New A-Level Physics AQA Revision Question Cards 1789085942, 9781789085945

Table of contents :
Intro
How To Use These Cards
Section 1 — Particles
1. Atomic Structure
Answers — 1. Atomic Structure
2. Stable and Unstable Nuclei
Answers — 2. Stable and Unstable Nuclei
3. Particles and Antiparticles
Answers — 3. Particles and Antiparticles
4. Forces and Exchange Particles (1)
Answers — 4. Forces and Exchange Particles (1)
5. Forces and Exchange Particles (2)
Answers — 5. Forces and Exchange Particles (2)
6. Classification of Particles (1)
Answers — 6. Classification of Particles (1)
7. Classification of Particles (2)
Answers — 7. Classification of Particles (2)
8. Quarks
Answers — 8. Quarks
Section 2 — Electromagnetic Radiation and Quantum Phenomena
9. The Photoelectric Effect
Answers — 9. The Photoelectric Effect
10. Energy Levels and Photon Emission
Answers — 10. Energy Levels and Photon Emission
11. Wave-Particle Duality
Answers — 11. Wave-Particle Duality
Section 3 — Waves
12. Progressive Waves
Answers — 12. Progressive Waves
13. Longitudinal and Transverse Waves
Answers — 13. Longitudinal and Transverse Waves
14. Superposition and Coherence
Answers — 14. Superposition and Coherence
15. Stationary Waves
Answers — 15. Stationary Waves
16. Diffraction
Answers — 16. Diffraction
17. Two-Source Interference
Answers — 17. Two-Source Interference
18. Diffraction Gratings
Answers — 18. Diffraction Gratings
19. Refractive Index
Answers — 19. Refractive Index
Section 4 — Mechanics
20. Scalars and Vectors
Answers — 20. Scalars and Vectors
21. Forces (1)
Answers — 21. Forces (1)
22. Forces (2)
Answers — 22. Forces (2)
23. Moments
Answers — 23. Moments
24. Mass, Weight and Centre of Mass
Answers — 24. Mass, Weight and Centre of Mass
25. Displacement-Time Graphs
Answers — 25. Displacement-Time Graphs
26. Velocity-Time Graphs
Answers — 26. Velocity-Time Graphs
27. Acceleration-Time Graphs
Answers — 27. Acceleration-Time Graphs
28. Motion With Uniform Acceleration
Answers — 28. Motion With Uniform Acceleration
29. Acceleration Due to Gravity
Answers — 29. Acceleration Due to Gravity
30. Projectile Motion
Answers — 30. Projectile Motion
31. Newton’s Laws of Motion
Answers — 31. Newton’s Laws of Motion
32. Drag, Lift and Terminal Speed
Answers — 32. Drag, Lift and Terminal Speed
33. Momentum and Impulse
Answers — 33. Momentum and Impulse
34. Work and Power
Answers — 34. Work and Power
35. Conservation of Energy and Efficiency
Answers — 35. Conservation of Energy and Efficiency
Section 5 — Materials
36. Properties of Materials
Answers — 36. Properties of Materials
37. Stress and Strain
Answers — 37. Stress and Strain
38. The Young Modulus
Answers — 38. The Young Modulus
39. Interpreting Graphs for Materials
Answers — 39. Interpreting Graphs for Materials
Section 6 — Electricity
40. Current, Potential Difference and Resistance
Answers — 40. Current, Potential Difference and Resistance
41. I/V Characteristics
Answers — 41. I/VCharacteristics
42. Resistivity and Superconductivity
Answers — 42. Resistivity and Superconductivity
43. Electrical Energy and Power
Answers — 43. Electrical Energy and Power
44. E.m.f. and Internal Resistance
Answers — 44. E.m.f. and Internal Resistance
45. Conservation of Energy and Charge
Answers — 45. Conservation of Energy and Charge
46. The Potential Divider
Answers — 46. The Potential Divider
Section 7 — Further Mechanics
47. Circular Motion
Answers — 47. Circular Motion
48. Simple Harmonic Motion
Answers — 48. Simple Harmonic Motion
49. Simple Harmonic Oscillators
Answers — 49. Simple Harmonic Oscillators
50. Free and Forced Vibrations
Answers — 50. Free and Forced Vibrations
Section 8 — Thermal Physics
51. Thermal Energy Transfer
Answers — 51. Thermal Energy Transfer
52. Gas Laws
Answers — 52. Gas Laws
53. Ideal Gas Equation
Answers — 53. Ideal Gas Equation
54. The Pressure of an Ideal Gas
Answers — 54. The Pressure of an Ideal Gas
55. Kinetic Energy and the Development of Theories
Answers — 55. Kinetic Energy and the Development of Theories
Section 9 — Gravitational and Electric Fields
56. Gravitational Fields
Answers — 56. Gravitational Fields
57. Gravitational Potential
Answers — 57. Gravitational Potential
58. Orbits and Gravity
Answers — 58. Orbits and Gravity
59. Electric Fields (1)
Answers — 59. Electric Fields (1)
60. Electric Fields (2)
Answers — 60. Electric Fields (2)
61. Electric Potential and Work Done
Answers — 61. Electric Potential and Work Done
62. Comparing Electric and Gravitational Fields
Answers — 62. Comparing Electric and Gravitational Fields
Section 10 — Capacitors
63. Capacitors
Answers — 63. Capacitors
64. Charging and Discharging
Answers — 64. Charging and Discharging
65. More Charging and Discharging
Answers — 65. More Charging and Discharging
Section 11 — Magnetic Fields
66. Magnetic Fields
Answers — 66. Magnetic Fields
67. Charged Particles in a Magnetic Field
Answers — 67. Charged Particles in a Magnetic Field
68. Electromagnetic Induction
Answers — 68. Electromagnetic Induction
69. Induction Laws and Alternators (1)
Answers — 69. Induction Laws and Alternators (1)
70. Induction Laws and Alternators (2)
Answers — 70. Induction Laws and Alternators (2)
71. Alternating Currents
Answers — 71. Alternating Currents
72. Transformers
Answers — 72. Transformers
Section 12 —Nuclear Physics
73. Rutherford Scattering and Atomic Structure
Answers — 73. Rutherford Scattering and Atomic Structure
74. Nuclear Radius and Density
Answers — 74. Nuclear Radius and Density
75. Radioactive Emissions (1)
Answers — 75. Radioactive Emissions (1)
76. Radioactive Emissions (2)
Answers — 76. Radioactive Emissions (2)
77. Investigations of Radioactive Emissions
Answers — 77. Investigations of Radioactive Emissions
78. Exponential Law of Decay
Answers — 78. Exponential Law of Decay
79. Nuclear Decay
Answers — 79. Nuclear Decay
80. Nuclear Fission and Fusion
Answers — 80. Nuclear Fission and Fusion
81. Binding Energy
Answers — 81. Binding Energy
Section 13: Option A —Astrophysics
82. Optical Telescopes (1)
Answers — 82. Optical Telescopes (1)
83. Optical Telescopes (2)
Answers — 83. Optical Telescopes (2)
84. Non-Optical Telescopes
Answers — 84. Non-Optical Telescopes
85. Distances and Magnitude
Answers — 85. Distances and Magnitude
86. Stars as Black Bodies
Answers — 86. Stars as Black Bodies
87. Spectral Classes and the H-R Diagram
Answers — 87. Spectral Classes and the H-R Diagram
88. Stellar Evolution — Low Mass Stars
Answers — 88. Stellar Evolution — Low Mass Stars
89. Stellar Evolution — High Mass Stars
Answers — 89. Stellar Evolution — High Mass Stars
90. The Doppler Effect and Red Shift
Answers — 90. The Doppler Effect and Red Shift
91. Quasars and Exoplanets
Answers — 91. Quasars and Exoplanets
92. The Big Bang Model of the Universe
Answers — 92. The Big Bang Model of the Universe
Section 13: Option B —Medical Physics
93. Physics of the Eye
Answers — 93. Physics of the Eye
94. Defects of Vision
Answers — 94. Defects of Vision
95. Physics of the Ear
Answers — 95. Physics of the Ear
96. Intensity and Loudness
Answers — 96. Intensity and Loudness
97. Electrocardiography (ECG)
Answers — 97. Electrocardiography (ECG)
98. Ultrasound Imaging
Answers — 98. Ultrasound Imaging
99. Endoscopy
Answers — 99. Endoscopy
100. X-Ray Production
Answers — 100. X-Ray Production
101. X-Ray Imaging Techniques
Answers — 101. X-Ray Imaging Techniques
102. Magnetic Resonance (MR) Imaging
Answers — 102. Magnetic Resonance (MR) Imaging
103. Medical Uses of Radiation
Answers — 103. Medical Uses of Radiation
Section 13: Option C —Engineering Physics
104. Inertia and Kinetic Energy
Answers — 104. Inertia and Kinetic Energy
105. Rotational Motion
Answers — 105. Rotational Motion
106. Torque, Work and Power
Answers — 106. Torque, Work and Power
107. Flywheels
Answers — 107. Flywheels
108. Angular Momentum
Answers — 108. Angular Momentum
109. The First Law of Thermodynamics (1)
Answers — 109. The First Law of Thermodynamics (1)
110. The First Law of Thermodynamics (2)
Answers — 110. The First Law of Thermodynamics (2)
111. p-V Diagrams
Answers — 111. p-V Diagrams
112. Four-Stroke Engines
Answers — 112. Four-Stroke Engines
113. Using Indicator Diagrams
Answers — 113. Using Indicator Diagrams
114. Engine Efficiency
Answers — 114. Engine Efficiency
115. Reversed Heat Engine
Answers — 115. Reversed Heat Engine
Section 13: Option D —Turning Points in Physics
116. Specific Charge of the Electron
Answers — 116. Specific Charge of the Electron
117. Millikan’s Oil-Drop Experiment
Answers — 117. Millikan’s Oil-Drop Experiment
118. Light — Particles vs Waves (1)
Answers — 118. Light — Particles vs Waves (1)
119. Light — Particles vs Waves (2)
Answers — 119. Light — Particles vs Waves (2)
120. The Photoelectric Effect and Photons (1)
Answers — 120. The Photoelectric Effect and Photons (1)
121. The Photoelectric Effect and Photons (2)
Answers — 121. The Photoelectric Effect and Photons (2)
122. Wave-Particle Duality
Answers — 122. Wave-Particle Duality
123. The Speed of Light and Relativity
Answers — 123. The Speed of Light and Relativity
124. Special Relativity
Answers — 124. Special Relativity
Practical and Investigative Skills
125. Experiment Design and Errors
Answers — 125. Experiment Design and Errors
126. Uncertainty and Errors
Answers — 126. Uncertainty and Errors
127. Presenting and Evaluating Data
Answers — 127. Presenting and Evaluating Data

Citation preview

A-Level

Physics

CGP

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P G C

A-Level

Gain momentum in A-Level Physics with CGP’s cards...

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Revision

Question Cards ISBN 978 1 78908 594 5

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Physics

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A-Level AQA

A-Level AQA

Physics

Revision Question Cards

Exam Board: AQA

Revision Question Cards

Physics

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A-Level

Physics Exam Board: AQA

PAF71

Revision Question Cards

How To Use These Cards Cards covering all the core content, plus Year 2 options 9-12. (Option 13: ‘Electronics’ isn’t covered in these cards.)

Questions on all the key course content, as well as on practicals and maths skills – you’ll need a calculator for most of the maths questions.

23. Moments

Section 4 — Mechanics

Quick Questions 1) State the principle of moments. 2) Which arrangement below shows a couple acting on the seesaw? F A. B. C. F F F

F

3) A student is assembling a flat-pack desk. They need to screw in different bolts A using a hex key, like the one shown on the right. Either end of the key B can be inserted into the bolt, and then the key is turned to tighten the bolt. The student inserts end A into a bolt, and applies a force of 14 N to produce a moment of 0.73 Nm. At what perpendicular distance from the pivot was the force applied? Use the equation moment = Fd. 4) Explain why the student finds it harder to tighten a bolt when they insert end B of the hex key compared to when they insert end A. 5) To attach the second set of legs to the desk, the student asks a friend to hold up one end of the desk, as shown on the right. The desk is pivoted around the already attached set of legs. What force must the friend use to keep the desk still? Use the equation moment = Fd.

Section 4 — Mechanics

F

Now try these:

135.0 cm 65.0 cm 352 N

a lanky student

23. Moments

ANSWERS

Complete answers to each question.

1) The principle of moments states that for a body to be in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about the same point. 2) B. A couple consists of two parallel forces acting in opposite directions, to produce a turning effect. 3) 0.052 m. Rearrange equation for d, d = moment ÷ F = 0.73 ÷ 14 = 0.05214... = 0.052 m (to 2 s.f.) 4) If the student inserts end B into a bolt, the maximum perpendicular distance from the pivot at which they can apply a force is smaller than if they insert end A into the bolt. Since moment = Fd, this means they need to apply a larger max. perp. distance for force to produce the same moment when end B is in the bolt than when end B inserted end A is in the bolt, and so it is harder to turn the bolt and tighten it.

max. perp. distance for end A inserted

B

A 5) 169 N. Using the principle of moments, clockwise (C) moments = anticlockwise (A) moments. This means FCdC = FAdA , so: FA = (FCdC ) ÷ dA = (352 × 65.0) ÷ 135.0 = 169.4814... = 169 N (to 3 s.f.) (Moments are usually given in the units Nm. However, there's no need to convert the units of distance from cm to m here, since all the distances are in the same units, and the distance units cancel out.)

TIP

When dealing with turning effect questions, take a moment (haha) to check where the pivot is. Any forces that act through the pivot of the system won't produce a moment, since their distance from the pivot will be zero.

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1. Atomic Structure

Section 1 — Particles

Quick Questions 1) What are the relative charges of a proton, a neutron and an electron? 2) What particles make up the nucleus of an atom? 3) What number is the same for two isotopes of the same element?

Now try these: 4) Give the nuclide notation of deuterium, an isotope of hydrogen (H) with one proton and one neutron. 5) The nuclide notation of chlorine-35 is 3517 C . How many protons, neutrons and electrons are in a chlorine-35 ion with a relative charge of –1? 6) All living things contain the same percentage of radioactive carbon-14 taken in from the atmosphere. After they die, the amount of carbon-14 inside them decreases over time as it decays to stable elements. A group of archaeologists dig up a wooden spoon. Suggest how isotopic 35 years of archaeology data can help them to determine the approximate age of this spoon. and all I find is a spoon. Just... no one talk to

7) Calculate the specific charge of the nucleus of one carbon atom, 126 C . me right now... The magnitude of the charge of an electron is 1.60 × 10–19 C, the mass of a proton is 1.67 × 10–27 kg and the mass of a neutron is 1.67 × 10–27 kg.

TIP

Remember to give numerical answers to the smallest number of significant figures used in the question. And I’m not talking about notable people here — if there’s any mention of Cleopatra, you might be in the wrong exam.

Section 1 — Particles

1. Atomic Structure

ANSWERS 1) Proton = +1. Neutron = 0. Electron = –1. 2) Protons and neutrons. 3) The proton number. Two isotopes of the same element have the same number of protons but a different number of neutrons. 4)

2 A 1 H . Nuclide notation is of the form Z X where X is the element, A is the nucleon number (total number of protons and neutrons) and Z is the proton number (number of protons). Deuterium has a proton number of 1 and a nucleon number of 2 so its nuclide notation is 21 H .

5) 17 protons, 18 neutrons, 18 electrons. The proton number is 17. Overall charge is –1 so it must have one more electron than the number of protons. No. of neutrons = nucleon number – proton number = 35 – 17 = 18 neutrons. 6) As the spoon is made out of wood, the archaeologists can use data on the amount of each isotope present in the spoon to find the percentage of carbon-14 which remains, and therefore its approximate age. 7) 4.79 × 107 Ckg –1. A 126 C nucleus has 6 protons and 12 – 6 = 6 neutrons. A neutron has no charge and the magnitude of the charge of a proton is the same as an electron. Charge of a 126 C nucleus = no. of protons × charge of one proton = 6 × 1.60 × 10–19 = 9.6 × 10–19 C Mass = no. of protons and neutrons × mass of one proton/neutron = 12 × 1.67 × 10–27 = 2.004 × 10–26 kg charge 9.6 × 10 -19 = 4.790... × 107 = 4.79 × 107 Ckg–1 (to 3 s.f.) Specific charge = mass = 2.004 × 10 -26

2. Stable and Unstable Nuclei

Section 1 — Particles

Quick Questions 1) Which particle was hypothesised in order to explain why energy didn’t appear to be conserved in beta decay? 2) What would happen to a nucleus if there was no strong nuclear force in the nucleus?

Now try these: 3) If two protons are separated by a distance of 0.3 fm, will they be attracted or repelled by each other? 4) Uranium-238 ( 23892 U ) decays to thorium (Th) via alpha decay. Give the balanced nuclear equation for this reaction. 5) An atom of carbon-14 (A = 14, Z = 6) decays to nitrogen via beta-minus decay. What is the balanced nuclear equation for this reaction? The chemical symbol for carbon is C and the chemical symbol for nitrogen is N. 6) If all the atoms in a radioactive sample were to decay, would the sample’s percentage mass decrease be larger if it underwent alpha or beta-minus decay? Explain your answer. You can assume any particles released during the decay leave the sample and that any change in mass due to mass defects is negligible.

Although she had a lot of mass, Lucy was feeling very stable.

Section 1 — Particles

2. Stable and Unstable Nuclei

ANSWERS 1) The neutrino. (We now know that the particle actually released in beta decay is an antineutrino.) 2) Nucleons would fly apart due to the repulsive electrostatic force between protons. 3) Repelled. For distances below around 0.5 fm the strong nuclear force is repulsive. 4)

238 234 4 92 U $ 90 Th + 2a . An alpha particle containing 2 neutrons and 2 protons is emitted so the nucleon number goes down by 4 and the proton number goes down by 2.

5)

14 14 0 6 C $ 7 N + -1b + u e . A neutron changes into a proton, so the nucleon number stays the same and the proton number increases by one. A beta-minus particle / an electron is emitted, and an antineutrino is also emitted to ensure energy conservation.

6) Alpha decay. During beta-minus decay, an atom loses no nucleons as a neutron changes to a proton, so the percentage mass decrease would be zero. During alpha decay, one atom loses 4 nucleons, so there would be a percentage mass decrease greater than zero.

TIP

The strong nuclear force is odd — the distance between two particles dictates whether it produces an attractive or repulsive force. But without it atoms wouldn’t exist, and the universe would be unimaginably different...

3. Particles and Antiparticles

Section 1 — Particles

Quick Questions 1) Give two properties that are the same for both a particle and its antiparticle. 2) What are the relative charges of a positron and an antiproton? 3) What is meant by annihilation between particles?

Now try these:

hc 4) A photon has 8.15 × 10–14 J of energy. What is the wavelength of this photon? Use the equation E = l where h = 6.63 × 10–34 Js and c = 3.00 × 108 ms–1. 5) A photon with a frequency of 3.74 × 1020 Hz produces an electron-positron pair, each with the same amount of energy. How much energy does each particle have? Use the equation E = hf where h = 6.63 × 10–34 Js. 6) What is the minimum energy (in J) that a photon must have in order to produce a proton-antiproton pair? Rest energy of a proton = 938.257 MeV. The magnitude of the charge on an electron = 1.60 × 10–19 C. 7) Calculate the minimum frequency of one of the photons produced when a neutron and an antineutron annihilate each other. Use the equation E = hf  where h = 6.63 × 10–34 Js. The rest energy of a neutron = 939.551 MeV and the magnitude of the charge on an electron is 1.60 × 10–19 C.

TIP

You’ll be given the rest energy of particles in MeV in the exam, so you’ll need to convert this to joules before you jump into calculating f or l. Missing out on marks because you’ve got the wrong units just won’t do.

Section 1 — Particles

3. Particles and Antiparticles

ANSWERS 1) E.g. mass and rest energy. 2) Positron = +1. Antiproton = –1. Antiparticles have an opposite charge to their particles and the relative charges of an electron and proton are –1 and +1 respectively. 3) Annihilation is when a particle meets its antiparticle and the combined mass gets converted to energy in the form of two photons. 6.63 × 10 -34 × 3.00 × 10 8 hc 4) 2.44 × 10–12 m. l = E = = 2.4404... × 10–12 = 2.44 × 10–12 m (to 3 s.f.). 8.15 × 10 -14 5) 1.24 × 10–13 J. Ephoton = hf = 6.63 × 10–34 × 3.74 × 1020 = 2.47962 × 10­–13 J . This energy is divided equally Ephoton 2.47962 × 10 -13 = 1.23981 × 10­–13 = 1.24 × 10–13 J (to 3 s.f.). between the two particles, so Eparticle = 2 = 2 6) 3.00 × 10–10 J. For this pair production, the minimum energy (Emin) is the total rest energy of the proton and antiproton. The rest energy of an antiproton is the same as a proton (E0), so Emin = 2E0 = 2 × 938.257 = 1876.514 MeV. To convert Emin into eV, multiply by 106. To convert into joules, multiply by 1.60 × 10–19. So Emin = 1876.514 × 106 × 1.60 × 10–19 = 3.002... × 10–10 = 3.00 × 10–10 J (to 3 s.f.) 7) 2.27 × 1023 Hz. A pair of photons are produced when the particles annihilate, each with the same energy. The rest energy of a neutron is the same as an antineutron (E0). For energy to be conserved, both photons need to have a minimum energy (Emin) which when added together equals at least the total rest energy of the particles (2E0), so 2Emin = 2E0 and therefore Emin = E0 = 939.551 × 106 × 1.60 × 10–19 = 1.503... x 10–10 J. Emin = hfmin. So fmin =

Emin 1.503... × 10 -10 23 23 h = 6.63 × 10 -34 = 2.267... × 10 = 2.27 × 10 Hz (to 3 s.f.)

4. Forces and Exchange Particles (1)

Section 1 — Particles

Quick Questions 1) Which exchange particles are exchanged in electromagnetic interactions? 2) True or false? As exchange particles, pions are exchanged between hadrons only. 3) How are exchange particles represented in particle interaction diagrams?

Now try these:

e–

4) The particle interaction diagram for beta-minus decay is shown on the right. What are the missing labels? 5) A K0 particle is a hadron with no charge. A K0 particle and a proton are attracted towards each other. Why isn’t it possible for this interaction to be due to the electromagnetic force? p

p – p

p

n

6) The diagram on the left shows the particle interaction diagram for two protons repelling due to their positive charge. Give two errors in this diagram. 7) A tau particle is a negatively-charged lepton that decays into a tau-neutrino, an electron antineutrino and an electron. Determine which force causes this interaction. Suggest an exchange particle for this interaction.

Section 1 — Particles

4. Forces and Exchange Particles (1)

ANSWERS 1) Virtual photons. 2) True. Pions act as the exchange particle in strong interactions, and the strong interaction affects hadrons only.

e– p

3) With a wiggly (rather than straight) line. 4) The completed diagram is shown on the right. The exchange particle is a W– boson travelling from the neutron. The neutron changes into a proton, and an electron and antineutrino are emitted.

W–

ne

n

5) The electromagnetic force only affects charged particles and the K0 particle has no charge. 6) The exchange particle should be labelled with a virtual photon, not a pion, as this is the correct exchange particle for electromagnetic repulsion. The arrows for the two protons on the right hand side should be pointing upwards, not downwards, because incoming particles start at the bottom of the diagram and move upwards. 7) The weak nuclear force and a W+ or W– boson. It can’t be the electromagnetic force as neutrinos are uncharged and the electromagnetic force only affects charged particles. It can’t be the strong nuclear force as electrons and neutrinos aren’t hadrons and the strong nuclear force only affects hardrons. So it must be the weak nuclear force. The weak nuclear force uses W– or W+ bosons as the exchange particle.

TIP

If there was an exchange particle between you and exam success, it would be revision. And remember now, not all forces can affect all particles — the electromagnetic force, for example, can only affect charged particles.

5. Forces and Exchange Particles (2)

Section 1 — Particles

Quick Questions 1) Which exchange particles are exchanged in strong interactions? 2) Which type of particles are virtual photons exchanged between? 3) Which two of the following aren’t fundamental forces in nature? A. gravity B. friction C. weak nuclear force D. tension

Now try these: 4) The particle interaction diagram for an electron-proton collision is shown on the right. How would the diagram be different for electron capture? 5) A metal plate is placed into an ionic solution. The ions in the solution are attracted towards the metal plate through the exchange of virtual photons. What force is acting on the ions?

n

ne

n p

6) A particle interaction diagram is shown on the left. State what interaction this is. What are the missing labels in this diagram? 7) Describe the particle interaction diagram for two S+ particles repelling each other.

W–

ne e–

Section 1 — Particles

5. Forces and Exchange Particles (2)

ANSWERS 1) Pions (p+, p–, p0). 2) Charged particles. 3) B and D. The four fundamental forces in nature are gravity, the weak nuclear force, the strong nuclear force and the electromagnetic force. 4) The diagram on the right shows the differences. For electron capture the exchange particle would be a W+ boson travelling in the opposite direction.

n

W+

p

ne e–

5) Electromagnetic force. The virtual photon is the exchange particle for the electromagnetic force. 6) Beta-plus decay. The incoming particle should be p (a proton), the interaction particle should be labelled W+ and the outgoing arrow needs to be labelled e+ (a positron). 7) Two straight lines with arrows pointing upwards, angled towards each other, both S+ S+ g labelled S+ indicate two incoming S+ particles. A wiggly line connecting them labelled g indicates a virtual photon being exchanged between them. Two straight S+ S+ lines with arrows pointing upwards, angled away from each other, both labelled S+ indicate two outgoing S+ particles. The diagram on the right shows this. (Don’t panic if you’ve not heard of S+ particles before. All repulsion interaction diagrams look the same.)

TIP

You may have noticed that despite being a fundamental force in nature, gravity takes a back seat in particle physics — it’s more important when there’s lots of mass involved. Like with planets and what have you.

6. Classification of Particles (1)

Section 1 — Particles

Quick Questions 1) What do all baryons eventually decay into? 2) Which two of these particles are mesons? A. Pion B. Proton C. Lepton

D. Kaon

3) State the symbol and charge of a muon-antineutrino. 4) Name a particle that can decay into a pion.

Now try these: 5) Give one reason why the reaction n + n $ p + p is not possible. 6) An atom of cobalt-60 ( 6027 Co) undergoes beta-minus decay. How many baryons are left in the atom once it has decayed? 7) After many years studying particles, I’ve finally worked out how to unlock the secrets of the universe — it turns out they’re just kept in a safe with a simple combination lock. The code is the number of leptons in a neutral atom of gold, 19779 Au , plus the number of hadrons in a neutral atom of silver, 10747 Ag . What number will unlock the safe? 8) Which one of these interactions takes place via the strong nuclear force? Explain your answer. B. K+ $ m+ + n m C. p $ n + e+ + n e A. p - + p $ p 0 + n

Section 1 — Particles

6. Classification of Particles (1)

ANSWERS 1) Protons. Protons are the only stable baryons. 2) A and D. Protons are baryons. Leptons are another class of particle that includes electrons and muons. 3) A muon-antineutrino has the symbol n m and a charge of 0. 4) E.g. a kaon. Kaons have a very short lifetime and can decay into pions. 5) The baryon number is not conserved. Both neutrons and protons are baryons. Baryons have a baryon number of +1 and antibaryons have a baryon number of –1, so the total baryon number at the start is 1 + 1 = 2 and the total baryon number at the end is –1 + 1 = 0. 6) 60. During beta-minus decay a neutron changes into a proton, so the number of nucleons stays the same. Neutrons and protons are both baryons with a baryon number of +1, so overall the baryon number is 60. 7) 186. Protons and neutrons are both hadrons, and electrons are leptons. A neutral atom of gold has the same number of electrons as protons, so it contains 79 leptons. A neutral atom of silver has a nucleon number (total number of protons and neutrons) of 107, so it contains 107 hadrons. 79 + 107 = 186. (What’s in the safe, you ask? Ah... well... it wouldn’t be much of a secret if I told you...) 8) A. This is because the strong force only acts on hadrons, and A is the only equation that contains just hadrons (electrons, muons, neutrinos and their antiparticles are leptons).

TIP

Call me a hadron, because I’m feeling a strong force to fall asleep right now. But keep calm and baryon — you’ll be feeling pretty smug in the exam if you can identify a kaon as a meson almost as fast as it can decay.

7. Classification of Particles (2)

Section 1 — Particles

Quick Questions 1) True or false? Muons don’t feel the strong nuclear force. 2) True or false? Electron lepton number and muon lepton number need to be counted separately when balancing equations. 3) What do muons eventually decay into?

Now try these: 4) Lekan is having a whale of a time learning about L0 particles. L0 particles are baryons. What will a L0 particle eventually decay into? 5) Late one night, Lekan writes down the following decay equation of a L0 particle: L 0 $ n + p 0 . Determine whether baryon number is conserved in this interaction or not. 6) A molecule of benzene contains six atoms of carbon, 126 C , and six atoms of hydrogen, 11 H . What is the total baryon number of one molecule of benzene? 7) The muon can decay via the weak interaction as shown: m - $ e - + n e + ? What is the missing particle? (Hint: for leptons, particles have a lepton number of +1 and antiparticles have a lepton number of –1.)

TIP

Baryon number and lepton number are both examples of quantum numbers — these numbers can only take a certain set of values. And remember, antiparticles have the opposite numbers to their corresponding particle.

Section 1 — Particles

7. Classification of Particles (2)

ANSWERS 1) True. Muons are leptons rather than hadrons, and only hadrons feel the strong nuclear force. 2) True. Each lepton, along with its lepton-neutrino, has its own lepton number so must be counted separately. 3) Electrons. Muons are unstable and decay into electrons which are stable. 4) A proton. All baryons (except the proton) are unstable, so all baryons eventually decay to a proton. 5) L0 particles and neutrons are both baryons so they have a baryon number B = +1. Pions are not baryons (they’re mesons) and therefore have a baryon number B = 0. The baryon number on both sides is equal to 1 and so the baryon number is conserved. 6) 78. Protons and neutrons are both baryons with a baryon number B = +1. One atom of carbon contains 12 baryons. One atom of hydrogen contains 1 baryon. So the total baryon number is (12 × 6) + (1 × 6) = 78. 7) A muon-neutrino. Lepton numbers must balance, and electron and muon lepton numbers need to be counted separately. The electron lepton number is 0 at the start and 1 + –1 = 0 at the end, so it balances. The muon lepton number is +1 at the start so the missing particle must have a muon lepton number of +1 for it to balance. Since e - and m - have charges of –1 and n e is unchanged, the missing particle must also be uncharged in order for charge to be conserved. Therefore the particle is a muon-neutrino.

I don’t like the sound of these party culls. You really leapt on that pun there. Also, did you hear they’re planning to elect Ron? Please stop.

8. Quarks

Section 1 — Particles

Quick Questions 1) Give three properties that are conserved in all particle interactions. 2) Which property can change in weak interactions, and by how much? 3) Give an example of a particle that is made up of three antiquarks.

Now try these: 4) Describe, in terms of quarks, what happens during beta-plus decay.

Quark u

5) Give the total charge of the K+ meson and the quark composition of its antiparticle. Use the table on the right to help you.

u

6) A X0 particle is a baryon with a strangeness of –2 and no charge. Determine the quark composition of a X0 particle. Use the table on the right to help you.

d

7) The kaon decays via the weak interaction. What is the missing particle in the following decay? K - $ p0 + m - + ?

s

d

s

Charge 2 +3 2 –3 1 –3 1 +3 1 –3 1 +3

e e e e e e

Baryon number

Strangeness

1 +3 1 –3 1 +3 1 –3 1 +3 1 –3

0 0 0 0 –1 +1

Section 1 — Particles

8. Quarks

ANSWERS 1) Any three from: charge / baryon number / electron lepton number / muon lepton number / energy / momentum. 2) Strangeness can change by –1, 0 or +1. 3) E.g. an antiproton/an antineutron. 4) An up quark turns into a down quark, changing the proton (uud) into a neutron (udd). 2 5) K+ has a quark composition of us . The charge of an up quark is + 3 e, and the charge of an anti-strange 1 quark is + 3 e, so the total charge is +1 e. The quark composition of the K+ antiparticle (K–) is su . 6) uss. A X0 particle is a baryon, so must be made up of three quarks. For a strangeness of –2, it must contain 2 strange quarks. To get a zero charge, the third quark must be an up quark. 7) n m . The table to the right shows the values of the properties that must be conserved in order to work out the missing particle. The charges for K– and p0 come from their quark compositions — su for K– and uu (or dd ) for p0. The required particle must be uncharged and have a muon lepton number of –1. This particle is the muon-antineutrino. (Although strangeness is not conserved, this is allowed in the weak interaction.)

TIP

K– " p0

+

m–

+

?

charge:

–1

0

–1

0

Muon lepton number:

0

0

+1

–1

Experiments to confirm predictions of new particles need lots of funding and resources. It’s only due to scientists working together that we’re able to make these discoveries. Three cheers for collaborative working. Hip hip...

9. The Photoelectric Effect

Section 2 — Electromagnetic Radiation and Quantum Phenomena

Quick Questions 1) True or false? The stopping potential is the potential difference required to stop the fastest moving photoelectrons emitted from a metal by the photoelectric effect. 2) What is the threshold frequency of a metal?

Now try these: 3) The diagram below shows the set-up for an experiment investigating the photoelectric effect. A photocell is connected in a series circuit with an ammeter. A lamp, emitting UV light with a frequency of 7.0 × 1015 Hz, is positioned so that the light is incident on the photoelectric plate in the photocell. This gives an ammeter reading of 0.5 A. Explain what would happen to the UV light reading on the ammeter if the intensity of the light is increased. 4) The work function of the photocell plate is 4.2 eV. Calculate the maximum velocity of the emitted photoelectrons, using the equations hf = f + Ek(max) and Ek = ½mv2. Use h = 6.63 × 10–34 Js, me = 9.11 × 10–31 kg and e = 1.60 × 10–19 C. 5) The UV lamp is replaced with an infrared lamp and the reading on the ammeter drops to, and stays at, zero. Suggest why this happens and how this observation would be different if light was behaving as a wave.

photocell

A

Section 2 — Electromagnetic Radiation and Quantum Phenomena

9. The Photoelectric Effect

ANSWERS 1) True. The work done by the potential difference in stopping the fastest photoelectrons is equal to their kinetic energy. This can be expressed in the equation eVS = Ek(max), where e is the magnitude of the electron charge, VS is the stopping potential and Ek(max) is maximum kinetic energy of the photoelectrons. 2) The frequency of incident radiation below which no photoelectrons are emitted. 3) The reading on the ammeter would increase because increasing the intensity of light means increasing the number of photons per second incident on the plate, which causes more photoelectrons to be emitted per second. This increases the rate of charge that is flowing across the photocell, so the current increases. 4) 3.0 × 106 ms–1. Convert 4.2 eV into J: 4.2 × 1.60 × 10–19 = 6.72 × 10–19 J. Rearrange the first equation for Ek(max): Ek(max) = hf – f = (6.63 × 10–34 × 7.0 × 1015) – 6.72 × 10–19= 3.969 × 10–18 J. Then rearrange the second equation for the velocity: 2Ek 2 × 3.969 × 10 -18 = 2.951... × 106 = 3.0 × 106 ms–1 (to 2 s.f.). vmax = m = 9.11× 10 -31 5) The ammeter drops to zero because there is no current flowing, as infrared radiation does not have a high enough frequency to liberate electrons from the photoelectric plate. If light was behaving as a wave, electrons would gradually gain energy from the incident infrared waves and then be emitted eventually (so there would be a time delay, but eventually, a current reading would be observed on the ammeter).

TIP

Increasing the intensity of light doesn’t mean the electrons get more energetic, as wave theory suggests. Instead, it increases the rate at which electrons are released as more photons are incident on the metal per second.

10. Energy Levels and Photon Emission

Section 2 — Electromagnetic Radiation and Quantum Phenomena

Quick Questions 1) Give the definition of the electronvolt. 2) What name is given to the process of an atom losing an electron? 3) How can you tell if an absorption spectrum and an emission spectrum are from the same element?

Now try these: 4) When excited electrons in mercury atoms return to their ground state, they emit ultraviolet (UV) photons. Explain how a fluorescent lamp uses the UV photons produced by the excited atoms in mercury vapour to create visible light. 5) The ground state of an atom is at an energy level of –11.3 eV. What would happen to the atom if one of its electrons in the n = 2 state absorbed a photon with an energy of 12 eV? Explain your answer. 6) The diagram on the right shows some of the energy levels for an atom of singly ionised helium. The electron is in the ground state of the helium atom when it absorbs a photon, and moves up to the n = 3 state. What was the frequency of the photon absorbed? Use the equation hf = E1 – E2, where h = 6.63 × 10–34 Js and e = 1.60 × 10–19 C.

n=5 n=4 n=3

–2.18 eV –3.40 eV –6.04 eV

n=2

–13.6 eV

n=1

–54.4 eV

Section 2 — Electromagnetic Radiation and Quantum Phenomena

10. Energy Levels and Photon Emission

ANSWERS 1) The electronvolt is the kinetic energy carried by an electron after it has been accelerated from rest through a potential difference of 1 volt. 2) Ionisation. 3) The black lines in the absorption spectrum would match up in frequency to the bright lines in the emission spectrum. 4) Fluorescent lamps have a phosphor inner coating that absorbs UV photons produced by the excited mercury atoms, exciting the phosphor electrons to higher orbits. These electrons then cascade down the energy levels, emitting lower energy photons in the form of visible light. 5) The atom would be ionised. An electron sat in the ground state would need to absorb a photon with a minimum of 11.3 eV of energy in order to be removed from the atom (ionisation). Any electron sat in other energy levels would need less energy than this to be completely removed from the atom. So as the photon has enough energy to remove an electron in the ground state from the atom, it will also have enough energy to remove an electron in the n = 2 sate. 6) 1.17 × 1016 Hz. The energy difference is 54.4 – 6.04 = 48.36 eV. Convert to J: 48.36 × 1.60 × 10–19 = 7.7376 × 10–18 J. E -E 7.7376 × 10 -18 Using the equation, f = 1 h 2 = = 1.167...× 1016 Hz = 1.17 × 1016 Hz (to 3 s.f.). 6.63 × 10 -34

TIP

Remember that the energies of energy levels are quoted as negative because of how ‘zero energy’ is defined — in calculations it may be easier to just deal with the magnitude of the energies and ignore those pesky minus signs.

11. Wave-Particle Duality

Section 2 — Electromagnetic Radiation and Quantum Phenomena

Quick Questions 1) When accelerated electrons are passed through a graphite crystal, they produce a diffraction pattern. What does this suggest about the nature of electrons? 2) Describe the process that new scientific theories, such as the theory of wave-particle duality, have to go through before they are validated by the scientific community.

Now try these: 3) An electron diffraction tube is shown on the right. It contains an electron gun electron gun that fires beams of electrons at a thin graphite crystal in a vacuum. A diffraction pattern is produced on a screen at the back of the tube. If the electrons in the electron beam are travelling at a electron beam speed of 4.0 × 106 ms–1, calculate their de Broglie wavelength. h Use l = mv , me = 9.11 × 10–31 kg and h = 6.63 × 10–34 Js. 4) The electrons in the experiment are slowed down. Explain what happens to the diffraction pattern on the screen.

5) Calculate the speed that a neutron would need to be travelling at in order to have the same de Broglie wavelength as the electrons in Q3. h Use the formula l = mv and mn = 1.67 × 10–27 kg.

vacuum tube

graphite crystal

screen

Section 2 — Electromagnetic Radiation and Quantum Phenomena

11. Wave-Particle Duality

ANSWERS 1) This suggests that electrons have wave-like properties, as only waves can undergo diffraction. 2) Other scientists have to evaluate the theory by peer review before it is published, and then the theory has to be tested with experiments to find enough evidence to back it up before it is accepted as valid. h 6.63 × 10 -34 3) 1.8 × 10–10 m. l = mv = 9.11× 10 -31 × 4.0 × 10 6 = 1.819... × 10–10 m = 1.8 × 10–10 m (to 2 s.f.). 4) The diffraction pattern gets more widely spaced out. This is because the momentum of the electrons h has decreased so their de Broglie wavelength has increased since l = mv (and the larger the wavelength, the more spaced out the pattern). 5) 2.2 × 103 ms–1. Rearrange the de Broglie wavelength equation for speed: 6.63 × 10 -34 h v = ml = = 2182.0... = 2.2 × 103 ms–1 (to 2 s.f.). 1.67 × 10 -27 × 1.819... × 10 -10

TIP

Chocolate or peanut butter? Sweet or salty popcorn? Why not just have both — that’s the way physicists like to do things around here... EM radiation and electrons can be both waves and particles, the greedy so-and-sos.

12. Progressive Waves

Section 3 — Waves

Quick Questions 1) True or false? Progressive waves always carry energy from one place to another by oscillating particles. 2) What is meant by the phase difference between two waves? 3) Give two pieces of evidence that show waves carry energy.

Now try these: 4) Calculate the wavelength of a 99.5 MHz radio signal when it is travelling in a vacuum. Use the formula c = f l and c = 3.00 × 108 ms–1. 5) A man sat in a boat on the sea measures that the boat rises and falls 8 times 1 in 20 seconds. Calculate the frequency of the water waves. Use f = T . 6) The graph below shows the displacement-distance graph for a water wave. What is the amplitude of the wave? 7) Between which two points is there an exact number of wavelengths? 8) Calculate the speed of the wave if the time period of the wave is 20.0 ms. 1 Use the formulas c = f l and f = T .

Displacement (cm)

D

3 0

C

A

F

1.0

G Distance (m)

E –3 B

2.0

Section 3 — Waves

12. Progressive Waves

ANSWERS 1) False. Although waves like water waves and sound waves oscillate particles, electromagnetic waves don’t — they oscillate fields. 2) The amount one wave lags behind the other. 3) Any two from: e.g. electromagnetic waves can cause things to heat up. / X-rays and gamma rays can ionise atoms. / Loud sounds can cause vibrations. / Wave power can be used to generate electricity. 3.00 × 10 8 c 4) 3.02 m. Rearrange the equation to get l = f = = 3.015... = 3.02 m (to 3 s.f.). 99.5 × 10 6 5) 0.4 Hz. The time period is the time taken for one wavelength (one rise and fall), so T = 20 ÷ 8 = 2.5 s. 1 1 The frequency is given by f = T = 2.5 = 0.4 Hz. 6) 3 cm. This is the maximum magnitude of the displacement of the wave. 7) A and F. These points are exactly in phase, and are two wavelengths apart. 8) 29 ms–1. Over a distance of 2.0 m, there are 3.5 wavelengths, so one wavelength must be 1 1 2 ÷ 3.5 = 0.5714... m. If the time period is 20.0 ms, the frequency of the wave is T = 0.02 = 50 Hz. The speed of the wave is then given by c = f  l = 50 × 0.5714... = 28.5714... = 29 ms–1 (to 2 s.f.).

TIP

There are quite a few terms thrown around when it comes to defining the properties of waves — like cycle, phase and amplitude — but you might be able to bag yourself a few marks in the exam if you can reel them off.

