Quantum : physics : third year of secondary cycle two : student textbook [2-3] 9782765214441, 2765214441

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Third Year of Secondary Cycle Two

PHYSICS

Student Textbook

Ahmed Bensaada Benoît Ouellette

Third Year of Secondary Cycle Two

PHYSICS

Student Textbook

Ahmed Bensaada Benoît Ouellette in collaboration with

Raynald Pepin

Translators: Valerie Buchanan, Jacquie Charlton, Cristina Cusano, Michael Varga

Quantum – Physics Third Year of Secondary Cycle Two

Acknowledgements

Student Textbook Ahmed Bensaada, Benoît Ouellette © 2011 Chenelière Education Inc. Editors: François Moreault, Isabel Rusin Coordinators: Samuel Rosa, Denis Fallu, Christiane Gauthier Copy-editors: Nicole Blanchette, Ginette Gratton Proofreaders: Danielle Maire Graphic design: Dessine-moi un mouton Cover design: Josée Brunelle Layout: Interscript Visual research: Marie-Chantal Laforge Printer: Imprimeries Transcontinental

The Editor would like to thank the following people for their valuable consulting work: Julie Anctil, teacher, École secondaire des Rives, CS des Affluents; Mario Cossette, teacher, École secondaire Georges-Vanier, CS de Montréal; Christian Drapeau, teacher, École secondaire de St-Marc, CS de Portneuf; Guy Lapointe, retired teacher; Charles Métivier, teacher, Polyvalente de Black Lake, CS des Appalaches; Steven Monzerol, teacher, Séminaire de Sherbrooke; Michel Paré, teacher, École secondaire des Pionniers, CS du Chemin-du-Roy; Jean-Marc Pétel, teacher, Collège Charlemagne Inc.; Yves-Alain Peter, Associate Professor, Department of Engineering Physics, École Polytechnique de Montréal; Fikry Rizk, retired teacher; Martin Trudeau, teacher, École Jacques-Rousseau, CS Marie-Victorin; Lucie Vallières, teacher, École secondaire du Versant, CS des Draveurs. The Editor would also like to thank the following person for his valuable expertise and scientific revisions: Richard Leonelli, Full Professor, Department of Physics, Université de Montréal.

English Version Editor: Colleen Ovenden Science Consultant: Geneviève Cormier Copy-editor: Dana Zidulka Proofreaders: Sarah Bild, Tamara Kramer Layout: Interscript Printer: Transcontinental Printing

The Editor of the English version would like to thank Geneviève Cormier for her valuable expertise and constructive review during the translation of this textbook. The Editor would like to thank the following people for their valuable contribution to the text: Nicholas Desrosiers, teacher, CS de la Seigneurie des-Milles-Îles (features: p. 161, 162, 178 and 179); Agence Science-Presse (features: p. 35, 37, 126 and 144).

7001 Saint-Laurent Blvd. Montréal (Québec) Canada H2S 3E3 Telephone: 514 273-1066 Fax: 450 461-3834 / 1 888 460-3834 [email protected]

ALL RIGHTS RESERVED. No part of this book may be reproduced in any form or by any means without written permission from the Publisher. ISBN 978-2-7652-1444-1 Legal deposit: 1st quarter 2011 Bibliothèque et Archives nationales du Québec Library and Archives Canada Printed in Canada 1

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We acknowledge the financial support of the Government of Canada through the Book Publishing Industry Development Program (BPIDP) for our publishing activities. Government of Québec – Tax credit program for book publishing – SODEC These programs are funded by Québec’s Ministère de l’Éducation, du Loisir et du Sport and through contributions from the Canada-Québec Agreement on Minority-Language and Second-Language Instruction.

Laboratoris and Dmonstrations                  viii Organization of th Txtbook                      ix Rviw                                          2 1 Waves                                        11 Types o waves                             12 Characteristics o waves                      13 Electromagnetic spectrum                     14 Electric eld                                15 Magnetic eld                             

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2 Properties o light                               8 21 Refection o light                            8 22 Reraction o light                           9 23 Converging lenses and diverging lenses         10 24 The eye                                   11 3 Force and motion                               31 Relationship between constant speed, distance and time                          32 Characteristics o orce                      33 Gravitational orce                          34 Relationship between mass and weight        35 Force o riction                            36 Equilibrium between two orces               37 Eective orce                             4 Energy                                        41 Relationship between work, orce and displacement                      42 Relationship between work and energy         43 Kinetic energy                             44 Potential energy                            45 Law o conservation o energy                46 Relationship between power and electrical energy                        47 Solar energy                              

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GeOMeTRIC OpTICS

UNIT INTRODUCTION AND ORGANIZATION CHART   20 CHApTeR 1 WAVeS                              23 11 Characteristics o waves                        24

For a detailed list of laboratories and demonstrations, see page VIII

111 Types o waves                         112 Periodicity o waves                     Practice makes perfect – Section 1.1          12 Light waves                                  121 Electromagnetic spectrum                1 and 2          122 Propagation o light 123 Light propagation in media                124 Light passage rom one medium to another   Practice makes perfect – Section 1.2          Applications                                      A brie history o…                                Review – Chapter 1                               Practice makes perfect – Chapter 1               

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CHApTeR 2 ReFLeCTION OF LIGHT                3  21 Types o refection 211 Specular refection                      212 Diuse refection                        Practice makes perfect – Section 2.1          22 Geometry o refection                         221 Incident ray                            222 Normal                                223 Angle o incidence (θi)                   224 Refected ray                           225 Angle o refection (θr)                   Practice makes perfect – Section 2.2          23 Refection on a plane mirror: laws o refection                             231 First law o refection                    4 232 Second law o refection  Practice makes perfect – Section 2.3          5 24 Refection on spherical mirrors  241 Concave mirrors                        242 Convex mirrors                         243 Spherical aberration                     Practice makes perfect – Section 2.4          25 Images                                      251 Image characteristics                    252 Images ormed 6 by a camera obscura  253 Images ormed 7  by plane mirrors 254 Images ormed by spherical 8 and 9                     mirrors Practice makes perfect – Section 2.5         

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Table of Contents

46 46 46 47 48 49 52 53 54 55 55 56 57 59 70

III

Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A brief history of… . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Review – Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Practice makes perfect – Chapter 2 . . . . . . . . . . . . . . .

71 72 73 76

CHAPTER 3 REFRACTION OF LIGHT . . . . . . . . . . . . . . . 3.1 Phenomenon of refraction . . . . . . . . . . . . . . . . . . . . . . 10 . . . . . . . . . . . . . . . . . . . . . 3.2 Index of refraction Practice makes perfect – Section 3.2. . . . . . . . . . 3.3 Geometry of refraction . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 Incident ray . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.2 Normal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.3 Angle of incidence (θi) . . . . . . . . . . . . . . . . . . 3.3.4 Refracted ray and angle of refraction (θR) . . . . . . . . . . . . . . . . . . . . . . . Practice makes perfect – Section 3.3. . . . . . . . . . 3.4 Laws of refraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4.1 First law of refraction . . . . . . . . . . . . . . . . . . . 11 and 12 ... 3.4.2 Second law of refraction Practice makes perfect – Section 3.4. . . . . . . . . . 3.5 Total internal reflection . . . . . . . . . . . . . . . . . . . . . . . . Practice makes perfect – Section 3.5. . . . . . . . . . Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A brief history of… . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Review – Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Practice makes perfect – Chapter 3 . . . . . . . . . . . . . . .

77 78 79 81 82 82 82 82 82 83 84 84 84 86 87 88 89 90 91 93

CHAPTER 4 LENSES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 13 4.1 Different types of lenses . . . . . . . . . . . . . . . 96 4.1.1 Converging lenses . . . . . . . . . . . . . . . . . . . . . . 96 4.1.2 Diverging lenses . . . . . . . . . . . . . . . . . . . . . . . 97 Practice makes perfect – Section 4.1. . . . . . . . . . 98 4.2 Refraction in lenses. . . . . . . . . . . . . . . . . . . . . . . . . . . 99 4.2.1 Refraction in converging 14 . . . . . . . . . . . . . . . . . . . . . . . . 99 lenses 4.2.2 Refraction in diverging lenses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 Practice makes perfect – Section 4.2. . . . . . . . . 105 4.3 Optical power of lenses . . . . . . . . . . . . . . . . . . . . . . 106 4.3.1 Optical power of a lens . . . . . . . . . . . . . . . . 106 4.3.2 Sign convention . . . . . . . . . . . . . . . . . . . . . . . 107 4.3.3 Optical power of a system 15 . . . . . . . . . . . . . . . . . . . . . 108 of lenses 4.3.4 Lens-maker’s equation . . . . . . . . . . . . . . . . . 109 Practice makes perfect – Section 4.3. . . . . . . . . 112 4.4 Images formed by lenses . . . . . . . . . . . . . . . . . . . . . 113

IV

Table of Contents

4.4.1 Geometric construction of images 16 and 17 . . . . . . . . . formed by lenses 4.4.2 Mathematical formalism of lenses . . . . . . . . . . . . . . . . . . . . . . . . . . . . Practice makes perfect – Section 4.4. . . . . . . . . 4.5 Optical aberrations of lenses . . . . . . . . . . . . . . . . . . 4.5.1 Chromatic aberrations . . . . . . . . . . . . . . . . . . 4.5.2 Spherical aberrations . . . . . . . . . . . . . . . . . . Practice makes perfect – Section 4.5. . . . . . . . . Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A brief history of… . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Review – Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Practice makes perfect – Chapter 4 . . . . . . . . . . . . . . CHAPTER 5 APPLIED GEOMETRIC OPTICS. . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 The camera . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Practice makes perfect – Section 5.1. . . . . . . . . 5.2 The human eye . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2.1 Geometric optics of vision . . . . . . . . . . . . . . 5.2.2 Visual disorders and their 18 correction .................... Practice makes perfect – Section 5.2. . . . . . . . . 5.3 The light microscope . . . . . . . . . . . . . . . . . . . . . . . . . Practice makes perfect – Section 5.3. . . . . . . . . 5.4 The telescope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4.1 The reflecting telescope . . . . . . . . . . . . . . . . 5.4.2 The refracting telescope . . . . . . . . . . . . . . . . Practice makes perfect – Section 5.4. . . . . . . . . A brief history of… . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Review – Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Practice makes perfect – Chapter 5 . . . . . . . . . . . . . .

113 115 122 123 123 124 124 125 126 127 130

131 132 133 134 134 136 138 139 140 141 141 142 143 144 145 147

PRELIMINARY NOTIONS OF MECHANICS UNIT INTRODUCTION AND ORGANIZATION CHART. . CHAPTER 6 FRAMES OF REFERENCE . . . . . . . . . . . . 19 . . . . . 6.1 The purpose of a frame of reference Practice makes perfect – Section 6.1. . . . . . . . . 6.2 Inertial frames of reference . . . . . . . . . . . . . . . . . . . Practice makes perfect – Section 6.2. . . . . . . . .

148 151 152 154 155 157

For a detailed list of laboratories and demonstrations, see page VIII

63 Coordinate systems                           631 The Cartesian coordinate system           632 The polar coordinate system             633 Relationship between Cartesian coordinates and polar coordinates         Practice makes perfect – Section 6.3         Applications                                     A brief history of…                               Review – Chapter 6                              Practice makes perfect – Chapter 6              

158 158 158 159 160 161 162 163 164

CHApTeR 7 QUANTITIeS AND UNITS             71 Quantity, measurement and unit of measure     72 The International System of Units               Practice makes perfect – Section 7.2         73 Fundamental standards of the basic units of mechanics                           731 The metre                            732 The kilogram                          733 The second                           Practice makes perfect – Section 7.3         74 Derived units in the International System         Practice makes perfect – Section 7.4         75 Multiples and submultiples of units              Practice makes perfect – Section 7.5         76 Scalar quantities and vector quantities                            761 Scalar quantities                       762 Vector quantities                       Practice makes perfect – Section 7.6         Applications                                     A brief history of…                               Review – Chapter 7                              Practice makes perfect – Chapter 7              

165 166 167 168

CHApTeR 8 VeCTORS                           81 Properties of vectors                          811 Scalars and vectors                     812 Magnitude and direction of a vector                            813 Components of a vector                 814 Vectors and the Cartesian plane          Practice makes perfect – Section 8.1         82 Adding vectors                               821 Graphical method                      822 Component method                    

183 184 184

169 170 170 170 171 172 172 173 175 176 176 176 177 178 179 180 182

184 186 188 189 190 190 190

For a detailed list of laboratories and demonstrations, see page VIII

823 Adding in the Cartesian plane            Practice makes perfect – Section 8.2         83 Subtracting vectors                           831 Graphical method                      832 Component method                     833 Subtracting in the Cartesian plane         Practice makes perfect – Section 8.3         84 Multiplying a vector by a number                841 Multiplying a vector by a scalar           842 Multiplying in the Cartesian plane         Practice makes perfect – Section 8.4         Review – Chapter 8                              Practice makes perfect – Chapter 8              

191 192 193 193 193 194 195 196 196 196 197 198 199

KINeMATICS UNIT INTRODUCTION AND ORGANIZATION CHART                                         CHApTeR 9 UNIFORM ReCTILINeAR MOTION                           91 Position, displacement and distance             911 Position                              912 Displacement                         913 Distance                             Practice makes perfect – Section 9.1         92 Graphical representation of position 20                   as a function of time Practice makes perfect – Section 9.2         93 Velocity                                    Practice makes perfect – Section 9.3         94 Graphical representation of velocity as a function of time                          Practice makes perfect – Section 9.4         Applications                                     A brief history of…                               Review – Chapter 9                              Practice makes perfect – Chapter 9              

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CHApTeR 10 UNIFORMLY ACCeLeRATeD ReCTILINeAR MOTION             221 101 Characteristics of uniformly accelerated rectilinear motion                  222 Table of Contents

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1011 Instantaneous velocity and average velocity                    1012 Determining instantaneous velocity using a graph                   1013 Acceleration                          1014 Distinguishing between velocity and acceleration                1015 Surace area under the curve in a graph o velocity as a unction o time                               1016 Surace area under the curve in a graph o acceleration as a unction o time                               Practice makes perfect – Section 10.1        102 Equations or uniormly accelerated rectilinear motion                           1021 A body’s motion on an 21                  inclined plane 1022 Equations or uniormly accelerated rectilinear motion                      1023 Analyzing uniormly accelerated rectilinear motions                     Practice makes perfect – Section 10.2        22 and 23                       103 Free all 1031 Axis o reerence or analyzing ree all                              1032 Free all analysis                       Practice makes perfect – Section 10.3        Applications                                     A brie history o…                               Review – Chapter 10                             Practice makes perfect – Chapter 10              CHApTeR 11 THe MOTION OF pROJeCTILeS      111 Describing the motion o projectiles             1111 Independence o horizontal and vertical motions                    1112 Equations o projectile motion           1113 The two-dimensional velocity vector                        Practice makes perfect – Section 11.1        112 The motion o objects launched 24 and 25                    horizontally

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Practice makes perfect – Section 11.2        113 The motion o objects launched at an angle 1131 Range                               1132 Maximum range                       Practice makes perfect – Section 11.3        Applications                                     A brie history o…                               Review – Chapter 11                             Practice makes perfect – Chapter 11             

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DYNAMICS

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UNIT INTRODUCTION AND ORGANIZATION CHART                                         CHApTeR 12 DIFFeReNT TYpeS OF FORCeS       26 and 27              121 The notion o orce 122 Gravitational orce                            1221 Law o universal gravitation              1222 Gravitational acceleration                28               1223 Mass and weight Practice makes perfect – Section 12.2        123 Normal orce                                Practice makes perfect – Section 12.3        29                       124 Force o riction 1241 The nature o riction                   1242 The coefcient o riction                Practice makes perfect – Section 12.4        125 Tension                                    Practice makes perfect – Section 12.5        126 Centripetal orce                             1261 Specifc circular motion terminology                           1262 Uniorm circular motion                 1263 Centripetal orce equation               Practice makes perfect – Section 12.6        Applications                                     A brie history o…                               Review – Chapter 12                             Practice makes perfect – Chapter 12             

262 265 266 267 267 268 270 272 273 274 275 275 275 277 278 278 279 279 279 280 281 282 283 284 286

For a detailed list of laboratories and demonstrations, see page VIII

CHApTeR 13 BODIeS SUBJeCT TO A NUMBeR OF FORCeS          131 Free-body diagram                            Practice makes perfect – Section 13.1        132 Resultant o several orces                     Practice makes perfect – Section 13.2        30                           133 Equilibrium 1331 Static equilibrium or dynamic equilibrium   1332 Equilibrant orce                       1333 Inclined plane                         Practice makes perfect – Section 13.3        Applications                                     A brie history o…                               Review – Chapter 13                             Practice makes perfect – Chapter 13             

