for the stability of the nucleus to halve D. A period during which the isotope is not very active Radioisotopes used in nuclear medicine usually have short half-lives. Why is this? A. They are cheaper. B. They are more active. C. The radiation is dangerous to the body so it is good to minimise the time of exposure. D. The more stable isotopes are more toxic. Alpha decay involves the ejection from the nucleus of which of the following? A. A tightly bound package of two protons and two neutrons B. An electron and an antineutrino C. A positron and a neutrino D. An energetic photon

238 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Cobalt-60 is often used to as a gamma source for sterilising medical equipment. It is made by bombarding stable cobalt-59 with neutrons in nuclear reactors. To answer the following think about the effect on the stability of a relatively small nucleus of adding a neutron. When cobalt-60 decays, it emits gamma rays and which of the following? A. Neutrons B. Alpha particles C. Beta minus particles and antineutrinos D. Beta plus particles and neutrinos 10. In a series of decays, the mass number drops by eight and the atomic number drops by two. Which sequence of decays could result in this transmutation? A. An alpha, a beta minus, a beta minus and another alpha B. An alpha, a beta plus, a beta minus and an alpha C. A beta plus, an alpha, a beta minus and an alpha D. An alpha, an alpha, a beta minus and another alpha 9.

7.5 Exercise 2: Short answer questions 1.

2. 3. 4.

5.

6.

7.

8.

Design baryons with a charge of: a. +2 b. −2 c. zero. Some mesons are their own antiparticle. Explain with an example. Why do you think it has taken so long and been so difficult to find neutral particles such as the neutron, neutrino and the Higgs boson? How many protons and neutrons are in the following atoms? a. 66 30 Zn b. 230 90 Th c. 45 20 Ca d. 31 14 Si Write the symbols for isotopes containing the following nucleons. a. Two neutrons and two protons b. Seven protons and 13 nucleons c. Ninety-one protons and 143 neutrons Determine the particle, X, that has been released in each of the following decay equations. 14 a. 146 C → 7N + X + energy 90 b. 90 38 Sr → 39Y + X + energy 234 c. 238 92 U → 90Th + X + energy Write a decay equation to show the following. a. 𝛼 decay of: i. radium-226 ii. polonium-214 iii. americium-241. b. 𝛽 decay of: i. cobalt-60 ii. strontium-90 iii. phosphorus-32. How many 𝛼 particles are released by one atom of uranium-238 as it becomes lead-206? How many 𝛽 particles are released? (Hint: Look at the change in the proton number and the change in the nucleon number.) Check your answer by using figure 7.22.

TOPIC 7 Particles in the nucleus 239

9.

What is the half-life of the substance represented in the following graph?

Portion remaining

1 3 – 4

1 – 2

1 – 4

0 1

10.

2

3 4 5 6 Time (hours)

7

Assume the half-life of carbon-14 is 5730 years. If you had 1 gram of carbon-14, how many years would it take for one-eighth of it to remain?

7.5 Exercise 3: Exam practice questions Question 1 (1 marks) Atomic nuclei consist of particles that can be classified as which of the following? A. Hadrons B. Mesons C. Baryons D. Leptons Question 2 (3 marks) PET scans provide doctors with information about how organs are functioning. PET scans use positron emission from sources introduced into the patient’s body. a. Compare the mass and charge of a positron with those of an electron. 2 marks b. The PET scan detects gamma rays rather than positrons. How are these gamma rays produced? 1 mark Question 3 (3 marks) a. Provide an example of a particle whose existence was predicted before it was discovered. b. How can the existence of an as yet unobserved particle be predicted scientifically?

1 mark 2 marks

Question 4 (6 mark) Consider the graph of nuclear stability in figure 7.15. Patterns can be seen where 𝛽 and 𝛼 decay occur in relation to the line of stability. a. Would adding a proton or a neutron to a stable nucleus make it more likely to undergo 𝛽 − decay? 1 mark b. Explain why 𝛽 − decay is likely to occur in this scenario. 2 marks c. Explain why 𝛼 emission is restricted to very large nuclei by referring to the forces involved in the nucleus. 3 marks Question 5 (5 marks) Targeted alpha particle therapy (TAT) is a relatively new treatment for some cancers. One of the radioisotopes used in TAT is actinium-225 ( 225 89A c), which decays to an isotope of francium (Fr). a. Write the decay equation for this. 3 marks b. The half-life of actinium-225 is 10 days. What fraction of the dose of actinium-225 administered would remain after 30 days? 2 marks

240 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

c.

The daughter nucleus of actinium-225 following 𝛼 decay is not stable. In fact, a sequence of 4 𝛼 decays and 2 𝛽 − decays occurs before stability is reached with an isotope of bismuth. Determine the atomic number and mass number for this isotope of bismuth. Show your working. 2 marks

7.5 Exercise 4: studyON topic test Fully worked solutions and sample responses are available in your digital formats.

Test maker Create unique tests and exams from our extensive range of questions, including practice exam questions. Access the assignments section in learnON to begin creating and assigning assessments to students.

TOPIC 7 Particles in the nucleus 241

AREA OF STUDY 3 WHAT IS MATTER AND HOW IS IT FORMED?

8

Energy from the atom

8.1 Overview Numerous videos and interactivities are available just where you need them, at the point of learning, in your digital formats, learnON and eBookPLUS at www.jacplus.com.au.

8.1.1 Introduction The universe can be considered to be made of matter and energy. In this topic we explore how the particles of matter that shape the universe cannot be considered separately from energy. This discovery has led to dangerous weapons and new sources of power for machines and electricity supply. Nuclear power plants harness the energy from matter to create large amounts of energy from very little raw material. It is the hope of many that further developments will provide an almost limitless source of energy with very little pollution. The relationship between mass and energy has supplied a range of light sources for various applications in medicine and many other fields. FIGURE 8.1 Nuclear power plants use nuclear fission to create roughly 13% of the world’s electricity supply.

242 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

8.1.2 What you will learn KEY KNOWLEDGE After completing this topic you will be able to: • explain nuclear energy as energy resulting from the conversion of mass: E = mc2 • compare the processes of nuclear fusion and nuclear fission • explain, using a binding energy curve, why both fusion and fission are reactions that produce energy • explain light as an electromagnetic wave that is produced by the acceleration of charges • describe the production of synchrotron radiation by an electron radiating energy at a tangent to its circular path • model the production of light as a result of electron transitions between energy levels within an atom. Source: VCE Physics Study Design (2016–2021) extracts © VCAA; reproduced by permission.

PRACTICAL WORK AND INVESTIGATIONS Practical work is a central component of learning and assessment. Experiments and investigations, supported by a Practical investigation logbook and Teacher-led videos, are included in this topic to provide opportunities to undertake investigations and communicate findings.

Resources Digital documents Key science skills — VCE Units 1–4 (doc-31856) Key terms glossary (doc-32266) Practical investigation logbook (doc-33011)

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0033).

8.2 Energy from mass KEY CONCEPT • Explain nuclear energy as energy resulting from the conversion of mass: E = mc2 .

The most famous physics equation of all is E = mc2 . The is an equation that Albert Einstein derived as a result of his work on relativity. We will now explore what this equation means and how it can be used. In the previous topic we learnt that some nuclei are unstable. They eventually decay by releasing particles including beta particles, neutrinos, alpha particles and gamma rays. Which particles are released depends on the particular isotope involved. Before 1905, scientists would have said that these decays must obey two conservation laws: conservation of mass and conservation of energy. Conservation of mass states that there is always the same amount of mass in a closed system — one where no mass is entering or leaving. This resulted from early understandings of chemical reactions, where the mass of the products is always the same as the mass of the reactants. Conservation of energy states that there is always the same amount of energy in an isolated system. Energy is never created or destroyed, it is transformed to different forms or transfered from one system to another.

FIGURE 8.2 Albert Einstein

TOPIC 8 Energy from the atom 243

However, in 1905, Albert Einstein stunned the scientific world by showing that these two conservation laws are not strictly correct. He demonstrated that only one conservation law was needed because mass and energy are essentially the same thing. What is conserved is the combination of mass and energy, which is sometimes called mass–energy. Einstein’s equation was a result of theory. Like the prediction of particles including the neutrino, positron, neutron and Higgs boson, there was no experimental evidence for it at the time and obtaining that evidence was not going to be easy. E = mc2 Where: E is the energy in joules m is the mass in kilograms c is the symbol for the speed of light (c = 2.997 924 58 × 108 m s−1 ). Note: The speed of light is usually rounded to 3.00 × 108 m s−1 in calculations. This equation is a statement that energy and mass are equivalent. If we need to know how much energy a certain mass is equivalent to, we use the equation. For example, for a mass of 1 kilogram: E = mc2 ( )2 = 1.0 × 3.00 × 108 = 9.0 × 1016 J That is a tremendous amount of energy. However, the Sun produces four billion times that every second. Einstein’s E = mc2 calculates that the Sun is losing energy at a rate that is equivalent to 4 billion kilograms every second! Why do we not see this loss of mass in everyday life when a body radiates energy? Let’s look at the example of an object cooling down, like a hot water bottle in your bed on a winter’s night. The hot water bottle would transfer approximately 100 000 J of energy to your bed as it cools. Using E = mc2 : E = mc2 m=

E c2

=(

100 000 3.00 × 108

)2

= 1.1 × 10−12 kg That is one-millionth of one-millionth of a kilogram, far too tiny to notice or measure. As far as typical events are concerned, energy is conserved and mass is conserved. Even the mass lost by the Sun per second is such a tiny percentage of the Sun’s mass that the Sun can continue losing mass at this rate for billions of years. However, Einstein’s equation has been tested in many ways. In 2008, researchers confirmed that E = mc2 explained the mass of protons and neutrons. In the previous topic, we learnt that protons comprise two up quarks and one down quark. Similarly, neutrons are made up of two down quarks and one up quark. These quarks are held together by the strong nuclear force, which involves the exchange of a boson called a gluon. However, the mass of a proton or a neutron cannot be determined by adding up the masses of the quarks they contain (gluons are massless).

244 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Most of the mass of the proton and neutron derives from the energies of the quarks within each particle. To account for a proton having a mass of 1.67 × 10−27 kilograms we need to add up the mass of the quarks it contains and the mass equivalent of the energy involved by using E = mc2 . Nuclear decays bring a nucleus to a state of lower energy and greater stability. The nucleus prior to the decay, or decay series, has a greater mass than the nucleus after decay, even when the masses of the particles emitted in the decay are accounted for. In the decay, mass is not conserved. Similarly, if the energies prior to the decay were compared with the energies of all of the particles after the decay, we find some energy that appears to have come from nowhere. The difference in mass (m) is equivalent to the difference in energy (E) as predicted by E = mc2 . For example, consider the alpha decay of uranium-238, shown in table 8.1.

FIGURE 8.3 Alpha decay of uranium-238 Uranium-238

TABLE 8.1 The decay of uranium-238 to thorium-234 and helium-4 Nucleus

Mass (kg)

Parent

Uranium-238

3.952 93 × 10−25

Daughter

Thorium-234

3.886 39 × 10−25

Helium-4

6.645 758 × 10−27

He-4 + Th-234

3.952 85 × 10−25

Mass defect

8.242 × 10−30

Decay particle Mass after decay Change of mass resulting from decay

Thorium-234

The thorium nucleus and the 𝛼 particle produced by the decay of uranium-238 have a slightly lower mass than the parent uranium nucleus. This change in mass is called the mass defect. Mass is not conserved here. Where is the missing mass? The mass can be accounted for by applying E = mc2 . The mass is equivalent to 7.42 × 10−13 J. This energy is the kinetic energy of the remaining particles, most of it going to the 𝛼 particle, which moves away from the thorium nucleus at approximately 10 000 kms−1 . Energy is often better expressed in a unit called the electronvolt (eV) when dealing with nuclear physics. 1 eV is the kinetic energy an electron or proton would gain if accelerated across a potential difference of 1 V. 1 eV = 1.6 × 10−19 J. The energy produced in the 𝛼 decay of uranium-238 is then 7.42 × 10−13 J = 4.63 MeV. Helium-4

SAMPLE PROBLEM 1

Calculate the energy released when thorium-232 undergoes alpha decay. Thorium-232 has a mass of 3.853 08 × 10−25 kg and radium-228 has a mass of 3.786 55 × 10−25 kg. An 𝛼 particle has a mass of 6.64 648 × 10−27 kg. Give your answer in MeV. Teacher-led video: SP1 (tlvd-0065)

TOPIC 8 Energy from the atom 245

THINK

WRITE

Determine the mass of the parent nucleus. 2. Determine the mass of the daughter nucleus and the 𝛼 particle. 3. Determine the change in the mass due to the decay.

mparent = 3.853 08 × 10−25 kg

1.

4.

Determine the equivalent energy.

mproducts = 3.786 55 × 10−25 kg + 6.64 648 × 10−27 kg = 3.85 301 × 10−25 kg mparent − mproducts = 3.85 308 × 10−25 kg − 3.85 301 × 10−25 kg E = mc2

= 7.26 × 10−30 kg

( )2 = 7.26 × 10−30 × 3.00 × 108 = 6.5 × 10−13 J 6.5 × 10−13 MeV 1.602 176 × 10−13

5.

Convert to MeV.

E=

6.

State the solution.

= 4.1 MeV 4.1 MeV of energy is released when thorium-232 undergoes alpha decay.

PRACTICE PROBLEM 1 An 𝛼 particle has a kinetic energy of 4.2 MeV as it leaves a nucleus. Calculate the mass defect of this reaction. (Assume that all the energy released went into the kinetic energy of the 𝛼 particle.)

8.2 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. What does the equation E = mc2 express about mass and energy? 2. What is the energy equivalent of 100 grams of mass in joule? 3. A positron and an electron (each of mass 9.1 × 10−31 kg) annihilate each other. How much energy is released? 4. What is the energy content of a 4500 kJ meal in eV? 5. Calculate the mass equivalent of 4500 kJ.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

246 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

8.3 Energy from the nucleus KEY CONCEPTS • Compare the processes of nuclear fusion and nuclear fission. • Explain, using a binding energy curve, why both fusion and fission are reactions that produce energy.

8.3.1 Binding energy The amount of energy needed to overcome the strong nuclear force and pull apart a nucleus is known as the binding energy. This is the amount of energy that has to be added to a nucleus to split it into its individual nucleons — that is, to reverse the binding process. For example, it would take 2.23 MeV of energy to split a ‘heavy’ hydrogen nucleus into a separate proton and neutron. Each isotope has its own specific binding energy. Nuclei with high binding energies are very stable as it takes a lot of energy to split them into separate protons and neutrons. Nuclei with lower binding energies are easier to split. Of course, it is difficult to supply sufficient energy to cause a nucleus to split totally apart. It is much more common for a nucleus to eject a small fragment, such as an 𝛼 or 𝛽 particle, to become more stable. To compare the binding energies of various nuclei, and therefore their stability, it is necessary to look at the average binding energy per nucleon. The average binding energy per nucleon is calculated by dividing the total binding energy of a nucleus by the number of nucleons in the nucleus. It can be seen from figure 8.4 that iron-56 has the highest binding energy per nucleon. This means it is the most stable of all nuclei. In order to become more stable, other nuclei tend to release some of their energy. Releasing this energy would decrease the amount of energy they contained, and therefore increase the amount of energy that must be added to split them apart. FIGURE 8.4 This graph of binding energy versus mass number peaks at nickel-62, although the much more common iron-56 is very close behind. 9 Average binding energy per nucleon (MeV)

14 7N

56 26 Fe

8

238 92 U

141 56 Ba

4 2He

7 6 6 3 Li

5 4 3

3 2 He

2 2 1H

1

1 1H

0 0

30

60

90

120 150 Mass number

180

210

240

270

The binding energy is not only the amount of energy required to separate a nucleus into its component parts, but also the amount of energy released when those parts are brought together to form the nucleus; that is, when a proton and a neutron collide to form a ‘heavy’ hydrogen nucleus, 2.23 MeV of energy is released (twice the binding energy per nucleon).

TOPIC 8 Energy from the atom 247

The curve of the graph in figure 8.4 indicates that if two nuclei with low mass numbers could be joined together to produce a single nucleus, then a lot of energy would be released. Similarly, if a nucleus with a very high mass number could split into two fragments with greater binding energy per nucleon, then once again a lot of energy would be released. These two possibilities were realised in the 1930s. The released energy can be calculated from Einstein’s equation E = mc2 , where m is the difference between the total nuclear mass before and after the event, and c is the speed of light. The joining of two nuclei is nuclear fusion and the splitting of a single nucleus is nuclear fission.

Resources Interactivity Making nuclei (int-6393)

8.3.2 Nuclear fission In 1934, Enrico Fermi investigated the effect of firing neutrons at uranium. The products had half-lives different from that of uranium. He thought that he had made new elements with atomic numbers greater than 92. Others repeating the experiment got different half-lives. In 1939, Otto Hahn and Fritz Strassmann chemically analysed the samples and found barium, which has atomic number 56, indicating that the nuclei had split. Lise Meitner and Otto Frisch called this process ‘fission’ and showed that neutrons could also initiate fission in thorium and protactinium. Further chemical analysis revealed a range of possible fission reactions, each with a different combination of fission fragments including bromine, molybdenum or rubidium (which have atomic numbers around 40) and antimony, caesium or iodine (which have atomic numbers in the 50s). Cloud chamber photographs showing two heavy nuclei flying off in opposite directions confirmed that fission had occurred. Meitner and Frisch also calculated from typical binding energies that the fission of one uranium-235 nucleus would produce about 200 MeV of energy, mainly as kinetic energy of the fission fragments. This is a huge amount of energy to be released by one nucleus, as can be seen when it is compared to the burning of coal in power plants. Each atom of carbon used in coal burning releases only 10 eV of energy — about 20 million times less than the energy released in the fission of uranium-235. FIGURE 8.5 The first chain reaction in a nuclear reactor was on a squash court at the University of Chicago during World War II.

248 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 8.6 Nuclear fission reactions involve a large nucleus splitting into two smaller nuclei and a couple of neutrons.

LISE MEITNER (1878–1968) Lise Meitner was the physicist who coined the term ‘fission’ and, along with her FIGURE 8.7 Lise nephew Otto Frisch, explained the splitting of uranium nuclei into barium and Meitner (1878–1968) lanthanum. was the first to describe Born in Vienna, Lise was fascinated by the world around her from an early how a nucleus could age. A talented student, she wanted to understand the things she observed undergo fission. in nature. Having decided that she would like to pursue her interest in physics and mathematics, Lise engaged a private tutor to prepare her for the university entrance exams, as schools that taught such subjects would not accept girls at that time. She was the second woman to be granted a Doctorate in Physics from the University of Vienna, conferred in 1906. Lise then moved to the Institute of Experimental Physics in Berlin to work with Otto Hahn. Initially, this proved difficult. Lise was forced to work in a converted workshop as females were not permitted to use the facilities available to male students. As the place of women in the institute became more accepted, Lise was given positions of responsibility, finally being made a professor in 1926. During her time at the institute, Lise made many important contributions to atomic and particle physics, including the co-discovery with Otto Hahn of the radioactive element protactinium. In 1938 Berlin became a dangerous place for Jews, and Lise moved to Sweden. It was there she and Otto Frisch interpreted the results of experiments conducted by Otto Hahn and Fritz Strassman to come up with the first explanation of the fission process. In doing so, Lise was the first person to use Einstein’s theory of mass–energy equivalence to calculate the energy released during fission. Her international reputation led to an invitation to join the Manhattan Project in 1941 and work on the development of the atomic bomb. Lise objected to the project and declined the offer. She continued to work in Sweden until moving to England in 1960, finally retiring at the age of 82.

Also in 1939, Frédéric Joliot, Irène Joliot-Curie and their team confirmed that two or three fast neutrons were emitted with each fission reaction. This allowed for the possibility of a chain reaction, where one fission triggers further fissions, which could potentially release enormous amounts of energy. FIGURE 8.8 The neutrons produced by a fission reaction can go on to produce a chain reaction.

TOPIC 8 Energy from the atom 249

Some possible equations for the fission of uranium-235 set off by the absorption of a neutron are: 235 92U

+ 10n →

236 92U

→

148 57L a

1 + 85 35B r + 3 0n + energy

235 92U

+ 10n →

236 92U

→

141 56B a

1 + 92 36K r + 3 0n + energy

235 92U

+ 10n →

236 92U

→

140 54X e

1 + 94 38S r + 2 0n + energy

The data in figure 8.4 and Einstein’s equation E = mc2 can be used to calculate the amount of energy released in each of these fission reactions. TABLE 8.2 Masses and binding energies of atoms Nucleus

Symbol

Mass (kg)

Total binding energy (MeV)

Uranium-236

236 92 U

3.919 629 × 10–25

1790.415 039

Lanthanum-148

148 57 La

2.456 472 × 10–25

1213.125 122

Bromine-85

85 35 Br

1.410 057 × 10–25

737.290 649

Barium-141

141 56 Ba

2.339 939 × 10–25

1173.974 609

Krypton-92

92 36 Kr

1.526 470 × 10–25

783.185 242

Xenon-140

140 54 Xe

2.323 453 × 10–25

1160.734 009

94 38 Sr

1.559 501 × 10–25

807.816 711

1 0n

1.674 746 × 10–27

Strontium-94 Neutron

Speed of light, c, = 2.997 924 58 × 108 m s–1 ; 1 MeV = 1.602 176 × 10–13 joules

FIGURE 8.9 Graph of energies in a fission reaction. The sum of the binding energies of La-148 and Br-85 is greater than the binding energy of U-236. The difference is released as kinetic energy of the neutrons and the fission fragments. The total energy after fission is the same as the energy before.

Kinetic energy

Before fission

After fission Neutron KE

0

La-148 Binding energy

=

U-236 +

Br-85

250 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Total energy after

SAMPLE PROBLEM 2

Answer the following questions about the fission of uranium-236 producing lanthanum-148 and bromine-85. Use table 8.2 for data on masses and binding energies. a. What is the difference between the binding energy of the uranium-236 nucleus and the sum of the binding energies of the two fission fragments? b. What is the difference between the mass of the uranium-236 nucleus and the sum of the masses of all the fission fragments, including neutrons? c. Use E = mc2 to calculate the energy equivalent of this mass difference in joules and MeV. Teacher-led video: SP2 (tlvd-0066)

THINK a. 1. 2. 3.

4.

5.

Write out the fission equation. Use table 8.2 to find the binding energy of uranium-236. Use table 8.2 to calculate the sum of the binding energies of the fragments. Calculate the difference between the binding energy of the uranium nucleus and its fission fragments. State the solution.

Use table 8.2 to find the mass of uranium-236. 2. Use table 8.2 to find the sum of the masses of the fragments.

b. 1.

Calculate the difference between the mass of the uranium nucleus and its fission fragments. 4. State the solution. 3.

c. 1.

Use E = mc2 to calculate the energy difference in joules. (The full value for c is used here, but you may use the rounded value.)

2.

Convert the energy to MeV.

3.

State the solution.

WRITE 236

85 1 → 148 57L a + 35B r + 3 0n + energy Binding energy of uranium-236 = 1790.415 039 MeV

a. 92U

Binding energy of lanthanum-148 = 1213.125 122 MeV Binding energy of bromine-85 = 737.290 649 MeV 1213.125 122 + 737.290 649 = 1950.415 771 MeV Energy difference = 1950.415 771 − 1790.415 039 = 160.000 732 MeV

The difference between the binding energy of the uranium-236 nucleus and the sum of the binding energies of the two fission fragments is 160.001 MeV. b. Mass of uranium-236 = 3.919 629 × 10–25 kg Mass of lanthanum-148 = 2.456 472 × 10–25 Mass of bromine-85 = 1.410 057 × 10–25 Mass of a neutron = 1.674 746 × 10–27 2.456 472 × 10−25 + 1.410 057 × 10−25 + 3 × 1.674 746 × 10−27 = 3.916 771 × 10−25 kg Mass difference = 3.919 629 × 10−25 − 3.916 771 × 10−25 = 2.857 62 × 10−28 kg The difference between the mass of the uranium-236 nucleus and the sum of the masses of all the fission fragments is 2.857 62 × 10−28 kg. c. E = mc2 ( )2 = 2.857 62 × 10−28 × 2.997 924 58 × 108 = 2.568 301 × 10−11 J 2.568 301 × 10−11 = 160.300 789 MeV 1.602 176 × 10−13 The energy equivalent of this mass difference is 2.568 301 × 10−11 J or 160.300 789 MeV.

TOPIC 8 Energy from the atom 251

PRACTICE PROBLEM 2 Answer the following questions about the fission of uranium-236 to barium-141 and krypton-92. Use table 8.2 for data on masses and binding energies. a. What is the difference between the binding energy of the uranium-236 nucleus and the sum of the binding energies of the two fission fragments? b. What is the difference between the mass of the uranium-236 nucleus and the sum of the masses of all the fission fragments, including neutrons? c. Use E = mc2 to calculate the energy equivalent of this mass difference in joules and MeV.

Resources Weblink Fission animation

8.3.3 Nuclear fusion Nuclear fusion is the process of joining two smaller FIGURE 8.10 Nuclear fusion of hydrogen-2 nuclei together to form a larger more stable nucleus. and a proton. The resulting nucleus is in an This was first observed by Australian physicist Mark excited state, which releases a gamma ray. Oliphant in 1932 when he was working with Ernest Rutherford. He was searching for other isotopes of hydrogen and helium. Heavy hydrogen (one proton and one neutron) was already known, but Oliphant discovered tritium (one proton and two neutrons) and helium-3 (with only one neutron). In his investigation he fired a fast heavy hydrogen nucleus at a heavy hydrogen target to produce a nucleus of tritium plus an extra neutron. This fusion reaction was to become the basis of the hydrogen bomb, but Oliphant was interested only in the structure of the nucleus and did not realise the energy implications. For fusion to occur more extensively, high temperatures and pressures are needed, such as those that exist inside the Sun or in a fission bomb explosion. The Sun’s core has a temperature of more than 15 million K, just perfect for fusion to occur! Inside the Sun, hydrogen nuclei fuse together to form helium. As helium is more stable than hydrogen, the excess nuclear energy is released. This energy is emitted from the nuclei as 𝛾 radiation, and is eventually received on Earth as light and heat. Fusion reactions also take place in other stars. Stars that are bigger than the Sun have such severe conditions that larger, more stable nuclei such as silicon and magnesium can be produced from the fusion of smaller nuclei. A star about 30 times more massive than the Sun would be needed to produce conditions that would enable the formation of iron by fusing smaller nuclei.

Our Sun The chain of events occurring in the Sun is quite complex. The major component of the Sun is 11 H; that is, nuclei consisting of only one proton and no neutrons. When collisions occur between 11 H nuclei, they have a chance of fusing together but a change needs to occur for this pairing of positively charged particles to become a stable nucleus. One of the protons is changed into a neutron (in much the same way as a neutron is changed into a proton and an electron during 𝛽 − decay). This forms a 21 H nucleus. The by-products of this process are a positron and a neutrino.

252 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Positrons are also produced when some FIGURE 8.11 Our sun is powered by fusion of small artificially produced isotopes undergo radioactive nuclei. decay. As we learnt previously, positrons are the antiparticle of the electrons; they have the same mass but carry a positive charge. When a positron and an electron collide, they immediately annihilate each other. The only thing that remains of either is a 𝛾 ray. Neutrinos are produced when protons change into neutrons, and vice versa. They have no charge, are considered massless and travel at close to the speed of light. Fifty trillion neutrinos from the Sun pass through the human body every second. When a 11 H nucleus and a 21 H nucleus collide, in the conditions found in the centre of the Sun, they form a more stable 32 He nucleus and release the extra energy as a 𝛾 ray. If two 32 He nuclei collide, they complete the process of turning hydrogen into helium. The collision results in the formation of a 42 He nucleus and two 11 H nuclei. Again, energy is released. The energy released during nuclear reactions inside the Sun provides energy for life on Earth.

FORMATION OF LARGER ATOMS Most of the atoms that make up your body (and the rest of the atoms in the Earth) were originally produced in a star. Fusion in stars produced most of the atoms other than hydrogen and helium. Those nuclei with atomic numbers up to that of iron were produced in regular stars. However, when large stars stop producing energy from fusion of elements up to iron, they implode, or collapse in on themselves. This causes conditions in which even larger atoms will fuse to produce very heavy elements such as gold, lead and uranium. These stars explode as supernovas, spreading the elements they have made out into space. The Earth was formed from a cloud consisting of remnants of old supernovas. Other atoms form from radioactive decay processes as described in the previous topic, and from high energy collisions of cosmic rays.

The main sequence of nuclear equations that occur in the Sun to convert hydrogen to helium is: 1 1H

+ 11H → 21H + 01e + v

2 1H

+ 11H → 32H e + 𝛾

3 2H e

+ 32H e → 42H e + 11H + 11H

The binding energy can be used to determine the amount of energy released in a fusion reaction. The difference between the binding energies of the reactants and the products is equivalent to the mass defect.

SAMPLE PROBLEM 3

In the final nuclear equations that occur in the Sun, shown above, two helium-3 nuclei collide to produce a helium-4 nucleus and two hydrogen-1 nuclei, that is, two protons. Use the data in the following table to calculate the: a. difference between the binding energy of the helium-4 nucleus and sum of the binding energies of the two helium-3 nuclei

TOPIC 8 Energy from the atom 253

difference between the sum of masses of the helium-4 nucleus and the two protons, and mass of two helium-3 nuclei c. energy equivalent of this mass difference in joules and MeV.

b.

Particle

Symbol

Mass (kg)

Total binding energy (MeV)

Helium-3

3 He 2

5.006 942 × 10–27

7.864 501

Helium-4

4 2 He

6.645 758 × 10–27

28.295 673

1 1 1 p or 1 H

1.673 351 × 10–27

Proton

THINK

Use the table to find the binding energy of the helium-4 nucleus. 2. Use the table to find the binding energy of the helium-3 nuclei. 3. Calculate the difference in binding energies. 4. State the solution.

a. 1.

b. 1.

2.

3.

4.

c. 1.

Use the table to find the mass of the reactants. Use the table to find the mass of the products. Calculate the difference in mass. State the solution.

Use E = mc2 to calculate the energy difference in joules.

WRITE a.

Binding energy of helium-4 nucleus = 28.295 673 MeV

Binding energy of two helium-3 nuclei = 2 × 7.864 501 = 15.729 002 MeV Difference = 28.295 673 − 15.729 002 = 12.566 671 MeV The difference between the binding energy of the helium-4 nucleus and sum of the binding energies of the two helium-3 nuclei is 12.566 671 MeV. b. Mass before fusion = 2 × 5.006 942 × 10−27 = 1.001 388 × 10−26 kg Mass after fusion = 6.645 758 × 10−27 + ( ) 2 × 1.674 746 × 10−27 = 9.995 25 × 10−27 kg Mass difference = 10.013 88 × 10−27 − 9.995 25 × 10−27 = 1.863 × 10−29 kg The difference between the sum of masses of the helium-4 nucleus and the two protons, and mass of two helium-3 nuclei is 1.863 × 10–29 kg. c. E = mc2 = 1.863 × 10−29 × (2.997 924 58 × 108 )2 = 1.674 381 × 10−12 J

2.

Convert the energy to MeV.

3.

State the solution.

1.674 381 × 10−12 = 10.45 MeV 1.602 176 × 10−13 The energy equivalent of this mass difference is 1.674 × 10−12 J or 10.45 MeV.

254 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

PRACTICE PROBLEM 3 A fusion reaction in the Sun of hydrogen-1 and hydrogen-2 produces helium-3. Use the data in the following table to calculate the: a. difference between the binding energy of the helium-3 nucleus and sum of the binding energies of the hydrogen-1 and hydrogen-2 nuclei b. difference between the masses of the helium-3 nucleus and the sum of the masses of the hydrogen-1 and hydrogen-2 nuclei c. energy equivalent of this mass difference in joules and MeV. Particle Helium-3 Hydrogen-2

Symbol

Mass (kg)

3 He 2

5.006 942 × 10–27

7.864 501

3.344 132 × 10

2.224 573

2 1H 1 1 p or

Proton

Total binding energy (MeV)

–27

1 H 1

1.673 351 × 10–27

Resources Digital document

Investigation 8.1 Chain reaction of dominoes (doc-32621)

Teacher-led video

Investigation 8.1 Chain reaction of dominoes (tlvd-0845)

Weblink

Fusion animation

8.3 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. 2. 3. 4.

5. 6. 7. 8.

Define the terms ‘fusion’ and ‘fission’. Which of these reactions occurs in our Sun? Explain why splitting uranium-235 nuclei releases energy, but joining hydrogen atoms also releases energy. Use the graph of binding energy per nucleon (see figure 8.4) to estimate the amount of energy released when a uranium-235 nucleus is split into barium-141 and krypton-92. Think carefully about the number of significant figures in your answer. How well does your answer agree with the measured value of 200 MeV? Why is energy released in the process of fusing two small nuclei together? What aspect of the binding energy per nucleon curve shows that fusion of light elements releases more energy than fission of heavy ones? When two light elements fuse to form a single heavier one, is the product lighter or heavier than the sum of the masses of the two light atoms? Explain. When uranium-236 undergoes fission to form two lighter nuclei plus some neutrons, is the mass of the uranium-236 greater than, equal to or less than the sum of the masses of the two nuclei plus the neutrons? Explain.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

TOPIC 8 Energy from the atom 255

8.4 Energy from accelerating charges KEY CONCEPTS • Explain light as an electromagnetic wave that is produced by the acceleration of charges. • Describe the production of synchrotron radiation by an electron radiating energy at a tangent to its circular path. • Model the production of light as a result of electron transitions between energy levels within an atom.

8.4.1 Matter and light The word ‘light’ is often used as a shorthand collective for all the FIGURE 8.12 Hundreds of different types of electromagnetic radiation from radio waves to mobile phone towers around the gamma rays, including infra-red, visible, ultraviolet and X-rays. state receive and transmit radio It is used here in that context. waves from mobile phones. So far in this text, matter has produced light in a variety of different ways. The glow of hot objects, such as the Sun, was explained by collisions between the outer electrons of atoms. The electrons’ collisions with each other caused a change in direction or speed, or often both. This acceleration of the electron produced light. If the temperature was high and the accelerations sudden, then the object glowed. The same situation applies when electrons are free from their atoms or free in space. A common example of where electromagnetic waves are produced is the mobile phone. A mobile phone turns the sound signal of the person speaking into it into an electrical signal. This electrical signal moves electrons back and forth in the antenna. The antenna is a conductor inside the mobile phone. The electrons moving back and forth along the antenna produce electromagnetic waves in the radio wave portion of the electromagnetic spectrum, that move outwards from the mobile phone at the speed of light. They are received by another antenna on a mobile phone tower in the neighbourhood. The electrons in the antenna experience a force in the presence of the electromagnetic waves causing them to move up and down the antenna, turning the radio wave signal back into an electrical signal. In this way the signals can be spread around the world. Antennas work in both directions, turning electrical signals into radio waves or radio waves into electrical signals, depending on whether they are transmitting or receiving a signal. In particle accelerators such as the Large FIGURE 8.13 Synchrotron radiation is emitted along Hadron Collider at CERN in Switzerland and the tangent of the circular path. the Australian Synchrotron, charged particles are accelerated by electric fields to speeds close to the speed of light, then deflected into a circular path by magnetic fields. The circular path means a Electron N constantly changing direction, so the particles are beam continually being accelerated and consequently giving off radiation. At CERN, this radiation and loss of energy is an unavoidable nuisance as the researchers using it want very fast particles, such S as protons, to smash into each other. Synchrotron radiation

256 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

However, at the Australian Synchrotron scientists use electrons accelerated to almost the speed of light. When electrons reach this speed, the radiation comes off in a very narrow beam along the tangent to the circular path. The radiation beam is also very intense and includes all the wavelengths across a large range. The beams at the Australian Synchrotron have a variety of uses including scientific research such as investigating the structure of proteins, medical uses such as high resolution imaging and cancer radiation therapy, as well as the analysis of mineral samples, forensic analysis and the investigation of advanced materials. The name synchrotron radiation comes from the fact that it was first observed in 1946 when the first synchrotron was built. Since then this type of radiation has been observed in galaxies, when electrons travelling at the speed of light spiral through an intense magnetic field. FIGURE 8.14 The Australian Synchrotron in Clayton Victoria produces intense beams of synchrotron radiation that are used to analyse the composition and structure of materials, right down to the atomic scale.

Source: ANSTO

8.4.2 Light from atoms Electrons within atoms are also able to produce light and electromagnetic radiation from other parts of the spectrum. They can do this because there are different energy states that the electrons within atoms can take. This model of the electrons in atoms differs from the common one that imagines electrons as circling the nucleus like planets revolving around the Sun. This planetary model of the atom cannot work because of synchrotron radiation. Those charged electrons orbiting the nucleus would radiate energy, which is not observed. Also, as the electrons radiated energy they would lose energy, spiralling into the nucleus. This does not happen. Instead, we need to think of electrons in atoms very differently. An example of the evidence for electron behaviour can be seen in the spectra from different materials. Figure 8.15 shows a typical absorption spectrum, where light is shone through a gas and specific colours are absorbed. The figure also shows an emission spectrum, where a gas is heated to a high temperature and gives a characteristic colour. The inverse of one pattern is almost identical to the other.

TOPIC 8 Energy from the atom 257

FIGURE 8.15 A continuous spectrum and two different ways of producing an element’s fingerprint Continuous spectrum

Emission line spectrum

Hot gas

Cold gas

Absorption line spectrum

Each line has a specific wavelength and frequency. Consequently, the light of that colour in the spectrum has a specific energy. Each line of light in the emission spectrum has come from an electron inside the atom that has jumped from one energy level down to a lower one. The difference in energy between the two levels is emitted as light energy. The existence of several lines tells us about the different energy levels that electrons in this atom can have and gives a picture of the structure inside the atom. These emission and absorption spectra are unique to each element. Each arrow in figure 8.16 shows possible energy transitions for electrons in hydrogen. The electron would normally be in level n = 1, known as the ground state, but collision with another charged particle or a photon could give the electron enough energy to enter one of the higher energy states. When this happens, it is only a matter of time before the electron gives up that additional energy in the form of a photon of light. Photons are packets of energy that make up light waves and all parts of the electromagnetic spectrum. The higher the energy of the photon, the further its light is towards the gamma end of the spectrum. Lower energy photons are towards the radio end of the spectrum. The colour of the light depends on the size of the energy transition. The arrows in figure 8.16 show energy transitions that would result in an emission spectrum. Only photons of energies that correspond to the changes in energy levels can occur, so the light emitted from excited hydrogen gas has very specific energies. Most of these lines have energies outside the visible spectrum of light, but the Balmer series are in the visible spectrum. This feature of gases is used to great effect in lighting. Gas discharge tubes like fluorescent tubes that probably light up your Physics classroom contain a gas at low pressure. The gas in fluorescent tubes is mercury vapour. When a high voltage is put across the ends of the tube, a current of electrons flows through the tube. These electrons hit electrons in the mercury atoms, giving them energy to rise to a higher energy level. When the electron falls back to its ground state it emits light of a particular frequency. With mercury, this tends to be light in the ultraviolet part of the spectrum. The glass tube is covered with a material that absorbs the ultraviolet light and re-emits it in the visible part of the spectrum. The different frequencies of this light can be viewed by observing the fluorescent tubes through a spectroscope. 258 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 8.16 Electron transitions for the hydrogen atom n=7 n=6 n=5

n=4 Brackett series E(n) to E(n=4) n=3 Paschen series E(n) to E(n=3) n=2 Balmer series E(n) to E(n=2)

n=1 Lyman series E(n) to E(n=1)

FIGURE 8.17 The emission spectrum of hydrogen UV

λ

200

IR

400

600

800

1000

1200

1400

1600

1800

2000

2200

2400

2600 nm

Other gas discharge tubes are used in advertising, with different gases used in the tubes, like neon, to produce light of different colours. Liquids and solids produce spectra that resemble a continuous spectrum due to the proximity of the atoms resulting in the energy levels that have little energy difference between them. Electrons in these ‘bands’ can take on a nearly continuous range of energies. As a result, glowing solids, like the filament in an incandescent light bulb, produce a continuous spectrum.

TOPIC 8 Energy from the atom 259

FIGURE 8.18 The skipping girl sign in Richmond is a Melbourne landmark using tubes of different gases to produce the effect of a girl skipping.

SAMPLE PROBLEM 4

Identify three processes in which electrons produce electromagnetic radiation. THINK 1

Recall three processes in which electrons produce electromagnetic radiation.

WRITE

Any three of the following: • Electrons in atoms in a hot metal move between energy levels and emit electromagnetic radiation. • Electrons moving in circles in a synchrotron produce intense electromagnetic radiation tangent to the path. • Electrons passing through a gas at high voltage excite the electrons in the gas, which then emit electromagnetic radiation as they return to ground state. • Electrons accelerate in an antenna.

PRACTICE PROBLEM 4 Explain why electrons cannot move around the nucleus like planets around the Sun.

260 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Resources Digital documents Investigation 8.2 The spectrum of hydrogen (doc-31903) Investigation 8.3 Spectroscopes (doc-32622) Teacher-led video

Investigation 8.3 Spectroscopes (tlvd-0254)

8.4 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. What situation leads to the production of synchrotron radiation? Electrons are said to exist in energy levels in atoms. What is some evidence for this claim? What is the source of electromagnetic radiation? What can you say about the electrons in the transmitter’s antenna at a radio station of frequency 621 kHz, compared with one of frequency 774 kHz? 5. Compare the energy levels of electromagnetic radiation emitted from the nucleus (gamma rays) with that emitted by the electrons in atoms. 1. 2. 3. 4.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

8.5 Review 8.5.1 Summary • •

• • • • • • •

Mass and energy are equivalent and related by the equation E = mc2 . The nuclei of different atoms have varying degrees of stability. The binding energy of a nucleus is the energy required to completely separate a nucleus into individual nucleons. Therefore, the binding energy is a measure of the stability of a nucleus. Iron-56 is commonly thought to be the most stable of all nuclei, however, the rarer iron-58 and nickel-62 have a slightly higher binding energy per nucleon. In order for a nucleus to become more stable, it emits energy called nuclear energy. The amount of energy released corresponds to the difference between the mass of the particles before and after the release of energy, according to E = mc2 . Fusion reactions generally occur between small nuclei which form nuclei that are smaller than iron. Fusion occurs in our Sun, where hydrogen nuclei are converted into helium nuclei, resulting in large amounts of energy being released. Fission reactions occur when a large nucleus is split into smaller, more stable fission fragments that have higher binding energy per nucleon. Electromagnetic radiation is produced when electric charges are accelerated. Electric charges can be accelerated in a number ways to produce electromagnetic radiation. Very fast electrons moving in a circular path produce synchrotron radiation. Electrons moving between energy levels inside an atom produce light in parts of the electromagnetic spectrum determined by the difference between the energy levels.

TOPIC 8 Energy from the atom 261

Resources

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0033).

8.5.2 Key terms The number of protons in a nucleus is called the atomic number of the atom. The binding energy of a nucleus is the energy that is required to split a nucleus into individual nucleons. Fission fragments are the products from a nucleus that undergoes fission. The fission fragments are smaller than the original nucleus. Atoms containing the same number of protons but different numbers of neutrons are called isotopes of an element. The mass defect is the difference in the mass of the products and reactants in a nuclear reaction. Mass number describes the total number of nucleons in an atom. Nuclear fission is the process of splitting a large nucleus to form two smaller, more stable nuclei. Nuclear fusion is the process of joining together two nuclei to form a larger, more stable nucleus. Synchrotron radiation is the electromagnetic radiation produced when electric charges move in circular motion. It is radiated at a tangent to the curve.

Resources Digital document Key terms glossary (doc-32266)

8.5.3 Practical work and investigations Investigation 8.1 Chain reaction of dominoes Aim: To model controlled and uncontrolled chain reaction in nuclear fission using dominoes Digital document: doc-32621 Teacher-led video: tlvd-0845

Investigation 8.2 The spectrum of hydrogen Aim: To observe the spectral lines of hydrogen, measure their wavelengths and compare these values with the theoretical values Digital document: doc-31903

Investigation 8.3 Spectroscopes Aim: To use a spectroscope to observe the spectrum of a light source Digital document: doc-32622 Teacher-led video: tlvd-0254

262 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Resources Digital document Practical investigation logbook (doc-33011)

8.5 Exercises To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au.

8.5 Exercise 1: Multiple choice questions 1.

2.

3.

4.

5.

6.

7.

8.

What does the equation E = mc2 describe? A. The equivalence of mass and energy B. That energy is much more significant than mass by a factor of c2 C. That energy has mass D. That the speed of light is constant When does fission occur? A. Two small nuclei merge. B. A small nucleus splits. C. Two large nuclei merge. D. A large nucleus splits. The products of fission have which of the following? A. Greater binding energy per nucleon than the original nucleus B. Less binding energy per nucleon than the original nucleus C. Equal binding energy per nucleon as the original nucleus D. It depends on the nucleus. What do current commercial nuclear plants run on? A. Fusion B. Fission C. Alpha decay D. A mixture of fusion and fission Electromagnetic radiation is not produced by accelerating which of the following? A. Protons B. Electrons C. Neutrons D. Positrons Which of the following is a source of the Sun’s energy? A. Fission of uranium-235 B. Fusion of carbon C. Exothermic chemical reactions D. Fusion of protons Which of the following is not a product of the fusion of protons? A. Electron B. Positron C. Antimatter D. Neutrino Radio waves from a radio station are produced by which of the following? A. Synchrotron B. Electrons moving between energy levels C. Electrons moving back and forth along an antenna D. Electrons moving in energy bands in the metal of an antenna TOPIC 8 Energy from the atom 263

Which of the following is one of the conditions required for a nucleus to undergo fission? A. The binding energy per nucleon must be less in the products that the parent nucleus. B. The binding energy per nucleon must be greater in the products that the parent nucleus. C. The mass of the products must be greater than the mass of the parent nucleus. D. The nucleus must be smaller than that of iron-56. 10. Which of the following is not a unit of energy? A. eV B. MeV C. J D. kW 9.

8.5 Exercise 2: Short answer questions 1. 2. 3. 4. 5. 6. 7.

Why might different elements produce different absorption and emission spectra? A gas absorbs some wavelengths of light to produce an absorption spectrum, but the light is very quickly re-emitted. Why doesn’t the re-emitted light fill in the dark line? The emission spectra shown in figure 8.17 is for hydrogen, which has only one electron. Why do you think there are so many lines in the spectrum? Why are neutrons good at initiating nuclear reactions? In what form does the energy released from a nuclear fusion reaction appear? What conditions are required before two protons will fuse to form a single nucleus? Why are this conditions required? Einstein’s theories suggested that energy and mass are not conserved. a. Describe what he found. b. Describe an example of where it can be observed that mass and energy are not conserved. c. Why is the problem with the laws of conservation and mass so difficult to observe?

8.5 Exercise 3: Exam practice questions Use table 8.3 to answer the questions that follow. TABLE 8.3 Masses and binding energies Nucleus

Symbol

Mass (kg)

Total binding energy (MeV)

240 94 Pu

3.985 755 × 10–25

1813.454 956

90 38 Sr

1.492 791 × 10–25

782.631 470

147 56 Ba

2.439 632 × 10–25

1204.158 203

Uranium-234

234 92 U

3.885 920 × 10–25

1778.572 388

Zirconium-95

95 40 Zr

1.575 810 × 10–25

821.139 160

Tellurium-136

136 52 Te

2.256 760 × 10–25

1131.440 918

1 0n

1.674 746 × 10–27

Plutonium-240 Strontium-90 Barium-147

Neutron Proton

1 1p

or 11 H

1.673 351 × 10–27

Hydrogen-2

2 1H

or 21 D

3.344 132 × 10–27

2.224 573

Hydrogen-3

3 1H

3 1T

5.007 725 × 10

–27

8.481 821

or

Helium-4

4 2 He

6.645 758 × 10–27

28.295 673

Lithium-6

6 3 Li

9.987 263 × 10–27

31.994 564

264 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Question 1 (6 marks) A plutonium-239 nucleus absorbs a neutron to become plutonium-240, which splits to form strontium-90, barium-147 and three neutrons. a. What is the difference between the binding energy of the plutonium-240 nucleus and the sum of the binding energies of the two fission fragments? 2 marks b. What is the difference between the mass of the plutonium-240 nucleus and the sum of the masses of all the fission fragments, including neutrons? 2 marks 2 c. Use E = mc to calculate the energy equivalent of this mass difference in joules and MeV. 2 marks Question 2 (6 marks) A uranium-233 nucleus absorbs a neutron to become uranium-234, which splits to form zirconium-95, tellurium-136 and three neutrons. a. What is the difference between the binding energy of the uranium-234 nucleus and the sum of the binding energies of the two fission fragments? 2 marks b. What is the difference between the mass of the uranium-234 nucleus and the sum of the masses of all the fission fragments, including neutrons? 2 marks c. Use E = mc2 to calculate the energy equivalent of this mass difference in joules and MeV. 2 marks Question 3 (9 marks) A fusion reactor could not feasibly use the same reactions as the Sun. A reactor on Earth would have to use a different reaction, preferably a one-step reaction with only two reactants. Three possible reactions for a terrestrial fusion reactor are as follows. 2 3 4 a. 1H + 1H → 2He + 10n 2 2 3 1 b. 1H + 1H → 1He + 1H 6 2 4 c. 1H + 3Li → 2 2He Using data from table 8.3, calculate the: i. difference between the binding energy of the products and the sum of the binding energies of the reactants ii. difference between the sum of masses of the products and of the reactants iii. energy equivalent of this mass difference in joules and MeV.

3 marks 3 marks 3 marks

Question 4 (3 marks) Compare the processes of fission and fusion. Question 5 (2 marks) List two reasons fusion would be a preferable way of generating power to fission.

8.5 Exercise 4: studyON topic test Fully worked solutions and sample responses are available in your digital formats.

Test maker Create unique tests and exams from our extensive range of questions, including practice exam questions. Access the assignments section in learnON to begin creating and assigning assessments to students.

TOPIC 8 Energy from the atom 265

UNIT 1 | AREA OF STUDY 3 REVIEW

AREA OF STUDY 3 What is matter and how is it formed? OUTCOME 3 Explain the origins of atoms, the nature of subatomic particles and how energy can be produced by atoms.

PRACTICE EXAMINATION STRUCTURE OF PRACTICE EXAMINATION Section

Number of questions

Number of marks

A

20

20

B

4

20 Total

40

Duration: 50 minutes Information:

• • •

This practice examination consists of two parts. You must answer all question sections. Pens, pencils, highlighters, erasers, rulers and a scientific calculator are permitted. You may use the VCAA Physics formula sheet for this task.

Resources Weblink VCAA Physics formula sheet

SECTION A — Multiple choice questions All correct answers are worth 1 mark each; an incorrect answer is worth 0. 1. Which of the following phenomena does Hubble’s Law refer to? A. The red shift that is observed in distant galaxies, due to the expansion of the universe B. That the more distant a galaxy is from us, the faster it is moving away from us C. That the red shift of distant galaxies is due to the Doppler effect D. The limit to the maximum possible speed that distant galaxies are receding from us 2. The currently accepted theory of the Big Bang origin of the universe is not supported by which of the following observation? A. The cosmic microwave background is the temperature the universe cooled to after the Big Bang. B. Distant galaxies are observed to be red shifted as they are receding from us. C. The further a galaxy is from us, the more red shifted it is. D. Distant galaxies are all red shifted by the same amount.

266 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

3. The light from a star near the Sun within the Milky Way galaxy is blue shifted. Which of the following best describes the current explanation for this change in wavelength? A. The star is moving away from us. B. The star is moving towards us. C. The space between the star and the Sun is contracting. D. The star is being drawn towards the centre of the Milky Way away from the Sun. 4. The Voyager 1 probe is the furthest human-made object from our solar system. It is travelling at a speed of approximately 17 km s–1 . What is the distance it will travel in one year, in metres? A. 2.2 × 108 m

B. 5.4 × 108 m

C. 2.2 × 1011 m D. 5.4 × 1011 m

5. The Hubble constant (H) could be used to estimate the age of the universe. Using the best observational data, Edwin Hubble estimated the constant to have a value 1.6 × 10−17 s−1 . Which of the following is the best estimate of the age of the universe? A. 1.6 × 1016 s−1

B. 6.3 × 1016 s−1

C. 1.6 × 1017 s−1 D. 6.3 × 1017 s−1

6. Which of the following best describes the reason for the ratio of protons to neutrons in the early universe as 7:1? A. There was insufficient matter to form neutrons. B. Seven is the ideal ratio for the protons to fuse with neutrons to form helium nuclei. C. The universe cooled to a point that protons do not interact with electrons with sufficient energy. D. There were too many protons that interfered with neutron formation. 7. According to the current theory of matter, which of the following groups of particles comprise the basic particles that form all other matter? A. Protons, neutrons and electrons with their antiparticles B. Hadrons and leptons with their antiparticles C. Baryons and mesons with their antiparticles D. Quarks and leptons with their antiparticles 8. An isotope of hydrogen is tritium. A neutrally charged atom of tritium comprises a proton and two neutrons. How many quark and leptons make up this atom? A. 3 quarks and 1 leptons B. 6 quarks and 3 leptons C. 3 quarks and 3 leptons D. 9 quarks and 1 leptons 9. Which of the following cannot be classified as a fermion? A. A beta particle B. A hydrogen nucleus C. A quark D. A gamma ray 10. Which of the following does not describe the hadron family of particles? A. They include the positron. B. They include baryons and mesons. C. They are composed of two or three quarks. D. They have an internal structure.

UNIT 1 Area of Study 3 Review 267

Use the following information to answer questions 11 and 12. Study the following table of quarks. Type Up (u) Down (d) Strange (s)

Charge + − −

2 3 1 3 1 3

Type Charm (c) Bottom (b) Top (t)

Charge + − +

2 3 1 3 2 3

11. Which of the following best describes the charge of the combination of a charm quark and an antibottom quark? 1 A. − 3 1 B. + 3 C. –1 D. +1

12. This combination of quarks results in what sort of particle? A. Baryon B. Lepton C. Meson D. Neutrino Use the following information to answer questions 13 and 14. When a matter particle and an antimatter particle collide, they annihilate each other and all matter and antimatter is transformed into energy in the form of electromagnetic radiation. Mass of electron: 9.1 × 10−31 kg Mass of proton: 1.67 × 10−27 kg Speed of light: 3.0 × 108 m s−1

13. Using the data provided, what is the best estimate of the energy released when an electron and a positron meet, at rest? A. 1.6 × 10−13 J

B. 8.2 × 10−14 J

C. 5.5 × 10−22 J

D. 2.7 × 10−22 J

14. When a proton and an antiproton meet, at rest, the energy released is 3.0 × 10−10 J. What is the approximate equivalent amount of this energy in MeV? A. 1.9 MeV B. 19 MeV C. 190 MeV D. 1900 MeV 15. How many protons and neutrons are there in an alpha particle 24𝛼 ? A. 4 protons and 2 neutrons B. 2 protons and 4 neutrons C. 2 protons and 2 neutrons D. 2 protons and 6 neutrons

268 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Use the following information to answer questions 16 and 17. 90 Strontium-90, 38S r, is a beta emitter with a half-life of 29 years. 16. Element X is the daughter element of strontium-90. Which of the following best describes the nucleus of X following the beta decay of strontium-90? A. B. C. D.

90 39X 86 36X 90 38X 86 39X

17. A chemical mixture from the last century contains strontium-90. A sample of it has been analysed and it is found that there is only 12.5% of the original amount of strontium-90 left. How old is this chemical mixture? A. 29 years old B. 58 years old C. 87 years old D. 116 years old 18. The binding energy per nucleon for the deuteron 21 H (or D) is 1.11 MeV. Which of the following best describes the binding energy of this nucleus? A. 0.556 MeV B. 1.11 MeV C. 2.22 MeV D. 3.33 MeV

19. Helium fusion occurs in stars where three nuclei of helium fuse to form a carbon nucleus. This process releases 1.16 × 10−12 J of energy per carbon nucleus formed. What is the best estimate of the amount of mass lost for each carbon nucleus formed? A. 3.9 × 10−21 kg

B. 1.3 × 10−29 kg

C. 3.9 × 10−29 kg

D. 1.3 × 10−21 kg

20. A radioactive source has 240 000 active nuclei initially. After 32 hours, there are 15 000 active nuclei remaining. Which of the following best describes the half-life of this source? A. 2 hours B. 4 hours C. 6 hours D. 8 hours SECTION B — Short answer questions Question 1 (6 marks) a. Within a second of the Big Bang, protons and neutrons were formed in equal amounts as the protons interacted with electrons to form neutrons, which decayed in turn to form protons and electrons. Explain why within the next few minutes the ratio of neutrons to protons dropped to 1:7. 2 marks b. Within minutes of the Big Bang, nearly all the neutrons had fused to form helium-4, resulting in helium-4 making up 25% of the mass in the universe. Provide a detailed explanation of how the mass of helium-4 is 25% that of the universe. 3 marks c. It took a further 380 000 years before neutral atoms were formed and the universe became transparent to electromagnetic radiation. What was the condition that led to the formation of neutral atoms? 1 mark

UNIT 1 Area of Study 3 Review 269

Question 2 (5 marks) Americium-241, 241 95A m, is an alpha emitter that is commonly used in smoke detectors. The following graph shows the activity curve of a sample of americium-241. This sample had an initial strength of 6.0 × 105 Bq. 700 000 600 000

Activity (Bq)

500 000 400 000 300 000 200 000 100 000 0 0

200

400

600 800 1000 Age of source (year)

1200

1400

a. Use the graph to estimate the half-life of americium-241. Outline the two points on the graph that you used to arrive at your estimate. 2 marks b. Describe what is meant by the term alpha decay (which occurs to an alpha emitter). 1 mark c. Americium-241 radioactively decays and its daughter element is neptunium, Np. Write the full decay equation of americium. 2 marks Question 3 (6 marks) Plutonium-239 is a fissile radioisotope which undergoes a fission reaction according to the following equation: 1 n 0

+ 239 94P u →

134 54X e

+ 103xZ r + y 10n

2 marks

a. What is the value of x in the equation?

b. How many neutrons, y, were produced in each reaction event? 2 marks −28 c. Each reaction event results in a mass deficit of 3.1 × 10 kg. What is the amount of energy produced per reaction event, expressed in MeV? 2 marks Question 4 (3 marks) Particle colliders have been used to create hadrons to study and discern the physics of the early universe. One such hadron is found to be made up of three quarks, with a charge of +2. a. Is the hadron a meson or a baryon? 1 mark b. Two of the quarks are known to be up quarks. Which of the following quarks might be the third quark? Justify your response. 2 marks Type Up (u) Down (d) Strange (s)

Charge

+ − −

Type 2 3 1 3 1 3

270 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Charm (c) Bottom (b) Top (t)

Charge

+ − +

2 3 1 3 2 3

PRACTICE SCHOOL-ASSESSED COURSEWORK ASSESSMENT TASK – A MODELLING ACTIVITY In this task, you will be required to explain the origins of atoms, the nature of subatomic particles and how energy can be produced by atoms. • A scientific calculator is permitted for this task. • Access to a spreadsheet program such as Excel will assist. Total time: One lesson to complete the simulation; 50 minutes for the completion of the three listed tasks (5 minutes reading and 45 minutes writing) Total marks: 35 marks

A MODELLING ACTIVITY FOR RADIOACTIVE SAMPLES The radioactive decay of a single nucleus is a random event. However, for a large enough population of identical radioactive nuclei, the half-life can be found for the sample. Even a small radioactive sample contains many billions of nuclei. Thus, the overall pattern of decay will appear quite ordered with the source getting less intense over time. 1 Dice, when thrown, have a probability of landing with a 6 face up equal to 6 . Like radioactive material, the outcome of each die, or nucleus, is unpredictable but a large sample will follow an ordered statistical behaviour. In this experimental assessment task you will produce a radioactive decay curve by simulating the radioactive decay process using dice. You will analyse this curve to determine the ‘half-life’ of your set of dice. The analogy is simple. Each trial corresponds to a set interval of time. If a die falls with 6 face up it is said to have decayed. If not, it is deemed to be still radioactive in the sense that it has not decayed. By recording the number of dice remaining at the end of each successive trial you will trace the ‘radioactive decay’ of the sample of dice.

Materials A box of dice, appropriate table (either by paper or on a digital spreadsheet) and graph paper

Method Firstly, count how many dice you have to start with; a good number would be 50 dice. Create a table similar to the following. The first row has been done assuming there were 50 dice to begin with. This table could be completed in a spreadsheet. Number of trial n

Number of dice remaining in sample

0

50

1 2

1. Toss all the dice randomly. 2. Count the number of 6s in the sample and remove them from the sample. Record the number of dice left after the first trial where n = 1. 3. Toss all the remaining dice and count the number of 6s. Again remove them from the sample and record the number of dice left after the second trial, this time n = 2. 4. Repeat this process a further 13 times making a total of 15 trials. 5. If all the dice have been removed before the 15th trial, record the dice remaining in the last trials as 0. 6. Now perform the process another two times to obtain three independent simulations in total. You will need to make two more columns for your simulation data.

UNIT 1 Area of Study 3 Review 271

Results and discussion Task 1 Create a spreadsheet or a table with column B recording the change in the sample size for trials 0 to 15 in your first simulation. In columns C and D record the results of your second and third simulations. In column E obtain the sum of each of the rows in each of the other three simulations to get the total results for your three simulations. Task 2 Plot a graph of the total number of dice remaining (summed over three trials (column E)) versus the number of trials, 1 through 15. Draw a curve of best fit. Task 3 Answer the following questions related to this investigation. 1. If there are 100 dice and they are tossed, what fraction on average would you expect to land with 6 face up? 2. Consequently, from 100 dice how many would you expect to be removed? 3. How many dice would be left if those dice were removed? 4. From those remaining after the first toss, repeat steps 1, 2 and 3 each time calculating the number of dice left after those with a face up value of 6 have been removed. Repeat this until there are less than 50 dice remaining. 5. Use this sequence of numbers (called a geometric sequence) to estimate the half-life. 6. Use your graph of your simulation to estimate the half-life for your population of dice? Ensure you show this on your graph and identify the result you obtained. 7. Describe how you used your graph to estimate this half-life. 8. What is meant by the half-life of a radioactive material? 9. What happens to the decay rate — that is, the number of atoms decaying — after each half-life. Explain. 10. 1 kilogram of Tc-99 (technetium-99) contains approximately 6.0 × 1024 atoms. The half-life of Tc-99 is 6 hours and it is a beta-emitter. What does the term beta-emitter mean? 11. How long would it take for the Tc-99 sample to be reduced to a mass of: (a) 500 grams (3.0 × 1024 atoms) (b) 125 grams (1.5 × 1024 atoms)? 12. Tim has 2 kilograms of Tc-99. It takes 6 hours for the sample to be reduced to 1 kilogram. Tim states that every 6 hours, 1 kilogram will decay, so after 12 hours no sample will be left. Explain why the statement made by Tim is incorrect. 13. Compare the processes of nuclear fusion and nuclear fission. Explain why both fusion and fission are processes that can produce energy. 14. The process of nuclear fusion was fundamental during the development of the universe. Explain how this process lead to the formation of atoms, with specific reference to atoms and subatomic particles present during the formation of the universe and early nucleogenesis. 15. There are two types of forces holding the nucleus together: the strong nuclear force and the weak nuclear force. Describe these two forces and identify which is responsible for the decay simulated in this investigation.

Resources Digital document School-assessed coursework (doc-32275)

272 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

UNIT 2 WHAT DO EXPERIMENTS REVEAL ABOUT THE PHYSICAL WORLD? In this unit, you will explore the power of conducting experiments and investigations to understand and develop models and theories. This unit allows you to investigate motion, exploring the effects of balanced and unbalanced forces both theoretically and practically. You will conduct further research in an area of interest, selecting from one of 12 options. This will allow for an in-depth application of physics, focusing on topics such as astrophysics, medical physics and sports science. You will also conduct a practical investigation related to the knowledge and skills developed across Units 1 and 2. AREA OF STUDY

OUTCOME

TOPICS

1. How can motion be described and explained?

Investigate, analyse and mathematically model the motion of particles and bodies.

9. Analysing motion 10. Forces in action 11. Energy and motion

2. OPTIONS: Observation of the physical world

Twelve options are available for selection in Area of Study 2. Each option is based on a different observation of the physical world. Students are required to apply and explain concepts for the one option selected.

Topics 12–23

3. Practical investigation

Design and undertake an investigation of a physics question related to the scientific inquiry processes of data collection and analysis, and draw conclusions based on evidence from collected data.

24. Practical investigation

Source: VCE Physics Study Design (2016–2021) extracts © VCAA; reproduced by permission.

AREA OF STUDY 1 HOW CAN MOTION BE DESCRIBED AND EXPLAINED?

9

Analysing motion

9.1 Overview Numerous videos and interactivities are available just where you need them, at the point of learning, in your digital formats, learnON and eBookPLUS at www.jacplus.com.au.

9.1.1 Introduction The study of the motion of objects relies upon an accepted and reliable means for describing and analysing motion. In this topic you will explore the key quantities used to describe motion, their characteristics and interrelationships. By the end of this topic you should be able to distinguish between scalar and vector quantities. You will be able to describe motion in terms of position, displacement, speed, velocity and acceleration. You will analyse motion using the most suitable method, including numerical calculations, estimations, algebraic analysis and graphical analysis. This area of study is often broadly referred to as kinematics. The skills developed through this topic provide the foundation for studies of the role of forces and energy in the understanding motion in the topics that follow. FIGURE 9.1 When driving a vehicle, the driver needs to be aware of their movements in terms of position, speed, direction and acceleration.

TOPIC 9 Analysing motion 275

9.1.2 What you will learn KEY KNOWLEDGE After completing this topic, you will be able to: • identify parameters of motion as vectors or scalars • analyse graphically, numerically and algebraically, straight-line motion under constant acceleration: v = u + at, v2 = u2 + 2as, s =

1 2

(u + v) t, s = ut +

1 2

at2 , s = vt −

1 2

at2

• graphically analyse non-uniform motion in a straight line. Source: VCE Physics Study Design (2016–2021) extracts © VCAA; reproduced by permission.

PRACTICAL WORK AND INVESTIGATIONS Practical work is a central component of learning and assessment. Experiments and investigations, supported by a Practical investigation logbook and Teacher-led videos, are included in this topic to provide opportunities to undertake investigations and communicate findings.

Resources Digital documents Key science skills — VCE Units 1–4 (doc-31856) Key terms glossary (doc-33010) Practical investigation logbook (doc-32267)

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0034).

9.2 Describing movement KEY CONCEPTS • Identify parameters of motion as vectors or scalars. • Analyse numerically and algebraically, straight-line motion under constant acceleration.

The study of the way in which an object moves is the starting point for developing an understanding of the nature of forces and their relationship with motion.

9.2.1 Distance and displacement Distance is a measure of the length of the path taken during the change in position of an object. Distance is a scalar quantity. It does not specify a direction. Displacement is a measure of the change in position of an object. Displacement is a vector quantity. In order to describe a displacement fully, a direction must be specified as well as a magnitude. The path taken by the fly in figure 9.2 as it escapes the swatter illustrates the difference between distance and displacement. The displacement of the fly is 60 centimetres to the right, while the distance travelled is well over 1 metre. In a 100-metre sprint, the runners travel in a straight line, therefore the magnitude of the displacement is the same as the distance. However, it is the displacement that fully describes the change in position of the runner because it specifies the direction.

276 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 9.2 Distance and displacement are different quantities.

FIGURE 9.3 Runners in the 100-metre sprint travel in a straight line.

The displacement of an object that has moved from position x⃗ 𝟏 to position x⃗ 𝟐 is expressed as: displacement = change in position = final position − initial position Δ x⃗ = x⃗ 𝟐 − x𝟏⃗ Displacement can also be represented by the symbols x⃗ or s ⃗ . (Vectors can be represented using either bold text or a right arrow.)

SAMPLE PROBLEM 1

A hare and a tortoise decide to have a race along a straight 100-metre stretch of highway. They both head due north. However, at the 80-metre mark, the hare realises that he dropped his phone at the 20-metre mark. He dashes back, grabs his phone and resumes the race, arriving at the finishing line at the same time as the tortoise. (It was a very fast tortoise!) a. What was the displacement of the hare during the entire race? b. What was the distance travelled by the hare during the race? c. What was the distance travelled by the tortoise during the race? d. What was the displacement of the hare during his return to collect his phone? Teacher-led video: SP1 (tlvd-0069) THINK

Recall the relationship Δx = x2 − x1 . 2. Identify the variables. Using the start as the reference point, the displacement was 100 m north. In symbols, this calculation can be done by denoting north as positive and south as negative. 3. Substitute into the relationship to find Δx.

a. 1.

4.

State the solution.

WRITE a.

Δx = x2 − x1 x1 = 0, x2 = 100 m north

Δx = 100 − 0 = 100 m north The displacement of the hare is 100 metres north.

TOPIC 9 Analysing motion 277

b. 1.

2.

The distance is the length of the path taken.

State the solution.

The distance travelled by the tortoise is the length of the path taken. 2. State the solution.

c. 1.

Recall the relationship Δx = x2 − x1 . The hare returns from a position 80 m north of the reference point (or start) back to a position 20 m north of the reference point. 3. State the solution.

d. 1.

2.

The hare travels a total distance of 80 m (before noticing his phone) + 60 m (running back to the 20 m mark) + 80 m (from the 20 m mark to the finishing line). Distance = 80 + 60 + 80 = 220 m The total distance that the hare travels is 220 metres. c. Length of the path taken = 100 m b.

The tortoise travels a total distance of 100 metres. d. Δx = x2 − x1 = 20 m − 80 m = −60 m The displacement is 60 metres south.

PRACTICE PROBLEM 1 A cyclist rides 4 kilometres due west from home, then turns right to ride a further 4 kilometres due north. She stops, turns back and rides home along the same route. a. What distance did she travel during the entire ride? b. What was her displacement at the instant that she turned back? c. What was her displacement from the instant she commenced her return journey until she arrived home?

9.2.2 Speed Speed is a measure of the rate at which an object moves over a distance. When you calculate the speed of a moving object, you need to measure the distance travelled over a time interval. The average speed of an object can be calculated by dividing the distance travelled by the time taken: average speed =

distance travelled time interval

The speed obtained using this formula is the average speed during the time interval. Speed is a scalar quantity as it does not include direction. The unit of speed is m s−1 if SI units are used for distance and time. However, it is often more convenient to use other units such as cm s−1 or km h−1 .

278 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

SNAIL’S PACE A snail would lose a race with a giant tortoise! A giant tortoise can reach a top speed of 0.37 km h−1 . However, its ‘cruising’ speed is about 0.27 km h−1 . The world’s fastest snail covers ground at the breathtaking speed of about 0.05 km h−1 . However, the common garden snail is more likely to move at a speed of about 0.02 km h−1 . Both of these creatures are slow compared with light, which travels through the air at 1080 million km h−1 , and sound, which travels through the air (at sea level) at about 1200 km h−1 . How long would it take the snail, giant tortoise, light and sound respectively to travel once around the equator, a distance of 40 074 km?

9.2.3 Converting units of speed It is often necessary to convert units that are not derived from SI units (such as km h−1 ) to and from units that are derived SI units, such as m s−1 . TABLE 9.1 Converting units of speed To convert 60 km h−1 to m s−1

To convert 30 m s−1 to km h−1 30 m

60 km

=

1s 0.030 km = 1 h 3600 3600 × 0.030 km = 1h = 108 km h−1

1h 60 000 m

3600 s = 16.7 m s−1

The speed in km h−1 has been multiplied by

The speed in m s−1 has been multiplied by

1000

3600

3600

, or divided by 3.6.

1000

, that is, by 3.6.

To quickly convert between speeds in m s−1 and km h−1 you can simply multiply or divide by 3.6 depending on which way you are converting.

÷ 3.6

km h−1

m s−1

× 3.6 TOPIC 9 Analysing motion 279

SAMPLE PROBLEM 2

A plane carrying passengers from Melbourne to Perth flies at an average speed of 250 m s−1 . The flight takes 3 hours. Use this information to determine the approximate distance by air between Melbourne and Perth. THINK 1.

Recall the relationship and rearrange to find distance.

Identify the values, and convert into the similar units. Note: Alternatively, the distance could be calculated in metres and converted to kilometres. Multiply the time interval by 3600 to convert h to s. 3 Substitute into the relationship to find distance. 4 State the solution.

2.

WRITE

distance travelled time interval Distance travelled = average speed × times time interval Time interval = 3 h ( ) Average speed = 250 m s−1 × 3.6 to convert to km h−1 Average speed =

= 900 km h−1

Distance travelled = 900 km h−1 × 3 h = 2700 km The approximate distance by air between Melbourne and Perth is 2700 kilometres.

PRACTICE PROBLEM 2 A car takes 8 hours to travel from Canberra to Ballarat at an average speed of 25 m s−1 . What is the road distance from Canberra to Ballarat?

9.2.4 Velocity In everyday language, the word velocity is often used to mean the same thing as speed. In fact, velocity is not the same quantity as speed; velocity is a measure of the rate of displacement, or rate of change in position, of an object. Because displacement is a vector quantity, velocity is also a vector quantity. The velocity has the same direction as the displacement. The symbol v is used to denote velocity. (Unfortunately, the symbol v is often used to represent speed as well, which can be confusing.) To make a distinction between vectors and scalars, this text mostly displays vector symbols in bold. When bold can’t be used to distinguish a vector (for example, in( the ) following(pink ) box or when writing by hand), it is common practice to place a vector symbol above v ⃗ or below v⃯ vector quantities. The average velocity of an object, v ⃗av , during a time interval t can be expressed as: vav ⃗ =

∆ x⃗ ∆t

where ∆ x⃗ represents the displacement (change in position).

280 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

For motion in a straight line in one direction, the magnitude of the velocity is the same as the speed. The motion of the fly in figure 9.4 illustrates the difference between velocity and speed. If the fly takes 2 seconds to complete its flight, its average velocity is: ∆x ∆t 60 cm to the right = 2s = 60 cm s−1 to the right

FIGURE 9.4 When motion is not in a straight line, in one direction, velocity is different to speed.

vav =

The path taken by the fly is about 180 cm. Its average speed is: Displacement

distance travelled time travelled 180 cm = 2s = 90 cm s−1

Start

average speed =

Finish 60 cm

Resources Digital document Investigation 9.1 Going home (doc-31874)

SAMPLE PROBLEM 3

Calculate the (a) average speed and (b) average velocity of the hare in Sample problem 1 if it takes 20 seconds to complete the race. THINK

WRITE

Average speed =

distance travelled time interval

Recall the relationship.

a.

2.

Identify the values.

3.

Substitute into the relationship to find the average speed.

4.

State the solution.

Distance travelled = 220 m Time interval = 20 s 220 m Average speed = 20 s = 11 m s−1 The average speed of the hare is 11 m s–1 . Δx b. Average velocity: vav = Δt Displacement = 100 m north Time interval = 20 s 100 m north vav = 20 s = 5.0 m s−1 north The average velocity of the hare is 5 m s–1 north.

a. 1.

b. 1.

Recall the relationship.

2.

Identify the values.

3.

Substitute into the relationship to find vav .

4.

State the solution.

TOPIC 9 Analysing motion 281

PRACTICE PROBLEM 3 During the final 4.2-kilometre run stage of a triathlon, a participant runs 2.8 kilometres east, then changes direction to run a further 1.4 kilometres in the opposite direction, completing the stage in 20 minutes. What was the participant’s: a. average speed b. average velocity? Express both answers in m s−1 to two significant figures.

9.2.5 Instantaneous speed and velocity Neither the average speed nor the average velocity provide information about movement at any particular instant of time. For example, when Jamaican athlete Usain Bolt broke the 100-metre world record in 2009 with a time of 9.58 seconds, his average speed was 10.4 m s−1 . However, he was not travelling at that speed throughout his run. He would have taken a short time to reach his maximum speed and would not have been able to maintain it throughout the run. His maximum speed would have been much more than 10.4 m s−1 . FIGURE 9.5 During his 100-metre world record in 2009, Usain Bolt’s average speed was 10.4 m s−1 .

The speed at any particular instant of time is called the instantaneous speed. The velocity at any particular instant of time is, not surprisingly, called the instantaneous velocity. If an object moves with a constant velocity during a time interval, its instantaneous velocity throughout the interval is the same as its average velocity.

282 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

9.2.6 Acceleration When the velocity of an object changes, it is helpful to describe it in terms of the rate at which the velocity is changing. In everyday language, the word accelerate is used to mean ‘speed up’. The word decelerate is used to mean ‘slow down’. However, if you wish to describe motion precisely, these words are not adequate. The rate at which an object changes its velocity is called its acceleration. Acceleration is a vector quantity. A car starting from rest and reaching a velocity of 60 km h−1 north in 5 seconds has an average acceleration of 12 km h−1 per second or 12 (km h−1 )s−1 north. This is expressed in words as 12 km per hour per second. In simple terms, it means that the car increases its speed in a northerly direction by an average of 12 km h−1 each second. The average acceleration of an object, a⃗ av , can be expressed as: aav ⃗ =

∆v⃗ ∆t

The change in velocity during the time interval ∆t can be expressed as: ∆ v ⃗ = final velocity − initial velocity This is commonly also written as: ∆ v ⃗ = v ⃗ − u⃗ The direction of the average acceleration is the same as the direction of the change in velocity. SAMPLE PROBLEM 4

Spiro leaves home on his bicycle to post a letter. He starts from rest and reaches a speed of 10 m s−1 in 4 seconds. He then cycles at a constant speed in a straight line to a letterbox. He brakes at the letterbox, coming to a stop in 2 seconds, posts the letter and returns home at a constant speed of 8 m s−1 . On reaching home he brakes, coming to rest in 2 seconds. The direction away from home towards the letterbox is assigned as positive. a. What is Spiro’s average acceleration before he reaches his ‘cruising speed’ of 10 m s−1 on the way to the letterbox? b. What is Spiro’s average acceleration as he brakes at the letterbox? c. What is Spiro’s average acceleration as he brakes when arriving home? d. During which two parts of the trip is Spiro’s acceleration negative? e. Does a positive acceleration always mean that the speed is increasing? Explain. Teacher-led video: SP4 (tlvd-0072) THINK a. 1.

Recall the relationship.

WRITE a.

Average acceleration: aav =

2.

Identify the values.

Δv = 10 m s−1 Δt = 4 s

3.

Substitute into the relationship to find the average acceleration.

Average acceleration =

Δv Δt

10 m s−1 4s = 2.5 m s−2

TOPIC 9 Analysing motion 283

4.

b. 1.

State the solution.

Recall the relationship.

2.

Identify the values.

3.

Substitute into the relationship to find the average acceleration.

4.

State the solution.

c. 1.

Recall the relationship.

2.

Identify the values.

3.

Substitute into the relationship to find the average acceleration.

4.

State the solution.

d. 1. 2.

e. 1. 2.

Recall the positive and negative directions.

b.

c.

d.

Acceleration is negative when the velocity in the direction towards the house is increasing.

Recall the definition of acceleration. As speed does not take into account direction, an increasing speed does not necessarily equate to a positive acceleration.

e.

Spiro’s average acceleration on his journey to cruising speed is 2.5 m s−2 towards the letterbox. Δv Average acceleration: aav = Δt Δv = −10 m s−1 Δt = 2 s −10 m s−1 Average acceleration = 2s = −5 m s−2 Spiro’s average acceleration when braking at the letterbox is 5 m s−2 away from the letterbox. Δv Average acceleration: aav = Δt Δv = 8 m s−1 Δt = 2 s 8 m s−1 Average acceleration = 2 s = 4 m s−2 Spiro’s average acceleration when braking at the end of his journey is 4 m s−2 towards the letterbox. Positive direction is away from the house, negative direction is towards the house. The two sections of the trip where acceleration is negative are braking at the letterbox and accelerating back towards home from the letterbox. Acceleration is equal to change in velocity. A decreasing speed in the negative direction is a positive change in velocity, and hence a positive acceleration.

PRACTICE PROBLEM 4 a. A cheetah (the fastest land animal) takes 2 seconds to reach its maximum speed of 30 m s−1 . What is the magnitude of its average acceleration? b. A drag-racing car reaches a speed of 420 km h−1 from a standing start in 6 seconds. What is its average acceleration in: i. km h−1 s−1 ii. m s−2 ?

284 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

CHANGING DIRECTION A non-zero acceleration does not always result from a change in speed. Consider a car travelling at 60 km h−1 in a northerly direction turning right and continuing in an easterly direction at the same speed. Assume that the complete turn takes 10 seconds. The average acceleration during the time interval of 10 seconds is given by: aav =

∆v ∆t

The change in velocity must be determined first. Thus:

FIGURE 9.6 A change in acceleration can occur even if there is no change in speed.

∆v = v − u = v + −u The vectors v and −u are added together to give the resulting change in velocity. The magnitude of the change in velocity is calculated using Pythagoras’ theorem or trigonometric ratios to be 85 km h−1 . Alternatively, the vectors can be added using a scale drawing and then measuring the magnitude and direction of the sum. The direction of the change in velocity can be seen in figure 9.6 to be south-east. aav = =

u 60 km h−1 north

v 60 km h−1 east v 45°

∆v ∆t

−u Δv

45°

85 km h−1 south-east

10 s = 8.5 km h−1 s−1 south-east

In fact, the direction of the average acceleration is the same as the direction of the average net force on the car during the 10-second interval. The steering wheel is used to turn the wheels to cause the net force to be in this direction.

60 km h−1 south

Δv = v + (−u) N

W

E

S

9.2 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. State whether each of the following is a vector quantity or a scalar quantity. (a) Distance (b) Displacement (c) Speed (d) Velocity (e) Acceleration 2. The speed limit on Melbourne’s suburban freeways is 100 km h−1 . Express this speed in m s−1 . 3. Leisel Jones’s average speed while swimming a 100-metre breaststroke race is about 1.5 m s−1 . Calculate what her average speed would be in km h−1 . 4. The speed limit on US freeways is 55 miles per hour. One mile is approximately 1.6 km. Express this speed in: (a) km h−1 (b) m s−1 . 5. A jogger heads due north from his home and runs 400 metres along a straight footpath before realising he has forgotten his sunscreen and runs straight back to get it. (a) What distance has the jogger travelled by the time he gets back home? (b) What was the displacement of the jogger when he started to run back home? (c) What was his displacement when he arrived back home to pick up the sunscreen?

TOPIC 9 Analysing motion 285

6. The world records for some women’s track events (as at early 2019) are listed in the table 9.2. TABLE 9.2 World records for women’s track events

7.

8.

9.

10.

11.

Athlete

Event

Time

Florence Griffith Joyner (USA)

100 m

10.49 s

Florence Griffith Joyner (USA)

200 m

21.34 s

Marita Koch (German Democratic Republic)

400 m

47.60 s

Jarmilia Kratochvilova (Czechoslovakia)

800 m

1 min 53.28 s

Genzebe Dibaba (Ethiopia)

1 500 m

3 min 50.07 s

Wang Junxia (China)

3 000 m

8 min 06.11 s

Tirunesh Dibaba (Ethiopia)

5 000 m

14 min 11.15 s

Almaz Ayana (Ethiopia)

10 000 m

29 min 17.45 s

(a) Calculate the average speed (to three significant figures) of each of the athletes listed in the table. (b) Why is there so little difference between the average speeds of the world-record holders of the 100-metre and 200-metre events despite the doubling of the distance? (c) How long would it take Genzebe Dibaba to complete the marathon if she could maintain her average speed during the 1500-metre event for the entire 42.2 km course? (The world record for the women’s marathon, set on 13 April 2003, is 2 h 15 min 25 s.) (d) Which of the athletes in table 9.2 has an average speed that is the same as the magnitude of her average velocity? Explain. (a) A jogger takes 30 minutes to cover a distance of 5 km. What is the jogger’s average speed in: i. km h−1 ii. m s−1 ? (b) How long does it take for a car travelling 60 km h−1 to cover a distance of 200 metres? In 2018, cyclist Chloe Dygert, of the USA, set a world record of 3 min, 20.060 seconds for the 3000-metre pursuit. (a) What was her average speed? (b) How long would it take her to cycle from Melbourne to Bendigo, a distance of 151 km, if she could maintain her 3000-metre pursuit average speed for the whole distance? (c) How long does it take a car to travel from Melbourne to Bendigo if its average speed is 80 km h−1 ? (d) A car travels from Melbourne to Bendigo and back to Melbourne in 4 hours. i. What is its average speed? ii. What is its average velocity? Once upon a time, a giant tortoise had a bet with a hare that she could beat him in a foot race over a distance of 1 kilometre. The giant tortoise can reach a speed of about 7.5 cm s−1 . The hare can run as fast as 20 m s−1 . Both animals ran at their maximum speeds during the race. However, the hare was a rather arrogant creature and decided to have a little nap along the way. How long did the hare sleep if the result was a tie? An unfit Year 11 student arrives at school late and attempts to run from the front gate of the school to the physics laboratory. He runs the first 120 metres at an average speed of 6 m s−1 , the next 120 metres at an average speed of 4 m s−1 and the final 120 metres at an average speed of 2 m s−1 . What was the student’s average speed during his attempt to arrive at his class on time? While on holidays, a physics teacher drives her old Volkswagen from Melbourne to Wodonga, a distance of 300 kilometres. Her average speed was 80 km h−1 . She trades in her old Volkswagen and purchases a brand-new Toyota Prius. She proudly drives her new car back home to Melbourne at an average speed of 100 km h−1 . Calculate the average speed for the entire journey and explain any difference between the predicted and calculated average speed.

286 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

12. Determine (i) the change in speed and (ii) the change in velocity of each of the following situations. (a) The driver of a car heading north along a freeway at 100 km h−1 slows down to 60 km h−1 as the traffic gets heavier. (b) A fielder catches a cricket ball travelling towards him at 20 m s−1 . (c) A tennis ball travelling at 25 m s−1 is returned directly back to the server at a speed of 30 m s−1 . 13. A car travelling east at a speed of 10 m s−1 turns left to head south at the same speed. Has the car undergone an acceleration? Explain your answer with the aid of a diagram. 14. Estimate the acceleration of a car in m s−2 as it resumes its journey through the suburbs after stopping at traffic lights. 15. Use the data in table 9.2 (question 5) to help you estimate the average acceleration of a world-class 100-metre sprinter at the beginning of a race.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au.

studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

9.3 Analysing motion graphically KEY CONCEPTS • Analyse graphically, numerically and algebraically, straight-line motion under constant acceleration. • Graphically analyse non-uniform motion in a straight line.

The motion of an object often varies with time. To help analyse the motion of objects it can be very useful to construct graphs of the characteristics of that motion against time.

9.3.1 Position-versus-time graphs Bolter Beryl and Steady Sam decide to race each other on foot over a distance of 100 metres. They run due west. Timekeepers are instructed to record the position of each runner after each 3-second interval. TABLE 9.3 The progress of Bolter Beryl and Steady Sam Position (distance from starting line) in metres Time (seconds)

Bolter Beryl

Steady Sam

0.0

0

0

3.0

43

20

6.0

64

40

9.0

78

60

12.0

90

80

15.0

100

100

TOPIC 9 Analysing motion 287

FIGURE 9.7 The position-versus-time graph of the race provides valuable information about the motion of the two runners.

100

er yl rB

60

dy

Sa

m

Bo lte

Position (m) west

80

St

ea

40

20

0

3

6

9

12

15

Time (s)

The points indicating Bolter Beryl’s position after each 3-second interval are joined with a smooth curve. It is reasonable to assume that her velocity changes gradually throughout the race. A number of observations can be made from the graph of position versus time. • Both runners reach the finish at the same time. The result is a dead heat. Bolter Beryl and Steady Sam each have the same average speed and the same average velocity. • Steady Sam, who has an exceptional talent for steady movement, maintains a constant velocity throughout the race. In fact, his instantaneous velocity at every instant throughout the race is the same as his average velocity. Steady Sam’s average velocity and instantaneous velocity are both equal to the gradient of the position-versus-time graph since: ∆x ∆t 100 m west vav = 15 s rise vav = run vav = gradient vav =

Steady Sam’s velocity throughout the race is 6.7 m s−1 west. • Bolter Beryl, in her usual style, makes a flying start; however, after her initial ‘burst’, her instantaneous velocity decreases throughout the race as she tires. Her average velocity is also 6.7 m s−1 west. A more detailed description of Bolter Beryl’s motion can be given by calculating her average velocity during each 3-second interval of the race (see table 9.4).

288 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

TABLE 9.4 Bolter Beryl’s changing velocity Time interval (s)

Displacement Δx⃗ (m west)

Average velocity during interval ) Δx ⃗ ( m s−1 west v⃗av = Δt

0.0−3.0

43 − 0 = 43

14.0

3.0−6.0

64 − 43 = 21

7.0

6.0−9.0

78 − 64 = 14

4.7

9.0−12.0

90 − 78 = 12

4.0

12.0−15.0

100 − 90 = 10

3.3

The average velocity during each interval is the same as the gradient of the straight line joining the data points representing the beginning and end of the interval. An even more detailed description of Bolter Beryl’s run could be obtained if the race was divided into, say, 100 time intervals. The average velocity during each time interval (and the gradient of the line joining the data points defining it) would be a very good estimate of the instantaneous velocity in the middle of the interval. In fact, if the race is progressively divided into smaller and smaller time intervals, the average velocity during each interval would become closer and closer to the instantaneous velocity in the middle of the interval. The gradient of the tangent of a position-versus-time graph gives the instantaneous velocity of the object.

Position (m) west

Figure 9.8 shows how this process of using smaller time intervals can be used to find Bolter Beryl’s instantaneous velocity at an instant 4 seconds from the start of the race. Bolter Beryl’s instantaneous FIGURE 9.8 The first 9 seconds of Bolter Beryl’s run velocity is not the same as the average velocity during the 3-second to Tangent 6-second time interval shown in table 9.4. However, it can be estimated by Q drawing the line AD and finding its 80 gradient. The gradient of the line BC would provide an even better estimate of the instantaneous velocity. If you continue this process of decreasing C D 60 the time interval used to estimate the instantaneous velocity, you will eventually obtain a line which is a B tangent to the curve. The gradient of 40 P the tangent to the curve is equal to the instantaneous velocity at the instant A represented by the point at which it meets the curve. 20 The gradient of the tangent to the curve at 4 seconds in figure 9.8 can be determined by using the points P and Q. 0 1

2

3

4

5

6

7

8

9

Time (s)

TOPIC 9 Analysing motion 289

rise run (84 − 36) m = (8 − 2) s 48 m = 6s = 8 m s−1

gradient =

Bolter Beryl’s instantaneous velocity at 4 seconds from the start of the race is therefore 8 m s−1 west. Just as the gradient of a position-versus-time graph can be used to determine the velocity of an object, a graph of distance versus time can be used to determine its speed. Because Bolter Beryl and Steady Sam were running in a straight line and in one direction only, their distance from the starting point is the magnitude of their change in position. Their speed is equal to the magnitude of their velocity. The gradient of the tangent of a distance-versus-time graph gives the instantaneous speed of the object.

Resources Digital document eModelling: Numerical model of motion 1: finding speed from position-time data (doc-0048)

9.3.2 Velocity-versus-time graphs

TABLE 9.5 Beryl’s velocity during the race Time (s)

Velocity (m s−1 west)

0.0

18.0

2.0

12.0

4.0

8.0

6.0

5.4

8.0

4.7

10.0

4.2

12.0

3.5

14.0

3.1

Velocity (m s−1 west)

The race between Bolter Beryl and Steady Sam described by the position-versus-time graph in figure 9.7 can also be described by a graph of velocity versus time. Steady Sam’s velocity is 6.7 m s−1 due west throughout the race. The curve describing Bolter Beryl’s motion can be plotted by determining the instantaneous velocity at various times during the race. FIGURE 9.9 Velocity-versus-time graph for This can be done by drawing tangents at a number of the race points on the position-versus-time graph in figure 9.7. Table 9.5 shows the data obtained using this method. The velocity-versus-time graph in figure 9.9 describes 20 the motion of Bolter Beryl and Steady Sam.

15

‘Bolter’ Beryl

10

‘Steady’ Sam 5

0 2

4

6

8 Time (s)

290 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

10

12

14

The velocity-versus-time graph confirms what you already knew by looking at the position-versus-time graph, namely that: • Steady Sam’s velocity is constant and equal to his average velocity • the magnitude of Bolter Beryl’s velocity is decreasing throughout the race. The velocity-versus-time graph allows you to estimate the velocity of each runner at any time. It provides a much clearer picture of the way that Bolter Beryl’s velocity changes during the race, namely that: • the magnitude of her velocity decreases rapidly at first, but less rapidly towards the end of the race • for most of the duration of the race, she is running more slowly than Sam. In fact Bolter Beryl’s speed (the magnitude of her velocity) drops below that of Steady Sam’s after only 4.7 seconds.

Displacement from a velocity-versus-time graph In the absence of a position-versus-time graph, a velocity-versus-time graph provides useful information about the change in position, or displacement, of an object. Steady Sam’s constant velocity, the same as his average velocity, makes it very easy to determine his displacement during the race. ) ( ∆x ∆x = vav ∆t since vav = ∆t −1 ∆x = 6.7 m s west × 15 s ∆x = 100 m west This displacement is equal to the area of the rectangle under the graph depicting Steady Sam’s motion. area = length × width = 15 s × 6.7 m s−1 west = 100 m west Because the race was a dead heat, Bolter Beryl’s average velocity was also 6.7 m s−1 . Her displacement during the race can be calculated in the same way as Steady Sam’s. ∆x = vav ∆t = 6.7 m s−1 × 15 s = 100 m west However, Bolter Beryl’s displacement can also be found by calculating the area under the velocityversus-time graph depicting her motion. This can be done by ‘counting squares’ or by dividing the area under the graph into rectangles and triangles. The area under Beryl’s velocity-versus-time graph is, not surprisingly, 100 metres. In fact, the area under any part of the velocity-versus-time graph is equal to the displacement during the interval represented by that part. The area under a velocity-versus-time graph gives the displacement (or change in position) of the object during that time interval. To determine the position of the object you must know its starting position. Note: If the velocity-versus-time graph is in the negative velocity range, you can get a negative value for the area. Although this may seem counterintuitive, it indicates a negative displacement. The area under a speed-versus-time graph gives the distance travelled by the object during that time interval.

TOPIC 9 Analysing motion 291

CALCULATING THE AREA UNDER A GRAPH WITH TIME INTERVALS When an object travels with a constant velocity, it is obvious that the displacement of the object is equal to the area under a velocity-versus-time graph of its motion. However, it is not so obvious when the motion is not constant. The graphs below describe the motion of an object that has an increasing velocity. The motion of the object can be approximated by dividing it into time intervals of ∆t and assuming that the velocity during each time interval is constant. The approximate displacement during each time interval is equal to ∆ x = vav ∆t, which is the same as the area under each rectangle. The approximate total displacement is therefore equal to the total area of the rectangles.

0

Time (s)

0

Time (s)

etc.

0

Time (s)

Velocity (m s−1)

Velocity (m s−1)

Δt

Velocity (m s−1)

Velocity (m s−1)

FIGURE 9.10 By dividing the velocity-versus-time graph into rectangles representing small time intervals, the displacement can be estimated.

0

Time (s)

To better approximate the displacement, the graph can be divided into smaller time intervals. The total area of the rectangles is approximately equal to the displacement. By dividing the graph into even smaller time intervals, even better estimates of the displacement can be made. In fact, by continuing the process of dividing the graph into smaller and smaller time intervals, it can be seen that the displacement is, in fact, equal to the area under the graph.

SAMPLE PROBLEM 5

At what time in the race between Bolter Beryl and Steady Sam did Bolter Beryl’s speed drop below Steady Sam’s speed? What is Steady Sam’s displacement at this point in time? THINK

WRITE

See the velocity-versus-time graph (figure 9.9) to determine when Bolter Beryl’s speed dropped below Steady Sam’s. 2. Determine Steady Sam’s displacement by calculating the area under the graph.

The time at which Bolter Beryl’s speed dropped below Steady Sam’s is approximately 4.7 seconds. Area under the graph = 4.7 × 6.7 = 3.1 m Steady Sam’s displacement was 31 metres at 4.7 seconds.

1.

292 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

PRACTICE PROBLEM 5 The following graph is of a section of figure 9.9 showing the first 4.7 seconds of Bolter Beryl’s motion.

20

Velocity (m s−1 west)

Bolter Beryl 15

10

5

0 1

2

3

4

5

Time (s)

a. Estimate Bolter Beryl’s displacement after 2 seconds. b. At 4.7 seconds into the race, determine how far ahead Bolter Beryl was of Steady Sam (whose velocity was a constant 6.7 m s–1 ).

Acceleration from a velocity-versus-time graph The graph in figure 9.11 describes the motion of an elevator as it moves from the ground floor to the top floor and back down again. The elevator stops briefly at the top floor to pick up a passenger. For convenience, any upward displacement from the ground floor is defined as positive. The graph has been divided into seven sections labelled A to G. Finding the acceleration The acceleration at any instant during the motion can be determined by calculating the gradient of the graph. This is a consequence of the definition of acceleration. The gradient of a velocity-versus-time graph is a measure of the rate of change of velocity just as the gradient of a position-versus-time graph is a measure of the rate of change of position. The gradient of a velocity-versus-time graph gives the acceleration of the object. Throughout interval A (see figure 9.11), the acceleration, a, of the elevator is: rise run +8.0 m s−1 = 5.0 s = +1.6 m s−2 or 1.6 m s−2 up

a=

TOPIC 9 Analysing motion 293

FIGURE 9.11 The motion of an elevator

15

10

B

Velocity (m s−1)

A

C

5 D

0 5

10

15

−5

20

25

30

35

E

−10

40

Time (s)

G

F

−15

During intervals B, D and F, the velocity is constant and the gradient of the graph is zero. The acceleration during each of these intervals is, therefore, zero. Throughout interval C, the acceleration is: −8.0 m s−1 2.5 s = −3.2 m s−2 or 3.2 m s−2 down

a=

Throughout interval E, the acceleration is: −12 m s−1 2.5 s = −4.8 m s−2 or 4.8 m s−2 down

a=

Throughout interval G, the acceleration is: +12 m s−1 a= 5.0 s = +2.4 m s−2 or 2.4 m s−2 up Notice that during interval G the acceleration is positive (up) while the velocity of the elevator is negative (down). The direction of the acceleration is the same as the direction of the change in velocity. Finding the displacement As explained previously, the area under the graph is equal to the displacement of the elevator. Dividing the area into triangles and rectangles and working from left to right yields an area of:

294 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

) ( ) ( ) 1 1 −1 −1 −1 × 5.0 s × 8.0 m s + 12.5 s × 8.0 m s + × 2.5 s × 8.0 m s + Area = 2 2 ( ) ( ) ) ( 1 1 × 2.5 s × −12 m s−1 + 7.5 s × −12 m s−1 + × 5.0 s × −12 m s−1 2 2 = 20 m + 100 m + 10 m − 15 m − 90 m − 30 m = −5 m (

This represents a downwards displacement of 5 metres, which is consistent with the elevator finally stopping two floors below the ground floor.

Resources Digital documents. eModelling: Numerical model of motion 2. Finding position from speed–time data (doc-0049) Investigation 9.2 Let’s play around with some graphs (doc-31875) eModelling: Numerical model for acceleration (doc-0050) Video eLessons

Motion with constant acceleration (eles-0030) Ball toss (eles-0031)

Teacher-led video Investigation 9.2 Let’s play around with some graphs (tlvd-0820)

9.3.3 Acceleration-versus-time graphs The graph in figure 9.12 represents the acceleration of the elevator described in the previous section. FIGURE 9.12 An acceleration-versus-time graph for the elevator

4.8 4.0 3.2 2.4 G Acceleration (m s−2 )

1.6 A 0.8 B

F

D

0 5

10

15

20

25

30

35

40

Time (s)

−0.8 −1.6 −2.4 C −3.2 −4.0 E −4.8

TOPIC 9 Analysing motion 295

The area under an acceleration-versus-time graph gives the change in velocity of the object during that time interval. To determine the velocity of the object you must know its starting velocity. Just as the area under a velocity-versus-time graph is equal to the change in position of an object, the area under an acceleration-versus-time graph is equal to the change in velocity of an object. The area under the part of the graph representing the entire upwards part of the journey is given by: Area A + Area B + Area C = 5.0 s × 1.6 m s−2 + 0 + 2.5 s × −3.2 m s−2 = + 8.0 m s−1 + −8.0 m s−1 =0 This indicates that change in velocity during the upward journey is zero. This is consistent with the fact that the elevator starts from rest and is at rest when it reaches the top floor. Similarly, the area under the whole graph is zero. The change in velocity during intervals C, D and E is given by the sum of areas C, D and E. Thus: ( ) ( ) Area C + Area D + Area E = 2.5 s × −3.2 m s−2 + 0 + 2.5 s × −4.8 m s−2 = 8.0 m s−1 + −12 m s−1 = −20 m s−1 The change in velocity is −20 m s−1 , or 20 m s−1 down. At the beginning of time interval C, the velocity was 8 m s−1 upwards. A change of velocity of −20 m s−1 would result in a velocity at the end of interval E of 12 m s−1 downwards. This is consistent with the description of the motion in the velocity-versus-time graph in figure 9.11. The change in velocity during interval G is 5 s × 2.4 m s−1 , or 12 m s−1 upwards, which is consistent with the elevator coming to rest at the end of its journey.

9.3.4 Working with motion graphs Position-versus-time graphs

• •

The instantaneous velocity of an object can be obtained from a graph of the object’s position versus time by determining the gradient of the curve at the point representing that instant. This is a direct consequence of the fact that velocity is a measure of the rate of change of position. Similarly, the instantaneous speed of an object can be obtained by determining the gradient of a graph of the object’s distance travelled from a reference point versus time.

Velocity-versus-time graphs

• • •

The displacement of an object during a time interval can be obtained by determining the area under the velocity-versus-time graph representing that time interval. The actual position of an object at any instant during the time interval can be found only if the starting position is known. Similarly, the distance travelled by an object during a time interval can be obtained by determining the corresponding area under the speed-versus-time graph for the object. The instantaneous acceleration of an object can be obtained from a graph of the object’s velocity versus time by determining the gradient of the curve at the point representing that instant. This is a direct consequence of the fact that acceleration is defined as the rate of change of velocity.

Acceleration-versus-time graphs

•

The change in velocity of an object during a time interval can be obtained by determining the area under the acceleration-versus-time graph representing that time interval. The actual velocity of the object can be found at any instant during the time interval only if the initial velocity is known.

296 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

TABLE 9.6 Summary of motion graphs Position-versustime graphs

Velocity-versustime graphs

Accelerationversustime graphs

Quantities that you can read directly from the graph

Horizontal axis

Time

Time

Time

Vertical axis

Position

Velocity

Acceleration

Quantities that you can calculate from the graph

Gradient of tangent

Instantaneous velocity

Instantaneous acceleration

Area under the graph

Change in position (displacement)

Change in velocity

Resources Digital document Investigation 9.3: On your bike or on your own two feet (doc-31876) Weblink

Constant acceleration app

9.3 EXERCISE

1. The position-versus-time graph shown describes the motion of five different objects, labelled A to E. (a) Which two objects start from the same position, but at different times? (b) Which two objects start at the same position at the same time? (c) Which two objects are travelling at the same speed as each other, but with different velocities? (d) Which two objects are moving towards each other for the whole period shown on the graph? (e) Which of the five objects has the lowest speed?

Position

To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au.

B

A

C D

Time

E

2. (a) Describe in words the motion shown for each of scenarios A, B and C shown on the right. (b) Copy the incomplete graphs for each scenario into your workbooks and then complete each graph.

A

B

x

0

t

v

0

C

x

x

0

0

v

v

0

0

a

a

0

0

t

a

0 t

TOPIC 9 Analysing motion 297

3. Sketch a velocity-versus-time graph to illustrate the motion described in each of the following situations. (a) A bicycle is pedalled steadily along a road. The cyclist stops pedalling and allows the bicycle to come to a stop. (b) A parachutist jumps out of a plane and opens his parachute midway through the fall to the ground. (c) A ball is thrown straight up into the air and is caught at the same height from which it was thrown. 4. Sketch a position-versus-time graph for each scenario in question 3. 5. The graph in the following figure is a record of the straight-line motion of a skateboard rider during an 80-second time interval. The time interval has been divided into sections labelled A to E. The skateboarder initially moves north from the starting point. (a) During which section of the Position (m) interval was the skateboard rider stationary? 80 (b) During which sections of the interval was the skateboarder travelling 60 north? (c) At what instant did the skateboard 40 rider first move back towards the starting line? 20 (d) What was the displacement of the skateboarder during the 80-second C D Starting 0 interval? point 10 20 30 40 50 60 70 80 Time (s) (e) What distance did the skateboarder A B −20 travel during the 80-second interval? E (f) During which section of the interval −40 was the skateboard rider speeding up? (g) During which section of the interval was the skateboard rider slowing down? (h) What was the skateboarder’s average speed during the entire 80-second interval? (i) What was the velocity of the skateboarder throughout section C? (j) Estimate the velocity of the skateboarder 65 seconds into the interval. 6. The figure that follows is a record of the motion of a battery-operated toy robot during an 80-second time interval. The interval has been divided into sections labelled A to G. (a) During which sections is the Velocity (m s−1) acceleration of the toy robot zero? 1.0 (b) What is the displacement of the toy robot during the 80-second 0.75 interval? (c) What is the average velocity of 0.50 the toy robot during the entire interval? 0.25 (d) At what instant did the toy robot first reverse direction? Starting 0 (e) At what instant did the toy robot point first return to its starting point? −0.25 (f) During which intervals did the toy robot have a negative acceleration? −0.50 (g) During which intervals did the toy robot decrease its speed? (h) Explain why your answers to (f) and (g) −0.75 are different from each other. (i) What is the acceleration of the toy robot −1.0 throughout section E? (j) What is the average acceleration during the first 20 seconds? (k) Describe the motion of the toy robot in words.

298 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

D

C A

10

B

20

30

40

E 50

F 60

G 70

80

Time (s)

Acceleration (m s−2)

4

r

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au.

6

Ca

7. The figure that follows compares the straight-line motion of a jet ski and a car as they each accelerate from an initial speed of 5 m s−1 . (a) Which is first to reach a constant speed — the jet ski or the car — and when does this occur? (b) What is the final speed of: i. the jet ski ii. the car? (c) Draw a speed-versus-time graph describing the motion of either the jet ski or the car.

Je

ts

ki

2

0 4

2

6

8 10

Time (s)

studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

9.4 Equations for constant acceleration KEY CONCEPT • Analyse numerically and algebraically, straight-line motion under constant acceleration.

In the absence of a graphical representation, a number of formulae can be used to describe straight-line motion as long as the acceleration is constant. These formulae are expressed in terms of the quantities used to describe such motion. The terms are: • initial velocity, u • final velocity, v • acceleration, a • time interval, t • displacement, s. Because the formulae describe motion along a straight line, vector notation is not necessary. The displacement, velocity and acceleration can be expressed as positive or negative quantities.

9.4.1 Deriving the equations algebraically The first formula is found by restating the definition of acceleration. a=

∆v ∆t

Where: ∆v = the change in velocity ∆t = the time interval. Thus: v−u t ⇒ v − u = at ⇒ v = u + at a=

[1] TOPIC 9 Analysing motion 299

The second formula is found by restating the definition of average velocity. vav =

∆s ∆t

Where: ∆s = the change in position. But:

Thus:

vav =

u+v 2

s u+v = 2 t 1 ⇒ s = (u + v) t 2

[2]

Three more formulae are obtained by combining formulae [1] and [2]. 1 (u + u + at) t 2 1 = (2u + at) t 2 ) ( 1 = u + at t 2 1 2 ⇒ s = ut + at 2 s=

1 (v − at + v) t 2 1 = (2v − at) t 2 ) ( 1 = v − at t 2 1 2 ⇒ s = vt − at 2 s=

(substituting v = u + at from formula [1] into formula [2])

[3] (substituting u = v − at from formula [1] into formula [2])

[4]

A final formula can be found by eliminating t from formula [2].

But:

s=

1 (u + v) t 2

t=

v−u . a

300 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

(formula [2])

(rearranging formula [1])

( ) 1 v−u ⇒ s = (u + v) a 2 1 v2 − u2 (expanding the difference of two squares) = 2 a ⇒ 2as = v2 − u2 ⇒ v2 = u2 + 2as

[5]

In summary, the formulae for straight-line motion under constant acceleration are: v = u + at

[1]

1 (u + v) t [2] 2 1 s = ut + at2 [3] 2

s=

1 s = vt − at2 2

[4]

v2 = u2 + 2as [5]

9.4.2 Deriving the equations graphically Each of the five formulae derived here allow you to determine an unknown characteristic of straight-line motion with a constant acceleration as long as you know three other characteristics. Although the formulae have not been derived from graphs, they are entirely consistent with a graphical approach.

C

D

vf

Velocity

acceleration = gradient of velocity-versus-time graph rise = run v−u = t v−u ⇒ a= t ⇒ v = u + at [1]

FIGURE 9.13 A velocity-versus-time graph for an object travelling in a straight line with constant acceleration

v−u

ui

E

B

A

displacement = area under graph = area of trapezium ABDF 1 = (u + v) t 2

F Time

t

[2]

TOPIC 9 Analysing motion 301

displacement = area under graph = area of rectangle ABEF + area of triangle BDE 1 (v − u = at from [1]) = ut + t × at 2 1 = ut + at2 2

[3]

displacement = area under graph = area of rectangle ACDF − area of triangle BCD 1 (v − u = at from [1]) = vt − × t × at 2 1 = vt − at2 2

[4]

Formula [5] can be derived by combining formula [1] with any of formulae [2], [3] or [4].

9.4.3 Problem-solving steps The following steps may help you when solving problems using the constant acceleration equations. 1. Sketch the situation. 2. Identify known quantities (converting to consistent units if necessary). 3. Identify the quantity to be determined. 4. Select an appropriate formula. 5. Substitute values and solve. It is possible to rearrange each of the equations to make different variables the subject. Table 9.7 summarises all possible versions of the equations and may be useful when solving problems. TABLE 9.7 Equations for solving problems

Variable that is to be calculated

Variables that are involved in the problem

u

v

uvat

uvas

u = v − at

u2 = v2 − 2as

u=

v = u + at

v2 = u2 + 2as

v=

a

a=

v−u t

t

t=

v−u a

s

–

a=

v 2 − u2 2s –

s=

uvts

v2 − u2 2a

302 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

2s t 2s t

−v

s=

(u + v) (u + v) t

vats

at s − t 2

–

v=

–

a= 2s

1 2

u=

−u

–

t=

uats

2 (s − ut) t2

a=

at s + t 2

2 (vt − s) t2

Find v then solve

Find u then solve

s = ut + 12 at2

s = vt − 12 at2

SAMPLE PROBLEM 6

Ying drops a coin into a wishing well and takes 3.0 seconds to make a wish. The coin splashes into the water just as she finishes making her wish. The coin accelerates towards the water at a constant 9.8 m s−2 . a. What is the coin’s velocity as it strikes the water? b. How far does the coin fall before hitting the water? Teacher-led video: SP6 (tlvd-0074)

THINK

WRITE

Recall the appropriate constant acceleration formula. 2. Identify the values. 3. Substitute into the formula to find the final velocity. 4. State the solution.

a. 1.

Recall the appropriate constant acceleration formula. 2. Identify the values.

b. 1.

3.

Substitute into the formula to find the distance.

4.

State the solution.

a.

v = u + at

u = 0, a = 9.8 m s−2 , t = 3.0 s v = 0 + 9.8 × 3.0 = 29 m s−1 The coin had a velocity of 29 m s−1 when it hit the water. 1 b. s = ut + at2 2 u = 0, a = 9.8 m s−2 , t = 3.0 s 1 s = 0 × 3.0 + × 9.8 × 3.02 = 44 m 2 The coin fell 44 metres before hitting the water.

Resources Digital document eModelling: Solving problems with a graphics calculator (doc-0051)

PRACTICE PROBLEM 6 A parked car with the handbrake off rolls down a hill in a straight line with a constant acceleration of 2 m s−2 . It stops after colliding with a brick wall at a speed of 12 m s−1 . a. For how long was the car rolling? b. How far did the car roll before colliding with the wall?

SAMPLE PROBLEM 7

The driver of a car was forced to brake in order to prevent serious injury to a neighbour’s cat. The car skidded in a straight line, stopping just 2 centimetres short of the startled but lucky cat. The driver (who happened to be a physics teacher) measured the length of the skid mark to be 12 metres. His passenger (also a physics teacher with an exceptional skill for estimating small time intervals) estimated that the car skidded for 2 seconds. a. At what speed was the car travelling as it began to skid? b. What was the acceleration of the car during the skid? Teacher-led video: SP7 (tlvd-0075)

TOPIC 9 Analysing motion 303

THINK

Recall the appropriate constant acceleration formula. 2. Identify the values.

a. 1.

3.

Substitute into the relationship to find the initial velocity.

4.

State the solution.

Recall the appropriate constant acceleration formula. 2. Identify the values.

b. 1.

3.

4.

WRITE a.

s=

1 (u + v) t 2

v = 0, t = 2 s, s = 12 m 1 12 = (u + 0) × 2 2 u = 12 m s−1 The initial speed of the car was 12 m s−1 . 1 b. s = vt − at2 2

Substitute into the relationship to find the acceleration.

v = 0, t = 2 s, s = 12 m 1 12 = 0 × 2 − a × 22 2 12 = −2a

State the solution.

a = −6 m s−2 The acceleration during the skid was −6 m s−2 ; that is, the car was decelerating at 6 m s−2 during the skid.

PRACTICE PROBLEM 7 A car travelling at 24 m s−1 brakes to come to a stop in 1.5 seconds. If its acceleration (deceleration in this case) was constant, what was the car’s: a. stopping distance b. acceleration? It is worth noting that sample problems 6 and 7 could both have been solved without the use of the constant acceleration formulae. Both examples could have been completed with a graphical approach and a clear understanding of the definitions of velocity and acceleration. Go ahead and try to answer both problems without the formulae.

9.4 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. How long does it take for: (a) a car to accelerate on a straight road at a constant 6 m s−2 from an initial speed of 60 km h−1 (17 m s−1 ) to a final speed of 100 km h−1 (28 m s−1 ) (b) a downhill skier to accelerate from rest at a constant 2 m s−2 to a speed of 10 m s−1 ? 2. In Acapulco, on the coast of Mexico, professional high divers plunge from a height of 36 metres above the water. (The highest diving boards used in Olympic diving events are 10 metres above the water.) Assuming that throughout their dive, the divers are falling vertically from rest with an acceleration of 9.8 m s−2 , estimate: (a) the length of the time interval during which the divers fall through the air (b) the speed with which the divers enter the water. 3. A skateboard rider travelling down a hill notices the busy road ahead and comes to a stop in 2.0 seconds over a distance of 12 metres. Assume a constant negative acceleration. (a) What was the initial speed of the skateboarder? (b) What was the acceleration of the skateboarder as they came to a stop? 304 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

4. A car is travelling at a speed of 100 km h−1 when the driver sees a large fallen tree branch in front of her. At the instant that she sees the branch, it is 50 metres from the front of her car. The car travels a distance of 48 metres after the brakes are applied before coming to a stop. (a) What is the average acceleration of the car while the car is braking? (b) How long does the car take to stop once the brakes are applied? (c) What other information do you need in order to determine whether the car stops before it hits the branch? Make an estimate of the missing item of information to predict whether or not the car is able to stop in time. 5. A dancer in a school musical is asked to leap 80 centimetres into the air, taking off vertically on one beat of the music and landing with the next beat. If the music beats every 0.5 seconds, is the leap possible? The acceleration of the dancer during the leap can be assumed to be 9.8 m s−2 downwards. 6. A brand-new Rolls Royce rolls off the back of a truck as it is being delivered to its owner. The truck is travelling along a straight road at a constant speed of 60 km h−1 . The Rolls Royce slows down at a constant rate, coming to a stop over a distance of 240 metres. It is a full minute before the truck driver realises that the precious load is missing. The driver brakes immediately, leaving a 25-metre skid mark on the road. The driver’s reaction time (time interval between noticing the problem and depressing the brake) is 0.5 seconds. How far behind is the Rolls Royce when the truck stops? 7. A girl at the bottom of a 100-metre high cliff throws a tennis ball vertically upwards. At the same instant a boy at the very top of the cliff drops a golf ball so that it hits the tennis ball while both balls are still in the air. The acceleration of both balls can be assumed to be 9.8 m s−2 downwards. (a) With what speed is the tennis ball thrown so that the golf ball strikes it at the top of its path? (b) What is the position of the tennis ball when the golf ball strikes it?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

9.5 Review 9.5.1 Summary • • •

Scalar quantities have a magnitude (size) only. Vector quantities have both a magnitude and a direction. Displacement is a measure of the change in position of an object. Displacement is a vector quantity. Speed is the rate at which distance changes over time and is a scalar quantity. Velocity is the rate at which displacement changes over time and is a vector quantity. distance travelled time interval displacement average velocity = time interval average speed =

The average velocity of an object, vav , during a time interval, t, can be expressed as: vav =

•

∆s ∆t

Instantaneous speed is the speed at a particular instant of time. Instantaneous velocity is the velocity at a particular instant of time. TOPIC 9 Analysing motion 305

•

• • •

• •

The change in velocity during the time interval ∆t can be expressed as:

∆v = final velocity − initial velocity = v − u

The instantaneous velocity of an object can be found from a graph of its displacement versus time by calculating the gradient of the graph. Similarly, the instantaneous speed can be found from a graph of distance versus time by calculating the gradient of the graph. The displacement of an object during a time interval can be found by determining the area under its velocity-versus-time graph. Similarly, the distance travelled by an object can be found by determining the area under its speed-versus-time graph. Acceleration is the rate at which an object changes its velocity. Acceleration is a vector quantity. The ∆v where ∆v = the change in average acceleration of an object, aav , can be expressed as aav = ∆t velocity during the time interval ∆t. The instantaneous acceleration of an object can be found from a graph of its velocity versus time by calculating the gradient of the graph. When acceleration of an object is constant, the following formulae can be used to describe its motion. v = u + at s=

1 (u + v) t 2 1 s = ut + at2 2 1 s = vt − at2 2 v2 = u2 + 2as

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0034).