13. Longitudinal and Transverse Waves Quick Questions 1) What is the difference between longitudinal and transverse waves? 2) State what types of wave a water wave and a sound wave are. 3) Which of the following types of wave can’t be polarised? A. Visible light B. Sound C. Gamma D. Infrared

Now try these: 4) Isaac is trying to tune his radio. As he moves his radio aerial around, the signal gets clearer and then more faded. Explain why this happens. 5) What is the minimum number of polarising filters you could pass unpolarised light through before the light is completely blocked? Explain your answer. 6) When light is reflected from a water surface, most of the reflected light waves have their electric fields aligned in one direction. Rita is attempting to take a photograph of some fish in a clear pond on a bright sunny day. Explain why using a polarising filter on the camera could help Rita to create a clearer photograph.

Section 3 — Waves

Section 3 — Waves

13. Longitudinal and Transverse Waves

ANSWERS 1) The direction of the vibrations in longitudinal waves is the same as the direction of the energy transfer, but the direction of the vibrations in transverse waves is perpendicular to the direction of the energy transfer. 2) A water wave is a transverse wave and a sound wave is a longitudinal wave. 3) B. Sound is a longitudinal wave, and only transverse waves can be polarised. 4) Radio signals are polarised in the orientation of the rods on the broadcasting aerial. For a strong signal to be received, Isaac needs to align the receiving aerial to the signal transmitted — so the signal comes and goes as he moves the radio aerial, as the transmitting and receiving aerials go in and out of alignment. 5) Two. Passing light through one polarising filter would cause all the light to be polarised in one direction. Another filter aligned at a right angle to the first would then block all the incoming light from getting through. 6) When sunlight is reflected from the surface of the pond, the electric fields of the light waves will mostly be aligned in one direction, i.e. the reflection will mostly be polarised. The polarising filter on the camera filters out this polarised light, so most surface reflections don’t pass through to the camera. Light from below the water’s surface isn’t polarised, so a lot of light from beneath the surface will pass through the polarising filter. This will allow Rita’s camera to capture a clear photograph of the fish in the pond, rather than the reflections from the surface.

TIP

If you’re ever asked for evidence that EM waves are transverse, polarisation is what you want to say. If EM waves were actually longitudinal (like sound waves) and you tried to polarise them, they’d be having none of it.

14. Superposition and Coherence

Section 3 — Waves

Quick Questions 1) State the principle of superposition. 2) State the smallest possible phase difference between two points on a sinusoidal wave that are exactly in phase, as a multiple of p. 3) Two waves are travelling along a string in opposite directions, as shown on the right. Describe what happens when the two waves meet.

Now try these: 4) An oscillator produces a wave on a string with a speed of 4 ms–1 and a wavelength of 0.8 m. Two points on the wave, A and B, have a phase difference of 3060°. Are points A and B in or out of phase? 5) A student sets up two speakers emitting coherent sound waves. He places a microphone on a slider at a set distance away from the speakers, and slides the microphone back and forth on the slider. The student finds that there are some places where the sound is loud and others where the sound is quiet. Explain why this occurs. 6) At a point where a loud sound is picked up by the microphone, the student measures the path difference between the sound waves to be 1.2 m. The student moves the microphone along the slider until he finds the next quiet region. He measures the path difference at this point to be 1.5 m. Determine the wavelength of the sound waves emitted from the speakers.

Section 3 — Waves

14. Superposition and Coherence

ANSWERS 1) The principle of superposition says that when two or more waves cross, the resultant displacement equals the vector sum of the individual displacements. 2) 2p. Two points are in phase if they’re a whole number of wavelengths apart, and so the shortest distance between two points in phase is one wavelength — which is 2p radians. 3) Both waves interfere with (non-total) destructive interference. At the meeting point the waves would form a trough smaller than the original trough. 4) Out of phase. 3060° ÷ 360° = 8.5, so A and B are 8.5 wavelengths apart, which means they’re exactly out of phase. 5) The sound waves from each speaker are interfering. Where the sound is loud, the sound waves have interfered constructively, as the waves have arrived at the same point in phase and reinforced each other. Where the sound is quiet, the sound waves have interfered destructively, as the waves have arrived at the same point out of phase and cancelled each other out. 6) 0.6 m. If there is a loud sound at a path difference of 1.2 m, there must be constructive interference here, and destructive interference at 1.5 m, where the quiet region is detected. The difference between these path differences is 1.5 – 1.2 = 0.3 m. As these points are next to each other, the difference between these points must be equal to half a wavelength. So one wavelength = 0.3 × 2 = 0.6 m.

TIP

I used to hate waves when I was younger, but as it turns out, it was just a phase — they seem much more coherent now... Anyway, don’t let me interfere with your revision, keep cycling through these cards.

15. Stationary Waves

Section 3 — Waves

Quick Questions 1) What is a stationary wave?

X

Y

2) What is the amplitude of vibration at a node of a stationary wave? 3) Name the points labelled X and Y on the stationary wave shown in the diagram above.

Now try these: 4) Jill sets up the apparatus shown on the right and switches on the microwave transmitter. The probe doesn’t detect any microwaves, but when Jill removes the metal plate, the probe does detect microwaves. Explain these observations.

metal plate

microwave transmitter

probe in fixed position

5) A stationary wave is created on a string fixed at both ends. The frequency of the first harmonic is 10 Hz and its wavelength is 3 m. What is the frequency and wavelength of the third harmonic? 6) A student is investigating the factors affecting the resonant frequencies of a piece of string. He measures the mass of the string to be 510 g and its length to be 3.0 m. He attaches one end of the string to a signal generator, and passes the string over a pulley which is 2.0 m from the signal generator. He attaches a hanging mass of 2.5 kg to the end of the string. What frequency should the signal generator be set 1 T to to obtain the first harmonic of the string? Use the formula f = 2l m and g = 9.81 Nkg–1. 7) If the student increases the distance between the oscillator and the pulley, what would you expect to happen to the frequency of the first harmonic? Explain your answer.

Section 3 — Waves

15. Stationary Waves

ANSWERS 1) A superposition of two waves, progressing in opposite directions, with same the frequency and wavelength. 2) 0. A node is a point on a stationary wave that doesn’t move, so the amplitude of vibration at a node is zero. 3) X — node, Y — antinode. 4) A stationary wave is created when the microwaves reflect back off the metal plate. The probe doesn’t detect any microwaves as it must be placed where a node is formed. When the plate is removed, there is no longer a standing wave, and so there’s no node, and the probe detects the microwaves that pass it. 5) f = 30 Hz, l = 1 m. The frequency of the third harmonic is just three times the frequency of the first harmonic, so it’s 3 × 10 = 30 Hz. The first harmonic displays a half wavelength, so if the wavelength is 3 m, then the length of the string must be 1.5 m. The third harmonic fits 1.5 wavelengths on the string, so length of string ÷ number of wavelengths = 1.5 ÷ 1.5 = 1 m. 0.51 M 6) 3.0 Hz. The mass per unit length of the string is given by m = L = 3.0 = 0.17 kgm–1. The tension in the string from the hanging mass is T = mg = 2.5 × 9.81 = 24.525 N. 1 T 1 24.525 The frequency of the first harmonic is f = 2l m = 2 × 2.0 0.17 = 3.002... = 3.0 Hz (to 2 s.f.). 7) The frequency of the first harmonic would decrease. This is because the length of string that the wave is on would be longer, and so the half wavelength at the resonant frequency would be longer. An increase in wavelength decreases the frequency.

TIP

For a stationary wave on a fixed string, harmonic number = number of half wavelengths on the stationary wave. Harmonic number of a stationery wave = number of half-pencils. Aha... ahahaha... HAHAHA... Ahem. Sorry.

16. Diffraction

Section 3 — Waves

Quick Questions 1) Explain why both blue and red light can be seen in the diffraction pattern of white light. 2) The diagram on the right shows a typical diffraction pattern produced when laser light is shone through a slit. Explain why there are some bright patches and some patches where there is no light.

Now try these: 3) One source emits sound waves with a wavelength of 0.5 m, and another source emits light waves with a wavelength of 600 nm. Compare the diffraction of the waves from the two sources when they pass through a gap of 1 m. 4) Describe how the diffraction pattern of a monochromatic red light source differs from that of a white light source, when light from each source gets passed through identical apertures. 5) A student passes microwaves from a transmitter through a gap. He uses a probe on the other side of the gap to detect a diffraction pattern. He expects the amplitude of the microwaves detected in the central region of the diffraction pattern to decrease when he shrinks the gap, but instead the probe no longer detects microwaves. Suggest why. 6) Holly lives in a small village in a hilly area. Her house is located several miles from the nearest radio transmitters. Suggest and explain why Holly is able to tune her radio to listen to Bangin’ Tunes Radio, which transmits radio signals at 945 kHz, but not Dull Talk Radio, which transmits at 97.5 MHz.

Section 3 — Waves

16. Diffraction

ANSWERS 1) White light is a mixture of different colours, which all have different wavelengths. When white light is shone through a narrow slit, all of the wavelengths are diffracted by different amounts, which produces spectra of colours — including blue and red light. 2) The bright patches are produced where the light waves from the slit are interfering constructively, and the patches where there is no light is where the light waves are interfering destructively. 3) For the sound waves, the gap is two wavelengths wide, so there will be noticeable diffraction of the sound and the waves will spread out. The light waves have a wavelength that is much, much smaller than the gap, so the diffraction of the light will be unnoticeable. 4) The red light would give a diffraction pattern with a central bright red fringe, and then dark and bright fringes of red alternating on either side, whereas the white light would form a central white fringe and then dark and bright fringes of rainbow light. 5) E.g. the gap is now much smaller than the wavelength of the microwaves, so the waves can’t pass through. 6) E.g. Bangin’ Tunes (BT) Radio transmits signals at a much lower frequency, and so a longer wavelength than Dull Talk (DT) Radio. So the gaps between the hills are probably of a similar size to the BT wavelengths but much larger than the DT wavelengths, causing the BT wavelengths to diffract more. So the BT wavelengths will spread out round the hills and reach Holly’s radio. But the diffraction of the DT wavelengths will be very small, so they won’t spread out as much and can’t reach Holly.

TIP

The amount of a wave diffracts depends on the wavelength of the waves and the slit width. Reducing the slit width increases the amount of diffraction, as long as it’s still wide enough for the waves to pass through.

17. Two-Source Interference

Section 3 — Waves

Quick Questions 1) Give three safety precautions you should take whilst working with lasers. 2) What kind of interference is taking place in the dark fringes of an interference pattern? 3) Explain how Young’s double-slit experiment gave evidence for the wave nature of light.

Now try these: 4) A student is investigating the interference of light using the set-up shown on the right. Explain why using a laser instead of a bulb would be better. 5) The student replaces the bulb with a laser and observes an interference pattern on the screen with a fringe spacing of w. What would the fringe spacing of the interference pattern be if the student halved lD the distance between the two slits? Use the formula w = s . 6) A scientist uses laser light projected at two slits to produce an interference pattern on a screen. Part of this pattern is shown on the right. The distance between the centres of the slits is 6.3 mm, and the distance from the slits to the screen is 1.2 m. lD Calculate the wavelength of the light in nm using the formula w = s .

screen slits bulb

0.66 mm

Section 3 — Waves

17. Two-Source Interference

ANSWERS 1) Any three from: e.g. don’t shine the laser towards a person. / Wear laser safety goggles. / Avoid shining the laser beam at a reflective surface. / Have a warning sign on display. / Turn the laser off when it’s not needed. / Never look directly into the laser. 2) destructive interference 3) E.g. the experiment proved that light could diffract and interfere, which are both unique properties of waves. 4) The light bulb isn’t a coherent light source, and it emits light of lots of different wavelengths, so no clear interference pattern would be produced on the screen. A laser does produce coherent and monochromatic light, so it would produce a clear interference pattern. 5) 2w. The equation shows that w (the fringe spacing) and s (the distance between the slits) are inversely proportional to each other. So as the distance between the slits is halved, then the spacing between the fringes doubles. 6) 580 nm. There are 7 bright fringes and so there is a spacing of 6 fringe widths in the pattern shown. So w = 0.66 ÷ 6 = 0.11 mm = 1.1 × 10–4 m. Rearranging the equation gives: 1.1× 10 -4 × 6.3 × 10 -3 ws = 5.775 × 10–7 l= D = 1.2 = 5.8 × 10–7 m (to 2 s.f.) = 580 nm

TIP

If you tried a two-source interference experiment with sound or water waves, you’d find pretty similar results. You’ve got to use coherent sources though, otherwise you’ll end up with a jumbled mess instead of a nice pattern.

18. Diffraction Gratings

Section 3 — Waves

Quick Questions 1) Give two applications of diffraction gratings. 2) Explain the appearance of the zero order line in the diffraction pattern produced when white light is passed through a diffraction grating.

Now try these: 3) A student finds a grey bird feather and holds it up to a white light source. She observes a rainbow pattern on the feather. Suggest why she observes this. 4) The light beam from a tunable laser (a laser that can be set to different wavelengths) is passed through a diffraction grating of 4.0 × 105 lines per metre. Calculate the wavelength of the light produced by the laser if the angle between the first order maximum and the zero order is 16°. Use the equation nl = dsinq. 5) How many maxima are produced by the diffraction grating? Use the equation nl = dsinq. 6) Give one way that the set-up could be altered to produce more maxima. Explain your answer.

TIP

You need to know how to derive that equation given in Q4 and 5. If you don’t know it, go and look it up and learn it. Go on, I’m watching you. (I’m not actually watching you, that would be creepy. Nice top by the way.)

Section 3 — Waves

18. Diffraction Gratings

ANSWERS 1) E.g. the production of spectra to help identify elements, and X-ray crystallography. 2) The zero order line is white because all the wavelengths of the light just pass straight through the diffraction grating. 3) The fine strands in the feather act as a diffraction grating, and so as the white light passes through the feather, the different wavelengths within the white light are spread out by different amounts, creating a rainbow pattern. d sin q 4) 6.9 × 10–7 m. Rearrange the equation for wavelength: l = n . d = 1 ÷ 4.0 × 105 = 2.5 × 10–6 m and n = 1 for the 1st order maximum. Put the numbers into the equation: l = 2.5 × 10–6 × sin(16°) = 6.89... × 10–7 = 6.9 × 10–7 m (to 2 s.f.). nl 5) 7. Rearrange the equation to get sin q = d . The largest possible value of sin q is 1, d d nl 2.5 × 10 -6 so d  ≤ 1 which gives n ≤ l . l = = 3.627..., so n ≤ 3.627... 6.89... × 10 -7 This means the that largest value of n is 3, as n is an integer. So the total maxima is 7 (the zero order maximum, plus a fringe either side of the zero order maximum for n = 1, n = 2 and n = 3). 6) Any one from: decrease the wavelength of incident light / replace the diffraction grating for one that has a larger distance between the slits. Both of these decrease how much the pattern spreads out (as q is decreased), meaning more maxima will be produced.

19. Refractive Index

Section 3 — Waves

Quick Questions 1) What would happen to the optical fibre signal shown on the right if it lost some energy due to absorption by the optical fibre material? 2) What is the purpose of the cladding in a step-index optical fibre?

Now try these: 3) Explain why the refractive index of a substance can never be less than 1. 4) Compare the differences and similarities between modal and material dispersion in optical fibres. Give one way for each that can reduce the effects of the dispersion.

B

A

C



5) Three rectangular blocks of transparent material — A, B and C — are placed next to each other so that they are touching, as shown in the 78° diagram on the right. A laser beam is shone into block A and the light beam follows the path shown in the diagram. Calculate q if the refractive index of block A is 1.52 and the refractive index of block B is 2.15. Use the equation n1sinq1 = n2sinq2. n 6) Explain whether block C has a higher or lower refractive index than block A. You can use sinqc = n2 . 1

TIP

Remember, if an incident light beam hits a boundary at its critical angle, it will refract along that boundary. Light undergoes total internal reflection for all angles greater than this critical angle, so there’s no refraction.

Section 3 — Waves

19. Refractive Index

ANSWERS 1) The amplitude of the signal would be reduced. 2) To allow total internal reflection within the optical fibre so that a signal can be transmitted from one end to the other. Cladding also protects the fibre from scratches that could allow light to escape. 3) The refractive index of a substance is the ratio between the speed of light in a vacuum and the speed of light in that substance. Since light can’t travel any faster than it does in a vacuum, the refractive index is always greater than (or approximately equal to) 1. 4) Both types of dispersion mean some light rays take longer to reach the end of the fibre than others, and both types can lead to pulse broadening. Modal dispersion is caused when light rays enter the fibre at different angles so take different paths, meaning some rays have longer to travel, whereas material dispersion occurs because light of different wavelengths travel at different speeds in the fibre. To reduce the effects, you can e.g. use a single-mode fibre to only let light take one path, stopping modal dispersion, and use monochromatic light to stop material dispersion. 5) 63°. The light ray has been reflected at the boundary between B and C, so the angle of incidence on block C is 78 ÷ 2 = 39°. By simple geometry, this is equal to the angle of refraction of the ray going from block A to n sin q 2.15 × sin 39° = 0.8901... block B. Rearranging the formula gives sinq1 = 2 n 2 = 1.52 1 –1 so q1 = sin (0.8901...) = 62.892... = 63° (to 2 s.f.). 6) E.g. if an angle of incidence of 39° on C produces total internal reflection, then the critical angle must be 39° or lower. Assuming it’s 39°, rearrange the equation for n2 = n1sinqc = 2.15 × sin39° = 1.353... = 1.4 (to 2 s.f.), so the refractive index of C must be 1.4 or lower, which is lower than the refractive index of A.

20. Scalars and Vectors

Section 4 — Mechanics

Quick Questions 1) Give two examples of scalar quantities. 2) True or false? When drawing a scale diagram to determine the resultant vector of two vectors, the two vectors should be drawn tip-to-tail.

Now try these: 3) A balloon has a displacement from the horizontal ground of 1.8 m at an angle of 65° to the horizontal. Which of the following gives the vertical component of the balloon’s displacement? C. 1.8 sin(65°) D. 1.8 cos(65°) A. 1.8 tan(65°) B. 652 + 1.82 4) An arrow is fired towards a dragon. Its velocity can be resolved into a northwards component of 67 ms–1 and an eastwards component of 12 ms–1. Determine the magnitude of the velocity of the arrow. 5) A knight runs towards the dragon at a velocity of 2.4 ms–1, on a bearing of 032°. Determine the northward component of the knight’s velocity. 6) The dragon takes off from the ground and flies 4.8 m vertically upwards. It then flies a distance of 6.3 m horizontally. What is the total displacement of the dragon from its starting position? Give the direction as an angle to the horizontal.

TIP

I bet you thought all that trigonometry in GCSE maths would never come in handy. Make sure you remember how to use the sine, cosine and tangent functions — SOHCAHTOA is a helpful mnemonic for that...

Section 4 — Mechanics

20. Scalars and Vectors

ANSWERS 1) Any two from: e.g. speed / mass / distance or length / temperature / time / energy. 2) True. You then draw the resultant vector as a straight line connecting the first tail to the last tip.

5) 2.0 ms–1. The bearing tells you the angle between north and the velocity, clockwise from north. Since we know the hypotenuse (the resultant velocity, see diagram), and adjacent want the adjacent (the north component), use cosine. cosq = hypotenuse so: adjacent = hypotenuse × cosq = 2.4 cos 32° = 2.0353... = 2.0 ms–1 (to 2 s.f.) 6) 7.9 m, 37° above the horizontal. Use Pythagoras to find the magnitude of the resultant displacement, sr. sr = 4.8 2 + 6.3 2 = 7.920... = 7.9 m (to 2 s.f.) opposite The angle from the horizontal can be found using tanq = adjacent (see diagram), opposite 4 . 8 n = tan -1 b l = 37.3... = 37° (to 2 s.f.). so q = tan–1 d 6.3 adjacent

2.4

32°



not to scale

not to scale

N

ms –1

1.8 m 3) C. The resolved components form a right-angled triangle with the displacement as the hypotenuse, as shown on the right. The vertical component is opposite the 65° angle. opposite 65° Rearrange sinq = hypotenuse for opposite = hypotenuse × sinq = 1.8 sin 65°. 4) 68 ms–1. The two components of velocity and the resultant velocity form a right-angled triangle, so use Pythagoras' theorem to calculate the hypotenuse, to give the magnitude of the resultant, vr. vr = 67 2 + 12 2 = 68.0661... = 68 ms–1 (to 2 s.f.)

top down view, not to scale

6.3 m 4.8 m

sr 

21. Forces (1)

Section 4 — Mechanics

Quick Questions 1) True or false? An object with a resultant force of 0 N acting on it must be at rest. 2) Look at the box on the slope on the right. Which of the following is the component of the box's weight parallel to the slope? B. W sinq C. W cosq D. W – (sinq + cosq) A. W tanq

Now try these:

N

3) Two people are moving a bed. The diagram on the right shows the forces the people exert on the bed. Determine the resultant force acting on the bed. Give the direction of the force as a bearing. 551 N

529 N

35°

42°

not to scale

670 N

TIP



W

25.0 N 55.0 N

not to scale

4) An actor is suspended from two wires to allow them to fly across the stage. The forces acting on them at a certain point are shown in the diagram on the left. Determine whether the actor is in equilibrium.

For your physics A-level, you'll only be expected to deal with forces acting on an object that are coplanar — that is, they all act along a single 2D plane. Sadly nothing to do with being the copilot of a Spitfire.

Section 4 — Mechanics

21. Forces (1)

ANSWERS 1) False. An object that's moving at a constant velocity will also have a 0 N resultant force acting on it. perpendicular 2) B. The weight forms a right-angled triangle with its  component components and angle q (as shown on the right). W opposite parallel Rearrange sinq = hypotenuse for opposite = hypotenuse × sinq = W sinq. component 3) 60.4 N at a bearing of 294°. The two forces drawn tip-to-tail form a right-angled triangle with the resultant force. Use Pythagoras' theorem to determine the magnitude of the resultant force, F. F = 55.0 2 + 25.0 2 = 60.4152... = 60.4 N (to 3 s.f.) Use the tangent function to determine the angle of the resultant force to the 55.0 N force. opposite 25.0 tan q = , so q = tan–1 b = 24.44... °. The 55.0 N force is at a bearing of 270°, 55.0 l adjacent so the bearing of the resultant force is 270 + 024.44... = 294.44... = 294° (to 3 s.f.)

4) The actor is not in equilibrium. For the actor to be in equilibrium, there has to be zero resultant force on the actor. Resolve the 551 N and 529 N forces into vertical and horizontal components, FV and FH. For the 551 N force: FV = 551 sin(35°), FH = 551 cos(35°) For the 529 N force: FV = 529 sin(42°), FH = 529 cos(42°) Vertical resultant force = 551 sin(35°) + 529 sin(42°) – 670 = 0.010... ≈ 0 N Horizontal resultant force = 551 cos(35°) – 529 cos(42°) = 58.229... N to the left. Since there is a horizontal resultant force to the left, the actor cannot be in equilibrium. (If you just checked all the horizontal components first and found that there was a non-zero resultant force, you won’t need to do the vertical forces as well, since you already know the actor isn’t in equilibrium.)

22. Forces (2)

Section 4 — Mechanics

Quick Questions 1) A free-body diagram of a ball, drawn to scale, is shown on the right. In which direction does the resultant force act? 2) Three coplanar forces are acting on an object. Describe how you could use a scale diagram to show whether or not the object is in equilibrium.

Now try these: 3) A wasp is caught in a spider web. It tries to fly away, and experiences the forces shown in the diagram on the right. The wasp is in equilibrium. Determine the size of force X. Give your answer to 3 significant figures. 6.5 N

not to scale

TIP



2.9 N W

0.54 N 45° 26° 0.87 N

not to scale

X

4) The diagram on the left shows a boot with weight W at rest on an inclined plane. A normal contact force of 6.5 N and a frictional force of 2.9 N up the slope act on the boot. Determine the angle, q, of the inclined plane.

For most equilibrium and resultant forces problems, you can use trigonometry or scale diagrams to solve them. When drawing scale diagrams, make sure you're armed with a ruler, a protractor and a really sharp pencil.

Section 4 — Mechanics

22. Forces (2)

ANSWERS 1) To the left. The up and down force arrows are of equal length, and act in exactly opposite directions, so they cancel out. The right force arrow is shorter than the left force arrow, so there is a resultant force to the left. 2) Draw the forces to scale and ‘tip-to-tail’ (by having the next force arrow start where the previous one ended). If the forces form a closed triangle (the last force arrow ends at the start of the first force arrow) then the object is in equilibrium. If the forces don't form a closed triangle, it's not in equilibrium. 3) 1.16 N. Resolve the two forces along the line of force X (the dashed line in the diagram). As the wasp is in equilibrium, the forces along this line must cancel out. Resolving parallel to force X: for the 0.54 N force, F1 = 0.54 cos(45°), for the 0.87 N force, F2 = 0.87 cos(26°). Since the wasp is in equilibrium, F1 + F2 – X = 0, so: X = F1 + F2 = 0.54 cos(45°) + 0.87 cos(26°) = 1.1637... = 1.16 N (to 3 s.f.) 4) 24°. Since the boot is at rest, the weight W must form a closed triangle with the two other forces acting on the boot. These forces act at right angles to each other, so they will form a right-angled triangle with W. The angle of the incline is equal to the angle formed between the weight and the 6.5 N force. So, use the tangent function to determine the angle: opposite 2.9 tan q = , so q = tan–1 b l = 24.04... = 24° (to 2 s.f.) 6.5 adjacent

 W

6.5 N

2.9

N

23. Moments

Section 4 — Mechanics

Quick Questions 1) State the principle of moments. 2) Which arrangement below shows a couple acting on the seesaw? F A. B. C. F F F

F

F

Now try these: 3) A student is assembling a flat-pack desk. They need to screw in different bolts A using a hex key, like the one shown on the right. Either end of the key B can be inserted into the bolt, and then the key is turned to tighten the bolt. The student inserts end A into a bolt, and applies a force of 14 N to produce a moment of 0.73 Nm. At what perpendicular distance from the pivot was the force applied? Use the equation moment = Fd. 4) Explain why the student finds it harder to tighten a bolt when they insert end B of the hex key compared to when they insert end A. 5) To attach the second set of legs to the desk, the student asks a friend to hold up one end of the desk, as shown on the right. The desk is pivoted around the already attached set of legs. What force must the friend use to keep the desk still? Use the equation moment = Fd.

135.0 cm 65.0 cm 352 N

a lanky student

Section 4 — Mechanics

23. Moments

ANSWERS 1) The principle of moments states that for a body to be in equilibrium, the sum of the clockwise moments about any point equals the sum of the anticlockwise moments about the same point. 2) B. A couple consists of two parallel forces acting in opposite directions, to produce a turning effect. 3) 0.052 m. Rearrange equation for d, d = moment ÷ F = 0.73 ÷ 14 = 0.05214... = 0.052 m (to 2 s.f.) 4) If the student inserts end B into a bolt, the maximum perpendicular distance from the pivot at which they can apply a force is smaller than if they insert end A into the bolt. Since moment = Fd, this means they need to apply a larger max. perp. distance for force to produce the same moment when end B is in the bolt than when end B inserted end A is in the bolt, and so it is harder to turn the bolt and tighten it.

max. perp. distance for end A inserted

A 5) 169 N. Using the principle of moments, clockwise (C) moments = anticlockwise (A) moments. This means FCdC = FAdA , so: FA = (FCdC ) ÷ dA = (352 × 65.0) ÷ 135.0 = 169.4814... = 169 N (to 3 s.f.) (Moments are usually given in the units Nm. However, there's no need to convert the units of distance from cm to m here, since all the distances are in the same units, and the distance units cancel out.)

TIP

When dealing with turning effect questions, take a moment (haha) to check where the pivot is. Any forces that act through the pivot of the system won't produce a moment, since their distance from the pivot will be zero.

B

24. Mass, Weight and Centre of Mass

Section 4 — Mechanics

Quick Questions 1) True or false? An object suspended from a string will always hang so that its centre of mass is vertically in line with the string. 2) Why is the weight of an object on the Moon different to its weight on Earth?

Now try these: 3) A trading card with a uniform density is shown on the right. Describe how you could determine the centre of mass of the trading card using only a ruler and pencil. 4) Camille travels into space in a spaceship with a mass of 2.03 × 106 kg. When it lands on an unknown planet, the weight of the spaceship is 1.75 × 107 N. Determine the gravitational field strength on the unknown planet. 5) A student has four figurines, as shown to scale below. The centre of mass of each figurine is shown by a white X. Which figurine is the most stable? Explain your answer. A.

B.

C.

D.

1

1

Section 4 — Mechanics

24. Mass, Weight and Centre of Mass

ANSWERS 1) True. You can use this to find the centre of mass of an irregular 2D object. Hang it from a string, and trace a line from the string vertically downwards on the object. Then hang it from a different point, and trace the vertical line for that position. The point where these lines intersect gives the location of the centre of mass. 2) The weight of an object depends on the strength of the gravitational field it's in, and the gravitational field strength on the Moon is different to the gravitational field strength on Earth. 3) E.g. measure the height of the card, mark the midpoint, and use it to draw a line of symmetry. Then measure the width of the card, mark the midpoint, and draw another line of symmetry. The point where the lines of symmetry intersect gives the geometric centre of the card, which is also where the centre of mass is (since the card has a uniform density). 4) 8.62 Nkg–1. weight = mass × g, so: g = weight ÷ mass = 1.75 × 107 ÷ 2.03 × 106 = 8.6206... = 8.62 Nkg–1 (to 3 s.f.) 5) Figurine D is the most stable. A figurine will topple over if it is tilted enough that a line drawn directly down from its centre of mass falls outside its base. At this point, the centre of mass causes a moment about the edge of the base which will cause the figurine to fall over. Figurine D has the widest base, and the lowest centre of mass, which sits close to the middle of its base. This means it will need to be tipped the furthest in order for its centre of mass to cause a moment that would cause it to topple over.

TIP

Centres of mass can be a bit weird — sometimes they can even be located outside the object. For example, for a uniform ring, the centre of mass will actually be in the centre of the gap in the middle of the ring. Bonkers.

25. Displacement-Time Graphs

Section 4 — Mechanics

Quick Questions Where do you want them putting?

1) True or false? If an object is undergoing a constant acceleration, the rate of change of the gradient of its displacement-time graph will be constant. DIS

Now try these: 3) Renee goes on a run in a straight line. The displacement-time graph of the first half of her run is shown on the right. Determine the velocity at which Renee was running at t = 50 s. Give your answer to 2 significant figures. 4) Calculate Renee’s average velocity for the part of her run Ds shown on the graph. Use the equation v = and Dt give your answer to 2 significant figures.

DIS

2) Describe how you could determine the instantaneous velocity of an accelerating object at time t from its displacement-time graph. s/m 240 160 80

5) On the second half of her run, she runs back to the start, still travelling in a straight line. She starts off running at a constant 0 20 40 60 velocity for 30 s, before decelerating for 20 s, stopping for 30 s, and then running at a constant velocity for the next 20 s, until she reaches her starting position. Describe how the rest of the displacement-time graph would look for the second half of her run.

80 t/s

Section 4 — Mechanics

25. Displacement-Time Graphs

ANSWERS 1) True. The gradient gives the velocity, and acceleration is the rate of change of velocity, so if the rate of change of the gradient is constant, then the acceleration is constant. 2) Draw a tangent to the curve at time t and calculate the gradient of the tangent. 3) 3.4 ms–1. The velocity is given by the gradient of the displacement-time graph at 50 s. This is equal to the gradient between 25 s and 60 s, as velocity is constant in this period, so use these points to find the gradient. Dy = 220 - 100 = 3.4285... = 3.4 ms–1 (to 2 s.f.) gradient = 60 - 25 Dx 4) 3.3 ms–1. To find the average velocity, use the total change in displacement (Ds) and time (Dt). 260 Ds Ds = 260 m, Dt = 80 s, so: average velocity = = 80 = 3.25 = 3.3 ms–1 (to 2 s.f.) Dt 5) For the first 30 s, the displacement-time graph will be a straight line, starting from a displacement of 240 m, with a negative gradient. For the next 20 s, the line will curve so the steepness of the gradient decreases. The graph will then be a horizontal line for the next 30 s. For the final 20 s, s / m 240 the graph will again show a straight line with a negative gradient, ending at a displacement of 0 m. (All the gradients of sloped lines on the graph should be negative. This is because Renee’s on her return journey, so her displacement from her starting point will always be decreasing, and the velocity (equal to the gradient) must be negative, since the velocity 0 30 s 20 s 30 s 20 s t of the outwards journey was positive.) The graph is shown on the right.

TIP

A graph speaks a thousand words, and s-t graphs are no exception. The gradient of a slope on an s-t graph gives the velocity, and the shape of the line shows if the object is accelerating or not, and how it's accelerating.

26. Velocity-Time Graphs

Section 4 — Mechanics

Quick Questions 1) What type of motion does a horizontal line on a velocity-time graph represent?

v A

B

2) What is represented by the area under a velocity-time graph?

C

3) Which line on the velocity-time graph on the right has the largest magnitude of acceleration?

D t

Now try these: 4) An object is decelerating, and its deceleration is getting smaller with time. Describe how this motion would be represented on a velocity-time graph. 5) A child drops a ball and it bounces on the floor. The velocity-time graph of the ball’s motion is shown on the right. Any energy losses are negligible. Determine the time at which the ball hit the floor. Explain your answer. 6) Calculate the acceleration of the ball at time t = 1.0 s. 7) Calculate the height the ball was dropped from.

v / ms

-1

4.0 2.0 0.0 -2.0 -4.0

0.5

1.0

t/s

Section 4 — Mechanics

26. Velocity-Time Graphs

ANSWERS 1) Motion with a constant velocity. 2) The displacement of the object travelling. 3) A. This is the steepest line, and the steeper the line, the greater the magnitude of acceleration (as gradient equals acceleration). A and D have negative gradients, so they show negative accelerations (deceleration). 4) A curved, downwards sloped line which gets shallower as time increases (as the acceleration equals the gradient). Like this:

v

t

5) 0.4 s. When the ball bounces, it hits the ground with its maximum speed (having accelerated due to gravity). After bouncing it immediately moves from the ground at the same speed, but in the opposite direction — its velocity changes sign. This can be seen to happen at 0.4 s on the graph. 6) –10 ms–2. To find the acceleration, find the gradient of the line at t = 1.0 s. This is equal to the gradient between 0.4 s and 1.2 s (as the gradient is constant over this period), so use these points to find the gradient. Dy = - 4.0 - 4.0 = –10 ms–2 Gradient = 1.2 - 0.4 Dx 7) 0.8 m. The ball hits the ground at 0.4 s, so the area under the graph in the first 0.4 s gives the displacement of the ball from the child's hand to the ground. The graph forms a triangle, so: change in displacement = 0.5 × base × height = 0.5 × 0.4 × (–4.0) = –0.8 m. Since we're looking for the height, not a displacement, you can ignore the minus sign, so height = 0.8 m.

TIP

You can find the area under a v-t graph by splitting it into basic shapes. To find the area under a curved graph though, you'll need to use the counting squares method. So make sure you're happy working it out both ways.

27. Acceleration-Time Graphs

Section 4 — Mechanics

Quick Questions 1) How could you find an object’s change in velocity from an acceleration-time graph of its motion? 2) How is decelerating at an increasing rate shown on an acceleration-time graph?

Now try these: 3) The graph on the right shows the acceleration-time graph for an object which starts at rest and moves along a straight line. Calculate the change in velocity of the object during the first 4.0 s of its motion.

a / ms-2 6 4 2

0 4) During its journey, the object spends some time in 2 4 6 8 10 12 14 t / s motion with a constant speed, and some time at rest. -2 Determine the time period in which the object is at rest. -4 Explain your answer. v 5) A velocity-time graph of a car’s journey is shown on the left. Describe what the acceleration-time graph for this car’s journey would look like. You can assume that for any periods of changing acceleration, the acceleration changes at a constant rate. 0 10 20 30 40 50 t/s

Section 4 — Mechanics

27. Acceleration-Time Graphs

ANSWERS 1) Calculate the area under the graph. 2) A line below the time axis with a negative gradient. 3) 12 ms–1. Change in velocity = area under graph in first 4.0 s = ½ × base × height = 0.5 × 4.0 × 6.0 = 12 ms–1 4) Between 12 s and 14 s. E.g. for an object travelling at a constant velocity, or at rest, acceleration = 0 ms–2. So the object must be at rest between 4 s and 8 s, or 12 s and 14 s. The graph shows the object starts with an increasing, positive acceleration, so the object is speeding up. At 4 s, the acceleration drops to 0 ms–2, so it must now be travelling at a constant velocity. At 8 s, the object decelerates until its acceleration becomes 0 ms–2 again. This must be where the object is at rest. (You could also have worked out the area under the graph for 8 – 12 s to show that the velocity decreased the same amount as it increased between 0 and 4 s.) 5) The acceleration-time graph will start with a straight, horizontal line with a positive value of acceleration to show a constant acceleration. This will then drop instantly at 10 s to 0 ms–2, as the car starts travelling at a constant velocity. At 20 s there will be an instant jump to a negative value of acceleration, followed by a straight line with a positive gradient showing a decreasing deceleration that a decreases at a constant rate. The line reaches 0 ms–2 at 40 s. The line will –2 then be horizontal at 0 ms as the car is stationary. From 50 s, there will 0 10 20 30 40 50 t / s be a straight line with a positive gradient, as the car's acceleration increases at a constant rate. This graph is shown on the right.

TIP

If you're given a motion graph in the exam, make sure you pay close attention to the labels on the axes — you don't want to go calculating the area under a displacement-time graph to find the change in an object's velocity.

28. Motion With Uniform Acceleration

Section 4 — Mechanics

Quick Questions 1) Define all the variables in the equation s = ut + ½at2. 2) Which one of the following equations can be used to determine the final velocity of an object dropped from a known height under freefall? ]u vg C. s = ut + ½at2 D. v2 = u2 + 2as A. v = u + at B. s = + × t 2

Now try these: 3) A whale-watching ferry, initially at rest, moves away from a jetty with a constant acceleration of 2.84 ms–2. Determine the time taken for the ferry to reach a speed of 15.4 ms–1. Use the equation v = u + at. 4) A seal at rest spots the ferry, and swims away from it with a constant acceleration. ]u vg Calculate the speed of the seal after it has travelled 54.0 m in 41.0 s. Use the equation s = + × t. 2 5) A passenger throws some bait directly upwards from the edge of the ship so it will land in the water. It has an initial velocity of 3.10 ms–1 upwards, and accelerates due to gravity. When the bait hits the water, its displacement from its starting position is –1.31 m. Calculate the speed at which the bait hits the water. Assume air resistance is negligible. Use g = 9.81 ms–2 and the equation v2 = u2 + 2as. 6) A gannet flying over the ocean dives straight down towards the bait. It accelerates due to gravity and travels 7.45 m to the water in 0.730 s. What was the gannet’s initial velocity? Assume air resistance is negligible. Use g = 9.81 ms–2 and the equation s = ut + ½at2.

Section 4 — Mechanics

28. Motion With Uniform Acceleration

ANSWERS 1) s = displacement, u = initial velocity, t = time and a = acceleration. 2) D. You know the initial velocity (u) is 0 ms–1 and you know the acceleration (a) and displacement (s). D is the only equation that includes the final velocity (v), and doesn't include a variable you don't know (time, t). 3) 5.42 s. The boat starts at rest, so u = 0 ms–1. Rearrange v = u + at for t, so: t = (v – u) ÷ a = (15.4 – 0) ÷ 2.84 = 5.42253... = 5.42 s (to 3 s.f.)

]u vg 4) 2.63 ms–1. The seal starts at rest, so u = 0 ms–1. Rearrange the equation s = + × t for v, so: 2 2 s 2 × 54 . 0 v= t –u= – 0 = 2.6341... = 2.63 ms–1 (to 3 s.f.) 41.0 5) 5.94 ms–1. Taking upwards as the positive direction, u = 3.10 ms–1, a = –g = –9.81 ms–2, s = –1.31 m. So, v2 = 3.102 + (2 × (–9.81) × (–1.31)) = 35.3122. So, v = 35.3122 = 5.9424... = 5.94 ms–1 (to 3 s.f.) (You were asked for the speed, so you don’t need to worry about the sign here, but if you were looking for the velocity you’d take the negative root, since you know the bait is travelling downwards when it hits the water).

6) 6.62 ms–1 downwards. Taking downwards as the positive direction, a = g = 9.81 ms–2, s = 7.45 m, t = 0.730 s. 2 1 7.45 - (0.5 × 9.81× 0.730 2) Rearrange the equation for u: u = s - t 2 at = = 6.6248.... = 6.62 ms–1 (to 3 s.f.) 0.730 As downwards is positive, the initial velocity is downwards. (If you took upwards as the positive direction, you'll get – 6.62 ms–1 as your answer.)

TIP

Don't get too trigger-happy with these equations. Make sure the situation you're dealing with definitely has a constant acceleration before sticking your variables in. It's best to think before sticking anything anywhere really.

29. Acceleration Due to Gravity

Section 4 — Mechanics

Quick Questions 1) True or false? An object is in free fall if the only force acting on it is air resistance. 2) Would it be better to use a small and heavy object, or a large and light object to determine acceleration due to gravity? light–gates

Now try these: 3) An experimental set-up is shown on the right. A weighted card can be dropped so that it passes through two light-gates a fixed distance apart. Each light-gate will measure the speed of the card as it passes through it, and the time at which the card passed through. Describe how you could use this set-up to carry out an experiment to determine the value of g. 0.0 -0.5 -1.0 -1.5 d/m

0.1

0.2

0.3

0.4

0.5

0.6

t/s

4) A scientist plots a displacement-time graph for a falling ball, as shown on the left. They draw a tangent to the curve at t = 0.5 s, shown in red. They determine that the velocity at t = 0.2 s is –1.96 ms–1. Use this information and the graph to determine a value of g. Use the equation a = Dv . Dt

Section 4 — Mechanics

29. Acceleration Due to Gravity

ANSWERS 1) False. An object is in free fall if the only force acting on it is a gravitational force/weight. 2) The smaller, heavier object would be better. A smaller object will have less air resistance acting on it than a larger object, and a heavier object will have a larger weight than a lighter object, so the size of the air resistance on a small, heavy object will be much smaller than the size of its weight. This means the value of g calculated from the motion of the smaller, heavier object will be more accurate, as the air resistance has less of an effect on the results. 3) Drop the weighted card through the light-gates. Using the speeds and times recorded by the light-gates, calculate the change in velocity and change in time between the two light-gates. Divide the change in velocity by the change in time to get a value of the acceleration of the card, the magnitude of which can be taken as g (assuming air resistance is negligible). Repeat the measurements with the light-gates set at difference distances apart, and then take an average value for g from the values of g obtained from each distance. 4) 10 ms–2. Determine the velocity at t = 0.5 s from the tangent, - 2.0 - (- 0.2) change in y velocity at 0.5 s = gradient of tangent = = = –5.0 ms–1 change in x 0.66 - 0.30 Substitute these velocities and times into the equation for acceleration: - 5.0 - (- 1.96) a = Dv = = –10.1333... = –10 ms–2 (to 1 s.f.) 0.5 - 0.2 Dt The value is negative as downwards was taken as negative, so g = 10 ms–2.