287 288 289 290 292 293 293 295 297 298 299 300 301 302

CHApTeR 14 NeWTON’S LAWS                  303 141 Newton’s frst law and inertia                  304 1411 The frst law and the practice o everyday lie                        305 1412 Inertia and its practical consequences      307 Practice makes perfect – Section 14.1        308 142 The notion o orce and Newton’s second law     309 31              309 1421 The notion o orce 32            310 1422 Newton’s second law Practice makes perfect – Section 14.2        314 33                    315 143 Newton’s third law Practice makes perfect – Section 14.3        317 Applications                                     318 A brie history o…                               319 Review – Chapter 14                             320 Practice makes perfect – Chapter 14              321

eNeRGY AND ITS TRANSFORMATIONS UNIT INTRODUCTION AND ORGANIZATION CHART                     322 CHApTeR 15 WORK AND MeCHANICAL pOWeR    325 34                               326 151 Work 1511 Work, orce and displacement            326 1512 Particular values o work                328 Practice makes perfect – Section 15.1        330 For a detailed list of laboratories and demonstrations, see page VIII

152 Mechanical power                            Practice makes perfect – Section 15.2        Applications                                     A brie history o…                               Review – Chapter 15                             Practice makes perfect – Chapter 15             

331 334 335 336 337 338

CHApTeR 16 MeCHANICAL eNeRGY              35                        161 Kinetic energy Practice makes perfect – Section 16.1        162 Work-energy theorem                         Practice makes perfect – Section 16.2        163 Gravitational potential energy                  Practice makes perfect – Section 16.3        164 Conservation o mechanical energy              Practice makes perfect – Section 16.4        165 Conservation o total energy                    Practice makes perfect – Section 16.5        Applications                                     A brie history o…                               Review – Chapter 16                             Practice makes perfect – Chapter 16             

339 340 341 342 346 347 349 350 353 354 356 357 358 359 361

CHApTeR 17 eLASTIC pOTeNTIAL eNeRGY       171 Behaviour o constrained helical springs          1711 Helical springs                         36                    1712 Hooke’s law Practice makes perfect – Section 17.1        37            172 The energy stored in a spring Practice makes perfect – Section 17.2        Applications                                     A brie history o…                               Review – Chapter 17                             Practice makes perfect – Chapter 17             

363 364 364 365 368 370 373 374 375 376 377

AppeNDICeS                                   APPENDIX 1 Laboratory Saety                      APPENDIX 2 Methods Used in Physics                APPENDIX 3 Presenting and Analyzing Scientifc Results APPENDIX 4 Interpreting Measurement Results         APPENDIX 5 Mathematics in Physics                 APPENDIX 6 Reerence Tables                      APPENDIX 7 Nobel Laureates in Physics              

378 380 385 389 396 400 413 415

Glossary                                       417 Indx                                          421 Sourcs                                        429 Table of Contents

VII

CHApTeR 1

WAVeS

D e MO 1

Behaviour o light beams

D e MO 2

Impact o various actors on the sharpness o a shadow

CHApTeR 10 UNIFORMLY ACCeLeRATeD ReCTILINeAR MOTION LA B 21 D e MO 22 LA B 23

CHApTeR 2

Motion on an inclined plane Impact o various actors on the all o objects Analyzing the position o a body in ree all

ReFLeCTION OF LIGHT CHApTeR 11 THe MOTION OF pROJeCTILeS

LA B 8

Observing a laser beam based on the type o refection Mathematical relationships between the angle o incidence and the angle o refection and between the angle o rotation o a mirror and the angle o deviation o a refected ray The principle points and rays o spherical mirrors Describing the trajectory o light Image ormation in a plane mirror Images ormed by a concave mirror

LA B 9

Mirror equations

CHApTeR 13 BODIeS SUBJeCT TO A NUMBeR OF FORCeS

D e MO 3 LA B 4

LA B 5

LA B 6 LA B 7

CHApTeR 3

D e MO 24 LA B 25

CHApTeR 12 DIFFeReNT TYpeS OF FORCeS D e MO 26 D e MO 27 LA B 28 LA B 29

ReFRACTION OF LIGHT LA B 30

D e MO 10 LA B 11 D e MO 12

CHApTeR 4 LA B 13 LA B 14 D e MO 15 LA B 16 LA B 17

CHApTeR 5 D e MO 18

Dispersion o white light Second law o reraction Apparent position and real position LeNSeS Optical properties o converging and diverging lenses Determining ocal length Optical power o a lens system Images ormed by a thin converging lens Images ormed by a thin diverging lens AppLIeD GeOMeTRIC OpTICS

Falling time Range o a projectile launched horizontally

Dierent eects o the same orce Eect o a constant orce on an object’s motion Relationship between mass and weight o an object Impact o various actors on riction

Forces exerted on a body at equilibrium

CHApTeR 14 NeWTON’S LAWS LA B 31 LA B 32 D e MO 33

Relationship between quantity o orce and acceleration Relationship between mass and acceleration The orces o action and reaction

CHApTeR 15 WORK AND MeCHANICAL pOWeR D e MO 34

Work along an inclined plane

CHApTeR 16 MeCHANICAL eNeRGY LA B 35

Impact o various parameters on the quantity o kinetic energy

Eye disorders and vision correction CHApTeR 17 eLASTIC pOTeNTIAL eNeRGY

CHApTeR 6

FRAMeS OF ReFeReNCe

D e MO 19

Observing trajectories rom dierent view points

CHApTeR 9

UNIFORM ReCTILINeAR MOTION

LA B 20

Analyzing uniorm rectilinear motion

VIII

Laboratoris and Dmonstrations

LA B 36 LA B 37

Hooke’s law Calculating the elastic potential energy o a spring

L’organisation du manueL th QUANTUM xbk c v units: gc opc, Ply n  mchc, Kc, dyc  ey   t. th   bv  17 chapters, ch wh  c b  sections p  h ccp b . Appendices pv   cc , h  chq,  lb  pbl-lv chq. th ppc l c l -

Review th Review  fc  p ccp c  h   c y  scy Cycl tw. th ccp l  h  h th Y  scy Cycl tw Phyc P.

c bl. a h   h xbk, h   glossary wh   h ccp  y  p    cc. a index  h   h xbk pv  l l  ky w, wh h p b  whch h w pp.

ech concept  clly .

Diagrams  illustrations k  y  vw ccp.

th In practice bx pv q  hlp  vw h ccp p  h rvw.

At the beginning of each unit th title  h .

a organization chart v  qck vw  ll h chp  c  h .

a introduction p h l ccp c  h .

Organization of the Textbook

IX

Within a chapter

the title o he chpe. the introductory text peses he geel coceps esse i he chpe. the Review box povies  quick eeece o he coceps pesee i he review secio  he begiig o he exbook.

the mes o he concepts  scientifc terms e pie i blue i he ex. defiios o hese ems e ou i he glossy.

a summary iouces he secios o he chpe.

Difcult words e mke wih  seisk*  e efe i he mgi.

Reerences o ohe coceps o ohe scieifc ems povie moe exesive iomio whe eee.

this picogm iices h hee is  appendix wih moe exesive iomio o he subjec beig iscusse. Important elements e highlighe i  she blue box.

noios h e impo o uesig he concepts ppe i he subheigs.

Formulas  equios e highlighe i  me box. Calculation examples e lwys pesee i  isicive box. the History highlights eue peses the Ino-science eue oes supplemey impo hisoicl fgues om he iomio ele o he meil ue suy. wols o sciece  echology.

X

Organization o the Textbook

the Furthering your understanding eure iroduces ddiiol ides beyod wh is prescribed i he progrm.

the Applications eure preses  echologicl pplicio reled o  scieifc cocep ddressed i he chper. A brief history of… preses he hisory o  subjec reled o he coceps ddressed i he chper.

numerous diagrams, photographs d illustrations mke i esier o udersd he coceps.

a he ed o mos secios, he Practice makes perfect segme provides quesios d exercises or beer udersdig d prcice o he secio’s coceps.

the Review summrizes he mos impor the Practice makes perfect pges or specs o ech o he coceps sudied i he he chper propose quesios d exercises chper. h help o beer udersd d pply ll o he kowledge cquired i he chper. Picogrms idice he degree o difculy: esy, medium, difcul, chllegig.

Appendices provide iormio o scieifc sregies, mehods d echiques, d elbore o problem-solvig echiques. the ppedices lso coi useul reerece bles.

Organization of the Textbook

XI

2

CON T E N T S 1

4

Waves                                     

4

Energy                                      16

1.1 Types o waves                          

4

1.2 Characteristics o waves                  

5

4.1 Relationship between work, orce and displacement                    16

1.3 Electromagnetic spectrum                 

6

4.2 Relationship between work and energy        16

1.4 Electric eld                            

6

4.3 Kinetic energy                            17

1.5 Magnetic eld                           

7

4.4 Potential energy                          17 4.5 Law o conservation o energy               18

2

Properties of light                          

8

2.1 Refection o light                        

8

2.2 Reraction o light                        

9

4.6 Relationship between power and electrical energy                       19 4.7 Solar energy                             19

2.3 Converging lenses and diverging lenses         10 2.4 The eye                                 11

3

Force and motion                            12 3.1 Relationship between constant speed, distance and time                         12 3.2 Characteristics o orce                     13 3.3 Gravitational orce                        13 3.4 Relationship between mass and weight       13 3.5 Force o riction                           14 3.6 Equilibrium between two orces             14 3.7 Eective orce                            15

3

1 Waves

REVIEW

1

A wave is defned as a disturbance that propagates through a vacuum or through a medium that contains matter. A wave carries energy rom one point to another without, however, carrying any matter along with it.

1.1

Types of waves

Waves can be distinguished based on their propagating medium and the angle ormed between the direction o the disturbance and the direction o propagation. • Mechanical waves require a material medium in order to propagate. Some examples o material media through which waves move include water (wave), air (sound) and the Earth (seismic wave). • Electromagnetic waves do not necessarily require a material medium; they can travel through a vacuum. Radio waves, light waves, UV rays and X-rays are examples o electromagnetic waves. • Waves can be transverse or longitudinal. • Transverse waves produce a disturbance that is perpendicular to the direction o propagation (see Figure 1). • Longitudinal waves produce a disturbance that is parallel to the direction o propagation (see Figure 2).

Motion of the spring (disturbance)

Crest

Position of the spring at rest

In Enpractice pratique pr Direction of propagation

Trough

Figure 1 A transverse wave created with a spring

Rarefaction

Position of the spring at rest

Compression Trough (rarefaction)

Crest (compression)

Direction of propagation

Figure 2 A longitudinal wave created with a spring

In Enpractice pratique pr 1. True or false? a) A wave carries energy from one point to another b) A wave carries matter from one point to another c) Mechanical waves and electromagnetic waves can move through material media d) Mechanical waves and electromagnetic waves can move through a vacuum

4

REVIEW 1 Waves

1.2

Characteristics of waves

REVIEW

1

All waves (see Figure 3) share four common characteristics: a wavelength, a frequency, an amplitude and a period.

Wavelength 1 cycle

1 cycle Crest

Amplitude A

B

C

Equilibrium position

Trough

Scale: 1 square = 1 cm

Figure 3 Diagram of a wave The time required for the wave to travel the distance between point A and point C is 10 seconds

• Wavelength (l) is the distance between two identical disturbances, that is, the length of a complete cycle. In Figure 3, l 5 8 cm. Wavelength can also be calculated by dividing the total distance travelled by the number of cycles completed. l5

Total distance Number of cycles

• Frequency (f ) is the number of disturbances (or cycles) produced per unit of In Enpractice pratique pusually r time, per second. It is measured in hertz (Hz). Frequency can be calculated by dividing the number of cycles by the total time it takes a wave to travel a distance equal to one cycle. f5

Number of cycles Total time

• Amplitude (A) is the maximum height reached by a wave from its equilibrium position. In Figure 3, the amplitude A 5 2 cm. • The period (T ) is the time it takes a wave to complete a full cycle. The period can be calculated by dividing the total time it takes the wave to travel this distance by the number of cycles completed. T5

Total time Number of cycles

The period of the wave shown in Figure 3 is T 5 5 s because it takes the wave 10 s to travel from point A to point C, and there are two cycles between A and C.

In Enpractice pratique pr 2. What is the frequency of the wave shown in Figure 3?

REVIEW 1 Waves

5

1.3

REVIEW

1

Electromagnetic spectrum

The electromagnetic spectrum consists o the entire range o electromagnetic radiation. • Electromagnetic waves are classiied into dierent categories according to their wavelength and requency (see Figure 4).

Long wavelength Low requency

Radio waves

Short wavelength High requency

Microwaves

Inrared rays

Visible light

Ultraviolet rays

X-rays

Gamma rays

Figure 4 The electromagnetic spectrum

• Visible light is made up o all o the waves in the electromagnetic spectrum that have wavelengths between 400 and 700 nanometres (nm). The rays with the longer wavelengths, and thereore the lower requencies, are red; those that have the higher requencies are violet. The other colours that make up white light all between these two extremes (see Figure 5). 700 nm

Red

600 nm

Orange

500 nm

Yellow

Green

400 nm

Blue

Violet

Figure 5 The wavelengths that make up visible light

1.4

Electric feld

An electric feld is a region o space where an electric orce created by a charged body can act on another charged body. • An electric feld is represented schematically by feld lines that diverge away rom a charged object or converge toward it (see Figure 6). • Field lines illustrate the direction o the electric orce acting against a positive charge located near the feld.

a) Positive charge

b) Negative charge

Figure 6 By convention, electric feld lines diverge away rom a positive charge (point toward the outside) and converge toward a negative charge (point toward the inside)

6

REVIEW 1 Waves

In Enpractice pratique pr • An electric feld is a region o space that transmits the electric orce rom a charged object to another charged object. A charged object that passes through an electric feld is subjected to an electric orce.

REVIEW

1

• The magnitude o an electric ield produced by an electric charge at a given point is directly proportional to the quantity o charge and inversely proportional to the square o the distance between this point and the charge. This relationship is expressed as ollows:

E5

kq

r2 where E 5 magnitude o the electric feld, expressed in newtons per coulomb (N/C) k 5 Coulomb’s constant, which is equal to 9 3 109 Nm2/C2 q 5 quantity o electric charge, expressed in coulombs (C) r 5 distance between the charge and a given point, expressed in metres (m)

In Enpractice pratique pr 3. What is the magnitude o an electric feld created by a negative electric charge o 5 3 10-7 C at a point located 10 cm rom the charge?

1.5

Magnetic feld

A magnetic feld is an invisible space that surrounds a magnet and inside which magnetic orces can act on other magnets or on erromagnetic substances. • The magnetic feld is represented by feld lines (or lines o orce). The spacing o the feld lines indicates the relative magnitude o the magnetic feld. • The magnetic feld lines o a bar magnet can be illustrated using iron flings (see Figure 7). • Magnetic feld lines have a specifc direction: they run rom the north pole to the south pole (see Figure 8).

Figure 7 Iron flings help to visualize magnetic feld lines

Arrows indicate the north-south direction o the feld

Figure 8 Magnetic feld o a bar magnet

REVIEW 1 Waves

7

2 Properties o light

REVIEW

2

Light moves in a straight line. However, it can be deviated when it encounters certain media. Images are ormed when the trajectory o light changes.

2.1

Refection o light

Refection is the phenomenon whereby light rays bounce o an obstacle. Specic terms are used to describe refection (see Figure 9). Refective surace

Refected ray

θr Angle o refection

θi Angle o incidence

Incident ray

Normal

Figure 9 Terms used to describe the refection o light rays

• Two laws help to predict what will happen to a ray when it refects o a surace. • The rst law o refection states that the incident ray, the normal and the refected ray all exist in the same plane (see Figure 10). • The second law o relection states that the angle o relection qr, which is ormed by the refected ray and the normal, is always equal to the angle o incidence qi, which is ormed by the incident ray and the normal (see Figure 11). • A mirror with a fat surace is called a plane mirror. A plane mirror produces a virtual image o an object where the light rays refected in the mirror intersect (see Figure 12 on the following page). Normal Normal

Plane o incidence

Incident ray

θi

Refected ray

30° Incident ray

θr 30° Refected ray

Refective surace Refected surace

Figure 10 First law o refection: the incident ray, the normal and the refected ray all exist in the same plane

8

REVIEW 2 Properties o light

Figure 11 Second law o refection: the angle o refection is always equal to the angle o incidence

Observer

REVIEW

Plane mirror

2

Virtual image of the object

Object

Figure 12 How a virtual image is ormed

• The object and its image ormed by a plane mirror are symmetrical in relation In practice Enthe pratique prplane o the mirror (see Figure 13). to A

Object

B

Extension of mirror

A’

Virtual image of the object B’

Figure 13 To represent the image o an object ormed by a plane mirror, each point o the image must be drawn at the same distance rom the mirror as the object

In Enpractice pratique pr 4. The angle between an incident ray and the surace o a plane mirror is 30° What is the angle o refection?

2.2

Refraction of light

Reraction is a phenomenon whereby light rays deviate rom their trajectory when they pass rom one medium to another with dierent optical properties. Specifc terms are used to describe reraction (see Figure 14).

Incident ray

Normal Angle of incidence

θi Air

• Media that have the ability to reract light rays are called reractive media.

θR

Refracted ray

Glass

Angle of refraction

Figure 14 Terms used to describe the reraction o light rays

REVIEW 2 Properties o light

9

• When a light ray passes from one medium to another that is more optically dense, it bends toward the normal. In the opposite situation, the light ray bends away from the normal (see Figure 15).