9.5.2 Key terms Acceleration is the rate at which an object changes its velocity. Acceleration is a vector quantity. Displacement is a measure of the change in position of an object. It is a vector quantity. Distance is a measure of the length of the path taken by an object. It is a scalar quantity. Instantaneous speed is the speed at a particular instant of time. Instantaneous velocity is the velocity at a particular instant of time. Scalar quantities specify magnitude (size) but not direction. Speed is the rate at which distance changes over time. Speed is a scalar quantity. A vector quantity specifies direction as well as magnitude (size). Velocity is the rate at which displacement changes over time, or the rate of change in position. Velocity is a vector quantity.

Resources Digital document Key terms glossary (doc-33010)

306 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

9.5.3 Practical work and investigations Investigation 9.1 Going home Aim: To analyse your journey from home to school Digital document: doc-31874

Investigation 9.2 Let’s play around with some graphs Aim: To demonstrate the motion represented by position-versus-time graphs and velocity-versus-time graphs Digital document: doc-31875 Teacher-led video: tlvd-0820

Investigation 9.3 On your bike or on your own two feet Aim: To record and analyse the motion of a bicycle or runner over a distance of 100 metres on a straight track Digital document: doc-31876 Other practical work ideas: • Place a small ruler with one end sitting over the edge of a desk. Hit the end so that it flies through the air. How far does it travel horizontally? What factors might affect this, and how? Investigate. • How does the initial acceleration of a sprinter depend on the spacing between their feet on the blocks? • Fill a bottle with some liquid. Lay it on its side and give it a push. The bottle may first move forward and then oscillate before it comes to rest. Investigate the bottle’s motion. • Make a small parachute from a piece of cloth, lengths of cotton and Blu-Tack. Hold it in the air and drop it with the canopy open. The parachute accelerates, then maintains a steady speed. Investigate the motion and what factors affect the initial acceleration and the final speed.

Resources Digital document Practical investigation logbook (doc-32267)

9.5 Exercises To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au.

9.5 Exercise 1: Multiple choice questions Which of the following does a vector quantity have? A. Size (magnitude) only B. Direction only C. Both size (magnitude) and direction D. None of the above 2. A track cyclist warming up for a race completes three laps of a 250-metre velodrome track. What is their displacement? A. 0 m B. 250 m C. 500 m D. 750 m 1.

TOPIC 9 Analysing motion 307

The Airbus A380 aircraft has a cruising speed 1060 km h–1 . How many metres does it travel in 1 second when cruising at this speed? A. 3816 m B. 1060 m C. 294 m D. 106 m 4. A hairy-nosed wombat that is feeling threatened runs as fast as its legs can take it. At its top speed it travels 21.0 metres in 1.90 seconds. What is its average velocity during this interval? A. 0.09 m s−1 B. 39.9 m s−1 C. 1.11 m s−1 D. 11.1 m s−1 5. A car is travelling at 12 m s-1 . How fast is this in km h-1 ? A. 3.3 km h−1 B. 43.2 km h−1 C. 33.3 km h−1 D. 4.3 km h−1 6. A Tesla Model S accelerates from a standstill to a speed of 97 km h−1 in 2.28 seconds. What is its average acceleration during this time interval? A. 12 m s−2 B. 43 m s−2 C. 153 m s−2 D. 27 m s−2 3.

Use the following graph, describing the motion of an elevator, to answer questions 7 and 8.

15

10

B

Velocity (m s−1)

A

C

5 D

0 5

10

15

−5

20

25

30 E

−10

−15

7.

The acceleration of the elevator is zero at which intervals? A. Intervals E, F and G B. Interval D C. Intervals B, D and F D. Intervals B and F

308 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

35

40 G

F

Time (s)

The accelerator reaches the maximum speed for the time period shown during which interval? A. Interval B B. Interval C C. Interval E D. Interval F 9. A student drops a stone into a wishing well (starting at rest). It falls at a rate of 9.80 m s−2 . The student hears the stone hit the water 2.00 seconds after they dropped it. How far down the well did the stone travel from the student’s hand until it hit the water? A. 9.80 m B. 19.6 m C. 39.2 m D. 78.4 m 10. A shopping trolley rolls down a hill in a straight line with a constant acceleration of 3.00 m s−2 . It starts at rest beside a car and travels 100 metres before it collides with a picket fence. How fast is it travelling the instant before it collides with the fence? A. 17.3 m s−1 B. 600 m s−1 C. 24.5 m s−1 D. 300 m s−1 8.

9.5 Exercise 2: Short answer questions A student is doing a ‘shuttle run’ activity where they run backwards and forwards along a straight line that runs north–south. They run 10 metres north, then 5 metres south, then 7 metres north, then 9 metres south. What distance have they travelled? What is their displacement? 2. An engineer takes a direct flight from Munich to Detroit on an Airbus A350. The flight takes 10 hours and 30 minutes and covers a distance of 6980 kilometres. What is the average speed of the aircraft during this flight? 3. A shuttlecock in a game of badminton is travelling at 330 km h−1 when it strikes the racket of an opposing player and rebounds at 264 km h−1 . Assuming that the collision takes 0.25 seconds. Determine the average acceleration that it experiences during the collision with the racket. 4. The following graph records the straight-line motion of a skateboarder. Calculate the velocity of the skateboarder during section A. 1.

Position (m)

80 60 40 20 Starting 0 point

C 10 A

20

30 B

40

50

D 60

70

80

Time (s)

−20 E −40

TOPIC 9 Analysing motion 309

5.

The following velocity-versus-time graph was sketched by a physics student from data collected during a practical investigation. Determine the change in position of the object during the 20-second motion.

v (m s–1)

30 20 10 0 5

10

20 t (s)

15

−10

The student is analysing their data presented in question 5. Use the graph to estimate the largest magnitude acceleration that the object experiences during this motion. 7. A physics student records the acceleration versus time for a dynamics trolley colliding with a bumper. Use the following graph to estimate the change in velocity of the trolley during the collision. 6.

a (m s–2 )

1.5 1.0 0.5 0 0.1

0.2

0.3

0.4

0.5

0.6

−0.5 −1.0 −1.5 −2.0 −2.5 −3.0 −3.5

310 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

0.7

0.8

0.9

1.0

1.1

1.2

1.3

1.4

1.5

1.6 t (s)

A physics student has their very first experience of cross-country skiing on a school camp. They tentatively ski in a straight line down a gentle slope, with constant acceleration, starting from rest. They travel 50 metres in 40 seconds. What was their final speed? 9. A model rocket is launched directly upwards from a school oval. The engine switches off when it is 45 metres above the ground and travelling at 90 m s−1 directly upwards. Assuming that it continues to travel directly upwards with a constant downwards acceleration due to gravity of 9.8 m s−2 . What is the total height above the oval that the rocket reaches? 10. A car travelling in a 40 km h−1 speed limit zone brakes hard to avoid hitting a kangaroo that hops onto the road. It comes to a complete stop 4.5 seconds after the brakes were applied and left a skid mark that was 27 metres long. Was the driver travelling faster than the speed limit when they braked? Use a calculation to support your answer. 8.

9.5 Exercise 3: Exam practice questions Question 1 (1 mark) Jo is jogging around a track at a leisurely constant speed. He is joined by a friend who matches his speed and challenges him to a race to the finish. Jo accelerates to run as fast as he can and then continues at this speed until the finish line. Sketch a position-versus-time graph of Jo’s motion. Question 2 (1 mark) A cyclist makes a left turn around a corner, keeping their speed constant. Do they experience an acceleration during this motion? Justify your response. Question 3 (5 marks) Two runners, Alex and Bo, are exchanging a baton in a 4 × 100-metre track relay. Their motion is shown in the following velocity-versus-time graph. Baton exchange v (m s–1) Alex

Bo

8 7

0

a. b.

1

2

3

4

5

6

7

t(s)

Calculate how far ahead Bo is of Alex at the instant Bo starts to run. 3 marks The rules of the race require that the baton exchange takes place over a maximum of 20 metres distance and that the second runner can start running a maximum of 10 metres before the exchange. Use the graph to determine if Alex and Bo are likely to have complied with these rules. 2 marks

Question 4 (2 marks) A diver is standing at the top of a 75-metre tall cliff. They leap off the cliff with an initial vertical velocity of 3 m s−1 . Assume that their acceleration throughout the dive is 9.8 m s−1 directly downwards. How long will it be from the beginning of their leap until the instant they hit the water below?

TOPIC 9 Analysing motion 311

Velocity (m s−1)

Question 5 (5 marks) During the filming of a new movie, a stuntman has to chase a moving bus and jump into it. The stuntman is required to stand still until the bus passes him and then start chasing it. The velocity-versus-time graph in the figure that follows describes the motion of the stuntman and the bus from the instant that the bus door passes the stationary stuntman. a. At what instant did the stuntman reach the same speed as the bus? 1 mark b. What is the magnitude of the acceleration of the stuntman during the first 4 seconds? 1 mark c. At what instant did the stuntman catch up with the bus door? 2 marks d. How far did the stuntman run before he reached the door of the bus? 1 mark

Stuntman

10

Bus

8 6 4 2 0 2

9.5 Exercise 4: studyON topic test Fully worked solutions and sample responses are available in your digital formats.

Test maker Create unique tests and exams from our extensive range of questions, including practice exam questions. Access the assignments section in learnON to begin creating and assigning assessments to students.

312 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

4

6

8 10 12

Time (s)

AREA OF STUDY 1 HOW CAN MOTION BE DESCRIBED AND EXPLAINED?

10

Forces in action

10.1 Overview Numerous videos and interactivities are available just where you need them, at the point of learning, in your digital formats, learnON and eBookPLUS at www.jacplus.com.au.

10.1.1 Introduction Have you ever seen somebody parachuting down from a skydive or BASE jump? They can land without injury as the force of the air resisting the motion of the parachute is large enough to slow them to a safe speed. This explanation stems from the concept of forces, which are central to our understanding and analysis of motion. In his groundbreaking book Philosophia Naturalis Principia, published in 1687, Sir Isaac Newton proposed three laws of motion. These laws accurately explain the motion of objects on Earth and throughout the universe. In this topic we will use Newton’s laws to explore the nature of forces and their relationship to motion. FIGURE 10.1 BASE jumpers use a high point such as a cliff to launch themselves. The forces involved in the jump must be carefully calculated to allow the parachute to open in time.

TOPIC 10 Forces in action 313

10.1.2 What you will learn KEY KNOWLEDGE After completing this topic you will be able to: ∆p ∆t • model the force due to gravity, F g , as the force of gravity acting at the centre of mass of a body, F g = mg, where g is the gravitational field strength (9.8 N kg−1 near the surface of Earth) • model forces as vectors acting at the point of application (with magnitude and direction), labelling these forces using the convention ‘force on A by B’ or F on A by B = −F on B by A F net , F on A by B = −F on B by A • apply Newton’s three laws of motion to a body on which forces act: a = m • apply the vector model of forces, including vector addition and components of forces, to readily observable forces including the force due to gravity, friction and reaction forces • calculate torque: 𝜏 = r⊥ F • investigate and analyse theoretically and practically translational forces and torques in simple structures that are in rotational equilibrium • apply concepts of momentum to linear motion: p = mv. • explain changes in momentum as being caused by a net force: F net =

Source: VCE Physics Study Design (2016–2021) extracts © VCAA; reproduced by permission.

PRACTICAL WORK AND INVESTIGATIONS Practical work is a central component of learning and assessment. Experiments and investigations, supported by a Practical investigation logbook and Teacher-led videos, are included in this topic to provide opportunities to undertake investigations and communicate findings.

Resources Digital documents Key science skills — Units 1–4 (doc-31856) Key terms glossary (doc-32268) Practical investigation logbook (doc-32269)

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0035).

10.2 Forces as vectors KEY CONCEPTS • Apply the vector model of forces, including vector addition and components of forces, to readily observable forces including the force due to gravity, friction and reaction forces. • Model forces as vectors acting at the point of application (with magnitude and direction), labelling these forces using the convention ‘force on A by B’ or F on A by B = −F on B by A . • Model the force due to gravity, F g , as the force of gravity acting at the centre of mass of a body, F g = mg, where g is the gravitational field strength (9.8 N kg−1 near the surface of Earth).

10.2.1 Describing a force A force is a push or a pull applied by one object on another. Forces can start things moving, stop them, or change their speed or direction. Forces can spin objects or change their size or shape. Some types of force require contact. For example, the force applied by your hand on a netball or basketball when shooting a goal requires contact between your hand and the ball. The friction applied by the road on a bicycle or car 314 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

requires contact between the road surface and the tyres. Some forces do not require contact. For example, the force of gravity applied on your body by the Earth is present even when you are not in contact with the Earth. A magnet attracts certain materials without being in contact with them. Note that descriptions of forces indicate both the object that the force is applied on and the object that applies the force. FIGURE 10.2 Contact and non-contact forces at work (a)

Non-contact force

(b)

(c)

Non-contact force

Non-contact force

Contact force

In diagrams showing forces, such as figure 10.3, a FIGURE 10.3 Force is a vector quantity. labelled arrow should be drawn from the centre of mass A vector quantity in this text is represented of the object upon which the force acts (this was not by symbols in bold italic font. included in figure 10.2 for the sake of simplicity). The length of the arrow should indicate the relative size of the force. To fully describe a force, you need to specify its direction as well as its magnitude or size. A quantity that can be fully described only by specifying a direction as well as a magnitude is called a vector quantity. Force is a vector quantity. A vector quantity can be described in writing or by a labelled arrow. If a symbol is used to represent a vector quantity, it should have a halfAir resistance (R) arrow above it (some people use a ‘squiggly’ line below R upwards the symbol instead). In this text, vector quantities are = Fon apple by air represented by symbols in bold italic font. When bold Force due to can’t be used to distinguish a vector (for example, writing gravity (Fg ) by hand), a right-facing arrow may be used instead. Fg downwards = Fon apple by Earth When labelling forces, it helps to describe the force as Fon A by B ; for example, the arrow representing the force due to gravity of the apple is labelled Fon apple by Earth . The SI unit of force is the newton (N). The force of gravity on a 100 gram apple is about 1 N downwards. A medium sized car starting from rest is subjected to a forward force of about 4000 N. Quantities that can be described without specifying a direction are called scalar quantities. Mass, energy, time and temperature are all examples of scalar quantities. The present understanding of our universe posits that there are four fundamental forces or interactions: • strong nuclear force • electromagnetic force • weak nuclear force • gravitational force. While these categories are fundamental to our understanding of the physics of interactions in the universe, they are not wholly useful for describing and analysing everyday motion at a human scale. The following sections introduce some everyday categories of forces. TOPIC 10 Forces in action 315

10.2.2 Force due to gravity (F g ) The apple in figure 10.3 is attracted to the Earth by the force due to gravity. Even before it falls, the force due to gravity is pulling it down. However, before it falls, the tree branch is pulling it up with a force of equal magnitude. The force due to gravity is a force of attraction that exists between any pair of objects that have mass. Gravity is such a small force that, unless at least one of the objects is as massive as a planet or a natural satellite like the Moon, it is too small to measure. The force due to gravity is a force that acts ‘at a distance’ in that the items that it acts on do not need to be in contact with each other. →

The force on an object due to the pull of gravity is usually given the symbol Fg . In simple scenarios where objects of relatively small mass (e.g. humans, buildings, whales) are in close proximity to a very massive object (e.g. the Earth) the force due to gravity can be represented by: → → Fg = m g Where: m = the mass of the smaller object in kg → g = the gravitational field strength in N kg−1 due to the larger object. The gravitational field strength is defined as the force of gravity on a unit of mass. The magnitude of g at the surface of the Moon is approximately one-sixth that at the surface of the Earth. Gravitational field strength is a vector quantity. The direction of the force due to gravity on an object is towards the centre of the source of attraction (e.g. the centre of the Earth). The magnitude of gravitational field strength at the Earth’s surface is, on average, 9.8 N kg−1 . The magnitude of g decreases as altitude (height above sea level) increases. It also decreases as one moves from the poles towards the equator. Table 10.1 shows the magnitude of g at several different locations. TABLE 10.1 Variation in gravitational field strength Location

Altitude (m)

Latitude

Magnitude of g (N kg−1 )

Equator

0

0°

9.780

Sydney

18

34°S

9.797

Melbourne

12

37°S

9.800

1609

40°N

9.796

New York

38

41°N

9.803

North Pole

0

90°N

9.832

Denver

The magnitude of g at the Earth’s surface will be taken as 9.8 N kg−1 throughout this text. At the surface of the Moon, the magnitude of g is 1.60 N kg−1 . SAMPLE PROBLEM 1

What is the force due to gravity by the Earth acting on a 50-kilogram student: on the Earth b. on the Moon? a.

THINK

Recall the formula for force due to gravity. 2. Substitute the values to determine the force due to gravity.

a. 1.

316 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

WRITE a.

Fg = mg Fg = 50 × 9.8 = 490 N downwards

3.

State the solution.

Recall the formula for force due to gravity. 2. Substitute the values to determine the force due to gravity. 3. State the solution.

b. 1.

The force due to gravity by the Earth acting on a 50-kilogram student on the Earth is 490 N downwards. b. Fg = mg Fg = 50 × 1.60 = 80 N downwards The force due to gravity by the Earth acting on a 50-kilogram student on the Moon is 80 N downwards.

PRACTICE PROBLEM 1 a. What is the difference between the force due to gravity by the Earth acting on a 70-kilogram person at the North Pole and at the equator? b. A hospital patient is very accurately measured to have a mass of 64.32 kilograms and the force due to gravity by the Earth acting on them is 630.08 N. In which of the locations in table 10.1 could the patient be?

WEIGHING IN Bathroom scales are designed for use only on Earth. Fortunately (at this point in time), that’s where most of us live. If a 60-kilogram student stood on bathroom scales on the Moon, the reading would be only about 10 kilograms. Yet the mass of the student remains 60 kilograms. Bathroom scales measure force, not mass. However, scales are designed so that you can read your mass in kilograms. Otherwise, you would have to divide the measured force by 9.8 to determine your mass. The manufacturer of the bathroom scales saves you the trouble of having to do this. The 60-kilogram student experiences a force due to gravity from Earth of about 588 N. However, on the Moon the force due to gravity from the Moon is only about 100 N. The reading on the scales will be 100 N divided by 9.8 N kg−1 , giving the result of 10.2 kilograms.

The term weight is often used in high school physics texts in order to draw a distinction between the force due to gravity on an object and its mass. However, the accepted definition of the term weight in physics is more complex than this and there are different conventions for how it is defined. Confusion that this may cause is further compounded by inconsistent use of the term weight in everyday life. As such, the term weight is not used in this text or the VCE Physics Study Design.

Resources Digital document Investigation 10.1 The relationship between mass and the force due to gravity (doc-31879)

TOPIC 10 Forces in action 317

10.2.3 Friction (F fr ) Friction is a force that surfaces exert on each other when FIGURE 10.4 This scanning electron they ‘rub’ together. The magnitude of the friction force micrograph shows a magnified metal depends greatly on the nature of each of the two surfaces. surface. Smooth surfaces experience smaller friction forces than rough surfaces. However, even very smooth surfaces are rough on a microscopic scale. It is this roughness that is mostly responsible for the resistance to motion that we call friction. As two surfaces move across each other, they intermesh, resisting the motion. Friction can seem to be a real nuisance at times. It makes doors squeak. It causes wear and tear in car engines and can make them overheat. Friction is also a necessary force in many situations. If you have ever walked on a banana peel or ice, you know the importance of friction to walking. When you walk, you push backwards against the ground so that the ground pushes forward on your foot. Without a significant friction force, your foot slides backwards and you fall. Friction is also needed for safe driving in a car. A large friction force is needed to start moving, change direction and stop. The rubber tyres of a car have a deep tread to ensure that the friction force is still present on a wet road when water forms a lubricating film between the road and the tyres. Smooth tyres would slide across a wet road, making it difficult to stop, turn or accelerate. The deep grooves in tyres pick up the water from the road and throw it backwards, providing a drier road surface and greater friction. The lubricating effect of water is also evident in ice skating. The moving blade melts the surface of the ice beneath it, creating a thin lubricating film of water on which it glides.

Resources Digital document Investigation 10.2 Friction (doc-31877) Teacher-led video Investigation 10.2 Friction (tlvd-0822) Video eLesson

Friction as a driving force (eles-0032)

Interactivity

Friction as a driving force (int-0054)

10.2.4 Forces from fluid motion The motion of an object through fluids such as air or water results in forces acting on the object. One component of this is a force acting in the opposite direction of the motion. This resistance to the motion of objects through air and water is given a variety of names including fluid friction, fluid resistance, drag or air resistance. Olympic swimmers, cyclists and track athletes are becoming increasingly aware of the effects of fluid resistance as the accuracy of timing improves. The desire to win gold medals has resulted in swimmers shaving their heads, cyclists shaving their legs, new bicycle designs and tighter fitting costumes. Because fluid friction increases with the speed of the object, streamlining is particularly important in cars, 318 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 10.5 Racing cyclists reduce the effects of fluid friction (air resistance) by wearing streamlined costumes and using aerodynamically designed bicycles.

planes, watercraft and bicycles. Streamlining involves creating a shape that reduces the slowing effect of collisions with particles of the fluid. While air resistance can be a seen as a problem, it is an absolute necessity for parachutists and paragliders, who rely on it to slow their descent and land safely. The forces arising from motion through fluids is very complex and has been very simply introduced here. To learn more about this, you may choose to study the option in topic 16 ‘How do heavy things fly?’ where more detail on these forces is provided.

10.2.5 The normal force (F N ) At this moment you are probably sitting on a chair with your feet on the floor. The material in the chair, whether it is plastic, timber, foam or steel springs, has been compressed and is pushing back up. This force that is pushing up is called a reaction force because if you were not sitting on the chair, there would be no force. This force is more properly called a normal force, FN , as it acts at right angles to the surface. The normal force and the friction force comprise the two perpendicular components of forces that arise from the interactions between two objects or surfaces. In both instances the fundamental force involved is the electromagnetic interaction between the atoms in the materials.

10.2.6 Compression and tension materials

FIGURE 10.6 The normal force acts perpendicular to the surface of the chair.

FN

Forces acting on materials, such as a concrete pillar or a steel cable, can cause the material to experience compression or tension. These are types of loadings on a material. In a material that is experiencing compression, the atoms and molecules are pushed closer together. In the photo of the pole house in figure 10.7, the concrete pillar is in compression. The house is pushing down on the pillar and the rocky hillside below is pushing up on it, causing the pillar to be squashed in compression. An object that is in compression will be pushing against the objects that are compressing it. Some materials are best suited for withstanding compression, such as concrete, stone, brick and bone. FIGURE 10.7 The Pole House located on the Great Ocean Road is supported by a concrete column that goes several metres into the ground. Fon house by Earth Fon house by pillar Fon pillar by house

Fon pillar by rock Fon rock by pillar

TOPIC 10 Forces in action 319

Similarly, when atoms and molecules in a material are pulled further apart by a loading, it experiences tension. In the photo of the crane in figure 10.8, the steel cable is in tension. The crane is pulling up on the cable and the load that it is supporting is pulling down on the cable, causing it to be stretched in tension. An object that is in tension will be pulling against the objects that are stretching it. Metals are a common example of materials that are well suited to withstanding tension loading. FIGURE 10.8 Cranes use metal cables in tension to lift large loads Fby crane on cable Fby cable on crane

Fby cable on mass

Fby Earth on mass

10.2.7 Free body diagrams A free body diagram or force diagram is used to depict all external forces acting upon an object (see figures 10.7 and 10.8). The ability to accurately and effectively represent the forces on an object through a free body diagram is a critical skill in analysing forces and motion. The key to developing this skill is practice. To create a free body diagram consider the following steps. • Represent the object with a suitable simplified shape or, if the problem is simple enough, a single point (e.g. a car could be represented very simply as a rectangular block). • Identify each interaction that the object is experiencing and the force that represents that interaction (e.g. gravitational interaction with the Earth represented by the force due to gravity, Fg ). • Represent each force with an arrow drawn at the point where it acts on the object (e.g. a friction force on a car by the road will act where the car tyres contact the road; the force due to gravity will act at the centre of mass of an object). • The direction of the arrow should be in the expected direction of the force. If the direction is not known, make an assumption (e.g. the force in a connection between a car and trailer could be either forwards or backwards). • The length of the arrow should represent the expected magnitude (size) of the force. If unknown, make an assumption. SAMPLE PROBLEM 2

Draw a forces diagram for each of the following. An ice skater moving at a constant velocity b. Ball hanging straight down on a string c. Trolley slowing down d. Block stationary on slope a.

Teacher-led video: SP2 (tlvd-0077)

320 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

THINK a.

The only forces acting on the ice skater are the normal force and the force due to gravity.

WRITE a. Normal force, FN

Force due to gravity, Fg

b.

The forces acting on a ball hanging on a string are the force due to gravity and a tension force.

b. Tension, T

Force due to gravity, Fg

c.

The forces acting on a trolley that is slowing down are the force due to gravity, the normal force and a friction force.

c. Normal force, FN Friction Force due to gravity, Fg

d.

The forces acting on a stationary block on a slope are the force due to gravity, the normal force and a friction force in the direction up the slope.

d. Friction

Normal force, FN

Force due to gravity, Fg

PRACTICE PROBLEM 2 Draw a forces diagram for each of the following. a. Falling stone b. Mass being pulled at a steady speed c. Piece of iron on a string, which is hanging near a strong magnet.

10.2.8 The net force (F net ) or sum of forces What happens if more than one force acts on an object? It is rare for just one force to be acting on an object. For example, a moving car has friction forces, the normal force and the force due to gravity acting on it. A suspended magnet will experience the force due to gravity, a tension force and a magnetic force. The net force is the result of combining all the forces into one force. You may also see this referred to as the sum of forces or the resultant force. Sometimes all the forces can balance leaving a net or overall force of zero. At other times there is a non-zero force left, and because it is a vector, it has a particular direction, which may be different from any of the individual forces acting on the object. However, because force is a vector, combining forces is not just simple arithmetic, but is a vector sum as the direction of the force has to be taken into account. TOPIC 10 Forces in action 321

When more than one force acts on an object, the net force is found by the vector addition of the forces. If an object has two forces acting on it, one of 30 N and another of 40 N, the sum of the two forces, or net force, is 70 N only if both forces are acting in the same direction. Figures 10.9b, c and d show these two forces acting in different directions; figure 10.9a shows them acting in the same direction. The net force is indicated in each of the three examples. The net force is usually denoted by the symbol Fnet . It is critical to distinguish that the net force is not a particular type or category of force that acts on an object. It is simply the combination of all forces. To distinguish the net force vector from the vectors for the actual forces, the arrow for Fnet is often drawn in another colour or as a dashed line. A geometric method of adding vectors uses a parallelogram, where the vector sum is the diagonal. This is shown in figure 10.9d. FIGURE 10.9 Adding forces together 30 N

Fnet = 70 N

(a)

30 N

(b)

40 N Fnet = 10 N

40 N (c)

(d)

Fnet = 50 N

30 N

FB

40 N

FA + FB

FA

SAMPLE PROBLEM 3

The free body diagram on the right shows the forces acting on an object. Calculate the: a. component of F2 that acts in the x direction b. component of F2 that acts in the y direction c. magnitude of the net force on the object.

F2 = 25 N F1 = 35 N

y

37º

Teacher-led video: SP3 (tlvd-0078) THINK a. 1.

2. b. 1.

2.

The force F2 acts at an angle of 37° from the x-axis. Its x component can be found using trigonometry.

State the solution. Similarly the y component of F2 can be found using sin.

State the solution.

WRITE

adjacent hypotenuse F2x cos(37°) = 25 F2x = 25 cos(37°) ≈ 20 N The component of F2 that acts in the x direction is 20 N. opposite b. sin 𝜃 = hypotenuse F2y sin(37°) = 25 F2y = 25 sin(37°) ≈ 15 N The component of F2 that acts in the y direction is 15 N. a.

cos 𝜃 =

322 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

x

c. 1.

2.

To find the net force, consider the force components in the x and y directions separately.

Combining these two forces, Pythagoras’ theorem can be used to fine the resultant net force.

c.

In the x direction: Fnet, x = F2x − F1 = 20 − 35 = −15 N In the y direction: Fnet, y = F2y = 15 = 15 N Fnet

15 N

15 N

c 2 = a2 + b2 F2net = 152 + 152 F2net = 450 √ Fnet = 450 3.

State the solution.

Fnet ≈ 21.2 N The magnitude of the net force on the object is 21.2 N. Note: This problem only requires the magnitude of the net force. If the direction was required it would be necessary to calculate the angle that the force is acting at.

PRACTICE PROBLEM 3 The free body diagram on the right shows the forces acting on an object. Calculate the magnitude of the net force acting on the object.

70 N y 45º x

30º 40 N

Resources Digital document Investigation 10.3 Force as a vector (doc-32308)

10.2 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Describe the difference between a vector quantity and a scalar quantity. 2. Which of the following are vector quantities? (a) Mass (b) Force due to gravity (c) Gravitational field strength (d) Time (e) Energy (f) Temperature

TOPIC 10 Forces in action 323

3. A car has a mass of 1400 kilograms with a full petrol tank. (a) What is the magnitude of the force due to gravity acting on it at the surface of the Earth? (b) What would be the magnitude of the force due to gravity acting on it on the surface of Mars where the magnitude of the gravitational field strength is 3.6 N kg−1 ? (c) What is the mass of the car on the surface of Mars? 4. Estimate the magnitude of the force due to gravity acting at the surface of the Earth on: (a) an apple (b) a textbook (c) your physics teacher. 5. Estimate your own mass in kilograms and determine: (a) the magnitude of the force due to gravity acting on you at the surface of the Earth (b) the magnitude of the force due to gravity acting on you at the surface of Mars where the magnitude of the gravitational field strength is 3.6 N kg −1 (c) your mass on the planet Mars. 6. Draw force diagrams for: (a) an open parachute falling slowly to the ground. (b) a thrown basketball approaching its maximum height before coming down into the basket (c) a car approaching a red light rolling slowly to a stop (d) a rocket during liftoff. 7. Draw force diagrams for figures a–d. For figures a, b and c, show the forces acting on the rock. For figure d, show the force acting on the block. (a)

(b)

(c)

(d) Stretched spring Swinging

8. A person is standing on a horizontal floor. Draw and label in the form Fon A by B all of the forces acting on the person, the floor and the Earth. 9. Determine the net force in the situations illustrated in diagrams a and b. (a)

(b) 2N

3N

100 N 4N 45° 45°

N W

E 100 N S

324 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

10. In the following diagrams, the net force is shown along with all but one of the contributing forces. Determine the magnitude and direction of the missing force. (a)

(b) Fnet = 200 N 200 N N 400 N

200 N 60°

200 N

W

30° 30°

S

200 N

Fnet = 0

E

11. A car is moving north on a horizontal road at a constant speed of 60 km h−1 . Draw a diagram showing all of the significant forces acting on the car. Show all of the forces as if they were acting through the centre of mass. 12. Determine the magnitude of the horizontal components of each of the forces shown in the following diagrams. (a)

(c)

(b) 200 N

200 N

200 N 25°

60°

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

10.3 Newton’s First Law of Motion KEY CONCEPT • Apply Newton’s three laws of motion to a body on which forces act: a =

F net , F on A by B = −F on B by A . m

It is difficult to explain the motion of objects without an understanding of force. The ancient Greek philosopher Aristotle (384–322 BCE) concluded from his observations that a moving object would come to rest if no force was pushing it. Aristotle thought that steady motion required a constant force and that ‘being at rest’ was the natural state of matter. This view held sway for almost 2000 years, although contrary views were expressed by philosophers such as Epicurus and Lucretius. It was not until Galileo (1564–1642) that this explanation of motion was seriously challenged. Galileo argued that if a ball rolled down an inclined plane gained speed and a ball rolled up an inclined plane lost speed, a ball rolled along a horizontal plane should neither gain nor lose speed. Galileo knew that this did not really happen. He claimed that if there was a lot of friction, the ball slowed down quickly; if there was little friction, the ball slowed down more gradually. However, he predicted that if there were no friction at all, the ball would continue to move with a constant speed forever unless something else caused it to slow down or stop. Galileo introduced the concept of friction as a force and concluded that objects retain their velocity unless a force, often friction, acts upon them. Galileo stated in his Discorsi (1638): A body moving on a level surface will continue in the same direction at constant speed unless disturbed. TOPIC 10 Forces in action 325

Sir Isaac Newton (1643–1727) was able to refine Galileo’s ideas about motion. In 1687, he published his Philosophia Naturalis Principia, which included three laws of motion. Newton’s First Law of Motion: Every object continues in its state of rest or uniform motion unless made to change by a non-zero net force.

A coin flicked across a tabletop changes its motion because the net force on it is not zero. In fact, it slows down because the direction of the net force is opposite to the direction of motion. The vertical forces, gravity and the support force of the table balance each other. The only ‘unbalanced’ force is that of friction. A coin pushed steadily across a tabletop moves in a straight line at constant speed as long as the net force is zero (that is, as long as the magnitude of the pushing FIGURE 10.10 The changing net force on this force is equal to the magnitude of friction). The coin bungee jumper determines their state of motion will speed up if you push horizontally with a force and whether or not they will stop in time. greater than the friction. It will slow down if the force of friction is greater than the horizontal pushing force. The motion of a bungee jumper can be explained in terms of Newton’s First Law of Motion. Until the bungee cord tightens, the net force is downwards and the speed of the bungee jumper increases. As the jumper’s speed increases, so does the air resistance. However, the air resistance is quite small compared with the force due to gravity on the jumper. Until the cord begins to tighten, the tension pulling the jumper up is zero. As the rope tightens, the tension increases. Until the tension and air resistance forces together balance the force due to gravity, the bungee jumper continues to speed up. The tension continues to increase, eventually resulting in an upwards net force which allows the bungee jumper to slow down, stop (just in time) and rise again.

SIR ISAAC NEWTON Sir Isaac Newton was one of many famous scientists who were not outstanding students at school or university. Newton left school at 14 years of age to help his widowed mother on the family’s farm. He turned out to be unsuited to farming and spent much of his time reading. At the age of 18, he went to Cambridge University, where he showed no outstanding ability. When Cambridge University was closed down in 1665 due to an outbreak of the plague, Newton went home and spent the next two years studying and writing. During this time, he developed the laws of gravity that explain the motion of the planets, and his three famous laws of motion. Over the same period, he put forward the view that white light consisted of many colours and he invented calculus. Newton’s laws of gravity and motion were not published until about 20 years later. Newton later became a member of Parliament, a warden of the Mint and President of the Royal Society. After his death in 1727, he was buried in Westminster Abbey, London, alongside many English kings, queens, political leaders and poets.

326 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 10.11 Sir Isaac Newton

10.3 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. With reference to Newton’s First Law of Motion explain why a coin flicked across a table does not move at a constant speed. 2. When you are standing on a bus, train or tram that stops suddenly, you lurch forwards. Explain why this happens in terms of Newton’s first law. 3. If the bicycle that you are riding runs into an obstacle such as a large rock, you may be flung forwards over the handlebars. Explain in terms of Newton’s first law why this happens.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

10.4 Newton’s Second Law of Motion KEY CONCEPT • Apply Newton’s three laws of motion to a body on which forces act: a =

F net , F on A by B = −F on B by A . m

Casual observations indicate that the acceleration of a given object increases as the net force on the object increases. It is also clear that lighter objects change their velocity at a greater rate than heavier objects when the same force is applied. It can be shown experimentally that the acceleration, a, of an object is: • proportional to the net force, Fnet , applied to it • inversely proportional to its mass, m. a ∝ Fnet

a∝

1 m

Thus: Fnet m kFnet ⇒a= m a∝

Where: k = a constant of proportionality. The SI unit of force, the newton (N), is defined such that a net force of 1 N causes a mass of 1 kilogram to accelerate at 1 m s−2 . The value of the constant, k, is 1. It has no units. Thus: Fnet m Fnet = ma a=

TOPIC 10 Forces in action 327

This equation describes Newton’s Second Law of Motion.

Newton’s Second Law of Motion: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object. → →

a =

→

Fnet m →

Fnet = m a

This statement of Newton’s second law allows you to: • determine the net force acting on an object without knowing any of the individual forces acting on it. The net force can be deduced as long as you can measure or calculate (using formulae or graphs) the acceleration of a known mass • determine the mass of an object. You can do this by measuring the acceleration of an object on which a known net force is exerted predict the effect of a net force on the motion of an object of known mass. •

SAMPLE PROBLEM 4

A 65-kilogram physics teacher, starting from rest, glides gracefully down a slide in the local playground. The net force on her during the slide is a constant 350 N. How fast will she be travelling at the bottom of the 8-metre slide?

THINK

WRITE

Recall Newton’s Second Law of Motion. 2. Substitute the mass and force to find the acceleration.

Fnet = ma Fnet = ma

1.

350 = 65a 350 65 ≈ 5.38 m s−2 2 v = u2 + 2as = 0 + 2 × 5.38 × 8 = 86.08 √ v = 86.08 a=

3.

4.

As the acceleration is constant, a constant acceleration formula can be used.

State the solution.

328 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

≈ 9.3 m s−1 She will be travelling at approximately 9.3 m s–1 at the bottom of the slide.

PRACTICE PROBLEM 4 a. What is the magnitude of the average force applied by a tennis racquet to a 58-gram tennis ball during service if the average acceleration of the ball during contact with the racquet is 1.2 × 104 m s−2 ? b. A toy car is pulled across a smooth, polished horizontal table with a spring balance. The reading on the spring balance is 2 N and the acceleration of the toy car is measured to be 2.5 m s−2 . What is the mass of the toy car? (Note: Because the table is described as smooth and polished, friction can be ignored.)

Resources Video eLesson Newton’s second law (eles-0033)

10.4.1 Applying Newton’s second law in real life Of course, the practice problems presented above do not reflect what really happens. When a tennis ball is served, the force applied by the tennis racquet is not the only force acting on the ball. However, the force due to gravity acting on the ball and the air resistance on the ball are negligible; that is, they are so small compared with the force of the racquet on the ball that they can be ignored while the racquet is in contact with the ball. The table surface in Practice problem 4b is smooth. This description was deliberately included so that you would know the force of friction on the toy car was negligible. If the table were not described as smooth, you could not have answered the question without taking friction into account. The event described in Sample problem 4 was also simplified. It is unlikely that the net force on the teacher gliding down the slide would be constant. The assumptions made to answer each of the questions asked in the sample problem and practice problem are called idealisations. However, caution is needed when making idealisations. For example, it would be unreasonable to ignore the air resistance on a tennis ball while it was soaring through the air at 150 km h−1 (42 m s−1 ) after the serve was completed. Most applications of Newton’s second law are not as simple as those given above. Some more typical examples are presented in the following sample problems.

SAMPLE PROBLEM 5

When the head of an 80-kilogram bungee jumper is 24 metres from the surface of the water below, her velocity is 16 m s−1 downwards and the tension in the bungee cord is 1200 N. Air resistance can be assumed to be negligible. a. What is her acceleration at that instant? b. If her acceleration remained constant during the rest of her fall, would she stop before hitting the water? Teacher-led video: SP5 (tlvd-0080)

THINK a. 1.

Calculate force due to gravity, then draw a diagram to show the forces acting on the bungee jumper.

WRITE a.

Fg = mg = 80 × 9.8 = 784 N

TOPIC 10 Forces in action 329

FT = 1200 N

Fg = mg = 784 N

2.

Calculate the net force.

3.

Use Newton’s second law to calculate the acceleration.

Fnet = 1200 − 784 = 416 N upwards Fnet = ma 416 = 80a a=

State the solution. b. 1. If the jumper’s acceleration were constant, one of the constant acceleration formulae could be used to answer this question. Assign down as positive for this part of the question as the bungee jumper has a downwards initial velocity and displacement during the time period being considered. 2. Substitute the values into the formula and solve for s. 4.

3.

State the solution.

330 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

416 80

= 5.2 m s−2 upwards Her acceleration at that instant is 5.2 m s–2 . b. v2 = u2 + 2as u = 16 m s−1 , v = 0, a = −5.2 m s−2 , s = ?

v2 = u2 + 2as 0 = 162 + 2 × −5.2 × s 0 = 256 − 10.4s 10.4s = 256 256 s= 10.4 ≈ 24 m The bungee jumper will not stop in time. However, don’t be upset. In practice, the acceleration of the bungee jumper would not be constant. The tension in the cord would increase as she fell. Therefore, the net force on her would increase and her upwards acceleration would be greater in magnitude than the calculated value. She will therefore almost certainly come to a stop in a distance considerably less than that calculated.

PRACTICE PROBLEM 5 A 1200-kilogram sports car is testing its brakes by driving at a constant speed of 100 km h−1 and then braking hard. To pass the test it needs to come to a complete stop in a distance of 50 metres. If the friction is a constant 1000 N, what force do the brakes need to apply for the sports car to pass the test?

SAMPLE PROBLEM 6

A waterskier of mass 80 kilograms, starting from rest, is pulled in a northerly direction by a horizontal rope with a constant tension of 240 N. After 6 seconds, he has reached a speed of 12 m s−1 . a. What is the net force on the skier? b. If the tension in the rope were the only horizontal force acting on the skier, what would his acceleration be? c. What is the sum of the resistance forces on the skier? Teacher-led video: SP6 (tlvd-0081) THINK a. 1.

A good first step is to draw a labelled force diagram. The force due to gravity can be calculated by the formula Fg = mg.

WRITE a.

Fg = mg = 80 × 9.8 = 784 N N Normal force = 784 N

Resistance forces

Tension = 240 N

Force due to gravity = mg = 784 N

The net force cannot be determined by adding the individual force vectors because the resistance forces are not given, nor is there any information in the question to suggest that they can be ignored. To calculate the net force we will first need to find the acceleration. 3. Use Newton’s second law to calculate the net force.

2.

4.

State the solution.

v = u + at 12 = 0 + a × 6 12 a= 6 = 2 m s−2 north

Fnet = ma = 80 × 2 = 160 N north The net force on the skier is 160 N north.

TOPIC 10 Forces in action 331

As the vertical forces cancel out, the net force is determined by the horizontal forces. If the tension were the only horizontal force acting on the skier it would be the net force. 2. Use Newton’s second law to calculate the acceleration.

b. 1.

3.

State the solution.

The sum of the resistance forces (friction caused by the water surface and air resistance) on the skier is the difference between the net force and the tension. 2. State the solution.

c. 1.

b.

Fnet = 240 N north

Fnet = ma 240 = 80a 240 a= 80 = 3 m s−2 north If the tension in the rope were the only horizontal force acting on the skier, his acceleration would be 3 m s–2 north. c. sum of resistance forces = Fnet − tension = 160 N north − 240 N north = 80 N south The sum of the resistance forces on the skier is 80 N south.

PRACTICE PROBLEM 6 A loaded sled with a mass of 60 kilograms is being pulled across a level snow-covered field with a horizontal rope. It accelerates from rest over a distance of 9 metres, reaching a speed of 6 m s−1 . The tension in the rope is a constant. The frictional force on the sled is 200 N. Air resistance is negligible. a. What is the acceleration of the sled? b. What is the magnitude of the tension in the rope?

SAMPLE PROBLEM 7

Velocity (m s−1)

The following velocity-versus-time graph describes the motion of a 45-kilogram girl on rollerblades as she rolls on a horizontal concrete path for 6 seconds before crossing onto a rough horizontal gravel path for the remaining 4 seconds. a. What was the magnitude of the net force on the girl on the concrete surface? 8 b. If the only horizontal force acting on the blades is the friction force applied by the path, what is the 6 value of the following ratio? 4 friction force of gravel path on rollerblades friction force of concrete path on rollerblades 2

2

4

6 Time (s)

332 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

8

10

THINK a. 1.

The acceleration of the girl while she was on the concrete surface is given by the gradient of the corresponding section of the velocity-versus-time graph.

2.

Use Newton’s second law to calculate the magnitude of the net force on the girl while on the concrete surface.

3.

State the solution.

b. 1.

If the only horizontal force acting on the rollerblades is friction, the net force on the girl is the same as the friction force on the blades.

2.

Apply Newton’s second law and cancel out common terms.

3.

Use the gradient of the graph to evaluate the ratio of accelerations.

WRITE

rise run −2 = 6 −1 m s−2 = 3 Fnet = ma −1 = 45 × 3 = −15 N The magnitude of the net force on the girl on the concrete surface is –15 N. friction force of gravel path on rollerblades b. friction force of concrete path on rollerblades a.

a=

=

Fnet on girl while on gravel Fnet on girl while on concrete

ma on gravel Fnet on girl while on gravel = Fnet on girl while on concrete ma on concrete a (during last 4 s) = a (during first 6 s) a (during last 4 s) gradient (for last 4 s) = a (during first 6 s) gradient (for first 6 s) ( −6 ) 4 = ( −1 ) 3

=

4.

State the solution.

36 8

= 4.5 The value of the ratio is 4.5.

PRACTICE PROBLEM 7 If the velocity-time graph in Sample problem 7 was applied to a car of mass 1200 kg on two road surfaces, what net force (in magnitude) acts on the car during: a. the first 6 seconds b. the final 4 seconds?

10.4.2 Falling down Objects that are falling (or rising) through the air in the atmosphere near the surface of Earth are subjected to two forces — the force due to gravity and air resistance. The force due to gravity of the object is effectively constant. The magnitude of the air resistance, however, is not constant. It depends on many TOPIC 10 Forces in action 333

factors, including the object’s speed, surface area and density. It also depends on the density of the body of air through which the object is falling. The air resistance is always opposite to the direction of motion. The net force on a falling object of mass m and force due to gravity Fg can therefore be expressed as: Fnet = ma (where a is the acceleration of the object) Fg − air resistence = ma When dense objects fall through small distances near the surface of the Earth it is usually quite reasonable to assume that the air resistance is negligible. Thus: Fg = ma ⇒ mg = ma (where g is the gravitational field strength) ⇒g=a The acceleration of a body in free fall in a vacuum or where air resistance is negligible is equal to the gravitational field strength. At the Earth’s surface, where g = 9.8 N kg−1 , this acceleration is 9.8 m s−2 . The units N kg−1 and ms−2 are equivalent. 1 N = 1 kg m s−2 ⇒ 1 N kg−1 = 1 kg m s−2 kg−1 (multiplying both sides by kg−1 ) ⇒ 1 N kg−1 = 1 m s−2

TERMINAL VELOCITY Terminal velocity is reached when the FIGURE 10.12 Skydivers accelerate until the drag force equals forces acting on a falling object are the force due to gravity, at terminal velocity. balanced and it stops accelerating. When considering objects falling vertically downwards in the atmosphere near the surface of the Earth, the two FD forces are air resistance or drag, FD acting to oppose the motion of the object and the force due to gravity, Fg , acting vertically downwards. The force due to gravity and air resistance in balance when an object is at terminal velocity. The force due to gravity, Fg , depends Fg only on the mass of the object, m, and the gravitational field strength, g, which can both be assumed to be roughly constant during the fall of an object near the surface of Earth. The air resistance or drag, FD , acting on an object depends on a number of factors: • CD is the drag coefficient, which measures the ease with which air can move over an object. In simple terms this indicates how streamlined an object is. This is a constant value for a particular object shape and orientation and is often determined experimentally or via computer analysis.

334 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

−3 • 𝜌 is the air density measured in kg m , which can be assumed to be constant, however, in reality it will vary with altitude, temperature, humidity and pressure. −1 • v is the velocity that the object is moving through the air measured in m s . • A is the cross-sectional or reference area that is perpendicular to the direction of motion of the object measured in m2 . For an object falling vertically this will be the horizontal cross-section. When terminal velocity occurs these two forces must be in balance (net force is zero, acceleration is zero).

FD = Fg 1 2

CD 𝜌v2 A = mg

FIGURE 10.13 Forces acting on a falling object at terminal velocity

FD = 1 CD ρv2A 2

This can be rearranged to determine a relationship for the terminal velocity. √ 2mg v= CD 𝜌A Provided that reasonable estimates can be made of the quantities involved, this relationship can be used to calculate the theoretical terminal velocity of falling objects. This can provide an interesting focus for an extended practical investigation.

Fg = mg

If a bowling ball, a golf ball and a table tennis ball were dropped at the same instant from a height of 2 metres in a vacuum, they would all reach the ground at the same time. Each ball would have an initial velocity of zero, an acceleration of 9.8 m s−2 and a downward displacement of 2 metres. If, however, the balls are dropped either in a classroom or outside, the table tennis ball will reach the ground a moment later than the other two balls. The acceleration of each of the balls is: FIGURE 10.14 A bowling ball, a golf ball and Fnet a table tennis ball dropped from a height of a= 2 metres. Which one would you expect to m reach the ground first? mg − FD = m mg FD (where FD is air resistance) − = m m FD =g− m FD is very small for the bowling ball and the golf ball. m Even though the air resistance on the table tennis ball is F small, its mass is also small and D is not as small as it is m for the other two balls.

WARNING: Do not drop a bowling ball from a height of 2 metres indoors. If you wish to try this experiment, replace the bowling ball with a medicine ball and keep your feet out of the way!

TOPIC 10 Forces in action 335

Resources Digital document Simulation of basketball throw (doc-0052) Video eLesson

Air resistance (eles-0035)

10.4 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. What is an idealisation? Provide an example of an idealism that could be used to simplify a physics problem. 2. When a space shuttle takes off, its initial acceleration is 3.0 m s−2 . It has an initial mass of about 2.2 × 106 kg. (a) Determine the magnitude of the net force on the space shuttle as it takes off. (b) What is the magnitude of the upward thrust as it takes off? 3. A 6 kilogram bowling ball and a 60 kilogram gold bar are dropped at the same instant from the third floor of the Leaning Tower of Pisa. Use Newton’s second law to explain why: (a) they both reach the ground at the same time (b) a 6 kilogram doormat dropped from the same location at the same time takes significantly longer to reach the ground. 4. A bungee jumper with a mass of 70 kilograms leaps from a bridge. (a) What is the force due to gravity acting on the bungee jumper? (b) During which part of the jump is: i. the upwards force on the jumper due to the tension in the bungee cord greater than force due to gravity on the jumper ii. the force due to gravity on the jumper greater than the upward pull of the bungee cord? (c) What tension in the bungee cord is needed for the jumper to travel at a constant speed? Does this occur at any time during the jump? Explain. 5. A car of mass 1200 kilograms starts from rest on a horizontal road with a forward driving force of 10 000 N. The resistance to motion due to road friction and air resistance totals 2500 N. (a) What is the magnitude of the net force on the car? (b) What is the magnitude of the acceleration of the car? (c) What is the speed of the car after 5 seconds? (d) How far has the car travelled after 5 seconds? 6. A train of mass 8.0 × 106 kilograms travelling at a speed of 25 m s−1 is required to stop over a maximum distance of 360 metres. What frictional force must act on the train when the brakes are applied if the train is to do this? 7. A short-sighted skier of mass 70.0 kilograms suddenly realises while travelling at a speed of 12.0 m s−1 that there is a steep cliff 50.0 metres straight ahead. What frictional force is required on the skier if he is to stop just before he skis off the edge of the cliff? 8. A physics teacher decides, just for fun, to stand on some bathroom scales (calibrated in newtons) in a lift. The scales provide a measure of the force with which they push up on the teacher. When the lift is stationary, the reading on the bathroom scales is 700 N. What will be the reading on the scales when the lift is: (a) moving upwards at a constant speed of 2.0 m s−1 (b) accelerating downwards at 2.0 m s−2 (c) accelerating upwards at 2.0 m s−2 ? 9. The cable holding a lift would break if the tension in it were to exceed 25 000 N. If the 480 kilograms lift has a load limit of 24 passengers whose average mass is 70 kilograms, what is the maximum possible upwards acceleration of the lift without breaking the cable? 10. A ball of mass 0.50 kilograms is thrown vertically upwards. (a) What is the velocity of the ball at the top of its flight? (b) What is the magnitude of the ball’s acceleration at the top of its flight? (c) What is the net force on the ball at the top of its flight?

336 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

10.5 Newton’s Third Law of Motion KEY CONCEPTS F net , F on A by B = −F on B by A . m • Model forces as vectors acting at the point of application (with magnitude and direction), labelling these forces using the convention ‘force on A by B’ or F on A by B = −F on B by A . • Apply Newton’s three laws of motion to a body on which forces act: a =

10.5.1 Forces in pairs To explain the motion of everyday objects, Isaac Newton devised some basic principles. A key starting point is the nature of force. Newton said that forces always act in pairs. Forces explain the interaction between two objects. Each object acts on the other, so where there is one force by one object, there is always another force by the other object. Examples of forces in pairs include: • bicycle back tyre. The action of the pedals push the tyre backwards against the road surface. The frictional interaction between the tyre and the road results in a forward force by the road on the tyre. • two suspended magnets. In the top diagram in figure 10.15b, each magnet is pushing the other away. In the bottom diagram, each is pulling the other towards it. • sitting in a chair. The normal force from the chair pushes up on the bones in the pelvis and the compressed bone pushes down. This compression is what you feel when you are sitting. FIGURE 10.15 Forces act in pairs (a)

(c)

(b)

Fon bone by chair

A

N

Fon A by B

A

S

N

B

Fon B by A

N

B

Fon tyre by road Fon road by tyre

Fon A by B

Fon B by A

Fon chair by bone

TOPIC 10 Forces in action 337

The pairing of these forces is apparent in the symmetry of their labels. If the label for one force is Fon B by A , then the label for the other is Fon A by B . It is important to note that in these pairs of forces, the forces act on different objects. Newton not only identified forces acting in pairs, he also said that these two forces act in opposite directions and are equal in magnitude or size. This statement became Newton’s Third Law of Motion. Figure 10.15 Newton’s Third Law of Motion: If object B applies a force to object A, then object A applies an equal and opposite force to object B. →

→

Fon A by B = −Fon B by A

This symmetry between the pair of forces can be used to identify the other of the pair if only one is given.

Resources Digital document eModelling: Skydiver spreadsheet (doc-0054)

REACTION OR NORMAL FORCE? Some texts summarise Newton’s third law as ‘For every action, there is an equal and opposite reaction’. This version is not preferred. The word ‘reaction’ here has a different meaning to its use in ‘normal force’. The statement also implies one force in the pair is a response to the other, which is incorrect.

There are four forces acting as two force pairs when you sit on a comfortable chair. The Earth pulls down on you, compressing the springs and foam, and the compressed springs and foam push up on you, compressing the bones in your pelvis. So, one force pair is the upward push by the springs on you and the downward push by the bones in your pelvis on the chair. The second force pair is the Earth pulling down on you and you pulling the Earth upwards. The net force on a person sitting in a chair is the vector sum of all the forces acting on the person. The net force is zero because the upward push by the chair, Fon student by chair , is balanced by the downward, and of equal size, pull of the force of gravity, Fon chair by student .

FIGURE 10.16 Forces while sitting Fon student by chair Chair pushes student.

Fon chair by student Student pushes chair.

338 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

SAMPLE PROBLEM 8

Two ice skaters, Jack and Jill, push each other away. The figure on the right shows the force on Jill by Jack with a black arrow. Draw an arrow and label it to show the force on Jack by Jill.

Fon Jill by Jack

THINK

WRITE

Recall Newton’s Third Law of Motion. 2. Draw the force on Jack by Jill as an arrow of equal length in the opposite direction.

Fon A by B = −Fon B by A

1.

Fon Jack by Jill

PRACTICE PROBLEM 8 Draw and label arrows for the other forces in the following force pairs. a.

b. Fon boy by girl

Fon gas by rocket

TOPIC 10 Forces in action 339

c.

d.

Fon apple by Earth Fon ball by club

10.5.2 Moving forward The rowing boat in figure 10.17 is propelled FIGURE 10.17 This rowing team relies on a reaction forward by the push of water on the oars. As the force to propel itself forward. face of each oar pushes back on the water, the water pushes back with an equal and opposite force on each oar. The push by the oars, which are held tightly by the rowers, propels the rowers and their boat forward. A greater push on the water results in a greater push on the oar. In fact, none of your forward motion, whether you are on land, water or in the air, could occur without a third law force pair. • When you swim, you push the water backwards with your hands, arms and legs. The water pushes in the opposite direction, propelling you forwards. • In order to walk or run, you push your feet backwards and down on the ground. The ground pushes in the opposite direction, pushing forwards and up on your feet. The forward driving force on the wheels of a car is the result of a push back on the road by the wheels. • • A jet or a propeller-driven plane is thrust forwards by air. The jet engines or propellers are designed to push air backwards with a very large force. The air pushes forward on the plane with an equally large force.

Resources Video eLesson Newton’s laws (eles-0036) Interactivity

Newton’s laws (int-0055)

340 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

10.5 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Copy the following table into your workbook. Describe fully the missing half of the following force pairs. Force 1

Pair of force 1

You push on a wall with the palm of your hand. Your foot pushes down on a bicycle pedal. The ground pushes up on your feet while you are standing. The Earth pulls down on your body. You push on a broken-down car to try to get it moving. A hammer pushes down on a nail. 2. Label all of the forces in the following figures in the form Fon A by B . (a)

N Normal force

(b) Normal force Friction

Tension Resistance forces

30° Fgy Fgx

Force due to gravity

30° Force due to gravity

3. Identify the force pairs in the following figure.

4. Explain, in terms of Newton’s first and third laws, why a freestyle swimmer moves faster through the water than a breaststroke swimmer. 5. A student says that the friction forces on the front and back tyres of a car are an example of Newton’s Third Law of Motion. Is the student correct? Explain.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

TOPIC 10 Forces in action 341

10.6 Forces in two dimensions KEY CONCEPTS F net , F on A by B = −F on B by A . m • Model forces as vectors acting at the point of application (with magnitude and direction), labelling these forces using the convention ‘force on A by B’ or F on A by B = −F on B by A . • Apply Newton’s three laws of motion to a body on which forces act: a =

10.6.1 Vehicles on horizontal surfaces The forces acting on a car being driven along a straight horizontal road are: • force due to gravity. The force applied by the Earth on the car acts through the centre of mass, or balancing point, of the car. This is normally closer to the front of the car than the back due to the significant mass of the engine. A medium sized sedan containing a driver and passenger will experience a force due to gravity of about 1.5 × 104 N. • the normal force. The force applied on the car by the road is a reaction to the force applied on the road by the car. A normal force pushes up on all four wheels. Its magnitude is normally greater at the front wheels than the rear wheels. On a horizontal road, the sum of these normal forces has the same magnitude as the force due to gravity. What do you think would happen if this were not the case? • the driving force. This is provided by the road and is applied to the driving wheels. The driving wheels are turned by the motor. In most cars, either the front wheels or the rear wheels are the driving wheels. The motor of a four-wheel-drive vehicle turns all four wheels. As a tyre on a driven wheel pushes back on the road, the road pushes forward on the tyre, propelling the car forward. The forward push of the road on the tyre is a type of friction commonly referred to as traction, or grip. If the tyres do not have enough tread, or the road is icy, there is not enough friction to push the car forward and the tyre slides on the road. The wheel spins and the car skids. The car cannot be propelled forward as effectively. Skidding also occurs if the motor turns the driving wheels too fast. • road friction. This is the retarding force applied by the road on the tyres of the non-driving wheels. The non-driving wheels of front-wheel-drive cars roll as they are pulled along the road by the moving car. In older cars, the non-driving wheels are usually at the front. They are pushed along the road by the moving car. Rolling friction acts on the non-driving wheels in a direction opposite to the direction of the car’s movement. When the driving wheels are not being turned by the motor, rolling friction opposes the forward movement of all four wheels. When the brakes are applied, the wheels to which the brakes are attached are made to turn too slowly for the speed at which the car is moving. They are no longer rolling freely. This increases the road friction greatly and the car eventually stops. If the brakes are applied hard enough, the wheels stop completely, or lock, and the car goes into a skid. The sliding friction that exists when the car is skidding is less than the friction that exists when the wheels are rolling just a little. • air resistance. The drag, or air resistance acting on the car, increases as the car moves faster. As a fluid friction force, air resistance can be reduced by streamlining the vehicle. The net force acting on the car in figure 10.18 is zero. It is therefore moving along the road at constant speed. We know that it is moving to the right because both the air resistance and road friction act in a direction opposite to that of motion. If the car were stationary, neither of these forces would be acting at all. • If the driving force were to increase, the car would speed up until the sum of the air resistance and road friction grew large enough to balance the driving force. Then, once again, the car would be moving at a constant, although higher, speed. • If the driver stopped pushing down on the accelerator, the motor would stop turning the driving wheels and the driving force would become zero. The net force would be to the left. As the car slowed down,

342 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

the air resistance and road friction would gradually decrease until the car came to a stop. The net force on the car would then be zero until such time as the driving force was restored. FIGURE 10.18 Forces acting on a car. The state of motion of a front-wheel-drive car on a horizontal road depends on the net force acting on it.

Air resistance, Fon car by air Normal force, Fon car by road

Normal force, Fon car by road

Driving force, Fon tyre by road

Road friction, Fon tyre by road

Force due to gravity, Fon car by Earth

COMPUTER CONTROLLED BRAKE SYSTEMS An anti-lock brake system (ABS) allows the FIGURE 10.19 ABS on a motorcycle wheels to keep rolling no matter how hard the brakes are applied. A small computer attached to the braking system monitors the rotation of the wheels. If the wheels lock and rolling stops, the pressure on the brake pads (or shoes) that stops the rotation is reduced briefly. This action is repeated up to 15 times each second. Antilock brake systems are most effective on wet roads. However, even on a dry surface, braking distances can be reduced by up to 20%. When brakes in an older car without ABS are applied too hard, as they often are when a driver panics, the wheels lock. The resulting sliding friction is less than the friction acting when the wheels are still rolling. The car skids, steering control is lost and the car takes longer to stop than if the wheels were still rolling. Before ABS became commonplace, drivers were often advised to ‘pump’ the brakes in wet conditions to prevent locking. This involves pushing and releasing the brake pedal in quick succession until the car stops. This, however, is very difficult to do in an emergency situation. Since the advent of ABS, a myriad additional computer controlled systems have been developed to improve safety when braking. One such system, often referred to as Brake Assist monitors the application of force to the brake pedal to detect an emergency and apply additional brake pressure. More advanced systems use a range of technology to monitor conditions on the road ahead and can warn the driver to apply the brakes and in some systems even apply the brakes automatically if there is no driver response. The development of systems such as these over recent decades has been a vital precursor to the rapid development of autonomous cars that is underway at present.

Resources Digital document Investigation 10.4 Static, sliding and rolling friction (doc-32309)

TOPIC 10 Forces in action 343

10.6.2 Vehicles on inclined planes A car left parked on a hill will begin to roll down the hill with increasing speed if it is left out of gear and the handbrake is off. Figure 10.20a shows the forces acting on such a car. In order to simplify the diagram, all of the forces are drawn as if they were acting through the centre of mass of the car. The forces on the car can then be modelled as acting on a single point. The direction of net force acting on the car is down the hill. It is clear that the force of gravity is a major contributor to the downhill motion of the car. FIGURE 10.20 (a) A simplified diagram of the forces acting on a car rolling down a slope (b) Vectors can be resolved into components. In this case, the force due to gravity has been resolved into two components. The net force is parallel to the slope. (b)

(a) Normal force

Normal force

Road friction and air resistance

Road friction and air resistance Force due to gravity

Force due to gravity

Components of the force due to gravity

It is often useful to divide vectors into parts called components. Figure 10.20b shows how the force due to gravity can be broken up, or resolved, into two components — one parallel to the slope and one perpendicular to the slope. Notice that the vector sum of the components is the force due to gravity. By resolving it into these two components, two useful observations can be made: 1. The normal force is balanced by the component of force due to gravity that is perpendicular to the surface. The net force has no component perpendicular to the road surface. This must be the case because there is no change in motion perpendicular to the slope. 2. The magnitude of the net force is simply the difference between the magnitude of the component of the force due to gravity that is parallel to the surface and the sum of the road friction and air resistance. SAMPLE PROBLEM 9

A car of mass 1600 kilograms left parked on a steep but rough road begins to roll down the hill. After a short while it reaches a constant speed. The road is inclined at 15° to the horizontal. The car’s speed is sufficiently slow that the air resistance is insignificant and can be ignored. Determine the magnitude of the road friction on the car while it is rolling at constant speed.

Normal force

15° Road friction Fgy = Magnitude of normal reaction force

15° Force due to gravity

Fgx = Magnitude of road friction

Teacher-led video: SP9 (tlvd-0084)

344 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

THINK

WRITE

Because the car is rolling at constant speed, the net force acting on it is zero. 2. As the net force is zero, the magnitude of the friction must be equal to the magnitude of the force due to gravity that is in the direction of the slope.

Fnet = 0

The force due to gravity in the direction of the slope can be calculated using the trigonometric ratio sine. 4. The force due to gravity can be found using the formula Fg = mg.

sin15° =

1.

3.

5.

Substitute the force due to gravity into the formula for Fgx .

Friction = Fgx Fgx

Fg Fgx = Fg sin(15°)

Fg = mg = 1600 × 9.8 = 15 680 N Fgx = Fg sin(15°) = 15 680 sin(15°) ≈ 4058 N The road friction on the car is 4058 N. Note: It is useful to consider the effect on the net force of the angle of the incline to the horizontal. If the angle is greater than 15°, the component of the force due to gravity parallel to the slope increases and the net force will no longer be zero. The speed of the car will therefore increase. The component of the force due to gravity perpendicular to the slope decreases and the normal force decreases by the same amount.

PRACTICE PROBLEM 9 a. A 5000-kilogram truck is parked on a road surface inclined at an angle of 20° to the horizontal. Calculate the component of the force due to gravity on the truck that is: i. down the slope of the road ii. perpendicular to the slope of the road. b. In the case of the car in Sample problem 9, what is: i. the component down the road surface of the normal force acting on it ii. the normal force?

TOPIC 10 Forces in action 345

SAMPLE PROBLEM 10

A loaded supermarket shopping trolley with a total mass of 60 kilograms is left standing on a footpath that is inclined at an angle of 30° to the horizontal. As the tired shopper searches for his car keys, he fails to notice that the trolley is beginning to roll away. It rolls in a straight line down the footpath for 9 seconds before it is stopped by an alert (and very strong) supermarket employee. Find the: a. speed of the shopping trolley at the end of its roll b. distance covered by the trolley during its roll. Assume that the footpath exerts a constant friction force of 270 N on the runaway trolley. Teacher-led video: SP10 (tlvd-0085)

THINK a. 1.

Draw a diagram to show the three forces acting on the shopping trolley. Air resistance is not included as it is negligible. The forces should be shown as acting through the centre of mass of the loaded trolley as in the figure on the right. The components of the force due to gravity, which are parallel and perpendicular to the footpath surface, should also be shown on the diagram.

WRITE a. Normal force Friction

30° Fgy Fgx 30° Force due to gravity = mg = 588 N

2.

Calculate the net force.

3.

Use Newton’s second law to calculate the acceleration.

4.

5.

b. 1.

2.

Use a constant acceleration formula to calculate the final speed. State the solution. Use a constant acceleration formula to calculate the distance covered by the trolley.

State the solution.

346 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Fnet = Fgx − friction = mg sin(30°) − 270 N = 588 N sin(30°) − 270 N = 294 N − 270 N = 24 N Fnet = ma 24 = 60a 24 a= 60 = 0.4 m s−2 down the slope v = u + at = 0 + 0.4 × 9 = 3.6 m s−1 The speed of the trolley at the end of its roll is 3.6 m s−1 . 1 b. s = ut + at2 2 1 = 0 + × 0.4 × 92 2 = 16.2 m The trolley travels 16.2 metres before it is stopped.

PRACTICE PROBLEM 10 A cyclist rolls freely from rest down a slope inclined at 20° to the horizontal. The total mass of the bicycle and cyclist is 100 kilograms. The bicycle rolls for 12 seconds before reaching a horizontal surface. The surface exerts a constant friction force of 300 N on the bicycle. a. What is the net force on the bicycle (including the cyclist)? b. What is the acceleration of the bicycle? c. What is the speed of the bicycle when it reaches the horizontal surface?

Resources Video eLesson Motion down an inclined plane (eles-0034) Weblink

Inclined plane

10.6.3 Connected objects In many situations, Newton’s laws need to be applied to more than one body. Figure 10.21 shows a small dinghy being pulled by a larger boat. The forces acting on the larger boat are labelled in red, while the forces acting on the small dinghy are labelled in green. Newton’s second law can be applied to each of the two boats separately. FIGURE 10.21 The forces acting on each of the two boats Normal force

Normal force Resistance forces

Tension

Resistance forces Thrust

Tension

Force due to gravity

Force due to gravity

Figure 10.22 shows only the forces acting on the whole system of the two boats and the rope joining them. When Newton’s second law is applied to the whole system, the system is considered to be a single object. FIGURE 10.22 The forces acting only on the whole system. The system consists of the two boats and the rope joining them. Normal force on both boats Resistance forces on both boats

System

Thrust

Force due to gravity

TOPIC 10 Forces in action 347

The thrust that acts on the larger boat and the system is provided by the water. The propeller of the larger boat pushes back on the water and the water pushes forwards on the propeller blades. The only force that can cause the small dinghy to accelerate forward is the tension in the rope. If the tension in the rope is greater than the resistance forces on the dinghy, the dinghy will accelerate. If the tension in the rope is equal to the resistance forces on the dinghy, it will move with a constant velocity. If the tension in the rope is less than the resistance forces on the dinghy, it will slow down. That is, its acceleration will be negative. The rope pulls back on the larger boat with the same tension that it applies in a forward direction on the small dinghy. This is consistent with Newton’s third law. Through the rope, the larger boat pulls forward on the small dinghy with a force that is equal and opposite to the force with which the small dinghy pulls on the large boat. SAMPLE PROBLEM 11

A car of mass 1400 kilograms towing a trailer of mass 700 kilograms accelerates at a constant rate on a horizontal road. A thrust of 5400 N is provided by the forward push of the road on the driving wheels of the car. The road friction on the car is 800 N, while that on the trailer is 400 N. The air resistance on both the car and the trailer is negligible. Determine the: a. acceleration of both the car and trailer b. force with which the trailer is pulled by the car (labelled P in the figure in the solution). Teacher-led video: SP11 (tlvd-0086) THINK a. 1.

Draw a diagram to show the forces acting on the car and trailer.

WRITE a.

Direction of motion

Normal force

Normal force P

P Driving force

Road friction

Road friction Force due to gravity

Force due to gravity

Consider the car and trailer as a system. The acceleration of the car and trailer can be calculated using Newton’s second law if the net force on the system is known. 3. Use Newton’s second law to calculate the acceleration. 2.

4.

State the solution.

348 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Fnet = driving force − road friction (car) − road friction (trailer) = 5400 N − 800 N − 400 N = 4200 N Fnet = ma 4200 = 2100a 4200 a= 2100 = 2 m s−2 to the right The acceleration of both the car and trailer is 2 m s−2 to the right.

Calculate the force with which the trailer is pulled by the car, P, by considering the net force on the trailer. 2. State the solution.

b. 1.

b.

Fnet = ma P − 400 = 700 × 2 P = 700 × 2 + 400 P = 1800 N The force with which the trailer is pulled by the car is 1800 N.

PRACTICE PROBLEM 11 A boat of mass 2000 kilograms tows a small dinghy of mass 100 kilograms with a thick rope. The boat’s propellers provide a forward thrust of 4700 N. The total resistance forces of air and water on the boat and dinghy system amount to 400 N and 100 N respectively. a. What is the acceleration of the boat and dinghy? b. What is the net force on the dinghy? c. What is the magnitude of the tension in the rope?

10.6 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. A ball rolls down a hill with an increasing speed. (a) Draw a diagram to show all of the forces acting on the ball. (b) What is the direction of the net force on the ball? (c) What is the largest single force acting on the ball? (d) When the ball reaches a horizontal surface, it slows, eventually coming to a stop. Explain, with the aid of a diagram, why this happens. 2. When you try to push a broken-down car with its handbrake still on, it does not move. What other forces are acting on the car to produce a net force of zero? 3. Redraw figure 10.18 for a car with rear-wheel drive. Normal force 4. A cyclist of mass 60 kilograms is riding up a hill inclined at 30° to the horizontal at a constant speed. The mass of the bicycle is 20 kilograms. D The figure on the right shows the forces acting on the bicycle–cyclist 10 N system. (a) What is the net force on the bicycle–cyclist system? (b) What is the magnitude of the component of the weight of the 800 N system that is parallel to the road surface? (c) The sum of the magnitudes of the road friction and air resistance 30° on the system is 10 N. What is the magnitude of the driving force D? Not to scale (d) What is the magnitude of the normal force on the bicycle–cyclist system? 5. An experienced downhill skier with a mass of 60 kilograms (including skis) is moving down a slope inclined at 30° with increasing speed. She is moving in a straight line down the slope. (a) What is the direction of the net force on the skier? (b) Draw a diagram showing the forces acting on the skier. Show all of the forces as if they were acting through her centre of mass. (c) What is the magnitude of the component of the skier’s force due to gravity that is parallel to the slope? (d) If the sum of the forces resisting the movement of the skier down the slope is 8 N, what is the magnitude of the net force on her? 6. A skateboarder of mass 60 kilograms accelerates down a slope inclined at an angle of 30° to the horizontal. Her acceleration is a constant 2.0 m s−2 . What is the magnitude of the friction force resisting her motion? 7. A roller-coaster carriage (and occupants) with a total mass of 400 kilograms rolls freely down a straight track inclined at 40° to the horizontal with a constant acceleration. The frictional force on the carriage is a constant 180 N. What is the magnitude of the acceleration of the carriage?

TOPIC 10 Forces in action 349

Speed (m s−1)

8. A skateboarder of mass 56 kilograms is rolling freely down a straight incline. The motion of the skateboarder is described in the graph on the right. 6 (a) What is the magnitude of the net force on the skateboarder? (b) If the friction force resisting the motion of the 4 skateboarder is a constant 140 N, at what angle is the slope inclined to the horizontal? 2 9. What force provides the forward force that gets you moving when you are: (a) ice skating 0 (b) downhill skiing 2 4 6 8 (c) waterskiing Time (s) (d) skateboarding (e) swimming (f) rowing? 10. Front-wheel-drive cars have a number of advantages over rear-wheel-drive cars. Compare and comment on the forces acting on the tyres in the two different types of car while being driven at a constant speed on a horizontal road. 11. The magnitude of the force due to air resistance, R, on a car of mass 1200 kilograms can be approximated by the formula R = 0.6v2 , where R is measured in newtons and v is the speed of the car in m s−1 . (a) Design a spreadsheet to calculate the magnitude of the force of air resistance and the net force on a car for a range of speeds as it accelerates from 20 km h−1 to 50 km h−1 on a horizontal road. Assume that, while accelerating, the driving force is a constant 1800 N and the road friction on the non-driving wheels is a constant 300 N. (b) Use your spreadsheet to plot a graph of the net force versus speed for the car. (c) Modify your spreadsheet to show how the net force on the car changes when the same acceleration from 20 km h−1 to 60 km h−1 is undertaken while driving down a road at an angle of 10° to the horizontal. 12. A well-coordinated rollerblader is playing with a yoyo while accelerating on a horizontal surface. When the yoyo is at its lowest point for several seconds, it makes an angle of 5° with the vertical. Determine the acceleration of the rollerblader. 13. A student argues that since there are friction forces on the front and back wheels of a bike that act in opposite directions, the bike cannot move. Explain how the bike moves. 5° 14. Two loaded trolleys of masses 3.0 kilograms and 4.0 kilograms, which are joined by a light string, are pulled by a spring balance along a smooth, horizontal laboratory bench as shown in the following figure. The reading on the spring balance is 14 N.

3.0 kg

4.0 kg

(a) What is the acceleration of the trolleys? (b) What is the magnitude of the tension in the light string joining the two trolleys? (c) What is the net force on the 4.0 kilogram-trolley? (d) What would be the acceleration of the 4.0 kilogram trolley if the string was cut? 15. A warehouse worker applies a force of 420 N to push two crates across the floor as shown in the figure on the right. The friction force opposing the motion of the crates is a constant 2.0 N for each kilogram. (a) What is the acceleration of the crates? (b) What is the net force on the 40-kilogram crate? (c) What is the force exerted by the 40-kilogram crate on the 30-kilogram crate?

350 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

(d) What is the force exerted by the 30-kilogram crate on the 40-kilogram crate? (e) Would the worker find it any easier to give the crates the same acceleration if the positions of the two blocks were reversed? Support your answer with calculations.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

10.7 Momentum and impulse KEY CONCEPTS • Apply concepts of momentum to linear motion: p = mv. • Explain changes in momentum as being caused by a net force: F net =

∆p . ∆t

10.7.1 Momentum How difficult is it to stop a moving object? How difficult is it to make a stationary object move? The answer to both of these questions depends on two physical characteristics of the object: • the object’s mass • how fast the object is moving, or how fast you want it to move. The product of these two physical characteristics is called momentum. The momentum, p⃗ , of an object of mass m with a velocity v ⃗ is defined as: p⃗ = m v ⃗ Momentum is a vector quantity and has SI units of kg m s−1 . SAMPLE PROBLEM 12

What is the momentum of a train of mass 8 × 106 kilograms that is travelling at a speed of 15 m s−1 in a northerly direction? THINK

WRITE

Recall the formula for momentum. 2. Substitute the mass and velocity in to find the momentum.

p = mv p = mv

1.

3.

State the solution.

= 8 × 106 × 15 = 1.2 × 108 kg m s−1 The momentum of the train is 1.2 × 108 kg m s−1 north.

TOPIC 10 Forces in action 351

PRACTICE PROBLEM 12 A car of mass 1200 kilograms travels east with a constant speed of 15 m s−1 . It then undergoes a constant acceleration of 3 m s−2 for 2 seconds. What is the momentum of the car: a. before it accelerates b. at the end of the 2 seconds acceleration?

10.7.2 Impulse Making an object stop, or causing it to start moving, requires a non-zero net force. The relationship between the net force applied to an object and its momentum can be explored by applying Newton’s second law to the object. Fnet = ma ( ) ∆v ⇒ Fnet = m ∆t ⇒ Fnet ∆t = m∆v The product Fnet ∆t is called the impulse of the net force. The impulse of any force is defined as the product of the force and the time interval over which it acts. Impulse is a vector quantity with SI units of N s. m∆v = m(v − u) = mv − mu = p f − pi Where: pf = the final momentum of the object pi = the initial momentum of the object. Thus, the effect of a net force on the motion of an object can be summarised by the statement: impulse = change in momentum. In fact, when translated from the original Latin, Newton’s second law reads: The rate of change of momentum is directly proportional to the magnitude of the net force and is in the direction of the net force.

This is expressed algebraically as: ∆p ∆t ∆p ⇒ Fnet = k ∆t Fnet ∝

The SI unit of force, the newton, has been defined so that the constant of proportionality, k, equals 1. Thus: ∆p ⇒ Fnet = ∆t. The effect of a net force on the motion of an object can be summarised by the statement: impulse = change in momentum ⃗ F net Δt = mΔ v ⃗ 352 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

SAMPLE PROBLEM 13

A 30-gram squash ball hits a wall horizontally at a speed of 15 m s−1 and bounces back in the opposite direction at a speed of 12 m s−1 . It is in contact with the wall for an interval of 1.5 × 10−3 seconds. a. What is the change in momentum of the squash ball? b. What is the impulse on the squash ball? c. What is the magnitude of the force exerted by the wall on the squash ball? Teacher-led video: SP13 (tlvd-0088)

THINK a. 1.

Change in momentum is calculated from the initial and final momentums. Consider towards the wall as the positive direction. (Note: This decision is arbitrary, you could choose the positive direction to be away from the wall, your answer will have the same magnitude but opposite sign.)

2. b. 1.

2.

c. 1.

State the solution. Impulse of the net force on the squash ball = change in momentum of the squash ball. State the solution. Recall the formula for the net force.

2.

Substitute the change in momentum and change in time into the equation to find the net force.

3.

State the solution.

WRITE a.

pi = mu = 0.03 × 15 = 0.45 kg m s−1 pf = mv = 0.03 × −12 = −0.36 kg m s−1 Δp = pf − pi = −0.36 − 0.45

= −0.81 kg m s−1 The change in momentum of the squash ball is 0.81 kg m s –1 away from the wall. b. I = Δp = −0.81 N s−1 The impulse on the squash ball is 0.81 N s−1 away from the wall. Δp c. Fnet = Δt Δp Fnet = Δt −0.81 = 1.5 × 10−3 = 540 N The magnitude of the force exerted by the wall on the squash ball is 540 N.

PRACTICE PROBLEM 13 During a crash test a 1400-kilogram car travelling at 16 m s−1 collides with a steel barrier and rebounds with an initial speed of 4.0 m s−1 before coming to rest. The car is in contact with the barrier for 1.4 seconds. What is the magnitude of: a. the change in momentum of the car during contact with the barrier b. the impulse applied to the car by the barrier c. the force exerted by the barrier on the car?

TOPIC 10 Forces in action 353

Resources Digital document Investigation 10.5 Impulse, momentum and Newton’s Second Law of Motion (doc-31880)

10.7.3 Impulse from a graph The force that was determined in Sample problem 13 was actually the average force on the squash ball. In fact, the force acting on the squash ball changes, reaching its maximum magnitude when the centre of the squash ball is at its smallest distance from the wall. The impulse (I) delivered by a changing force is given by: I = Fav ∆t If a graph of force versus time is available, the impulse can be determined from the area under the graph. (You might recall that the displacement of an object can be determined by calculating the area under its velocityversus-time graph — and displacement = vav ∆t. Similarly, the change in velocity of an object can be determined by calculating the area under its acceleration-versus-time graph — and change in velocity = aav ∆t.) SAMPLE PROBLEM 14

The following graph describes the changing horizontal force on a 40-kilogram ice skater as she begins to move from rest. Estimate her velocity after 2 seconds.

400

Force (N)

300

C

200 A

B

100

0

0.5

1.0 Time (s)

354 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

1.5

2.0

THINK 1.

The magnitude of the impulse on the skater can be determined by calculating the area under the graph.

2.

Use the formula for impulse to calculate the change in velocity.

3.

State the solution.

WRITE

magnitude of impulse = area A + area B + area C 1 1 = × 1.1 × 400 + 0.9 × 200 + × 0.9 × 200 2 2 = 220 + 180 + 90 = 490 N I = Δp I = mΔv 490 = 40Δv 490 Δv = 40 = 12.25 m s−1 As the skater started at rest, her velocity after 2 seconds will be equal to the change in velocity. The skater’s velocity after 2 seconds is 12.25 m s−1 .

PRACTICE PROBLEM 14 Consider the motion described in Sample problem 14. a. Estimate the velocity of the skater after 1.1 seconds. b. What is the acceleration of the skater during the first 1.1 seconds? c. What constant force would produce the same change in velocity after 2.0 seconds?

10.7.4 Follow through Players of ball games are often advised to ‘follow through’. The force is then applied to the ball by the bat, racquet, club, stick or arm for a larger time interval. The impulse, F∆t, is larger and the change in momentum, ∆p, is therefore larger. Consequently, the change in velocity of the ball as a result of the applied force is greater.

FIGURE 10.23 Golfers are advised to ‘follow through’. The force is applied to the ball for a longer time, giving it more momentum.

10.7.5 Protecting that frail human body The human body does not cope very well with sudden blows. The skeleton provides a fairly rigid frame that protects the vital organs inside and, with the help of your muscles, enables you to move. A sudden impact to your body, or part of your body, can: • push or pull the bones hard enough to break them • tear or strain the ligaments that hold the bones together • tear or strain muscles or the tendons that join muscles to bones • push bones into vital organs like the brain and lungs • tear, puncture or crush vital organs like the kidneys, liver and spleen. The damage that is done depends on the magnitude of the net force and the subsequent acceleration to which your body is subjected. In any collision, the net force acting on your body, or part of your body, can be expressed as: ∆p Fnet = ∆t TOPIC 10 Forces in action 355

The symbol ∆p represents the change in momentum of the part FIGURE 10.24 Gloves make it possible of your body directly affected by that net force. The magnitude for wicketkeepers to catch a solid cricket of your change in momentum is usually beyond your control. ball travelling at high speed without severe pain and bruising. For example, if you are sitting in a car travelling at 100 km h−1 when it hits a solid concrete wall, the magnitude of the change in momentum of your whole body during the collision will be your mass multiplied by your initial speed. When you land on a basketball court after a high jump, the magnitude of the change in momentum of each knee will be its mass multiplied by its speed just as your feet hit the floor. You have no control over your momentum. You do, however, have control over the time interval during which the momentum changes. If ∆t can be increased, the magnitude of the net force applied to you will be decreased. You can do this by: • bending your knees when you land after jumping in sports such as netball and basketball. This increases the time interval over which your knees change their momentum, and decreases the likelihood of ligament damage • moving your hand back when you catch a fast-moving ball in sports such as cricket. The ball changes its momentum over a longer time interval, reducing the force applied to it by your hand. In turn, the equal and opposite force on your hand is less • wearing gloves and padding in sports such as baseball, softball and gridiron. Thick gloves are essential for wicketkeepers in cricket, who catch the solid cricket ball while it is travelling at speeds up to 150 km h−1 • wearing footwear that increases the time interval during which your feet stop as they hit the ground. This is particularly important for people who run on footpaths and other hard surfaces. Indoor basketball and netball courts have floors that, although hard, bend a little, increasing the period of impact of running feet.

DON’T BE AN EGGHEAD After bicycle helmets became compulsory in Victoria in July 1990, the number of head injuries sustained by cyclists decreased dramatically. Bicycle helmets typically consist of an expanded polystyrene liner about 2 centimetres thick, covered in a thin, hard, polymer shell. They are designed to crush on impact. In a serious bicycle accident, the head is likely to collide at high speed with the road or another vehicle. Even a simple fall from a bike can result in the head hitting the road at a speed of about 20 km h−1 . Without the protection of a helmet, concussion is likely as the skull decelerates and collides with the brain because of the large net force on it. If the net force and subsequent deceleration is large enough, the brain can be severely bruised or torn, resulting in permanent brain damage or death. The effect is not unlike that of dropping a soft-boiled egg onto a hard floor. Although a helmet does not guarantee survival in a serious bicycle accident, it does reduce the net force applied to the skull, and therefore increases the chances of survival dramatically. The polystyrene liner of the helmet increases the time interval during which the skull changes its momentum. Helmets used by motorcyclists, as well as in horse riding, motor racing, cricket and many other sports, all serve the same purpose — to increase the time interval over which a change in momentum takes place.

356 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 10.25 Helmets save lives and prevent serious injury in many activities.

BUCKLE UP Seatbelts are a relatively new introduction to vehicles, with Victoria being one of the first places in the world to make wearing them compulsory in 1970. Their introduction and the increasing emphasis on vehicle safety over recent decades has drastically reduced the number of fatalities. In a collision a car comes to a stop rapidly. An occupant not wearing a seatbelt continues at the original speed of the car (as described by Newton’s first law) until acted on by a non-zero net force. An unrestrained occupant therefore moves at speed until: • colliding with part of the interior of the car, stopping even more rapidly than the car itself, usually over a distance of only several centimetres • crashing through the stationary, or almost stationary, windscreen into the object collided with, or onto the road • crashing into another occupant closer to the front of the car. An occupant properly restrained with a seatbelt stops with the car. In a typical suburban crash, the acceleration takes place over a distance of about 50 centimetres. The rate of change of the momentum of a restrained occupant is much less. Therefore, the net force on the occupant is less. The addition of airbags to complement seatbelts has further improved occupant safety. Airbags provide an additional means of restraining the occupant and reducing the force experienced. As well as increasing the time interval over which the occupant comes to a stop, the combination of seatbelts and airbags: • spreads the force over a larger area of the body • reduces the likelihood of a collision between the body and the interior of the vehicle • keeps the occupant in an aligned position, reducing injuries to vulnerable areas such as the neck, as well as stopping them from crashing through the windscreen.

COMPUTER CRASH MODELLING Automotive engineers use computer modelling during FIGURE 10.26 Engineers use computers to the design and development of new vehicles to model collisions to design and develop features investigate the effectiveness of safety features. This that improve the safety characteristics of cars. sort of crash modelling takes place long before the Computer modelling takes place long before the first prototype is built and the first physical crash tests real crash tests are performed. take place. Computer crash modelling has resulted in improvements to front and side structural design and to internal safety features such as seatbelt and airbag systems. Modelling crashes allows the investigation of a wide range of collision types, including full frontal, offset frontal, angled frontal and pole or barrier; collisions between trucks and cars; and rear impacts. The possibilities are endless. The computer models are then verified with the crash testing of real vehicles. During side-impact modelling, the computer is used to test thousands of combinations of seatbelt, cushioning and airbag designs. For each test, the computer can be set up to calculate the forces acting on occupants, estimate the severity of injuries and compare results with other design solutions. One aspect of design that can be tested is the sensor that triggers airbags to inflate. Complex calculations and comparisons are performed by a microprocessor within the sensing module before it ‘decides’ whether or not to trigger the airbags. The crash events that are modelled to develop the airbag sensors include high- and low-speed collisions, full-frontal and angled-frontal impacts and pole- or tree-type collisions.

TOPIC 10 Forces in action 357

CARS THAT CRUMPLE Modern cars are designed to crumple at the front FIGURE 10.27 A vehicle that has undergone a roof and rear. This provision increases the time interval crush test during which the momentum of the car changes in a collision, further protecting its occupants from death or serious injury. Even though the front and rear of the car crumple, the passenger compartment is protected by a rigid frame. The engine is also surrounded by rigid structures that prevent it from being pushed into the passenger compartment. The tendency of the roof to crush is currently being reduced by increasing the thickness of the windows, using stronger adhesives and strengthening the structure of the roof panel and pillars. The inside of the passenger compartment is also designed to protect occupants. Padded dashboards, collapsible steering wheels and airbags are designed to reduce the rate of change of momentum of occupants in the event of a collision. Interior fittings like switches, door knobs and handbrakes are sunk so that the occupants do not collide with them.

10.7 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. A 1400-kilogram car travels at 60 km h−1 east. Calculate the momentum of the car. 2. Make an estimate to one significant figure of the magnitude of each of the following. (a) The average net force on a car while it is accelerating from 0 to 40 km h−1 in 3.2 seconds (b) The magnitude of the air resistance on an 80 kg skydiver who has reached a terminal velocity of 200 km h−1 (c) The momentum of an Olympic class athlete participating in the 100-metre sprint event (d) The momentum of a family car travelling at the speed limit along a suburban street (e) The impulse that causes a 70-kilogram football player who is running at top speed to stop abruptly as he collides with a goal post that he didn’t see (f) The impulse applied to a netball by a goal shooter as she pushes it up towards the goal at a speed of 5 m s−1 (g) The change in momentum of a tennis ball as it is returned to the server in a Wimbledon final 3. A 60 gram tennis ball is bounced vertically onto the ground. After reaching the ground with a downwards velocity of 8.0 m s−1 , the ball rebounds with a velocity of 6.0 m s−1 vertically upwards. (a) What is the change in momentum of the tennis ball? (b) What is the impulse applied by the tennis ball to the ground? Explain how you obtained your answer without any information about the change in momentum of the ground. (c) Does the ground actually move as a result of the impulse applied by the tennis ball? Explain your answer. (d) If the tennis ball is in contact with the ground for 2.0 × 10−3 s, what is the average net force on the tennis ball during this interval? (e) What is the average normal force during this time interval? 4. A 75-kilogram basketballer lands vertically on the court with a speed of 3.2 m s−1 . (a) What total impulse is applied to his feet by the ground? (b) If the basketballer’s speed changes from 3.2 m s−1 to zero in 0.10 seconds, what total force does the ground apply to his feet? (c) Estimate the height from which the basketballer fell to the court.

358 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Upward push of court floor (N)

Horizontal force on occupant (N)

5. A car with a total mass of 1400 kilograms (including occupants) travelling at 60 km h−1 hits a large tree and stops in 0.080 seconds. (a) What impulse is applied to the car by the tree? (b) What force is exerted by the tree on the car? (c) What is the magnitude of the deceleration of the 70-kilogram driver of the car if he is wearing a properly fitted seatbelt? 6. Airbags are fitted to the centre of the steering wheel of many new cars. In the event of a sudden deceleration, the airbag inflates rapidly, providing extra protection for a driver restrained by a seatbelt. Explain how airbags reduce the likelihood of serious injury or death. 7. Joggers are advised to run on grass or other soft surfaces rather than concrete paths or bitumen roads to reduce the risk of knee and other leg injuries. Explain why this is so. 8. The graph on the right shows how the horizontal force on the upper body of Occupant without seatbelt Occupant with seatbelt 10 000 each of two occupants of a car changes as a result of a head-on collision. One 8000 occupant is wearing a seatbelt while 6000 the other is not. Both occupants are stationary 0.10 seconds after the initial 4000 impact. 2000 (a) What is the horizontal impulse on the occupant wearing the seatbelt? 0 (b) If the mass of the occupant wearing the 0.02 0.04 0.06 0.08 0.10 seatbelt is 60 kilograms, determine the Time (s) speed of the car just before the initial impact. (c) Is the occupant who is not wearing the seatbelt heavier or lighter than the other (more sensible) occupant? (d) Write a paragraph explaining the difference in shape between the two curves on the graph. 9. The graph on the right shows how the upward push of the court floor changes as a 60-kilogram basketballer jumps vertically upwards to complete a 1600 slam dunk. (a) What is the impulse applied to the basketballer 1200 by the floor? (b) With what speed did the basketballer leave 800 the ground? (c) What was the average force exerted on the basketballer by the floor during the 400 0.10 second interval? (d) Explain why the initial upward push of the floor 0 is not zero. 0.02 0.04 0.06 0.08 0.10 10. A well-known politician makes the suggestion that if Time (s) cars were completely surrounded by rubber ‘bumpers’ like those on dodgem cars, they would simply bounce off each other in a collision and passengers would be safer. Discuss the merits of this suggestion in terms of Newton’s laws of motion.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

TOPIC 10 Forces in action 359

10.8 Torque KEY CONCEPT • Calculate torque: 𝜏 = r⊥ F.

10.8.1 Torque or the turning effect of a force So far the explanation of forces and motion has treated objects as if they are a single point, or as if the force acts through the middle of the object; that is, its centre of mass. However, nature is more complicated than this. Friction acts at the rim of the front tyre of a bike to make it roll, the thigh muscle straightens the leg, a billiard cue hits the bottom edge of a ball to make it spin backwards, the wind blows over a tree, a pull on a handle opens the door. All these actions involve rotation, and a force has made the object turn. The turning effect of a force is called a torque. The symbol for torque is 𝜏, the Greek letter tau. Torque is always measured about a particular point. In some situations, the point about which an object will rotate is obvious (such as in a see-saw), and it may be referred to as a pivot or fulcrum. FIGURE 10.28 Examples of torque being applied

r⊥

Wheel rolls forward

Friction resists sliding

The size of a torque about a point or pivot is determined by the product of two factors: F • the size of the force, ⃗ • the perpendicular distance between the line of action of the force and the pivot, r⟂ . 𝜏 ⃗ = r⊥ ⃗ F

As a product of force and distance, torque has the units of newton metre (N m). It is also a vector, but because its effect is rotation, the direction of the vector is set by a rule. The rule is: If the rotation in the plane of the page is clockwise, the direction of the vector is into the page.

360 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

SAMPLE PROBLEM 15

A torque wrench is used to tighten nuts onto their bolts to a specific tightness or force. A torque wrench has a handle (black in the photo on the right) on one end and a socket that fits over a nut on the other end. In between is a scale that gives a reading in Newton metres. The scale on a torque wrench has a reading of 30 Newton metres. If the hand applying the force is 30 centimetres from the end, what is the size of the force by the hand on the wrench? Teacher-led video: SP15 (tlvd-0090) THINK

WRITE

Recall the formula for calculating torque. 2. Substitute the torque and perpendicular distance into the equation to find the force applied.

𝜏 = r⊥ F 30 = 0.3 × F 30 F= 0.3 F = 100 N The force by the hand on the wrench is 100 N.

1.

3.

State the solution.

PRACTICE PROBLEM 15 The handle of a torque wrench is hollow so an extension rod can be inserted. If you can exert only 30 N of force, how far along the extension rod from the handle should you place your hand to achieve a torque of 30 N m?

10.8 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. (a) A mechanic applies a force of 200 N to a wheel nut using a shifter. The perpendicular distance from where they apply the force to the nut is 25 centimetres. What torque are they applying to the nut? (b) The mechanic is unable to loosen the nut using the torque applied. Suggest two ways that they could increase the torque in this situation. 2. A lever is used to apply a torque of 20 N m about a pivot point. The perpendicular distance is 0.25 metres between the application of the force and the pivot point. What is the applied force? 3. There are myriad examples of everyday situations where we use devices that have a lever of some form to increase the torque that we apply. Examples include door handles, car steering wheels, electric motors, push bike pedals, wrenches, wheelbarrows, bottle lids. For one of these examples, or another that you can identify, estimate the force applied and the perpendicular distance to calculate an estimate of the torque involved. 4. Sam is standing at the right-hand end of the seesaw shown in the following figure. He places a bag on the seesaw and then begins walking up the plank to the left. Describe what happens as he walks towards, and then beyond, the fulcrum.

TOPIC 10 Forces in action 361

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

10.9 Equilibrium KEY CONCEPT • Investigate and analyse theoretically and practically translational forces and torques in simple structures that are in rotational equilibrium.

10.9.1 Equilibrium or keeping still Earlier in this topic, ‘keeping still’ meant not moving. If the net force was zero and the object was at rest, it would stay still. The forces were considered as acting on a single point. However, if the forces act at different points on the object, it is possible to have a net force of zero, but the object can still spin. In figure 10.29 the force upwards equals the force downwards, so the net force is zero, but the sphere rotates. In this case there is a net torque. The torques of the two forces about the centre add together. In cases such as car engines and electric motors, the production of a torque is essential for rotation and movement. But torque, and the rotation and movement it causes, can be detrimental. In bridges and building, the torque effect of a force can’t be avoided, but needs to be controlled if the structure is to remain standing. Such structures need to be designed so that not only is the net force equal to zero, but the net torque is also zero, and importantly this is true about every point in the structure. For a structure to be in equilibrium, two conditions need to apply. 1. Translational equilibrium: net force = zero 2. Rotational equilibrium: net torque about any point = zero

FIGURE 10.29 A zero net force can cause rotation. Nozzle

Pivot

Steam

Steam rises through tubes Water heated

10.9.2 Strategy for solving problems involving torque Questions regarding torque will often involve determining the value of two forces, so the solution will require generating two equations, which can then be solved simultaneously. First, draw a diagram with all the forces acting on the structure. Label each force. If its size is given in the question, write the value, for example, 10 N. If the size of the force is unknown, use a symbol such as F or R.

362 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

1. Using translational equilibrium: net force = zero It is easier to break this into two simpler tasks such as: a. sum of forces up = sum of forces down b. sum of forces left = sum of forces right 2. Using rotational equilibrium: net torque about any point = zero Choose a point about which to calculate the torques. Any point is acceptable, however, it can make solving the problem easier if you choose a point through which an unknown force acts. The torque of this force about that point will be zero as its line of action passes through the point, so that unknown will be effectively eliminated from this equation. sum of clockwise torques = sum of anticlockwise torques Now you will have two equations with two unknowns: one equation from the net force and one from the net torque. You can then solve the equations simultaneously to determine the unknown quantities. SAMPLE PROBLEM 16

Where should person 1 sit to balance the seesaw? R Person 1

Fg1 800 N

Person 2

Fg2 600 N 2m

d

THINK

WRITE

To satisfy equilibrium, both the sum of the forces acting on the seesaw and the sum of the torques must equal zero. 2. Consider the net force equilibrium. The sum of the upwards forces must equal the sum of the downwards forces. 3. Consider the net torque, taking the torques about the fulcrum at the centre.

Fnet = 0

1.

𝜏 net = 0 R = 800 + 600 R = 1400 N upwards

𝜏 net = 0 ⇒ sum of clockwise torques = sum of anticlockwise torques 600 × 2 = 800 × d 1200 800 = 1.5 m

d=

4.

State the solution.

To balance the seesaw person 1 must sit 1.5 metres to the left of the fulcrum.

TOPIC 10 Forces in action 363

PRACTICE PROBLEM 16 The following seesaw is balanced. Calculate the mass of person 1. R Person 1

Person 2

Fg2 600 N

Fg1 3.2 m

2.0 m

SAMPLE PROBLEM 17

Consider the painter’s plank supported between two trestles shown. The plank behaves as a simple bridge or beam and the weight of the painter must be transferred through the plank to the two trestles. The mass of the beam is 40 kilograms, the mass of the painter is 60 kilograms and she is a quarter of the distance from trestle 1. What is the magnitude of the reaction forces R1 and R2 ?

R1 R2

Fgp Fgb Trestle 1

L

Trestle 2

Teacher-led video: SP17 (tlvd-0092) THINK

WRITE

For the structure to be stable, the sum of the forces and the sum of the torques must both equal zero. 2. Consider the net force. The sum of the upwards forces must equal the sum of the downwards forces. 3. Consider the net torque, taking the torques about trestle 1.

Fnet = 0

1.

𝜏 net = 0 R1 + R2 = 40 × 9.8 + 60 × 9.8 = 980 N 𝜏 net = 0 ⇒ sum of clockwise torques = sum of anticlockwise torques 1 1 40 × 9.8 × L + 60 × 9.8 × L = R2 × L 2 4 1 1 40 × 9.8 × + 60 × 9.8 × = R2 2 4 196 + 147 = R2 R2 = 343 N

364 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

4.

Substitute the value of R2 into the equation for the net force.

R1 + R2 = 980 R1 + 343 = 980 R1 = 980 − 343

5.

R1 = 637 N The magnitude of R1 is 637 N and R2 is 343 N.

State the solution.

PRACTICE PROBLEM 17 A eucalyptus tree, 15 metres high and with a 200 centimetre diameter, was pulled over until it failed. The applied load was 6.0 kN m about the base of the tree. a. If the root ball of the tree has an average depth of 0.80 metres, what is the size of the force by the soil on the root ball at the point of failure? b. If the rope pulling on the tree was attached halfway up the tree, calculate the size of the force at the point of failure: i. in the rope (assuming the rope is horizontal) ii. by the ground at the base of the tree.

10.9.3 Types of structures: cantilevers A cantilever is a beam with one end free to move. A diving board, flagpole and a tree are examples of cantilevers. The diving board in figure 10.30 is supported by an upward force, R1 , from the bracket. The force due to gravity acts down through the middle of the board at a point further out. If these were the only forces on the diving board, the board would rotate anticlockwise. To prevent this rotation, the other end of the bracket pulls down on the diving board. The board is bolted to each end of the bracket. At which end are the bolts not needed? The tree in figure 10.31 is buffeted by winds from FIGURE 10.31 A tree buffeted by winds is a the left. The soil on the right at the base of the tree cantilever. is compressed and pushes back to the left. The Force on tree by wind combination of these two forces pushes the roots of the tree to the left, and the soil to the left of the roots pushes back to the right.

FIGURE 10.30 A diving board is an example of a cantilever. R1 (compression)

Tension Force on tree by surface Force due to gravity, mg

Force on tree (roots) by soil

TOPIC 10 Forces in action 365

10.9 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. A truck crosses a concrete girder bridge (as shown in the 20 m figure on the right). The bridge spans 20 metres and is supported at each end on concrete abutments. (a) Describe what happens to the reaction at each abutment as the truck moves across the bridge from left to right. (b) The truck weighs 12 tonne. Calculate the reaction at each support when the centre of gravity of the truck is 4 metres from the right abutment.

12 t

2. A person standing on the outside edge of the cantilevered balcony shown in the following figure walks inside.

(a) Explain what forces are necessary to support the balcony. (b) As the person walks across the balcony, describe what happens to the reaction at the support. 3. The truck crane in the following figure is able to lift a 20-tonne load at a radius of 5 metres. If the weight of the truck is evenly distributed, how heavy must the truck be if it is not to tip over? Assume the weight of the truck is uniformly distributed. 5m

1.8 m

20 t

366 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

1.8 m

4. The pedestrian bridge spanning the creek in the following figure weighs 2 kN. Calculate the reaction at each end when the three people are in the positions shown. 5.0 m 4.0 m 2.0 m

800 N

600 N

200 N

6.0 m

5. How far beyond the edge of the boat can Pirate Bill walk along the plank before it tips and he falls into the water? The 6-metre long plank weighs 800 N and Pirate Bill weighs 500 N. 6m 4m

500 N 800 N

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

TOPIC 10 Forces in action 367

10.10 Review 10.10.1 Summary • • • • • • • • • • •

• • • • • • • • • •

Forces are used to model interactions between objects. Force is a vector quantity. Force due to gravity is a measure of the force on an object due to the pull of gravity. The force due to gravity on an object is directly proportional to its mass. A free body diagram depicts all external forces acting upon an object. The vector sum of the forces acting on an object is called the net force. The velocity of an object can change only if a non-zero net force acts on it. This statement is an expression of Newton’s First Law of Motion. When a non-zero net force acts on an object, the object accelerates in the direction of the net force. Newton’s Second Law of Motion describes the relationship between the acceleration of an object, the net force acting on it and the object’s mass. It can be expressed as Fnet = ma. When an object applies a force to a second object, the second object applies an equal and opposite force to the first object: Fon A by B = –Fon B by A . This statement is an expression of Newton’s Third Law of Motion. The forces acting on a moving vehicle are: – force due to gravity, acting down through the centre of mass – the normal force, applied perpendicular to the surface of the road – the driving force, applied in the direction of motion by the road – road friction, applied to the non-driving wheels opposite to the direction of motion – air resistance, applied opposite to the direction of motion. Newton’s second law can be applied to a single object or a system of multiple bodies that are in contact or connected together. The momentum of an object is the product of its mass and its velocity, p = mv. The impulse delivered to an object by a force is the product of the force and the time interval during which the force acts on the object, I = F∆t. The impulse delivered by the net force on an object is equal to the change in momentum of the object. This can be expressed as Fnet ∆t = m∆v. The impulse delivered by a force can be found by determining the area under a graph of force versus time. The damage done to the human body during a collision depends on the magnitude of the net force and subsequent acceleration the body is subjected to. The net force on a human body during a collision can be decreased by increasing the time interval during which the body’s momentum changes. Increasing the time interval to reduce the net force, and hence damage, to the human body is used to advantage in many sports and in road safety. Torque is a measure of the turning effect of a force and is calculated as 𝜏 = r⊥ F. Translational equilibrium occurs when all forces acting on an object are in balance. Rotational equilibrium occurs when all torques acting on an object are in balance.

Resources

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0035).

368 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

10.10.2 Key terms Air resistance is the force applied to an object, opposite to its direction of motion, by the air through which it is moving. The centre of mass of an object is the point at which all of its mass can be considered to be. Components are parts. Any vector can be resolved into a number of components. When all the components are added together, the result is the original vector. A force is a push or a pull. Forces model interactions between objects. Force is a vector quantity. The force due to gravity is the force applied to an object, due to gravitational attraction. Friction is the force applied to the surface of an object when it is pushed or pulled against the surface of another object. Gravitational field strength (g) is the force of gravity on a unit of mass. An idealisation makes modelling a phenomenon or event easier by assuming ideal conditions that don’t exactly match the real situation. The impulse of a force is the product of the force and the time interval over which it acts. Impulse is a vector quantity. Momentum is the product of the mass of an object and its velocity. It is a vector quantity. A quantity that is negligible is so small that it can be ignored when modelling a phenomenon or an event. The vector sum of the forces acting on an object is called the net force. The normal force is a force that acts perpendicularly to a surface as a result of an object applying a force to the surface.

Resources Digital document Key terms glossary (doc-32268)

10.10.3 Practical work and investigations Investigation 10.1 The relationship between mass and the force due to gravity Aim: To examine the relationship between the force due to gravity and mass Digital document: doc-31879

Investigation 10.2 Friction Aim: To observe differences in friction when a wooden block is pulled across a surface Digital document: doc-31877 Teacher-led video: tlvd-0822

Investigation 10.3 Force as a vector Aim: To show that force is a vector and that the net force is the vector sum of all the forces acting on an object; To analyse the forces acting on an object by resolving the forces into components Digital document: doc-32308

Investigation 10.4 Static, sliding and rolling friction Aim: To compare the relative sizes of different forms of friction Digital document: doc-32309

TOPIC 10 Forces in action 369

Investigation 10.5 Impulse, momentum and Newton’s Second Law of Motion Aim: To measure and record the velocity of the trolley (or glider) at two separate instants as the load is falling Digital document: doc-31880 Other practical work ideas: • Compare the accuracy of different methods of determining g. • Investigate the friction of running shoes or of a towel falling off a rack. • Investigate the compression and tension behaviour of plasticine. • A wet rag is hard to drag when it is spread out and pulled across the floor. What does the resistive force depend on? • Design and test a safety helmet for a papier mâché ‘egg’.

Resources Digital document Practical investigation logbook (doc-32265)

10.10 Exercises To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. In answering the following questions, assume the magnitude of the gravitational field strength near the Earth’s surface is 9.8 N kg−1 .

10.10 Exercise 1: Multiple choice questions Which of the following is not a vector quantity? A. Force B. Mass C. Acceleration D. Momentum 2. Which of the following forces would be acting on a basketball falling through the air during a game of basketball? A. Gravity B. Normal force C. Net force D. Air resistance 3. Which of the following is not at rest or in a uniform state of motion? A. A car travelling along a highway at a constant velocity using cruise control B. A rocket sitting motionless on the launch pad C. A ladder leaning against a wall D. A train accelerating uniformly as it departs a station 4. An object of mass 43 kilograms has a net force of 86 N applied to it. Calculate its acceleration. A. 9.8 m s−2 B. 0.50 m s−2 C. 2.0 m s−2 D. 1.6 m s−2

1.

370 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

5.

6.

7.

8.

9.

10.

Which of the following is not a Newton’s third law pair of forces? A. The force of a book on a table and the force of the table on the book B. The normal force on someone sitting in a chair and the force of gravity on them C. The force of a car tyre on the road and the force of the road on the car tyre D. The force due to gravity of the Earth on you and the force due to gravity of you on the Earth A car of mass 2250 kilograms is travelling at 20 m s−1 . Calculate its momentum. A. 45 000 kg m s−1 B. 4500 kg m s−1 C. 1125 kg m s−1 D. 112.5 kg m s−1 A tennis ball of mass 58 grams experiences a change in velocity of 155 km h−1 when struck by a racquet during a serve. Calculate the change in momentum it experiences. A. 90 kg m s−1 B. 9 kg m s−1 C. 25 kg m−1 D. 2.5 kg m s−1 A car of mass 1850 kilograms is travelling at 8 m s−1 around a carpark when it collides with a parked car. The collision lasts 0.2 seconds, after which the car has come to a complete stop. Determine the magnitude of the average force acting on the car. A. 9250 N B. 74 000 N C. 2960 N D. 14 800 N Some engineering students are discussing ways to reduce the force acting on a car during a collision. Based on your understanding of impulse and momentum, which of the following will not be effective in reducing the force acting on the car? A. Reduce the initial speed of the car (using more effective braking or lowered speed limits). B. Decrease the mass of the car (through better structural design and material selection). C. Reduce the duration (time taken) for the collision. D. All of the above will be effective. A student sits at the end of pipe attached to a shifter to apply a force of 735 N at a perpendicular distance of 125 centimetres from a wheel nut that they are struggling to undo. Calculate the torque applied to the nut. A. 93.8 N m B. 9380 N m C. 919 N m D. 91 900 N m

10.10 Exercise 2: Short answer questions As part of a practical investigation a physics student rolls objects horizontally off the edge of a table and records the motion of their fall onto the floor below. Identify the forces that will be acting on these objects during their fall to the floor. For each force you identify, describe the interaction it represents and the likely direction of the force. 2. A rear-wheel-drive car is accelerating forwards along a horizontal road surface. Draw and label a complete diagram of all external forces acting on the car. 3. A child pulls along a toy on a piece of string at an angle of 30 degrees to the horizontal. They apply a force of 25 N along the string. Determine the horizontal and vertical components of this force. 1.

TOPIC 10 Forces in action 371

4.

The following forces are acting on ropes used in a four-way tug of war (as viewed from above) as part of a school athletics carnival. Resolve all of the forces into north–south or east–west components and determine the size and direction of the net force. 3400 N North

15° 2500 N

10° 3000 N

2400 N 5°

A sports car of mass 1645 kilograms accelerates at 9.6 m s−2 . Determine the net force that would be required to produce this acceleration. 6. The Falcon Heavy rocket produces approximately 2.2 × 106 N of thrust during the first moments of liftoff. If the acceleration of the rocket at this instant is 1.57 m s−2 , calculate its mass. 7. A piano is falling through the air near the surface of Earth. It has a mass of 410 kilograms. a. At a particular instant during its fall, the force of air resistance acting to oppose its fall is 2400 N. Calculate its acceleration at this instant. b. Determine the magnitude of air resistance force that would be required for the piano to reach a constant velocity during its fall. 8. A removalist is pushing two heavy boxes across the floor of their truck. The horizontal forces acting on the boxes are shown in the following diagram. a. Calculate the acceleration of the boxes. b. Calculate the compression force acting between the two boxes. 5.

Box A 25 kg

Box B 50 kg

Push by removalist 175 N

Total friction from floor 40 N

A tennis ball is travelling at 65 m s−1 towards a player’s racquet an instant before it collides with it. A moment later, the ball leaves the racquet travelling at 33 m s−1 in the opposite direction. Assume that the ball has a mass of 57 grams. a. Determine the change in momentum that the ball experiences during this collision. b. Assuming that the collision duration is 0.0020 seconds, determine the average force applied by the racquet on the ball during the collision. 10. The following graph is a simplified representation of the force applied by a trampoline on an acrobat over the duration of a single rebound. Use this graph to determine the impulse applied to the acrobat. 9.

Force (N)

1200

0

372 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

0.15 Time (s)

0.3

10.10 Exercise 3: Exam practice questions Question 1 (1 mark) Consider the example of a textbook at rest on a table. Explain why the normal force acting on the book and the force due to gravity are not a Newton’s third law pair of forces. Question 2 (3 marks) A cycle tourist is towing all of their camping equipment, clothes and food behind their pushbike in a bike trailer. Whilst pedalling at a reasonable pace they produce a driving force of 172 N acting forwards from the rear wheel of their bike. The road friction and air resistance opposing the motion of the bike and trailer is equal to 34 N acting backwards (20 N on the bike and 14 N on the trailer). The total mass of the cyclist and bike is 95 kilograms. The total mass of the trailer and all of its payload is 20 kilograms. Determine the tension force in the link between the trailer and the bike. Question 3 (3 marks) A small rocket of mass 2500 kg is launched up along an inclined ramp at an angle of 42° from the horizontal. During the launch, the rocket engine provides a constant thrust of 18 000 N. Determine the acceleration of the rocket during its launch. It is reasonable to consider air resistance and friction to be negligible in this situation. Question 4 (2 marks) During a standardised car crash test, a vehicle of mass 1980 kilograms is travelling at 64.0 km h−1 when it strikes a barrier. The car rebounds and is travelling at 12.0 km h−1 in the opposite direction immediately after the collision. The duration of the collision with the barrier is 160 milliseconds. Determine the average force exerted by the car on the barrier during the collision. Question 5 (3 marks) A truck of mass 14 500 kilograms is crossing a bridge over a small river. The bridge span is 40.0 metres between the two supports. The bridge has a total mass of 46 000 kilograms, with its centre of gravity exactly in the middle of its span. Determine the reaction at each of the supports when the centre of mass of the truck is 9.00 metres from the right-hand support. 40 m 9m

10.10 Exercise 4: studyON topic test Fully worked solutions and sample responses are available in your digital formats.

Test maker Create unique tests and exams from our extensive range of questions, including practice exam questions. Access the assignments section in learnON to begin creating and assigning assessments to students.

TOPIC 10 Forces in action 373

AREA OF STUDY 1 HOW CAN MOTION BE DESCRIBED AND EXPLAINED?

11

Energy and motion

Numerous videos and interactivities are available just where you need them, at the point of learning, in your digital formats, eBookPLUS and learnON at www.jacplus.com.au.

11.1 Overview 11.1.1 Introduction What happens when a ball is dropped from a height of 1 metre? Depending on the type of ball, and the surface it is bounced on, there will be a variety of different answers, but could a ball ever rebound to a greater height than it started at? The answer to this question is based on the conservation of energy, a law of nature which is inherent to the world we live in. In this topic we will learn about the different forms of energy and how it can be transferred from one form to another. These transfers allow energy from the food we eat to be turned into mechanical energy, powering the body throughout the day.

FIGURE 11.1 In this collision the kinetic energy of a car is transferred into many other forms of energy. Energy from the car is used up as the car’s front is deformed and a large amount of energy is transferred to the object it collides with, the road and the particles of air around the car. Some of the energy is transferred into sound and heat.