TIP

Whenever you take a measurement, there'll be some uncertainty in the value. Devices like light-gates take recordings automatically, so reduce the uncertainty in measurements by removing the influence of human error.

30. Projectile Motion

Section 4 — Mechanics

Quick Questions 1) True or false? An object undergoing projectile motion is accelerating in both the horizontal and vertical directions. 2) An object is travelling at v ms–1 at an angle q above the horizontal. What equation should be used to determine the horizontal component of its velocity?

Now try these: 3) A girl fires a pebble from a slingshot. The pebble leaves the slingshot at a velocity of 7.21 ms–1, exactly along the horizontal, at a height of 1.64 m above the ground. Determine the time taken for the pebble to hit the ground. Assume air resistance is negligible. Use g = 9.81 ms–2 and the equation s = ut + ½at2 . 4) Calculate the horizontal distance travelled by the pebble. Use g = 9.81 ms–2 and the equation v = Ds . Dt 5) A darts player throws a dart towards a 0.25 m tall target that is 2.37 m away, as shown below. The –1 dart has an initial velocity of 10.2 ms at an angle of 12° to the horizontal. Will the dart hit the target? Assume air resistance is negligible. Use g = 9.81 ms–2 and the equations v = Ds and s = ut + ½at2. Dt 2.37 m

10.2

not to scale 0.25 m

-1

ms

12°

Section 4 — Mechanics

30. Projectile Motion

ANSWERS 1) False. In projectile motion, the object is only accelerating in the vertical direction.

v

2) v cos q. The velocity forms a right-angled triangle with its components, as shown on the right  adjacent horizonal component So cos q = hypotenuse = , so horizontal component = v cos q. v 3) 0.578 s. Deal only with the vertical components of the velocity, and take upwards as the positive direction. Since the pebble is fired directly along the horizontal, the initial vertical velocity, u = 0 ms–1. s = –1.64 m, a = –g = –9.81 ms–2. Rearrange the equation for t (ignoring the ut term, since u = 0 ms–1), 2 × (- 1.64) t = 2as = = 0.57823... = 0.578 s (to 3 s.f.) (- 9.81) 4) 4.17 m. Rearrange the equation for Ds: Ds = vDt = 7.21 × 0.57823... = 4.16905... = 4.17 m (to 3 s.f.) 5) The dart will hit the target. First find the time it takes the dart to travel 2.37 m horizontally. The horizontal component of the darts' velocity is given by v cosq = 10.2 cos(12°) 2.37 Rearrange the velocity equation for Dt: Dt = ∆v s = = 0.237... s 10.2 cos (12°) Find the vertical displacement in this time, taking upwards as positive. u = 10.2 sin(12°), a = –g = –9.81 ms–2 s = (10.2 × sin(12°) × 0.237...) + (0.5 × (–9.81) × 0.237...2) = 0.22698... m. This is positive and less than 0.25 m, the height of the target, so the dart will hit the target.

TIP

When it comes to projectile motion, the first thing you'll want to do is resolve the initial speed into horizontal and vertical components. So don't forget those trigonometric functions. They really get about don't they?

31. Newton’s Laws of Motion

Section 4 — Mechanics

Quick Questions 1) Which of Newton’s laws of motion can be written as the equation F = ma? 2) State Newton’s other laws of motion.

Now try these: 3) The diagrams below show some objects moving at a constant speed. The dashed lines show the paths followed by each object. Which object must have a resultant force acting on it? Explain your answer.

A.

B.

C.

4) An interaction pair is a pair of equal and opposite forces that act due to Newton's third law. Give an example of an interaction pair that act when a bauble is hanging from a string. 5) A tennis ball collides with a wall and decelerates at a rate of 22.5 ms–2 under a force of 1.31 N. Determine the mass of the tennis ball in grams. Use the equation F = ma. 6) A train of mass 4.66 × 105 kg sets off from rest with an initial acceleration of 0.446 ms–2­. It experiences a constant driving force and initially experiences no resistive forces. After a period of time, the train reaches a constant velocity. Determine the size of the total resistive force acting against the motion of the train when it is moving at a constant velocity. Use the equation F = ma.

Section 4 — Mechanics

31. Newton’s Laws of Motion

ANSWERS 1) Newton’s second law. 2) Newton’s first law — the velocity of an object will not change unless a resultant force acts on it. Newton’s third law — if an object A exerts a force on object B, then object B exerts an equal and opposite force on object A. 3) B. The other two objects are moving in a straight line, and as their speed is constant, so is their velocity, and according to Newton's first law the resultant force acting on them must be 0 N. The bird travels in a curved path, so its direction, and therefore velocity, must be changing, so there must be a resultant force acting on it. 4) E.g. the gravitational force of the Earth acting on the bauble (the bauble's weight), and the gravitational force of the bauble acting on the Earth. / The upwards tension force acting on the bauble and the downwards tension force acting on the string. (Watch out. The weight of the bauble and the tension force acting on the bauble don't make a valid interaction pair for Newton's third law. They may be the same size, and act in opposite directions, but they're different types of force, and are acting on the same object.) 5) 58.2 g. Rearrange F = ma for mass: m = F ÷ a = 1.31 ÷ 22.5 = 0.05822... kg = 58.2 g (to 3 s.f.) 6) 208 000 N. The driving force remains constant, and when the train is travelling at a constant velocity, it must have a zero resultant force acting on it. This means that the total resistive force at this time must be equal to the driving force. Use F = ma to determine the size of the driving force when the train started moving, and so the size of the final total resistive force: F = ma = 4.66 × 105 × 0.446 = 207 836 = 208 000 N (to 3 s.f.)

TIP

Newton's second law gave a mathematical proof that all falling objects undergo the same acceleration due to gravity, which backed up the experimental evidence from Galileo 100 years earlier. Aaah, the interplay of science.

32. Drag, Lift and Terminal Speed

Section 4 — Mechanics

Quick Questions 1) True or false? The faster an object is moving through air, the lower the air resistance acting on it. 2) True or false? The drag force felt by an object moving in a fluid depends on the object's shape. 3) What is meant by an object's terminal speed?

Now try these: 4) The diagram on the right shows a child sliding down a playground slide. State the direction of the frictional force acting on the child, and describe the effect it will have on the child's kinetic energy. 5) Explain how the air flow around an aeroplane wing produces lift. 6) A skydiver is falling towards the ground, and reaches a terminal speed. Shortly after reaching their terminal speed, the skydiver opens their parachute. Explain how the forces acting on the skydiver change after they open their parachute, and how this will affect the motion of the skydiver as they fall. 7) Taking a downwards velocity as positive, describe the velocity-time graph of the skydiver from the point at which their parachute is first fully open to the point at which they hit the ground.

TIP

Try not to get drag (fluid friction) and the friction between two surfaces mixed up. In general, the friction from a surface will stay roughly the same as an object moves across it, but drag will change based on the object's speed.

Section 4 — Mechanics

32. Drag, Lift and Terminal Speed

ANSWERS 1) False. Air resistance increases with the speed of the object it’s acting on. 2) True. The larger the area of the object pushing against the fluid, the greater the drag force. 3) The speed at which the driving force(s) acting on the object is (are) equal to the frictional force(s) acting on it. 4) The frictional force will act up the slope of the slide (in the opposite direction to the child's motion). It will cause some of the child's kinetic energy to be converted to heat and sound energy. 5) As the aeroplane wing moves through the air, it pushes down on the air (and changes the direction of the air flow so that air moves beneath the wing). Due to Newton’s third law, this downwards force on the air causes an equal and opposite upwards force from the air on the wing. This upwards force is lift. 6) Before opening the parachute, the weight of the skydiver is equal to the air resistance acting on them. When the skydiver opens their parachute, the surface area of the skydiver increases. This means the air resistance acting on the skydiver will increase, so that it is bigger than their weight. This creates an upwards resultant force that causes them to decelerate. This decrease in speed will cause the air resistance on the skydiver to decrease. The skydiver will continue to decelerate until the air resistance is equal to their weight again, and the skydiver reaches a new terminal speed at the point when no resultant force acts on them. 7) The velocity-time graph (as shown on the right) will show a downwards sloping curve, starting from a non-zero positive velocity. The gradient of the curve will get gradually more shallow with time, until the curve becomes a horizontal line at the new value of terminal speed. It then drops abruptly to 0 when the skydiver hits the ground.

v

t

33. Momentum and Impulse

Section 4 — Mechanics

Quick Questions 1) What quantity is given by the area under a force-time graph for an interaction? 2) What is the difference between an elastic collision and an inelastic collision?

Now try these: 3) A gymnast dismounts from a balance beam, lands on a crash mat and comes to rest. She has a mass of 51 kg and hits the crash mat with an initial velocity of 1.85 ms–1. Determine the magnitude of the impulse that acts on the gymnast as she collides with the crash mat. Use the equation FDt = D(mv). 4) Explain, with regards to Newton’s second law, why landing on the crash mat is less likely to cause injury than the gymnast landing directly onto the floor. 5) Two magnets, A and B, are held together at rest with their north poles facing each other. Magnet A has a mass of 124 g and magnet B has a mass of 186 g. When the magnets are released, they move away from each other. Magnet A moves away with an initial velocity of 0.121 ms–1. Ow... Determine the initial velocity at which magnet B moves away. 6) During a rugby match, Ewan tackles Hamish, who is at rest. Ewan has a mass of 76 kg and Hamish has a mass of 83 kg. Immediately after the tackle, the two players move through the air together with a velocity of 1.35 ms–1 to the right. Determine the velocity at which Ewan tackled Hamish.

Section 4 — Mechanics

33. Momentum and Impulse

ANSWERS 1) The impulse on the object the force acts on / the change in momentum of the object the force acts on. 2) Kinetic energy is conserved in an elastic collision, but it isn’t conserved in an inelastic collision. 3) 94 Ns. The gymnast comes to rest, so the magnitude of her change in momentum is equal to her initial momentum. Impulse = FDt = D(mv) = 51 × 1.85 = 94.35 = 94 Ns (to 2 s.f.) 4) From Newton’s second law, force is equal to the rate of change of momentum. The crash mat increases the time over which the gymnast is brought to rest from her initial velocity. This reduces the gymnast's rate of change of momentum. Due to Newton's second law, this means that the force acting on her is less than it would be if she landed directly on the floor, and this smaller force is less likely to cause injury. 5) –0.0807 ms–1. The initial momentum of the system, when the magnets are held at rest, is 0 kgms–1. Momentum is conserved, so the total momentum after the magnets have been released must also equal 0 kgms–1. 0 = pA + pB which means 0 = mAvA + mBvB , so: vB = –(mAvA) ÷ mB = –(124 × 0.121) ÷ 186 = 0.080666... = –0.0807 ms–1 (to 3 s.f.) (You didn't need to convert the mass from g to kg here, because the mass units cancel out.) 6) 2.8 ms–1 to the right. Total momentum before the collision is equal to the total momentum after the collision. Hamish was at rest before the collision, so he had no momentum. So, the momentum of Ewan and Hamish after the collision equals the momentum of Ewan before the collision. mEvE = mEHvEH, so vE = (mEHvEH) ÷ mE = ((76 + 83) × 1.35) ÷ 76 = 2.8243... = 2.8 ms–1 (to 2 s.f.)

TIP

From splitting nuclei to colliding planets, momentum is always conserved — it's a fundamental principle of physics. And remember, momentum is a vector, so take care with those + and – signs in your calculations.

34. Work and Power

Section 4 — Mechanics

Quick Questions 1) In physics, what does it mean to say 'work is done'? 2) In which of these situations is no work being done by a force? A. a car moving at a constant velocity B. a pear at rest on a table C. a skydiver who has just jumped from a plane

Now try these: 8N

3) A trolley is pulled along a flat surface by a cord, at an angle to the horizontal, as shown on the right. The trolley is pulled across a distance of 58.0 m with a constant force of 108 N. A total of 3510 J of work is done on the trolley by this force. Determine the angle of the cord to the horizontal. Use the equation W = F s cosq. force / kN 4) A gardener pushes a wheelbarrow with a constant horizontal force of 51.5 N. The wheelbarrow travels at a constant speed of 1.42 ms–1. 12 Determine the work done on the wheelbarrow in 12.5 seconds. 8 Use the equations P = DW and P = Fv. 4 Dt 5) The graph on the right shows how the force acting on a lift changes with distance as it moves between floors. 0 Calculate the total work done when the lift moves through the first 16 m.

4

10 

8

12 16 20 24 distance / m

Section 4 — Mechanics

34. Work and Power

ANSWERS 1) Energy is transferred when an object is moved by a force. 2) B. The pear is not moving, so no work is being done on it. In A., work is being done against resistive forces to keep it travelling at a constant velocity. In C., work is being done by gravity as the skydiver falls. 3) 55.9°. Rearrange the equation for q: q = cos–1(W ÷ (Fs)) = cos–1(3510 ÷ (108 × 58.0)) = 55.92... = 55.9° (to 3 s.f.) 4) 914 J. First find the power, P = Fv = 51.5 × 1.42 = 73.13 W Rearrange P = DW for work done: DW = PDt = 73.13 × 12.5 = 914.125 = 914 J (to 3 s.f.) Dt 5) 156 000 J. Total work done is equal to the area under a force-distance graph. force / kN Up to 16 m, the graph can be split into two trapeziums. 12 Area of trapezium 1 = 0.5 × (6 + 10) × 12 000 = 96 000 J Area of trapezium 2 = 0.5 × (4 + 6) × 12 000 = 60 000 J 8 So, total work done = 96 000 + 60 000 = 156 000 J 1 2 4 (You could also have split the graph into three 2 m by 12 kN triangles and two rectangles. Either way, you should get the same answer.) 0 4 8 12 16 20 24 distance / m

TIP

Don't be fooled, there doesn't have to be a resultant force acting on an object for work to be done. As long as there's a component of a force acting in line with the direction of the object's motion, work is being done.

35. Conservation of Energy and Efficiency

Section 4 — Mechanics

Quick Questions 1) What is the conservation of energy principle? 2) Describe the energy transfers that occur as a box begins sliding down a rough slope.

Now try these:

6) Calculate the maximum speed that the child could reach during their swing. Assume resistive forces are negligible. Use the equation Ek = ½mv2.

2.6

4) A microwave has an efficiency of 57.0%, and a useful output power of 852 W. useful output power What is its input power? Use the equation efficiency = . input power 5) The diagram on the right shows a child on a swing, at the top of their swing. Calculate the change in the gravitational potential energy of the child as they move from the top of their swing to the swing’s lowest point. Use g = 9.81 Nkg–1 and the equation DEp = mgDh.

m

3) A bird with a mass of 1.21 kg takes off from the ground. When it reaches its maximum height, it has gained 192 J of gravitational potential energy. Determine the maximum height of the bird's flight. Use g = 9.81 Nkg–1 and the equation DEp = mgDh.

mass = 42 kg

34°

not to scale

Section 4 — Mechanics

35. Conservation of Energy and Efficiency

ANSWERS 1) Energy cannot be created or destroyed. Energy can be transferred from one form to another but the total amount of energy in a closed system will not change. 2) The box loses gravitational potential energy as its height above the ground decreases, which is transferred to kinetic energy as the box accelerates down the slope. Some of the box's kinetic energy is transferred to heat energy as work is done against the friction of the slope and air resistance acting on the box. 3) 16.2 m. Rearrange the equation for Dh: Dh = DEp ÷ mg = 192 ÷ (1.21 × 9.81) = 16.17509... = 16.2 m (to 3 s.f.) 4) 1490 W. 57.0 % efficiency expressed as a decimal = 0.570. Rearrange the equation for input power: useful output power = 852 = 1494.736... = 1490 W (to 3 s.f.) input power = 0.570 efficiency 5) 180 J. First determine the change in the vertical height of the swing. This is equal to swing length, 2.6 m, minus the vertical component of its length at the position shown. So: Dh = 2.6 – 2.6 cos(34°) = 0.4445... m DEp = mgDh = 42 × 9.81 × 0.4445... = 183.143... = 180 J (to 2 s.f.) 6) 3.0 ms–1. Maximum speed occurs at the bottom of the swing. At this point, all of the gravitational potential energy lost by the child as they move from the top of their swing to the bottom has been transferred to kinetic energy. Rearrange the Ek equation for v: v = 2mEk = 2 × 183.144... = 2.9531... 42 = 3.0 ms–1 (to 2 s.f.)

TIP

Just like momentum, energy is always conserved. A specific type of energy might not be conserved — e.g. kinetic energy may be transferred to heat energy — but the total energy before and after an event will always be equal.

36. Properties of Materials

Section 5 — Materials

Quick Questions 1) Name the law that says the extension of a stretched object is proportional to the force applied. 2) What’s the difference between elastic deformation and plastic deformation?

Now try these: 3) 22 N is applied to a spring with a spring constant of 1800 Nm–1. Calculate its extension using F = kDL. 4) A box has a density of 650 kgm–3. A sample of oil has a mass of 1.5 kg and a volume of 1.63 × 10–3 m3. m Use r = V to determine if the box would float on the oil. 5) By considering the energy transfers that take place during a car crash, explain why having a crumple zone at the front of a car is safer for passengers than not having one.

6) The diagram on the right shows a bowl of frogs. The total 8 cm mass of all the frogs in the bowl is 2200 g. Using the dimensions given on the diagram, estimate the average density of a frog. m Use r = V and area of a circle = pr², and give your answer in gcm–3.

20 cm

7) A mass is hanging on a spring. 0.12 J of energy is used to pull the mass down, but when the mass is released, only 0.09 J of energy is transferred to the kinetic energy of the mass. Suggest why. You can ignore the effects of friction and air resistance in this question.

Section 5 — Materials

36. Properties of Materials

ANSWERS 1) Hooke’s law. 2) If a material undergoes elastic deformation, the material will return to its original shape and size once the forces acting on it are removed, whereas if a material undergoes plastic deformation it won’t return to its original shape and size once the forces acting on it are removed. 3) 0.012 m. F = kDL, so DL = F ÷ k = 22 ÷ 1800 = 0.0122... = 0.012 m (to 2 s.f.) 1.5 m 4) Density of the oil: r = V = = 920.2... = 920 kgm–3 (to 2 s.f.) 1.63 × 10 -3 The density of the box is lower than the density of the oil, so yes, it would float on the oil. 5) During a crash, the crumple zone deforms plastically, so some of the car’s kinetic energy is used to plastically deform the crumple zone, so less is transferred to the passengers. Without a crumple zone, more kinetic energy is transferred to the passengers, so the chances of injury are higher, and injuries would be more severe. 6) 0.9 gcm–3. First estimate the volume of the bowl (roughly a cylinder shape with a radius equal to 10 cm): V ≈ pr2 × l ≈ p × 102 × 8 ≈ 2513.2... cm3 2200 m Use this to estimate the average density of the frogs: r = V ≈ 2513.2... ≈ 0.875... ≈ 0.9 gcm–3 (to 1 s.f.) 7) E.g. the spring may have been stretched past its elastic limit, and so 0.03 J of the 0.12 J of energy is used to plastically deform the spring and then dissipated as heat. So only 0.09 J of energy is stored as elastic strain energy, so when the spring is released, only 0.09 J is converted to the kinetic energy of the mass.

TIP

The things I do for science, eh? Because I tell you what, it took quite a long time to convince those frogs to get in the bowl and then weigh them... Don’t worry, I’m joking, of course I’m joking. It was actually a doddle.

37. Stress and Strain

Section 5 — Materials

Quick Questions 1) How can you calculate the elastic strain energy stored by a material from its force-extension graph, assuming the material hasn’t been stretched past its elastic limit? 2) Where is the breaking stress found on a stress-strain graph for a material?

Now try these:

5) The graph on the right shows the force-extension relationship for a copper wire with a cross-sectional area of 1.3 × 10–5 m2. Calculate the tensile stress in the wire when it has an extension F of 0.34 mm, using tensile stress = A .

6) Use the graph to find the work done on the wire to extend it by 0.18 mm.

force (N)

3) A 2.5 m long aluminium wire has a stiffness constant of 8.6 × 104 Nm–1. A force is applied to each end of DL the wire so it extends by 1.8 mm. Calculate the strain of the wire as a percentage. Use strain = L . 4) Calculate the elastic strain energy stored in this aluminium wire. Use F = kDL and E = ½FDL. 200 150 100

50 7) An artist wants to hang a large sculpture with a weight of 2450 N from 0 the ceiling using a metal cable. The artist has a cable with a breaking stress of 5.1 × 108 Pa and a cross-sectional area of 7.8 × 10–3 m2. F Use tensile stress = A to determine if this steel cable is suitable for the artist to use.

0.1 0.2 0.3 0.4 extension (mm)

Section 5 — Materials

37. Stress and Strain

ANSWERS 1) By calculating the area under the graph. 2) At the very end of the line. 1.8 × 10 -3 DL = 0.00072. 0.00072 × 100 = 0.072% L = 2.5 4) 0.14 J. Substitute F = kDL into E = ½FDL to get E = ½kDL2. E = ½kDL2 = ½ × 8.6 × 104 × (1.8 × 10–3)2 = 0.13932 = 0.14 J (to 2 s.f.) 3) 0.072%. strain =

5) 1.2 × 107 Nm–2. The force applied to the wire to give it an extension of 0.34 mm is 150 N. F 150 tensile stress = A = = 1.153... × 107 = 1.2 × 107 Nm–2 (to 2 s.f.) 1.3 × 10 -5 6) 0.0072 J. The work done on the wire is equal to the elastic strain energy stored in the wire, which is equal to the area under the graph up to an extension of 0.18 mm. Area under the graph = ½ × force × extension = ½ × 80 × 0.18 × 10–3 = 0.0072 J 7) The stress in the cable when the sculpture is hung from it will be: 2450 F = 3.141... × 105 = 3.1 × 105 Pa (to 2 s.f.) tensile stress = A = 7.8 × 10 -3 This stress is lower than the breaking stress of the steel cable, so yes, the cable is suitable to use as it won’t break when the sculpture is hung from it.

TIP

You may also hear about the ‘ultimate tensile stress’ of a material. Beyond this stress, the material will continue to stretch without any more force being applied, until it eventually reaches its breaking stress and it fractures.

38. The Young Modulus

Section 5 — Materials

Quick Questions 1) True or false? The stress-strain graphs of two different sized pieces of nickel have the same gradient. 2) Why is it better to use a longer wire when using a wire to find the Young modulus?

Now try these: 3) Billy uses the apparatus shown on the right to find the Young modulus of a wire. He measures the wire diameter to be 0.86 ± 0.005 mm. Calculate the percentage uncertainty in this measurement.

clamp

pulley

test wire

ruler marker

weights

6) Calculate the elastic strain energy per unit volume stored by the wire when the stress in the wire is 1.4 × 108 Pa.

stress (× 108 Pa)

4) Calculate the stress in the wire when Billy hangs a 150 N weight on the end of the wire. F Area of a circle = pr2 and tensile stress = A . 2.0 5) From his results, Billy produces the graph shown on 1.0 the right. Calculate the Young modulus of the wire. 0

0.4

0.8

strain 1.2 (× 10–3)

7) A wire has a Young modulus of 4.55 × 1010 Nm–2, a cross-sectional area of 1.13 mm2 and an initial length of 0.700 m. Calculate the force that needs to be applied to the wire to extend the wire to 0.702 m. F tensile stress DL Young modulus = tensile strain , tensile stress = A and tensile strain = L .

Section 5 — Materials

38. The Young Modulus

ANSWERS 1) True. The gradient of a stress-strain graph is equal to the Young modulus of the material. Both pieces are made from the same material and so will have the same Young modulus and therefore the same gradient. 2) The longer the wire, the bigger the extension for a given force. A bigger extension is easier to see and measure, and it reduces the uncertainty in the measurements. 0.005 3) 0.6%. Percentage uncertainty = 0.86 × 100 = 0.581... = 0.6% (to 1 s.f.) 4) 2.6 × 108 Pa. Radius = diameter ÷ 2 = 0.86 ÷ 2 = 0.43 mm Cross-sectional area = pr2 = p × (0.43 × 10–3)2 = 5.8088... × 10–7 m2 F 150 tensile stress = A = = 2.5822... × 108 = 2.6 × 108 Pa (to 2 s.f.) 5.8088... × 10 -7 2.0 × 10 8 Dstress 5) 1.7 × 1011 Pa. Young modulus = gradient of graph = = = 1.724... × 1011 Dstrain 1.16 × 10 -3 = 1.7 × 1011 Pa (to 2 s.f.) 6) 56 000 Jm–3. Elastic strain energy stored per unit volume = area under the graph = ½ × base × height = ½ × 0.8 × 10–3 × 1.4 × 108 = 56 000 Jm–3 0.702 - 0.700 DL 7) 147 N. Tensile strain = L = = 2.85... × 10–3. 0.700 Tensile stress = Young modulus × tensile strain = 4.55 × 1010 × 2.85... × 10–3 = 1.3 × 108 Nm–2. F = tensile stress × A = 1.3 × 108 × 1.13 × 10–6 = 146.9 = 147 N (to 3 s.f.)

TIP

Boy that’s a lot of equations. Remember — if you’re doing more than one calculation in a question, don’t round any numbers until the end, otherwise you could end up with the wrong answer. (That’s a bad thing, by the way.)

39. Interpreting Graphs for Materials

Section 5 — Materials

Quick Questions 1) What is meant by the yield point of a material? 2) Which point on the graph opposite is the yield point?

stress

B A

C

D strain

Now try these:

stress

3) The graph on the right shows the stress-strain relationship for three different materials. Rajesh has a sample of each material. Each sample is the same shape and size. If Rajesh applies an increasing force to each sample, which one, X, Y or Z, will snap first?

X

Y

Z strain

force (N)

6.0

E

4.0 2.0 0

0.5 1.0 1.5 2.0 2.5 extension (cm)

4) The graph on the left shows the force-extension relationship for a spring. The elastic limit of the spring is marked as point E. If a force of 6.0 N is applied to the spring, and then the force is slowly removed, how much work is done to permanently deform the spring? Hint: the total number of small squares under the graph between 0 and 6 N = 182.5.

Section 5 — Materials

39. Interpreting Graphs for Materials

ANSWERS 1) The stress at which a large amount of plastic deformation takes place with a constant or reduced load. 2) B 3) Z. Material Z has a lower breaking stress than the other two materials, so a smaller force is needed to snap a sample of material Z.

force (N)

4) 0.028 J. The spring has been stretched past its limit of proportionality, so as the force is removed the unloading curve appears as shown on the right. (The unloading curve must be parallel 6.0 E to the loading curve.) The work done to permanently deform 4.0 the wire is equal to the area between the two curves. First calculate the area under the loading curve using 2.0 the total number of small squares: –2 Area of one square = base × height = 0.1 × 10 × 0.4 = 0.0004 J 0 So value of total area under loading curve 0.5 1.0 1.5 2.0 2.5 = value of one square × no. of squares = 0.0004 × 182.5 = 0.073 J extension (cm) Then find value of area under the unloading curve: –2 Area under unloading curve = ½ × base × height = ½ × ((2.0 – 0.5) × 10 ) × 6.0 = 0.045 J So work done = area under loading curve – area under unloading curve = 0.073 – 0.045 = 0.028 J

TIP

Brittle materials have very boring stress-strain and force-extensions graphs — for both of them it’s just a straight line and then it stops when it fractures. Materials like wires have curves and bends... much more exciting.

40. Current, Potential Difference and Resistance

Section 6 — Electricity

Quick Questions 1) What measuring device in a circuit can be assumed to have zero resistance? 2) What is Ohm’s law? 3) Which of the following units is equivalent to 1 JC–1? A. 1 V B. 1 W C. 1 A

D. 1 s

Now try these: 4) A lamp is switched on for 50.0 s. While the lamp is switched on, a total of 175 C of charge passes through the lamp. What is the average current through the lamp while it is switched on? DQ . Use the equation I = Dt 5) A component in a circuit has a potential difference of 2.7 V across it, and a resistance of 4.5 W. What is the current through the component? Use the equation R = V . I 6) A battery has an e.m.f. of 6.80 V and negligible internal resistance. How much energy is transferred by the battery when a constant current of 2.75 A passes through it for 56.0 s? DQ . Use the equations V = W and I = Q Dt

Section 6 — Electricity

40. Current, Potential Difference and Resistance

ANSWERS 1) An ammeter. It has to have close to zero resistance so that connecting it won’t change the current through the circuit. 2) Ohm’s law states that, provided physical conditions remain constant, the current through an ohmic conductor will be directly proportional to the potential difference across it. 3) A. The equation for potential difference is V = W . Work done, W, is measured in joules, J, Q and charge, Q, is measured in coulombs, C, so the unit for potential difference, V, must be equivalent to JC–1. DQ 4) 3.5 A. I = = 175 = 3.5 A. 50.0 Dt 5) 0.6 A. R = V , so I = V = 2.7 = 0.6 A. Resistance is futile! I R 4.5 DQ I think you’ll find it’s 6) 1050 J. Rearrange I = for DQ, Dt potential difference divided by current. so DQ = IDt = 2.75 × 56.0 = 154 C. Rearrange V = W , for W : NOBODY ASKED Q YOU STEVE. W = VQ = 6.80 × 154 = 1047.2 = 1050 J (to 3 s.f.)

TIP

Remember, Ohm’s law is a special case with specific conditions that need to be true for it to hold. If variables like the temperature aren’t kept constant, Ohm’s law won’t apply – so keep an eye out for changing conditions.

41. I/V Characteristics

Section 6 — Electricity

Quick Questions 1) Name a circuit component which has an I/V characteristic that is a straight line through the origin. 2) Look at the characteristic of a filament lamp on the right. State the variables that are represented by the x and y axes.

Now try these:

I/A 0.8

3) An I/V characteristic of an NTC thermistor is shown on the right. Explain how the graph shows that the resistance of the thermistor decreases as the current increases.

0.4 4) Determine the resistance of the thermistor when there is a V current of 0.6 A through it. Use the equation R = . I 0.0 5) A student constructs the circuit shown below on the right, which 0.4 0.0 contains a variable power supply. He varies the potential difference across the circuit in order to determine the I/V characteristic of the circuit. Explain which of the graphs below shows the correct shape of this I/V characteristic. A.

I

B. V

I

C. V

I V

0.8

1.2 V/V -

+

A

V

Section 6 — Electricity

41. I/V Characteristics

ANSWERS 1) E.g. a fixed resistor (or any ohmic conductor) at a constant temperature. (Remember that temperature point. Ohmic conductors only have linear graphs through the origin when physical conditions are constant.) 2) Potential difference is represented by the y-axis and current is represented by the x-axis. (It’s a V/I characteristic.) 3) Resistance is equal to potential difference divided by current (R = V ÷ I ). By reading values for V and I at various points on the graph and putting them into the equation for R, you can see that as the current increases, the resistance decreases. 4) 2 W. The value of potential difference when the current is 0.6 A is 1.2 V. R = V = 1.2 = 2 W. I 0.6 5) C. The presence of the diode in the circuit will mean that little to no current will flow for a negative potential difference. The only graph that shows this behaviour is graph C.

TIP

Be sure to pay close attention to the axes of the graphs you’re given. In the exam, you may have to interpret graphs with current on the y-axis and p.d. on the x-axis, or the other way around. Don’t let it catch you out.

42. Resistivity and Superconductivity

Section 6 — Electricity

Quick Questions 1) Define resistivity. 2) Give one factor which can affect the resistivity of a material.

The lacklustre characters and shaky plot line were an unwelcome addition to the unrealistic scenes and overbearing symbolism.

3) What is meant by the ‘critical temperature’ of a material?

Now try these: 4) Explain why it isn’t practical to use superconducting metals in everyday electronics. 5) A student has a wire that is cylindrical, with a diameter of 0.64 mm. Calculate the cross-sectional area of the wire in m2 using the equation A = pr2.

Resistivity (Wm)

4.9 ×10–7

1.1 × 10–6

6.9 × 10–7

1.6

Resistance (Ω)

6) The student carries out an experiment to see how the resistance of this wire changes with length. A graph of the student’s results is shown on the right. The gradient of the line of best fit is 3.6 Wm–1. The table below shows the resistivity of three different metals at room temperature. Determine which of these three metals the wire is most likely to be made from. Use the equation r = RA . l Metal Constantan Nichrome Stainless Steel

1.2 0.8 0.4 0.0

0.1

0.2 0.3 0.4 Length (m)

0.5

Section 6 — Electricity

42. Resistivity and Superconductivity

ANSWERS 1) The resistivity of a material is the resistance of a 1 m length with a 1 m2 cross-sectional area. 2) E.g. temperature / light intensity. Don’t be fooled, the variables in the resistivity equation don’t affect resistivity — resistivity is a material constant that’s true for that material no matter what its shape or size. 3) The critical temperature of a material is the temperature below which the material will become a superconductor (its resistivity will become 0). 4) E.g. metals only become superconducting when they are cooled to very low temperatures (usually less than 10 K). Cooling materials to these temperatures is very difficult and expensive, which means it would not be practical to use for everyday electronics. 5) 3.2 × 10–7 m2. Radius of wire = 0.64 ÷ 2 = 0.32 mm = 3.2 × 10–4 m. A = pr2 = p × (3.2 × 10–4)2 = 3.216... × 10–7 = 3.2 × 10–7 m2 (to 2 s.f.) 6) The graph has resistance on the y-axis and length on the x-axis, so the gradient of the line of best fit is equal to R . Substituting this into the equation gives: l r = gradient × A = 3.6 × 3.216... × 10–7 = 1.158... × 10–6 = 1.2 × 10–6 Wm (to 2 s.f.) This is closest to the value of resistivity given in the table for nichrome, so the wire is most likely to be made from nichrome.

TIP

Superconductivity is a pretty mind-blowing property — below the critical temperature, the resistivity of a superconductor doesn’t just get close to zero, it is zero, so there’s no resistance at all. Chilling stuff.

43. Electrical Energy and Power

Section 6 — Electricity

Quick Questions 1) True or false? 1 W is equal to 1 kJ of energy transferred per hour. 2) The equation E = VIt can be formed by combining which two of the following equations? A. V = IR B. P = Et C. V = W D. P = VI E. r = RA l Q

Now try these: 3) There is a potential difference of 1.4 V across a bulb. The current through the bulb is 0.85 A. What is the power of the bulb? Use the equation P = VI. 4) An electric heater is supplied with a constant current of 11.4 A for 2.0 minutes. The potential difference across the heater is 230 V. Calculate the total energy transferred by the heater while the current flows through it. Use the equations P = Et and P = VI.

5) A circuit device has a power of 54 W, and a resistance of 3.2 W. 2 Calculate the potential difference across the device. Use the equation P = V . R 6) An electrical fault causes a fuse to be supplied with a power of 18.8 mW. The fuse has a resistance of 7.5 × 10–4 W. What is the current through the fuse? Use the equation P = I 2 R.

Section 6 — Electricity

43. Electrical Energy and Power

ANSWERS 1) False. 1 W is equal to 1 J per second. (1 kJ per hour is 1000 J every 3600 s, which works out to about 0.28 J per second.) 2) B and D. Simply substitute one into the other, and rearrange the equation, and you get E = VIt. 3) 1.2 W. P = VI = 1.4 × 0.85 = 1.19 W = 1.2 W (to 2 s.f.) 4) 310 000 J. 2.0 minutes = 2.0 × 60 = 120 s. Substitute P = VI into P = Et and rearrange to get E in terms of V, I and t. E = VIt = 230 × 11.4 × 120 = 314 640 = 310 000 J (to 2 s.f.) 2 5) 13 V. P = V , so V = PR = 54 × 3.2 = 13.145... = 13 V (to 2 s.f.) R -3 6) 5.0 A. P = I 2 R, so I = P = 18.8 × 10-4 = 5.0066... = 5.0 A (to 2 s.f.) R 7.5 × 10

P = V² P = I²R P = VI R There are a lot of different equations you can use for calculations involving power and electrical energy. In the TIP exam, it’s all about choosing the right equation for the variables you’re given in the question. Choose wisely.

44. E.m.f. and Internal Resistance

Section 6 — Electricity

Quick Questions 1) How do you find the total e.m.f. of multiple cells connected in series? 2) What is meant by the ‘terminal potential difference’ of a power supply? 3) True or false? Internal resistance is caused by a battery having an existential crisis.

Now try these: 4) A battery in a circuit has an e.m.f. of 12.0 V. The current in the circuit is 1.63 A and the load resistance in the circuit is 6.75 W. What is the internal resistance of the battery? Use the equation e = I(R + r) 5) A student has a circuit containing a power supply and a variable resistor. They use the variable resistor to change the current through the circuit. They measure the potential difference across the circuit and the current through the circuit, for different resistances of the V/V variable resistor, and plot the graph shown on the right. 0.8 Determine the internal resistance and the e.m.f. of the power supply. V 0.6 Use the equations e = I(R + r) and R = to help you. I 6) A battery with an internal resistance of 0.44 W is connected in 0.4 a circuit with a total resistance of 6.3 W and a current of 3.2 A. 0.2 How much energy does the battery supply to 5.5 C of charge? E Use the equations e = and e = I(R + r) 0.0 0.4 0.8 1.2 1.6 2.0 I / A Q

Section 6 — Electricity

44. E.m.f. and Internal Resistance

ANSWERS 1) Add the e.m.f.s of each cell together. 2) The potential difference between the two terminals of a power supply. This is equal to e.m.f. when there is no internal resistance. 3) False. It’s caused by the electrons that carry the current colliding with other particles inside the battery. Batteries aren’t capable of contemplating their role in the universe. At least, as far as we know... 4) 0.612 W. Rearrange the equation for internal resistance, r, so r = (e ÷ I) – R = (12.0 ÷ 1.63) – 6.75 = 0.6119... = 0.612 W (to 3 s.f.) 5) Internal resistance = 0.3 W and e.m.f. = 0.7 V. The equation e = I(R + r) can be combined with R = V I to get e = V + Ir, and rearranged to V = e – Ir. By comparing this to the straight line equation y = mx + c, this shows that the gradient of the line, m = –r. change in y So, r = – = – 0.2 - 0.6 = – - 0.4 = 0.266... = 0.3 W (to 1 s.f.) change in x 1.9 - 0.4 1.5 From comparison of the equations, e = c, the y-intercept of the graph. So e = 0.7 V. 6) 120 J. e = I(R + r) = 3.2 × (6.3 + 0.44) = 21.568 V Rearrange e = E for E, and substitute e into the equation, so: Q E = eQ = 21.568 × 5.5 = 118.624 = 120 J (to 2 s.f.)

TIP

You’ll be told in exam questions if you can ignore the internal resistance of a power supply — if it doesn’t say it’s negligible, you’ll need to account for it in your calculations. Electricity questions just got a bit more complicated.

45. Conservation of Energy and Charge

Section 6 — Electricity

Quick Questions 1) True or false? The total current leaving a junction is always less than the total current entering it. 2) A student makes the circuits below containing a cell and three identical resistors. In which circuit will there be the lowest potential difference across the blue resistor? A.

B.

C.

Now try these: 3) A scientist sets up the circuit shown on the right. The battery has negligible internal resistance. A The current at point Y is 1.57 A. Determine the potential difference you would Y expect to see on the voltmeter. Use any of the equations R = V , I 1 1 1 1 RT = R1 + R2 + R3 + ... (in series) and = + + + ... (in parallel). RT R1 R2 R3 4) Calculate the total resistance of the circuit. 5) Determine the current that you would expect to see on the ammeter.

15 V

6.0 Ω 4.0 Ω 2.5 Ω V

8.0 Ω

Section 6 — Electricity

45. Conservation of Energy and Charge

ANSWERS 1) False. The total current leaving a junction is always the same as the total current entering it. This is conservation of charge in action. 2) A. In each circuit, each branch has the full source potential difference, and it is shared between the components connected in series on that branch. Since the resistors are identical, they’ll share the potential difference equally when connected in series. Circuit A has the blue resistor connected in series with the most other resistors, so it will have the lowest potential difference across it in this circuit. 3) 2.4 V. Find the total resistance of the two resistors in parallel 1 1 1 1 RT = 1 ÷ b + l = 1 ÷ b = 1.538... W + R1 R2 4.0 2.5 l V R = , so V = IR = 1.57 × 1.538... = 2.415... = 2.4 V (to 2 s.f.) I 4) 3.7 W. Total resistance on bottom branch, RT = R1 + R2 = 8.0 + 1.538... = 9.538... W 1 1 1 1 Total resistance RT = 1 ÷ b + l = 1 ÷ b = 3.683... = 3.7 W (to 2 s.f.) + R1 R2 9.53... 6.0 l V V 15 5) 4.1 A. R = , so I = = = 4.072... = 4.1 A (to 2 s.f.) (You could also have answered this question I R 3.683... by finding the current through the first branch, and adding it to the value from Q3).

TIP

You might’ve thought conservation of charge was about how long you can keep your phone on at 10% battery, but sadly that’s not what this card is about. Conservation of charge underpins many essential rules for circuits.

46. The Potential Divider

Section 6 — Electricity

Quick Questions 1) The circuit on the right shows a potential divider with a Vout that varies depending on light intensity. What is the circuit symbol for the device at point Z?

Z V

2) The device at point Z has a resistance R1. Give an equation that can be used to calculate Vout.

R2

Vout

Now try these: 3) A student builds a potential divider to serve as a dimmer switch for a bulb. The circuit is shown below. The battery has negligible internal resistance. The maximum resistance of R1 is 2.5 W and the bulb has a resistance much higher than R2. What resistance should R1 be set to so that the bulb is at its brightest? 4) The student sets the resistance of R1 to 0.55 W. Calculate the potential difference across the bulb (Vbulb). 5) What resistance does R1 have when Vbulb = 7.8 V? 6) The student replaces the bulb. The new bulb has an average operating resistance of 0.42 W. R1 is kept at the same value as in Q5. Will the value of Vbulb still be 7.8 V? Explain your answer.

R1 12 V R2 = 0.35 Ω

Section 6 — Electricity

46. The Potential Divider

ANSWERS 1)

This is the symbol for a light-dependent resistor (LDR). An LDR’s resistance changes with light intensity. 2) Vout = R2 V. The source potential difference, V, is shared in proportion to the resistances of the two R1 + R2 resistors, so to find Vout, find the ratio of R2 to the total resistance (R1 + R2) and multiply it by V. 3) 0 W. When the variable resistor has no resistance, the second resistor, and so the bulb, will receive the full source potential difference and so the bulb will be at its brightest. R2 V = 0.35 4) 4.7 V. Vout = × 12 = 4.666... = 4.7 V (to 2 s.f.) R1 + R2 0.55 + 0.35 5) 0.19 W. Rearrange the equation from Q4 for R1, so R1 = VR2 - R2 = 12 × 0.35 - 0.35 = 0.1884... = 0.19 W (to 2 s.f.) Vout 7.8 6) No, it will not. The bulb has a resistance similar to R2, so the bulb and resistor will act as two resistors in parallel. This means they will have a total resistance that is lower than R2, and so Vout will not still have a value of 7.8 V (it will be lower than 7.8 V).

TIP

Potential dividers are all about controlling and adjusting the p.d. supplied to a component. And they’re a key part of many electronic devices — from dimmer switches for lights, to the volume control for your banging tunes.