REVIEW

2

Normal

Normal

θR

θi Air

Air

Water

Water

Bends away from the normal

θi

θR Bends toward the normal

a) From air to water

b) From water to air

Figure 15 Trajectory of refracted rays

2.3

Converging lenses and diverging lenses

Lenses are curved elements manufactured from translucent materials, such as glass or plastic, that cause the light rays that pass through them to deviate from their trajectory. They are classified into two categories, based on how they affect light rays: converging lenses and diverging lenses. • A converging lens causes the light rays that travel through it to bend toward its principal axis. Converging lenses come in many forms, but they all have at least one convex side, that is, a side that curves outward (see Figure 16).

a) A biconvex lens

b) A planoconvex lens

c) A positive meniscus lens

• All rays that pass through a converging lens deviate toward the lens’s principal axis (which corresponds to the normal). The point on the axis where the rays converge is called the focal point of the lens. The higher the curvature of a converging lens, the closer the focal point will be to the lens (see Figure 17).

Figure 16 Three types of converging lenses

Focal point of the lens

Focal point of the lens Principal axis of the lens

a) Lens with a low curvature

Figure 17 Focal point of converging lenses

10

REVIEW 2 Properties of light

Principal axis of the lens

b) Lens with a high curvature

• A diverging lens causes the light rays that travel through it to bend away rom the principal axis. Diverging lenses come in many dierent shapes, but they all have at least one side that is concave, that is, a side that curves inward (see Figure 18). • All light rays that travel through a diverging lens bend away rom the principal axis o the lens. When the reracted rays are extended, they appear to originate rom the same point on the axis o the lens. This is the ocal point o the diverging lens. The higher the curvature o the diverging lens, the closer the ocal point is to the lens (see Figure 19).

Focal point of the lens

REVIEW

2

a) A biconcave lens

c) A negative meniscus lens

Figure 18 Three types of diverging lenses

Principal axis of the lens

a) Lens with a low curvature

b) A planoconcave lens

Focal point of the lens

Principal axis of the lens

b) Lens with a high curvature

Figure 19 Focal points of diverging lenses

2.4

The eye

The eye is the body’s receptor organ or light. It is a complex, spherical organ with a diameter o about 2.5 cm. • The eye consists o several transparent media that allow light to reach the retina, where images are ormed (see Figure 20). • Light deviates when it passes through transparent media with dierent indices o reraction and dierent curvatures. This occurs in the eye, and more specifcally in the lens. The lens o the eye is lexible and roughly biconvex in shape, allowing an image to orm on the retina. • When we look at an object that is ar away, the lens does not have to change shape in order or the rays to converge and orm an image on the retina. However, i we are looking at an object that is very close, the lens must become more curved so that the rays will converge and ocus on the retina. The process through which the lens adapts to the light rays that reach the retina is called accommodation.

Retina Vitreous humor

Lens Cornea

Aqueous humor

Figure 20 Anatomy of the eye

REVIEW 2 Properties of light

11

3 Force and motion

REVIEW

3

Force and motion are important concepts in physics. They are particularly useful in technology, where they are used in the design of transportation mechanisms, in hydraulic systems and in machines with moving parts.

3.1

Relationship between constant speed, distance and time

Speed is the ratio between the distance a body travels and the amount of time it takes to travel this distance. • The relationship between constant speed (v), distance (d) and the time interval (Dt) is represented by the following equation: v5 d Dt where v 5 constant speed, expressed in metres per second (m/s) d 5 distance, expressed in metres (m) Dt 5 time interval, expressed in seconds (s) • When the speed is not constant over the entire trajectory, we use average speed. It in the same way as constant speed and is expressed by the folIn Enispractice pratique pcalculated r lowing equation: vave 5 d Dt where vave 5 average speed, expressed in metres per second (m/s) d 5 total distance, expressed in metres (m) Dt 5 time interval, expressed in seconds (s)

In Enpractice pratique pr 5. The cabin of an aerial tram moves at a constant speed of 8 m/s over a distance of 60 m How long will it take the cabin to cover this distance? 6. Calculate the average speed in each of the following examples: a) A remote-control car travels 50 m in 13 s b) A person is standing on a moving sidewalk in an airport This person travels 255 m in 90 s c) It takes a driver 2 h and 55 min to travel 300 km between Saint-Jérôme and Québec City 7. A car travels 210 km in 2 h and 15 min What is the car’s average speed: a) in km/h? b) in m/s?

12

REVIEW 3 Force and motion

3.2

Characteristics of force

REVIEW

A orce is a push or pull applied to a body that can change the body’s motion or shape. • A orce can act at a distance, like the gravitational orce, or by contact (by pushing or pulling the object). • In physics, orce is represented schematically by an arrow, which indicates its magnitude, point o application and direction (see Figure 21). • The magnitude o the orce is represented by the length o the arrow or by a number indicating its value. • In the International System, orce (F) is measured in newtons (N). • The point o application o a orce is the specifc point at which the orce is applied to a body.

3

Direction

F51

0N

Point of application

Magnitude of the force

Figure 21

Schematic representation of a force

• The direction o a orce is indicated by a line segment with an arrow at the end.

3.3

Gravitational force

The gravitational orce (Fg ) is an attractive orce between two bodies. It depends on the mass o each body and the distance between them. • The greater the mass o the bodies, the greater the gravitational orce. • The greater the distance between the bodies, the weaker the attraction between them. • In The Enpractice pratique pmost r common occurrence o gravitational orce is weight, that is, the attraction between a celestial body and the objects on its surace.

3.4

Relationship between mass and weight

The relationship between mass (m) and weight (Fg ) is expressed by the ollowing equation: Fg 5 mg where Fg 5 object’s weight, expressed in newtons (N) m 5 object’s mass, expressed in kilograms (kg) g 5 gravitational acceleration, whose average value on Earth is 98 N/kg • Mass is the amount o matter a body contains, whereas weight is equal to the product o a body’s mass and the gravitational acceleration. • The greater a body’s mass, the greater the gravitational orce (weight).

In Enpractice pratique pr 8. The Canadarm used on space shuttles has a mass of 410 kg How much does it weigh on Earth?

REVIEW 3 Force and motion

13

REVIEW

3

3.5

Force of friction

The force of friction (Ff ) is a force resisting the motion of an object that is in contact with a surface. • The force of friction depends on the type of surface and the magnitude of the contact force between the surface and the object (see Figure 22). • The plane in which the friction occurs is parallel to the surfaces that are in contact, and the direction of the force is opposite to the direction of the motion.

Ff

Ff

Ff

Figure 22 The forces of friction (Ff ) between the car’s tires and the road, and between the front of the car and the air

3.6

Equilibrium between two forces

Two forces are in equilibrium when the resultant force is zero and the body’s motion does not change. • The resultant force (FR ) is the sum of all forces applied to a body. • If an object is subjected to two forces of the same magnitude but in the opposite direction, the resultant force is zero; the object stays at rest if it was initially at rest. Therefore, the two forces are in equilibrium (see Figure 23). • A body that is in equilibrium stays at rest if it is stationary but continues to move at a constant speed if it is already in motion.

F1 5 20 N

F2 5 20 N

FR 5 0 N

Figure 23 The resultant force (FR ) is equal to F1 2 F2 5 20 N 2 20 N 5 0 N Since the resultant force (FR) is zero, the bookcase is in equilibrium and remains at rest

14

REVIEW 3 Force and motion

3.7

Effective force

REVIEW

3

When the force applied to an object is not parallel to the direction in which it is moving, only one component of the force is responsible for the motion. The component that is parallel to the direction of the displacement is called the effective force (see Figure 24).

F

F

θ θ Displacement

Displacement

a) The applied In practice En pratique prforce forms an angle with the direction of the displacement

Feff

b) The effective force (Feff ) is the component of the applied force that is parallel to the displacement

Figure 24 Effective force and the direction of displacement

• Effective force (Feff ) is expressed as a function of force (F ) and of the cosine of the angle q, as illustrated by the following equation: Feff 5 F cosq where Feff 5 effective force, expressed in newtons (N) F 5 applied force, expressed in newtons (N) q 5 value of the angle between the direction of the applied force and the direction of the displacement

In Enpractice pratique pr 9. A person pulls a sled with a force of 50 N The rope the person is pulling forms an angle of 37° with the direction of the sled’s displacement What is the effective force? 10. A young child pulls on a rope tied to a toy with a force of 20 N The rope forms an angle of 22° with the toy What is the effective force?

REVIEW 3 Force and motion

15

4 Energy

REVIEW

4

Energy is a physical quantity measured in joules (J). It is defned as a system’s capacity to produce work.

In Enpractice pratique pr 4.1

Relationship between work, force and displacement

The relationship between work (W), orce (F ) and displacement (d ) is expressed by the ollowing equation: W 5 Fd where W 5 work, expressed in joules (J) F 5 force, expressed in newtons (N) d 5 object’s displacement, expressed in metres (m) The ollowing conditions must be met or work to be done on an object: • the object must move • a orce must be applied to the object • the object’s displacement must be in the same direction as the orce applied to the object, or in the same direction as a component o the orce

In Enpractice pratique pr In Enpractice pratique pr

11. Calculate the work done on an object by a force of 20 N over a displacement of 4 m The force is applied in the direction of the object’s displacement

4.2

Relationship between work and energy

The concepts o energy and work are closely related: work is defned as the transer o energy between two bodies, two objects or two systems. • Work is the process o transerring energy. • The relationship between work (W ) and energy (E ) is expressed by the ollowing equation: W 5 DE where W 5 work, expressed in joules (J) DE 5 change in the object’s energy, expressed in joules (J)

In Enpractice pratique pr 12. To draw water, a horse pulls horizontally on a rope that runs through a pulley and is tied to a bucket If the horse pulls with a force of 100 N and covers a distance of 5 m, how much energy is transferred to the bucket by this force? (Do not consider the forces of friction)

16

REVIEW 4 Energy

In Enpractice pratique pr

4.3

Kinetic energy

REVIEW

4

Kinetic energy is energy related to an object’s motion. The faster the object moves, the greater its kinetic energy. • Kinetic energy is proportional to the mass of a moving body and the square of its speed. • The relationship between kinetic energy (Ek ), mass (m) and speed (v) is expressed by the following equation: Ek 5 1 mv 2 2 where Ek 5 kinetic energy, expressed in joules (J) m 5 object’s mass, expressed in kilograms (kg) v 5 object’s speed, expressed in metres per second (m/s)

In Enpractice pratique pr 13. What is the kinetic energy of a 05 kg billiard ball that moves at 2 m/s?

4.4

Potential energy

Potential energy is energy that is stored in a body and that can be transformed into another form of energy. • There are different forms of potential energy, including elastic potential energy and gravitational potential energy. • Elastic potential energy is energy stored in elastic materials that are comIn Enpractice pratique pr or stretched. pressed • Gravitational potential energy is the energy a body possesses due to its position in space. • The gravitational potential energy stored in a body is equal to the product of the mass, gravitational acceleration and height of a body in relation to a reference point. • The relationship between gravitational potential energy (Epg ), mass (m), gravitational acceleration (g) and height (h) is expressed by the following equation: Epg 5 mgh where Epg 5 gravitational potential energy, expressed in joules (J) m 5 object’s mass, expressed in kilograms (kg) g 5 gravitational acceleration, whose average value on Earth is 98 m/s2 h 5 object’s height in relation to a reference point, expressed in metres (m)

In Enpractice pratique pr 14. An amusement park ride that moves vertically has an empty passenger seat with a mass of 100 kg What is the gravitational potential energy stored in the seat when it is raised to a height of 10 m?

REVIEW 4 Energy

17

4.5

REVIEW

4

Law of conservation of energy

The law of conservation of energy states that energy cannot be created or destroyed; it can only be transformed from one form to another. According to this law, these transformations occur with no loss of energy if they take place in an isolated system. • An isolated system is a system that does not exchange matter or energy with its surroundings. • In an isolated system, mechanical energy is conserved (and is therefore constant). • Mechanical energy (Em) is the sum of a system’s gravitational potential energy (Epg ) and kinetic energy (Ek). It is expressed by the following equation: Em 5 Epg 1 Ek where Em 5 mechanical energy, expressed in joules (J) Epg 5 gravitational potential energy, expressed in joules (J) Ek 5 kinetic energy, expressed in joules (J) • Any decrease in gravitational potential energy must be compensated for by an increase in its kinetic energy and vice versa (see Figure 25). Epg maximum

Ek minimum

In Enpractice pratique pr

Epg

Epg Em

Ek Ek Epg

Ek

Figure 25 The snowboarder’s maximum gravitational potential energy at her highest point in the air This energy is transformed into kinetic energy, which reaches its maximum at the lowest point of the half-pipe, and so on This illustration demonstrates what would happen if this were an isolated system

In Enpractice pratique pr 15. A rock falls from a cliff at a height of 50 m Without taking friction into account, determine whether the kinetic energy and gravitational potential energy are at a maximum, zero or equal: a) when the rock is at the top of the cliff b) when the rock is at a height of 25 m as it is falling

18

REVIEW 4 Energy

4.6

Relationship between power and electrical energy

REVIEW

4

Electrical power is the amount o energy consumed or supplied by an electrical device in a given unit o time. • Electrical power is the rate at which electrical energy is supplied or consumed. • Electrical power (P) is calculated by dividing the quantity o energy (E ) by the time interval (Dt) during which the energy is supplied or consumed. This relationship is expressed by the ollowing equation:

In Enpractice pratique pr P5 E Dt where P 5 electrical power, expressed in watts (W) E 5 electrical energy, expressed in joules (J) Dt 5 time interval, expressed in seconds (s) • The relationship between the watt (unit o power) and the joule (unit o energy) is expressed as ollows: 1 W 5 1 J/s

In Enpractice pratique pr 16. What is the power of an electric light bulb that consumes 108 MJ in 5 h? 17. How much energy is used when a 348-W television is on from 6 pm until 8 pm? 18. The electric power of a toaster is 965-W How much energy is used if the toaster works for 4 min? 19. How much energy would an 875-W electric coffee maker consume after being used for 10 min?

4.7

Solar energy

Solar energy is radiation that comes rom the Sun in the orm o electromagnetic waves with various wavelengths. • The electromagnetic spectrum is made up o the various wavelengths o solar radiation. • The amount o energy associated with these waves depends on their wavelength. • Solar energy that reaches the Earth’s surace consists mostly o visible light. • Many o the other types o rays, such as ultraviolet rays and inrared rays, are mostly absorbed or refected by the Earth’s atmosphere.

REVIEW 4 Energy

19

20

COnTenTs ChApTer 1

Wav . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .23 ChApTer 2

rfctio o Ligt. . . . . . . . . . . . . . . . . .41 ChApTer 3

ractio o Ligt . . . . . . . . . . . . . . . . .77 ChApTer 4

L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .95 ChApTer 5

Alid Gomtic Otic . . . . . . . .131

O

tic i o o t ld o cic tat a g­ atd t mot itt ic atiquity. T Gk, Aab ad Ci w itigud by ligt ad tid to ulock it may ct. Ti ot vd a a bai o lat, mo laboat toi. Gomtic otic l u to tudy ligt ad li o t coct tat ligt tavl i a taigt li i t om o ay. T advatag o ti modl li i t cotuctio o gomtic gu, wic a lativly iml to aalyz. Fo ti ao, ti bac o otic i calld gomtic otic.

Dit it imlicity, gomtic otic l to xlai comlx oma uc a t omatio o aibow, t aaac o miag ad ow t y wok. Uig iml quatio, it i oibl to dtmi t fctio o actio o ligt ay ad to actai t caactitic o imag omd by mio ad l. Aid om it totical valu, gomtic otic a layd a imotat ol i t dig ad mauactu o actical otical ytm uc a micoco, tl­ co, cama, gla ad cotact l.

21

ChApTer

1

1.1 Characteristics o waves WAVes

1.2 Light waves

2.1 Types o refection 2.2 Geometry o refection ChApTer

2

reFLeCTiOn OF LiGhT

2.3 Refection on a plane mirror: laws o refection 2.4 Refection on spherical mirrors 2.5 Images

3.1 Phenomenon o reraction

UniT

1

3.2 Index o reraction ChApTer

GeOmeTriC OpTiCs

3

reFrACTiOn OF LiGhT

3.3 Geometry o reraction 3.4 Laws o reraction 3.5 Total internal refection

4.1 Dierent types o lenses 4.2 Reraction in lenses ChApTer

4

Lenses

4.3 Optical power o lenses 4.4 Images ormed by lenses 4.5 Optical aberrations o lenses

5.1 The camera ChApTer

5

AppLieD GeOmeTriC OpTiCs

5.2 The human eye 5.3 The light microscope 5.4 The telescope

22

Wavs

W

hen a pebble is cast onto the surace o a calm body o water, undulations appear at the point o impact and travel out in all directions. This pheno­ menon is similar to that o the sound produced by a guitar string or the light emitted by a projector. In act, although they appear to be dierent, in each case they are physical phenomena that occur in a point in space and subsequently travel toward other points in space. Moreover, they have one characteristic in common: they can be described as waves.