374 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

11.1.2 What you will learn KEY KNOWLEDGE After completing this topic you will be able to: • apply the concept of work done by a constant force using: • work done = constant force × distance moved in direction of force: W = Fs • work done = area under force-distance graph • investigate and analyse theoretically and practically Hooke’s Law for an ideal spring: F = −k∆x • analyse and model mechanical energy transfers and transformations using energy conservation: • changes in gravitational potential energy near Earth’s surface: Eg = mg∆h 1 • potential energy in ideal springs: Es = k∆ x2 2 1 2 • kinetic energy: Ek = mv 2 E • analyse rate of energy transfer using power: P = t useful energy out • calculate the efficiency of an energy transfer system: 𝜂 = total energy in • analyse impulse (momentum transfer) in an isolated system (for collisions between objects moving in a straight line): I = ∆p • investigate and analyse theoretically and practically momentum conservation in one dimension. Source: VCE Physics Study Design (2016–2021) extracts © VCAA; reproduced by permission.

PRACTICAL WORK AND INVESTIGATIONS Practical work is a central component of learning and assessment. Experiments and investigations, supported by a Practical investigation logbook and Teacher-led videos, are included in this topic to provide opportunities to undertake investigations and communicate findings.

Resources Digital documents Key science skills — VCE Units 1–4 (doc-31856) Key terms glossary (doc-32270) Practical investigation logbook (doc-32271)

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0036).

11.2 Impulse and momentum KEY CONCEPTS • Analyse impulse (momentum transfer) in an isolated system (for collisions between objects moving in a straight line): I = ∆p. • Investigate and analyse theoretically and practically momentum conservation in one dimension.

11.2.1 Impulse and momentum in collisions When two objects, A and B, collide with each other, each object exerts a force on the other. These two forces are an example of Newton’s Third Law of Motion. The two forces act on different objects, are equal in magnitude and act in opposite directions. TOPIC 11 Energy and motion 375

The collision starts at some point in time and finishes at another point in time. The symbol Δt represents the duration of the collision. From Newton’s point of view, both objects measure the same duration. Multiplying Fon B by A by ∆t gives Fon B by A ∆t, which is the impulse on B by A. This impulse produces a change in momentum, but whose momentum? A or B? Because it is the impulse by A on B, it is B’s momentum that is changed. ⃗ F on

B by A

Δt = Ion

B by A

= Δ pB⃗

Similarly, the force on A by B produces a change in A’s momentum. ⃗ F on

A by B

Δt = Ion

A by B

= Δ pA⃗

Forces are actions by one object on another, but momentum can be said to be a quantity an object has, even if it is a vector. So a force acting for a time changes how much momentum an object has. In a collision, the two forces are equal in magnitude and opposite in direction. So the changes in momentum of A and B are also equal in magnitude and opposite in direction. Δ p⃗ B = −Δ pA⃗ Because momentum is a quantity, this statement can be interpreted as ‘the momentum that B gains equals the momentum that A loses’. This is the basis of a conservation principle. Total momentum is conserved. The ‘change’ in something is always the ‘final’ value minus the ‘initial’ value, or what is added to the ‘initial’ value to get the ‘final’. Δp = pfinal − pinitial or pfinal = pinitial + Δp For object B: ΔpB = pB final − pB initial For object A: ΔpA = pA final − pA initial So, from the above relationship: ΔpB = −ΔpA ( ) pB final − pB initial = − pA final − pA initial

Putting the initial momenta together gives:

pA final + pB final = pA initail + pB initial sum of momentum after = sum of the momentum before Total momentum is conserved. From this analysis it can be seen that the conservation of momentum is a logical consequence of Newton’s third law. The interaction between objects A and B can be summarised as follows. • The total momentum of the system remains constant. • The change in momentum of the system is zero. • The change in momentum of object A is equal and opposite to the change in momentum of object B.

376 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 11.2 The net force on this system of two blocks is zero; therefore, momentum is conserved. pB

pA

Before the collision A B

pA + pB

During the collision

FA

A FB B

pAB After the collision

A B

The total momentum of the system pAB after the collision is the same as the total momentum of the system before and during the collision.

SAMPLE PROBLEM 1

Consider the collision illustrated in figure 11.2. Block A has a mass of 5 kilograms, block B has a mass of 3 kilograms and each block has a speed of 4 m s−1 before the collision. After the collision, the blocks move off together. Friction may be ignored. a. Determine the velocity of the blocks after the collision. b. What is the change in momentum of each of the blocks? c. What is the impulse on block A during the collision? d. Determine the final velocity of block B if, instead of moving off together, block A rebounds to the left with a speed of 0.5 m s−1 .

Teacher-led video: SP1 (tlvd-0093)

THINK

Recall the formula for conservation of momentum. 2. Calculate the momentum of each block to find pA + pB . Take right as the positive direction.

a. 1.

3.

Use the formula for conservation of momentum to find the velocity of the blocks after the collision.

WRITE a.

pA + pB = pA+B pA + pB = mA vA + mB vB = 5 × 4 + 3 × −4 = 20 − 12 = 8 kg m s−1 to the right pA+B = pA + pB mA+B vA+B = 8 (5 + 3)v = 8 v = 1 m s−1

TOPIC 11 Energy and motion 377

4. b. 1.

State the solution. Consider block A only.

The velocity of the blocks after the collision is 1 m s−1 to the right. b. ΔpA = mA ΔvA = 5(1 − 4) = 5 × −3 = −15 kg m s−1

2.

3.

ΔpB should be 15 kg m s−1 to the right since the change in momentum of the whole system is zero. To confirm this, consider block B alone. State the solution.

Impulse is equal to the change in momentum. d. 1. By conservation of momentum pA + pB = 8 kg m s−1 .

c. 1.

= 15 kg m s−1 to the left ΔpB = mB ΔvB = 3(1 − (−4)) =3×5 = 15 kg m s−1 to the right The change in momentum of each of the blocks is 15 kg m s−1 . c. The impulse on block A is equal to the change in momentum of block A: 15 kg m s−1 to the left. d. pAf + pBf = 8 5 × −0.5 + 3vBf = 8 −2.5 + 3vBf = 8

2.

State the solution.

3vBf = 10.5 10.5 vBf = 3 vBf = 3.5 m s−1 to the right The final velocity of block B if, instead of moving off together, block A rebounds to the left with a speed of 0.5 m s−1 is 3.5 m s−1 to the right.

PRACTICE PROBLEM 1 Consider a collision in which a model car of mass 5 kilograms travelling at 2 m s−1 in an easterly direction catches up to and collides with an identical model car travelling at 1 m s−1 in the same direction. The cars lock together after the collision. Friction can be assumed to be negligible. a. What was the total momentum of the two-car system before the collision? b. Calculate the velocity of the model cars as they move off together after the collision. c. What is the change in momentum of the car that was travelling faster before the collision? d. What is the change in momentum of the car that was travelling slower before the collision? e. What was the magnitude of impulse on both cars during the collision? f. How are the impulses on the two cars different from each other?

378 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

11.2.2 Modelling real collisions The Law of Conservation of Momentum makes it possible to predict the consequences of collisions between two cars or between two people on a sporting field. For example, if a 2000-kilogram delivery van travelling at 30 m s−1 (108 km h−1 ) collided with a small, stationary car of mass 1000 kilograms, the speed of the tangled wreck (the two vehicles locked together) could be predicted. However, you would need to assume that the frictional forces and driving force acting on both cars were zero after the collision. A reasonably good estimate can be made of the speed of the tangled wreck immediately after the collision in this way. The initial momentum p⃗ i of the system is given by: p⃗ van + pcar ⃗ = 2000 kg × 30 m s−1 + 0 kg m s−1 = 60 000 kg m s−1 Where the initial direction of the van is taken to be positive. The momentum of the system after the collision pf is the momentum of just one object — the tangled wreck: pf = 3000 kg × v where v is the velocity of the tangled wreck after the collision. But since pf = pi : 3000 kg × v = 60 000 kg m s−1 60 000 kg m s−1 3000 kg = 20 m s−1

⇒v=

The speed of the small car changes a lot more than the speed of the large van. However, the change in the momentum of the car is equal and opposite to that of the van. Δpcar = 1000 kg × 20 m s−1 − 0 kg m s−1 = 20 000 kg m s−1 Δpvan = 2000 kg × 20 m s−1 − 2000 kg × 30 m s−1 = 40 000 kg m s−1 − 60 000 kg m s−1 = −20 000 kg m s−1 If the small car hit the stationary van at a speed of 30 m s−1 , and the two vehicles locked together, the speed of the tangled wreck would be less than 20 m s−1 . Apply the Law of Conservation of Momentum to predict the speed of the tangled wreck immediately after this collision.

Resources Digital document Investigation 11.1 Simulating a collision (doc-31878) Teacher-led video Investigation 11.1 Simulating a collision (tlvd-0823)

TOPIC 11 Energy and motion 379

11.2 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Explain in terms of the Law of Conservation of Momentum how astronauts walking in space can change their speed or direction. 2. A physics student is experimenting with a low-friction trolley on a smooth horizontal surface. Predict the final velocity of the 2-kilogram trolley in each of the following two experiments. (a) The trolley is travelling at a constant speed of 0.6 m s−1 . A suspended 2 kilograms mass is dropped onto it as it passes. (b) The trolley is loaded with 2 kilograms of sand. As the trolley moves with an initial speed of 0.6 m s−1 , the sand is allowed to pour out through a hole behind the rear wheels. 3. Two stationary ice skaters, Catherine and Lauren, are facing each other and use the palms of their hands to push each other in opposite directions. Catherine, with a mass of 50 kilograms, moves off in a straight line with a speed of 1.2 m s−1 . Lauren moves off in the opposite direction with a speed of 1.5 m s−1 . (a) What is Lauren’s mass? (b) What is the magnitude of the impulse that results in Catherine’s gain in speed? (c) What is the magnitude of the impulse on Lauren while the girls are pushing each other away? (d) What is the total momentum of the system of Catherine and Lauren just after they push each other away? (e) Would it make any difference to their final velocities if they pushed each other harder? Explain. 4. Nick and his brother Luke are keen rollerbladers. Nick approaches his stationary brother at a speed of 2.0 m s−1 and bumps into him. As a result of the collision Nick, who has a mass of 60 kilograms, stops moving and Luke, who has a mass of 70 kilograms, moves off in a straight line. The surface on which they are ‘blading’ is smooth enough that friction can be ignored. (a) With what speed does Luke move off? (b) What is the magnitude of the impulse on Nick as a result of the bump? (c) What is the magnitude of Nick’s change in momentum? (d) What is the magnitude of Luke’s change in momentum? (e) How would the motion of each of the brothers after their interaction be different if they pushed each other instead of just bumping? (f) If Nick held onto Luke so that they moved off together, what would be their final velocity? 5. An unfortunate driver of mass 50 kilograms travelling on an icy road in her 1200-kilogram car collides with a stationary police car of total mass 1500 kilograms (including occupants). The tangled wreck moves off after the collision with a speed of 7.0 m s−1 . The frictional force on both cars can be assumed to be negligible. (a) At what speed was the unfortunate driver travelling before her car hit the police car? (b) What was the impulse on the police car due to the collision? (c) What was the impulse on the driver of the offending car (who was wearing a properly fitted seatbelt) due to the impact with the police car? (d) If the duration of the collision was 0.10 seconds, what average net force was applied to the police car? 6. The Law of Conservation of Momentum can be written as ∆pB = −∆pA , which equates the change in momentum of two different objects, A and B. Newton’s third law is often expressed as Fon B by A = −Fon A by B . Although this equation looks very similar to a law of conservation, no such law exists for forces. Explain why this is the case.

380 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

11.3 Work and energy KEY CONCEPTS • Apply the concept of work done by a constant force using: • work done = constant force × distance moved in direction of force: W = Fs • work done = area under force-distance graph.

11.3.1 The concept of energy A lot has already been said about energy in previous topics of this text. The word energy is used often in everyday language. Try writing down your own definition of energy. Look up a dictionary — any dictionary — to find out what it means. Most dictionaries and some physics textbooks define energy as the capacity to do work. The term work is defined as the quantity of energy transferred when an object moves in the direction of a force applied to it. These definitions do not, however, provide a clear understanding of what energy actually is. Energy is a concept — an idea — that is used to describe and explain change. The following list of some of the characteristics of energy provides some clues as to what it really is. • All matter possesses energy. • Energy takes many different forms. It can therefore be classified. Light, internal energy, kinetic energy, gravitational potential energy, chemical energy and nuclear energy are some of the different forms of energy. The names given to different forms of energy sometimes overlap. For example, light is an example of radiant energy. Gamma radiation could be described as nuclear energy or radiant energy. Sound energy and electrical energy involve kinetic energy of particles. • Energy cannot be created or destroyed. This statement is known as the Law of Conservation of Energy. The quantity of energy in the universe is a constant. However, nobody knows how much energy there is in the universe. • Energy can be stored, transferred to other matter or transformed from one form into another. • Some energy transfers and transformations can be seen, heard, felt, smelt or tasted. • It is possible to measure the quantity of energy transferred or transformed. However, many energy transfers and transformations are not observable and therefore cannot be measured. The concept of energy allows us to keep track of the changes that take place in a system. The system could be the universe, Earth, your home, the room you are in, your body or a car.

TOPIC 11 Energy and motion 381

11.3.2 Getting down to work Energy can be transferred to or from matter in several different ways. It can be transferred by: • emission or absorption of electromagnetic radiation or nuclear radiation • heating or cooling as the result of a temperature difference • the action of a force on an object resulting in movement. An interaction that involves the transfer of energy by the action of a force is called a mechanical interaction. When mechanical energy is transferred to or from an object, the amount of mechanical energy transferred is called work. The work (W) done when a force ( ⃗ F ) causes a displacement ( s ⃗ ) in the direction of the force is defined as: work = magnitude of the force × magnitude of displacement in the direction of the force W=⃗ F s⃗ Work is a scalar quantity. The SI unit of work is the joule. One joule of work is done when a force of magnitude of 1 newton (N) causes a displacement of 1 metre in the same direction of the force. That is: 1J = 1N × 1m = 1Nm It is important to remember that work is always done by a force acting on something. SAMPLE PROBLEM 2

A shopper pushes horizontally on a loaded supermarket trolley of mass 30 kilograms with a force of 150 N to move it a distance of 5 metres along a horizontal, straight path. The friction force opposing the motion of the trolley is a constant 120 N. How much work is done on the trolley by: a. the force applied by the shopper b. the net force c. the shopper to oppose the friction force? Teacher-led video: SP2 (tlvd-0094) THINK a. 1.

2. b. 1.

2.

Recall the formula for work done.

State the solution. Calculate the net force on the trolley, then use the formula for work done.

State the solution.

382 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

WRITE

W = Fs = 150 × 5 = 750 J 750 J of work is done on the trolley by the force applied by the shopper. b. Fnet = 150 − 120 = 30 N a.

W = Fnet s = 30 × 5 = 150 J 150 J of work is done on the trolley by the new force.

The force applied to oppose the friction is equal to the friction force. Use this force to calculate the work done. 2. State the solution.

c. 1.

c.

W = Fs = 120 × 5 = 600 J 600 J of work is done on the trolley by the shopper to oppose the friction force.

PRACTICE PROBLEM 2 A warehouse worker pushes a heavy crate a distance of 2 metres across a horizontal concrete floor against a constant friction force of 240 N. He applies a horizontal force of 300 N on the crate. How much work is done on the crate by: a. the warehouse worker b. the net force?

11.3.3 Force-versus-distance graphs The work done by a force can be calculated from a force-versus-displacement, or force-versus-distance graph. For this to be applicable the force presented in the graph and the displacement must be in the same direction. The work done is equal to the area under the force-versus-distance graph.

SAMPLE PROBLEM 3

A filing cabinet is pushed in a straight line across the floor of an office during some spring cleaning. The force applied in the direction of its motion and its displacement are recorded in the graph provided. Determine the work done.

Force (N)

150

0

1

2 Distance (m)

3

THINK

WRITE

The work done can be calculated by the area under the graph. 1 2. The area of a trapezium is (a + b)h where a and b 2 are the lengths of the horizontal sides and h is the height. 3. State the solution.

W = area under graph

1.

1 (3 + 1) 150 2 = 300 J

W=

300 J of work is done.

TOPIC 11 Energy and motion 383

PRACTICE PROBLEM 3 A spring is used in an old fashioned pinball machine to launch the ball from a rest position into the arcade game. The following graph shows the force applied to the ball by the spring during this motion against the distance travelled in the direction of the force. Determine the work done.

Force (N)

50

0 5

10 Distance (cm)

15

11.3 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. A box is pushed across the floor with force as shown in the following force-versus-displacement graph. Calculate the work done to push the box 25 metres.

Force (N)

300 200 100

0

5

10

15

20

25

Distance (m)

2. How much work is done on a 4-kilogram brick as it is pushed a distance of 1.5 metres by a net force of 40 N? 3. Imagine that you are trying to push-start a 2000-kilogram truck with its handbrake on. How much work are you doing on the truck? 4. A toddler swings her fluffy toy by a string around in circles at a constant speed. How much work does she do on the toy in completing: (a) one full revolution (b) half of a full revolution?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice questions Fully worked solutions and sample responses are available in your digital formats.

384 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

11.4 Energy transfers KEY CONCEPTS • Investigate and analyse theoretically and practically Hooke’s Law for an ideal spring: F = −k∆x. • Analyse and model mechanical energy transfers and transformations using energy conservation: • changes in gravitational potential energy near Earth’s surface: Eg = mg∆h 1 • potential energy in ideal springs: Es = k∆x2 2 1 2 • kinetic energy: Ek = mv . 2

11.4.1 Kinetic energy Kinetic energy is the energy associated with the movement of an object. By imagining how much energy it would take to stop a moving object, it can be deduced that kinetic energy depends on the mass and speed of the object. A formula for kinetic energy can be deduced by recalling Newton’s First Law of Motion: Every object continues in its state of rest or uniform motion unless made to change by a non-zero net force.

The kinetic energy of an object can change only as a result of a non-zero net force acting on it in the direction of motion. It follows that the change in kinetic energy of an object is equal to the work done on it by the net force acting on it. If an object initially at rest is acted on by a net force of magnitude Fnet and moves a distance s (which will necessarily be in the direction of the net force), its change in kinetic energy ∆Ek can be expressed as: ∆Ek = Fnet s The quantity of kinetic energy it possesses is: Ek = Fnet s because the initial kinetic energy was zero. Applying Newton’s second law (Fnet = ma) to this expression: Ek = mas where m is the mass of the object. The movement of the object can also be described in terms of its initial velocity v and its final velocity u. The magnitudes of the quantities a, s, v and u are related to each other by the equation: v2 = u2 + 2as If the object acquires a speed v as a result of the work done by the net force: v2 = 2as since u = 0 ⇒ as =

v2 2

Substituting this into the expression for kinetic energy: Ek = mas ⇒ Ek =

mv2 2 TOPIC 11 Energy and motion 385

The kinetic energy of an object of mass m and speed v ⃗ can therefore be expressed as: 1 𝟐 Ek = m v ⃗ 2 𝟐 Note that the momentum ( p⃗ = m v ⃗ ) is a vector quantity, whereas kinetic energy Ek = 21 m v ⃗ is a scalar quantity. SAMPLE PROBLEM 4

Compare the kinetic energy of an Olympic track athlete running the 100-metre sprint with that of a family car travelling through the suburbs. Assume the athlete is 70 kilograms and travelling at 10 m s–1 and the car is 1500 kilograms and travelling at 60 km h–1 . Teacher-led video: SP4 (tlvd-0096) THINK 1.

Use the estimated mass and speed to calculate the kinetic energy of the athlete.

2.

Convert the speed into m s−1

3.

Use the estimated mass and speed to calculate the kinetic energy of the family car.

4.

Interpret the results.

5.

State the solution.

WRITE

1 Ek = mv2 2 1 = × 70 × 102 2 = 3500 J Speed ≈ 60 km h−1 ≈ 17 m s−1 1 Ek = mv2 2 1 = × 1500 × 172 2 ≈ 217 000 J 217 000 = 62 3500 The family car has approximately 62 times more kinetic energy than the athlete.

PRACTICE PROBLEM 4 a. Calculate the kinetic energy of a 2000-kilogram elephant charging at a speed of 8.0 m s−1 . b. Estimate the kinetic energy of: i. a cyclist riding to work ii. a snail crawling across a footpath.

SAMPLE PROBLEM 5

A shopper pushes horizontally on a loaded supermarket trolley of mass 30 kilogram with a force of 150 N to move it a distance of 5 metres along a horizontal, straight path. The friction force opposing the motion of the trolley is a constant 120 N. If the trolley starts from rest what is its final speed? Teacher-led video: SP5 (tlvd-0097)

386 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

THINK

WRITE

The change in kinetic energy of the trolley is equal to the work done on it by the net force acting on it. 2. As the trolley was initially at rest the change in kinetic energy is equal to the final kinetic energy.

ΔEk = Fnet s

1.

3.

4.

Use the formula for kinetic energy to calculate the velocity.

Ek = Fnet s = 30 × 5.0 = 150 J 1 Ek = mv2 2 1 150 = × 30 × v2 2 150 v2 = 15 = 10 √ v = 10

≈ 3.2 m s−1 The trolley’s final speed is 3.2 m s−1 .

State the solution.

PRACTICE PROBLEM 5 A gardener pushes a loaded wheelbarrow with a mass of 60 kilograms a distance of 4 metres along a straight horizontal path against a constant friction force of 120 N. He applies a horizontal force of 150 N on the wheelbarrow. If the wheelbarrow is initially at rest, what is its final speed? If the net force is in the opposite direction to that in which the object is moving, the object slows down. For example, the work done by the net force to stop a 70-kilogram athlete running at a speed of 10 m s−1 is given by: work done by net force = ∆Ek

1 = 0 − mv2 2 ( )2 1 = − × 70 kg × 10 m s−1 2 = −3500 J

The negative sign indicates that the direction of the net force is opposite to the direction of the displacement.

SPEED KILLS The truth of the slogan ‘Speed kills’ can be appreciated by comparing the kinetic energy of a 1500-kilogram car travelling at 60 km h−1 (16.7 m s−1 ) with that of the same car travelling at 120 km h−1 (33.3 m s−1 ). At 60 km h−1 , the car’s kinetic energy is: Ek = =

1 2 1

mv2 ( )2 × 1500 kg × 16.7 m s−1

2 = 2.1 × 105 J

TOPIC 11 Energy and motion 387

At 120 km h−1 , its kinetic energy is: Ek = =

1 2 1

mv2 ( )2 × 1500 kg × 33.3 m s−1

2 = 8.3 × 105 J

In other words, a 100% increase in speed produces a 400% increase in the kinetic energy and therefore four times as much work needs to be done on the car to stop it during a crash with a solid object.

11.4.2 Potential energy Energy that is stored is called potential energy. Objects that have potential energy have the capacity to apply forces and do work. Potential energy takes many forms. • The food that you eat contains potential energy. Under certain conditions, the energy stored in food can be transformed into other forms of energy. Your body is able to transform the potential energy in food into internal energy so that you can maintain a constant body temperature. Your body transforms some of the food’s potential energy into the kinetic energy of blood, muscles and bones so that you can stay alive and move. Some of it is transformed into electrochemical energy to operate your nervous system. • Batteries contain potential energy. • An object that is in a position from which it could potentially fall has gravitational potential energy. The gravitational potential energy of an object is ‘hidden’ until the object is allowed to fall. Gravitational potential energy exists because of the gravitational attraction of masses towards each other. All objects with mass near the Earth’s surface are attracted towards the centre of the Earth. The further away from the Earth’s surface an object is, the more gravitational potential energy it has. • Energy can be stored in objects by compressing them, stretching them, bending them or twisting them. If the change in shape can be reversed, energy stored in this way is called strain potential energy. Strain potential energy can be transformed into other forms of energy by allowing the object to reverse its change in shape.

11.4.3 Gravitational potential energy When an object is in free fall, work is done on it by the force of gravity, transforming gravitational potential energy into kinetic energy. When you lift an object, you do work on it by applying an upwards force on it greater than or equal to its weight. Although the gravitational field strength, g, decreases with distance from the Earth’s surface, it can be assumed to be constant near the surface. The increase in gravitational potential energy ∆Egp by an object of mass m lifted through a height ∆h can be found by determining how much work is done on it by the force (or forces) opposing the force of gravity. W = Fs = mg∆h (substituting F = mg and s = ∆h) ⇒ ∆Eg = mg∆h This formula only provides a way of calculating changes in gravitational potential energy. If the gravitational potential energy of an object is defined to be zero at a reference height, a formula for the quantity of gravitational potential energy can be found for an object at height h above the reference height. ∆Eg = mg∆h ⇒ Eg − 0 = mg(h − 0) ⇒ Eg = mgh 388 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Usually the reference height is ground or floor level. Sometimes it might be more convenient to choose another reference height. However, it is the change in gravitational potential energy that is most important in investigating energy transformations. Figure 11.3 shows that the gain in gravitational potential energy as a raw egg is lifted from the surface of a table is mgd. When the raw egg is dropped to the table, the result will be the same whether you use the height of the table or ground level as your reference height. The gravitational potential energy gained will be transformed into kinetic energy as work is done on the egg by the force of gravity. FIGURE 11.3 The choice of reference height does not have any effect on the change in gravitational potential energy. ΔEg = mgΔh = mg(he − 0) = mgd

ΔEg = mgΔh = mg(H − ht) = mgd

h = he

h=H

d

d

h=0

h = ht

h=0

THE FOSBURY FLOP High jumpers use a technique called the Fosbury Flop that allows them to clear the bar while keeping their centre of mass as low as possible. The gravitational potential energy needed to clear the bar is minimised. Thus, with their maximum kinetic energy at take-off, high jumpers can clear those extra few centimetres. Incidentally, you might like to estimate just how much energy is needed to clear the bar in the high jump. Start by working out the change in height of an athlete’s centre of mass during a jump of about 2 metres.

FIGURE 11.4 The Fosbury Flop places a high jumper’s centre of mass below the bar.

TOPIC 11 Energy and motion 389

11.4.4 Strain potential energy Work must be done on an object by a force in order to change its strain potential energy. However, when objects are compressed, stretched, bent or twisted, the force needed to change their shape is not constant. For example, the more you stretch a rubber band, the harder it is to stretch it further. The more you compress the sole of a running shoe, the harder it is to compress it further. The amount of strain potential energy gained by stretching a rubber band or by compressing the sole of a running shoe can be determined by calculating the amount of work done on it. The amount of work done by a changing force is given by: W = Fav s It can be determined by calculating the area under a graph of force versus displacement in the direction of the force. In the case of a simple spring, a rubber band or the sole of a running shoe, the work done (and hence the change in strain potential energy) can be calculated by determining the area under a graph of force versus extension, or force versus compression. SAMPLE PROBLEM 6

Force applied to spring (N)

The following figure shows how the force required to compress a spring changes as the spring is compressed. How much strain potential energy is stored in the spring when it is compressed by 25 centimetres?

20

15

10

5

0 5

10

15

20

25

Compression (cm)

THINK

WRITE

The amount of strain potential energy added to the spring when it is compressed is equal to the amount of work done to compress it. 2. Work is equal to the area under the graph.

Es = W

1.

3.

State the solution.

390 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

W = area under graph 1 = × 20 × 0.25 2 = 2.5 J 2.5 J strain potential energy is stored in the spring when it is compressed by 25 centimetres.

PRACTICE PROBLEM 6 How much strain potential energy is stored in the spring described in Sample problem 6 when it is compressed by a distance of: a. 10 centimetres b. 20 centimetres?

Hooke’s Law The spring in Sample problem 6 is an example of an ideal spring. For an ideal spring, the force required to compress (or extend) the spring is directly proportional to the compression (or extension): F ∝ ∆x Where: F is the force exerted by the spring ∆x is the displacement of the spring. This relationship is expressed fully by Hooke’s Law, which states: F = −k∆x where k is known as the spring constant. The negative sign in Hooke’s Law is necessary because the direction of the force applied by the spring is always opposite to the direction of the spring’s displacement. For example, if the spring is compressed, it pushes back in the opposite direction. If the spring is extended, it pulls back in the opposite direction. Hooke’s Law is more conveniently expressed without vector notation as: ⃗ F = kΔ x⃗ Where: ⃗ F is the magnitude of the force applied by the spring Δ x⃗ is the magnitude of the extension or compression of the spring.

FIGURE 11.5 Graphs showing the force applied by an ideal spring versus (a) compression and (b) extension

Force applied to spring

F = kΔx

Force applied to spring

(b)

(a)

0 0

Δx Compression

F = kΔx

Δx Extension

TOPIC 11 Energy and motion 391

The strain energy stored in a spring that is changed in length by Δ x⃗ , whether it is compressed or extended, is equal to the area under the force-versus-compression graph or the force-versus-extension graph. That is: 1 × kΔ x⃗ × Δ x⃗ 2 1 = k(Δ x⃗ )2 2

strain potential energy (Es ) =

SAMPLE PROBLEM 7

A wooden block is pushed against an ideal spring of length 30 centimetres until its length is reduced to 20 centimetres. The spring constant of the spring is 50 N m−1 . a. What is the magnitude of the force applied on the wooden block by the compressed spring? b. How much strain potential energy is stored in the compressed spring? c. How much work was done on the spring by the wooden block? Teacher-led video: SP7 (tlvd-0099) THINK

Recall Hooke’s Law. 2. Substitute the values into the equation.

a. 1.

3.

b. 1.

State the solution. Recall the formula for strain potential energy.

2.

Substitute the values into the equation.

3.

State the solution.

The work done on the spring is equal to the elastic potential energy. 2. State the solution.

c. 1.

WRITE

F = kΔx Δx = 0.30 − 0.20 = 0.10 m = 50 × 0.10 = 5.0 N 5 N is applied on the wooden block by the compressed spring. 1 b. Es = k(Δx)2 2 a.

1 × 50 × 0.102 2 = 0.25 J 0.25 J of strain potential energy is stored in the compressed spring. c. W = Es = 0.25 J 0.25 J of work was done on the spring by the wooden block. =

PRACTICE PROBLEM 7 a. An object hanging from the end of a spring extends the spring by 20 centimetres. The spring constant is 60 N m−1 . i. What upwards force is applied to the object by the spring? ii. How much strain potential energy is stored in the spring when it is extended by 50 centimetres? iii. What is the mass of the object? b. What is the spring constant of the spring described in Sample problem 6?

392 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

11.4.5 Conservation of energy Conservation of energy is a law of the universe. It states that the total energy in an isolated system is constant. Considering the universe is an isolated system, the total amount of energy is constant, meaning energy cannot be created or destroyed. Energy can, however, be transferred from one form FIGURE 11.6 Energy is transferred into different to another. A single bounce of a tennis ball onto forms when a tennis ball is bounced. a hard surface involves the following mechanical energy transformations. • As the ball falls, the force of gravity does work on the ball, transforming gravitational potential energy into kinetic energy. • As soon as the bottom of the tennis ball touches the ground, the upwards push of the ground does work on the tennis ball, transforming kinetic energy into strain potential energy. A small amount of gravitational potential energy is also transformed into strain potential energy. This continues until the kinetic energy of the ball is zero. • As the ball begins to rise and remains in contact with the ground, the upwards push of the ground does work on the tennis ball, transforming strain potential energy into kinetic energy and a small amount of gravitational potential energy until the ball loses contact with the ground. • As the ball gains height, the force of gravity does work on the ball, transforming kinetic energy into gravitational potential energy. SAMPLE PROBLEM 8

A skateboarder of mass 50 kilograms, starting from rest, rolls from the top of a curved ramp, a vertical drop of 1.5 metres (see the following figure). What is the speed of the skateboarder at the bottom of the ramp? (The frictional force applied to the skateboarder by the ramp is negligible.) Height of centre of mass

1.5 m

1.5 m Height of centre of mass

Teacher-led video: SP8 (tlvd-0100)

TOPIC 11 Energy and motion 393

THINK

WRITE

It can be assumed in this case that the total mechanical energy is conserved. The only transformation that takes place is that from gravitational potential energy to kinetic energy. The gain in kinetic energy of the skateboarder is therefore equal to the magnitude of the loss of gravitational potential energy. 2. Substitute the values into the equation.

ΔEk = ΔEgp 1 2 mv = mgΔh 2

1.

ΔEk = ΔEgp 1 2 mv = mgΔh 2 1 × 50v2 = 50 × 9.8 × 1.5 2 25v2 = 735 735 25 √ 735 v= 25 v ≈ 5.4 m s−1 The speed of the skater at the bottom of the ramp is 5.4 m s−1 . v2 =

3.

State the solution.

PRACTICE PROBLEM 8 A toy car of mass 0.5 kilograms is pushed against an ideal spring so that the spring is compressed by 0.1 metres. The spring constant of the spring is 80 N m−1 . a. How much strain potential energy is stored in the spring when it is compressed? b. After the toy car is released, what will be its speed at the instant that the spring returns to its natural length?

AS A MATTER OF FACT Kangaroos have huge tendons in their hind legs that store and return elastic potential energy much more efficiently than do those of other mammals of comparable size. This allows them to hop for very large distances without tiring. An adult red kangaroo can jump over obstacles up to 2 metres in height. A young 50-kilogram kangaroo is capable of storing about 360 J of energy in each of its hind legs. A typical four-legged animal of the same mass stores about 55 J in each of its hind legs while running.

394 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 11.7 An adult red kangaroo.

11.4 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Use the formulae for work and kinetic energy to show that their units are equivalent. 2. Estimate the kinetic energy of: (a) a car travelling at 60 km h−1 on a suburban street (b) a 58 gram tennis ball as it is returned to the server in a Wimbledon final. 3. Estimate the amount of work done on a 58 gram tennis ball by the racquet when the ball is served at a speed of 200 km h−1 . 4. Estimate the change in gravitational potential energy of: (a) a skateboarder riding down a half-pipe (b) a child sliding from the top to the bottom of a playground slide (c) you at your maximum height as you jump up from rest. 5. A truck driver wants to lift a heavy crate of books with a mass of 20 kilograms onto the back of a truck through a vertical distance of 1 metre. The driver needs to decide whether to lift the crate straight up, or push it up along a ramp.

(a)

6. 7.

8.

9.

(b)

(a) What is the change in gravitational potential energy of the crate of books in each case? (b) How much work must be done against the force of gravity in each case? (c) If the ramp is perfectly smooth, how much work must be done by the truck driver to push the crate of books onto the back of the truck? (d) In view of your answers to (b) and (c), which of the two methods is the best way to get the crate of books onto the back of the truck? Explain your answer. World-class hurdlers raise their centre of mass as little as possible when they jump over the hurdles. Why? Two ideal springs, X and Y, have spring constants of 200 N m−1 and 100 N m−1 respectively. They are each extended by 20 centimetres by pulling with a hook. For each of the springs, determine: (a) the magnitude of the force applied to the hook (b) the strain potential energy. A tourist on an observation tower accidentally drops her 1.2-kilogram camera to the ground 20 metres below. (a) What kinetic energy does the camera gain before shattering on the ground? (b) What is the speed of the camera as it hits the ground? The following figure shows part of a roller-coaster track. As a fully loaded roller-coaster car of total mass 450 kilograms approaches point A with a speed of 12 m s−1 , the power fails and it rolls freely down the track. The friction force on the car can be assumed to be negligible.

A

Who switched off the lights?

20 m

C

8m Ground level

B

D

(a) What is the kinetic energy of the loaded car at point A? (b) Determine the speed of the loaded car at each of points B and C. (c) What maximum height will the car reach after passing point D?

TOPIC 11 Energy and motion 395

Driving force (N)

10. The following graph shows how the driving force on a 1200-kilogram car changes as it accelerates from rest over a distance of 1 kilometre on a horizontal road. The average force opposing the motion of the car due to air resistance and road friction is 360 N.

2000 1500 1000 500 0

250

500 750 Distance (m)

1000

(a) How much work has been done by the forward push (the driving force) on the car? (b) How much work has been done on the car to overcome both air resistance and road friction? (c) What is the speed of the car when it has travelled 1 kilometre? 11. A toy truck of mass 0.5 kilograms is pushed against a spring so that it is compressed by 0.1 metres. The spring obeys Hooke’s Law and has a spring constant of 50 N m−1 . When the toy truck is released, what will be its speed at the instant that the spring returns to its natural length? Assume that there is no frictional force resisting the motion of the toy truck. 12. A pogo stick contains a spring that stores energy when it is compressed. The following graph shows how the upwards force of a pogo stick on a 30-kilogram child jumping on it changes as the spring is compressed. The maximum compression of the spring is 8 centimetres. Assume that all of the energy stored in the spring is transformed to the mechanical energy of the child. The mass of the pogo stick itself can be ignored.

Upwards force (N)

1500

1000

500

2

4

6

8

Compression (cm)

(a) (b) (c) (d)

How do you know that the spring in the pogo stick is an ideal spring? What is the spring constant of the spring? How much work is done on the child by the pogo stick as the spring expands? What is the kinetic energy of the child at the instant that the compression of the pogo stick spring is zero? (e) How high does the child rise from the ground? Assume that the child leaves the ground at the instant that the compression of the pogo stick spring is zero. 13. Describe the mechanical energy transformations that take place when a child jumps up and down on a trampoline.

396 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

14. Discuss the mechanical energy transformations that take place when a diver uses a springboard to dive into the water, from the time that the diver is standing motionless on the springboard until the time she reaches her lowest point in the water. Use a graph describing the energy transformations in both the springboard and the diver to illustrate your answer. 15. Discuss the mechanical energy transformations that take place when a skateboard rider gets airborne off the end of a ramp (see the figure on the right). (a) Use a graph to describe the energy transformations that occur during the time interval between starting at one end of the ramp, getting airborne at the other end and returning to the starting point. (b) Explain in terms of the energy transformations how it is possible for the rider’s feet to remain in contact with the skateboard while in the air.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

11.5 Efficiency and power KEY CONCEPTS • Calculate the efficiency of an energy transfer system: 𝜂 = • Analyse rate of energy transfer using power: P =

E

useful energy out . total energy in

.

t

11.5.1 Efficiency Along with kinetic energy, gravitational potential energy and strain potential energy are referred to as forms of mechanical energy. Transformation to or from each of these forms of energy requires the action of a force. Of course, if mechanical energy were conserved, a ball would return to the same height from which it was dropped. In fact, mechanical energy is not conserved. During each of the mechanical transformations that occur during a bounce, some of the ball’s mechanical energy is transformed to thermal energy of the air, ground and ball, resulting in a small temperature increase. Some mechanical energy is also lost as sound, and as the ball undergoes permanent deformation. Mechanical energy losses to thermal energy, sound, and so on are largely permanent. It is very difficult to convert this lost energy back into mechanical energy and so it is not considered useful.

The efficiency, 𝜂, of an energy transfer is calculated from the ratio: 𝜂=

useful energy out total energy in

where 𝜂 is the Greek letter ‘eta’.

TOPIC 11 Energy and motion 397

SAMPLE PROBLEM 9

A ball dropped from 1.5 metres rebounds to 1.2 metres. What is the efficiency? THINK 1.

Recall the formula for efficiency.

2.

Calculate the total energy in.

3.

Calculate the useful energy out.

4.

Substitute these values into the formula to find the efficiency.

5.

State the solution.

WRITE

useful energy out total energy in The total energy in is the initial gravitational potential energy of the ball. Eg = mgh = mg × 1.5 The ‘useful energy out’ is the gravitational potential energy of the ball at its rebound height of 1.2 metres. Eg = mgh = mg × 1.2 useful energy out 𝜂= total energy in 1.2mg = 1.5mg = 0.8 = 80% The efficiency is 80%. 𝜂=

PRACTICE PROBLEM 9 A basketball is pumped up to give an efficiency of 80% when dropped. If this basketball is dropped from a height of 2 metres, to what height does it rebound after the fourth bounce?

11.5.2 Power Power is the rate at which energy is transferred or transformed. P=

E t

Where: E = the energy transformed, in joules t = the time taken, in seconds. In the case of conversions to or from mechanical energy or between different forms of mechanical energy, power, P, can be defined as the rate at which work is done. P=

W Δt

Where: W = the work done, in joules ∆t = the time interval during which the work is done, in seconds.

398 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

The SI unit of power is the watt (W), which is defined as 1 J s−1 . The power delivered when a force, F, is applied to an object can also be expressed in terms of the object’s speed v. W Fx = ∆t ∆t x =F× ∆t = Fv

P=

SAMPLE PROBLEM 10

A student of mass 40 kilograms walks briskly up a flight of stairs to climb four floors of a building, a vertical distance of 12 metres in a time interval of 40 seconds. a. At what rate is the student doing work against the force of gravity? b. If energy is transformed by the leg muscles of the student at the rate of 30 kJ every minute, what is the student’s power output? Teacher-led video: SP10 (tlvd-0102) THINK a. 1.

The work done by the student against the force of gravity is equal to the gain in gravitational potential energy.

2.

Recall the formula for power.

3.

State the solution.

b. 1.

Recall the formula for power.

WRITE

W = mgΔh = 40 × 9.8 × 12 = 4704 J W P= Δt 4704 = 40 ≈ 118 W The student is doing work against the force of gravity at a rate of 118 W. energy transferred b. P = time taken = 30 kJ min−1 a.

30 000 J 60 s = 500 W If energy is transformed by the leg muscles of the student at the rate of 30 kJ every minute, the student’s power output is 500 W. =

2.

State the solution.

PRACTICE PROBLEM 10 a. If all of the 720 J of energy stored in the hind legs of a young 50-kilogram kangaroo were used to jump vertically, how high could it jump? b. What is the kangaroo’s power output if the 720 J of stored energy is transformed into kinetic energy during a 1.2 second interval?

TOPIC 11 Energy and motion 399

Resources Digital document Investigation 11.2 Climbing to the top (doc-31881)

WHICH IS EASIER — RIDING A BIKE OR RUNNING? A normal bicycle being ridden at a constant speed of 4 m s−1 on a horizontal road is subjected to a rolling friction force of about 7 N and air resistance of about 6 N. The forward force applied to the bicycle by the ground must therefore be about 13 N. The mechanical power output required to push the bicycle along at this speed is: P = Fv = 13 N × 4 m s−1 = 52 W Running at a speed of 4 m s−1 requires a mechanical power output of about 300 W. Even walking at a speed of 2 m s−1 requires a mechanical power output of about 75 W. Riding a bicycle on a horizontal surface is less tiring than walking or running for two reasons. 1. Less mechanical energy is needed. The body of the rider does not rise and fall as it does while walking or running, eliminating the changes in gravitational potential energy. 2. Because the rider is seated, the muscles need to transform much less chemical energy to support body weight. The strongest muscles in the body can be used almost exclusively to turn the pedals. Once you start riding uphill or against the wind, the mechanical power requirement increases significantly. For example, in riding along an incline that rose 1 metre for every 10 metres of road distance covered, the additional power needed by a 50-kilogram rider travelling at 4 m s−1 would be: P= =

∆Egp ∆t mg∆h

1

of 4 m = 0.4 m. 10 50 kg × 10 N kg−1 × 0.4 m

In a time interval of 1 seconds, the vertical climb is ⇒P=

∆t

1s

= 200 W

11.5 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. When a cricket ball bounces on a hard surface, 32% of the kinetic energy is stored as elastic potential energy. If a 160 gram cricket ball is dropped from a height of 2 metres onto a hard surface, calculate: (a) the kinetic energy of the ball as it hits the ground (b) the maximum amount of elastic potential energy stored in the ball (c) the height to which it will rebound. 2. A car of mass 1500 kilograms travelling at 50 km h−1 collides with a concrete barrier. The car comes to a stop over a distance of 60 centimetres as the front end crumples. (a) What is the average net force on the car as it stops? (b) What is the average acceleration of the car and its occupants? Assume that the occupants are wearing properly fitted seatbelts. (c) What would be the average acceleration of properly restrained passengers in a very old car with no crumple zone if it stopped over a distance of only 10 centimetres? (The maximum magnitude of acceleration that humans can survive is about 600 m s−2 .)

400 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

(d) Explain in terms of mechanical energy transformations why the front and rear ends of cars are designed to crumple. (e) In a collision with a rigid barrier, would you feel safer in a large or a small car? Use some sample calculations to illustrate your answer. 3. A girl of mass 50 kilograms rollerblades freely from rest down a path inclined at 30° to the horizontal. The following graph shows how the magnitude of the net force on the girl changes as she progresses down the path.

Magnitude of net force (N)

300

200

100

0

4

8

12

16

20

Distance (m)

24 22 20 18 16 14 12 10 8 6 4 2

127 mm

Load (kN)

(a) What is the kinetic energy of the girl after rolling a distance of 8 metres? (b) What is the sum of the friction force and air resistance on the girl over the first 8 metres? (c) What is the kinetic energy of the girl at the end of her 20-metre roll? (d) How much gravitational potential energy has been lost by the girl during her 20-metre roll? (e) Account for the difference between your answers to (c) and (d). 4. The following graph shows the results of a roof crush test conducted in the laboratories of the Department of Civil Engineering at Monash University.

0 20

40 60 80 100 Displacement of ram (mm)

120

(a) How much work has been done on the roof when the ram has reached its maximum displacement? (b) If the car has a mass of 1400 kilograms, from what height would it need to be dropped on its roof to crush it by 127 millimetres? 5. Jo and Bill are conducting an experimental investigation into the bounce of a basketball. Bill drops the ball from various heights and Jo measures the rebound height. They also use an electronic timer with thin and very light wires attached to the ball and to alfoil on the floor to measure the impact time. A top-loading balance measures the mass of the ball. What physical quantities can they calculate using these four measurements? 6. A tractor engine has a power output of 80 kW. The tractor is able to travel to the top of a 500-metre high hill in 4 minutes and 30 seconds. The mass of the tractor is 2.2 tonnes. What is the efficiency of the engine?

TOPIC 11 Energy and motion 401

7. Human muscle has an efficiency of about 20%. Take a heavy mass, about 1–2 kilograms, in your hand. With your hand at your shoulder, raise and lower the mass 10 times as fast as you can. Measure the mass, your arm extension and the time taken, and calculate the amount of energy expended and your power output. 8. A pile driver has an efficiency of 80%. The hammer has a mass 500 kilograms and the pile a mass of 200 kilograms. The hammer falls through a distance of 5 metres and drives the pile 50 millimetres into the ground. Calculate the average resistance force exerted by the ground. 9. Estimate the average power delivered to a 58 gram tennis ball by a racquet when the ball is served at a speed of 200 km h−1 and the ball is in contact with the racquet for 4 milliseconds. 10. At what average rate is work done on a 4-kilogram brick as it is lifted through a vertical distance of 1.5 metres in 1.2 seconds? 11. In the sport of weightlifting, the clean-and-jerk involves bending down to grasp the barbell, lifting it to the shoulders while squatting and then jerking it above the head while straightening to a standing position. In 1983, Bulgarian weightlifter Stefan Topurov became the first man to clean and jerk three times his own body mass when he lifted 180 kilograms. Assume that he raised the barbell through a distance of 1.8 metres in a time of 3.0 seconds. (a) How much work did Stefan do in overcoming the force of gravity acting on the barbell? (b) How much power was supplied to the barbell to raise it against the force of gravity? (c) How much work did Stefan do on the barbell while he was holding it stationary above his head? 12. A small car travelling at a constant speed of 20 m s−1 on a horizontal road is subjected to air resistance of 570 N and road friction of 150 N. What power provided by the engine of a car is used to keep it in motion at this speed? 13. While a 60-kilogram man is walking at a speed of 2 m s−1 , his centre of mass rises and falls 3 centimetres with each stride. At what rate is he doing work against the force of gravity if his stride length is 1 metre? 14. A bicycle is subjected to a rolling friction force of 6.5 N and an air resistance of 5.7 N. The total mass of the bicycle and its rider is 75 kilograms. Its mechanical power output while being ridden at a constant speed along a horizontal road is 56 W. (a) At what speed is it being ridden? (b) If the bicycle was ridden at the same speed up a slope inclined at 30° to the horizontal, what additional mechanical power would need to be supplied to maintain the same speed? Assume that the rolling friction and air resistance are the same as on the horizontal road.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

11.6 Review 11.6.1 Summary • • • • •

If the net force acting on a system is zero, the total momentum of the system does not change. This statement is an expression of the Law of Conservation of Momentum. If there are no external forces acting on a system of two objects when they collide, the change in momentum of the first object is equal and opposite to the change in momentum of the second object. The Law of Conservation of Energy states that energy cannot be created or destroyed. Work is done when energy is transferred to or from an object by the action of a force. The work done on an object by a force is the product of the magnitude of the force and the magnitude of the displacement in the direction of the force. All moving objects possess kinetic energy. The kinetic energy of an object can be expressed as: Ek = 12 mv2

402 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

• •

The work done on an object by the net force is equal to the object’s change in kinetic energy. The change in gravitational potential energy of an object near the Earth’s surface can be expressed as: ∆Eg = mg∆h

• •

where ∆h is the object’s change in height. The change in the strain potential energy stored in an object can be found by determining the area under a force-versus-compression or force-versus-extension graph. The force F applied by an ideal spring when it is compressed or extended is proportional to its displacement ∆ x. This relationship is expressed by Hooke’s Law: F = −k∆x where k is known as the spring constant.

• • • •

•

1 The strain potential energy stored in an ideal spring is expressed as Es = k (∆x)2 . 2 Kinetic energy, gravitational potential energy and strain potential energy are referred to as forms of mechanical energy. During a mechanical interaction, it is usually reasonable to assume that total mechanical energy is conserved. The efficiency of an energy transfer is calculated from the ratio: useful energy out efficiency, 𝜂 = total energy in Power is the rate at which energy is transferred or transformed. In mechanical interactions, power is also equal to the rate at which work is done. E P= t The power delivered by a force is the product of the magnitude of the force and the velocity of the object on which the force acts.

Resources

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0036)

11.6.2 Key terms Gravitational potential energy is the energy stored in an object as a result of its position relative to another object to which it is attracted by the force of gravity. Kinetic energy is the energy associated with the movement of an object. Like all forms of energy, kinetic energy is a scalar quantity. A mechanical interaction is one in which energy is transferred from one object to another by the action of a force. Power is the rate of doing work, or the rate at which energy is transformed from one form to another. Strain potential energy is the energy stored in an object as a result of a reversible change in shape. It is also known as elastic potential energy. Work is done when energy is transferred to or from an object by the action of a force. Work is a scalar quantity.

Resources Digital document Key terms glossary (doc-32270)

TOPIC 11 Energy and motion 403

11.6.3 Practical work and investigations Investigation 11.1 Simulating a collision Aim: To show that momentum is conserved in a collision in which there are no unbalanced external forces acting on the system Digital document: doc-31878 Teacher-led video: tlvd-0823

Investigation 11.2 Climbing to the top Aim: To investigate the difference that extra load makes to the work done against gravity and the power developed Digital document: doc-31881

Other practical work ideas: • Investigate the magnetic collision between two air track gliders. Hint: Use an ultrasound motion detector or a high-speed digital camera. • Investigate the force of impact on a bouncing ball.

Resources Digital document Practical investigation logbook (doc-32271)

11.6 Exercises To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au.

11.6 Exercise 1: Multiple choice questions Two identical toy cars are travelling directly towards each other at 5 m s−1 . They have a head on collision. They are stuck together after the collision. Assume friction is negligible. What is their combined speed after the collision? A. 0 m s−1 B. 2.5 m s−1 C. 5 m s−1 D. 10 m s−1 2. Two physics students, Steve and Terri, are standing at rest next to each other on skateboards. They push against each other and move off in opposite directions. The friction on the floor is negligible. Steve has a mass of 75 kilograms and was moving at a speed of 5 m s−1 immediately after he lost contact with Terri. Terri has a mass of 60 kilograms. What is the impulse that Steve exerted on Terri. A. 15 N s−1 B. 300 N s−1 C. 375 N s−1 D. 6.25 N s−1

1.

404 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

3.

4.

5.

6.

7.

8.

9.

Trish applies a constant horizontal force of 125 N to push a 95-kilogram rock a distance of 5 metres horizontally across her lawn. Calculate the work done. A. 11 875 J B. 625 J C. 59 375 J D. 475 J Emily kicks a football high into the air before she is tackled. The ball strikes the ground at a speed of 25 m s−1 . Assuming that the ball is of regulation size and has a mass of 500 grams, calculate its kinetic energy the instant before it struck the ground. A. 12.5 J B. 156 J C. 313 J D. 6.25 J Kate creates a toy for her cats to play with by hanging a fluffy object at the bottom of a spring. The spring constant is 5 N m−1 . One of the cats is able to stretch the spring by 10 centimetres. How much energy has the cat stored in the spring? A. 0.5 J B. 0.05 J C. 0.25 J D. 0.025 J Rohan rides his pushbike home from work. At the end of his journey he has gained 25 metres in altitude. Assuming that his mass is 80 kilograms, determine how much gravitational potential energy he has gained over this journey. A. 2000 J B. 19 600 J C. 245 J D. 784 J Louise stretches a spring, causing it to extend in length by 17 centimetres. The spring constant is 125 N m−1 . Determine the force that Louise has applied to the spring at this extension. A. 21.25 N B. 2125 N C. 7.36 N D. 736 N To assist them in their studies of motion in physics, Grover and Bailey are riding on a roller coaster. The combined mass of the two students and the carriage is 369 kilograms. As part of the ride they stop momentarily at the highest point in the track. The cart then accelerates down a steep slope, dropping 40 metres in vertical height at the end of the slope. What is the total kinetic energy of the cart and students at the end of the slope. A. 200 295 J B. 295 200 J C. 648 144 J D. 144 648 J Scott is building a small robot that uses a number of electrical engines to run its wheels and arms. One of the engines is specified to be 90% efficient. If the engine is provided with 42 J of energy to perform a manoeuvre, how much useful energy will it output. A. 34 J B. 37.8 J C. 42 J D. 46.7 J

TOPIC 11 Energy and motion 405

10.

Matt is building a wall out of stone. This requires a lot of stone to be lifted into place. Matt lifts a 3.7-kilogram stone 47 centimetres in 0.8 seconds. What is the average rate that he is doing work on the stone during this motion? A. 2130 W B. 1704 W C. 21.3 W D. 17 W

11.6 Exercise 2: Short answer questions 1.

2.

3.

4. 5.

6.

7.

8.

9.

10.

John is practising for a tennis tournament by hitting a ball against a wall. The ball strikes the wall at 27 m s−1 and rebounds in the opposite direction at 19 m s−1 . The ball has a mass of 58.5 grams. a. Calculate the change in momentum of the ball. b. Calculate the average force exerted by the wall on the ball during this collision. A large car of mass 1980 kilograms travelling at 11 m s−1 collides head on with a small car of mass 970 kilograms. Immediately after they collide, the cars are stuck together and have come to a complete stop. Calculate the speed of the smaller car immediately before the collision. A teacher applies a constant vertical force 40 N to lift a 4-kilogram box of physics exam papers 0.8 metres directly upwards from the floor onto a bench. Assume that the motion is at a constant speed. Calculate the work done by the teacher on the box of exam papers. An Airbus A380 of mass 560 000 kilograms lands on a runway at a speed of 250 km h−1 . Determine its kinetic energy at this instant. As part of a practical investigation a physics student hangs various objects on an ideal spring and measures its extension. The spring constant is 90 N m−1 . One of the objects that the student hangs on the spring causes it to extend by 12 centimetres. a. What upwards force is applied to the object by the spring? b. What is the mass of the object? c. How much strain potential energy is stored in the spring? A rock climber climbs a cliff, gaining a vertical height of 185 metres above where they started by the end of their climb. Assuming that their mass is 72 kilograms, how much gravitational potential energy have they gained as a result of their climb? A vase of flowers, is at rest 175 metres vertically above the surface of the Earth. It falls freely, smashing onto the ground below. Assume that it has a mass of 420 grams and air resistance is negligible. Considering the instant immediately before it smashes on the ground, determine its: a. kinetic energy b. speed. A car of mass 1500 kilograms is travelling along a level road at a constant speed of 62 km h−1 . The driver applies the accelerator and the force on the driven wheels does 550 000 J of work to accelerate the car to a higher speed. Determine what the new speed of the vehicle is. A softball of mass 180 grams strikes a hard stationary surface while travelling at a speed of 23 m s−1 . It rebounds in the opposite direction at 10 m s−1 . a. Calculate the kinetic energy of the ball before and after this collision. b. Determine the efficiency of the collision. A car travelling at a constant speed of 30 m s−1 on a horizontal stretch of highway experiences a combined resistance force of 1150 N from the air and the road. Determine the power that the engine of the car must deliver to the driven wheels to keep it in motion at this constant speed.

406 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

11.6 Exercise 3: Exam practice questions Question 1 (1 mark) As part of a practical investigation, physics students are measuring the energy transformations of objects sliding down a ramp that they have constructed. A heavy box is allowed to slide down the ramp. The students find that the kinetic energy gained is less than the change in gravitational potential energy. Explain why this is the case. Assume that there are no significant errors in their methodology or measurement. Question 2 (3 marks) A runaway tram of mass 26 000 kilograms is travelling at a speed of 54 km h−1 when it collides with a stationary tram of mass 35 000 kilograms. After an impact of duration 0.40 seconds the two trams are stuck together and continue to move along the track. a. Calculate the speed of the trams after the collision. 2 marks b. Determine the average force acting on the runaway tram during the collision. 1 mark Question 3 (2 marks) A physics student is waiting tables at a restaurant. They have been doing this work for some time and have perfected the ability to carry a tray of drinks across the room in a purely horizontal motion at a constant speed. Is the student doing work on the tray of drinks during this constant speed motion? Refer to relevant physics principles in your response. Question 4 (2 marks) A new energy storage system is proposed as an alternative to using chemical batteries. It stores energy as gravitational potential energy instead. A prototype of this system uses an old mine shaft with a 450-metre vertical drop to reach the bottom. A 2500-kilogram mass is attached to an electrical motor/generator. The mass is lowered to the bottom of the shaft. When there is electricity available it is used to run the motor to raise the mass up. When electricity is needed, the mass is dropped and the motor is used to generate electricity. At peak capacity, the system takes 520 seconds for the mass to travel the entire 450-metre drop. Calculate the power output of the system. Question 5 (3 marks) A physics student is investigating the energy transformations involved in the workings of pinball machines. In particular they decide to focus on the launch of the ball. The following is a diagram they made of the key details of the machine. B Spring k = 300 Nm–1

A

Change in height 6 cm

Ball m = 80 g

At the start of the game the spring is compressed by 10 centimetres by the machine and the ball sits motionless against it at point A. When the spring is released it pushes on the ball, transferring its energy to the ball and travels up the machine to point B. At point B the ball has gained 6 centimetres in height. Calculate the speed of the ball when it reaches B. Assume that friction is negligible.

11.6 Exercise 4: studyON topic test Fully worked solutions and sample responses are available in your digital formats.

Test maker Create unique tests and exams from our extensive range of questions, including practice exam questions. Access the assignments section in learnON to begin creating and assigning assessments to students.

TOPIC 11 Energy and motion 407

UNIT 2 | AREA OF STUDY 1 REVIEW

AREA OF STUDY 1 How can motion be described and explained? OUTCOME 1 Investigate, analyse and mathematically model the motion of particles and bodies.

PRACTICE EXAMINATION STRUCTURE OF PRACTICE EXAMINATION Section

Number of questions

Number of marks

A

20

20

B

4

20 Total

40

Duration: 50 minutes Information:

• • •

This practice examination consists of two parts. You must answer all question sections. Pens, pencils, highlighters, erasers, rulers and a scientific calculator are permitted. You may use the VCAA Physics formula sheet for this task.

Resources Weblink VCAA Physics formula sheet

SECTION A — Multiple choice questions All correct answers are worth 1 mark each; an incorrect answer is worth 0. Use the following information to answer questions 1 and 2. A body travels 25 metres north in a time of 8.0 seconds. It stops for 2.0 seconds, then moves 7.0 metres south in a time of 2.0 seconds. 1. Which of the following is the best estimate of its average velocity? A. 2.7 m s–1 B. 1.5 m s–1 C. 3.3 m s–1 D. 3.5 m s–1 2. Which of the following is the best estimate of its average speed? A. 2.7 m s–1 B. 1.5 m s–1 C. 3.3 m s–1 D. 3.5 m s–1

408 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

3. A dodgem car was travelling at 5.0 m s−1 south when it collided head-on with another dodgem car, and its velocity changed to 3.0 m s−1 north over a time interval of 0.80 seconds. What is the average acceleration of the first dodgem car? A. 2.5 m s–2 north B. 2.5 m s–2 south C. 10.0 m s–2 north D. 10.0 m s–2 south Use the following information to answer questions 4 and 5. The following is a velocity-versus-time graph of a car moving away from a traffic stop. v (m s–1) 10 8 6 4 2

0

1

2

3

4

5

t (s)

4. What is the average acceleration of the car in the first two seconds? A. 2 m s–2 B. 4 m s–2 C. 6 m s–2 D. 8 m s–2 5. What is the distance covered by the car in the first four seconds? A. 24 m B. 32 m C. 40 m D. 48 m 6. Consider the following four descriptions for the motion of a body. i. Velocity is positive and acceleration is positive. ii. Velocity is positive and acceleration is negative. iii. Velocity is negative and acceleration is positive. iv. Velocity is negative and acceleration is negative. Which of the following best describes the two scenarios in which the body is slowing down? A. i. and ii. B. ii. and iii. C. i. and iii. D. i. and iv.

UNIT 2 Area of Study 1 Review 409

Use the following information to answer questions 7 and 8. A truck travelling on a road changes its velocity from 3.0 m s−1 to 9.0 m s−1 and covers a distance of 24 metres during this period of acceleration. 7. What is the magnitude of the truck’s acceleration? A. 0.25 m s–2 B. 0.50 m s–2 C. 1.5 m s–2 D. 4.0 m s–2 8. Which of the following is the best estimate for the time taken to accelerate the truck? A. 2 seconds B. 4 seconds C. 6 seconds D. 8 seconds 9. A passenger is not wearing a seatbelt in a car moving forward. The car runs head-on into a barrier and the passenger hits her head against the windscreen in front of her. The best explanation for her motion immediately after the car hit the barrier is which of the following? A. She experienced a force that pushed her forward towards the windscreen. B. She experienced no force on her and kept moving forward. C. She experienced a force from the barrier that pushed her towards the windscreen. D. She experienced no force on her by the barrier but one from the car, which pushed her forwards. 10. A man is in a lift travelling upwards at constant speed. He stands on a spring balance that reads 750 N. What is the magnitude of the force exerted on the man by the floor of the lift? A. Zero B. Slightly less than 750 N C. 750 N D. Slightly more than 750 N

11. A small car of mass 800 kilograms is accelerating at 2.5 m s−2 . What is the magnitude of the net force on the car? A. 320 N B. 1200 N C. 1600 N D. 2000 N 12. A truck of mass 3600 kilograms is carrying a load of mass M. A net driving force of 9600 N causes an acceleration of 2.0 m s−2 . What is the mass M? A. 6000 kg B. 4800 kg

C. 2400 kg D. 1200 kg 13. A body, J, travelling to the right at a high velocity, collides with a stationary body, K. Which of the following best describes the forces in the interaction between J and K? A. F on J by K = F on K by J

B. −F on J by K = −F on K by J

C. −F on J by K = F on K by J

D. F on J by K − F on K by J = 0

14. A window-washer with a mass of 60 kilograms is standing on a platform that is accelerating upwards at a constant rate of 1.2 m s–2 . What is the force exerted on the platform by the person? A. 660 N down B. 660 N up C. 588 N up D. 588 N down

410 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Use the following information to answer questions 15 and 16. A box is at rest on a ramp inclined at 40º to the horizontal, as shown in the following diagram. The force due to gravity is 98 N.

98 N

40°

15. What is the magnitude of the normal force that the surface of the ramp is exerting on the box? A. 63 N B. 75 N C. 82 N D. 98 N 16. What is the magnitude of the friction force that the surface of the ramp is exerting on the box? A. 63 N B. 75 N C. 82 N D. 98 N Use the following information to answer questions 17 and 18. A jogger with mass of 80 kilograms is moving east at 7 m s−1 . He then slows down to a velocity of 2 m s−1 east over a period of 2 seconds. 17. Which of the following best describes the change in momentum of the jogger? A. 560 kg m s–1 east B. 560 kg m s–1 west –1 C. 400 kg m s east D. 400 kg m s–1 west 18. What is the magnitude of the average force exerted on the jogger to slow him down? A. 280 N B. 200 N C. 160 N D. 80 N 19. A ball of mass 0.42 kilograms is at rest on the field. It is kicked and reaches a maximum speed of 13.5 m s–1 . What is the work done on the ball when it was kicked? A. 12 J B. 24 J C. 38 J D. 77 J 20. A spring has an unstretched length of 0.7 metres. After a mass of 2.5 kilograms is attached to it, its length is 1.2 metres. What is the spring constant of this spring? A. 20 N m–1 B. 35 N m–1 –1 C. 49 N m D. 98 N m–1 SECTION B — Short answer questions Question 1 (4 marks)

A cyclist is pedalling along a straight stretch of road at a constant velocity of 6.0 m s−1 . She then accelerates at a constant rate for 4.0 seconds, reaching a final velocity of 14 m s−1 . a. What is the magnitude of the cyclist’s acceleration?

1 mark

b. How far did she travel during the period that she was accelerating?

1 mark

While travelling at 14 m s

−1

the cyclist applied constant braking, coming to a stop over a distance of 24.5 m.

c. What is the magnitude of the cyclist’s acceleration while she was braking?

1 mark

d. How long did the bicycle slow down to a stop?

1 mark

UNIT 2 Area of Study 1 Review 411

Question 2 (6 marks) A van with mass 150 kilograms tows a small trailer with mass of 50 kilograms. They are connected by a towing rod that can be considered to be massless. The driving force is entirely provided by the electric motor on the van. The total resistive force on the van is 30 N, while the total resistive force on the trailer is 20 N. The van and trailer are accelerating at 1.9 m s–2 .

150 kg

50 kg rod 20 N

30 N

a. Calculate the net force acting on the van and trailer.

1 mark

b. Determine the size of the driving force provided by the electric motor on the van.

2 marks

c. What is the size of the tension force of the rod on the trailer?

3 marks

Question 3 (4 marks) Suzy, an ice skater with a mass of 55 kilograms, is skating at a constant velocity of 4.8 m s–1 east. She collides with Kai, another ice skater, who is stationary. They move off together at a constant velocity of 3.0 m s–1 east. The collision may be considered as isolated. a. What was Suzy’s initial momentum before the collision? 1 mark b. Considering that the collision is isolated, what is the mass of Kai?

3 marks

Question 4 (6 marks) A stationary cart, mass 600 kilograms, is at the top of a sloping ramp at point A. It is released and rolls down the ramp and attains its maximum speed of 14.4 m s–1 at point B. u = 0 m s–1

m = 600 kg

A

h v = 14.4 m s–1

B

C

a. Calculate the kinetic energy of the cart at point B.

2 marks

b. Assuming that friction is negligible and there is no energy lost, what is the height, h, of the ramp?

2 marks

c. The cart hits a spring at point C, which compresses and slows the cart down to a stop. The spring compresses a distance of 7.40 metres. What is the spring constant of the spring?

2 marks

412 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

PRACTICE SCHOOL-ASSESSED COURSEWORK ASSESSMENT TASK — DATA ANALYSIS In this task, you will be required to investigate, analyse and mathematically model the motion of particles and bodies. • A scientific calculator is permitted. You will need graph paper for this task. Total time: 50 minutes (5 minutes reading and 45 minutes writing) Total marks: 34 marks

TRACKING THE POSITION OF ATHLETES The motion of an athlete can be studied using a radar system that can track their position on a course. The data obtained (shown in table 1 below) gives the position of the runner down a straight athletics track in half-second intervals for the first three seconds of the race and one-second intervals thereafter. The athlete is training for a 100-metre race. At time t = 0, the athlete is in the starting blocks and the gun goes off. ∆x ∆t

USEFUL FORMULAE FOR THIS TASK Average speed: v =

Average acceleration: a = Momentum: p = mv

Kinetic energy: Ek = 21 mv2

∆v ∆t

1. Use the data in table 1 to plot a distance-versus-time graph. TABLE 1 Distance versus time Time (s)

Distance (m)

0.00

0.00

0.50

0.60

1.00

2.40

1.50

5.40

2.00

9.60

2.50

15.0

3.00

21.0

4.00

32.9

5.00

44.7

6.00

56.4

7.00

67.8

8.00

79.0

9.00

89.9

10.0

100.7

UNIT 2 Area of Study 1 Review 413

2. By taking pairs of data, complete a table for the average speed of the runner for each pair of data points in table 1. The first data point, in which the average speed of the athlete in the first interval is 1.2 m s–1 (determined through dividing the change in distance by the change in time), has been done for you. It occurs at the midpoint in time of the first frame. This will be at t = 0.25 seconds. TABLE 2 Average speed versus time Time (s)

Average speed (m s–1 )

0.25

1.20

0.75 1.25

3. Using your data from table 2, plot a graph of the athlete’s average speed versus time. Now we are going to find the acceleration of the athlete during the course of the race. 4. Complete table 3 for acceleration versus time. Again, your time should be the midpoint between each set of data points in table 2. TABLE 3 Acceleration versus time Time (s) 0.50

414 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Average acceleration (m s–2 )

5. Provide a definition of acceleration and explain, using an example from table 3, how you used this definition to determine the average acceleration between two time frames. 6. Plot a graph of the athlete’s acceleration versus time. 7. Give a detailed summary of the athlete’s motion, making reference to their speed and acceleration during the race. 8. Use your velocity-versus-time graph to estimate the distance run after 7.0 seconds, showing all workings. How does your estimate compare to the data in table 1? 9. By appealing to Newton’s Third Law of Motion, explain why the athlete is able to accelerate out of the starting blocks. 10. The mass of the athlete is 75 kg. Draw a free body force diagram to illustrate the three significant forces acting on the athlete during the time they are accelerating. (Hint: There is one vertical force, one horizontal force and one force acting on the athlete that has both horizontal and vertical components.) 11. During the period of acceleration, what is the magnitude of the horizontal component of the net force acting on the athlete? 12. From your graph of acceleration versus time, approximately when is the net force acting on the athlete zero? 13. Use your results from one of your graphs to estimate both the momentum and the kinetic energy of the athlete at t = 4.

Resources Digital document School-assessed coursework (doc-32276)

UNIT 2 Area of Study 1 Review 415

“Unit_2_AOS_1_Reviews_print” — 2019/10/21 — 5:56 — page 416 — #9

UNIT 2 | AREA OF STUDY 2 OPTIONS

OBSERVATION OF THE PHYSICAL WORLD The twelve option topics available for selection in Unit 2, Area of Study 2, are all based on a different observation of the physical world. The Option topics 12–23 can be accessed through your digital formats, learnON and eBookPLUS, or you can download a PDF at www.jacplus.com.au. Students will study only one option topic. Topic

Option

Outcome

12

What are stars?

Apply concepts of light and nuclear physics to describe and explain the genesis and life cycle of stars, and describe the methods used to gather this information.

13

Is there life beyond Earth’s solar system?

Apply concepts of light and atomic physics to describe and analyse the search for life beyond Earth’s solar system.

14

How do forces act on the human body?

Analyse the physical properties of organic materials including bone, tendons and muscle, and explain the uses and effects of forces and loads on the human body.

15

How can AC electricity charge a DC device?

Construct, test and analyse circuits that change AC voltage to a regulated DC power supply, and explain the use of transducers to transfer energy.

16

How do heavy things fly?

Apply concepts of flight to investigate and explain the motion of objects through fluids.

17

How do fusion and fission compare as viable nuclear energy power sources?

Apply the concepts of nuclear physics to describe and analyse nuclear energy as a power source.

18

How is radiation used to maintain human health?

Use nuclear physics concepts to describe and analyse applications of electromagnetic radiation and particle radiation in medical diagnosis and treatment.

19

How do particle accelerators work?

Apply the principles related to the behaviour of charged particles in the presence of electric and magnetic fields to describe and analyse the use of accelerator technologies in high energy physics.

20

How can human vision be enhanced?

Apply a ray model of light and the concepts of reflection and refraction to explain the operation of optical instruments and the human eye, and describe how human vision can be enhanced.

21

How do instruments make music?

Apply a wave model to describe and analyse the production of sound in musical instruments, and explain why particular combinations of sounds are more pleasing to the human ear than others.

22

How can performance in ball sports be improved?

Apply concepts of linear, rotational and fluid mechanics to explain movement in ball sports.

23

How does the human body use electricity?

Explain the electrical behaviour of the human body and apply electricity concepts to biological contexts.

Source: VCE Physics Study Design (2016–2021) extracts © VCAA; reproduced by permission.

UNIT 2 Area of Study 2 OPTIONS 417

AREA OF STUDY 2 OPTIONS OBSERVATION OF THE PHYSICAL WORLD

12

What are stars?

12.1 Overview Numerous videos and interactivities are available just where you need them, at the point of learning, in your digital formats, learnON and eBookPLUS at www.jacplus.com.au.

12.1.1 Introduction Astrophysics is the area of science in which the laws of physics are applied to the universe. It is quite remarkable that we can learn anything about the universe beyond the solar system as most stars are so far away that we cannot observe any features on them — they are mere points of light — and we cannot yet travel to the stars to explore them. If we ever travel to the stars, the journey will take so long that generations will pass before word of the discoveries comes back to Earth. Yet in the past hundred years, astrophysics has flourished and revealed much about our universe. Astrophysicists now can determine what stars are made of, how they generate their energy, their age and distance from us. This topic explores scientist’s understanding of stars and how they came to gain this knowledge. We have barely left the surface of the Earth, so how have we gained such understanding? FIGURE 12.1 Astrophysics is the field of science in which the laws of physics are applied to the universe.

TOPIC 12 What are stars? 1

12.1.2 What you will learn KEY KNOWLEDGE After completing this topic you will be able to: Astronomical measurement • explain the use of electromagnetic radiation in collecting information about the universe • identify all electromagnetic waves as travelling at the same speed, c, in a vacuum 1 • calculate wavelength, frequency, period and speed of light: c = f𝜆, T = f • identify spectroscopy as a tool to investigate the light from stars, and interpret and analyse spectroscopic data with reference to the properties of stars • apply methods used for measurements of the distances to stars and galaxies (standard candles, parallax, red shift) to analyse secondary data Classification of stars • describe the Sun as a typical star, including size, mass, energy output, colour and information obtained from the Sun’s radiation spectrum • identify the properties of stars, including luminosity, radius and mass, temperature and spectral type, and explain how these properties are used to classify stars 2 • explain nuclear fusion as the energy source of a star including: E = mc • distinguish between the different nuclear fusion phenomena that occur in stars of various sizes Stellar life cycle • apply the Hertzsprung–Russell diagram as a tool to describe the evolution and death of stars with differing initial mass • relate the formation of stars to the formation of galaxies and planets • describe future scenarios for a star, including white dwarfs, neutron stars and black holes 2GM to calculate the Schwarzschild radius • explain the event horizon of a black hole and use rs = c2 • describe the effects of the gravitational fields of black holes on space and time • compare the Milky Way galaxy to other galaxies with different shape, colour or size • explain and analyse how the chemical composition of stars and galaxies is used to determine their age • investigate selected aspects of stellar life cycles by interpreting and applying appropriate data from relevant databases. Source: VCE Physics Study Design (2016–2021) extracts © VCAA; reproduced by permission.

PRACTICAL WORK AND INVESTIGATIONS Practical work is a central component of learning and assessment. Experiments and investigations, supported by a Practical investigation logbook and Teacher-led videos, are included in this topic to provide opportunities to undertake investigations and communicate findings.

Resources Digital documents Key science skills — VCE Units 1–4 (doc-31856) Key terms glossary (doc-32279) Practical investigation logbook (doc-32280)

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0037).

2 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

12.2 BACKGROUND KNOWLEDGE Changing views of the universe BACKGROUND INFORMATION Observations of the night sky have changed over time from using just the naked eye to the use of sophisticated instruments.

12.2.1 Galileo’s telescope Galileo pointed his telescope at the heavens in 1609, only to find that they bore an unexpected resemblance to the Earth. The laws of physics that Galileo had begun to formulate would not only work down here, but out there as well; we were a part of the universe. On the other hand, as astrophysicists applied physics to the universe, they discovered that it is a very strange place indeed. By exploring space we have learnt that our experience on Earth is somewhat limited. This topic outlines how we have come to understand so much about the nature of stars since Galileo’s time.

FIGURE 12.2 Galileo was the first person to observe the heavens through a telescope.

12.2.2 The universe By ‘the universe’ we mean all matter, all space and all time. Anything that you can name or point out is part of the universe. Stars, planets, galaxies, people, light, even the space between the stars and the time between events are parts of our universe. Astrophysicists have found that much of the mass of the universe, as measured using the laws of physics, is hidden. So the universe, also sometimes called the cosmos, is everything that we know exists, and all that is yet to be found.

Ancient peoples’ view of the stars Imagine living in ancient times. You were often outside at night, under a sky unpolluted by the lights that have ‘fogged’ our night view in the past century. You saw stars in the sky, some bright and some dim. Some were clustered together; some were red, while others were white or blue. These differences enabled you to distinguish one star from another. You named them and grouped stars in an area of the sky to form images of things important to you: animals, heroes, gods and mythical creatures. We know these groupings as constellations, and different cultures organised the stars into their own sets of constellations. The arrangement of stars in a constellation does not noticeably change in a human lifetime, but people did notice that, as the night went on, the constellations moved across the sky, similar to the Sun moving across the sky in the daytime. As days passed, people saw that the constellations overhead slowly shifted. They realised that the same constellations were overhead in the cold months each year, and different ones were overhead in the warm months. This cycle repeated and people had a way to measure the passing of time and predict seasons, thus knowing when particular food would be available and, eventually, when to plant crops. Religious festivals were linked to particular events in this astronomical calendar; we continue to mark some of these today when we celebrate Christmas and Easter. The movement of the Sun and stars across the sky also gave important navigational information. It led to our understanding of north, south, east and west long before magnetic compasses were discovered.

TOPIC 12 What are stars? 3

Luckily in Australia, if we have the chance to get away from the lights of the large cities and view the stars, we can see them almost as our ancient ancestors did. Different cultures have viewed the universe differently, but until the seventeenth century most people believed that the Earth did not move, the stars were points of light fixed on a sphere that revolved around the Earth and the world of the heavens was made of different materials to those on Earth. Many believed it was the gods who caused things to happen, like the Sun rising and setting. The ancient Greeks developed several models of the universe; in some the Earth moved but most philosophers believed it stayed still, consistent with their everyday experience. The stars were fixed points of light on a sphere that was higher than a mountain top, but maybe not much more distant than that. These models that place the Earth at the centre of the universe are called geocentric. In the sixteenth and seventeenth centuries, Copernicus, Kepler and Galileo developed new models of the universe with the Sun at the centre and the Earth revolving around it — that is, heliocentric models. These models required the stars to be much further away than previously thought; otherwise, their relative positions would change as the Earth revolved around the Sun. Over the following hundred years, the idea that the stars were just points of light, all at the same distance from the centre of the universe, ceased. Stars were seen as suns like our own, at great distances from Earth. What else could be so far away, yet be so bright? FIGURE 12.3 (a) One ancient geocentric depiction of the universe (b) Copernicus’s heliocentric model, which had the planets revolving around the Sun (a)

(b)

Saturn Jupiter

Saturn

Stars

Mars Moon

Earth Venus

Venus Mercury

Sun

Earth Mercury

Jupiter

Sun

Moon

Mars

12.2.3 The discovery of galaxies Apart from the stars and planets, other objects are visible in the night sky, including a bright band known as the Milky Way. Galileo’s telescopic investigations showed this to be an enormous concentration of stars too faint for the naked eye to distinguish. In the southern hemisphere two large fuzzy objects are visible. These are the Large Magellanic Cloud and the Small Magellanic Cloud, named in honour of the explorer Magellan, although they would have been observed tens of thousands of years earlier by Indigenous Australians and other peoples of the southern hemisphere. Another object, fainter than the Magellanic clouds and about the size of the full Moon, was first recorded in the writings of the Persian astronomer Al-Sufi in 964 BCE. This ‘small cloud’ was in the constellation Andromeda. With the invention of the telescope, many more of these extended fuzzy objects were discovered. Due to their appearance they were named nebulae (meaning ‘clouds’). 4 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 12.4 An optical image showing part of the Milky Way. It shows an area of extremely dense star clusters and dark and bright (pink) nebulae. Dark nebulae are formed by clouds of interstellar dust that scatter and absorb the radiation from stars situated behind them. Bright nebulae are clouds of interstellar gas that is ionised and illuminated by the ultraviolet radiation of nearby hot and young stars. The Large Magellanic Cloud is the smudge of light near the bottom right corner and Canopus is the very bright star above that.

In the eighteenth century, William Herschel and his sister Caroline Hershel also shook the ancient perceptions of the universe. William built the largest telescope of his time and made some very significant discoveries — the most famous being his discovery of a new planet, Uranus, in 1781. This planet was so far away that its discovery doubled the size of the known solar system. He also noticed that some stars, called binary stars, occurred in pairs that orbited a common point. When William analysed their orbits he found that they obeyed Isaac Newton’s law of gravitation that described how masses attract each other. This was confirmation that the physical laws that applied on Earth also applied in the far reaches of space. Herschel also found that the Sun, with its planets, was moving through space. This meant that the Sun, like the Earth, was not fixed at the centre of the universe. Moreover, the stars did not seem to be evenly distributed throughout space but were gathered together in an elongated ‘island’ of stars. William and Caroline also discovered many new nebulae. With their large telescope they noticed that some of these ‘clouds’ appeared to contain distant stars. The large and small Magellanic clouds and the large nebula in Andromeda were part of this group. Many other nebulae were nearly devoid of stars, and it is these that are still called nebulae today.

TOPIC 12 What are stars? 5

FIGURE 12.5 These two nebulae are just visible to the naked eye and now known to be galaxies: (a) the Andromeda galaxy (b) the Large Magellanic Cloud. (a)

(b)

Source: (a) Adam Evans / Creative Commons (b) NASA / JPL-Caltech / M. Meixner STScI & the SAGE Legacy Team

Some of the nebulae the Herschels were seeing were actually galaxies. However, a good understanding of exactly what these objects were came only after another hundred years. The problem was that there was no way of measuring the distance to the nebulae. In the 1920s, debate raged about the size of the ‘island of stars’ we live in and whether some of the nebulae that had been discovered, particularly some in the shape of spirals, were within Herschel’s island of stars or outside it. By this time, an accurate method of measuring distance to these objects had been discovered using a particular type of variable star, known as a Cepheid variable. An astronomer by the name of Edwin Hubble applied this technique to the Andromeda nebula and found it to lie well beyond our own island of stars. The implication was that we lived in one galaxy (island of stars), the Milky Way, while the Andromeda nebula was an enormous galaxy of stars in its own right. It is actually a galaxy much bigger than the Milky Way, and is one of the closest to it at a distance of over two million light-years. There are several small galaxies, including the Large and Small Magellanic Clouds, that are closer than Andromeda. The closest yet discovered is a small galaxy found in 2003, named the Canis Major Dwarf Galaxy. The distance to the Andromeda nebula was the first clear evidence that the universe was not just the group of stars and gas clouds that form the Milky Way. The universe was clearly much, much larger. Our understanding of the stars, and their distribution throughout the universe, was once more dramatically changed. 6 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 12.6 Astronomers noted that some of the nebulae looked like spirals, but these were later found to be spiral galaxies. This photograph shows the large spiral galaxy known as the Whirlpool Galaxy.

Source: S. Beckwith STScI, Hubble Heritage Team, STScI/AURA, ESA, NASA

12.2.4 Where we live In the 1920s and 1930s Hubble and other astronomers surveyed the sky in search of galaxies and found tens of thousands that were roughly evenly distributed throughout the sky. Even the closer galaxies were so far away that it took the largest telescopes to identify that they were made up of individual stars. However, through a technique known as spectroscopy scientists were able to distinguish galaxies from clouds of gas when the stars could not be resolved into individual points of light. This is because stars emit a continuous spectrum of light, but some of it is absorbed by atoms in their atmospheres to form absorption spectra. However, the clouds of gas in space — nebulae — can be excited by radiation from nearby stars, which causes them to radiate light at particular frequencies. As galaxies are composed of stars and gas, they have spectra that include absorption lines within a continuous spectrum as well as some emission lines. Nebulae, on the other hand, have spectra dominated by emission lines. FIGURE 12.7 (a) This spectrum of the Sun shows the characteristic continuous colour spectrum produced by stars, crossed by absorption lines resulting from the gases in the Sun’s atmosphere. (b) A nebula is revealed by the emission lines in its spectrum, like this one showing the emission spectrum of hydrogen. (a) KH

390

h

g

G

fe

d

450

400

h

F

c

D

b E h 4-1

3-1

550

500

C

a

B

650

600

A

700

750

Wavelength in nm

(b)

IR

UV

λ

200

400

600

800

1000

1200

1400

1600

1800

2000

In the 1960s, astronomers conducted more detailed surveys of galaxies, mapping over 1 million. These maps showed that not only were stars clumped into galaxies, but galaxies were clumped into clusters. There are even clusters of clusters, known as superclusters. Today, estimates suggest that there are a hundred billion galaxies in the universe, each containing millions, if not billions or hundreds of billions, of stars. The size of the universe is beyond comprehension.

2200

2400

2600

nm

FIGURE 12.8 The Hubble Ultra-deep Field. How many galaxies can you count in this photograph of a tiny section of the sky?

Source: ESA / H.Teplitz and M.Rafelski IPAC/Caltech, A. Koekemoer STScI, R. WindhorstASU, Z. Levay STScI / NASA

TOPIC 12 What are stars? 7

FIGURE 12.9 The solar system is a tiny spot in a vast galaxy that is just one of billions in the universe.

Uranus

But one star of 100 billion in our galaxy Saturn

Jupiter

Mars Sun

Earth Mercury

Venus

But one galaxy of 100 billion thought to exist

The model of the universe as we know it today is vastly different to the universe that people thought they knew 2000 years ago. It is interesting that most people still think like the ancients. We talk of the Sun rising and setting, think of the Earth as stationary and that our lives and homes are the centre of the universe! The universe is in fact so large that if we were anywhere else, even within our own galaxy, the Earth would barely feature as a speck of dust worthy of investigation. Except, that is, for one thing: Earth is the only place in the universe currently known to host life. Even if we find that millions of worlds have life, these places would still make up a tiny fraction of the universe. The more astronomers learn about the universe, the less significant our place in it can seem. Yet it may also have heightened our sense of how extraordinary Earth and its inhabitants really are.

8 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

12.2.5 Information from the stars Scientists now have a very good idea about stars — what they are, how they work and what will happen to them over time. This has been one of the great triumphs of physics. As discussed previously, Galileo accumulated evidence that what he was discovering about physics on Earth also applied to the physics of objects in the sky. He saw that the Moon had mountains and craters and that its changing shape was simply the result of half of it being lit by the Sun and half being in shadow. He saw that other planets were also spheres and Jupiter had moons orbiting it. This discovery was tremendously important because it showed that moons could orbit a body that was itself in motion. An argument against the heliocentric model was that the Moon clearly orbits the Earth and pre-Galilean views of physics dating back to the ancient Greeks claimed that it was impossible for something to orbit a moving body — it would be left behind! Galileo, unlike Aristotle, claimed that a body would continue in motion unless a force acted to change its motion. He found support for this in the motion of the moons around Jupiter.

12.2 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. How has science changed our understanding of: (a) the distance to stars (b) the nature of stars (c) the number of stars (d) our position in relation to the stars? 2. What tool did Galileo introduce to the study of stars? What were three discoveries he made with it? 3. What did William Herschel discover about the Sun and its planets? 4. What technique enabled galaxies of stars to be distinguished from nebulae of gas? 5. Identify one piece of evidence that the laws of physics operate beyond Earth in the same way they do on Earth. Fully worked solutions and sample responses are available in your digital formats.

12.3 Using electromagnetic radiation to collect information KEY CONCEPTS • Explain the use of electromagnetic radiation in collecting information about the universe. • Identify all electromagnetic waves as travelling at the same speed, c, in a vacuum. 1 • Calculate wavelength, frequency, period and speed of light: c = f𝜆, T = . f

12.3.1 Collecting information about the universe Through the telescope he had invented, Gallileo found that the stars, unlike the planets, remained as points of light. When he pointed his telescope at the band of light across the sky known as the Milky Way, he could see that it was made from countless stars, too dim to be seen individually with the naked eye. Galileo’s simple telescope collected light coming from stars and focused it on his eye. Over the next 400 years, successive developments in telescope technology led us to the understanding we have of stars today. To understand what telescopes do and what they can teach us about stars, we need to understand more about light.

TOPIC 12 What are stars? 9

12.3.2 Electromagnetic radiation and the speed of light Virtually all of our information about stars comes to us in the form of electromagnetic radiation. The tiny points of light we see as stars provide so little light that with a clear sky and no moon, the thousands of stars visible in the sky provide barely any light to see by. However, we can learn so much about a star from the tiny amount of light that reaches us. The gap we experience between seeing lightning and hearing thunder shows that sound travels relatively slowly. Light seems to travel so fast that to us its speed seems infinite; that is, we seem to observe events at the instant they happen. Galileo was not convinced of this. He attempted to determine the speed of light by measuring the time delay between the flash of his lamp to an assistant on a distant mountain and the return flash from his assistant’s lamp. No detectable delay was observed and Galileo concluded that the speed of light was very high. A greater distance was needed. Olaus Roemer was a Danish astronomer born two years after Galileo’s death. He observed that the time between eclipses of Jupiter’s moons by Jupiter decreased as the Earth moved closer to Jupiter and increased as the Earth moved away. Roemer reasoned that this was because the distance the light travelled from Jupiter to Earth from one eclipse to the next became greater as the Earth’s orbit took it further from Jupiter (see figure 12.10). Roemer used this time and the known diameter of the Earth’s orbit around the Sun to estimate the speed of light. The value he obtained was 2.7 × 108 m s−1 . FIGURE 12.10 Measured from Earth, the time between successive eclipses of Jupiter’s moon Io increases as the Earth moves from A to B. (The diagram is not to scale.)

Io (moon) B Sun

A

Jupiter Earth’s orbit

Eventually, in the nineteenth century, with stronger light sources and more precise timing devices, Galileo’s method could be revisited, but the assistant was replaced by a mirror. The values obtained then for the speed of light were about 3.0 × 108 m s−1 . Early in the twentieth century, US scientist Albert A. Michelson (1852–1931) used a rapidly rotating eight-sided mirror. The light was reflected to a distant mirror about 35 kilometres away then reflected back to the rotating mirror. For some particular rotation rates, this light is reflected by one of the sides of the rotating mirror directly to the observer. The rotation rate can be used to calculate the speed of light. The value Michelson obtained was 2.997 96 × 108 m s−1 .

10 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 12.11 Light from the source reflects off one of the sides of the rotating mirror towards a mirror 35 kilometres away. The returning beam hits the rotating mirror. If one of the sides of the mirror is in the right position, the light enters the eyepiece and can be seen by the observer. By measuring the speed of rotation when the beam enters the eyepiece, the speed of light can be calculated. Observer

Fixed mirror

Rotating mirror with eight sides

Source of light 35 km

In 1973, a laser beam was used to measure the speed of light at 299 792.4574 ± 0.001 km s−1 . In 1983, the value was set internationally at 299 792.458 km s−1 and used to define the length of a metre. In the nineteenth century, it was well established that light in many ways behaved like a wave. James Clarke Maxwell provided a theoretical description of electricity and magnetism and his formulas predicted the existence of electromagnetic waves that all travelled at a particular speed. He noticed that this speed was essentially equal to the speeds experimenters were measuring for light. So he proposed that light was an electromagnetic wave.

Looking back in time All of the electromagnetic radiation we use to gather information travels at the speed of light. This means that all of this information refers to the past. On Earth, the distances are relatively small, so the immense speed of light means that transmission of information is almost instantaneous. The universe, however, is vast. Even light from the Moon takes over a second to reach us. Light from the Sun takes more than 8 minutes. Light from the nearest star, Alpha Centauri, takes 4.2 years to reach us. As we look out to this brighter of the ‘Pointers’ in the southern hemisphere sky, we are seeing it as it was over four years ago. It does not matter how good our telescopes are, the images and data will still represent Alpha Centauri as it was when the light left it, 4.2 years ago.

Resources Weblink The speed of light

12.3.3 Properties of electromagnetic waves The frequency of a periodic wave is the number of times that it repeats itself every second. Frequency is measured in hertz (Hz) and 1 Hz = 1 s−1 . Frequency can be represented by the symbol f. The period of a periodic wave is the time it takes a source to undergo one complete cycle. This is the same as the time taken for a complete wavelength to pass a given point. The period is measured in seconds and is represented by the symbol T.

TOPIC 12 What are stars? 11

The period of a wave is the reciprocal of its frequency. For example, if five complete waves pass every second, that is f = 5 Hz, then the period (the time for one complete wavelength to pass) is 15 = 0.2 seconds. 1 1 In other words, f = . It follows that T = . T f The amplitude of a wave is the size of the maximum disturbance of the medium from its normal state. The units of amplitude vary from wave to wave. For example, in sound waves the amplitude is measured in units of pressure, while the amplitude of a water wave would normally be measured in centimetres or metres. In electromagnetic waves, the amplitude represents the intensity of the electromagnetic fields. The wavelength is the distance between successive corresponding parts of a periodic wave. The wavelength is also the distance travelled by a periodic wave during a time interval of one period. For transverse periodic waves, the wavelength is equal to the distance between successive crests (or troughs). Wavelength is represented by the symbol 𝜆 (lambda). FIGURE 12.12 Transverse periodic waves in a piece of string

Wavelength

Crest

Direction of wave movement

Amplitude

Amplitude Position of undisturbed medium

Trough

Direction of particle motion

The speed, v, of a periodic wave is related to the frequency and period. In a time interval of one period, T, the wave travels a distance of one wavelength, 𝜆. Thus: speed =

distance 𝜆 = = f𝜆 time T

This relationship v = f𝜆 is sometimes referred to as the universal wave equation. The frequency of a periodic wave is determined by the source of the wave. The speed of a periodic wave is determined by the medium through which it is travelling.

Because the wavelength is a measure of how far a wave travels during a period, if it can’t be measured, it can be calculated using the formula: 𝜆=

12 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

v f

12.3.4 Frequency of light This periodic wave model can be applied to light, with each colour corresponding to a particular frequency. Table 12.1 shows some typical values for the frequencies and wavelengths of the colours of the spectrum. In fact, the frequency and wavelength steadily change across the spectrum but the eye and brain divide the spectrum into separate colours. Electromagnetic waves do not stop at the values of energies and wavelengths shown in table 12.1. These are merely the limits of what our eyes are able to detect. TABLE 12.1 Frequency and wavelength of colours Red

Orange

Yellow

Green

Blue

Violet

Frequency (× 1012 hertz)

430

480

520

570

650

730

Wavelength (nanometres)

700

625

580

525

460

410

SAMPLE PROBLEM 1

When light with a frequency of 5.6 × 1014 Hz travels through a vacuum, what is its: a. period b. wavelength (in nanometres)? The speed of light in a vacuum is 3.0 × 108 m s−1 . THINK a. 1.

Recall that the period is the reciprocal of the frequency.

WRITE a.

T=

1 f

1 5.6 × 1014 = 1.8 × 10−15 s The period of the light is 1.8 × 10−15 seconds. v b. 𝜆 = f 3.0 × 108 = 5.6 × 1014 = 5.4 × 10−7 m 1 nm = 1 × 10−9 m =

2. b. 1.

2.

3.

State the solution. Recall the relationship between speed, frequency and wavelength.

The wavelength of visible light is usually expressed in nanometres (nm). State the solution.

𝜆 = 5.4 × 102 nm The wavelength of the light is 540 nanometres.

PRACTICE PROBLEM 1 Find the frequency and period of light with a wavelength of 450 nm.

TOPIC 12 What are stars? 13

12.3.5 Electromagnetic waves At shorter wavelengths we find the higher energy electromagnetic radiation of ultraviolet light, X-rays and gamma rays. We use these shorter wavelengths in many medical and industrial processes. The longer wavelengths include infra-red (that we detect as radiated heat), microwaves and radio waves. These wavelengths are the key to our wireless communications. These frequencies of the electromagnetic spectrum are also valuable sources of information from stars as stars radiate across the broad range of wavelengths. Special telescopes have been built to detect these parts of the electromagnetic spectrum coming from space. There are infra-red telescopes, ultraviolet telescopes, X-ray and gamma ray telescopes, usually on a satellite orbiting the Earth because the atmosphere prevents most of the energy at these wavelengths from reaching the ground. Radio telescopes can be built on Earth because the radio waves largely pass through the atmosphere. FIGURE 12.13 Different telescopes have been built to detect the various parts of the electromagnetic spectrum coming from space.

14 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 12.14 Australia’s vast regions with low levels of radio pollution are suitable for radio telescopes, such as the Australia Telescope Compact Array at the Narrabri Observatory in NSW.

12.3 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

Describe the role of measuring distance in the discovery of galaxies. Why is light so important in our understanding of the universe? Given that speed equals distance divided by time, why was it so challenging to measure the speed of light? Describe Galileo’s attempt to measure the speed of light. What refinements were made to his experiment over the next couple of centuries to provide a precise measurement? What type of wave is light and who was the first to propose this? An electromagnetic wave with a period of 2.0 × 10–15 seconds is observed. Use the data in table 12.1 to determine its colour. An electromagnetic wave with a frequency of 626 kHz is detected. Calculate its wavelength and period. Is this wave visible to the human eye? The radio wave spectrum ranges in wavelengths from 1 mm to 100 km. Calculate the range of frequencies of radio waves. Why are X-ray and ultraviolet observatories placed in orbit? At what speed do radio waves travel? It is difficult to learn much about the universe without electromagnetic radiation. Other sources of information include gravitational waves and something that featured in topic 7. What is it?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

TOPIC 12 What are stars? 15

12.4 Spectroscopy KEY CONCEPT • Identify spectroscopy as a tool to investigate the light from stars, and interpret and analyse spectroscopic data with reference to the properties of stars.

12.4.1 Extracting information from the light Isaac Newton (1642–1727) published his book Opticks in 1704. In the first volume he demonstrated that light from the Sun can be dispersed into its constituent colours. Other theories about why rainbows formed, why prisms of glass produced a spectrum of colours and why soap bubbles appeared coloured involved the prism, raindrop or bubble altering the light. However, as Newton demonstrated, the prism, raindrop or bubble simply disperse the light according to its colour (wavelength), revealing information about the Sun. Newton’s prisms showed the colour spectrum from the Sun to contain red, orange, yellow, green, blue, indigo and violet. FIGURE 12.15 Forms of radiation and their place in the electromagnetic spectrum. The visible portion of the spectrum is shown enlarged in the upper part of the diagram. 750

(Nanometres)

700

650

600

550

500

450

400

Visible spectrum (white light) Power and telephone Electrical generator

108

Radio waves AM

FM

Klystron Incandescent tubes, lamps, heat microwave lamps lasers

Electronic tubes and semi-conductor devices

106

104

102

Ultraviolet rays

Infrared rays

Microwaves Radar

1

10−2

10−4

10−6

Lamps, fluorescent tubes, sparks, lasers

10−8

Gamma rays

X-rays

X-ray tubes

10−10

Cosmic rays, radioactive isotopes

10−12

10−14

Wavelength in metres 102

104

106

108

1010

1012

1014

1016

1018

1020

1022

Frequency in hertz

In 1802 William Wollaston (1766–1828) invented the spectroscope in an effort to explore the spectrum in more detail. He found the solar spectrum was not continuous but was crossed by a number of black lines. In 1814, Joseph von Fraunhofer (1787–1826) mapped the spectrum in much greater detail, finding 576 black lines. These have become known as Fraunhofer lines. John Herschel (1792–1871), William’s son, and W.H. Fox Talbot (1800–1877), a pioneer in photography, found that when chemical substances were heated in a flame and observed through a spectroscope, each chemical had a distinctive set of bright bands of colour forming its spectrum. This meant that scientists could identify chemicals simply by observing their spectra. Other scientists found that when sunlight is passed through gas before entering the spectroscope, it had extra black lines through its spectrum. This suggested that the black lines in the solar spectrum were due to light passing through gases in the Sun. These scientists had identified a method for determining the elements in stars. 16 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 12.16 A continuous spectrum produced by a spectroscope Continuous spectrum

Prism or diffraction grating

Slits

Light from source

In 1859 Gustav Kirchhoff (1824–1887) with his friend Robert Bunsen (1811–1899) used Bunsen’s burner to burn elements and clearly describe the cause of these spectral lines. They found that: • a continuous colour spectrum is produced by glowing solids or dense gaseous bodies like the Sun • if a gas is between the light source and the spectroscope, light is absorbed from the continuous spectrum at wavelengths or colours characteristic of the chemical components of the gas • a glowing gas produces bright lines on a dark background at wavelengths or colours characteristic of the chemical components of the gas. One of the first successes with this new tool of astrophysics was the spectroscopic analysis of planetary nebulae by William Huggins (1824–1910). Working in London in 1864, he found that these nebulae produced the bright line spectra of glowing gas, showing that they were clouds of gas rather than groups of very distant stars (see table 12.2). Some of the nebulae documented by Hershel, however, showed that they were collections of stars, as they emitted continuous spectra interrupted by black lines. Huggins’s investigations also provided very convincing evidence that the stars in the sky really are distant suns. TABLE 12.2 Types of spectra and the celestial bodies that produce them Type of spectrum

Generally produced by

Celestially produced by

Continuous

Hot solids, liquids, gases under pressure

Galaxies, inner layers of stars

Emission

Incandescent low-density gases

Emission nebulae, quasars

Absorption

Cool gases in front of continuous spectrum

The atmospheres of stars

12.4.2 Helium In 1868, Joseph Norman Lockyer (1836–1920), detected some Fraunhofer lines in the solar spectrum that did not correspond to any known element on Earth. He predicted that there must be an as yet undiscovered element in the Sun and called it helium, after the Greek helios, meaning Sun. William Ramsay (1852–1916) confirmed this in 1895, when he isolated the gas helium in the laboratory. Although the second most common element in the universe, helium is rare on Earth because it has so little mass that, even at normal

TOPIC 12 What are stars? 17

temperatures, it has sufficient energy to escape the atmosphere, a property that makes it useful in blimps and party balloons. Helium is also extremely unreactive with other chemicals so it forms few compounds; unlike hydrogen, which is even less massive but occurs in many compounds on Earth including water and organic compounds.

12.4.3 Composition of the Sun Early in the twentieth century, no one understood what could power the Sun. Chemical reactions did not produce sufficient energy. Scientists considered that it may have been some form of nuclear reaction that could release some of the Sun’s mass in the form of energy, as Einstein described in 1905 in the equation E = mc2 . However, scientists thought the Sun was made largely of iron, the most stable nucleus of all. This conclusion was based on measurements of tidal effects on the Earth that showed the Earth was mostly made of iron and on analysis of meteorites from space that showed that they too were composed largely of iron. Spectroscopic results seemed to confirm the iron content of the Sun. However, reading spectrographs of the Sun is not easy as many spectra of different elements overlap. In 1925, astrophysicist Cecilia Payne (1900–79) interpreted the spectra as showing that the Sun was mostly hydrogen, not iron at all. The scientists had all been reading the Fraunhofer lines incorrectly, which is easy to do when the lines of so many elements overlap. It was not long before the scientific community accepted that hydrogen was the main element found in the Sun. Hydrogen not only fitted the spectra, it explained the Sun’s energy, which iron could not. SAMPLE PROBLEM 2

Describe how astronomers can determine the elements that are present in the Sun. THINK

WRITE

Recall that astronomers use spectroscopes to disperse the light and observe Fraunhofer lines. 2. Recall that the Fraunhofer lines are unique for different elements.

Astronomers observe the Sun’s light through a spectroscope. The solar spectrum contains absorption lines that are characteristic of the elements found in the Sun’s atmosphere.

1.

PRACTICE PROBLEM 2 How was the element helium predicted before it was detected on Earth?

12.4.4 What produces the spectra? When the atoms are part of a solid, or when the material is very dense, like the gases in a star, the electrons in the atoms can exist with a range of energies. When the material is heated, the electrons gain energy. They then re-emit this energy in the form of packets of light energy called photons and fall back to a lower energy state. This process results in the emission of light with a range of energies or frequencies. We see this in the light emitted by hot wires in incandescent globes. The light produced has a continuous spectrum when passed through a spectroscope. The relative intensities of the different wavelengths of the spectrum determine the colour of the object. The hotter the object, the more the highest intensity wavelengths shift towards the blue end of the spectrum.

18 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

12.4.5 Absorption spectra Atoms and molecules in gaseous form allow their electrons to have only certain specific energies. When light with a continuous spectrum passes through a gas, most of the light will pass straight through. This is because most of the light is not at the specific energies that the electrons in the gas can absorb. The particular energies that are absorbed are unique to each element and molecule in the gas. When light with a continuous spectrum passes through a gas, these energies are absorbed and so are removed from the spectrum. When the light is passed through a spectroscope, it forms an absorption spectrum. The energies that have been absorbed indicate the elements that the light has passed though. FIGURE 12.17 An absorption spectrum produced by shining light with a continuous spectrum through a cool gas Absorption spectrum

Cool gas absorbs certain wavelengths and re-emits them in all directions. This light is deficient in certain wavelengths.

Incandescent bulb producing continuous spectrum

This is important for our understanding of stars, because the hot dense plasma at the centre of a star produces light with a continuous spectrum. As the light moves out from the centre, the gases in the star’s cooler atmosphere absorb specific frequencies and produce light with an absorption spectrum. Analysis of this spectrum on Earth shows us what elements are present in the atmosphere of the star without having to go there to take samples, which we have no way of doing.

FIGURE 12.18 The absorption spectrum for hydrogen

Absorption lines

TOPIC 12 What are stars? 19

12.4.6 Emission spectra When a gas is heated, the electrons increase their energies to higher energy states. These higher energy states are temporary and energy can be released at any time by an electron dropping back down to one of the lower energy states allowed by that element. This process produces light of very specific frequencies — an emission spectrum. Sometimes it is difficult to determine what fuzzy light sources in space are. Are they galaxies too distant to distinguish the individual stars, or are they clouds of gas or nebulae, glowing because they are being energised by nearby stars? Passing the light through a spectroscope answers this question. If the spectrum is an absorption spectrum, we know that it is produced by stars. If it is an emission spectrum, we know that it is produced by diffuse clouds of gas. We have seen how this technique was important in the discovery of galaxies. FIGURE 12.19 The spectra of 13 different stars. The stars increase in temperature from bottom to top. The top star is bright at the blue end of the spectrum, but emits very little light at the red end of the spectrum. It is a blue star. The star at the bottom emits very little light in the blue wavelengths. It is a cool red star. The absorption lines are visible in the spectra.

O6.5

HD 12993

B0

HD 158659

B6

HD 30584

A1

HD 116608

A5

HD 9547

F0

HD 10032

F5

BD 61 0367

G0

HD 28099

G5

HD 70178

K0

HD 23524

K5

SAO 76803

M0

HD 260655

M5

Yale 1755

F4

HD 94028

M4.5 B1

SAO 81292 HD 13256

FIGURE 12.20 The spectrum of light from a nebula. Notice that it is an emission spectrum, distinguishing it from the absorption spectra of stars.

Source: © Shelyak Instruments / Olivier Thizy

20 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

12.4 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

What instrument is used for the analysis of the spectra of light? What information can the frequencies of the absorption lines tell astronomers about stars? What part of stars produces the absorptions lines? Why? Faint light patches in the sky could be glowing clouds of gas or collections of very distant stars. How can stellar spectra be used to help distinguish one from the other? Spectroscopy helped to explain the Sun’s source of power. Explain. What process is occurring within an atom when absorption spectra are produced? Contrast the processes within an atom that produce absorption spectra and emission spectra. If you needed to determine the elemental composition of a star, what process would you use. Why? What would be different about the process within an atom that produces a blue Fraunhofer line compared to that which produces a red Fraunhofer line. Absorption lines can be used to determine the proportions of elements in stars. How might that work?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

12.5 Measuring distances to stars and galaxies KEY CONCEPT • Apply methods used for measurements of the distances to stars and galaxies (standard candles, parallax, red shift) to analyse secondary data.

12.5.1 Methods used to measure distances Hubble’s conclusion that the Andromeda nebula was actually a separate galaxy depended on a measurement of its distance from Earth and knowledge of the size of our galaxy, the Milky Way. How could such enormous distances be measured? Some of the methods use parallax, moving star clusters and the relationship between star brightness and distance.

12.5.2 Parallax Parallax is an effect that is noticed whenever an object is viewed from two different locations. A simple example is what you see when you stretch your right arm out with the index finger on your right hand pointing up. Point the index finger of your left hand up in a similar way about halfway between your face and your outstretched right hand. With one eye shut, move your hands until the finger from your right hand is hidden behind the finger from your left hand. If you now move your head left or right you will notice that the fingers are no longer in line. This effect is parallax. Another way of changing the position the fingers appear to be at is to swap viewing eyes. If the fingers are in line when viewed with one eye, they will not be if you close that eye and view your fingers with the other eye. Notice that it is the finger closest to your eye that changes position the most in relation to the background.

TOPIC 12 What are stars? 21

FIGURE 12.21 Observing parallax

Parallax angle when viewed through the right eye

If you hold your head very still with one eye closed, there is no parallax, so you will find it very difficult to judge distances, especially if you are unfamiliar with the objects in your field of view. Using two eyes, however, we get a slightly different view through each eye. The distance between our eyes is called the baseline. Ancient astronomers were unable to discern any change in the relative position of the stars at different times of the year. This supported the idea that the Earth was fixed in the centre of a sphere of stars. However, the accuracy of parallax measurements greatly increased with the invention of the telescope and many astronomers attempted to measure the parallax that must occur if the Earth is in motion. This is because our view of the sky in six months’ time will be from a position one diameter of the Earth’s orbit away from where we are today. Surely this very large baseline would be enough to observe changes in the positions of the stars relative to each other? In spite of this large baseline, it took until 1838 for the first stellar parallax to be observed by Friedrich Bessel. He noticed that the star known as 61 Cygni moved about a small ellipse when its position was measured relative to two adjacent stars. The parallax for a full year meant that 61 Cygni moved a mere 0.3 arc seconds relative to these two stars, which must have been much further away. To get an idea of how small this angle is, consider that one arc second is one-sixtieth of an arc minute, which is one-sixtieth of 1 degree, and there are 360 degrees in a full revolution. The full Moon has a diameter of about 30 arc minutes in the sky, so 0.3 arc seconds is roughly one six-thousandth of the diameter of the Moon, or alternatively the angle marked out by your little finger viewed from 7 kilometres away! No wonder parallax was not observed sooner. 61 Cygni is one of the closest stars to Earth and the closer the star, the larger the parallax. The closest star to the Sun is Proxima Centauri, which has a parallax of only 0.764 arc seconds. All other stars have smaller parallaxes than that.

22 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 12.22 A nearby star appears to change position relative to distant stars when viewed from different places in the Earth’s orbit. Earth — 2nd position

1st apparent position 1 AU

Earth’s orbit

Sun Close star

2nd apparent position

Earth — 1st position

Modern techniques to measure parallax include taking photographs of the stars through telescopes at various times during the Earth’s orbit about the Sun. Distant stars will not appear to change position relative to each other but near stars will. These measurements take years as the true movement of the star relative to the Sun is usually much more significant than the parallax. The parallax will be observed to come and go every six months, whereas the motion of the star against the background stars will continue unchanged.

12.5.3 The Gaia spacecraft Parallax is the only direct method of measuring distances in space; however, it is limited to the nearest stars only. Variation in the refraction of light through the atmosphere (like heat haze) is one of the reasons it is difficult to measure parallax. The Hubble Space Telescope, in orbit outside the Earth’s atmosphere, avoids this problem and has made a significant contribution to measuring the distances of our local neighbourhood of stars. A satellite called Hipparcos made an enormous contribution, gathering data from 1989–93, which enabled astronomers to accurately measure the distance to over 2.5 million of the nearest stars. A new space-based instrument called Gaia has replaced Hipparcos. This ambitious project FIGURE 12.23 The Gaia spacecraft provides data for astronomers to understand much more about our galaxy’s formation, structure and evolution. It will survey a billion stars in our galaxy, which is 1% of the total number of stars in the Milky Way, greatly expanding the data obtained from Hipparcos. Gaia will measure the position, motion, colour and brightness of stars, giving astronomers precise measurements of distance to each star, chemical composition and changes in each star’s brightness over a five-year period. Much of the work of astrophysicists involves working with the enormous databases of information that observatories such as Hubble, Source: ESA / ATG medialab Hipparcos and Gaia collect. TOPIC 12 What are stars? 23

The formula for small parallax angles, where sinp ≈ p (parallax angles for all stars are small), is: 1 d where p is expressed in arc seconds and is the parallax angle calculated as half the maximum parallax measured in one year. That is, the parallax corresponding to a baseline of one astronomical unit (1 AU); d is the distance to the star in parsecs (pc). This method of parallax measurement is limited to objects about 100 pc away using ground-based technology. One parsec is equal to 3.086 × 1016 m. p=

FIGURE 12.24 When p = 1 arcsec, d = 1 parsec

1 parsec 1 AU

1 arcsec

SAMPLE PROBLEM 3

How far is a light-year in metres? THINK 1.

WRITE 8

−1

Light travels at 3.00 × 10 m s .

1 year = 365 × 24 × 60 × 60 seconds = 3.15 × 107 s Distance = speed × time = 3.00 × 108 m s−1 × 3.15 × 107 s

2.

State the solution.

= 9.46 × 1015 m One light-year is a distance of 9.46 × 1015 metres.

PRACTICE PROBLEM 3 How many light-years are there in a billion metres (1.00 × 109 m)?

SAMPLE PROBLEM 4

What is the furthest distance that could be measured by parallax with the naked eye, using 1 AU as the baseline? It is estimated that the smallest parallax angle detectable with the naked eye is 30 arc seconds. Give your answer in parsecs, metres and light-years.

24 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

THINK 1.

Determine the distance that would result in a parallax angle of 30 arc seconds.

2.

Calculate the distance in metres and light-years.

WRITE

1 d 1 d= p 1 = 30 = 0.033 pc 1 parsec = 3.086 × 1016 m 0.033 pc = 1.0 × 1015 m 1 light-year = the distance light travels in 1 year = speed × time p=

= 3.0 × 108 × 1 × 365 × 24 × 60 × 60 = 9.5 × 1015 m

3.

State the solution in all the required units.

1.0 × 1015 = 0.11 light-years 9.5 × 1015 The maximum distance for which a naked eye could see parallax is 0.033 pc, 1.0 × 1015 metres and 0.11 light-years. This is less than one-fortieth of the distance to the nearest star.

PRACTICE PROBLEM 4 The parallax angle of the star Sirius is 0.379 arc seconds. How far is Sirius from Earth in: a. parsecs b. metres c. light-years?

SAMPLE PROBLEM 5

How many light-years make 1 parsec? One light-year is a distance of 9.46 × 1015 m. THINK 1.

Convert the parsec to metres.

2.

Divide the number of metres in a parsec by the number of metres in a light-year.

3.

State the solution.

WRITE

1 parsec = 3.086 × 1016 m 3.086 × 1016 1 pc = 9.46 × 1015 = 3.26 light-years 1 parsec is equal to 3.26 light-years.

TOPIC 12 What are stars? 25

PRACTICE PROBLEM 5 A star is measured to be 25 pc from Earth. What is its distance from Earth in light-years?

12.5.4 Star brightness and distance The methods discussed so far are not useful beyond our galaxy, because the huge distances make parallax unmeasurable. As a result, the discovery that stars are also gathered in other galaxies outside our own did not occur until the mid-1920s. This discovery required understanding a special type of star, called a Cepheid variable, that was bright enough to be observed in distant galaxies. The luminosity of a star is the total energy that the star radiates per second. As this energy moves away from the star at the speed of light, it becomes spread over a sphere of surface area 4𝜋r2 , where r is the distance to the centre of the star. The brightness of a star is defined as the energy per second per square metre at the place where the star is observed. b=

L 4πr2

Where: b = brightness L = luminosity. The brightness of a star is measurable from Earth. If we know the luminosity of the star, we can determine its distance using the formula. But how can the luminosity of a star be determined? In 1912 Henrietta Leavitt (1868–1921) of Harvard University found that a Cepheid variable (a large pulsating star) can tell us how luminous it is. When studying Cepheid variables in the two small galaxies called the Large and Small Magellanic Clouds, Leavitt noticed that the brightness of Cepheid variables was proportional to their period of pulsation. She was able to determine this because all of the stars in each galaxy are approximately the same distance from Earth. She noticed that the brighter the star was, the longer its period of pulsation. Harlow Shapley (1885–1972), of the Mount Wilson Observatory in California, recognised that if he could accurately determine the distance to some Cepheids he would be able to determine how luminous they were. He achieved this using Cepheid variable stars close enough to measure the distance using parallax. Once he knew the luminosity of those stars, he could calculate the distance to other Cepheids, too distant to measure using parallax methods. He used the Sun’s motion through the galaxy to produce a baseline to determine the distance to some Cepheids as no Cepheid variables were found close enough to use parallax due to the Earth’s annual motion around the Sun. It was by using Cepheid variables that Hubble was able to measure the distance to the Andromeda nebula and determine that it was a galaxy separate from our own. Other types of stars have also been found that work as ‘standard candles’ of known luminosity, such as the Cepheid variables. The accuracy of this method depends on the accurate determination of the distance of nearby examples. As methods and data have improved, the distances to stars and galaxies calculated by this method have been revised.

26 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

SAMPLE PROBLEM 6

The luminosity of a Cepheid variable star is known to be 8.0 × 1029 W. Astronomers measure its brightness as 8.6 × 10−10 W m−2 . What is its distance from us? THINK 1.

Write the relationship between luminosity and brightness.

2.

Rearrange the formula so that it gives the distance in terms of luminosity and brightness.

Insert the values of luminosity and brightness in the formula. 4. Calculate the distance.

3.

5.