47. Circular Motion

Section 7 — Further Mechanics

Quick Questions 1) How do you convert an angle in degrees to an angle in radians? 2) How do you know an object travelling in a circle at a constant speed has a resultant force acting on it?

Now try these: 3) A cheetah is moving in a circle with a radius of 4.65 m. The cheetah’s linear velocity 2 v is 2.4 ms–1. Calculate the angular acceleration of the cheetah. Use the equation a = r . 4) Alice and Nick are sat on a merry-go-round. Alice is sat 2.5 m from the centre of the merry-go-round and Nick is sat closer to the centre. Alice has a linear speed of 4.8 ms–1 and Nick has a linear speed v of 3.7 ms–1. How far is Nick sat from the centre of the merry-go-round? Use the equation w = r .

5) A toy car is driving round a circular track with a radius of 57 cm with an angular acceleration of 1.6 ms–2. q Calculate how long it takes for the car to go round half the track using the equations a = w2r and w = t . 6) A 1500 kg car travels in a circle round a roundabout. The centripetal force needed to keep the car travelling in the circle is 750 N. What size centripetal force would be needed to keep a 4500 kg tractor travelling round the same roundabout at the same speed and radius as the car’s motion? mv2 Use the equation F = r .

TIP

‘Don’t displease and use degrees, radians are the potion for circular motion!’ I can tell you’re unimpressed... but don’t underestimate the power of making up stupid rhyming phrases to help you remember tricky physics facts.

Section 7 — Further Mechanics

47. Circular Motion

ANSWERS p 1) Multiply the number by 180 . 2) An object moving in a circular motion is constantly changing direction, so its velocity is constantly changing. The object must therefore have an acceleration, and, according to Newton’s laws, any accelerating object must have a resultant force acting on it. v2 2.4 2 3) 1.2 ms–2. a = r = 4.65 = 1.2387... = 1.2 ms–2 (to 2 s.f.) 4) 1.9 m. The angular velocity of both Alice and Nick are the same (as they’re covering the same angle in the same amount of time). Calculate the angular velocity using Alice’s linear velocity and distance from the centre v 4.8 of the merry-go-round: w = r = 2.5 = 1.92 rads–1 v Use this angular velocity to calculate Nick’s distance from the centre. w = r , so 3.7 v r = w = 1.92 = 1.927... = 1.9 m (to 2 s.f.) a 1.6 5) 1.9 s. Rearrange a = w2r for w = = 1.6754... rads–1. The toy travels half of the track, r = 57 ×10 -2 q q p so q = p. w = t rearranges to give t = w = 1.6754... = 1.8751... = 1.9 s (to 2 s.f.) mv2 6) 2300 N. F = r so F µ m as v and r are constants. F F F 750 So mcar = mtractor which rearranges to Ftractor = mcar × mtractor = 1500 × 4500 = 2250 = 2300 N (to 2 s.f.) car tractor car

48. Simple Harmonic Motion

Section 7 — Further Mechanics

Quick Questions 1) In order for an object to be undergoing SHM, what must its acceleration be directly proportional to? 2) True or false? The total energy of a mass on a pendulum reaches zero at its maximum displacement.

Now try these: 3) A duck is oscillating up and down on the surface of a pond with simple harmonic motion. The duck has an angular frequency of 0.45 rads–1 and a maximum displacement of 2.5 cm. The displacement of the duck is given by x = Acos(wt). Calculate the displacement of the duck from its central position at t = 5.2 s. x/m 4) Calculate the acceleration of the duck at t = 5.2 s using a = –w2x. 0.04 X 5) A mass is oscillating on a spring. The displacement-time graph of 0 this oscillation is shown on the right. Which of the graphs below, 0.2 0.4 0.6 t / s A, B or C, shows the correct acceleration-time graph for this mass? –0.04 a / ms 10

-2

A

a / ms 10

-2

B

a / ms 10

-2

C

0 0 0 0.2 0.4 0.6 t / s 0.2 0.4 0.6 t / s 0.2 0.4 0.6 t / s –10 –10 –10 6) The mass has an angular frequency of 16 rads–1. Use vmax = wA to find the maximum speed of the mass. 7) Calculate the velocity of the mass at t = 0.27 s (point X on the x-t graph) using v = ±w (A 2 - x 2) .

Section 7 — Further Mechanics

48. Simple Harmonic Motion

ANSWERS 1) Negative displacement. (a µ –x) 2) False. The total energy of a mass on a pendulum remains constant during oscillation (assuming no energy is lost to the surroundings). At its maximum displacement, the gravitational potential energy of the mass is at its maximum and its kinetic energy reaches zero. 3) –1.7 cm. x = Acos(wt) = 2.5 × cos(0.45 × 5.2) = –1.738... = –1.7 cm (to 2 s.f.) 4) 3.5 × 10–3 ms–2. a = –w2x = –(0.452) × (–1.738... × 10–2) = 3.521... × 10–3 = 3.5 × 10–3 ms–2 (to 2 s.f.) 5) B. The velocity-time graph for this oscillating mass is derived from the gradient of the displacement-time graph, and the acceleration-time graph is derived from the gradient of the velocity-time graph. So the velocity-time graph would be a quarter of a cycle in front of the displacement (i.e. it would have the same shape as graph C) and the acceleration-time graph is a quarter of a cycle in front of the velocity (graph B). 6) 0.64 ms–1. Read the amplitude of the oscillation from the displacement-time graph, A = 0.04 m. vmax = wA = 16 × 0.04 = 0.64 ms–1 7) –0.55 ms–1. Displacement of the mass at 0.27 s (point X on the displacement-time graph) = 0.02 m v = ±w (A 2 - x 2) = ±16 × (0.04 2 - 0.02 2) = ±0.5542... = ±0.55 ms–1 (to 2 s.f.) The gradient of the displacement-time graph is equal to the velocity of the mass. The gradient at 0.27 s is negative, so v = –0.55 ms–1

TIP

Visualise the object to help you work out the correct sign of a quantity. For displacement, it depends on which side of equilibrium the object is, for velocity it’s direction of motion, and acceleration is always towards the centre.

49. Simple Harmonic Oscillators

Section 7 — Further Mechanics

Quick Questions 1) True or false? If the amplitude of a mass oscillating freely on the end of a spring is doubled, the time period halves.

x / cm 1

2) The graph on the right shows the displacement of an oscillating mass 0 against time. What is the time period of this mass’s oscillation? –1

0.3

Now try these: 3) Aisha sets up the experiment shown on the right to investigate the simple harmonic motion of a pendulum. She uses the stopwatch to time how long it takes for the pendulum to complete one oscillation. Explain how Aisha could improve the accuracy of this measurement using the same apparatus.

0.6 t / s

clamp and clamp stand

pendulum

stopwatch 4) Aisha’s pendulum has a length l and time period T. Aisha sets up a second pendulum with half the length of the first pendulum. What is the time period of this second pendulum? l Use the equation T = 2p g and give your answer in terms of T. 5) A horizontal spring has one end fixed to a wall, and a trolley attached to the other. The trolley is pulled to the right by 4.9 cm with a force of 2.8 N and then released. It oscillates with simple harmonic motion with m a time period of 0.44 s. Calculate the trolley’s mass. Use the equations F = kDL and T = 2p k .

6) What would a graph of T2 against k look like for this mass oscillating on a spring?

Section 7 — Further Mechanics

49. Simple Harmonic Oscillators

ANSWERS 1) False. The time period of a mass oscillating on a spring doesn’t depend on the amplitude of the oscillation, so m doubling the amplitude won’t change the time period. (You can also see this from T = 2p k .) 2) 0.4 s. It takes 0.6 s to complete 1.5 oscillations. So one oscillation (the time period) = 0.6 ÷ 1.5 = 0.4 s. 3) E.g. she could time how long it takes the pendulum to complete 10 oscillations and divide that by 10 to find how long one oscillation takes. This reduces the random error in the time measurement for one oscillation. T l . T = 2p g where p and g are constants, so T µ l , and so T = k l where k is a constant. 2 T T l This can be written as 1 = 2 . Substituting T1 = T, l1 = l and l2 = 2 into the equation gives l1 l2 2 T2 lT T T = which can be rearranged to give T2 = = l l 2 l 2 2 T 2.8 F –1 5) 0.28 s. F = kDL so k = = = 57.142... Nm DL 4.9 × 10 -2 T 2 0.44 2 m T = 2p k so m = b 2p l k = b 2p l × 57.142... = 0.2802... = 0.28 s (to 2 s.f.) m m 1 6) If you rearrange T = 2p k for T2 = 4p2 k , you can see that T2 µ k , so the graph would be an inverse proportion relationship — a negative gradient becoming less steep, like this graph. 4)

TIP

When tackling questions like Q4, find the proportionality relationship first, then set up the algebraic equation. It’s tempting to try to find the answer by just looking at the equation, but you’re more likely to make mistakes.

k

50. Free and Forced Vibrations

Section 7 — Further Mechanics

Quick Questions 1) True or false? A free vibration is an oscillation where no energy is transferred to or from the surroundings. 2) What’s the difference between critical damping and overdamping?

Now try these: 3) A mass is attached to the end of a spring. When the mass is pulled down and released, it oscillates with a frequency of 0.7 Hz. At which of the following driving frequencies would the amplitude be greatest? A. 0.35 Hz B. 0.7 Hz C. 1.4 Hz

5) A wooden block is sat on some sandpaper and is fixed between two horizontal springs. One of the springs is attached to a vibration generator. A graph of the block’s amplitude for a given driving frequency is shown on the right. Describe how the graph would be different if the block was sat on a sheet of ice.

amplitude

4) A set of bathroom weighing scales has an analogue measuring scale. When a person steps on the scales, a needle moves to point at the person’s mass on the measuring scale. A speedometer needle acts in a similar way, moving to point to the speed a car is moving. State which needle is heavily damped and which needle is critically damped. Explain your answer.

driving frequency

Section 7 — Further Mechanics

50. Free and Forced Vibrations

ANSWERS 1) True. 2) Both types of damping return an object to equilibrium without oscillation. Critical damping reduces the amplitude in the shortest possible time, overdamping takes longer to reduce the amplitude. 3) B. The amplitude is at its maximum when the driving frequency is equal to the resonant frequency.

5) The amplitude would be higher at all frequencies, except for at very low driving frequencies, and the peak would be at a higher amplitude and a slightly higher driving frequency, as shown opposite. This is because the amount of damping has been reduced.

amplitude

4) E.g. the bathroom weighing scales needle is heavily damped and the speedometer needle is critically damped. On the set of weighing scales, once a person has stepped on the scales, the value of the mass won’t change, and so the mass can be read from the scale once the needle has stopped oscillating. In a car, the speed can vary a lot, and the driver needs to be able to read the speed using the speedometer needle, which would be difficult to do if the needle was oscillating. Critically damping the needle ensures the needle won’t oscillate, so the driver is always able to accurately read off the speed.

driving frequency

TIP

If you’re asked about an unfamiliar situation PANIC. Oh wait... hang on... I’ve just been told you should NOT PANIC. Take your time to work out the physics being tested, then apply what you know to the situation.

51. Thermal Energy Transfer

Section 8 — Thermal Physics

Quick Questions 1) What quantity is equal to the sum of the kinetic energies and potential energies of an object’s particles? 2) True or false? Specific heat capacity is the thermal energy required to change the state of 1 kg of a substance.

Now try these: 3) State the change (if any) in a gas’s internal energy as the gas condenses. Explain your answer. 4) Using Q = ml, prove that it takes around seven times more energy to boil any mass of water at 100 °C than to melt the same mass of ice at 0 °C. lfusion for water = 3.34 × 105 Jkg­–1, lvaporisation for water = 2.26 × 106 Jkg–1. 5) In a simple electric shower, water flows at a rate of 0.015 kgs–1 over a heating element in a tank. The temperature of the water is raised from 8 °C to 41 °C as it passes through the tank. The water then leaves the shower through the showerhead. Calculate the total thermal energy transferred to the water during a 4 minute shower. Use the equation Q = mcDq. cwater = 4180 Jkg–1K–1. 6) Maude decides to liven up a Tuesday night by melting a 2.0 kg bar of chocolate. The initial temperature of the chocolate is 21 °C. The chocolate melts at 36 °C. Maude heats the chocolate continuously and it takes 6.5 minutes to completely melt the bar. Use the equations Q = ml and Q = mcDq to calculate the rate at which thermal energy was transferred to the chocolate. Assume heat losses are negligible and the chocolate remains at 36° once melted. cchocolate = 1.30 × 103 Jkg–1K–1, lfusion for chocolate = 4.5 × 104 Jkg–1.

Section 8 — Thermal Physics

51. Thermal Energy Transfer

ANSWERS 1) The internal energy of the object. 2) False. Specific heat capacity is the amount of energy needed to raise the temperature of 1 kg of a substance by 1 K (or 1 °C). The definition given is for specific latent heat. 3) It decreases. Internal energy is the sum of the potential energies and kinetic energies of the gas particles. While the kinetic energy of the particles stays the same, the potential energy of the particles decreases as the gas changes to a liquid, so the internal energy must decrease. 4) The energy needed to boil a mass, m, of water is Qboil = mlvaporisation for water = 2.26 × 106 m. The energy needed to melt the same mass of water is Qmelt = mlfusion for water = 3.34 × 105 m. Qboil ÷ Qmelt = 2.26 × 106 m ÷ 3.34 × 105 m = 6.76... = 7 (to 1 s.f.) 5) 5 × 105 J. Dq = 41 – 8 = 33 °C. The thermal energy transferred to the water per second = Q = mcwaterDq = 0.015 × 4180 × 33 = 2069.1 J The thermal energy transferred in 4 minutes = 2069.1 × 4 × 60 = 4.96584 × 105 = 5 × 105 J (to 1 s.f.) 6) 330 Js–1. Dq = 36 – 21 = 15 °C. The energy required to heat the chocolate from 21 °C to 36 °C, Qheat = mcchocolateDq = 2.0 × 1.30 × 103 × 15 = 39 000 J. The energy required to melt the chocolate at 36 °C, Qmelt = mlfusion for chocolate = 2.0 × 4.5 × 104 = 90 000 J. So the total energy required to melt the chocolate, Qtotal = 39 000 + 90 000 = 129 000 J. It took 6.5 minutes to transfer this energy, so the rate of thermal energy transfer = 129 000 ÷ (6.5 × 60) = 330.7... = 330 Js–1 (to 2 s.f.).

TIP

Well, if all of that hasn’t got you hot under the collar, I don’t know what will. Make sure you can wield the equations for specific latent heat and specific heat capacity like a grand master thermal energy... er... ninja.

52. Gas Laws

Section 8 — Thermal Physics

Quick Questions 1) What name is given to the temperature –273 °C (to 3 s.f.)? What is special about this temperature? 2) At a constant temperature, the pressure and volume of a gas are inversely proportional. What is the name of this gas law? 3) Which of the following is true for a fixed mass of gas at a constant volume? A. Pressure = absolute temperature B. Pressure µ absolute temperature C. Pressure µ 1/absolute temperature D. Pressure is independent of absolute temperature

Now try these: 4) Convert 21 °C into kelvins. 5) A gas is initially at a pressure of 1.2 × 105 Pa and a volume of 0.55 m3. The temperature of the gas is kept constant as its volume is increased to 0.80 m3. Calculate the new pressure of the gas.

7) The graph on the right shows how pressure varies with volume for an ideal gas kept at a constant temperature. Explain the difference in pressure at volumes V1 and V2 in terms of the gas particles.

Pressure

6) A student investigates how the length of a column of air, trapped in a capillary tube, varies with temperature at a constant pressure. Which gas law is the student investigating? Describe the relationship you would expect the student to see between length and temperature. Explain your answer.

V1

V2

Volume

Section 8 — Thermal Physics

52. Gas Laws

ANSWERS 1) Absolute zero, it’s the lowest temperature theoretically possible. 2) Boyle’s law. 3) B. This is the Pressure law. 4) 294 K. K = C + 273 = 21 + 273 = 294 K 5) 8.3 × 104 Pa. At a constant temperature pV = constant, so p1V1 = p2V2. p1 V1 1.2 × 105 × 0.55 Rearrange to give p2 = V = = 8.25 × 104 = 8.3 × 104 Pa (to 2 s.f.) 0.80 2 6) Charles’s law. The student should find the length of the air column increases proportionally with temperature. The volume of the column of air is proportional to its length (V = pr2l), and volume is proportional to the temperature (Charles’s law). That means length must also be proportional to temperature. 7) At V2, the pressure is lower than at V1. This is because the particles are further apart since the volume at V2 is higher. This means they’ll collide less often with each other and the container walls, so the pressure is lower.

TIP

It’ll impress no one if you get Charles’s and Boyle’s laws mixed up. So remember, Boyle sounds like boil, and both the law and boiling happen at a constant temperature. Oh boy(le), that’s revision gold right there.

53. Ideal Gas Equation

Section 8 — Thermal Physics

Quick Questions 1) pV = NkT is known as the ideal gas equation of state. What quantity does N represent? 2) True or false? Avogadro’s constant is equal to the number of particles in one mole of any gas.

Now try these: 3) Nitrogen has the chemical symbol

14 7

N . What is the relative molecular mass of nitrogen gas, N2?

4) Helium, He , exists as a single atom, not as part of a molecule. What is the mass of 3.6 moles of helium? 4 2

5) 6.0 kJ of work is done as a gas expands at a constant pressure from an initial volume of 0.20 m2 to a volume of 0.40 m2. Calculate the pressure of the gas. 6) A scuba-diving cylinder stores air at a pressure of 1.65 × 104 kPa and at a temperature of 21 °C. The internal volume of the cylinder is 0.16 m3. Calculate the number of moles of air in the cylinder using the equation pV = nRT. R = 8.31 Jmol–1K–1. 7) Calculate the number of air molecules in my 32 m3 bedroom. Use the equation pV = NkT. You can assume I’m a minimalist and that the volume of everything apart from the air in the room is negligible. Atmospheric pressure = 101.3 kPa, bedroom temperature = 16 °C, k = 1.38 × 10–23 JK–1.

Section 8 — Thermal Physics

53. Ideal Gas Equation

ANSWERS 1) The number of gas molecules. 2) True. 3) 28. The relative atomic mass is 14. The gas molecule has the formula N2, which shows 2 atoms make up one molecule. So the relative molecular mass = 14 × 2 = 28. 4) 14 g. The molar mass of helium = 4 g. So the total mass of 3.6 moles of helium = 3.6 × 4 = 14.4 = 14 g (to 2 s.f.) 5) 3.0 × 104 Pa. Work done = pDV, so p = work done ÷ DV = 6.0 × 103 ÷ (0.40 – 0.20) = 3.0 × 104 Pa. 6) 1100 mol. T = 21 + 273 = 294 K, 1.65 × 104 kPa = 1.65 × 107 Pa. pV 1.65 × 107 × 0.16 = 1080.57... = 1100 mol (to 2 s.f.) pV = nRT, so n = RT = 8.31× 294 26 7) 8.1 × 10 molecules. T = 16 + 273 = 289 K. pV 101.3 × 103 × 32 pV = NkT, so N = kT = = 8.127... × 1026 = 8.1 × 1026 molecules (to 2 s.f.) 1.38 × 10 -23 × 289

The good news is that you’ll get given the ideal gas equations and the values of k and R in the exam, so you TIP don’t need to learn them. The bad news is that you’ll need to learn work done = pDV, and sit A-levels. Sorry.

54. The Pressure of an Ideal Gas

Section 8 — Thermal Physics

Quick Questions 1) True or false? The internal energy of an ideal gas is equal to the total kinetic energies of the gas particles. 2) List five assumptions that are used in kinetic theory.

Now try these: 1 3) The cubic box on the right can be used to derive the equation pV = 3 Nm (crms) 2 for an ideal gas. It contains N particles of gas, each with a mass m. The particle shown travels directly towards the right-hand wall of the cube with a speed u. Derive the equation for the total force Nmu 2 acting on one wall of the cube due to the particles in the box, F = l .

l

l

m

u

l

Nmu 4) Use this equation for F to derive p = V , where p = the pressure acting on that wall due to the particles and V = the volume of the cube. 2

5) Hydrogen gas is stored in a 2.0 m3 box at a pressure of 1.2 × 106 Pa. There are 1.9 × 1022 gas particles in the box and each particle has a mass of 1.67 × 10-27 kg. 1 Use the equation pV = 3 Nm (crms) 2 to calculate the root mean square speed of the particles.

TIP

Make sure you can derive the equation for the pressure of an ideal gas, and quote all the kinetic theory assumptions. Just keep writing it all down over and over and over again until it gets firmly stuck in your brain.

Section 8 — Thermal Physics

54. The Pressure of an Ideal Gas

ANSWERS 1) True. For an ideal gas, the assumption that the particles have no potential energy is made. 2) E.g. the molecules continually move about randomly. The motion of the molecules follows Newton’s laws. Collisions between molecules themselves or at the walls of a container are perfectly elastic. Except for during collisions, the molecules always move in straight lines. Any forces that act during collisions last for much less time than the time between collisions. 3) Assume the particle has a perfectly elastic collision with the right-hand wall. The particle’s change in momentum is –mu – mu = –2mu. Assuming the particle doesn’t collide with other particles, the time between collisions of the particle and the wall is 2l ÷ u. So the number of collisions per second = u ÷ 2l, so the rate of change of momentum is –2mu × u ÷ 2l. Force = rate of change of momentum, so the force exerted by the wall on this particle = –2mu2 ÷ 2l = –mu2 ÷ l. Each particle in the cube will have a different component of velocity u1, u2 etc. towards the wall. The total force, F, of all these particles on the wall is  m (u12 + u22 + ...) . This force is now positive as we’re talking about the force on the wall. The mean F= l u 2 + u 2 + ... Nmu 2 square speed of the particles u 2 = 1 N2 . Substituting this into the force equation gives F = l . force Nm u 2 ' l Nmu 2 Nmu 2 = 3 = V 4) pressure = area = l2 l 1 5) 4.8 × 105 ms–1. Rearrange pV = 3 Nm (crms) 2 to give: 3pV 3 × 1.2 × 10 6 × 2.0 = 4.763... × 105 = 4.8 × 105 ms–1­ (to 2 s.f.) crms = Nm = 1.9 × 10 22 × 1.67 × 10 -27

55. Kinetic Energy and the Development of Theories

Section 8 — Thermal Physics

Quick Questions 1) True or false? Kinetic theory has been a widely accepted scientific theory for over 500 years. 2) What is an empirical law? 3) Is kinetic theory an empirical law? Explain your answer.

Now try these: 4) Explain how Brownian motion provides evidence for kinetic theory. 1 3 5) Use the equation 2 m (crms) 2 = 2 kT to calculate the average kinetic energy of oxygen gas molecules at 300 K. k = 1.38 × 10–23 JK–1. 6) The temperature of a gas is increased by a factor of 4. Determine the factor by which the root mean square speed of the 1 3 gas molecules will increase. Use the equation 2 m (crms) 2 = 2 kT .

Ain’t no party like a

kinetic theory revision party... 7) A party balloon is filled with helium gas. The mass of a helium atom is 6.64 × 10–27 kg. Calculate the root mean square speed 1 3 of a helium atom in the balloon at 27 °C. Use the equation 2 m (crms) 2 = 2 kT . k = 1.38 × 10–23 JK–1.

Section 8 — Thermal Physics

55. Kinetic Energy and the Development of Theories

ANSWERS 1) False. It wasn’t widely accepted until the 1900s. 2) A law based on observations and evidence. This means that it can predict what will happen but it doesn’t explain why it happens. 3) No — kinetic theory is a theory based on assumptions (that help to explain observations). 4) Brownian motion is the random movement of particles suspended in a fluid, e.g. pollen grains in water. It can be explained by kinetic theory, which is the theory that all substances are made from tiny, moving particles. Large, heavy particles suspended in a fluid move randomly as a result of collisions with lots of smaller, fast, randomly-moving particles in a fluid. Therefore Brownian motion provides evidence for kinetic theory. 1 1 3 5) 6 × 10–21 J. Average kinetic energy = 2 m (crms) 2 and 2 m (crms) 2 = 2 kT , so 3 3 average kinetic energy = 2 kT = 2 × 1.38 × 10 -23 × 300 = 6.21 × 10-21 = 6 × 10–21 J (to 1 s.f.) 1 3 3kT 2 6) 2. E.g. 2 m (crms) = 2 kT rearranges to give crms = m . k and m are constants, so crms µ T . So, if T increases by a factor of 4, crms will increase by a factor of 4 = 2. 1 3 7) 1400 ms–1. T = 27 + 273 = 300 K. 2 m (crms) 2 = 2 kT . -23 3kT 3 × 1.38 × 10 × 300 So crms = = 1.87... × 10 6 = 1367.6...= 1400 ms–1 (to 2 s.f.) m = 6.64 × 10 –27

TIP

Brownian motion is pretty awesome — who doesn’t like to see particles jig about? Make sure you understand what’s causing the particles to move about in random directions and how it provides evidence for kinetic theory.

56. Gravitational Fields

Section 9 — Gravitational and Electric Fields

Quick Questions 1) State what makes a gravitational field a type of force field. 2) What do parallel gravitational field lines show? g A

Now try these: 3) The opposite graph shows the Earth’s gravitational field strength. Which value (A-D) is equal to 9.81 Nkg–1? F 4) Using g = m and the graph in Q3, explain why a person would weigh slightly less at the top of a mountain than at the bottom.

B C D

r

5) A Mars rover has a weight of 8.829 kN on Earth, where g = 9.81 Nkg . On Mars, the rover collects samples and increases its total mass by 5%. F The rover now weighs 3.497 kN on Mars. Use g = m to show that g = 3.7 Nkg–1 on Mars. GM 6) The rover is on the surface of Mars. Use g = 2 to calculate the mass of Mars. r G = 6.67 × 10–11 Nm2kg–2, radius of Mars = 3390 km. –1

7) The Earth is 81 times more massive than the Moon. An object directly in between the Earth and the Moon is Gm1 m2 to explain why the object must be 9 stationary and does not move towards either of them. Use F = r2 times closer to the centre of the Moon than the centre of the Earth. Assume no other gravitational forces act.

Section 9 — Gravitational and Electric Fields

56. Gravitational Fields

ANSWERS 1) In a gravitational field, an object with mass experiences an (attractive) non-contact force, so it is a force field. 2) Parallel gravitational field lines show a uniform gravitational field (it’s the same everywhere). 3) Point A. 9.81 Nkg–1 is the value of g at the surface of the Earth. 4) For a person at the top of a mountain, the value of r in the Earth’s radial gravitational field is greater than at F 1 the bottom, so the value of g is lower (since g \ 2 , see the graph in Q3). g = m , so F = gm. The force on r a person due to gravity is their weight, so their weight will be lower at the top of the mountain since mass is constant and g is lower at the top of the mountain. 8829 F 5) Its mass on Earth was m = g = 9.81 = 900 kg. Its mass on Mars is 5% larger, so 1.05 × 900 = 945 kg. F 3497 So the gravitational field strength on Mars is g = m = 945 = 3.700... Nkg–1 = 3.7 Nkg–1 (to 2 s.f.) gr 2 3.7 × ]3390 × 103g2 GM so M = G = = 6.374... × 1023 = 6.37 × 1023 kg (to 3 s.f.) r2 6.67 × 10 -11 7) E.g. the gravitational forces acting on the object due to the Earth and the Moon must be equal (and opposite) GmE mo GmM mo = where mE, mM and mO since the object isn’t moving towards either. This means that rE2 rM2 are the masses of the Earth, Moon and object, and rE and rM are the distances from the Earth and Moon. m m Cancelling G and mo , 2E = M2 , and since mE = 81 mM, rE2 = 81rM2, so rE = 81 rM = 9rM rE rM The equation g = F ÷ m may look familiar — it’s the same as W = mg, weight is just the force due to gravity. TIP All planets can be assumed to have radial fields, so make sure you can use those r2 formulas.

6) 6.37 × 1023 kg. g =

57. Gravitational Potential

Section 9 — Gravitational and Electric Fields

Quick Questions 1) True or false? Heavier objects have a greater escape velocity. 2) How would you find the work done against gravity to move a mass, m, between two points in a gravitational field from a graph of g against r for the field and the equation DW = mDV? 3) Describe the shape of a graph of V against r for the gravitational field of a spherical mass.

Now try these: 4) Calculate the gravitational potential on the surface of Earth, GM using V = – r , mass of Earth = 5.97 × 1024 kg, radius of Earth = 6.37 × 106 m and G = 6.67 × 10–11 Nm2kg–2.

A.

B.

C.

5) Which of the diagrams of the Earth opposite shows lines along which no work is done by gravity to move a mass in the Earth’s gravitational field. 2GM - GM and Ek = 1 mv 2 . 6) Show that the formula for escape velocity is v = r , using V = r 2 7) After the 1969 Moon landing, the 4500 kg Eagle lunar module left the Moon’s surface and travelled back to the Columbia spacecraft orbiting above. The work done against gravity was 792 MJ. Calculate the height of the Columbia spacecraft’s orbit above the Moon’s surface, DV , DW = mDV and g on the Moon = 1.6 N kg–1. using g = – Dr Assume g is constant between the Moon’s surface and the Columbia spacecraft.

Section 9 — Gravitational and Electric Fields

57. Gravitational Potential

ANSWERS 1) False. Escape velocity doesn’t depend on mass. 2) Find the area under the graph between the two points and multiply by m (DW = mDV and the area under the graph is equal to DV). 3) On the surface of the mass, V is negative. As r increases, V decreases at a decreasing rate, tending to 0.

]6.67 × 10 -11g × ]5.97 × 10 24g GM 4) –6.25 × 107 J kg–1. V =– r = – = –62 511 616.95 = –6.25 × 107 J kg–1 (to 3 s.f.) ]6.37 × 10 6g 5) A. No work is done by gravity to move a mass along the equipotential lines. For a radial field like the Earth’s, equipotentials are circles around the centre of the mass. 6) The escape velocity is the velocity at which the object’s kinetic energy (Ek) is equal to minus its gravitational potential energy. Gravitational potential, V, is gravitational potential energy per unit mass, 1 GMm GMm so Ep = Vm = - r . Putting Ek = –Ep gives 2 mv2 = r . 2GM Cancelling the m and rearranging gives v = r . - 792× 10 6 - DV - DW 7) 110 km. Dr = g = mg = 4500 × 1.6 = –110 000 m = –110 km (Ignore the – as you just need distance.)

TIP

Don’t forget that gravitational potential (V) is not energy – it’s the energy per unit mass. If you need to find the gravitational potential energy of an object you need to multiply V by its mass. V is not in joules — that’s the clue.

58. Orbits and Gravity

Section 9 — Gravitational and Electric Fields

Quick Questions 1) True or false? The orbital period of a satellite depends on its mass. A. B. 2) Which of the planes shown opposite could a geostationary satellite orbit in?

C.

D.

3) Is a geostationary satellite in a synchronous orbit?

Now try these: 4) A comet orbits the Sun in an elliptical orbit, as shown. Explain how the total 1 energy, kinetic energy (given by 2 mv2 ) and gravitational potential energy GMm (given by - r ) of the comet vary during the orbit.

Comet

Sun

5) At which part of the comet’s orbit is it moving fastest? 4p 2 r 3 6) To derive T2 = GM , you need the equations for four other quantities. Which four quantities? 7) Briefly describe the main steps in deriving the formula in Q6. 8) Use the formula in Q6 to show that the orbital radius of a geostationary satellite is 42 000 km. G = 6.67 × 10–11 Nm2kg–2 and mass of Earth = 5.97 × 1024 kg.

TIP

That derivation goes on a bit, but if you know the four formulas and the main steps it should all come together, so focus on that and you’ll have it memorised. That formula is all the maths you need to explain orbits, really.

Section 9 — Gravitational and Electric Fields

58. Orbits and Gravity

ANSWERS 1) False. The orbital period only depends on the mass of the object it’s orbiting, and the radius of the orbit. 2) B. A geostationary orbit always orbits in the plane of the equator. 3) Yes. A synchronous orbit is one where the orbital period of the orbiting object is the same as the rotational period of the orbited object. A geostationary orbit is a type of synchronous orbit where the satellite is above the same point on Earth all the time. 4) The total energy stays the same throughout the orbit. Total energy = kinetic energy + gravitational potential 1 GMm energy = 2 mv2 - r . As the comet moves closer to the Sun, r decreases, so the gravitational potential energy decreases (it gets more negative). As the total energy must stay the same, when the potential energy decreases, the kinetic energy increases. The reverse is true when the comet moves away from the Sun. 5) The comet moves fastest when closest to the Sun, as the kinetic energy is at a maximum at that point. 6) E.g. the force acting on an object in circular motion, the force of attraction due to gravity, speed, and the circumference of a circle. 7) E.g. equate the force on an object in circular motion and the force of attraction due to gravity. Rearrange for v. Use speed = distance ÷ time, where the distance is the circumference of a circle, to find the time taken for one orbit (T). Substitute the expression you found for v into the formula for T and rearrange. -11 24 3 4p 2 r 3 GMT2 3 ]6.67 × 10 g × ]5.97 × 10 g × ]24 × 60 × 60g2 8) T2 = GM so r = = 4p 2 4p 2 = 42 226 910.18 m = 42 000 km (to 2 s.f.)

59. Electric Fields (1)

Section 9 — Gravitational and Electric Fields

Quick Questions 1) What is meant by the electric field strength, E, of a charge Q? 2) True or false? The electrostatic force between two charges depends on the material between them.

Now try these:

4) Sadiq measures the mass of each balloon as 4 g before the experiment. He measures the angle shown, q, to be 8° and the distance between the balloons’ centres to be 35 cm. mg He assumes the balloons are equally charged and calculates the charge on 1 Q1 Q2 . Calculate the charge on a balloon. one balloon using F = 4pe 0 r2 –1 Woo-hoo. g = 9.81 Nkg , e0 = 8.85 × 10–12 Fm–1. 5) Suggest why Sadiq’s results are not very precise. 6) Suggest how Sadiq could improve the accuracy of his result.

18

3) Sadiq is desperate to know the charge on a balloon that he’s rubbed on his jumper at a party. He sets up the experiment shown with two balloons that he’s rubbed on his jumper for T a while to try to calculate the electrostatic force on a balloon. Each balloon is suspended  using an equal length of thread. He assumes the balloons behave as charged spheres and works out that the force between them is equal to mg tan q. Using the diagram, explain why the force of repulsion between the balloons is F = mg tan q.

Section 9 — Gravitational and Electric Fields

59. Electric Fields (1)

ANSWERS 1) It’s the force per unit positive charge that a charge experiences in the electric field of Q. 2) True. For charges not in a vacuum, the force depends on the permittivity of the material between them. 3) F acts horizontally on each balloon. The force is repulsive (because the balloons are pushing each other apart), so the forces are as shown in the left-hand F diagram. The forces are in equilibrium, so drawing a closed triangle of forces (right-hand diagram) shows that F = mg tan q. 4) 3 × 10–7 C. Assume the balloons are equally charged, so Q = Q1 = Q2.



F

T

mg T mg



Rearranging Coulomb’s Law gives Q =  4pe 0 r2 F = 2r pe 0 mg tan q = 2 × (35 × 10–2) × p × (8.85 × 10 -12) × (4 × 10 -3) ×9.81× tan 8 = 2.74... × 10–7 C = 3 × 10–7 C (to 1 s.f.). 5) E.g. measuring the small angle precisely with a protractor would be very difficult and introduce a lot of random error / measuring the distance between the centres of the balloons precisely would be difficult and introduce a lot of random error / he won’t know the exact point at which the charge can be assumed to be concentrated, which will introduce random error to his measurement of distance between charges. 6) E.g. he could use identical balloons charged in the same way for the same amount of time to ensure they have the same charge, as assumed. He could also try to use balloons that are as spherical as possible, since he has assumed the balloons act as point charges, which can only be assumed for spherical objects. He could repeat the experiment and take an average of his results.

TIP

There are two cards on electric fields — why not have a thrilling race with your study buddy. Grab a card each and answer all the questions as fast as you can. You’ll need a timer. What could be more fun? You’re welcome.

60. Electric Fields (2)

Section 9 — Gravitational and Electric Fields

Quick Questions 1) What law can be used to calculate the electrostatic force between two point charges? 2) What is Cole’s Law? 3) How does the value of F between two point charges tell you whether they are like or opposite charges?

Now try these: 4) The diagrams opposite show four particles, each travelling at a constant horizontal speed between two charged plates. Only the particle’s charge and the potential difference of the plates are different in each set-up. Which particle has the greatest charge? F V Use the formulas E = Q and E = d to explain your answer.

+V A

+2V B

0

0

+V

+½V D

C 0

0

5) Describe the field lines between the plates. How will they differ in the diagrams of charges C and D? 6) One particle has a charge Q =  e. Calculate the electric field strength due to this charge at a distance of 1 Q 1.0 mm from the centre of the charge, using E = 4pe 2 , e0 = 8.85 × 10–12 Fm–1 and e = 1.60 × 10–19 C. 0 r 7) Explain why, as the particle travels between the plates, the combined electric field strength at a fixed radius from the particle is greater below the particle than above it.

TIP

Remember, field lines point in the direction that a positive charge would move in the field. Keep that in your head and things should fall into place. You’ll be given the equations, so you just need to be comfortable using them.

Section 9 — Gravitational and Electric Fields

60. Electric Fields (2)

ANSWERS 1) Coulomb’s law. 2) Thinly sliced cabbage with a splash of mayo. Ha. Gotcha. 3) F is negative for opposite charges, and a positive value for like charges. F F 4) D. E = Q , so Q = E . The biggest value of charge (Q) comes from the set-up with the smallest electric field strength (E) and the biggest F. The more the path curves, the greater the force (F), so charges C and D have the greatest force acting on them. The smaller the potential difference between the plates, the smaller the E V between them (since E = d ). Charge D has the greatest F and the smallest E, so it has the largest charge. 5) The field lines between the plates will be straight lines pointing from the top (positively-charged) plate to the bottom plate (i.e. vertical, parallel lines). The field lines in the set-up for charge D will be spaced further V apart than for charge C — the potential difference is smaller, so the electric field strength E = d is weaker. -19 Q 1 60 × 10 . 1 6) 1.4 ×10–3 NC–1. E = 4pe 2 = = 1.43... × 10–3 = 1.4 × 10–3 NC–1 (to 2 s.f.) 0 r 4p ]8.85 × 10 -12g × ]1.0 × 10 -3g2 7) Electric field strength, E, of the particle is a vector acting away from field strengths cancel the centre of the positively-charged particle. E for the field between the plates acts away from the positive plate. So below the charge, the fields both act downwards and combine to make a greater electric field strength. Above the charge, the electric fields act in field strengths combine opposite directions and therefore cancel, meaning the field above the particle has a smaller electric field strength.

61. Electric Potential and Work Done

Section 9 — Gravitational and Electric Fields

Quick Questions 1) Describe the E-r graph for a field between two parallel charged plates, if r is the distance from one plate. 2) True or false? For the field in Q1, no work is done when moving a charge anywhere on a flat plane at 90° to the plates because this plane is an equipotential.

Now try these: 3) The two graphs shown opposite show the absolute electric potential as you move a distance (r) along a line directly between two point charges. One graph is for identical like charges, the other is for identical opposite charges. Explain which is which. 4) Explain why graph B reaches zero. 5) For which graph can the work done on a charge moved between the two point charges be equal to zero? Use DW = QDV to explain your answer. 6) Calculate the work done to bring an electron and a proton from infinity to 1.0 fm 1 Q apart. Use the equations V = 4pe r and DW = QDV. e0 = 8.85 × 10–12 Fm–1, 0 charge on a proton =  e = 1.60 × 10–19 C, charge on an electron = – e.

V

Graph A

0 V 0

r

Graph B

r

7) Explain why the answer to Q6 is negative. You need to know how to derive that QDV is equal to work done. Just equate the two equations for electric field TIP strength, E, in a uniform field. You’re given both formulas in the exam so there’s literally no excuse. Learn it.

Section 9 — Gravitational and Electric Fields

61. Electric Potential and Work Done

ANSWERS 1) The graph would be a horizontal line. The electric field between the plates is uniform, so E is constant. 2) False. No work is done on equipotential surfaces, but the equipotentials for this field are flat planes that are parallel to the plates rather than perpendicular. 3) Graph A shows the absolute electric potential between two like charges, and graph B shows it for two opposite charges. V is the electric potential energy that a unit positive charge has. For like charges, V will stay either positive (if the charges are positive) or negative (if the charges are negative) for all values of r. For opposite charges, V will be positive close to the positive charge and negative close to the negative charge. 4) Graph B reaches zero because the total electric potential between opposite charges is the sum of the positive electric potential due to the positive charge and the negative electric potential due to the negative charge. So the graph reaches zero when these values have equal magnitudes and therefore cancel out. 5) For graph A, the work done can be zero. DW = QDV and looking at the graphs, the only graph for which DV can be zero is graph A, as the curve starts and stops at the same value of V. If you think about it, you’d need to put in work against the electric field to move a particle for half the journey, but for the other half the electric field will do the work to move the particle. ^ - 1.60 × 10 -19h 1 1 Qe 6) –2.3 × 10–13 J. DV = 4pe r = = –1.438... × 106 JC–1 -12 0 1× 10 -15 4p ]8.85 × 10 g –19 6 –13 DW = QpDV = 1.60 × 10 × –1.438... × 10 = –2.30... × 10 = –2.3 × 10–13 J (to 2 s.f.). 7) DW is the work you need to put in. Electrons and protons are oppositely charged, so they attract. That means that it takes work to separate them, rather than to bring them together. Or you can think of it as the electric field is doing work to bring them together.

62. Comparing Electric and Gravitational Fields

Section 9 — Gravitational and Electric Fields

Quick Questions 1) True or false? A main difference between gravitational fields and electric fields is that gravitational forces are always attractive, but electrostatic forces are not. 2) Give five (yes, five) similarities between electric and gravitational fields.

Now try these: 3) An alpha particle with charge +2e, and a proton with charge +e, are separated by a distance of d. 1 Q1 Q2 Use the formula F = 4pe to find the distance, in terms of d, from the proton to the point directly 0 r2 between the two particles where the electrostatic force on a charge is zero. 4) An alpha particle is made up of two protons and two neutrons. Assume mp = mn. Show that the point where the gravitational force on a mass directly between them is zero isn’t the same as the point where absolute electrostatic force is zero. Gm1 m2 5) A proton’s charge (in C) is 108 times bigger than its mass (in kg). Without calculating forces, use F = r2 1 Q1 Q2 and F = 4pe to explain why attractive gravitational forces between two protons are negligible compared 2 0 r 1 to repulsive electrostatic forces, at a fixed distance, r. G = 6.67 × 10–11 Nm2kg–2 and 4pe = 8.99 × 109 mF–1. 0

TIP

It’s two fields for the price of one in this section. As long as you’re clear on the differences between them, a lot of the physics and maths is the same. It’s like Physics is putting a gentle hand on your exam-stressed shoulder.