In this chapter, you will discover types o waves, their periodic behaviour and characteristics. In parti­ cular, the ocus will be on light waves, how they trav­ el, the electromagnetic spectrum and the behaviour o these waves when the medium through which they travel is modifed.

rviw Types o waves                                           4 Characteristics o waves                                    5 Electromagnetic spectrum                                   6 Electric felds                                             6 Magnetic felds                                            7 Reection o light                                          8 Reraction o light                                          9

1.1 1.2

Caactistics of wavs . . . . . . . . . . . . . . 24 Ligt wavs . . . . . . . . . . . . . . . . . . . . . . . . . . 29 ChApTer 1 Waves

23

1.1 Chaactestcs of waves A wave is a travelling disturbance that carries energy from one point to another. Certain waves can only travel in a material medium and thereore belong to the category o mechanical (or material) waves. This is the case with sound waves, which cannot travel through a vacuum. Their propagation, despite closely relying on the presence o the medium’s atoms, does not transport matter. In order to gain a better understanding o this, let us consider the example o the “wave” produced by the ans o a sports team in the stands o a stadium. The wave travels in a medium ormed by the ans, but each o the ans making up the wave remains in his or her place (see Figure 1). In contrast to mechanical waves, electromagnetic waves do not need a medium in order to propagate: they can travel through a vacuum. This is how light reaches Earth rom space.

1.1.1

Figure 1 The propagation of a “wave” in the stands of a stadium: the wave travels, but each person remains in his or her place

Tyes of waves

There are two main types o waves: transverse waves and longitudinal waves. In order to distinguish these two types o waves, one has to compare the angle be­ tween the direction o the disturbance and the direction in which the waves travel.

Wave

Tasvese waves

pot roe

The propagation o a wave in a fexible rope, which one causes to move up and down at one o its ends, provides an illustra­ tion o a transverse wave (see Figure 2). The disturbance moves each point o the rope up and down, whereas the wave travels along the length o the rope. Since the direction o movement o the points on the rope is perpendicular to the direction o propagation, the wave is transverse.

t 5 t0

t 5 t1

t 5 t2

t 5 t3

Electromagnetic waves, which are discussed in Section 1.2, are transverse waves.

t 5 t4

Logtudal waves

Figure 2 The direction of propagation of a transverse wave is perpendicular to the direction of movement of each point on the rope

raefacto

Comesso

Wave

pot o col

Figure 3 The direction of propagation of a longitudinal wave is parallel to the direction of movement of each point on the spring coil

24

A wave is said to be transverse when the direction in which it travels is perpendicular to the direction of the disturbance.

UniT 1 Geometric Optics

A wave is said to be longitudinal when its direction of propagation is parallel to the direction of the disturbance. One can produce a longitudinal wave with a coil, in which one initiates a movement rom one end o the coil to the other (see Figure 3). Certain spring coils are compressed (compression zone), whereas others are not (rareaction zone). The two zones correspond to the wave’s crests (maximums) and troughs (mini­ mums) respectively. The compression zone then moves along the coil. Because the direction o movement o each point on the spring coils is parallel to the compression’s direction o propaga­ tion, the wave is longitudinal. Sound waves, which allow sound to travel rom one place to another, are longitudinal waves.

1.1.2

piodicity of wavs

Although waves can be very complex, using simple examples will allow us to easily become amiliar with the mathematical processing o wave movements. Figure 2 on the previous page shows that the movement o each point on the rope is cyclical (or periodic). The successive posi­ tions o the red point describe a complete cyclical movement. I one continues to move the rope at the same speed and with an upward movement that is identical to the downward movement, the wave is perectly sinusoidal, meaning that it is possible to accurately describe it using a sine unction. This type o unction is characterized by two pulses, one positive and one negative (see Figure 4). Figure 4 clearly shows that the wave cyclically recurs in space. Its unction is positive between O and B (positive pulse) and negative between B and C (negative pulse). This dual pulse, bounded by O and C, represents a complete cycle that then repeats itsel between C and E and between E and G. This behav­ iour is said to be periodic. Based on fgures 2 and 4, it is possible to describe the characteris­ tics o waves. These characteristics are wavelength (l), period (T), amplitude (A) and requency (f ).

l

f (x)

l

l

1A

B

O

C

D

e

F

G x (m)

2A OB 5 positiv uls BC 5 ngativ uls

Figure 4 The sinusoidal behaviour o a wave travelling along a rope This fgure shows three complete cycles

Wavlgt (l) The wavelength (l) is the distance between two points of a wave separated by a complete cycle. For example, in Figure 4, the distance between points O and C constitutes a wavelength. However, the wavelength can be measured between any two points separated by a complete cycle, in particular between two successive crests or troughs. The unit o wavelength is expressed in metres (m), and its symbol is the Greek letter l (lambda). Figure 4 comprises three complete cycles.

piod (T )

f (t) 1A

Wav iod (T ) The wave period (T) corresponds to the time required for the wave to complete a cycle. Figure 2 on the previous page makes it possible to see the variation o the wave created in a rope as a unction o time. The position o the red point may be represented as a unction o time (t). At time t0, the point is at its equilibrium position. Then, when the wave reaches the point, it moves to a minimal posi­ tion (t 5 t1), returns to its equilibrium position (t 5 t2), goes to a maximum position (t 5 t3) and fnally returns to its equilibrium position (t 5 t4). At this moment, the cycle is complete. Any urther disturbance would have the eect o orcing the point to repeat the same movement. The period (T) is expressed in seconds (s). The period is the time required to cover the distance equivalent to the wavelength (l) [see Figure 5].

O

t (s)

2A t0

t1

t2

t3

t4

Figure 5 A wave period (T ) is the time required to cover the distance equivalent to the wavelength (l)

ChApTer 1 Waves

25

The relationship between the period (T ) and the wavelength (l) is written as follows: l5v3T where l 5 Wavelength, expressed in metres (m) v 5 Wave velocity, expressed in metres per second (m/s) T 5 Wave period, expressed in seconds (s) f (t )

exampl A wave that travels at a velocity of 500 cm/s has a period (T ) equal to 00250 s What is its wavelength (l)?

1A Ampltud (A)

O

Data: v 5 500 cm/s T 5 00250 s l5?

t (s)

Solution: l 5 v 3 T 5 500 cm/s 3 00250 s 5 125 cm 5 00125 m

Wav ampltud (A) 2A

t0

t1

t2

t3

The amplitude (A) of a wave is the maximum displacement of a point from its equilibrium position (see Figure 6).

t4

The unit of measurement of a wave’s amplitude (A) depends on the nature of the wave. In the case of a wave travelling along a rope, the amplitude is a length and is measured in metres (m).

Figure 6 The amplitude (A) of a wave is the maximum displacement of a point from its equilibrium position

Wav frqucy (f ) The frequency ( f ) of a wave is equal to the number of cycles completed by one wave per second (see Figure 7). The frequency (f ) of this wave is calculated by dividing the number of cycles com­ pleted by the wave by the total time, as conveyed by the following relationship:

f5

Number of cycles Total time

Figure 7 shows a wave completing three cycles in six seconds. Therefore, the frequency (f ) of this wave is: f5

f (t) 1A

O

2

4

Since the unit of frequency (f ) is hertz (Hz), which is related to one second by the following relationship: 1 Hz 5 1 s-1

2A

Figure 7 Representation of a wave during a period of time of six seconds

26

6 t (s)

Number of cycles 3 1 05 05 5 5 5 5 5 05 s-1 Total time 6s 2s 1s s

UniT 1 Geometric Optics

the frequency of the wave represented in Figure 7 is f 5 0.5 Hz.

In accordance with its defnition, the period (T) is equal to the time required by the wave to complete a cycle. It can be obtained by dividing the total time by the number o cycles.

T5

Total time Number of cycles

Thereore, the period o the wave represented in Figure 7 on the previous page is: T5

Total time 6s 5 52s Number of cycles 3

It should be noted that the requency (f ) is the inverse o the period (T) and vice versa.

f (Hz) 5

1 1 or T (s) 5 T (s) f (Hz)

Given that l 5 v 3 T, one can iner the relationship that connects v, l and f . l l5v3T⇒v5 T 1 T5 f

⇒v5 l 5l3f 1 f

The mathematical expression derived rom this is called the universal wave equation. The term “universal” means that the expression applies to all types o waves. v5l3f where v 5 Wave velocity, expressed in metres per second (m/s) l 5 Wavelength, expressed in metres (m) f 5 Wave frequency, expressed in hertz (Hz)

examl Water drops fall from a pipe at regular intervals onto the surface of a calm body of water Given that 150 drops fall each minute and the distance between 2 successive crests of the sinusoidal wave created in the process measures 60 cm, calculate the velocity of the wave Data: 1 minute → 150 drops Distance between two successive crests 5 Wavelength (l) l 5 60 cm v5?

Solution: 1. Calculation of wave frequency: Since 150 drops fall in one minute, the wave frequency (f ) is written: f5

150 150 5 5 25 s-1 5 25 Hz 1 min 60 s

2. Calculation of wave velocity: v 5 lf 5 60 cm 3 25 s-1 5 0060 m 3 25 s-1 5 015 m/s

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seCTiOn 1.1

Caractrtc of wav 5. The ollowing graph represents a wave created on a calm body o water: 0.5 hgt of wav (cm)

1. A tuning ork produces sound waves. Calculate the period (T ) and requency (f ) o a tuning ork whose prongs vibrate 375 times in 3.00 seconds.

0.3 0.1 ­0.1 ­0.3 ­0.5

0

5

10 Dtac (cm)

15

20

a) What is this wave’s amplitude (A)? b) What is its wavelength (l)? c) I the requency (f ) o this wave is equal to 20 Hz, what is its velocity (v)? 6. Several waves are represented in the ollowing graph: 2. A longitudinal wave is created by pushing a coil orward and then pulling it backward at a requency (f ) o 2.00 Hz. I the velocity o the waves in the coil is 5.40 m/s, what is their wavelength (l)? 3. To tune their instruments, musicians use the A note, which has a requency (f ) o 440 Hz. I the speed o sound in an auditorium is 350 m/s, what is the wavelength (l) corresponding to this requency? 4. A hummingbird can fap its wings at a maximum o 4800 faps per minute. a) What is the requency (f ) o the faps? b) What is the period (T ) o the faps?

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UniT 1 Geometric Optics

A B C D e Tm ()

a) b) c) d) e)

What can you say about the shape o these waves? Which wave has the smallest period (T )? Classiy the waves by increasing order o period. Which wave has the smallest requency (f )? Classiy the waves by increasing order o requency.

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

1.2 Ligt wav A light wave is an electromagnetic disturbance whose propagation allows the transportation of light energy. In contrast to mechanical waves, light waves can travel equally well through a vacuum as in certain media composed o atoms or molecules. But how can light waves move through a vacuum? In order to understand this phenome­ non, it is necessary to know the composition o this type o wave. In act, light waves are composed o an electric feld and a magnetic feld that are perpen­ dicular to each other (see Figure 8). This is the reason why light waves are called electromagnetic waves.

JAmes CLerk mAxWeLL Scottish physicist (1831–1879)

E

B Diction o wav oogation

E 5 elctic fld B 5 magntic fld

Figure 8 Light waves are composed o an electric feld and a magnetic feld that are perpendicular to each other They are electromagnetic waves

Furthering

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

James Clerk Maxwell mathematically ormalized the connection between electric and magnetic felds as well as electromagnetic wave propagation He is considered to be one o the world’s greatest physicists Einstein said that his work led to the most proound changes in physics since Newton In act, his work made possible the appearance o another major innovation in physics in the 20th century: quantum theory

your understanding

mauing t d o ligt Measuring the speed o light has ascinated scientists or centuries We know that Galileo (1564–1642), although unsuccessul, conducted experiments using lanterns and his pulse to measure time It was necessary to wait or the Danish astronomer Ole Christensen Rømer (1644– 1710) beore methods became more refned Rømer was interested in the revolutions o Io, one o the moons o Jupiter He noted that the period o its eclipse behind Jupiter depended on the distance between Earth and Jupiter From this, he inerred that the time dierence between the maximum and minimum periods could be explained by the dierent amounts o time it took light to travel rom Jupiter to Earth when the Earth moved away rom or approached Jupiter He proved that the speed o light is not infnite, as had been believed until then Later, technological progress led to the frst direct measurements o the speed o light In 1848, with the help o an experimental apparatus, the French physicist Hippolyte Fizeau (1819–1896) obtained the value c 5 315 000 km/s, which was only 5% more than the currently known value In 1878, Albert Abraham Michelson (1852–1931) constructed an intererometer, a device composed o a series o rotating mirrors (see Figure 9 ) His experiments enabled him to obtain the value c 5 300 140 ± 480 km/s He then refned his measurements and in 1926 determined that c 5 299 796 ± 4 km/s Subsequently, owing to the development o electronics and laser, assessments o the speed o light became more accurate In 1978, Woods, Shotton and Rowley determined the most accurate value measured to date: c 5 299 792458 98 ± 00002 km/s Figure 9 Michelson’s intererometer

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As Figure 8 on the previous page shows, electric and magnetic felds are variable. By oscillating each on a plane, they create an electromagnetic wave that travels in a direction that is perpendicular to their planes o oscillation. Electromagnetic waves are thus transverse waves and move through a vacuum at speed (c).

See Fudatal stadads of t basc uts of cacs, p 169

c 5 299 792 458 m/s However, in order to simpliy matters, the value c 5 3.00 3 108 m/s is used in­ stead. This value can also be used when the electromagnetic wave moves through air since the resulting error is negligible.

1.2.1

elctoagtc spctu

In the case o electromagnetic waves moving through a vacuum, the universal wave equation is written as ollows: c5l3f ⇒ l5

8 c 5 3 00 3 10 m/s f f

As this equation indicates, the wavelength (l) is inversely proportional to the requency (f ). The shorter the wavelength, the more substantial the energy transported by the electromagnetic wave.

29

*  Nanometre (1 nm 5 10

By varying the requency (f ) [or the wavelength (l)] o electromagnetic waves, the electromagnetic spectrum is obtained (see Figure 10). The fgure shows that visible light represents only a minute part o the electromagnetic spectrum. In act, the human eye can only detect the wavelengths (l) contained between 400 nm* (violet) and 700 nm (red).

m)

The requencies (f ) [or wavelengths (l)] o visible electromagnetic waves determine their colours. The colours visible to humans are the ones that make up a rainbow. Table 1 on the ollowing page indicates the wavelengths o visi­ ble colours.

Wav­ lgt () 103

104

102

101

rado wavs

105

106

107

10­2

10­3

mcowavs

108

Fqucy (hz)

Figure 10 Electromagnetic spectrum

30

10­1

1

UniT 1 Geometric Optics

109

1010

1011

10­5 10­4 ifad adato

1012

1013

10­6 10­7 Vsbl lgt

1014

1015

10­8 10­9 Ultavolt adato

1016

1017

10­10

10­11

x­ays

1018

1019

10­12 Gaa ays

1020

Tabl 1 Visible light spectrum

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

Violet

Blue

Green

Yellow

Orange

Red

Wavlgt (l) [m]

400 to 450

450 to 490

490 to 570

570 to 585

585 to 620

620 to 700

Fqucy (f ) [1014 h]

75 – 67

67 – 61

61 – 53

53 – 51

51 – 48

48 – 43

White light is a combination of colours that make up the visible light spectrum. It can be broken up with a prism. examl Calculate the wavelength (l) o orange light with a requency o f 5 500 3 1014 Hz Data: f 5 500 3 1014 Hz c 5 300 3 108 m/s l5?