Convert the answer to light-years.

6.

State the solution.

WRITE

L 4πr2 √ L r= 4πb √ 8.0 × 1029 = 4π × 8.6 × 10−10

b=

= 8.6 × 1018 m 8.6 × 1018 = (convert metres to light-years) 9.461 × 1015 = 909 light-years The Cepheid variable star is 909 light-years from us.

PRACTICE PROBLEM 6 The distance to a star cluster has been measured using the Cepheid variable technique to be 4000 light-years. A star in that cluster has a measured brightness of 2.1 × 10−11 W m−2 . What is the luminosity of that star?

The period of a Cepheid variable star with a brightness of 5.1 × 10−13 W m−2 is measured to be 10 days. What is the distance to this star? The luminosity of the Sun is 3.8 × 1026 W. Use the following graph to determine your answer. (Note the logarithmic scales on the graph axes.)

Luminosity (Lsun)

SAMPLE PROBLEM 7 104

103

102 10 0

THINK 1.

Use the graph to determine the luminosity of a Cepheid variable with a period of 10 days.

2.

Calculate this luminosity in watts.

1

3

10 30 Period (days)

100

WRITE

From the graph, a period of 10 days corresponds to a luminosity of about 103.4 (or 2.5 × 103 ) times that of the Sun. L = 3.8 × 1026 × 2.5 × 103 = 9.5 × 1029 W

TOPIC 12 What are stars? 27

√ 3.

4.

Use the relationship between distance, luminosity and brightness to determine the distance.

State the solution.

r=

L 4πb

√

9.5 × 1029 4π × 5.1 × 10−13 = 3.85 × 1020 m

=

= 4 × 104 light-years The Cepheid variable star is 40 000 light-years from us.

A note on reading logarithmic scales: You should notice that the scales on this graph are not linear. It is known as a logarithmic graph. The scales are equally divided by powers of 10. Halfway between 103 and 104 is 103.5 = 3.2 × 103 .

PRACTICE PROBLEM 7 a. What is the luminosity of a Cepheid variable with a period of 30 days? b. If this Cepheid variable was located in the Large Magellanic Cloud (at a distance of 157 000 light-years), how much energy would we receive from it per square metre on Earth?

Resources Digital documents Investigation 12.1 Brightness versus distance (doc-31882)

12.5.5 The red shift method We learnt in topic 6 that Edwin Hubble had plotted the red shift of galaxies against their distance. For distant galaxies, he showed that there was a linear relationship between red shift and distance. Hubble’s work has been extended to very distant galaxies using different standard candles. A type of exploding star called a Type 1a supernova always emits the same amount of light. These are rare events, but when they occur, they outshine the light in the rest of the galaxy for a period of time, making it possible to determine the distance to galaxies far beyond the reach of the Cepheid variable technique. As a result of this work, we know the relationship between red shift and distance to sufficient precision that we can determine how far away distant galaxies are by measuring their red shift alone. This is important when no standard candles like Cepheid variables or Type 1a supernova are observed in the galaxy we are interested in. We have learnt about the techniques used to determine the chemical composition and distance to stars in the sky. In the next subtopic we examine stars more closely. Where better to start than with the star we depend on the most.

28 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 12.25 The relationship between red shift and distance

Increasing distance

Increasing red shift

12.5 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. A star’s parallax is 0.155 arc seconds using a baseline of 1 AU. How far is this star from the Sun in: (a) parsecs (b) metres (c) light-years? 2. The Hubble Space Telescope (HST) can measure parallax angles as small as 0.05 arc seconds. What is the furthest object for which the HST could be used to measure its distance using the method of parallax? 3. What restricts the use of parallax to measure the distance to relatively nearby stars? 4. The orbit of Jupiter has a radius of more than five times that of the Earth. Explain why an observatory at Jupiter would be able to measure the parallax of more distant stars than those in orbit around Earth. 5. What makes Cepheid variable stars so useful in measuring distances to galaxies? 6. A Cepheid variable star in one galaxy has the same period as a Cepheid variable star in a second galaxy. However, the first star has a brightness that is eight times that of the second. What can you say about the distance of the second galaxy compared to the first? 7. Explain why the term ‘standard candle’ is sometimes used to describe stars like Cepheid variables. 8. Brightness and the period of change in brightness of stars are directly measurable from Earth. (a) Explain why luminosity cannot be directly measured. (b) Explain why, despite this, astronomers can measure the luminosity of Cepheid variables. 9. What methods can be used to measure the distance to galaxies too distant to observe Cepheid variables? 10. Explain why measuring red shift can be used to determine the distance to more distant galaxies, but not to galaxies neighbouring our own.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

TOPIC 12 What are stars? 29

12.6 The properties of stars KEY CONCEPTS • Describe the Sun as a typical star, including size, mass, energy output, colour and information obtained from the Sun’s radiation spectrum. • Identify the properties of stars, including luminosity, radius and mass, temperature and spectral type, and explain how these properties are used to classify stars.

12.6.1 The Sun We have seen that as scientists moved away from the idea of a geocentric universe, their understanding of the nature of stars changed. No longer were stars just points of light on a sphere revolving around the Earth. It seemed highly likely that they were like the Sun. The Sun is a typical star, but the closest to us. It is not the centre of the universe, just one of billions of stars in a universe that has no centre. It is easy to think that the Earth is of great significance in the universe, with a few bright objects like the Sun, Moon and stars passing in the sky overhead. We need to look only as far as the Sun to get a better sense of perspective. The mass of the Earth may seem huge at 5.97 × 1024 kg, but this is small in astronomical terms. The Sun’s mass is 1.99 × 1030 kg, or 333 000 times the mass of Earth. The mass of the Sun is a useful unit for comparing the masses of stars, and is called a solar mass. The total mass of our galaxy is roughly 100 billion times the mass of the Sun (100 billion solar masses), and there are about 100 billion galaxies in the universe, some smaller than ours and some much larger. FIGURE 12.26 The Sun

Source: NASA / SDO

30 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 12.27 The Earth is very small compared to the Sun.

Sun Earth

The radius of the Earth is an impressive 6.378 × 106 m, but the Sun has a radius of 109 times this at 6.955 × 108 m. Nearly all of the energy available to Earth comes from the Sun, whose energy output is 3.86 × 1026 J s−1 (3.86 × 1026 W). This is known as its luminosity. A tiny portion of this energy hits the Earth, heating and lighting it. Alpha Centauri, the brighter of the two Pointers is the third brightest star in the sky. Through a small telescope Alpha Centauri appears as two bright stars, Alpha Centauri A and B, but it is actually a system of three stars! The third star is a very dim type of star called a red dwarf and is the closest star to our solar system, hence its name Proxima Centauri. Many red dwarfs can be found in our part of the galaxy but they are not visible to the naked eye because they have low luminosity. Alpha Centauri A has a luminosity slightly greater than that of the Sun. From Alpha Centauri the Sun would appear as a star slightly dimmer than Alpha Centauri is when viewed from Earth. FIGURE 12.28 Energy from a star radiates in a sphere around it. The star’s brightness is an indication of how much of the light it emits passes through a square metre each second.

ino Lum sity

r Brightness

TOPIC 12 What are stars? 31

SAMPLE PROBLEM 8

A star, located 6 light-years away, is equal in luminosity to a star 18 light-years away. How much light energy from the first star enters the eye of an observer on Earth, compared with the light energy entering the eye from the second star? THINK

WRITE

What is the ratio of the distance to the first star to the distance to the second star? 2. Square the ratio. This gives the fraction of light entering the eye from the first star compared to the second star.

The first star is one-third of the distance from the observer. As light intensity diminishes with the square of the distance, its light will be nine times as intense as the light from the second star. (However, due to the nature of light sensitivity of the eye, this does not correspond to a nine-fold increase in how bright we perceive the star to be.)

1.

PRACTICE PROBLEM 8 Alpha Centauri A is 267 000 times as far from Earth as Earth is from the Sun. Assuming it is the same luminosity as the Sun, how bright is it compared to the Sun?

Resources Digital documents Investigation 12.2 Solar observation (doc-31883) Investigation 12.3 Cardboard solar observatory (doc-31884) Weblink

The nearest stars applet

12.6.2 The mass of stars To understand the processes occurring inside a star, we need to know its mass. The mass of the Sun can be determined from the orbits of the planets around it.

Newton’s laws of gravity and the analysis of the motion of masses in a circle lead directly to a formula for the mass of the Sun: M=

4π2 r3 GT 2

Where: r = the radius of the orbit T = the period or time to complete the orbit G = 6.67 × 10−11 N m2 kg−2 and is the universal constant of gravitation. This formula applies to all masses orbiting a central body of mass M.

32 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Log L / L

The distance between the Earth and the Sun was measured FIGURE 12.29 The mass– using parallax by observing a transit of Venus from different luminosity relationship, L ∝ M3.5 . Note that the scales are not locations on Earth. A transit of Venus is like an eclipse of the linear. M⊙ and L⊙ are the mass Sun, except that Venus looks much smaller than the Moon when and luminosity of the Sun, so it is viewed from Earth. When Venus passes between the Earth and at (0, 0). the Sun it can be viewed by projection (never look at the Sun directly or through binoculars or a telescope) as a small disc 5 passing across the face of the Sun. This method of measuring the distance between the Earth and the Sun was first suggested 4 by a contemporary of Newton’s, Edmund Halley (1656–1742), in 1716. Transits of Venus are rare but two occurred in the 1760s, 3 after Halley’s death, and the distance to the Sun was measured. It was on a journey to the Pacific Ocean to measure the second of 2 these transits that Captain James Cook landed at Botany Bay. The mass of distant stars can be found in a similar way. Many 1 stars, called binary stars, come in pairs and orbit a common centre of mass. Using the formula above, astronomers can calculate the 0 total mass of the two stars. To work out the individual masses, they apply centre of mass calculations. These are based on the –1 fact that the heavier body must be closer to the centre of mass 0.5 1.0 1.5 0 than the lighter one by an amount proportional to the masses. Log M / M Using this technique, astronomers are able to measure the mass The mass-luminosity relationship of many stars. Source: © Dr. Mike Guidry In 1924, Arthur Eddington made a very significant discovery. When he graphed the masses of stars versus their luminosity he found a clear relationship. The more luminous the star, the greater its mass. This meant that it was possible to determine the mass of most stars whose luminosity was known. The luminosity of a star can be determined using the brightness and the distance from Earth, if it is known. This method works for stars in their hydrogen-burning phase, but not for white dwarfs or red giants. SAMPLE PROBLEM 9

The radius of Earth’s orbit about the Sun is 1.5 × 1011 m and the period is 365 days. Calculate the mass of the Sun. THINK 1.

2.

Use the formula for mass of a star and substitute the values for Earth’s orbit provided.

State the solution.

WRITE

M= =

4π2 r3 GT2 4π2 (1.5 × 1011 m)3

(6.67 × 10−11 N m2 kg−2 × (365 × 24 × 60 × 60 s)2 ) = 2.0 × 1030 kg The mass of the Sun is 2 × 1030 kg. This agrees with the accepted value for the mass of the Sun.

PRACTICE PROBLEM 9 How would you describe the mass of the Sun compared to the masses of the other stars plotted on the graph in figure 12.29?

TOPIC 12 What are stars? 33

12.6.3 Diameters of stars Stars other than the Sun are too far away to see as discs in a telescope. The Earth’s atmosphere distorts the images of stars, making the smallest object that can be seen directly much larger than even the nearest stars. The Hubble Space Telescope can resolve objects one-tenth of the diameter of those resolved by groundbased telescopes because it does not have to deal with the problems introduced by the atmosphere. It has been able to image Betelgeuse, a very large red giant over 600 light-years from Earth, as a disc. FIGURE 12.30 A Hubble Space Telescope photograph of Betelgeuse showing the size of this enormous star relative to parts of our solar system. Betelgeuse is the bright star at top left in the well-known constellation Orion, shown in the photograph on the right.

Size of star Size of earth’s orbit Size of jupiter’s orbit

Atmosphere of betelgeuse • Alpha orionis Hubble space telescope • Faint object camera

Source: NASA / A. Dupree CfA, R. Gilliland STScI

Other methods can be used to measure the diameters of nearby stars; for example, when a star passes behind the Moon it fades, very quickly but detectably. Knowing the speed that we should observe it passing behind the Moon, its diameter can be determined by timing how long the star takes to fade. Another method involves superimposing the light from two mirrors separated by many metres. An interference pattern is produced that depends on the diameter of the star.

12.6.4 Classifying stars The information gathered about stars through the light that reaches our telescopes and spectroscopes can be used to classify stars into types. Classification helps astronomers to see that not all stars are the same and prompts the questions that reveal the processes that occur in stars. Luminosity has been found to vary greatly between stars. The stars that radiate the least energy have a luminosity of around 0.0001 of the luminosity of the Sun. The most luminous stars are millions of times as luminous as the Sun. The Sun is neither particularly dim nor particularly powerful in the realm of stars. So, what gives stars their power, and why are some so much more luminous than others?

34 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 12.31 Stars vary widely in size; Rigel is a supergiant. Most stars are smaller than the Sun but some of the larger stars are enormous. The Sun is about average in size.

Rigel

Sun

A clue to this might be the range of radii of stars. Stars are known to exist with a radius of around 10% of the radius of the Sun. That is not much bigger than the planet Jupiter. The largest stars known have a radius of around 2000 times that of the Sun. In the context of the solar system, the star’s surface would be about the radius of Saturn. So the largest known star is about 20 000 times the radius of the smallest. Not surprisingly, the masses of stars also vary tremendously. The star with the smallest radius may not be much bigger than Jupiter but it has much more mass, about 100 times as much. The least massive known star has a mass of about 7% that of the Sun, the largest is about 350 times that of the Sun. When astronomers plot a graph of luminosity versus mass, they observe a very strong correlation for most of the stars, as shown in the graph in figure 12.32. This suggests that luminosity might be due to the mass of these stars. There are many other stars that do not follow this relationship. The ones that do are known as main sequence stars. The radius of these main sequence stars also increases with mass.

FIGURE 12.32 Graph showing that the radius of main sequence stars increases with mass.

Radius (solar units)

10

3

Sun

1

0.5

1

2

5

10

20

Mass (solar units)

Stars also vary greatly in temperature. The surface of the coolest stars is around 3500 K and the hottest over 40 000 K. The Sun, for comparison, is around 5700 K. For main sequence stars, as the temperature increases, so does the luminosity. Perhaps unsurprisingly, large, more massive stars are hotter and more luminous on the whole. However, there are exceptions to this main sequence trend. Stars exist that are hot but small. They are known as white dwarfs. They do not fit the trends of the main sequence. Also, there are some stars with a large radius and luminosity but average mass. Their temperature is not as high as found in main sequence stars of that radius, so they appear red. These stars are known as red giants. Stars that deviate from the main sequence reveal a lot to astronomers about how stars change over time. This will be explored in a later section. An important part of the story of classifying stars is the consideration of their spectra. The spectra of different stars emphasise different absorption lines differently. Using the spectra, scientists in the early twentieth century classified stars by their spectral type. Annie Jump Cannon was a Harvard University astronomer who devised a system of spectral class that is used in a modified form today. Stars are classified as of spectral type O, B, A, F, G, K or M, according to the intensity of absorption lines in their spectra. TOPIC 12 What are stars? 35

12.6 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. 2. 3. 4.

5. 6. 7. 8. 9.

What provides most of the Earth’s energy? Stars appear as pinpoints of light in the night sky. List evidence that the Sun is also a star. The Sun dwarfs everything else in the solar system. Name two stars that dwarf the Sun. The luminosity of the Sun is 3.86 × 1026 W, the radius of Earth’s orbit is 1.50 × 1011 m and the Earth has a radius of 6.37 × 106 m. (a) Find the surface area of a sphere whose surface lies at the Earth’s orbit and is centred on the Sun. (b) Find the fraction of that surface that is taken up by the Earth. (c) Calculate the total energy that the Earth receives from the Sun each day. (d) Calculate the total energy received by one solar panel of area 1 m2 per day (assume the solar panel receives the equivalent of six hours sunlight directly overhead per day). The volume of the Sun is how many times that of the Earth? (Radius of Earth = 6.378 × 106 m, radius of Sun = 6.955 × 108 m) Identify five characteristics of stars that can be used to classify them. How is a star’s spectral type determined? What relationship defines stars classified as main sequence stars? Which types of stars are not classified as main sequence stars?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

12.7 Nuclear fusion KEY CONCEPTS 2 • Explain nuclear fusion as the energy source of a star including: E = mc . • Distinguish between the different nuclear fusion phenomena that occur in stars of various sizes.

12.7.1 The energy source of a star An understanding of how the Sun generates its energy did not come until well into the twentieth century. What could be the source of energy that seems to last for millions and even billions of years without any signs of running out? All particles exert a force of gravity on other particles, and it is the force of gravity that dominates on the large scale of the universe. Clouds of hydrogen and helium gas that were present in the early universe eventually collapsed under the force of gravity. The larger and more massive the cloud, the stronger the gravitational force that pulled the gas particles in the cloud together. Temperature is directly linked to the average kinetic energy of atoms. The cloud of gas shrinks because the force of gravity accelerates the atoms inward, the average speed of the atoms increases, therefore the temperature rises. The high temperatures in the centre of these collapsing gas clouds ensure that the cloud remains in a gaseous state and does not condense into liquid or solid. The more the cloud collapses, the more the gas particles will collide, which eventually provides enough resistance to prevent further collapse.

36 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

For some time this was the view of how stars worked — clouds of gas heated by gravitational collapse. The collisions slow the motion of the gas particles and release energy in the form of electromagnetic radiation, in a similar way to billiard balls colliding and producing sound. As particles slow down they are more easily drawn in by gravity, again increasing the temperature, the rate of collision and the release of electromagnetic radiation. Astrophysicists calculated how long the Sun would continue to produce the energy that it does if gravity were its only source of energy. The answer was about 15 million years. Yet evidence on Earth suggested that the Sun had been a reliable source of energy for much longer than that. The Earth has been dated to about 4500 million years old using the radioactive decay of uranium isotopes and other nuclei with long half-lives. This age is consistent with theories of how the solar system formed, the evolution of life on Earth and the time required for the geological formations that we see in the Earth’s crust to develop. Fifteen million years is clearly nowhere near long enough! Consider the Sun’s composition. Hydrogen makes up about 92% of the nuclei in the Sun. The rest is mostly helium, about 7.8%. Carbon, nitrogen and oxygen are the next most common elements, in total making up less than 0.1% of the atomic nuclei in the Sun. The Sun is so hot in its centre that the electrons are not attached to atoms; the nuclei and electrons swarm around separately, forming what is known as plasma. When talking of hydrogen in stars we usually mean individual protons. Some of these protons will be attached to one or two neutrons, forming the different isotopes of hydrogen — deuterium and tritium. In the same way, helium-3 and helium-4 isotopes are formed. FIGURE 12.33 (a) The proportion of atomic nuclei in the Sun (b) The most common isotopes present (a)

(b) 92% Hydrogen +

+

Hydrogen

Deuterium

Tritium

+

+ +

+

Other 0.2%

+

Helium 7.8% Helium-3

Helium-4

Arthur Eddington (1882–1944) was famous for his measurement of the bending of starlight around the Sun, which helped verify Einstein’s General Theory of Relativity. Around 1926, Eddington proposed fusion to be the energy source of the Sun. The nuclei of atoms are held together by a force known as the strong nuclear force, which acts over only very short distances within atoms. In the centre of the Sun, the pressure is so great and the temperature so high that protons are pushed together with enormous force. However, the electrostatic repulsive force that exists between all positive charges resists this coming together. Eddington calculated that the centre of the Sun would reach temperatures of about 15 million degrees Celcius simply by contraction due to gravity. Physicists at the time thought this was not high enough for the protons to get sufficiently close for the strong nuclear force to overcome the electrostatic force, allowing the protons to fuse.

TOPIC 12 What are stars? 37

In 1928, George Gamow solved the problem of how large nuclei can eject alpha particles in alpha decay. According to pre-twentieth century physics, alpha particles would not have enough energy to escape the strong nuclear force holding them to the nucleus. However, studies in the new field of quantum mechanics showed that there was a small probability that the alpha particles would escape certain nuclei. Physicists realised that if alpha particles were able to ‘tunnel’ their way out of a nucleus, protons would have a probability of tunnelling their way in. This probability is high only when the protons have a lot of kinetic energy (which they do when the temperature is very high) and becomes significant at about 15 million degrees Celsius! Some of the protons come close enough to fuse into a single nucleus, releasing energy in the process. This fusion of protons does not happen easily, otherwise we would have a wonderful source of energy to replace fossil fuels and nuclear fission. Maybe one day a practical fusion reactor will be produced, but so far attempts to achieve controlled fusion on Earth have used more energy to bring the nuclei together than was released by the fusion. The thermonuclear (hydrogen) bomb is the exception, but it requires a rather large ‘match’ to light the fuse. These bombs use a fission (atom) bomb to achieve the necessary temperatures. In the Sun, the gases are pulled closer together by gravity until the temperature is high enough for fusion to occur. The energy released by the fusion reactions prevents the Sun from contracting any further. In 1938, Hans Bethe (1906–2005) at Cornell University was trying to understand what happens to protons when they are forced close together in the Sun. One of the reaction chains that Bethe found is known as the PP1, or proton–proton chain, in which two protons are forced together. A nucleus of two protons represents helium, but a nucleus of two protons alone is very unstable. One of the protons undergoes positive beta decay; that is, it releases a positive beta particle (called a positron or positively charged electron), another particle called a neutrino and about 1.44 MeV of energy. This can be summarised in the following nuclear equation: 1 H 1

+ 11 H → 21 H + 𝛽 + + v + 1.44 MeV

where v is the symbol for the neutrino. The isotope of hydrogen formed is called deuterium. Deuterium has a chance of undergoing fusion with another proton in the following reaction: 2 H 1

+ 11 H → 23 He + 𝛾 + 5.49 MeV

where 𝛾 is a gamma ray. Two of these helium nuclei can fuse to form the much more stable helium-4 isotope: 3 He 2

+ 32 He → 42 He + 211 H + 12.9 MeV

This is the most common of several chains of nuclear reactions that occur in the Sun; all of them start with protons and end up with helium nuclei.

REMEMBER In nuclear equations the mass numbers on each side of the equation have to balance. The atomic numbers must also balance.

38 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 12.34 The proton–proton chain, PP1 v

e+

γ 2 H 1

1 H 1

1 H 1 3 He 2 4 He 2

1 H 1 1 H 1

γ

3 He 2

1 H 1

2 H 1

γ 1 H 1

v

e+

The mass of the products of each of the chains of fusion reactions is less than the total mass of the protons that undergo fusion in the chain. Einstein’s statement about the equivalence of mass and energy, described by his famous equation E = mc2 , explains what happens to the ‘missing’ mass. Each reaction results in a loss of mass as energy is released. The mass loss can be calculated using E = mc2 , where E is the energy in joules, m is the mass loss in kilograms and c is the speed of light in metres per second.

REMEMBER The electron volt (eV) is a unit of energy that is often used by physicists when dealing with energy of subatomic particles. 1 eV = 1.6 × 10−19 J

How can we be sure that this theoretical model of the Sun describes how it actually works? Experiments with particle accelerators and the hydrogen bomb have shown that fusion of hydrogen does occur and that a lot of energy is produced in the process. The neutrinos and gamma rays produced in the reactions interact very weakly with matter and therefore travel out of the Sun into space. Their presence can be detected in laboratories on Earth and in orbit, but because neutrinos interact so weakly with matter their detection is difficult. Photons of light take millions of years to make their way from the centre to the surface of the Sun because of their continual absorption and re-emission. Neutrinos, on the other hand, leave the Sun’s surface in seconds. If we could see neutrinos during the day we would see them radiating from the Sun. At night we would see them coming through from the other side of the Earth, as most of them travel straight through it uninterrupted! The levels of neutrinos predicted by the fusion models agree with what has been detected in experiment, further validating the already well-established theory of fusion in the Sun.

TOPIC 12 What are stars? 39

SAMPLE PROBLEM 10

When two hydrogen nuclei in the centre of a star fuse, 1.44 MeV of energy is released. How much mass is lost by the Sun in this reaction? THINK 1.

Convert the energy released from MeV to joule.

WRITE

1.44 MeV = 1.44 × 106 eV

= 2.3 × 10−13 J E 2. Use E = mc2 to calculate the mass equivalent. m= 2 c 2.3 × 10−13 J = (3.0 × 108 ms−1 )2 = 2.56 × 10−30 kg The Sun loses 2.56 × 10−30 kg in this 3. State the solution. reaction. Note: Both the mass and energy involved in each fusion reaction may seem tiny. However, the Sun contains in the order of 1057 protons that can fuse in its lifetime to form helium.

PRACTICE PROBLEM 10 How much mass is lost by the Sun each second if there are 1038 fusions of protons per second?

SOLAR NEUTRINO PROBLEM When physicists measured the neutrinos arriving on Earth from the Sun they found less than half the number predicted using the theoretical model of how the Sun works. This became known as the Solar Neutrino Problem. This problem puzzled physicists for over 30 years but the evidence for the hydrogen fusion model was so strong that they held on to it. In 2001 the problem was considered solved. The detectors that had been used detected only one type (flavour) of neutrino, but neutrinos change flavour on their journey from the Sun so that many of them went undetected. When this was taken into account, the neutrino count agreed with the theoretical prediction.

12.7.2 What happens next? After about 10 billion years, most of the hydrogen in the core of the Sun will have fused to form helium. Once the hydrogen in the Sun’s core is exhausted, it will cool and contract a little. The temperature at the core’s edge will rise, due to the extra contraction, causing the hydrogen at the edge to undergo fusion. This fusion closer to the surface of the star will heat its outer layers, causing them to expand and then cool. At this point the Sun will become a red giant. The Sun as a red giant will be about 1000 times as bright as it is now and have a radius about 100 times its current size. Its core will be small and extremely dense and hot, while the outer layers will have very low density and will be quite cool. After a time, the collapsing core will reach a temperature where helium begins to fuse. This will produce enormous amounts of energy throughout the whole core. The rapid heating that occurs in this time is known as a helium flash. The core will reach an enormous 350 million degrees Celsius, causing the star to expand and cool. Then the Sun will continue burning helium in the core and hydrogen on the edge of the core.

40 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

The helium fusion will involve the following reactions: 4 He 2

+ 42 He ↔ 84 Be + γ then 84 Be + 42 He →

12 6 C

+γ

FIGURE 12.35 The development of a star from the hydrogen fusion stage through to the helium fusion stage Non-reacting H Non-reacting H

Fusing H

Non-reacting He

Core H runs ou

t

Main sequence star

ing

urn

ll b he ,s s s t ac gin ntr be

End of main sequence life

o

c re

Co

Fusing H

Fusing H Hot, non-reacting He

Sh

ell

Fusing H

ex

pa

nd

s

Fusing He

He Core

ins

g beg

burnin

Red giant Hot, non-reacting He Red giant (Possible pulsating variable)

The fusion of helium to carbon is known as the triple alpha process because each helium nucleus is an alpha particle and three are required to make the carbon-12 nucleus. The double-headed arrow in the first equation, which indicates that the reaction occurs in both directions, is included because beryllium-8 is not very stable and tends to disintegrate into two helium-4 nuclei if it does not fuse quickly with another helium-4 nucleus. In contrast, carbon-12 is a very stable nucleus and eventually it will replace the hydrogen and helium that currently form the Sun’s core. Helium will continue to fuse on the edges of the carbon core, again causing the outer layers to expand. The Sun will become a bit unstable at this time, its size pulsing in and out every 10 000 years or so due to the sensitivity of the rate of helium fusion to temperature. As the outer layers expand, the temperature of the Sun drops, causing fusion to stop. The outer layers then cool and contract until the temperature again reaches the point where helium fusion can occur. Eventually this pulsing will throw off the outer layers of the Sun, which will form a ring of hot gases called a planetary nebula.

FIGURE 12.36 The Ring Nebula, a planetary nebula

Source: NASA

TOPIC 12 What are stars? 41

The Sun is not large enough to cause carbon to undergo fusion. The mainly carbon remnant is known as a white dwarf and this gradually cools over a few billion years to form a black dwarf. All of this happens over billions of years and so has not been observed. However, astrophysicists have considered what would happen to the Sun over time if it is powered by hydrogen fusion. They have observed stars at all the stages that they predict the Sun will go through. Red giants, planetary nebulae and white dwarfs are all present in our galaxy.

12.7 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. What is the source of energy in a star? 2. What two elements make up the vast majority of the nuclei in the universe? 3. The Sun radiates energy at 3.86 × 1026 J s−1 . Assume that this is the result of the fusion of hydrogen to helium following the chain of reactions presented in this topic. We can summarise the fusion of these reactions using the following equation. 411 H → 42 He + 2𝛽 + + v + 𝛾 + 26.76 MeV (1 MeV = 1.6 × 10−13 J) (a) How many of these fusion reactions would occur in the Sun per second? (b) Given that the Sun has about 1057 hydrogen nuclei but only about 10% of those will fuse in the core, how long do you predict the Sun will continue to fuse hydrogen? (Give your answer to one significant figure.) 4. One of the reactions that occurs in the Sun is the fusion of helium-3 and helium-4. (a) Complete the following equation. 3 He 2

5. 6. 7. 8. 9. 10.

+ 42 He → ___ + 𝛾 + 1.59 MeV

(b) Use the energy released by the reaction to determine the mass difference between the nuclei on the left- and right-hand sides of the equation in part (a). (c) Where could the helium isotopes for this reaction have come from? (d) This is an intermediate reaction in a chain of reactions that occurs in the Sun. What is the final product of this chain of reactions, given that the Sun is a main sequence star? (a) What are three arguments for the case that fusion is the source of the Sun’s energy? (b) Why is gravity not a good explanation of the source of the Sun’s power? Explain why, if hydrogen and helium make up the vast majority of the nuclei in the universe, they are relatively rare on Earth. Research another nuclear reaction chain that occurs in the Sun, other than the proton–proton chain. What is the product of fusion in main sequence stars? In red giants, fusion of larger nuclei occur to form heavier elements. Explain why the radius of the star grows significantly from its time as a main sequence star. Why do red giants not fit the patterns for surface temperature versus mass and luminosity versus mass observed for main sequence stars?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

42 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

12.8 The Hertzsprung–Russell diagram KEY CONCEPTS • Apply the Hertzsprung–Russell diagram as a tool to describe the evolution and death of stars with differing initial mass. • Relate the formation of stars to the formation of galaxies and planets. • Describe future scenarios for a star, including white dwarfs, neutron stars and black holes. • Explain the event horizon of a black hole and use to calculate the Schwarzschild radius. • Describe the effects of the gravitational fields of black holes on space and time.

12.8.1 The temperature of stars

Intensity

In the 1920s, astronomers like Cecilia Payne determined how stellar temperatures related to their spectrum. The spectrum of a star is not of equal intensity for all colours. They found that hot stars radiate more energy at short wavelengths than cooler stars. Short wavelengths correspond to the blue end of the visible spectrum, while longer wavelengths correspond to the red end of the spectrum. We are all familiar with this relationship between colour and temperature. In the school laboratory you have probably used a Bunsen burner. These burners have two settings, one a cool, yellow flame and the other a hot, blue flame. The flames emit more than just the colours we see. If you hold your hand in the air about 30 cm from the flame you can feel heat, indicating that the flame is emitting infra-red radiation that we can feel but cannot see. Astronomers generally like to use the kelvin FIGURE 12.37 Emission at different wavelengths for objects of different temperatures temperature scale. Every degree in temperature difference is the same as the Celsius scale, but 0 K 8000 K is set at the coldest possible temperature, which corresponds to −273.15 °C. This means that 0 °C is 273.15 K. On a clear night it is possible to see some 6000 K variation in the colour of stars, but the rods in the retinas of our eyes that are responsible for distinguishing colours are not very sensitive to dim light. A photograph will show the colours much more clearly. We notice that some stars are red and some are white or blue. The Sun is a yellow star, indicating that it is neither particularly hot nor cool in the range of star temperatures. The colour of a star indicates the area of the spectrum of the star that is most intense. Some stars are so hot that they emit most of their radiation at very short wavelengths of ultraviolet light, making them invisible from Earth. 3000 K They must be observed using UV telescopes in orbit because the atmosphere absorbs most UV radiation, preventing it from reaching ground-based telescopes. 1000 0 2000 3000 The temperature of a star’s outer layers determines UV visible IR its colour. The core of a star is much hotter than the Wavelength (nm) outer layers, due to fusion reactions and gravitational energy.

TOPIC 12 What are stars? 43

FIGURE 12.38 A portion of the night sky including the constellation of Orion. Four of the brightest stars in the sky can be seen. Notice the different colours: Betelgeuse is a red supergiant and Rigel is a blue supergiant.

Source: NASA

SAMPLE PROBLEM 11

Describe what happens to the wavelength and intensity of light from a star as its temperature increases. THINK 1.

Refer to figure 12.37. The graph shows the intensity of different wavelengths for different temperature stars.

WRITE

As temperature increases the wavelength of the light emitted becomes shorter and the intensity increases.

PRACTICE PROBLEM 11 Describe how astronomers can determine the surface temperature of a star.

12.8.2 Spectral type When the spectra of stars were first observed in detail in the nineteenth century, it seemed that the spectrum of every star was different. Gradually, some sense was made of the multitude of lines that crossed the spectra and stars were classified into spectral types. The system developed by Annie Jump Cannon (1863–1941) has been used since 1910. It classes stars as O, B, A, F, G, K or M according to the relative intensity of various absorption lines in their spectra. For example, for type A stars the lines of the hydrogen spectrum are very clear. The spectral classes are arranged in order of temperature from O, the hottest with a spectrum peaking in the ultraviolet, to M, the coolest with a spectrum peaking in the infra-red. The Sun is a type G star and these are yellow. A full description of the spectral classes is given in table 12.3. 44 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

TABLE 12.3 Spectral classifications and their corresponding features. Spectral class

Surface temperature (K)

Colour

Spectral features

O

28 000–50 000

Blue

Ionised helium lines Strong UV component

B

10 000–28 000

Blue

Neutral helium lines

A

7500–10 000

Blue-white

Strong hydrogen lines Ionised metal lines

F

6000–7500

White

Strong metal lines Weak hydrogen lines

G

5000–6000

Yellow

Ionised calcium lines Metal lines present

K

3500–5000

Orange

Neutral metals dominate Strong molecular lines

M

2500–3500

Red

Molecular lines dominate Strong neutral metals

Note: In astronomy the term ‘metal’ refers to any element other than hydrogen or helium.

Resources Digital documents Investigation 12.4 The colour of stars (doc-31885) Investigation 12.5 Hertzsprung–Russell diagrams (doc-31886) Teacher-led video Investigation 12.4 The colour of stars (tlvd-0830)

12.8.3 Hertzsprung–Russell diagrams

Luminosity

Absolute magnitude

In 1911 in Denmark, Ejnar Hertzsprung FIGURE 12.39 An H–R diagram showing various star types (1873–1967) plotted star luminosities (or equivalently, absolute magnitudes) Temperature (K) versus their spectral types (or 25000 3000 6000 10000 –10 equivalently, temperatures). In 1913, Supergiants Henry Norris-Russell (1877–1957) did –5 104 the same thing at Princeton. The result is known as a Hertzsprung–Russell 0 (H–R) diagram. H–R diagrams provide 102 Giants a wonderful synthesis of the data that we +5 1 have about stars. Main sequence Stars on the left side of the H–R diagram are hot and those on the right +10 10–2 are cool. Those towards the top of the White dwarfs diagram are very luminous, while those +15 10–4 towards the bottom are dim. Most stars K M F B O A G Spectral class cluster along a diagonal line from the bright and hot down to the dim and cold. These are the main sequence stars. They are all doing much the same thing, fusing hydrogen in their cores, just like the Sun. The hotter, brighter stars are more massive than the cooler, dimmer stars.

TOPIC 12 What are stars? 45

Many stars do not lie on the main sequence. Above the main sequence are stars that are brighter than their spectral class would predict for a main sequence star. These stars are very large and are known as giants and supergiants. Many are red because of their cooler surface temperature, but are still very bright because they have such a large radiating surface. Below the main sequence is a cluster of hot, dim stars known as white dwarfs. White dwarfs are made mainly of carbon and fusion has ceased in their cores. They are simply cooling down. During World War II Walter Baade discovered that there appeared to be yet another classification of stars. He called these classes Population I and Population II stars. Population I stars contain a greater variety of elements, while Population II stars contain little more than hydrogen and helium. Population I stars tend to be hot and blue, and all fit on the main sequence, whereas Population II stars are much more varied. Many are red giants and the top left corner of the main sequence in the H–R diagram is nearly empty of Population II stars. The Sun is a Population I star. The explanation for these two types of star is that Population I stars are young. They have more heavy elements in them because they are formed from the remnants of older stars, which produced heavy elements (helium and heavier) during the processes of fusion. Population I stars are all on the main sequence because they are still fusing hydrogen in their cores. The most massive stars are found in the top left corner of the main sequence. This area of the H–R diagram is nearly empty of Population II stars. Massive stars have a much greater pull of gravity on the gases that they are made from, so equilibrium between gravitational collapse and the radiation pressure from the core is reached at much higher temperatures than in the smaller stars. This accelerates the rate at which fusion occurs and so these stars ‘live fast and die young’. The most massive Population II stars have had time to complete the fusion of hydrogen in the core and move on to fusing helium. Few Population I stars, however, are old enough to have completed this stage of their lives. By identifying the two star populations, Baade had discovered a way for astrophysicists to view star populations at two points in time. This opens a window of research as stellar evolution takes billions of years, so it is not possible to just sit and wait to see what happens. FIGURE 12.40 This H–R diagram shows the ‘path’ of a star of one solar mass throughout its lifetime. 106 Supergiants 104

L (Lsun)

102

Possible shell He burning Shell H burning

Core He burning, possible variable

Main sequence

Red giants 1 solar mass

1

White dwarfs

10–2

10–4

40 000

20 000

10 000 T (K)

46 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

5000

2500

When the hydrogen in the core is exhausted, the star moves off the main sequence as it cools and expands into a red giant. Fusion of helium in the core begins. Once the helium in the core is exhausted, fusion of helium around the carbon core will cause the star to expand. In larger stars, fusion of nuclei can continue up to iron-56, but beyond iron it takes more energy to cause fusion than the fusion reaction produces, so the cycle stops. While fusion is occurring in shells around the core, the star pulses as a variable star. Due to its size and temperature it remains high above the main sequence on the H–R diagram. When a star about the size of the Sun throws off most of the remaining hydrogen and helium to form a planetary nebula, it can no longer sustain fusion and cools as a white dwarf, now below the main sequence as it is dim but still quite hot. White dwarfs are only about the size of Earth, but are much denser. They slowly cool, moving to the right in the H–R diagram as they become cold black dwarfs. These processes are summarised in table 12.4. TABLE 12.4 Fusion in different star types Star group

Energy-producing reactions

Main sequence

Nuclear fusion of H to He in core

Red giants

Nuclear fusion of He to C in core, with H fusion continuing in shell

Supergiants

Multiple nuclear fusions possible in shells, forming elements up to iron in core

White dwarfs

No nuclear fusion reactions occurring

SAMPLE PROBLEM 12

A star is observed to have a luminosity 1000 times that of the Sun and a surface temperature of about 10 000 K. Use an H–R diagram to predict what stage of evolution this star is in. THINK 1.

Move along the temperature axis of an H–R diagram to 10 000 K. Move up the luminosity axis to 1000 solar luminosities to locate the star on the H–R diagram. Name which group it is in.

WRITE

This star is on the main sequence, so it is fusing hydrogen in its core.

PRACTICE PROBLEM 12 Two stars have luminosities 10 000 times that of the Sun. One is a main sequence star and the other is a supergiant. Compare the temperatures, masses and ages of the stars. More massive stars have a violent end. When fusion ends in these stars, they start to collapse very rapidly. This process leads to enormously high temperatures and more fusion, including the formation of elements heavier than iron-56. The outer layers come crashing inwards and bounce off the core in a supernova, blasting a rich soup of elements into space. A supernova can outshine a whole galaxy for a period of time. Stars between 8 and 50 solar masses end in this way and are called Type II supernovae. They are mostly found in the spiral arms of galaxies, where stars form. These massive stars have lifetimes in the millions of years, rather than the billions of years of solar mass stars. Their death provides the material for new stars in the galaxy that contain elements other than hydrogen and helium; the stars that could form planets like those in the solar system. Type I supernovae are found in all parts of all types of galaxies. They form from old, low-mass stars. Stars with less than 8 solar masses lose a lot of their mass as they form a planetary nebula following the red giant phase. They are too small to go supernova at this stage, so begin to cool as white dwarfs. Many

TOPIC 12 What are stars? 47

stars, however, are in a binary system. If a white dwarf has a partner that is a red giant in a close orbit, the white dwarf will accrete hydrogen gas from the giant star. As the hydrogen falls onto the white dwarf, it undergoes fusion, heating the surface of the star rapidly and causing a nova, where the star glows a million times as brightly as the Sun. If the white dwarf is able to accrete sufficient hydrogen to have a total mass of more than 1.4 solar masses, it will collapse. Tremendous heat and pressure is generated, causing fusion reactions on a massive scale throughout the star and releasing incredible amounts of energy. These Type I supernovae are thousands of times brighter again than the nova. With this energy source rapidly depleted, the star collapses. This time the mass is too great for protons and electrons to remain apart. In reverse beta decay, the electrons and protons combine to form neutrons. The density of the resulting neutron star is the same as the density of atomic nuclei. The whole star has a diameter of little more than 10 kilometres, but it has a mass greater than that of the Sun. FIGURE 12.41 Stellar evolution

Note: M = Solar mass

Main sequence

Red giant (Number of burning shells depends on mass)

Planetary nebula (Original mass < 8 M ) Leaving behind a White dwarf (Core < 1.4 M )

Supernova (Original mass > 8 M ) Leaving behind a

Neutron star (Core < 3 M )

It is easy to distinguish Type I and Type II supernovae by their spectra. Type I supernovae have very little remaining hydrogen so the lines for hydrogen are missing from their spectra, while hydrogen lines are clear in Type II. Supernovae are rare, occurring at a rate of only about one per century per galaxy, with the last in the Milky Way recorded by Kepler in 1604. In 1987, a supernova was observed in the Large Magellanic Cloud. Photos of the cloud before the supernova showed the star before the massive explosion. It offered a wonderful opportunity to study supernovae and featured in the popular press. If the core of the star that remains following a supernova has a mass of more than about 3 solar masses, gravity is unstoppable. Without the energy supply of fusion to hold the star up, no known force can support the matter against the intense gravity and the remnant collapses to form a black hole. These exotic objects have a gravitational pull so great that even light cannot escape.

Black hole (Core > 3 M )

FIGURE 12.42 The Crab Nebula, the remnant of a supernova observed in 1054; at the centre a neutron star sweeps a beam of radiation past us as it rotates. These types of neutron stars are called pulsars.

Source: NASA

48 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 12.43 Stars of different masses evolve in different ways. 106 Supergiants

Core burning Shell burning

10

10 solar masses

4

Shell He burning Shell H burning

L (Lsun)

102

Core He burning, possible variable

Main sequence

Red giants Shell H burning

1

1 solar mass

White dwarfs

10–2

0.1 solar mass 10–4

40 000

20 000

10 000 T (K)

5000

2500

12.8.4 Schwartzchild radius Black holes are the result of sufficient mass being squeezed into a small enough space. You can imagine if we could squeeze the Earth, we could make it into a ball that has a smaller radius. As we reduce the volume of the Earth, the mass not ) change, so the density increases. For example, if we halve the radius of the ( does 4 3 Earth, its volume V = 3 πr will be only an eighth of what is now. The same mass squeezed into an eighth of the volume means that the density would be eight times as great. Gravity on the surface of this compressed Earth would be four times as strong as we know it. That means that gravity would pull inwards on the material that makes the Earth much more strongly. We could imagine squeezing the Earth smaller and smaller so that gravity gets larger and larger. Eventually we would reach the point where electrons would merge with protons to form an Earth of neutrons. At this point, the Earth would have a radius of around 200 metres! But what if we kept squeezing so that the neutrons could no longer withstand the pressure? The gravitational force on the neutrons would be so great that the matter would collapse completely and form a black hole. The radius at which this collapse occurs is called the Schwartzchild radius. The Schwartzchild radius is determined by the formula: rs =

2GM c2

Where: G = the universal constant of gravitation (6.67 × 10−11 N m2 kg−2 ) M = the mass of the star c = the speed of light (3.0 × 108 m s−1 )

TOPIC 12 What are stars? 49

The Schwartzchild radius for the Earth is: 2GM c2 2 × 6.67 × 10−11 × 6.0 × 1024 = (3.0 × 108 )2 ≈ 9 mm

rs =

This is a bit unrealistic because there is no clear way of squeezing the Earth into a ball with a radius of 9 mm. What is needed to make this happen is such an enormous mass that gravity itself does the work; that is what you have when a giant star collapses as a result of consuming all of its nuclear fuel. The smallest star that would collapse to form a black hole is about 25 times the mass of the Sun. However, a lot of this material is lost in the supernova that occurs at the end of the star’s life. The remaining mass would be about three times the mass of the Sun or 6 × 1030 kg. 2GM c2 2 × 6.67 × 10−11 × 6.0 × 1030 = (3.0 × 108 )2 ≈ 9 km

rs =

The Schwartzchild radius for this enormous star is around 9 km. About three times the mass of the Sun would be within this 9 km radius sphere, and the gravity would be so powerful that no forces could stop it collapsing. Anything that moved within this radius could not escape. No events inside this radius can be known to observers outside — they would be over the boundary of the black hole, known as the event horizon.

12.8.5 Black holes Black holes are predicted by Einstein’s General Theory of Relativity. In this theory, Einstein described changes in gravitational fields as curves in space-time. Space-time is required to locate events. It is not enough to ask where an event happens, we also need to ask when. Einstein’s theory was that mass produced curves in space-time and the stronger the gravity, the tighter the curves. The curves explain why it is difficult to escape a massive body. It is like going uphill on the curves of space-time. Fascinatingly though, gravity also affects time. Observers see time passing more slowly for objects in an intense gravitational field. In the case of a black hole, space-time is affected so much that it is impossible to climb the hill to escape once inside the event horizon. As you watch something fall into a black hole, perhaps a clock, you would notice it ticking more and more slowly as it approached the event horizon. The light from the clock finds it increasingly difficult to reach anyone observing the event and it would appear to never cross the event horizon. Unfortunately for anyone falling into the black hole, time would pass as normal and tidal forces — the difference in gravitational pull (or the shape of space-time) between your head and your toes — would stretch you out in a process fittingly called ‘spaghettification’ and you would rapidly die. Black holes are difficult to detect because they do not emit light, or any form of electromagnetic radiation. However, the motion of objects nearby can reveal their presence. A star might be detected moving as though it was orbiting another star. If that other star is not visible and the strength of its gravity is sufficient, then it may be a black hole. Black holes also cause material that is falling into them to release radiation. Once the material has reached the event horizon, no further information of any type can escape. However, X-rays released by matter in its last moments before disappearing beneath the event horizon can indicate the presence of a black hole.

50 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Another form of black hole has been found at the centre of some galaxies, including the Milky Way. These have the mass of millions of stars and are known as galactic black holes.

12.8 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. (a) (b) (c) (d)

Sketch an H–R diagram and circle and label the main sequence. Is it normal for stars to move along the main sequence during their life spans? Will the Sun ever become a supergiant? On the diagram you produced in part (a), circle and label the stars that have mainly fusion of helium and heavier elements in their cores. (e) In what circumstances is it possible to have a very massive star positioned at the right-hand side of the diagram? (f) What section of the H–R diagram contains remnants of stars that no longer have fusion reactions as a source of energy? (g) Andre measures the composition and temperature of two stars to be the same. He expects to place them in the same region of the H–R diagram. One of the stars is brighter than the other. What does this tell him about the two stars? (h) A star lies on the main sequence. What does this tell you about the star? (i) What colour are stars on the left-hand side of the diagram? 2. Would you expect to find planets like Earth around Population II stars? Explain. 3. Compare and contrast the spectra of type O and type A stars. (a) A type O star

(b) A type A star

4. The Sun is a yellow star. Estimate its surface temperature and give its spectral class. 5. You are an astronomer with a research project to search for black holes. How will you look for them? 6. Use the internet to identify two black hole ‘candidates’ and explain why astrophysicists believe them to be black holes. 7. Write a description of the life cycle of a one solar mass star. 8. A massive star sheds a lot of material in a supernova explosion and the remaining mass, equal to four solar masses, collapses to form a black hole. What is the Schwarzchild radius of this black hole? 9. The black hole from question 8 consumes more mass over time, stripping it from a neighbouring star. What happens to the Schwarzchild radius? 10. You are in a spacecraft just outside the event horizon of a black hole. (a) Do you notice anything different about the rate that time passes on the spacecraft? (b) Later you return to Earth. What do you notice about how you have aged compared with those who remained on Earth?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

TOPIC 12 What are stars? 51

12.9 Galaxies KEY CONCEPTS • Compare the Milky Way galaxy to other galaxies with different shape, colour or size. • Explain and analyse how the chemical composition of stars and galaxies is used to determine their age. • Investigate selected aspects of stellar life cycles by interpreting and applying appropriate data from relevant databases.

12.9.1 Comparing the Milky Way galaxy to other galaxies Galaxies come in several forms. Understanding how they came to take these forms is a major area of research. Galaxies range in size from hundreds of thousands of stars to hundreds of billions of stars. The Milky Way, our galaxy, is a large galaxy with approximately 100 to 200 billion stars. Galaxies are classified by their shape as either spiral, elliptical, irregular or dwarf. The Milky Way is a spiral galaxy. Galaxies are classified by their shape as either spiral, elliptical, irregular or dwarf. The Milky Way is a spiral galaxy. FIGURE 12.44 Galaxies are classed as (a) spiral, (b) elliptical, (c) irregular or (d) dwarf, depending on their shape. (a)

(c)

(b)

(d)

Source: (a) and (b) ESA / Hubble & NASA (c) NASA, ESA, and The Hubble Heritage Team STScI/ AURA / NASA

52 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Most galaxies formed in the early years of the universe. A clue to the age of a galaxy, and the stars within it, comes from its chemical composition. Stars and galaxies that have few ‘metals’ — that is, elements other than hydrogen and helium — formed from material that is little changed since the beginning of the universe, where around 75% of the mass of atoms in the universe was hydrogen and 25% helium. Atoms of higher atomic number were formed in stars that have since exploded as supernovae, which pushed the elements out into the surrounding space. If a star or galaxy contains elements such as oxygen, carbon and iron, that are plentiful on Earth, we know they have been formed from material made in earlier stars. The galaxies containing these elements must be relatively old and the stars containing them relatively new. The solar system, for example, is understood to be over four and a half billion years old. It contains significant quantities of elements formed in stars that exploded before it formed. The Milky Way is understood to be one of the older of the universe’s galaxies, or at least some of its oldest stars are, dating back to when galaxies first started to form. A method of dating stars is to measure the quantity of beryllium-9. This isotope does not form in fusion in stars and was not formed in the big bang. It forms from the collision of cosmic rays (fast-moving protons) with heavier elements. The more beryllium-9 that is found, the older the star that contains it and the longer it has been exposed to cosmic rays. Some of the oldest stars in the Milky Way are dated at around 13.6 billion years old using this technique, close to the 13.7 billion years age of the universe.

Spiral galaxies A spiral galaxy, like the Milky Way, has arms that spiral around a central ‘bulge’. Young stars are found within the spiral arms and new stars are continually being formed in some of the interstellar gas clouds in the arms. New stars are bluer on average than old stars (which are red). This means that the spiral arms of these galaxies tend to be blue. In contrast, stars in the central bulge are usually old and not much star formation occurs. As a result, the bulge is redder than the spiral arms. As we look away from the spiral arms, only old stars are found. Astronomers know they are old, because their spectra reveal them to be Population II stars. They are found in a sphere or halo around the galaxy, with many of them grouped in globular clusters. The stars, or clusters of stars, in the halo are so widely spaced that they have little effect on the appearance of a galaxy, which looks like a flat disc. The fact that these old stars form a sphere around spiral galaxies suggests that these galaxies collapsed into rotating discs after the stars were formed. Stars formed from gas clouds once most of the galaxy had collapsed into a disc would continue to orbit around the centre of the galaxy within the disc, as they have no component of velocity to move them outside it; however, the early stars must have had a component of their velocities that took them outside the plane of the galaxy. Spiral galaxies are flat discs because the stars within them revolve around the centre of mass of the galaxy, due to the initial motion of the gas cloud from which the galaxy formed. As this gas cloud contracted, the rotation rate increased in the same way that an ice-skater spins more quickly when she pulls in her arms. Eventually, the spinning and the gravitational collapse reach a balance, just as there is a speed where the ice-skater is not strong enough to pull in her arms any further. However, no force acted to create this balance parallel to the axis of rotation, so the galaxies flattened into discs.

TOPIC 12 What are stars? 53

FIGURE 12.45 One model for the formation of a spiral galaxy

(a) The hydrogen/helium cloud contracts due to gravity.

N e t ro t a t i o n

(b)

Axis of rotation

R o t a ti o n

The cloud contracts, mostly parallel to the axis of rotation, forming a disc.

(c)

Tidal forces from nearby galaxies cause spiral arms to form.

Elliptical galaxies Elliptical galaxies contain only Population II stars, unlike spiral galaxies, which contain very old stars as well as young and still-forming stars. Due to the dominance of old stars, eliptical galaxies tend to have a reddish tinge. It seems that something about the formation of elliptical galaxies causes a quick burst of star formation. Elliptical galaxies tend to be found within large clusters of galaxies, in contrast to spiral galaxies, which formed where the density of galaxies was lower. Elliptical galaxies are composed of stars that revolve randomly around the centre of mass, so there is no cause for the galaxy to flatten into a disc. This random motion may have come about because two galaxies with different planes of rotation collided in the early universe. The interaction caused by the collision could also explain the sudden burst of star formation early in the existence of elliptical galaxies.

Irregular galaxies Irregular galaxies tend to be small, like the Magellanic clouds. In contrast to the stars in spiral and elliptical galaxies, the types of stars found in irregular galaxies do not follow any particular pattern. The irregular shape is probably due to the gravitational disturbances of large neighbouring galaxies. Collisions between galaxies are also thought to cause the formation of some irregular galaxies. Irregular galaxies are often bright and blue due to the star formation occurring within them.

Unanswered questions It is still not known how galaxies form. Astrophysicists use computer simulations to attempt to discover why galaxies form particular shapes, such as why spiral galaxies have spiral arms. One attempt at explaining the spiral arms has been to consider tidal effects. Just as the Moon exerts tidal effects on the Earth, most notably in the rise and fall of the oceans, satellite galaxies like the Large Magellanic Cloud

54 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

exert tidal forces on the Milky Way. The result is that the stars do not revolve about the centre of the galaxy in circular orbits. It has been suggested that this might cause the stars to bunch together in arms that form spirals, but why would such a regular feature appear in one galaxy while the other galaxy, in this case the Large Magellanic Cloud, remains so irregular? Clearer answers to how and when the galaxies formed will probably be found in the next few decades as data from the many new instruments are analysed.

The Milky Way Our galaxy, the Milky Way, contains over 100 billion stars. It is a spiral galaxy whose colour appears to be white. It is difficult to study the Milky Way’s properties because we are inside it. There is no way of stepping outside to get a view of what it looks like overall, so astronomers need to piece together an understanding from the information they have. There is a lot of gas and dust clouds that block the views to parts of the galaxy. This includes our view of the centre of the galaxy where a supermassive black hole is understood to be. Due to our obscured view, scientists have only been able to produce an image of a black hole at the centre of a different galaxy, even though the one in our galaxy is relatively much closer.

12.9 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. 2. 3. 4. 5. 6.

7. 8. 9.

10.

There are no photographs of the Milky Way galaxy from the outside. Explain why this is. There are three major types of galaxy: spiral, elliptical and irregular. Which one describes the Milky Way? Describe the colours of different parts of the Milky Way galaxy. What causes these colours? Approximately how many stars make up the Milky Way galaxy? How does this compare to other galaxies? One galaxy is observed to contain a relatively high number of red stars and another has more blue stars. What does this tell you about the two galaxies? Two red dwarfs are observed. One’s spectrum reveals virtually no elements other than hydrogen and a small amount of helium. The other contains significant quantities of other elements like oxygen, carbon and nitrogen. What can you say about these two stars from this information? You observe an enormous blue giant in our galaxy. What would we expect to observe in its chemical composition? Compare the chemical compositions of galaxies that are relatively nearby to those at the most distant reaches of observation. Use www.atlasoftheuniverse.com/12lys.html (or a similar database of nearby stars). At the bottom of the page there is a list of all of the known stars within 12.5 light-years. (a) How many stars are within this distance? (b) How many white dwarfs are there? (c) How many red giants are there? (d) How many main sequence stars are there? (e) How many of these stars are in multiple systems (two or three stars)? Use the data on www.atlasoftheuniverse.com, or similar, to plot the closest 20 stars on an H–R diagram. The vertical axis can be in absolute magnitude if luminosity is not provided. Note, absolute magnitude is a scale that represents the dimmest stars as a large positive number that diminishes as the stars get more luminous. The most luminous stars will have a large negative absolute magnitude! The horizontal axis should be spectral type (class).

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

TOPIC 12 What are stars? 55

12.10 Review 12.10.1 Summary • • • • • • • • • • • • • • • • •

• • •

The information we have about the universe is gathered from the electromagnetic radiation that has reached us. Spectroscopy can be used to analyse the chemical composition of stars, which provides the key to determining the age of the star. From electromagnetic radiation, astrophysicists are able to determine the temperatures of stars and the elements that make up the stars. Standard candles, such as Cepheid variables whose period of varying luminosity is related to their average luminosity, enable us to measure distances to distant stars. Parallax is the only direct method that astrophysicists have to measure the distance to stars, and it only works for stars that are relatively nearby due to the small parallax of such distant objects. The spectra of more distant galaxies are all red shifted by an amount that increases with distance, so red shift is a very useful technique for measuring how far away distant galaxies are. The Sun is a typical star, enormous by Earth standards, but dwarfed by many larger stars. The Sun is made mostly of ionised hydrogen and helium. The total energy output of a star is called its luminosity, which varies with the age and mass of the star. More massive stars shine more brightly and are hotter than low mass stars; they also pass through their life cycle more quickly. The Hertzsprung–Russell (H–R) diagram is a graph of the luminosity of stars against their temperature or colour. In an H–R diagram, most stars are found on a diagonal line called the main sequence, from hot and luminous down to cool and dim. Main sequence stars are fusing hydrogen into helium in their cores. Fusion is the source of energy in stars and involves a mass loss in accordance with E = mc2 . Stars above the main sequence in an H–R diagram have consumed all of the hydrogen in their cores and have expanded in size to form red giants due to fusion heating outer layers of the star. Stars below the main sequence in an H–R diagram are ending their life cycle; fusion has finished and most of their material has been shed forming a planetary nebula. They are remnants of stars cooling down as white dwarfs. The event horizon is the name given to the sphere around a black hole beyond which no matter or light can escape. The radius of the sphere is called the Schwarzchild radius and can be calculated using: 2GM r= 2 . c To observers at a distance from a black hole, time is measured to pass more slowly for objects near the black hole. Black holes distort the shape of space around them to the point that it collapses on itself. The Milky Way, our galaxy, is a large spiral galaxy. Other galaxies vary vastly in size and shape.

Resources

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0037).

56 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

12.10.2 Key terms When a body, such as a star or a planet, accretes it gains mass as material is attracted to it by gravity. Amplitude of a wave is the size of the maximum disturbance of the medium from its normal state. Astrophysics is the area of science in which the laws of physics are applied to the universe. An astronomical unit (AU) is the average distance from the Earth to the Sun (1 AU = 1.496 × 1011 m). Baseline is the distance between the two observation points in a parallax measurement. Binary stars are stars that occur in pairs which orbit a common point. A black dwarf is a white dwarf that has had time to cool. A black hole is an object whose gravity is so strong that nothing can escape it. Brightness is the energy per second per square metre at the place where the star is observed. A Cepheid variable is a large star, near the end of its life, that pulses at a rate that is related to its luminosity. A constellation is a group of stars forming recognisable patterns in the sky. Frequency refers to the number of times of a periodic wave repeats itself every second. Frequency is measured in hertz (Hz) and 1 Hz = 1 s−1 . Fusion is the process where two nuclei combine to form a larger nucleus and release energy. A galaxy is an enormous group of stars held together by gravity. A geocentric model of the universe is one that has the Earth as its centre. A heliocentric model of the universe is one in which the Sun is at the centre. A light-year is the distance that light travels through a vacuum in one year (1 light-year = 9.46 × 1015 m). Luminosity is the total energy that the star radiates per second. Main sequence stars are clustered on a diagonal line through an H–R diagram. They are in the phase of their cycle where they are fusing hydrogen in their cores. A nebula (plural = nebulae) is a vast cloud of gas in space. Parallax is an effect that is noticed whenever an object is viewed from two different positions. A parsec (pc) is the distance to a star that would have a parallax of one arc second using the radius of the Earth’s orbit about the Sun as a baseline (1 pc = 3.086 × 1016 m). Period refers tothe time it takes a periodic wave source to produce a complete wave. A planetary nebula is a ring of gas that has been thrown off a star in the latter stages of its life. A red giant is a star that has completed hydrogen fusion in its core and swelled to an enormous size. The expanded surface of the star is cool so it glows red. Revolving means to move around a body in an orbit. Solar mass refers to the mass of the Sun. The solar system is made up of the eight planets, their moons and other smaller bodies in orbit around the sun. Spectroscopy is the science of examining spectra consisting of a range of electromagnetic radiation wavelengths. A supernova is an explosion of a massive star once fusion reactions in its core cease. A transit occurs when one body passes across the face of another, for example, when Venus passes between the Earth and the Sun. Triple alpha process is the process of three alpha particles producing a carbon-12 nucleus. The universe refers to everything, including all matter, all space and all time. Wavelength is the distance between successive corresponding parts of a periodic wave. A white dwarf is the hot remains of a star in which fusion has ceased. It is small and dense and made mostly of carbon.

Resources Digital document Key terms glossary (doc-32279)

TOPIC 12 What are stars? 57

12.10.3 Practical work and investigations Investigation 12.1 Brightness versus distance Aim: To model the decrease in brightness of a star with distance Digital document: doc-31882

Investigation 12.2 Solar observation Aim: To observe the Sun and sunspots in a safe manner Digital document: doc-31883

Investigation 12.3 Cardboard solar observatory Aim: To make a safe solar observatory to observe the Sun Digital document: doc-31884

Investigation 12.4 The colour of stars Aim: To model the colour of stars by connecting an incandescent globe to different voltages Digital document: doc-31885 Teacher-led video: tlvd-0830

Investigation 12.5 Hertzsprung–Russell diagrams Aim: To construct a Hertzsprung–Russell diagram using provided secondary data Digital document: doc-31886

Resources Digital document Practical investigation logbook (doc-32280)

12.10 Exercises To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au.

12.10 Exercise 1: Multiple choice questions 1.

Which wave does not travel at c, the speed of light? A. Radio wave B. Microwave C. Gamma ray D. Sound wave

58 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

2.

3.

4.

5.

6.

7.

8.

9.

10.

What is the only direct method we have of measuring distances to stars? A. Parallax B. Standard candles C. Redshift D. Spectroscopy The radiation spectrum from the Sun when passed through a spectroscope forms which of the following? A. Emission spectrum B. Absorption spectrum C. Continuous spectrum D. Blackbody spectrum What does the absorption spectrum from a star reveal? A. The chemical composition of the star B. The mass of the star C. The radius of the star D. The luminosity of the star The difference between a nebula and a galaxy can be determined using spectroscopy. A. A nebula has an emission spectrum and a galaxy an absorption spectrum. B. A nebula has an absorption spectrum and a galaxy an emission spectrum. C. A nebula has beryllium in its emission spectrum, a galaxy does not. D. A galaxy produces a continuous spectrum. The mass of a star can be determined if it has a companion star orbiting it. The more massive the star: A. the longer the period of orbit of the companion for a given radius of orbit. B. the shorter the period of orbit of the companion for a given radius of orbit. C. the smaller the radius of orbit of the companion. D. the larger the radius of orbit of the companion. What is the major energy producing process in the core of a red giant? A. Hydrogen is fusing to form helium. B. Helium is fusing to form carbon. C. Carbon is fusing to form iron. D. There is fission of helium to form hydrogen. The more massive the star, the more luminous. This relationship is true for which of the following? A. All stars B. White dwarf stars C. Red giants D. Main sequence stars Which stars last longest? A. Blue giants B. Red dwarfs C. Yellow stars like the Sun D. It has nothing to do with colour. In which of the following is fusion taking place? A. Black hole B. White dwarf C. Red dwarf D. Neutron star

TOPIC 12 What are stars? 59

12.10 Exercise 2: Short answer questions 1. 2. 3.

4. 5. 6.

7. 8. 9. 10.

Explain the difference between the luminosity and the brightness of a star making use of how they each change when the observation distance doubles. Light from a star enters a spectroscope with wavelength 550 nm. Calculate its period and frequency. a. Explain how parallax can be used to measure the distance to stars. b. How can the distance to ’standard candles’ be measured if they are too far away to measure their parallax? Describe the Sun’s mass and temperature compared with other stars. Explain how protons in the core of the Sun undergo fusion to form helium-4 nuclei. Include the number of protons required and any other particles produced in the process in your answer. a. Use figure 12.39 to list three types of stars that could have the spectral class G. b. You observe a star with a spectral type of G. What other measurement needs to be made to determine which type of star it is? Does the spectral class of a star change in its life cycle? Give an example. What is the main factor determining the life cycle that a star will take? Give two examples to show how they differ when this factor is different. List the three main types of galaxies and the differences between them. Access a list of the nearest stars such as the list of nearest stars and brown dwarfs. a. Complete a table showing each spectral type (stellar class) O, B, A, F, G, K and M to determine the percentage of nearby stars in each class. b. What proportion of these stars are on the main sequence (indicated by a V in their class)?

12.10 Exercise 3: Exam practice questions Question 1 (3 marks) What source of data can astronomers use to determine each of the following? a. The chemical composition of a star b. The distance to a star

1 mark 2 marks

Question 2 (1 mark) Figure 12.19 shows the spectra of different spectral type stars. What is the difference between stars of different spectral types? Question 3 (5 marks) a. Use figure 12.29 to predict the luminosity of a main sequence star that is observed to have a mass 1.5 times that of our Sun. 3 marks b. A star has a companion star orbiting at a radius of 3 × 1012 km. The period of revolution of the companion is 1 years. Calculate the mass of the star. 2 marks Question 4 (3 marks) The oldest stars in the Milky Way tend to be in globular clusters. What difference in chemical composition of the stars in globular clusters would you notice compared with stars in the rest of the galaxy? Explain your reasoning. Question 5 (3 marks) In spiral galaxies, young stars are much more likely to be found in the spiral arms than in the central bulge. What colour difference would you see between the spiral arms and the central bulge? Explain.

60 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

12.10 Exercise 4: studyON topic test Fully worked solutions and sample responses are available in your digital formats.

Test maker Create unique tests and exams from our extensive range of questions, including practice exam questions. Access the assignments section in learnON to begin creating and assigning assessments to students.

TOPIC 12 What are stars? 61

AREA OF STUDY 2 OPTIONS OBSERVATION OF THE PHYSICAL WORLD

13

Is there life beyond Earth’s solar system? 13.1 Overview Numerous videos and interactivities are available just where you need them, at the point of learning, in your digital formats, eBookPLUS and learnON at www.jacplus.com.au.

13.1.1 Introduction Understanding how life could exist in our universe is the ultimate quest for astronomers. The investigation of our solar system using spacecraft has revealed much new information and stimulated extensive debate on the origins of life on Earth and beyond. But how do we investigate outside our solar system? How can we discover the composition and origin of other planets and stars when they are too far away to physically visit? In earlier science studies, you may have discovered that visible light is just a small part of a much bigger electromagnetic spectrum. This spectrum identifies light through different wavelengths and frequencies — that is, light modelled as a wave. This model of light is the key to our understanding of the universe, so it is important to look at what the model tells us. FIGURE 13.1 Observations from ALMA (Atacama Large Millimeter/submillimeter Array) reveal extraordinarily fine detail that has never been seen before in the planet-forming disc around a young star.

Source: ALMA ESO/NAOJ/NRAO / NASA / ESA

TOPIC 13 Is there life beyond Earth’s solar system? 1

13.1.2 What you will learn KEY KNOWLEDGE After completing this topic you will be able to: Information from beyond Earth’s solar system • identify the spectrum of electromagnetic radiation as the basis for all observations of the universe • explain how emission and absorption line spectra are produced with reference to the transition of electrons between energy levels of the atom • identify spectroscopy as a tool to investigate the light from stars, and interpret and analyse spectroscopic data with reference to information from beyond our solar system • describe how planets can be identified by using the common centre of mass and the gravitational effect of a planetary system on a star Locating extrasolar planets • compare methods of exoplanet detection including astrometric, radial velocity, transit method and microlensing, referring to databases that differentiate for size, eccentricity and radius v ∆𝜆 = • explain and apply Doppler shift including spectral shift and ‘wobble’ of planetary systems using: 𝜆0 c • investigate how the composition of an exoplanet can be determined using spectral analysis Conditions for life beyond Earth’s solar system • explain the presence of liquid water as determining the habitable zones of a star system and the most likely place for life • explain the origins of life in the universe as having come from organic molecules in space, or as originating on Earth or an Earth-like planet through reactions of elements and compounds Possibility of life beyond Earth’s solar system • explain the use of the Fermi paradox to question the possibility of life outside Earth’s solar system and identify its counter arguments ∗ • apply the Drake equation: N = R fp ne f1 fi fc L, as a way of predicting the likelihood of life existing in the universe by making reasonable assumptions based on evidence and speculation • distinguish between targeted and untargeted searches for extra-terrestrial intelligence (ETI), and describe optimising strategies including where to look and how to ‘listen’ with reference to choice of frequency and bandwidth • explain why the radio spectrum is the best section of the electromagnetic spectrum to search the sky for possible ETI signals, including the cosmic radio window and the use of radio astronomy in the search • explain the nature of information that humans transmit beyond Earth to signal that intelligent life exists on Earth. Source: VCE Physics Study Design (2016) extracts © VCAA; reproduced by permission.

PRACTICAL WORK AND INVESTIGATIONS Practical work is a central component of learning and assessment. Experiments and investigations, supported by a Practical investigation logbook and Teacher-led videos, are included in this topic to provide opportunities to undertake investigations and communicate findings.

Resources Digital documents Key science skills — VCE Units 1–4 (doc-31856) Key terms glossary (doc-32277) Practical investigation logbook (doc-32278)

To access key concept summaries and past VCE exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0038).

2 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

13.2 Spectroscopic analysis of light KEY CONCEPTS • Identify the spectrum of electromagnetic radiation as the basis for all observations of the universe. • Explain how emission and absorption line spectra are produced with reference to the transition of electrons between energy levels of the atom. • Identify spectroscopy as a tool to investigate the light from stars, and interpret and analyse spectroscopic data with reference to information from beyond our solar system.

13.2.1 Electromagnetic radiation In order to understand how light can be used to detect exoplanets we must first understand what light is and how it behaves. Light is a form of electromagnetic radiation which is a transverse periodic wave that travels at a constant speed of 3.00 × 108 m s−1 in a vacuum. Electromagnetic radiation consists of an electric component perpendicular to a magnetic component as shown in figure 13.2. FIGURE 13.2 A transverse wave model E

B v=c

The frequency of a periodic wave is the number of times that it repeats itself every second. Frequency is measured in hertz (Hz) and 1 Hz = 1 s−1 . Frequency can be represented by the symbol f. The period of a periodic wave is the time it takes a source to produce a complete wave. This is the same as the time taken for a complete wave to pass a given point. The period is measured in seconds and is represented by the symbol T. The period of a wave is the reciprocal of its frequency: T=

1 f

The wavelength is the distance between successive corresponding parts of a periodic wave. The wavelength is also the distance travelled by a periodic wave during a time interval of one period. For transverse periodic waves, the wavelength is equal to the distance between successive crests (or troughs). For longitudinal periodic waves, the wavelength is equal to the distance between two successive compressions (regions where particles are closest together) or rarefactions (regions where particles are furthest apart). Wavelength is represented by the symbol 𝜆 (lambda). The speed, v, of a periodic wave is related to the frequency and period. In a time interval of one period, T, the wave travels a distance of one wavelength, 𝜆. speed =

distance 𝜆 𝜆 = = 1 = f𝜆 time T f

TOPIC 13 Is there life beyond Earth’s solar system? 3

This relationship, v = f𝜆, is sometimes referred to as the universal wave equation. Recall that for light (and all electromagnetic radiation) the speed is a constant in a vacuum. The colour of light is determined by the frequency (or wavelength). Red light has a lower frequency than violet light, and therefore a longer wavelength. The frequency and wavelengths of some colours are summarised in table 13.1. TABLE 13.1 Frequency and wavelength of colours Red

Orange

Yellow

Green

Blue

Violet

Frequency ( × 1012 hertz)

430

480

520

570

650

730

Wavelength (nanometres)

700

625

580

525

460

410

REMEMBER The visible spectrum of light contains red, orange, yellow, green, blue, indigo and violet. The spectrum continues into invisible forms of radiation, including infra-red at lower frequencies than red and ultraviolet at higher frequencies than violet.

FIGURE 13.3 Forms of radiation and their place in the electromagnetic spectrum. The visible portion of the spectrum is shown enlarged in the upper part of the diagram.

(Nanometres)

750

700

650

600

550

500

450

400

Visible spectrum (white light) Power and telephone Electrical generator

108

Radio waves AM

FM

Microwaves

104

Ultraviolet rays

Incandescent lamps, heat lamps

lamps, fluorescent tubes, sparks, lasers

Radar

Klystron tubes, microwave lasers

Electronic tubes and semi-conductor devices

106

Infrared rays

102

1

10−2

10−4

10−6

10−8

X-rays

X-ray tubes

10−10

Gamma rays

Cosmic rays, radioactive isotopes

10−12

10−14

Wavelength in metres 102

104

106

108

1010

1012

1014

Frequency in hertz

4 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

1016

1018

1020

1022

13.2.2 Production of light It is important to understand why objects emit light. Think about the sources of light you come into contact with everyday. Can you explain why they make light? Electromagnetic radiation is produced when charged particles accelerate. When an object gets hot, the particles that make up the object gain kinetic energy; that is, they move about. When an object gets really hot, charged particles such as electrons start to move so much they can oscillate very quickly and sometimes collide with nuclei or other charged particles around them. Thus, we have charged particles experiencing acceleration, and so we have light! The hotter the object, the more kinetic energy and the more energetic movement of the charge, resulting in higher energy light emissions. This is how an incandescent source emits light. Incandescent light sources use heat (through electrical resistance) to create light; fluorescent light sources use electrical energy to stimulate transitions of electrons within the energy levels of an atom (see figure 13.4). All incandescent sources produce a continuous spectrum. This is because the interactions and collisions between charged particles are random. This means any size collision is possible and can occur, from a very small glancing bump to a head-on smash. The size of the acceleration (or deceleration) results in different types of electromagnetic emissions or wavelengths produced. Cooler objects tend to be red in colour as red wavelengths represent lower energy emissions. As objects get hotter, their charged particles experience more frequent, higher energy collisions resulting in higher energy emission and the colours we see move towards yellow, white and then to blue (figure 13.5). The intensity or brightness of light increases because there are many more collisions as the object gets hotter, which in turn means more electromagnetic radiation is emitted. FIGURE 13.4 An incandescent light source and fluorescent light source. Both involve the excitement of charged particles within a material.

FIGURE 13.5 When iron such as this iron pipe is heated, the hottest section is the brightest and is yellow in colour. The cooler sections are less bright and red.

The relationship between heat and the electromagnetic emission of an object is often represented as a blackbody curve (figure 13.6). This curve enables us to examine the spread of an object’s spectrum as well as how intense (bright) different parts of the spectrum are. The blackbody curve shows that as temperature increases, the shape of the curve tends to ultraviolet and gets more peaked. You can see how the wavelength of maximum brightness moves up and towards the purple end of the visible spectrum. Remember, all the other parts of the spectrum are still being produced, this is why the curve slopes down on both sides.

TOPIC 13 Is there life beyond Earth’s solar system? 5

FIGURE 13.6 The blackbody curve

Intensity (relative scale)

1.0

6000 K

0.5

5000 K

4000 K 3000 K 0 0

5000

Ultraviolet

Visible

10 000

15 000

20 000

Infra-red Wavelength (× 10−10 m)

As all stars are very hot, they all emit a continuous spectrum of incandescent light. On a clear night it is possible to see some variation in the colour of stars but the rods in the retinas of our eyes that are responsible for distinguishing colours are not very sensitive to dim light. A photograph will show the colours much more clearly. We notice that some stars are red and some are white or blue. The Sun is a yellow star, indicating that it is neither particularly hot nor cool in the range of star temperatures. The colour of a star indicates the area of the spectrum of the star that is most intense. SAMPLE PROBLEM 1

Describe what happens to the wavelength and intensity of light from a star as its temperature increases. THINK 1.

Consider the blackbody curves in figure 13.6.

WRITE

The peak wavelength decreases and the intensity increases as the temperature increases.

PRACTICE PROBLEM 1 Describe how astronomers can determine the surface temperature of a star.

13.2.3 Dispersion: producing colour from white light White light is a part of the electromagnetic spectrum and can be separated into colours using a narrow beam of light and a glass triangular prism. This phenomenon is called dispersion. It was first analysed by Isaac 6 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Newton in 1666, although René Descartes had sought an explanation for rainbows in 1637 by working with a spherical glass flask filled with water. As light enters a triangular glass prism, it is bent. It then travels through the prism to the other side where it is bent again. The colours spread as they enter the glass and travel on different paths through the triangular prism (figure 13.7). They are spread even more as they leave the glass. Violet light is bent the most and red the least. The order of the colours, from the colour that bends least to the colour that bends most, is red, orange, yellow, green, blue, indigo, violet. You would know these colours as the colours of a rainbow. They are also known as the visible spectrum. FIGURE 13.7 The colours in white light separate as they enter the glass and separate even more when they leave. At each edge, violet light is deflected more than red light. Glass prism

White light

13.2.4 Spectroscopy By analysing the different spectra of light from distant stars we can determine their temperature, chemical composition and age. We can tell how fast they are moving and whether they have planets. This process of examining the spectra of light is called spectroscopy.

The solar spectrum Isaac Newton (1642–1727) published his book Opticks in 1704. In the first volume he demonstrated that light from the Sun can be dispersed into its constituent colours. Other theories about why rainbows formed, why prisms of glass produced a spectrum of colours and why soap bubbles appeared coloured involved the prism, raindrop or bubble altering the light. However, as Newton demonstrated, the prism, raindrop or bubble simply disperse the light according to its colour (wavelength), revealing information about the Sun. Newton’s prisms showed the colour spectrum from the Sun to contain red, orange, yellow, green, blue, indigo and violet (figure 13.8). In 1802, William Wollaston (1766–1828) invented the spectroscope in an effort to explore the spectrum in more detail. He found the solar spectrum was not continuous but was crossed by a number of black lines. In 1814, Joseph von Fraunhofer (1787–1826) mapped the spectrum in much greater detail, finding 576 black lines. These have become known as Fraunhofer lines. John Herschel (1792–1871) and W.H. Fox Talbot (1800–1877), a pioneer in photography, found that when chemical substances were heated in a flame and observed through a spectroscope, each chemical had a distinctive set of bright bands of colour forming

FIGURE 13.8 A spectrum of colours is produced when white light is passed through a prism

TOPIC 13 Is there life beyond Earth’s solar system? 7

its spectrum. This meant that scientists could identify chemicals simply by observing their spectra. Other scientists found that when sunlight is passed through gas before entering the spectroscope it had extra black lines through its spectrum. This suggested that the black lines in the solar spectrum were due to light passing through gases in the Sun. These scientists had identified a method for determining the elements in stars. In 1859, Gustav Kirchhoff (1824–1887) with his friend Robert Bunsen (1811–1899) used Bunsen’s burner to burn elements and clearly describe the cause of these spectral lines. They found that: • a continuous colour spectrum is produced by glowing solids or dense gaseous bodies like the Sun • if a gas is between the light source and the spectroscope, light is absorbed from the continuous spectrum at wavelengths or colours characteristic of the chemical components of the gas. A spectrum of this type is called an absorption spectrum (figure 13.9) • a glowing gas produces bright lines on a dark background at wavelengths or colours characteristic of the chemical components of the gas. A spectrum of this type is called an emission spectrum. TABLE 13.2 Types of spectra and the celestial bodies that produce them Type of spectrum

Generally produced by

Celestially produced by

Continuous

Hot solids, liquids, gases under pressure

Galaxies, inner layers of stars

Emission

Incandescent low-density gases

Emission nebulae, quasars

Absorption

Cool gases in front of continuous spectrum

The atmospheres of stars

FIGURE 13.9 An absorption spectrum produced by shining light with a continuous spectrum through a cool gas Absorption spectrum

Cool gas absorbs certain wavelengths and re-emits them in all directions. This light is deficient in certain wavelengths.

Incandescent bulb producing continuous spectrum

8 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

One of the first successes with this new tool of astrophysics was the spectroscopic analysis of planetary nebulae by William Huggins (1824–1910). Working in London in 1864, he found that these nebulae produced the bright line spectra of glowing gas, showing that they were clouds of gas rather than groups of very distant stars. Some of the nebulae documented by Herschel, however, showed that they were collections of stars, as they emitted continuous spectra interrupted by black lines. Huggins’s investigations also provided very convincing evidence that the stars in the sky really are distant suns.

Composition of the Sun Early in the twentieth century, no one understood what could power the Sun. Chemical reactions did not produce sufficient energy. Scientists considered that it may have been some form of nuclear reaction that could release some of the Sun’s mass in the form of energy, as Einstein described in 1905 in the equation E = mc2 . However, scientists thought the Sun was made largely of iron, the most stable nucleus of all. This conclusion was based on measurements of tidal effects on the Earth that showed the Earth was mostly made of iron and on analysis of meteorites from space that showed that they too were composed largely of iron. Spectroscopic results seemed to confirm the iron content of the Sun. However, reading spectrographs of the Sun is not easy as many spectra of different elements overlap. In 1925, a young astrophysicist named Cecilia Payne (1900–79) wrote one of the most highly regarded PhD theses in astronomy on the spectra of stars. She interpreted the spectra as showing that the Sun was mostly hydrogen, not iron at all. The scientists had all been reading the Fraunhofer lines incorrectly, which is easy to do when the lines of so many elements overlap. It was not long before the scientific community believed that hydrogen was the main element found in the Sun. Hydrogen not only fitted the spectra, it explained the Sun’s energy, unlike iron. FIGURE 13.10 The absorption spectrum for hydrogen

Absorption lines

SAMPLE PROBLEM 2

Describe how astronomers can determine the elements that are present in the Sun. THINK 1.

Each element has its own unique absorption spectrum.

WRITE

Astronomers observe the Sun’s light through a spectroscope. The solar spectrum contains absorption lines that are characteristic of the elements found in the Sun’s atmosphere.

PRACTICE PROBLEM 2 The Sun is a hot ball of gas. Explain why it does not produce an emission spectrum.

TOPIC 13 Is there life beyond Earth’s solar system? 9

13.2.5 Spectral type When the spectra of stars were first observed in detail in the nineteenth century, it seemed that the spectrum of every star was different. Gradually, some sense was made of the multitude of lines that crossed the spectra and stars were classified into spectral types. The system developed by Annie Jump Cannon (1863–1941) has been used since 1910. It classes stars as O, B, A, F, G, K or M according to the relative intensity of various absorption lines in their spectra. For example, for type A stars the lines of the hydrogen spectrum are very clear. The spectral classes are arranged in order of temperature from O, the hottest with a spectrum peaking in the ultraviolet, to M, the coolest with a spectrum peaking in the infra-red. The Sun is a type G star and these are yellow. A full description of the spectral classes is given in table 13.3. TABLE 13.3 Spectral classifications and their corresponding features. Spectral class

Surface temperature (K)

Colour

Spectral features

O

28 000–50 000

Blue

Ionised helium lines Strong UV component

B

10 000–28 000

Blue

Neutral helium lines

A

7500–10 000

F

Blue-white

Strong hydrogen lines Ionised metal lines

6000–7500

White

Strong metal lines Weak hydrogen lines

G

5000–6000

Yellow

Ionised calcium lines Metal lines present

K

3500–5000

Orange

Neutral metals dominate Strong molecular lines

M

2500–3500

Red

Molecular lines dominate Strong neutral metals

Note: In astronomy the term ‘metal’ refers to any element other than hydrogen or helium.

Using the properties of light, astronomers are able to investigate the temperature and chemical composition of stars. There is one more important tool that is used to investigate the universe. This relates to the way we model light as a wave.

Resources Digital document Investigation 13.1 The colour of stars (doc-32317) Teacher-led video Investigation 13.1 The colour of stars (tlvd-0830)

13.2 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Define the following terms: refraction, reflection, dispersion, spectrum, spectroscopy. 2. (a) White light enters a crown glass rectangular prism. Sketch the path of red and deep blue light through the glass and back into air. How does the direction of the emerging coloured rays compare with that of the incoming white ray? (b) Suggest why a glass triangle is used to observe the visible spectrum, rather than a glass rectangle. 3. Explain the difference between absorption and emission spectra. 4. Explain how incandescent light is produced.

10 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

5. 6. 7. 8.

What type of spectrum does an incandescent source have? Why? Describe how the shape of the emission curve of a black body changes as the object gets hotter. How can we use our understanding of blackbody emission curves to determine the temperature of stars? Explain how the spectra from a star can enable us to determine the chemicals present in the star.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

13.3 Locating extrasolar planets (exoplanets) KEY CONCEPTS • Describe how planets can be identified by using the common centre of mass and the gravitational effect of a planetary system on a star. • Compare methods of exoplanet detection including astrometric, radial velocity, transit method and microlensing, referring to databases that differentiate for size, eccentricity and radius. ∆𝜆 v = . • Explain and apply Doppler shift including spectral shift and ‘wobble’ of planetary systems using: 𝜆0 c • Investigate how the composition of an exoplanet can be determined using spectral analysis.

13.3.1 The search for exoplanets So how can we use this understanding of light to detect life in our universe? Our quest starts with looking for other planets: extrasolar planets or exoplanets. The detection of extrasolar planets or exoplanets has long been heralded as a way to justify the likelihood that life is not unique to Earth. Since 1995, ground- and satellite-based surveys have detected many planets orbiting other suns. To date, over 3900 planets have been confirmed with more than 3400 candidates pending — none an Earth analogue. The planets that have been detected range considerably in orbits and size. Finding exoplanets is very difficult. They are near impossible to observe directly due to the fact that they are tens of thousands of times fainter than the stars they orbit. The three restricting factors to direct observation are: 1. Planets do not make their own light; they reflect their star’s light. 2. The star they orbit is so bright the planet cannot be seen through the star’s glare. 3. They are very, very far away. Very early on astronomers realised that directly observing exoplanets was unlikely to give results. Other methods of detecting needed to be considered using our understanding of how planetary systems behave. Astronomers used the fact that the gravity of planets orbiting their star would cause the star to ‘wobble’.

13.3.2 The astrometric method In our solar system, the planets orbit the Sun at different orbits and different rates. The reason they maintain their orbits is due to the gravitational force between the Sun and the planets. This force acts on both objects and means that a planet’s motion around a star pulls the star slightly from the centre of the system; as the planets rotate around the star, the star is pulled ever so slightly towards the planets. This creates a very regular, simple wobble in the star. Figure 13.11 illustrates this in a single planetary system. Our solar system, with its eight planets and many other orbiting bodies, has a much more TOPIC 13 Is there life beyond Earth’s solar system? 11

complicated wobble. However, the shape and size of the wobble is a direct result of the many and varied orbits of bodies in our planetary system. Knowing this, we can use mathematical processes to breakdown a complicated wobble to infer the size and orbits of the bodies around a star. FIGURE 13.11 The orbit of a planet about a star causes the star to wobble because both objects assert a gravitational force on each other.

Knowing that a star wobbles is only half the process; the biggest hurdle is detecting the wobble. Observing this wobble directly is very difficult. This process, known as astrometric measurement, attempts to identify the wobble of a star by measuring tiny variations in the star’s position over time. This process relies on very accurate instruments and to date no exoplanet has been identified using this method. This is not to say that planets cannot be detected this way. In the future, with very large telescopes like the Keck telescopes in Hawaii or proposed space projects such as NASA’s Space Interferometry Mission (SIM), exoplanets may be identified with this method.

13.3.3 The Doppler effect On the surface of the Earth, when fast moving objects go by, the sound they make drops in pitch. You can hear this when trains, fire engines or racing cars speed past you. This is known as the Doppler effect, after Christian Doppler, who predicted its existence in 1842, before it had been observed. As a result of the Doppler effect, the pitch of the sounds that we hear depends on the motion of the sound source and the observer. For example, a siren that has a frequency of 500 Hz will have a frequency higher than 500 Hz when it is approaching you and a pitch lower than 500 Hz when it is moving away from you. Sound travels as a wave made up of a series of compressions and rarefactions. The distance between one compression and the next is the wavelength of the sound. In still air, a sound wave travels at a constant speed that is independent of the speed of the source. However, if the source is moving towards the listener, the distance between one compression and the next will be shorter than if the source is at rest, making the wavelength of the sound shorter. The number of compressions that the listener hears per second will be greater than if the source was at rest and this higher frequency will be heard as a higher pitch. Similarly, if the source is moving away from the listener, the motion of the source increases the distance between compressions. This results in a lower frequency and a lower pitch being heard. Doppler realised that as light behaved like a wave, it should also experience a Doppler effect. While with sound the frequency of the wave determines the pitch, with light the frequency determines the colour. When a light source is travelling away from an observer, the wavelength lengthens and the light appears redder in colour. This phenomenon is an example of the Doppler effect and is called red shift. Similarly, light that is travelling towards an observer appears bluer in colour; this is called blue shift.

Resources Digital document Investigation 13.2 The Doppler effect (doc-31887)

12 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 13.12 The Doppler effect: as the source moves to the right, the wavelengths in front of it are smaller than those behind it.

S

S

= Source

13.3.4 Measuring radial velocity As we have seen, when an object emitting waves is in motion, the Doppler effect causes a change in the detected frequency and wavelength. When a star wobbles due to a planetary system, we can observe the wobble as a red shift (as the star moves away from us) or a blue shift (as the star moves towards us) of the star’s light. Bigger shifts indicate larger movements of the star. Variations in the wavelength of a star’s light over a period have enabled the identification of planetary systems. This form of exoplanet detection is known as the radial velocity (RV) method, a name which relates to the fact that to observe the change in wavelength the observer must be parallel to the motion, or the line of sight is along the radius connecting Earth to the star (figure 13.13). The size and complexity of the Doppler shift can indicate the size of single or multiple planet orbits about a star. FIGURE 13.13 The RV method uses Doppler shift to detect the wobble of a star caused by a planetary system. The size and variation of the shifts over time indicates the size and number of planets in a system.

Host star

Exoplanet

Source: ESO TOPIC 13 Is there life beyond Earth’s solar system? 13

Michel Mayor and Didier Queloz of the Geneva Observatory in Switzerland made the first exoplanet discovery in 1995 using the radial velocity method. They found a planet orbiting the star 51 Pegasi, 47.9 light-years from us in the constellation Pegasus (figure 13.14). Since then, thousands of stars have been identified as having planetary systems. FIGURE 13.14 Data from multiple RV readings of 51 Pegasi illustrated the telltale sine graph indicating a close orbiting body. With a period of 4.231 days and a mass of just less than half that of Jupiter, it was the first verified exoplanet to be found.

Source: © Jason Wright / Exoplanet

The speed at which a star wobbles can be determined by the shift in wavelength from a known (stationary) spectrum. The wave equation applied to light can be used to determine how fast an object is moving if we know the change in wavelength: Δ𝜆 v = c 𝜆0 Where: ∆𝜆 = change in wavelength 𝜆0 = stationary wavelength v = velocity of moving object c = speed of light. 14 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Astronomers use the lines of the hydrogen spectrum to compare spectra from different stars (figure 13.15). This is because hydrogen is the most common element in the universe and is the main constituent of stars. The hydrogen spectrum measured in a laboratory on Earth is compared to that observed from a star. The shift is measured and then the above relationship is used to determine the speed of the moving object. The negative indicates the shift was to a shorter wavelength, that is, a blue shift, suggesting the star is moving towards Earth. FIGURE 13.15 The visible spectrum of hydrogen. The positions of the spectral lines are known to high precision, which enables astronomers to compare them to starlight to identify any shift in the spectrum. 410 nm 434 nm

486 nm

656 nm

SAMPLE PROBLEM 3

A star is observed to have a wavelength of H𝛼 = 656.280 07 nm, the known H𝛼 for a stationary source is 656.280 10 nm. Determine the radial velocity at which the star is moving relative to Earth. THINK 1.

Calculate the change in wavelength.

2.

Rearrange to make v the subject and substitute in the known values.

3.

State the solution.

WRITE

Δ𝜆 = 𝜆 − 𝜆0 = 656.280 07 − 656.280 10 nm = −0.000 03 × 10−9 m Δ𝜆 v = 𝜆v c Δ𝜆c v= 𝜆0 −0.000 03 × 10−9 × 3.0 × 108 = 656.2 801 × 10−9 = −13.7 m s−1 The radial velocity at which the star is moving relative to Earth is –13.7 m s–1 .

PRACTICE PROBLEM 3 A star has a known wavelength of 484.56 028 nm. When it was observed it was found to have a wavelength of 484.56 023 nm. Determine the radial velocity at which the star is moving relative to Earth. The beauty of the radial velocity method is that the shift in the spectrum of a star due to its wobble is the same no matter how far away that star is from us. However, the changes in wavelength are very small because the movements due to planetary systems are small. The Sun moves at about 12.5 m s–1 around the solar system’s centre. If we observed this change in speed on a spectrum, it would give a shift in wavelength of about 2.6 × 10–5 nm. This change accounts for our entire solar system, not an Earth-sized planet on its own. If the Earth were the only planet in our system, the Sun would only move by about 9 cm s–1 . This TOPIC 13 Is there life beyond Earth’s solar system? 15

means that to detect an Earth-like planet, we need to be able to make detections within 1 cm s–1 . This is a very high level of precision. The best ground-based telescopes are currently able to achieve a precision of about 1 m s–1 . The RV method has been the most successful Earth-based technique to date but it does have several drawbacks. To identify the variation in the speed of a star, the system is best observed ‘edge on’. Of course, not all systems are in this position relative to Earth; however, a radial velocity component will be detected unless a system is ‘flat on’ or perfectly perpendicular to the Earth. This problem means that the RV method can’t identify an accurate mass of the planets under observation. The RV method can only give a minimum mass value. This leads to errors where possible planets could in fact be a binary star system. In general, planets discovered using the RV method tend to be ‘hot Jupiters’ — named because they are large gas planets that orbit very close to their star — because these types of planets have a greater gravitational effect on their star, and thus cause a larger wobble. Precision of Earth-based telescopes is another limit of the RV method. The quest for radial velocity precision significantly less than 1 m s–1 is being made with the development of the Next Generation Adaptive Optics (NGAO) system at the WM Keck Observatory and the Echelle SPectrograph for Rocky Exoplanets and Stable Spectroscopic Observations (ESPRESSO) at the Very Large Telescope (VLT) in Cerro Paranal, Chile, run through the European Space Agency. It is the hope that both projects will achieve precisions to a few cm s–1 .

13.3.5 The transit method Transit photometry, commonly known as the transit method, is the most successful technique for discovering exoplanets (figure 13.16). This method involves detecting the small variation in brightness when an exoplanet passes in front of its star. This technique is usually used in conjunction with the RV method to verify any discovery. This is because the dimming effect observed could also be caused by other factors. The most significant contributor to the transit method of detecting exoplanets is the Kepler space telescope, launched in 2009, which has found over 5000 planet candidates. Most detections from Kepler are termed ‘candidates’ until they are verified by ground-based observations. The Kepler data is double-checked through a Follow-up Observation Program (FOP) using the Lick, Las Cumbres and Keck observatories. In addition, an online project known as Planet Hunters enables thousands of online users to assist in identifying possible exoplanet transits from the thousands of star candidates identified by Kepler. FIGURE 13.16 A transit

As well as being a good method of detecting an exoplanet, the transit method also gives a much better picture of the exoplanet. As the planet transits the star, the degree to which the star’s visible light is diminished can give an indication of the planet’s diameter. During an exoplanet’s transit, additional absorption lines due to starlight travelling through the planet’s atmosphere can be detected to identify an atmosphere and its make-up. When the planet moves behind the star, a drop in infra-red light can be used to determine the temperature of the planet. Space-based transit method observations have revealed broader and more detailed data on exoplanet properties as technologies improve. The transit method does have drawbacks. The most significant is that for us to see a transit, the system must be very close to edge on from our point of view. This means the transits we observe represent a very small number of the total systems in our universe. A second difficulty is that the time of a transit is only a very small fraction of the total orbit time of an exoplanet. This means that the likelihood of observing the transit at all is very low. In addition, multiple transits need to be observed to verify a sighting. Finally, the transit method can produce false readings where a binary system can be mistaken for a planetary system. 16 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

In order to find transits, our telescopes need to be continually observing the sky. This is achieved by using both space-based and ground-based telescopes. In addition to Kepler, the Spitzer space telescope, launched in 2003, uses the transit method to gather detailed spectra in the infra-red and has sensitivity three orders of magnitude greater than any other infra-red observatory. Spitzer’s replacement, the James Webb Space Telescope (JWST), which is also the optical replacement for the Hubble space telescope, has suffered delays and funding concerns and is working towards a 2021 launch date.

13.3.6 Microlensing In 1913 Einstein’s General Theory of Relativity described how light would be affected by gravity. Einstein’s theory predicted that gravitational effects of large bodies such as stars would cause light to bend around them. Astronomers realised that they could use Einstein’s theory and the variations of gravity caused by a planetary system on a star to identify exoplanets. Stars in our sky move relative to each other, with closer stars appearing to move faster than more distant background stars. When the light from a distant star travels past a closer star, some of its light is bent around the closer star, causing the distant light to appear brighter than it actually is. This phenomenon is called microlensing as the foreground star (known as the lens star) bends the light in the same way a magnifying glass does (see figure 13.17). If the lens star has a planetary system, it changes the effects of the lensing in a mathematically predictable way. This method was first proposed to discover binary star systems in 1992. The first successful exoplanet discovery using microlensing was made in 2002. FIGURE 13.17 As the lens star and planet move in between Earth and a source star, the light curve created can be used to identify a planetary system.

Source Observer

star Planet Lens star

Planet Lens Source Observer

star star

Microlensing has several advantages over the other methods for exoplanet detection. It can identify planets of much smaller masses and larger orbital radii than the ‘hot Jupiters’ discovered through RV and transit methods. This means it can detect planets that are rocky and much more like Earth. A second advantage is that microlensing can detect systems at much larger distances than RV and transit methods. The biggest drawback of microlensing is that it measures unique events. Unlike RV and transit events that can be measured and verified by repeat observations, microlensing events occur once, when background and foreground stars line up in the perfect position to create the lensing effect. Other methods of detecting exoplanets, such as direct imaging and timing (with pulsars, eclipsing binaries or stellar oscillations), have achieved a small number of planetary discoveries. TOPIC 13 Is there life beyond Earth’s solar system? 17

13.3.7 Trends in the data Looking at the planets discovered so far we might think that Earth and our solar system are very rare and different to what the rest of the universe holds. However, we need to be careful about how we interpret our discoveries. Each method of exoplanet discovery tends to favour planets with a particular type of period and mass. This indicates the limits in our detection methods rather than limits in the types of planets out there. Exoplanet masses are given as a ratio of Jupiter’s mass because so many of the exoplanets discovered thus far are very large. The period indicates not only how quickly the planet moves around its star, but also its orbital radius. Smaller periods indicate planets that are closer to their star. The distribution of masses, periods, radii and eccentricity of discovered exoplanets have not been what astronomers expected. The planets and systems do not look like our solar system, nor do they fit simply into the accretion model, the model used to explain the formation of our solar system. Many exoplanets discovered have been described as ‘hot Jupiters’. Explaining how and why these planets ended up so close to their star became a very important task for astronomers, as it was so unexpected. It is now understood that the systems we are observing are much older that our solar system. Astronomers believe that over time the large gas planets have migrated (drifted very slowly) towards their star. Smaller, rocky planets like our Earth, may have already fallen into their sun or been knocked out of orbit through the gravitational effects of such migration. This migration model also supports another oddity of so many exoplanets discovered: their highly eccentric orbits. FIGURE 13.18 Exoplanet mass (in Jupiter masses) versus period in years by detection method. The clusters illustrate how the various detection methods seem to bias particular types of exoplanets. This scatterplot was generated using data from www.exoplanet.eu. The data from microlensing appears limited here as many of the discoveries do not have periods and as such do not appear on this plot. 100 28.5 10

1 0.0001

0.001

0.01

0.1

Orbital period (years)

1.5 1

10

100

1000

10000

Imaging Transit

0.1

Radial velocity Microlensing Pulsar

0.01

TTV

0.0001

0.00001

Mass (Jupiter masses)

Astrometry 0.001

Our solar system

13.3.8 Finding the right star In the search for life-sustaining exoplanets, the stars themselves are the first target of interest. To find an Earth analogue, we would first seek stars like our Sun; that is, a main sequence star of spectral type F, G or K. A main sequence star is a star that can be described as in mid-life. These stars have featured strongly in exoplanet searches because they are numerous and their properties suit the RV method. They have had long lives, which would enable enough time for a biosphere to form and they have not experienced the cataclysmic events of a post–main sequence star that would sterilise any life. Type M stars have not featured widely in past exoplanet searches, mainly due to their lower luminosities and more complicated 18 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

spectral emissions. However, type M stars are numerous and space-based observatories and advancement in instrumentation means they are now considered excellent candidates for RV or transit methods.

13.3 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Why is finding exoplanets so difficult? 2. How does our understanding of gravity help us find exoplanets? 3. Using the last line on the spectrum in the following image calculate the radial velocity of objects A and B. Standard spectrum

400

500

600

700

(nm)

500

600

700

(nm)

500

600

700

(nm)

Object A

400

Object B

400

4. Would you expect the velocities calculated in question 4 to be due to the movement about a star or something else? Justify your answer. Use figure 13.18 to answers questions 5 to 8. 5. Why is the mass of the exoplanets measured in Jupiter masses? 6. What is the relationship between the period of an exoplanet’s orbit and its radius? 7. Why would we conclude that certain methods of detection favour particular types of planets? 8. Does the data in figure 13.18 support the idea that our solar system is unique and we will never find another one like it? Justify your answer.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

13.4 Conditions for life beyond Earth’s solar system KEY CONCEPTS • Explain the presence of liquid water as determining the habitable zones of a star system and the most likely place for life. • Explain the origins of life in the universe as having come from organic molecules in space, or as originating on Earth or an Earth-like planet through reactions of elements and compounds.

TOPIC 13 Is there life beyond Earth’s solar system? 19

13.4.1 Life based on carbon The molecules that make up life on Earth are carbon based, thus, these compounds are called organic molecules. The extraordinary thing about carbon is that it is able to form long chains and intricate structures. This property enables a huge array of molecules, some of which perform complex, selfregulating chemical reactions. The constituents of organic molecules — carbon, hydrogen, nitrogen and oxygen, with trace elements of sulfur and phosphorus — are readily abundant in the universe. With this abundance and the versatility of carbon to bond into complicated structures, many scientists believe life’s origin lies in organic chemistry. Organic molecules are not unique to the Earth or our solar system. Spectroscopy of planetary nebula, interstellar dust and circumstellar envelopes (gas and dust around stars) shows organic molecules. Hundreds of organic compounds have been discovered FIGURE 13.19 Carbonaceous chondrite throughout our galaxy, and their presence within stellar nurseries suggests that at least some of the organic compounds found on Earth may have already existed when the solar system was created. Spectroscopy is not the only evidence of organic molecules beyond Earth. Meteorites known as carbonaceous chondrite, believed to be leftover from the formation of our solar system, have also been found to contain organic molecules. FIGURE 13.20 Many organic molecules can be found in space. This image shows infra-red spectroscopic data of the Orion Nebula from the European Space Agency’s Herschel Space Observatory.

Sulphur dioxide

Hydrogen sulphide Methanol

Carbon monoxide

Formyl cation

Water Dimethyl ether

Sulphur dioxide

Methanol

Formaldehyde

Deuterium cyanide

Water Methanol

Hydrogen cyanide Sulphur dioxide

HIFI spectrum of water and organics in the orion nebula

20 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Acrylonitrile Methyl formate

© ESA, HEXOS and the HIFI consortium E.Bergin

13.4.2 The importance of water The other molecule important to life that is found throughout the universe is water. Most water in space is not liquid, but ice. Ice is a main constituent of interstellar dust, but when that dust condenses in stellar nebular, temperatures can rise to a level that enables pockets of liquid water within the dust grains. Liquid water, even in miniscule amounts, enables organic chemicals to interact to form more complicated molecules. On Earth, liquid water is integral to life. It is the major constituent of living tissue, accounting for around 70% of tissue mass. Water’s polar nature means that it is an excellent solvent of many minerals and compounds, which in turn enables chemical processes to occur. Water is required for metabolic processes within the cell. Most scientists agree that without liquid water, life on Earth would not have evolved. So the question becomes, how could organic molecules form life? To date, no experiment in a laboratory has created life from organic molecules. However, many experiments have been able to produce complex organic molecules. By replicating the conditions of primordial Earth, scientists have been able to create molecules such as nucleic acids, which are the building blocks of molecules such as RNA and DNA. In addition, experiments on organic molecules found in meteorites have demonstrated their ability to form more complex molecules when exposed to liquid water (figure 13.21). Water is crucial to life on Earth, so when we are searching for exoplanets, the existence of liquid water is a key search parameter. FIGURE 13.21 Organic molecules, when extracted from carbonaceous chondrite, such as the Murchison, meteorite and then exposed to water form self-assembled vesicular structures that could be a precursor to living cells. (Magnification 400x)

Source: (a) David Deamer

13.4.3 The habitable zone Liquid water will only exist on the surface of a planet if the right conditions exist. The main requirement is a temperature at which water remains a liquid. Astronomers can calculate the region in which a planet needs to orbit a star to ensure liquid water is possible. This region is known as the habitable zone (HZ), or sometimes the ‘Goldilocks zone’, and it varies depending on the type of star. The habitable zone is a significant factor for any life-sustaining exoplanet. This zone describes the distance from the sun that an exoplanet needs to be if it is to have liquid water on its surface. The size of this zone is approximated using the size and temperature of the star, which is limiting because albedo (how reflective the planet is), greenhouse effect, eccentricity, obliquity, rotation rate and geologic age all contribute to the surface temperature of a planet. To determine HZ, astronomers make assumptions about these other factors. TOPIC 13 Is there life beyond Earth’s solar system? 21

The HZ is a clear focus for searching for life-sustaining exoplanets. One of the problems in our quest to find a habitable planet is that so many of the exoplanets found so far have highly eccentric orbits. This means their orbits are far from circular. An eccentric orbit often results in the planet moving in and out of the predicted habitable zone. The likelihood of an exoplanet sustaining life is determined by its ability to maintain liquid water on its surface. However, moving in and out of the HZ may not prevent life. If an atmosphere is thick enough, and depending on the period of orbit, a planet may be able to maintain liquid water on its surface while outside the HZ. Astronomers analyse the amount of time eccentric orbiting exoplanets spend within and outside the HZ in attempt to identify better targets for further examination. These studies use models to estimate the surface temperatures and atmospheric conditions of exoplanets. In the search for habitable exoplanets, these studies are imperative to reduce the number of candidates that are examined directly.

13.4.4 Planet size The last property to consider when examining a viable candidate for life on another planet is mass. The very existence of large planets orbiting close to their suns makes the prospect of finding smaller terrestrial planets in the same system unlikely. As the vast majority of exoplanets discovered are more similar in mass to Jupiter than Earth, we may need to consider the prospects of life outside terrestrial planets. If there is a possibility of life forming on these giant planets, and certainly many are within their habitable zones, it may not resemble life as we know it on Earth. A planet with a minimum mass of more than Jupiter would have gravitational effects and pressures like nothing comparable on Earth. This does not rule the possibility of life out; life on Earth is found in very extreme environments. Microbiological life on Earth is found in many different environments. Extremophiles, which are organisms found in extreme environments, can survive and sometimes thrive in environments of extreme temperature by Earth standards (–75 °C to 113 °C), radiation, salinity, pH, pressure (up to 130 MPa), vacuum, variations in oxygen and other chemical extremes. It is very difficult to predict the environments that exoplanets could provide for life, but with the variability of life on Earth, it is not unimaginable that life could evolve on planets such as hot Jupiters. The prospect of finding Earth analogues is unlikely if we focus on the data collected so far. However, missions such as Kepler are increasing the number of candidate exoplanets exponentially, giving a much broader picture of the types of systems that exist and moving towards discovering smaller terrestrial planets. The likelihood of life existing in some manner somewhere in the universe and us finding it seems high.

13.4 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. What are the chemical constituents of life? 2. How do we know that these constituents are relatively abundant in the universe? 3. How do meteorites help us understand the origins of life? 4. Why is liquid water so important in the search for life? 5. What factors influence the habitable zone of a star? 6. What is an eccentric orbit, and why is this an issue for life on an exoplanet? 7. Other than the habitable zone, what factors affect the prospect for life on another planet?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

22 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

13.5 The search for extraterrestrial intelligence KEY CONCEPTS • Explain the use of the Fermi paradox to question the possibility of life outside Earth’s solar system and identify its counter arguments. ∗ • Apply the Drake equation: N = R fp ne f1 fi fc L, as a way of predicting the likelihood of life existing in the universe by making reasonable assumptions based on evidence and speculation. • Distinguish between targeted and untargeted searches for extraterrestrial intelligence (ETI), and describe optimising strategies including where to look and how to ‘listen’ with reference to choice of frequency and bandwidth. • Explain why the radio spectrum is the best section of the electromagnetic spectrum to search the sky for possible ETI signals, including the cosmic radio window and the use of radio astronomy in the search. • Explain the nature of information that humans transmit beyond Earth to signal that intelligent life exists on Earth.

13.5.1 The Fermi paradox The most fantasised and romanticised search for life in our universe is the search for intelligent life. Science fiction writers have for many years deliberated on the possibility that we are not the only intelligent form of life. By just considering the sheer number of stars and galaxies out there, many conclude we cannot be alone. Scientists and philosophers have also considered the question of life and intelligent life in our universe. Many would like to believe that we are not the only civilised beings, that there has to be many technologically advanced beings out there, some of which, must be more evolved than us and be able to travel between the stars. When this idea was put to physicist Enrico Fermi in a 1950 interview, he asked that if intelligent life is common in the Milky Way, and given the millions of years another life form may have had to evolve their technology, then why are we not inundated with space ships and visiting aliens? This is called the Fermi paradox, and in essence it demonstrates the contradiction between the probability of intelligent life in the universe and our lack of contact with that life. Of course, it is only a paradox if you accept its two premises: 1. Life in our universe is not rare. 2. Intelligent life and technological advancement are an expected outcome of billions of years of evolution. Neither of these premises seem implausible given this is what happened on Earth. So was Fermi right? There are several arguments that attempt to explain the Fermi paradox; some considerably more radical than others: • Life on Earth is the first to evolve in the universe. • Intelligent life on Earth is the first to evolve in the universe. • Other intelligent civilisations may have risen and fallen in the billions of years the universe has existed. • Other intelligent civilisations may have risen, but the technology to travel or communicate between the stars is impossible. • Other intelligent civilisations may have risen and have the technology to travel or communicate between the stars, but we don’t understand them. • Other intelligent civilisations may have risen and have the technology to travel or communicate between the stars, but choose not to. • The aliens have made contact, but no one believes me.

TOPIC 13 Is there life beyond Earth’s solar system? 23

13.5.2 The Drake equation Several of these ideas were formulated into an equation by radio astronomer Frank Drake. In 1961, Drake developed an equation that tried to estimate the number of civilisations that exist in our galaxy with the means to communicate with us. Drake stipulated that the number of such civilisations, N, could be determined as: N = R∗ f p ne f l f i f c L Where: R* = the rate at which suitable stars are formed in our galaxy fp = the fraction of those with planets ne = the number of planets in each planetary system capable of bearing life fl = the fraction of those life-bearing planets on which life actually evolves fi = the fraction of planets whose life forms evolve intelligence fc = the fraction of those intelligent life forms who choose to communicate L = the length of time for which communicating life forms exist on a planet. Clearly this equation could not be solved with any kind of accuracy, but Drake’s equation clarified the many factors involved in searching for intelligent life in our universe, which in turn has significantly contributed to the design of targeted search projects.

SAMPLE PROBLEM 4

Given that the Milky Way galaxy is estimated to have 300 billions stars and the oldest known star in the galaxy is 13.6 billion years old, give an estimate for the rate at which stars might form in our galaxy. THINK 1.

State the formula for the rate of star formation.

2.

State the solution.

WRITE

number of stars 300 billion = age of oldest known star 13.6 billion years = 22 stars per year The rate at which stars might form in our galaxy is 22 stars per year.

PRACTICE PROBLEM 4 The Andromeda galaxy is estimated to have 1 trillion stars, with the oldest star in the galaxy approximately 8 billion years old. Estimate the rate at which stars might form in this galaxy.

13.5.3 The search for extraterrestrial intelligence (SETI) The search for extraterrestrial intelligence (SETI) outside our solar system has been conducted in earnest since the 1960s. Frank Drake was one of the pioneers of applying our scientific understanding in a systematic and targeted search to find evidence of extraterrestrial intelligence. He used radio astronomy as his tool, and this remains the main tool used by SETI today.

24 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

13.5.4 Radio telescopes Optical telescopes extend our vision by capturing more light and magnifying images. But many astronomical objects emit electromagnetic radiation outside the visible range. It has been said that to view the stars only in the visible part of the spectrum is like listening to a symphony with an ear that can hear only middle C and the two notes either side of it. The first telescopes to observe frequencies outside the visible range were radio telescopes. In 1931 Karl Jansky was trying to remove the noise that was interfering with early radio transmissions when he realised that some of the noise was coming from space. The first radio telescope was made in 1937 when Grote Reber, a radio enthusiast, built one in his back yard. It was a parabolic dish nearly 10 metres in diameter that could be directed at different parts of the sky. Reber spent five years using his telescope to locate radio sources in the sky before publishing his work. He found that the Sun emitted radio waves, as did many stars. He also found that some sources of radio waves could not be seen in optical telescopes, making radio astronomy a particularly exciting prospect. You cannot look through a radio telescope and see a picture in the way that you can through an optical telescope. The information comes in the form of numbers, which show the intensity of the radio sources at different wavelengths. Modern radio telescopes use computers to turn the numbers into pictures that are easier for people to analyse and appreciate. In 1957, using materials from World War II FIGURE 13.22 The Parkes radio telescope battleships, the giant 76 m radio telescope at Jodrell Bank in England was completed. It was ready just in time to track the first spacecraft, the Russian Sputnik 1. The science of radio astronomy took off and in 1961 the Parkes radio telescope (featured in the film The Dish) was built in New South Wales (figure 13.22). In 1963 in Puerto Rico, the Americans built the enormous 305-metre diameter Arecibo radio telescope. It is fixed to the ground and relies on the rotation of the Earth to point in different directions. One of the main purposes of the Arecibo dish was to explore the chemistry and dynamics of the atmosphere. It has also discovered the first planet around a star other than the Sun, determined the rate of rotation of Mercury, mapped the distribution of galaxies in the universe, found ice at the poles of Mercury and made many other major astronomical discoveries in the past five decades. Radio telescopes can be made to send out radio signals as well as receive them. In 1974, astronomers searching for intelligent life in the universe used the Arecibo dish to send radio signals into space hoping to hear a signal in return, like the one featured in the 1997 film Contact. This was very controversial, largely because people were concerned about a small group of scientists speaking to unknown beings on behalf of the whole planet. Whether we deliberately transmit radio signals or not may make no difference. If there are technologically advanced beings on other planets in our part of the galaxy, they will be able to detect the many transmissions we send to each other that are continually leaking away from the planet at the speed of light. Radio telescopes can also be used as radar telescopes. This involves sending out radio waves to nearby objects such as the Moon, Mars, Mercury and Venus. The radio signal is reflected back to the radio telescope, providing a very accurate measurement of the distance of these objects. This has enabled a very accurate measurement of the solar system. The surfaces of the planets have also been mapped in this way. Radar is particularly useful for mapping the surface of Venus because of the thick clouds that cover that planet. The radio waves pass through the clouds while light simply reflects off. Much of the sky is invisible to optical telescopes because of clouds of gas and dust, or nebulae, in the universe. Some radio frequencies pass through these clouds, enabling radio telescopes to detect objects that will never be visible with an optical telescope.

TOPIC 13 Is there life beyond Earth’s solar system? 25

SAMPLE PROBLEM 5

A radio telescope is used to measure the distances to nearby planets. In one test a reflected pulse is received by the telescope 12 minutes after the initial radio transmission. How far away was the planet? THINK

Recall the speed of light. 2. As it is a reflected pulse received, it has travelled to the planet and back.

1.

3.

4.

3.0 × 108 m s−1 The radio signal has travelled to the planet and returned in 12 minutes, so it took 6 minutes to reach the planet. Six minutes is 360 seconds. d = vt

WRITE

Recall the equation to calculate distance with a known velocity and time. Substitute in known values with the correct units. State the solution.