Section 9 — Gravitational and Electric Fields

62. Comparing Electric and Gravitational Fields

ANSWERS 1) True. Electric forces can be attractive or repulsive. 2) E.g. the field lines for radial and uniform electric and gravitational fields are the same shape. Absolute electric 1 potential and gravitational potential are both proportional to r . The forces due to gravitational and electric 1 fields are both proportional to 2 . Electric field strength and gravitational field strength for a radial field is r 1 proportional to 2 . The equipotentials for a point mass and a point charge are both spherical surfaces. r d 3) . Both particles are positively charged so the forces act in different 1+ 2 directions. So the point where the force is zero is the point where the forces, 1 Q1 Q2 F = 4pe on a charge due to the proton and the alpha particle are equal. alpha alfalfa 0 r2 field field 1 Qp Qq 1 Qa Qq = 4pe , where Qp, Qa and Qq are the charges on the proton, So 4pe 0 0 rp2 ra2 alpha particle and charge respectively, and rp and ra are the distance of the charge from the proton Qp Qa 1 and alpha particle respectively. Cancelling Qq and 4pe gives 2 = 2 . Qa = 2Qp so ra² = 2rp², 0 rp ra d therefore ra = 2 rp. The distance is d, so if d = ra + rp, then d = (1 + 2 )rp or rp = . 1+ 2 m p ma Gm1 m2 d 4) ma = 4mp. F = , using the same reasoning as above, 2 = 2 so ra² = 4rp² and rp = 3 . rp r2 ra 20 8 5) The constant is about 10 times bigger for electric fields. And the value of each charge is 10 times bigger than the value of each mass (of which there are two), so the force caused by the electric field is around 1036 times bigger than the force caused by the gravitational field, thus it’s negligible in comparison.

63. Capacitors

Section 10 — Capacitors

Quick Questions 1) True or false? Capacitance is the amount of current through a capacitor per unit potential difference. 2) What is permittivity? Give the equation for relative permittivity. 3) Which properties of a Q-V graph are equal to the energy stored on a capacitor and the capacitance of a capacitor?

4) The graph on the right shows the charge stored on a capacitor against the potential difference across it as it's charged. Calculate the total energy stored by this capacitor. 5) A plates of a capacitor each have an area of 25 cm2. The capacitor contains a dielectric with a permittivity of 4 × 10–11 Fm–1. The capacitance is measured to be 50 pF. Ae e Using C = d0 r , work out the separation of the plates. e0 = 8.85 × 10–12 Fm–1. 6) Explain why the addition of a dielectric between the plates of a capacitor increases its capacitance.

Q (mC)

Now try these:

40 35 30 25 20 15 10 5 0 0 1 2 3 4 5 6 p.d. (V)

Section 10 — Capacitors

63. Capacitors

ANSWERS Q 1) False. Capacitance is the amount of charge stored by a capacitor per unit potential difference (C = V ). 2) Permittivity is a measure of how difficult it is to generate an electric field in a certain material. e Relative permittivity (also known as the dielectric constant) er = e 1 , 0 where e1 is the permittivity of a material and e0 is the permittivity of free space. 3) The area under the graph gives you the energy stored. The gradient of the graph gives you the capacitance. 1 4) 9 × 10–5 J. Energy stored = area under the graph. Area of a triangle = 2 × base × height so 1 –6 –5 –5 E = 2 × 5 × 35 × 10 = 8.75 × 10 J = 9 × 10 J (to 1 s.f.). e 4 × 10 -11 5) 2 ×10–3 m. er = e1 = = 4.519... 0 8.85 × 10 -12 -4 -12 Ae e 25×10 × 8.85 × 10 × 4.519... d = C0 r = = 2 × 10–3 m 50 × 10 -12 6) A dielectric material contains polar molecules, with a positive and negative end. In the presence of an electric field generated by a capacitor these molecules align themselves in the direction of the field. The positive ends are attracted to the negatively charged plate and the negative ends are attracted to the positively charged plate. Each molecule has its own electric field which now opposes the electric field generated by the capacitor. This means, with the addition of a dielectric, the overall electric field is reduced and the potential difference required to charge the capacitor is reduced too. Therefore the capacitance increases.

TIP

Like a best friend with a load of equations scrawled up her arm, the A-level physics formula booklet will give you most of the capacitor equations you'll need in the exam. Just make sure you're comfortable using them all.

64. Charging and Discharging

Section 10 — Capacitors

Quick Questions 1) Describe a set-up you could use to measure how the current through a capacitor changes with time as the capacitor is charged. 2) Describe the I-t graph for a charging capacitor. What is the area under the graph equal to? 3) True or false? The Q-t graph for charging a capacitor will be identical to the Q-t graph for discharging the same capacitor.

Now try these: 4) A graph showing the potential difference across a capacitor over time is shown to the right. State whether the capacitor was charging or discharging. What determines the maximum value of the potential difference across a capacitor? 5) A 450 nF capacitor has a charge of 3.24 mC. Using the equation 2 1Q E = 2 C , calculate the energy stored by this capacitor. 6) A capacitor is being charged. By what factor does the energy stored by the capacitor change as the potential difference across the capacitor increases 1 from 5 V to 10 V? Use the equation E = 2 CV 2 .

V

0

t

64. Charging and Discharging

Section 10 — Capacitors

ANSWERS 1) E.g. construct a circuit containing a power supply, high resistance resistor, a capacitor and an ammeter in series. Connect a data logger to the ammeter. The data-logger will record values of the current given by the ammeter as the capacitor charges. 2) The I-t graph starts at a maximum value and decreases exponentially. This graph is I shown to the right. The area under the graph gives you the charge, since Q = I × t. 3) False. The Q-t graphs for charging and discharging are given below. Q

Q Charging 0

t

Discharging 0

0

t

t

4) The graph is for a discharging capacitor. The maximum value of potential difference = the voltage of the power supply used to charge the capacitor. -6 2 2 1 (3.24 × 10 ) 1Q 5) 1.2 × 10–5 J. E = 2 C = 2 = 1.1664 × 10–5 = 1.2 × 10–5 J (to 2 s.f.) 450 × 10 -9 1 6) The energy increases by a factor of 4. From E = 2 CV 2 you can see that the energy stored is proportional to V 2. If the potential difference increases by a factor of 2, the energy stored increases by a factor of 22 = 4.

TIP

The graphs for a capacitor discharging all have the same shape, but when it comes to charging, the current-time graph is the only one with a downwards slope — the other two have the same shape but are flipped vertically.

65. More Charging and Discharging

Section 10 — Capacitors

Quick Questions 1) True or false? The larger the capacitance, the longer it will take to fully charge a capacitor. 2) What is meant by the time constant?

3) A capacitor is connected to a power supply in a circuit and fully charged. It is then discharged through a resistor. A potential difference of 4.44 V is measured across the capacitor by a voltmeter after one time constant has passed. What is the voltage of the power supply? t The time constant t = RC and V = V0 e - RC . 4) This capacitor has a capacitance of 15 mF and the resistor has a resistance of 650 W. How much energy is stored by this capacitor 8.0 ms after it t 1 began discharging? Use the equations V = V0 e - RC and E = 2 CV 2. 5) A plot of ln(Q) against t for a different capacitor is shown. The equation t for Q is Q = Q0 e - RC . Work out the time constant, in ms, from this graph.

ln(Q)

Now try these:

4.0 3.0 2.0 1.0

- t RC

6) Using I = I0 e , calculate the time taken for the current through a discharging capacitor to reach 25% of its initial value in terms of R and C.

0.0

1.0

2.0 3.0 t (ms)

Section 10 — Capacitors

65. More Charging and Discharging

ANSWERS 1) True. 2) t, the time constant, is the time taken for the charge, potential difference or current of a discharging capacitor to fall to 37% of its value when fully charged. (t = RC) t

3) 12.1 V. If you substitute t = t = RC into V = V0 e - RC you get V = V0e–1. So V0 = V × e = 4.44 × e = 12.069... = 12.1 V (to 3 s.f.). f-

8.0 × 10 -3

-6 p

4) 2.1 × 10–4 J. V = 12.069... × e 650 × 15 × 10 = 5.312... V 1 1 E = 2 CV 2 = 2 × 15 × 10–6 × (5.312...)2 = 2.117... × 10–4 = 2.1 × 10–4 J (to 2 s.f.)

t t 1 5) 0.56 ms. Q = Q0 e - RC , so ln(Q) = ln(Q0 e - RC ). Applying log rules and rearranging: ln(Q) = b - RC lt + ln(Q0). 1 1 1 This equation is in the format y = mx + c and the gradient, m = - RC = - t , so t = - m . - 3.2 1 = 0.5625 = 0.56 ms (to 2 s.f.) The gradient of this line, m = 1.8 = – 1.777... so t = - 1.777... t 1 1 1 - t 6) 1.4RC. You want the time when I = 4 I0 so I = 4 I0 = I0 e RC . Therefore 4 = e - RC . t 1 1 t Taking the natural log of both sides gives ln b 4 l = ln ^e - RC h = - RC . So t = –ln b 4 l RC. 1 –ln b 4 l = 1.386... so t = 1.4RC (to 2 s.f.).

TIP

Exponential and log relationships are very useful and crop up a lot in A-level physics. So like a magnificent beaver building a dam, make sure you take the time to really get to grips with logs and their fancy rules.

66. Magnetic Fields

Section 11 — Magnetic Fields

Quick Questions 1) A straight current-carrying wire is in an external magnetic field. At what angle must the wire be to the magnetic field lines in order to feel no force? 2) Give the definition of magnetic flux density.

Now try these:

slab magnets with

opposite poles 3) A student sets up the apparatus as shown on the right. facing each other He is investigating how the force felt by the metal cradle varies with current through the wire. The strength of the magnetic field is 30 mT and the length of wire in the magnetic field is 15 cm. The student zeroes the balance and then switches on the power supply. Calculate the magnitude of the force felt by the metal cradle when there is a current of 2 A through the wire. Use F = BIl.

4) When a current flows in the wire, the reading on the mass balance is negative. In which direction is the current flowing?

stiff copper wire clamped at both ends (connected to variable resistor and power supply) metal cradle

top pan balance

5) Suggest how the student could change the set-up to show a positive reading. 6) The student varies the current through the wire and takes pairs for readings for current and mass. He uses the mass to calculate the force for each reading and then plots a graph of force against current. He draws a straight line of best fit. What value would you expect as the gradient of the graph? Use F = BIl. Give your answer to 2 significant figures.

Section 11 — Magnetic Fields

66. Magnetic Fields

ANSWERS 1) 0°. At 0° there will be no component of the magnetic field perpendicular to the current, so the wire will feel no force. At all other angles from 0° to 90° there will be some component of the field perpendicular to the current, so there will be some force, and the force will be at a maximum at 90°. 2) The force on one metre of wire carrying a current of one amp at right angles to a magnetic field. 3) 9 ×10–3 N. Magnitude of force on cradle = magnitude of force on wire, so F = BIl = 30 × 10–3 × 2 × 15 × 10–2 = 9 × 10–3 N. 4) The current acts from left to right (looking at the diagram). If the mass balance reads a negative value, then the force on the cradle must act upwards. Due to Newton's 3rd Law, the force on the wire is equal and opposite, so it acts downwards. Using Fleming's left-hand rule on the wire, if the force acts downwards and the field lines go from north to south, the current acts from left to right.

Though his physics teacher wasn’t convinced he’d grasped it, Brandon insisted the lesson was ‘Totally Fleming awesome, dude.’

5) The student could reverse the direction of the current e.g. by swapping the connections to the wire OR reverse the direction of the magnetic field e.g. by turning the cradle round. This swaps the direction of the force acting on the wire and therefore the cradle, so the force on the cradle will be a downwards force, causing a positive mass reading. 6) 4.5 × 10–3 Tm (the units NA–1 are also correct). F = BIl is in the format y = mx + c where y = F, x = I and m = Bl. So the gradient is equal to Bl = 30 × 10–3 × 15 × 10–2 = 4.5 × 10–3 Tm

TIP

Make you Look, suremay you're think totally you clued look stupid up about contorting how to your use Fleming's hand intoleft-hand weird positions rule because to findchances directions are, ofyou're fieldsgoing and to forces,tobut need usethat's it a fair physics bit. for Ohyou. physics, we love you... where else can you have so much fun with hand contortion?

67. Charged Particles in a Magnetic Field

Section 11 — Magnetic Fields

Quick Questions 1) When using Fleming's left-hand rule on a positive charge moving in a magnetic field, what does the second finger represent? 2) Describe the structure of a cyclotron.

Now try these: 3) Calculate the force acting on a particle with a charge of 3.2 × 10–19 C, travelling at 3.0 × 106 ms–1 perpendicular to a magnetic field with a flux density of 8.0 T. Use F = BQv. 4) The diagram on the right shows the path of a particle entering a magnetic field. Determine whether the particle is positively or negatively charged. 5) The particle enters the field with a velocity of 5.71 × 107 ms–1. The force acting on the particle causes it to move in a circular path with a radius of 6.21 × 10–4 m. Calculate the charge on the particle if the magnetic flux density of the field is 0.523 T and the mass of the particle is 9.11 × 10–31 kg. mv2 Use F = r and F = BQv.

particle

magnetic field out of card

6) A proton and an electron both enter a magnetic field at right angles to the field lines. They both enter at the same speed and then each move in a circular path. Explain the differences in their circular paths.

Section 11 — Magnetic Fields

67. Charged Particles in a Magnetic Field

ANSWERS 1) The direction of motion of the positive charge. 2) A cyclotron is made up of two hollow semicircular electrodes, with an alternating potential difference applied between the electrodes. A magnetic field is also applied perpendicular to the plane of the electrodes. 3) 7.7 × 10–12 N. F = BQv = 8.0 × 3.2 × 10–19 × 3.0 × 106 = 7.68 × 10–12 = 7.7 × 10–12 N (to 2 s.f.). 4) Negatively charged. Using Fleming's left hand rule (with the magnetic field coming out of the card and the force acting upwards), the 'current' (or the motion of a positively charged particle) flows towards the left. As the particle is actually travelling towards the right, the particle must be negatively charged. 9.11× 10 -31 × (5.71× 107) 2 mv2 = 4.782... × 10–12 N r = 6.21× 10 -4 F 4.782... × 10 -12 Rearrange the second formula for charge: Q = Bv = = 1.601... × 10–19 0.523 × 5.71× 107 = 1.60 × 10–19 C (to 3 s.f.)

5) 1.60 × 10–19 C. F =

6) A proton has more mass than an electron, so the radius of the proton's circular path is larger as the proton is deflected less. The charges on a proton and an electron are opposite, so one path will be clockwise while the other is anticlockwise.

TIP

When using Fleming's left-hand rule, your middle finger points in the direction that a positive charge is moving. For a negative charge, your middle finger points in the opposite direction to its motion. Get it? Got it? Good.

68. Electromagnetic Induction

Section 11 — Magnetic Fields

Quick Questions 1) What are the units of magnetic flux linkage? 2) What is meant by electromagnetic induction?

Now try these: 3) Determine whether or not an e.m.f. is being induced in each of the following diagrams. A.

B.

N S

C. N

N

S

S

4) A single circular loop of wire has a magnetic flux of 188 mWb through it when it is perpendicular to a magnetic field of 1.5 T. Calculate the diameter of the loop of wire. Use F = BA and area of a circle = pr². 5) A copper wire with a length of 90 cm is twisted into a rectangular coil, with sides measuring 5 cm and 4 cm. The coil is placed in a magnetic field of 2.3 T. There is an angle of 30° between the normal to its plane and the magnetic field lines. Calculate the magnetic flux linkage in the coil. Use NF = BANcosq.

TIP

Electromagnetic induction can also be demonstrated by the search coil experiment, something you should have done in class. Here, e.m.f. is induced because current is alternating — it wouldn't work if the current was direct.

Section 11 — Magnetic Fields

68. Electromagnetic Induction

ANSWERS 1) Webers, Wb. This is the same as the units for magnetic flux (as magnetic flux linkage is just magnetic flux multiplied by an integer, N, which has no units). 2) When an e.m.f. is induced in a conductor due to the conductor experiencing a change in magnetic field. 3) An e.m.f. is only induced when the magnetic field passing through the wire or coil changes. An e.m.f is not induced in A — the wire moves in the direction of the field lines, so the field around the wire doesn't change. An e.m.f. is not induced in B — as the magnet turns, the magnetic field around the coil stays the same (this one can be a bit easier to visualise if you imagine the coil of wire turning rather than the magnet, and consider that we're looking at the magnetic field around the whole coil, not just a small section of wire). An e.m.f. is induced in C — the magnet's movement cuts magnetic field lines, changing the field around the coil. 188 × 10 -3 F 4) 0.40 m. Rearrange the first equation for A: A = B = = 0.12533... m². 1.5 A 0.12533... Rearrange the second equation for r: r = = 0.1997... m, p = p so the diameter is twice this value: 0.1997... × 2 = 0.39947... = 0.40 m (to 2 s.f.). 5) 0.02 Wb. The area of the rectangular coil is A = 5 × 4 = 20 cm² = 0.002 m². The perimeter of the coil is 5 + 5 + 4 + 4 = 18 cm. The wire is 90 cm long so it must have been wound 90 ÷ 18 = 5 times, so N = 5. Putting all these values into the equation: NF = 2.3 × 0.002 × 5 × cos30° = 0.01991... = 0.02 Wb (to 1 s.f.).

69. Induction Laws and Alternators (1)

Section 11 — Magnetic Fields

Quick Questions 1) What is the name of the law that states that induced e.m.f. is directly proportional to the rate of change of flux linkage?

Now try these:

4

3) A straight wire is moved in a magnetic field. The graph on the right shows the magnetic flux linkage in the wire against time. Explain how you can tell that the e.m.f. induced across the wire is constant. 4) Calculate the magnitude of the e.m.f. induced across the wire. Axis of rotation

N (Wb)

2) How would you find the direction in which an e.m.f. is induced in a conductor moving in a field, given the direction of the motion that causes it, and the direction of the field lines?

Magnetic field into page.

Coil

3

2 1 0

0

1

2

3

4 5 time (s)

5) A coil of wire with 65 turns and an area of 0.047 m² is rotating uniformly in a magnetic field on an axis perpendicular to the field, as shown in the diagram on the left. If the e.m.f. induced in the coil at a time of 9.5 ms is 420 V, and the coil rotates 1800 times a minute, calculate the magnetic flux density of the magnetic field that the coil is in. Use e = BANwsinwt. 6) Calculate the e.m.f. induced in the coil at a time of 5.91 s. Use e = BANwsinwt.

Section 11 — Magnetic Fields

69. Induction Laws and Alternators (1)

ANSWERS 1) Faraday's law. 2) Use Fleming's left-hand rule, with your first finger pointing in the direction of the field lines and your thumb pointing in the opposite direction to the motion (the direction of the force that opposes the motion). Your second finger then points in the direction of the induced e.m.f. (This is Lenz's law.) 3) The magnitude of the induced e.m.f. is equal to the magnitude of the rate of change of flux linkage. Therefore it's equal to the magnitude of the gradient of the graph, which is constant. 4-0 4) 0.8 V. e = gradient of the graph = - = 0.8 V 5 0 5) 0.75 T. First, work out the angular speed from the revolutions per minute. 1800 revolutions per minute = 1800 ÷ 60 = 30 revolutions per second. Convert this to radians: 30 × 2p = 60p rad s–1. Then rearrange the equation for magnetic field strength: 420 e = 0.7473... = 0.75 T (to 2 s.f.) B = ANw sin wt = 0.047 × 65 × 60p × sin (60p × 9.5 × 10 -3) 6) 410 V. e = BANwsinwt = 0.7473... × 0.047 × 65 × 60p × sin(60p × 5.91) = 409.3... = 410 V (to 2 s.f.)

TIP

For those of you who have come across differentiation, the equation for e.m.f. is what you get when you differentiate the equation for flux linkage with respect to time. If that sounds like gibberish to you, feel free to not give a hoot.

70. Induction Laws and Alternators (2)

Section 11 — Magnetic Fields

Quick Questions 1) True or false? The area under a graph of NF against t gives the magnitude of an induced e.m.f.. DF 2) Name and state the law that is responsible for the minus sign in the equation e = – N . Dt

Now try these:

4) The graph on the right shows how the magnetic flux linkage of a coil rotating in a magnetic field varies with time. At which point on the graph, X, Y or Z, is the e.m.f. induced equal to zero?

motion

N

3) A conducting rod is moved in a magnetic field, as shown in the diagram opposite. State the direction of the induced e.m.f.

magnetic field (into the card)

Z X Y t 5) A rectangular coil is set up as shown on the left, so that the coil's area is perpendicular to the magnetic field lines. The coil is then set spinning on the axis shown by the dotted line at 9p rad s–1. What is the minimum amount of time it takes for the magnetic flux linkage to halve? Give your answer to 2 s.f.. Use NF = BANcosq to help you.

6) A 4.2 m long, straight conducting rod is moved through the Earth's magnetic field, perpendicular to the direction of the field. It's moved at a speed of 3.6 ms–1 for 0.50 s. The Earth's magnetic field strength at the rod's location is 45 mT. Calculate the magnitude of the e.m.f. induced across the ends of the rod. DF Ds , F = BA and v = . Use e = N Dt Dt

Section 11 — Magnetic Fields

70. Induction Laws and Alternators (2)

ANSWERS 1) False. The magnitude of the gradient of a graph of NF against t gives the magnitude of an induced e.m.f.. 2) Lenz's law. The induced e.m.f. is always in such a direction as to oppose the change that caused it. 3) Left to right. From Fleming's left-hand rule — first finger points into the card, and thumb points up (direction of force from the induced e.m.f., opposing the rod's downward motion), so middle finger points left to right. 4) Y. E.m.f is directly proportional to the rate of change of flux linkage, and the rate of change of flux linkage is represented by the gradient of the graph. The gradient is zero at point Y (and non-zero at X and Z). 5) 0.037 s. The coil's starting position is at q = 0. As B, A and N are all constants, and cos(0) = 1, this means the magnetic flux linkage is at a maximum in the starting position. The magnetic flux linkage reaches half its value 1 when cosq = ½, so the angle at this point is q = cos–1(½) = 3 p . So the minimum amount of time it takes 1 for the magnetic flux linkage to halve is equal to the time it takes to move through an angle of 3 p . q 1 –1 As q = wt, for an angular speed of 9p rad s , t = w = 3 p ÷ 9p = 0.037037... = 0.037 s (to 2 s.f.). 6) 6.8 × 10–4 V. The distance the rod travels is Ds = vDt = 3.6 × 0.50 = 1.8 m. The area of flux that the rod cuts, A, is equal to the rod's length multiplied by the distance it travels, so A = lDs = 4.2 × 1.8 = 7.56 m2. This means the change in magnetic flux, DF = BA = 45 × 10–6 × 7.56 = 3.402 × 10–4 Wb. 3.402 × 10 -4 DF DF Since N = 1 the e.m.f. is e = N = = = 6.804 × 10–4 = 6.8 × 10–4 V (to 2 s.f.). 0.5 Dt Dt

TIP

Make sure that you're really comfortable converting between units for angles, and always, ALWAYS double check your calculator is in the right units before you start doing calculations with trigonometric functions.

71. Alternating Currents

Section 11 — Magnetic Fields

Quick Questions 1) Describe the oscilloscope display for the voltage of a dc supply when the time base is switched on. 2) Describe the oscilloscope display for the voltage of an ac supply when the time base is switched off.

Now try these: 3) A country's mains electricity supply has a root mean square (rms) value of 120 V. Calculate the peak-to-peak voltage of the mains electricity supply. V Use Vrms = 0 . 2 4) The image on the right shows the display on an oscilloscope screen when an alternating voltage supply is connected. The oscilloscope is set up so that the time base is 2.5 ms per division. 1 Calculate the frequency of the voltage supply. Use f =  T .

V0 . 2 6) State what would happen to the waveform shown if the time base was doubled to 4 ms per division. 5) The Y-gain is 6 V per division. Calculate the rms voltage of the voltage supply to 3 s.f.. Use Vrms =

7) A device has a peak current of 2.5 A when connected to an ac supply that has a peak voltage of 4.0 V. An identical device has a current of 2.5 A when connected to a dc supply that has a voltage of 3.0 V. V I Determine which supply has the highest average power output. Use Vrms = 0 , Irms = 0 and P = VI. 2 2

Section 11 — Magnetic Fields

71. Alternating Currents

ANSWERS 1) A horizontal line straight across the screen. This is because a dc source is always at the same voltage. 2) A vertical line in the middle of the screen. This just shows the peak-to-peak voltage, and since the time base is turned off, the trace of the voltage just goes up and down on the spot. 3) 340 V. Rearrange the formula for the peak voltage: V0 = Vrms × 2 = 120 × 2 = 169.705... V Then double this value to find the peak-to-peak voltage: 169.705... × 2 = 339.4... = 340 V (to 2 s.f.) 4) 100 Hz. One full wavelength is completed in 4 divisions (squares), and each division is 2.5 ms, 1 1 so the time period = 4 × 2.5 = 10 ms = 0.01 s. f =  T = 0.01 = 100 Hz. 5) 12.7 V. The peak voltage is 3 divisions high and the Y-gain is 6 V per division, so the peak voltage is 3 × 6 = 18 V. V 18 Vrms = 0 = = 12.727... = 12.7 V (to 3 s.f.) 2 2 6) Doubling the time base would mean the waves would get more squashed together so that one wavelength is two squares wide. The height of the peak would stay the same since the Y-gain hasn't changed. 7) The dc supply has the highest average power. V0 I 4.0 2.5 The ac average power is P = VrmsIrms = × 0 = =5W × 2 2 2 2 The average power of the dc supply is P = 3.0 × 2.5 = 7.5 W

TIP

Nobody ever said that physics was easy, but at least the equations in this bit are fairly straight forward. And it's cool to know all about how mains electricity works I guess... well, some people would say so... I'm sure of that.

72. Transformers

Section 11 — Magnetic Fields

Quick Questions 1) Which of the following describes how electricity is transmitted across the country in power cables? A. low voltage & high current B. high power & low current C. high voltage & low power 2) What are eddy currents in transformers, and how do they reduce the magnetic field strength?

Now try these: 3) Describe the basic structure of a step-up transformer, and explain how it increases the voltage. 4) A transformer has an efficiency of 72%. When there's an input current of 0.5 A, there is an output power of 36 W. IV Calculate the input voltage. Use efficiency = I s Vs and P = VI. p p 5) Suggest and explain two ways to increase the efficiency of the transformer in Q4.

Stinkbrute certainly saw

6) The primary coil of a transformer has 23 times the number of turns on it than a transformation when on the secondary coil. State the type of transformer this is and calculate the he looked in the mirror. Ns Vs output voltage if the input voltage is 55 V. Use the equation N = V . p p 7) Two cables, X and Y, are transmitting electricity. Cable X transmits electricity at 700 A and cable Y transmits electricity at 2100 A. If the cables have the same resistance at these currents, by what factor is the power lost in cable Y bigger than in cable X? Use P = I²R.

Section 11 — Magnetic Fields

72. Transformers

ANSWERS 1) B. Electricity needs to be transmitted at a high power and low current to minimise energy losses. 2) Eddy currents are looping currents induced by the changing magnetic flux in the core. They reduce magnetic field strength by creating a magnetic field that acts against the field that induced them. 3) It consists of an iron core with two coils wrapped around either side (as shown opposite) — the primary and secondary coils. The secondary coil has more turns than the primary coil. An alternating input voltage in the primary coil creates a changing magnetic field, which passes through the iron core and into the secondary coil. A voltage is induced in the secondary coil, which is greater then the input voltage since there are more turns on the secondary coil. Is Vs 36 4) 100 V. Rearrange for Vp: Vp = I × efficiency . P = VI, so output power = IsVs, so Vp = 0.5 × 0.72 = 100 V p 5) E.g. laminate the core to reduce the effect of eddy currents. Eddy currents create a magnetic field that acts against the field that induced them and dissipate energy by generating heat. Use thick copper wire for the coils to lower the resistance. A lower resistance causes less energy to be dissipated by generating heat. Reducing dissipated energy increases the efficiency. 6) Step-down transformer with Vs = 2.4 V. More turns on the primary coil than the secondary coil means it's N 1 a step-down transformer. Rearrange the equation so: Vs = Vp Ns = 55 × 23 = 2.39... = 2.4 V (to 2 s.f.) p 7) 9 times. For a constant resistance, power is proportional to the current squared. Cable Y transmits electricity at 3 times the current of cable X (2100 ÷ 700 = 3), so cable Y has 32 = 9 times the power lost.

TIP

Transformers — they bring together loads of concepts in this section. Currents creating fields... fields inducing currents... ac... mains electricity... they're the best real world example of using magnetism I can think of right now.

Section 12 — Nuclear Physics

73. Rutherford Scattering and Atomic Structure Quick Questions 1) What were alpha particles fired at in the Rutherford alpha scattering experiment? 2) What is the name of the model of the atom that was disproved by Rutherford’s alpha scattering experiment?

detector

Now try these: 3) A group of scientists fire a beam of alpha particles at a thin copper foil. A circular detector is set up around the source and foil, as shown on the right, and the scientists record where alpha particles are detected. Suggest what the results of the experiment will be.

alpha source

copper foil

4) An alpha particle with a kinetic energy of 5.3 MeV travels straight towards the nucleus of a copper atom. A copper nucleus contains 29 protons. Determine the distance of closest approach of the alpha particle. Q . Use e = 1.60 × 10–19 C, e0 = 8.85 × 10–12 Fm–1 and the equations DW = QDV and V = 1 4πε 0 r 5) Explain how the value calculated in Q4 is related to the radius of a copper nucleus.

TIP

The Rutherford alpha scattering experiment wasn’t the end for our understanding of the atom. Throughout the 20th century, many more experiments were done, which revealed new things like neutrons and electron orbits.

Section 12 — Nuclear Physics

73. Rutherford Scattering and Atomic Structure

ANSWERS 1) A (piece of ) thin gold foil. 2) The plum pudding model. So sadly, there’s no longer an excuse to bring dessert into your physics class. 3) The scientists should find that the majority of the alpha particles that are fired at the copper foil will pass straight through the foil. Some of the alpha particles will be deflected, most of them by small angles but a small number will be deflected by large angles, sending them back in the direction of the alpha source. 4) 1.6 × 10–14 m. At the distance of closest approach, all the kinetic energy of the alpha particle is converted to electric potential energy. So the kinetic energy of the alpha particle is equal to the work done to move the particle to its position of closest approach, DW. Converting the kinetic energy of the alpha particle to J, Ek = DW = 5.3 × 106 × 1.60 × 10–19 = 8.48 × 10–13 J. DW 8.48 × 10 -13 Using DW = QDV and Qalpha = 2e, DV = Q = = 2.65 × 106 J alpha 2 × 1.60 × 10 -19 so the electric potential at the distance of closest approach is equal to 2.65 × 106 J. Q Rearrange V = 1 for r to find the distance of closest approach using Qnucleus = 29e. 4πε 0 r Qnucleus ]29 × 1.60 × 10 -19g r= = = 1.574... × 10–14 = 1.6 × 10–14 m (to 2 s.f.) 4πε 0 V 4π × 8.85 × 10 -12 × 2.65 × 106 5) The value calculated in Q4 is a maximum possible value for the radius of a copper nucleus, and so serves as an estimate for the radius of the nucleus.

74. Nuclear Radius and Density

Section 12 — Nuclear Physics

Quick Questions 1) True or false? The typical radius of a nucleus is around 5 × 10–11 m. 2) True or false? The nucleon number of a nucleus is directly proportional to the nuclear radius cubed.

Now try these: 3) A bismuth (Bi) nucleus has a nucleon number of 209, and an estimated radius of 7.84 × 10–15 m. Determine the nuclear radius of an atom of palladium (Pd) with nucleon number 106. Use the equation R = R0 A . relative 4) Estimate the density of a bismuth-209 nucleus. intensity m 4 –27 3 and V = pr . Use mnucleon = 1.67 × 10 kg and the equations r = V 3 5) A scientist fires a beam of high-energy electrons at a thin film of tantalum. The de Broglie wavelength of an electron in the beam is 7.2 fm. The scientist plots a graph of the relative intensity of the diffracted electron beam against diffraction angle, shown on the right. Determine the radius of a tantalum nucleus, R. –36 0 36 Use the equation sin q = 1.22λ , where q is the angle of the 2R angle of diffraction (°) first minimum of the diffraction pattern, l is the de Broglie wavelength of a diffracted electron. 1

3

74. Nuclear Radius and Density

Section 12 — Nuclear Physics

ANSWERS 1) False. The typical radius of a nucleus is about 1 × 10–15 m. 5 × 10–11 m is the typical radius of an atom. 2) True. Nuclear radius is directly proportional to the cube root of the nucleon number, so the nucleon number is directly proportional to the nuclear radius cubed. 3) 6.25 × 10–15 m. First, determine the constant R0 from the values for bismuth: -15 R0 = RBi = 7.84 × 10 = 1.321... × 10–15. Then calculate the radius of a palladium nucleus: (209) ABi RPd = R0 APd = 1.321... × 10–15 × (106)⅓ = 6.252... × 10–15 = 6.25 × 10–15 m (to 3 s.f.) 1

1

3

1

3

3

4) 1.73 × 1017 kgm–3. mass of a bismuth-209 nucleus = 209 × mnucleon = 209 × 1.67 × 10–27 = 3.4903 × 10–25 kg volume of bismuth-209 nucleus = 4 πRBi3 = 4 π × (7.84 × 10–15)3 = 2.0185... × 10–42 m3 3 3 So, density of bismuth-209 nucleus: -25 r = m = 3.4903 × 10 -42 = 1.729... × 1017 = 1.73 × 1017­ kgm–3 (to 3 s.f.) V 2.0185... × 10 5) 7.5 fm. From the graph, the angle of first diffraction minimum is 36°. Rearrange the equation for R, so: 1.22 × 7.2 × 10 -15 R = 1.22λ = = 7.47... × 10–15 m = 7.5 fm (to 2 s.f.) 2 × sin (36°) 2 sin θ

TIP

The density of nuclear matter is much larger than atomic density, which suggests that an atom contains lots of empty space. Just like this tip.

75. Radioactive Emissions (1)

Section 12 — Nuclear Physics

Quick Questions 1) Which type of radioactive emission has the greatest mass? 2) Name the type of radioactive emission that is negatively charged.

To: @007 Mission: Become more stable.

3) State the type of radioactive emission that won’t be affected by a magnetic field.

Now try these: 4) A student is doing an experiment to determine the radiation emitted by an unknown source. They set up the equipment shown in the diagram on the right. They record the count rate detected by the detector when sheets of different materials are placed at position X. Suggest why this would not be a suitable experimental set-up to test whether the source emits beta-plus radiation.

source 1 cm X

5) The student finds there is no change in the count rate recorded when there is a sheet of paper placed at position X compared to when there is nothing placed at position X. What can you conclude about the type of radiation emitted by the source? Explain your answer.

1 cm Geiger counter

6) The student places a 5 mm sheet of aluminium at position X. What would you expect the Geiger counter to detect now if the source emits only beta-minus radiation? Explain your answer.

Section 12 — Nuclear Physics

75. Radioactive Emissions (1)

ANSWERS 1) Alpha particle. 2) Beta-minus particle. All the other radioactive emissions are positively charged (alpha and beta-plus) or uncharged/neutral (gamma radiation). 3) Gamma radiation. All the other radioactive emissions have an electric charge, and so will interact with a magnetic field. 4) Beta-plus particles are positrons, which are a form of antimatter. They have a very short range in air because they are quickly annihilated by their matter counterpart (an electron), so it is likely they would not be able to reach the detector, regardless of what material was placed at position X. This means the experimental set-up cannot be used to determine if the source emits beta-plus radiation. 5) The source doesn’t emit alpha radiation. If it did, there would have been a decrease in the radiation detected when a sheet of paper was placed at position X. This is because the paper would have absorbed the alpha particles emitted from the source. 6) They would observe the count rate detected dropping to equal to the background radiation count rate, because the beta-minus particles emitted by the source would all be absorbed by the aluminium sheet. (Remember, the Geiger counter will always pick up background radiation. The student should correct their results for it though).

TIP

You need to know four different types of nuclear radiation, each with distinct properties. Make sure you know the mass, charge, ionising power, and penetration of each type, so you can identify them from experimental results.

76. Radioactive Emissions (2)

Section 12 — Nuclear Physics

Quick Questions 1) True or false? Alpha sources are more dangerous if ingested than when they’re outside the body. 2) Why should radioactive isotopes used as medical tracers have short half-lives?

Now try these: 3) An engineer is making sheets of aluminium by pressing some aluminium through rollers. As the sheets are produced, they pass between a beta-minus source and a Geiger counter. When the sheets are the correct thickness, the Geiger counter records a count-rate of 100 counts per second (cps). Describe what the engineer should program the machine to do if the count rate is higher than 100 cps. What about if the count rate is lower than 100 cps? 4) Another engineer suggests that they could modify the machine to make it more precise by swapping the beta-emitter for an alpha-emitter. Is this a sensible idea? Explain your answer. 5) In a PET scan, a patient is injected with a beta-plus emitting isotope. The gamma rays emitted when the beta-plus particles annihilate are detected outside the body and used to produce a detailed image. Explain why gamma radiation is suitable for this use. 6) Suggest why a doctor may prioritise other, less-detailed scanning techniques before using a PET scan.

TIP

Even the dangerous aspects of radiation can have uses. Irradiating an object with gamma radiation can sterilise it — this ionises atoms in the bacteria on the object to kill them, while leaving the object intact and safe to use.

Section 12 — Nuclear Physics

76. Radioactive Emissions (2)

ANSWERS 1) True. Alpha particles have a very short range but are highly ionising. So if an alpha source gets into the body it can do a lot of damage to a concentrated area around the source. Outside the body the alpha particles are absorbed either before they reach the skin, or by the skin. 2) They should have short half-lives because the ionising radiation can cause damage to the human body, and the shorter the half-life, the less time the tracer will be radioactive for. This will minimise the amount of radiation that the patient is exposed to, and so minimise the risk of them being harmed by the radiation. 3) If the count rate is higher than 100 cps, the engineer should program the machine to roll the aluminium thicker (as too much beta radiation is passing through). If the count rate is lower than 100 cps, they should program the machine to roll it thinner. (Beta-minus particles are absorbed by aluminium, and the thicker the aluminium, the more beta-minus particles it will absorb.) 4) No, this isn’t a sensible idea. Alpha particles will be completely absorbed by all thicknesses of aluminium, and so alpha radiation cannot be used to monitor the thickness of the sheet. 5) Gamma radiation is highly penetrating, so most of the radiation can pass out of the human body to be detected by an external detector. Gamma radiation is also weakly ionising, so it’s less likely to do damage to the human body as it passes through it. 6) The beta-plus and gamma radiation will ionise healthy cells in the patient’s body, which is potentially harmful to them. So a doctor may try to make a diagnosis with a less-detailed scan that poses less risk to the patient before resorting to a PET scan, in order to minimise the risk to the patient’s health.

Section 12 — Nuclear Physics

77. Investigations of Radioactive Emissions Quick Questions 1) Give two reasons why background radiation depends on location. 2) Why is it important to calculate the count rate due to background radiation when you’re doing an investigation into radiation?



3) True or false? The intensity of gamma radiation 1 m from a gamma source will be twice the intensity 2 m away from the source.

Now try these:

5) A student wants to investigate how the intensity of gamma radiation changes with distance by measuring the count rate detected from a source of gamma radiation at different distances. Explain why count rate is proportional to intensity for a given experimental set-up, if the same Geiger counter is used throughout. 6) The graph of the student’s results is shown on the right. At what distance from the source would the count rate be 166 counts per second? Use the equation I = k2 . x

count rate (s-1)

4) A student is doing an experiment involving a source of gamma radiation. During the experiment, the student stays several paces away from the source whenever possible. Why is this good practice for safety?

150 100 50 0

20 40 60 80 100 distance (cm)

Section 12 — Nuclear Physics

77. Investigations of Radioactive Emissions

ANSWERS 1) Any two from: e.g. the amount of certain sources of background radiation (e.g. radon gas) varies from place to place. / Some sources of background radiation are localised to a specific location (e.g. nuclear power plants). / Some locations are more shielded from types of background radiation (e.g. cosmic rays). 2) Because the background radiation will add a systematic error to all your measurements of count rate (they’ll all be off by an amount equal to the background count rate). 3) False. Intensity of gamma radiation follows an inverse-square law. The intensity 2 m away from the source will be 4-times smaller than the intensity 1 m away from the source — the distance has doubled and 22 = 4. 4) The intensity of gamma radiation follows the inverse square law, so the intensity decreases with the square of the distance from the source. By staying a few paces away from the source whenever possible, the student reduces their exposure to gamma radiation, and so the risk of damage from it is significantly reduced. 5) Intensity is the radiation per unit area. Because the same detector is being used throughout the experiment, the area over which the gamma radiation is detected is constant. Since the area is constant, the amount of radiation detected (the count rate) will be proportional to the intensity. 6) 19 cm. Since count rate and intensity are proportional, you can use count rate in the place of intensity in the equation. Find the constant k using a pair of values from the graph: e.g. k = Ix2 = 150 × 202 = 60 000. Rearrange the equation for x, x = k = 60 000 = 19.01... = 19 cm (to 2 s.f.) I 166

TIP

Experimental results for gamma intensity won’t always show the inverse-square law exactly. Look out for errors in an experiment that might cause some deviation, such as the source emitting other types of radiation.

78. Exponential Law of Decay

Section 12 — Nuclear Physics

Quick Questions 1) A 900 g pure sample of a particular radioactive isotope has a half-life of 10 hours. What is the half-life of a 50 g pure sample of the same isotope? 2) Why is it important that isotopes used in radioactive dating have a long half-life? 3) Describe how to determine the half-life of an isotope using an activity-time graph.

Now try these:

lnN

4) A sample of the radioactive isotope scandium-48 contains 1.2 × 1024 scandium-48 nuclei. It has an activity of 5.3 × 1018 Bq. Calculate the decay constant of scandium-48. Use the equation A = lN. 5) Determine the half-life of scandium-48 in hours. Use the equation T½ = ln 2 . λ 7.0 6) The graph on the right shows the natural logarithm of the number of nuclei of a particular radioactive isotope plotted against time. Work out the probability per second of any given nucleus of the isotope undergoing nuclear decay. 7) The initial activity of another sample of this isotope was 6.7 × 1016 Bq. Determine the activity after 3 hours. Use the equations A = lN and N = N0e–lt.

TIP

6.0 0.0

6.0 12.0 time (hours)

Take care when using the decay constant — for A = lN, the decay constant needs to be in s–1 if activity’s in Bq. In the exponential decay equation, it can be in any (time unit)–1 as long as your t is in the same time unit.