Solution: According to the universal wave equation: c c5l3f⇒l5 f 8 c 300 3 108 m/s Thereore: l 5 5 5 300 3 10 m/s 14 14 f 500 3 10 Hz 500 3 10 s21 26 5 0600 3 10 m 5 600 nm The wavelength (l) o orange light with a requency o f 5 500 3 1014 Hz is l 5 600 nm

Cll os A cellular telephone is an amazing device As soon as it is switched on, a cellular telephone connects to a cellular central ofce through a numeric user identifer The telephone then remains connected to a network o stations on the ground called base stations (which are generally located on electrical towers, church steeples or any other place that is ree o obstacles and intererence) A cellular telephone communicates with the base station through electromagnetic waves at requencies o around 109 Hz (see Figure 11) All inormation sent and received by the cellular telephone is relayed by the base station to another subscriber o the same network or to a public telephone network that uses cables The cellular telephone converts the analogue signal o the voice into a ow o digital data sent to the closest base station, which may be located at a distance o 08 to 13 km rom the telephone Each base station sends out a low-strength signal that cellular telephones pick up When a user moves, his or her cellular telephone manages the signals that come rom the dierent base stations nearby It moves rom one station to the next so as to maintain the greatest possible signal strength The digital data low is unique to each technology, and each Figure 11 The cellular network has its own requency, allowing many networks to main- telephone has become a comtain a presence in the same place simultaneously without inter- munication device with a neverending growth in applications ering with one another

HISTORY HISTORY h e i n r i HIGHLIGHTS CHIGHLIGHTS h rUDOLF herTz German engineer and physicist (1857–1894) Heinrich Rudol Hertz studied in Berlin where he took classes with German scientists Gustav Kirchho and Hermann Ludwig von Helmholtz In 1887, with the help o an astute apparatus, he demonstrated the existence o electromagnetic waves He thus provided an unequivocal experimental confrmation o the theoretical work o Scottish physicist J C Maxwell The research Hertz undertook during his short lie (he died at the age o 37) cleared the way or telegraphy and radio This is why radio waves are also called Hertzian waves The SI unit o requency was named in his honour

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HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

phOTOns

The word “photon” comes from the Greek phôtos, meaning “light” In physics, photons represent the elementary particles that constitute visible light as well as all other electromagnetic waves These massless particles possess energy that is dependent on the electromagnetic wave frequency In fact, the higher the wave frequency, the greater the related photon’s energy For example, a photon in the ultraviolet region has more energy than a visible light photon The photon concept was introduced by Albert Einstein (1879–1955) in order to explain physical phenomena that could not be understood based on the wave properties of light The use of this concept allowed for considerable advances in various areas of physics

* Cocetrc Having the same centre Wave­ frot

Wave ray Wave­ legt (l)

Figure 13 The wave’s direction of propagation can be represented by an arrow that is perpendicular to the wavefronts

32

UniT 1 Geometric Optics

1.2.2

proagato of lgt

In order to better understand the phenomenon of the propagation of light, it is necessary to imagine the movement of a light wave. To do this, it is interest­ ing to consider the example of the wave created on the surface of a calm lake after a pebble has been cast into it. Even though this wave is not electromag­ netic, it presents simple analogies that can be transposed to the case of light. When the pebble strikes the water’s surface, it creates circular ripples that are perfectly centred on the point of impact. These concentric* ripples move away from the source of the disturbance. It is possible to recreate this disturbance in a laboratory with the help of an oscillating vibrator that periodically strikes the surface of water contained in a sink. When a sinusoidal oscillation is applied to the vibrator, the waves created will, in turn, be sinusoidal. If an imaginary vertical orthogonal cut is made on the surface of the water containing the source (S) of the disturbance, it is possible to visualize the disturbed area (see Figure 12). Wave tra

Vbrator

Wave Wavefrot A s source

B W

x

l

Figure 12 The formation of a wave train on a calm lake

When the vibrator strikes the water’s surface, a wave is created. At a given time, the movement reaches point W, which forms part of a circle of radius SW. This circle, centred on point S, is called the wavefront. But as the vibra­ tor continues to disturb the water’s surface, new wavefronts are created. Two consecutive wavefronts are separated by a distance that is equal to the wave’s wavelength, as indicated in Figure 12. The periodic disturbance imposed on the water’s surface generates a series of consecutive waves that represent a wave train. A wavefront is an imaginary line that connects all the points touched by the wave at the same moment in time. A wave train is a consecutive series of waves that travel in the same direction. Since the wave moves in a direction that is perpendicular to the wavefronts, it is possible to represent its direction of propagation with an arrow that itself is also perpendicular to the wavefront (see Figure 13). This arrow is called the wave ray. Since light is a wave, it travels in a manner similar to the one just described, with wavefronts and wave trains also playing a part. In this case, the wave rays are light rays (see Figure 14 on the following page). The use of light rays to represent the propagation of light forms the basis of geometric optics. It allows for the explanation of many optic phenomena using simple mathematical formulas based on geometry. Moreover, rays make it possible to study a number of these phenomena in the following chapters.

Wavfont Wav­ font Ligt ay

Ligt ay

Wavlngt (l)

Ligt ay

Figure 14 A cross-section representation o the light produced by an electric bulb The light rays are perpendicular to the waveront o the electromagnetic wave that constitutes the light

Figure 15 When electromagnetic waves are plane, the light rays are parallel to one another

It should be noted that the bulb in Figure 14 produces spherical concentric electromagnetic waves. However, as the waves move away from the source, the curvature of their wavefronts dwindles. If the distance is very great, the wavefronts can be considered as planes. In this case, they are plane waves and the light rays become parallel to one another (see Figure 15).

a) Parallel beam

Ty of bam A light beam is a set of light rays. When the beam’s rays are parallel, the beam is said to be parallel. When the rays appear to come from the same point, it is a diverging beam. If the rays move toward the same point, the beam is converg­ ing (see Figure 16).

b) Diverging beam

Lina oagation of ligt Many phenomena of everyday life demonstrate that light travels in a straight line. One of the clearest examples is that of umbra formation. If light did not travel in a straight line, but instead followed curves, a shadow would not form. A point source* contributes to umbra formation, whereas an extended light source* causes the formation of umbra and penumbra (see Figure 17).

c) Converging beam

Figure 16 Types o beams point ouc A source o light that is * very small or that is so distant that it can be classifed as a point extndd ligt ouc A source o light with well-defned physical dimensions so that it cannot be classifed as a point

* sadow

sadow



Ligt ay

point ouc

Obtacl scn

a) Point source

extndd ligt ouc

b) Extended light source

Obtacl scn

pnumba

Figure 17 The shadow and the penumbra are created by the linear propagation o light

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1.2.3

Lght popagato  meda

In contrast to mechanical waves, light can travel through a vacuum, meaning in the absence o matter. But when it encounters a material medium, light can behave in dierent ways depending on the nature o the medium. Opaque media block the passage o light completely so that it is impossible to see through them (see Figure 18a). This is the case, or example, with stone, wood and metal. Transparent media allow light to pass. As a result, it is possible to see through these media. Water, glass and air are some examples (see Figure 18b). Translucent media represent an intermediate case between the two preceding ones. They allow or the passage o light but diuse the light in all directions. This diusion prevents one rom clearly distinguishing objects when looking through these media. Frosted glass and tracing paper are good examples (see Figure 18c).

a) Opaque medium

b) Transparent medium

c) Translucent medium

Figure 18 Types of propagation in media

1.2.4 The surface of separation * iteace between two different media

See refecto o lght, p 41

See reacto o lght, p 77

34

UniT 1 Geometric Optics

Lght passage om oe medum to aothe

Light travels in a straight line in a given medium until it encounters a new me dium. This change o medium can be associated with three phenomena: refection, transmission or absorption. It should be noted that these phenom­ ena do not necessarily happen in isolation but may also be simultaneously produced. However, only refection and transmission occur at the interace* between the two media. Refection occurs when light reaches the surace o certain materials with the property o modiying the trajectory o light rays and sending them back into the medium they came rom. This property can be observed in shiny metals, calm bodies o water and mirrors, among others. Owing to this property, it is possible to obtain true images o the objects mirrored in the refecting devices. Transmission takes place when light passes through certain transparent or trans­ lucent materials. This transmission can modiy the initial trajectory o light rays. When this occurs, we are dealing with a phenomenon called reraction. Absorption occurs when a light wave travels through a substance and its energy diminishes over the course o its journey. This phenomenon may produce dier­ ent changes in the substance passed through, including an increase in tempera­ ture. When exposed to light radiation, dark­coloured materials heat up more quickly than those that are light­coloured. In act, dark­coloured materials have a much greater ability to absorb light energy than light­coloured materials.

La The acronym LASER comes rom the phrase light amplification by stimulated emission of radiation A laser produces a beam o photons that all have the same energy (requency and wavelength) and travel in phase As such, it is a very thin but intense light beam (see Figure 19 ) Ever since Planck and Einstein ormulated quantum theory, physicists have sought to produce a powerul beam o photons o visible light From the end o the 1950s onward, scientists in several countries worked on this and on July 7, 1960, the American physicist Theodore Maiman developed the very frst laser

Lasers make it possible to take measurements with a high degree o precision As such, the dimensions o an object can be measured with the precision o one millionth o a centimetre, which is about 1/250 o the diameter o a hair Using a laser, the distance between the Earth and the Moon (384 000 km) was measured with a precision o 5 cm! There are plenty o laser applications in many other felds For example, in medicine, lasers are used in the treatment o certain skin cancers In the industrial sector, they are used to carry out specialized welding In the business world, lasers are used to read the bar codes printed on the packaging o goods

seCTiOn 1.2

Figure 19 An optical bench apparatus using a green laser ray

Lgt wav

1. Calculate the period (T ), in seconds, of the following periodic events: a) 5 classes every 375 minutes b) 10 oscillations of a pendulum in 6.7 seconds c) 1000 engine turns in 1 minute d) 68 sit-ups in 57 seconds

8. A local radio station transmits its broadcasts at a frequency of 95.1 MHz. What is the length of the wavelength emitted by the antenna? 9. When an obstacle is placed between an extended light source and a screen, the regions of umbra and penumbra appear on the screen.

2. Calculate the frequency (f ), in hertz (Hz), of the following periodic events: a) 120 oscillations in 2.0 seconds b) 1200 engine turns in 1 minute c) 40 pulsations in 1.2 hours d) 65 words typed every 48 seconds

A B

C

3. Convert the periods in question 1 into frequencies. 4. Convert the frequencies in question 2 into periods. 5. Visible light is the region in the electromagnetic spectrum that is detectable by the eye. Name the immediately adjoining regions in the spectrum: a) toward the lower frequencies b) toward the upper frequencies 6. The radius of the Earth is 6400 km. Calculate the time it takes light to travel the distance equal to the circumference of the Earth. 7. How much time does it take sunlight to reach the Earth’s surface, given that the distance between the Sun and the Earth is about 150 million kilometres?

Obtacl Lgt ouc

D e sc

a) Where are the regions of umbra and penumbra on the screen? b) What would become of these regions if the light source were a point source? 10. During hot weather, is it preferable to wear black or white clothing? Explain your answer.

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APPLICATIONS x­rays German physicist Wilhelm Röntgen (1845–1923) dis­ covered X­rays at the end o the 19th century. He did this by studying the properties o cathode rays in a Crookes tube, an instrument widely used in exper­ iments at the time. A Crookes tube has a negative metallic cathode and a positive anode between which a potential dierence o several thousand volts exists. The partial vacuum created in the tube and the elec­ tric voltage generate a fow o electrons rom the cathode to the anode, orming cathode rays. On November 8, 1895, Röntgen sealed a Crookes tube in a thick, black cardboard cover beore running a current through it. He observed that a paper screen covered with barium platinocyanide, a salt, became fuorescent i placed close to the tube. Since this salt emits light only when it is stimulated by light, Röntgen deduced that the tube produced a orm o radiation. Röntgen called this unknown invisible radiation “X­rays.” Ater many experiments, Röntgen was awarded the 1901 Nobel Prize or Physics or his discoveries (see Figure 20). Unlike cathode rays, which are streams o electrons, X­rays are photons produced by electrons striking an anode. X­rays are commonly used in medical imaging

Figure 20 One o the frst radiographs taken by Röntgen

36

UniT 1 Geometric Optics

(see Figure 21). The sot tissues o the human body contain small atoms such as carbon, oxygen and hydrogen. These tissues decrease the X­rays’ inten­ sity to a lesser degree than bones do (bones contain calcium and phosphorous, which are larger atoms). The intensity o the X­rays detected ater they pass through the body is thereore dierent rom one area to another, which provides inormation about the inside o the body. There are also some disadvantages associated with X­rays. Since the X­ray photons have a lot o energy, they can strip electrons rom the atoms that absorb them, creating harmul ions. This can have negative health consequences or individuals. However, this disadvantage has been put to good use in some cases. For instance, in radiation therapy, X­rays are directed at tumours to destroy cancer cells while causing mini­ mal damage to surrounding tissues. Also, X­ray telescopes in space measure the radiation emitted by celestial objects such as neutron stars, black holes, active galaxies and hot gas clouds. The study o cosmic rays using these telescopes has led to major advances in astrophysics.

Figure 21 X-rays are a orm o radiation routinely used in medical imaging

radio boadcasting Shortly ater discovering the existence o radio waves, Heinrich Rudol Hertz, in response to one o his stu­ dents, who had asked what purpose a radio wave might serve, replied, “It’s o no use whatsoever!” Nevertheless, electromagnetic waves revolutionized the world o communications. They have multiple uses today, including radio, television, cellular tele­ phones, wireless Internet, GPS receivers, satellite communication, remote­controlled objects, and many others. Very soon ater their discovery, the frst uses o radio waves were discovered. In 1893, in the United States, the Serbian engineer Nikola Tesla (1856–1943) described the principle o a radio broadcast and constructed the frst transmitter. In 1895, the Italian Guglielmo Marconi (1874–1937) set up the frst radio link over 1.5 km and six years later, the frst transat­ lantic transmission (see Figure 22). However, he only sent Morse code signals. Nevertheless, his eorts made him the pioneer o wireless telegraphy. In 1906, the Canadian Reginald Fessenden (1866­ 1932) achieved the frst transmission o the human voice and o music using radio waves. He is considered the inventor o radio broadcasting (see Figure 23). The frst broadcast used the AM band width and reached across the Atlantic Ocean.

Figure 22 In 1901, Guglielmo Marconi succeeded in making the frst transatlantic radio transmission between England and Newoundland

The frst users o wireless telegraphy and then radio were ship crews who could rom then on communi­ cate ship­to­shore or ship­to­ship while on the open seas. For example, in 1912, when the Titanic sank, the survivors owed their lives to wireless telegraphy (an S.O.S signal). Radio is an application o electromagnetism. During a radio transmission, a sound is picked up by a micro­ phone and converted into alternating electrical cur­ rents that oscillate in the radio station’s antenna. These oscillating currents emit electromagnetic waves. The requency o the currents in the antenna determines the requency o the radio waves. Radio waves have a requency below 1 GHz. They do not carry sound, but they transmit inormation that radio receivers can reconstruct. This inormation can be written into the radio waves by adjusting their amplitude (AM) or their requency (FM). Radio waves travel at the speed o light. When they reach the antenna o a receiver set, they induce weak electrical currents. The currents are analyzed, and then the inormation is decoded and then sent to the loudspeaker, which transorms the electrical impulses it receives into mechanical waves that can be heard.

Figure 23 During the frst hal o the 20th century, radio broadcasting expanded rapidly and radios became a common household item, numbering in the thousands ChApTer 1 Waves

37

chapter

1

Wvs

1.1 Charactrstcs o wavs • A wave is a travelling disturbance that carries energy. • A wave is transverse when its direction o propagation is perpendicular to the direction o the disturbance. • A wave is longitudinal when its direction o propagation is parallel to the direction o the disturbance. • The wavelength (l) is the distance between two points o a wave separated by a complete cycle. • The wave period (T) corresponds to the time required by the wave to complete a cycle. • The amplitude (A) o a wave is the maximum displacement o a point rom its equilibrium position. • The requency (f ) o a wave is equal to the number o cycles completed by a wave in one second. • The universal equation o waves is: v5l3f

1.2 Lght wavs • A light wave is an electromagnetic disturbance whose propagation allows the transportation o light energy. • Electromagnetic waves are transverse waves that travel through a vacuum at a speed o 299 792 458 m/s. • The electromagnetic spectrum represents all electromagnetic waves classifed according to their requencies (f ) or their wavelengths (l). • The electromagnetic spectrum can be divided into several regions. In order rom least to greatest amount o energy, there are: radio waves, microwaves, inrared radiation, visible light, ultraviolet radiation, X­rays and gamma rays. E

• The region o visible light is contained between the regions o inrared and ultraviolet radiation. • A waveront is an imaginary line that connects all the points touched by the wave at the same moment in time.

B

E 5 elctrc fld B 5 magtc fld

38

UniT 1 Geometric Optics

Wav’s drcto o propagato

(cont.) • A wave train is a consecutive series o waves that travel in the same direction. • A wave ray is an arrow perpendicular to the waveront that represents the wave’s direction o propagation. • A light ray is an arrow that represents the direction o propagation o an electromagnetic wave and that is perpendicular to its waveront. • A light beam is a set o light rays. • The ormation o shadows is an indication o the linear propagation o light. • Opaque media block the passage o light completely so that it is impossible to see through them. • Transparent media allow light to pass. It is possible to see through these media. • Translucent media allow the passage o light but diuse the light into all directions. This diusion prevents one rom clearly distinguishing objects when looking through these media. • Three phenomena can occur when light encounters a new medium: refection, transmission or absorption. • Refection occurs when light reaches the surace o materials with the property o modiying the trajectory o light rays and sending them back into the medium they came rom. • Transmission takes place when light passes through certain transparent or translucent materials. • Absorption occurs when a light wave travels through a substance and its energy diminishes over the course o its journey.

ChApTer 1 Waves

39

ChApTer 1

Wavs

1. The cardiac ultrasound was developed using sound waves (ultrasounds) with a requency o f 5 2 MHz. The velocity (v) o these waves in the heart tissue is v 5 1.5 km/s. a) Calculate their wavelength (l). b) In your opinion, is the velocity o these waves the same when they move through the air as when they move through heart t issue? 2. During certain thunderstorms, a ash o lightning can be seen, and ater a certain period o time thunder can be heard. These two wave phenomena occur simultaneously but are perceived with a time-lag. This can be explained by the dierence in their velocity. The sound o thunder has a velocity o approximately 330 m/s; it is heard with a delay in time, whereas one can almost instantaneously perceive the light o the ash o lightning (c 5 3.00 3 108 m/s). Taking this inormation into account, calculate the distance (d) at which the ash o lightning occurred i the thunder can be heard our seconds later. 3. An electromagnetic wave with a requency o 5.2 3 1014 Hz reaches your eyes. a) Is it possible to know the colour o this wave? I so, what is the colour? b) What is the period (T ) o this wave? c) Is it possible to calculate its wavelength (l)? I so, what is its value? d) What occurs i this wave reaches a shimmering surace?