= 3.0 × 108 × 360

= 1.08 × 1011 m The planet is 1.08 × 1011 m from the Earth at the time of the measurement. (This is comparable to the distance between the Earth and the Sun, 1.5 × 1011 m.)

PRACTICE PROBLEM 5 Give two examples of situations in which a radio telescope reveals information that is not detectable with optical telescopes. One of the uses of radio telescopes is to map the occurrence of neutral hydrogen atoms in the sky (figure 13.23). They are able to do this because hydrogen atoms emit radio waves of 1420 MHz when the proton and electron switch from spinning in the same direction to spinning in opposite directions. Hydrogen in hot stars is largely ionised, so tuning radio telescopes to 1420 MHz enables them to detect clouds of cool hydrogen without detecting stars. As these clouds do not emit light, radio astronomy is a great way of detecting where they are, thus improving our knowledge of the galaxy. All types of astronomical bodies can be studied using radio telescopes. FIGURE 13.23 A radio view of hydrogen atoms in the universe

Source: NASA / J. Dickey UMn, F. Lockman NRAO, SkyView

26 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Radio telescopes need to be much larger than optical telescopes to achieve equivalent resolution. One method of improving their resolution is to link them to produce a telescope that has a much greater effective size. A series of radio antennas connected in this way is called an interferometer and can resolve much smaller objects than a single telescope. The Very Large Array (VLA) in New Mexico, where 27 dish antennas each 25 metres in diameter are laid out in a Y shape, is an example. These can be used together to achieve the resolution of a single 36-kilometre antenna. The Australia Telescope Compact Array in New South Wales is another example (figure 13.24). It uses six 22-metre dishes at Narrabri on a 6 kilometre long railway track. It can also link up with the 64-metre Parkes dish and another 22-metre dish near Coonabarabran to form the Australian Long Baseline Array. FIGURE 13.24 The Australia Telescope Compact Array

Source: John Masterson / CSIRO Science Image

The SKA, or Square Kilometre Array, is a huge radio telescope planned by many different teams in countries around the world, including Australia. With the collecting power of this telescope, astronomers hope to be able to analyse radiation emitted when galaxies were forming, discover planets around other stars and gain a greater understanding of the early universe. It will have 100 times the sensitivity of any existing radio telescope.

Why radio? Since it was first discovered, radio has been recognised as a medium for communication. SETI’s focus on the radio part of the electromagnetic spectrum comes from several important properties of the radio emission. Radio waves are low energy, so they require less power to be produced by an extraterrestrial

TOPIC 13 Is there life beyond Earth’s solar system? 27

civilisation trying to communicate. Radio waves experience much less galactic absorption from the interstellar medium, and most stars are quiet in the radio part of the electromagnetic spectrum, so the host star (or our Sun) would not interfere with a transmission. There is a natural radio window that allows a large range of radio frequencies through the Earth’s atmosphere, and some significant naturally occurring frequencies lie within the radio window. The radio window illustrated in figure 13.25 is very large (1–10 GHz), which is wonderful because so many different wavelengths are able to reach Earth’s surface to be analysed. The challenge of such a large window is that it becomes extremely difficult to pinpoint one specific broadcast frequency, and it is highly unlikely that an alien race, however advanced, would have the power required to broadcast a broadband signal across even one tenth of this spectrum. In 1959, physicists Guiseppe Cocconi and Phillip Morrison suggested, as many others have since then, that the frequency of 1420 MHz would be a likely candidate for a narrowband alien broadcast. FIGURE 13.25 The cosmic radio window between 1 GHz and 10 GHz is a gap in the radio spectrum that enables clear signal to be measured from Earth. 1000 k Atmospheric window

Galactic noise

100 k

H OH 10 k Quantum noise

3K Background

0 Hz

1 GHz

10 GHz

100 GHz

1 THz

Source: SETI League image (www.setileague.org), used by permission.

Also known as the 21-centimetre line, the 1420 MHz frequency represents the emission made by neutral hydrogen when its electron ‘flips’. Similar emissions are measured for the hydroxyl radicals (OH) at 1612 MHz, 1665 MHz, 1667 MHz and 1720 MHz. Combined with the neutral hydrogen frequency this describes a window (termed the water hole), which has been the focus of much SETI work. The modern SERENDIP program and SETI@home projects focus on the neutral hydrogen frequency. The first dedicated search for extraterrestrial intelligence, Project Ozma, was conducted by Drake in 1960 with a targeted search of the Sun-like stars Tau Ceti and Epsilon Eridani at the 1420 MHz (21 cm) frequency.

28 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 13.26 The magnetic energy between the two states of an electron and proton in hydrogen-1 results in the emission of a photon at the 21-centimetre wavelength when the electron ‘flips’ to the lower energy state.

Transition from the higher energy state of an electron in a hydrogen atom to the lower energy state releases a photon at 1420 MHz.

Resources Weblink Gallery of radio telescopes with links

13.5.5 Deciding where and how to search There are two types of signal we would look for as evidence of intelligent life elsewhere in the universe: 1. unintentional signal, which is the noise created by a technologically advanced civilisation 2. intentional signal, which assumes the extraterrestrial intelligent life forms are trying to reach us. There are many difficulties with searching for unintentional signals. The most significant is that because the signals are not directed at us they will be weak and difficult to identify. The weakness of any indirect signal means that only very close stars would be worth investigating, which gives a very small target group. Searching for an intentional signal is considered a more likely way of identifying extraterrestrial intelligence. To some extent, this is based on our ability to communicate. At present we have the technology to send a narrow beam and bandwidth signal into the centre of the galaxy. So we assume that a civilised alien race could do likewise. However, this assumption also has its issues. Firstly, deciding which bandwidth to search. We would assume an intentional signal would be broadcast on a wavelength that is common to the universe. It is this reasoning that led to the focus on the 21-centimetre line. Secondly, a narrow signal is very difficult to identify in our very large sky. Surveys of significant areas of sky are needed. There are two main search methods in astronomy: 1. untargeted search that sweeps large portions of the sky 2. targeted search that focuses on selected stellar candidates. Throughout the 1960s, the Soviets also developed SETI projects; however, their searchers spanned large areas of the sky, assuming several advanced civilisations would be attempting to contact us with enormously powerful transmissions. The significance of the Russian search was that in their attempts to eradicate the significant interference from omnidirectional searches, they started using a process of correlating signals from different antennae. In the 1970s, NASA joined the SETI community through the Ames Research Centre in California, starting with Project Cyclops, a proposed telescope array of 1000 dishes. The project was never funded but the analysis of the technical and scientific issues involved became a significant reference for further SETI projects and was often referred to as the ‘SETI bible’. NASA was forced to abandon its involvement in SETI due to funding cuts, and since 1989 SETI projects have been primarily privately funded.

TOPIC 13 Is there life beyond Earth’s solar system? 29

The SERENDIP (Search for Extraterrestrial Radio Emissions from Nearby Developed Intelligent Populations) project was initiated at University of California, Berkeley in 1976, originally using the 26-metre Hat Creek antenna and the 64-metre Goldstone antenna in California. The project moved to the 92-metre antenna at Greenbank in 1986. In 1992, it moved again to the 302-metre Arecibo telescope, the world’s largest single-dish radio telescope, in Puerto Rico. The initial project used a 100 channel receiver, when it moved to Greenbank the receiver was capable of 65 000 channels. The move to the Arecibo telescope has seen three upgrades, the latest is a 128 million channel digital spectrometer covering 200 MHz of bandwidth. This receiver uses the 7-beam, Arecibo L-band Feed Array (ALFA), piggybacking the telescope while it is being used for other projects. It is able to do real-time spectrometry over the 200 MHz bandwidth, searching for narrow band signals. FIGURE 13.27 Arecibo is the largest radio telescope in the world, with a diameter of 305 metres. It is built into the hills of Arecibo, Puerto Rico and is the most sensitive Earth-based radio telescope.

Source: Photo courtesy of the NAIC - Arecibo Observatory, a facility of the NSF

Piggybacking involves mounting a receiver onto a telescope permanently and then making use of the observation plan set for other projects. The data collection equipment is autonomous and silent (does not interfere with the ongoing data of the host project). This enables the collection of large amounts of highquality data at a lower cost. Piggyback systems do have several disadvantages, the main one being that there is no control over where in the sky observations are made. This makes it difficult to get full and uniform sky

30 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

coverage. In addition, not all astronomers are convinced that the piggyback equipment is silent, and it will often be switched off if they suspect it of interfering with the host project.

13.5.6 Dealing with the data The amount of data generated from sky surveys is huge. The process of sorting through signal data to find possible extraterrestrial signals is bigger than the collection of data itself. In 1997, a team at UC Berkeley proposed using the personal computing power of 100 000 home users to perform the massive computational task of analysing a continuous recording of a 2.5 MHz bandwidth signal centred on the 21 centimetre-line. The signal was tapped off the SERENDIP receiver and SETI survey operating at Arecibo. Since its initial launch in 1999, SETI@home has had over 5.2 million users, and at the time of writing had 1.525 million current users. The latest version of the BOINC (Berkeley Open Infrastructure for Network Computing) software was developed for SETI@home by UC Berkeley and is now used in over 100 projects ranging from searching for a cure for Alzheimer’s to predicting global temperatures. In May 2011, UC Berkeley announced they would be using the SETI@home community to analyse data collected from a targeted search of 86 Earth-like exoplanets discovered by the Kepler space observatory. Using the Robert C Byrd Green Bank radio telescope, data was collected on 86 planets identified by Kepler to be within the habitable zone of their stars. This was the first time that SETI@home users were part of a targeted SETI survey. SETI@home has been hugely successful as a large scale multi-user platform and has led the way for many other projects requiring large-scale processing. It has not yet been successful in detecting extraterrestrial intelligence, but has made significant contributions to radio astronomy through the wealth of technological developments and scientific understanding that have been pioneered for the project. SETI has developed a long history of scientific endeavour and human ingenuity.

13.5.7 Shouting out If we expect extraterrestrial intelligent life to send messages to us, we need to send messages out as well. In 1974, Frank Drake and Carl Sagan sent a signal from the Arecibo telescope. Its target was the M13 globular cluster. It was sent in binary code and contained information on human DNA, the Earth’s population at the time and the atomic numbers of carbon, nitrogen, oxygen and phosphorus. Several scientists at the time questioned the wisdom of the act, and Drake and Sagan did not ask permission of any governing body before sending the signal. Most of the scientific community were not concerned as they understood the message would take 25 000 years to reach the cluster. The ‘Cosmic Call’ messages were a set of messages created by Canadian scientists Yvan Dutil and Stephane Dumas, sent from the RT-70 radio telescope in Ukraine in 1999 and 2003. They repeated the message sent previously from Arecibo, but also included text, audio, video and other image files submitted by everyday people from around the world. The messages were sent to five different star clusters and will start to arrive from 2036 through to 2059. Throughout history humans have sought new knowledge and understanding of our purpose. The search for life outside our solar system continues that quest. Although imagined many times by science fiction writers, nothing will prepare us for the moment when a signal of life is confirmed.

TOPIC 13 Is there life beyond Earth’s solar system? 31

FIGURE 13.28 The Pioneer spacecrafts included messages to any potential extraterrestrial intelligent life who happened to find them. The prospect of them being found is ridiculously small, and yet it was considered important to include them.

13.5 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Explain the Fermi paradox. 2. If the Drake equation is impossible to calculate accurately, why is it valuable? 3. Why is the radio part of the spectrum thought to be the best frequency to use in SETI? 4. What part of the spectrum is defined as radio? 5. Show why the 1420 MHz frequency of neutral hydrogen is called the 21-centimetre line. 6. What are the wavelengths for the hydroxyl radicals at 1665 MHz and 1720 MHz? 7. Describe two types of search methods employed in SETI. 8. Why is processing the data from sky surveys difficult and how is this problem solved? 9. Do you think it is a good idea to be ‘shouting out’ to the universe by sending messages into space?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

32 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

13.6 Review 13.6.1 Summary • • • • • • • • • • • • • • • • •

Our understanding of the composition of the universe is based on our ability to interpret the light from stars. The electromagnetic spectrum describes the range of energies light can have. White light can be separated into its constituent colours through the process of dispersion. Spectroscopy is the process of analysing the dispersion of light. Spectra can be continuous or discrete, absorption or emission. Spectra are used to determine the chemical composition of stars by observing the distinct lines that signify particular elements. Electromagnetic radiation is produced when electrons accelerate. Incandescent sources of light make their light through heat and have a continuous spectrum. A blackbody curve can be used to determine the temperature of stars, which are then classified into spectral types. The Doppler effect enables astronomers to determine how fast stellar objects are moving in relation to Earth. The search for exoplanets uses several methods including radial velocity, transit, direct imaging and microlensing. The exoplanets found so far challenge our understanding of planet formation; many are very large gas giants orbiting very close to their suns and with eccentric orbits. The way in which exoplanets are discovered can be biased towards particular types of planets. The search for life on other planets is based on a search for liquid water, which can be found in the habitable zone around a star. The Fermi paradox and the Drake equation both express the probabilities associated with the search for intelligent life in the universe. The search for extraterrestrial intelligence (SETI) is based in radio astronomy due to the signature of the 21-centimetre neutral hydrogen line and the ground-based radio window. Multiple projects using targeted and untargeted methods have, to date, not revealed intelligent life outside of Earth.

Resources

To access key concept summaries and past VCE exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0038).

13.6.2 Key terms Amplitude of a wave is the size of the maximum disturbance of the medium from its normal state. Astrometric measurement is a technique for finding exoplanets by observing the minute periodic motion of a star caused by the gravitational effects of an orbiting body. Blue shift is the decrease in wavelength that results from a light source moving towards an observer. A continuous spectrum is a light spectrum that is smoothly spread over a wide range of wavelengths. Dispersion is the separation of light into different colours as a result of refraction. Frequency refers to the number of times of a periodic wave repeats itself every second. Frequency is measured in hertz (Hz) and 1 Hz = 1 s−1 . An incandescent source isproduced by heating a solid or gas under high pressure. Longitudinal waves are those for which the disturbance is parallel to the direction of propagation.

TOPIC 13 Is there life beyond Earth’s solar system? 33

Microlensing refers to when a foreground (lens) star bends the light in the same way a magnifying glass does. Period refers tothe time it takes a periodic wave source to produce a complete wave. Periodic waves are disturbances that repeat themselves at regular intervals. A radar telescope sends out radio waves and bounces them off surfaces of nearby planets to measure distances and map surface features. The radial velocity (RV) method is a technique to find exoplanets that measures the Doppler shift of a star’s spectrum to identify the gravitational effects of an orbiting body. A radio telescope detects faint, low-frequency electromagnetic radiation from the sky. Red shift is the increase in wavelength that results from a light source moving away from the observer. Spectroscopy is the analysis of light by dispersion. The transit method looks for dips in a star’s brightness over time, which indicates a body moving in front of the star. Transverse waves are those for which the disturbance is at right angles to the direction of propagation. A wave is a transfer of energy through a medium without any net movement of matter. Wavelength is the distance between successive corresponding parts of a periodic wave.

Resources Digital document Key terms glossary (doc-32277)

13.6.3 Practical work and investigations Investigation 13.1 The colour of stars Aim: To model the colour of stars by connecting an incandescent globe to different voltages Digital document: doc-32317 Teacher-led video: tlvd-0830

Investigation 13.2 The Doppler effect Aim: To observe the effects of wind on the frequency of sound Digital document: doc-31887

Resources Digital document Practical investigation logbook (doc-32278)

13.6 Exercises 13.6 Exercise 1: Multiple choice questions 1.

What is it called when light is bent due to travelling from one medium to another medium of different density? A. Dispersion B. Reflection C. Refraction D. Diffraction

34 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

2.

3.

4.

5.

6.

7.

8.

9.

10.

Which colour bends the most when white light is passed through a triangular prism? A. Red B. Blue C. Yellow D. Violet What can an absorption spectra reveal about a star? A. Its intensity B. Elements in its atmosphere C. Its temperature D. Its frequency When observing a star, what is blue shift? A. When the star is moving towards us B. When the star is stationary C. When the star is spinning fast D. When the star is moving away from us Which colour light travels the fastest in air? A. Yellow B. Red C. Violet D. They all travel at the same speed. Which of the following are not one of the chemical constituents of life? A. Hydrogen B. Helium C. Carbon D. Nitrogen Which of the following frequencies fit in the spectrum defined as radio? A. 5.2 × 1014 Hz B. 4.8 × 102 Hz C. 6.9 × 106 Hz D. 1.5 × 1011 Hz What do astronomers use the Doppler effect to determine? A. The speed of a stellar object B. The colour of a stellar object C. The distance of a stellar object D. The temperature of a stellar object The electromagnetic spectrum (see figure 13.3) increases from right to left in relation to which of the following? A. Wavelength B. Frequency C. Amplitude D. Height An excited state of an atom is a state where its potential energy is which of the following? A. Equal to the ground state B. Higher than the ground state C. Lower than the ground state D. Zero

TOPIC 13 Is there life beyond Earth’s solar system? 35

13.6 Exercise 2: Short answer questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

What is the main element found in the Sun? What is the hotter part of a star? The core or the outer layer? Explain. Which physics principle was used to explain the movement of Sirius (the brightest star in the sky)? Using this principle explain what was discovered. What is an exoplanet? What units are used to describe the mass of exoplanets and why? In searching for life-sustaining exoplanets, what do we first search for and why? What is the habitable zone? How has the Hubble Space Telescope contributed to the discovery of many exoplanets? If a radio telescope sends out a signal to a planet and receives the signal back 18 minutes later, how far away is the planet? List three methods used to find exoplanets.

13.6 Exercise 3: Exam practice questions Question 1 (3 marks) A star is observed to have a wavelength of 356.978 66 nm. The known wavelength for a stationary source is 356.978 72 nm. Determine the redial velocity at which the star is moving relative to Earth. Question 2 (3 marks) A radio telescope sent out two separate signals, one to planet A and the other to planet B. The planet B signal takes 13 minutes to return while the planet A signal takes 21 minutes to return. How much further is planet A from Earth than planet B? Question 3 (1 mark) A star emits visible colour at 550 nm. If the star is moving away from a stationary observer, which of the following cannot be the wavelength observed? A. 555 nm B. 580 nm C. 530 nm D. 575 nm Give reasoning for your response. Question 4 (3 marks) Comparing blue light and yellow light: a. Which has the greatest frequency? b. Which has the greatest wavelength? c. Which travels faster in air? Question 5 (5 marks) List five factors that affect the prospect for life on another planet.

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Test maker Create unique tests and exams from our extensive range of questions, including practice exam questions. Access the assignments section in learnON to begin creating and assigning assessments to students.

36 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

1 mark 1 mark 1 mark

AREA OF STUDY 2 OPTIONS OBSERVATION OF THE PHYSICAL WORLD

14

How do forces act on the human body? 14.1 Overview Numerous videos and interactivities are available just where you need them, at the point of learning, in your digital formats, learnON and eBookPLUS at www.jacplus.com.au.

14.1.1 Introduction We work, we play, we move. Lift an object and you feel it pull on your arm. Throw or catch a ball and you feel it push on your hand. You should also feel something happening inside your body. Perhaps you’ll feel a particular muscle when pushing off to run or when jumping to catch a ball. Forces generated by our muscles enable our bones to move relative to each other; that’s how we move (figure 14.1). This happens when we bend, run or sit, whether we are exercising vigorously, sitting down or turning around to look over our shoulder. Even when we are standing still our muscles work to help us keep our balance. Humans must be strong enough to create these internal forces and safely respond to the external forces that act on us. Try to think of your body as a structure made of bone, tendons, muscles and skin. To find out what the body is capable of, we will look at the forces and loads on these structures. FIGURE 14.1 Forces generated by our muscles enable our bones to move relative to each other.

TOPIC 14 How do forces act on the human body? 1

14.1.2 What you will learn KEY KNOWLEDGE After completing this topic you will be able to: Forces in the human body • identify different types of external forces including gravitational forces, that can act on a body to create compression, tension and shear x1 m1 + x2 m2 + ... + xn mx • apply centre of mass calculations to a body or system: xm = m1 + m2 + ... + mn ( ) • investigate and apply theoretically and practically translational forces and torques 𝜏 = r⊥ F in simple lever models of human joints under load Materials in the human body • calculate the stress and strain resulting from the application of compressive and tensile forces and loads to ∆l F and 𝜀 = materials in organic structures including bone and muscle using: 𝜍 = A l • compare the behaviour of living tissue under load with reference to extension and compression, including 𝜍 Young’s modulus: E = 𝜀 • investigate how the behaviour of living tissue under load compares with common building materials, including wood and metals • investigate the suitability of different materials for use in the human body, including bone, tendons and muscle, by comparing tensile and compressive strength and stiffness, toughness, and flexibility under load • calculate the potential energy stored in a material under load (strain energy) using area under stress versus strain graph • investigate the elastic or plastic behaviour of materials under load, for example skin and membranes Materials used to replace body parts • investigate the development of artificial materials and structures for use in prosthetics, including external prostheses for the replacement of lost limbs, and internal prostheses such as hip or valve replacements • identify the difficulties and problems with implanting materials within the human body • compare the performance of artificial limbs with natural limbs with reference to function and longevity. Source: VCE Physics Study Design (2016–2021) extracts © VCAA; reproduced by permission.

Resources Digital documents Key science skills — Units 1–4 (doc-31856) Key terms glossary (doc-32286)

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2 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

14.2 Forces in the human body KEY CONCEPTS • Identify different types of external forces, including gravitational forces, that can act on a body to create compression, tension and shear. • Investigate and apply theoretically and practically translational forces and torques 𝜏 = r⊥ F in simple lever models of human joints under load. x1 m1 + x2 m2 + ... + xn mn . • Apply centre of mass calculations to a body or system: xm = m1 + m2 + ... + mn

14.2.1 Equilibrium — staying in place The net external force acting on a stationary object must be zero because its acceleration is zero. Whether we are stationary or moving, the primary external force we experience is caused by gravity. For us to stand still, the gravitational force acting on us, mg, must be balanced by the force from the ground, R, which is equal in magnitude and opposite in direction to our weight (figure 14.2).

FIGURE 14.2 Force diagram for a person standing still

mg

Resolving forces vertically: ΣF = 0 R − mg = 0 ∴ R = mg When suspended from the horizontal bar, as shown in figure 14.3, the person is balanced. The force, T, in each of the arms can be calculated using the principles of equilibrium.

R

FIGURE 14.3 Force diagram for a person hanging from a bar

60º

60º

T

T

Resolving forces vertically: ΣF = 0 2Tsin 60∘ − mg = 0 mg T= 2sin 60∘ T = 0.58 mg

mg

TOPIC 14 How do forces act on the human body? 3

14.2.2 Rotational equilibrium If an object is in equilibrium, the torque caused by one force must be balanced by the turning effect of one or more other forces; that is, the sum of the torques acting on the object must equal zero. The effect of torque can be seen if you stand with your heels touching a wall (figure 14.4a). Keeping your legs straight and without moving your feet, try to bend forwards to touch your toes or ankles. You will lose your balance and fall forwards. Now stand clear of the wall and any furniture or other objects (figure 14.4b). While keeping your legs straight, bend forwards to try and touch your ankles. This time you should be able to keep your balance. FIGURE 14.4 When your centre of gravity is not above the normal force it creates a torque, which causes you to lose balance. (a)

(b)

mg

mg R R

When you stood against the wall, two forces were acting on you: your weight, mg, and the reaction from the floor, R. Although translational equilibrium was satisfied, that is, R + mg = 0, with your straight legs and heels touching the wall, the force, mg, could not act along the same line as R. Rotational equilibrium was not satisfied and the pair of forces caused a rotation, meaning you lost your balance. In contrast, when you were not restrained by the wall, you could adjust your body position so that your centre of gravity was over your feet. You were able to keep your balance when your weight and the reaction from the floor acted along the same vertical line. Under these conditions, rotational equilibrium was satisfied.

14.2.3 Centre of mass When applying Newton’s laws, it is often convenient to consider the mass of an object to be concentrated at a single point, and that the net force acting on the object acts through that single point. We refer to that concentration of mass as the centre of mass (CM). The centre of mass of a symmetrical object with uniformly distributed mass is at the centre of the object. For example, the centre of mass of an orange is at its centre. If the object is symmetrical in two dimensions, the CM will be at the intersection of the two lines of symmetry, as shown in figure 14.5. A symmetrical object behaves as if its mass were concentrated in the middle, where its lines of symmetry intersect. If you support an object under its centre of mass, it will balance. Applying a force through the CM of an object will not cause rotation.

4 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 14.5 Centre of mass occurs at the intersection of the lines of symmetry. Line of symmetry

CM

Line of symmetry

AS A MATTER OF FACT When an object is hung from a single point, the CM will be on the same vertical line as the support. The CM of an object like a table tennis bat can be found by hanging it successively from more than one position and using the gravitational field and a plumb bob as shown in figure 14.6. The CM is where the lines intersect. FIGURE 14.6 Using a plumb bob to find the centre of mass of a table tennis bat (a)

(b)

Pivot

Previous construction line

Pivot CM

The CM is where the lines intersect. CM lies somewhere on this line.

Plumb bob

Plumb bob

Try using this method to find the CM of a map of Victoria or Australia.

TOPIC 14 How do forces act on the human body? 5

The CM for certain shapes are shown in figure 14.7. The CM for some objects, such as a boomerang or a doughnut, are not within the object. FIGURE 14.7 The centre of mass of these objects may occur within, or outside of the object.

CM

CM CM

CM CM CM

The position for the CM can be determined mathematically by using the equation for rotational equilibrium, ∑𝜏 = 0. Rotational equilibrium has been used to locate the CM of the tray and contents being carried by the waiter in figure 14.8.

FIGURE 14.8 The CM of the plate can be found using the formula for rotational equilibrium.

R

∑F = 0 4N + 2N + 2N − R = 0 R = 8 N upwards The force from the waiter’s hand acts along the same vertical line as the CM of the tray and its contents. ∑𝜏 = 0 take torques about the CM:

2N

2N

4N

2 N × (0.2 + d)m + 2 N × d m − 4 N × (0.2 − d) m = 0 8 × d = 0.4 m d = 0.05 m = 5 cm To balance the tray, the waiter should place his hand 5 centimetres from the centre of the tray.

d 0.2 m 0.2 m

The location of the CM of a system of two or more objects can be found using the formula: xCM =

x1 m1 + x2 m2 + … + xi mi (where i is the number of objects) m1 + m2 + … + mi

6 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

14.2.4 Centre of gravity Centre of gravity (CG) is the point at which gravity FIGURE 14.9 Centre of mass of a standing person acts on an object. Unlike centre of mass, it is not constant and depends on the gravitational field acting on it. For most objects on or near the Earth’s surface, it is reasonable to assume that the CG is at the same point as the CM. Even though each part of our body — torso, head, arms and legs — contributes its own weight to our total weight, it is usually convenient to consider the CM CM total weight of a person as acting through a single point, our CG, which corresponds to our CM. When looked at from the front, the standing person in figure 14.9 is symmetrical about the vertical line passing through the middle of their body, and their CM is at approximately waist level. Looked from the side, a standing person is not symmetrical, but their CM is still approximately at waist level and in the middle of their body. When we move a limb or bend, as shown in figure 14.10, we change our body shape and there is a corresponding change in our centre of mass. Moving a limb moves our CM in the same direction. When we tuck during a dive, curl up to pole vault or arch our backs during a high jump we change the location of our centre of mass. Bend far enough and our CM will move to a point outside our body. This enables us to move in different ways. FIGURE 14.10 The centre of mass moves around as a person moves their limbs and body. Bending into a tuck, the diver performing a somersault relocates their CM to outside their body. The change in position of the CM enables the diver to rotate about their centre of mass. Before entering the water, the diver straightens their body and stops the somersault.

CM CM

TOPIC 14 How do forces act on the human body? 7

14.2.5 Lever models of human joints Systems of muscles, bones and tendons in our body work together to enable us to move. They also work to keep us balanced when we are stationary. When Jo is standing and stationary, two forces are acting on him: his weight, mg, and the reaction from mg , and acts downwards. However, what the ground, R. The force in each leg is equal to half his weight, 2 happens when Jo stands on his toes? The external forces, his weight and the reaction from the ground have not changed, but he can feel his calf muscles stretch and he will eventually tire. Let’s look at the forces in Jo’s feet in isolation from the FIGURE 14.11 Force diagram of a person rest of his body. Using a diagram to consider the equilibrium standing on their toes of his feet, all of the forces acting on his feet need to be shown, whether the forces are known or unknown. While B T it might not be initially clear whether an unknown force is pushing or pulling, this will become apparent after the calculations have been completed. There is a force caused by the rest of Jo’s body pushing down on his ankles, primarily through his talus, the bone in his ankle at the base of his leg. This force is shown in figure 14.11 as force B. Jo’s calf muscles are pulling upwards through the Achilles tendon at the back of his ankle, shown as force T, acting upwards. There is also the reaction from the ground, R, which also acts upwards. Considered as isolated objects, Jo’s feet are in mg R= 2 equilibrium, so the net force acting on his feet must be zero. 30 mm 120 mm For translational equilibrium to be satisfied B − T − R = 0. mg . We know that on each foot R = 2 Taking torques about the line of effort of one of the unknown forces, in this case B, and assigning clockwise torque as positive and anti-clockwise torques as negative, we find: 0.030T −

0.120 mg =0 2 0.120 mg T= 0.060 T = 2.00 mg in each leg

By substitution, the force in the talus, B, can be calculated. B=T+R

mg 2 B = 2.50 mg in each leg

B = 2.00 mg +

Jo’s feet act as levers with their fulcrum at the ball of his feet. To stand on his toes, Jo creates a force in his muscles, and his tendons pull upwards at the back of his ankles. This in turn increases the magnitude of the force in the bones in Jo’s legs. In this example, the force needed to lift Jo’s legs is less than the force in the bones of the leg. The levers, his feet, are said to have provided a mechanical advantage, which is fundamental to our walking and being able to quickly accelerate when running.

8 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

There are many other muscle, tendon and bone systems in our bodies forming levers that might be necessary to stabilise us, enable us to move — sometimes quickly or through relatively large distances — or to perform fine movements as might be required by a neurosurgeon. Of the three types of levers shown in figure 14.12, examples of each can be found in the human body. In a first-class lever, the load and force are either side of the fulcrum. In a second-class lever, the load is between the fulcrum and the force and in a third-class lever, the force is between the fulcrum and the load. When we nod up and down, as shown in the figure 14.13, our head rotates about the top of our spine. This is an example of a first-class lever in which our skull is the lever. The centre of mass of our head is forward of our spine and the muscles at the back of our neck provide the force to counter the head’s weight, rotating about the top of the spine. This lever mechanism is important as it stabilises our head and neck.

FIGURE 14.12 (a) A first-class lever (b) A second-class lever (c) A third-class lever (a) Force

Load

Fulcrum (b)

Load Force

Fulcrum (c) Load Force

Fulcrum

FIGURE 14.13 Nodding your head is an example of a first-class lever.

mg

Pivot

Force from muscles at back of neck

TOPIC 14 How do forces act on the human body? 9

Splinting is another example of a first-class lever that is used to support parts of our body. The wrist splint in figure 14.14 might be used to protect a wrist following a fracture or sprain, while also enabling functional use of the hand. With the fulcrum at the heel of the palm of the hand, a force acting up on the hand, U, is resisted by the force R at the end of the splint to the right. FIGURE 14.14 A splint is an example of a first-class lever that helps to stabilise a joint.

R

U

Fulcrum

Jo standing on his toes was an example of a second-class lever in which the force he created was able to lift a larger force, therefore, providing him with a mechanical advantage. Third-class levers are more common in our body. If you lie face down and then flex your knee to raise your foot, your lower leg, knee joint and hamstring act as a third-class lever, as shown in figure 14.15. In such cases, the muscle is not necessarily pulling at right angles to the bone. The efficiency of the force generated by the muscles depends on the angle of its insertion to the bone. The efficiency also changes as you flex your knee because the perpendicular distance from your knee to the centre of gravity of your lower leg decreases. FIGURE 14.15 An example of a third-class lever action. Efficiency of the movement decreases as the CM gets closer to the knee. E E

mg

mg

Although the force needed by your hamstrings is large compared to the weight of your lower leg, this type of lever enables fast and extensive movement of your leg. Contracting your muscles a small amount will cause a larger movement in your leg.

Combinations of lever actions While the fundamental principles of levers can be applied to a system, identifying which muscles are acting can be complex, particularly as an apparently simple action might involve more than one muscle, or group of muscles, at different stages of the movement. For example, when using your forearm to lift a cup you might engage two different muscle groups, the brachialis (biceps) and the brachioradialis (see figure 14.16). Using your brachialis muscles is an example of a third-class lever with your elbow as the fulcrum. However, when engaging the brachioradialis you create a second-class lever.

10 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 14.16 Lifting your arm using the (a) brachialis or bicep and (b) brachioradialis (right) (a)

(b)

Humerus

Humerus

Brachialis Brachioradialis

Radius

Ulna

Radius

Ulna

Whereas one group of muscles might provide the force necessary to produce or resist movement, another muscle group might control the movement by opposing it. For example, while the biceps might act to lift an object, the triceps might control the action by resisting the biceps. An analysis of these lever systems depends on a knowledge of both anatomy and the principles of mechanics as it applies to human movement.

14.2.6 Types of forces Tension and compression When standing on your toes, the tendons at the FIGURE 14.17 Tension in arterial walls back of your ankles are pulled. The muscles and tendons are in tension, whereas the bones in the leg are in compression. Objects respond to compressive and tensile forces by deforming. This is evident in objects such as springs and rubber bands when under T load. How much an object deforms depends T on the magnitude of the applied force and the physical properties of the material it is made from. T The bones, muscles and tendons in our bodies also respond to forces by deforming. A change in length of our muscles and tendons is more noticeable than changes in our bones, except T when a bone fractures. However, our bones do compress and bend in response to the forces that act on them. T T Tension is also created in the walls of our arteries from the blood being circulated around our body (figure 14.17). The blood travelling along our arteries is like water travelling in a hose. Under pressure, the blood pushes out against the wall of the artery, causing circumferential tension. The skin is the largest organ in the human body. Push on your skin and it will depress and stretch. When you pushed on your skin you caused a multidirectional tensile force in it (see figure 14.18). This is a special example of tension in that the primary deformation is transverse to the force you applied. This might be

TOPIC 14 How do forces act on the human body? 11

likened to a tight rope stretching when walked on, the difference being that the skin is a membrane and the resulting tension will occur in several directions.

FIGURE 14.18 Pressing something with your finger creates multidirectional tension forces.

Bending and shear There are forces that cause deformations different to those caused by tensile and compressive forces. Bending and shear are two such forces. When bending, an object tends to curve. Some sections of the object move closer together and other sections move further apart, causing compression on one side and tension on the other side, as shown in figure 14.19. When you stand in the middle of a plank that is supported at both ends, it bends and you might feel it sag. When it is bent, the top of the plank is compressed and the underside is stretched and is in tension.

T

T

T

T

T

T T

FIGURE 14.19 An object is bent, resulting in compression on the top and tension on the bottom.

C

C C

C

C

T

T T

T

Shear forces create a tendency for part of an object to slide over itself; the effect is similar to a sideways force acting on a thick book, shown in the figure 14.20. The top of the book displaces sideways, the bottom remains stationary, and there is a proportional displacement of the pages in between as they slide over each other. FIGURE 14.20 A shear force acts on a book.

12 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

When a person grinds their teeth, they create shear forces that erode the surface of the tooth and sometimes cause the tooth to fracture. The famous tennis player, Roger Federer, in figure 14.21 is also creating shear forces in their foot when they slide across or try to push off the court surface, risking injury to their ankle. FIGURE 14.21 Roger Federer creates a shear force as he slides on a clay court surface.

TOPIC 14 How do forces act on the human body? 13

14.2 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. A 65-kilogram gymnast is completing a vertical handstand. To make the degree of difficulty greater they have their arms spread at an angle of 60°. Calculate the force through each arm. 2. The skull shown in the figure on the right acts as a 30 mm 50 mm lever with its fulcrum at the top of the spine. (a) Calculate the magnitude of the force in the muscles at the back of the neck. During a car crash, the shoulders of a person are restrained by a seatbelt. However, their head continues moving forward, to be resisted only by the CM muscles in the back of the neck. (b) On a diagram of a skull, show the forces that are 75 mm now acting. mg Force from (c) Calculate the force in the muscles at the back muscles at of the neck when the acceleration is 4g backwards. back of neck (d) What is the force in the spine when the acceleration is 4g backwards? 3. Your jaw acts as a lever when you eat, hinging at the temporomandibular joint. If the force needed to bite a carrot is 120 N, what force is required from the muscles shown in the following figure?

Temporomandibular joint 120 N

90 mm

30 mm

4. When holding an arm horizontally and to the side as shown in the following figure, Ben’s arm acts as a lever that rotates at the shoulder joint. Deltoid muscle De

lto

B Shoulder joint (pivot point)

mg

id 16o

Elbow B

Centre of hand

mg

170 mm

170 mm 400 mm

400 mm

680 mm

680 mm

(a) If the mass of Ben’s arm is 4 kilograms with its centre of mass at 400 millimetres from his shoulder joint, what force is needed at B to keep his arm horizontal? The deltoid muscle group, which is the primary muscle group acting, inserts at the bone at approximately 16° to the horizontal. (b) What force must be provided by the deltoids? (c) When Ben also holds a 2-kilogram mass in his hand, what force must be provided by his deltoids?

14 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

5. (a) For the circus routine shown in the following figure, draw a diagram that shows the forces acting on the performer. (b) The performer’s mass is 64 kilograms. Calculate the direction and magnitude of the force in each arm. (Assume the vertical forces are equally distributed between each arm.)

CM 1m

1.6 m

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

14.3 Stress and strain KEY CONCEPTS • Calculate the stress and strain resulting from the application of compressive and tensile forces and loads to ∆l F . materials in organic structures including bone and muscle using 𝜍 = and 𝜀 = A l • Compare the behaviour of living tissue under load with reference to extension and compression, including 𝜍 Young’s modulus: E = . 𝜀

To survive a bungee jump, people rely on the suitable choice of bungee cord. The cord must stop the freefall before the jumper hits the ground, and must cushion the fall by absorbing the deceleration. An unsuitable choice of a cord could be fatal. If the bungee jumper is heavier, instinct and experience would suggest that a thicker cord should be used. The thicker cord will be stronger and have a greater resistance to deformation. If a person is jumping from a greater height, then a longer cord will be needed, but the longer cord will also stretch further. Similarly, it is important that our muscles and tendons do not break, or stretch too much, when resisting external forces. They also need to tolerate the internal forces necessary for balance and movement. There are many factors that make some materials strong or flexible, and the physical quantities stress and strain are helpful in classifying and predicting the behaviour of materials, including those in our body. To compare meaningfully the performance of different sized forces on different sized objects, we measure stress and strain.

TOPIC 14 How do forces act on the human body? 15

14.3.1 Stress We use stress to compare the strength of materials — a strong material will sustain a larger stress than a weak material. Stress is defined as the force per unit area of cross-section supporting a force and is a vector quantity. 𝜎 = stress =

F force = area A

Stress is measured in pascals, where 1 Pa = 1 N m−2 . However, for many materials such as those used in the construction of buildings, it is more convenient to use megapascals, MPa, where 1 MPa = 1 × 106 Pa. FIGURE 14.22 The force, F, acts on the cross-sectional area, A.

A

F

SAMPLE PROBLEM 1

The femur in the following figure must support 340 N. 340 N

340 N

Assuming that the femur is homogenous and at mid-length is cylindrical with a diameter of 26 millimetres, what is the stress in the bone?

16 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

THINK

WRITE

1.

Recall the relationship between stress and area.

2.

Using the)formula for the area of a circle ( A = 𝜋r2 , calculate the area of the mid-point of the femur.

3.

Substitute the force and area into the equation to find the stress.

4.

State the solution.

F A A = 𝜋 × 132 mm2 𝜎=

= 530 mm2 = 5.3 × 10−4 m2 F 𝜎= A 340 N = 5.3 × 10−4 m2 = 6.4 × 105 Pa = 0.64 MPa The stress in the bone is 0.64 MPa.

PRACTICE PROBLEM 1 A person is doing a push-up and in doing so they are applying 220 N through the ulna. Assuming the ulna is homogeneous and at mid-length it is cylindrical with a diameter of 1.6 centimetres, what is the stress on the ulna?

14.3.2 Strain When a stress is applied to a material, it changes shape. Strain is a measure of the magnitude of this deformation. The amount a material deforms depends on its initial length in the direction of the force. For example, when pulling on a short piece of fishing line, you could not stretch it as far as when applying the same force to a longer piece of the same fishing line. Similarly, when under the same stress, the change in length of a shorter tendon would be different from that of a longer piece of the same tendon. For this reason, we use strain to compare the fractional change in length of materials. Strain, 𝜀, is the ratio of the change in length to the original length. 𝜀=

∆l l

Where: ∆l = the change in length measured in metres l = the original length measured in metres. Note: Strain does not have any units; it is dimensionless! In many materials strain is so small that it is hardly noticeable, but it is always there in response to stress. The strain of a material when it breaks is often expressed as a percentage and called the percentage elongation. Strain affects every material. Every solid material will deform, even though the deformation is often so small that it is not visible. Excessive or permanent deformation could make an object unserviceable. Fortunately, many objects recover their original shape when the force is removed.

TOPIC 14 How do forces act on the human body? 17

SAMPLE PROBLEM 2

Camille’s Achilles tendon is 200 millimetres long. When Camille stretches her tendon to 208 millimetres, what is the strain? b. Camille ruptured her Achilles tendon when the strain reached 10%. How much had the tendon stretched when it ruptured? a.

THINK a. 1. 2.

Recall the formula for strain. Find l.

3.

Substitute the values into the formula.

4.

State the solution.

b. 1.

Recall the formula for strain.

Rearrange to make l the subject. 3. Substitute the values into the formula. 2.

4.

State the solution.

WRITE a.

Δl l Δl = 208 − 200 𝜀=

= 8 mm Δl 𝜀= l 8 = 200 = 0.04 or 4% The strain when Camille stretches her tendon to 208 millimetres is 0.04, or 4%. Δl b. 𝜀 = l Δl = 𝜀 × l Δl = 𝜀l = 0.1 × 200 = 20 mm The tendon had stretched 20 millimetres when it ruptured.

PRACTICE PROBLEM 2 Noah is stretching before his basketball game. His Achilles tendon is 212 millimetres long naturally and 223 millimetres long when he stretches it. a. What is the strain on Noah’s Achilles tendon when he is stretching it? b. Noah can injure his Achilles tendon if the strain reaches 15%. How much would the tendon need to stretch for him to injure his Achilles tendon?

14.3.3 Stiffness When a force deforms a material, it does work against internal forces. Whether it is a bone being compressed or a tendon stretching, work is done by moving tissues and cells either closer together or further apart. If the material is stiffer, the amount of work required to deform it will be greater. Knowing the stiffness of materials enables us to predict their behaviour when acted on by a force. Stress and strain are used to compare the properties of materials because they are independent of the size and shape of the material. Scientists and engineers gather this information by conducting laboratory tests on small samples of materials. Information from these tests enables engineers to choose the most appropriate material, and the required amount of material, for a particular use.

18 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Tension tests are typically carried out on small bars that are stretched until the specimen breaks. Compression tests are performed by crushing blocks or cylinders. The stress applied and the subsequent deformation of a material may be displayed as a stress–strain graph. The strength of a material is the largest stress applied before it fails and is often available from the stress–strain graph. Tensile tests can be undertaken using apparatus similar to that shown in figure 14.23. These tests assume that the composition and behaviour of the material being tested are the same throughout; that is, the material is homogenous. It is also assumed that the material is isotropic — its properties are the same in each direction. While the assumptions of homogeneity and isotropy might not apply to biological materials, such as skin, the tensile tests can still be used to inform us of their mechanical properties. The mechanical properties of skin (figure 14.24) vary according to many factors, including where on the body the skin comes from, its age, level of hydration and how quickly the force is applied to the test sample. In the case of living tissue such as that found in the human body, medical scientists might obtain stress and strain information by testing in vivo — alive and on the body — or in vitro — after removing the material from the body. An understanding of the behaviour of different materials, including those occurring naturally in the human body, has helped develop artificial devices to help people with implants, prostheses and orthoses. FIGURE 14.24 Stress–strain relationship for a skin sample

FIGURE 14.25 The shape of the stress–strain graph for some different materials

25

Stress (MPa)

500

20 Stress (MPa)

FIGURE 14.23 A tensile testing apparatus

15

10

Mild steel 400

300

200

Glass

Aluminium Rubber

100

5

0

10

20

30 40 Strain (%)

50

60

0

Strain

TOPIC 14 How do forces act on the human body? 19

SAMPLE PROBLEM 3 a. b.

Which of the two materials in the following graph is stronger? Which elongates the most before breaking?

Stress

Material 1

Material 2

0

Strain

THINK a. 1.

WRITE

Recall the definition of strength.

Strength = maximum stress before breaking Material 1 is stronger because a greater stress was needed to break it. b. Strain = distance stretched a.

From the graph determine which material is stronger. b. 1. Recall that strain is the measure of how far a material has stretched. 2. From the graph determine which material stretched more. 2.

Material 2 elongated more before breaking.

Stress

PRACTICE PROBLEM 3 a. Which material in the following graph is stronger? b. Which material is more ductile?

Material 2 × Material 1 ×

0

20 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Strain

SAMPLE PROBLEM 4

Consider the two materials represented in the following stress–strain graph. Of the two materials, which would be more difficult to stretch?

× Material 1

Stress

σ1

× Material 2

σ2

0

ε Strain

THINK

WRITE

Recall that strain is the measure of how far a material has stretched. 2 From the graph determine which material stretches further under the same stress.

Strain = distance stretched

1

Material 2 stretches further than material 1 under the same stress. Therefore, material 1 would be more difficult to stretch.

PRACTICE PROBLEM 4 Which material in the following stress–strain graph would be more difficult to stretch?

Stress

σA

Material A ×

σB × Material B

0

Strain

TOPIC 14 How do forces act on the human body? 21

14.3.4 Young’s modulus of elasticity For many materials, the stress–strain graph for a tensile test is straight, or nearly straight, near the origin. Here, the stress in the material is proportional to the strain and the constant of proportionality is called the modulus of elasticity or Young’s modulus, E.

FIGURE 14.26 Young’s modulus of elasticity can be calculated from the gradient of a stress–strain curve.

A material with a higher Young’s modulus is stiffer; it is more resistant to deformation as shown by the steeper gradient of the stress–strain curve. Steel is used in large and small structures because it is stiff and needs a large stress to change its shape. Materials that have a low Young’s modulus would not be suitable for buildings because they would deform too much. The modulus for most materials is large, so it is usually expressed in gigapascals, where 1 GPa = 109 Pa. For the material represented graphically in figure 14.26: 𝜍 𝜀 ∆𝜍 = ∆𝜀 180 MPa = 3 × 10−3 = 6 × 104 × MPa

E=

So for this material, E = 60 GPa. Some typical values of E are listed in table 14.1. Our bones are not homogenous. They comprise a hard outer layer called cortical bone and a spongy interior named trabecular bone. Before it breaks, the bone exhibits a linear stress–strain relationship (shown in the figure 14.27). ∆𝜍 ∆𝜀 120 MPa = 0.009 = 1.3 × 104 MPa

E=

So, for this sample of bone E = 13 GPa.

22 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Fracture ×

∆σ = 180 MPa

Young’s modulus is calculated from the gradient of the 𝜎 stress–strain curve, E = . This relationship is an 𝜀 expression of Hooke’s Law and can be written in the form 𝜍 = E × 𝜀.

Stress (MPa)

200

100

0

1

2

3

4

∆ɛ = 3 ×10–3 Strain (× 10–3)

TABLE 14.1 Some typical values of E Material Skin

Young’s modulus (GPa) 2 × 10−4

Tendon

2

Polyethylene

7

Bone

7–30

Wood

14

Concrete

18

Aluminium

70

Glass

100

Titanium

100

Copper

110

Steel

210

Carbon fibre Diamond

410 1200

FIGURE 14.27 The stress–strain graph of a bone sample. The stress–strain graph is linear and Young’s modulus can be calculated from the graph.

Stress (MPa)

150

100

50

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

1.2

Strain (%)

Even in materials that do not exhibit a linear stress–strain relationship, the slope of the curve is often used to describe the stiffness of the material. For the skin sample shown in figure 14.28, the slope of the linear part of the graph provides an estimate of Young’s modulus — a measure of the stiffness of the skin within its elastic range. FIGURE 14.28 Stress–strain graph of a skin sample

Stress (MPa)

0.02

0.01

0

10

20

30

40

50 Strain (%)

60

70

80

TOPIC 14 How do forces act on the human body? 23

From the graph, Young’s modulus can be calculated. For example, when subject to larger stress: ∆𝜍 ∆𝜀 0.02 MPa = (0.80 − 0.70) = 0.2 MPa

E=

So, for this sample of skin E = 0.2 MPa = 200 kPa. Note that the skin initially deforms without carrying much stress. After it deforms, it behaves more stiffly. When under stress the skin will deform, something which is important if the body is to accommodate movement at joints. The ability of the skin to stiffen and resist larger stresses then protects other parts of the body, such as vital organs, from penetration.

SAMPLE PROBLEM 5

Using the test data in the following table, plot a stress–strain curve and calculate Young’s modulus for each material. Which material is stiffer? Which is stronger? Material A Stress (MPa)

0

35

70

110

140

145

150

Strain (× 10−3 )

0

0.5

1.0

1.5

2.0

2.3

2.5

Stress (MPa)

0

30

60

90

110

130

140

Strain (× 10−3 )

0

0.5

1.0

1.5

2.0

3.0

4.0

Material B

THINK 1

WRITE

Place strain on the x-axis and stress on the y-axis and plot the points to form the graph.

Material A 150 Material B Stress (MPa)

120

90

60

30

0

24 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

1.0

2.0 3.0 –3 Strain (x 10 )

4.0

2

Use the straight region of the graphs to calculate Young’s modulus.

3

Recall that stiffness is measured by the gradient of the stress–strain curve.

4

Strength is defined as the maximum stress reached before a material breaks.

Material A: Δstress E= Δstrain 140 = 2.0 × 10−3 = 7 × 104 MPa = 70 GPa Material B: Δstress E= Δstrain 90 = 1.5 × 10−3 = 6 × 104 MPa = 60 GPa From the graph, it can be seen that the slope of the curve for A is steeper than that for B; therefore, A is stiffer than B. The maximum stress that material A can sustain before fracture is greater than that for material B: 150 MPa compared with 140 MPa. Therefore, material A is stronger than material B.

PRACTICE PROBLEM 5 Using the experimental data in the following table, plot a stress–strain curve and calculate Young’s modulus for each material. Which material is stiffer? Which material is stronger? Material 1 Stress (MPa)

0

20

40

60

80

100

110

–3

0

0.2

0.4

0.6

0.8

1.0

1.2

Strain (×10 )

Material 2 Stress (MPa)

0

16

32

48

64

85

100

Strain (×10–3 )

0

0.2

0.4

0.6

0.8

1.2

1.8

14.3 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. What is the value of Young’s modulus for the polymer in the following stress–strain graph?

TOPIC 14 How do forces act on the human body? 25

Stress (MPa)

5

0 2.0 –3 Strain (x 10 )

1.0

2. What stress would create a strain of 0.0005 in a bone with a Young’s modulus of 18 GPa? 3. At its mid-length, Belinda’s femur is approximately circular in cross-section with a diameter of 22 millimetres. If her mass is 56 kilograms, calculate the stress in her femur. 4. (a) What stress will cause a strain of 0.04 in a tendon with a Young’s Force (kN) Length (mm) modulus of 0.25 GPa? (b) If the original length of the tendon was 200 millimetres, how much 0 50.80 has it stretched under this load? 72.4 50.90 5. A 2-millimetres diameter steel cable 5 metres long lifts a 15 kN load. Assuming Young’s modulus for the steel is 200 GPa, how much 108.6 50.95 will the cable stretch? 144.8 51 6. A 12.7-millimetres square bar 50.8 millimetres in length is loaded in tension. The data collected are shown in the following table. (a) Convert the data to stress and strain. (b) Plot the data and calculate Young’s modulus. (c) What stress would cause a strain of 0.5% in the 12.7-millimetre square bar? 7. Estimate the stiffness of the material described in the following graph.

181.0

51.05

189.5

51.10

100 × Fracture

Stress (MPa)

80

60

40

20

0

0.001

0.002 Strain

0.003

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

26 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

14.4 Elastic and plastic behaviour of materials KEY CONCEPTS • Calculate the potential energy stored in a material under load (strain energy) using area under stress versus strain graph. • Investigate the elastic or plastic behaviour of materials under load, for example skin and membranes.

14.4.1 Strain energy Bungee cords are selected for their good energy-absorbing characteristics. Otherwise, a bungee jumper would stop too quickly, which might cause injury. Similarly, our tendons, muscles and cartilage absorb large amounts of energy to help avoid injury. The energy is absorbed during deformation caused by the forces doing work on that part of the body. We use strain energy to describe the energy stored in a material when it is deformed. Strain energy is equal to the work done to deform a material. In general, the strain energy per unit volume stored in a material for any strain can be determined from the area under the stress–strain graph up to that particular strain. The strain energy per unit volume is usually measured in joules per metre cubed. The unit J m−3 is equivalent to N m−2 , as: 1 J m−3 = 1 N m m−3 = 1 N m−2 If a material returns to its original shape when the force is removed, the energy that was stored is called elastic strain energy. SAMPLE PROBLEM 6

×

280 240 Stress (MPa)

The stress–strain characteristics of a particular material are shown in the following graph. a. Calculate how much energy per unit volume is needed to strain the material 0.2% (2 × 10−3 ). b. The test specimen was 200 mm long and 12 mm in diameter. If the specimen fractured at a strain of 3 × 10−3 , estimate the energy required to fracture it.

200

100

0

THINK a. 1

The energy per unit volume is found from the area under the curve up to a strain of 2 × 10−3 . As the shape is a triangle use the formula for the area of a triangle.

0.1

0.2

WRITE a.

0.3 0.4 Strain (%)

0.5

1 × 2 × 10−3 × 240 × 106 2 = 2.4 × 105 J m−3

TOPIC 14 How do forces act on the human body? 27

2

b. 1

State the solution. Energy per unit volume found between 2 × 10−3 and 3 × 10−3 is given by the area between 2 × 10−3 and 3 × 10−3 . As the shape is roughly trapezoidal use the formula for the area of a trapezoid.

The energy per unit volume needed to strain the material is 2.4 × 10 Jm−3 . 1 1 b. ≈ × 2 × 10−3 × 240 × 106 + (240 + 280) 2 2) ( ×106 × 3 × 10−3 − 2 × 10−3 = (2.4 + 2.6) × 105 = 5.0 × 105 J m−3 Volume = 𝜋 × (0.006)2 × 0.200 = 2.26 × 10−5 Total energy = 5.0 × 105 × 2.26 × 10−5 ≈ 11.3 J The energy required to fracture the specimen at a strain of 3 × 10−3 is 11.3 J.

Calculate the volume of the specimen. 3 Total energy = energy per unit volume × volume 4 State the solution. 2

PRACTICE PROBLEM 6 The stress–strain characteristics of a particular material are shown in the following graph. a. Calculate how much energy per unit volume is needed to strain the material 0.8%. b. The test specimen was 150 millimetres long and 16 millimetres in diameter. If the specimen fractured at a strain of 1.2%, estimate the energy required to fracture it. ×

Stress (MPa)

100 80 60 40 20

0

0.2

0.4

0.6

0.8 1.0 1.2 Strain (%)

1.4

1.6

14.4.2 Toughness We use the term toughness to describe the ability of a material to store energy up to the point of fracture. A material with a greater strain energy up to fracture is said to be tougher than one requiring less energy. Toughness can be determined from the total area under the tensile stress–strain curve for a material up to the fracture strain.

28 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

SAMPLE PROBLEM 7

Consider the following stress–strain graph of two materials. Which material is: a. tougher b. stronger c. stiffer?

×A

Stress

×B

0

Toughness = total area underneath the stress–strain curve b. Strength = maximum stress before breaking c. Stiffness = gradient of the stress–strain curve a.

WRITE

Material B has more area under its stress–strain curve, thus, is tougher. b. Material A reaches a higher stress before breaking than material B, thus, is stronger. c. Material A has a greater gradient in its stress–strain curve than material B, thus, is stiffer. a.

PRACTICE PROBLEM 7 Consider the following stress-strain graph of two materials. Which material is: a. stiffer b. tougher c. stronger?

Material 1 × × Material 2

100 Stress (MPa)

THINK

Strain

80 60 40 20 0

0.2

0.4

0.6

0.8 1.0 1.2 Strain (× 10–3)

1.4

1.6

1.8

14.4.3 Elastic behaviour A cyclist pulls on shorts to stretch them over the hips, but the deformation of the shorts is not permanent. When the cyclist stops pulling, the shorts assume the shape of the body. If they did not, the shorts would fall down. This is an example of reversible deformation that occurs in many materials, including metals used in structures like bridges and buildings. When the force stops acting, the material returns to its original shape (figure 14.29). Deformation that disappears when the stress is removed is called elastic deformation. The maximum stress for which a material behaves elastically is called the elastic limit or yield stress. When stretched beyond its elastic limit, a material will not return to its original shape.

TOPIC 14 How do forces act on the human body? 29

Visco-elastic behaviour Naturally occurring materials in our body — blood vessels, skin, tendons and bones — exhibit another type of elasticity. The stress–strain graph of skin in figure 14.28 shows that skin does not strain linearly when stress is initially applied; the strain is time dependent. As the stress increases, the skin assumes linear elastic properties. This time-dependent behaviour is largely attributed to the response of elastin fibres in the skin. When a load is removed and the skin or other material returns to its original shape, the energy might not be returned, which shows up as a loop on in figure 14.30. This loop is called a hysteresis loop, which is a characteristic behaviour for tendons and skin. The unreturned energy is typically used to do work within the system, and might be evident in the form of heat. For some materials loaded and unloaded within the elastic range, the energy absorbed during loading may not be completely recovered. This is called hysteresis and the discrepancy appears as a hysteresis loop on a stress–strain graph. FIGURE 14.29 The stress–strain behaviour of a test specimen. When the load was removed, the specimen behaved elastically, recovering its original shape.

FIGURE 14.30 Stress–strain graph containing a hysteresis loop

Stress

Stress

Unload Unload Load

Load 0

Strain

0

Strain

14.4.4 Plastic behaviour Deformation that does not completely recover when the force is removed is called plastic deformation. The panels of a car, dented in an accident, have undergone plastic deformation. When you pull on a piece of chewing gum it stretches, but the deformation is not reversible. We also plastically deform aluminium foil and cling wrap to the shape of various containers. The plastic properties of materials are very useful for manufacturing and shaping items, including some prosthetics. For example, wire is bent into shape for dental braces without breaking, and polymers are used to form complex shapes when making a hand splint.

Ductile Materials that deform plastically before they fracture are said to be ductile. Most metals and polymers are ductile. The steel tensile test piece shown in figure 14.31 started as a uniform shape. As it was loaded, it elongated and the cross-sectional area decreased. When unloaded, the test piece did not return to its original shape because it had been loaded beyond its elastic limit; it had deformed plastically.

30 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 14.31 The test specimen was loaded beyond its elastic limit, then there was some elastic recovery but the energy used to stretch the specimen was not fully recovered.

Elastic limit Stress

Unload

Plastic

Elastic

deformation

strain

(permanent)

recovery Strain

Necking is the noticeable thinning that occurs before fracture of many metals, and occurs where the material is weakest. After ductile fracture, the broken pieces do not fit together well. The percentage elongation of a material at fracture is often used as a measure of its ductility.

Brittle Materials that are not ductile are called brittle. Brittle materials experience little or no plastic deformation before fracture — less than about 5% strain. In general, the surface of a brittle failure is flat and the broken pieces could be fitted back together to resemble the original item. The glass in figure 14.32 is an example of a brittle failure. Note that the pieces can be put back together to look like the original object. Concrete, plaster and glass are examples of brittle construction materials. FIGURE 14.32 Although brittle materials are usually not as tough as ductile materials, the maximum stress to fracture may be large. Consequently, brittle materials are sometimes strong.

Source: Jef Poskanzer / Flickr

TOPIC 14 How do forces act on the human body? 31

14.4 EXERCISE

𝜍 (MPa)

0

4

8

12

16

20

24

𝜀 (%)

0

2

4

6

8

10

12

𝜍 (MPa)

22

20

19

20

20

20

18

𝜀 (%)

15

30

50

80

110

160

210

×A

×B Stress

To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. The stress–strain characteristics for two different materials, A and B, tested to fracture are shown. (a) Which material has the greater Young’s modulus? (b) At the elastic limit, which material has the greater strain? (c) Which material is tougher? (d) Which material is more ductile? (e) Which material is stronger? 2. The results of a tensile test are given in the following table.

0

Strain

(a) Plot the stress–strain graph for this material. (b) Is the material brittle or ductile? Explain. (c) Calculate Young’s modulus for the material. (d) How much energy is stored in the material when it is stretched to twice its original length? 3. Grant states that strain energy exists only when a material is in the process of being deformed. Is Grant correct? Discuss your reasoning. 4. Gaby is testing the properties of materials. Based on her observations, explain the properties of the following materials. (a) Material A has a large strain value. (b) Material B breaks before reaching its elastic limit.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

14.5 Materials for prostheses KEY CONCEPTS • Investigate how the behaviour of living tissue under load compares with common building materials, including wood and metals. • Identify the difficulties and problems with implanting materials within the human body.

Steel is used in large and small structures such as buildings and bridges because it is stiff and needs a large stress to change its shape. Similarly, metals and ceramics are used in orthopaedic and dental prostheses because of their strength, stiffness and bearing properties. When producing a prostheses, it is important to select materials with similar mechanical properties to the body part it is going to replace. These materials

32 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

must also be biocompatible, non-toxic and not susceptible to decay or changes in temperature. They must also be readily secured mechanically or chemically to hold them in place.

14.5.1 The behaviour of living tissues Nature successfully uses materials with appropriate properties for a variety of purposes. For example, an eggshell is strong enough to withstand the compressive stresses caused by the weight of brooding parents. However, a hatching chick can easily break the shell. This is because the shell is brittle. Once the chick has pecked a hole in the shell, cracks spread easily throughout the brittle material, causing it to break. Similar to the shell of an egg, our skull must be strong in compression and be hard enough to resist penetration. However, because it must resist larger forces, the skull needs to be both stronger and tougher with a larger elastic limit than the brittle eggshell.

14.5.2 Common artificial materials The materials that we use for construction can be classified as coming from one of four groups — metals, ceramics, polymers or composites; composite materials are composed of two or more different materials. Strength and stiffness are two important properties necessary for structures, whether they are for a building or for an artificial limb. Typical strength and stiffness values of these groups of materials are shown in figure 14.33. FIGURE 14.33 The range of strength and stiffness for different materials Stiffness

Ceramics

Metals

Polymers

Composites

10–3

10–2

10–1

101

1

102

103

104

105

Elastic moduli (GPa)

Strength

Ceramics

Metals

Polymers

Composites

10–1

1

101

103

102 Yield strength (MPa)

TOPIC 14 How do forces act on the human body? 33

Another useful way to display this information is by grouping the typical range of values of two properties of materials together. This type of chart is often referred to as an Ashby chart (figure 14.34). Materials with identical chemical composition do not always behave in the same way. Their behaviour might depend on other factors such as ambient temperature, the duration and frequency of loading, and the environment in which the material is used.

FIGURE 14.34 An Ashby chart showing strength and stiffness of different groups of materials

1000

Ceramics Metal alloys

Young’s Modulus E (GPa)

100

10

Composites 1 Polymers

0.1

0.01 0.1

1

10

100

1000

10000

Strength (MPa)

14.5.3 Metal in prosthetic limbs Limb prostheses are functional replacements for a missing limb. Records show that steel and iron were used in the manufacture of prosthetic arms as far back as the sixteenth century. Prior to that, wood and copper are known to have been used for leg prostheses. The strength and stiffness of steel make it a good substitute for bone, but it is significantly heavier for the user. Stainless steel, aluminium and titanium are metals that are now used for manufacturing prosthetic limbs and joints. These metals are readily available, although titanium is more expensive. Both stainless steel and titanium can equal bone for strength and stiffness when acted on by axial forces as well as when bending or in torsion. The ductility of these metals is an important consideration given the cyclic loading and unloading that occurs in our body. However, the weight of metal prosthetics is larger than bone, and therefore they require more energy to use, causing the user to tire sooner. For this reason, aluminium might be preferred over stainless steel and titanium when considering large components such as the pylon in a leg prosthesis. The development of materials such as carbon fibre and Kevlar, both strong and durable, has enabled lighter prostheses to be built.

34 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

In general, limb prostheses include a socket to join the prosthetic to the body, an attachment mechanism and, for a dynamic prosthetic, a control mechanism. For a lower leg prosthesis like that shown in figure 14.35, the pylon is the main internal structure, preferably made from carbon fibre, titanium or aluminium. Metals are commonly used when joining components, particularly if they are designed to move relative to each other. Some wood might be used to structure the foot, but plastics and foams are suitable for the rest of the prosthesis.

FIGURE 14.35 An above-knee leg prosthesis

14.5.4 The effect of temperature on materials The properties of materials depend on temperature. For example, a decrease in temperature might cause a ductile material to become brittle. At high temperatures, materials can deform plastically even when the stress is less than the material’s elastic limit determined at room temperature. FIGURE 14.36 The mechanical properties of polymethyl methacrylate vary with temperature.

80 4°C

Stress (MPa)

60

20°C

40 40°C

20

60°C

0 0.1

0.2

0.3

Strain

TOPIC 14 How do forces act on the human body? 35

The temperature-dependent behaviour of materials can be used as an advantage, as in the shaping of some metals and plastics. A splint such as that shown in figure 14.37 was shaped to the profile of the person’s hand while the thermoplastic was warm. As the thermoplastic cooled, it became stiffer and stronger.

FIGURE 14.37 Splints can be shaped at high temperatures to fit the shape of a person’s body.

14.5.5 Limitations of ductile materials Repetition of low-stress movements Materials subjected to cyclic loads can fail at lower stresses than their yield strength. This failure is common in metals and is called fatigue. For example, a paperclip can be broken by repeatedly bending and straightening it in opposite directions. The maximum magnitude of the force you apply does not change, but the cyclic loading and unloading causes cracks to spread and eventually fractures the clip. Similarly a tooth that is repeatedly subjected to compressive stresses might eventually fracture. Cyclic loads are important to consider when selecting materials for joint replacements. When metal — a ductile material — is used in an implant, it is more prone to fatigue than the biological material it replaces. In their early years, metal-on-metal hip implants were known to last up to two years. The current generation of hip implants uses other materials and the hips can be reasonably expected to last more than 10 years.

Given enough time FIGURE 14.38 Typical strain–time curve for a material showing creep under constant stress and temperature

Fracture

Strain

Creep is the time-dependent and permanent deformation of a material under a constant load. Both the temperature and the size of the applied stress affect the rate at which creep occurs. The visco-elastic materials in our body, particularly our skin, exhibit creep. The strain increases with time even when the stress is constant. When stretched and kept at a constant deformation, the stress in our skin will decrease over time and it is said to have relaxed. Creep is evident in most materials at higher temperatures; however, lead and some polymers will readily creep at room temperature. Although some materials take years to show significant signs of creep, some soft metals and polymers show signs of creep in a short time.

0 Instantaneous deformation

36 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Time

14.5 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. 2. 3. 4. 5.

Name the four groups that materials we use for construction can be classified as. Which of the four classified materials demonstrates the greatest strength? Which of the four classified materials displays the least stiffness? Why might aluminium be preferred for a leg prosthesis over stainless steel or titanium? Explain what creep is.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

14.6 Performance of artificial limbs KEY CONCEPTS • Investigate the suitability of different materials for use in the human body, including bone, tendons and muscle, by comparing tensile and compressive strength and stiffness, toughness, and flexibility under load. • Investigate the development of artificial materials and structures for use in prosthetics, including external prostheses for the replacement of lost limbs, and internal prostheses such as hip or valve replacements. • Compare the performance of artificial limbs with natural limbs with reference to function and longevity.

When choosing materials to make a prosthesis, it is necessary to consider the stresses each component must resist. Will each component be subject to an axial tensile or compressive stress? Bending or torsion? Does the component move, rendering it susceptible to wear and tear from frictional forces? The properties of the available materials also need to be determined according to their function. For example, the pylon in a below-knee prosthetic must be strong in compression and able to resist some bending and torsion. It should also be stiff, ductile and preferably with a weight similar to that of a natural leg. Consideration should also be given to how the pylon can be made. Can it be readily cast or machined? In contrast, a cover over the pylon is most likely cosmetic and other materials such as polyurethane foam and plastic might be appropriate because of their low weight and the ease with which they form irregular shapes.

14.6.1 Composite materials Composite materials are materials made by combining two or more different materials to create a single material with enhanced performance characteristics, such as stiffness, strength and toughness. Fibre reinforced composites (FRCs) such as carbon fibre have become competitive substitutes for traditional materials, not just because of their weight, but also for the shapes that they can create. Fibre composites can also exhibit good energy absorbing and storage properties.

TOPIC 14 How do forces act on the human body? 37

PROSTHETIC ADVANTAGE Could someone with lower leg prostheses have an advantage over athletes with natural legs? This question was asked when Paralympic athlete Oscar Pistorius applied to compete at the 2012 London Olympics. When running and responding to impact forces, Pistorius’s prosthetic leg deformed, primarily about the curve at the bottom, and absorbed the stress. In an athlete with a natural leg, the stress would also be absorbed but in the joints of their body: ankle, knee, hip and lower back. However, it was alleged that unlike an athlete with natural legs, the carbon-fibre prostheses would rebound off the ground, returning a larger portion of the stored energy, to the advantage of the runner. Is this possible?

FIGURE 14.39 An athlete using a composite prosthetic leg

COMPOSITES Both timber and bone are naturally occurring composites. Timber consists of hollow cellulose fibres bound together in a matrix of lignin. Bone comprises a hard outer layer, cortical, which surrounds and protects a softer inner region, trabecular. Perhaps the earliest artificial composites developed for construction were mud bricks in which straw was mixed with clay. The composite mix reduced cracking that occurred as the bricks dried and shrank. Reinforced concrete is an artificial composite widely used for the past century in the construction industry. Cellular reinforcement, such as honeycombs and foams, is used in the construction of aircraft and train carriages, but most structural composites are a plastic matrix with fibre.

14.6.2 Arm prostheses Like our legs, our arms need to resist large forces when lifting objects or withstanding impacts. Our arms must do this while also enabling the hands to perform delicate and complex tasks such as cleaning our teeth or playing a musical instrument. Prostheses that need to move about a joint might be operated by cables or even small motors to do the job that muscles and tendons in a natural arm would perform. A dynamic hand splint may use external metal cables to transfer forces across the joints of the hand and fingers to assist in the rehabilitation of an injured hand. Traditionally, the body of the splint would be made from plaster. Now it is more likely to be a thermoplastic or fibre reinforced composite such as fibreglass.

14.6.3 Joint replacements Sometimes people need a replacement joint or other internal body part because of a congenital issue, injury, disease such as osteoarthritis, or wear and tear from overuse or ageing. Stainless steel and titanium are used for implants such as hip and knee replacements. These metals are also commonly used in dental work for braces and implants. Stainless steel and titanium are generally biocompatible, hard and can be prepared so that they can integrate into bone. There are different types of hip implants. However, as a ball and socket joint, they typically comprise a tapered metal peg inserted into the head of the femur, to which is attached a ball to facilitate movement of the joint, and a socket and cup fixed to the pelvis (figure 14.40).

38 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 14.40 A disassembled prosthetic socket joint. Metals are well suited for attachment to the femur because their mechanical properties are comparable to healthy bone. Typically the femoral peg is stainless steel or titanium and the other components are mostly made from ceramics or hard polymers. The peg and cup might be mechanically fixed or cemented in place. These parts are often coated with materials such as hydroxylapatite to facilitate the growth of bone and the eventual integration of the replacement part with the existing bone.

As with all joint implants, the surgical insertion of a hip replacement comes with risks and complications. The risks can include fracture of the femur, particularly if the bone is not healthy, and loosening of the peg or cup where they are attached to the bone. As it is subject to frequent and heavy loads, the movement of the replacement hip can also cause wear of the ball and socket. If debris created from wear and tear is not biocompatible, this can cause complications. In Britain in 2010, blood poisoning was reported from certain metal-on-metal implants that had worn and deposited potentially toxic metal into the bloodstream. Ceramic and polymer ball-and-socket joints now provide an alternative wearing surface to earlier metal-on-metal implants. However, these materials also create their own debris from wear and tear, which the body must deal with.

14.6.4 Materials in dentistry and stents The advantages of the ductility of metals is also evident in their use in dental braces and in stents, which are inserted into arteries (figure 14.42). Shaped like a tube, stents are placed in diseased arteries to keep them open and to facilitate blood flow. Usually made of metal mesh, a stent is implanted in the artery. A synthetic balloon already within the stent is then inflated, deforming the stent to its intended size. The balloon is then removed to enable a healthier blood flow.

TOPIC 14 How do forces act on the human body? 39

FIGURE 14.41 X-ray of a hip replacement

The malleability of metals is a useful behaviour that makes them tolerant to compressive forces. Softer metals such as lead and gold can be easily pushed and hammered into shape. Dentists use materials that can be easily shaped to fill cavities in teeth. Once in place, a filling must be resistant to biochemical attack from bodily fluids. Gold has been used for more than a century for filling tooth cavities. It is malleable, but is now used less frequently in part because of its cost. Amalgam (an alloy of mercury, silver, tin and copper) is a more cost-effective mixture of metals, which forms a malleable putty at time of mixing. After shaping in situ by the dentist, the amalgam sets and becomes a hard, strong and stiff filling for tooth cavities. The properties of ceramics have also made them suitable for some dental work including caps. Ceramics are typically hard, resistant to corrosion and can usually withstand high temperatures. Some ceramics are naturally occurring and some are artificial substances. There have been significant advances in the materials used in the manufacture of prostheses, particularly with the development of composite materials for the structural components and electronic systems for controlling movement. The human body is not built to last forever; however, will prostheses ever be as desirable as a natural body part?

40 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 14.42 A stent is inserted, expands into its place and then allows blood to run freely through the artery. Coronary artery

Narrowed artery

Plaque

Plaque

Artery cross-section Closed stent

Catheters

Expanded balloon

Widened artery

Compressed plaque

Expanded stent

Increased blood flow

Widened artery

TOPIC 14 How do forces act on the human body? 41

14.6 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Explain why composite materials are used in so many different types of prosthetics. 2. When a prosthetic needs to move about a joint what is used to replace muscles and tendons? 3. What characteristic of a metal is beneficial when used for braces and stents? 4. The dental implant shown in the following figure comprises a titanium alloy replacement for the Implant tooth root inserted into the jaw. An abutment is inserted then added to which a ceramic crown is fitted. into bone (a) What are the properties of the titanium alloy Abutment and the ceramic tooth that make them suitable for this use in the human body? Crown (b) What factors might affect the success of this implant? 5. To manufacture a 30-millimetre diameter pylon for a below-the-knee limb prostheses, three potential materials are being examined. The mechanical properties of each material are shown in the following table. Lower jaw

Density (kg m−3 )−3

Strength (MPa)

Young’s modulus (GPa)’

Bone

2000

200

20

Material A

4430

970

110

Material B

1400

190

30

Material C

2200

140

20

Material

List and describe the advantages and disadvantages of each of the materials being considered relative to natural bone.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

42 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

14.7 Review 14.7.1 Summary • • • • • • • • • • • • • • • •

A structure is in equilibrium if the net force acting on it is zero and the net torque acting on it is zero. Each part of a stationary structure must be in translational and rotational equilibrium. A force that pulls on something is called a tensile force. A force that pushes on something is called a compressive force. When a force is applied to a material, the material deforms. The amount of deformation depends on the size of the force and the stiffness of the material. When a material is bent, part of it is in compression and part is in tension. Deformation that is reversed when the load is removed is called elastic deformation. Deformation that is not completely reversed when the load is removed is called plastic deformation. Stress, 𝜍, is the force per unit area of cross-section under load. Strain, 𝜀, is the ratio of the change in length, ∆l, to the original length of the material under load, l. When the stress–strain curve is a straight line, the gradient is called Young’s modulus, E. The relationship between stress and strain is an expression of Hooke’s Law and is written in the form 𝜍 = E × 𝜀. Young’s modulus is a measure of the stiffness of a material. The strain energy of a material is equal to the work done to deform it. The area under the stress–strain curve measures the strain energy per unit volume of material. Toughness is a measure of the energy required to fracture a material and is determined from the total area under the stress–strain curve up to the fracture point. Brittle materials show little or no plastic deformation before breaking. Ductile materials show plastic deformation before breaking. Composite materials are made by combining two or more different materials to create a single material with enhanced characteristics.

Resources

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0039).

14.7.2 Key terms A brittle material experiences little or no plastic deformation before it fractures. The centre of gravity is the point at which gravity acts on an object. The centre of mass of an object is the average position of the object’s mass. We often consider forces acting through an object’s centre of mass. Creep is the time-dependent and permanent deformation of a material under a constant stress. A ductile material deforms plastically before it fractures. Deformation that disappears when stress is removed is called elastic deformation. Necking is a noticeable thinning in a material that occurs when it is under tension. Deformation that does not completely recover when stress is removed is called plastic deformation. Stiffness is a measure of how much work is required to deform a material. Strain is the amount by which a material is deformed by a force. Strain energy is the energy stored in a material that is deformed. It is equal to the work done to deform a material. Stress is a measure of the strength of a material. It is defined as the force per unit area. Torque is the turning effect of a force. Toughness describes the ability of a material to store energy up to the point of fracture. A material that holds a large amount of strain energy before fracturing is said to be tough. Young’s modulus, E, is the ratio of stress to strain in a material.

TOPIC 14 How do forces act on the human body? 43

Resources Digital document Key terms glossary (doc-32286)

14.7 Exercises 14.7 Exercise 1: Multiple choice questions 1.

2.

3.

4.

5.

6.

7.

What is the normal reaction force supplied by the ground to an 80-kilogram person standing still? A. 80 N B. 80 Kg C. 784 N D. 712 N A bricklayer is supporting several bricks on his head as he is hanging by his arms from a construction frame. Which of the following statements is correct? A. His neck joint is in compression; his arm bones are in compression. B. His neck joint is in compression; his arm bones are in tension. C. His neck joint is in tension; his arm bones are in compression. D. His neck joint is in tension; his arm bones are in tension. A basketballer slides slightly before coming to a stop once their shoes grab the court. Which of the following statements best describes the type of deformation the rubber sole of the shoe undergoes? A. The particles of the rubber sole will undergo shear. B. The particles of the rubber sole will undergo tension. C. The particles of the rubber sole will undergo compression. D. The particles of the rubber sole will undergo rotation. A bone with a cross-section of radius 1.2 centimetres is supporting a load of 420 N. Which of the following values is the best estimate of the stress in the bone? A. 9.28 × 104 Pa B. 9.28 × 105 Pa C. 9.28 × 101 Pa D. 1.1 × 104 Pa A prosthetic leg is made from aluminium of length 50 centimetres and a diameter of 40 millimetres. If this prosthetic leg is placed under tension with stress of 350 MPa, what force was applied to the leg to provide this stress? A. 1.8 × 106 N B. 1.8 N C. 4.4 × 103 N D. 4.4 × 105 N A tendon of length 12 centimetres stretches 8 millimetres under load. Which of the following values is the best estimate of the strain in the tendon? A. 66.7 B. 0.67 C. 0.07 D. 15 If a material is stretched sufficiently, it does not return to its original length when the stress is removed. Which of the following statements does not correctly describe the behaviour of this material? A. It has become permanently deformed. B. It has shown ductile failure. C. It has exceeded its elastic limit. D. It has shown plastic behaviour.

44 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

The following stress–strain graph represents two polymers. The full scale is not shown. Which statement concerning the properties of the polymers is true? A. Acrylic is tougher than polyethylene. B. Acrylic is more brittle than polyethylene. C. Acrylic is more flexible than polyethylene. D. Acrylic breaks more easily than polyethylene.

* Acrylic

60 Stress (MPa)

8.

40

Polyethylene 20

* 0

2

4

6

8

10

500

Strain (%)

A test specimen is stretched to the point P, as shown on the following stress–strain graph, and then released. Which value is the best estimate of the permanent strain on the test specimen? A. 4.0 × 10−2 B. 2.0 × 10−2 C. 1.5 × 10−2 D. Zero

P

5

Stress (x 106 Pa)

9.

4

3

2

1

0

10.

1

2

3 4 Strain (x 10–2)

5

A test specimen, with Young’s modulus equal to 2.0 × 1010 Pa, is deformed within its elastic region, up to a strain of 0.50%. Which of the following values is the best estimate of the energy stored, per cubic metre of specimen? −3 A. 2.5 × 105 Jm −3 B. 5.0 × 107 Jm −3 8 C. 1.0 × 10 Jm −3 D. 2.5 × 109 Jm

14.7 Exercise 2: Short answer questions A 30-kilogram student is hanging from the monkey bars at school with each arm making an angle of 70° from the horizontal monkey bar. Calculate the tension in each arm. (Assume both arms are straight.) 2. A person is lifting a 15 kilogram weight at the gym. When the arm is 30° from the horizontal, calculate the torque.

1.

Weight cm

35 30o

B Elbow

TOPIC 14 How do forces act on the human body? 45

3.

4.

5.

6.

7.

8.

A stress–strain graph of a sample material in the elastic region has a strain of 0.6% when a stress of 180 MPa is applied. a. Calculate Young’s modulus. b. Calculate the energy stored in the sample material. a. Determine the loads that each of the following is most likely to experience: tension or compression. i. Strings of a tennis racquet ii. Human bone iii. A lift cable iv. Bicycle tube v. Human tendon vi. Sole of a shoe vii. Kitchen cling wrap viii. The leg of a table A dental technician is stretching a 40-millimetres long piece of stainless steel wire 1.2 millimetres in diameter. Stainless steel has the following mechanical properties: • E = 195 GPa • yield strength (stress applied before permanent deformation) = 215 MPa • tensile strength (stress applied before failure occurs) = 505 MPa. a. What is the minimum force needed to deform the wire permanently? b. How much must the wire be stretched to create a permanent change in length? c. What is the minimum force required to break the wire? The data sheet provided by a manufacturer for its fibre product stated that it had a Young’s modulus of 75 GPa and an ultimate strength of 90 MPa. If the stress–strain curve for the fibre is linear, by how much would a 1 metre long fibre elongate before fracture? A 0.5 metre long piece of wire stretched 0.4 millimetres when a 6 kN force was applied. The wire had a diameter of 3 millimetres. Assuming that it was behaving in a linear elastic manner, what was: a. the maximum stress b. the maximum strain c. the Young’s modulus d. the strain energy e. the total energy absorbed by the wire? An aluminium bar with diameter of 12.5 millimetres was tested in tension. The bar fractured at a length of 55.13 millimetres. The following table shows the data collected up to the point of fracture. a. Convert the data to stress versus strain. Load (kN) Length (mm) b. Plot the stress–strain graph. 0 50.00 c. On the graph, label: i. elastic limit 4.5 50.02 ii. breaking strength. 13.4 50.07 d. Calculate: i. elastic limit 22.3 50.13 ii. breaking strength.

46 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

31.2

50.18

33.4

50.75

35.2

52.00

35.7

53.00

35.7

54.00

33.8

55.13

9.

The stress–strain relationship for skin is shown in the following graph. XH σ

0

ɛ

Describe how the skin will behave when it is acted on by a force from zero to H. The stress–strain relationship for another material, A, is shown in the following graph. a.

σ X

H

0

ɛ

If material A was used to make synthetic skin, what would happen as it was loaded from zero to a stress H? c. Would you consider this an adequate substitute for skin?

b.

14.7 Exercise 3: Exam practice questions Question 1 (3 marks) The human jaw acts as a lever when you eat, hinging at the temporomandibular joint. If the force needed to bite a mint is 180 N, what force is required from the muscles shown in the following figure?

Temporomandibular joint 180 N

80 mm

25 mm

Question 2 (3 marks) At its mid-length, Zali’s femur is approximately circular in cross-section with a diameter of 18 millimetres. If her mass is 66 kilograms, calculate the stress on her femur. TOPIC 14 How do forces act on the human body? 47

Question 3 (4 marks) Which material on the following stress–strain graph: a. has the greatest Young’s modulus 1 mark b. is more ductile 1 mark c. is tougher 1 mark d. is stronger? 1 mark

×A

Stress

×B

0

Strain Question 4 (3 marks) Brad, who is 75 kilograms, is hanging from the gymnastic horizontal bar before his routine. Calculate the tension in each of his arms.

60º

60º

T

T

Question 5 (2 marks) Use the stress–strain graph to calculate the value of the following ratio. Young’s modulus of material A Young’s modulus of material B

Tensile Stress (x 106 N m–2)

mg

300

Material A

×

200

× 100

0

0.01

Material B

0.02 Strain

48 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

0.03

14.7 Exercise 4: studyON topic test Fully worked solutions and sample responses are available in your digital formats.

Test maker Create unique tests and exams from our extensive range of questions, including practice exam questions. Access the assignments section in learnON to begin creating and assigning assessments to students.

TOPIC 14 How do forces act on the human body? 49

AREA OF STUDY 2 OPTIONS OBSERVATION OF THE PHYSICAL WORLD

15

How can AC electricity charge a DC device? 15.1 Overview Numerous videos and interactivities are available just where you need them, at the point of learning, in your digital formats, learnON and eBookPLUS at www.jacplus.com.au.

15.1.1 Introduction We live in a world surrounded by electronic devices. Many of us feel the need to be connected constantly to the internet and the telephone network. To be connected we need mobile communication devices such as smart phones, tablets and laptops. These devices need batteries to provide electrical energy. Rather than using batteries that are thrown away and replaced, these devices are fitted with rechargeable batteries that need to be recharged from the mains power network. The mains power network supplies electricity to houses and buildings at 230 V AC (previously, in Australia, this was 240 V AC). Batteries are charged by providing a reverse voltage of 6 V DC. This topic looks at how to convert a 230 V AC voltage to a DC voltage to charge your devices and keep you connected. It also looks at the use of transducers to transfer energy in long distance telecommunications systems. In this topic you will investigate the processes involved in transforming the alternating current delivered by the electrical supplier into low-voltage direct current for use with small-current electrical devices. You will investigate a variety of circuits to explore processes including transformation, rectification, smoothing and regulation. You will then investigate how data is transferred from one point to another using optical fibre networks. FIGURE 15.1 Smartphones are charged using a reverse voltage of 6 V DC.

TOPIC 15 How can AC electricity charge a DC device? 1

This involves transducer circuits that: • transform energy, such as heat and light, into electrical signals • transform electrical signals into light signals that can be transmitted along optical fibres • transform light signals back into electrical signals. In this topic, you will design, build and test a low-voltage AC to DC regulated voltage power supply system that could be used to charge a device. To help you achieve this goal, you will study theory relating to: • using diagnostic equipment such as a cathode ray oscilloscope (CRO) and a multimeter • converting an AC voltage signal to a DC voltage signal (voltage rectification) using diodes • smoothing the output of a rectifier using a capacitor • selecting and using a voltage regulator that will give a constant voltage output. (The voltage rectifier will do this even though there are changes to the voltage provided by the rectifier and there are changes to the load.)

15.1.2 What you will learn KEY KNOWLEDGE After completing this topic you will be able to: • analyse the role of the transformer in the power supply system including the analysis of voltage ratio: N1 V1 = (not including induction or its internal workings) N2 V2 • explain the use of diodes in half-wave and full-wave bridge rectification • explain the effect of capacitors with reference to voltage drop and current change when charging and discharging (time constant for charging and discharging, 𝜏 = RC) leading to smoothing for DC power supplies • describe the use of voltage regulators including Zener diodes and integrated circuits • analyse systems, including fault diagnosis, following selection and use of appropriate test equipment • interpret a display on an oscilloscope with reference to voltage as a function of time. • apply the use of heat and light sensors such as thermistors and light-dependent resistors (LDRs) to trigger an output device such as lighting or a motor • evaluate the use of circuits for particular purposes using technical specifications related to potential difference (voltage drop), current, resistance, power, temperature and illumination • compare different light sources (bulbs, LEDs, lasers) for their suitability for data transfer • explain the use of optical fibres for short and long distance telecommunications. Source: VCE Physics Study Design (2016–2021) extracts © VCAA; reproduced by permission.

PRACTICAL WORK AND INVESTIGATIONS Practical work is a central component of learning and assessment. Experiments and investigations, supported by a Practical investigation logbook and Teacher-led videos, are included in this topic to provide opportunities to undertake investigations and communicate findings.

Resources Digital documents Key science skills — VCE Units 1–4 (doc-31856) Key terms glossary (doc-32288) Practical investigation logbook (doc-32289)

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (doc-sonr-0040).

2 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

15.2 Voltage, current and resistance KEY CONCEPTS • Interpret a display on an oscilloscope with reference to voltage as a function of time. • Analyse the role of the transformer in the power supply system including the analysis of voltage ratio: N1 V1 = (not including induction or its internal workings). N2 V2

15.2.1 AC voltage and current An AC current is one that periodically changes direction. Figure 15.2 shows a graph of how voltage varies with time for an AC voltage signal. Note that the voltage is sinusoidal, which means that the graph has the same form as a sine curve. FIGURE 15.2 The variation of voltage with time for a sinusoidal AC signal

Vpeak VRMS Voltage

Vpeak

Vp–p

2T

T

t

–Vpeak

An AC signal is described in terms of: • period, T ( ) 1 • frequency, f f = T amplitude. If the voltage cycle is centred on 0 V, the voltage amplitude is Vpeak . Similarly, if the • current cycle is centred on 0 A, the current amplitude is Ipeak • VRMS . This is the value of the DC voltage that generates the same power as the AC voltage when placed across a resistor. Vpeak VRMS = √ 2 Similarly, IRMS is the value of the DC current that generates the same power when it flows through a resistor. Ipeak IRMS = √ 2

The maximum variations in the voltage and current are called the peak-to-peak voltage, Vp-p , and the peak-to-peak current, Ip-p , respectively. TOPIC 15 How can AC electricity charge a DC device? 3

Figure 15.3 shows how the current through a resistor varies with time when an AC voltage is placed across it. FIGURE 15.3 The variation of current with time for a sinusoidal AC signal

Ipeak IRMS Ip–p

Current

Ipeak

2T

T

t

–Ipeak

15.2.2 Cathode ray oscilloscope A cathode ray oscilloscope (CRO) is an instrument that shows how voltage drop across a device varies with time. A cathode ray consists of a beam of electrons that strike a phosphorescent screen at the front of the instrument, creating a spot. The beam can be made to sweep across the screen at different speeds and the voltage across the device affects the vertical position of the beam. By adjusting the controls, a picture, called a trace, of how voltage varies with time is displayed. The trace is identical to a graph where the y-axis gives the voltage and the x-axis gives time. The scales for the x- and y-axes may be altered using the controls. This is explained in the next section.

How to use a cathode ray oscilloscope Figure 15.4 shows (a) the basic controls of a single trace CRO and (b) how to connect it to a device — in this case a signal generator that you may use when studying sound. The ground terminal of the CRO is held at 0 V. The other terminal (marked ‘input’) samples the voltage of the signal generator output. Before connecting the CRO to the device, turn it on and use the focus and intensity knobs to adjust the trace. The horizontal and vertical deflection knobs are used to position the trace at the centre of the screen. The screen has a centimetre grid marked on it to assist in centring the trace and in interpreting the values that the trace represents.

Time scale

FIGURE 15.4 (a) The front panel of a single trace CRO (b) Connecting a signal source to a CRO (a)

Focus knob Intensity knob Screen

Locate Horizontal deflection knob Vertical deflection knob

Timebase control on–off switch Input Trigger Vertical amplifier (b) on–off

• x1 • x10 • x100 • x1000

Frequency

CRO

Output

Ground

Ground (0 V) Input The timebase control is adjusted so that at least one complete waveform is displayed on the screen. This enables the period of the signal to be read. The period is the time it takes for the signal to complete one cycle. Measure the length of one complete cycle of the trace from the screen and multiply this value by the factor indicated on the timebase control.

4 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Voltage scale The vertical scale of the trace is set by the vertical amplifier. This should be set so that the top and bottom of the trace fall within the screen grid. The amplitude of the trace is then multiplied by the factor indicated by the vertical amplifier control.

Dual trace CROs A dual trace CRO has two inputs, each with separate vertical deflection controls, that enable the signal at different parts of a circuit to be compared. SAMPLE PROBLEM 1

What are the period, frequency and amplitude of the trace shown if the following timebase and voltage settings are used?

Timebase 5 ms cm−1 , voltage 1 V cm−1 b. Timebase 0.1 ms cm−1 , voltage 5 mV cm−1 a.

THINK a. 1.

Each line on the CRO represents 1 cm.

2.

Period is four lines in length.

3.

Frequency is the inverse of period.

4.

Amplitude is two lines high.

b. 1.

2.

Each line on the CRO represents 1 cm.

Period is four lines in length.

WRITE

Each line represents 5 ms on the horizontal axis and 1 V on the vertical axis. T = 4 × 5 = 20 ms The period is 20 milliseconds. 1 f= 2.0 × 10−2 s = 50 Hz The frequency is 50 Hz. A = 2 × 1 = 2V The amplitude is 2 V. b. Each line represents 0.1 ms on the horizontal axis and 5 mV on the vertical axis. T = 4 × 0.1 = 0.4 ms The period is 0.4 miliseconds. a.

TOPIC 15 How can AC electricity charge a DC device? 5

3.

Frequency is the inverse of period.

4.

Amplitude is two lines high.

1 4 × 10−4 s = 2.5 × 103 Hz The frequency is 2.5 × 103 Hz. A = 2 × 5 mV = 10 mV The amplitude is 10 mV. f=

PRACTICE PROBLEM 1 With a time base and voltage setting of 2 ms cm–1 and 5 mV cm–1 respectively, what are the period, frequency and amplitude for each of the following traces? a.

b.

15.2.3 Using a multimeter The cathode ray oscilloscope discussed in the preceding section is a FIGURE 15.5 Switched benchtop diagnostic tool. It is a good device for measuring the period and range multimeter amplitude of AC signals, or the output of voltage rectifiers. A multimeter is a hand-held diagnostic device. It is useful for analysing DC circuits that you might construct. Multimeters have a digital display of the quantities being measured. A multimeter can be used as an ammeter or a voltmeter in either DC or AC circuits, or as an ohmmeter. Figure 15.5 shows the front of a typical multimeter. This type is known as a ‘switched range multimeter’ because you can select its function and range of readout by rotating the central control dial. Always switch off the circuit’s power supply before connecting a multimeter to a circuit. You must never use a multimeter to measure mains voltages. This will blow the fuse, damage other internal circuitry and could injure or kill you. Choose the voltage, current or resistance you wish to measure with the selection knob. Estimate the magnitude of the reading you expect to find and select the appropriate range. If you have absolutely no idea what the reading might be, start on the greatest range — for example, the 10 A range for current. This will reduce the risk of damaging the multimeter. Figure 15.6a shows a simple DC circuit consisting of a power source, a switch and two resistors. We will now consider how to use a multimeter to analyse this circuit. When using a multimeter as an ammeter to measure current through a component or branch of a circuit, the ammeter must be part of the circuit. The ammeter must therefore be connected in series with the circuit components whose current is to be measured. It is necessary to break the circuit at an appropriate point and connect the meter, as shown in figure 15.6b. In this mode, the multimeter has very little resistance so that it 6 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

has a minimal effect on the current flowing through that branch of the circuit. The resistance of the ammeter will cause a slight reduction in the current. FIGURE 15.6 (a) The original circuit. Connecting the multimeter as (b) an ammeter, (c) a voltmeter and (d) an ohmmeter (a)

(d)

(c)

(b) R1

R1

R1

R2

R2

R2

R1 Ω V

R2

A

If there is no current detected flowing through the circuit branch you are testing check for breaks, loose connections and open switches. If there is still no current detected after doing this try using a lower range on the multimeter. When using a multimeter as a voltmeter to measure the voltage drop between points in the circuit, the voltmeter is connected across the two points (see figure 15.6c). If measuring the voltage drop across a component, the multimeter is connected in parallel with the component. In this mode the multimeter has a very high resistance, of the order of half a million ohms. A negligible current will still flow through the multimeter. When using a multimeter to measure current or voltage in a DC circuit, connect the common terminal of the multimeter to the part of the circuit closest to the ground or negative terminal of the power supply. The ohmmeter is not used when the circuit is connected to the power supply. It uses the internal battery of the meter to send a current through the desired part of the circuit and shows the resistance by analysing the current (see figure 15.6d). If the ohmmeter indicates that there is an infinite resistance in that part of the circuit, it means there is either a break in the circuit, a faulty connection or that a switch is open. Ohmmeters are difficult to use with circuits that have already been constructed because it is not always apparent when other components are connected in parallel with the component being tested.

Transformers A transformer is a device that increases or decreases the voltage from an AC supply. A transformer consists of two coils of wire linked by a soft iron core. The two coils have different numbers of turns of wire wrapped around the core. If there are more turns of wire on the input side than the output side of the transformer, the input voltage will be greater than the output voltage. This type of transformer is called a step-down transformer. The input and output voltages (V1 and V2 ) are related to the number of turns of wire on the input and output sides (N1 and N2 ) by the relationship:

FIGURE 15.7 Circuit symbol for a transformer

N1 V = 1 N2 V2 Transformers operate using AC voltages. The input and output voltages are alternating currents.

TOPIC 15 How can AC electricity charge a DC device? 7

SAMPLE PROBLEM 2

A transformer is used to step down a 230 V voltage to 6 V. What is the ratio of the input turns to output turns of wire in this transformer? THINK 1.

WRITE

V1 = 230

The initial voltage is 230 V and the final voltage is 6 V.

V2 = 6

Recall the relationship between the number of turns of wire on the input and output sides with the input and output voltages. 3. Substitute the voltages to find the ratio of turns in the wires.

2.

4.

N1 V1 = N2 V2 N1 230 = = 38.3̇3̇ N2 6 The ratio of the input turns to output turns of wire in this transformer is 38.3̇3.̇

State the solution.

PRACTICE PROBLEM 2 What is the ratio of input turns to output turns on a transformer used to step down a voltage of 110 V to 2.5 V?

15.2 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. The variation of voltage drop with time for a 100 Ω resistor is shown here.

Voltage drop (V)

5

t (ms)

0 1

2

–5

What is the value of the following quantities? (a) Vpeak (b) Vp-p (c) VRMS (d) T (e) f (f) IRMS

8 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

3

4

5

6

7

8

9

2. The following trace was obtained on a CRO with the voltage scale set at 1 V cm–1 and the timebase set at 1 s cm–1 .

Calculate the: (i) period (ii) frequency (iii) peak voltage (iv) RMS voltage. 3. A signal generator is connected to a CRO as shown in the following figure. The CRO trace obtained is also shown. The timebase setting is 5 ms cm−1 and the voltage setting is 1 V cm−1 . (a)

(b)

x1 x10 x100 x1000

Signal generator

Input ground CRO

(a) What are the values of the following quantities? i. T ii. f iii. Vpeak iv. VRMS

TOPIC 15 How can AC electricity charge a DC device? 9

(b) Sketch the CRO trace obtained if a 1.5 V cell is connected between the signal generator and the CRO as shown in the following figure.

x1 x10 x100 x1000

Signal generator

Input ground CRO

(c) Sketch the CRO trace obtained if the cell in part (b) were reversed. 4. (a) Describe how you would use a multimeter as a voltmeter. (b) Why must you never use a hand-held multimeter to measure the mains voltage? 5. For a transformer, the ratio: number of turns on the transformer secondary number of turns on the transformer primary

=

21 1200

If the input voltage is 120 V AC, what is the output voltage? 6. A transformer is used to provide an AC voltage of 6 V from a mains supply of 230 V AC. What is the value of the following ratio? number of turns on the transformer secondary number of turns on the transformer primary

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

10 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

15.3 Converting AC to DC KEY CONCEPTS • Explain the use of diodes in half-wave and full-wave bridge rectification. • Explain the effect of capacitors with reference to voltage drop and current change when charging and discharging (time constant for charging and discharging, 𝜏 = RC) leading to smoothing for DC power supplies.

Many electronic devices such as tablets and mobile phones require a DC supply to operate. Electric power is usually supplied to buildings in AC form. It is therefore necessary to convert or ‘rectify’ the voltage supplied to the electronic circuits inside many appliances from AC to DC. Many portable digital radios operate using four 1.5 V cells in series as a power supply. This means that the circuits inside the radio are designed to run from a 6 V supply. When the player is plugged into the mains, the voltage must be converted from 230 VRMS AC to 6 V DC. The first step in such a conversion is to use a transformer to reduce the AC voltage to a suitable level. At this stage the current is still AC. What is needed next is a device that allows current to pass through it in only one direction. Such a device is called a diode.

15.3.1 Diodes A diode is a device that allows current to flow through it in only one direction. Diodes act like a valve in a car tyre. They effectively provide little resistance to a current flowing in one direction, but a very large resistance to a current flowing in the other direction. The circuit symbol for a diode is shown in figure 15.8a, with different types of diode commonly used in electronics shown in figure 15.8b. Diodes are usually made from two semi-conductor materials fused together as two layers, the most common materials being silicon and germanium that have been doped with atomic impurities. The doping of these materials facilitates the movement of charge in one direction. Diodes have a positive connection, the anode, and a negative connection, the cathode. The anode is connected to a p-type semi-conductor and the cathode to a n-type semi-conductor. FIGURE 15.8 (a) Circuit symbol for a diode (b) Different types of diode (b)

(a) Bar

Anode

Cathode

A diode conducts current in the direction indicated by the arrow in the circuit diagram. The bar indicates that current will not flow in the opposite direction. The ends of a diode are known as the cathode and the anode. Current flows from the anode to the cathode. Diodes are marked with a band around the cathode. This corresponds to the bar on the circuit symbol. The characteristic curve of a device is a graph showing how a particular device behaves electrically. The characteristic curve for a typical silicon diode is shown in figure 15.9.

TOPIC 15 How can AC electricity charge a DC device? 11

Note that the scales on the graph are different on the FIGURE 15.9 The current-versus-voltage positive and negative sides of both axes. When the voltage characteristic of a typical silicon diode across the diode is positive, the diode is said to be forward Current (mA) biased. Before a silicon diode starts to conduct, or behave like it has low resistance, it has to have a voltage of about 20 0.7 V across it. When a negative voltage is placed across a diode, it is 10 said to be reverse biased; a small leakage current of only Voltage (volts) Reverse voltage (volts) a few microamperes will flow. If the negative voltage is 0 –100 –50 0.5 1.0 large enough, the semiconductor will break down and a –100 large current will flow. Diodes are usually placed in series with other circuit –200 elements such as resistors. Reverse current (μA) In the circuit shown in figure 15.10a, when the Note: The scales are different on both sides of both axes. voltage from the supply is increased from 0 V to 0.6 V, little current flows and the voltage drop is almost entirely across the diode because it has a very large resistance. As the voltage of the supply is further increased, the voltage drop across the diode stays at about 0.7 V, and the rest of the voltage drop is across the resistor. The size of the current through the circuit is then largely determined by the resistor. Note that when a diode is reverse biased, as shown in figure 15.10b, very little current flows through the diode and the resistor. The voltage drop across the resistor will be negligible and all the voltage drop will be across the diode. Diodes have ratings for voltage and current that should be observed to avoid damage to the diode and other elements in a circuit. Three special types of diodes, Zener, LED (light emitting diodes) and photodiodes will be dealt with later in the topic. FIGURE 15.10 (a) A forward-biased diode connected in series with a variable voltage supply and a resistor.(b) The voltage drop across a reverse-biased diode (b)

(a)

6.0 V R

R 6.0 V

0V

SAMPLE PROBLEM 3

A diode and resistor are connected in series with a power supply. If the emf of the supply is 10 V, and the value of the resistor is 100 Ω, estimate the current flowing through the diode. THINK 1.

The diode can be assumed to have a voltage drop of about 0.7 V across it.

12 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

WRITE

Voltage drop across the resistor: 10 − 0.7 = 9.3 V

2.

The current through the resistor can be found using Ohm’s Law.

3.

State the solution.

V R 9.3 = 100 = 0.093 A, or 93 mA I = 93 mA As the diode and resistor are in series, the current through the diode is also 93 mA. I=

PRACTICE PROBLEM 3 A forward-biased diode and a 600 Ω resistor are connected in series with an 18 V power supply. Estimate the current flowing through the diode.

15.3.2 Capacitors Capacitors are devices comprising parallel metal plates that store FIGURE 15.11 An RC circuit electric energy. They are used to produce a DC voltage and deliver a more constant voltage when there is a variable output. Switch The value of a capacitor, it’s capacitance, is measured in farads, where 1 farad equates to one coulomb of charge being stored by one volt, 1 F = 1 C V−1 . As the farad is a large unit of measure, R smaller units such as mF and nF are typically used. E It is important to understand what happens when capacitors charge and discharge, something we will do by looking at an RC C circuit, one that contains a resistor and capacitor in series. When the switch in the RC circuit is closed, a current flows through the circuit. As charge builds up on the capacitor, the current decreases until the voltage across the plates is equal to the emf of the cell. Remember that a current flows from a high voltage point to a low voltage point. When the voltages of the cell and the capacitor are equal, the current ceases. The magnitude of the current is affected by the FIGURE 15.12 How voltage varies with time resistance of the resistor. The bigger the resistance, the when charging a capacitor smaller the current and the slower the capacitor will Voltage (V ) across capacitor charge. E The amount of time it takes to fully charge the capacitor depends on the capacitance of the capacitor and the resistance of the resistor. The greater the capacitance, 0.63 E the more charge can be stored on the plates per unit 0.5 E voltage across the plates. Figure 15.12 shows how the voltage across the capacitor varies with time. The time constant, 𝜏 or t, for an RC circuit is the time in seconds it takes for the capacitor to reach 63% (or approximately two-thirds) of its final voltage, E, when t 1𝜏 3𝜏 5𝜏 2𝜏 0 4𝜏 6𝜏 charging. The time constant can be calculated by finding the product of the resistance in ohms and capacitance in farads for the RC circuit. 𝜏 = RC

TOPIC 15 How can AC electricity charge a DC device? 13

When a voltage source is removed from a fully charged RC circuit the capacitor will discharge back through the resistor. The circuit shown in figure 15.13a will discharge the capacitor when the switch is closed. The way voltage varies with time for discharging the capacitor is shown in figure 15.13b. In this case, the time constant is the amount of time it takes for the voltage across the capacitor to fall to 37% of its original value. A capacitor is considered to be fully charged or discharged after five time constants have elapsed. FIGURE 15.13 (a) A circuit for discharging a capacitor (b) How voltage varies with time when discharging a capacitor (a)

(b)

Switch

V0

C

R 0.37 V0

𝜏 = RC

t

AS A MATTER OF FACT An electronic heart pacemaker is an example of an RC circuit. Pacemakers provide regular voltage pulses that start and control the frequency of the heartbeat. They can be worn externally or implanted beneath the skin. Pacemakers contain a resistor and a capacitor with electrodes implanted in or near the heart. The charge on the capacitor builds up and then discharges through the electrodes. Then it starts to build up again. The time between pulses depends on the values of R and C.

SAMPLE PROBLEM 4

A camera flash charging circuit consists of a 100 𝜇F capacitor in series with a 220 kΩ resistor as shown.

C = 100 μF E = 1.5 V

R = 220 kΩ

The emf of the cell is 1.5 V. Approximately how long will it take for the voltage across the capacitor to reach 1 V? THINK

WRITE

The time constant for a circuit is equal to: resistance × capacitance, 𝜏 = RC 2. Referring to figure 15.13b, one time constant is the time it takes to the capacitor to reach approximately two-thirds of the final voltage.

𝜏 = 220 × 103 × 100 × 10−6 = 22 seconds

1.

3.

One volt is two-thirds of the final voltage (1.5 V).

14 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

1.5 ×

2 = 1V 3

4.

State the solution.

The capacitor will reach 1 volt at approximately one time constant, that is, 22 seconds.

PRACTICE PROBLEM 4 How long will it take the RC circuit in Sample problem 4 to fully charge the capacitor to 1.5 V?

Resources Digital document eModelling: Charging a capacitor (doc-0039)

15.3.3 Half-wave rectification A rectifier is a circuit element that is connected to the output of a transformer and converts the AC voltage into a DC voltage. The AC voltage from the transformer gives a sine wave when graphed against time. A half-wave rectifier uses a diode to pass only one direction of the AC voltage from the transformer onto another part of the device that is sometimes called the load. The remaining part of the voltage signal is across the diode, since the diode has a very large resistance when reverse biased. A half-wave rectifier circuit is shown in figure 15.14a, and how the input voltage varies with time is shown in figure 15.14b. Since the load has an effective resistance, it is represented as a resistor. FIGURE 15.14 (a) A half-wave rectifier circuit (b) How the input voltage varies with time (a)

(b)

Vin (V) 9

Vin (from a transformer)

Rload

t (ms) 0 20

40

60

–9

If the input voltage from the transformer has a peak voltage of +9 V, the maximum voltage drop across the load resistance will be +8.3 V, because silicon-based diodes have a voltage drop of 0.7 V when they are forward biased. When the input voltage is negative, the diode effectively prevents current from flowing through the load resistance and the voltage drop across the load will be zero. The load now has only a positive voltage drop across it and the current, which is now said to be clipped, flows through the load in only one direction. A DC voltage has been achieved but this voltage is far from being steady. This rectifier is called a half-wave rectifier because only one-half of the sine wave has been allowed to pass through it.

Capacitor smoothing Most electronic circuits require a steady DC voltage to operate effectively. The process of converting the DC voltage into a steady DC voltage is called smoothing. Smoothing is achieved by placing a capacitor in parallel with the load resistance as shown in figure 15.15. When the positive part of the transformer voltage passes through the diode, it produces a voltage drop across the load resistance and charges the capacitor. When the diode blocks the negative part of the transformer voltage, the capacitor discharges through the load. The voltage drop across the load is shown in figure 15.16.

FIGURE 15.15 A halfwave rectifier with a smoothing capacitor

Vin

+ _

C

Rload

TOPIC 15 How can AC electricity charge a DC device? 15

The voltage drop shown in figure 15.16 is still not constant. The variations in voltage are called ripples. The size of the ripples depends on the capacitance of the capacitor, the load resistance and the frequency of the waveform from the transformer. The bigger the time constant when the capacitor is discharging through the load, the smoother the voltage drop across the load. It is for this reason that good quality DC power supplies require smoothing capacitors that have large capacitance. FIGURE 15.16 The voltage drop across the load for a half-wave rectifier with a smoothing capacitor Voltage drop across load

Voltage drop (V)

8.3

0

Time (s) Discharging

Charging

Resources Digital document Investigation 15.1 Constructing a half-wave rectifier (doc-31889)

15.3.4 Full-wave rectification A half-wave rectifier makes use of only half the input AC current or voltage. A full-wave rectifier makes use of both half-cycles of an AC waveform to supply a current that flows in one direction through a load.

Bridge rectifier A bridge rectifier uses four diodes to produce full-wave rectification. Figure 15.17 shows a circuit diagram for a bridge rectifier with input and output signals. FIGURE 15.17 A bridge rectifier Vout

Vout

Vin

0

0

t

Vin

Figure 15.18 shows the pathway taken by the current through the load of a bridge rectifier. Remember that there will be an approximate voltage drop of 0.7 V across each diode that the current passes through. Each half of the AC cycle passes through two diodes in a bridge rectifier. This means that 16 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

the peak voltage from a bridge rectifier will be approximately 1.4 V less than the peak voltage of the AC signal. FIGURE 15.18 Direction of current through the load of a bridge rectifier when the AC current (e.g. from the transformer) is (a) positive and (b) negative (a)

Vout +

Vin

RL

–

(b)

Vout –

Vin

RL

+

Capacitor smoothing Full-wave rectifier output signals can be smoothed in the same way as half-wave rectifier signals (see figure 15.19a). The main difference is that for a given RC combination, the capacitor has less time to discharge before it is recharged by the output signal, as shown in figure 15.19b. FIGURE 15.19 Capacitor smoothing of a bridge rectifier: (a) circuit diagram and (b) comparison of signals (a) VL

C

Vout RL

Vin

(b)

Vin (V)

VL (V)

VL (V) 8.6 V

8.6 V 10 8 6 4 2 0 –2 –4 –6 –8 –10

10 8 6 4 2

10 8 6 4 2 t Input signal (without capacitor)

0 –2 –4 –6 –8 –10

t Rectified signal (without capacitor)

Smoothed signal

0

TOPIC 15 How can AC electricity charge a DC device? 17

t

Centre-tap full-wave rectifier The centre-tap full-wave rectifier uses only two diodes. The diodes are connected to either end of the secondary coil of a transformer. The centre of the coil is connected (tapped) to the earth. The arrangement is shown in the figure 15.20. In this case there is also an approximate voltage drop of 0.7 V across each diode that the current passes through. Each half of the AC cycle passes through only one diode in the centre-tap rectifier. This means that the peak voltage from a centre-tap rectifier will be approximately 0.7 V less than the peak voltage of the AC signal. The output of a centre-tap rectifier can also be smoothed using a capacitor, as shown in figure 15.21.

FIGURE 15.20 Centre-tap rectifier circuit diagram

RL

FIGURE 15.21 Capacitor smoothing of a centre-tapped rectifier: (a) circuit diagram and (b) comparison of signals (b)

(a)

Input signal

VL

Vin (V)

10 8 6 4 2 0 –2 –4 –6 –8 –10

VL (V)

VL (V) with capacitor

10 8 6 4 2

9.3 V

0

t Input signal (without capacitor)

t Rectified signal (without capacitor)

0

Smoothed signal t

Resources Weblink Rectifiers

15.3.5 Ripple voltage Ripple voltage is the periodic variation in a DC voltage that results from the rectification of an AC voltage. The peak-to-peak value of the ripple voltage is: Vr(p−p) = Vmax − Vmin

18 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

The value of the peak-to-peak ripple voltage depends on the following factors: • the maximum value of the voltage (Vmax ) supplied by the diode(s) of the rectifier • the period (T) of the unsmoothed voltage supplied by the diode(s) of the rectifier. This is half the period of the input AC voltage supplied to a full-wave rectifier, or the same as the period of the input AC voltage supplied to a half-wave rectifier. In Australia, the frequency of the AC power supply is 50 Hz. This means that the AC power supply has a period of 0.020 seconds • the time constant (𝜏) of the capacitor and load resistance: 𝜏 = RC. Vr(p−p) =

Vmax T RC

15.3 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Convert the following capacitances to farads. (a) 0.1 𝜇F (b) 220 𝜇F (c) 100 pF (d) 140 nF 2. (a) Define the term ‘time constant’ for an RC circuit. (b) A capacitor is charged to 10 V. What will be the voltage drop across the capacitor after one time constant if it is discharged through a resistor? 3. Calculate the time constants for the following RC combinations. (a) 390 Ω, 100 𝜇F (b) 33 kΩ, 100 𝜇F (c) 68 kΩ, 10 𝜇F (d) 470 kΩ, 0.47 𝜇F 4. The variation of voltage across a capacitor with time for an RC circuit is shown in the following graph. V (V) 10

8

6

4

2

0

5

10

15

20

25 t (ms)

(a) What is the voltage drop across the capacitor when t = 1 ms? (b) What is the time constant for this RC circuit? (c) If R = 100 Ω, what is the value of C? (d) If C = 0.1 𝜇F, what is the value of R? 5. Define rectification. 6. Describe the output of a half-wave rectifier for a sinusoidal input voltage. 7. (a) What is the maximum voltage drop possible across a silicon diode? (b) How many diodes are needed for half-wave rectification?

TOPIC 15 How can AC electricity charge a DC device? 19

8. (a) (b) (c) 9. (a) (b)

Describe the effect of placing a capacitor in parallel with the load resistor of a half-wave rectifier. Why does this effect occur? What happens to the output voltage as the value of the capacitor is increased? Sketch the circuits of a bridge rectifier and a centre-tap rectifier. If the peak voltage entering the bridge and centre-tap rectifier is 12 V, calculate the peak voltage of the output of each rectifier. 10. Calculate the peak-to-peak ripple voltage of a voltage signal if the maximum voltage is 6.3 V and the minimum voltage is 5.8 V. 11. A full-wave rectifier has an output with a peak value of 12.5 V and a frequency of 100 Hz. Calculate the ripple voltage when the following combinations of load resistance and smoothing capacitor are used. Load resistance (kΩ)

Smoothing capacitor (𝜇F)

a.

10

100

b.

3.3

100

c.

10

5.0

d.

25

20

e.

100

25

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

15.4 Voltage regulators KEY CONCEPTS • Describe the use of voltage regulators including Zener diodes and integrated circuits. • Analyse systems, including fault diagnosis, following selection and use of appropriate test equipment.

Most electronic circuits operate using a steady DC voltage. This voltage generally needs to remain constant. There might be variations in the input voltage — for example, from the AC voltage supplied to your house by the power lines or from a battery that is slowly ‘going flat’. There might also be variations in the effective resistance of the load circuit. Most digital logic circuits and processors require a 5 V DC power supply. The diode- and capacitor-based rectifier circuits discussed so far do not provide a steady DC voltage. Attaching a load across the output of a rectifier, as shown in figure 15.22, will reduce the effective resistance of the rectifier and therefore affect the time constant of the smoothing capacitor. The solution to this problem is to use a voltage regulator. A voltage regulator is an integrated circuit (IC) device that delivers a steady terminal voltage despite variations in the input voltage or variations in the current drawn by the load. In contrast, the voltage drop across the output terminals of a school power supply varies as its current changes.