Section 12 — Nuclear Physics

78. Exponential Law of Decay

ANSWERS 1) 10 hours. The half-life of a particular isotope is always the same for that isotope, no matter the amount. 2) Radioactive dating is used to date very old specimens based on the activity of certain isotopes, so the isotopes need to still have a detectable activity after a long period of time. 3) Choose a value of activity from the y-axis, and read from the graph the value of time at that activity. Use the graph to find the time when the activity is half of your initial value of activity. Find the difference between these two values of time — this will be equal to the half-life of the isotope. 4) 4.4 × 10–6 s–1. Rearrange the equation for the decay constant, l = A ÷ N = 5.3 × 1018 ÷ 1.2 × 1024 = 4.41... × 10–6 = 4.4 × 10–6 s–1 (to 2 s.f.) ln 2 5) 44 hours. T½ = ln 2 = = 156 938.98... s λ 4.416... × 10 –6 T½ = 156 938.98 ÷ (60 × 60) = 43.5... = 44 hours (to 2 s.f.) 6) 1.2 × 10–5 s–1. The probability of a given nucleus decaying per second is the decay constant. The decay constant of an isotope is equal to the negative of the gradient of a ln(N)-time graph. Convert the time to seconds and determine the gradient. - 0.6 6.4 - 7.0 gradient = = 50 400 = –1.19... × 10–5 = –1.2 × 10–5 s–1 (to 2 s.f.) (14.0 × 60 × 60) - 0 So the decay constant l = 1.2 × 10–5 s–1 (to 2 s.f.) 7) 5.9 × 1016 Bq. Combine the equations to get the decay equation in terms of A. N = A , so A = A0 e–lt, λ λ λ -5 so A = A0e–lt = (6.7 × 1016) × e -1.19...×10 ×3×60×60 = 5.89... × 1016 Bq = 5.9 × 1016 Bq (to 2 s.f )

79. Nuclear Decay

Section 12 — Nuclear Physics

Quick Questions R

1) How many neutrons are there in a nucleus of

N

I?

131 53

Q

N=Z

2) Which line, Q, R or S, correctly shows the line of stability on the N-Z graph on the right?

S

3) The emission of which type of radiation causes a neutron in the nucleus to turn into a proton? Z

Now try these: 4) Describe the process of electron capture. Give an equation for the process that takes place. 5) Give a nuclear equation for the beta-plus decay of

Hg to gold-195 (symbol Au).

Fr into astatine-217 (symbol At).

7) The graph on the right shows an incomplete energy level diagram for a nuclear decay. The total energy of the decay is 1.174 MeV. Determine the type of radioactive emission represented by the transition labelled X, and determine the energy change during this transition.

Energy

6) Give a nuclear equation for the decay of

195 80

221 87

137 55

Cs 

-

DE = 0.512 MeV X

137 56

Ba

Section 12 — Nuclear Physics

79. Nuclear Decay

ANSWERS 1) 78. Just take away the proton number from the nucleon number. 2) R. (Protons repel each other by the electrostatic force. The stability line is above N = Z as you need more neutrons than protons to increase the distance between protons so they don’t repel each other as strongly. The more protons a nucleus has, the stronger the electrostatic repulsion between them, and so the more neutrons are needed to keep the nucleus stable.) 3) Beta-minus particles. The neutron changes into a proton and releases a beta-minus particle (an electron) and an antineutrino. 4) In electron capture, an atomic nucleus absorbs one of its own orbiting electrons. A proton in the nucleus interacts with the electron, changing the proton into a neutron, and emitting a neutrino and gamma radiation. p + e $ n + ne + g 0 Hg $ 195 79 Au + +1β + ν e

5)

195 80

6)

Fr $ 21785At + 42α . The decay is an alpha decay, since the francium-221 nucleus has lost 4 nucleons to become astatine-217, so it must have emitted an alpha particle. 221 87

7) Transition X is a gamma emission with energy 0.662 MeV. After the beta-minus decay, the nucleus has 137 nucleons and 56 protons. So the second stage must be a gamma decay as the nucleon and proton number stay the same during this transition. The energy of the transition is 1.174  – 0.512 = 0.662 MeV.

TIP

Just like a carefully constructed house of cards, your nuclear equations have to be balanced. And don’t forget those pesky neutrinos in beta decays. Honestly, without them, all of particle physics would come crashing down.

80. Nuclear Fission and Fusion

Section 12 — Nuclear Physics

Quick Questions 1) What part of a nuclear fission reactor minimises the amount of radiation that escapes from the reactor in normal operation? 2) Other than being radioactive, how else is the waste from a nuclear reactor dangerous? How should this danger be reduced? 3) Which of the following nuclei is most likely to undergo spontaneous fission? B. 25298 Cf C. 1432 Si A. 6830 Zn

Now try these: 4) Explain what would happen in a nuclear fission reactor if no control rods were used. 5) How does the electrostatic force between nuclei affect the conditions required for nuclear fusion? 6) Water is commonly used as a moderator in nuclear fission reactors. Explain the purpose of a moderator, and suggest why water is suitable for this purpose. 7) The isotope cadmium-113 is a good absorber of neutrons, known to absorb all neutrons below a particular energy level. Suggest how cadmium-113 could be used in a nuclear reactor and why it may be suitable for that use.

TIP

Waste is the big problem with nuclear fission — material from a fission reactor plant can be dangerously radioactive. It needs to be stored well away from living organisms for years until its activity drops to safe levels.

Section 12 — Nuclear Physics

80. Nuclear Fission and Fusion

ANSWERS 1) Shielding such as a concrete case around the reactor. It absorbs radioactive emissions released by the reactor 2) The waste is initially very hot. It should be placed in cooling ponds until its temperature drops to safe levels. 3) B. 25298 Cf has the higher nucleon number and so is the most unstable. The less stable an isotope is, the more likely it is to be able to undergo spontaneous fission. 4) There would be a runaway chain reaction. Nuclear reactors use a supercritical mass, meaning each fission reaction would quickly trigger multiple other fission reactions without controls rods absorbing neutrons. This would cause a huge release of energy, causing the reactor to overheat and explode. 5) The electrostatic force between two nuclei is repulsive, since all nuclei are positively charged. The nuclei need enough kinetic energy to overcome the repulsive forces between them. This means that for fusion to occur, conditions have to be such that the nuclei have very high kinetic energies (e.g. very high temperatures). 6) A moderator is used to slow down high-energy neutrons emitted from fission reactions to a low enough speed that they can trigger further fission reactions. Water contains hydrogen, which has a similar mass to a neutron. Collisions between neutrons and the moderator are elastic, and the closer the masses of the particles are, the more kinetic energy transferred from one to the other in an elastic collision. Neutrons need to be slowed down a lot to cause a fission, so water is suitable as a moderator. 7) Cadmium-113 could be used in control rods. The purpose of control rods is to absorb low energy/thermal neutrons in the reactor (which would otherwise trigger nuclear fission reactions). Cadmium-113 may be able serve this purpose as it absorbs neutrons below a certain energy level (provided that energy level is higher than the energy of a thermal neutron).

81. Binding Energy

Section 12 — Nuclear Physics

Quick Questions 1) True or false? The mass of a nucleus is less than the total mass of its nucleons. 2) What term is used for the energy required to separate all of the nucleons in a nucleus?

Now try these: 4) The graph on the right shows a plot of average binding energy per nucleon against nucleon number, A. Two nuclei, X and Y, are marked on the graph. Would energy be released if nuclei X and Y combined in a nuclear fusion reaction? Explain your answer.

Average binding energy per nucleon

3) How does the average binding energy per nucleon change when two hydrogen nuclei fuse into a helium nucleus? X

Y

0 A 5) A phosphorus-31 nucleus has a mass defect of 0.28225 u. Determine the binding energy per nucleon of phosphorus-31 in J. Use u = 1.661 × 10–27 kg, c = 3.00 × 108 ms–1 and the equation E = mc2. 6) A carbon-12 nucleus fuses with a hydrogen-2 nucleus to form a nitrogen-14 nucleus. Use the binding energies per nucleon below to estimate the energy released by this fusion reaction in MeV. Binding energy per nucleon (12C) = 7.68 MeV, binding energy per nucleon (2H) = 1.11 MeV, binding energy per nucleon (14N) = 7.48 MeV.

Section 12 — Nuclear Physics

81. Binding Energy

ANSWERS 1) True. The difference between these two values is called the mass defect. 2) The binding energy of the nucleus. 3) It increases. Average binding energy per nucleon increases with nucleon number up to A = 56. 4) No, energy wouldn’t be released. If nuclei X and Y were fused together, the nucleon number of the product would be higher than Y, so the binding energy per nucleon would be lower than for Y. Since the binding energy per nucleon decreased from reactants to products, no energy will be released (in fact, energy would be required to cause them to fuse). 5) 1.36 × 10–12 J. The mass defect is equivalent to the binding energy, according to E = mc2. Convert the mass defect from u to kg, mass defect = 0.28225 × 1.661 × 10–27 = 4.688... × 10–28 kg. Total binding energy, E = mc2 = 4.688... × 10–28 × (3.00 × 108)2 = 4.21935... × 10–11 J Binding energy per nucleon = 4.21935... × 10–11 ÷ 31 = 1.361... × 10–12 = 1.36 × 10–12 J (to 3 s.f.) 6) 10.3 MeV. Find the total binding energy of each nucleus. Binding energy of 12C = 7.68 × 12 = 92.16 MeV. Binding energy of 2H = 2 × 1.11 MeV = 2.22 MeV. Binding energy of 14N = 7.48 × 14 = 104.72 MeV. Energy released = binding energy of products – binding energy of reactants = 104.72 – (92.16 + 2.22) = 10.34 MeV = 10.3 MeV (to 3 s.f.)

TIP

Fusion reactions release so much energy because, for small nuclei, binding energy increases dramatically for a small increase in nucleon number. You can see this by how steep the curve of the graph in Q4 is for small A.

82. Optical Telescopes (1)

Section 13: Option A — Astrophysics

Quick Questions 1) Which is meant by the quantum efficiency of a CCD? 2) Give two advantages that a reflecting telescope has over a refracting telescope. 3) True or false? All light rays that pass through a converging lens will converge on the principal focus.

Now try these: 4) What causes chromatic aberration? 5) In a refracting telescope's normal adjustment, the objective lens and the eye lens are 24 cm apart. fo If the eye lens has a focal length of 15 mm, calculate the magnification of the telescope. (M = f ) 6) A large, reflecting telescope with a parabolic mirror is constructed. When the first images from the telescope are taken, scientists find that they are much more blurry than expected. Suggest and explain what could have caused this. What name is given to this phenomenon? 7) Sirius is a binary star system, that appears as one star when viewed with refracting telescope A, but two stars when viewed through refracting telescope B. Describe how the structures of the two telescopes are different. l Use the Rayleigh criterion, q ≈  D , to help explain your answer.

e

Section 13: Option A — Astrophysics

82. Optical Telescopes (1)

ANSWERS 1) The proportion of incident photons hitting the CCD that are detected (which is typically 80% or more). 2) Any two from: e.g. large mirrors in reflecting telescopes are cheaper to build than large lenses in refracting telescopes. / Mirrors in reflecting telescopes can be supported from underneath so they don’t distort as much as lenses in refracting telescopes. / Reflecting telescopes don’t suffer from chromatic aberration like refracting telescopes can. 3) False. Only light rays initially travelling parallel to the principal axis will converge on the principal focus. 4) A glass lens refracts different colours/wavelengths of light by different amounts, causing the image for each colour to be formed in a slightly different position. 22.5 5) 15. fe = 15 mm = 1.5 cm. fo = 24 – 1.5 = 22.5 cm. Putting these into the equation: M = 1.5 = 15. (You could have used any units of length here, just as long as they're the same for fe and fo.) 6) E.g. the shape of the mirror in the telescope. If the shape of the mirror wasn't perfectly parabolic, the incoming parallel light rays that were reflected from the mirror wouldn't all converge to the same point, so the images would appeared blurred. This is known as spherical aberration. 7) Telescope A does not have the resolving power to distinguish the two stars (so they appear as one star) whereas telescope B does. The higher the resolving power, the smaller the minimum angular resolution, q, l so q must be smaller for telescope B than telescope A. The Rayleigh criterion q ≈  D , shows that D, the objective lens diameter, must therefore be larger for telescope B than telescope A.

TIP

Refracting telescopes may have revolutionised astronomy, but larger and fancier telescopes tend to use mirrors these days — it's much easier to make them bigger to get smaller resolutions. Plus it's cheaper. It's win win.

83. Optical Telescopes (2)

Section 13: Option A — Astrophysics

Quick Questions 1) The diagram on the right shows part of a reflecting telescope set-up. Name this telescope arrangement. 2) The diagram shows a single light ray incident on the primary mirror. Describe the path that the light ray takes after it hits the primary mirror.

incoming ray

hole

eye lens

primary mirror

convex secondary mirror

3) State in which type of telescope chromatic aberration takes place.

Now try these: 4) Compare the convenience of using a CCD with the convenience of using the human eye as an image detector for a telescope. 5) Two light sources have an angular separation in the sky of 0.10° and both emit light with a wavelength of 650 nm. Calculate the smallest aperture size that a l telescope would require to resolve the two sources. Use q ≈  D .

6) A refracting telescope in normal adjustment is set up so that the focal length of the eye lens is 5.0 cm. When an object is viewed with the naked eye, it subtends an angle of 0.20°. When the object is viewed with the telescope, the image subtends an angle of 3.40°. Calculate the focal length of the objective lens. angle subtended by image at eye fo Use the formulas, M =  angle subtended by object at unaided eye and M = f . e

Section 13: Option A — Astrophysics

83. Optical Telescopes (2)

ANSWERS 1) The Cassegrain arrangement. 2) The ray is reflected from the primary mirror onto the secondary mirror. It is then reflected from the secondary mirror through the hole in the primary mirror to then refract through the eye lens. 3) Chromatic aberration occurs in refracting telescopes. 4) E.g. CCDs are more convenient in that they produce digital images that can be stored and shared globally. However the human eye is more convenient in that it doesn’t need any extra equipment — you can just look down a telescope to see an image. p 5) 3.7 × 10–4 m. Convert the angle to radians: 0.10 × 180 = 1.745...× 10–3 rad l 650 × 10 9 Rearrange the equation for the diameter of the aperture: D ≈ q = 1.745...× 10 -3 = 3.724...× 10–4 = 3.7 × 10–4 m (to 2 s.f.) 3.40 6) 85 cm. Using the first formula, M = 0.20 = 17 Rearranging the second formula for fo gives: fo = Mfe = 17 × 5.0 = 85 cm

TIP

Fun fact — in the 1600s, chromatic aberration was minimised by using lenses with very long focal lengths — known as aerial telescopes. Sometimes, these telescopes were so long that they needed cranes to hold them up.

84. Non-Optical Telescopes

Section 13: Option A — Astrophysics

Quick Questions 1) Give one way that the structure of a radio telescope differs from that of an optical reflecting telescope. 2) Give one reason why an infrared telescope captures better images in space than it would on the Earth’s surface.

Now try these: 3) Explain how the set-up of a grazing X-ray telescope brings X-rays to a focus. 4) Radio telescopes often exist in large arrays, with the data from each telescope linked together to provide a single image. Explain why arrays are needed for radio telescopes to obtain a similar resolving power to that of optical telescopes. 5) Telescope A, a radio telescope, has a collecting dish with a diameter that is three times the size of the parabolic mirror in telescope B, an optical telescope. Compare the resolving power and collecting power l of telescope A to those of telescope B. Use q ≈  D and the fact that radio waves have a wavelength at least 103 times longer than visible waves.

TIP

UV telescopes work in a similar way to optical reflecting ones, but they have to be more precisely made. This is because UV light has a shorter wavelength than visible light, so is more affected by imperfections in the mirror.

Section 13: Option A — Astrophysics

84. Non-Optical Telescopes

ANSWERS 1) Any one from: e.g. it's made up of one reflecting dish rather than two reflecting mirrors. / A wire mesh dish can be used instead of a mirror. / An antenna is used instead of an eye or camera. 2) E.g. most infrared wavelengths are absorbed by the Earth’s atmosphere, so having a telescope in space allows more infrared wavelengths to be observed, creating better images. 3) A grazing X-ray telescope contains a series of nested mirrors. As the X-rays enter the telescope, they reflect off the mirrors by grazing their surfaces, causing the path of the X-rays to be gradually altered until they focus on a detector. 4) Radio waves have much longer wavelengths than visible light waves, so the collecting dish in radio telescopes would need to be impossibly bigger than that of optical telescopes, in order to have the same resolving power. Linking the data from many radio telescopes means the effective size of the dish is much larger (it's the size of the separation of the telescopes), increasing the resolving power. l 5) The resolving power of a telescope depends on the Rayleigh criterion, q ≈  D . The smaller the minimum angular resolution, q, the better the resolving power. The minimum angular resolution of telescope A is 103 at least 3 ≈ 300 times bigger than telescope B. So telescope B has a much higher resolving power. Collecting power µ diameter². Telescope A is 3 times larger, so it has a collecting power of 32 = 9 times the collecting power of telescope B.

85. Distances and Magnitude

Section 13: Option A — Astrophysics

Quick Questions 1) True or false? The brightness of a star measured by different people could produce different results. 2) How is one light-year defined? 3) How is one parsec defined?

Now try these: 4) What condition is required for the apparent magnitude of a star to be equal to its absolute magnitude? 5) Star A has an apparent magnitude of 2.0 and star B has an apparent magnitude of 5.0. Determine which star appears brighter, and calculate by how much one star appears to be brighter than the other. 6) The Tarantula Nebula is at a distance of about 50 kpc from Earth, and has an absolute magnitude of around –10.5. By calculating its apparent magnitude, determine whether the d Tarantula Nebula is visible with the naked eye. Use the formula m – M = 5log b 10 l . 7) The brightness of star X appears to be six times that of star Y. The absolute magnitude of star X is 2.3 and the apparent magnitude of star Y is 0. Calculate the d distance, in parsecs, from Earth to star X. Use the formula m – M = 5log b 10 l . Give your answer to 2 s.f..

TIP

The log in the absolute magnitude formula is always to the base 10. Most calculator’s log buttons are set to base 10 anyway, so I tend not to worry about it. In fact, I sleep like a log*. *I keep waking up in the fireplace...

Section 13: Option A — Astrophysics

85. Distances and Magnitude

ANSWERS 1) True. Brightness is a subjective scale of measurement. 2) The distance that electromagnetic waves travel through a vacuum in one year.



3) The distance between an object and the Earth when the angle of parallax (q in the diagram on the right) is equal to one arcsecond.

SUN

Object EARTH

4) The star must be 10 parsecs away from Earth. The absolute magnitude of a star is defined as what the apparent magnitude would be if it were 10 parsecs from Earth. 5) Star A appears to be 16 times brighter than star B. Star A has a lower apparent magnitude than star B, I I so it appears to be brighter. I2  = 2.51m –m so IA = 2.515–2 = 15.813... = 16 (to 2 s.f.) 1 B 6) No, it isn't visible with the naked eye. Rearrange the formula for apparent magnitude and substitute d 50 × 103 in the numbers: m = M + 5log b 10 l = –10.5 + 5log b 10 l = 7.99... = 8 (to 1 s.f.) So the Tarantula Nebula would not be visible with the naked eye because the limit of the naked eye corresponds to an apparent magnitude of 6. 1

2

7) 1.4 pc. If star X appears to be 6 times brighter than star Y, the intensity ratio of star X to star Y is 6. I I Rearrange IX = 2.51mY – mX for mx : mx = my – log2.51( IX ) = 0 – log2.51(6) = –1.946... Y

Then rearrange the formula for d: d = 10 × 10

Y

m- M 5

= 10 × 10

-1.946... - 2.3 5

= 1.4145... = 1.4 pc (to 2 s.f.)

86. Stars as Black Bodies

Section 13: Option A — Astrophysics

Quick Questions 1) What is a black body? 2) True or false? If the distance between a star and the Earth was halved, the intensity of radiation received on Earth from the star would double.

Now try these:

Power radiated

3) Proxima Centauri is the nearest star to the Sun, at a distance of 1.3 pc. Calculate the intensity of the radiation received on Earth from Proxima Centauri if its luminosity is 6.51 × 1023 W. 1 pc = 3.08 × 1016 m. You can assume Proxima Centauri is a sphere. The surface area of a sphere = 4pr². 4) Proxima Centauri has a diameter of 2.15 × 108 m. Determine which of the black body radiation curves in the graph on the right corresponds to Proxima Centauri. Use the equations P = sAT4, lmaxT = 2.9 × 10–3 mK and the surface area of a sphere = 4pr². s = 5.67 × 10–8 Wm–2K–4. 5) The Pistol Star is one of the most luminous stars in the Milky Way, with a temperature 2.0 times that of the Sun. The surface area of the Pistol star is 9.3 × 104 times bigger than the surface area of 0 the Sun. How many times bigger is the power output of the Pistol Star compared to the Sun? Use P = sAT4.

A

B C 1.0 1.5 0.5 Wavelength (μm)

Section 13: Option A — Astrophysics

86. Stars as Black Bodies

ANSWERS 1) A black body is a body that absorbs and can emit all wavelengths of electromagnetic radiation. 2) False. The energy emitted from the star is assumed to be emitted from a point source, so intensity and distance obey an inverse square law. So if distance halves, intensity increases by a factor of four. 3) 3.2 × 10–11 Wm–2. Luminosity is the same as power. The intensity is the power per square metre, which is P I= when the power is emitted from a sphere, where d is the distance from the source. Convert the 4pd 2 6.51× 10 23 distance to metres: 1.3 × 3.08 × 1016 = 4.004 × 1016 m, so: I = 4p × (4.004 × 1016) 2 = 3.23... × 10–11 = 3.2 × 10–11 Wm–2 (to 2 s.f.) 4) Curve C. The surface area of Proxima Centauri is 4pr² = 4p × ((2.15 × 108) ÷ 2)² = 1.452... × 1017 m² 4 4 P 6.51× 10 23 b sA l = Rearrange the first equation: T = 5.67 × 10 -8 × 1.452... × 1017 = 2981.8... K 2.9 × 10 -3 2.9 × 10 -3 Rearrange the second equation: lmax = = 2981.8... = 9.725... × 10–7 = 0.97 mm (to 2 s.f.). T This corresponds to curve C, which has a peak just below 1.0 mm. P sA T 4 5) 1.5 × 106 times bigger. Using P = sAT4, PPistol = Pistol Pistol4 . The sigmas cancel to give: Sun sASun TSun PPistol APistol TPistol4 APistol b TPistol l4 4 = = = 9.3 × 10 × 2.04 = 1.488 × 106 = 1.5 × 106 (to 2 s.f.) × 4 PSun TSun ASun ASun TSun Remember that the shorter the peak wavelength emitted by a star, the hotter that star is. And the hotter the TIP stars get, the more power they radiate, creating black body curves that tower over those with lower temperatures.

87. Spectral Classes and the H-R Diagram

Section 13: Option A — Astrophysics

Quick Questions 1) What are the hydrogen Balmer absorption lines? 2) State the two spectral classes that show the strongest neutral metal atom absorption lines. 3) State the temperature range of stars in spectral class B.

Now try these: 4) Describe stars in spectral class M in terms of their colour, temperature and absorption lines. 5) The lines in a star’s absorption spectrum show strong hydrogen Balmer lines as well as weak spectral lines for ionized calcium. State which spectral class this star belongs to and a possible temperature for this star.

7) Match the following stars to their correct number on the H-R diagram: Aldebaran (a red giant), Rigel (a blue supergiant), Sirius A (a blue main sequence star), Sirius B (a white dwarf ), and the Sun. 8) Explain why the hottest stars only show very weak hydrogen Balmer lines in their absorption spectra.

B

O

absolute magnitude

6) The H-R diagram on the right is missing scales on both its axes. Describe the scale on each axis. Give values in your answer.

A

G

F

K

M

3 2 5 4 1

temperature (K)

Section 13: Option A — Astrophysics

87. Spectral Classes and the H-R Diagram

ANSWERS 1) Lines in spectra of hydrogen due to electrons moving between higher energy levels and the n = 2 state. 2) G and K. 3) 11 000 to 25 000 K. 4) Spectral class M stars are red, with a temperature of less than 3500 K. They show absorption lines in their spectra from neutral atoms and molecules like titanium oxide. 5) Spectral class A. You could have said any temperature within the range 7500 K to 11 000 K. 6) The x-axis (temperature) goes from 50 000 K on the left down to a value of around 2500 K on the right, in a decreasing, non-linear scale. The y-axis (absolute magnitude) starts at +15 on the bottom and goes to –10 at the top (from dimmest to brightest). 7) Aldebaran — 2, Rigel — 3, Sirius A — 5, Sirius B — 1, the Sun — 4. Red giants lie in the top right, and white dwarfs are in bottom left corner. Blue stars (like Rigel) must lie somewhere in the O, B or A region. The bit that runs through the middle shows main sequence stars. The Sun is a G star, so Sirius A must be the A star. 8) Hydrogen Balmer lines are only observed in absorption spectra when electrons in hydrogen atoms, existing in the n = 2 state, are excited to higher energy levels. In really hot stars, the temperature is so high that the majority of electrons exist in the n = 3 state, which don't give Balmer transitions when they are excited to higher energy levels.

TIP

There’s a common mnemonic for the spectral classes, but it might stick in your head even better if you come up with your own like, um... Older Bears Are Fantastically Good Knitting Machines... or, you know. Something else.

88. Stellar Evolution — Low Mass Stars

Section 13: Option A — Astrophysics

Quick Questions 1) At the end of the Sun’s life, what two elements will its core mostly be made from? 2) What stops a white dwarf from collapsing due to gravity?

3) Describe the life cycle of the Sun from birth to present day. 4) Describe the path the Sun will take on the H-R diagram on the right, from its current position to the end of its life. Describe how the temperature of the Sun will change as it changes position on the H-R diagram.

Absolute magnitude

Now try these: red giants

main sequence white dwarfs

5) Star X is a main sequence star with a similar mass to the Sun. After it leaves the main sequence stage, it M O K B G F A becomes a red giant and shell hydrogen burning takes Spectral class place within the star. Describe the process that star X will undergo as it changes from being a main sequence star to being in the shell hydrogen burning stage. 6) Describe the rest of the process that star X will go through during its red giant phase.

TIP

There's no actual motion of the star involved when it 'moves' on the H-R diagram — it's just changing its position based on its temperature and luminosity. The path of the Sun is the only one you need to know about.

Section 13: Option A — Astrophysics

88. Stellar Evolution — Low Mass Stars

ANSWERS 1) Carbon and oxygen — in low mass stars such as the Sun, the carbon-oxygen core produced at the end of the red giant stage isn't hot enough to undergo any further fusion. 2) Pressure exerted by electrons (electron degeneracy pressure). 3) The Sun was formed in a cloud of dust and gas, when denser clumps of the cloud contracted under the force of gravity. Eventually, a clump got dense enough to form a protostar, which continued to contract and heat up until the core temperature was hot enough for hydrogen nuclei to fuse into helium nuclei. This created enough radiation pressure to stop the gravitational collapse — and the Sun reached its main sequence stage, the stage it is at currently. 4) The Sun is currently in the main sequence region of the H-R diagram, in spectral class G. When the Sun expands into a red giant, it will move into the red giant area in the top right as it has become cooler and more luminous. As the red giant phase ends, the path of the Sun will move to the left of the H-R diagram and then down to the bottom left region as it becomes a white dwarf. It is now very hot but not very luminous. 5) At the end of star X's main sequence stage it will have used up the hydrogen in its core, which will cause the outward pressure to stop and the helium core to contract and heat up under the weight of the star. The heat produced by the contracting core will raise the temperature of the hydrogen surrounding the core enough for it to undergo fusion. This is the shell hydrogen burning stage. 6) The helium core will contract until it is hot and dense enough to undergo the fusion of helium into carbon and oxygen, releasing large amounts of energy and pushing the outer layers of the star outwards. When the helium runs out, the carbon-oxygen core contracts and heats a shell around it, allowing helium shell burning to take place.

89. Stellar Evolution — High Mass Stars

Section 13: Option A — Astrophysics

Quick Questions 1) During which stage of a high mass star's life will it emit a large burst of gamma rays? 2) What is meant by the event horizon of a black hole?

5) Which of the graphs on the right shows the light curve for a type 1a supernova? 6) Explain why type 1a supernovae are useful for determining distances. 7) A black hole is formed after the collapse of a massive star. The surface area of its event horizon is 2.8 × 1011 m2. 2GM Calculate the mass of the black hole. Use Rs = , c2 –11 2 –2 8 –1 G = 6.67 × 10  Nm kg , c = 3.00 × 10  ms and the surface area of a sphere = 4pr².

A.

B.

C.

Absolute magnitude

4) A star explodes in a supernova. The core remnant it leaves behind isn't massive enough to become a black hole. State what astronomical object this core remnant is and describe its composition and density.

Absolute magnitude

Now try these:

Absolute magnitude

3) True or false? The energy output of a type 1a supernova is roughly the same as the energy output of the Sun over its entire lifetime.

0 30 days Time since peak magnitude

0 2 years Time since peak magnitude

0 10 seconds Time since peak magnitude

Section 13: Option A — Astrophysics

89. Stellar Evolution — High Mass Stars

ANSWERS 1) During a supernova. 2) The boundary of the region in which nothing, not even light, can escape the gravitational pull of the black hole. / The boundary of the region in which the escape velocity is greater than the speed of light. 3) True. Both the Sun over its entire lifetime and a type 1a supernova release around 1044 J. 4) A neutron star. It is composed of neutrons and is incredibly dense (about 4 × 1017 kg m–3). 5) A. Type 1a supernova light curves have a sharp initial peak and then a gradually decreasing curve. The peak only lasts for a few days. 6) Type 1a supernovae are standard candles — their absolute magnitude is known because they all have the same absolute magnitude curve. The apparent magnitude (measured on Earth) and absolute magnitude of a type 1a supernova can be used to calculate the distance to the supernova. 7) 1.0 × 1032 kg. Rearrange the equation for the surface area of a sphere to find the radius: 2.8 × 1011 surface area r= = = 1.492... × 105 m. This is equal to the Schwarzschild radius. 4p 4p 1.492... × 105 × (3.00 × 10 8) 2 R c2 Rearrange the Schwarzschild radius equation for mass: M = 2sG = 2 × 6.67 × 10 -11 = 1.007... × 1032 = 1.0 × 1032 kg (to 2 s.f.)

TIP

You need an awful lot of mass to achieve a sizeable Schwarzschild radius — if you (somehow) transformed the Earth into a black hole, it would have a Schwarzschild radius of less than 9 mm. I'd rather you didn't though...

90. The Doppler Effect and Red Shift

Section 13: Option A — Astrophysics

Quick Questions 1) If electromagnetic radiation has been emitted from an object moving towards Earth, what will have happened to the radiation by the time it is observed on Earth? 2) What is the Hot Big Bang theory?

Now try these: 3) The Andromeda galaxy is the closest galaxy to the Milky Way. A line is observed in Andromeda’s absorption spectrum at a frequency of 545.5 THz. On Earth, this line occurs at a frequency of 545.3 THz. Calculate the observed Doppler shift of light from Andromeda, and Andromeda's velocity relative to the Df Dl v v Milky Way galaxy. Use f = - l = c , z = – c and c = 3.00 × 108 ms–1. 4) Suggest how the velocity of Andromeda would be different from the Day 0 value calculated in Q3 if Andromeda was actually a very distant galaxy. 5) The stars in a binary star system complete one revolution every twenty days. The absorption spectrum of the star system is viewed along its orbital plane and recorded daily from Earth. Spectra showing the hydrogen Balmer series from the star system on day 0, day 5 and day 10 are shown on the right. Explain the variations in the absorption lines between the three days shown.

Day 5

Day 10

Section 13: Option A — Astrophysics

90. The Doppler Effect and Red Shift

ANSWERS 1) It will have been blue shifted (the wavelengths of radiation decrease, so it shifts towards the blue end of the spectrum). 2) The universe started off very hot and very dense (perhaps as an infinitely hot, infinitely dense singularity) and has been expanding ever since. 3) z = –3.668 × 10–4 and v = 1.10 × 105 ms–1. Combine the two equations to find the Doppler shift: (545.5 × 1012) - (545.3 × 1012) Df fobserved - femitted z=– f = = = –3.6677...× 10–4 = –3.668 × 10–4 (to 4 s.f.) femitted 545.3 × 1012 Then rearrange the second equation for velocity: v = –zc = –(–3.6677... × 10–4) × 3.00 × 108 = 1.1003... × 105 = 1.10 × 105 ms–1 (to 3 s.f.) The negative sign for z, and positive sign for v show that Andromeda is moving towards the Earth. 4) E.g. if Andromeda were a very distant galaxy, it is likely that it would be moving away from us due to the expansion of the universe. So the value for the velocity would be negative (showing that it is receding). 5) On day 0, there's no Doppler shift in the star's absorption lines, as they aren't moving towards or away from us, but instead are moving at right angles to our line of sight. On day 5, each absorption line is split into two, showing that one star is blue shifted and moving towards Earth and the other is red shifted and moving away from Earth. On day 10, there's again no Doppler shift in the star's absorption lines, as they are again moving at right angles to our line of sight, having completed half a revolution in 10 days.

TIP

My sister never used to understand the Doppler effect, but she changed her tune when I ran it past her... it is a weird one to be fair — it depends on perspective, so what you observe changes depending on where you are.

91. Quasars and Exoplanets

Section 13: Option A — Astrophysics

Quick Questions 1) What are the most distant measurable objects in the universe? 2) Give one limitation of using variations in Doppler shift from the light of a star for detecting the presence of exoplanets.

Now try these: 3) A telescope has a minimum angular resolution smaller than the angle subtended by a star and its known exoplanet. Suggest one reason why the telescope may be unable to form a clear image of the exoplanet.

5) An exoplanet is orbiting a star. The light curve for this star is shown on the right. Explain how the dip in the light curve is caused by the exoplanet orbiting the star. 6) Explain how the dip in the light curve would be different if the exoplanet had a lower orbital angular velocity and a larger volume.

TIP

apparent magnitude

4) A quasar and a star have the same recorded intensity on Earth. If the quasar is 300 000 times further away from Earth than the star, by what factor is the power of the quasar greater than the star?

time

Quasars are so-called because they're quasi-stellar — they're sort of, in a way, stars... and yet, not quite stars. In the same way that this section is quasi-fun. It's sort of, in a way, fun... and yet, still revision. Quasfun.

Section 13: Option A — Astrophysics

91. Quasars and Exoplanets

ANSWERS 1) Quasars. 2) E.g. the orbit of an exoplanet around a star has to be aligned with the observer’s line of sight for any Doppler shift to be detected. 3) E.g. the star is likely to be much brighter than the exoplanet, so the light from the star would drown out any light from the exoplanet, making it difficult to produce a clear image of the exoplanet. 4) The quasar emits 9 × 1010 times the power emitted by the star. Intensity is the power of radiation per square P metre. As light from a star spreads out, the intensity follows an inverse square law: I = , so P µ Id². 4pd 2 2 Pquasar dquasar 2 m = 300 000 = 9 × 1010 = c If the intensity of the quasar and the star is the same, then P dstar star 5) As the exoplanet crosses in front of the star X, some of the light from star X is blocked from Earth’s view causing a dip in the apparent magnitude.

apparent magnitude

6) The lower orbital angular velocity would mean that the exoplanet would take longer to cross in front of the star, so the dip would be wider. The start and end of the dip would also be more angled because of this, as the exoplanet would take longer to move from not covering the star to fully covering the star. The larger volume of the exoplanet would mean that it would block out more of the star's light. So the apparent magnitude from the star would be lower and the dip would be deeper. The dip would look like the graph on the right. time

92. The Big Bang Model of the Universe

Section 13: Option A — Astrophysics

Quick Questions 1) What has been theorised to cause the accelerated expansion of the universe? 2) What is the approximate ratio of hydrogen to helium in the universe? A. 1:1 B. 2:1 C. 3:1 D. 4.1

Now try these: 4) State what cosmic microwave background radiation (CMBR) is and explain how it provides evidence for the Hot Big Bang model. 5) The graph on the right shows a line of best fit for a plot of recessional velocity against distance from Earth for different type 1a supernovae. Calculate the value of the Hubble constant in units of kms–1Mpc–1 from this graph. Use v = Hd. 6) Use the value calculated in Q5 to estimate the age of the universe in years to 2 s.f.. 1 pc = 3.08 × 1016 m. 

TIP

recessional velocity (kms–1)

3) How does the abundance of helium in the universe provide evidence for the Hot Big Bang model? 40 000 30 000 20 000 10 000 0 0 100 200 300 400 500 distance (Mpc)

Your unit conversion skills are your best friend when it comes to converting the Hubble constant. I'm sure they had a reason for making the units tricky, even if it does mean more work for you. You'd better get practising.

Section 13: Option A — Astrophysics

92. The Big Bang Model of the Universe

ANSWERS 1) Dark energy. 2) C. The universe is approximately 74% hydrogen and 24% helium. 3) The Hot Big Bang model says that the universe used to be much hotter. The large abundance of helium provides evidence for this as the universe must have been hot enough for the fusion of hydrogen into helium to occur. 4) CMBR is the microwave radiation all around us that is nearly isotropic and homogenous. The Hot Big Bang model predicts that a lot of electromagnetic radiation was produced in the early universe, that has now been stretched to microwave wavelengths due to the expansion of the universe. This is the CMBR, which therefore provides evidence for the Hot Big Bang model. v 5) 70 kms–1Mpc–1. Rearrange the equation for the Hubble constant to give H = d . This shows that the Dy 35 000 - 0 Hubble constant is given by the gradient of the graph. Gradient = = = 70 kms–1Mpc–1. Dx 500 - 0 6) 1.4 × 1010 years. Convert the Hubble constant into units of s–1: 1 Mpc = 3.08 × 1016 × 106 = 3.08 × 1022 pc, so 70 kms–1Mpc–1 = (70 × 1000) ÷ 3.08 × 1022 = 2.2727... × 10–18 s–1. 1 1 The universe age estimate is given by H = = 4.4 × 1017 s 2.2727... × 10 -18 17 4.4 × 10 Convert this to years: 60 × 60 × 24 × 365 = 1.395... ×1010 = 1.4 × 1010 years (to 2 s.f.) (14 billion years)

93. Physics of the Eye

Section 13: Option B — Medical Physics

Quick Questions 1) What are rods and cones types of ? 2) True or false? The image distance doesn’t change when the eye focuses on different distances.

Now try these: 3) A man is looking lovingly at his cat. For this example, describe the 1 1 1 distances that represent u, v and f In the equation f = u + v .

B

4) What quantity should the vertical axis of the graph opposite be labelled with? What do curves A, B and C represent?

5) A condition called complete achromatopsia results in individuals seeing everything in black and white. Suggest which cells don’t function in individuals with complete achromatopsia.

A

300

400

C

500 600 700 wavelength (nm)

6) The fovea in the human eye contains only cones in high concentration. Outside the fovea, rods and cones are found, but in much lower concentrations. Explain why individuals with complete achromatopsia might also suffer with reduced spatial resolution. 7) An unaccommodated eye is an eye that is focussed at the far point. A young person has a far point of infinity, and the power of their unaccommodated eye is 59 D. Calculate the length of their eyeball in cm. 1 1 1 1 Use the equations f = u + v and P = f .

Section 13: Option B — Medical Physics

93. Physics of the Eye

ANSWERS 1) Rods and cones are types of photoreceptor (light-sensitive) cells in the retina. 2) True. The image distance is the distance from the lens to the retina, which doesn’t change. The focal length changes to make sure the image distance is always the same regardless of object distance. 3) u is the distance from the man’s eye lens to the cat. v is the distance from the man’s eye lens to his retina. f is the focal length of the man’s eye (lens and cornea combined). 4) The vertical axis should be labelled with the relative light absorption of cones in the eye. A is the curve for the blue cone, B for the green cone, and C for the red cone. 5) Cones. These are the cells responsible for reacting to colour. 6) Spatial resolution is the ability to resolve or distinguish two objects. To resolve objects, beams of light from each object need to hit two photoreceptor cells separated by at least one other photoreceptor cell. With complete achromatopsia, the cones, and therefore the fovea, can’t be used. This means the eye has to focus images onto other parts of the retina, where there are rods, but in much lower concentrations. Since cells are much further apart here, the individual’s eye won’t be able to resolve objects as easily. 1 1 1 1 1 1 7) 1.7 cm. P = f , f = P = 59 = 0.0169... m. f = u + v where the eyeball length is equal to the 1 1 1 image distance, v, and u = 3, so u = 0. So f = v , so v = f = 0.0169... m = 1.7 cm (to 2 s.f.)

TIP

So there’s more to the eye than meets... the eye. Seriously though, this is physics at its best — just think, because our eyes are so powerful at focussing light we can see many things, from moths to mountain tops, in great detail.

94. Defects of Vision

Section 13: Option B — Medical Physics

Quick Questions 1) What is the name for the vision defect in which the near point of the eye is too far away? 2) True or false? Diverging lenses have a positive focal length. 3) Describe how the power of a converging lens differs from that of a diverging lens.

Now try these:

SPH

CYL

Axis

4) On a glasses prescription like the one opposite, SPH tells you Right Eye – 4.50 0 n/a the power in D required to correct the patient’s myopia or Left Eye – 4.00 + 0.50 60 hypermetropia, CYL represents the added cylindrical power required and Axis tells you the angle to the horizontal of the plane that doesn’t need correcting . For the prescription shown, what two vision defects is the patient being corrected for? 1 5) Use P = f to calculate the patient’s uncorrected far point in their right eye, in cm. v 6) The patient’s right eyeball is 1.8 cm long. Use m = u to calculate the magnification of the patient’s eye when viewing an object without corrective lenses at their uncorrected far point. 7) The ray diagram opposite shows how the right eye is corrected by the lens prescribed. What do distances A, B, C and D represent?

A

B C

D

Section 13: Option B — Medical Physics

94. Defects of Vision

ANSWERS 1) Hypermetropia (this is just long sight, but make sure you use the correct scientific terminology). 2) False. Diverging lenses have a negative focal length. 3) A converging lens has a positive power, whereas a diverging lens has a negative power. 4) Myopia (both eyes) and astigmatism (left eye only). A negative power in dioptres indicates myopia/short-sight, and a cylindrical lens indicates an astigmatism. 5) 22.2 cm. The right eye lens corrects myopia, so its focal length is equal to the patient’s uncorrected far point. 1 1 1 Its power is –4.50 and P = f , so uncorrected far point = f = P = = –0.2222... m = –22.2 cm (to 3 s.f.) 4.50 (You can ignore the minus sign, that’s just because it’s a diverging lens.) 6) 0.081. For an object at the patient’s far point of 0.2222... m, the eye can focus the image onto the retina at v 0.018 an image distance of 0.018 m. So m = u = 0.2222... = 0.081. 7) A is the distance of the object from the eye. B is the distance of the patient’s uncorrected far point from the eye. C is the distance from the corrective lens to the eye lens. D is the distance from the eye lens to the retina.

TIP

If you wear glasses, dig out your prescription and calculate your near or far point. It’s a little depressing but good practice. I used my prescription for these questions and honestly I’m a bit gutted I can only see to 22 cm.

95. Physics of the Ear

Section 13: Option B — Medical Physics

Quick Questions 1) Which structure separates the outer ear from the middle ear? 2) Briefly describe the basic structure of the outer ear, and state the main purpose of it. 3) Which of the following would a sound wave meet last when passing through the ear? A. oval window B. basilar membrane C. tympanic membrane D. stapes/stirrup bone

Now try these:

P 4) Using I = A and surface area of a sphere = 4pr2, calculate the power of a siren that has an intensity of 0.20 Wm–2 at a distance of 2.0 m away from the siren.