40

UniT 1 Geometric Optics

4. Sound and light are waves. In spite o this, sound waves do not appear in the electromagnetic spectrum. Explain why not. 5. The fgure below represents a sinusoidal wave. y (cm) 2 0

1.25

t (s)

This type o wave can be described mathematically with the ollowing unction: y (t) 5 A 3 sin (ω 3 t) where A 5 Wave amplitude ω 5 Pulsation related to the period (T ) by the ollowing relationship: ω 5 a) b) c) d)

2π T

What is the period (T ) o this wave? What is its requency (f )? What is its wavelength (l)? Write the equation o this wave in the orm shown above, indicating the numerical values o the dierent variables.

Refection o Light

T

he relection o light is a phenomenon that is easy to observe. Examples include the refection o a landscape on the surace o a calm lake and the image a driver sees in a car’s rear-view mirror. Did you know that without the refection o light, you would not be able to see the objects around you? In this chapter, you will discover how light interacts with certain types o suraces. This will help you to understand the laws governing the phenomenon o refection. You will also study how images are ormed and learn about their characteristics.

2.1 2.2 2.3

Types o refection                       42

Review

2.4

Electromagnetic spectrum                                   6 Refection o light                                          8 The eye                                                 11

Refection on spherical mirrors                                   48

2.5

Images                                   55

Geometry o refection                   44 Refection on a plane mirror: laws o refection                        46

CHAPTER CHAPTER22 Refection RefectionooLight Light

41 41

2.1 Tpes o refection Refection reers to the change in the direction o light ater it meets a surace that returns it to its original medium.

See Propagation o light, p 32

See Wavelength (l), p 25

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

Despite the complexity o the nature o light, it is helpul to illustrate it schematically by a light ray. The light ray simply represents the light’s direction o propagation. This simplifcation allows us to use mathematical tools such as geometry to explain many optical phenomena. The scientifc approach used in this unit is called geometric optics. Light travels in a straight line in a homogeneous medium and is undisturbed until it meets the surace o a new medium. Reection occurs when light returns to its original medium ater meeting a surace. The rough surace o a piece o wood does not reect light in the same way as polished metal does. The nature o the surace on which light is reected is very important because it determines the type o reection, i.e. specular or diuse.

2.1.1

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY I b N A L - H A y T H A m, A.K.A. ALHAZEN Arab mathematician, physicist and philosopher (965–1039) Ibn al-Haytham, also known as Alhazen, carried out important experiments on the propagation o light, reection, atmospheric reraction and colour He was the frst to suggest that light travels in a straight line He also conducted experiments on curved mirrors and discovered spherical aberration This learned man wrote over a hundred works His Kitâb al-Manâzir (Book of Optics) served as a reerence in the feld o optics and experimental physics or several centuries This book is considered one o the most signifcant in the history o physics

Specular refection

At the atomic level, all suraces possess imperections or irregularities; there is no such thing as a perectly plane surace. However, i the wavelength o the light is longer than the size o the irregularities on the surace, specular (regular) reection will occur (see Figure 1). This means that i the light directed toward the surace consists o parallel rays, the reected light will consist o parallel rays as well. Specular reection occurs on smooth suraces such as mirrors, calm bodies o water and polished metals. This type o reection produces discernable images o objects.

2.1.2

Diuse refection

When the size o the surace irregularities is greater than or approximately equal to the light’s wavelength, the rays are reected in a disorderly manner, in dierent directions. This is reerred to as diuse or irregular reection (see Figure 2). Rough, matte and dull suraces give rise to diuse relection. This happens with many amiliar suraces such as brick walls, asphalt, wood urniture and sheets o paper. Diuse reection does not produce discernable images like a mirror does. However, it allows us to see objects around us because our eyes detect the light rays, even i the rays are reected in a disorderly manner.

Figure 1 Specular or regular reection

42

UNIT 1 Geometric Optics

Figure 2 Diuse or irregular reection

Furthering

your understanding

The wavelength and type o refection o visible light The wavelength o visible light ranges rom 07 µm 1 (red) to 04 µm (violet) Thereore, a surace with irregularities that are greater than or approximately equal to 04 µm in height will refect visible light in a diuse manner and will not produce a clear image This is what happens with the surace o a sheet o paper Surace irregularities can be observed using a light microscope, which has a resolution limit o about 02 µm In contrast, i the surace irregularities do not exceed one-tenth o a micrometre, the surace will appear specular and behave like a mirror For example, the mirror-like surace o a silicon pellet used in microelectronics is not ree o imperections at the atomic level It has surace irregularities in the order o tens o nanometres (nm2), but this has no eect on the specular nature o the surace or visible light (see Figure 3 ) Figure 3 The irregularities on the surace o a silicon pellet are too small to aect its specular nature

1 1 µm (micrometre) 5 1026 m 2 1 nm (nanometre) 5 1029 m

SECTION SECTION2.12.1

Types o refection

1. Explain the dierence between specular reection and diuse reection 2. A man is polishing his kitchen oor He takes a break midway through and notices that hal o the oor is shiny but the other hal still looks dull Explain this phenomenon based on what you know about reection 3. Under bright light, why is it easier to read a book printed on porous paper with a matte fnish rather than on smooth, glossy paper? 4. Why does the image o the Sun on the rough sur­ ace o a lake appear distorted?

5. Indicate whether each o the ollowing examples involves diuse reection or specular reection a) a ashlight beam on a brick wall b) a boat’s reection on the calm waters o a lake c) the blank page o a notebook d) the Sun’s reection on a car’s metallic paint e) a person’s reection in a mirror 6. Green light (wavelength o 550 nm) is shone on a metallic surace that has surace irregularities whose average size is 50 nm a) Will the reection be specular or diuse? Explain your answer b) The surace is then scored using a tool that creates numerous grooves with an average depth o 2 µm What type o reection occurs ater the surace is scored?

CHAPTER 2 Refection o Light

43

2.2 Geometry o refection Reection, whether it is diuse or specular, ollows certain laws. However, it is more difcult to use geometry to analyze diuse reection, where light is reected in many dierent directions, than it is to analyze specular reection. The fgures at the bottom o the page illustrate the trajectory o a light ray travelling toward a reective surace. We will use these fgures to defne the terminology o reection.

Incident ray

2.2.1

The incident ray is the light ray that travels toward the refective surace (see Figure 4).

Normal

2.2.2

The normal is an imaginary line that is perpendicular to the reective surace and originates rom the point o incidence, that is, the intersection o the incident ray and the reective surace (see Figure 4). The plane defned by the incident ray and the normal is called the plane o incidence. In optics, the normal is used as a reerence or measuring angles. It is oten represented by a dotted line.

Angle o incidence (θi)

2.2.3

The angle ormed by the incident ray and the normal is called the angle o incidence and is represented by the symbol θi.

Refected ray

2.2.4

The refected ray is the light ray that travels away rom the refective surace (see Figure 5).

2.2.5

Angle o refection (θr)

The angle ormed by the refected ray and the normal is called the angle o refection and is represented by the symbol θr. Normal Incident ray

Plane of incidence Angle of incidence (θi)

Normal Angle of reflection (θr)

Reflected ray

Point of incidence Reflective surface

Figure 4 The plane o incidence is a plane that Figure 5 The refected ray travels away rom the contains the normal and the incident ray The angle o inci- refective surace The angle o refection (θr) is ormed by dence (θi) is ormed by the incident ray and the normal the refected ray and the normal

44

UNIT 1 Geometric Optics

metallic paint All paint available on the market is made up o three essential components: a binder, a solvent and one or more pigments The pigments are very fne powders that diuse visible light and determine the paint colour The pigment concentration as well as the grain size and shape also inuence the brightness o the painted surace Metallic paint, as its name suggests, contains metallic pigments, which behave like microscopic mirrors and reect light (see Figure 6) These pigments are made up o metals such as aluminum or o metallic alloys such as bronze Although the light that relects against these metal particles provides a “metallic” fnish, the surace is not perectly specular because the pigments are not regularly oriented and the rays are not always reected in the same direction, as in the case o a mirror The paint used on most cars is metallic

SECTION 2.2

Geoetry o refection

1. The angle o reection (θr) is ormed by: A the normal and the incident ray B the normal and the reected ray C the incident ray and the reected ray D the normal and the reective surace 2. On a sheet o paper, reproduce the fgure shown below and draw a normal on the outside ace o each o the three mirrors (M1, M2 and M3) in the fgure M1

M2 M3

Figure 6 Metallic paint reects light due to its metallic pigments

3. A light ray is directed toward a mirror at an angle o incidence o 30° The plane o incidence is perpendic­ ular to the mirror a) What is the value o the angle ormed by the incident ray and the normal? b) What is the value o the angle ormed by the incident ray and the mirror? The mirror is tilted clockwise by 20° c) What is the value o the angle ormed by the incident ray and the normal? d) What is the value o the angle ormed by the incident ray and the mirror? 20°

CHAPTER 2 Reection o Light

45

2.3 Refection on a plane mirror: laws o refection A plane mirror is an object that orms images using the specular reection o light rays on a plane surace. When a light ray strikes a plane mirror, it behaves like a ball that hits the edge o a billiard table. According to this analogy, which applies only to the trajectory, both the ball and the light ray return in very specifc directions.

2.3.1

First law o refection

The frst law o reection states that the incident ray, the reected ray and the normal are all located in the same plane. This law implies that the incident ray determines the plane in which reection will occur. In other words, the reected ray is located in the plane o incidence. Thereore, i the incident ray is rotated by a given angle in relation to the normal, then the plane and the reected ray will be rotated by the same angle as well (see Figure 7). The billiard ball analogy works here, too. The trajectory o the ball that hits the edge o the table is located entirely in the plane ormed by the surace o the billiard table (see Figure 8).

2.3.2

Second law o refection

The second law o reection states that the angle o incidence is equal to the angle o reection: θi 5 θr (see Figure 9).

Figure 7 The incident ray, the reected ray and the normal are always located in the same plane

This law, which is simple to test experimentally, is valid or any angle o incidence (θi). When the angle is zero, the reected ray ollows the same path as the incident ray but in the opposite direction. In this case, the two rays coincide with the normal.

θi θr

Figure 8 As it travels, the ball does not leave the plane ormed by the surace o the billiard table

46

UNIT 1 Geometric Optics

Figure 9 Experimental verifcation o the second law o reection: the angle o incidence (θi) and the angle o reection (θr) are both equal to 45°

SECTION 2.3

Refection on a plane mirror: laws o refection

1. I the angle separating the incident ray and the refect­ ed ray is 60°, what is the angle o incidence (θi)? 2. Light rom a fashlight is directed toward a plane mir­ ror The angle o incidence (θi) is 20° a) What is the angle o refection (θr)? b) What angle is ormed by the refection o the light ray and the mirror’s surace?

7. In the ollowing diagram, two plane mirrors (M1 and M2) are set up perpendicular to each other Use a diagram to show that the incident ray on mirror M1 and the refected ray on mirror M2 are parallel M1 θi

3. The angle ormed by a light ray and the surace o a plane mirror is 40° What is the value o the angle o refection (θr)? 4. The suraces in the ollowing diagram are plane mirrors Reproduce the diagram on a sheet o paper, and use the laws o refection to help you draw the pathway that the light ray ollows until it exits the set­up

5. Reproduce the ollowing diagram on a sheet o paper, and draw the trajectory o the refected ray

M2

8. To help scientists accurately measure the dis­ tance between the Earth and the Moon, optical devices capable o refecting light were placed on the Moon’s surace by American missions Apollo 11 (1969), 14 and 15 (1971) and by the Russian probe Lunokhod 2 (1973) These devices, called retrorefectors, consist o an array o cube corners Each corner is made up o three mirrors that are perectly perpendicular to each other Light rom a strong laser beam directed at them rom stations on Earth is refected back to its source By measuring the time it takes or the laser beam to travel to and rom the devices (the beam travels at the speed o light), scientists have been able to determine the distance between the Earth and the Moon with great accuracy In the ollowing diagram, three mirrors are assembled to orm the corner o a perect cube Explain how this set­up can refect a light ray back to where it came rom

6. A light ray is refected on a plane mirror at a certain angle o incidence (θi) The mirror is then turned clockwise at an angle equal to θ ’ What is the angle o refection? Show your calculations CHAPTER 2 Refection o Light

47

2.4 Refection on spherical mirrors Unlike plane mirrors, which have a two-dimensional plane surace, curved mirrors have a three-dimensional curved surace. There are several types o curved mirrors. They are named ater the geometric fgures that are used to produce them. Thereore, we have spherical, cylindrical, parabolic, elliptical and hyperbolic mirrors, which are generated by respectively rotating a circle, rectangle, parabola, ellipse and hyperbola around a given axis (see Figure 10). In all cases, only a portion o the geometric fgure is used as a mirror.

* Spherical cap Portion o a sphere HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

Parabola

Cylinder

Rectangle

Cylindrical mirror

Ellipsoid of revolution

Paraboloid of revolution

Parabolic mirror

Hyperboloid of revolution

ISAAC NEWTON

HISTORY HIGHLIGHTS HISTORY HIGHLIGHTS English mathematician, physicist and astronomer (1642–1727) Isaac Newton’s discoveries revolutionized optical physics With his theory o colour, he established that white light consists o an infnite number o coloured rays To prove this he used prisms, which can break down light into a spectrum o visible colours His research on the reraction and reection o light helped him to perect the telescope, which he equipped with parabolic and second ary mirrors By reecting light, these mirrors corrected chromatic aberrations and, as a result, the blurred and iridescent images produced by previous telescopes Another advantage o his innovation was the ease with which the telescope could then be built

48

UNIT 1 Geometric Optics

Ellipse

Elliptical mirror

Hyperbola

Hyperbolic mirror

Figure 10 Curved mirrors generated rom various two-dimensional geometric fgures

Spherical mirrors are the most commonly used because they are simple to manuacture. They consist o a spherical cap* cut rom a refective hollow sphere (see Figure 11).

Centre (C)

A refective hollow sphere

A spherical cap

Figure 11 A spherical mirror is a portion o a reective hollow sphere

A spherical cap has two aces: interior and exterior. The interior ace, which is hollow, corresponds to the inside o the sphere. The exterior ace, which is domed, corresponds to the outside o the sphere. I the hollow side is used as the relective surace, the mirror is concave (see Figure 12a). However, i the domed ace is used as the relective surace, the mirror is convex (see Figure 12b). Thereore, in a concave mirror, the centre (C) o the sphere is located on the side o the reective surace. In a convex mirror, the centre is located on the opposite side o the reective surace. Although spherical mirrors are shaped dierently than plane mirrors, the laws o reection still apply. In a plane mirror, all normals to the reective surace are parallel to each other. This is not the case or spherical mirrors, which behave as i they were made up o many tiny mirrors that are tilted in relation to each other. Thereore, to fnd the normal at a point on a spherical mirror, frst draw a tangent to the mirror at this point. Next, draw a line that is perpendicular to Centre (C) the tangent passing through this point (see Figure 13).

Tangent

Tangent

Tangent

Tangent

θi

θi

θi

θi

θr

θr

θr

θr

a) Concave mirror

Centre (C)

Centre (C

Figure 12a Concave mirror

Centre (C)

Figure 12b Convex mirror

b) Convex mirror

Figure 13 The laws o refection apply to a spherical mirror

2.4.1

Concave mirrors

The surface of a concave mirror converges the light rays that are parallel to its principal axis (which is the mirror’s axis of symmetry and is labelled P) toward a point located inside its curvature and referred to as the focal point (F) (see Figure 14). For this reason, concave mirrors are also called converging mirrors.

F

P

Figure 14 In a concave mirror, the light rays that are parallel to its principal axis (P) converge toward its ocal point (F)

CHAPTER 2 Refection o Light

49

The principal points of a spherical mirror

C

P

F

V

To understand the geometry o reection in spherical mirrors, it is important to have a good grasp o the terminology. Begin with the three principal points o spherical mirrors or, more generally, o curved mirrors. All o these points are located on the mirror’s principal axis (P) (see Figure 15). 1. Centre of curvature (C) The centre o curvature (C) is the point at the centre o the sphere rom which the spherical mirror originates.

Figure 15 The principal points o a spherical mirror

2. Focal point (F) The ocal point (F) is the point at which the light rays that are parallel to the principal axis (P) converge. 3. Vertex (V) The vertex (V) is the geometric centre o the mirror’s surace.

Characteristic lengths of a spherical mirror Each o these three principal points defnes characteristic lengths o a spherical mirror (see Figure 16). C

P

F

V

1. Focal length (f ) The ocal length (f ) is the distance between the vertex (V) and the ocal point (F) o the spherical mirror.

f R

Figure 16 The characteristic lengths o a spherical mirror

2. Radius of curvature (R) The radius o curvature (R) is the distance between the centre o curvature (C) and the vertex (V). Since the spherical mirror derives rom a sphere, the radius o curvature is also equal to the distance between the centre o curvature and any other point on the surace o the mirror.

Concave mirrors of solar furnaces The concave mirror is a basic element in many solar urnaces The substance to be heated, usually water, passes through a cylinder located at the mirror’s ocal point The refective mirrors concentrate the solar energy at this ocal point, heating the water, which is then stored or used immediately, or instance, to heat a building Solar urnaces concentrate the Sun’s energy at a speciic point using a large concave mirror and thereore produce very high temperatures (several thousand degrees) The largest solar urnace is located in southwest France (see Figure 17) It is made up o many plane mirrors assembled to orm a concave mirror Smaller urnaces, such as those used in cooking, rely on this same principle

50

UNIT 1 Geometric Optics

Figure 17 The Odeillo solar urnace, in the Pyrénées-Orientales (France), is made up o approximately 20 000 mirrors

Geometric considerations Certain relationships can be deduced from the symmetry of spherical mirrors. 1. Normal

R

In a sphere, all radii are perpendicular to a tangent to the surface passing through a point of intersection between the radius and the surface of this sphere. This property is important because it means that the normal of a spherical mirror is simply a line that connects the centre of curvature (C) and the point of incidence of a light ray (see Figure 18).