20 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 15.22 Attaching a load resistor in parallel with the resistor of a halfwave rectifier reduces the effective resistance of the rectifier.

Vin

C

R

RL

1 = 1 + 1 Reff R RL

Many DC power supplies used in school laboratories consist of a transformer connected to a rectifier circuit. Changes to the input AC voltage of the transformer can alter the magnitude of the output voltage of the supply. Voltage regulators are usually integrated circuits that are placed between the power supply and the load. Figure 15.23 shows a 7805 three-pin voltage regulator and a circuit diagram showing how a voltage regulator can be used in a practical situation. Note that voltage regulators provide an output voltage that is less than their input voltage. FIGURE 15.23 (a) A 7805 voltage regulator (b) How a voltage regulator is placed in a circuit to give a constant output voltage (a)

Output

(b)

8 V to 20 V

Common Input

Input Poorly regulated power supply

Output 7805

+5 V

Common 0V

Resources Digital document Investigation 15.2 Constructing a voltage regulator (doc-31890)

15.4.1 Internal resistance Students are often required to carry out experiments to verify that when resistors are connected in parallel, the voltage across them remains constant. When they do the practical work, students usually find that adding extra resistors in parallel results in the voltage across them falling. The drop in voltage occurs because power supplies are not ideal. They also offer resistance to the current drawn from the supply. This resistance is known as ‘internal resistance’. Even though a student connects resistors in parallel, they are also in series with the internal resistance of the power supply. When the student adds an extra resistor to the parallel array, the total resistance of the circuit is reduced. The current through the power supply increases and the voltage drop across the ‘internal resistance’ increases. This results in the terminal voltage of the supply decreasing.

USING VOLTAGE REGULATORS Voltage regulators are used in portable sound systems. The amplifying circuits need a constant voltage so that sound is reproduced accurately. Voltage regulators are used because the output voltage of batteries and dry cells falls as their internal resistance increases. (That is, the batteries go flat!)

Voltage regulators, such as the 7805, are available in three-pin integrated circuit (IC) packages. One pin connects to the unregulated input voltage, one pin provides the regulated output voltage and the other pin connects to the ground. There are two types of IC voltage regulators. A fixed regulator gives a set output voltage, and an adjustable regulator can give a range of output voltages.

TOPIC 15 How can AC electricity charge a DC device? 21

15.4.2 Choosing a suitable voltage regulator Voltage regulators come in different types and sizes. They are able to operate with different input and output voltages as well as current ranges. If you are going to use a voltage regulator in a circuit, you will need to choose the best one for your requirements. You can choose a voltage regulator by studying data sheets. Another way to choose a voltage regulator is to examine the current-versus-voltage characteristics of the regulators. These are graphs that show how the regulator’s voltage varies as its output current changes. Figure 15.24 shows the characteristics for a 7805 voltage regulator. FIGURE 15.24 The current-versus-voltage characteristic curve for a 7805 voltage regulator

6

Output voltage (V)

Iout = 0 A

4 Iout = 500 mA

Iout = 1 A 2

0

2

4

6

8

10

Input voltage (V)

If you required a steady supply voltage of 5 V operating from a 12 V supply, you would choose a 7805 voltage regulator because it is designed to give a steady 5 V DC output. A 7809 voltage regulator is designed to give a steady 9 V DC output from an unregulated supply with an output voltage greater than 10 V.

15.4.3 Voltage regulator using a Zener diode A Zener diode is a diode that breaks down at a specific voltage when it is FIGURE 15.25 Circuit reverse biased. This means that a Zener diode will stop a reverse current from diagram symbol for a flowing through it until the reverse voltage applied across it reaches a fixed value Zener diode known as the breakdown voltage. Zener diodes are designed to break down in a reliable and non-destructive way so that they can be used in reverse to maintain a fixed voltage across their terminals. Zener diodes are used to provide a stable output voltage from an unstable source. One application is to provide a stable voltage from the storage battery of a solar cell array. The output of such a battery might vary depending on the amount of charge stored. Zener diodes are also used in circuits to maintain a fixed voltage across a load. Figure 15.26 shows the Zener diode’s voltage versus current characteristics and figure 15.27 shows how it can be connected to an unstable source. The resistor is placed in series to limit the current from the unstable source. 22 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 15.26 Zener diode voltage-versus-current diagram

FIGURE 15.27 Circuit diagram of a Zener diode

Current +

R

D

Unstable voltage IN

Forward current Breakdown voltage V

Stable voltage OUT

–

Leakage current

0

Voltage

Avalanche current Reverse voltage

Zener diodes can be distinguished from ordinary diodes by their code and breakdown voltage, which are printed on them. Zener diode codes begin BZX or BZY. Their breakdown voltage is printed with V in place of a decimal point, so 4V7 means 4.7 V. The minimum voltage Zener diode available is 2.7 V. If a Zener diode is placed in parallel with one of the resistors in a voltage divider circuit (see figure 15.28), no current flows through the Zener diode when the voltage across the resistor is less than the breakdown voltage. When the voltage is equal to the breakdown voltage, some current passes through this resistor, the rest passes through the diode. The result is that the voltage across the diode and the resistor R2 remains constant. If the input voltage increases, the current through the voltage divider increases and any increase in voltage happens across the other resistor, R1 .

FIGURE 15.28 A Zener diode is placed in parallel with a resistor.

R1 Vin R2

Vout

15.4.4 Power dissipation In electric and electronic circuits, power is the rate at which electrical energy is transformed into other forms of energy, usually heat. When this happens, power is said to be dissipated. The amount of power dissipated in a circuit element can be calculated using the relationship: P = VI Where: V = the potential difference (or voltage drop) across the circuit element I = the current flowing through the element. If the circuit element has a fixed resistance, R, the above formula can be expanded to include the resistance using Ohm’s Law: V = IR So, if you only know the current I and the resistance R, you can use the relationship: P = I2 R

TOPIC 15 How can AC electricity charge a DC device? 23

If you only have the potential difference V and the resistance R, you can use the relationship: P=

V2 R

SAMPLE PROBLEM 5

A 100 Ω resistor in series with a diode carries a current of 5 mA. Calculate the power dissipated in the resistor. THINK

WRITE

Recall the formula for power dissipation. 2. Substitute the values for current and resistance into the formula. 3. State the solution.

P = I2 R P = (5 × 10−3 )2 × 100 = 2.5 × 10−3 W

1.

2.5 × 10−3 W is dissipated in the resistor.

PRACTICE PROBLEM 5 A 5 kΩ resistor is connected in series with a diode. Calculate the power dissipated in the resistor when there is a 2 mA current flowing through it.

15.4.5 Controlling voltage in circuits It is often necessary to control the voltage at various points in the electronic circuits that make up electronic systems. The electrical potential, or voltage, is defined as the amount of electrical potential energy at a point in a circuit per unit charge. The potential difference, or voltage drop, across a device is the difference in potential on either side of the device. The SI unit for voltage is the volt (V). A voltage of 1 volt at a point in a circuit means that every coulomb of charge passing that point has the ability to transfer 1 joule of energy to the rest of the circuit. The electrical potential energy is provided to a circuit by a source of electromotive force (emf). Emf is also measured in volts. An emf of 1 volt means that the source of emf provides the circuit with 1 joule of energy for every coulomb of charge passing through the source. Examples of sources of emf include cells, batteries, generators, photovoltaic cells and thermocouples. The source of voltage is sometimes represented in circuit diagrams by two horizontal lines known as supply rails. These indicate the voltage that the source supplies to the circuit. They are shown in figure 15.29a. Figure 15.29b shows the traditional way of showing a DC source. Supply rails are used in circuit diagrams when the actual nature of the source is not important, other than being a DC supply of a particular value. The voltage drop, or potential difference, across a device in a circuit is a measure of the amount of energy transformed in that device by the source of emf for every coulomb of charge that passes through it. If a resistor has a voltage drop of 4.5 V across it, 4.5 J of energy are transformed by the source of emf for every coulomb of charge passing through the resistor. The current flowing in the circuit is the means by which energy is transferred from the source of emf to the devices in the circuit.

24 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 15.29 (a) Supply rail circuit (b) Equivalent traditional circuit (a)

(b)

+5 V

5V

0V

15.4.6 Basic voltage dividers When two resistors are placed in series with a supply, the supply voltage is shared, or divided, between the resistors. The way that the voltage is divided depends on the resistance values of the resistors. Such an arrangement of resistors, as shown in figure 15.30, is known as a voltage divider and is used when it is necessary to provide a useful voltage to a part of the circuit that is smaller than the supply voltage. The effective resistance of the voltage divider is R1 + R2 . The current flowing through each resistor is: I=

Vin R1 + R2

FIGURE 15.30 The basic voltage divider Vin R1 Vin R2

Vout

0V

Therefore, the value of Vout is the current times the resistance: Vout =

R2 Vin R1 + R2

When using voltage dividers, it is assumed that the resistance of the part of the circuit being provided with the output voltage (Vout in figure 15.30) will be so great that it will not affect the current flowing through R2 and the calculated value of Vout . This is because when a resistance is placed in parallel with another resistor, the effective resistance is reduced. SAMPLE PROBLEM 6

Variable resistors are often used in voltage dividers to give an adjustable output voltage. Reducing the size of one of the resistances increases the current flowing through the divider and therefore increases the voltage drop across the other resistor. Calculate the output voltage in the following voltage divider circuit.

+5.0 V R1 = 2.2 kΩ

R2 = 3.3 kΩ

Vout

0V

THINK 1.

Recall the formula for voltage drop over one part of a voltage divider.

WRITE

Vout =

R2 Vin R1 + R2

TOPIC 15 How can AC electricity charge a DC device? 25

2.

Substitute the values into the formula.

3.

State the solution.

3.3 × 5.0 2.2 + 3.3 = 3.0 V The output voltage is 3 V. =

PRACTICE PROBLEM 6 Calculate the output voltage in the following voltage divider circuit. 9V 1.5 Ω 3.0 Ω

Vout

0V

SAMPLE PROBLEM 7

What would the resistance of the variable resistor have to be to give an output voltage of 4 V in the following circuit? +5.0 V

R1

R2

2.2 kΩ

Vout = 4.0 V

0V

THINK

WRITE

1.

Recall the formula for voltage drop over one part of a voltage divider.

Vout =

2.

Rearrange the formula to make R1 the subject.

R1 =

3.

Substitute the values into the formula.

4.

State the solution.

26 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

R2 Vin R1 + R2

R2 Vin − R2 Vout 2.2 × 5.0 = − 2.2 4.0 = 0.55 kΩ The resistance of the variable resistor to give an output voltage of 4 V is 0.55 kΩ.

PRACTICE PROBLEM 7 For the following circuit what size resistor is needed to give an output voltage of 5 V? 9V R

12.5 kΩ

Vout = 5.0 V

0V

Resources Digital document eModelling: Voltage divider (doc-0036)

15.4 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. A simple RC circuit is shown here. (a) How long does it take for the voltage drop across the capacitor to reach approximately 1 V after R = 1 MΩ the switch is closed? (b) How long will it take for the capacitor to be fully charged? (c) What will be the voltage drop across the capacitor when it 1.5 V C = 0.1 μF is fully charged? (d) When the voltage drop across the capacitor is 1 V, what is the voltage drop across the resistor? 2. A half-wave rectifier circuit using (a) a silicon diode with (b) an input voltage signal is shown in the following figure. (a)

(b) Vin (V) 4 2 Vin

1 kΩ

Vout

0 –2

10

20

30

40 t (ms)

–4

(a) What is the peak output voltage for this circuit? (b) Sketch the output voltage for this circuit. (c) Sketch the output voltage if the following capacitors are placed in parallel with the load resistor: i. 10 𝜇F ii. 100 𝜇F.

TOPIC 15 How can AC electricity charge a DC device? 27

3. A half-wave rectifier circuit is shown. The capacitor can be engaged by closing the switch. The input voltage to the rectifier is 9 V (RMS) at 50 Hz.

Vout

R = 500 Ω

Vin

C = 20 μF

(a) What is the peak input voltage? (b) What is the period of the output voltage waveform when the switch is open? (c) Sketch the waveform of Vout when the switch is open, paying attention to the maximum voltage and the period. (Assume that the maximum voltage drop across the diode is 0.7 V.) (d) Sketch the waveform of VD , the voltage across the diode, when the switch is open. (e) What is the time constant if the capacitor discharges through the resistor? (f) Sketch the waveform of Vout when the switch is closed. (g) List two changes that would give a smoother output voltage. 4. A square waveform of period 2 seconds is fed into a resistor capacitor circuit as shown in the following figure. (a)

(b) Vin (V) 20 kΩ

VR

10 Vin

10 μF

1

2

3

4

5

6

7

VC

t (s)

(a) What is the time constant for this RC combination? (b) Will the capacitor fully charge before the input voltage falls to zero? Justify your answer. (c) Sketch the waveform for VC , the voltage drop across the capacitor. (d) Sketch the waveform for VR , the voltage drop across the resistor. 5. List three factors that affect the variation in the output of DC power supplies. 6. (a) What is meant by voltage regulation? (b) Why are voltage regulators necessary in circuits? (c) Where are voltage regulators placed in circuits? (d) Compare the output voltage of a voltage regulator with its input voltage. 7. Find the output voltage for the voltage divider shown in the following circuits. (a)

(b) R1 = 2.2 kΩ

R1 = 2.2 kΩ

Vin = 6.0 V

Vin = 9.0 V R2 = 2.2 kΩ

R2 = 4.4 kΩ

Vout

8. Find the value of R2 in the following voltage divider that would give an output voltage of 2 V. R1 = 3.0 kΩ Vin = 6.0 V R2

28 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Vout = 2.0 V

Vout

9. Describe a circuit containing a Zener diode that can provide a constant voltage across a load from a rectifier. Use the following circuit diagram to answer questions 10 and 11. +

R1

RL

Vin

–

10. R1 is 4 kΩ, RL is 10 kΩ and the breakdown voltage of the Zener diode is 30 V. The input voltage ranges between 70 V and 100 V. (a) What is the range of voltage drops across R1 ? (b) Calculate the maximum and minimum currents through R1 ? (c) What is the current in the load resistor? (d) Calculate the maximum and minimum currents through the Zener diode. 11. The voltage drop across RL is 12 V, Vin has values between 20 V and 35 V. The minimum current through RL is 100 mA, and the minimum current through the Zener diode is 8 mA. (a) What is the breakdown voltage of the Zener diode? (b) What is the minimum voltage drop across R1 ? (c) What is the minimum current through R1 ? (d) Calculate the value of RL . (e) Calculate the value of R1 .

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

15.5 Transducers and diodes KEY CONCEPTS • Apply the use of heat and light sensors such as thermistors and light-dependent resistors (LDRs) to trigger an output device such as lighting or a motor. • Evaluate the use of circuits for particular purposes using technical specifications related to potential difference (voltage drop), current, resistance, power, temperature and illumination.

Transducers are devices that transform energy from one form to another. They are used in electronic communications, instrumentation and control systems. Transducers are the interface between the environment and the electronic system. There are two types of transducer: input and output. In electronic systems, a transducer that changes non-electrical energy into electrical energy is called an input transducer. For example, photocells or solar cells convert light energy into electricity. They are used for automatic exposure control in cameras. Microphones transform sound energy into an electrical signal. Another example of an input transducer is a thermocouple, which consists of two different metals such as copper and iron that are joined together. They are used in electronic thermometers because the size of the voltage they produce is related to the temperature. Input transducers are used in sensing devices in household and industrial systems.

TOPIC 15 How can AC electricity charge a DC device? 29

15.5.1 Thermistors A thermistor is an input transducer made from a mixture of semiconductors. The resistance of a thermistor varies with temperature. Thermistors are considered to be input transducers because they convert thermal energy into a voltage when they are used in potential dividers. Thermistors are used in fire alarms, temperature circuits in car engines, electronic thermometers and thermostats for controlling the temperature in household and industrial equipment. They are also included in protection circuits used to protect equipment from sudden current and voltage surges. You can investigate the way in which the resistance of a thermistor varies with temperature by doing Investigation 15.3. Figure 15.31a shows the circuit symbol for a thermistor. Figure 15.31b and c are examples of graphs showing how the resistance of different thermistors varies with temperature. Note that in figure 15.31c the graph shows the characteristic curves for three separate thermistors and the vertical scale is logarithmic. Each vertical line represents the next whole number multiple of the previously stated power of ten. For example, on the vertical scale the lines above that labelled 10 are 20, 30, 40, and so on. Similarly, the lines above that labelled 100 are 200, 300, 400, and so on. FIGURE 15.31 (a) Circuit symbol for a thermistor (b) and (c) Resistance-versus-temperature characteristics for different thermistors (a)

(c) 100 000

Resistance (Ω)

10 000

1000

100 Resistance (kΩ)

(b) 2 10 1

0

0 20

40

60

Temperature (°C)

50

100 150 200 Temperature (°C)

Thermistors are produced in two forms. The most common is the negative temperature coefficient thermistor, which has a resistance that decreases as the temperature increases. This type has a resistance-versus-temperature graph like that shown in figure 15.31b and c. Thus, an increase in temperature leads to an increase in the current in a temperature-sensing circuit. The other type of thermistor is the positive temperature coefficient thermistor, which has a resistance that increases as the temperature increases. For positive coefficient thermistors, an increase in the temperature will lead to a decrease in the current in a temperature-sensing circuit. Thermistors are usually connected to variable resistors in voltage dividers. 30 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

250

300

FIGURE 15.32 The two common forms of thermistor

SAMPLE PROBLEM 8

A thermistor has the temperature-resistance characteristic shown by the bottom curve in figure 15.31c. It is placed in the following voltage divider. a. Is this a positive or negative coefficient thermistor? b. What is the resistance of the thermistor when the temperature is 150 °C? c. What is the value of the variable resistor if the temperature is 200 °C and Vout is 6 V? +9.0 V

R1

R2 = thermistor

Vout

0V

THINK

WRITE

Recall that a positive coefficient thermistor has a resistance that increases as the temperature increases. b. Use the graph to find the resistance when the temperature is 150 °C. c. 1. From the graph, when the temperature is 200 °C, the resistance is 10 Ω. Substitute the known values into the formula for voltage drop across a voltage divider.

a.

a.

Since this thermistor’s resistance decreases as the voltage increases it is a negative coefficient thermistor. b. When the temperature is 150 °C the resistance is 20 Ω. c. Vin = 9 V, Vout = 6 V, R2 = 10 Ω RV Vout = 2 in R1 + R2 10 Ω × 9 V 6V = R1 + 10 Ω 6 × R1 + 60 = 90 6 × R1 = 30

2.

State the solution.

R1 = 5 Ω The value of the variable resistor if the temperature is 200 °C and Vout is 6 V is 5 Ω.

TOPIC 15 How can AC electricity charge a DC device? 31

PRACTICE PROBLEM 8 A thermistor has the temperature-resistance characteristic shown by the top curve in figure 15.31c. It is placed in the following voltage divider. 4.5 V R1 = thermistor

R2

Vout

a. What is the resistance of the thermistor when the temperature is 200 °C? b. What is the value of the variable resistor needed to give an output voltage of 1.5 V when the temperature is 200 °C?

Resources Digital documents Investigation 15.3 Thermistors (doc-31888)

THERMISTORS IN FIRE ALARMS AND THERMOSTATS Fire alarm sensors are used in areas where smoke detectors FIGURE 15.33 A fire alarm sensor are not suitable, for instance, in smoky or dusty environments. They use a sensing circuit that employs a thermistor that is in contact with the air and is placed on the ceiling of the room. The circuit gives a voltage output that is related to the temperature of the air because the resistance of the thermistor varies with temperature. When the air temperature reaches a predetermined value, a control system is activated. This may sound alarms, turn on sprinklers and cause a light-emitting diode to flash. The thermistor is used as an input transducer that transforms thermal energy into electrical energy. A thermostat is a device that controls the heating or cooling of an object in order to keep it at a constant temperature. It consists of a temperature-sensing device connected to a switching device. As in fire alarm systems, thermistors are used in thermostats as transducers to convert thermal energy into electrical energy. They are used in voltage dividers so that when a predetermined temperature is reached, the voltage output from the divider causes a switch to be triggered. This activates a heating or cooling device. Thermostats are used in ovens, hot-water systems, space-heating systems and refrigerators.

32 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

15.5.2 Photonics Photonics is the branch of science that studies or uses the manipulation of photons (the basic unit of light) to manipulate, encode, transmit, decode and/or store information. It involves: • using electronic systems to convert information into light (or other electromagnetic radiation) energy signals • transmitting the light — for example, through optical fibres • converting the light signal back into useful information — for example, sound • storing information using light — for example, by burning CDs (compact discs) or DVDs (digital video discs). Photonics does not just involve the use of visible light. It also involves infra-red and ultraviolet sections of the electromagnetic spectrum. A television remote control transmits information to the television set using infra-red radiation. An integrated circuit chip detects when you press a button on the remote. It then produces a coded signal that is something like Morse code. There is a different code for each button. The coded signal is amplified and sent to an infra-red LED that sends invisible radiation into the room. The code varies the intensity of the radiation. This varying of intensity is known as modulation. FIGURE 15.34 Block diagram of a On the television set there is a photodetector that converts the television remote control system infra-red radiation back into an electric signal. The detector might be a photocell — such as a photovoltaic (solar) cell — Code or a photodiode or phototransistor. The magnitude of the current Amp. generator produced by the photodetector is proportional to the intensity of the infra-red radiation. The infra-red LED is an example of a transducer that transforms electrical energy to electromagnetic radiation. The infra-red radiation is the carrier of the signal. The photodetector is an example of a transducer that transforms electromagnetic Television radiation to electrical energy. This photonic system is illustrated Photodetector control in figure 15.34. system

Resources Weblink Photonics resources for teachers

15.5.3 Light-emitting diodes (LEDs) You will have seen light-emitting diodes (LEDs) on many electronic devices. An LED is a small semiconductor diode that emits light when a current passes through it. LEDs are used in electronic displays. They can be made to emit any colour (red, green and yellow are the most common) by the choice of impurities added to the base semiconductor used in their construction. They are used on amplifiers to indicate the amplitude of the sound signal. They also appear in cash register displays and as indicators on modems and routers. Light-emitting diodes have the following advantages over incandescent light globes: • LEDs are very durable and can be used in situations where excessive vibrations would damage the filaments of ordinary light globes — for example, in aircraft cockpits, in the space shuttle and on the control panels of vibrating machinery. • LEDs emit light using very low voltages, so they are useful indicator lights for electronic systems. • An LED will operate for over 500 000 hours (depending on the operating current and temperature) compared with about 1000 hours for a light globe.

TOPIC 15 How can AC electricity charge a DC device? 33

•

LEDs are much faster to respond than light globes. A light globe takes many milliseconds to turn on and off, while red or yellow LEDs take about 75 nanoseconds to turn on and off. • LEDs consume much less power than light globes. Before considering how LEDs operate, you should revise the information on diodes covered previously in the topic.

The structure of LEDs An LED consists of a small semiconductor chip mounted on, and in electrical contact with, a metal base. The base is reflective to ensure much of the radiation is projected forwards through the top of the LED. The cathode lead is connected to the metal base. A fine wire is connected to the upper surface of the chip and leads to the anode lead of the LED. This assembly is covered with an epoxy resin that acts as a lens to intensify the light passing through the top of the LED. This structure is shown in figure 15.35a. The circuit diagram symbol for an LED is shown in figure 15.35b. The colour of the LED determines the voltage at which the LED reaches saturation or the maximum voltage across the LED when it is conducting a current. The current-versus-voltage characteristic curve for a red LED is shown in figure 15.35c. FIGURE 15.35 (a) The structure of a typical LED (b) The circuit diagram symbol for an LED (c) current-versusvoltage characteristic curve for a red LED (a)

LED chip

Magnifying epoxy lens

(b)

(c) + Anode

Flat edge on cathode side

Radiation

5 mA

Cathode Anode

– +

30 mA

– Cathode 0

1.58 V 1.66 V

VLED

Commercial LEDs are designed to release visible light or infra-red radiation. The chip is based on a semiconducting material made from a gallium–arsenic–phosphorus compound. Adjusting the ratio of phosphorus to arsenic enables different wavelengths (and colours) of light to be emitted. As with other diodes, the chip of an LED is made by fusing a p-type semiconductor to an n-type semiconductor. The p-type material forms the anode and the n-type material forms the cathode. An LED will emit light when a current passes through it; that is, when it is forward biased. It is important that the current is not too high, as the LED might be damaged, or not too low as not enough light will be emitted. The forward bias voltage drop across an LED is between 1.6 V and 2.8 V, depending on the colour of the LED and the current through it. A red LED has a forward bias voltage drop of approximately FIGURE 15.36 Limiting resistor in series 1.6 V, whereas for a green or yellow LED the value is with a forward-biased LED approximately 2.3 V. To ensure that the current passing Limiting resistor through the LED is suitable, the LEDs are protected by being placed in series with resistors (known as limiting resistors). The value of the resistor is calculated to reduce the potential 0V +5 V drop across the LED to the required value. This arrangement is shown in figure 15.36. Anode

34 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Cathode

SAMPLE PROBLEM 9

The following LED has a voltage drop across it of 1.8 V and carries a current of 20 mA. a. What is the voltage drop across the limiting resistor? b. What is the value, R, of the limiting resistor?

I = 20 mA

R

6.0 V

1.8 V

THINK

WRITE

The emf of the source equals the sum of the voltage drops around the circuit. 2. State the solution.

a. 1.

b. 1.

2.

The current over the resistor is equal to the current over the LED. Use Ohm’s Law to find the resistance.

State the solution.

a

Voltage drop = 6 − 1.2 = 4.8 V

The voltage drop across the limiting resistor is 4.8 V. b V = IR V R= I 4.2 = 0.020 = 210 Ω The value of the limiting resistor is 210 Ω.

PRACTICE PROBLEM 9 When an LED is connected in series with an unknown limiting resistor to a 9.0 V battery, a current of 30 mA flows and there is a voltage drop of 2.4 V across the LED. What is the value of the limiting resistor?

LED NUMBER DISPLAYS The seven-segment number display is used in clock radios, microwave ovens and many other appliances. Figure 15.37 shows a unit with the number seven activated. Each segment of the display consists of an LED and a lens. When the LED conducts a current, it emits light that passes through the top of the lens. By activating different segments, different numbers can be formed.

FIGURE 15.37 The seven-segment number display Surface of segment

Connecting pins

Plastic outer

Each segment contains an LED chip.

Diffusing plastic

TOPIC 15 How can AC electricity charge a DC device? 35

Infra-red LEDs emit electromagnetic radiation with a longer wavelength than that of visible light. They are used widely in optical communications, especially in television and audio system remote controls. Laser diodes are made by polishing the ends of suitable p–n junction crystals. They are used in DVD players and laser pointers.

HOW DO LEDS PRODUCE LIGHT? When an electron at the bottom of the conduction band of a semiconductor falls into a hole at the top of a valence band, an amount of energy, Eg , is released, where Eg is the energy gap width. In silicon-based semiconductors this energy is transformed into thermal energy in the vibrating lattice of the atoms. In some semiconducting materials, the released energy can appear as electromagnetic radiation. The wavelength of this radiation can be calculated using the formula: 𝜆=

c f

=

hc Eg

Where: 𝜆 = wavelength of the electromagnetic radiation c = speed of light f = frequency of the radiation h = Planck’s constant.

AS A MATTER OF FACT Optical-fibre telecommunication systems usually use LEDs and laser diodes that operate in the infra-red region of the electromagnetic spectrum. Laser diodes produce a much smaller range of wavelengths than LEDs. Certain wavelengths of infra-red radiation are used because they travel better through the optical fibres than visible light.

Resources Digital document Investigation 15.4 Light-emitting diodes (LEDs) (doc-31891) Teacher-led video Investigation 15.4 Light-emitting diodes (LEDs) (tlvd-0836) Weblink

LEDs

15.5.4 Light-dependent resistors (LDRs) Photonic transducers that transform light energy to electrical energy can be used in electronic circuits to produce current or voltage signals. A light-dependent resistor (LDR) is a semiconductor device that has a resistance which decreases with the amount of light falling on it. Figure 15.38b gives an example of a graph that shows the relationship between resistance and light intensity for an LDR. Note that in this example, the scales on both axes are logarithmic. LDRs are used in cameras and in a wide range of automatic devices controlled by light. They are often found in voltage dividers in conjunction with variable resistors; the output voltage is used to switch another device on or off. The variable resistor is used to control the light intensity at which the switching device is activated.

36 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 15.38 (a) Circuit symbols for an LDR (b) Resistance-versus-light intensity graph for an LDR (c) An LDR (b) 10 000

(a)

(c) Cadmium sulfide track

LDR resistance (Ω)

or

1000

100

10 0.1

1

10

100

1000

Illumination (lux)

SAMPLE PROBLEM 10

A shop minder circuit consists of a beam of light that shines onto the following voltage divider circuit. The LDR has the same characteristic curve as that shown in figure 15.38b.

+ 12 V –

R1 Switching device

R2 = LDR

Buzzer

Vout

What is the resistance of the LDR when the light intensity is 1 lux? If the variable resistor is set at 500 Ω, calculate the value of Vout when the light intensity is 1 lux. c. What is the value of the variable resistor if the light intensity is 0.1 lux and Vout is 6 V?

a. b.

THINK

WRITE

a. 1.

Use the graph to find the resistance when the light intensity is 1 lux.

a.

b. 1.

R1 = 500 Ω

b.

R2 = 1000 Ω Use the formula to find Vout .

2.

State the solution.

When the intensity is 1 lux, R2 = 1 kΩ. R2 Vin R1 + R2 1000 × 12 = 500 + 1000 12 000 = 1500 = 8.0 V When the variable resistor is set at 500 Ω, the value of Vout when the light intensity is 1 lux is 8 V. Vout =

TOPIC 15 How can AC electricity charge a DC device? 37

c. 1.

When the light intensity is 0.1 lux, R2 = 10 kΩ. Use the formula to determine R1.

c.

R2 Vin R1 + R2 10 000 Ω × 12 V 6V = R1 + 10 000 Ω 6 × R1 + 60 000 = 120 000

Vout =

6 × R1 = 60 000 2.

State the solution.

R1 = 10 000 Ω When the light intensity is 0.1 lux and Vout is 6 V the value of the variable resistor is 10 000 Ω.

PRACTICE PROBLEM 10 An outdoor sensor light switches on when the light intensity drops below 1 lux. The circuit comprises a voltage divider in which the input voltage is 12 V and the switch is triggered by the output voltage across the LDR. The LDR has the same characteristics as shown in figure 15.38b. The light switches on when the voltage across the LDR is 1.5 V. a. What is the value of the resistor that is in series with the LDR? b. To make the light switch on when there is more light would you need a resistor with a smaller or larger resistance? Explain your answer.

LDRS IN CAMERAS To get the best results from a camera FIGURE 15.39 LDRs are used in camera lenses it is necessary to control the amount of light that falls on the detector. This is achieved by changing the aperture of the lens diaphragm and the shutter speed. High shutter speeds are necessary when photographing fastmoving objects. If you use a high shutter speed you have to use a wide aperture opening so that enough light falls on the detector to get proper resolution. Similarly, in dull conditions you need a wide aperture opening so that enough light enters the camera. Many professional-level cameras have a light meter so that photographers can adjust the shutter speed and aperture opening. These meters are either connected to a photocell that converts light directly into an electric current or they are connected to a voltage divider that has an LDR in it. In a camera that uses a voltage divider, part of the light passing through the lens system is reflected onto the LDR by a small mirror. The output voltage of the voltage divider is fed to the meter that converts the reading to the appropriate units for light intensity. It is then up to the photographer to make the necessary adjustments.

38 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Resources Digital documents Investigation 15.5 Light-dependent resistor (doc-31892) Investigation 15.6 Light-dependent resistor: voltage divider method (doc-31893)

15.5.5 Photodiodes Photodiodes are used to detect light and to convert light energy to electrical energy. A photodiode consists of an active p–n junction that is operated in reverse bias. When a photon of sufficient energy strikes an electron in the junction, the photon transfers its energy to the electron allowing the electron to escape the crystal lattice. This creates a free electron and a hole that move in opposite directions under the influence of the electric field provided by the source of electromotive force. The intensity of light depends on the number of photons present. The current produced by a photodiode is therefore proportional to the intensity of light incident on the photodiode. Figure 15.40 shows a circuit with a reverse-biased photodiode and a response characteristic curve. FIGURE 15.40 Photodiode: (a) circuit symbol, (b) reverse biased and (c) sample characteristic curves (b)

(c) I (μA) L4

125 100 75 50 25 0

L3 L2

10

20

40

60

80

L1 Dark V

Increasing light intensity

(a)

Photodiodes are fast to respond to light. It takes only a few nanoseconds (10−9 s) for a photodiode to produce a current after the photons arrive. This makes them useful for converting light signals from fibreoptic communication systems back into electrical signals. Although the response time of a photodiode seems to be extremely rapid, it does limit the bandwidth (amount of information carried per second) of an optical-fibre communications system. The response time determines both the minimum duration of pulses and the maximum number of pulses that can be transmitted down the fibre each second. Different types of photodiode have different optimum wavelengths of light to which they respond. Silicon-based photodiodes operate best with wavelengths between 700 nm and 900 nm. Germanium-based photodiodes operate best with wavelengths between 1400 and 1500 nm.

15.5.6 Phototransistors A phototransistor is similar to an ordinary transistor except that incident light on the p–n junction controls the response of the device. Phototransistors have a built-in gain and are more sensitive than photodiodes. A phototransistor is usually an n–p–n transistor. The base region is large and usually does not have a terminal attached to it. The circuit symbol for a phototransistor is shown in figure 15.41. The collector–base junction is sensitive to light in the same way as a photodiode. In fact, the collector–base junction is also a reverse-biased p–n junction, just like a photodiode. When light shines on it, a base current flows. The base current is proportional to the light intensity and produces a collector current that is also proportional to the light intensity. Phototransistors are more sensitive than photodiodes because transistors amplify the collector–base current. They are not as quick to respond to light as photodiodes. This reduces the bandwidth of photonic systems that use phototransistors rather than photodiodes.

FIGURE 15.41 Phototransistor circuit symbol

TOPIC 15 How can AC electricity charge a DC device? 39

Optical-fibre systems that have a bandwidth of 43 Tbps (terabits per second) have been trialled. This means that the system can transmit up to 43 × 1012 bits of data per second. By comparison, coaxial cable, which is used to carry telephone calls between cities and is also used to transmit data over shorter distances, has a bandwidth of up to 100 Mbps.

15.5 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. A thermistor has the following characteristic curve.

8 7

Resistance (kΩ)

6 5 4 3 2 1

0

10

20

30

40 50 60 Temperature (°C)

70

80

90

(a) What is the resistance of the thermistor at: i. 20 °C +9 V ii. 80 °C? (b) What is the temperature when the thermistor has resistance of: i. 4 kΩ R ii. 1.5 kΩ? 2. The thermistor from question 1 is placed in a voltage divider as shown. (a) If the variable resistor is set at 4 kΩ and the temperature is 40 °C, Vout what is the output voltage? (b) An output voltage of 6 V is required when the temperature is 80 °C. What must the value of the variable resistor be? 3. A temperature sensing system in an oven consists of a thermistor connected into a voltage divider as shown in the following figure (a). The thermistor has the characteristic curve shown in figure (b). (a) What is the resistance of the thermistor when the temperature in the oven is 100 °C? (b) What is the temperature in the oven when the resistance of the thermistor is 400 Ω? (c) Calculate the resistance of the variable resistor when the temperature in the oven is 200 °C and the output voltage, Vout , is 8 V.

40 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

12.0 V

(a)

To sensor Vout

(b)

100 000

Resistance (Ω)

10 000

1000

100

10

0

50

100 150 200 Temperature (°C)

250

300

4. The following is a current-versus-voltage characteristic curve of a diode. 60

Current (mA)

50 40 30 20 10 0 0.2 0.4 0.6 0.8 1.0 Voltage (volts)

TOPIC 15 How can AC electricity charge a DC device? 41

The diode is placed in the following circuit.

12 V 1.0 kΩ

Calculate the current flowing in the circuit. Express your answer in mA. 5. What is a transducer? 6. Explain the relationship in a light-dependent resistor (LDR) between the resistance and the amount of light falling on the LDR. 7. What is the role of LDRs in cameras? 8. An LDR is used in a voltage divider circuit as shown in the following figure (a). The characteristic curve of the LDR is given in figure (b). (a)

(b) 10 000

+ 6.0 V

3000 Ω

LDR resistance (Ω)

LDR

Vout

0V

1000

100

10 0.1

1

10

100

1000

Illumination (lux)

(a) (b) (c) 9. (a) (b)

10. 11. 12. 13.

What is the illumination when the LDR has a resistance of 2 kΩ? Calculate Vout for this illumination. Describe what happens to Vout as the illumination increases. What is the meaning of the terms ‘forward biased’ and ‘reverse biased’? Draw circuit diagrams to show how to: i. forward bias an LED ii. reverse bias an LED. Give three examples of a situation where an LED would be preferable to an ordinary light source. Analyse each case to justify your choices. Why is it that not all diodes emit light? Why are limiting resistors placed in series with LEDs? Consider the circuit shown in the following figure (a). The characteristic curve of the LED is shown in figure (b). (a)

I (mA) 6 5 4 3 2 1

(b)

A

6.0 V R

–6.0

(a) If the current is measured to be 4 mA: i. What is the voltage drop across the LED? ii. What is the voltage drop across the resistor? iii. What is the resistance of the resistor?

42 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

–3.0

0

V (V) 0.2 0.4 0.6 0.8

(b) The LED is now reversed, as shown in the following figure. A 6.0 V R

i. What is the voltage drop across the LED? ii. What is the current flowing in the circuit? iii. What is the voltage drop across the resistor? 14. A student carries out a practical activity on an LED. She initially sets up the circuit shown in the following figure (a). She is able to measure the voltage drop across the diode and the resistor as well as the emf supplied by the variable DC supply. This enables her to construct the current-versus-voltage characteristic curve for the diode, as shown in figure (b). (a)

(b) A

Current (mA) 60 R = 500 Ω 40

20 Voltage (V) –6

–3

0

1.0

2.0

When the current in the circuit is 60 mA, calculate the: (a) voltage drop across the diode (b) voltage drop across the resistor (c) emf of the supply. 15. If the student from question 14 reverses the polarity of the supply, very little current flows through the circuit. (a) Explain why this occurs. (b) If the emf of the supply is −2.5 V, what is the voltage drop across the diode? Justify your answer. 16. The following LED has a voltage drop across it of 1.8 V and carries a current of 40 mA. I = 40 mA

9.0 V

R

1.8 V

(a) What is the voltage drop across the limiting resistor? (b) What is the value, R, of the limiting resistor? 17. Describe the purpose of a photodetector. 18. (a) Describe the operation of a photodiode. (b) Describe the operation of a phototransistor. (c) Which device has the fastest response time? (d) Which device is the most sensitive? (e) Describe the effect of response time on bandwidth.

TOPIC 15 How can AC electricity charge a DC device? 43

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

15.6 Data transfer KEY CONCEPTS • Compare different light sources (bulbs, LEDs, lasers) for their suitability for data transfer. • Explain the use of optical fibres for short- and long-distance telecommunications.

15.6.1 Communication carriers There are different types of communication carriers involved in carrying telephone calls and data between users. Most houses are still connected to the telephone network by twisted pairs of wires.

Twisted-pair copper cable Bundles of copper cables are commonly used throughout all telecommunications networks. Each pair of cables carries one call. A twisted-pair copper cable consists of two independently insulated wires twisted around one another. One wire carries the signal while the other wire is grounded and absorbs signal interference. Twisted-pair cables are used by older telephone networks. They are the least expensive type of local area network (LAN) cable. Most networks still contain some twisted-pair cabling at some point along the network.

Coaxial cable Coaxial cable consists of a centre wire surrounded by insulation and then a grounded shield of braided wire. The shield minimises electrical and radio frequency interference. It was once the main cable used in the trunk network. It carried calls between major cities. Each cable could carry up to 2700 calls at a time. Coaxial cabling is widely used in the cable television industry and computer networks. It is much less susceptible to interference than twisted-pair cable and can carry much more data. Coaxial cables are gradually being replaced by optical fibres and microwave systems.

Optical-fibre cable An optical fibre is a thin filament of glass or plastic that has a central core and a cladding with a lower refractive index, so that total internal reflection will happen. FIGURE 15.42 Light reflections in an optical fibre

n2 < n1 n1 n2 < n1

44 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Cladding θθ

Core Cladding

An optical-fibre cable consists of several optical fibres in a common package. A strength member, such as a steel wire, supports the fibres. This protects the delicate optical fibres from stretching. A single optical fibre can carry 30 000 telephone calls at a time or hundreds of television signals. Some optical-fibre systems in development have demonstrated a capacity of over 40 million telephone calls. FIGURE 15.43 Communications carriers: (a) twisted pair (b) coaxial cable and (c) optical fibres

Attenuation Attenuation is the loss of power of a signal along a communicating channel. Attenuation is measured in decibels (dB). Telephone systems that use electrical signals travelling through copper cables experience attenuation. The strength of the signal decreases with the distance travelled. These systems use amplifiers to increase the power of the signals. One reason for the attenuation of signals in metal wire carriers is the skin effect. The skin effect is the uneven distribution of current carriers throughout the cross-section of a conductor when it is carrying an alternating current. The current density becomes greater at the surface than it is at the centre. The skin effect is caused by electromagnetic effects, and it becomes more significant as the frequency of the alternating current increases. One consequence of the skin effect is that the effective resistance of a conductor is greater when it is carrying an alternating current than its true or direct current resistance.

Bandwidth Bandwidth is the amount of data that can be transmitted in a fixed amount of time. For digital devices, the bandwidth is usually expressed in bits per second (bps) or bytes per second. For analog devices, the bandwidth is expressed in cycles per second, or hertz (Hz). The bandwidth of an information system is essentially the highest frequency or rate at which the data can be transmitted. System bandwidth is the range of frequencies that are carried by the transmission system. In practice, a commercial quality speech transmission system need carry only frequencies between 300 Hz and 3400 Hz. This is a bandwidth of 3400 Hz because the maximum frequency it can carry is 3400 Hz. Fibre-optic cables have a much greater bandwidth than metal cables. This means that they can carry more data. The transmission rate of optical fibres is in the order of 1 trillion bits (1 terabit) per second per fibre. The transmission rate of telephone wire pairs is in the range of 16–100 million bits per second (16–100 Mbps).

TOPIC 15 How can AC electricity charge a DC device? 45

15.6.2 Modulation Modulation is the changing of one wave train caused by another wave. An example of this is amplitude modulation (AM) used in radio transmission systems. In this case, the amplitude of the carrier signal is changed by superimposing the waveform of a sound such as music or voice. FIGURE 15.44 Amplitude modulation of a radio signal Information to be transmitted, e.g. speech or music

Carrier radio signal of higher frequency

Carrier wave amplitude modulated by speech or music information

Time

Optical intensity modulation in photonic communications systems involves varying the output intensity of a carrier light source (for example, an LED or laser diode) by using the electric signal from a sound-toelectricity transducer such as a microphone, or from a sound playback device. Figure 15.45 shows an analog modulation system where the output from an LED is modulated using a microphone and amplifier. FIGURE 15.45 (a) Optical intensity modulation system block diagram (b) Adding the signals (a) Information Driver Source

Bright

(b)

Dim Information Encoding process

Intensity modulated carrier Carrier

The output of the amplifier is an amplified version of the information signal. The carrier is the light from the LED or laser. This light is very close to being monochromatic. Monochromatic light has a constant

46 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

wavelength. The output of the amplifier makes the light produced by the output transducer brighter or dimmer. The output transducer can respond to variations in intensity very rapidly.

LEDS IN DATA TRANSMISSION LEDs are widely used in data transmission systems to transmit data at rates below 270 Mbps (2.7 × 108 bits per second). This means they can be switched on and off at this rate. Laser diodes are used when faster transmission rates are required.

15.6.3 Decoding signals Once the modulated signal has been transmitted through the carrier system, it needs to be changed back to its original form so that it can be used. The first step in this process is known as demodulation.

Demodulation Demodulation is the process whereby the transmitted signal is converted back into an electrical signal. This can then be amplified and used to drive a loudspeaker (for example, in an AM radio) or used by a controlling system to control another device (for example, a television set). In photonic systems, the light signal falls on a transducer called a detector. The most common devices used as detectors are the photodiode or phototransistor. The detector produces an electrical current whose magnitude is proportional to the intensity of the light. This produces an identical electrical signal to that which modulated the light carrier waveform. FIGURE 15.46 (a) Optical intensity demodulation system block diagram (b) Restoring the electrical signal Amplifier

(a)

Output circuit

Detector

(b) Demodulation system

Modulated carrier

Original information signal

15.6 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. (a) What is the bandwidth of an information system? (b) State the bandwidth of a communications system that can carry a maximum frequency of 15 000 Hz. (c) Compare the bandwidths of metal wires and optical fibres. (d) Why does the skin effect occur?

TOPIC 15 How can AC electricity charge a DC device? 47

2. (a) What is optical intensity modulation? (b) Describe how an analog signal — for example, the electrical output signal from a microphone — can be used to alter the intensity of light produced by a light-emitting diode. 3. The following diagram shows part of an optical-fibre telephone system. Driver

Amplifier

Waveguide Microphone

Modulator

Modulator

Speaker

(a) Explain the terms ‘modulation’ and ‘demodulation’ as they apply to the transmission of sound by this system. (b) State a device that could be used as a modulator in this system. (c) State three devices that could be used to demodulate the light signal.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

15.7 Review 15.7.1 Summary • • • • • • • • • • • • • •

A cathode ray oscilloscope shows how voltage varies with time. A multimeter is a diagnostic tool that can be used as an ammeter, an AC or DC voltmeter, or an ohmmeter. DC currents flow in one direction; AC currents periodically change direction. A diode is a semiconductor device that allows current to pass through it in one direction only. Capacitors store charge. The time constant (𝜏) of a resistor–capacitor (RC) circuit is the time it takes the voltage across the capacitor to reach 63% of its final value when charging, or to fall to 37% of its original value when discharging. A rectifier converts alternating currents to direct currents. A half-wave rectifier uses one diode to allow half an AC signal to produce a DC signal. A full-wave rectifier uses an array of diodes to make both directions of an AC signal flow in one direction. Capacitors are used to smooth the outputs of rectifiers. A voltage regulator is a device that keeps the terminal voltage of a voltage supply within required limits despite variations in the input voltage or the load. A Zener diode breaks down at a specific voltage when reverse biased. It can be used as a voltage regulator. Ripple voltage is the periodic variation in a DC voltage that results from the rectification of an AC voltage. A voltage divider consists of two or more resistors arranged in series to produce a smaller voltage at its output.

48 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

•

The output of a basic voltage divider can be calculated using the equation: Vout =

• • • • • • • • • • • • • • • • • •

R2 Vin R1 + R2

A transducer is a device that can be affected by, or affect, the environment. An input transducer transforms non-electrical energy into electrical energy. A thermistor is a semiconductor device whose resistance varies with temperature. The resistance of a negative coefficient thermistor falls as the temperature increases. The resistance of a positive coefficient thermistor rises as the temperature increases. The relationship between resistance and temperature for a thermistor is usually shown graphically. Photonics involves the use of photons to manipulate, encode, transmit, decode and/or store information. A photonic transducer transforms light energy into electrical energy, or electrical energy into light energy. A light-emitting diode (LED) is a semiconductor diode that emits light when a current passes through it. A light-dependent resistor (LDR) is a semiconductor device that has a resistance that decreases as the amount of light falling on it increases. The relationship between resistance and light intensity for an LDR is shown graphically. Photodiodes produce a current when they are reverse biased and light falls on the junction. The current is proportional to the intensity of the light. Phototransistors generally have two terminals. A photosensitive collector–base junction provides the base current. Photodiodes are faster at responding to light than phototransistors, but phototransistors are more sensitive. An optical fibre can carry 30 000 telephone calls at a time. Attenuation is the loss of power of a signal along a communicating channel. The bandwidth of an information system is essentially the highest frequency or rate at which the data can be transmitted. Optical intensity modulation in photonic communications systems involves varying the output intensity of a carrier light source by using the electric signal from a transducer.

Resources

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0040).

15.7.2 Key terms Attenuation is the loss of power of a signal along a communicating channel. Bandwidth is a measure of the amount of data that can be transmitted in a fixed amount of time without the signals becoming so distorted that the data cannot be recovered. A characteristic curve is a graph that shows how one quantity varies against another quantity — for example, voltage and current — for an electric or electronic component. The current-versus-voltage characteristics graph shows how voltage varies with output current for voltage regulators. A decibel is the standard unit used to express gain or loss and relative power levels.

TOPIC 15 How can AC electricity charge a DC device? 49

A diode is a two-terminal semiconductor device that allows current to pass through it in one direction. A half-wave rectifier uses one half only of an AC signal to produce a DC signal. A light-emitting diode (LED) is a small semiconductor diode that emits light when a current passes through it. A light-dependent resistor (LDR) is a device which has a resistance that varies with the amount of light falling on it. Modulation is the process by which a characteristic of one wave (a carrier) is modified by another wave (the signal). A multimeter is a hand-held diagnostic device that can be used as an ammeter, voltmeter or ohmmeter. A photodetector is a device that produces an electric current when subjected to infra-red, visible or ultraviolet electromagnetic radiation. A photodiode is a p–n junction diode designed such that when light interacts with the junction, it greatly increases the reverse bias leakage current. Photodiodes are used in fibre-optic systems to convert light signals into electric signals. A phototransistor is a light-sensitive transistor whose response is controlled by the intensity of incident light on its base. An RC circuit consists of a resistor and a capacitor. A rectifier is a device that converts an AC signal into a DC signal. Ripple voltage is the periodic variation in a DC voltage that results from the rectification of an AC voltage. Ripples are change in the amplitude of a rectified signal. Ripples can be observed using a CRO. Smoothing is the process of reducing or removing variations in the amplitude of a rectified signal. A supply rail is a wire or line in a circuit that is connected to the positive or negative terminal of a voltage supply. Circuit elements are connected between the supply rails. A thermistor is a device that has a resistance that changes with temperature. The time constant for an RC circuit is the amount of time it takes for the circuit to either discharge to 37% of the original voltage across the capacitor, or to charge to 63% of the final voltage across the capacitor. The timebase control affects the speed at which the trace moves across the screen. Trace is the visible path of a moving spot on the screen of a CRO. A transducer is a device that converts energy from one form to another. It can be affected by, or can affect, the environment. The vertical amplifier controls the number of centimetres the trace is detected vertically for every volt of input signal. A voltage divider is a series arrangement of two or more resistors connected across a voltage source. A desired or smaller voltage is obtained at a junction of the resistors. A voltage regulator a device that keeps the output voltage of a power supply relatively constant for a range of operating conditions. Zener diode breaks down at a specific voltage when it is reverse biased.

Resources Digital document Key terms glossary (doc-32288)

15.7.3 Practical work and investigations Investigation 15.1 Constructing a half-wave rectifier Aim: To investigate the output waveform using half-wave rectifiers and capacitors Digital document: doc-31889

Investigation 15.2 Constructing a voltage regulator Aim: To connect a voltage regulator to a rectifier circuit to produce a steady-value voltage supply Digital document: doc-31890

50 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Investigation 15.3 Thermistors Aim: To construct a resistance-versus-temperature characteristic for a thermistor Digital document: doc-31888

Investigation 15.4 Light-emitting diodes (LEDs) Aim: To determine how an LED behaves when it is forward and reverse-biased, and to explore the relationship between current and voltage in an LED Digital document: doc-31891 Teacher-led video: tlvd-0836

Investigation 15.5 Light-dependent resistor Aim: To examine the way in which the resistance of a light-dependent resistor varies with light intensity Digital document: doc-31892

Investigation 15.6 Light-dependent resistor: voltage divider method Aim: To produce a graph and examine how the resistance of an LDR varies with light intensity and distance Digital document: doc-31893

Resources Digital document Practical investigation logbook (doc-32289)

TOPIC 15 How can AC electricity charge a DC device? 51

15.7 Exercises 15.7 Exercise 1: Multiple choice questions Use the following input signal to answer questions 1 and 2. The oscilloscope has a vertical setting of 5 V cm–1 and a horizontal setting of 4 ms cm–1 .

What is the VRMS of this signal? A. 0 V B. 11 V C. 15 V D. 30 V 2. What is the signal frequency, f ? A. 1 Hz B. 25 Hz C. 63 Hz D. 83 Hz 3. An oscilloscope connected across the output of a transformer traces the voltage as shown in the following figure. 1.

The oscilloscope has a vertical setting of 4 V cm–1 and a horizontal setting of 1 ms cm–1 . A multimeter is used to measure the voltage. Which of the following is the reading on the multimeter? A. 8 V B. 12 V C. 17 V D. 24 V 52 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

A student builds a transformer which they will connect to a 24 VRMS AC power. If 800 turns are wound on the primary side, how many turns must there be on the secondary side to deliver 1.5 VRMS ? A. 16 B. 50 C. 12 800 D. 28 800 5. A camera lamp is connected to a battery and a capacitor as follows. 4.

S1

S2

Flash lamp

Which of the following describes what happens in this circuit? A. When switch 1 is closed and switch 2 is open the capacitor charges. B. When switch 1 is open and switch 2 is closed the lamp flashes. C. The brightness of the flash will diminish over time. D. All of the above 6. The following graph shows the voltage of a capacitor over time while it is charging. Voltage (V) 12.0 10.0 8.0 6.0 4.0 2.0 0

0.5

1.0

1.5 2.0 t ( × 10–5 s)

2.5

3.0

Which of the following describes the characteristics of the capacitor? A. It’s time constant, 𝜏 = 5 𝜇s. B. When discharging the capacitor, the voltage after 5 𝜇s would be 4.4 V. C. The capacitor is fully charged after 25 𝜇s. D. All of the above 7. Which of the following best explains why Zener diodes can be used as voltage regulators. A. Zener diodes stop reverse currents flowing until the voltage reaches a known fixed value. B. With a forward current flowing through it there is typically a fixed 0.7 V drop across a Zener diode. C. Zener diodes break down when subject to a reverse current. D. Zener diodes only allow current to flow in one direction.

TOPIC 15 How can AC electricity charge a DC device? 53

A 120 Ω resistor and a 2700 Ω resistor load are connected in series with a 9 V battery. A Zener diode with a breakdown voltage of 6 V is then connected with a reverse bias and in parallel with the 2700 Ω resistor load. What is the current through the Zener diode? A. 0 mA B. 2 mA C. 23 mA D. 25 mA 9. To reduce the ripple voltage supplied to a load the following circuit is connected across the output of a bridge rectifier. 8.

R

C

Which of the following would best reduce the ripple voltage? A. A Zener diode with a larger cut-off voltage B. A resistor with a larger resistance C. A capacitor with a larger capacitance D. None of the above 10. When a handheld remote control is used to switch on a television, the diode on the television changes colour and the TV switches on. Which of the following explanations are most likely true? A. The TV has a built-in motion sensor that switches on when you move the remote control. B. A photodiode in the remote control senses light that is emitted by an LED in the TV. C. A voltage divider circuit with an LDR switches on the TV when it detects the signal from the remote control. D. The LED in the TV is faulty.

15.7 Exercise 2: Short answer questions 1.

A capacitor is connected in series with a resistor and a switch as shown.

60 Ω 12 V

C

54 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

To charge the capacitor should the switch be open or closed? Explain your answer. When fully charged the capacitor is then discharged and its voltage plotted against time as shown.

Capacitor voltage (V)

a.

12

9

6

3

0

0.01 0.02 Time (S)

0.03

What is the value of the capacitor in 𝜇F? What current will flow through the resistor after 0.03 seconds? 2. The resistor in the RC circuit in question 1 is replaced with a 500 Ω resistor. A different capacitor is then charged in the circuit. During charging, the voltage is plotted and graphed as shown. b.

c.

V 12

10

8

6

4

2

0

a. b.

t (s) 0.01

0.02

0.03

0.04

0.05

What is the value of the capacitance in 𝜇F? When the capacitor is initially being charged, what current will flow through the 500 Ω resistor? Explain your answer.

TOPIC 15 How can AC electricity charge a DC device? 55

3.

A diode with the following characteristics is connected in series with a 10 V battery and a 200 Ω resistor. I(mA) 200

100

–3.0

a.

–2.0

1.0

–1.0

2.0

3.0

V

What current will flow through the resistor? A 50 Ω resistor is now added to the circuit as shown. 200 Ω

50 Ω

10 V

What current will now flow through the 200 Ω resistor? 4. In the following circuit, Vin is 40 V, R1 is 50 Ω, RL is 100 Ω and the breakdown voltage of the Zener diode is 20 V. b.

R1 40 V RL

What is the voltage drop across the load resistance, RL ? Calculate the current through the load resistor, RL . c. What is the voltage drop across R1 ? d. Calculate the current through R1 . e. What is the current through the Zener diode? 5. Consider the circuit and character following figure. a.

b.

(a)

(b) Current (mA) 30

A

6.0 V 20

R

10

–6.0

56 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

–3.0 0 1.0 Voltage (V)

2.0

When the current is 15 mA: a. What is the voltage drop across the diode? b. What is the voltage drop across the resistor? c. What is the resistance of the resistor? d. Calculate the power dissipated in the resistor. 6. The diode in the circuit in question 5 is now reversed. a. What is the voltage drop across the diode? b. What is the current flowing in the circuit? c. What is the voltage drop across the resistor? 7. A stair light connected to a sensor circuit turns on when it is dark. The circuit, which is shown, comprises an LDR with the characteristics shown in the graph. R 12 V

1.5 V

Stair light

Switch 106

105

R (Ω)

104

103

102

10 0

0.1

1

102

10

103

104

Illumination (lux)

The switch for the stair light turns on when the voltage across the LDR is 1.5 V. a. If the stair light must turn on when the light level is 40 lux or less, what value of resistance, R, is needed in this circuit? b. To switch the stair light on when the light level is larger than 40 lux, should the value of R be increased or decreased? Explain your answer.

TOPIC 15 How can AC electricity charge a DC device? 57

8.

A light bank is connected in series with a 6 V battery supply and a 50 Ω resistor. The light bank is built using four LEDs, each with current-versus-voltage characteristics shown in the following figure. I (mA) 100

50

–1.5 –1.0 –0.5

0

0.5

1.0

1.5

V

If the light bank is built with the four LEDs in series, what current will flow through the 50 Ω resistor? b. If the light bank is built with the four LEDs in parallel, what current will flow through the 50 Ω resistor? c. Explain the advantages and disadvantages of the options described in parts (a) and (b). 9. An environmental monitoring station transmits data back to base using an optical fibre. Measurements are taken by sensors that trigger voltages when regulation limits are exceeded. Using the voltages from the sensors as input signals, explain the electronic process for delivering and interpreting the information at base. 10. An AC supply is connected to an electronic circuit with unknown components. A CRO shows the following trace when connected across the output from the circuit. a.

3V

0

–3 V

Explain what electronic components have most likely have been used in the circuit and show how they would be connected.

15.7 Exercise 3: Exam practice questions Question 1 (3 marks) A capacitor is connected in series with a resistor as shown.

200 Ω

9V

1000 μF

After the capacitor is fully charged it is discharged.

58 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

What will be the value of the voltage drop over the capacitor 0.2 seconds after discharge has commenced? b. What is the minimum time taken for the capacitor to fully discharge? c. On a set of axes sketch a curve showing the capacitor voltage versus time. a.

1 mark 1 mark 1 mark

Question 2 (11 marks) A student is asked to measure the output of a transformer that is connected to a 240 VRMS AC supply. a. If there are 4800 turns on the primary, how many turns must there be on the secondary to provide 12 VRMS . 1 mark –1 An oscilloscope set with a vertical scale of 5 V cm and a horizontal scale of 5 ms cm is used to trace the voltage output on the secondary. b. Use the grid to draw the trace to scale, showing the value of the maximum and minimum voltages. Assume the frequency of the current is 50 Hz. 3 marks

The student is given four diodes to build a bridge rectifier. Draw a circuit that shows how the four diodes should be connected. 1 mark The rectifier is then connected to the secondary terminals of the transformer and the oscilloscope connected to the output of the rectifier. d. Use the following grid to draw the trace the oscilloscope will now show. 1 mark c.

e.

The student is given a capacitor to add to ‘smooth’ the circuit. On a circuit diagram show where the capacitor should be connected. 1 mark

TOPIC 15 How can AC electricity charge a DC device? 59

f.

Use the following grid to draw the trace the oscilloscope will show when the capacitor is included in the circuit. 1 mark

g.

Explain the effect of decreasing the value of the capacitor in this circuit. 1 mark When one of the four diodes in the bridge rectifier fails it does not allow current to pass through it. Use the grid below to draw the trace that the oscilloscope will now show. 2 marks

h.

Question 3 (6 marks) You are given a thermistor with the characteristics shown in the following graph.

3000

Resistance (Ω)

2500

2000

1500

1000

500

0

5

10

15

20

25

T (°C)

What is the resistance of the thermistor when the temperature is 24 °C? 1 mark You have been asked to build a circuit that will activate a switch when the ambient temperature is 24 °C or more. The switch is activated when the input is 3V, starting a cooling fan. a.

60 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Draw a circuit diagram to show how the thermistor could be connected in a circuit with a 9 V battery to provide a 3 V output to the switching circuit. On this diagram show the value of any components you use 2 marks in your circuit. c. What modifications would be needed in your circuit if the fan is to be switched on at a higher temperature? Explain your answer. 3 marks

b.

Question 4 (6 marks) Chris and Jo have been asked to build two different electronic circuits that can be used to convert a 230 V AC supply to a 6 V DC output. One circuit must be built using the least number of components possible, and the other circuit built to provide the most constant output. With the assistance of circuit diagrams, describe two circuits that would satisfy these criteria. For each circuit also explain the effect of varying the value of the components in these circuits. Question 5 (2 marks) In some intranet systems computers are linked using an optical network. Explain the key steps involved in transferring information from electrical impulses to light signals and then to electrical signals, and the type of devices that might be used.

15.7 Exercise 4: studyON topic test Fully worked solutions and sample responses are available in your digital formats.

Test maker Create unique tests and exams from our extensive range of questions, including practice exam questions. Access the assignments section in learnON to begin creating and assigning assessments to students.

TOPIC 15 How can AC electricity charge a DC device? 61

AREA OF STUDY 2 OPTIONS OBSERVATION OF THE PHYSICAL WORLD

16

How do heavy things fly?

16.1 Overview Numerous videos and interactivities are available just where you need them, at the point of learning, in your digital formats, learnON and eBookPLUS at www.jacplus.com.au.

16.1.1 Introduction The science and technology of flight originated over 2000 years ago, when the Chinese began making and experimenting with kites. The idea of people taking to the air was just a dream until the early sixteenth century, when Leonardo Da Vinci produced detailed drawings of flying machines. However, his machines were not built or tested. The first recorded flight driven and controlled by an engine took place in 1903, by the Wright brothers. That first flight followed many experiments with gliders and using a wind tunnel. This topic deals with the study of aeronautics — the science of flight through the Earth’s atmosphere. It includes the study of the forces that enable flight through the air, aircraft design and jet propulsion.

FIGURE 16.1 The Wright Flyer, 1903 — the first controlled powered aircraft to carry a person

TOPIC 16 How do heavy things fly? 1

Resources Weblink Timeline of aviation

16.1.2 What you will learn KEY KNOWLEDGE After completing this topic, you will be able to: Aerodynamics • model the forces acting on an aircraft in flight as lift, drag, the force due to gravity and thrust • identify aerodynamic forces as arising from the movement of fluid over an object • explain the production of aerodynamic lift with reference to: • Bernoulli’s principle and pressure differences • conservation of momentum and downwash • compare contributions to aerodynamic drag, including skin friction, form and lift-induced • explain the changes in aerodynamic behaviour at supersonic speeds, including compressibility, shock wave formation and increase in drag • explain the production of thrust with reference to Newton’s laws of motion • investigate how it is possible for an aircraft to generate lift when flying upside down Manipulating flight • calculate lift and drag forces acting on an aircraft: 2 1 • lift: F L = 2 CL 𝜌v A • • • • •

•

2 1 • drag: F D = 2 CD 𝜌v A investigate theoretically and practically the variation of lift coefficient with angle of attack, including identification of stall model aerodynamic forces as acting at the centre of pressure and the force due to gravity as acting at the centre of mass calculate the torque applied by a force acting on an aircraft: 𝜏 = r⊥ F describe the roles of the rudder, elevator and ailerons as the primary control surfaces on an aircraft apply balance of forces and torques with reference to Newton’s laws of motion to: • controlling an aircraft in roll, pitch and yaw • stages of flight, including takeoff, climb, cruise, descent, landing and manoeuvres explain the possible advantages and difficulties in designing an unconventional aircraft, such as a flying wing

Applications of flight • apply aerodynamics principles beyond conventional aircraft to investigate practically and/or theoretically at least one of: • strategies to improve the efficiency of cars by reducing drag area (CD A) • the design and use of aerofoil shapes to produce forces in propellers, wind turbines, racing cars or submarines • improving lift in boomerangs, kites or helicopters • the production of thrust using propellers, jet engines and rockets. Source: VCE Physics Study Design (2016–2021) extracts © VCAA; reproduced by permission.

PRACTICAL WORK AND INVESTIGATIONS Practical work is a central component of learning and assessment. Experiments and investigations, supported by a Practical investigation logbook and Teacher-led videos, are included in this topic to provide opportunities to undertake investigations and communicate findings.

2 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Resources Digital documents Key science skills — Units 1–4 (doc-31856) Key terms glossary (doc-32997) Practical investigation logbook (doc-32998)

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0041).

16.2 Forces acting on an aircraft KEY CONCEPTS • Model the forces acting on an aircraft in flight as lift, drag, the force due to gravity and thrust. • Model aerodynamic forces as acting at the centre of pressure and the force due to gravity as acting at the centre of mass.

16.2.1 Forces in flight The main forces acting on an aircraft in level flight (shown in the figure 16.2) can be identified as vertical and horizontal pairs. The vertical force pair is lift and force due to gravity. The horizontal force pair is thrust and drag. Lift acts upwards and at right angles to airflow direction. The lift force is generated over the entire wing, although it can be modelled as acting at one position along a wing. This position is known as the centre of lift or the centre of pressure (CP). The wing is the primary source of lift on a conventional aircraft; however, the fuselage and tail also contribute to the total lift generated. In figure 16.2, the total lift force is represented as a single arrow. FIGURE 16.2 The forces acting on an aircraft in level flight and with constant speed. For the sake of simplicity, all forces are shown as acting through the centre of gravity.

Lift (FL)

Thrust (FT)

Drag (FD)

Force due to gravity (FG)

TOPIC 16 How do heavy things fly? 3

The force due to gravity acting on an aircraft is considered to act through the centre of gravity. This is the point at which all the aircraft’s mass can be considered to be concentrated. It is also the point of balance — if an aircraft were hung from a cable attached to its centre of gravity, it would hang level and in perfect balance. The location of an aircraft’s centre of gravity depends upon two factors: the load it carries (for example, fuel, passengers and cargo) and the positioning of the load within the aircraft. In steady level flight, the lift force and the force due to gravity are equal in size and opposite in direction. The thrust force that causes an aircraft to move through the air comes from the propeller blades or jet engines pushing the air backwards (an action). The air pushed backwards therefore pushes the plane forwards with a force of equal magnitude (a reaction). As an aircraft moves through the air in flight, it experiences air friction or drag. The faster the aircraft moves, the greater the resulting drag force. In figure 16.2, the arrow that represents drag is the result of all the drag forces that act on every part of the aircraft. There are several different types of drag force that act when an aircraft flies. If the net force acting on an aircraft in flight is zero, it maintains a constant velocity. If the net force acting on the aircraft is not zero, the magnitude and direction of the net force determine: • the magnitude and direction of the acceleration of the aircraft • the rate of change of momentum of the aircraft.

16.2.2 Newton’s laws of motion 1. The velocity of an object can change only if there is a non-zero net force acting on it. 2. The relationship between the acceleration of an object, the net force acting on it, and the object’s mass can be expressed as Fnet = ma. The relationship can also be expressed in terms of momentum. The net m∆v force acting on an object is proportional to the time rate of change of momentum; that is, Fnet = . ∆t 3. When an object applies a force (an action) to a second object, the second object applies an equal and opposite force (called a reaction) to the first object.

Resources Digital document Investigation 16.1 Power and thrust (doc-31894)

16.2 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Explain the difference between the centre of pressure and the centre of gravity. 2. Describe the resulting motion, if an aircraft has the following forces acting on it in flight. (a) Lift = 6000 N, drag = 500 N, force due to gravity = 5900 N, thrust = 500 N (b) Lift = 4000 N, drag = 600 N, force due to gravity = 4000 N, thrust = 500 N (c) Lift = 7000 N, drag = 300 N, force due to gravity = 6800 N, thrust = 310 N (d) Lift = 6600 N, drag = 450 N, force due to gravity = 6800 N, thrust = 460 N 3. A business jet is travelling at a constant speed of 200 m s−1 while its engines provide a total thrust of 25 kN. (a) If it is in level flight, what is the magnitude of the total drag on the jet? (b) Assuming that all of the energy delivered by the engines is used to provide thrust, what is the power output of the engines?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

4 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

16.3 Aerodynamic lift KEY CONCEPTS • Identify aerodynamic forces as arising from the movement of fluid over an object. • Explain the production of aerodynamic lift with reference to: • Bernoulli’s principle and pressure differences • conservation of momentum and downwash. • Investigate how it is possible for an aircraft to generate lift when flying upside down. • Apply aerodynamics principles beyond conventional aircraft to investigate practically and/or theoretically: • the design and use of aerofoil shapes to produce forces in propellers, wind turbines, racing cars or submarines • improving lift in boomerangs, kites or helicopters. • Investigate theoretically and practically the variation of lift coefficient with angle of attack, including identification of stall.

16.3.1 Movement of fluids Aeronautics is principally concerned with the motion of aircraft through gases and, in particular, the air. All liquids and gases are fluids. Fluids, like solid objects, are composed of small particles. However, the particles that make up fluids do not behave in the same way as those that make up solid objects because they are able to move more freely.

The Equation of Continuity In the early 1700s, Swiss mathematician Daniel Bernoulli developed the Equation of Continuity: ‘All material that enters a pipe will leave the pipe’. This statement is based on Newton’s mechanics and the idea that mass cannot be created or destroyed.

FIGURE 16.3 Daniel Bernoulli

The Equation of Continuity can be expressed as: Q = v1 A1 = v2 A2 Where: Q = flow rate, measured in cubic metres per second (m3 s−1 ) v = fluid speed, measured in metres per second (m s−1 ) A = cross-sectional area of the pipe, measured in square metres (m2 ). If, as illustrated in figure 16.4, A2 is smaller than A1 , v2 must be greater than v1 . In order to increase the velocity of the fluid through the pipe, work must be done on the fluid travelling through the pipe to increase its kinetic energy. This can occur only if the pressure of the fluid entering the pipe is greater than the pressure of the fluid leaving the pipe. In short, it means that faster moving fluids have lower pressure.

Source: Wellcome Library, London

TOPIC 16 How do heavy things fly? 5

FIGURE 16.4 The flow of fluid through changing cross-sectional areas within a pipe A2 V2 A1

V1

SAMPLE PROBLEM 1

Air flows into a house through an open window. The airspeed is a gentle 30 cm s−1 . It is a sliding window of dimensions 1200 mm × 400 mm. Inside the room, the far door is opened 10 cm. The door is a standard height of 2050 mm. What is the speed of the air as it flows through the door? THINK 1.

Find the area of the window.

2.

Find the fluid speed through the window.

WRITE

Window area = A1 = 1.2 × 0.4 = 0.48 Fluid speed through the window = v1 = 0.30 m s−1

3.

4.

5.

Find the area of the door.

Recall that the rate of air flow is a constant.

State the solution.

Door area = A2 = 0.10 × 2.05 = 0.205 m2 Flow rate in through the window = flow rate out through the door v1 A1 = v2 A2 0.30 × 0.48 = 0.205 × v2 v2 = 0.70 m s −1 The air flows through the door at a speed of 70 cm s−1 .

PRACTICE PROBLEM 1 a. Water flows into one end of a pipe at a speed of 1 m s−1 . The radius of the pipe at the point of entry is 5 cm. The pipe narrows so that the water leaves through an opening that is only 2 cm in radius. With what speed does the water flow out of the pipe? b. If a fluid flows at a speed of 1.2 m s−1 through a pipe of cross-sectional area 0.45 m2 , at what speed will it flow when the cross-sectional area: i. narrows to only 0.32 m2 ii. widens to 0.60 m2 ?

6 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

16.3.2 Fluid speed and pressure By applying the Law of Conservation of Energy to fluid flow, Bernoulli derived the following equation. 1 2 𝜌v + 𝜌gh + P = constant 2 Where: 𝜌 = the fluid density (measured in kg m−3 ) v = the speed of the fluid (measured in m s−1 ) g = acceleration due to gravity (measured in m s−2 ) h = the vertical displacement of the fluid (measured in m) P = the static pressure of the fluid (in N m−2 or Pa). According to this principle, the total energy of the fluid is constant throughout the flow. In simplest terms, this energy comprises the kinetic energy due to motion, the gravitational potential energy due to changes in height and the potential energy associated with the pressure of the fluid. This equation does not apply to aircraft moving at supersonic speeds due to changes in the behaviour of the fluid and significant heating of the wing. In 1738, the Swiss mathematician Leonhard Euler derived an equation that related the speed of a fluid to its static pressure. He found that an increase in speed, created when a liquid flows through a narrower section of a pipe, produced a decrease in static pressure. Likewise, a decrease in speed (by enlarging the pipe) produced an increase in static pressure.

16.3.3 The Bernoulli principle The statement that the pressure of a fluid decreases as its velocity increases is known as the Bernoulli principle, in honour of Euler’s mentor, Daniel Bernoulli. Euler applied Bernoulli’s equation to a fluid at a constant height, producing the equation: 1 2 𝜌v + P = constant 2 The cross-section of an aircraft wing forms a shape known as an aerofoil. This shape is designed so that air travelling over it will speed up. According to the Bernoulli principle, this reduces the air pressure above the wing. The opposite happens on the lower surface, with a lower speed resulting in an increased pressure. The result is an upwards force known as the lift force. FIGURE 16.5 How lift is developed by an aerofoil Lift High speed, reduced pressure

Aerofoil

Direction of motion

Low speed, increased pressure

TOPIC 16 How do heavy things fly? 7

Formula One racing cars are fitted with aerofoils at the front and back, called the front and rear ‘wings’. They are actually inverted wing shapes that help push the tyres onto the road and hence increase their grip. This also increases the drag force on the car, but the improved handling outweighs the effect of the increased drag. The descriptions of the movement of smooth surfaces in air can be extended to surfaces travelling through other fluids such as water. Ships must not pass within several metres of each other as they will be drawn together and may collide. Boats approaching a wharf may experience the same effect if they come in too fast.

FIGURE 16.6 Aerofoils are not always used to provide lift. In racing cars, aerofoils are used to oppose lift and increase the car’s grip on the track.

Resources Digital document Investigation 16.2 Bernoulli effects (doc-31895) Teacher-led video Investigation 16.2 Bernoulli effects (tlvd-0840)

16.3.4 Aerofoil characteristics There are two key factors in the ability of an aerofoil to generate lift. The first is its shape. The front of the aerofoil is called the leading edge. This is usually quite rounded so it deflects the airflow above and below the wing. The rearmost point of the aerofoil is called the trailing edge. At the trailing edge, the upper and lower surfaces of the wing come to a sharp point to reintroduce the two airstreams with minimal disturbance. The theoretical line directly between the leading and trailing edge is called the chord line. Many simple aerofoils are symmetrical about the chord line. However, the majority of aerofoil shapes used today are not symmetrical and feature a curvature that is called camber, which is represented by the camber line.

8 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 16.7 Terminology used to describe the shape of an aerofoil Chord line Camber line Upper surface Leading edge

Lower surface Trailing edge

The second factor in an aerofoil’s ability to generate lift is its orientation to the undisturbed airflow. This is known as the angle of attack and is measured as the angle between the undisturbed airflow and the chord line of the wing. Simple symmetrical aerofoils only generate lift when placed at an angle to the airflow, whereas a cambered aerofoil can generate lift at an angle of attack of zero degrees. FIGURE 16.8 The angle of attack is the angle between the wing of an aircraft and the direction of airflow. Trailing edge

Angle of attack

Direction of airflow across aerofoil

As the angle of attack is increased from zero degrees, the difference between the pressure on the upper and lower surfaces increases and more lift is generated. This is accompanied by an increase in drag due to the greater disturbance in the airflow. At large angles of attack, the airflow on the top of the wing can separate and a swirling pattern of turbulence can be created. This reduces the lift and increases the drag and is commonly referred to as stall. FIGURE 16.9 The effect of increasing the angle of attack; if the angle of attack becomes too great, turbulence is created, lift decreases and the drag increases. Lift Lift

Lift

Drag Aerofoil

Direction of airflow

Drag

Direction of airflow

Drag

Direction of airflow

Turbulence

TOPIC 16 How do heavy things fly? 9

WIND TURBINES The design of a wind turbine is very similar to an aircraft propeller engine, but in reverse. The wind moves over the blades of the turbine, generating a force that causes the blades to rotate around the central hub. Within the hub this rotation is transformed into electrical energy using a generator.

FIGURE 16.10 Wind turbines create energy from wind.

16.3.5 Newton’s Third Law of Motion The mechanism by which aerofoils and wings generate lift can also be explained using Newton’s third law. The overall effect of the movement of the wing through the air is to push the air downwards, creating what is called downwash. This can be thought of as an action, with the wing pushing the air downwards. The Newton’s third law reaction pair of this is the air pushing the wing upwards, or in other words the generation of lift. This is not an additional source of lift, but rather another approach to understanding the generation of lift by a wing. FIGURE 16.11 Newton’s third law can be used to explain the lift force on a wing. Force on wing by air Direction of airflow

Force on air by wing

16.3.6 Flying upside down The wing of an aircraft is specifically designed to be efficient and effective in steady level flight as well as a range of manoeuvres. It did not take long in the history of aviation before pilots extended normal flight to flying aircraft upside down. In a very simple design with a symmetrical aerofoil for the wing, the lift produced is directly related to the angle of attack, so once such an aircraft has been manoeuvred into an upside down position, it can still generate lift, provided it is oriented correctly. This requires the aircraft controls and engine to be capable of operating in such an orientation — and the same must be said of the pilot. The design of efficient modern passenger aircraft has produced more complex aerofoil shapes that cannot simply be flipped upside down and expected to respond in the same way to a change in angle of attack. Not to mention the thrust required to manoeuvre into an upside down position and the associated loading on an aircraft structure. This is one of many reasons that specialised aircraft designed for acrobatics are used by those intending to fly upside down. 10 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 16.12 A stunt plane is able to fly upside down due to the design and manoeuvrability of its wings.

16.3 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. What basic difference between fluids and solids causes them to behave differently in terms of their motion? 2. If a fluid flows through a pipe of cross-sectional area 61 cm2 at a speed of 9.3 cm s−1 , what must the cross-sectional area be to make it speed up to 13 cm s−1 ? 3. If a fluid flows at a speed of 2.1 m s−1 through a pipe of diameter 0.15 metres, what speed will it flow at when the pipe widens to a diameter of 0.45 metres? 4. Air flows through a wind tunnel with a circular cross-section. (a) How would you change the cross-sectional area of the wind tunnel in order to double the speed of the air passing through it? (b) By what factor would the radius of the wind tunnel change to achieve the doubling of airspeed? 5. Assuming that everything else remains constant, what change in diameter of a wind tunnel would produce a 10-fold increase in the speed of the air moving through it? 6. Describe what happens to the pressure in a fluid as its speed increases. 7. Explain in terms of Bernoulli’s principle how an aerofoil develops lift. 8. On the following aerofoil: (a) Draw and label an arrow to represent the lift force acting on the aerofoil. Direction of airflow (b) Draw and label the angle of attack. (c) Label the trailing edge of the aerofoil.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

TOPIC 16 How do heavy things fly? 11

16.4 Thrust and drag KEY CONCEPTS • Explain the production of thrust with reference to Newton’s laws of motion. • Compare contributions to aerodynamic drag, including skin friction, form and lift-induced. • Apply aerodynamics principles beyond conventional aircraft to investigate practically and/or theoretically: • the production of thrust using propellers, jet engines and rockets.

16.4.1 Producing thrust In order to move through the air, an aircraft must produce thrust. This is generally done with one or more propellers, or with jet engines. A propeller-driven aircraft is actually pulled through the air. The propeller itself is an aerofoil turned by the engine and each blade of the propeller generates its own ‘lift’ force. A large component of the ‘lift’ force on the propellers is in the direction of motion of the aircraft and provides the thrust. Regardless of the specific design of the engine, the production of thrust can also be understood simply in terms of Newton’s third law. The engine has the effect of pushing the air backwards, and in response the air pushes the engine and, therefore, the aircraft forwards. The power output, P (in W), when a force, F (in N), is applied to an object causing the object to move with a speed, v (in ms−1 ), is given by the equation: P = Fv The thrust and speed of an aircraft can be related to the mechanical power output of the engines using the same equation. Thus, for an aircraft: mechanical power output = thrust × speed

16.4.2 Types of drag There are several types of drag that are created when an aircraft moves through the air; the two main types are induced drag and parasite drag.

Induced drag Induced drag occurs due to the three-dimensional nature of an aircraft wing. At the tip of the wing the difference in pressure between the upper and lower surfaces draws air around the tip towards the upper surface, creating a large whirl of air called a vortex. This changes the effective angle of attack across a significant length of the wing span, resulting in a proportion of the lift force acting to oppose the forward motion of the aircraft. This component of the lift is called induced drag.

DESIGNING WINGS TO REDUCE DRAG Modern passenger aircraft commonly feature a vertical surface at the wing tip known as a winglet or wing tip fence. This is designed to decrease the interaction between the upper and lower surfaces, reducing the size of the vortex and in turn reducing the induced drag. This reduced drag leads to lower fuel consumption and decreased operating costs.

12 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 16.13 A winglet on a commercial plane

Parasite drag Parasite drag is the name given to the combined effect of skin friction drag and form drag. Neither of these forces, in contrast to induced drag, contributes towards the lift force, so they are given the collective name of parasite drag. Skin friction drag arises from the friction force caused by the contact between air and the surface of the aircraft, and it can be reduced by keeping the surface of the aircraft clean and polished. Form drag is the drag due to the shape or ‘form’ of a surface. Improving the streamlining of an aircraft (or a car, truck or train) reduces the amount of form drag it experiences. Smooth covers called fairings are used to enclose, or partially enclose, shapes such as wheels or joints between different parts of an aircraft. Parasite drag increases roughly proportionally to the square of the speed. For example, if the airspeed is doubled, the parasite drag will increase by roughly a factor of four. It is therefore in the best interests of aircraft designers to create aircraft that are as aerodynamically streamlined as possible. FIGURE 16.14 A wide body airliner with vortexes and condensation above the wing

16.4.3 Lift-to-drag ratio In the final analysis, it is the total drag on an aircraft that determines the necessary thrust required for it to achieve a given airspeed. If we try to get more lift from an aircraft by increasing its speed, for example, we usually get more drag. For every aircraft, there is a particular airspeed that generates the minimum drag, and this is also the airspeed that will produce the maximum lift-to-drag ratio. The higher the lift-to-drag ratio, the less power is required from the engine. The ratio of lift to drag can be found from the ratio of horizontal distance travelled (glide distance) to loss of altitude when gliding. This is known as the glide ratio. The glide ratio and the lift-to-drag ratio have the same value. A gliding aircraft with a glide ratio of 9:1 can fly 9 units of distance forward through the air for every 1 unit of height lost. Some modern gliders used for long distance racing have a glide ratio of more than 50:1. Large commercial passenger planes have glide ratios of about 10:1. The lift-to-drag ratio changes with the angle of attack. As the angle of attack increases from zero, the lift-to-drag ratio increases. Once the angle reaches about 15–20°, turbulence occurs — the lift decreases and the drag increases. When a plane slows down, the lift decreases. The pilot can restore the lift, and the lift-to-drag ratio, by slightly increasing the angle of attack. TOPIC 16 How do heavy things fly? 13

FIGURE 16.15 For a gliding aircraft, the lift-to-drag ratio is equal to the glide ratio. Drag

Force due to gravity

Lift

The two right-angled triangles are similar. glide distance lift = glide ratio Therefore, = drag loss of altitude

θ

Loss of altitude

Glide path

θ Glide distance

SAMPLE PROBLEM 2

After the engines of an aircraft fail, it glides in a straight line to a safe landing in a field 5.6 kilometres from the point directly below where the engines failed. The glide ratio of the aircraft is 12:1. a. How much altitude is lost by the aircraft for every 1 kilometre of ground distance covered? b. What was the altitude of the aircraft when its engines failed? Assume that the ground over which the aircraft glides is level and that there is insignificant thermal activity in the air. c. What is the lift-to-drag ratio while the aircraft is gliding?

THINK a. 1.

2.

b. 1.

2.

Recall the formula for glide ratio and rearrange to make loss of altitude the subject.

State the solution.

Recall the formula for glide ratio and rearrange to make loss of altitude the subject.

State the solution.

Recall the lift-to-drag ratio is equal to the glide ratio. 2. State the solution.

c. 1.

14 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

WRITE

glide distance loss of altitude glide distance Loss of altitude = glide ratio 1000 = 12 = 83 m The aircraft loses 83 metres of altitude for every 1 kilometre of ground distance covered. glide distance b. Loss of altitude = glide ratio 5.6 × 103 = 12 = 4.5 × 102 m The aircraft had an altitude of 450 metres when its engines failed. c. Lift-to-drag = glide ratio = 12:1 The lift-to-drag ratio while the aircraft is gliding was 12:1. a.

Glide ratio =

PRACTICE PROBLEM 2 a. A glider loses 120 m in altitude over a ground distance of 2.5 kilometres during an exhibition of straight line flight. i. What is the glide ratio of the glider? ii. What is the lift-to-drag ratio of the glider? b. During a safety drill, a pilot must glide a small plane in a straight line to a safe landing from an altitude of 500 metres after the engines are switched off. The glide ratio of the plane is 24:1. What ground distance should the pilot allow for the landing after switching off the engines?

Resources Digital documents Investigation 16.3 Investigating gliders (doc-31898) Investigation 16.4 Turbulence (doc-31896)

16.4.4 Faster than sound As an aircraft approaches the speed of sound, the behaviour of the airflow over the aircraft changes. The air becomes compressible, rendering the standard theories of fluid flow behaviour outlined in the preceding sections inadequate. At such high speeds, the rate at which the aircraft is disturbing the air forms shock waves. A shock wave is a propagating disturbance associated with abrupt changes in temperature, pressure and density. The formation of a shock wave on the surface of a wing creates significant turbulence in its wake, which leads to a dramatic increase in drag. The Mach number is the ratio of speed of an aircraft to the speed of sound. Mach number =

vaircraft vsound

The speed of sound is approximately 343 m s–1 or 1235 km h–1 ; an aircraft travelling at this speed is said to be moving at Mach 1. The speeds at which aircraft travel are broken into four broad areas: • Subsonic speeds are significantly less than Mach 1. The effects of compressibility can be ignored. • Transonic speeds are around Mach 0.8 to 1. The entire aircraft is moving slower than the speed of sound; however, as the air speeds up over the top of the wing it may exceed the speed of sound, causing shock waves to form and significantly increasing drag. • Supersonic speeds are Mach 1 to 5. The majority of air interacting with the aircraft is travelling faster than the speed of sound. Shockwaves are present and drag is significantly increased. Heating becomes increasingly prominent at higher speeds. • Hypersonic speeds are Mach 5 and above. In addition to the supersonic effects, it becomes necessary to consider the chemistry of the air molecules. The majority of passenger aircraft travel at subsonic speeds to avoid the significant increase in drag associated with speeds in the transonic region and above. Aircraft designed to travel at supersonic speeds require a thinner and more swept back wing, significantly more thrust capability, carefully designed control surfaces, and specialised engine intakes and outlets. Supersonic travel remains elusive to the general public due primarily to much higher costs from higher fuel consumption and limited flight paths due to the disruption caused by sonic booms affecting populated areas.

TOPIC 16 How do heavy things fly? 15

16.4 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. The Airbus A380 has four jet engines, each capable of producing a maximum thrust of 370 kN. If the aircraft is travelling at 270 km h–1 and each engine is running at maximum thrust, what is the total mechanical power output of the aircraft? 2. What is the cause of wing-tip vortices? 3. The following graph shows how the parasite drag and induced drag acting on a particular aircraft change as the airspeed changes.

Parasite drag

Drag (N)

Induced drag

0

20

40

60

80 100 120 140 Air speed (km h–1)

160

180

200

(a) On the graph, draw and label the curve representing the total drag (that is, the sum of the parasite and induced drags). (b) At what airspeed does the maximum lift-to-drag ratio occur? (c) Explain the importance of a high lift-to-drag ratio. (d) Which region of the graph represents the conditions under which a stall is likely to occur? (e) Explain what happens to the air around an aircraft wing when a stall occurs. 4. An aircraft has a glide ratio of 9:1 and is at an altitude of 1200 metres when its engine cuts out. (a) How far could it travel before landing, assuming minimal thermal activity? (b) What is the lift-to-drag ratio for this aircraft while it is gliding? 5. A glider loses 800 metres in altitude while it covers a ground distance of 12 kilometres. Calculate its: (a) glide ratio (b) lift-to-drag ratio. 6. A glider with a glide ratio of 40:1 glides in a straight path over a ground distance of 3.6 kilometres to make a perfect landing. What was its initial altitude?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

16 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

16.5 Manipulating flight KEY CONCEPTS • Calculate lift and drag forces acting on an aircraft: 2 1 • lift: F L = 2 CL 𝜌v A 2 1 • drag: F D = 2 CD 𝜌v A. Calculate the torque applied by a force acting on an aircraft: 𝜏 = r⊥ F. • Apply aerodynamics principles beyond conventional aircraft to investigate practically and/or theoretically: • • strategies to improve the efficiency of cars by reducing drag area (CD A).

16.5.1 Lift and drag coefficients The lift coefficient (CL ) and drag coefficient (CD ) are used by aircraft designers to provide a comparable measure of the amount of lift and drag generated by a particular shape as it moves through the air. The lift coefficient and drag coefficient are defined as: F F CL = 1 L2 and CD = 1 D2 𝜌v A 𝜌v A 2 2 Where: CL = the lift coefficient CD = the drag coefficient FL = the lift force generated (measured in N) FD = the drag force generated (measured in N) 𝜌 = the fluid density (measured in kg m–3 ) v = the speed of the fluid (measured in m s–1 ) A = the reference area in m2 . (For lift this is the area of the wing viewed from above, for drag this is the area of the wing viewed from the front.) The values of the lift and drag coefficients for different designs are determined by varying the shape, angle of attack and speed, and taking measurements of the forces generated. This can be done experimentally in a wind tunnel or using computer-based analysis tools. The lift and drag coefficient formulae can be rearranged to give the following formulae for calculating the lift and drag forces. FL = CL 12 𝜌v2 A and FD = CD 21 𝜌v2 A These relationships can be very useful for estimating the effect that a change in the design or configuration of an aircraft will have on the lift and drag forces produced. For example, it can be seen that doubling the velocity of the aircraft will quadruple the lift and drag force generated.

AERODYNAMICS OF CARS Reducing the drag force is crucial for designing energy efficient cars. To minimise drag, it is necessary to design for the lowest value of drag coefficient multiplied by frontal area (commonly referred to as drag area) or CD A. The frontal area can be minimised by designing more compact vehicles, and the drag coefficient can be reduced by streamlining the vehicle shape to reduce disruption to the airflow. Modern vehicles have drag coefficients between 0.25 and 0.35, with the Tesla Model S setting the benchmark at 0.24. The drag area of the first mass-produced electric car, the General Motors EV1, was 0.37 m2 compared to 2.47 m2 for the Hummer H2. The Tesla Model S has a drag area of 0.57 m2 .

FIGURE 16.16 A Tesla Model S car being charged

TOPIC 16 How do heavy things fly? 17

Wind tunnels A great deal of research into the effects of airflow is done FIGURE 16.17 A scale model is used in a in wind tunnels. Models of aircraft (and cars, trucks, trains, wind tunnel to assist in aircraft design. and so on) are fixed in the wind tunnel. A large fan creates an artificial airflow that is effectively the same as if the air were still and the aircraft was moving. Sensors can be used to measure lift and drag and special cameras can detect variations in the airflow caused by changing the position or dimensions of the control surfaces. Advances in computer technology have led to significant advances in aerodynamic modelling using computational fluid dynamics (CFD); however, due to the complex nature of aerodynamic behaviours there is still widespread use of wind tunnels alongside simulations. Simple wind tunnels can now be built, leased or even bought in kit form. Plans for building your own wind tunnel Source: NASA can be downloaded from the internet. Simulations of wind tunnel investigations are also available on the internet. FIGURE 16.18 A replica of the wind tunnel built and used by the Wright brothers in 1901 to help in the design of the gliders they built before their first powered flight in 1903. The data obtained from their 1901 wind tunnel was used in the design of the propellers for their powered Flyer.

Resources Digital document Investigation 16.5 DIY wind tunnel (doc-31899) Weblink

Online wind tunnel simulation

16.5.2 Turning effect of a force For an aircraft to maintain level, steady flight — that is, constant speed at a constant altitude — the vertical force pair of lift and the force due to gravity must be balanced, and the horizontal force pair of thrust and drag must also be balanced. This results in the whole aircraft being balanced or in equilibrium. However, it is not only the size of the forces that must be taken into account, but also the distance between the line of action of the force and centre of gravity of the aircraft.

18 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

The location where the lift force on the wing acts, commonly called the centre of pressure (CP), is not in the same location as the centre of gravity of the aircraft. If unbalanced, this would cause the aircraft to rotate. Similarly, the drag and thrust forces do not necessarily act along the same line, which can also cause an aircraft to rotate. This tendency for the aircraft to rotate is due to the turning effect of the forces. The turning effect of a force is called a torque, or moment. In this topic, the term torque will be used.

The magnitude of the torque created by a force about a given pivot point can be calculated by: 𝜏 = F × r⊥ Where: 𝜏 = magnitude of the torque, measured in units of newton metres (N m) F = magnitude of the force, measured in units of newton (N) r = perpendicular distance between the line of action of the force and the pivot or reference point, measured in metres (m). When an object is not rotating, it is said to be in rotational equilibrium. In order to achieve rotational equilibrium about an axis, the clockwise torques must be balanced by the anticlockwise torques. In other words, the net torque, 𝜏 net , must be zero. Imagine two children sitting on opposite sides of a seesaw, as shown in figure 16.19. If they want to be in rotational equilibrium they each must sit at a distance from the centre of the seesaw such that the net torque is zero. Thus, 𝜏 clockwise = 𝜏 anticlockwise . FIGURE 16.19 Two children sitting on opposite sides of a seesaw R

Child 1

Child 2

m2 g m1 g

d2 1.5 m

They would normally find the right distances by trial and error, but if they felt inclined to use calculations they would take torques about the centre of the seesaw. The only forces that need to be considered are the respective force due to gravity on the children. The normal force applies no torque because it acts at the pivot point. So if child 1 was 1.5 metres from the centre, the equation becomes: m2 g × d2 = m1 g × 1.5 m The distance d2 can be calculated as long as their masses are known. When an aircraft flies normally in a straight line, the net torque on it about any axis is zero. It is in rotational equilibrium. Although the forces on an aircraft are ideally kept in balance, it is always advisable to have an extra force available that can be called into action when required to correct any out-of-balance occurrence. TOPIC 16 How do heavy things fly? 19

In a conventional aircraft design this force is provided by the tail plane. The tail assembly of an aircraft comprises a vertical tail fin and a horizontal tail plane, called the vertical and horizontal stabilisers, respectively. The tail plane provides a downwards force called the tail lift. The balance of the aircraft can be corrected by adjustments to the tail plane to make the net torque equal zero. The load in an aircraft must be ‘balanced’ to achieve level flight. When luggage is loaded on commercial aircraft the location of the total mass has been carefully considered. Luggage is weighed at passenger check-in to allow accurate calculation of location of baggage. The mass and location of the fuel for the flight must also be considered. It must be positioned so that it doesn’t produce an unwanted turning effect. SAMPLE PROBLEM 3

An aircraft in level flight has a total wing lift of 3.6 × 106 N. The centre of pressure is 12 metres from the centre of gravity and the tail is a distance of 45 metres from the centre of gravity as shown in the following figure. Calculate the tail lift. dW

FLW dT

Tail fin

Tail plane

Centre of gravity

THINK 1

2

3

In level flight there must be no rotation about the centre of gravity. Take torques about the centre of gravity. Recall that the centre of pressure is the point at which the lift is considered to be acting on the wings.

Substitute in the known values. State the solution.

FLT WRITE

𝜏 net = 0 𝜏 clockwise = 𝜏 anticlockwise LT × dT = Lw × dw Where: Lw = wing lift LT = tail lift (The torque due to the weight is zero because the weight acts through the centre of gravity.) LT × 45 = 3.6 × 106 × 12 LT = 9.6 × 105 N The tail lift is equal to 9.6 × 105 N in a downwards direction.

PRACTICE PROBLEM 3 a. Calculate the tail lift required to balance the wing lift of 480 000 N for a light plane in which the wing distance, dw , is 3 metres behind the centre of gravity and the tail distance, dT , is 8 metres behind the centre of gravity. The plane is in level flight. b. A tail lift of 2 × 105 N balances a wing lift of 6.5 × 105 N for a plane in level flight. The centre of pressure is 4 metres behind the centre of mass. How far is the tail from the centre of mass?

20 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

SAMPLE PROBLEM 4

Excess baggage of mass 120 kilograms is loaded into a plane. The turning effect of the excess baggage is balanced by loading extra fuel into a tank located 2.4 metres in front of the centre of gravity of a plane before it takes off. The excess baggage has been loaded 1.6 metres behind the centre of gravity. What mass of extra fuel must be added? Assume that g = 9.8 N kg−1 . THINK 1

Take torques about the centre of gravity.

2

State the solution.

WRITE

𝜏 net = 0 𝜏 fuel = 𝜏 baggage mfuel g × dfuel = mbaggage g × dbaggage mfuel × 9.8 × 2.4 = 120 × 9.8 × 1.6 120 × 9.8 × 1.6 mfuel = 9.8 × 2.4 = 80 kg A mass of 80 kilograms of extra fuel must be added to the tank.

PRACTICE PROBLEM 4 a. Unexpected cargo of 600 kilograms is loaded onto a plane in a hold 2.8 metres in front of the centre of mass of the plane. In order to compensate for the turning effect of this cargo, the pilot loads extra fuel into a reserve tank located 4.2 metres behind the centre of mass. What mass of extra fuel should the pilot add? b. Two containers with masses of 2000 and 3000 kilograms must be loaded onto a cargo plane so that they have no net turning effect. The lighter container is placed 6 metres in front of the centre of mass. Where must the heavier container be placed?

16.5 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. The lift equation is given as: F L = CL 21 𝜌v2 A What would be the overall effect of the lift produced by an aircraft in each of the following scenarios? (a) Decreasing the wing area by a factor of 1.5 by retracting the flaps (b) Increasing the aircraft speed by a factor of 2 (c) Increasing the lift coefficient by a factor of 2 by deploying the flaps (d) Halving the aircraft speed 2. An aircraft in level flight has a wing lift force of 15 000 N. The centre of pressure is 1.1 metre behind the centre of gravity. The tail lift acts at a distance of 9.3 metres from the centre of gravity. (a) What is the size and direction of the tail lift? (b) Why is it not necessary to know the mass of the plane to answer part (a) of this question? 3. If the tail lift in question 2 was reduced while the wing lift remained the same, the centre of gravity of the aircraft would need to be shifted. How can this be achieved?

TOPIC 16 How do heavy things fly? 21

4. A cargo plane is loaded with two large containers. The first container, which has a mass of 10 000 kilograms, is loaded into a hold located 12 metres behind the plane’s centre of gravity. The second container, which has a mass of 15 000 kilograms, is loaded so that it compensates for the torque applied by the first container. Where should the second container be located? 5. A passenger plane’s rear fuel tank has been filled to allow for the usual amount of baggage in the rear hold, which is located 3.6 metres behind the plane’s centre of mass. The rear fuel tank is 4 metres behind the centre of mass. However, an extra 200 kilograms of cargo has been placed in the hold. What mass of fuel must the pilot release from the rear tank before taking off to compensate for the turning effect of the extra baggage?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

16.6 Controlling an aircraft KEY CONCEPTS • Apply balance of forces and torques with reference to Newton’s laws of motion to: • controlling an aircraft in roll, pitch and yaw • stages of flight, including takeoff, climb, cruise, descent, landing and manoeuvres. • Describe the roles of the rudder, elevator and ailerons as the primary control surfaces on an aircraft. • Explain the possible advantages and difficulties in designing an unconventional aircraft, such as a flying wing.

16.6.1 Flight controls The primary control surfaces on an aircraft are called the elevator, rudder and ailerons. These components can be moved in either direction to change the camber and angle of attack of the aerofoil. This changes the size, and sometimes the direction, of the force generated, which in turn generates a torque that causes the aircraft to rotate in a particular direction. Movement around the lateral axis is known as pitch and is controlled by moveable areas on the rear edge of the horizontal tail plane called elevators. Raising the elevators decreases the camber of the tail, creating more downwards force, in turn raising the nose of the aircraft. Moving the elevators down produces the opposite effect. Movement about the vertical axis is known as yaw and is controlled by a single moveable area along the rear edge of the vertical tail known as the rudder. This controls the direction, left or right, in which the aircraft is pointing. Movement about the longitudinal axis is known as roll, which is controlled by moveable areas on the trailing edges of both wings, known as ailerons. The ailerons work in opposite directions, one up and the other down. The aileron that moves up reduces the lift, so that the wing will drop. The aileron that moves down causes more lift, so that the wing will rise. The resultant turning effect causes the plane to roll, or bank.

22 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 16.20 Pitch, yaw and roll

Yaw Longitudinal

Pitch

Roll

Centre of gravity

Lateral

Vertical

16.6.2 Stages of flight The typical flight of an aircraft contains a number of stages, each of which require a difference balance between the forces on the aircraft. • During takeoff the thrust must be significantly greater than the drag so that the aircraft will accelerate down the runway. Once the aircraft reaches the appropriate speed, the pilot will use the elevator to raise the nose, increasing the lift. Once the lift is greater than the force due to gravity the aircraft will accelerate upwards. • During climb the aircraft will be flying upwards at an angle (figure 16.21a). As a consequence of this incline, a component of the force due to gravity acts to oppose the motion of the aircraft, requiring greater thrust to balance this out. • The cruise stage is where aircraft are designed to spend most of their operating life. During this stage the forces are in balance; lift equals force due to gravity and thrust equals drag. Minimising drag in this configuration will minimise fuel consumption. • When an aircraft makes a turn or manoeuvre to the left or right you’ll notice that it also rolls ‘into’ the turn (figure 16.21b). This is to re-orient the lift force so that a component of it provides the required force left or right to cause the aircraft to change direction. • During the glide or descent phase of flight the nose of the aircraft is pointing slightly downwards (figure 16.21c). At this angle a component of the force due to gravity acts in the direction of motion of the aircraft, allowing the thrust to be reduced. • In preparation for landing the pilot will decrease the speed of the aircraft by reducing the thrust. The required lift is maintained using moveable devices on the rear of the wing called flaps that increase the area and camber of the wing, producing more lift at a lower speed. FIGURE 16.21 Forces during climb, turn and glide (a)

(b)

Lift

(c)

Lift

Lift Thrust Component of force due to gravity opposing motion

Component of lift to create turn

Drag Fg

Drag

Component of force due to gravity in direction of motion Thrust

Fg

Fg

TOPIC 16 How do heavy things fly? 23

Unconventional aircraft The basic configuration of the majority of aircraft in service today is fundamentally the same as almost a century ago: a cylindrical fuselage, a wing attached towards the front of the fuselage, and a tail with horizontal and vertical stabilisers at the rear. A number of alternate configurations have been trialled with varying success. The primary challenge faced by many alternate approaches is creating an aircraft that is stable and can be controlled. In recent years the development of sophisticated computer-based flight control systems has made it possible to consider aircraft designs that would never have been viable with manual controls. A canard configuration aircraft has the horizontal stabiliser and control surface, called a canard, in front of the wing. This was the configuration of the Wright Flyer. The primary advantage of this configuration is that both the canard and the wing can contribute to the lift generated, whereas a typical horizontal tail opposes the lift of the wing. The interaction between the canard and the wing is complex and was not well understood in the early decades of aviation. FIGURE 16.22 A canard in flight

Source: © Brandon Thetford

A flying wing has no fuselage and is in effect a very large wing with all payload carried inside it. There is potential for a very streamlined and aerodynamically efficient design, however, with significant complexities in stability, control and engine placement.

24 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 16.23 A flying wing aircraft

Source: Transferred from the US Air Force

A v-tail aircraft has two rear surfaces in a v-shape. Advantages of this configuration include reduced drag as well as enabling different engine placement. This is currently rarely used in passenger aircraft but is quite common in unmanned aerial vehicles (UAVs). FIGURE 16.24 A v-tail aircraft

TOPIC 16 How do heavy things fly? 25

Additional alternate configurations that have been explored include: pivoting oblique wings, lifting bodies, blended wing and body, joined or box wing, multiple fuselage aircraft and wing-in-ground effect vehicles. Most have been explored in the quest to improve performance, payload or efficiency. With future pressures to make aviation more efficient and transition to zero emissions transport, it will be interesting to see if more unconventional configurations come to the fore in future decades. FIGURE 16.25 Boeing’s X-48B uses blended wing–body technology.

Source: Tony Landis for NASA

High-lift devices The design of an aircraft is a series of tradeoffs. A typical passenger aircraft spends most of its time at cruise, and this dominates the basic configuration. When landing, an aircraft needs to travel much slower, so needs to generate a lot more lift at a lower speed. To do this, aircraft use a range of high-lift devices that alter the shape and operation of the wing, significantly increasing lift. This also increases drag, however, this can be overcome by engine thrust and is only for a relatively small period of time. Flaps are moveable areas at the rear of the wing that increase the wing area and change its camber. This significantly increases the lift coefficient of the wing. Slats and slots are devices at the front of the wing that improve the stall performance of the aircraft, meaning it can fly slower or at higher angles of attack. FIGURE 16.26 An Airbus A380 preparing to land at Melbourne Airport. To generate enough lift at the slow speeds required for landing it has its high lift devices extended.

Source: Michael Rosenbrock

26 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

16.6 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. A pilot wishes to raise the nose of an aircraft. What is the name for this type of motion and what is the primary control surface they will need to use to achieve this? 2. Referring to the balance of forces on an aircraft, explain why aircraft require more thrust during takeoff. 3. Explain how the movement of a control surface causes a change in the motion of an aircraft. 4. Explain what flaps do to the characteristics of the wing in order to provide an increase in lift upon landing.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

16.7 Review 16.7.1 Summary • • • •

The primary forces acting on an aircraft in flight are lift, the force due to gravity, thrust and drag. The force due to gravity acting on an aircraft is thought of as acting at one position, the centre of gravity. When a force acting on an aircraft in flight is drawn as an arrow, the arrow represents the resultant of all the component forces that contribute from various parts of the aircraft. The Equation of Continuity can be expressed as: Q = v1 A1 = v2 A2

•

The Bernoulli principle was expressed as an equation and states: 1 2 𝜌v + 𝜌gh + P = constant 2

• • • • • • • • • •

Another way to express the Bernoulli principle in the context of flight is that ‘faster moving fluids have lower pressure’. The generation of lift by a wing can also be explained using Newton’s third law. The wing pushes the air downwards and the air pushes the wing upwards. There are several types of drag that are created when an aircraft moves through the air. The two main types are induced drag and parasite drag. Parasite drag is the combined effect of skin friction drag and form drag. The total drag acting on an aircraft in flight determines the necessary thrust required for the aircraft to maintain a given airspeed. The behaviour of airflow changes when travelling at or beyond the speed of sound. This can lead to the formation of shockwaves and significant increases in drag. Torque is the turning effect of a force about a pivot or reference point. An aircraft in flight can move around three different axes: the lateral (known as pitch), the longitudinal (known as roll) and the vertical (known as yaw). The primary control surfaces on an aircraft are the elevator, rudder and ailerons. There are six main stages of flight: take-off, climb, cruise, turn, glide and landing. The balance of forces on the aircraft is different in each stage.

TOPIC 16 How do heavy things fly? 27

• •

Wind tunnels are used extensively to test the aerodynamics of aircraft designs. An aircraft’s performance can be judged from its lift-to-drag ratio, also known as the glide ratio.

Resources

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0041).

16.7.2 Key terms An aerofoil is any surface that extends into the air so as to create a useful force. An aircraft’s wing is an aerofoil so are the blades of a propeller and an aircraft’s tail surfaces. The angle of attack is the angle between the undisturbed airflow and the chord line of the wing, which allows an aerofoil to generate lift. The Bernoulli principle states that the pressure of a fluid decreases as its velocity increases. A canard aircraft has a small wing in front of of the main wing. The centre of gravity of an object is the point at which the weight can be considered to act. The centre of pressure (CP) or centre of lift is the point on an aircraft’s wing at which the lift forces are considered to be acting. Control surfaces are moving parts on the surface of an aircraft that are used to change the direction of motion of the aircraft. Drag is the rearward-acting force that resists the motion of an aircraft through the air. Flaps are moveable areas at the rear of the wing that increase the wing area and change its camber. Fluids are substances that can flow. All liquids and gases are fluids. Force due to gravity is the force applied to an object with mass due to gravitational attraction. Form drag is the drag due to the shape or ‘form’ of an object. A flying wing aircraft does not have a separate fuselage or horizontal tail. Glide ratio is the ratio of horizontal distance travelled (glide distance) to loss of altitude when gliding. Induced drag is the component of lift that opposes the motion of the aircraft. It arises from wing tip vortices which alter the effective angle of attack of the wing. Lift is the upward-acting force created by a wing moving through the air. Mach number is the ratio of speed of an aircraft to the speed of sound. The mechanical power output of a device is the rate at which it does work on an object. Parasite drag is the name given to the combined effect of skin friction drag and form drag (drag due to the shape or ‘form’ of a surface). Pitch refers to the rotation of an aircraft in flight about the lateral axis. It makes the nose move upwards or downwards. Roll refers to the rotation of an aircraft in flight about the longitudinal axis. This rotation is commonly referred to as banking. Rotational equilibrium is when an object is not rotating. A shock wave is a propagating disturbance associated with abrupt changes in temperature, pressure and density. Skin friction drag is the drag produced by contact between the air and a surface. Slats and slots are devices at the front of the wing that improve the stall performance of the aircraft. Stall occurs when the angle of attack is very large. When this happens, turbulence is large, lift decreases and drag increases. Static pressure is the pressure caused by the random motion of molecules in a fluid. Thrust is the forward force that drives an aircraft through the air. Torque (or moment ) is the turning effect of a force. A v-tail aircraft has two tail surfaces in a v-shape in place of a conventional horizontal and vertical tail. A vortex (plural vortices) is a roughly circular whirlpool in a fluid, often occurring in regions of separated flow. Cyclones and tornadoes are examples of very large vortices. Yaw refers to rotation of an aircraft in flight about the vertical axis. This rotation points the nose of the aircraft to the left or right.

28 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Resources Digital document Key terms glossary (doc-32997)

16.7.3 Practical work and investigations Investigation 16.1 Power and thrust Aim: To explore the relationship between the power delivered to a propeller driven by a small electric motor, and the thrust the propeller delivers to a model plane Digital document: doc-31894

Investigation 16.2 Bernoulli effects Aim: To investigate how Bernoulli’s principle applies to moving air Digital document: doc-31895 Teacher-led video: tlvd-0840

Investigation 16.3 Investigating gliders Aim: To construct simple gliders to investigate and model various forces associated with flight Digital document: doc-31898

Investigation 16.4 Turbulence Aim: To examine the impact of the velocity and size of an object on causing turbulence Digital document: doc-31896

Investigation 16.5 DIY wind tunnel Aim: To construct a simple wind tunnel and explore fluid flow, lift and drag Digital document: doc-31899

16.7 Exercises 16.7 Exercise 1: Multiple choice questions 1.

What is the name given to the location where the lift force acts? A. Centre of gravity B. Centre of lift C. Centre of pressure D. Centre of wing

TOPIC 16 How do heavy things fly? 29

2.

3.

4.

5.

6.

7.

8.

9.

10.

The air flowing down through a constriction in a wind tunnel decreases to one-quarter its initial speed. By what factor does the cross-sectional area of the wind tunnel increase by? A. 2 B. 4 C. 12 D. 16 If the speed of an airflow is increased by a factor of three, according to the Bernoulli principle, what will happen to the pressure? A. Increase to three times the original value B. Increase to nine times the original value C. Decrease to one-third of the original value D. Decrease to one-ninth of the original value Which type of drag is related to the formation of vortices at the tops of the wings? A. Parasite drag B. Form drag C. Induced drag D. Skin friction drag In which of the speed regions can the effects of compressibility be reasonably ignored? A. Hypersonic B. Subsonic C. Supersonic D. Transonic Which of the following would not cause an increase in lift? A. Extending the flaps B. Increasing the angle of attack C. Increasing the air speed D. Decreasing the thrust In a conventionally configured aircraft where the centre of gravity is forward of the centre of pressure of the wing, which direction must the force on the horizontal tail be acting to ensure that the aircraft remains in rotational equilibrium? A. Forwards B. Backwards C. Upwards D. Downwards What is the balance of forces required during climb? A. Thrust = drag B. Thrust < drag C. Thrust > drag D. The balance of forces is irrelevant during climb. Which of the primary flight controls on the aircraft is used to control yaw? A. Elevator B. Flaps C. Ailerons D. Rudder According to the lift force equation, FL = 12 CL 𝜌v2 A, what will happen to the lift force generated if you halve the speed of the aircraft? A. Increase to double its initial value B. Increase to four times its initial value C. Decrease to half its initial value D. Decrease to one-quarter its initial value

30 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

16.7 Exercise 2: Short answer questions 1. 2.

3. 4. 5.

6.

7.

8. 9.

10.

A fluid flows at 6.2 m s–1 in a pipe of diameter 0.48 metres, calculate the speed it will it flow at if the diameter of the pipe is constricted to 0.12 metres? Above a certain angle of attack, an aerofoil will experience a stall. a. What happens to the airflow over the top of the aerofoil when stall occurs? b. What happens to the lift and drag forces acting on the aerofoil once stall occurs? Explain briefly how induced drag is caused and identify one solution used by aircraft designers to address this issue in modern passenger aircraft designs What can be done to reduce the parasite drag acting on an aircraft? The lift force can be calculated as FL = 12 CL 𝜌v2 A. a. Which factor in this equation depends on the environmental conditions and cannot be controlled directly by the pilot or changed by design? b. Which factors in this equation can be influenced by the design of an aircraft? c. Which factors in this equation can be controlled by a pilot during flight? Explain how they are controlled. An aircraft experiences a total drag force of 38 kN while cruising at a speed of 168 m s–1 . Assuming that all of the energy delivered by the engines is used to provide thrust, what is the total power output of the engines of the aircraft? A conventionally configured aircraft has the centre of pressure of the wing 5 metres behind the centre of gravity and the centre of pressure of the tail 16 metres behind the centre of gravity. When it is cruising in steady level flight, the total upwards lift generated by the wing is 1.45 × 105 N. What is the magnitude and direction of the force provided by the horizontal tail to keep the aircraft in rotational balance? The elevator of a conventionally designed aircraft is typically part of the horizontal tail. Compare the role of the elevator to that of the horizontal tail. An aircraft is flying at a speed of 728 m s–1 . a. Calculate the Mach number associated with this speed? b. What speed region is it flying within? c. What effects, if any, of faster-than-sound travel would you expect to occur at this speed? A glider covers a ground distance of 1.2 kilometres in a straight line while losing 45 metres in altitude. Determine its glide ratio and lift-to-drag ratio.

16.7 Exercise 3: Exam practice questions Question 1 (4 marks) The use of external air flowing through a house is a common means of providing ventilation and cooling. In one such design, air flows in a door opening that is 820 millimetres by 2040 millimetres in size, at a speed of 3.2 m s–1 . It crosses the house and flows out a window with an opening of dimensions 600 millimetres by 600 millimetres. a. Will the velocity of the air be slower or faster through the window? Justify your prediction. 2 marks b. Calculate the velocity of the air through the window. 2 marks Question 2 (3 marks) Use your knowledge of either the Bernoulli principle or Newton’s third law to explain in brief and simple terms how an aerofoil shape can generate lift. Question 3 (3 marks) Explain how the ailerons on an aircraft cause it to roll.

TOPIC 16 How do heavy things fly? 31

Question 4 (6 marks) An aircraft has been designed in a canard configuration with a single horizontal stabiliser and control surface ahead of the main wing. The total lift generated by the main wing during cruise is 250 kN. The force due to gravity on the aircraft is 275 kN. a. What must the upwards force on the canard wing be for the aircraft to be in translational equilibrium? 2 marks b. The distance between the centre of pressure of the main wing and the canard is 9 metres. Assuming that the aircraft is in rotational equilibrium, determine the location of the centre of gravity. 3 marks c. Identify one advantage of this canard configuration over a conventional aircraft. 1 mark Question 5 (3 marks) Designers in many industries work to reduce the drag experienced by objects such as aircraft, cars, boats and trucks. Picking one example, explain why this is important, and identify at least two things which can be addressed in design that can reduce drag.