5) The diagram opposite shows the middle ear. Describe the role of the ossicles in the transmission of sound waves through the ear. 6) In the middle ear, pressure variations are increased by a factor of around 23 between the tympanic membrane and the oval window. The area of the oval window is around 3.0 mm2. The tympanic membrane has an area of about 45 mm2. Prove that the ossicles in the middle ear increase the force F of the vibrations by a factor of about 1.5. Use P = A .

ossicles

auditory canal

tympanic membrane (eardrum)

oval window

Section 13: Option B — Medical Physics

95. Physics of the Ear

ANSWERS 1) The tympanic membrane (eardrum). 2) The outer ear consists of the pinna/external ear and the auditory canal. It reflects and funnels sound towards the eardrum/tympanic membrane at the end of the auditory canal. 3) B. This is the part of the cochlea with hair cells attached. The hairs are vibrated causing electrical impulses to be sent to the brain. The other three structures have all been reached by this point. P 4) 10 W. I = A , where A is the surface area of a sphere of radius 2.0 m, as the sound waves spread out equally in all directions. So P = I × A = 0.20 × 4 × p × 2.02 = 10.05... = 10 W (to 2 s.f ) 5) The ossicles are tiny bones in the middle ear that transmit sound waves through the middle ear. A sound wave entering from the outer ear causes the tympanic membrane to vibrate. The tympanic membrane is connected to one of the ossicles (the malleus) and the vibration of the sound wave is transmitted into the bone. The vibration is then passed from the first ossicle along to the other two (the incus and stapes). The last ossicle (the stapes) is connected to the oval window, where the vibration is passed into the inner ear. The ossicles also amplify the sound signal and reduce the energy reflected back from the inner ear. F P A F 6) Rearrange P = A for F: F = PA, so Foval = Poval × Aoval . Substitute in the values: tm tm tm Foval 3.0 = 23 × = 1.533..., so the force increases by a factor of about 1.5. Ftm 45

TIP

There’s lots to remember about the ear. All sorts of physics too, including pressure, waves and forces. It’s amazing to think that everything you hear is going through this complicated process in such a tiny space.

96. Intensity and Loudness

Section 13: Option B — Medical Physics

Quick Questions 1) True or false? The threshold of hearing is the same at all frequencies. 2) On the graph shown below, what do the solid lines show?

Now try these:

Intensity Level (dB)

120

3) A 1000 Hz sound at 0 dB is perceived to be as loud as a 100 Hz sound. Use the graph on the right to determine the intensity level of the 100 Hz sound. 80 4) If the vertical axis showed the dBA scale instead, how would the solid lines look? 5) Most smoke alarms produce sound at 3000 Hz. A scientist is trialling an alarm that 40 sounds at 520 Hz. The scientist’s new alarm sounds with the same intensity level, in dB, as a standard 3000 Hz smoke alarm. Explain whether or not this is sensible. 0

100 1000 10 000 6) The scientist claims that people with age- or noise-related hearing loss will be Frequency (Hz) more likely to hear the new alarm than a standard alarm. Explain his reasoning. I 7) A whisper has an intensity level of 30 dB. Using intensity level = 10 log b I l , determine which of the 0 following noises has an intensity 100 times greater than a whisper. A. very quiet talking at 40 dB B. rainfall at 50 dB C. a jet taking off at 130 dB.

TIP

The human ear’s response to sound isn’t straightforward. Not only is its response to intensity logarithmic, but the ear works best around 3000 Hz, so those sounds seem louder than other frequencies of the same intensity.

Section 13: Option B — Medical Physics

96. Intensity and Loudness

ANSWERS 1) False. The threshold of hearing is the minimum intensity of sound a human can hear at a given frequency, and it’s different for different frequencies. 2) Curves of equal loudness (each curve shows a range of sounds that all have the same perceived loudness). 3) 40 dB. The curve of equal loudness shows that a 1000 Hz sound at 0 dB is on the same line as a 100 Hz sound at 40 dB. 4) The lines would be horizontal, as the dBA scale gives the same value of intensity level to sounds of the same perceived loudness. 5) It is not a sensible idea. If the 520 Hz alarm had the same intensity level as a 3000 Hz alarm, it would have a different perceived loudness because perceived loudness also depends on frequency. The curves of equal loudness show that perceived loudness is always lower for a 520 Hz sound than for a 3000 Hz sound at the same intensity level. So a 520 Hz alarm would sound quieter than a 3000 Hz alarm at the same intensity level in dB, so it may not be heard as easily. 6) With age-related hearing loss, higher frequencies are most affected, so being able to hear the low frequency alarm would be less affected by this type of hearing loss. For noise-related hearing loss, frequencies closer to 4000 Hz are usually more affected. 3000 Hz may therefore not be heard very well by people with this problem, and the low frequency alarm might work better. I 7) B. Intensity level of a whisper = 10 log b Iw l = 30 dB. If I is 100 times greater than lw, l = 100lw, the 0 100I I I larger intensity level = 10 log b I w l = 10 log b100 × Iw l = 10 log100 + 10 log b Iw l = 20 + 30 = 50 dB. 0 0 0 So a sound with 100 times the intensity of a whisper has an intensity level of 50 dB.

97. Electrocardiography (ECG)

Section 13: Option B — Medical Physics

Quick Questions 1) What does an ECG trace show? 2) Why does the signal for an ECG trace need to be amplified?

Now try these: 3) Look at the ECG trace of a patient shown opposite. How many T waves are shown and what do they represent?

p.d. at electrodes (mV) 1

4) What is the patient’s heart rate (number of beats per minute)? 5) Describe what is happening in the patient’s heart between 0.0 s and 0.6 s.

0.0 0.2 0.4 0.6 0.8 1.0 1.2 time (s)

6) Give two ways in which the procedure being used to obtain an ECG trace can be modified to improve the accuracy of the trace and explain why each way works. 7) In a healthy heart, an electrical signal reaches the right and left ventricles of the heart at the same time. A ‘left bundle branch block’ is a condition that causes the signal to travel more slowly to the left ventricle than to the right ventricle. Which sections of an ECG trace would you expect to be affected for someone with this condition?

Section 13: Option B — Medical Physics

97. Electrocardiography (ECG)

ANSWERS 1) The electrical signals of the heart that cause the atria and ventricles to beat each heartbeat. 2) The signal measured at the surface of the skin is very faint as the signal is attenuated (absorbed and weakened) by the body, so it has to be amplified. 3) One T wave is shown (at around 0.7 s). A T wave shows the electrical signal when the ventricles relax ready for the next heartbeat. 4) 75 beats per minute. The pattern repeats after 0.8 s, so there will be 60 ÷ 0.8 = 75 beats in a minute. 5) At 0 s, the heart has finished its last beat and the ventricles are relaxed. At 0.2 s, a generated signal causes the atria to contract. At about 0.35 s, the signal reaches the ventricles and causes them to contract. By 0.6 s, the ventricles have finished contracting but have not yet begun to relax. 6) E.g. avoid interference by removing any nearby ac sources / by shielding the apparatus / by using shielded leads. Nearby electrical signals can be picked up by the equipment and show up on the trace, reducing its accuracy. Improve conductivity between the skin and the electrodes by using a conducting gel and removing dead skin or hairs. This reduces resistances and so improves conduction at the surface of the skin, meaning the signal quality is improved and therefore the signal received is more accurate. 7) The QRS wave and T wave. These represent the signals that cause the ventricles to contract, and then relax, so if the signal to one ventricle is delayed, it will show up in these parts of the trace.

TIP

Well, this feels awfully like a biology lesson. Not that you should be complaining... you were the one that wanted to study medical physics... or maybe it was your teacher... anyway it’s too late to turn back now. Learn it.

98. Ultrasound Imaging

Section 13: Option B — Medical Physics

Quick Questions 1) Give two advantages of using an ultrasound scan to see inside a patient. 2) True or false? Ultrasound scans work best if the acoustic impedances of the different tissues at a boundary being imaged are identical.

Now try these: 3) Describe how piezoelectric crystals in an ultrasound transducer can create and detect ultrasound pulses. 4) A diagram and trace of a foetal ultrasound scan is shown opposite. What type of scan has been used?

ultrasound transducer coupling gel skin uterus lining outline of foetus’s skull uterus lining one square = 2 × 10–5 s

5) Explain why a coupling gel is used during the scan, as shown in the diagram. 6) For bone, Z = 7.8 × 106 kgm–2s–1 and for soft tissue Z = 1.6 × 106 kgm–2s–1. Calculate the 2 I percentage of ultrasound reflected at the foetus’s soft tissue-bone boundary. Use Ir = c Z2 Z1 m . Z2 + Z1 i 7) The two peaks on the oscilloscope trace show ultrasound reflections from either side of the foetus’s skull. Brain tissue has a density of 1025 kg m–3 and an acoustic impedance of 1.6 × 106 kgm–2s–1. Use Z = rc and the oscilloscope trace to calculate the width of the foetus’s skull in cm.

TIP

Learning your ultrasound scans is as easy as A, B... Make sure you can use the equations and understand how acoustic impedance affects the amount of ultrasound reflected — it’ll help you understand the results of scans.

Section 13: Option B — Medical Physics

98. Ultrasound Imaging

ANSWERS 1) Any two from: e.g. there are no known risks / it’s relatively cheap / it’s quick/convenient. 2) False. Although too much of a difference in impedance leads to most of the signal being reflected back, and therefore being blocked from imaging any deeper, identical impedances would lead to no reflection at all, so the scan wouldn’t detect the boundary between the two tissues. 3) When a potential difference (p.d.) is applied across piezoelectric crystals, they stretch and squash, which can be used to create an ultrasound pulse. When an ultrasound pulse hits piezoelectric crystals, it causes them to stretch and squash, which causes a p.d. to be generated across them, which can be detected as an electrical signal. 4) An A-scan. An A-scan produces an oscilloscope trace (whereas a B-scan produces an image). 5) Soft tissue has a very different acoustic impedance from air (as does the ultrasound transducer), so if an air gap is left between the transducer and the skin, almost all of the ultrasound is reflected from the surface of the body. A coupling gel has a similar acoustic impedance to soft tissue and is used to fill the gap between the transducer and the skin to minimise the amount of ultrasound reflected back. 2 I 7.8 × 10 6 - 1.6 × 10 6 m2 = 0.435... = 44% (to 2 s.f.) 6) 44%. Ir = c Z2 Z1 m = c Z2 + Z1 i 7.8 × 10 6 + 1.6 × 10 6 7) 7.8 cm. There are 5 squares between the two peaks, so ultrasound takes 5 × 2 × 10–5 = 1 × 10–4 s to travel across the skull and back, so 5 × 10–5 s to travel across it once. Z 1.6 × 10 6 The speed at which it travels is c = r = 1025 = 1560.97... ms–1. Speed = distance ÷ time, so distance = speed × time = 1560.97... × 5 × 10–5 = 0.0780... m = 7.8 cm (to 2 s.f.)

99. Endoscopy

Section 13: Option B — Medical Physics

Quick Questions 1) For total internal reflection to occur at a core-cladding boundary, what must be true about the core and cladding materials? 2) Give two advantages of diagnosing a patient using optical fibres for internal imaging instead of performing diagnostic surgery.

Now try these:

eyepiece

3) A patient is having a procedure to look at their small intestine. A gastroscope tube containing two optical fibre bundles is passed down their throat as shown. One bundle is attached to the eyepiece and one has a light source directed down it. Explain how a gastroscope allows the user to see inside the patient’s body, including the use of coherent and non-coherent fibre bundles.

gastroscope tube

light source

4) The cladding and core of the fibres have refractive indices of 1.45 and 1.48 respectively. n Calculate the critical angle using sin qc = n2 .

light from endoscope

1

5) The patient has to lie in a particular position in the procedure to ensure the gastroscope tube is as straight as possible. Explain why this is necessary.

Section 13: Option B — Medical Physics

99. Endoscopy

ANSWERS 1) The refractive index of the cladding must be lower than the core. (Otherwise total internal refraction will not occur.) 2) E.g. there is less risk for the patient / the patient has much quicker recovery times / it is less costly. 3) A light is sent down one fibre from outside the body to illuminate the area at the other end of the tube (inside the body). This fibre is non-coherent, meaning the relative position of each fibre can change along the fibre. The other fibre carries an image formed by an objective lens from inside the body back to the eyepiece. This fibre has to be coherent so that the image can be viewed at the eyepiece. n 1.45 4) 78.4 °. Since sin qc = n2 , qc = sin–1 b 1.48 l = 78.44... = 78.4° (to 3 s.f.). 1 Ooh. Fancy. 5) Total internal reflection will only occur if the angle of incidence is greater than the critical angle. If the tube is bent too much, then this won’t always be the case and some light will escape the fibres. This will mean less light will get to the area being illuminated, and also some of the light formatting the image going back to the eyepiece will be lost, leading to a decrease in image quality.

TIP

Optical fibres have loads of uses outside of medicine too. These days they carry high speed internet around. And back in the 90s we used them to wow our friends with space-age, rainbow-coloured fibre-optic lamps.

100. X-Ray Production

Section 13: Option B — Medical Physics

1) What are the names of the two spectra that combine to give the overall spectra of X-rays produced by an X-ray tube? 2) Why does the anode of an X-ray tube have to rotate? 3) On the graph opposite, which letter represents a value equal to the potential difference of the X-ray tube multiplied by the charge on an electron?

Relative Intensity

Quick Questions A B

D C Photon Energy

4) How does an intensifying screen help to minimise the radiation dose to a patient in an X-ray scan? 5) Two X-ray production spectra for the same X-ray tube are shown on the graph opposite. What changes have been made to the X-ray tube to get from spectrum A to spectrum B? Explain your answer.

Relative Intensity

Now try these: A

6) Explain how the clusters of spikes on the curves are produced.

B Photon Energy

7) Explain why there is only one cluster of spikes on spectrum A, but two on spectrum B.

TIP

You might have heard the continuous spectra of X-ray radiation also being referred to as bremsstrahlung. It means ‘braking radiation’ — a nice little reminder that it’s the energy released when the electrons decelerate.

Section 13: Option B — Medical Physics

100. X-Ray Production

ANSWERS 1) Continuous spectra and characteristic spectra. 2) To prevent overheating. Most of the energy of the accelerated electrons is converted to heat — overheating is prevented by rotating the anode. 3) D. This is the maximum photon energy, which is equal to the potential difference multiplied by e. 4) An intensifying screen consists of crystals that fluoresce, meaning they absorb X-rays and re-emit the energy as visible light, which helps to develop the photograph quicker. A shorter exposure time is needed, keeping the patient’s radiation dose lower. 5) The voltage has been increased because the maximum photon energy is higher and the current has also been decreased, because the overall intensity is lower. 6) They are produced when beam electrons knock out other electrons from the inner shells of the anode material’s atoms. Electrons in the atoms’ outer shells move into the vacancies in the lower energy levels, and emit energy in the form of X-ray photons. (The energies of these X-rays, called characteristic X-rays, are fixed for a given metal as they relate to the energy between electron shells in the anode.) 7) E.g. the spikes are the characteristic spectra, which are present at energies corresponding to the energy level transitions of the anode material. The tube voltage used to produce spectrum B was higher than that used to produce spectrum A, so the electrons hitting the anode had more kinetic energy. These higher energy electrons could knock out electrons from shells deeper within the anode atoms, so outer-shell electrons dropped further to fill vacancies which emitted higher energy X-rays, leading to more spikes on the graph.

101. X-Ray Imaging Techniques

Section 13: Option B — Medical Physics

Quick Questions 1) What technique allows moving images to be captured by X-ray scans?

I believe there’s a joke in here somewhere.

2) Describe how a flat panel detector turns X-rays into an image. 3) Give two differences between an X-ray scan and a CT scan.

If only CGP editors could pull their finger out and think of one.

Now try these: 4) A patient is having an X-ray scan to diagnose a problem with their stomach. The stomach is surrounded by other organs. The patient ingests a ‘barium meal’ before the scan. Explain why a barium meal is needed to produce an X-ray image of the stomach. 5) Bone has a density of around 1900 kgm–3 and a linear attenuation coefficient of 58 m–1. m Use m m = r to find the mass attenuation coefficient of bone. 6) Starting with I = I0e–mx, derive the equation for half-value thickness, x½. 7) Use I = I0e–mx to calculate the thickness of bone required to reduce the intensity of the X-ray beam by 99%.

TIP

Wishing you had X-ray vision to help you cheat on the answers? I feel ya. The equation for x½ isn’t given in the exam, but don’t worry if you can’t remember it — you can derive it from the equation for intensity (see Q6).

Section 13: Option B — Medical Physics

101. X-Ray Imaging Techniques

ANSWERS 1) Fluoroscopy. 2) X-ray photons excite scintillator material in the detector producing light with intensity proportional to the photon’s energy. The light hits photodiode pixels which generate a voltage proportional to the light’s intensity. Thin-film transistors are then used to read the voltages and create a digital image. 3) Any two from: e.g. a CT scan produces an image of a slice through the body, and an X-ray scan produces an image of looking from one side of the body to the other. / In a CT scan, the X-ray beam is rotated around the body and an array of detectors pick up the signal, and in a standard X-ray, the beam doesn’t move so there’s just one detector. / A CT scan gives the patient a higher radiation dose than an X-ray. 4) Contrast in X-ray scans is caused by different materials having different attenuation coefficients, and therefore attenuating an X-ray beam by different amounts. Soft tissues such as the stomach and other organs have similar attenuation coefficients, so there would be little contrast between them on an X-ray image and the stomach would be hard to see. Barium has a high atomic number compared to soft tissue, so it absorbs more X-rays. If a patient ingests it into their stomach, their stomach will be clearly contrasted in the scan from other surrounding organs due to the barium absorbing X-rays, so the stomach is seen on the X-ray image. m 58 5) 0.031 m2 kg–1. m m = r = 1900 = 0.0305... m2 kg–1 = 0.031 m2 kg–1 (to 2 s.f.) 1 1 6) x½ is the thickness when I = 2 I0. Substitute this into I = I0e–mx, 2 I0 = I0e–mx½ , and divide by I0, 1 ln ( 2 ) 1 1 1 so 2 = e–mx½. Using log laws, ln( 2 ) = ln(e–mx½), so ln( 2 ) = –mx½. Dividing by –m, x½ = – = ln (2) m m 1 I 7) 0.079 m. Rearranging I = I0e–mx, x = – m ln I . 0 1 I For a 99% reduction in intensity, I = 1% = 0.01 , so x = – 58 ln 0.01 = 0.0793... m = 0.079 m (to 2 s.f.) 0

102. Magnetic Resonance (MR) Imaging

Section 13: Option B — Medical Physics

Quick Questions 1) What is the term used to describe the ‘wobble’ of a proton as it rotates about an axis that is moving with circular motion. 2) Which element in the body is imaged by an MR scan?

Now try these: 3) Which method of medical imaging, out of an ultrasound, MR scan and CT scan, would be suitable for diagnosing a broken bone in a patient with a metal plate in their hip? 4) Doctors will sometimes use ultrasound to image soft tissue, even though the resolution is much worse than an image from an MR scanner. Give two possible reasons for this. 5) Explain what happens to the protons in a person’s body when they are exposed to the large magnetic field in an MR scanner. 6) Explain why a radio frequency (RF) pulse of a certain frequency will cause an MR signal to come from one part of the body but not another in an MR scan. Explain how this can be used to create an image of a single ‘slice’ of the body.

TIP

It’s complicated, but hey, there’s no maths (woo). Simply put, the field determines which protons will be excited by the RF pulse, and therefore emit an MR signal. Right, time for a precess (a break to wobble like a proton).

Section 13: Option B — Medical Physics

102. Magnetic Resonance (MR) Imaging

ANSWERS 1) Precession. 2) Hydrogen. MR scans affect protons, which is what a hydrogen nucleus is. 3) A CT scan. MR scanning usually can’t be used on patients with metal in their body and MR scanning has a low resolution for bone. Ultrasound cannot penetrate bone.

Dave’s centrepiece dessert was undeniably the epitome of precision and precession.

4) Any two from: e.g. an ultrasound scanner is much cheaper than an MR scanner. / An ultrasound scan is a lot quicker. / There’s more likely to be an ultrasound scanner locally to the patient. / An MR scanner can make patients feel claustrophobic so a patient might find it difficult to have an MR scan. / An MR scan can’t usually be performed on someone with metal in their body, so the doctor might choose an ultrasound instead. 5) The protons have a property called spin, which causes them to behave like tiny magnets. They’re originally randomly orientated, but align parallel with the magnetic field when it’s turned on and precess (wobble) about the magnetic field lines.

6) In an MR scan, field coils cause a magnetic field strength gradient across the body. The precession frequency of a proton is proportional to the magnetic field strength, so the precession frequency of the protons is different in different parts of the body. When an RF pulse is sent across the body, only the protons with the same precession frequency as the frequency of the RF pulse will absorb a radio wave, get excited, and change their spin state. An excited proton then relaxes and emits an MR signal, so only protons in certain parts of the body will emit an MR signal for certain RF pulses. By using the field coils to create a certain magnetic field strength in the slice of the body the doctor is interested in, this gives the protons in that slice a certain precession frequency. So they are the only protons that emit MR signals and only that slice is imaged.

103. Medical Uses of Radiation

Section 13: Option B — Medical Physics

Quick Questions 1) Pick your favourite radioisotope tracer and give its half-life and the energy of the gamma radiation it emits. 2) What does a photomultiplier tube in a gamma camera do? 3) Give one way that the risks from using high-energy X-rays to treat cancer can be reduced.

Now try these: 4) Explain why technetium-99m cannot be delivered to hospitals ready to use as a tracer. How do hospitals get around this problem? 5) FDG is a medical tracer used in PET scans. It contains positron-emitting fluorine-18 bound to a glucose substitute. The body uses the tracer wherever it would normally use glucose. Explain how a PET scan uses this tracer to create an image of metabolic activity in the body. 6) The half-life of fluorine-18 is 110 minutes. The effective half-life of FDG depends on the amount of water the patient drinks to flush the isotope out of their body. Explain why this is the case. 7) Calculate the effective half-life of FDG for a patient whose biological half-life is 14 minutes. 1 1 1 Use the formula T = T + T . E

B

P

Scientists have come up with some pretty nifty ways to limit a patient’s exposure to radiation during a procedure. TIP But radiation exposure always carries a risk, so the pros and cons of using it on a patient must still be considered.

Section 13: Option B — Medical Physics

103. Medical Uses of Radiation

ANSWERS 1) E.g. technetium-99m: half-life = 6 hours, energy of g radiation = 140 keV. / Iodine-131, half-life = 8 days, energy of g radiation = 360 keV. / Indium-111: half-life = 2.8 days, energy of g radiation = 170 or 250 keV. 2) A photomultiplier tube turns the flashes of light (that are emitted by the scintillating crystal when X-rays hit it) into electrical signals by releasing electrons via the photoelectric effect. Each electron released is then multiplied into a cascade of electrons. 3) E.g. use a focussed beam to ensure radiation mostly hits the tumour. / Use shielding to shield other parts of the body. / Rotate the beam around the patient to minimise radiation dose to healthy tissue. 4) Technetium-99m has a very short half-life of 6 hours. For it to be used as a tracer, the activity can’t get too low, so long transport times and storing technetium-99m on site is not possible. Instead, hospitals are delivered a Molybdenum-Technetium generator. The generator contains molybdenum, which has a much longer half-life (of 66 hours). This then decays into Technetium-99m which can be collected for use on site. 5) The positrons emitted by fluorine-18 annihilate with electrons in the body, producing gamma rays. Detectors measure where the gamma rays came from, and so the FDG can be tracked around the body. The FDG is used by the body as glucose would be — it’s used more by cells that are doing more work. This means the PET scan detects which cells in the body are working most, and therefore it can create an image of metabolic activity. 6) Effective half-life is determined by physical half-life, which is constant, and biological half-life, which measures how quickly the body gets rid of the isotope, i.e. by using it up or removing it from the body. So drinking more water to flush the isotope out will reduce the biological half-life and therefore the effective half-life. 1 1 1 1 1 1 7) 12 minutes. T = T + T = 14 + 110 = 0.080... So TE = 0.080... = 12.41... = 12 minutes (to 2 s.f.) E B P

104. Inertia and Kinetic Energy

Section 13: Option C — Engineering Physics

Quick Questions 1) True or false? An object will have different moments of inertia if it is rotated about different points. 2) How would you work out the total moment of inertia for an extended object, given that the moment of inertia for a point mass is given by I = mr2?

Now try these: 3) A solid snooker ball has a mass of 160 g and a moment of inertia of 4.3 × 10–5 kgm2. 2 What is its diameter? For a solid sphere, I = 5 mr2. 4) Shannon drops a 12 g coin with a radius of 1.42 cm and it rolls away from her with an angular velocity of 6.2 rads–1. Calculate the rotational kinetic energy of the coin. 1 1 Use the equations Ek = 2 Iw 2 and I = 2 mr2 for a solid disc.

5) A 3.6 m long barrier to a car park rotates about its end to allow cars to pass. The barrier can be treated as a uniform rod with a mass of 6.0 kg. A sign of mass 560 g is attached halfway along. Assuming the sign is a point mass, calculate the total moment of inertia of the barrier 1 and sign. I = mr2 for a point mass and I = 3 mL2 for a rod of length L rotated about its end.

6) A 15 g spherical sugar shell with a radius of 0.050 m is completely full of solid chocolate. The moment of inertia of the shell and chocolate is 5.88 × 10–4 kgm2. Assuming the shell’s thickness is 2 2 negligible, calculate the mass of the chocolate. I = 3 mr2 for a hollow sphere and I = 5 mr2 for a solid sphere.

Section 13: Option C — Engineering Physics

104. Inertia and Kinetic Energy

ANSWERS 1) True. Moment of inertia depends on the distribution of the mass around the axis of rotation, so if the position of the axis of rotation changes (i.e. from the centre of an object to the edge), the distribution of mass around the axis is now different, and so the moment of inertia will be different. 2) By adding together all the individual moments of inertia of the point masses that make up the object (i.e. I = Smr2 since each point mass has a moment of inertia of I = mr2). 5 I 5 4.3 × 10 -5 2 3) 0.052 m. Rearrange I = 5 mr2 to get r = 2 m = 2 160 × 10 -3 = 0.0259... m. d = 2r = 2 × 0.0259... = 0.0518... = 0.052 m (to 2 s.f.) 1 1 4) 2.3 × 10–5 J. I = 2 mr2 = 2 × 12 × 10–3 × (1.42 × 10–2)2 = 1.20... × 10–6 kgm2 1 2 1 Ek = 2 Iw = 2 × 1.20... × 10–6 × 6.22 = 2.32... × 10–5 = 2.3 × 10–5 J (to 2 s.f.) 1 1 1 5) 28 kgm2. Ibarrier = 3 mL2 = 3 × 6.0 × 3.62 = 25.92 kgm2. r = 2 × 3.6 = 1.8 m, so 2 –3 2 2 Isign = mr = 560 × 10 × 1.8 = 1.8144 kgm . Itotal = Ibarrier + Isign = 25.92 + 1.8144 = 27.7344 = 28 kgm2 (to 2 s.f.) 2 2 6) 0.56 kg. Sugar shell acts as a hollow sphere, Ihollow = 3 mr2 = 3 × 15 × 10–3 × 0.0502 = 2.5 × 10–5 kgm2. Ichocolate = Itotal – Ihollow = (5.88 × 10–4) – (2.5 × 10–5) = 5.63 × 10–4 kgm2. The chocolate is a solid sphere so 5 I 5 5.63 × 10 -4 2 = 0.563 = 0.56 kg (to 2 s.f.) Ichocolate = 5 mr2 = 5.63 × 10–4 kgm2. Then m = 2 2 = 2 r 0.050 2

TIP

I had a moment of inertia when I tried to get out of bed this morning... Inertia tells you all about how difficult it is to get something to move. The larger the mass, and the further it is from the axis of rotation, the harder it will be.

105. Rotational Motion

Section 13: Option C — Engineering Physics

Quick Questions 1) Which quantity is given by the gradient of an angular displacement-time graph? 2) What is meant by angular velocity? 3) True or false? A larger angular acceleration gives a smaller gradient on an angular velocity-time graph.

Now try these: 4) The wheel of a bike starts from rest and accelerates with a uniform angular acceleration of 0.73 rads–2. What is its angular velocity after 3.0 seconds? Use the equation w2 = w1 + at.

6) Two spinning toys, A and B, are set spinning. Toy A starts spinning with an angular velocity of 12 rads–1. It has a constant angular deceleration and spins for 10.0 seconds before stopping. How many revolutions did toy A make while it was spinning? (w1 + w 2) t . Use the equation q = 2 6) The graph on the right shows the angular velocity against time for toy B from the point it starts spinning. What is its total angular displacement, to 2 s.f.?

angular velocity (rads–1)

5) What is the wheel’s angular displacement from rest when it reaches an angular velocity of 3.5 rads–1? Use w22 = w12 + 2aq. 20 18 16 14 12 10 8 6 4 2 0 0 1 2 3 4 5 6 7 8 9 10

time (s)

Section 13: Option C — Engineering Physics

105. Rotational Motion

ANSWERS 1) The angular velocity. 2) Angular velocity is a vector quantity describing the angle a point rotates through per second (given in rads–1). 3) False. It will give a larger gradient, as the gradient is equal to the angular acceleration. 4) 2.2 rads–1. Substituting the values in, w2 = w1 + at = 0 + (0.73 × 3.0) = 2.19 = 2.2 rads–1 (to 2 s.f.) 5) 8.4 rad. We want to find q, and we know that w1 = 0 rads–1, w2 = 3.5 rads–1 and a = 0.73 rads–2. w2 - w2 Rearrange w22 = w12 + 2aq to get q = 2 2a 1 . Substituting the values in: 2 2 3.5 - 0 q = 2 × 0.73 = 8.390... = 8.4 rad (to 2 s.f.) (w1 + w 2) t (12 + 0) × 10.0 6) 9.5 revolutions. Substitute the values into q = = = 60 rad 2 2 60 One revolution = 2p rad, so number of revolutions = 2p = 9.54... = 9.5 revolutions (to 2 s.f.) 7) 68 rad. Angular displacement is given by the area under the graph. Each square represents 2 × 1 = 2 rad. Number of squares under the curve ≈ 34 squares. So the total angular displacement = 34 × 2 = 68 rad (As long as you’ve counted between 33 and 35 squares, and your answer’s correct for your number of squares, you can give yourself the credit.)

TIP

It’s true what they say — angular velocity makes the world go round — so make sure you get lots of practice using and rearranging these angular versions of the equations of motion. It’ll pay off when it comes to the exam.

106. Torque, Work and Power

Section 13: Option C — Engineering Physics

Quick Questions 1) State the variables represented by each of the symbols in the formulas T = Fr and T = Ia. 2) Forces A, B and C in the diagram on the right are all of identical magnitude. Which of the forces would produce the largest torque on the wheel?

axis C

A B

Now try these: 3) A bolt requires a minimum torque of 18 Nm to turn. If you use a wrench to apply a perpendicular force 30.0 cm from the bolt, what is the minimum force you must apply to turn the bolt? Use the equation T = Fr. 4) The bolt turns 180° in 1.5 seconds when the minimum torque is applied. Using W = Tq and P = Tw, calculate the work done and power exerted. 5) A machine applies a 5.0 Nm torque to a rotating disc with a radius of 25 cm. A force of 1.2 N acts at the edge of the disc to oppose this motion. If the disc has an angular acceleration of 8.4 rads–2 at this point, what is its moment of inertia? Use T = Fr and T = Ia. 6) The torque applied to the disc is reduced so that it rotates at a constant angular velocity. The force opposing the motion is still 1.2 N. If the power used to rotate the disc is now 90 W, calculate the angular velocity of the rotating disc. Use the equation P = Tw.

Section 13: Option C — Engineering Physics

106. Torque, Work and Power

ANSWERS 1) T is the torque, F is the applied force, r is the perpendicular distance from the axis of rotation to the point of the applied force, I is the moment of inertia and a is the angular acceleration. 2) B. Since all forces are identical in magnitude, the biggest torque will come from the force applied at the greatest perpendicular distance from the axis of rotation. A is applied closest, and B and C are applied at the same point, but since B is applied perpendicular to the distance from the axis of rotation, the perpendicular distance is greatest and it results in the greatest torque. T 18 3) 60 N. T = Fr so F = r = = 60 N 30.0 × 10 -2 4) W = 57 J and P = 38 W. Converting degrees to radians, q = 180° = p rad p Dq = 1.5 = 2.09... rad s–1 W = Tq = 18 × p = 56.54... = 57 J (to 2 s.f.). w = Dt P = Tw = 18 × 2.09... = 37.69... = 38 W (to 2 s.f.) 5) 0.56 kgm2. The frictional torque, Tfrictional = Ffrictional × r = 1.2 × 25 × 10–2 = 0.3 Nm The net torque on the disc, Tnet = Tapplied – Tfrictional = 5.0 – 0.3 = 4.7 Nm 4.7 T T = Ia so I = a = 8.4 = 0.559... = 0.56 kgm2 (to 2 s.f.) 6) 300 rad s–1. Frictional force is 1.2 N, so Tfrictional = 0.3 Nm = Tapplied, since the disc rotates 90 P at a constant angular velocity. P = Tw, so w = T = 0.3 = 300 rad s–1 I had a joke about torque to put here, but I decided it was a bit forced. The work and power equations take the TIP same form for rotational and straight-line motion, but the variables are different so you need to learn what’s what...

107. Flywheels

Section 13: Option C — Engineering Physics

Quick Questions 1) Which of these factors doesn’t affect the energy stored by a flywheel? A. its mass B. its angular speed C. the direction it spins D. whether it is spoked or not 2) Explain why it is important that a flywheel is lubricated regularly. 3) What is a risk associated with increasing the angular velocity of a flywheel?

Now try these: 4) A potter’s wheel is operated by a foot pedal which provides a force each time it is pressed. The wheel needs to rotate at a constant speed to sculpt clay. Explain why a flywheel is used in a potter’s wheel. 5) An electric car contains a flywheel as part of its braking system. When the brakes are applied, some of the rotational kinetic energy from the car’s wheels is transferred to the flywheel. Explain how this helps to make the car more efficient. 6) The flywheel in a machine is replaced with one that is made to spin twice as fast and is half as heavy as the original flywheel. Compare the energy stored by the new flywheel to the energy 1 stored by the old flywheel. Use the equations Ek = 2 Iw 2 and I = mr2. Assume both flywheels have the same radius.

Something had clearly been lost in translation.

Section 13: Option C — Engineering Physics

107. Flywheels

ANSWERS 1) C. The energy stored by a flywheel depends on its angular speed, total mass, and mass distribution. 2) Flywheels lose energy due to friction between the wheel and the bearings. Lubricating the wheel reduces friction and means less energy is lost. 3) E.g. if the angular velocity of a flywheel is too high it could break apart. 4) The force supplied is variable, delivered in spurts rather than constantly. The flywheel uses each spurt of force to charge up, then it delivers the stored energy smoothly to the rest of the system so that the potter’s wheel spins at a constant angular velocity. 5) Some of the energy from the car’s wheels that would usually be dissipated during braking is instead stored by the flywheel until it’s needed. This energy can then be transferred from the flywheel, and used usefully in the car (e.g. to help it accelerate again). This reduces the amount of energy wasted by the car, and so the car is more efficient. 6) The energy stored in the new flywheel is double the energy stored in the old flywheel. Doubling the angular velocity means that the energy quadruples, as Ek is proportional to w2. Halving the mass means that the energy halves, as Ek is proportional to m (because Ek is proportional to I and I is proportional to m). The overall effect of quadrupling and halving the energy, is doubling the energy.

TIP

Flywheels are useful in many different devices — if you need to smooth out torque or store energy, a flywheel might be just the ticket. Make sure you’re aware of some of their disadvantages as well as advantages, though.

108. Angular Momentum

Section 13: Option C — Engineering Physics

Quick Questions 1) True or false? Objects with the same angular velocity always have the same angular momentum. 2) True or false? If no external forces are applied, the total angular momentum of a system is constant.

Now try these: 3) A ball is attached to a pole by a string, as shown on the right. The ball swings around the pole in a horizontal circle. As the balls swings around the pole, some of the string wraps around the pole, bringing the ball closer to the centre as it continues moving in a horizontal circle. Using angular momentum = Iw, explain how the angular velocity of the ball changes as the string wraps around the pole. 4) During a break from revision, Sammi entertains herself by spinning around on her desk chair. From rest, she provides a constant torque of 74 Nm for 0.30 seconds. The total moment of inertia of the system is 2.7 kgm2. What is her angular speed after applying the torque? Use the equation D(Iw) = TDt. 5) When she pulls her arms and legs in, her angular speed becomes 13 rads–1. What is the moment of inertia of the system when her arms and legs have been pulled in? Use the equation angular momentum = Iw. You can assume that resistive forces are negligible. 6) A car wheel is slowed by the constant application of 3.2 Nm of torque from a brake. The wheel experiences an angular impulse of 26 Nms as it comes to rest. Using D(Iw) = TDt, calculate how long it takes for the wheel to come to a stop.

Section 13: Option C — Engineering Physics

108. Angular Momentum

ANSWERS 1) False. The angular momentum of a rotating object depends on its moment of inertia, as well as its angular velocity. Moment of inertia will vary depending on the mass of an object and how its mass is distributed. 2) True. This is the law of conservation of angular momentum, which can be written as Iinitialwinitial = Ifinalwfinal. 3) The angular velocity of the ball will increase as the string wraps around the pole. As the string wraps around the pole, the ball is brought closer to the axis of rotation, reducing its moment of inertia. No external forces are applied, so according to conservation laws the angular momentum will remain constant. Angular momentum = Iw so if the moment of inertia decreases, the angular velocity will increase. 4) 8.2 rads–1. D (Iw) = TDt can be rewritten as Ifinalwfinal – Iinitialwinitial = TDt. We know she started at rest so 74 × 0.30 TDt = = 8.22... = 8.2 rads–1 (to 2 s.f.) Iinitialwinitial = 0. This means Ifinalwfinal= TDt and wfinal = I 2.7 final I w 2.7 × 8.22... 5) 1.7 kgm2. Iinitialwinitial = Ifinalwfinal and Ifinal = initialw initial = = 1.70... = 1.7 kgm2 (to 2 s.f.) 13 final 6) 8.1 s. The change in angular momentum of the wheel, D(Iw), is equal to the angular impulse experienced by the wheel, so D(Iw) = 26 Nms–1. D (Iw) 26 Rearrange D (Iw) = TDt to get Dt = = 3.2 = 8.125 = 8.1 s (to 2 s.f.) T

TIP

Remember, a change in angular momentum is called angular impulse and means there must be some kind of external force acting on the system. If there aren’t any external forces then angular momentum is conserved.

109. The First Law of Thermodynamics (1)

Section 13: Option C — Engineering Physics

Quick Questions 1) True or false? An adiabatic process is one where Q = 0. 2) True or false? No work is done on or by a gas during a process where its volume remains constant. 3) What is meant by a non-flow process?

Now try these: 4) The diagram on the right shows a cylinder with a diameter of 30.0 cm, containing gas at a pressure of 8.0 × 104 Pa. Energy is transferred to the gas in the cylinder. This causes the gas to expand, moving the piston up by 5.0 cm. The pressure of the gas remains constant during this expansion. Using W = pDV, calculate how much work is done by the system.

5.0 cm

5) The internal energy of the gas increases by 18 J. How much energy was supplied to the system? Use Q = DU + W. 6) A gas at a pressure of 1.34 × 105 Pa occupies a volume of 0.12 m3. The gas isothermally expands to a pressure of 1.0 × 105 Pa. What is the final volume of the gas? For isothermal processes, pV = constant. 7) There are 5.5 moles of gas in this system. Using pV = nRT, calculate the temperature of the gas. R = 8.31 JK–1mol–1.

30.0 cm

Section 13: Option C — Engineering Physics

109. The First Law of Thermodynamics (1)

ANSWERS 1) True. In an adiabatic process no heat is lost or gained by the system and therefore Q = 0. 2) True. For work to be done on or by the system, a change in volume is required. 3) A change in a system during which no matter is able to enter or escape the system (also called a closed system). 4) 280 J. The change in volume DV = pr2Dh, where Dh is the change in height. 30.0 × 10 -2 = 0.15 m, so DV = p × 0.152 × 5.0 × 10–2 = 3.53... × 10–3 m3 r= 2 W = pDV = 8.0 × 104 × 3.53... × 10–3 = 282.7... = 280 J (to 2 s.f.) 5) 300 J. Energy supplied = Q = DU + W = 18 + 282.7... = 300.7... = 300 J (to 2 s.f.) p1 V1 1.34 × 105 × 0.12 6) 0.16 m3. pV = constant, so p1V1 = p2V2 and V2 = p = = 0.1608 = 0.16 m3 (to 2 s.f.) 2 1.0 × 105 7) 350 K. As isothermal processes have a constant temperature, you can use the initial or final conditions to work this out. Using the initial conditions p1 and V1: p1 V1 1.34 × 105 × 0.12 = 351.8... = 350 K (to 2 s.f.) p1V1 = nRT so T = nR = 5.5 × 8.31

TIP

I wanted the first law of thermodynamics to be ‘do not talk about thermodynamics’, but I got outvoted. Luckily you’ll be given Q = DU + W in the exam, so focus on what those symbols mean and where they’ve come from.

110. The First Law of Thermodynamics (2)

Section 13: Option C — Engineering Physics

Quick Questions 1) If work is done on a system, is the value of W positive or negative? 2) In what type of process is the heat energy transferred to a gas equal to the work done by the gas?

Now try these: 3) A diatomic gas in a container is adiabatically compressed to half its original volume. The initial pressure inside the container was 1.4 × 105 Pa. Using the fact that for adiabatic processes pVg = constant, determine the pressure in the container after this compression. For a diatomic gas, g = 1.4. 4) Using Q = DU + W, explain how this will change the temperature of the gas. 5) An ideal gas in a closed system is initially at 280 K. 5100 J of work is done by the gas as it expands to a volume of 0.056 m3. The pressure is kept constant at 1.2 × 105 Pa. How many moles of gas does the system contain? Eleonora was doing a Use W = pDV and pV = nRT. R = 8.31 JK–1mol–1. lot of work to increase her volume.

TIP

In a closed system you know the number of moles doesn’t change, so pV = nRT will only depend on p, V and T. For a process where one of these values is kept constant, the other two only depend on each other. Simple... ish.