P

f

V

R

R

2. Mathematical relationship between the radius of curvature (R) and the focal length (f ) When a light ray hits a concave spherical mirror, it is reflected according to the laws of reflection. The reflected ray intersects the principal axis (P) at a specific point called the focal point (F), which is separated from the vertex (V) by the focal length (f ). The mathematical relationship between the radius of curvature (R) and the focal length (f ) is:

C

Figure 18 The radii of curvature (R ) are normals to the surface of spherical mirrors.

R 2

However, this relationship is only valid for small angles.

Furthering

your understanding

Mathematical relationship between the radius of curvature (R ) and the focal length (f ) To determine this mathematical relationship, it is necessary to use geometric concepts (see Figure 19 ). According to the laws of reflection, the angle of incidence is equal to the angle of reflection: θ1  θ2. In addition, since angles θ 1 and θ 3 are alternate interior angles, they are equal: θ 1  θ 3. We can therefore deduce that θ1  θ2  θ3  θ. Triangle ACF is therefore an isosceles triangle because it has two equal angles. As a result, it has two equal sides: CF  AF. If the angles are small, which is true when the size of the mirror is smaller than the radius of curvature (R ), then AF  VF. It follows that CF  VF, which means that the focal point (F) is located at the halfway point of segment CV. Since VF is the focal length (f ) and CV is the radius of curvature (R ), we can write: R f 2

θ1

P

C

3

A θ2

F

V

f R

Figure 19 Rays that are parallel to the principal axis (P) of a spherical concave mirror are reflected toward the focal point (F) if the angles of incidence (θi) are small.

CHAPTER 2 Reflection of Light

51

Principal rays of a concave mirror P

C

F

V

The principal points are used to defne the principal rays. 1. First principal ray The irst principal ray is an incident ray that is parallel to the principal axis (P). When this ray strikes a concave mirror, it is reected toward the mirror’s ocal point (F) (see Figure 20). This is consistent with the defnition o the ocal point o a concave mirror.

Figure 20 First principal ray

2. Second principal ray P

C

F

V

The second principal ray is an incident ray that travels through the ocal point (F). When the ray strikes a concave mirror, it is reected parallel to the mirror’s principal axis (P) (see Figure 21). 3. Third principal ray

Figure 21 Second principal ray

P

C

F

V

Figure 22 Third principal ray

The third principal ray is an incident ray that passes through the centre o curvature (C). When the ray meets the concave mirror, it is reected back on itsel as in the case o a plane mirror where the angle o incidence (θi) is zero. Since the radius o curvature (R) is also a normal in the case o a concave mirror, this type o ray reects back on itsel (see Figure 22).

2.4.2

Convex mirrors

The surface of a convex mirror causes light rays that are parallel to its principal axis (P) to diverge (see Figure 23). For this reason, convex mirrors are also called diverging mirrors.

C

F

V

P

Figure 23 A convex mirror, also known as a diverging mirror, causes light rays that are parallel to its principal axis (P) to diverge

Like concave mirrors, convex mirrors have a centre o curvature (C), a ocal point (F) and a vertex (V). However, some dierences exist with respect to their principal rays: none o them passes through the principal points because these points are located behind the mirror. In diagrams, it is customary to use dotted lines to extend the reected rays behind the convex mirror.

52

UNIT 1 Geometric Optics

Principal rays of a convex mirror 1. First principal ray Any incident ray that is parallel to the principal axis (P) is relected in a direction originating rom the mirror’s ocal point (F) (see ray 1 in Figure 24). 2. Second principal ray Any incident ray directed toward the virtual ocal point will reect back parallel to the principal axis (P) (see ray 2 in Figure 24). 3. Third principal ray Any incident ray directed at the centre o curvature (C) is reected back on itsel (see ray 3 in Figure 24).

Ray 1

Ray 3 Ray 2 V

F

C

Figure 24 Principal rays o a convex mirror P

2.4.3

Spherical aberration

Spherical aberration is an optical defect. It occurs when the rays that are parallel to the principal axis (P) and that strike a concave spherical mirror at large angles of incidence (θi) are not focused at one common focal point (F) (see Figure 25).

Figure 25 Spherical aberration Spherical mirror

This phenomenon is caused by a specifc characteristic o spherical mirrors. Figure 25 shows that i the angles o incidence are too large, the rays will not be ocused at the same point. In this case, the distance AF is not equal to VF, and the ocal length (f ) is R not equal to hal o the radius o curvature (R) : f   2 Spherical aberration can be minimized using spherical mirrors whose size is smaller than their radius o curvature (R). A parabolic mirror is the ideal shape or avoiding spherical aberration (see Figure 26). With this shape, all incident rays that are parallel to the principal axis (P) are ocused at a single point, which is the ocal point (F) o the parabolic mirror. However, these mirrors are more difcult to manuacture and are more expensive as a result.

Parabolic mirror

P

Figure 26 Unlike a spherical mirror, a parabolic mirror does not produce spherical aberration

CHAPTER 2 Refection o Light

53

SECTION 2.4

Refection on spherical mirrors

1. Identiy points 1 to 5 in the ollowing diagram: 5

1

2

3

4

7. Examine the ollowing diagrams and in each case explain whether the refected ray passes through V, F, C or in a direction parallel to the principal axis

A.

B. C

F

V

C

F

V

C

F

V

2. Name the two types o mirrors ound on a spoon 3. In the ollowing diagram, a light ray is directed toward a concave mirror What is the value o the angle between the incident ray and the refected ray?

C.

D. C

F

V

8. Draw a geometric representation o the principal rays or a diverging mirror and illustrate the ollowing terms: a) centre o curvature b) ocal point c) vertex 4. A converging mirror refects the light emitted by a dis­ tant object, ocusing it at a point 20 cm rom the mirror What is the radius o curvature (R ) o the mirror? 5. Parabolic mirrors have an advantage over spherical mirrors with respect to ocusing incident rays that are parallel to the principal axis (P) What is this advantage? 6. A light ray, travelling parallel to the principal axis (P), strikes the surace o a concave spherical mirror At what distance rom the vertex (V) o the mirror will the refected ray cross the principal axis i the radius o curvature (R ) o the concave mirror is equal to 30 cm?

54

UNIT 1 Geometric Optics

9. Examine the ollowing diagrams and or each one state whether the projection o the refected ray passes through V, F, C or in a direction parallel to the principal axis A.

B. V

F

C

V

F

C

C.

V

F

C

V

F

C

D.

2.5 Images The concept o an image has a very specifc meaning in the feld o optics. An image is a representation of an illuminated object produced by a series of points caused by the convergence of light rays originating from various points on the object or from the extension of these rays. The points that make up an image are called image points, and the points that make up an illuminated object are called source points. All illuminated objects can be broken down into many source points, and a beam o light rays travels away rom the object rom each source point. The image is ormed by an optical system such as a mirror or lens, which deviates the rays travelling rom the source points and orms image points. In the case o a camera obscura, light rays pass through a tiny pinhole (see Figure 29 on page 56). The image o an illuminated object is made up o a series o image points. Thereore, the light emitted by an object conveys the image o this object.

2.5.1

Image characteristics

Images obtained rom optical systems can be characterized according to type, orientation, size and position. The characteristics o an image vary based on the optical system used.

Image type An image can be real or virtual. It is real if the optical system deviates (or directs) the light rays originating from a source point by converging them toward an image point (see Figure 27). A real image can be projected onto a screen. The image o a flm on a movie theatre screen is an example o a real image (see Figure 28a). An image is virtual if its image points appear to originate from a fictitious extension of light rays that are deviated by the optical system (see Figure 28). A virtual image cannot be projected onto a screen because it is not ormed by the intersection o real rays. The image we see when we look at ourselves in a plane mirror is an example o a virtual image (see Figure 28b).

A

A′

Source point

Image point

Optical system

Figure 27 Image point A’ is the real image o source point A The confguration o the deviated rays and the position o the real image depend on the optical system

Image orientation Depending on the optical system and position o the illuminated object, an image is either upright or inverted. An image is upright i it is oriented in the same direction as the object; it is inverted i it is rotated by 180° with respect to the object.

Image size An image is not necessarily the same size as the object. It can be smaller or larger than the object, depending on the optical system and the object’s position in relation to the system.

a) A real image

b) A virtual image

Figure 28 The nature o images CHAPTER 2 Reection o Light

55

*

Note It is important not to conuse magnifcation (M ) and magniying power (MP) Magnifcation is the ratio between the linear dimensions o the image and the object Magniying power is the ratio between the angles o the object and the image: θi MP 5 θo

Magnifcation* (M) is the ratio between the image height (hi) and the object height (ho).

M5

hi ho

where M 5 Magnifcation hi 5 Height o the image, expressed in metres (m) ho 5 Height o the object, expressed in metres (m) Since magnifcation (M) is a ratio between two dimensions having the same units, the ratio itsel has no units.

Image position The image o an object can be ormed either in ront o or behind the optical system. This depends on the optical system and on the object’s position in relation to the system.

2.5.2

Images formed y a camera oscura

The camera obscura has been around since antiquity. It is an opaque box with a tiny hole in its ront side. The opposite interior wall is covered by a screen. The tiny opening is called a pinhole. Since light travels in a straight line, the light rays travelling rom the object enter the camera obscura and orm an inverted image on the screen (see Figure 29). In addition, there is a speciic ratio between the height o the image and the height o the object (see Figure 30). This fgure helps to illustrate that: h d B′A′ OH′ ⇒ i 5 i 5 AB OH ho do This equation demonstrates that the image height (hi) is inversely proportional to the distance between the illuminated object and the pinhole (do), but that it is directly proportional to the depth o the camera obscura (di).

A b′

ho

H′ hi

A′

Pinhole

Oject Camera oscura

Image

Figure 29 The image produced by a camera obscura is real, inverted and usually smaller than the object

56

O

H

UNIT 1 Geometric Optics

b

do

di

Figure 30 The geometry o image ormation in a camera obscura

Based on the defnition o magnifcation (M) provided above, we have:

M5

hi ho

5

di do

Furthermore, since the image in Figure 30 on the previous page is created by the intersection o real rays, the image is a real image. Like all real images, it can be projected onto a screen (in this case, the screen o a camera obscura). A photograph o the object may be obtained by securing photographic flm to the screen. The frst cameras were based on this principle. Example A pinhole camera with a depth o 30 cm is used to take a photograph o a person standing 5 m away I the height o the image in the photograph is 10 cm, what is the magnifcation (M ) o the pinhole camera, and how tall is the person? Data: di 5 30 cm 5 030 m do 5 50 m hi 5 10 cm 5 010 m ho 5 ?

Solution: d M5 i do

M5

030 m 5 006 50 m

The magnifcation (M ) is 006 We can use this result to determine the size o the illuminated object: h h 010 m M 5 i 5 006 ho 5 i 5 5 17 m M ho 006 The person is 17 m tall

Table 1 summarizes the characteristics o an image ormed by a camera obscura. Table 1 Characteristics o an image ormed by a camera obscura Image characteristics

Object position

Type

Orientation

Size

Position

Any position

Real

Inverted

Depends on the ratio between the depth o the camera obscura and the distance between the object and the pinhole

Behind the pinhole

2.5.3

Images formed by plane mirrors

Images formed by a single plane mirror Images ormed by a plane mirror result rom the specular reection o light rays originating rom source points on the mirror’s surace. The images, which appear to emanate rom behind the mirror, are upright and are the same size as the object (see Figure 31). To determine the position and types o images obtained with plane mirrors, it is important to understand how these images are ormed (see Figure 32 on the next page). The image obtained using a plane mirror is ormed by the fctitious extension o reected light rays. It is thereore a virtual image. Figure 32 also shows that the distance between the object and the mirror (do) is equal to the distance between the image and the mirror (di).

Figure 31 This fgure shows an image on a plane mirror The man is shaving with his right hand but his image appears to be using the let hand This phenomenon is called lateral inversion

CHAPTER 2 Reection o Light

57

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

Oject

Oject

Image

Ee

Ee

a) Draw two rays perpendicular to the mirror rom both ends o the object

d) Draw two lines rom the ends o the image to the eye

Oject

Oject

Image A b

DAGUERREOTyPE The principle o the pinhole was discovered long beore HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY the process o capturing images on a support Since antiquity, people have known that piercing a hole in a box produces an inverted image on the side o the box opposite the hole In the 19th century, French physicist Nicéphore Niépce (1765–1833) exposed a pewter plate soaked with bitumen to the light entering a pinhole Ater several hours o exposure, he obtained the frst negative in the history o photography Following Niépce’s death, Louis Daguerre (1787–1851) improved the process by using a copper plate covered with a thin layer o silver, which reduced the exposure time This process, which he called daguerreotype, is considered the frst photographic process intended or commercial purposes

Ee

Ee

b) Extend both rays an equal distance behind the mirror (The lines behind the mirror are dotted lines) Oject

e) Identiy the points o intersection o both lines with the mirror (see points A and B in the fgure) These points o intersection are the points o incidence o the light rays that originate rom the object

Image

Oject

Image

f) A b

Ee

Ee

c) Draw the image: it is symmetrical to the object in relation to the plane o the mirror

) Draw the light rays originating rom the ends o the object These rays reect on the mirror at the points o incidence (A and B) and reach the eye along the lines drawn in d The angle o incidence (θi) is equal to the angle o reection (θr) or both rays

Figure 32 How the eye sees image ormation in a plane mirror

Table 2 summarizes the characteristics of an image formed on a plane mirror. Tale 2 Characteristics o an image ormed on a plane mirror Oject position Any position

Image characteristics Tpe

Orientation

Size

Position

Virtual

Upright

Equal to size o object

Behind the mirror, at the same distance rom the mirror as that o the object

Images formed  two mirrors When two mirrors are used, several images are formed. The number of images formed (N) depends on the angle (θ) between the two mirrors and can be determined according to the following rule: N5

( 360°θ ) 2 1

where N 5 Number o images ormed θ 5 Angle between the mirrors, expressed in degrees (°)

58

UNIT 1 Geometric Optics

Note that when the mirrors are parallel, the angle is zero. In this case, the number o images produced is infnite. As the denominator in the equation used to calculate the number o images ormed decreases, the raction increases. When the angle between the mirrors approaches zero, the number o images approaches infnity. Exaple How many images (N ) are ormed by two mirrors set up at a right angle to each other? Data: θ 5 90° N5?

Solution: N5

21542153 ( 360°θ ) 2 1 5 ( 360° 90° )

When two mirrors are set up at a right angle to each other, three images are ormed (I1, I2 and I3) The third image is produced by double reection I2

I1

mirror

mirro

2

r1

mirror 1 I3 I1 O

I3

I2 O

mirror 2 How the eye sees point I3

2.5.4

How the eye sees point I3

Iages fored by spherical irrors

Spherical mirrors reect light rays by causing them to converge or to diverge. Images ormed by concave (or converging) mirrors will be discussed separately rom images ormed by convex (or diverging) mirrors.

Geoetric construction of iages fored by concave irrors When representing images ormed by a concave mirror, use the properties o its three principal rays. To illustrate the method or the geometric construction o images, we will consider a situation where the object is located along the mirror’s principal axis (P), at a distance greater than the distance rom its centre o curvature (C) (see Figure 33). mirror C

F

V

mirror

P

Object

See Concave irrors, p 49

C

F

V

P

Object Iage

c) Draw the image

a) Draw the frst principal ray mirror C

F

Object

V

P

mirror C

F

V

Object

P

Iage

b) Draw the second principal ray

d) Draw the third principal ray to veriy the image

Figure 33 Formation o an image by a concave mirror when the object is located at a distance greater than the distance rom its centre o curvature (C)

CHAPTER 2 Reection o Light

59

Figure 33 shows that we can obtain the image with just two principal rays. We can use the third ray to check whether the image is in the right place. The image obtained in this case is real, inverted, smaller than the object and located between the ocal point (F) and the centre o curvature (C). I the object is moved along the principal axis (P), the image characteristics change in certain positions. Depending on the object’s position, ive other situations are possible (see Figure 34). The characteristics o the various images ormed by a concave mirror are summarized in Table 3. mirror

mirror C

F

V

P C

F

P V

a) Object at infnity d) Object at F

mirror F

C

mirror

V

P F

Iage

V

P

b) Object at C mirror C

F

V

P

e) Object between F and V

Iage

c) Object between C and F

Figure 34 Images ormed by concave mirrors The object’s position determines the image’s characteristics

Table 3 Characteristics o various images ormed by a concave mirror Object's position

Iage characteristics Type

Orientation

Size Point image

Position

At infnity

Real

Beyond C

Real

Inverted

Smaller than the object

Between F and C

At C

Real

Inverted

Identical to the object

At C

Between C and F

Real

Inverted

Larger than the object

Beyond C

At F Between F and V

At F

No image Virtual

Upright

Larger than the object

Behind the mirror, arther rom the mirror than the object is

To summarize, concave mirrors can orm fve kinds o images. They can be real or virtual, inverted or upright, enlarged or reduced, and with positions that vary according to the object’s position. Concave mirrors do not orm an image when the object is at the ocal point (F).