16.7 Exercise 4: studyON topic test Fully worked solutions and sample responses are available in your digital formats.

Test maker Create unique tests and exams from our extensive range of questions, including practice exam questions. Access the assignments section in learnON to begin creating and assigning assessments to students.

32 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

AREA OF STUDY 2 OPTIONS OBSERVATION OF THE PHYSICAL WORLD

17

How do fusion and fission compare as viable nuclear energy power sources? Numerous videos and interactivities are available just where you need them, at the point of learning, in your digital formats, eBookPLUS and learnON at www.jacplus.com.au.

17.1 Overview 17.1.1 Introduction We use energy every day of our lives. As such, the need for sustainable sources of energy is continually growing. Fission and fusion are two ways of extracting energy from the nucleus of atoms. While the sources of nuclear energy, such as uranium, are classed as non-renewable, they can produce large quantities of energy from very small fuel quantities, making them fairly sustainable. This topic investigates the following methods of nuclear energy production: 1. A heavy nucleus splits into two medium-sized nuclei. This nuclear reaction is called fission. 2. Two very light nuclei are forced together to form one nucleus. This reaction is called fusion. After studying this topic you will be able to compare the two reactions and their associated technologies, understanding the benefits and risks associated with nuclear energy production.

FIGURE 17.1 The underwater core of the Reed Research Reactor, Reed College, Oregon, USA

Source: United States Nuclear Regulatory Commission

TOPIC 17 How do fusion and fission compare as viable nuclear energy power sources? 1

17.1.2 What you will learn KEY KNOWLEDGE After completing this topic you will be able to: Energy from the nucleus • explain nuclear fusion reactions of proton-proton and deuterium-tritium with reference to: • reactants, products and energy production • availability of reactants • energy production compared with mass of fuel 235 239 • explain nuclear fission reactions of U and Pu with reference to: • fission initiation by slow and fast neutrons respectively • products of fission including typical unstable fission fragments and energy • radiation produced by unstable fission fragments 238 239 • describe neutron absorption in U, including formation of Pu • explain fission chain reactions including: • the effect of mass and shape on criticality • neutron absorption and moderation Nuclear energy as a power source • compare nuclear fission and fusion with reference to: • energy released per nucleon and percentage of the mass that is transformed into energy • availability of reactants • limitations as a source of energy for electricity production • environmental impact • analyse fission and fusion with reference to their viabilities as energy sources • describe the energy transfers and transformations in the systems that convert nuclear energy into thermal energy for subsequent power generation • explain the risks and benefits for society of using nuclear energy as a power source. Source: VCE Physics Study Design (2016–2021) extracts © VCAA; reproduced by permission.

PRACTICAL WORK AND INVESTIGATIONS Practical work is a central component of learning and assessment. Experiments and investigations, supported by a Practical investigation logbook and Teacher-led videos, are included in this topic to provide opportunities to undertake investigations and communicate findings.

Resources Digital documents Key science skills — VCE Units 1–4 (doc-31856) Key terms glossary (doc-32999) Practical investigation logbook (doc-33000)

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0042).

2 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

17.2 Nuclear fusion KEY CONCEPT • Explain nuclear fusion reactions of proton-proton and deuterium-tritium with reference to: • reactants, products and energy production • availability of reactants • energy production compared with mass of fuel.

17.2.1 Proton–proton and deuterium–tritium fusion reactions As previously mentioned in topic 8, the Sun’s core has a temperature of more than 15 million K, which is perfect for fusion to occur. Inside the Sun, hydrogen nuclei fuse together to form helium. As helium is more stable than hydrogen, the excess nuclear energy is released. This energy is emitted from the nuclei as gamma radiation and is eventually received on Earth as light and heat. A part of this process is referred to as proton–proton fusion, where two protons fuse, then usually the pair instantly break apart, but in some instances one of the protons is transmuted into a neutron. The resulting proton–neutron pair is call deuterium (a form of hydrogen). Then another proton collides with deuterium to create helium-3. When two helium-3 nuclei collide, helium-4 is created plus two extra protons. The equations that demonstrate this process are: 1 1 2 0 1H + 1H → 1H + 1e + 2 1 3 1H + 1H → 2H e + 𝛾 3 2H e

v

+ 32H e → 42H e + 11H + 11H

Another fusion reaction is a deuterium–tritium fusion reaction, this is a reaction where two atomic nuclei (deuterium and tritium) are brought so close together that they fuse. They are both positively charged so to fuse they must overcome this mutual repulsion. This requires them to travel at high speeds, which is only achieved at high temperatures of about 100 million degrees. This fusion process is: 3 1H

+ 21H → 42H e + 10n + energy

( 21H is deuterium and 31H is tritium) The amount of energy required to overcome the strong nuclear forces is known as the binding energy. This was covered in topic 8 and is revised in the next section.

17.2.2 Binding energy The amount of energy needed to overcome the strong nuclear force and pull apart a nucleus is known as the binding energy. This is the amount of energy that has to be added to a nucleus to split it into its individual nucleons — that is, to reverse the binding process. For example, it would take 2.23 MeV of energy to split a ‘heavy’ hydrogen nucleus into a separate proton and neutron. Each isotope has its own specific binding energy. Nuclei with high binding energies are very stable as it takes a lot of energy to split them into separate protons and neutrons. Nuclei with lower binding energies are easier to split. Of course, it is difficult to supply sufficient energy to cause a nucleus to split totally apart. It is much more common for a nucleus to eject a small fragment, such as an 𝛼 or 𝛽 particle, to become more stable.

TOPIC 17 How do fusion and fission compare as viable nuclear energy power sources? 3

To compare the binding energies of various nuclei, and therefore their stability, it is necessary to look at the average binding energy per nucleon. The average binding energy per nucleon is calculated by dividing the total binding energy of a nucleus by the number of nucleons in the nucleus. It can be seen from figure 17.2 that iron-56 has one of the highest binding energies per nucleon. This means it is one of the most stable nuclei. In order to become more stable, other nuclei tend to release some of their energy. Releasing this energy would decrease the amount of energy they contained, and therefore increase the amount of energy that must be added to split them apart. FIGURE 17.2 This graph of binding energy versus mass number peaks at nickel-62, although the much more common iron-56 is very close behind. 9 Average binding energy per nucleon (MeV)

14 7N

56 26 Fe

8

238 92 U

141 56 Ba

4 2He

7 6 6 3 Li

5 4 3

3 2 He

2 2 1H

1

1 1H

0 0

30

60

90

120 150 Mass number

180

210

240

270

The binding energy is not only the amount of energy required to separate a nucleus into its component parts, but also the amount of energy released when those parts are brought together to form the nucleus; that is, when a proton and a neutron collide to form a ‘heavy’ hydrogen nucleus, 2.23 MeV of energy is released (twice the binding energy per nucleon). The curve of the graph in figure 17.2 indicates that if two nuclei with low mass numbers could be joined together to produce a single nucleus, then a lot of energy would be released. Similarly, if a nucleus with a very high mass number could split into two fragments with greater binding energy per nucleon, then once again a lot of energy would be released. These two possibilities were realised in the 1930s. The released energy can be calculated from Einstein’s equation E = mc2 , where m is the difference between the total nuclear mass before and after the event, and c is the speed of light. The joining of two nuclei is nuclear fusion and the splitting of a single nucleus is nuclear fission.

17.2.3 Fusion reactors Stars are giant fusion reactors, and many people have proposed that fusion reactors could be the answer to our energy needs. The fusion of hydrogen isotopes to form helium releases a large amount of energy, which is a result of the large difference in their respective binding energies as seen in figure 17.2. The practicalities of a fusion reactor, however, are proving difficult. To initiate fusion of hydrogen, the forces of repulsion between the positive nuclei must be overcome. One way to do this is to give the nuclei large amounts of kinetic energy by heating them to extremely high temperatures (millions of degrees). It is incredibly difficult to contain material at such temperatures. Currently, scientists are using magnetic fields

4 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

to contain the hydrogen, while laser beams are used to heat it. This method has proved successful, but for only very short periods of time. Viable production of energy using a fusion reactor is still many years away. A fusion reactor could not feasibly use the same reactions as the Sun. A reactor on Earth would have to use a different reaction, preferably a one-step reaction with only two reactants. Three possible reactions for a terrestrial fusion reactor are: 2 1H

+ 31H → 42H e + 10n

2 1H

+ 21H → 31H + 11H

2 1H

+ 63L i → 42H e + 42H e

SAMPLE PROBLEM 1

Use data from the following table. Nucleus

Symbol

Mass (kg)

Neutron

1 n 0

1.674 746 × 10–27

Total binding energy (MeV)

Hydrogen-2

2 1H

or 21D

3.344 132 × 10–27

2.224 573

Hydrogen-3

3 1H

or 31T

5.007 725 × 10–27

8.481 821

Helium-4

4 2H e

6.645 758 × 10–27

28.295 673

Lithium-6

6 3L i

9.987 263 × 10–27

31.994 564

For the third reaction of a terrestrial fusion reactor 21H + 63L i → 42H e + 42H e calculate the: a. difference between the binding energy of the products and reactants b. difference between the sum of the masses of the products and the reactants c. energy equivalent of this mass difference in joules and MeV d. energy released per nucleon in MeV. THINK

WRITE

Sum the binding energies of the reactants. 2. Sum the binding energies of the products. 3. Calculate the difference in the binding energies. 4. State the solution.

a. 1.

b. 1.

Sum the masses of the reactants.

Binding energy of reactants = 2.224 573 + 31.994 564 = 34.219 137 MeV Binding energy of products = 2 × 28.295 673 = 56.591 346 MeV Difference = 56.591 346 − 34.219 137 = 22.372 209 MeV The difference between the binding energy of the products and reactants is 22.372 209 MeV b. Mass of the reactants = 3.344 132 × 10−27 a.

+ 9.987 263 × 10−27 2.

Sum the masses of the products.

= 1.333 1395 × 10−26 kg Mass of products = 2 × 6.645 758 × 10−27 = 1.329 1516 × 10−26 kg

TOPIC 17 How do fusion and fission compare as viable nuclear energy power sources? 5

3.

4. c. 1.

Calculate the difference between the two masses. State the solution. Use the equation E = mc2 to calculate the energy difference.

Difference = 1.333 1395 × 10−26 − 1.329 1516 × 10−26 = 3.9879 × 10−29 kg The difference between the sum of the masses of the products and the reactants is 3.9879 × 10–29 kg. c. E = mc2 ( )2 = 3.9879 × 10−29 × 2.997 924 58 × 108 = 3.584 146 × 10−12 J

2.

Convert to eV by dividing by 1.602 176 × 10−19 .

3.

Convert to MeV by dividing by 1.0 × 106 .

4.

State the solution.

d. 1.

2.

3.584 146 × 10−12 1.602 176 × 10−19 = 22 370 487 eV 22 370 487 In MeV = 1.0 × 106 = 22 370 487 MeV The energy equivalent of this mass difference is 3.584 146 × 10–12 J or 22.370 487 MeV. 22 370 487 d. Energy released per nuleon = 8 = 2.796 311 MeV The energy released per nucleon is 2.796 311 MeV. In eV =

Calculate the energy per nucleon by dividing by the number of nucleons. State the solution

PRACTICE PROBLEM 1 Use data from the following table. Particle

Symbol

Mass (kg)

Neutron

1 n 0

1.674 746 × 10–27

Total binding energy (MeV)

Hydrogen-2

2 1H

or 21D

3.344 132 × 10–27

2.224 573

Hydrogen-3

3 1H

3 1T

5.007 725 × 10–27

8.481 821

6.645 758 × 10–27

28.295 673

Helium-4

or

4 2H e

For the first reaction between a deuterium and a tritium nucleus, which forms a helium nucleus and a neutron, calculate the: a. difference between the binding energy of the products and sum of the binding energies of the reactants b. difference between the sum of masses of the products and the reactants c. energy equivalent of this mass difference in joules and MeV d. energy released per nucleon of reactants in MeV.

6 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 17.3 The stars in Omega Centauri, like all other stars, are giant fusion reactors.

Source: Gordon Mandell / NASA

Resources Weblink Fusion basics

17.2.4 Availability of deuterium and tritium Deuterium is a stable isotope of hydrogen. It has one proton and one neutron. In nature, only about one in every 6400 hydrogen atoms is a deuterium atom. However, being twice as heavy as normal hydrogen it is quite easy to separate. In steam, water molecules with two deuterium atoms have a molecular weight of 20 compared to 18 for most water molecules. This 11% difference in mass means that the heavier molecules on average travel slower and can be separated in a distillation process. Water that was almost exclusively deuterium oxide could be made in large quantities by the 1930s. It was predominately used as a moderator in nuclear reactors. It can be easily extracted from sea water, but most industrial processes use glacial ice or mountain water. Tritium is radioactive. It beta decays into helium-3, with a half-life of 4500 days. It is extremely rare; very small amounts are produced by cosmic rays hitting atoms in the atmosphere. It can be produced inside nuclear reactors by using the neutrons to bombard targets such as lithium-6.

TOPIC 17 How do fusion and fission compare as viable nuclear energy power sources? 7

SAMPLE PROBLEM 2 a. b.

How many protons and neutrons are in deuterium? What is the difference in mass of deuterium compared to normal hydrogen?

THINK

WRITE

a.

Deuterium is a stable isotope of hydrogen.

b.

Deuterium has one proton and one neutron, both of which have very similar masses.

Deuterium has one proton and one neutron. b. Deuterium is twice as heavy as normal hydrogen. a.

PRACTICE PROBLEM 2 a. What is the decay equation for tritium? b. What is the product formed from bombarding a lithium-6 nucleus with a neutron?

17.2.5 Fusion reactor designs The key difficulties for fusion reactors are to confine the mixture of deuterium and tritium at a high enough temperature, at a great enough density and for a long enough time for fusion to occur on a sustainable basis.

Magnetic confinement In a doughnut-shaped machine a mixture of deuterium and tritium is constricted by a magnetic field. An electric current is passed through the gas, heating it up to beyond 10 million degrees Celsius. At these temperatures, electrons are stripped from atoms, and the gas becomes a plasma of ions. An example of magnetic confinement is the tokamak, which is a Russian-designed fusion reactor concept.

FIGURE 17.4 A photo of the Alcator C-Mod tokamak at MIT. The tokamak (the large upright cylinder at the centre) is encased within boron-doped concrete for neutron and X-ray shielding.

Laser fusion Over 190 laser beams simultaneously emit a strong pulse of light on a small pellet of frozen deuterium and tritium. About 1.8 million joules of energy hits the pellet in billionth-of-a-second pulses and causes fusion.

Source: Jim Irby, Head of Operations, MIT Plasma Science and Fusion Center

8 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

17.2 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. (a) Define the terms fusion and fission. (b) Which of these reactions occurs in our Sun? 2. Why does the splitting of uranium-235 nuclei release energy, but the joining of hydrogen atoms also releases energy? 3. Why is energy released in the process of fusing two small nuclei together? 4. In what form does the energy released from a nuclear fusion reaction appear? 5. Use data from the following table. Nucleus Proton

Symbol 1 p 1

Hydrogen-2

2 1H

Hydrogen-3

3 1H

Mass (Kg)

Total binding energy (MeV)

or

1 H 1

1.673 351 × 10

or

2 1D

3.344 132 × 10–27

2.224 573

or

3 1T

5.007 725 × 10

8.481 821

–27

–27

For the reaction: 2 1H

+ 21H → 31H + 11H

(a) What is the difference between the sum of binding energies of the products and the binding energies of the reactants? (b) What is the difference between the sum of masses of the products and the reactants? (c) What is the energy equivalent of this mass difference in joules and in MeV? (d) Calculate the energy released per nucleon of reactants in MeV.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

17.3 Nuclear fission KEY CONCEPTS • Explain nuclear fission reactions of 235 U and 239 Pu with reference to: • fission initiation by slow and fast neutrons respectively • products of fission including typical unstable fission fragments and energy • radiation produced by unstable fission fragments. • Describe neutron absorption in 238 U, including formation of 239 Pu. • Explain fission chain reactions including: • the effect of mass and shape on criticality • neutron absorption and moderation.

17.3.1 Fission fragments Topic 7 described three different fission reactions of uranium-236. The reactants in each case are a neutron and a uranium-235 nucleus, but there are many combinations of products. In topic 8 you saw three different reactions with six different products, but about 30 different elements can be produced, two at a time with about 100 isotopes among those 30 elements.

TOPIC 17 How do fusion and fission compare as viable nuclear energy power sources? 9

FIGURE 17.5 Fission of uranium-236; distribution of fission fragments by mass number

9% 8% 7% 6% 5% 4% 3% 2% 1% 0% 75

80

85

90

95 100 105 110 115 120 125 130 135 140 145 150 155 160 Mass number

SAMPLE PROBLEM 3

One of the fragments of the fission of uranium-236 has an atomic mass number of about 137. What is the species produced? THINK

WRITE

Use a periodic table to determine which element has a mass number of 137.

The element with the mass number of 137 is barium, which has a relative atomic mass of 137.3.

PRACTICE PROBLEM 3 a. From the graph in figure 17.5, what is the mass number of the most common fragment of uranium-236 fission? b. Look up the periodic table to find the most likely atomic number of this fragment. c. Uranium-236 has 92 protons. Determine the atomic number of the other fragment in the most common fission reaction.

10 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

17.3.2 Fission chain reactions There were also two or three neutrons emitted with each fission reaction. This allowed the possibility of a chain reaction with one fission producing two or three others, and so on. FIGURE 17.6 This is what happens if every free neutron goes on to produce another fission. A situation such as this quickly releases an enormous amount of energy. It is called an uncontrolled chain reaction and is what happens in a nuclear bomb.

Reless of energy

Neutron

Fission products

Resources Digital document: Investigation 17.1 Chain reaction with dominoes (doc-31900) Teacher-led video: Investigation 17.1 Chain reaction with dominoes (tlvd-1074)

17.3.3 Achieving a chain reaction There are three factors that make achieving a chain reaction very difficult. 1. Uranium has two main naturally occurring isotopes: uranium-235 (0.7%) and uranium-238 (99.3%). The isotopes’ responses to an incoming neutron are different and depend on the speed of the neutron (see table 17.1). 2. The high speed of the ejected neutrons: they have kinetic energies about 1 MeV, equivalent to speeds of 1.4 × 107 m s−1 . 3. If a chain reaction could be achieved, there needs to be a way of controlling or stopping it.

TOPIC 17 How do fusion and fission compare as viable nuclear energy power sources? 11

TABLE 17.1 Reactions of uranium isotopes to incoming neutrons Neutron speed and energy

235

238

Very fast neutrons (5 MeV)

Few nuclei fission

Most nuclei fission

Fast neutrons (1 MeV)

Many nuclei fission

Very few nuclei fission

Slow neutrons (200 eV)

Most nuclei fission

Nearly all nuclei absorb neutrons

Very slow neutrons (0.03 eV)

All nuclei fission

All nuclei absorb neutrons

U

U

A high-speed neutron from the fission of a uranium-235 nucleus is travelling too slowly to cause a uranium-238 nucleus to split. By the time successive collisions slow the neutrons down to a speed to cause most of the uranium-235 to split, the neutrons will have been gobbled up by the uranium-238 nuclei, which outnumber the uranium-235 nuclei by about 140 to 1 in naturally occurring uranium. Therefore, a chain reaction cannot occur in a block of pure natural uranium. Solutions have been developed for each of the three difficulties mentioned. 1. Increase the proportion of the uranium-235 isotope. This process is called enrichment. 2. Slow down the neutrons very quickly. This is done using a moderator. 3. To control the chain reaction, use a material that readily absorbs neutrons and takes them out of the reaction, and that can be quickly inserted at a moment’s notice. This is done with control rods.

17.3.4 Enriched uranium Natural uranium cannot be used for nuclear bombs or power plants because it contains only small amounts of fissionable uranium-235. The uranium-238 absorbs free neutrons and prevents a sustainable chain reaction from occurring. Enrichment must be carried out to increase the percentage of uranium-235 in the ore. Because uranium-235 and uranium-238 isotopes are chemically identical, the process of separating them is difficult. A number of enrichment methods have been developed, but all are complex and costly. Enriched uranium for nuclear power plants must contain between 1% and 4% uranium-235. For nuclear bombs the percentage of uranium-235 must be closer to 97%, because an uncontrolled chain reaction is required. FIGURE 17.7 One centrifuge cylinder Depleted uranium (increased 238U)

Liquid uranium hexafluoride Enriched uranium (increased 235U)

Rotating cylinder

Motor

12 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

One enrichment method, called the centrifuge system, uses a rotating cylinder that sends the heavier isotope (uranium-238) in liquid uranium hexafluoride to the outside of the cylinder, where it can be drawn off, while the uranium-235 diffuses to the centre of the cylinder. To effectively enrich the uranium-235, thousands of centrifuges are connected in series.

Moderators A fast neutron bouncing off a uranium nucleus is like a golf ball bouncing off a basketball. The large difference in mass means that the neutron does not lose much kinetic energy. To slow down quickly, the neutron needs to collide with something of similar mass, like one billiard ball hitting another billiard ball. Also, the moderating material should be a liquid or solid at room temperature, reasonably inexpensive, and not chemically reactive. These constraints mean that ordinary water is commonly used as a moderator. Other possible moderators are carbon (in the form of graphite) and ‘heavy’ water (water in which the hydrogen atoms are the deuterium isotope, which has one proton and one neutron).

Control rods The elements cadmium and boron are deficient in neutrons and readily absorb them. Cadmium and boron are usually encased within steel control rods that can be rapidly raised or lowered into the reactor. The first controlled fission reactor was constructed by Enrico Fermi at the University of Chicago in December 1942. He used graphite as the moderator and cadmium for the control rods.

The chain reaction To create an uncontrollable chain reaction, a large proportion of fissionable nuclei is necessary. If uranium is the energy source, weapons-grade enriched ore (about 97% uranium-235) is required. This means that there are very few nuclei present that can absorb the free neutrons without undergoing fission themselves. Therefore, more of the available neutrons cause an energy release.

Criticality A large ball of weapons-grade material makes a large nuclear bomb, but a small ball of weapons-grade material may not make a small nuclear bomb! The difference lies in the volume-to-surface-area ratio. A small ball has a small ratio of volume to surface area. This means a low proportion of the free neutrons stay inside the ball where they can initiate a fission. Large balls of weapons-grade material will have a higher ratio of volume to surface area; therefore, a higher proportion of the free neutrons stay inside the ball to produce further fissions. Critical mass is the smallest mass of a fissionable substance which, when formed into a sphere, will sustain an uncontrolled chain reaction. This mass or criticality varies according to the percentage of fissionable nuclei in the material and the fissionable isotope used. Any mass less than the critical mass is known as subcritical. If the shape of the fissionable material is changed from a sphere to a brick shape, then there is more surface area available for neutrons to escape the material before initiating another fission reaction. A brick shape therefore needs to be larger and heavier before it becomes critical. The sphere is the shape with the smallest surface area for a given volume. TABLE 17.2 Critical mass and diameter of a sphere of fissionable material Fissionable nucleus

Critical mass

Critical diameter

Uranium-235

52 kg

17 cm

Plutonium-239

10 kg

10 cm

Uranium-233

15 kg

11 cm

TOPIC 17 How do fusion and fission compare as viable nuclear energy power sources? 13

The critical mass can be reduced by using beryllium around the outside as a neutron reflector. Escaping neutrons are ‘reflected’ back into the fissionable material, increasing the chances of a fission reaction. The neutrons collide with the beryllium nuclei and a high proportion bounce back.

FIGURE 17.8 An uncontrolled chain reaction

(a)

(b)

A small volume compared to surface area does not allow neutrons to stay in the material, so a small sphere may not be able to sustain a chain reaction.

(c)

A large ball has a large volume-to-surface-area ratio. A large ball can sustain a chain reaction.

14 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

A flat sheet of weapons-grade material cannot sustain a chain reaction. A flat sheet has a very low volume-to-surface-area ratio.

17.3 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Explain why a large spherical mass of uranium may be able to sustain a chain reaction while the same mass spread into a flat sheet could not. 2. In what form does the released energy from a nuclear fission reaction appear? 3. Why are neutrons good at initiating nuclear reactions? 4. Describe a chain reaction. 5. What are the advantages and disadvantages of fusion power as compared to fission power?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

17.4 Nuclear bomb design BACKGROUND INFORMATION Explore the effects of nuclear weapons on the surrounding environment.

17.4.1 Fission bombs With more countries developing nuclear facilities as power sources, the production of nuclear weapons is growing. One type of nuclear weapon is the fission bomb, which releases massive amounts of energy by splitting large nuclei. A nuclear fission bomb is simply a lump of fissionable material in which an uncontrolled chain reaction is allowed to occur. Fortunately, this happens only under fairly specific conditions. All fission bombs rely on two or more subcritical masses being brought together to form a critical mass. The joining of the subcritical masses is usually instigated by the detonation of a small conventional bomb.

FIGURE 17.9 This gun-style bomb is similar to ‘Little Boy’, the bomb dropped on Hiroshima on 6 August 1945. It uses the fission of enriched uranium-235 as its energy source. (a) Before firing Separating tube

Two pieces of enriched uranium-235, both less than the critical mass

(c) Compressed piece of uranium created which exceeds critical mass

Explosive charge (d) Atomic explosion

(b) Detonation

TOPIC 17 How do fusion and fission compare as viable nuclear energy power sources? 15

FIGURE 17.10 This implosion bomb is similar to ‘Fat Man’, the bomb dropped on Nagasaki on 10 August 1945. It uses several subcritical pieces of plutonium-239 as its energy source. (a) Before firing

(b) Detonation

(d) Atomic explosion

Detonators

Small pieces of plutonium-239 (each less than the critical mass) and separating material

Beryllium shell Compressed small ball of plutonium-239 which exceeds critical mass

17.4.2 The effects of nuclear weapons The size of nuclear weapons is measured according to the mass of TNT needed to produce a similar blast. Both bombs dropped on Japan in 1945 were about 20-kiloton bombs, meaning they produced a blast similar to that of 20 thousand tons of TNT. Weapons currently stockpiled around the world are much larger than this, with 20-megaton bombs being common. The devastating damage caused by nuclear weapons can be classified in two ways: immediate and longterm effects.

17.4.3 Immediate effects Thermal flash The high temperatures generated by a nuclear blast cause a flash of thermal (heat) radiation to spread out from ground zero (the centre of the blast). The thermal radiation takes the form of an enormous fireball. A 10-megaton bomb would produce a fireball about 4.3 kilometres in diameter, and the thermal flash emitted would cause second-degree burns up to 32 kilometres away. Eyesight may also be damaged in any creatures looking at the fireball. When exposed to such large amounts of thermal radiation, dry grass and paper may spontaneously ignite, setting fires which may be fanned by the accompanying high winds. Close to ground zero, most substances are melted, burned or exploded.

Shock wave Immediately after the detonation of a nuclear bomb, a shock wave spreads out from the centre of the blast. This high-pressure wave moves out at speeds which may be greater than 3000 km h−1 . It is estimated that a 10-megaton bomb would irreparably damage houses within a 17.7 kilometre radius, and moderately damage homes up to 24 kilometres away.

Electromagnetic pulse The huge amounts of gamma (𝛾) radiation emitted by the nuclear explosion can ionise atoms in the air. The numerous free electrons produced form strong electromagnetic fields. These fields are capable of destroying electric and electronic systems, including power distribution systems, telecommunications and computer networks. Data stored electronically would be wiped from memory chips and flash drives.

Initial nuclear radiation Enormous amounts of 𝛾 radiation and free neutrons would irradiate everything near the blast. While the radius of damage due to nuclear radiation would not be as great as for the shock wave or thermal blast, the severity of the radiation received by people and animals would cause immediate death in most.

16 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 17.11 The immediate after-effects of the nuclear bomb explosion in Hiroshima, Japan, August 1945

Source: United States Department of Energy

17.4.4 Long-term effects Radioactive fallout In the days and weeks after a nuclear blast, the radioactive products of the fission reaction, floating as dust in the atmosphere, would start to return to Earth. Many of these isotopes have long half-lives, ensuring that their effects would remain for months and years. The fallout from fusion bombs is less than for fission bombs because radioisotopes are not produced by fusion reactions. (A fusion bomb does still require a small fission bomb to initiate the fusion reaction.) The radiation from the fallout would ensure that any remaining people would continue to receive a greater than normal radiation dose for many years.

Nuclear winter Some scientists have predicted that a so-called nuclear winter may follow large-scale nuclear detonations, such as in a nuclear war. The dust and smoke brought into the atmosphere by the explosions would gather as clouds covering great expanses of the sky. Sunlight would be blocked from the Earth’s surface, resulting in lower temperatures and the destruction of plant life until the clouds finally settled. The ozone layer could be damaged, leaving the Earth without protection from the Sun’s ultraviolet rays once the dust clouds cleared.

17.4.5 Fusion bombs The extreme conditions necessary for the fusion of hydrogen can be created by the detonation of a small fission bomb. This means a fusion (or hydrogen) bomb is actually two bombs — a small fission bomb that triggers a much larger fusion explosion. Such weapons are known as thermonuclear, because the initial explosion creates the intense heat necessary to overcome the repulsion between positive nuclei and allow them to get close enough for the fusion to occur. TOPIC 17 How do fusion and fission compare as viable nuclear energy power sources? 17

17.4 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. How can a nuclear bomb contain sufficient fissionable material for an explosion but be transported without exploding? (Use the term critical mass in your answer.) 2. What energy source was used for the ‘Fat Man’ bomb? 3. What form of radiation is emitted from a nuclear explosion? 4. Why are there long-term effects of a nuclear blast? 5. Why are some weapons referred to thermonuclear? Fully worked solutions and sample responses are available in your digital formats.

17.5 Nuclear energy as a power source KEY CONCEPTS • Compare nuclear fission and fusion with reference to: • energy released per nucleon and percentage of the mass that is transformed into energy • availability of reactants • limitations as a source of energy for electricity production • environmental impact. • Analyse fission and fusion with reference to their viabilities as energy sources. • Describe the energy transfers and transformations in the systems that convert nuclear energy into thermal energy for subsequent power generation. • Explain the risks and benefits for society of using nuclear energy as a power source.

17.5.1 Nuclear power plants Many countries use nuclear power to supply their domestic electric power needs. It is the world’s second largest source of low-carbon power (30% of the total in 2016). Four countries get more than 50% of their electricity from nuclear energy, another seven countries get between 50% and 30%. Overall 10.2% of the world’s electric power is supplied by nuclear energy. In 2019, 31 countries were producing nuclear energy for domestic use and energy importation. The amount of pure uranium required to generate this nuclear energy is about 67 000 tonnes per year. Uranium resources are the amount of uranium that is known to exist and can be mined. The amount of uranium resources for various countries shown in figure 17.14. The world’s current resources of uranium, approximately 5.9 megatonnes, would last about 90 years at existing rates of consumption. It is expected that further exploration and higher prices will yield further resources.

18 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 17.12 France uses nuclear power for 72% of its electricity.

FIGURE 17.13 Percentage of electricity supplied by nuclear power in 2018 Percentage of energy from nuclear power 2018

80 70 60 50 40 30 20 10

Fr a Sl nce ov U akia kr H a in un e g Sw ary ed B Sw elg en itz ium er Sl lan o d C ze B ven ch ul ia Re gar pu ia Fi blic nl a So Arm nd ut e h ni Ko a re Sp a ai n U Ru SA ss ia Ro U m K a C nia an G a e r da m a Ta ny i w Pa a ki n st Ja an p M an Ar exi So ge co ut nti h na Af ric C a hi n N et In a he d rla ia nd Br s az il Ira n

0

FIGURE 17.14 Uranium resources as in 2018. Data from OECD Nuclear Energy Agency and International Atomic Energy Agency.

Uranium resources (tonnes)

1 600 000 1 400 000 1 200 000 1 000 000 800 000 600 000 400 000 200 000

SA U

di a In

e kr ai n U

il

na C hi

Af ri h ut

Br az

ca

a si Ru s So

er N ig

ib ia am

an N

kh

st

ad a Ka za

an C

Au

st

ra lia

0

17.5.2 Generating electricity There are many energy sources that can readily be converted into electricity. In Australia, coal, gas, hydro and wind are used to generate commercial quantities of electricity. Australia does not have any nuclear power plants that produce electricity for use by the community. Whatever the energy source used to generate electricity, the techniques are surprisingly similar. By some means, the energy source is used to turn a turbine. The kinetic energy of the turbine is then converted into electrical energy by a generator. The electrical energy is distributed to the community via a series of transformers and transmission lines. Power plants that use coal are very similar to those using nuclear energy. In both cases, the energy is used to heat water and produce steam, which is then used to turn the turbine.

TOPIC 17 How do fusion and fission compare as viable nuclear energy power sources? 19

FIGURE 17.15 Various methods can be used to turn the turbines that allow electricity to be produced. (a) Thermal power stations. Burning fuel or nuclear reactions release energy. This heats water and products steam, which travels through pipes and spins the turbines. (i) Gas or coal Generator Turbine Steam

Water Burning fuel

(ii) Nuclear

Generator Turbine

Water

Steam

Reactor

(b) Hydro-electric power stations. Water rushing down from the dam spins the turbines. Generator

Dam Turbine

A NATURAL FISSION REACTOR In 1972 scientists discovered a site in Gabon, West Africa, which is believed to be the remnant of a natural fission reactor that operated about 2000 million years ago! Scientists suspected that a natural arrangement of uranium ore and water had acted like a modern fission reactor when they noticed lower than expected amounts of uranium-235 in the ore of a mine (0.44% instead of the usual 0.72%). It took some good scientific detective work to solve the puzzle. The main evidence for the existence of a natural reactor was the presence of fission products. The graphs below show the distribution by mass number of the isotopes of the fission product neodymium (Nd). This evidence leaves little doubt that a natural reactor really did exist: (a) and (c) are virtually identical. FIGURE 17.16 Evidence for a natural reactor in Gabon: (a) natural reactor site, (b) regular ore and (c) spent fuel rods from a nuclear power plant (b)

(a)

20 15 10

25 Percentage

Percentage

Percentage

(c)

25

25

20 15 10 5

5 0

15 10 5

0 142 143 144 145 146 148 150 Mass number

20

0 142 143 144 145 146 148 150 Mass number

20 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

142 143 144 145 146 148 150 Mass number

Nuclear reactors There are many different styles of nuclear reactor. The pressurised water reactor (PWR) is the most common. It uses the fission of uranium-235 to produce energy. Generally, nuclear fission reactors have the same basic components. Figure 17.17 shows the arrangement of a typical nuclear reactor.

17.5.3 Australia’s nuclear reactor Australia currently has one small uranium-235 reactor, OPAL (Open Pool Australian Light Reactor), which is used for industrial research and the production of medical isotopes, but not for power production. Australia had another reactor, HIFAR (High Flux Australian Reactor), which was closed in January 2007 and permanently decommissioned over the following ten years. OPAL is a 20-megawatt reactor that uses low enriched uranium fuel and is cooled by ordinary water. It has twice the power of HIFAR. The reactor core, which contains the nuclear fuel and the moderator, is about the size of a filing cabinet; it contains 30 kilograms of uranium, including 6 kilograms of uranium-235. It is designed to produce a stream of neutrons, which are used to irradiate materials to either identify their structure or to produce radioisotopes for medical or industrial purposes.

17.5.4 Fast breeder reactors Like coal and gas, the world’s supply of uranium-235 is limited. To overcome this problem, fast breeder reactors have been developed. These reactors use plutonium-239 as the energy source. Plutonium-239 is not a naturally occurring isotope; rather, it is produced when uranium-238 captures a neutron and becomes uranium-239. This unstable isotope releases a 𝛽 − particle and becomes neptunium-239. Another 𝛽 − particle is then released and the nucleus becomes plutonium-239, a fissionable isotope. 238 92U

+ 10n →

239 92U

→

239 93N p

+ β− →

239 94P u

+ β−

Fission reaction for plutonium-239 One possible fission reaction for plutonium-239 is: 239 94P u

+ 10n →

240 94P u

→

148 58C e

1 + 89 36K r + 3 0n + energy

Fuel rods in a fast breeder reactor consist of 20% plutonium-239 surrounded by 80% natural uranium. As the plutonium undergoes fission, it actually produces more plutonium when the uranium-238 is bombarded with neutrons. In this process more plutonium is produced than is used. This is why they are known as ‘breeder’ reactors. Uranium-238 is much more plentiful than uranium-235, so fast breeder reactors are an alternative to conventional reactors. There is another major difference between reactors that use enriched uranium-235 and fast breeder reactors. Plutonium-239 readily absorbs fast-moving neutrons (hence, the name ‘fast’ breeder). It is therefore not necessary for these reactors to have a moderator to slow down the free neutrons. Generally they use liquid sodium as a coolant. Fast breeder reactors have had a troubled history. France, the United Kingdom and the United States built a small number of such reactors during the 1950s and 1960s but, due to political factors and the relatively low cost of conventional uranium fuel, none of these are currently operational. Russia still has two fast breeder reactors running, and France, Japan, India and China are all planning the construction of new breeder reactors.

TOPIC 17 How do fusion and fission compare as viable nuclear energy power sources? 21

FIGURE 17.17 A nuclear power plant

Control roads

Centainment vessel

Fuel roads

Moderator

Steam

Steam generator (heat exchanger)

Turbine generator

Electricity

Pressure vessel Pump

Coolant Reactor core Pump

Reactore core The reactor core is where the fission place to produce energy. It contains the nuclear fuel, control rods, moderatoe and coolant. A shield of concrete and steel surrounds the core to prevent radiation from escaping Fuel rods Fuel rods consist of a number of thin tubes full of pellets of fissionable material. Usually the fissionable material is enriched uranium–235. Fission occurs inside the fuel rods, releasing nuclear energy. Because the uranium–235 is used up in the process of fission, the fuel rods must be replaced regularly. Control rods The purpose of these cadmium or boron control rods is to absorb neutrons. This allows the chain reaction to be controlled. When the reactor is just starting, the rods can be raised to allow the number of nuclei being split each second to increase. Once the desired level is reached, the rods are lowered part way into the reactore core until they absorb sufficient neutrons to maintain the fission at a constant rate. In case of emergency, safety rods drop fully into the core, abosorbing lots of neutrons, and shutting down the chain reaction. Moderator A moderator is used to slow down the neutrons released during fission. Collisions with the moderator material slow the neutrons down. Slow–moving, or ‘thermal’, neurons are more likely to be absorbed by a uranium–235 nucleus than fast–moving neutrons. Using a moderator means each neturon released during fission has a greater chance of causing another nucleus to underator. Sometimes graphite is used, but it may catch fire if an accident occurs inside the reactore core. For this reason, water–moderated reactors are the only type used in the Western world.

22 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Water

Coolant The coolant keeps the temperature constant throughout the reactor core, and is also used to transfer heat energy from the reactor core to the heat exchanger. Water is the most commonly used coolant. Often it is pumped under higbh pressure in between the rods and through the rest of the system. Water under high pressure does not turn into steam, even when the reactor operates at temperatures exceeding 300°C. In raectors that do not use graphite as a moderator, the water acts as both a coolant and a moderator. In graphite–moderated reactors, carbon dioxide (under enormous pressure) is used as a coolant. Containment vessel The portions of the nuclear power plant that contain radioactive material are housed in a containment vessel. This is an airtight, concrete and steel structure that isolates the reactore from the environment. It prevents any radioactive material escaping if an accident occurs. For safety reasons, containment vessels are designed to withstand a plane crashing into them! Heat exchanger In this section of the power plant, heat is transferred from the reactor coolant to a secondary water system used to turn the turbines. The purpose of the heat exchanger is to keep the water that runs through the reactor core separate from the water that travels outside the containment vessel. This is safety precaution as the coolant water itself can become radioactive.

TABLE 17.3 Mass and binding energies of some fission products and reactants Nucleus

Symbol

Mass (kg)

Total binding energy (MeV) −25

1813.454 956

236 U 92

3.919 629 × 10−25

1790.415 039

Cerium-148

148 Ce 58

2.455 916 × 10−25

1219.569 580

Barium-141

141 Ba 56

−25

1173.974 609

Krypton-92

92 Kr 36

1.526 470 × 10−25

783.185 242

Krypton-89

89 Kr 36

1.476 512 × 10−25

766.909 351

Plutonium-240 Uranium-236

240 Pu 94

Neutron Proton

1 P 1

3.985 755 × 10

2.339 939 × 10

−27

1 n 0

1.674 746 × 10

or 11H

1.673 351 × 10−27

SAMPLE PROBLEM 4

Use table 17.3 to answer the following questions about the fission of uranium-236 to barium-141 and krypton-92. a. What is the difference between the binding energy of the uranium-236 nucleus and the sum of the binding energies of the two fission fragments? b. What is the difference between the mass of the uranium-236 nucleus and the sum of the masses of all the fission fragments, including neutrons? c. Use E = mc2 to calculate the energy equivalent of this mass difference in joules and MeV. THINK

WRITE

Write out the fission equation. 2. Use table 17.3 to find the binding energy of uranium-236. 3. Calculate the sum of the binding energies of the fragments.

a. 1.

Calculate the difference between the binding energy of the uranium nucleus and its fission fragments. 5. State the solution. 4.

Use table 17.3 to find the mass of uranium-236. 2. Calculate the sum of the masses of the fragments.

b. 1.

236

92 1 → 141 56B a + 36K r + 3 0n + energy Binding energy of uranium-236 = 1790.415 039 MeV

a. 92U

Binding energy of barium-141 = 1173.974 609 MeV Binding energy of krypton-92 = 783.185 242 MeV 1173.974 609 + 783.185 242 = 1957.159 851 MeV Energy difference = 1957.159 851 – 1790.415 039 = 166.744 812 MeV The difference between the binding energy of the uranium nucleus and its fission fragments is 166.744 812 MeV. b. Mass of uranium-236 = 3.919 629 × 10−25 kg Mass of barium-141 = 2.339 939 × 10−25 Mass of krypton-92 = 1.526 470 × 10−25 Mass of a neutron = 1.674 746 × 10−27 2.339 939 × 10−25 + 1.526 470 × 10−25 + 3 × 1.674 746 × 10−27 = 3.916 651 × 10−25 kg

TOPIC 17 How do fusion and fission compare as viable nuclear energy power sources? 23

3.

4. c. 1.

Calculate the difference between the mass of the uranium nucleus and its fission fragments. State the solution. Use E = mc2 to calculate the energy difference in joules.

Convert to MeV by dividing by 1.602 176 × 10−13 . 3. State the solution. 2.

Mass difference = 3.919 629 × 10−25 − 3.916 651 × 10−25 = 2.978 × 10−28 kg The difference between the mass of the uranium nucleus and its fission fragments is 2.978 × 10–28 kg. c. E = mc2 ( )2 = 2.978 × 10−28 × 2.997 924 58 × 108 = 2.676 493 × 10−11 J 2.676 493 × 10−11 = 167.053 615 MeV 1.602 176 × 10−13 The energy equivalent of the mass difference is 2.676 493 × 10−11 J or 167.053 615 MeV.

PRACTICE PROBLEM 4 A plutonium-239 nucleus absorbs a neutron to become plutonium-240, which splits to form krypton-89, cerium-148 and three neutrons. Use the data in table 17.3 to answer the following questions. a. What is the difference between the binding energy of the plutonium-240 nucleus and the sum of the binding energies of the two fission fragments? b. What is the difference between the mass of the plutonium-240 nucleus and the sum of the masses of all the fission fragments, including neutrons? c. Use E = mc2 to calculate the energy equivalent of this mass difference in joules and MeV.

Resources Weblink Nuclear power plant simulation

17.5.5 Fukushima Daiichi nuclear disaster On Friday 11th March 2011, an earthquake, measuring 9.0 on the Richter scale, occurred about 70 kilometres off the east coast of Japan. It was the fourth most powerful earthquake since modern records began in 1900. As well as structural damage on the coast, the earthquake produced a tsunami that reached heights of 40 metres and travelled 10 kilometres inland. Fukushima Daiichi is a complex of six nuclear reactors. It was built on a bluff that was reduced to a height of 10 metres above sea level to give solid foundations. On the day of the earthquake only three units (1, 2 and 3) were operating; the others were shut down for maintenance. The earthquake produced a ground acceleration of 5.5 m s–2 at unit 2, which was above the design tolerance for that reactor of 4.4 m s–2 . Immediately after the earthquake, the control rods automatically shut down the fission reactions, but there was still heat in the reactor core, sufficient to require active cooling for several days to prevent the fuel rods melting.

24 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

The tsunami arrived 50 minutes after the earthquake. The tsunami height at Fukushima was 14 metres, which flooded the low-lying rooms and disabled the emergency diesel generators. The battery operated emergency generators took over, but ran out of power the next day and cooling of the reactor core stopped. Replacement batteries were delayed by poor road conditions. Over the next few days, with increasing reactor temperature, the zirconium coating of the fuel rods reacted with water to produce hydrogen in each of the three reactors. The hydrogen gas vented out of the reactor vessel, mixed with air and exploded in the outer secondary containment building causing a fire. It is estimated that most of the fuel in each of the reactor cores melted and is now on the concrete floor of the containment vessels. Radioactive material, consisting mainly of iodine-131, caesium-134 and caesium-137, was released from the reactors by the following mechanisms: • deliberate venting to reduce gas pressure in the containment vessels • deliberate discharge of coolant water into the sea • uncontrolled events. All three radioactive isotopes released undergo beta decay with the following half-lives: • iodine-131 — 8 days • caesium-134 — 2 years • caesium-137 — 30 years. Radioactive material vented to the atmosphere eventually falls out from the atmosphere downwind of the site (hence, the term ‘fallout’). If the material falls on pasture, it can enter the food chain. Iodine is found in cow’s milk and when consumed it concentrates in the thyroid. After the Chernobyl disaster, there was an increase in the incidence of thyroid cancer among children. In Japan after the Fukushima accident, care was taken to break the food chain and also to remove the caesium-contaminated topsoil. There was no increase in the incidence of thyroid cancer in Japan, but the psychological effects of relocation and uncertainty are now becoming noticeable. Caesium-137 is the main health concern in decontaminating land around Fukushima. In the first few months following the tsunami, about 8 kilograms of caesium-137 flowed into the ocean. Fortunately the currents off the coast are rapid and the caesium dispersed to low concentrations. Atmospheric releases of radioisotopes spread around the world, reaching the west coast of the United States two days after the explosions. Measurements across the globe suggested that the release of radiation was about 10–40% that of the Chernobyl disaster and covered an area about 10% the size. The World Health Organization (WHO) FIGURE 17.18 Damage to the roof of the outer containment indicated that people who were evacuated building at the Fukushima power plant due to the hydrogen air from the area around Fukushima were explosions in each of the three reactors exposed to so little radiation that radiation-induced health effects are likely to be below detectable levels. However, decommissioning the reactors remains a problem, with radioactivity seeming to still leak into the groundwater. The Fukushima Nuclear Accident Independent Investigation Commission found the nuclear disaster was manmade and that its direct causes were all foreseeable. The report also found that the plant was incapable of withstanding the earthquake and tsunami. The company, the regulators and the government body promoting the nuclear power industry all failed to meet the most basic safety requirements, such as assessing the probability of damage, preparing for containing collateral damage from such a disaster and developing evacuation plans.

TOPIC 17 How do fusion and fission compare as viable nuclear energy power sources? 25

DISASTER AT CHERNOBYL As in any other industry, accidents have occurred in the nuclear power industry. In most cases the damage has been minimal, and was contained inside the reactor without posing danger to nearby communities. This is due to the large number of safety features incorporated into reactor design. Unfortunately, on 26 April 1986, a major accident happened at Chernobyl, near Kiev — then part of the Soviet Union. The Chernobyl nuclear power plant reactors were graphite moderated and water cooled. Unlike those in Western countries, Soviet reactors were sometimes built without a containment vessel and the Chernobyl reactors were of this kind. On the day before the accident, one of the reactors at the power plant had been reduced to running at about 50% of its usual power output and the emergency core cooling systems had been switched off to allow tests to be carried out. The tests were to measure the effectiveness of modifications that had been made to the generators and were not due to concerns about the reactor core. Just after 1 am on 26 April, the operators had to make changes to the system in order to get the reactor to behave in a way that would allow the test to continue. When another change was made to set the system up for the tests, the reactor became more unstable and could no longer be controlled. The operator tried to insert the control rods fully to stop the reaction. In four seconds the power rose to 100 times its normal level and a steam explosion occurred. The temperature of the core reached about 5000 K, one-third of the fuel was destroyed by the explosion and about one-tenth of the core (mostly the graphite moderator) was burned, releasing about 4% of the fuel into the atmosphere because there was no containment vessel. Two staff were killed at the time of the accident — one was hit by a jet of steam, the other by a block of concrete. A further 300 reactor staff and emergency workers were treated in hospital, and 29 of them died of acute radiation poisoning. FIGURE 17.19 A giant confinement arch 150 metres long and 92 metres high is now being constructed at the Chernobyl site to prevent leakage of radioactive material. This is the second sarcophagus for the Chernobyl reactor. The first is beginning to break down. Components of the new shell are being assembled off site, as the air above the reactor is still too radioactive for humans. When completed, the shell will be rolled over the reactor to the left.

The International Nuclear Safety Group investigated the causes of the Chernobyl disaster. They concluded that there were problems with the design of the reactor and with the test procedures, which contributed to the accident. In addition, investigators felt that operators were not fully aware of safety issues, due to a poor ‘safety culture’, both locally and nationally. As a result of the investigation, much has been learned about design and safety of reactors in general, and about Soviet reactors in particular.

26 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

17.5.6 Nuclear wastes High-level waste Fission reactors do not produce as much waste as coal or gas power plants, but the waste they do produce is highly radioactive. When fission occurs inside the fuel rods, the fission fragments produced are unstable. These fragments then undergo a series of radioactive decays to become more stable. Commonly the fission fragments release 𝛽 − particles in order to increase the number of protons in the nucleus. Some of the isotopes produced have very short half-lives, while others have longer half-lives. Two frequently produced fission fragments and their decay chains are: 94 38S r 140 54X e

→

140 55C s

→

→

94 39Y

→

140 56B a

94 40Z r

→

140 57L a

(stable) →

140 58C e

(stable)

The used (or spent) fuel rods can be reprocessed. During this procedure any unused fuel is removed from the rods for reuse. Plutonium produced in the fuel rods is also separated. The remnants are classified as high-level waste. While such waste does not take up a lot of space, it is highly radioactive. It has been estimated that high-level waste will take about 1000 years to return to the same level of radioactivity as the uranium ore that was originally mined, and at least 5 million years before there is no longer any significant radiation. High-level waste must be stored in shielded containers to prevent radiation escaping into the environment. It must also be cooled to stop overheating. Australia does not produce any high-level waste.

SPENT FUEL RODS IN THE US The United States has so far refused to reprocess spent fuel rods. The reason is concern that the plutonium-239 that is removed during reprocessing may be illegally used to manufacture nuclear weapons.

Medium-level waste In addition to the fuel rods, all other material in the reactor core is exposed to a huge amount of radiation. Some of this radiation can be absorbed by the material, which itself becomes radioactive. The fuel containers, pipes, gauges and other reactor components are classified as medium-level waste. This waste does not require cooling, but still needs to be shielded. Radioactive isotopes used in medicine and industry are classified as medium-level waste once their useful life is over.

Low-level waste Used protective clothing, water from showers and the cleaning of protective gear, and old plant equipment all make up low-level waste. Often such waste contains levels of radiation that are just above safe levels, or isotopes with very short half-lives or low activities. Sometimes these wastes can be released into the environment after being diluted, or simply stored for a short time. Low-level waste does not require shielding during handling. Such waste has a dose rate of about 2 millisieverts per hour. If the dose rate is above this, then waste is classified as intermediate-level and needs shielding.

Waste disposal Because of the long life of high- and medium-level radioactive waste, careful consideration must be given to its storage. At present most waste is planned to be stored underground in geologically stable areas away from underground water. Some of the waste is simply placed in steel storage canisters. Another option is to fuse the waste into glass blocks — a process called vitrification. Australian scientists have developed a process by which wastes are encapsulated in an artificial rock dubbed ‘Synroc’. In this synthetic substance, the waste is incorporated into the crystal lattice, making it resistant to high temperatures and water. These properties

TOPIC 17 How do fusion and fission compare as viable nuclear energy power sources? 27

make Synroc ideal for underground storage of nuclear wastes. It can resist the high temperatures present very deep in the Earth’s crust, and heat produced by radioactive decay. Underground water supplies will not break down Synroc, thus avoiding the possibility that the waste will contaminate water supplies. Natural rocks with similar composition to Synroc have been known to survive harsh conditions for millions of years. It is hoped that Synroc will do the same. Storage of nuclear waste deep in the oceans has also been used by some countries. While this is mostly used for lower level wastes, it is still a source of concern for environmentalists because the metal storage canisters will corrode in time, allowing the radioactive contents to leak into the world’s oceans. Suggestions have been made that nuclear waste should be shot away from the Earth in rockets, possibly into the Sun’s core (although the amount of energy needed to do this makes it prohibitive). It is not known, however, what effect this might have on the stability of the Sun! In addition, the possibility of a rocket carrying nuclear waste exploding in our atmosphere, or crashing back to Earth, is not something most governments would be prepared to risk.

17.5.7 Energy released As we discovered in sample problem 4, the amount of energy released by the fission of uranium-236 to produce barium-141 an krypton-92 is approximately 167 MeV. There are two measures of the energy released that can be used to compare nuclear reactions, in particular fission and fusion reactions. They are: • energy released per nucleon • percentage of mass that is transformed into energy.

Energy released per nucleon In the fission of uranium-236, there are a total of 236 protons and neutrons involved. So: 167 MeV 236 = 0.71 MeV

energy released per nucleon =

SAMPLE PROBLEM 5

Uranium-236 undergoes a fission reaction releasing 172.646 963 MeV of energy. Calculate how much energy is released per nucleon in MeV. THINK

WRITE

1.

Nucleons are the total number of protons and neutron in uranium-236.

There are 236 nucleons in uranium-236.

2.

The energy per nucleon is calculated by dividing the energy by the number of nucleons.

Energy per nucleon =

3.

State the solution.

172.646 963 236 = 0.731 554 MeV 0.731 554 MeV of energy is released per nucleon in the fission reaction.

PRACTICE PROBLEM 5 Fission of uranium-236 to xenon-140 and strontium-94 releases 178.189 774 MeV of energy. Calculate the energy released per nucleon for this reaction.

28 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

17.5.8 Percentage of mass that is transformed into energy The total mass of the fission fragments including the neutrons is less than the mass of the starting uranium-236 nucleus. This missing mass has appeared as the 167 MeV of kinetic energy of the fragments. It is useful to know how much mass this 167 MeV represents. The amount of missing mass can be calculate using E = mc2 , but first the energy of 167 MeV needs to be converted into joules. 1 MeV = 1.602 176 × 10–13 joules. The number comes from the charge on the electron: 1.602 176 × 10–19 coulomb. The other number required is the speed of light, c = 2.997 924 58 × 108 m s1 . Energy released in MeV = 167 MeV Energy released in joules = 167 MeV × 1.602 176 × 10−13 J MeV−1 = 2.675 634 × 10−11 J 2.675 634 × 10−11 (2.997 924 58 × 108 )2 = 2.977 044 × 10−28 kg

Missing mass =

The percentage of the original uranium-236 nucleus converted into energy can now be determined. The mass of the uranium-236 nucleus is 3.919 629 × 10–25 kg. 2.977 044 × 10−28 × 100 3.919 629 × 10−25 = 0.076%

Percentage =

17.5.9 Environmental impact of a fusion reactor The advantages of a fusion reactor over fission are that there is no chain reaction that could possibly run away, and there is no radioactive waste. Any accident is not likely to require the evacuation of the surrounding area. Neutrons will be emitted from the reaction vessel, but they can be stopped by a thick containment vessel. Tritium is radioactive with a reasonably long half-life, but if accidentally released into water, it is impossible to remove and will flow through the planet’s water cycle.

17.5.10 Conclusion Having studied this topic, you are now in a position to compare fission and fusion on each of the following aspects: • the reactants — that is, the starting material for the reaction • how the reaction is initiated • the products at the end of the reaction • how plentiful and easily obtained the reactants are • how much energy is produced • the opportunities and difficulties of the technology being used as a source of society’s energy • the environmental impact of the technology • the risks and benefits for society of using nuclear energy as a power source. The thermodynamics topics also invited you to investigate solar thermal technology as a power source for society, so you may wish to include that technology as part of your comparison.

TOPIC 17 How do fusion and fission compare as viable nuclear energy power sources? 29

17.5 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Make a list of the similarities and differences between the way electricity is produced in a nuclear power plant and the way it is produced in a coal-burning plant. 2. How do control rods allow the fission chain reaction to be controlled? 3. Explain why fast breeder reactors are likely to be the main producers of nuclear power in the future. 4. Enriching uranium is difficult. Why? 5. After the explosion at the Chernobyl reactor, tonnes of lead, sand and boron were dropped into the reactor. Why was boron used? 6. Why are ‘thermal’ reactors so called? 7. What do control rods control? 8. What does the phrase ‘reprocessing of spent fuel rods’ mean?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

17.6 Review 17.6.1 Summary •

• • • •

Fission reactions occur when a nucleus is split into smaller, more stable fission fragments. If every neutron released in fission is free to initiate more fission reactions, an uncontrolled chain reaction occurs. Controlled chain reactions occur when some of the free neutrons are absorbed by non-fissionable substances. Nuclear reactors use the energy generated by controlled chain reactions to heat water. The steam produced turns the turbines that produce electricity. The fuel in some nuclear reactors is more likely to undergo fission when it absorbs slow-moving neutrons. Moderators are used in these reactors to slow down neutrons. Control rods start and stop the nuclear reaction by absorbing neutrons. The amount of uranium-235 in natural uranium is not enough to sustain a chain reaction. In order for uranium to be used in some types of nuclear reactors and nuclear weapons, the percentage of uranium-235 needs to be increased to 1–4% for nuclear reactors and 97% for weapons. The fuel in fast breeder reactors undergoes fission when it absorbs fast-moving neutrons. Its fuel does not need to be enriched, because it uses plutonium-239 derived from uranium-238 as the fuel source. The reaction is: 238 92U

• • •

+ 10n →

239 92U

→

239 93N p

+ β− →

239 94P u

+ β−

A critical mass is needed for a sustainable chain reaction. The nuclear fusion reaction between deuterium and tritium is a possible energy source. The processes of nuclear fission and fusion can be compared by a variety of measures including energy released per nucleon and percentage of mass lost.

30 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Resources

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0042).

17.6.2 Key terms A chain reaction occurs when neutrons, emitted from the decay of one atom, are free to initiate fission in surrounding nuclei. Control rods start and stop the nuclear reaction by absorbing neutrons. Critical mass of a fissionable substance is the smallest spherical mass that will sustain an uncontrolled chain reaction. Enrichment is the process of increasing the percentage of uranium-235 in a sample of uranium. Enrichment is important because naturally occurring uranium does not have a high enough percentage of uranium-235 to sustain a chain reaction. High-level waste does not take up a lot of space but is highly radioactive. High-level waste needs to be stored in shielded containers and cooled to stop overheating. Low-level waste contains levels of radiation that are just above safe levels, or isotopes with very short half-lives or low activities. Low-level waste is either stored for a short time and then released into the environment, or released immediately, without first being stored. Medium-level waste does not require cooling, but still needs to be shielded. A moderator is a material that slows down the speed of a neutron. A nuclear winter may follow large-scale nuclear detonations. It involves the blocking of sunlight by the smoke, dust and soot produced by a nuclear explosion. The reduction of sunlight would lead to a drop in the Earth’s temperature and interrupt vital processes such as photosynthesis A thermonuclear weapons are those whose initial fission explosion provides the intense heat necessary to overcome the repulsion between positive nuclei and allow them to get close enough for fusion to occur.

Resources Digital document Key terms glossary (doc-32999)

17.6.3 Practical work and investigations Investigation 17.1 Chain reaction with dominoes Aim: To model controlled and uncontrolled chain reactions in nuclear fission using dominoes Digital document: doc-31900 Teacher-led video: tlvd-1074

Resources Digital document Practical investigation logbook (doc-33000)

TOPIC 17 How do fusion and fission compare as viable nuclear energy power sources? 31

17.6 Exercises 17.6 Exercise 1: Multiple choice questions 1.

Which of the following is the isotope tritium? 2 A. 1H 3 B. 1H 1

C. 1H 2 D. 1D

Under what conditions can fusion occur? A. Electric field B. Bright light C. High temperature D. In water 3. Which of the following is one of the three possible reactions for a terrestrial fusion reactor? 3 6 3 A. 1H + 1H → 2H e + 10n 2.

4

3

B. 2H e + 1H

→ 43L i + 3 10n

+ 11H → 22H e + 10n + 31H → 42H e + 10n Tritium is radioactive and it decays into helium-3. What form of decay does it undergo? A. Alpha B. Beta C. Gamma D. X-ray When uranium-235 collides with a neutron, which of the following is a possible chemical equation? 235 141 92 A. 92U + 10n → 46P d + 36K r + 3 10n 87 146 235 B. 92U + 10n → 35B r + 67L a + 3 10n 141 92 235 C. 92U + 10n → 56B r + 36K r + 3 10n 147 89 235 D. 92U + 10n → 58C e + 36K r + 3 10n One of the fragments of fission of uranium-236 has an atomic mass number of about 96. What is the most likely species produced? A. Gold B. Barium C. Caesium D. Rubidium Which of the following is one of the processes used to overcome difficulties in causing a chain reaction? A. Engaging B. Enhancing C. Enrichment D. Evolving The process to slow down the neutrons very quickly is done using which of the following? A. A net B. A moderator C. Water D. Heat 1

C. 1H 2 D. 1H

4.

5.

6.

7.

8.

32 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Why are neutrons good for initialising a chain reaction? A. They are neutrally charged. B. They are light. C. They are heavy. D. They are large in size. 10. An element that has 105 protons and 125 neutrons releases 180.5465 MeV. What is the energy released per nucleon? A. 1.7195 MeV B. 0.7850 MeV C. 1.4444 MeV D. 0.8570 MeV 9.

17.6 Exercise 2: Short answer questions Explain the how a chain reaction works. Write the equations that make up the three-stage fusion process which occurs in the Sun to convert hydrogen to helium. 3. Given the general equation A + B → C + D and using the following table, calculate the: a. difference between the binding energy of products and reactants b. difference between the sum of masses of the products and reactants c. energy equivalent of this mass difference in joules and MeV.

1. 2.

Symbol

4.

5. 6. 7. 8. 9. 10.

Mass (kg)

Total binding energy (MeV) −25

A

3.610 422 × 10

B

1.956 117 × 10−27

C

2.307 189 × 10

−25

890.065 063

D

1.321 596 × 10−25

644.312 162

1467.164 791 0

Using the values from your solution to question 3, calculate the: a. energy released per nucleon of reactants in MeV, given the reactants have 72 protons and 106 neutrons combined b. percentage of mass transformed into energy. How and why are control rods used in nuclear reactors? How does a nuclear reactor generate electricity? What is the difference between a moderator and a control rod in a nuclear reactor? What is the critical mass of a fissionable substance? What are the advantages of a fusion reactor compared to a fission reactor? The mass difference of a fission reaction is 3.723 48 × 10−28 kg. If the mass of the original uranium-236 nucleus is 3.919 629 × 10−25 kg, calculate the percentage of the original uranium-236 nucleus that is converted into energy.

17.6 Exercise 3: Exam practice questions Question 1 (2 marks) 85 148 1 For the fission equation 236 92U → 35B r + 67L a + 3 0n , calculate the mass difference between the fragments and the uranium-236 nucleus given the following information. Mass of uranium-236 = 3.919 629 × 10−25 kg Mass of lanthanum-148 = 2.456 472 × 10−25 kg Mass of bromine-85 = 1.410 057 × 10−25 kg Mass of neutron = 1.674 746 × 10−27 kg TOPIC 17 How do fusion and fission compare as viable nuclear energy power sources? 33

Question 2 (3 marks) The fission of uranium-236 to barium-141 and krypton-92 releases 166.744 812 MeV of energy. Use the data in table 17.3 to calculate the percentage of mass transformed into energy. Question 3 (6 marks) Nuclear reactors do not produce as much waste as coal or gas power plants, but they do produce what is described as high, medium and low-level waste. Describe each level of waste from a nuclear reactor and how it must be handled or disposed of. Question 4 (3 marks) Describe the difference between a fission and fusion bomb. Question 5 (9 marks) Consider the fission reaction Use the following values. Nucleus

236 92U

→

137 52T e

1 + 97 40Z r + 2 0n .

Symbol

Mass (kg)

Total binding energy (MeV)

Uranium-236

236 U 92

−25

3.919 629 × 10

1790.415 039

Tellurium-137

137 Te 52

2.273 507 × 10−25

1139.760 337

Zirconium-97

97 Zr 40

Neutron

1 n 0

−25

1.609 304 × 10

837.065 893

−27

1.674 746 × 10

Calculate the: a. difference between the mass of the uranium-236 nucleus and the sum of the binding energies of the fission fragments 2 marks b. difference between the mass of the uranium-236 nucleus and the sum of the masses of the fission fragments, including neutrons 2 marks c. energy equivalent of this mass difference in joules and in MeV 2 marks d. energy released per nucleon in MeV 1 mark e. percentage of mass transformed into energy. 2 marks

17.6 Exercise 4: studyON topic test Fully worked solutions and sample responses are available in your digital formats.

Test maker Create unique tests and exams from our extensive range of questions, including practice exam questions. Access the assignments section in learnON to begin creating and assigning assessments to students.

34 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

AREA OF STUDY 2 OPTIONS OBSERVATION OF THE PHYSICAL WORLD

18

How is radiation used to maintain human health? 18.1 Overview Numerous videos and interactivities are available just where you need them, at the point of learning, in your digital formats, eBookPLUS and learnON at www.jacplus.com.au.

18.1.1 Introduction The models used by physicists to explain the behaviour of electromagnetic radiation and the nucleus of the atom have provided the basis for huge improvements in medical diagnosis and treatments during the last century. In this topic we will look at how electromagnetic radiation and particle radiation are used in medical imaging and diagnosis. FIGURE 18.1 Using computer analysis, the data from images of ‘slices’ through the body can be combined to produce a three-dimensional image of the area under investigation.

TOPIC 18 How is radiation used to maintain human health? 1

18.1.2 What you will learn KEY KNOWLEDGE After completing this topic you will be able to: Radiation and the human body • distinguish between electromagnetic radiation and particle radiation • describe how X-rays for medical use are produced including the distinction between soft and hard X-rays • describe how medical radioisotopes may be produced by neutron bombardment and high energy collisions • analyse decay series diagrams of medical radioisotopes with reference to type of decay and stability of isotopes • compare ionising and non-ionising radiation with reference to how each affects living tissues and cells • explain the effects of 𝛼, 𝛽 and 𝛾 radiation on humans, including: • different capacities to cause cell damage • short- and long-term effects of low and high doses • ionising impacts of radioactive sources outside and inside the body • calculations of absorbed dose (gray), equivalent dose (sievert) and effective dose (sievert) The use of radiation in diagnosis and treatment of human illness and disease • compare the processes of, and images produced by, medical imaging using two or more of X-rays, computed tomography (CT), 𝛾 radiation, magnetic resonance imaging (MRI), single photon emission computed tomography(SPECT) and positron emission tomography (PET) • describe applications of medical radioisotopes in imaging and diagnosis • explain the use of medical radioisotopes in therapy including the effects on healthy and damaged tissues and cells • relate the detection and penetrating properties of 𝛼, 𝛽 and 𝛾 radiation to their use in different medical applications • analyse the strengths and limitations of a selected contemporary diagnostic or therapeutic radiation technique. Source: VCE Physics Study Design (2016–2021) extracts © VCAA; reproduced by permission.

Resources Digital documents Key science skills — VCE Units 1–4 (doc-31856) Key terms glossary (doc-33001)

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0043).

2 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

18.2 Radiation and the human body KEY CONCEPTS • Distinguish between electromagnetic radiation and particle radiation. • Describe how X-rays for medical use are produced including the distinction between soft and hard X-rays. • Compare ionising and non-ionising radiation with reference to how each affects living tissues and cells. • Explain the effects of 𝛼, 𝛽 and 𝛾 radiation on humans, including: • different capacities to cause cell damage • short- and long-term effects of low and high doses • ionising impacts of radioactive sources outside and inside the body • calculations of absorbed dose (gray), equivalent dose (sievert) and effective dose (sievert).

18.2.1 Electromagnetic radiation and particle radiation

Ionising radiation is the collective name given to 𝛼 and 𝛽 particles, neutrons that have been released from the nucleus, 𝛾 rays and X-rays. These forms of radiation are grouped together because they have high energies and therefore similar effects on matter. Other forms of electromagnetic radiation, such as radio waves, microwaves and visible light, have lower energies and do not interact with matter in the same way. They are non-ionising radiations. X-rays are the only type of ionising radiation that are not formed by changes in the nucleus. They result from large energy losses by electrons. Ionising radiation has sufficient energy to knock an electron from its orbit around a nucleus. Once an electron has been knocked away from the nucleus, the atom has more positive charges (protons) than negative charges (electrons), giving the atom an overall positive charge. Atoms that have an overall charge are called ions, hence, the name ionising radiation. Sometimes the electron that is knocked from the atom is part of a bond between one atom and another. This causes the bond to be broken, and can result in the molecule being split in two. Each piece of the molecule would then have a charge. The charged pieces are called free radicals. Both ions and free radicals are chemically very reactive. This may result in new chemical reactions taking place inside the substance that was exposed to the ionising radiation.

18.2.2 What are X-rays? X-rays are electromagnetic waves of very high frequency and very short wavelength, in the range 0.001 nm to 10 nm. Because of their high frequency, and hence high energy, they can penetrate flesh and may cause ionisation of atoms they encounter on the way through. When X-rays pass through the body, the body tissue absorbs energy and the intensity of the beam is reduced. The more dense material, such as bone, absorbs more X-radiation.

FIGURE 18.2 An X-ray of the lungs showing damage due to tuberculosis

18.2.3 Production of X-rays X-rays are emitted from a cathode-ray tube when the cathode rays strike the glass of the tube. Similar principles are used to produce X-rays for medical diagnosis. A cross-section of an X-ray tube is shown in figure 18.3. The tube is highly evacuated and a very high voltage, from 25 000 to 250 000 volts, is applied between the anode and cathode. The cathode is a filament of wire through which a TOPIC 18 How is radiation used to maintain human health? 3

current is passed. Electrons are emitted from the hot filament and a metal focusing cup directs the electrons towards the anode. The very high voltage between the cathode and anode accelerates the electrons to the anode. The anode is usually made of tungsten that can withstand the high temperatures generated. When the electrons strike the tungsten they are absorbed and some of their energy is converted to X-rays. By placing the tungsten target at an angle to the incoming electron beam, the X-rays emitted from the tungsten can be sent in a predetermined direction. FIGURE 18.3 An X-ray tube Evacuated chamber Heated filament Anode mounting (copper)

Electron beam

Coolant circulates here

Metal target (tungsten)

X–rays

Window

Very high potential difference

Tungsten is usually used for the target as it has a very high melting point of about 3400 °C and emits X-rays when struck by electrons. This is not a very efficient way to produce X-rays as only about 1% of the energy reaching the target is converted to X-rays. The rest is converted to thermal energy in the target — enough to heat a cup of water to boiling point in one second. Hence, it is important to prevent the target from overheating or melting. Copper — a good conductor of heat — is used for the anode mountings and oil, circulating in the outer region near the anode, helps the cooling by convection. Rotating the target at a rapid rate, approximately 3600 revolutions per minute, also allows the heat produced to be distributed over a large area.

18.2.4 Hard and soft X-rays An X-ray beam consists of a range of frequencies. A thin sheet of material placed in the path of the X-ray beam acts as a filter and absorbs more low frequency rays than high frequency rays. The beam will then have a greater proportion of high frequency (high energy) rays, which are more penetrating than the low frequency rays. The beam is said to consist of hard X-rays. By contrast, a beam of X-rays with lower frequency rays has less energy, is less penetrating and is said to consist of soft X-rays. Note that the higher the frequency of the rays, the shorter their wavelength. Hard X-rays are preferred for imaging as they penetrate the body and are absorbed by material such as bone, allowing images of the bone to be observed. Soft X-rays are not useful for imaging as they will not penetrate the body. They expose the patient to additional useless and possibly harmful X-radiation.

18.2.5 Ionising radiation and living things The chemical changes resulting from the production of ions and free radicals in living cells can have a range of effects. The cytoplasm (the part of the cell that surrounds the nucleus) has a high water content, therefore, it is often water molecules that are broken. This results in the production of H+ and OH− ions, 4 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

which are chemically very reactive. The ions may react with important molecules, causing damage to DNA, the mechanisms for controlling cell division, or the production of molecules necessary for the life of the cell. Cells undergoing division when they are irradiated are particularly at risk. Often the cell is able to repair itself, but sometimes the chemical changes cause the cell to die. This is not a problem for a living thing unless a large number of cells are damaged. If the mechanism for cell division is damaged, the cell may begin to reproduce uncontrollably, forming a cancer. If DNA in the ovaries, the testes or in an unborn foetus is damaged, genetic mutations may be passed on. Usually these mutations are recessive and are not exhibited.

18.2.6 The effects of 𝛼, 𝛽 and 𝛾 radiation on humans How much radiation is safe? This seemingly simple question has a very complicated answer. The damage caused by radiation depends on the type of radiation, the rate at which it is received, the part of the body exposed, the general health of the individual and many other factors.

FIGURE 18.4 Visualisation of the damage caused by ionising radiations of different penetration powers in a block of carbon

Absorbed dose One measure of the amount of radiation received is the absorbed dose. This is the amount of energy absorbed by each kilogram of the tissue being irradiated. The unit of absorbed dose is the gray, which is given the symbol Gy (1 Gy = 1 J kg−1 ). energy absorbed absorbed dose = mass

α

β

Unfortunately, the number of grays absorbed by a person does not provide much information about the γ extent of the damage to that person. The penetrating power of the type of radiation is important. For example, alpha (𝛼) particles are stopped in a short distance. They pass on all their energy in a short space, causing a great deal of localised damage. Beta (𝛽) particles are more penetrating, so the damage they cause is less severe in any one area but is more widespread. Neutrons, 𝛾 rays and X-rays are far more penetrating than either 𝛼 or 𝛽 particles. They spread their energy over a large range.

Equivalent dose To take into account the different styles of damage caused by the various forms of ionising radiation, another measure of the amount of radiation, equivalent dose, was developed. The units for equivalent dose are sieverts (Sv). equivalent dose (Sv) = absorbed (Gy) × quality factor

The quality factor is determined by the type of radiation that delivered the energy. One sievert of radiation causes the same amount of biological damage, no matter what type of radiation is used.

TOPIC 18 How is radiation used to maintain human health? 5

TABLE 18.1 Quality factors for different types of radiation Type of radiation

Approximate quality factor

𝛾 rays and X-rays 𝛽 particles

1

Slow neutrons

3

Fast neutrons

10

1

𝛼 particles

10–20

SAMPLE PROBLEM 1

A 60-kilogram person absorbs 0.054 J of energy due to ionising radiation. a. Calculate the absorbed dose. b. What would be the equivalent dose if the energy was delivered by 𝛾 rays? c. What would be the equivalent dose if the energy was delivered by 𝛼 particles? (Take the quality factor to be 20.) d. Which would cause more biological damage to the person? THINK a. 1.

Recall the formula for the absorbed dose.

State the solution. b. 1. The quality factor of gamma radiation is 1. 2.

2. c. 1.

2. d. 1.

2.

State the solution. The quality factor of alpha radiation is 20.

State the solution. Recall that damage is measured by equivalent dose, in sieverts.

State the solution.

6 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Absorbed dose =

WRITE

energy absorbed mass 0.054 J = 60 kg = 9 × 10−4 Gy The absorbed dose is 9 × 10−4 Gy. b. Equivalent dose = absorbed dose × quality factor a.

= 9 × 10−4 Gy × 1

= 9 × 10−4 Sv = 0.9 mSv The equivalent dose if the energy was delivered by 𝛾 rays is 0.9 mSv. c. Equivalent dose = absorbed dose × quality factor = 9 × 10−4 Gy × 20

= 1.8 × 10−2 Sv = 18 mSv The equivalent dose if the energy was delivered by 𝛼 particles is 18 mSv. d. Equivalent dose of 𝛾 rays = 0.9 mSv Equivalent dose of 𝛼 particles = 18 mSv 18 = 20 0.9 The equivalent dose delivered by the 𝛼 particles would cause about 20 times more damage than that delivered by the 𝛾 rays.

PRACTICE PROBLEM 1 On average, each crew member on the Apollo space missions received a dose of 12 mSv while in space. The exposure was mainly electrons and 𝛾 rays. Estimate an astronaut’s mass and determine how much energy each astronaut absorbed.

Effective dose Radiation affects different parts of the body in different ways. Each organ or tissue in the body has a different sensibility to radiation doses. For example, the head is less sensitive than the chest. The effective dose is a number that is calculated for an individual patient. This number takes into account the absorbed dose, the quality factor (relative harm level) of the different types of radiation and the sensitivity of each organ or tissue type to the different types of radiation. It also takes into account the fact that different parts of the body will not receive the same amount of radiation when undergoing a medical procedure. The calculation of the effective dose helps to estimate the risk to a patient from a procedure. The actual risk to an individual patient will also depend on such factors as the size and age of the patient. The effective dose for a patient is the total of the equivalent doses for all the different parts of the body. Effective dose is measured in sieverts.

18.2 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Why can the formation of free radicals and ions be damaging to living cells? 2. X-rays can be classified as hard or soft. (a) How are hard X-rays different from soft X-rays? (b) Why are hard X-rays preferred for imaging the human body? 3. A 30-kilogram child receives 3 mGy of radiation. How much energy did the child absorb? 4. An adult (60 kg) absorbs 0.09 J of radiation. What is the adult’s absorbed dose? 5. What is the equivalent dose of 3 mGy of radiation, assuming the energy was delivered by 𝛾 radiation? 6. What is the equivalent dose of a 60-kilogram adult who receives 3 mGy of radiation, assuming the energy was delivered by 𝛼 radiation? Assume a quality factor of 18. 7. Why is 𝛼 radiation given a higher quality factor than 𝛾 radiation? 8. Why is equivalent dose often a more useful measure than absorbed dose? 9. It is more dangerous for pregnant women to be exposed to high radiation levels than for other people. Why? 10. A particularly concerned man is keen to minimise his exposure to background radiation. What advice could you give him on the lifestyle changes he should make? 11. Australians receive on average 2 mSv of radiation each year. Assuming this radiation is all 𝛽 particles with energy of 1 MeV, how many 𝛽 particles pass in or out of your body every second? (Hint: Estimate your body mass and find out how many joules of radiation you receive each year. Find out how many joules of energy there are in a 1 MeV 𝛽 particle, then find out how many 𝛽 particles pass through your body every year, then every second.) 12. Ionising radiation can cause cancer, yet it can also cure cancer. Explain this contradictory statement.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

TOPIC 18 How is radiation used to maintain human health? 7

18.3 The use of radiation in diagnosis and treatment KEY CONCEPTS • Describe applications of medical radioisotopes in imaging and diagnosis. • Explain the use of medical radioisotopes in therapy including the effects on healthy and damaged tissues and cells. • Relate the detection and penetrating properties of 𝛼, 𝛽 and 𝛾 radiation to their use in different medical applications. Describe how medical radioisotopes may be produced by neutron bombardment and high energy • collisions. • Analyse decay series diagrams of medical radioisotopes with reference to type of decay and stability of isotopes.

18.3.1 Radioactivity as a diagnostic tool The best known use of radioactive materials in medicine is probably in the ‘radiotherapy’ treatment of cancer. Less well known is the use of a radioactive material inside the body to diagnose disease. This use of radioactive material in the body may seem very risky because of the danger associated with radioactivity. In fact the use of radioisotopes and, more recently, PET (positron emission tomography) to image organs and study their function has become a very common, effective and safe means of diagnosis. For the purposes of medical diagnosis, radioactive substances may be introduced into the body and used to target areas of interest. The radiation produced is measured and used to determine the health of the organ or section of the body under investigation.

PRODUCING RADIOISOTOPES: THE MEDICAL CYCLOTRON The effectiveness of nuclear medicine for FIGURE 18.5 A cyclotron for radioisotope synthesis in diagnosis of disease has depended on the a clinical medical centre ability to: • produce radioisotopes • detect the gamma radiation produced. The production of radioisotopes became possible with the development of the cyclotron by E. O. Lawrence in 1931. From the mid-1940s, a range of radioisotopes from the United States and later from the United Kingdom was available. The radioisotopes needed in nuclear medicine in Australia are made at the nuclear research reactor based at Lucas Heights in the south of Sydney, or in a cyclotron under the control of ANSTO (Australian Nuclear Science and Technology Organisation). Cyclotrons are needed to make radioisotopes for positron emission tomography (PET), a diagnostic technique discussed later in this topic. PET facilities are presently found in hospitals in Brisbane, Melbourne and Sydney.

18.3.2 Choosing the right medical radioisotope When a radioisotope is introduced into the body, other factors in addition to the half-life of the radioisotope need to be considered. The radioisotope is removed from the patient’s body by processes such as respiration, urination and defecation. However, some patients metabolise the chemical to which the radioisotope is attached more quickly than others, so it is important that the characteristics of the particular patient are considered when dosages are being determined.

8 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

The half-life of the radioisotope must be long enough to allow useful readings to be taken after it has been taken up by the targeted organ. Generally, if the radioisotope remains in the patient’s body for a long period of time, its half-life should be comparable to the time taken to carry out the investigation, to minimise the dose to the patient. When the radioisotope is excreted in about the same time as is needed for the investigation, a longer half-life radioisotope can be safely used. Radioisotopes that emit 𝛼 particles are not used in the diagnosis of disease because the 𝛼 particles cause damaging ionisation inside the body. 𝛽 particles travel further than 𝛼 particles before they are absorbed but their ionisation damage is much less. They are used in therapy but not in diagnosis of disease. The most useful radioisotopes for nuclear medicine are those that emit gamma radiation only. Technetium-99m and iodine-123 are two such isotopes. A gamma-emitting radioisotope inside the body can be detected outside the body because gamma radiation is very penetrating. Common radioisotopes used in medical diagnosis are listed in table 18.2. TABLE 18.2 Radioisotopes used in medical diagnosis Radioisotope

Production site

Half-life

Chromium-51

Nuclear reactor

27.70 days

Iodine-131

Nuclear reactor

8 days

To diagnose and treat various diseases associated with the thyroid gland; used in the diagnosis of the adrenal medullary; used for imaging some endocrine tumours

Iodine-123

Cyclotron

13 hours

To monitor thyroid function, evaluate thyroid gland size and detect dysfunction of the adrenal gland; to assess stroke damage

Nuclear reactor

65.94 hours

‘Milked’ from

6 hours

Molybdenum-99

Technetium-99m

Function To label red blood cells and measure gastro-intestinal protein loss

Used as the ‘parent’ in a generator to produce technetium-99m, which is the most widely used isotope in nuclear medicine To investigate bone metabolism and locate bone disease; assess thyroid function; study liver disease and disorders of its blood supply; monitor cardiac output, blood volume and circulation clots; monitor blood flow in lungs; assess blood and urine flow in kidneys and bladder; investigate brain blood flow and function; estimate total body plasma and blood count

molybdenum-99

Thallium-201

Cyclotron

3.05 days

To detect the location of damaged heart muscles

SAMPLE PROBLEM 2

A 20-milligram sample of iodine-123 is to be used as a radioactive tracer in the body. The half-life of the iodine-123 is 13 hours. a. How long will it take for 17.5 milligrams to decay? b. Calculate how much iodine-123 will remain after 26 hours. THINK a. 1.

Determine the amount that decays in each half-life until the total amount has reached 17.5 mg.

WRITE a.

In one half-life, 10 mg of iodine-123 will decay. This will leave 10 mg of iodine-123. In the second half-life, 5 mg of iodine-123 will decay, leaving 5 mg of iodine-123. In the third half-life, 2.5 mg of iodine-123 will decay.

TOPIC 18 How is radiation used to maintain human health? 9

2. b. 1.

2.

State the solution. Recall that 26 hours is two half-lives (2 × 13 hours). State the solution.

17.5 mg (10 + 5 + 2.5) of iodine will have decayed in three half-lives, or 39 hours. b. After one half-life, 10 mg of iodine-123 will decay, leaving 10 mg of iodine-123.After two half-lives, 5 mg of iodine-123 will decay, leaving 5 mg of iodine-123. 5 mg of iodine-123 will remain after 26 hours.

PRACTICE PROBLEM 2 A radioisotope with a half-life of 13 hours is used in the diagnosis of a patient. A check 52 hours later reveals that 1 milligram of the radioisotope remains. a. What mass of the radioisotope was used in the diagnosis? b. How much of the radioisotope will remain after a further 52 hours?

SAMPLE PROBLEM 3

A sample of a radioisotope has a half-life of 10 minutes. Calculate the time it will take the radioisotope’s activity to drop from 8 MBq (megabecquerels) to 4 MBq. b. Calculate the time it will take for its activity to be 1 MBq. a.

THINK

WRITE

When half the sample has decayed, the activity will also halve. This assumes that the atoms formed are not radioactive. b. 1. Halving the activity each half-life means that three half-lives have passed before the activity is 1 MBq.

a.

The time needed to reduce the activity to 4 MBq is one half-life, or 10 minutes.

b.

The time taken is 30 minutes.

a. 1.

PRACTICE PROBLEM 3 A sample of a radioisotope with a half-life of 8 days has an activity of 8 MBq 16 days after it is placed in safe storage. a. What was the activity of the sample when it was placed in safe storage? b. What is the activity of the sample after a further 16 days? c. How long will it take after the sample is placed in safe storage for its activity to decrease to 1 MBq?

18.3.3 Labelling with isotopes The radioisotope chosen needs a way of getting to the target organ. First, it must be chemically attached to a compound that would normally be metabolised by the organ of interest. When this compound is

10 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

‘labelled’ with the radioisotope (they are chemically attached) it is called a radiopharmaceutical. The radiopharmaceutical that is used must not alter the functioning of the organ or the area being examined. It must be sterile, non-toxic and compatible with the body. For example, glucose is a compound that is readily absorbed by the brain. Hence, glucose is labelled to become a suitable radiopharmaceutical for imaging brain function.

RADIOISOTOPES EMITTING GAMMA RADIATION

Both iodine-123 and technetium-99m are valuable radioisotopes because they decay by the emission of 𝛾 radiation only. Iodine-123 is more expensive than iodine-131, which has been used in the investigation of the thyroid gland. However, iodine-131 emits both 𝛾 and 𝛽 radiation, leading to larger radiation doses than desirable. The half-life of 8 days for iodine-131 is relatively long, resulting in exposure of the patient to radiation well after the testing has been carried out. By contrast, iodine-123 has a short half-life of 13 hours. Technetium-99m has a short half-life of only 6 hours so it must be produced in the hospital where it is to be used. A purpose-built generator system is used to obtain the technetium-99m when it is needed. The generator contains the ‘parent’ isotope molybdenum-99, which decays to the excited ‘daughter’ isotope technetium-99m. The technetium-99m is flushed from the molybdenum using a saline solution and the molybdenum remains in the generator as it is chemically attached to a central column. (Figure 18.6 shows a cross-section through a technetium generator.) The technetium-99m is said to be ‘milked’ from the molybdenum. This operation usually happens daily, allowing the technetium sufficient time to build up. As the molybdenum has a half-life of approximately 66 hours it must be replaced weekly because, by that time, the rate of production of technetium is too low to be of value. Technetium-99m has the added advantage that it readily attaches to different compounds to form radioactive tracers. Technetium is produced by the following equation: 99 42M o

Its decay equation is:

→

99m 43T c

+ −10𝛽 + 𝜈

→

+𝛾

99m 43T c

99 43T c

FIGURE 18.6 A cross-section through a typical technetium generator used in hospitals to generate technetium-99m

Source: Gentech® Generator courtesy of ANSTO

It is a gamma emitter that is used as a diagnostic tool for finding and treating cancer.

18.3.4 Targeting body organs As indicated in table 18.2, particular radioisotopes are chosen to target particular organs. The radiopharmaceutical is injected into the bloodstream, inhaled or taken orally and its passage through the body can be traced by measuring the radiation it emits. Sometimes an image is taken after a period of up to several hours, to allow the radioisotope to accumulate in the target organ. This image measures the amount of radiation emitted from the target organ and shows where the radioisotope has accumulated. In other situations, a series of images are taken over a period of time, starting from when the radioisotope is first introduced. This type of investigation shows the distribution of the radioisotope and the rate of absorption or excretion by various organs. The images may be taken over a few minutes for a heart or lung study, or over a period up to half an hour for kidney or bladder investigation. In analysing the images, radiographers identify ‘hot spots’ with a higher than normal concentration of radioisotope and ‘cold spots’ showing a lack of radioisotope. TOPIC 18 How is radiation used to maintain human health? 11

Obtaining the image — SPECT The image is obtained by measuring the amount of gamma radiation coming out of the patient’s body using a gamma camera. The gamma camera is stationary and collects gamma radiation over a large area. It converts the gamma rays into light flashes (scintillations) that are transformed into amplified electrical signals. These signals are decoded and converted to an image on a computer screen. This process is known as single photon emission computed tomography (SPECT).

FIGURE 18.7 Radiopharmaceuticals can also be used for bone scans with images taken by a gamma camera.

18.3.5 Some applications of radioisotopes Thyroid investigations The thyroid gland metabolises iodine. A dilute solution of sodium iodide tagged with iodine-123 is given to the patient to drink and the radioisotope’s accumulation is measured over a 10-minute to 48-hour time interval. An image of the goitre may be obtained or the uptake of the isotope may be graphed and compared with a standard, as in figure 8.8. Thyroid investigations now commonly use technetium-99m, which is also taken up readily by the thyroid but is also more readily released than iodine-123. Source: Andreas Praefcke

FIGURE 18.8 Uptake of iodine-123 by the thyroid gland Hyperthyroid (overactive)

Uptake (%)

100

Normal

> 50%

50 30 – 50% Hypotyroid (underactive) 15 – 30% 0 24 Time (hours)

48

The heart Human serum albumen is labelled with technetium-99m and injected into the patient. The passage of the radiopharmaceutical through the heart chambers is monitored to measure the efficiency of the heart as a pump. Thallium-201, as part of thallium chloride, can be injected and monitored to assess damage caused by a stroke or to measure the effect on the heart of exercise or drugs (see figure 18.9).

12 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 18.9 Performance of heart muscle using thallium-201. A series of images produces ‘slices’ through a chamber in the heart. The top row is images taken during exercise and the bottom row is taken when at rest.

Source: US National Library of Medicine

Bones, lungs and brain Technetium-99m is used in imaging the bones, lungs and brain. Polyphosphate ions are labelled with technetium-99m and injected, accumulating in bone within an hour. The image shows the function of the bone. Areas of increased blood flow show up as ‘hot spots’. Such areas are frequently associated with disease. Bone imaging often shows up bone tumours and stress fractures earlier than standard X-rays, which show the structure of the skeleton (see figure 18.10). Brain studies using technetium-99m as a tracer measure blood flow through the brain, allowing dementia and stroke damage to be identified. To study the blood flow in the lungs, technetium-99m attached to albumen is mixed with saline solution and injected into the veins in the arm. It becomes trapped in the fine capillaries in the lung and allows a map to be made of the functioning capillaries. Any blockage in the lung, perhaps due to a clot, shows as a region without any radioactive tracer. This is called a perfusion study. FIGURE 18.10 (a) An X-ray of a broken leg (b) A bone scan showing (i) a healthy skeleton and (ii) a skeleton with tumours (Note: The white spot on the right arm shows where the isotope was injected.)

Source: © Westmead Hospital

TOPIC 18 How is radiation used to maintain human health? 13

To enable the health of the airways to be studied, the patient inhales an aerosol labelled with technetium-99m. This ventilation study shows, over about half an hour, ‘cold spots’ where the radioisotope has not accumulated because the airway is blocked (see figure 18.11). FIGURE 18.11 Lung studies. (a)(i) A normal perfusion study and (ii) ventilation study of the lungs (b) Front view of lung scans of a patient with a blockage in the left pulmonary artery: (i) the perfusion scan showing no blood flow to the left lung and (ii) the ventilation scan showing both lungs as the airway is not blocked

Source: © Westmead Hospital

Blood To determine the volume of blood in the body, a measured quantity of a radioisotope is administered and, after a period of time, a sample of blood is taken. If the activity of the tracer in the blood is measured, the dilution of the tracer and hence the volume of blood in the body can be calculated. This procedure, known as dilution analysis, is valuable in investigating disorders such as anaemia, assessing stroke damage and monitoring blood loss as a result of an accident.

FIGURE 18.12 The decay series of uranium-238 N α 145 234

59 27C o

+ 10n →

60 28N i

+ 𝛾 + −10𝛽

Recall that radioactivity is a result of radioactive decay, thus, the element cobolt-60 transmutes to nickel-60. This process along with many other radioactive decays can be shown in a decay series diagram, which demonstrates the change in mass

14 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

4.5×109 y Th

β 24 d 234Pa 1.2 min β 234 U α 2.5×105 y 230 Th

140

α

18.3.6 Production of medical radioisotopes

226

8×104 y Ra α 1600 y

222Rn

Number of neutrons

A radioisotope can be produced for medical purposes through the absorption of a neutron. This is achieved by a bombardment of thermal neutrons, thus, enabling a stable element to absorb a neutron. The process increases the element’s mass number by 1. After the neutron is absorbed by the nucleus, the energy that binds the neutron is released as radiation. Radioactivity is the result of radioactive decay. The following example looks at this process with cobolt-59:

238U

α

135 218

3.8 d

Po

β 3.0 mα 218 At α 214 Pb β 27 m 214Bi 20 m β 214 130 210 Po α Tl α 1.6×10−4 s β 210 Pb 1.3 m β 210 22 y Bi 5.0 d β 210 206Tl α Po α 125 138 d β 206 Pb

120 80

85

Key α alpha decay β beta decay y year(s) m month(s) d day(s) s second(s)

90 95 Number of protons

100

Z

and atomic number once an element has undergone 𝛼 or 𝛽 decay. Figure 18.12 shows the decay series of uranium-238, demonstrating the type of decay and how long it takes for uranium-238 to decay to lead-206.

FIGURE 18.13 A radiation warning sign. The trefoil is the internationally recognised sign indicating a controlled area.

18.3.7 Safety issues In hospitals, the general public, medical teams and patients must be protected from overexposure to radioactive material. Strict guidelines are implemented to control and monitor exposure to radiation. Areas where work is carried out with ionising radiation are clearly marked as controlled areas with limited access. Equipment is checked regularly to make sure it does not leak radioactive material. Personnel distance themselves from radioactive material where possible and wear monitors to measure their exposure to radioactive sources. These monitors are checked regularly.

18.3 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Carbon-11 has a half-life of 20 minutes while bromine-75 has a half-life of 100 minutes. If samples of these isotopes initially have the same activity, show on one graph how their activities vary with time. 2. A small amount of iodine-131, which has a half-life of 8 days, is used to treat a patient with a thyroid condition. Sixteen days later, an amount of 6 milligrams remains. (a) How much iodine-131 was used in the treatment? (b) How much of the radioisotope will remain after another 16 days? (c) When is iodine-123 preferred to iodine-131 even though it is more expensive? 3. A sample of a radioisotope has a half-life of 2 minutes. (a) Calculate the time it will take the activity to drop from 4 MBq to 1 MBq. (b) Calculate the time it will take for its activity to be 0.25 MBq. 4. A particular isotope has a half-life of 100 days. Discuss the suitability of this isotope for use in medical diagnosis. 5. Describe the problems associated with using a radioisotope of very short half-life for medical diagnosis. 6. (a) Choose two specific radioactive isotopes used in medical diagnosis and outline where they would be used in the body. Justify your answer. (b) Explain why 𝛼-emitting radioisotopes are not used for medical imaging. 7. Identify a radioactive tracer study in which the tracer: (a) mixes with the substance under investigation (b) is accumulated in the organ of interest. 8. Explain why technetium-99m is such an ideal radioisotope for medical imaging. 9. Figure 18.11 shows two different types of studies of lungs. (a) Contrast the studies. (b) Relate the type of study to the disease diagnosed. 10. Figure 18.10 shows an X-ray of a leg and a bone scan of the body. (a) Compare the X-ray image with the bone scan. (b) Explain why there are differences in the images obtained.

TOPIC 18 How is radiation used to maintain human health? 15

11. The function of the lungs can be studied using a radioactive gas. The choices are xenon-133 or krypton-81m and their properties are listed in the following table. Isotope Xenon-133 Krypton-81m

Emission products

Half-life

𝛾

5.3 days

𝛽, 𝛾

13 seconds

Evaluate the claim that Xenon should be used in preference to krypton for investigations of lung function.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

18.4 X-rays in medical diagnosis KEY CONCEPT • Compare the processes of, and images produced by, medical imaging using two or more of X-rays, computed tomography (CT), 𝛾 radiation, magnetic resonance imaging (MRI), single photon emission computed tomography (SPECT) and positron emission tomography (PET).

18.4.1 Medical imaging using X-rays X-rays are used frequently in medicine and dentistry. It is likely that you or someone you know has had an X-ray at some time, for example, to check the development of teeth at the dentist or at a hospital for a suspected broken bone.

Use and detection of X-rays Since X-rays cannot be focused, the images from X-rays are shadows of objects placed in the beam. To obtain a sharp image it is necessary to have an object that is as still as possible and illuminated by an X-ray beam of small cross-sectional area, with the detecting plate as close to the object as possible. In this way, blurring of the image is minimised and the shadow is sharper. X-rays may also be scattered by surrounding tissue, which will affect the sharpness of the image. This is illustrated, using light, in figure 18.14. FIGURE 18.14 Obtaining a sharp shadow image. (a) A narrow source produces a sharp shadow. (b) An extended source or large distance between object and screen results in a shadow that is less sharp. (c) Cloudy water scatters light and produces a blurry image. (a)

(b)

(c)

Cloudy water Pinhole opening Object

Image is sharp

Large penumbra causes image to blur

16 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Image is not sharp

A narrow beam of X-rays can be obtained by shaping the target to allow the beam of electrons to strike it over a reasonable area while significantly reducing the width of the X-ray beam, as illustrated in figure 18.15. FIGURE 18.15 Electrons hitting the target over a wide area produce a narrow beam of X-rays. Anode angle

Filament

Wide beam of electrons Rotating tungsten target

Narrow beam of X–rays

The X-ray beam is directed at the part of the patient being imaged. Some tissues absorb X-rays very well and cast a shadow on the detecting screen. Bone is more dense than soft tissue and absorbs X-rays. Consequently, bones produce a clear image when X-rayed. X-rays may be detected on a photographic film or by an image intensifier. The photographic film is used when only a record of the image is required. An image intensifier allows direct viewing of the X-ray image. X-rays strike a phosphor screen that produces light. This light stimulates a photocathode to produce electrons that are accelerated to strike an output phosphor screen, producing more light than was generated originally and intensifying the image up to 1000 times. The image produced can be viewed directly by the eye, a movie camera or a TV camera. The viewing area can be altered while the X-ray process is occurring.

18.4.2 Effect of X-radiation on the body If the intensity of X-radiation striking the body is great enough it may be absorbed and cause electrons to be removed from atoms or molecules (ionisation). The effect may be harmful, which is why X-radiation is often referred to as ‘harmful ionising radiation’. One reaction that may occur is the ionisation of water molecules in the body and the formation of hydroxyl and hydrogen free radicals. These free radicals may alter base structures and sequences in the DNA in chromosomes, causing mutations. This may affect not only the person exposed to the radiation, it may also be passed on to that person’s children. Radiation that can cause damage to the body includes alpha (𝛼), beta (𝛽) and gamma (𝛾) radiation as well as X-rays. Government bodies set dose limits that are considered to be safe, but they vary from country to country. For the general population, the limit may be 1 mSv per year. For children under 16, it is usually about 0.5 mSv per year. These appear to be conservative values as the limit for radiation workers is generally set at 20 mSv per year. These values are in addition to the background radiation from the Earth and cosmic

TOPIC 18 How is radiation used to maintain human health? 17

rays, which amounts to a value under 3 mSv. Approximate values for radiation from various sources are listed in table 18.3. TABLE 18.3 Radiation from various sources Source Dental X-ray Chest X-ray Pelvic X-ray Mammogram ‘Barium meal’ X-ray CT scan of head CT scan of chest Aircraft crew additional annual exposure

Radiation received < 10 𝜇Sv 20 𝜇Sv 70 𝜇Sv

< 4000 𝜇Sv 3000 𝜇Sv 2000 𝜇Sv 8000 𝜇Sv 2000 𝜇Sv

18.4.3 Using conventional X-rays as a diagnostic tool The effects that X-rays have on the tissues of the body depend on the frequency (and therefore energy) of the X-rays and the time of exposure to them. For diagnostic purposes, the optimal frequency is around 7 × 1018 Hz, resulting in the best contrast between different tissues. At this frequency the X-rays are absorbed by the tissues and electrons are released. The extent of the X-ray absorption depends on the cube of the number of protons in the nuclei of the atoms encountered. For example, bone that has a high atomic density (high number of protons in the nuclei), attenuates the beam about 11 times more than the surrounding tissue and hence produces a strong X-ray shadow and allows a very good image of the bone to be obtained. Atomic density values are high for bone, moderate for soft tissue and low for air. Hence, the skeleton is imaged very well by X-rays.

18.4.4 Imaging parts of the body To image soft tissue, an artificial contrast medium that absorbs X-rays readily may be introduced. For investigations of the circulatory system, iodine in a compound is introduced into the bloodstream. To X-ray the gastrointestinal tract, which is composed of soft tissue, a ‘barium meal’ consisting of a thick suspension of barium sulfate is swallowed by the patient or introduced into the intestines through the anus. The barium compound absorbs X-rays and gives a clear image, as shown in figure 18.16. A chest X-ray is the most common way of detecting lung cancer or tuberculosis. The X-ray must be taken from several different directions to overcome the problem that the heart sometimes obstructs a clear view of the lungs. The teeth and jaw are X-rayed to detect tooth decay and the position of crowded teeth or wisdom teeth before surgery or orthodontal treatment. Someone who has swallowed a foreign object may be X-rayed to locate its position.

18 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 18.16 X-ray of the abdomen with barium sulfate used to provide a white contrast to image the bowel

For imaging the breast, which is an area of continuous soft tissue, careful choice of the X-ray beam and film detector provides high resolution. Molybdenum targets in the tube and low voltage maximise photoelectric attenuation. High tube current and short exposure time minimise image blur due to movement by the patient. FIGURE 18.17 An X-ray image of a breast showing a tumour

A better technique for imaging soft tissue is computed axial tomography (CAT) scanning, which detects small differences in X-ray attenuation.

18.4 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. With the aid of a labelled diagram, give a description of the way in which X-rays are produced. 2. Explain why the X-rays usually pass through a thin filter before they are used to image the patient. 3. Outline how the attenuation of X-rays changes for different materials in the body. 4. Describe and account for the appearance of an X-ray image of part of the body containing bone, muscle and air spaces.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

TOPIC 18 How is radiation used to maintain human health? 19

18.5 Medical imaging with CT, PET, MRI and ultrasound

KEY CONCEPTS • Compare the processes of, and images produced by, medical imaging using two or more of X-rays, computed tomography (CT), 𝛾 radiation, magnetic resonance imaging (MRI), single photon emission computed tomography (SPECT) and positron emission tomography (PET). • Analyse the strengths and limitations of a selected contemporary diagnostic or therapeutic radiation technique.

18.5.1 CT scanning Computed axial tomography scanning (CAT scanning) uses X-rays to obtain an image of a cross-section through the body. Very slight differences in X-ray attenuation can be measured, so soft tissue can be accurately imaged. Sometimes the name of the technique is abbreviated to computed tomography scanning (or CT scanning).

How is a CT scan produced? A CT scanner consists of an X-ray tube that is rotated around the patient being imaged. The tube and detection mechanism are mounted on a frame called a gantry. The part of the patient’s body being scanned is positioned in a gap in the gantry. An image is obtained in the plane being examined. The patient, on a bed, is moved slowly through the gantry so that a series of images of ‘slices’ through the body may be obtained. The X-ray source must produce a very narrow beam so that the path of the X-rays can be carefully controlled. To produce the narrow beam, the tube voltages are high and consequently a lot of heat must be conducted away from the anode in the tube generating the X-rays. This requirement, coupled with the tube movement during scanning, means that tubes fail and have to be replaced after a few months of use. The cost of such replacement is high. The beam is filtered to remove some soft X-rays that are not needed. This ensures the beam is relatively uniform in frequency and the dose to the patient’s skin is reduced. The X-rays are detected by an array of several hundred detectors. The detectors convert the X-radiation directly into electrical signals that go to multiple integrated-circuit amplifiers. FIGURE 18.18 A modern CT scanner

20 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

The patient is accurately positioned in the gantry so that a plane of the body can be scanned. A beam of X-rays is sent through the patient and detected on the other side. The tube is then rotated, usually 1°, and another beam transmitted and detected. This process is repeated until an angle of 180° has been swept out. The data from the scan are collected, displayed and reconstructed using a powerful computer and software. The computer analyses the absorption of the X-rays at each measured point in the slice. For example, if X-ray beam absorption is measured at 160 distinct points along each scanning path and 1° increments in angle are used, approximately 29 000 distinct pieces of data about X-ray absorption are obtained. The reconstruction, which is explained in simplified form in figure 18.19, is the result of around one million computations. The image can be displayed on a TV screen or stored in the computer memory and used with other data to produce an image in a different plane. Using computer analysis, the data from images of ‘slices’ through the body can be combined to produce a 3D image of the area under investigation. FIGURE 18.19 Creating a CAT scan image X–rays

Attenuation = 5 + 5

10

9

7

9

10

9

8

6

8

9

7

6

4

6

7

4

9

8

6

8

9

5

10

9

7

10

10

5 4 2

X–rays

5

4

2

4

Attenuation = 5 + 2

5

(a) The attenuation is measured at many points (pixels) from different angles. (Here only 2 angles at 90° are recorded.)

(b) The total attenuation of the X–rays at each pixel is calculated.

(c) A shade of grey is assigned to each pixel and from this the image is created. (As an example assign the darkest shade to the smallest number.)

In recent years, full body CT scans have been advertised for those who want to detect problems before symptoms appear. The medical profession has criticised this offer on several grounds: people are exposed to unnecessary radiation; potential problems may not be detected; and abnormalities, which are harmless, may be found. People may either be given false security or false alarms. For further information about full CT body scans, enter ‘CT scans’ or ‘full body scans’ in a search engine.

Comparing CT scans and conventional X-rays CT scans are significantly more expensive than conventional X-rays. They are, however, a superior diagnostic tool to X-rays when fine detail is needed. CT scans provide detail to distinguish between areas where the density difference is quite small and a dense material shields the area. For example, in the brain the density range is only a few per cent but the bony skull is so dense that it absorbs most of the X-rays. A conventional X-ray will therefore provide an image of the dense skull rather than the brain tissue inside. However, by taking X-ray images from many angles in a CT scan, the material along the path of the X-ray beams can be distinguished clearly. The method of obtaining and analysing the image makes it possible to see behind bone using a CT scan.

FIGURE 18.20 CT scan of a shoulder showing the detail possible with this technique

TOPIC 18 How is radiation used to maintain human health? 21

Conventional X-rays are valuable when there is high natural contrast between the tissues to be viewed. The contrast is high for bone, moderate for soft tissue and low for air. Hence, X-rays are good for diagnosing bone problems such as fractures, dislocation and arthritis. Conventional X-rays can also be used to image the digestive tract if an artificial contrast medium, such as a barium meal, is introduced. Because very good soft tissue resolution is possible with CT scans, soft tissue damage due to bone fracture or ruptured spinal discs can be investigated. CT scans are also used to scan the liver and kidneys to obtain resolution better than 1 millimetre (meaning that differences separated by 1 mm can be detected). CT scans are preferred for imaging the lungs. Although conventional X-rays give adequate routine lung screening, CT scans provide clearer detail.

Resources Weblink CT scan applets

18.5.2 Ultrasound X-rays of sufficient intensity are a harmful ionising radiation, whereas ultrasound does not produce any ionisation. Hence, ultrasound is so safe it can be used with foetuses. Ultrasound is used to view the performance of organs, vessels and tissues without needing to make an incision. It is a medical test that doesn’t produce any ionisation as it uses high-frequency sound waves to capture images from inside the human body. As ultrasound uses sound waves, which are reflected by dense materials such as bone and are affected by gaps in a medium such as an air-tissue boundary, it is not able to be used in a large variety of situations. Ultrasound can be used many times with the same patient without any harmful effects. The specific advantage over a CT scan or X-ray would need to be considered in each case before exposing a patient to the X-ray doses involved. The less expensive, quick and portable imaging techniques using X-rays or ultrasound may give an initial diagnosis that could lead to further testing for tissue damage or internal bleeding by ordering a CT scan.

18.5.3 Positron emission tomography (PET) Positron emission tomography, known as PET, is used to diagnose and monitor brain disorders, investigate heart and lung functioning and detect the location and spread of tumours. Using particular radiopharmaceuticals, a cross-sectional image through an organ can be obtained or a region of the body can be imaged, allowing the function of an area to be determined. A PET image of the brain shows the patient’s responses to factors such as noise, illumination and changes in mental concentration.

Positron–electron interactions Certain radioisotopes decay by the emission of positrons — positively charged beta particles. That is, they are positively charged electrons formed when a proton disintegrates to form a neutron and a positron. Radioisotopes that are deficient in neutrons often decay in this way. For example, carbon-11 decays to boron-11 emitting a positron (𝛽 + ): 11 6C

→

11 5B

+ 𝛽+

22 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 18.21 A PET scan of a patient’s brain

When a positron meets an electron they ‘annihilate’ each other, converting their combined energy and mass into two 𝛾 rays. The energy of each of these 𝛾 rays is 0.51 MeV. This process is sometimes called ‘pair annihilation’.

How a PET scan is obtained To obtain a PET scan, a suitable pharmaceutical is labelled with a positron-emitting radioisotope. The radiopharmaceutical is usually injected into the patient, but sometimes the chemical is inhaled. After a short period of time the radiopharmaceutical will have accumulated in particular areas of the body and begun to decay by the emission of positrons. These positrons travel a short distance, of the order of a few millimetres, before they encounter electrons in the body. Pair annihilation takes place and two 𝛾 rays are produced. The 𝛾 rays travel in opposite directions from the site of annihilation and emerge from the body, where they are detected by gamma cameras. Gamma cameras surround the patient in the section being scanned, so that 𝛾 rays can be detected from all angles. Pairs of 𝛾 rays travelling in opposite directions are detected and their relative intensity measured. By comparing these measurements with known attenuation coefficients for 𝛾 rays passing through tissue, the position of the decaying radioisotope can be approximately determined. In this way, an image is produced showing where radioisotopes accumulate. A PET imaging system detecting emissions from a region of the brain is illustrated in figure 18.22.

Isotopes used in PET

FIGURE 18.22 (a) Some PET images of brain activity (b) Cross-section showing pairs of 𝛾 rays travelling in opposite directions and reaching detectors (a)

(b)

Photomultipliers Detector

γ

γ

γ

γ

γ γ

Annihilation of electron and positron produces two gamma rays

Common isotopes used in PET are listed in table 18.4. TABLE 18.4 Common isotopes used in PET As can be seen from the table, the half-lives of isotopes suitable for PET are very small. The isotopes must be Radioisotope Symbol Half-life created on the day of use and, except for fluorine-18, 11 Carbon-11 18.4 min C 6 must be made at the site of use. A cyclotron is needed 13 N Nitrogen-13 10.0 min for their production. This is a serious limitation, as the 7 18 cost of an on-site cyclotron and facility for producing Fluorine-18 109.8 min F 9 radiopharmaceuticals is extremely high. In Victoria, the 15 O Oxygen-15 2.13 min 8 Austin Hospital, the Royal Melbourne Hospital and the Peter McCallum have on-site cyclotrons for their own PET facilities. The largest cyclotron in Victoria, at Bundoora, is privately owned and produces radioisotopes for PET facilities in a number of hospitals, including some in other states. Fortunately the longer half-life of fluorine-18 means tracers can be labelled with fluorine-18 and shipped to nearby hospitals from a central location.

Resources Weblink PET scans

TOPIC 18 How is radiation used to maintain human health? 23

18.5.4 Magnetic resonance imaging (MRI) Magnetic resonance imaging (MRI) makes use of the effect of a strong magnetic field on nuclei in the body to obtain images of organs and tissue. The images show significant clear contrast between different types of soft tissue, making MRI scans suitable for examining the brain, spinal cord, muscle, tendons, cartilage and joints. Our bodies consist of a relatively small variety of elements combined in a large variety of compounds. The protons in the nuclei of hydrogen atoms and isotopes such as carbon-13, fluorine-19, sodium-23 and phosphorus-31 respond to a strong external magnetic field by aligning themselves so that they are parallel to the field. Hydrogen is the most commonly imaged element in MRI because it is the most abundant of these elements in the body. When a patient is placed inside an MRI machine, the machine generates a strong magnetic field. A pulse of lowenergy electromagnetic radiation in the radio frequency range is beamed into the patient. The aligned protons resonate with the radio frequency, absorbing a small amount of energy. When the pulse is turned off, the protons release the absorbed energy as a radio frequency pulse. The intensity and duration of the radio frequency pulse is analysed by computers, enabling an image of a ‘slice’ through the patient’s body to be obtained. The contrast between different tissues on the images exists because they have different concentrations of hydrogen and different hydrogen compounds. The greater the density of hydrogen protons in the tissue, the larger the signal and the brighter the image. Air and outer bone contain little or no hydrogen, so they appear dark in an MRI scan. Water, however, contains many mobile hydrogen protons, so it produces a strong signal. Cerebrospinal fluid in the brain and spinal cord has a large amount of water that is not bound to other molecules, so it shows up brightly on an MRI scan. The types of hydrogen compounds in the tissue also determine how easily the protons can release energy to neighbouring nuclei. Therefore, the type of tissue influences the intensity of the signal. The type of tissue can be identified by examining the time it takes for protons in the tissue to release the energy absorbed from the radio frequency pulse.

FIGURE 18.23 A patient having a scan using magnetic resonance imaging

FIGURE 18.24 Images using MRI and showing clear contrast of soft tissue: (a) MRI of the abdomen showing liver cancer (b) MRI image of the lumbar spine

Source: (b) © David Grabham

24 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

The use of MRI in medical diagnosis MRI is considered the best diagnostic technique for obtaining sharp and crisp images of tissues. It depicts soft tissue so well that it is the preferred choice for imaging the brain and spine, where it is able to show suspected tumours and slipped discs. Cancerous tumours contain different amounts of water from normal tissue or are surrounded by watery tissue; they can be distinguished in an MRI scan because of the different brightness. Functional MRI allows brain activity to be investigated while the patient is awake and able to think and respond to stimuli. There is an increased flow of oxygenated blood to areas that are stimulated. Knowledge of the magnetic properties of oxygenated blood allows the parts of the brain involved in the activity to be identified and studied (see figure 18.25). Parts of the brain may be able to be studied prior to surgery.

FIGURE 18.25 Brain activation showing increased blood flow to reward centres of the brain

Source: Powell K

CARDIAC MRI Cardiac MRI allows investigation of congenital abnormalities and coronary heart disease to be carried out. Improvements in the speed of MRI have made abdominal imaging possible. Early MRI machines took 10 minutes to scan 24 ‘slices’ of the body and this can now be done in under 1 second. Injection of a contrast agent into the blood, combined with rapid imaging techniques, now allows blood flow in the kidneys to be examined and narrowing of the arteries due to fatty plaques to be seen.

18.5.5 Strengths and limitations of imaging techniques Imaging methods working together Medical imaging to obtain both functional and structural images is often needed for adequate diagnosis. For example, CT scans are used to obtain structural images. Radioisotopes, on the other hand, allow functional information to be gathered. For example, a nuclear medicine image may show tumours but not very much normal tissue. Hence, it may be difficult to determine the position of the tumour relative to other structures. If a CT scan is obtained at the same time, the location of the tumour can be established precisely.

Comparison of imaging techniques Table 18.5 provides some comparison between imaging techniques. Improvements are being made in the machines used in all these imaging methods, and students are advised to search the internet for the latest advances.

TOPIC 18 How is radiation used to maintain human health? 25

TABLE 18.5 Comparing imaging techniques X-rays

CT

SPECT and PET

MRI

Cost of machine (capital cost)

Least expensive

Quite expensive

Quite expensive

Very expensive

Mobility of machines

Small portable machines available

Fixed machines

Fixed machines

Very few mobile machines

Spatial clarity (ability to see fine detail)

0.1 mm

0.25 mm

5–15 mm

0.3–1 mm

Time for examination

Very fast

Moderate

May be long, depending on tracer and procedure

Relatively long but some procedures are now quite short

Comfort and safety

Small dose of ionising radiation

Usually higher dose of ionising radiation than for X-rays

Moderate dose of ionising radiation from radioisotopes

Some claustrophobia from lying inside the bore containing the magnetic field. Patients with metallic implants cannot be scanned.