Section 13: Option C — Engineering Physics

110. The First Law of Thermodynamics (2)

ANSWERS 1) If work is done on the system, W will be negative (and if work is done by the system W is positive). 2) Isothermal processes. During an isothermal process the temperature stays constant so there is no change in internal energy. From Q = DU + W we can see that this means Q = W. 3) 3.7 × 105 Pa. pVg = constant which means that p1V1g = p2V2g. p1 V1g V g This can be rearranged to give p2 = V g = p1 b V1 l . 2 2 As V1 = 2V2, substitute this to get p2 = 1.4 × 105 × 21.4 = 3.69... × 105 = 3.7 × 105 Pa (to 2 s.f.) 4) In adiabatic processes no heat is lost or gained by the system, so Q = 0. Using Q = DU + W this means DU = –W. As the volume decreases, work is being done on the system so W is negative and therefore DU is positive, so internal energy increases. Internal energy only depends on temperature, so if internal energy increases the temperature must increase too. W 5100 5) 0.70 moles. W = pDV so DV = p = = 0.0425 m3. You know the final volume, V2, 1.2 × 105 and DV so you can find the initial volume, V1 = V2 – DV = 0.056 – 0.0425 = 0.0135 m3 Rearrange pV = nRT for n: pV 1.2 × 105 × 0.0135 = 0.696... = 0.70 moles (to 2 s.f.) n = RT = 8.31× 280

111. p-V Diagrams

Section 13: Option C — Engineering Physics

Quick Questions 1) True or false? After a system has completed one cycle of a cyclic process, it will have the same pressure and volume as it did at the start. 2) How do you tell whether a curve on a p-V graph represents a compression or an expansion? p

Now try these:

C

3) The p-V diagram on the right shows three curves, A B and C. How would you find the work done by the process represented by curve C?

p / 105 Pa

4) For the graph on the right, one curve shows an isothermal process at 320 K, one shows an adiabatic process with an initial temperature of 320 K, and one shows an isothermal process at 300 K. Which curve represents each process? 10 8 6 4 2 0

E F

H G

0 1 2 3 4 5 6 7

A

V 5) The p-V diagram on the left shows a cyclic process of a system made up of 6 processes, D-I. Which of these processes shows no work being done and an increase in temperature? Explain your answer.

D I

B

V / 10–3 m3

6) How much work is done by this system after 3 cycles have taken place? Give your answer to 2 s.f..

Section 13: Option C — Engineering Physics

111. p-V Diagrams

ANSWERS 1) True. One cycle of a cyclic process creates a closed loop on a p-V diagram, starting at a particular pressure and volume and returning to the same point on the graph at the end of the cycle. 2) The arrow on the curve. If the arrow points towards a smaller value of V, then it’s a compression. If it points towards a larger value of V, it’s an expansion. 3) Find the area beneath curve C. The work done by a process on a p-V diagram equals the area under a curve. 4) Curve A shows an isothermal process at 300 K, curve B shows an isothermal process at 320 K, and curve C shows an adiabatic process at 320 K. (Curves B and C start at the same pressure, that is higher than A, so A must be the 300K process. Curve B is similar to curve A, while Curve C gets much steeper at smaller volumes, so curve B is the isothermal process and curve C is the adiabatic process.) 5) I. I  is represented by a vertical line on the p-V diagram. This shows the volume isn’t changing / the area below the line is equal to zero, and so no work is being done. The direction of the arrow shows that the pressure is increasing, so the temperature must also be increasing. 6) 9600 J. The work done per cycle = the area of the loop. There are 16 squares enclosed by the loop, and each square has an area of (2 × 105) × (1 × 10­–3) = 200 J. This means the total work done over 3 cycles is 200 × 16 × 3 = 9600 J

TIP

Graphs can be great ways of visualising thermodynamic changes, so practise drawing some p-V diagrams until you’ve got the hang of how the different processes are represented. I promise it’s not just going round in circles.

112. Four-Stroke Engines

Section 13: Option C — Engineering Physics

Quick Questions 1) What defines a stroke in a four-stroke engine? 2) True or false? Four-stroke engines burn fuel during every stroke. 3) What enters the cylinder via the inlet valve during the induction stroke for a petrol engine? State how this is different for a diesel engine.

Now try these: 4) A diagram of a cylinder in a four-stroke petrol engine, and the indicator diagram for the engine, are shown on the right. State which stroke is represented by the red line on the indicator diagram and describe what happens during this stroke. 5) During which stroke does the spark plug ignite the gas? Describe how the temperature and pressure change at the end of this stroke, once the gas has been ignited.

inlet valve

exhaust valve

p

gas

piston

V

6) Explain how the engine causes a net output of work. State how this is shown on the graph. 7) A four-stroke diesel engine works in a similar way to a four-stroke petrol engine. Describe what happens in a cylinder of a diesel engine during the induction and compression strokes.

Section 13: Option C — Engineering Physics

112. Four-Stroke Engines

ANSWERS 1) A stroke happens every time a piston moves in or out of a cylinder. 2) False. Four-stroke engines burn fuel during only one of the four strokes. 3) A mixture of fuel and air. For diesel engines only air enters the cylinder during this stroke. 4) The expansion stroke. During this stroke, the gas mixture expands and the piston moves downwards. At the end of this stroke the exhaust valve opens and the pressure reduces. 5) The compression stroke. The temperature and pressure suddenly increase while the volume stays almost constant. 6) More work is done by the gas during the expansion stroke than was done on the gas during the compression stroke. This means overall more work is output that was input. This is shown on the graph by the expansion stroke curve having a greater area underneath it than the compression stroke curve. 7) During the induction stroke, the piston moves to increase the volume of gas in the cylinder, sucking in air through the open inlet valve. During the compression stroke, the inlet valve is closed, and the piston moves to compress the air. This does work on the air and increases its temperature and pressure. The air is compressed until it reaches a high enough temperature to ignite diesel fuel. The fuel Petrol: Diesel: I play sport, and sing, is then sprayed into the cylinder and ignites. sharp peak well rounded and speak French...

TIP

The differences between petrol and diesel engines can be seen in the shapes of their respective indicator diagrams. Petrol engines have a sharp peak at the very top of the cycle, while diesel engines have a more rounded peak.

113. Using Indicator Diagrams

Section 13: Option C — Engineering Physics

Quick Questions 1) Which of these is not an assumption that a theoretical model of an engine makes? A. The heat source is external. B. The gas is replaced every cycle. C. The engine is frictionless. 2) Why do real engine indicator diagrams have a smaller peak than theoretical engine indicator diagrams?

Now try these: 3) The diagram on the right shows the theoretical indicator diagram for a four-stroke diesel engine. The steps of the cycle are labelled A-D. Describe what happens to the gas during each step. 4) Which step differs between a theoretical petrol engine and a theoretical diesel engine? Describe how it is different.

p

B C A

D

5) How would the area of the p-V loop of a real engine be different from that of the theoretical engine shown? Give your reasoning.

V 6) A diesel engine has a torque of 110 Nm and its crankshaft rotates at an –1 angular velocity of 56p rads . The power needed to overcome friction is 5400 W. Calculate its indicated power. Use P = Tw and friction power = indicated power – brake power. 7) This engine has 4 cylinders and its p-V loop has an area of 442 J. How many cycles occur each second? Indicated power = (area of p-V loop) × (number of cycles per second) × (number of cylinders).

Section 13: Option C — Engineering Physics

113. Using Indicator Diagrams

ANSWERS 1) B. In a theoretical model it is assumed that the same gas is taken continuously around the cycle. 2) Real engines have an internal heat source, from the fuel being burnt inside them, whereas theoretical engines have an external one. You don’t get the maximum energy from the fuel in a real engine as not all of it is burned in the cylinder. This means a real engine can’t reach as high a temperature as a theoretical engine, and so can’t reach as high a pressure, so its indicator diagram has a lower peak. 3) During step A, the gas is adiabatically compressed. During B, heat is supplied to the gas and its pressure is kept constant. During C, the gas cools adiabatically. During D, the gas is cooled at a constant volume. 4) Step B is different. For theoretical petrol engines this step involves supplying heat to the gas while keeping the volume, rather than pressure, constant. (This results in a vertical line on the indicator diagram instead.) 5) The area of the p-V loop would be smaller for the real engine. A real engine has friction caused by moving parts so work must be done to overcome this and the overall net work is lower. As the area enclosed by the loop is equal to the net work, this means a smaller area. 6) 25000 W. Calculate brake power as P = Tw = 110 × 56p = 19 352.2... W Indicated power = friction power + brake power = 5400 + 19 352.2... = 24 752.2... = 25 000 W (to 2 s.f.) 7) 14 cycles. No. of cycles per second =

TIP

indicated power 24 752.2... = 442 × 4 = 14.0... = 14 cycles (to 2 s.f.) ^area of p–V loop h × ^no. of cylinders h

Doing these questions is a good indicator of how well you know your stuff... Theoretical indicator diagrams make quite a few assumptions — be sure to remember how these make theoretical models different to real life.

114. Engine Efficiency

Section 13: Option C — Engineering Physics

Quick Questions 1) Give a reason why the efficiency of a real heat engine is lower than its maximum theoretical efficiency. 2) What implication does the second law of thermodynamics have for the efficiency of a heat engine? 3) How do combined heat and power plants try to minimise the energy waste of an engine?

Now try these: 4) Daniel claims to have invented a robot that will carry out 70 J of work when provided with 200 J of energy from a heat source at 80 °C. Some energy is transferred to a heat sink at 20 °C. Using T -T W efficiency = Q and maximum theoretical efficiency = H T C , determine whether this claim is possible. H H 5) Diesel fuel has a calorific value of 45.5 MJkg–1. An engine uses 3.4 g of diesel per second. An indicator diagram shows that the engine has an indicated power of 50.0 kW. What is the thermal efficiency of the engine? Use the equation input power = calorific value × fuel flow rate. QH - QC W 6) A steam engine with an efficiency of 27% does 20.0 kJ of work. Use efficiency = Q = QH H to calculate how much heat energy the engine transfers to the heat sink. Give your answer in kJ.

TIP

It’s important to convert °C to K in these calculations — you can make the conversion by remembering that absolute zero (0 K) = –273 °C. You’re not given this in the exam so make sure it’s firmly in your brain, 0 K?

Section 13: Option C — Engineering Physics

114. Engine Efficiency

ANSWERS 1) Any one from: e.g. there are frictional forces inside the engine. / The fuel doesn’t burn entirely. / Energy is needed to move internal components of the engine. 2) No heat engine can be 100% efficient. The second law tells us that heat engines must operate between a heat source and a heat sink. If some heat energy (QC) must be transferred away at the lower temperature, then not all of the heat energy supplied is converted to useful work, and some energy is always wasted. 3) By ensuring that as much of the input heat energy (QH) as possible can be converted into useful work (W). They do this by using the energy transferred to the heat sink for other useful purposes, for example, heating local buildings. 4) It isn’t possible. First, convert the temperatures to kelvins: 80 °C = 353 K. 20 °C = 293 K T -T 353 - 293 = 0.169... Then work out the Then calculate the maximum theoretical efficiency = H T C = 353 H 70 W actual efficiency = Q = 200 = 0.35. This is higher than the maximum theoretical efficiency, so H Daniel’s dream of a robot army may need to be put on hold while he recalculates some numbers. 5) 32% (or 0.32). Input power = 45.5 × 106 × 3.4 × 10–3 = 154 700 W Thermal efficiency tells you how well heat energy is converted into work and is given by indicated power 50.0 × 103 input power = 154 700 = 0.323... = 0.32 (to 2 s.f ) = 32% (to 2 s.f.) 20 W 6) 54 kJ. Rearrange the equation for efficiency to get QH = efficiency = 0.27 = 74.07... kJ W = QH – QC so Qc = QH – W = 74.07... – 20 = 54.07... = 54 kJ (to 2 s.f.)

115. Reversed Heat Engine

Section 13: Option C — Engineering Physics

Quick Questions 1) True or false? The coefficient of performance can be greater than 1. 2) Why must work be done for a reverse heat engine to function?

Now try these:

27 °C

250 J 3) A refrigerator is maintained at a temperature of 2.0 °C. 60.0 J Energy transfers take place as shown in the diagram on the right. QC QC . What is its coefficient of performance? COPref = W = QH - QC 190 J 4) What would its coefficient of performance be if it QC 2.0 °C was operating at its maximum theoretical efficiency? COPref = . QH - QC 5) A heat pump transfers heat between hot and cold spaces at the same temperatures as the refrigerator. What is the difference between the coefficients of performance for the heat pump and the refrigerator QH . if they are both running at maximum theoretical efficiency? COPhp = QH - QC 6) A heat pump is running with a COP of 9.50. The heat pump transfers 8350 J to the hot space. QH . How much energy is transferred from the cold space? COPhp = QH - QC

Section 13: Option C — Engineering Physics

115. Reversed Heat Engine

ANSWERS 1) True. Coefficient of performance can be above 1 — it’s a measure of how much heat transfer can be done per unit of work. 2) Reverse heat engines transfer heat from a colder space to a hotter space. Heat naturally flows from a hotter to a colder space, so work must be done for this heat transfer to go in the opposite direction. QC 190 3) 3.2. From the diagram, W = 60.0 J and QC = 190 J. COPref = W = 60.0 = 3.16... = 3.2 (to 2 s.f.) QC 190 (You could also use to give COPref = = 3.16... = 3.2 (to 2 s.f.).) QH - QC 250 - 190 4) 11. First convert the temperatures to kelvins: 2.0 °C = 275 K, and 27 °C = 300 K TC 275 At maximum efficiency Q can be substituted by T so COPref = = = 11 TH - TC 300 - 275 TH 300 5) 1. At maximum efficiency Q can be substituted by T so COPhp = = = 12 TH - TC 300 - 275 The difference = COPhp – COPref = 12 – 11 = 1 QH QH 6) 7470 J. COPhp = so QC = QH – COP hp QH - QC eni gne 8350 t aeh = 8350 – 9.50 = 7471.05.... = 7470 J (to 3 s.f.)

TIP

The COP for a reversed heat engine depends on what heat transfers are useful — for refrigerators it’s all about the heat transferred away from the cold space, while transferring heat to the hot space is a heat pump’s goal.

116. Specific Charge of the Electron

Section 13: Option D — Turning Points in Physics

Quick Questions A

1) What particles make up a cathode ray? 2) True or false? The work done on an electron as it is accelerated from rest through a potential difference is equal to the electron’s final kinetic energy. 3) The set-up used for a specific charge experiment is shown on the right. Identify the equipment labelled A, B and C in the diagram.

Now try these: 4) The diagram on the right shows an electron gun. Describe how the electron gun produces a beam of electrons.

B electron beam

C

-

+

cylindrical anode with hole in it

heater

5) In a particle physics experiment, a charged particle travels cathode through a magnetic field with a magnetic flux density of 1.5 T. The magnetic field causes the particle to move in a circular path with radius 6.7 cm, with a linear speed of 1.5 × 107 ms–1. mv Determine the specific charge of the particle. Use the equation r = Be .

evacuated glass tube

6) It is known that the particle has the same charge as a proton, and that the specific charge of the particle is larger than the specific charge of a proton. Compare the mass off the particle to that of a proton.

Section 13: Option D — Turning Points in Physics

116. Specific Charge of the Electron

ANSWERS 1) electrons 2) True. All the work done by the electric field is transferred to the electron’s kinetic energy. 3) A = magnetic field coils B = electron gun C = glass bulb containing low pressure gas 4) The heater heats the cathode, which emits electrons via thermionic emission. These electrons are attracted to the anode, and are accelerated across the potential difference between the cathode and the anode. When they reach the anode, some electrons pass through the hole in the anode, forming a narrow beam of electrons. e e 5) 1.5 × 108 Ckg–1. The specific charge is equal to m , so rearrange the equation for m . 1.5 × 107 e v 8 8 –1 m = Br = 1.5 × 0.067 = 1.492... × 10 = 1.5 × 10 Ckg (to 2 s.f.) 6) The specific charge of a particle is the charge to mass ratio of the particle. So, if the particle has the same charge as a proton and a larger specific charge, it must have a smaller mass than a proton.

TIP

b lom ow ou Sh C se e Th Wi d n a

Cathode rays found a nifty application when TV sets were developed. The glow they produced in discharge tubes could be used to form images on the screen. So thank electrons for your Nan’s favourite classic comedies.

117. Millikan’s Oil-Drop Experiment

Section 13: Option D — Turning Points in Physics

Quick Questions 1) What does it mean to say something is ‘quantised’? 2) What two forces act on an object falling at a terminal speed through the air? 3) True or false? No isolated object can have an electric charge that is larger than 0 C, and smaller than 1.60 × 10–19 C.

Now try these:

top plate

4) A scientist is recreating Millikan’s oil-drop experiment using the equipment shown on the right. The electric field is initially switched off. A negatively-charged drop of oil with a weight of 2.0 × 10–13 N falls between the plates at a terminal speed of 3.2 × 10–4 ms–1. The viscosity microscope of air is 1.84 × 10–5 kgm–1s–1. What is the radius of the oil droplet? Use the equation F = 6phrv.

atomiser variable p.d. V

bottom plate

5) The electric field is turned on as the droplet is falling. The field exerts an upwards force on the droplet equal to its weight. Describe how the motion of the droplet and the forces acting on it change. 6) The scientist swaps the oil for another liquid which loses electrons when it passes through the atomiser. Explain how and why the scientist would need to change their experimental set-up to use this new liquid.

Section 13: Option D — Turning Points in Physics

117. Millikan’s Oil-Drop Experiment

ANSWERS 1) If something is quantised, it can only exist in whole number multiples of a set value. (You can think of it as coming in set packets.) 2) The weight of the object and the viscous drag force of the air. 3) True. Quarks can have a charge between these values, but you can’t find them in isolation. They’ll always be bonded together in particles with an overall charge that’s a whole number multiple of 1.60 × 10–19 C. 4) 1.8 × 10–6 m. Since it’s falling at its terminal speed, the viscosity force on the oil drop will be equal to its weight. Rearrange the equation for r : r = F ÷ 6phv = 2.0 × 10–13 ÷ (6 × p × 1.84 × 10–5 × 3.2 × 10–4) = 1.802... × 10–6 = 1.8 × 10–6 m (to 2 s.f.) 5) Since the droplet was in equilibrium before the field was switched on, the field causes a resultant upwards force on the oil droplet, causing it to decelerate. As the drop slows down, the upwards viscous force acting on the drop decreases. This decreases the resultant upwards force, and the deceleration. This continues until the droplet decelerates to rest, at which point the viscous force no longer acts and the droplet is held in equilibrium by its weight and the force from the electric field. 6) Droplets of the new liquid will have a positive charge. The electric field currently exerts an upwards force on a negatively-charged droplet, and so it will cause a downwards force on a positive charge. The scientist would need to change the direction of the electric field to cause an upwards force on the new droplets.

TIP

Millikan’s big revelation was that charge is quantised, and a quantum of charge is 1.60 × 10–19 C. Quantum of Charge is also the name of my ill-received film about electrical espionage, but that’s a story for another time.

118. Light — Particles vs Waves (1)

Section 13: Option D — Turning Points in Physics

Quick Questions 1) State Huygens’ principle. 2) True or false? Diffraction can’t be explained by Newton’s theory of light. 3) True or false? Both Newton’s theory of light and Huygens’ theory of light explain the refraction of light as being due to light travelling faster in denser materials.

Now try these:

light ray

4) A ray of light reflects from a metal plate, as shown on the right. How is this behaviour explained by Newton’s theory of light? metal plate

5) Briefly describe how diffraction and interference of light are demonstrated by Young’s double-slit experiment. 6) Explain how this provided evidence for Huygens’ theory of light. 7) Explain why Huygens’ theory of light took a long time to be accepted despite this evidence.

corpuscle core muscle

TIP

“Corpuscle” is a funny old word. It just means ‘particle’ and was used to describe electrons in the plum pudding model, as well as Newton’s version of particulate light. I’m glad we settled on some better names in the end.

Section 13: Option D — Turning Points in Physics

118. Light — Particles vs Waves (1)

ANSWERS 1) Every point on a wavefront may be considered to be a point source of secondary wavelets that spread out in the forward direction at the speed of the wave. The new wavefront is the surface that is tangential to all of these secondary wavelets. 2) True. 3) False. In Huygens’ theory, light travels slower in denser materials. 4) The ray of light consists of a stream of particles/corpuscles, that undergo an elastic collision with the metal plate. The reflected ray is formed as the particles are pushed away from the surface by a force that is due to the collision (just like a ball bouncing against a wall). 5) Diffraction is demonstrated by the light spreading out as it passes through the slits. Interference is demonstrated by the fringes that are displayed on the screen after the light has passed through the double‑slits, showing that the light from the two slits had undergone constructive and destructive interference. 6) Diffraction and interference couldn’t be explained by a particulate theory of light, as they were exclusively wave properties. So, demonstrating that light could undergo diffraction and interference provided evidence that light was a wave, not a particle. 7) Huygens’ theory took a long time to be accepted because Newton had revolutionised physics so much that the scientific community were more inclined to believe he was also correct about light. In addition, there were still some things that Huygens’ version of the wave theory of light couldn’t explain (e.g. double refraction), so since there were things that both theories couldn’t explain, scientists preferred to stick with the already established theory developed by Newton.

119. Light — Particles vs Waves (2)

Section 13: Option D — Turning Points in Physics

Quick Questions 1) Which scientist discovered radio waves, and determined they were electromagnetic waves? 2) True or false? e0 is a physical constant that relates the magnetic flux density produced by a wire to the current flowing through it.

Now try these: 3) Describe the experiment Fizeau performed to estimate the speed of light. radio wave 4) Explain how the results of this experiment supported source Maxwell’s theory that light is an electromagnetic wave.

radio waves

5) The diagram on the right shows the experimental set-up for an experiment investigating the nature of radio waves. Describe what will be observed at the point marked X, and explain how this shows that radio waves have a magnetic component.

metal loop with small gap

6) A standing radio wave is set up in a vacuum chamber, as shown below. Describe how you could use this experimental set-up to determine the speed of radio waves in a vacuum. radio transmitter of known frequency

reflector movable detector

Section 13: Option D — Turning Points in Physics

119. Light — Particles vs Waves (2)

ANSWERS 1) (Heinrich) Hertz. 2) False. That’s m0. e0 relates to the electric field strength due to the charge on the object producing it. 3) Fizeau passed a beam of light through the gap between two cog teeth to a reflector placed eau de fizz around 9 km away. The reflector reflected the beam back towards the cog. Fizeau rotated the cog at the correct speed to cause the reflected beam to be blocked by the next cog tooth. Using the frequency of rotation and the number of gaps in the cog, Fizeau determined the time it took for the light to travel to and from the reflector. Using this and the distance travelled by the reflector, he determined the speed of light. 4) Maxwell’s theory predicted that electromagnetic waves would move through a vacuum at a particular speed that depends only on physical constants. This speed (~2.998... × 108 ms–1) was very close to the speed of light calculated by Fizeau, which suggested that light was an electromagnetic wave. 5) A spark will jump across the gap in the wire loop at point X. In order for a current (the spark) to flow, a potential difference (p.d.) must have been induced across the wire loop. The loop must have experienced a changing magnetic field to induce this p.d., so the incident radio waves must have a magnetic field component. 6) Move the detector along the standing wave to detect the location and distance between the nodes. Use this information to calculate the wavelength of the standing wave (the distance between two nodes is equal to half a wavelength). Multiply the wavelength by the frequency of the transmitter to calculate the speed of the radio waves.

TIP

You can’t teach an old dog new tricks, but you can support a new theory with existing data. Maxwell calculated the theoretical speed of EM waves over 10 years after Fizeau estimated the speed of light to be about the same value.

Section 13: Option D — Turning Points in Physics

120. The Photoelectric Effect and Photons (1) Quick Questions

1) What is the name given to an object that absorbs all wavelengths of electromagnetic radiation and can emit all wavelengths of electromagnetic radiation? 2) What is the photoelectric effect?

Now try these:

4) Explain why, according to wave theory, light of any frequency should be able to trigger the photoelectric effect. 5) The graph on the right shows the graph of intensity against wavelength for radiation emitted by a perfect black body, as predicted by wave theory. Explain the feature of the graph that provides evidence against the wave theory of electromagnetic radiation. Describe how the real graph for a perfect black body differs to the graph shown.

TIP

intensity

3) Give the three main conclusions that can be drawn from investigations into the photoelectric effect. Identify the conclusions that cannot be explained by a wave theory of light.

wavelength

The photoelectric effect was really the breaking point for thinking about light purely as a wave. Hard as they tried, physicists just couldn’t explain it in terms of electromagnetic waves. Enter our saviour, the photon...

Section 13: Option D — Turning Points in Physics

120. The Photoelectric Effect and Photons (1)

ANSWERS 1) A black body. 2) The photoelectric effect is when electrons are released from the surface of a metal when light is incident on the metal. (The electrons released are known as photoelectrons.) 3) For a given metal, no photoelectrons are emitted if the radiation incident on the metal has a frequency below a certain value, called the threshold frequency. The photoelectrons are emitted with a variety of kinetic energies, ranging from zero to a maximum value, and the maximum kinetic energy increases with the frequency of the radiation, but is unaffected by radiation intensity. The number of photoelectrons emitted per second is directly proportional to the intensity of the radiation. The existence of the threshold frequency and the fact that kinetic energy only depends on the frequency cannot be explained by a wave theory of light.

intensity

4) In wave theory, the energy carried by a light wave would be spread evenly across the whole wavefront. So when a wave of any frequency hits the metal, each free electron in the surface of the metal would gain a little bit of energy from the wave. As more waves hit the metal, the energy of each electron should increase, so gradually each electron should gain enough energy to be able to leave the metal and be emitted as a photoelectron, regardless of the frequency of the incident light. 5) The graph shows that for small wavelengths, the intensity of radiation emitted by a black body tends to infinity. This is impossible, and does not agree with observations of real black bodies, which suggests that the wave theory of electromagnetic radiation cannot be correct. The real graph (see blue curve) matches the wave theory prediction for large wavelengths, but reaches a peak intensity at a particular value of wavelength, tending towards 0 intensity as the wavelength decreases below the peak value.

wavelength

121. The Photoelectric Effect and Photons (2)

Section 13: Option D — Turning Points in Physics

Quick Questions 1) What was the ‘ultraviolet catastrophe’? 2) What concept, introduced by Max Planck, provided a solution to the ultraviolet catastrophe? 3) Which of the following is equal to the minimum energy required to remove an electron from the surface of a metal via the photoelectric effect? A. the Planck constant, h B. the Boltzmann constant, k C. the work function, f

Now try these: 4) Explain the photoelectric effect in terms of the photon model. 5) Explain why the kinetic energy of photoelectrons is unaffected by an increase in the intensity of the light source. 6) A metal plate is illuminated by monochromatic light and emits photoelectrons. Explain why the photoelectrons have a range of kinetic energies.

TIP

Sometimes physicists can be a bit dramatic. I mean, ‘ultraviolet catastrophe’? That’s just a little extra, don’t you think? Even if it was a pretty big problem with our understanding of EM radiation. Thankfully, they solved it.

Section 13: Option D — Turning Points in Physics

121. The Photoelectric Effect and Photons (2)

ANSWERS 1) Wave theory predicted an infinitely high peak towards the ultraviolet region of an intensity-wavelength graph for a black body, which is impossible. 2) The idea that electromagnetic radiation could only be emitted in discrete packets — quanta. 3) C. The other two are physical constants. 4) When light illuminates a metal, the metal’s surface is bombarded by photons. A photon can interact with a free electron on the surface and be absorbed. The electron gains energy equal to the energy of the photon (hf). If the energy is greater than or equal to the work function, f, then the electron will have enough energy to break the bonds holding it in the metal and it will be emitted from the surface. 5) The intensity of the light source is a measure of the number of photons emitted per second from it. Each electron has to absorb a single photon to be emitted from the metal, and the kinetic energy of a photoelectron is equal to the energy of the absorbed photon minus the work function. Since the energy of each photon is unchanged, an increase in intensity, and so the number of photons per second, won’t increase the kinetic energy of the emitted photoelectrons. 6) All light incident on the metal plate has the same frequency (it’s monochromatic), so each electron absorbs the same amount of energy when it absorbs a photon. Each electron uses some energy to overcome the bonds holding it in the metal, and the rest becomes the electron’s kinetic energy. The distance of the electrons from the surface will vary, meaning electrons will have to use different amounts of this kinetic energy to escape the metal. This is what causes there to be a range of kinetic energies.

122. Wave-Particle Duality

Section 13: Option D — Turning Points in Physics

Quick Questions 1) Estimate the de Broglie wavelength an electron in an electron microscope needs to resolve an atom. 2) Describe the diffraction pattern for electrons that have passed through a graphite crystal. 3) How will increasing the speed of the incident electrons change the diffraction pattern?

Now try these: 4) Determine the momentum of an electron that has a de Broglie wavelength of 0.884 nm. h Use h = 6.63 × 10–34 Js and the equation l = p . 5) A crystal has an atom layer spacing of 9.2 × 10–10 m. Will the electron in Q4 diffract when it passes through this crystal? Explain your answer.

6) An X-ray has a wavelength of 4.96 × 10–11 m. Determine the potential difference that an electron would need to be accelerated by in order to have a de Broglie wavelength equal to the wavelength of the X-ray. h . Use h = 6.63 × 10–34 Js, me = 9.11 × 10–31 kg, e = 1.60 × 10–19 C and the equation l = 2meV 7) Describe how a scanning tunnelling microscope (STM) works.

TIP

If light could be a wave with particle-like properties, it would make sense that particles could show wave-like properties too. A bold statement for de Broglie to put in his PhD thesis, but it was correct. Clever lad.

Section 13: Option D — Turning Points in Physics

122. Wave-Particle Duality

ANSWERS 1) 1 × 10–10 m or smaller. This is approximately equal to the diameter of an atom. 2) The diffraction pattern is a circular point with high intensity, surrounded by concentric rings. The rings have a lower intensity and are spaced further apart as they get further from the central circle. 3) The concentric rings will get closer together. / The spacing between the rings will decrease. (This is because the de Broglie wavelength of the electrons decreases, which reduces the spacing of the diffraction pattern.) 4) 7.5 × 10–25 kgms–1. Rearrange the equation for momentum: h 6.63 × 10 -34 p= l = = 7.5 × 10–25 kgms–1 0.884 × 10 -9 5) Yes, it will diffract, because the de Broglie wavelength of the electron (8.84 × 10–10 m) is close in size to the spacing between the atoms in the crystal (9.2 × 10–10 m). 6) 613 V. Rearrange the equation for potential difference: (6.63 × 10 -34) 2 h2 V= = = 612.90... = 613 V (to 3 s.f.) 2mel 2 2 × 9.11× 10 -31 × 1.60 × 10 -19 × (4.96 × 10 -11) 2 7) A very fine probe is moved over the surface of a sample, and a voltage is applied between the probe and the surface. Electrons ‘tunnel’ from the probe to the surface, causing a weak electrical current between them. The smaller the distance between the probe and the surface, the greater the current. By scanning the probe over the surface and measuring the current, an image can be produced of the sample’s surface.

e

123. The Speed of Light and Relativity

Section 13: Option D — Turning Points in Physics

Quick Questions 1) True or false? One of Einstein’s postulates for special relativity was that light has different apparent speeds to different observers. 2) True or false? Einstein’s other postulate of special relativity states that physical laws have the same form in any inertial reference frame.

Fath e Nor r Christ th P mas ole

3) What conclusion was drawn from the Michelson-Morley experiment about absolute motion?

Now try these: 4) A scientist does an experiment to test Newton’s first law of motion in the back of a motor home. While he does the experiment, the motor home is moving along a perfectly flat, straight road at a constant speed. How are his results likely to compare to those obtained from the same experiment carried Mirror, M2 Partial reflector out in a lab? Explain your answer. glass 8 –1 5) A light wave travels through space at 3.00 × 10 ms towards a L block 6 –1 spaceship that is travelling at 2.45 × 10 ms in the same direction. Light Mirror, M1 What would an alien on the spaceship observe the speed of source the light wave to be? Explain your answer. L 6) Explain how the equipment shown on the right was intended to operate and detect absolute motion of the Earth through the ether.

Section 13: Option D — Turning Points in Physics

123. The Speed of Light and Relativity

ANSWERS 1) False. Einstein postulated that the speed of light was invariant — no matter where the observer is or how fast they’re moving, they’ll observe light moving at the same speed. 2) True. 3) It was concluded that it was impossible to detect absolute motion (and so the ether could not exist). 4) His results should agree with the results from the lab. Since the motor home is moving in a straight line at a constant speed, it isn’t accelerating, so it is an inertial reference frame. This means that Newton’s first law will be obeyed in the motor home, and the results of the scientist’s experiments should agree with the results from the same experiments carried out in a lab. 5) 3.00 × 108 ms–1. The second postulate of special relativity states that the speed of light is invariant — it will always be observed to have the same value, regardless of how the alien and their spaceship are moving. 6) Monochromatic light is sent from the source towards the partial reflector. The partial reflector splits the light into two beams at right angles to each other. One beam is reflected at mirror M1, and the other is reflected at M2. When the reflected beams meet each other again at the partial reflector, they form an interference pattern. If the Earth has absolute motion compared to the ether, light moving parallel to the Earth’s motion should take e.g. longer to travel to the mirror and back than light travelling at right angles to the Earth’s motion. Rotating the apparatus 90° should change the travel time in the two beams and cause a change in the interference pattern.

TIP

And the prize for the worst relativity joke goes to: I bought some new glasses and now whenever I read a news article, I can see where all the information originally came from. Turns out my glasses have reference frames...

124. Special Relativity

Section 13: Option D — Turning Points in Physics

Quick Questions 1) What is meant by the ‘proper time’ between two events? 2) How does the mass of an object change Ek as its speed approaches the speed of light? 3) Which of the graphs on the right correctly shows how the kinetic energy of an object varies with speed?

Ek

0

c v

Ek

0

A.

c v

0

c v

B.

C.

Now try these: 4) Which scientist’s work confirmed the relationship in Q3 experimentally? 5) Describe how observations of muons at the Earth’s surface provide evidence for time dilation. 6) A giant robot is travelling past a space station at a speed of 0.92c. An observer on the space station sees the robot as 4.7 m long. Determine the length of the robot in its reference frame. Use the equation l = l0

1-

7) A positron with a rest mass of 9.11 × 10–31 kg travels at a speed of 2.96 × 108 ms–1. Determine the energy of the positron. Use c = 3.00 × 108 ms–1 and the equation E =

v2 c2

.

m0 c 2 . v2 1- 2 c

Section 13: Option D — Turning Points in Physics

124. Special Relativity

ANSWERS 1) The time between two events as measured in the reference frame where the events are stationary. 2) It increases. 3) B. Since kinetic energy is proportional to speed-squared at non-relativistic speeds, the energy is 0 when the speed is 0. The shape of graph A shows how the mass of an object changes with speed. 4) (William) Bertozzi. (He measured the mass and energy of high energy particles in particle accelerators.) 5) Muons are produced in the upper atmosphere and move towards the ground at speeds close to the speed of light, c. Muons have a half-life in the laboratory of less than 2 ms, and should take multiple half-lives to reach the ground, so most muons should decay before then. However, when muons are measured at the ground, it is found that more make it to the ground than would have been expected. This is because the time of flight of the muons has been dilated for the observers on Earth. In the muon’s rest frame, much less time has passed, and so fewer muons have decayed. 4.7 l 6) 12 m. Rearrange the equation for l0: l0 = = = 11.992... m = 12 m (to 2 s.f.) 2 (0.92c) 2 v 1- 2 1 2 c c (9.11× 10 -31) × (3.00 × 10 8) 2 m0 c 2 –13 7) 5.04 × 10  J. E = = = 5.0376... × 10–13 = 5.04 × 10–13 J (to 3 s.f.) (2.96 × 10 8) 2 v2 1- 2 1 c (3.00 × 10 8) 2

TIP

Special relativity is where physics really starts to get weird. Times getting l o n g e r, lengths getting shorter, and masses getting bigger... All due to something getting closer and closer to that ultimate speed limit, c. Wild.

125. Experiment Design and Errors

Practical and Investigative Skills

Quick Questions 1) Define the terms ‘independent variable’ and ‘dependent variable’. 2) Identify a hazard associated with using electrical circuits as part of an experiment. 3) What is meant by calibration? What type of error can calibration reduce?

Now try these: 4) A student does an experiment to determine how the potential difference across a component changes as the current through it increases. Explain why the temperature of the circuit should be kept constant during the experiment. 5) A scientist heats a liquid at a constant power. She measures the temperature of the liquid every 20 s using a thermometer in the liquid. The thermometer has a maximum reading of 120 °C, and a scale with 1 °C divisions. What is the uncertainty in each of the scientist’s temperature measurements? 6) The scientist’s temperature measurements are shown in the table below. While taking one measurement, the thermometer was removed from the liquid, causing an anomalous result. Time (s) 0 20 40 60 Identify the anomalous result Temperature (°C) 16 18 21 24 in the table. Explain your answer.

80

100

120

24

29

32

7) The scientist repeats the experiment, using a digital thermometer that automatically measures temperature to the nearest 0.01 °C every 20 s. Explain how this will reduce the effect of random error in the results.

Practical and Investigative Skills

125. Experiment Design and Errors

ANSWERS 1) The independent variable is the thing that you change during an experiment. The dependent variable is the thing that you measure during an experiment. 2) E.g. you may get an electric shock from the circuit / the circuit could overheat and cause burns. 3) Calibration is testing the accuracy of a measuring device by measuring a known value, and adjusting the device to correct inaccuracy. Correctly calibrating a measuring device can reduce the systematic error in your results. 4) Changes in temperature could affect the resistance of the component, which will affect the potential difference at a given current, so the results will not be valid and the test will not be fair. 5) ±0.5 °C. The uncertainty in the measurement is equal to half the smallest division on the temperature scale. 6) The value recorded at 80 s is the anomalous result. Removing the thermometer from the liquid is likely to have caused the scientist to record a temperature lower than the true value. For all of the other results, there is a steady increase of 2-3 °C between recordings. The value at 80 s is the only result where no increase in temperature was recorded, despite a constant rate of energy being supplied to the liquid. This means this is likely to have been where the thermometer was removed from the liquid. 7) The automatic measurements at each time interval will remove the random error caused by the scientist manually reading the temperature at each time interval by looking at a stopwatch. The greater resolution, and the removal of any human error in the scientist reading the temperature, reduces the size of any random errors in the temperature measurements.

TIP

Remember, when planning experiments, you need to make sure everyone involved will behave safely and ethically. I heard some famous scientist had the bright idea to stick a cat in a box. Doesn’t sound very ethical to me...

126. Uncertainty and Errors

Practical and Investigative Skills

Quick Questions 1) What does an error bar on a graph show? 2) What is meant by the level of confidence of a result? 3) How do you find the uncertainty of a mean value calculated from repeated measurements?

Now try these: 4) When a capacitor with capacitance 8.0 ± 0.1 nF is fully charged, a student measures a potential difference of 2.4 ± 0.1 V across the capacitor. He calculates the energy stored by the capacitor using the equation E = ½CV 2. Determine the percentage uncertainty in the student’s value of energy stored.

R/

5) A 10.0 ± 0.1 cm length of current-carrying wire in a magnetic field experiences a force of 1.50 ± 0.02 N. A current of 10.50 ± 0.05 A flows through the wire. Determine the magnetic flux density of the magnetic field, and the fractional uncertainty of this value. Use the equation F = BIl. 6) The graph on the right shows the results of an experiment into how the resistance of a wire changes with length. The wire has a cross-sectional area of 7.84 × 10–7 m2. The gradient of the line of best fit of the graph is 0.500 Wm–1. The gradient of the blue worst line is 0.529 Wm–1. The gradient of the red worst line is 0.474 Wm–1. 0 RA Use the equation r = L to determine the resistivity of the wire, and the uncertainty in this value.

L/m

Practical and Investigative Skills

126. Uncertainty and Errors

ANSWERS 1) An error bar shows the uncertainty in a measurement by showing the range the measurement is likely to lie in. 2) A measure of the likelihood that the true value lies within the range given by the uncertainty of a result. 3) Find the range of the repeated measurements, and divide it by two. This is the uncertainty of the mean. 4) ±10%. Percentage uncertainty in potential difference = (0.1 ÷ 2.4) × 100 = 4.166...% Percentage uncertainty in capacitance = (0.1 ÷ 8.0) × 100 = 1.25% Percentage uncertainty in E = percentage uncertainty in capacitance + percentage uncertainty in V 2 = 1.25 + (2 × 4.166...) = 9.583... = 10% (to 1 s.f.) 59 F 1.50 5) 1.43 T, fractional uncertainty = ± 2100 . B = Il = = 1.4285... T = 1.43 T (to 3 s.f.). 10.50 × 10.0 × 10 -2 All the variables are being multiplied or divided, so the fractional uncertainty in B is equal to the sum of the 0.02 0.05 0.1 59 fractional uncertainties in each variable. Fractional uncertainty in B = 1.50 + 10.50 + 10.0 = 2100 R 6) 3.92 × 10–7 ± 0.23 × 10–7 Wm. The gradient of the line of best fit gives L , so: r = gradient × A = 0.500 × 7.84 × 10–7 = 3.92 × 10–7 Wm Use the worst line furthest from the line of best fit to determine the worst value of r. The blue worst line is further from the line of best fit than the red line (0.529 – 0.500 > 0.500 – 0.474), so: worst value of r = worst gradient × A = 0.529 × 7.84 × 10–7 = 4.14... × 10–7 Wm. Uncertainty in r = r – worst r = (4.14... × 10–7) – (3.92 × 10–7) = 2.2736 × 10–8 = 0.23 × 10–7 Wm (to 2 d.p.)

TIP

The two worst lines of a graph are the maximum and minimum possible slopes that fall within the error bars. The one furthest from the line of best fit is the true worst line and should be used to find the gradient’s uncertainty.

127. Presenting and Evaluating Data

Practical and Investigative Skills

Quick Questions 1) What does it mean if experimental results are said to be repeatable and reproducible? 2) Which of the scatter graphs below shows a negative correlation between variables M and N? A. M B. M C. M N

N

Repeats (GPa)

Now try these: 3) The table on the right shows some students’ results for finding the Young modulus of copper. The Young modulus is known to be 117 GPa. Which student’s results are the most accurate? Whose results are the most precise? Explain your answers. T/s 0.30 0.20 0.10 0.00

0.02

0.04

N

0.06

Student

1st

2nd

3rd

Mean (GPa)

Andre

115

101

105

107

Bijan

118

113

117

116

Carla

121

119

120

120

4) The graph on the left shows the results of a student’s investigation into how the period of oscillation of a mass-spring system changes with the mass hung from the spring. The student concludes that the period of oscillation is directly proportional to mass. Evaluate how confident the student should be in his conclusion.

0.08 m / kg

Practical and Investigative Skills

127. Presenting and Evaluating Data

ANSWERS 1) If results are repeatable, you’re able to get the same / very similar results by repeating the same experiment. If results are reproducible, another scientist is able to recreate the experiment using different equipment and/or methods, and get the same / very similar results. 2) A. This is the only graph which shows a consistent trend of M decreasing as N increases. 3) Bijan’s results are most accurate, because his final mean is closest to the known true value of the Young modulus (117 GPa). Carla’s results are the most precise, as her individual results have the smallest range/ least variation (having a range of only 2 GPa, compared to 5 GPa for Bijan and 14 GPa for Andre). 4) The student should not be very confident in their conclusion. The student has only taken 4 measurements, which is not enough to make a strong conclusion about the correlation between the two variables, or mathematical relationship between them. In addition, the error bars on each point are very large (representing as high as ±50% for one point) and overlap between successive points. This means that the student’s results are not very precise, and so the student’s line of best fit cannot be relied upon to give an accurate representation of the trend and relationship between the two variables. Sylvia was confident in her A-Level results.

TIP

Be sure to look at the whole graph when you’re evaluating results. There may be an impressive line of best fit and the points might seem to show a clear trend, but that means nothing if they have enormous error bars.

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