Imaging soft tissue of abdomen

Image poor — needs contrast medium

Good for whole abdomen scan

Good for growth of tumours and functional study of liver and kidneys

Good clarity for specific areas e.g. kidneys

Imaging soft tissue of joints

Poor contrast

Good — preferred to MRI when extra bone detail is needed

Poor clarity but good for functional information

Excellent for studying muscles, tendons and cartilage

Imaging heart and circulation

Contrast medium is needed

Limited use with digital imaging techniques

Good for blood flow studies

Good clarity and ability to measure blood flow

Imaging chest

Adequate for routine lung screening

Better detail than X-rays

Good for functional studies of blood and air flow

Not good for imaging air spaces

Imaging brain and spinal cord region

Limited use as bone blocks most waves

Good — preferred to MRI for details of bone of spine

PET scans are useful for showing function

Excellent for giving good contrast between tissues

Imaging bone

Given very good clarity

Good when more complicated structures must be viewed

Good for whole body bone cancer and early diagnosis of stress fractures

Weak signal, so of limited use

18.5 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Describe the differences between the ways in which CT scans and conventional X-ray images are produced. 2. Use a table to summarise situations in which CT scans are a superior diagnostic tool to X-rays and ultrasound. 3. How are the radioisotopes used in PET scans different from those not used in PET scans?

26 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

4. (a) What is a positron? (b) How are positrons obtained? (c) Identify issues associated with positron–electron interaction and describe how this interaction is used in medical diagnosis. 5. Describe how a radioisotope of your choice is used in a PET investigation. In your answer, name the isotope and state what radiation is emitted and how it is monitored. Describe what measurements are made and how they are used to obtain a result. You should also mention any precautions or safety procedures. 6. Describe how an external magnetic field influences a hydrogen proton. 7. Why are hydrogen nuclei imaged more than any other nuclei in MRI? 8. Describe two different pieces of information that can be analysed during an MRI scan when the low-energy radio frequency pulse is turned off. 9. Why is MRI useful for imaging cancerous tumours in the brain? 10. Compare the advantages and disadvantages of X-ray scans, CT scans, ultrasound and MRI scans for each of the following purposes: (a) imaging the brain (b) imaging bone (c) imaging the heart and circulation.

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

18.6 Review •

18.6.1 Summary • • • • • •

The effect of radiation exposure can range from nausea to death. The amount of radiation energy received by each kilogram of living tissue is measured in gray (Gy), but this value does not take into account the type of radiation that has been absorbed. Each type of radiation has a different effect because of its ionising power. X-rays are produced by the collision of electrons with a target material. Soft X-rays are less penetrating and have lower frequency than hard X-rays. Equivalent dose measures the radiation energy absorbed by each kilogram of biological tissue and its effect by taking into account the form of radiation energy absorbed. Equivalent dose is measured in sievert (Sv). The Australian average annual radiation dose is 2 mSv, most of which is from background radiation. Gamma radiation from radioisotopes is detected and used to make an image of an organ. This process is known as SPECT (single photon emission computed tomography). Radiopharmaceuticals to which radioisotopes have been attached are taken up by particular organs in the body. The rate at which the radioisotope accumulates in the target organ indicates the health of the organ. The half-life of the radioisotope and length of time needed for the procedure must be considered when choosing an appropriate radioisotope. A CT scan is produced by the computer analysis of the attenuation of X-rays moving around a slice of the body. CT scans can distinguish soft tissue with small differences in density and can produce an image of tissue behind bone. PET imaging uses radioisotopes that are positron emitters. Positrons and electrons annihilate each other in the body, producing two gamma rays. Detecting the position from which the gamma rays originate enables the position of the positron emitter to be mapped.

TOPIC 18 How is radiation used to maintain human health? 27

•

• •

PET scans indicate the biochemistry, metabolism and function of a particular area. PET scans are used for studying the brain and heart, detecting cancers at an early stage and monitoring cancers during treatment. MRI scans make use of the magnetic effects of a strong external magnetic field on certain nuclei, particularly hydrogen, together with pulses of radio waves, to produce images of internal body tissue. MRI scans show soft tissue clearly, making them suitable for imaging the brain and spinal cord.

Resources

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0043).

18.6.2 Key terms Absorbed dose, measured in gray (Gy), describes the amount of energy absorbed by each kilogram of tissue that is irradiated. To attenuate is to reduce the intensity of the beam. A cathode ray is a narrow beam of electrons emitted from a hot filament through which an electric current flows. Equivalent dose, measured in sievert (Sv), describes the biological effect of radiation that has been absorbed by living tissue. It takes into account the type of radiation absorbed. A free radical is an uncharged fragment of a molecule resulting from a covalent bond being broken. Hard X-rays have a higher frequency and more energy than soft X-rays and are therefore more penetrating. Ionising radiation is high-energy radiation that has the ability to change atoms by removing electrons and therefore giving the atom an overall charge. A positron is a positively charged beta particle; that is, it is a positively charged electron formed when a proton disintegrates to form a neutron and a positron. A radiopharmaceutical is a compound that has been labelled with a radioisotope. To resonate means to absorb energy when an applied external frequency matches the natural frequency of an object. Soft X-rays have a lower frequency and less energy, and are less penetrating than hard X-rays. X-rays are electromagnetic waves of very high frequency and very short wavelength.

Resources Digital document Key terms glossary (doc-33001)

18.6 Exercises To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au.

18.6 Exercise 1: Multiple choice questions 1.

Which of the following is the most penetrating radiation? A. Alpha B. Beta C. X-rays D. Gamma

28 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

2.

3.

4.

5.

6.

7.

8.

9.

10.

What are sieverts a measure of? A. Equivalent dose B. Quality factor C. Absorbed dose D. Energy absorbed A radioactive isotope has a half-life of 8 days. What percentage of substance is left after 24 days? A. 50% B. 12.5% C. 25% D. 87.5% A sample of a radioisotope has a half-life of 12 minutes. How long will it take the radioisotope’s activity to drop from 6.0 MBq to 1.5 MBq? A. 12 minutes B. 24 minutes C. 36 minutes D. 18 minutes You can get a clear image from an X-ray of a bone, due to which characteristic of the bone? A. Absorbs the X-rays B. Reflects the X-rays C. Is soft D. Refracts the X-rays A patient who has a chest X-ray taken is exposed to approximately how much radiation? A. 20 Sv B. 100 𝜇Sv C. 100 Sv D. 20 𝜇Sv A CT scan takes multiple images by rotating how much each time? A. 1° B. 10° C. 5° D. 45° During a PET scan, what happens when a positron meets an electron? A. They turn into a neutron. B. They repel each other. C. They annihilate each other. D. They both become electrons. What are half-lives of isotopes suitable for PET usually measured in? A. Minutes B. Hours C. Days D. Weeks What is the preferred imaging for the brain and spinal cord? A. X-ray B. PET C. CT D. MRI

18.6 Exercise 2: Short answer questions The X-ray tube is highly evacuated and a very high voltage runs between the anode and cathode. What does this high voltage do? 2. What is the definition of absorbed dose?

1.

TOPIC 18 How is radiation used to maintain human health? 29

3. 4. 5.

6. 7.

8. 9. 10.

Why are radioisotopes that emit 𝛼 particles are not used for diagnosis? Why are both iodine-123 and technetium-99m valuable radioisotopes? If the intensity of X-rays striking the body is high enough: a. What is one of the reactions that may occur? b. What effect can this reaction cause? What is the amount of radiation set by government bodies for the general population each year? When using X-rays as a diagnosis tool: a. What is the optimal frequency? b. What is the benefit of using X-rays at this frequency? What advantages do CT scans have over X-rays? What is the benefit of using ultrasound rather than X-rays in relation to safety? What causes an MRI to have a brighter image?

18.6 Exercise 3: Exam practice questions Question 1 (4 marks) A 75-kilogram person absorbs 0.068 J of energy due to ionising radiation. a. Calculate the absorbed dose. b. What would be the equivalent dose if the energy was delivered by 𝛾 rays? c. What would be the equivalent dose if the energy was delivered by 𝛼 particles? (Quality factor of 20) d. Which would cause more biological damage to the person?

1 mark 1 mark 1 mark 1 mark

Question 2 (6 marks) A 35-milligram sample of technetium-99m is used for medical diagnosis in the body. The half-life of technetium-99m is 6 hours. a. How long will it take for 26.25 milligrams to decay? 3 marks b. Calculate how much iodine-123 will remain after 18 hours? 3 marks Question 3 (3 marks) Nuclear medicine plays an important part in the health system. a. What is the most commonly used radioisotope in nuclear medicine? b. Where is this radioisotope produced from? c. Write the decay equation of this radioisotope.

1 mark 1 mark 1 mark

Question 4 (1 mark) Why can therapeutic radioisotopes be either 𝛼, 𝛽 or 𝛾emitters, but for diagnostic tracing purposes the radioisotopes can only be 𝛾?

Question 5 (1 mark) Breast cancer awareness is continuing to grow, and women are encouraged to self-check for any unusual lumps. If lumps appear the doctor is likely to order a mammogram. Which X-rays would be used, hard or soft? Explain.

18.9 Exercise 4: studyON topic test Fully worked solutions and sample responses are available in your digital formats.

Test maker Create unique tests and exams from our extensive range of questions, including practice exam questions. Access the assignments section in learnON to begin creating and assigning assessments to students.

30 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

AREA OF STUDY 2 OPTIONS OBSERVATION OF THE PHYSICAL WORLD

19

How do particle accelerators work? 19.1 Overview Numerous videos and interactivities are available just where you need them, at the point of learning, in your digital formats, learnON and eBookPLUS at www.jacplus.com.au.

19.1.1 Introduction Following the discovery of radioactivity, the newly identified particles were soon being used as investigative tools. They were fired at targets to see what the targets were made of. Rutherford fired alpha particles at thin gold foil to explore the structure of the atom. He also fired them at nitrogen gas, producing oxygen. Chadwick discovered the neutron by firing alpha particles at beryllium. These high-speed ‘bullets’ were fired at the nuclei of various atoms to break them apart. By looking at the fragments, an insight into the structure of the atoms and the nucleus and ultimately the structure of the nuclear particles themselves could be achieved. However, Rutherford, Chadwick and other early researchers were limited to particles from naturally occurring radioactive sources. They soon realised there were three limitations to these natural ‘bullets’: • The energy of the incoming particle could not be measured. • The energy could not be controlled or varied. • Higher energies were needed to investigate further. A machine was needed. FIGURE 19.1 The Large Hadron Collider at CERN

Source: ATLAS Experiment © 2014 CERN

TOPIC 19 How do particle accelerators work? 1

19.1.2 What you will learn KEY KNOWLEDGE After completing this topic you will be able to: Particle accelerators and the production of light • distinguish between the use of particle accelerators to produce synchrotron light and to collide particles • distinguish between the capabilities of a particle collider and the capabilities of the Australian Synchrotron • explain the general purpose of the electron linac, circular booster, storage ring and beamlines in the Australian Synchrotron • explain, using the characteristics of brightness, spectrum and divergence, why for some experiments synchrotron radiation is preferable to laser light and radiation from X-ray tubes Accelerator technology and the development of modern particle physics • explain the evolution of collider technology including: • particles involved in the collision event • the increasing energies attained since the 1950s • evaluate the role of colliders in the development of the Standard Model of particle physics, including reference to subatomic structure and processes • describe the products of collisions with reference to symbol, charge, rest energy and lifespan • compare the physical designs and purposes of particle detectors at the Large Hadron Collider including ATLAS, CMS, ALICE and LHCb Current and future applications of accelerator technology for society • explain how the immense amount of data collected by the Large Hadron Collider is stored and analysed, and the associated role particle detectors have had in the development of information processing technologies • describe at least one application of particle accelerators selected from: • materials analysis and modification which results in the improvement of consumer products such as heat shrinkable film and chocolate • implanting of ions in silicon chips to make them more effective in electronic products such as computers and smart phones • nuclear energy applications such as the use of thorium as an alternative fuel for the production of nuclear energy or the treatment of nuclear waste • pharmaceutical research involving the analysis of protein structure leading to the development of new pharmaceuticals to treat major diseases • DNA research involving the analysis of protein metabolism leading to the development of new antibiotics • medical applications such as the production of a range of radioisotopes for medical diagnostics and treatments or cancer therapy through the use of particle beams • use of spectrometry in environmental monitoring or the use of blasts of electrons in the treatment of pollution such as contaminated water, sewage sludge and gases from smokestacks • use of particle accelerators in a selected experiment or scientific endeavour • investigate current and proposed future directions of collider technologies. Source: VCE Physics Study Design (2016–2021) extracts © VCAA; reproduced by permission.

Resources Digital documents Key science skills — VCE Units 1–4 (doc-31856) Key terms glossary (doc-33008)

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0044).

2 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

19.2 Particle accelerators and the production of light KEY CONCEPTS • Distinguish between the use of particle accelerators to produce synchrotron light and to collide particles. • Distinguish between the capabilities of a particle collider and the capabilities of the Australian Synchrotron. • Explain the general purpose of the electron linac, circular booster, storage ring and beamlines in the Australian Synchrotron. • Explain, using the characteristics of brightness, spectrum and divergence, why for some experiments synchrotron radiation is preferable to laser light and radiation from X-ray tubes.

19.2.1 Synchrotron The Large Hadron Collider (LHC) (figure 19.1) is the most powerful accelerator in the world. It boosts the particles in a loop 27 kilometres in circumference at an energy of 6.5 TeV (teraelectronvolts), generating collisions of 13 TeV. Accelerated to a speed close to that of light, these particles collide with other particles. These collisions produce massive particles, such as the Higgs boson or the top quark. By measuring their properties, scientists increase our understanding of matter and of the origins of the universe. These massive particles only last for a very short time and cannot be observed directly. Almost immediately they transform (or decay) into lighter particles, which in turn also decay. The particles emerging from the successive links in this decay chain are identified in the layers of the detector. The type of particles, the energy of the collisions and the luminosity are among the important characteristics of an accelerator. The principles of a synchrotron are not new — aspects of them have been used for over 100 years. Synchrotrons were originally designed as particle accelerators — that is, machines that can accelerate charged particles such as electrons and protons up to high speeds (close to speed of light) and energies (around 7 trillion eV), using electromagnetic fields. One of the early problems with particle accelerators was when any charged particle was accelerated, it emitted electromagnetic radiation, UV, X-rays or visible light, depending on the speed and acceleration. This radiation is a loss of energy from the particle, but it became a significant and detectable problem only with the construction of synchrotrons. For this reason, the radiation is commonly called synchrotron radiation. The radiation produced is such an intense and narrow beam that synchrotron radiation became a powerful experimental research tool. In fact, synchrotrons ended up being built just to produce the synchrotron radiation! This name, synchrotron radiation, is now applied to the radiation produced by any accelerating particle. Synchrotron radiation is emitted as cosmic ray electrons enter Earth’s magnetic field, in the form of radio waves. Distant galaxies are also sources of radio waves as charged particles move through their strong magnetic fields. Synchrotron radiation from a synchrotron was first observed in 1948. Both the Australian Synchrotron and the LHC use synchrotrons to accelerate charged particles — it’s what happens next that makes the big difference. In the case of synchrotron as a light source, the electrons are accelerated close to the speed of light that gives off intense and very narrow beams of mainly X-rays synchrotron radiation, which is harnessed for beamline experiments in biological, chemical and materials science. Whereas, the LHC accelerates protons or lead ions close to the speed of light, then lets them collide. This is used to test the predictions of theories of particle physics, for example, the existence of the Higgs boson.

TOPIC 19 How do particle accelerators work? 3

Why do accelerating charges produce radiation? When a charged particle is accelerated, the electric field around the charged particle is disturbed. This changing electric field produces a changing magnetic field. These changing fields radiate (away from the charge) as electromagnetic radiation (at the speed of light). The transmission towers of TV and radio stations and the small aerial inside a mobile phone have electrons oscillating along a length of wire producing the electromagnetic signal. The radiation from a fast-moving charge is all emitted in the direction the particle is moving in. If a particle is moving in a circular path in a magnetic field, radiation is emitted progressively along the tangent to the circle, like water flung off a spinning wet ball, as shown in figure 19.2.

FIGURE 19.2 Radiation from synchrotron is emitted progressively along the tangent to the circle, like water flung off a spinning wet ball.

19.2.2 Design features of the Australian Synchrotron Opened in 2007 in Clayton, Victoria, the Australian Synchrotron (figure 19.3) is Australia’s largest and arguably most successful scientific user facility, benefiting over 5000 researchers from academia, medical research institutes, industry, government and other research organisations. FIGURE 19.3 The Australian Synchrotron in Melbourne has a suite of X-ray and infra-red beamlines with applications in health, medical, food, environment, nanotechnology, energy, mining, advanced materials, agriculture and cultural heritage.

Source: ANSTO

4 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Put simply, a synchrotron is a very large, circular, gigavoltage machine similar in size to a football field. The exterior of the Australian Synchrotron, for example, looks just like a roofed football stadium, but inside instead of a pitch and seats, there is a vast, circular network of interconnecting tunnels and high-tech equipment. Synchrotrons use electricity to produce intense beams of light a million times brighter than the Sun. The light is produced when high-energy electrons are forced to travel in a circular orbit inside the synchrotron tunnels by ‘synchronised’ application of strong magnetic fields. With this new knowledge that synchrotron science provides about the molecular structure of materials, researchers can invent ways to tackle diseases, make plants more productive and metals more resilient — among many other beneficial applications of synchrotron science. The design features of the Australian synchrotron are described below. FIGURE 19.4 Design features of the Australian Synchrotron at Clayton, Melbourne

Source: Australian Nuclear Science and Technology Organisation

Electron gun Electrons are produced at the electron gun by thermionic emission from a heated tungsten matrix cathode. Electrons are attracted out of this by 120 kilovolts. Applying a 500 MHz voltage signal to the gun as it fires means the electrons are generated in bunches two nanoseconds apart. The emitted electrons are then accelerated to 90 keV (kiloelectron volts) by a 90-kilovolt potential difference applied across the gun and move into the linear accelerator.

TOPIC 19 How do particle accelerators work? 5

Linear accelerator (linac) The linear accelerator (or linac) accelerates the electron beam to 100 MeV over about 10 metres (see figure 19.5). FIGURE 19.5 A schematic diagram of a linear accelerator

Electrons bunch

Electrons

Gate

Radio frequency source

Anode

Electron gun

Buncher

Linac

To booster synchrotron

Cathode Source: Courtesy of the Advanced Light Source, Lawrence Berkeley National Laboratory

The anode is a torus (doughnut) shaped to create an electromagnetic field that guides most of the electrons through the hole into the next part of the accelerator, called the buncher. The purpose of the buncher is to accelerate the pulsing electrons as they come out of the electron gun and pack them into bunches. This involves a series of radio frequency (RF) cavities operating at a frequency of 3 GHz. Due to the nature of the acceleration, the beam must be separated into discrete packets, or ‘bunches’, with a spacing consistent with the 3GHz acceleration frequency of the linac. This is done at the start of the linac, using several ‘bunching’ cavities. The linac can accelerate a beam once every second. After the first metre of acceleration in the linac, the electrons are already travelling at more than 99.99% of the speed of light. Linacs utilise travelling waves rather than standing waves. A vacuum is created to ensure that the electron beam is not impeded by other particles.

Circular booster (booster ring) The booster is an electron synchrotron 130 metres in circumference that takes the 100 MeV beam from the linac and increases its energy to 3 GeV (giga electron volts). The booster ring contains 60 combinedfunction (steering and focusing) electromagnets to keep the electrons inside the stainless-steel vacuum chamber and a single 5-cell RF cavity (operating at 500 MHz) to supply energy for acceleration. Electrons are bent into a circular path by a force produced by magnets at a right angle to its motion. The beam is accelerated by a simultaneous ramping of magnet strength and cavity fields. Each ramping cycle takes approximately 1 second for a complete ramp up and down. An electron spends about half a second in the booster ring and completes over one million laps.

Storage ring The accelerated electrons are transported to their final destination, the storage ring. It can hold 200 mA of stored current with a beam lifetime of over 20 hours. The storage ring is 216 metres in circumference, with a radius of 34.4 metres. The ring, however, is not really circular. It consists of 14 nearly identical sectors; each sector consists of a 4.4-metre straight section and an 11-metre arc. In the straight-line sections, the electrons are accelerated back up to speed to compensate for the loss of energy due to the emission of the radiation. Every arc contains two dipole ‘bending’ magnets where synchrotron light will be produced. The magnets force the electrons into a snake-like path, so that the light from all the curves adds together (see figure 19.6a).

6 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

These beams can be captured and focussed to a specific wavelength appropriate for a particular technique. Insertion devices increase the intensity of light in the straight sections of the ring (figure 19.6b). There are two types of insertion devices: wiggler and undulator. In the multipole wiggler (MPW), a cone of light is emitted at each bend in the wiggler so that the cones of light superimpose on each other, the intensity increasing with the number of bends. The undulator uses less powerful magnets to move the electrons up and down. In this case, the light cones just overlap and interfere with each other, so that certain wavelengths of light are enhanced approximately 10 000 times. These wavelengths can be changed by altering the gap between the component magnets so that the light is tunable to specific wavelengths (figure 19.6c). The synchrotron light emitted by the electrons is directed to beamlines through the round beam ports. FIGURE 19.6 Illustration of different sections of storage ring (a) Bending magnet. At each defection of the electron path a beam of light is produced. The effect is similar to the sweeping of a search light. (b) Multipole wiggler. At the peak of each wave a beam of light is emitted. These beams reinforce each other and appear as a broad beam of incoherent synchrotron light when viewed in the horizontal plane ahead of the wiggler. (c) Undulator. The poles produce less deflection of the electron beam. This results in a narrow beam of coherent synchrotron light, with certain frequencies amplified by up to 10 000 times.

(a) Illustration of a bending magnet

(b) Illustration of a multipole wiggler

(c) Illustration of an undulator

Source: Australian Nuclear Science and Technology Organisation

Beamlines Synchrotron radiation is directed into separate experimental stations (figure 19.7). Each beamline is usually set up for a particular application using a specific frequency range. The first section of every beamline is the photon delivery system (also called the ‘beamline optics’). It incorporates filters, monochromators, mirrors, attenuators and other devices to focus and select appropriate wavelengths for particular research techniques. The frequency for the beamline is selected by a monochromator, which means that only one frequency leaves the end of the beamline and hits the target. Table 19.1 shows some of the processes for which beamlines of different frequencies are used. Experiments employing synchrotron light are conducted in customised facilities called end-stations. Most of the end-stations are housed inside radiation shielding enclosures called ‘hutches’ to protect staff and visitors from potentially harmful X-rays. Each beamline utilises the synchrotron light to gather data in the form of images, chemical spectra, and/or scattered light. As research scientists cannot enter the hutches during data collection, much of the equipment is controlled remotely via motors and robotic devices.

TOPIC 19 How do particle accelerators work? 7

FIGURE 19.7 Generation of synchrotron light Bending magnets Synchrotron radiation (light) is generated when the electron beam is deflected between powerful magnets. One analogy is the manner in which a wet tennis ball will fling off drops off water when swung in a circle.

Numerous detectors can be positioned to record diffraction patterns.

Electron beam Synchroton light

Electron beam

Synchroton light

Slits define the beam.

Monitor detector Measures intensity of the beam.

Monochromator A single piece of silicon cut with high precision allows only a specific wavelengths to pass. Other wavelengths are absorbed as heat.

Analyser crystal Selects scattered radiation. Sample Target object is rotated to build comprehensive image.

TABLE 19.1 Beamline uses Photon energies

Process

Science and industry applications

Example

X-ray diffraction

Protein crystallography

Determining 3D structure of proteins and pharmaceuticals

2–20 keV

Single-crystal diffraction

Chemistry, materials science, nanotechnology, geochemistry, pharmacology, mining, polymers, pigments

Determining structure of very small samples

3.6–5.5 keV

Infra-red

Geology Forensic Metallurgy Cell biology

Detection of minerals Analysis of fibres, dyes, residues Study of corrosion Study of drug uptake

Soft X-rays

Surface analysis, mining, polymers, engineering chemical analysis

Spectroscopy of materials with low atomic number elements (Li to K), e.g. the coal and sulfur industries

0.1–4 keV

Highperformance microprobe

Mining Materials science

Determining trace elements in soil High-resolution surface mapping

2–20 keV

Generalpurpose microprobe

Geochemical Manufacturing Biotechnology Forensic

Mineral exploration Corrosion, stress/strain mapping Metal uptake by plants Analysis of biological samples

5–30 keV

10–104 cm−1 ; exit through a diamond window

(continued)

8 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

TABLE 19.1 Beamline uses (continued) Photon energies

Process

Science and industry applications

Example

Vacuum ultraviolet

Electronic and magnetic properties of atoms, molecules and solids

Electronic structure of alloys; chemical processes at the nanometre level

10–350 eV

Imaging and hard X-rays

Biological/biomedical, materials science, fundamental X-ray imaging studies

Obtain images at the organic or even cellular level for biomedical research and clinical medical practice

4–60 keV

Powder diffraction

Structural chemistry, materials science, nanotechnology, mineral processing

Structure of transition metal oxides and superconductors; measurement of fast chemical reactions

4–30 keV

X-ray absorption spectroscopy

Structural and chemical analysis of materials

Structure of photonic devices; structure of metal complexes in enzymes

4–65 keV

Lithography

Nanotechnology

Product manufacture, e.g. photonic devices, hearing implants, communication chips

2–25 keV

Surface scattering

Use X-rays to probe the structural properties of the first few nanometres of the surface of a material

Coatings, adhesion, corrosion, catalysts, surfactants

5–25 keV

Small angle X-ray scattering

Chemical, material and biological sciences, polymer science

Growth of crystals and polymers; structure of colloids and membranes

5–30 keV

There are currently 10 operational beamlines with plans to expand the suite with an additional seven beamlines, where separate experiments using specific wavelengths can be conducted simultaneously. Synchrotron techniques can generate images plus elemental, structural and chemical information from diverse sample types ranging from biological to industrial materials.

19.2.3 Technical features of the Australian Synchrotron The technical features of the Australian Synchrotron include the following. • Air pressure. The air needs to be sucked out of all the components where electrons are moving. If the electrons hit an air molecule, they will be deflected and hit the side of the tube. The pressure inside the synchrotron is reduced to 10−9 millibar — about one millionth of atmospheric pressure. • High voltage. The linac uses high voltages to accelerate the electrons. The magnets in the booster and storage rings are electromagnets with high voltage, producing large currents through many coils. Vacuum pumps also run off high voltages. • Magnetic field strength. The magnetic field to bend electrons around in a circle has a strength of 1.30 T. • Electric current. The electric current in the storage ring is 200 mA. Once injected into the storage ring, electrons are slowly lost from the beam as a result of collisions with the remaining air molecules. The current can be kept at 200 mA for approximately 20 hours before a top-up is needed; but usually electrons are injected about twice a day. • Energy loss. The electrons lose energy by synchrotron radiation; the energy lost each turn is 931.6 keV. This energy is made up by accelerators in the straight-line sections of the storage ring.

TOPIC 19 How do particle accelerators work? 9

19.2.4 Synchrotron radiation and X-rays Synchrotron storage rings are designed to produce synchrotron radiation, the electromagnetic radiation emitted when charged particles are accelerated. Synchrotron radiation is emitted as photons that form a narrow cone as they head towards the target. The main characteristics of synchrotron radiation are as follows. • Spectrum. Synchrotron radiation is mostly in the form of X-rays, as they are the most useful. However, radiation across the electromagnetic spectrum, from infra-red upwards, can be produced. The spectrum is also continuous, which means there are no gaps or missing frequencies. Any frequency can be found in the range (see figure 19.8).

FIGURE 19.8 A continuous band of electromagnetic spectrum including infra-red, visible light, ultraviolet, and X-rays produced by Synchrotron is shown here.

Size

Taipei 101 Mall

103

People

101

1

Football

10–1

(1 m)

(1 km)

10–2

Virus

Cell

10–3

(1 cm) (1 mm)

10–5

10–6

10–7

Atom

Nucleus

10–9 10–10

10–12

10–14 (m)

(1 nm) (1 Å)

(1 µm)

Wavelength Visible light Radio waves

Micro-waves

Infra-red

Ultra-violet

Soft X-rays Hard X-rays Gamma rays

Photon energy

10–9 Source

(1 neV)

10–7 10–6 10–5 (1 µeV)

Radio

10–3

10–1 1

(1 meV)

Microwave tube

(1 eV)

101

103

105 106 107

(1 keV)

Light bulbs Synchrotron light source

(eV)

(1 MeV)

Radioactive elements

Source: Permission to use from National Synchrotron Radiation Research Center

•

Brightness. The intensity of the beam is hundreds of thousands times greater than that of conventional X-ray tubes (see figure 19.9). Brightness can be understood as the number of photons every second. It is more properly measured as the number of photons emitted per second per square millimetre of source size, per square milliradian of cone angle within a specific frequency range. In the past 120 years there has been a considerable amount of growth in synchrotron brightness, as shown in figure 19.10.

10 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 19.9 Brigtness of difference sources in comparision to synchrotron radiation. Higher orders of brightness enables synchrotron to performe quick experiments on small samples.

FIGURE 19.10 Growth of synchrotron brightness over the past 120 years 1035

1016 1015

Synchrotron radiation with bend magnet

1013 1012

Brightness

1011 Sun

1010 109 108

X-ray tube

107 106

Brightness (Photons/sec/mm2/mrad/0.1 % bw)

1030

1014

105

4th generation light source (free electron lasers)

1025 3rd generation light source (Bending magnets, wigglers and undulators)

1020

1015

2nd generation light source (Bending magnets and wigglers) 1st generation light source (Bending magnet)

1010

60 W light bulb X-ray tubes

104 103

Candle

105 1900

102

•

1920

1940

1960 Year

1980

2000

2020

Source: (b) Amardeep Bharti and Navdeep Goyal February 8th 2019. Fundamental of Synchrotron Radiations, Synchrotron Radiation - Useful and Interesting Applications, Daisy Joseph, IntechOpen, DOI: 10.5772/intechopen.82202. Available from: https://www.intechopen.com/books/synchrotronradiation-useful-and-interestingapplications/fundamental-of-synchrotron-radiations

Divergence. The beam of radiation spreads out like a cone as it FIGURE 19.11 Cone of travels down the beamline (see figure 19.11). Typically a beam cone radiation showing cone would have a cone angle of a few microradians — that is, angle. The size of the cone angle is a few microradians, less than half of one thousandth of a degree. which is less than half of the • Polarisation. The radiation from a synchrotron is linearly polarised, one-thousandth of a degree. minimises background scattering and improves sensitivity. • Tunability. The frequency for the beamline is selected by a monochromator, which means that only one frequency leaves the end of the beamline and hits the target. • Collimation. The beam can be focused down to less than a micron (10–6 m) across, enabling chemical speciation to be mapped. • Duration. Synchrotron radiation comes in pulses, typically lasting about one billionth of a second. Cone angle These features allow the X-rays to be used to investigate the fine structure of many materials — that is, to locate specific atoms in a molecule, even a large molecule such as haemoglobin that is found in red blood cells. This information is of value to researchers across a range of fields, because it enables them to answer such questions (that are not possible to explore using visible light) as: • What are the differences between malignant and non-malignant brain tumors? • What is the structure of material, such as semiconductor nanocrystals, which may be used in the next generation of computers?

TOPIC 19 How do particle accelerators work? 11

•

What are the steps or dynamics of a chemical reaction, either an industrial situation or a biological one? To get some idea of how X-rays can answer these questions, we should go back in time to their discovery. FIGURE 19.12 A reconstruction of 3D structure of the active site of an influenza virus based on information revealed by synchrotron radiation. This collection of precise information can improve the knowledge of causes of infectious diseases like flu.

Source: © Dr. Jose Varghese

Discovery of X-rays In the late nineteenth century, scientists were keen to discover the exact nature of electricity, but they had no way of getting the electricity out of the wires. Two new technologies offered this possibility: the high DC voltage equipment invented by Heinrich Ruhmkorff, capable of producing 20 000 volts; and the mechanical pump that could reduce the air pressure inside a glass tube to very low, near vacuum, levels. Placing electrodes inside the glass tube and connecting them to high voltage enabled electricity to flow through about 20 centimetres of low-pressure air producing strange visual effects. Ribbons and coloured bands of light were produced inside the tube. These lights seemed to come from the terminal in the tube that was connected to the negative of the power supply, and so were called cathode rays. However, a clear answer to the question of what electricity is was not forthcoming. In October 1895 while experimenting with cathode rays, Wilhelm Röntgen, professor of Physics at the University of Wurzburg in Germany, discovered an unknown form of radiation, which he called X-rays. He had noticed that photographic film beyond the end of the tube was exposed when the cathode rays hit the end of the tube. Röntgen spent the next seven weeks carrying out a variety of experiments on the X-rays to determine their properties and to identify them. In that time, he compiled and published the following properties. 1. X-rays are produced when cathode rays (later to be identified as electrons) strike any solid object. 2. All substances are more or less transparent to X-rays. The more dense or thick the substance, the more the X-rays are absorbed. (Röntgen produced the first ‘X-ray’ of the bones in his hand, which attracted the immediate interest of the medical profession.) 3. X-rays make a number of materials fluoresce — that is, give off visible light. 4. X-rays expose photographic film. 5. X-rays cannot be deflected by electric or magnetic fields. 6. X-rays travel in straight lines, and had not been found to reflect or refract. 7. X-rays discharge electroscopes, regardless of whether they were charged positive or negative.

12 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Comparison between synchrotron, laser and X-rays As shown in table 19.2 the synchrotron is the most powerful tool to investigate complex issues. A conventional X-ray is less bright and produces radiation at a few different frequencies. Laser, on the other hand, does not spread and is a coherent light source, meaning all photos have the same frequency. However, being coherent does reduce laser’s versatility to be used in different investigations at different frequencies. Synchrotron are extremely bright, polarised, continuous and very narrow. The broad range of available wavelengths allow scientists to look at the size and shape of macromolecules and voids in bulk materials, peer into the biochemistry of single cells and delve all the way down to the bonds between atoms. TABLE 19.2 Comparisons of radiation a synchrotron, a laser and an X-ray tube Brightness

Spectrum

Divergence

Synchrotron

Extremely intense

Continuous and wide

Very narrow

Laser

Very intense

Single frequency

Narrow

X-ray tube

Intense

Narrow, continuous but not smooth

Wide

19.2 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. What is a particle accelerator? Give an example of a particle accelerator. 2. How is radiation generated in a synchrotron? 3. What is the difference between Australian Synchrotron and the LHC? 4. Label the design features of the Australian Synchrotron with the given options in the boxes provided and write the purpose for each of them. Electron gun, Linear accelerator, Booster ring, Storage ring, Beamline, End station

TOPIC 19 How do particle accelerators work? 13

5. Why does the storage ring have to be in an ultra-high vacuum? 6. The direction of the electron beam is controlled with the magnets in a synchrotron. How is the temperature of these magnets and the entire building regulated?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

19.3 Accelerator technology and the development of modern particle physics KEY CONCEPTS • Explain the evolution of collider technology including: • particles involved in the collision event • the increasing energies attained since the 1950s. • Evaluate the role of colliders in the development of the Standard Model of particle physics, including reference to subatomic structure and processes. • Describe the products of collisions with reference to symbol, charge, rest energy and lifespan. • Compare the physical designs and purposes of particle detectors at the Large Hadron Collider including ATLAS, CMS, ALICE and LHCb.

The first particle accelerator The first particle accelerator was an offshoot of equipment designed to investigate the nature of electricity. The development of high voltage devices and vacuum technology in the 1850s enabled research into how electricity flowed through gases at very low pressures. This ultimately led to the discovery of the electron and the identification of its properties. The deflection of the electron in a magnetic field showed that the electron was negatively charged and the radius of curvature gave a value to what is called the ‘chargeto-mass ratio’ of the electron. This principle is now used in mass spectrometer chemical analysis. FIGURE 19.13 A mass spectrometer. The ions move in a circular path in the magnetic field. The radius of the path depends on the charge-to-mass ratio of the ion. Particles are Ions are accelerated ionised into beam

Magnetic field causes deflection

Sample

Electromagnet

To vacuum pump

Ion stream hits Ion detector To amplifier To recorder 14 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

When the properties of the electron were being investigated in the 1890s, the radius of curvature in a magnetic field gave the charge-to-mass ratio of the electron, but neither the charge nor the mass separately. It was only after Robert Millikan, in his oil-drop experiment, had experimentally determined the charge on the electron that its mass could be calculated. Particle accelerators can be split into two fundamental types, electrostatic and oscillating field accelerators.

19.3.1 The cyclotron

FIGURE 19.14 Schematic diagram of a cyclotron

Magnet pole Dee

Dee Magnet pole

Vacuum

Dees

Ernest O. Lawrence invented the cyclotron in 1930, while at the University of California. In his design, the electron made repeated passes through the same electric field, getting an increase in speed each time. The design consisted of what are called ‘dees’, because they look like the letter D. The hollow metal dees were placed in a vacuum between the poles of a strong magnet, facing each other across a narrow gap (see figure B A 19.14). They were connected to a power supply, which could switch polarity very quickly — that is, one moment one dee would be positive, and the other negative; then a very short time later, the voltage would be reversed. A charged particle was released in the middle of the FIGURE 19.15 Ernest Orlando Lawrence, apparatus. It would be attracted to one of the dees and (1901–1958), US physicist, winner of the 1939 gain speed. Once inside the hollow of the dee, the Nobel Prize for Physics for his invention of the voltage would have no effect because the particle cyclotron — the first particle accelerator to was effectively ‘inside’ the conductor. However, the achieve high energies magnetic field would have an effect and curve the particle around in a semicircle and come back into the space between the dees. Meanwhile the polarity of the dees had been reversed, and the particle would be attracted across to the other dee, and so gain speed. Inside this dee the particle, now faster, would be curved around by the magnetic field, but in a larger radius because of the higher speed. If the polarity of the voltage on the dees could be reversed just at the right time as the particle emerges from one dee, then the particle would continue to get faster and faster as it spiralled outwards towards the edge of the dees. One of the difficulties with the cyclotron is that in this mode of operation it works only for relatively low speeds. We know from Einstein’s theory of relativity that the mass of particles increases with their speed, which means that at speeds close to the speed of light, the faster the particle, the longer the time for one revolution. The movement of the particle between the dees would get out of sync with the voltage reversals. This problem can be overcome by decreasing the cyclotron frequency as the particle gains in speed. These modified cyclotrons are called synchrocyclotrons. The first cyclotron built by Ernest Lawrence could accelerate protons to an energy of 80 keV and by 1939 to 8 MeV, using a magnet 150 cm in diameter (see figure 19.15). By 1953 variable energy cyclotrons were being designed, with one of the first built at the University of Melbourne. TOPIC 19 How do particle accelerators work? 15

19.3.2 Early synchrotrons The strategy used in the synchrotron to overcome the relativistic effect is to progressively increase the magnetic field as the particle gains speed, keeping it in a path of constant radius. To minimise operating costs, the radius needs to be large. In fact, the radius can be as large as 1 kilometre. Of course a giant magnet with a 1-kilometre radius has not been built. Rather, small magnets are placed on the circumference of the synchrotron (the ring), where the particles are. Between these magnets are high-voltage accelerators to top up the energy to the value requested by the experimenter. It would be wasteful to inject particles into a synchrotron at low speeds and let the ring of magnets force them up to almost the speed of light. Instead a linear accelerator outside the ring is used to accelerate the particles up to a significant fraction of the speed of light before they are directed into the ring along a tangential line. The early particle accelerators used electrons because they are easy to produce with an electron gun and because they are very light, which means they can be accelerated to very high speeds. However, their lightness also means they have very little momentum, so their capacity to affect a nucleus is very limited. Machines such as the proton synchrotron, mentioned earlier, were designed to produce protons, which are much heavier than electrons. In 1943, while working at the University of Birmingham during World War II, Australian physicist Marcus Oliphant suggested modifying the design of the cyclotron to produce the synchrotron. In 1946 he and his team began construction of a proton synchrotron. The Melbourne physicists ‘had in mind the urgent need of a local source of radioactive isotopes’ for medical research. Such a focus also allowed for tapping into the more generous funding possibilities associated with medical science. The design of the variable-energy cyclotron, by David Caro and John Rouse, was a world-first (figure 19.16) but a team of better resourced Americans managed to build one sooner. FIGURE 19.16 In the late 1950s at the University of Melbourne, David Caro and John Rouse designed the 35 MeV Betatron, the world’s first variable-energy cyclotron.

Source: 35MeV Betatron from School of Physics Museum The University of Melbourne

16 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

A US team at Brookhaven, who started on their synchrotron after the UK team, were the first to be operational. They produced 900 MeV protons in May 1952. The UK team were operational in July 1953 and went on to produce 2900 MeV in January 1954. Later in 1954, a second US synchrotron, rated at 6000 MeV, was built at Berkeley, California and in 1955 a Russian 10 000 MeV machine was built. By 1960 an internationally owned machine of 30 000 MeV was built in Geneva. Most particle accelerators developed since the 1950s have used the basic circular design described in the previous section, ‘Design features of a synchrotron’. They differ in the amount of energy they can deliver to the particles and also in their size, which is determined by the diameter of the storage ring.

European Council for Nuclear Research (CERN) In December 1951, at an intergovernmental meeting of UNESCO in Paris, the first resolution concerning the establishment of a European Council for Nuclear Research (in French Conseil Européen pour la Recherche Nucléaire) was approved. Two months later, an agreement was signed, and the acronym CERN was born. Today, our understanding of matter goes much deeper than the nucleus, and CERN’s main area of research is particle physics. Because of this, the laboratory operated by CERN is often referred to as the European Laboratory for Particle Physics. The first foundation stone was laid down in July 1955 by the CERN’s director-general, Felix Bloch. Since CERN began, it has helped to uncover what the universe is made of and how it works. In the 1980s, the discovery of the W and Z particles brought the first experimental evidence of the theories to explain weak and electromagnetic force. CERN researchers Simon van der Meer and Carlo Rubbia shared the 1984 Nobel Prize in Physics for this discovery. During the 1990s, CERN experiments designed in light of this discovery tested the so-called electroweak theory with extreme precision. In 2010, the LHC started to provide particle collisions in a new high-energy domain, leading to the discovery at CERN of a Higgs boson (figure 19.17) — long sought as the particle linked to the mechanism that gives mass to elementary particles. FIGURE 19.17 A Higgs boson in a large collider

The Higgs boson (Higgs particles) is a particle in the standard model of physics. In the 1960s Peter Higgs was the first person to think of it, and the particle was found in march 2013. It is one of the 17 particles in the standard model. The Higgs particle is a boson.

The best-known CERN technology is the World Wide Web, invented to allow large number of scientists to share information. For many of us today, life without the Web seems inconceivable. The Grid has been developed at CERN to process the vast amounts of data collected by the LHC experiments. Electronic particle detection techniques have revolutionised medical diagnosis. Detectors invented by Georges Charpak in 1968 allow X-ray images to be made using a fraction of the dose required by photographic methods. Crystals developed for CERN experiments in the 1980s are now ubiquitous in PET scanners. And today, developments for a new generation of CERN detectors are allowing PET and MRI imaging techniques to be combined in a single device. The transfer of CERN technologies and expertise to society is an integral part of these activities, providing novel solutions in many fields (see figure 19.18). TOPIC 19 How do particle accelerators work? 17

CERN uses 1.3 terawatt hours of electricity annually. That’s enough power to fuel 300 000 homes for a year in the United Kingdom. But the energy needed changes from month to month, as the seasons shift and the experimental requirements are adjusted. At peak consumption, usually from May to mid-December, CERN uses about 200 megawatts of power, which is about a third of the amount of energy used to feed the nearby city of Geneva in Switzerland. The Large Hadron Collider (LHC) runs during this period of the year, using the power to accelerate protons to nearly the speed of light. CERN’s power consumption falls to about 80 megawatts during the winter months. Particle accelerators can be designed to work with any charged particle or antiparticle. As the names in table 19.3 reflect, electrons, positrons and protons are commonly used. Particle accelerators use a straight-line section, called a linac, to accelerate the charged particles up to an extremely high speed and energy. Some may use a booster ring to take the energy even higher. The increasing energies attained since 1960s are shown in figure 19.19. Once the particles are at the maximum energy, they are redirected to a storage ring. The storage ring is not so much a ring as a series of short curved sections joined by short straight sections. In each curved section, magnets bend the beam of particles into a circular path. In each straight section, the beam is given a little extra push along to compensate for the loss FIGURE 19.18 How CERN’s various areas of expertise translate into impact across industries beyond CERN

Source: CERN

18 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

TABLE 19.3 Synchrotron developments

Name

Location

Year

Diameter

Maximum energy

Results

Cyclotron

University of California, Berkeley, USA

1932

10 cm

80 keV

Proof of concept

Cyclotron

University of California, Berkeley, USA

1939

150 cm

8 MeV

Discovered many isotopes

Cosmotron

Brookhaven National Laboratory, Brookhaven, USA

1952

23 m

900 MeV

Detected mesons only found in cosmic rays

Birmingham Synchrotron

Birmingham, UK

1953

20 cm

2900 MeV

First synchrotron to reach 1 GeV

Bevatron

Berkeley, USA

1953

36 m

6000 MeV

Discovered anti-proton and anti-neutron

Synchrophasotron

Joint Institute for Nuclear Research, Dubna, Russia

1957

57 m

10 GeV

First synchrotron to reach 10 GeV

Proton Synchrotron

CERN, Geneva, Switzerland

1960

200 m

28 GeV

Discovered many sub-atomic particles

Stanford Positron–Electron Accelerating Ring (SPEAR)

Stanford University, Stanford, USA

1968

3.2 km*

90 GeV

Found the charm quark in 1976 and quark structure of protons and neutrons in 1990

Tevatron

Fermilab, USA

1983

2.2 km

1000 GeV

Found the top quark in 1995; shut down in 2011 due to budget cuts

Large Electron–Positron Collider (LEP)

CERN, Geneva, Switzerland

1989

8.5 km

209 GeV

Precisely measured the mass of subatomic particles, supporting the Standard Model

Large Hadron Collider (LHC)

CERN, Geneva, Switzerland

2008

8.5 km

14 000 GeV

Produced a quark plasma, the densest matter besides black holes, in 2011; discovered Higgs Boson in 2012; discovered two new heavy baryons in 2014

* SPEAR uses the 3.2 km long Stanford Linear Accelerator (SLAC), which has no storage ring.

of energy due to synchrotron radiation emitted on the bend. As the beam of particles circles the storage ring, some particles stray off course and the number of particles in the beam diminishes. However, the beam can hold enough particles to be useful for hours at a time before it needs to be replenished. The diameter of the storage ring is determined partly by the energy of the particles, but also by the how strong the magnets are. As the magnets are electromagnets, the electricity cost of running the magnets is a significant factor in deciding the size of the accelerator.

TOPIC 19 How do particle accelerators work? 19

FIGURE 19.19 Livingston plot showing the centre of mass energy of particle accelerators (lepton and hadron colliders) versus time. In the case of the hadron machines, energies have been adjusted to account for quark and gluon constituents. This suggests that the advancements in accelerator technology have increased the energy records achieved by new machines by a factor of 10 every six years. Using these units, the energy of collisions at the Large Hadron Collider is nearly 100 000 TeV. Livingston plot – update 2009 – symmetry magazine 100 000 TeV LHC

10 000 TeV 1000 TeV 100 TeV 10 TeV 1 TeV

First accelarator Cyclotrons Cook or oft–Walton electrostatic accel Van be Graaff electrostatic accelerators Betatrons Synchrocyd otrons Linear accelarator Electron synchrotrons Proton synchrotrons Storage ring colliders Linear colliders

100 GeV 10 GeV 1 GeV 100 MeV 10 MeV 1 MeV 1930

1940

1950

1960

1970

1980

1990

2000

2010

Source: From 1954 ‘Energy of colliders is plotted in terms of the laboratory energy of particles colliding with a proton at rest to reach the some center of mass energy’.

SIR MARCUS OLIPHANT Marcus Oliphant was born and educated in Adelaide. After completing his Master of Science, he won a scholarship to go to the Cavendish Laboratory at Cambridge University to join the research team under Ernest Rutherford. Oliphant’s main contributions were the discovery of tritium, the third isotope of hydrogen, which has two neutrons and one proton, and the confirmation of the possibility of a fusion reaction between deuterons (the nuclei of the second isotope of hydrogen, which have one neutron). In 1939 he began building a 60-inch cyclotron, but World War II intervened and the cyclotron was not finished until after the war. During the war, Oliphant’s research focused on radar. The resonant cavity magnetron, developed by his team, enabled radar with a 10 centimetre wavelength instead of the existing 150 centimetre. This became a vital part of Britain’s defence during the 1940 Battle of Britain. He also played a role in the development of the atom bomb. In 1940, while working in England, Frisch and Peierls had showed by calculation that the critical mass for a chain reaction of uranium-235 was only a few kilograms. Oliphant brought the results to the attention of government authorities and a committee was set up to develop an atom bomb. In 1941 Oliphant visited the United States to talk about radar development. While there he discovered that, even

FIGURE 19.20 Marcus Oliphant (1901–2000)

Source: Bassano Ltd / Wikimedia Commons

20 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

though the reports from the UK committee on the atom bomb had been sent to Washington, no work was being done by the US on the atom bomb. The reports had been filed away and not discussed because the United States was not at war. He immediately told US physicists of the developments in the United Kingdom and shortly after the Manhattan Project was established to develop an atom bomb. Oliphant strongly opposed the use of the atom bomb on Japan. He never overcame his feelings of guilt about the part he played in the Manhattan Project. He subsequently took an active part in campaigns against nuclear weapons, being a founding member of the Pugwash Movement. The mission of Pugwash is to ‘bring scientific insight and reason to bear on threats to human security arising from science and technology in general, and above all from the catastrophic threat posed to humanity by nuclear and other weapons of mass destruction’. In 1972 he was appointed Governor of South Australia, a post he held for five years. Sir Marcus Oliphant died in 2000 at the age of 98.

19.3.3 The development of the Standard Model of particle physics The Standard Model of particle physics describes all known elementary particles of matter and three of the four known fundamental forces that describe their behaviour (see figure 19.21). There are 17 particles named in the Standard Model, all of which can be classified into two types of particles: fermions and bosons. FIGURE 19.21 The Standard Model of particle physics

Fermions describe the fundamental particles that make up all matter, while bosons are the mediators (cause) of interactions between other particles. A fundamental particle is one that is not made up of other smaller particles. The six quarks are paired in three generations: the ‘up quark’ and the ‘down quark’ form the first generation, followed by the ‘charm quark’ and ‘strange quark’, then the ‘top quark’ and ‘bottom (or beauty) quark’. Quarks also come in three different ‘colours’ and only mix in such ways as to form colourless objects. The six leptons are similarly arranged in three generations – the ‘electron’ and the ‘electron

TOPIC 19 How do particle accelerators work? 21

neutrino’, the ‘muon’ and the ‘muon neutrino’, and the ‘tau’ and the ‘tau neutrino’. The electron, the muon and the tau all have an electric charge and a sizeable mass, whereas the neutrinos are electrically neutral and have very little mass. Three of the fundamental forces result from the exchange of force-carrier particles, which belong to a broader group called ‘bosons’. Particles of matter transfer discrete amounts of energy by exchanging bosons with each other. Each fundamental force has its own corresponding boson – the strong force is carried by the ‘gluon’, the electromagnetic force is carried by the ‘photon’, and the ‘W and Z bosons’ are responsible for the weak force. The precise measurements of the mass and charge of all these various particles enabled physicists to confirm the theory called the Standard Model. Some of the exotic particles detected are listed in table 19.4. TABLE 19.4 Exotic particles detected in particle accelerator experiments Particle

Symbol

Charge (e)

Mass† (GeV/c2 )

Mean life‡ (s)

Spin (h)

5.11 × 10−4

Stable

1/2

Stable

1/2

2.20 × 10−6

1/2

Stable

1/2

Leptons: e−

Electron

−1

ve

0

< 2 × 10

Muon

μ−

−1

0.106

Muon neutrino

vμ

0

Electron neutrino

𝜏

Tau

−

v𝜏

Tau neutrino

−1 0

−9

< 2 × 10−9 1.78 < 2 × 10−9

2.90 × 10

−13

1/2

Stable

1/2

Hadrons: Baryons: Proton

p

+1

0.938

Stable

1/2

Neutron

n

0

0.940

880

1/2

Lambda

Λ

0

1.12

1/2

Ξ− 0

−1 0

2.63 × 10−10

1.32 1.31

Ω

−1

π+ , π − K+ , K−

{

Cascade

Ξ

Omega

1.64 × 10−10 2.90 × 10−10

1/2 1/2

1.67

8.2 × 10−11

3/2

+1, −1 0

0.140 0.135

2.60 × 10−8 8.5 × 10−17

0 0

+1, −1

0.494 0.498

Mesons: Pion

Kaon

{ {

J/psi

B

π

K

0

0

, K0

0

0

J/𝜓 {

B+ , B− B

0

, B0

+1, −1 0

3.10 5.28 5.28

1.24 × 10−8 ¶ 1.25 × 10−19

1.64 × 10−12 1.52 × 10−12

0 0 1 0 0 (continued)

22 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

TABLE 19.4 Exotic particles detected in particle accelerator experiments (continued) Particle

Symbol

Charge (e)

Mass† (GeV/c2 )

Mean life‡ (s)

Spin (h)

0

Stable

1

Gauge bosons: 𝛾

Photon

0

W+ , W−

+1, −1 0

80.4 91.2

1.59 × 10−25 1.32 × 10−25

1 1

Gluons

g

0

0

Stable

1

Graviton‖

G

0

0

Stable

2

126

1.56 × 10−22**

0

Weak bosons

{

Z

0

Higgs boson: Higgs

H0

0

† Antiparticle symbols are indicated by an overline. ‡ Where available, values are quoted to three significant digits, which is sufficient precision for most purposes. ¶ Due to mixing of neutral kaons with their antiparticles, these particles do not have definite lifetimes. Symmetric and antisymmetric linnear combinations of K0 and K0 , known as KS and KL , have lifetimes of 8.95 × 10−11 s and 5.29 × 10−8 s, respectively. ‖ The existence of this particle has been postulated but is not yet definitively established. ** Predicted; measurements are not yet precise enough to determine the Higgs lifetime.

The Higgs boson Since the positron was observed in 1932, confirming a prediction from theory, particle after particle that has been predicted has been observed. By the end of 2000, all predicted particles had been observed in experiment except for the Higgs boson. This particle had been predicted in 1964 to solve problems with existing theory. Scientists had developed a theory that made sense of what was known about the particles of matter that made up the known universe. However, it only worked if certain particles had zero mass. Some of those particles were known to have mass so something was missing. Peter Higgs, François Englert and three others proposed that there was a field through space, now known as the Higgs field, that particles with mass interacted with via a particle called the Higgs boson. If the Higgs boson was found, the theory of the Standard Model would match experimental results. To search for this elusive particle, the largest particle accelerator ever built was constructed at CERN on the boarder of Switzerland and France. More than 100 countries are involved in the Large Hadron Collider, which has cost more than US$13 billion. In July 2012 scientists announced that this enormous investment had paid off and the final piece in the Standard Model had been observed. Higgs and Englert were awarded the Nobel Prize for Physics in 2013, for the prediction they had made nearly fifty years earlier. The detection of these particles has posed significant challenges, as they do not stay around to be ‘stored in a bottle’; they rapidly decay into other particles, which themselves decay almost immediately. Achievements in particle physics owe as much to the design of the detectors as to the design of the machines that produce the collisions in the first place.

19.3.4 Features of the Large Hadron Collider The Large Hadron Collider (LHC) is the world’s most powerful particle collider. It mostly uses protons (hence its name, as protons are hadrons), but also uses ions and lead nuclei. The LHC was built underground, as the Earth’s crust provides a shield for the radiation produced. It generates about 1 billion particle collisions per second. Each proton beam flying around the LHC makes 11 245 circuits every second and could reach an energy of 7 TeV, so for two protons the collision energy is 14 TeV. In absolute terms, these energies, if compared to the energies we deal with every day, are not remarkable. In fact, about 1 TeV of energy is produced by a flying mosquito. What makes the LHC so extraordinary is that it squeezes energy into a space about one million times smaller than a mosquito.

TOPIC 19 How do particle accelerators work? 23

The detectors There are four main detectors at the LHC, each with a specific purpose and special design features. They are known by the acronyms ATLAS, CMS, ALICE and LHCb. For all of the particles produced in a reaction, particle detectors need to be able to measure their mass, momentum, kinetic energy, lifetime and charge.

ATLAS ATLAS is a general-purpose detector. The acronym stands for ‘A Toroidal LHC ApparatuS’. (The word ‘toroidal’ means ‘donut-shaped’.) It is a large construction, 45 metres long and 25 metres in diameter, with a mass of 7000 tonnes. Particle beams can be arranged to collide in the middle of ATLAS so that the numerous particles that are produced in a collision can be detected. ATLAS is made up of four layers of concentric detectors, each for different types of particles. 1. Inner detector: determines the exact path of the particles 2. Calorimeter: measures their energy 3. Muon spectrometer: measures the momentum of muons 4. Magnet system: measures the momentum of other particles The charge and momentum of particles are detected by their curvature in the magnetic fields in the layers. The only particles that cannot be detected by ATLAS are neutrinos.

FIGURE 19.22 The ATLAS detector

CMS The Compact Muon Solenoid, or CMS, is another general-purpose detector. It is smaller than ATLAS but Source: Justin Clements / Flickr heavier: 22 metres long, 15 metres in diameter, with a mass of 14 000 tonnes. It is also of a similar shape to ATLAS, although more cylindrical. The CMS has five layers that serve similar functions to those in ATLAS. The CMS is mainly used to detect photons, electrons, muons and different types of hadrons.

Why both? If the ATLAS and CMS do similar jobs, why is there a need for two detectors? ATLAS and CMS experiments each involve over 3000 physicists from almost 200 institutions from more than 40 countries. The LHC frequently runs two or more experiments at the same time. The benefits are healthy competition between teams, cross-checking of results, increased data and insurance against failure. The arrangement also allows the experiments to use the different expertise of physicists and explore different strategies to solve problems. ATLAS and CMS both confirmed the existence of the Higgs boson. ATLAS gave its mass as 126 ± 0.4 GeV c2 and CMS gave the result 125.3 ± 0.4 GeV c2 . (The Tevatron at Fermilab also produced results consistent with the Higgs boson but there was a 1-in-500 chance that those results could have been due to other events, so the Tevatron finding was not categorised as a confirmed result.)

ALICE ALICE stands for ‘A Large Ion Collider Experiment’. It is different from ATLAS and CMS in that it has a specific purpose. It is designed to analyse collisions between lead nuclei. The temperature of such a collision produces a quark plasma, the densest matter besides black holes, similar to conditions thought to have existed a fraction of the second after the big bang. In 2012 ALICE measured the highest temperature on Earth at 5.5 trillion degrees Celsius.

24 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

LHCb LHCb stands for ‘Large Hadron Collider beauty’. It is investigating matter and antimatter. At the beginning of the big bang, matter and antimatter were created in equal quantities, but now there is little antimatter to be found. The LHCb is seeking an explanation by colliding heavy hadrons that contain the bottom quark. This detector has about 800 people working together on it. The LHCb detected new baryons in 2014.

Putting the LHC together The LHC consists of accelerators and detectors. How are they all related? Figure 19.23 shows the different components of the LHC. FIGURE 19.23 p and Pb: The linear accelerators for protons and lead ions; where they start their journey CMS LHC ALICE

p Pb

• • • • • •

LHCb SPS

ATLAS

PS

PS: Proton Synchrotron, where the energy of the protons is increased from 1.4 GeV to 28 GeV SPS: Super Proton Synchrotron, where the energy is increased further, from 28 GeV to 450 GeV; particles are sent into the large ring in opposite directions to collide in one of the detectors ATLAS: A Toroidal LHC ApparatuS detector CMS: Compact Muon Solenoid detector ALICE: A Large Ion Collider Experiment LHCb: Large Hadron Collider beauty detector

19.3 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Who invented the first cyclotron and what was its purpose? 2. What can the LHC do that the Australian Synchrotron can’t? 3. What can the Australian Synchrotron do that the LHC can’t? 4. (a) What are the four main detectors at the LHC? (b) Which is the only particle that cannot be detected by ATLAS? 5. The ATLAS and CMS do similar jobs, why is there a need for two detectors? 6. What does LHCb stand for and what is its purpose? 7. What features does the LHC have that the early particle colliders did not? 8. In what way is the LHC crucial for our increased understanding of the universe?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

TOPIC 19 How do particle accelerators work? 25

19.4 Current and future applications of accelerator technology for society KEY CONCEPTS • Explain how the immense amount of data collected by the Large Hadron Collider is stored and analysed, and the associated role particle detectors have had in the development of information processing technologies. • Describe at least one application of particle accelerators selected from: • materials analysis and modification which results in the improvement of consumer products such as heat shrinkable film and chocolate • implanting of ions in silicon chips to make them more effective in electronic products such as computers and smart phones • nuclear energy applications such as the use of thorium as an alternative fuel for the production of nuclear energy or the treatment of nuclear waste • pharmaceutical research involving the analysis of protein structure leading to the development of new pharmaceuticals to treat major diseases • DNA research involving the analysis of protein metabolism leading to the development of new antibiotics • medical applications such as the production of a range of radioisotopes for medical diagnostics and treatments or cancer therapy through the use of particle beams • use of spectrometry in environmental monitoring or the use of blasts of electrons in the treatment of pollution such as contaminated water, sewage sludge and gases from smokestacks • use of particle accelerators in a selected experiment or scientific endeavour. • Investigate current and proposed future directions of collider technologies.

19.4.1 Data analysis In the ATLAS detector, two beams of protons FIGURE 19.24 Tim Berners-Lee, the CERN scientist, are aimed at each other. A ‘crossing’ is when made the World Wide Web inaugural page available at a particle from one beam crosses the path of a CERN in December 1990. However, the world’s first particle from the other beam; there are 40 million website was made public on 6 August 1991. crossings every second. Each impact produces 25 megabytes of raw data. In total, about one million gigabytes of raw data are produced every second. Analysing the data is an international exercise relying on high-speed data links around the world. The 230 000 processor cores and 15 000 servers run 24/7. Over 90% of the resources for computing in the Data Centre are provided through a private cloud based on OpenStack, an open-source project to deliver a massively scalable cloud operating system. It runs over 2 million tasks per day and, Source: © CERN 2019 at the end of the LHC’s Run 2, global transfer rates regularly exceeded 60 GB per second. These numbers will increase as time goes on and as computing resources and new technologies become ever more available across the world. With its high demands for data storage, data transfer and software design, it is no wonder that CERN was the birth place of the internet.

26 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

19.4.2 Applications of particle accelerators There are over 25 000 particle accelerators across the world. Not all of these are as big or as powerful as the Australian Synchrotron, let alone the LHC. Many have quite specific purposes, including: • medical uses — about 40% of the world’s particle accelerators are dedicated to medical uses such as: • radiation therapy with protons or electrons • sterilisation of equipment • ion implantation — about another 40% of particle accelerators are used this way. Ions are accelerated to high speeds, then fired at the surface of a material. This process is used in the manufacture of semiconductors and the altering of surface properties, such as toughening steel tools non-destructive testing to find out about the internal structure of an object such as an aircraft wing • without damaging it in any way. • blasts of electrons from a particle accelerator are an effective way to clean up dirty water, sewage sludge and polluted gases from smokestacks. • treating nuclear waste and allowing the use of an alternative fuel, thorium, for the production of nuclear energy. Accelerators are used in two ways for diagnostics and for therapy for medical purposes.

Radiotherapy Radioisotopes are an essential part of radiopharmaceuticals (consists of a radioisotope tracer attached to a pharmaceutical) and have been used regularly in medicine for more than 30 years. They can give exclusive biochemical and physiological information when injected into living creatures. The radioisotopes that have short half-lives are appropriate for diagnostic purposes; whereas, others with longer half-lives are used for therapeutic purposes. Radioisotopes are employed to observe anomalies of metabolism (changes in blood flow, oxygen utilization, glucose metabolism) or detect tumours. The aim of radiotherapy is to destroy the malignant cells without damaging the healthy tissues. After entering the body, the radio-labelled pharmaceutical will collect in a particular organ or tumour tissue. The radioisotope attached to the targeting pharmaceutical will undergo decay and produce specific amounts of radiation that can be used to diagnose or treat diseases and injuries. The amount of radiopharmaceutical given is carefully chosen to ensure the safety of each patient. The most common radiopharmaceuticals used in Australian’s nuclear medicine centres are technetium-99m and iodine-131. Technetium-99m has a half-life of six hours that make it ideal for imaging organs for disease detection without delivering a significant radiation dose to the patient as it emits weak gamma radiation. On the other hand, iodine-131 with a half-life of eight days and a higher energy beta particle decay, it is used in the treatment of thyroid cancer.

Sterilisation Sterilising medical equipment in hospitals is a critical process in medical facilities. This can be done by bombarding the equipment with a beam of electrons or X-rays produced by a particle accelerator. These high energy particles eradicate any microbes on the surface of the product. This technique can also be used in sterilisation of fresh food and industrial waste. In this process ionisation radiation is produced without using radioisotopes. Accelerators could be used to sterilise single objects or multiple objects quickly.

19.4.3 Current and proposed future of collider technologies At the time of writing the LHC was shut down for two years for upgrades. This planned improvement will bring the energy up to 14 trillion eV by 2021. It will also lay the groundwork for another expression of the collider further, known as the High-luminosity LHC. This is expected to be ready by 2026 and will increase the rate of proton collisions by at least a factor of five.

TOPIC 19 How do particle accelerators work? 27

CERN has announced the ambitious plan to build a future circular collider (FCC) in a circular tunnel stretching 100 kilometres near Geneva that could start operating in 2040 and take over from the existing 27-kilometre LHC. The goal of the FCC is to significantly push the energy and intensity frontiers of particle colliders, with the aim of reaching collision energies of 100 TeV, in the search for new physics. The project is the combined work of 1500 contributors across 150 universities, research institutions and industrial partners and is yet to be approved.

FIGURE 19.25 The proposed future circular collider would take decades to design and build.

Source: © CERN 2019

19.4 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. How much raw data is produced every second in the ATLAS detector? 2. How is the huge amount of data transferred and stored at CERN? 3. Why was CERN the birth place of ‘internet’? 4. Which charged particles has the LHC detected and how? 5. What is the future direction of collider technologies?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

19.5 Review 19.5.1 Summary • • • •

Synchrotrons accelerate electrons to produce light, called synchrotron radiation. Colliders accelerate electrons, protons and heavy ions to produce particle collisions. The maximum energies have grown exponentially since the 1950s. In the Australian Synchrotron the linac is where the electrons receive most of their energy gain. The circular booster ring raises the energy even further before the electrons are passed to the storage ring. At each bending magnet in the storage ring, synchrotron radiation is produced along the tangent into beamlines. Synchrotron radiation is characterised by its extreme brightness, wide spectrum and narrow spread. Laser light has a narrower spread.

28 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

• • •

Colliders have played a central role in detecting and measuring subatomic particles that support the Standard Model of particle physics. The LHC has several ways of detecting the particles produced in collisions. These include ATLAS, CMS, ALICE and LHCb, which have different purposes. The need to store massive amounts of data produced by the LHC, the software to analyse it and the networks to share it has had an impact of the development of information processing technologies.

Resources

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0044).

19.5.2 Key terms A beamline is the line along which the synchrotron radiation passes to reach the target. The booster ring is the part of a synchrotron that takes the already fast electrons from the linear accelerator and increases their speed and energy before injecting them into the storage ring. A cathode ray is a narrow beam of electrons emitted from a hot filament through which an electric current flows. A cyclotron was an early design for a particle accelerator that was replaced when speeds close to the speed of light were desired. A mass spectrometer uses a magnetic field to separate ions by their charge-to-mass ratio. It gives information about the presence or absence of particular ions and also their proportion. A monochromators is a device that allows radiation of only one frequency through. A nanocrystal is a very small crystal with only a few hundred to a thousand atoms. A particle accelerator is a large expensive machine that accelerates charged particles such as electrons, protons and atomic nuclei to speeds close to the speed of light, and aims them at a target to gain some understanding about the target or the ‘bullet’ itself. A photon is a discrete bundle of electromagnetic radiation. It can be thought of as a discrete packet of light energy with zero mass and zero electric charge. A proton synchrotron is designed to run with positively charged particles such as the proton. The storage ring of a synchrotron is where the electrons, now at maximum possible speed, produce radiation for use in the beamlines. A synchrotron is a particle accelerator in which the final path of the particle is a circle of constant radius. Synchrotron radiation is the electromagnetic radiation produced when electric charges are accelerated. X-rays are electromagnetic waves of very high frequency and very short wavelength.

Resources Digital document Key terms glossary (doc-33008)

19.5 Exercises To answer questions online and to receive immediate feedback and sample responses for every question, go to your learnON title at www.jacplus.com.au.

TOPIC 19 How do particle accelerators work? 29

19.5 Exercise 1: Multiple choice questions 1.

2.

3.

4.

5.

6.

7.

8.

What is a particle accelerator used for? A. Accelerating particles other than subatomic particles to very high velocities by means of electric or electromagnetic fields B. Accelerating particles other than subatomic particles to very high velocities by means of electric or electromagnetic fields C. Accelerating subatomic particles to very high velocities by means of electric or electromagnetic fields D. Accelerating subatomic particles to very high velocities by means of electric and electromagnetic fields Which statement is correct about the Australian synchrotron? A. It accelerates charged particles to very high velocities, to produce intense and narrow beams of electromagnetic radiation. B. It accelerates charged particles to velocities less than the speed of light, to produce intense and narrow beams of electromagnetic radiation. C. It accelerates charged particles to very high velocities and then let them collide. D. It accelerates charged particles to velocities less than speed of light and then let them collide. Where is synchrotron radiation produced? A. All the parts of synchrotron B. Storage ring C. Booster ring D. Beamlines If a fast-moving charge is moving in a circular path within a magnetic field, how is radiation emitted? A. Along the circular path B. Uniformly in all directions C. Along the tangent to the circle D. None of the above Which of the following is true in relation to the first particle accelerators? A. They were electromagnetic accelerators. B. They were electrostatic accelerators. C. They were either electromagnetic or electrostatic accelerators. D. They were neither electromagnetic nor electrostatic accelerators. What is the purpose of beamlines? A. To generate electrons B. To bend the electrons into a circular path C. To accelerate charged particle to the speed of light D. To bring the X-ray beam to the separate experimental stations Why is the synchrotron a more powerful tool than laser? A. Synchrotron produces an extremely intense, continuous and very narrow beam. B. Synchrotron produces an extremely intense, single frequency and very narrow beam. C. Synchrotron produces an extremely intense, continuous and wide beam. D. Synchrotron produces an extremely intense, single frequency and wide beam. What is the only particle that cannot be detected by ATLAS? A. Protons B. Neutrons C. Neutrinos D. Electrons

30 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Which of these is not a use of particle accelerators? A. Particle physics experiments B. A new type of weaponry C. Particle therapy to treat cancer D. Condensed matter physics 10. ALICE stands for ‘A Large Ion Collider Experiment’. It is designed to analyse collisions between which of the following? A. Potassium nuclei B. Carbon nuclei C. Uranium nuclei D. Lead nuclei 9.

19.5 Exercise 2: Short answer questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

What components do the LHC and the Australian Synchrotron have in common? Why is vacuum required in the particle accelerators? In the storage ring of a synchrotron, energy is lost by electrons in the form of synchrotron radiation. How is this lost energy replaced? What components do each of the LHC and the Australian Synchrotron have that the other does not? What is the benefit of using X-ray synchrotron over a laser source? What is the main difference between a cyclotron and a synchrotron? What is the significance of the word ‘hadron’ in the title of the LHC? What is the purpose of a wiggler and an undulator magnet? What does ALICE stand for and what is its specific purpose? What are some of the areas where technologies developed at CERN had an impact across society?

19.5 Exercise 3: Exam practice questions Question 1 (2 marks) What is the main difference in the purpose of the LHC and the Australian Synchrotron? Question 2 (4 marks) What is the benefit of using X-ray synchrotron over conventional X-ray sources? Question 3 (6 marks) List three main characteristics of synchrotron radiation that allow it to be a powerful tool for investigating the fine structure of materials. Question 4 (4 marks) Describe the purpose of each of the following components of the Australian Synchrotron: linac, circular booster, storage ring and beamlines. Question 5 (3 marks) Why did it take physicists so many years to discover the Higgs boson?

19.5 Exercise 4: studyON topic test Fully worked solutions and sample responses are available in your digital formats.

Test maker Create unique tests and exams from our extensive range of questions, including practice exam questions. Access the assignments section in learnON to begin creating and assigning assessments to students.

TOPIC 19 How do particle accelerators work? 31

AREA OF STUDY 2 OPTIONS OBSERVATION OF THE PHYSICAL WORLD

20

How can human vision be enhanced? 20.1 Overview Numerous videos and interactivities are available just where you need them, at the Invepoint of learning, in your digital formats, learnOn and eBookPLUS at www.jacplus.com.au.

20.1.1 Introduction Physics is the field of science that is based on making sense of the physical universe. To do this we use our senses, looking for patterns and connections to evaluate the data they provide. Our sense of sight is one of our most significant. Although it detects only a small part of the electromagnetic spectrum, it provides us with vast amounts of useful information. Our ability to enhance human vision is incredible — glasses and bionic eyes have been designed to treat vision impairment and blindness, and lenses in telescopes and microscopes allow us to see very small and very distant objects. We will explore these amazing innovations further in this topic. FIGURE 20.1 Sight is one of the senses we heavily rely on for collecting data and making observations. Consider how vision can be extended to gather even more information.

TOPIC 20 How can human vision be enhanced? 1

20.1.2 What you will learn KEY KNOWLEDGE After completing this topic you will be able to: Behaviour of light • identify that light travels in straight lines in a uniform medium • investigate and apply theoretically and practically the two laws of reflection at a plane surface: • the angle of incidence is equal to the angle of reflection • the incident ray, reflected ray and the normal at the point of incidence are coplanar • investigate theoretically and practically refraction using Snell’s Law, n1 sin(𝜃 1 ) = n2 sin(𝜃 2 ) Manipulating light for a purpose • describe image formation using pinhole cameras and convex and concave lenses • calculate image positions for thin lenses using either accurate ray tracing scale diagrams and/or the thin 1 1 1 lens equation: = + f u v u • calculate image sizes in pinhole and simple lens cameras: M = − v • explain the operation of simple two-lens telescopes and microscopes Light and the eye • model and explain human vision as refraction at a spherical surface with an adjusting lens • distinguish between short-sightedness and long-sightedness, and explain their correction by concave and convex lenses, respectively 1 • apply the power of a lens: P = to eye glasses f • explain accommodation in the human eye including the effects of ageing • investigate and explain the treatment of cataract blindness including the use of intraocular lenses • investigate the operation of the bionic eye. Source: VCE Physics Study Design (2016–2021) extracts © VCAA; reproduced by permission.

PRACTICAL WORK AND INVESTIGATIONS Practical work is a central component of learning and assessment. Experiments and investigations, supported by a Practical investigation logbook and Teacher-led videos, are included in this topic to provide opportunities to undertake investigations and communicate findings.

Resources Digital documents Key science skills — VEC Units 1–4 (doc-31856) Key terms glossary (doc-33002) Practical investigation logbook (doc-33003)

To access key concept summaries and practice exam questions download and print the studyON: Revision and practice exam question booklet (sonr-0045).

2 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

20.2 Behaviour of light KEY CONCEPTS • Identify that light travels in straight lines in a uniform medium. • Investigate and apply theoretically and practically the two laws of reflection at a plane surface: • the angle of incidence is equal to the angle of reflection • the incident ray, reflected ray and the normal at the point of incidence are coplanar.

20.2.1 Light travels in straight lines In this topic we will discuss the ray model for light. The ray model makes use of the fact that light travels in straight lines in uniform media. Rays are arrows drawn from a light source, as shown in figure 20.2. The direction of light moving out from its source is called the direction of propagation. Objects that emit light are called luminous objects. They include the Sun, the stars, light globes, computer screens (when switched on), lit candles and glow worms. Light propagates away from luminous objects. Luminous objects shine even when there is no other light source. Luminous objects that produce light as a result of being hot are described as incandescent. All other objects are described as non-luminous. They need a luminous object to shine light on them for us to be able to see them. They do not produce light of their own. Nonluminous objects include trees, tables and the Moon. It turns out that all objects are luminous to some extent, but ones we call non-luminous are radiating light that is invisible to the unaided eye. We know that very hot objects tend to glow, for example, a hot coal in a fire. What we see depends on the object’s temperature. When the coal cools down, it is still radiating but the light it emits has a longer wavelength than the human eye can see. We call this radiation infra-red, microwave or radio wave radiation, depending on how cool the object is.

FIGURE 20.2 Rays are straight lines indicating light propagating from a light source. The further they are apart, the dimmer the light.

Light bulb

FIGURE 20.3 The Sun is luminous; clouds, rocks, trees and water are all non-luminous.

TOPIC 20 How can human vision be enhanced? 3

Devices such as night-vision goggles can be used to detect some of these wavelengths that our unaided eyes cannot and turn them into wavelengths on a screen that are visible to the human eye. Night-vision devices are used by soldiers and journalists, in wildlife observation, navigation and surveillance and are useful in firefighting to find the hotspots behind the smoke.

FIGURE 20.4 An image seen through a night-vision device

Resources Digital documents Investigation 20.1 Luminous or not? (doc-32327) Investigation 20.2 The Galilean telescope (doc-31901)

20.2.2 Objects and images We can see luminous objects because the FIGURE 20.5 Light radiating from a flame with some light from them directly enters our eyes. entering an eye We draw rays from the luminous object to the retinas of our eyes, through the iris. Notice that in figure 20.5 rays of light leave the flame in all directions. Only some are drawn. Our eyes only see the rays that enter the eyes. All of the other rays play no part in our perception of the flame. Eye Some terminology will help us here and for the rest of this topic. In this example, the flame is called the object. It is the source of Object (candle) the rays. Optical devices such as our eyes collect the rays that enter them to form an image of the object. Without these images, we would not be able to see. Eyes do not always successfully form clear images. Many people cannot form clear images of letters on the page in front of them. Reading glasses help some people form clear images of the letters they read.

20.2.3 Looking in the mirror Most of what we see depends significantly on the phenomenon of reflection. We can see the light emitted by luminous objects, but we can see everything else around us because light reflects from non-luminous objects into our eyes. In the previous section we noted that rays of light radiate in all directions from a luminous object such as a flame. What happens to those other rays that do not enter the eye? Some of them might hit a book that is lying on the table beside the flame. When they hit the book, two things can happen: the ray can be absorbed or the ray can be reflected. 4 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Different materials absorb different amounts of light. Compare the sunlight striking snow and soil. The snow absorbs very little light but the soil absorbs a lot. This means that most of the light hitting the snow is reflected, but only a little of the light hitting the soil is reflected. The snow and soil have irregular surfaces, so the reflected light leaves them in many different directions. Any ray that reflects so that it enters your eye will help form an image of the snow or the dirt. Many more rays come from the snow, so it is very bright, whereas the soil has absorbed a lot of the light, so it appears dark. The book next to the candle is visible to us because some of the light that hits it reflects into our eyes, which use that light to form an image of the book (shown in figure 20.6). This is how we see all nonluminous objects. This type of reflection, in which the directions of the reflected rays appear random, is called diffuse reflection. The directions of rays in diffuse reflection are not really random; they are simply a result of the rays striking an irregular surface. Regular reflection, also referred to as specular reflection, is reflection from a smooth surface. The more polished a surface is, the less we see the surface itself and the more we see the reflection of other objects in the surface (for example, in a mirror). FIGURE 20.6 Light rays from a luminous object hitting a book and reflecting into an eye

Eye

Candle

Book

The ray model can explain diffuse reflection. On a microscopic scale, the surfaces of most objects are not flat or smooth. Rays hitting the surface only a tiny distance apart will therefore be reflected at different angles. Although the law of reflection is satisfied on the microscopic scale, the overall effect of light hitting that surface is for it to be scattered in all directions. This scattering of the light in random directions is called diffusion, hence, the term diffuse reflection (figure 20.7). FIGURE 20.7 This is diffuse reflection. Each of the incoming parallel rays meets the irregular surface at a different angle of incidence. The reflected rays will therefore go off in different directions, enabling observers in all directions to receive light from the surface; in other words, to see the surface.

Observer A

Observer B

Irregular surface

TOPIC 20 How can human vision be enhanced? 5

The law of reflection The law of reflection governs the reflection of light rays off smooth conducting surfaces, such as polished metal or glass mirrors. Consider the incident of a light ray on a plane mirror, as shown in figure 20.8. The ray of light approaching the surface is called the incident ray and the ray of light reflecting from the surface is called the reflected ray. The normal (N) is an imaginary line drawn perpendicular to the surface of the mirror. The angle that the incident ray makes with the normal to the surface is called the angle of incidence (i). The angle that the reflected ray makes with the normal to the surface is called the angle of reflection (r). (Note: The surface is not necessarily flat, so the angle with the surface can be different depending on the direction of the ray.) The law of reflection states that the angle of incidence (i) equals to angle of reflection (r). The incident ray, the reflected ray and the normal all lie in the same plane. FIGURE 20.8 Diagram showing incident and reflected rays for a plane surface Normal (N)

Incident ray

Reflected ray Angle of incidence

Angle of reflection

i

r

Mirror

Forming images in a mirror In flat, or plane mirrors, the image is virtual, upright, the same distance behind the mirror as the object is in front of the mirror and the same size as the object. We describe this image as having a magnification of 1 (it is the same size as the object) and it is upright. These images are also laterally inverted, which means that the left-hand side of the object is the right-hand side of the image, but the image is facing the object. So, if you wear a watch on your left hand (the object), the image will have the watch on its right hand. This is illustrated using the simple ray diagram in figure 20.9. FIGURE 20.9 In flat, or plane mirrors, images have a magnification of 1 and they are laterally inverted.

FIGURE 20.10 A ray diagram of image formation in a flat, or plane, mirror C N1

P

O

P

D

A

M

I 6 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

N2

B

M1

To understand how this happens, consider figure 20.10. Two rays emerge from point O (object), strike the mirror (M) and reflect into the observer’s eye. We use the law of reflection to construct the reflected rays. If the reflected rays are extended backward behind the mirror (shown by the dashed lines, as rays are imaginary not real light rays), they seem to originate from point I (image). This is where the image of point O is located. By forming images of all points of the object, we obtain an upright image of the object behind the mirror. The reflected rays appear to the observer to come directly from the image behind the mirror; in reality, these rays come from the points on the mirror where they are reflected. The image behind the mirror is called a virtual image because it cannot be projected onto a screen — the rays only appear to come from a common point behind the mirror, so that is where the virtual image is located (figure 20.11).

FIGURE 20.11 The image of your face in the mirror is an optical illusion caused by the reflection of light in the mirror. It is called a virtual image.

SAMPLE PROBLEM 1

Calculate the angles a, b and c in the following figure.

Mirror c°

50° a°

b°

THINK

WRITE

The angle of incidence (a) is measured from the normal line. 2. The law of reflection states that the angle of incidence (i) is equal to angle of reflection (r).

a = 90° – 50° = 40°

1.

The angle between the normal and the mirror is 90°. 4. State the answer. 3.

According to the law of reflection, the angle of incidence (i) is equal to angle of reflection (r). Therefore, b = 40°. c = 90° – b = 90° – 40° = 50° a = b = 40° and c = 50°

TOPIC 20 How can human vision be enhanced? 7

PRACTICE PROBLEM 1 A ray of light strikes the surface of a plane mirror so that the angle between it and the mirror is 35°. a. Determine: i. the angle of incidence ii. the angle of reflection. b. What do the incident ray, the normal and the reflected ray all have in common (other than being straight lines)?

Reflection in curved mirrors Images are not always the same size as their objects. FIGURE 20.12 In a car headlight, the light The effect of an optical device on the size of the image source (a globe or LED) is placed at the focus of is indicated by the magnification (discussed later in the curved reflective surface. This produces a section 20.4.1). As the image in a plane mirror is the narrow beam of light. same height as the object, the magnification is 1. In a device such as the bottom of a spoon, where the height of the image is smaller than the height of the object, the magnification is between 0 and 1. This is known as a diminished image. When the magnification is greater than 1, the image is said to be enlarged. If you look at the reflection of your eye in the concave (curved inwards) side of a polished spoon, with your eye very close to the spoon, you may see an enlarged image of your eye. Sometimes mirrors are used for purposes that require them to be curved. For example, a car headlight has a curved mirror behind its bulb (figure 20.12). This mirror collects the rays that would otherwise illuminate objects behind the light and reflects them forward. A plane mirror behind the light would cause the light to reflect in a forward direction but diverge (figure 20.13a). The curved mirror results in all of the rays reflecting in the forward direction without diverging, to create an intense beam of light (figure 20.13b). Mirrors curved like this are called concave mirrors. The edge of a concave mirror is closer to the object than the centre (figure 20.13c). Such mirrors are also used in many forms of telescope. FIGURE 20.13 (a) Rays from a light source in front of a plane mirror are reflected in all directions. (b) The curved circular mirror brings the reflected rays inwards. (c) When the curved mirror is in the shape of a parabola, the reflected rays become parallel to each other and the light source is said to be at the focus of the curved mirror. (a) Plane mirror

(b) Circular concave mirror

(c) Parabolic concave mirror

Convex mirrors are like the bottom side of a spoon: the middle is closer to the object than the edge. These mirrors can be used to help see around corners in driveways and small streets by producing a wider field of view (figure 20.14). The magnification of convex mirrors is always less than one, and the image is virtual.

8 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 20.14 Convex mirrors are often used to help improve vision around corners.

Resources Digital documents Model of a concave mirror (doc-0055) Investigation 20.3 Luminosity and temperature (doc-32329) Weblink

Concave mirror applet

20.2 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. Describe the light path from a light source to your eye in seeing an object. 2. Use the ray model and the sources of light to rephrase the statements (a) ‘I looked at a flower through the window’ and (b) ‘I watched the TV’. 3. Explain how early astronomers knew the Moon must have a rough surface. 4. Copy the following figure and draw the incident and reflected rays from the two ends of the object to the eye. Locate the image. Object

Plane mirror

TOPIC 20 How can human vision be enhanced? 9

5. A horizontal light ray strikes a mirror such that the reflected ray is directed at a 60° angle with respect to the horizontal as show in the following figure. Find the value of the angle, a, that the mirror makes with the horizontal?

60°

a°

6. The two arrowed lines in the following figures represent reflected rays. The line AB represents the plane mirror. Locate the image and the light source in each of the two figures. A A

B

B

7. A student argues that you cannot photograph a virtual image because light rays do not pass through the space where the image is formed. How would you argue against this statement? 8. Sketch the path of each of the rays entering each of the pair of joined mirrors in the following figure.

9. You are walking towards a plane mirror at a speed of 1 m s−1 . How fast is your image walking? How quickly are you and your image approaching each other? 10. (a) You are standing 2 metres in front of a plane mirror and you wish to take a sharp photograph of yourself in the mirror. At what distance do you set the camera lens? (b) Your friend is standing beside you, 1 metre away. At what distance do you set the camera lens for a sharp photograph of your friend?

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

10 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

20.3 Refraction KEY CONCEPT • Investigate theoretically and practically refraction using Snell’s Law, n1 sin(𝜃 1 ) = n2 sin(𝜃 2 ).

20.3.1 The behaviour of light when changing media So far we have considered opaque objects such as books, FIGURE 20.15 An example of refraction heads and mirrors. Light does not pass through these objects. Materials that light passes through are called transparent and translucent materials. Materials that are transparent include clear glass. The light passes through undeviated so that you can see the objects on the other side clearly. Translucent materials are used when you want some light but you do not want to be able to see through clearly, for example, glass in a bathroom window. The glass is etched so that the surface is not smooth, or it has some other material embedded in it that diffuses the light. Light that strikes the surface of transparent glass is partially reflected from the material and is partially transmitted through the material. When light moves from one medium to another (in this case from air into glass), the speed of the light tends to change. If the light strikes the surface at an angle other than 90 degrees, this usually results in bending of the light. This bending is called refraction. (The occasions when it does not bend are when the two materials are such that they convey light at the same speed.) You may have noticed some of the effects of bending of light. Hunters spearing for fish in the shallows must aim the spear below where the fish appears to be in the water. At the beach or in a pool, people standing in the shallows appear to have shorter legs. Our perception is distorted by light travelling from one medium into another. This may be more apparent when we set up a straight rod in a beaker of liquids that do not mix, as seen in figure 20.15. Each time the medium changes, the light changes direction, distorting our vision. This change in the direction of the light is refraction.

Resources Digital document Investigation 20.4 Seeing is believing (doc-32331)

20.3.2 Bending of light The ray model can help explain these observations, as shown in figure 20.16. If we try to spear a fish and the fish swims safely beneath where the spear passes, it is because a virtual image of the fish has formed above where the fish actually was. The rays of light from the fish that enter our eyes are bent when they leave the water. To our eyes, the rays seem to be coming from another direction. Given that light can travel both ways along a light path, the fish will see spear throwers stand taller above it than they in reality are. This effect was well known to Indigenous Australians in their traditional hunting methods, in which spears are used to hunt fish and sea life.

TOPIC 20 How can human vision be enhanced? 11

FIGURE 20.16 The rays from the fish bend when they enter the air. To the eye, the rays appear to come from a point closer to the surface.

Air water

Fish appears to be here. Fish is here.

The ray model not only gives us a way of describing our observations of the bending of light; it also helps us to take measurements. The angle that a ray of light makes with the normal (either the angle of incidence or the angle of refraction) can be measured and investigated.

20.3.3 Snell’s Law In 1621, the Dutch physicist Willebrand Snell investigated the refraction of light and found that the ratio of the sines of the angles of incidence (i) and angle of refraction (r) was constant for all angles of incidence. Figure 20.17 shows how an incident ray is affected when it meets the boundary between air and water. The normal is the line at right angles to the boundary, and all angles are measured from the normal. Some of the light from the incident ray is reflected back into air. The rest is transmitted into the water. The following ratio is a constant for all angles for light travelling from air to water: sin i = constant sin r

FIGURE 20.17 A graphical representation of Snell’s Law for any two substances. 𝜃 2 is the angle of refraction in medium 2. Angle of incidence

Normal

Angle of reflection Reflected ray

Incident ray

θi

θi

Boundary

n1 n2 θr

Angle of refraction

12 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Refracted ray

Snell repeated his experiments with different substances and found that the ratio was still constant, but it had a different value for different substances. This suggested that different substances bend light by different amounts. (Remember that some light is always reflected.) In fact, there is a different ratio for each pair of substances (for example, air and glass or air and water). A different ratio is obtained for light travelling from air into glass than for light travelling from air into water. The value of the ratio is called the relative refractive index, because it depends on the properties of two different substances. The refractive index is a measure of the bending of a ray of light as it passes from one medium to another. It is not possible to find the effect of a particular substance on the deflection of light without adopting one substance as a reference standard. Once you have a standard, every substance can be compared with it. A natural standard is a vacuum, the absence of any substance. The absolute refractive index of a vacuum is given the value of 1. From this, the absolute refractive index of all other substances can be determined. Some examples are given in table 20.1. (The word ‘absolute’ is commonly omitted and the term ‘refractive index’ usually refers to the absolute refractive index.) A medium that produces a large amount of the light path is said to have a large refractive index. TABLE 20.1 Values for absolute refractive index Material

Value

Vacuum

1.0 000

Air at 20 °C and normal atmospheric pressure

1.00 028

Water

1.33

Perspex

1.49

Quartz

1.46

Crown glass

1.52

Flint glass

1.65

Carbon disulfide

1.63

Diamond

2.42

The refractive index is given the symbol n because it is a pure number without any units. This enables a more useful restatement of Snell’s Law: nair sin 𝜃air = nwater sin 𝜃water Where: nair = refractive index of air nwater = refractive index of water 𝜃air = angle of incidence in air 𝜃 water = angle of incidence in water. More generally this would be expressed as: n1 sin 𝜃1 = n2 sin 𝜃2 Where: n1 = refractive index of incident medium n2 = refractive index of the refractive medium 𝜃1 = angle of incidence in medium 1 𝜃2 = angle of refraction in medium 2.

TOPIC 20 How can human vision be enhanced? 13

As with all formulas, it is important to enter the data carefully into Snell’s Law. The refractive index inserted for n1 must be the refractive index of the medium in which the light makes an angle of 𝜃 1 with the normal. Similarly, the refractive index inserted for n2 must be the refractive index of the medium in which the light makes an angle of 𝜃2 with the normal. SAMPLE PROBLEM 2

A ray of light strikes a glass block of refractive index 1.45 at an angle of incidence of 30°. What is the angle of refraction? THINK

WRITE

State the known and unknown quantities. 2. Recall Snell’s Law.

nair = 1.0, 𝜃air = 30°, nglass = 1.45, 𝜃glass = ? n1 sin 𝜃1 = n2 sin 𝜃2 nair sin 𝜃air = nglass sin 𝜃glass 1.0 × sin 30° = 1.45 × sin 𝜃glass sin 30° = 0.3448 sin 𝜃glass = 1.45 𝜃glass = sin−1 (0.3448) = 20° The angle of refraction is 20°.

1.

3.

Substitute known values in and solve for 𝜃glass .

4.

State the answer.

PRACTICE PROBLEM 2 A ray of light enters a plastic block at an angle of incidence of 40°. The angle of refraction is 30°. What is the refractive index of the plastic?

GRAVITATIONAL FIELDS AND BENDING LIGHT Light can be bent by a strong gravitational field, such as that near the Sun. The gravitational field can act like a convex lens. Light from a distant star that is behind and blocked by the Sun bends around the Sun so that astronomers on Earth see an image of the star to the side of the Sun.

Resources Video eLesson Refraction and Snell’s Law (eles-0037) Interactivity

Refraction and Snell’s Law (int-0056)

20.3.4 The speed of light in glass The speed of a wave depends on the properties of the medium through which it travels. During the seventeenth century, one model of light proposed that light travelled faster in glass. Another model of light proposed that light travelled slower in glass. Scientists at the time did not have the technology to measure the speed of light through materials such as water or glass. It was only in the nineteenth century that Augustin-Jean Fresnel and Jean Bernard Léon Foucault were able to measure the speed of light in water as being less than the speed in air. This helped to improve understanding of the nature of light. This gave the refractive index a physical meaning:

14 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

absolute refractive index of a medium =

speed of light in a vacuum speed of light in a medium

where the speed of light in a vacuum = 3.0 × 108 m s–1 . As light passes through the interface between two mediums of different refractive indices, it bends. The medium with greater refractive index is called the denser medium and the one with smaller refractive index is called the rarer medium. In figure 20.18a, the path of a light ray is bent towards the normal when the ray enters a medium with a refractive index higher (water) than the one from which it emerges (air). This is because the light travels slower in the medium with a higher refractive index. In figure 20.18b, the light ray bends away from the normal when the ray enters a medium with a refractive index lower (air) than the one from which it emerges (water). This is because the light travels faster in the medium with a lower refractive index. This property is very helpful in designing optical lenses as it allows the calculation of the bend angle of the beam of light when it passes from one transparent medium to another. FIGURE 20.18 A graphical depiction of Snell’s Law for any two substances (a)

(n1 < n2)

Rarer (n1)

(b)

Normal

Rarer

Denser (n2)

(n1 > n2)

Denser (n1)

i

Normal

i

r

(n2)

r

ri

Dispersion White light is split up into different colours as it passes through a triangular prism due to refraction. This phenomenon is known as dispersion (see figure 20.19). These colours are generally described as red, orange, yellow, green, blue, indigo and violet. The refractive index varies with the wavelength of light. Red light has a longer wavelength than violet light and therefore red light bends least as it passes through a refractive medium. This means that the red light is travelling faster than the violet light through the medium and therefore has a smaller refractive index than violet. It is important to use monochromatic light to prevent dispersion of light into different colours. The refractive index also depends on the temperature of a medium. A medium with a higher temperature means the liquid becomes less dense and less viscous, causing light to travel faster within the medium. This results in a smaller value for the refractive index due to a smaller

FIGURE 20.19 Dispersion

TOPIC 20 How can human vision be enhanced? 15

ratio. A lower temperature means the liquid becomes denser and has a higher viscosity, causing light to travel slower in the medium. This results in a larger value for the refractive index due to a larger ratio. SAMPLE PROBLEM 3

The refractive index of glass is 1.5. How fast does light travel in glass? THINK

WRITE

1.

State the known and unknown quantities.

2.

Recall the refractive index of a medium.

nglass = 1.5, speed of light in a vacuum = 3.0× 108 , speed of light in glass =? absolute refractive index of a medium speed of light in a vacuum speed of light in a medium 3.0 × 108 1.5 = speed of light in glass =

3.0 × 108 1.5 = 2.0 × 108 m s−1 The speed of light in glass is 2.0 × 108 m s−1 .

speed of light in glass = 3

State the answer.

PRACTICE PROBLEM 3 How fast does light travel in diamond?

APPARENT DEPTH As described earlier in this topic, spear throwers need to aim below a fish if they are to have a chance of spearing the fish. A similar phenomenon occurs when a spear thrower is directly above a fish. The fish appears to be closer to the surface than it actually is. This observation is known as apparent depth. Swimming pools provide another example of apparent depth: they look shallower than they actually are. The refraction of light combined with our two-eyed vision makes the pool appear shallower. The relationship is illustrated in figure 20.20 and can be expressed as: real depth = refractive index apparent depth FIGURE 20.20 The phenomenon of apparent depth

Apparent depth

Real depth

16 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Air water

Resources Digital document Investigation 20.5 Using apparent depth to determine the refractive index (doc-31904) Teacher-led video Investigation 20.5 Using apparent depth to determine the refractive index (tlvd-1075)

20.3.5 Total internal reflection and critical angle Light can play some strange tricks. Many of these involve refraction away from the normal and the effect on light of a large increase in the angle of incidence. It has already been mentioned that some light is reflected off a transparent surface, while the rest is transmitted into the next medium. This applies whether the refracted ray is bent towards or away from the normal. When light passes from a denser medium (such as water) to a rarer medium (air), the refracted ray is bent away from the normal (as illustrated in figure 20.21a). As the angle of incidence increases, the angle of refraction also increases. Eventually the refracted ray becomes parallel to the surface and the angle of refraction reaches a maximum value of 90° (see figure 20.21b). The corresponding angle of incidence is called the critical angle. If the angle of incidence is increased beyond the critical angle, all the light is reflected back into the water and no refraction takes place. This phenomenon is called total internal reflection (see figure 20.21c). As all angles are less than the critical angle (i), both refraction and reflection occur in varying proportions. FIGURE 20.21 Three stages of refraction leading to total internal reflection (a) Before critical angle

(b) At critical angle

(c) After critical angle (total internal reflection)

θr Air water θr θr

θc

Total internal reflection is a relatively common atmospheric phenomenon (as in mirages) and it has technological uses (for example, in optical fibres). Total internal reflection, together with a large index of refraction, is why diamonds sparkle more than other materials. The critical angle for a diamond-to-air surface is only 24.4°, so when light enters a diamond it can exit only if it makes an angle of less than 24.4° with the normal. Facets (flat surfaces) on diamonds are specifically intended to make this unlikely, so that the light can exit only in certain places. Good quality diamonds are very clear so that the light makes many internal reflections and is concentrated at the few places it can exit, which causes it to sparkle. Zircon is a natural gemstone that has a large index

θc

FIGURE 20.22 Total internal reflection, together with a large index of refraction, explains why diamonds sparkle more than other materials.

White light

Total internal reflection – all incident angles > 24.4°

Incident angles < 24.4° light is refracted

Total internal reflection – all incident angles > 24.4°

TOPIC 20 How can human vision be enhanced? 17

of refraction, but not as large as diamond so it is not as highly prized. Cubic zirconia is manufactured and has a refraction index of approximately 2.17. The colours emerging from a sparkling diamond are due to dispersion. A mirage creates the illusion of water or an inverted reflection as a result of the refraction of light through a non-uniform medium. Consider figure 20.23. The air near the surface of the road is hot and much cooler higher up. Light travels faster through the thin hotter air than it does through the denser cooler air. Therefore, the light travels in a curved line instead of travelling in a straight path —the effect on the road we observe is a reflection of the sky. As the light has different speeds in different media, the light bends due to refraction. FIGURE 20.23 (a) Atmospheric refraction when air near the surface of the road is hot and cooler above (b) A mirage when driving along a hot road (c) A mirage when traveling in dessert (a)

(b) Cool air (denser) Warm air (rarer)

Hot surface

(c)

OPTICAL FIBRES An optical fibre is a thin, transparent fibre, usually made of glass or plastic, for transmitting light. Fibres in bundles (figure 20.24a) are clad by a material that has a lower index of refraction than the core to ensure total internal reflection, even when fibres are in contact with one another. The structure of a single optical fibre is shown in figure 20.24b. An optical fibre consists of an outer protective material called cladding and an inner part called the core, through which light travels. The difference in the refractive index of the cladding and the core allows total internal reflection in the same way as happens at an air–water surface, as discussed earlier. If light is incident on a cable end with an angle of incidence greater than the critical angle, then the light will remain trapped inside the glass strand. In this way, light propagates very fast down the cable over a very long distance (tens of kilometres). The cladding prevents light from escaping out of the fibre, minimising the loss of signal and ensuring that a quality image is formed at the other end. Most telephone conversations and internet communications are now carried by laser signals along optical fibres, because information can be transported over long distances, with minimal loss of data.

18 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 20.24 (a) A bundle of optical fibres (b) A single fibre with its cladding (a)

(b) Total internal reflection

Cladding has lower refractive index

Core has higher refractive index Source: (b) Woodford, Chris. 2006/2019 Fiber optics. Retrieved from https://www.explainthatstuff.com/fiberoptics.html. [Accessed 14 November 2019]

Another common use can be found in medicine in endoscopes. Endoscopes are used to explore the body through various orifices or minor incisions. Light is propagated down one fibre bundle to illuminate internal parts, and the reflected light is transmitted back out through another to be observed. Arthroscopic surgery on the knee joint can be performed, using cutting tools attached to and observed with the endoscope. Optical fibres have made microsurgery and remote surgery possible where the incisions are small, and the surgeon’s fingers do not need to touch the diseased tissue.

SAMPLE PROBLEM 4

What is the critical angle for water given that the refractive index of water is 1.3? THINK

WRITE

1.

State the known and unknown quantities.

nair = 1.0, 𝜃air = 90°, nwater = 1.3, 𝜃water = ?

2.

Recall Snell’s Law.

n1 sin 𝜃1 = n2 sin 𝜃2 nair sin 𝜃air = nwater sin 𝜃water

3.

Solve for 𝜃water .

4.

State the answer.

1.0 × sin 90° = 1.3 × sin 𝜃water sin 90 = 0.7692 sin 𝜃water = 1.3 𝜃water = sin–1 (0.7692) = 50° The critical angle is 50°.

PRACTICE PROBLEM 4 A glass fibre has a refractive index of x and its cladding has a refractive index of y. What is the critical angle in the fibre in terms of x and y?

TOPIC 20 How can human vision be enhanced? 19

20.3 EXERCISE To answer questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. 1. What is the angle of refraction in water (n = 1.33) for an angle of incidence of 40°? If the angle of incidence is increased by 10°, by how much does the angle of refraction increase? 2. A ray of light enters a plastic block at an angle of incidence of 55° with an angle of refraction of 33°. What is the refractive index of the plastic? 3. A ray of light passes through a rectangular glass block with a refractive index of 1.55. The angle of incidence as the ray enters the block is 65°. Calculate the angle of refraction at the first face of the block, then calculate the angle of refraction as the ray emerges on the other side of the block. Comment on your answers. 4. Immiscible liquids are liquids that do not mix. Immiscible liquids will settle on top of each other, in the order of their density, with the densest liquid at the bottom. Some immiscible liquids are also transparent. (a) Calculate the angles of refraction as a ray passes down through immiscible layers as shown in the following figure.

Light ray

25˚

Air

n = 1.00

Acetone

n = 1.357

Glycerol

n = 1.4 746

Carbon tetrachloride

n = 1.4 601

n = 1.53

Glass beaker

(b) If a plane mirror was placed at the bottom of the beaker, calculate the angles of refraction as the ray reflects back to the surface. Comment on your answers. 5. Light rays are shown passing through boxes in the following figure. Identify the contents of each box from the options (a)–(g) given below. Option (b) is a mirror. All others are solid glass. Note: There are more options than boxes.

(i) (a)

(ii) (b)

(iii) (c)

20 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

(iv) (d)

(v) (e)

(f)

(vi) (g)

6. (a) To appear invisible you need to become transparent. What must your refractive index be if your movement is not to be detected? (b) The retina of your eye is a light-absorbing screen. What does that imply about your own vision if you are to remain invisible? (Hint: If you are invisible all light passes through you.) 7. Calculate the angle of deviation at a glass–air interface for an angle of incidence of 65° and refractive index of glass of 1.55. 8. Calculate the sideways deflection as a ray of light goes through a parallel-sided plastic block (n = 1.4) with sides 5 centimetres apart, as in the following figure.

30°

n = 1.4

5 cm

9. Calculate the angle of deviation as the light ray goes through the triangular prism shown in the following figure. n = 1.5 60°

? 40°

To answer practice exam questions online and to receive immediate feedback and sample responses for every question go to your learnON title at www.jacplus.com.au. studyON: Practice exam questions Fully worked solutions and sample responses are available in your digital formats.

TOPIC 20 How can human vision be enhanced? 21

20.4 Manipulating light for a purpose KEY CONCEPTS • Describe image formation using pinhole cameras and convex and concave lenses. • Calculate image positions for thin lenses using either accurate ray tracing scale diagrams and/or the thin 1 1 1 lens equation: = + . f u v u Calculate image sizes in pinhole and simple lens cameras: M = − . • v

20.4.1 Cameras Pinhole cameras The earliest cameras did not have lenses, just a pinhole for light to pass through onto a screen. Pinhole cameras help us to understand how modern cameras work; this knowledge also applies to the human eye. A pinhole camera is essentially a box (possibly an entire room) with a small hole made in one side for light to enter. The light passing through the hole shines on a screen on the other side of the box. Photographic film can be placed on the screen to preserve the images formed. Figure 20.25 shows a pinhole camera forming an image of a person. Rays from the top of the person shine in all directions. Because the pinhole is so small, very few rays pass through, but those that do make a small spot of light on the bottom of the screen. The rays from the feet of the person that make it through the pinhole shine on the top of the screen. You can draw rays from every other part of the object through the pinhole to the screen and you will see that an inverted image is formed on the screen. It is a real image because light rays form it, rather than it just being an illusion. The image is there even if we are not looking at it. FIGURE 20.25 (a) Clear upside-down (inverted) image with a small pinhole (b) Fuzzy out-of-focus image with a large pinhole (a)

Image

Object

(b)

Image

22 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Object

The size of this image depends on the distance of the pinhole from the object: the closer the object, the larger the image. The size of the object does not change; the size of the image depends on the distance to the object. It also depends on the size of the box: the longer the distance from the pinhole to the screen, the larger the image. This ability to form a clear image of objects at various distances from the camera is called depth of field. The depth of field is essentially infinite for a pinhole camera; however, in reality none of the images are particularly sharp. Pinhole cameras are a great way for safely viewing the Sun, particularly during an eclipse, as the event can be viewed on the screen using only the small amount of light that has passed through the pinhole being spread out to form an image of low intensity (figure 20.26). Before viewing the Sun, you must be very careful to check that your method is safe. It is very easy to permanently damage your eyes by observing the Sun. FIGURE 20.26 Stand with your back to the Sun and look through the view hole. Once you have lined up the pinhole with the Sun, you will be able to see an image of the Sun reflected from the paper screen at the back of the box. Using shadows of the box can help you line it up. Never turn to line it up by sighting directly.

White paper taped to inside end

Sunlight

Aluminum foil with pinhole

Small image of partially eclipsed sun Source: AAS Sky Publishing

Pinhole cameras have played a very important role in the history of the understanding of light, with Aristotle, Leonardo da Vinci and Johannes Kepler all using them to further their understanding of light and optics. The earliest cameras were pinhole cameras. Sometimes an early camera could be a room called a camera obscura, with the image formed on a table. The image could then be recorded by drawing on paper placed on the table. Only in relatively recent times (since 1850) has a sensitive material been placed on the screen to preserve the image, such as emulsions, film, photographic paper or the image sensors found in digital cameras (charge coupled devices). One of the features of a pinhole camera is that the images formed have fuzzy edges; the images are never in sharp focus. This is because the hole allows rays to enter from a small range of angles. Instead of a point on the object forming a point of light on the image, it forms a small disk. All of these small disks form the image. Making the hole larger has the advantage of forming a brighter image because more light is allowed through, but the image is more blurred. A smaller hole allows less light in, forming a clearer but duller image. If the hole is too small, another effect called diffraction distorts the image. Pinhole cameras have been very useful for exploring the behaviour of light but they have serious limitations, which has led to the exploration of more advanced cameras. TOPIC 20 How can human vision be enhanced? 23

Adding a lens To overcome the limitations another component needs to be added to the pinhole camera. This is a lens. A lens makes use of the refractive properties that we learned about earlier in this topic. What we need a lens to do is to take all of the rays that are coming from one point on the object and direct them to one point on the image. This will produce a sharp image. If we can do that using a larger opening (aperture), then we will also have more light to make a brighter image. To begin to understand a lens, we can start with a rectangular block of glass as in figure 20.27. Parallel rays from the left pass through the lens unaffected if they are normal to the lens (red lines). Parallel rays that are not normal to the lens are slightly displaced due to refraction when passing through the lens, but emerge parallel. This is essentially what happens with light passing through a pane of glass in a window. This would not help the function of our camera apart from keeping dust and rain out. In fact, the glass would absorb a little of the light entering the camera. FIGURE 20.27 Ray diagram of parallel rays normal to the lens (shown in red) and parallel rays not normal to the lens (shown in black) falling on a rectangular block of glass.

If we grind the lens down so that it is now shaped like figure 20.28, refraction converges (bends the rays to meet at a point) the three rays in each case. This works for these three rays from the two angles, but if we grind the lens so that it is a continuous curve in the arc of a circle (or better, a parabola), we find that the convergence occurs for all sets of parallel rays that hit the lens, as shown in figure 20.29. FIGURE 20.28 Refraction converges three rays when a lens is not curved but shaped as shown.

24 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

FIGURE 20.29 Refraction converges all rays when a lens is curved.

Now that we have some idea about how a lens works for parallel rays, we need to think of how that might be useful in a camera. If the object is distant from the camera, the light that enters the lens diverging from one point on the object will be nearly parallel. As an example, light from an object 5 metres from the camera enters a lens with a radius of 1 centimetre. The angle of the light passing through the edge of the lens only differs from a line parallel with light through the centre of the lens by 0.1 degrees. A pinhole camera with a pinhole of 1 centimetre radius and depth of 25 centimetres would produce a disk of light with a 1.05 centimetres radius on the screen for every point of light on the object. This would be very blurred, which highlights why it needs to be a pinhole camera. Placing a lens at the large pinhole that focuses that light onto a point on the screen enables the clear image to be produced, as shown in figure 20.30. FIGURE 20.30 Adding a lens to a camera ensures that the light rays converge to give a clear image, rather than diverging to form a blurry image.

Image disk Object Pinhole camera

Image point Object Lens

Simple cameras, like many small ‘point and shoot cameras and those in mobile phones, use a very small aperture so that they do not need to change focus for each photo taken. They have a large depth of field but do not function well in low-light environments or when the objects are moving quickly. More sophisticated cameras have lenses that can move to focus on near and far objects. This means that the aperture can be larger while still producing a focused image. A large aperture means more light is allowed into the camera, making it possible to take good photos in lower light levels, or to reduce exposure times so that clear images can be taken of moving objects. TOPIC 20 How can human vision be enhanced? 25

We will now look at lenses more generally before exploring how we can apply them to different situations.

Forming images with lenses The refraction of light at a boundary between two transparent media can be put to use if the boundaries are curved. There are two possibilities for curved boundaries — curving inwards or curving outwards. A convex lens has its faces curving outwards. A lens that curves inwards is a concave lens. The simple ray tracings in figure 20.31 illustrate what each lens does to the light rays. As rays enter the glass, they are bent towards the normal. When they reach the air on the other side of the lens, they are bent away from the normal. In the case of the convex lens, the emerging rays converge (come together) at a point called the focus (F). For the concave lens, the rays diverge (move apart) so that they appear to come from a point, also called the focus, on the other side of the lens. For these reasons, convex lenses are sometimes called converging lenses and concave lenses can be called diverging lenses. FIGURE 20.31 Refraction of rays through (a) a convex and (b) a concave lens (a) Convex lens

(b) Concave lens

F

F

In fact, the focus is more than just a point. It is a plane — a focal plane through the focus point. In figure 20.32, for example, parallel light rays from a distant object coming in at an angle to the lens are still brought to a focus, not at the same focus as for light coming in directly, but elsewhere in the focal plane. The distance from the lens to the focal point is called the focal length. The value of the focal length is positive for converging lenses and negative for diverging lenses. FIGURE 20.32 The convex lens has two focal points. Optical axis

Convex lens

Principal axis

Optical center Focus Focus

Focal length (f )

26 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

Convex lenses — locating images using ray tracing The features of a convex lens are illustrated in figure 20.33. The convex lens has two symmetrical curved surfaces, which means that it has two focus points. The straight line that goes through the centre of the lens at right angles to the lens surface is known as the principal axis. The vertical line passing through the centre of a lens and joining the two centres of curvature of its surfaces is known as optical axis of a lens. FIGURE 20.33 Rays converge at a point on the focal plane, which passes through the focus point (F).

Focal plane

F

Principal axis

Ray diagrams can be used to locate an image. The object and the image are usually represented by a vertical line with an arrow. In the case of an image, the arrow can be down if the image is inverted, and up if the image is upright. The three principal rays used to locate and size the image formed by a single lens (convex or concave) are as follows: • Ray 1 parallel to the principal axis passes through the principal focus on the other side of the lens. • Ray 2 passes through the principal focus of a lens and emerges parallel to the principal axis on the other side of the lens. Ray 3 passes through the optical centre of a lens and emerges undeviated on the other side of the lens. • (See table 20.2 for a convex lens and table 20.3 for a concave lens.) The ray diagram in figure 20.34 shows the location of the image of an object placed beyond the principal focus. All three rays converge on the same point after passing through the lens. This is where the image of the head of the object is located. Note that the image could have been located with any two of the rays, but the other can be used to confirm the location of the image.

TOPIC 20 How can human vision be enhanced? 27

TABLE 20.2 The three principal rays used to locate and size the image formed by a convex lens Light ray from object is 1

Ray diagram

How it a appears after refraction

Parallel to the principal axis

After refraction from a convex lens, passes through the principal focus on the other side of the lens

O F1

2

F2

Passing through a principal focus

After refraction from a convex lens, will emerge parallel to the principal axis O F1

3

F2

Passing through the optical centre of a lens

After refraction from a convex lens, will emerge without any deviation O F1

F2

TABLE 20.3 The three principal rays used to locate and size the image formed by a concave lens Light ray from object is 1

Ray diagram

How it a appears after refraction

Parallel to the principal axis

After refraction from a convex lens, the ray appears to diverge from the principal focus located on the same side of the lens

O F1

2

F2

Passing through a principal focus

After refraction from a convex lens, will emerge parallel to the principal axis O F1

3

F2

Passing through the optical centre of a lens

After refraction from a convex lens, will emerge without any deviation O F1

28 Jacaranda Physics 1 VCE Units 1 & 2 Fourth Edition

F2

The features of the image can now be described — its location, size, orientation and nature. There are four aspects of an image which may be determined. • Nature. Is the image real or virtual? • Orientation. Is the image upright or inverted? • Position. Is the image on the same side of the lens to the object or it is on the opposite side of the lens to the object? • Size or magnification. Is the image magnified (enlarged), diminished (reduced) or the same size as the object? In the example in figure 20.34, we have a real, inverted and magnified image on the opposite side of the lens to the object. FIGURE 20.34 The location of the image is determined according to the point where the three rays cross. All the rays that pass through the lens pass through the image.

3

Object

Convex lens

1

Image

I O

F

F 2

FLAT LENS A lens works by changing the direction of the light ray at the front surface and then again at the back surface. The glass in the middle is there to keep the two surfaces apart. Augustin-Jean Fresnel devised a way of making a lens without the need for all the glass in the middle. The glass surface of the lens is a series of concentric rings. Each ring has the slope of the corresponding section of the full lens, but its base is flat. The slopes of the rings get flatter towards the centre. FIGURE 20.35 A side view of a convex Fresnel lens showing how it is constructed

This design substantially reduces the weight of the lens, so lenses of this type are used in lighthouses. Their relative thinness means they are also used where space is at a premium, such as in overhead projectors, and as a lens to be used with the ground-glass screens in camera viewfinders. Flat lenses, or Fresnel lenses as they are called, are now attached to the rear windows of vans and station wagons to assist the driver when reversing or parking.

TOPIC 20 How can human vision be enhanced? 29

Convex lenses are used in a variety of applications. The image obtained depends on the placement of the object in relation to the focus. A range of these applications is given in table 20.4. TABLE 20.4 A variety of applications for convex lenses Object Ray diagram distance (u)

Type of image

u=∞

• • •

Inverted Real Diminished

•

• • •

Inverted Real Diminished

•

• • •

Inverted Real Same size

•

• • •

Inverted Real Magnified

•

• • •

Upright Virtual Magnified

• • •

Upright Virtual Magnified

Parallel rays from a distant object v

Object

F 2F

2F Image

F u

v

u = 2f Object

F

2F

2F Image

F u

v

F

Object

2F

F

Image u

u=f

v=f Opposite side of the lens

•

f < v < 2f Opposite side of the lens

• •

v = 2f Opposite side of the lens

•

v > 2f Opposite side of the lens

• •

•

•

Object lens of a telescope

Image

u > 2f

2F

Uses

F

F

f