60

UNIT 1 Geometric Optics

Geoetric construction of iages fored by convex irrors See Convex irrors, p 52

When representing images ormed by a convex mirror, we use the properties o its three principal rays. Since the principal points o these mirrors are located on the other side o the refective surace, only one possibility exists (see Figure 35). Here as well, the image can be obtained with just two principal rays and the third principal ray can be used to veriy the results. Table 4 lists the properties o images ormed by a convex mirror. Table 4 Characteristics o images ormed by a convex mirror

a) Draw the frst principal ray (with its extension)

Iage characteristics

Object's position

Type

Orientation

Size

Position

Any position

Virtual

Upright

Smaller than the object

Between F and V (behind the mirror, closer to the mirror than the object is)

To summarize, convex mirrors can only orm one kind o image. These images are always virtual, upright, reduced and located behind the mirror, but closer to the mirror than the object is.

matheatical foralis of spherical irrors

V

Object

b) Draw the second principal ray

ho

Image hi

F

F

C

miroir mirror Iage V

Object C

C

c) Draw the image

Mirror Object

F

mirror

V

Object

In addition to geometric constructions, it is possible to establish mathematical relationships between the parameters that characterize spherical mirrors (see Figure 36).

mirror

F

C

V

d) Draw the third principal ray to veriy the image

mirror

f di

Iage Object

V

F

C

do

Figure 36 Parameters that characterize spherical mirrors: ocal length (f ), image distance (di) and object distance (do)

Figure 35 Formation o an image by a convex mirror

The mathematical relationship between ocal length (f ), image distance (di) and object distance (do) is: 1 1 1 5 1 f do di This ormula is called the mirror equation.

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61

When the object is located at ininity, that is, as d o approaches ininity 1 1 (do → ∞), the raction approaches zero ( → 0). do do In this case, f 5 di, which means that the image is located at the ocal point. This result was obtained in the frst diagram o Figure 34 on page 60. Magnifcation (M ) is also expressed as a unction o these parameters:

M5

hi d 5 i ho do

It is relatively easy to understand where these equations come rom. Think about the similarities o certain triangles observed in the ormation o images by a concave mirror (see Figure 37).

D

O ho

C

F

A

V

hi B

I f di do

Figure 37 Parameters in the ormation o images by a concave mirror

First, triangles DOF and BIF are similar We can thereore write: BV BI 5 DV DO Considering the parameters defned in Figure 37, we can deduce that: hi d 5 i ho do This equation is the magnifcation (M) equation that applies to camera obscuras and spherical mirrors Second, triangles AOF and BVF are similar as well We can thereore write: BV FV 5 AO AF Thereore: hi f 5 ho do 2 f Taking into account the magnifcation relationship determined above, we obtain: di f 5 do do 2 f By inverting this equation we obtain: do di

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UNIT 1 Geometric Optics

5

d do 2 f f 5 o 2 f f f

Next, divide both sides o the equation by do: do do 21 d di d d 1 1 1 1 1 1 1 1 f ⇒ o 3 5 o 21 3 ⇒ 5 o 3 2 13 ⇒ 5 2 5 do do d i do do di do d i f do do f f

(

)

(

) ( )

Thereore: 1 1 1 1 5 f do di Note that a variation o this equation exists We know that i we take into account certain approximations: R 2 where R 5 mirror’s radius o curvature f5

It ollows that: 2 1 1 5 1 R do di

Sign convention As mentioned at the beginning o this section, images obtained using optical systems can be characterized according to type, orientation, size and position. They can be real or virtual, upright or inverted, and enlarged or re duced. To take all o these possibilities into account, it is important to allocate signs (1 or –) to the parameters in the ormulas involving spherical mirrors. We use these signs so that the mathematical calculations will be consistent with the experimental results and geometric constructions o the images ormed by the mirrors. The allocation o these signs is a convention that provides specifc meaning to a negative or positive parameter. There are several sign conventions, all o which can be used as long as they are used consistently. The sign convention used throughout this textbook is the ollowing: Distances are measured rom the vertex (V) o spherical mirrors Focal lengths (f ) are positive or converging mirrors and negative or diverging mirrors Radii o curvature (R ) are positive or converging mirrors and negative or diverging mirrors The image distances (di) and object distances (do) are positive or real images and real objects The image distances (di) and object distances (do) are negative or virtual images and virtual objects Image heights (hi) and object heights (ho) are positive when they are upright and negative when they are inverted 7. To abide by this convention, magnifcation (M ) must be written: h d M5 i 5- i ho do

1. 2. 3. 4. 5. 6.

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63

In addition to providing image characteristics, this sign convention also helps to predict image position when applied to the mirror equation. Example A Concave mirror, object beyond C An object with a height o 4 cm is located 30 cm rom a converging mirror whose radius o curvature (R ) is 10 cm Determine: a) the image position (di) b) the magnifcation (M ) c) the image height (hi) Data:

C 110 cm Because the mirror is a converging mirror, f and C are positive: f 5 5 5 15 cm 2 2 • The object is real: d 5 130 cm o • The object is upright: h 5 14 cm o di 5 ? M5? hi 5 ? •

Solution: a) Calculation o image position: 1 1 1 1 1 1 1 5 1 ⇒ 5 2 ⇒ di 5 f do di di f do 1 1 2 f do Numerical application: di 5

1 1 1 2 f do

5

1 1 1 2 5 cm 30 cm

5

1 6 1 2 30 cm 30 cm

5

1 5 6 cm 5 30 cm

The image distance is positive, which means that the image is real The numerical value tells us that the image is located between F and C b) Calculation o magnifcation: h d 6 cm 1 M5 i 5- i 55 - 5 - 02 ho do 30 cm 5 The magnifcation is negative; the image is inverted The absolute value o the magnifcation is less than 1; the image is smaller than the object The numerical value shows that the image is fve times smaller than the object c) Calculation o image height: h M 5 i ⇒ hi 5 M 3 ho ho Numerical application: hi 5 M 3 ho 5 - 02 3 4 cm 5 - 08 cm The image height is negative; the image is inverted (this result was obtained in the previous question) These results are consistent with those obtained using the geometric construction in Figure 33 on page 59 and are summarized in Table 3 on page 60

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UNIT 1 Geometric Optics

Example b Concave mirror, oject at C An object with a height o 4 cm is located 10 cm rom a converging mirror whose radius o curvature (R ) is 10 cm Determine: a) the image position (di) b) the magnifcation (M ) c) the image height (hi) Data: C 110 cm 5 5 15 cm 2 2 • d 5 110 cm o • h 5 14 cm o di 5 ? M5? hi 5 ? •

f5

Solution: a) Calculation o image position: di 5

1 1 1 2 f do

5

1 1 1 2 5 cm 10 cm

5

1 2 1 2 10 cm 10 cm

5

1 5 10 cm 1 10 cm

The image distance is positive, which means that the image is real The numerical value indicates that the image is located at C b) Calculation o magnifcation: h d 10 cm M5 i 5- i 55 -1 ho do 10 cm The magnifcation is negative; the image is inverted The absolute value o the magnifcation is equal to 1; the image is the same size as the object c) Calculation o image height: hi 5 M 3 ho 5 - 1 3 4 cm 5 - 4 cm The image height is negative; the image is inverted (this result was obtained in the previous question) These results are consistent with those obtained using the geometric construction in Figure 34b and are summarized in Table 3 on page 60

Example C Concave mirror, oject etween C and F An object with a height o 4 cm is located 6 cm rom a converging mirror whose radius o curvature (R ) is 10 cm Determine: a) the image position (di) b) the magnifcation (M ) c) the image height (hi) Data: •

f5

C 110 cm 5 5 15 cm 2 2 CHAPTER 2 Reection o Light

65

Example C (cont.) • d 5 16 cm o • h 5 14 cm o di 5 ? M5? hi 5 ? Solution: a) Calculation o image position: di 5

1 1 1 2 f do

5

1 1 1 2 5 cm 6 cm

5

1 6 5 2 30 cm 30 cm

5

1 5 30 cm 1 30 cm

The image distance is positive, which means that the image is real The numerical value indicates that the image is located beyond C b) Calculation o magnifcation: h d 30 cm M5 i 5- i 55 -5 ho do 6 cm The magnifcation is negative; the image is inverted The absolute value o the magnifcation is greater than 1; the image is larger than the object The numerical value shows that the image is fve times larger than the object c) Calculation o image height: hi 5 M 3 ho 5 - 5 3 4 cm 5 - 20 cm The image height is negative; the image is inverted (this result was obtained in the previous question) These results are consistent with those obtained using the geometric construction in Figure 34c and are summarized in Table 3 on page 60

Example D Concave mirror, object at F An object with a height o 4 cm is located 5 cm rom a converging mirror whose radius o curvature (R ) is 10 cm Determine: a) the image position (di) b) the magnifcation (M ) c) the image height (hi) Data: C 110 cm •f5 5 5 15 cm 2 2 • The object is real: d 5 15 cm o • The object is upright: h 5 14 cm o di 5 ? M5? hi 5 ?

Solution: a) Calculation o image position: di 5

1 1 1 2 f do

5

1 1 1 2 5 cm 5 cm

5

1 0

The equation is indeterminate No image is ormed, which is consistent with Figure 34d and summarized in Table 3 on page 60 b) Does not apply because there is no image c) Does not apply because there is no image

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UNIT 1 Geometric Optics

Example E Concave mirror, object between F and S An object with a height o 4 cm is located 10 cm rom a converging mirror whose radius o curvature (R ) is 10 cm Determine: a) the image position (di) b) the magnifcation (M ) c) the image height (hi) Data: C 110 cm 5 5 15 cm 2 2 • The object is real: d 5 14 cm o • The object is upright: h 5 14 cm o di 5 ? M5? hi 5 ? •

f5

Solution: a) Calculation o image position: di 5

1 1 1 1 5 5 5 5 - 20 cm 1 1 1 1 4 5 -1 2 2 2 f do 5 cm 4 cm 20 cm 20 cm 20 cm

The image distance is negative, which means that the image is virtual The numerical value indicates that the image is located beyond C b) Calculation o magnifcation: h d - 20 cm M5 i 5- i 55 15 4 cm ho do The magnifcation is positive; the image is upright The absolute value o the magnifcation is greater than 1; the image is larger than the object The numerical value shows that the image is fve times larger than the object c) Calculation o image height: hi 5 M 3 ho 5 5 3 4 cm 5 20 cm The image height is positive; the image is upright (this result was obtained in the previous question) These results are consistent with those obtained using the geometric construction in Figure 34e and are summarized in Table 3 on page 60

Example F Convex mirror An object with a height o 4 cm is located 5 cm rom a diverging mirror whose radius o curvature (R ) is 10 cm Determine: a) the image position (di) b) the magnifcation (M ) c) the image height (hi) Data: • Because

the mirror is a diverging mirror, f and C are negative: f 5

C - 10 cm 5 5 - 5 cm 2 2 CHAPTER 2 Reection o Light

67

Example F (cont.) • The object is real: d 5 15 cm o • The object is upright: h 5 14 cm o di 5 ? M5? hi 5 ? Solution: a) Calculation o image position: di 5

1 1 1 5 5 5 - 25 cm 1 1 1 1 -2 2 2 f do - 5 cm 5 cm 5 cm

The image distance is negative, which means that the image is virtual The numerical value indicates that the image is located between V and F (the image is closer to the mirror than the object is) b) Calculation o magnifcation: h d - 25 cm 1 M5 i 5- i 55 1 5 105 5 cm ho do 2 The magnifcation is positive; the image is upright The absolute value o the magnifcation is less than 1; the image is smaller than the object The numerical value shows that the image is hal the size o the object c) Calculation o image height: hi 5 M 3 ho 5 05 3 4 cm 5 2 cm The image height is positive; the image is upright (this result was obtained in the previous question) These results are consistent with those obtained using the geometric construction in Figure 35 and as summarized in Table 4 on page 61

Furthering

your understanding

Another sign convention Problems involving image ormation can also be solved correctly using other sign conventions The sign convention below is commonly used 1. All negative values are associated with virtual images, and all positive values are associated with real images 2. For concave mirrors, di, hi and M are negative i the image is virtual They are positive i the image is real 3. For convex mirrors, di, hi and M are always negative 4. The ocal length (f ) and radius o curvature (R ) o a convex mirror are negative They are positive or a concave mirror 5. Magnifcation is written as ollows: M5

hi ho

5

di do (continued on the following page)

68

UNIT 1 Geometric Optics

Another sign convention (cont.) With this convention, the image is upright when the magnification (M) is negative and vice versa. The following example applies this convention. Example Concave mirror, object between F and V (Compare with Example E on page 67) An object with a height of 4 cm is located 4 cm from a converging mirror whose radius of curvature (R ) is 10 cm. Determine: a) the image position (di) b) the magnification (M ) c) the image height (hi) Data: •

Because the mirror is a converging mirror, f and C are positive: f 

C 10 cm   5 cm 2 2

The object is real: do  4 cm The object is upright: ho  4 cm di  ? M? hi  ?

• •

Solution: a) Calculation of image position: 1 1 1 1 1 1 1   ⇒   ⇒ di  f do di di f do 1 1  f do Numerical application: di 

1 1 1 1     - 20 cm 1 1 1 1 4 5 -1    f do 5 cm 4 cm 20 cm 20 cm 20 cm

The image distance is negative, which means that the image is virtual. The numerical value indicates that the image is located beyond C. b) Calculation of magnification: h d - 20 cm M i  i   -5 ho do 4 cm The magnification is negative; the image is upright. The absolute value of the magnification is greater than 1; the image is larger than the object. The numerical value shows that the image is five times larger than the object. c) Calculation of image height: h M  i ⇒ hi  M  ho ho Numerical application: hi  M  ho  - 5  4 cm  - 20 cm The image height is negative; the image is virtual (this result was obtained in the first question). These results are consistent with those obtained with the sign convention used in the textbook. . CHAPTER 2 Reflection of Light

69

SECTION 2.5

Images

1. Calculate the height o a building located 50 m rom a camera obscura The depth o the camera obscura is 20 cm, and the height o the image in the camera obscura is 40 cm

b) Using a ray diagram, show the position o the mirror’s ocal point (F) in each case o

2. An 30­m tree orms an inverted image with a height o 15 cm on the screen o a pinhole camera What is the magnifcation? 3. When we look at ourselves in a plane mirror, we see an image Describe the image characteristics (type, orientation and height) 4. The hands o a clock with no numbers, when seen on a plane mirror at various times, appear to indicate the ollowing times: 8 am, 8:30 am and 6 pm What is the actual time in each case?

i

o

i

5. The ollowing photo shows an ambulance with the word “ambulance” written backwards Which image characteristic comes into play here?

8. An outside diverging rear­view mirror on a truck has a ocal length (f ) o - 60 cm A truck driver sees an image in this mirror o a car passing the truck I the height o the car is 15 m and it is located at a distance o 6 m, what are the image height and image position? 9. A truck driver sees her ace in her rear­view mirror I the mirror’s ocal length (f ) is 50 cm and its magni­ fcation (M ) is 010, what are the image distance and object distance?

6. How many images (N ) are ormed by two mirrors i the angle between the mirrors is 60°? 7. The ollowing diagrams represent an object (o) and its image (i) ormed by a converging mirror and a diverging mirror a) Copy both diagrams on a sheet o paper

10. A dentist holds a converging mirror with a ocal length (f ) o 20 mm at a distance o 15 mm rom a tooth What is the magnifcation (M ) o the image o the tooth flling? 11. A person is standing 2 m rom a distorting mirror and observes that her image is three times larger than she is What is the mirror’s radius o curvature (R )? 12. A child looks at his reection in a spherical Christmas ornament with a diameter o 8 cm and sees that his image is reduced by hal What is the distance between the ornament and the child’s ace?

70

UNIT 1 Geometric Optics

APPLICATIONS blu-ray optical disc Optical discs are circular media used to store data. The most common are 12 cm in diameter and 1.2 mm thick. They all work based on the same principle: the reection o light rom a laser on a thin metallic layer. This layer has surace irregularities made up o microscopic pits o dierent sizes (see Figure 38). Light rom the laser, which reects dierently when it meets a pit, is directed toward an optoelectronic device1 that detects the variation in the reected light. The two most common types o optical discs are the CD (compact disc) and the DVD (digital versatile disc). The dierence between a CD and a DVD lies in the size o the pits. These pits are much smaller and closer together in a DVD, which thereore allows it to store much more inormation. A CD can hold approximately 700 MB (megabytes)2 o inormation, whereas a DVD can hold 4.7 GB (gigabytes). Thereore, the storage capacity o a DVD is seven times greater than that o a CD. Also, or both the CD and the DVD, the light rom the laser is either red (wavelength o 635 to 650 nm) or inrared (wavelength o 780 nm).

Blu-ray discs are a response to this new requirement. Although it is based on the same principle as the CD and DVD, Blu-ray technology does present signifcant dierences (see Figure 39). First, the name Bluray comes rom the blue-violet light (wavelength o 405 nm) emitted by the laser, which has a smaller wavelength than a red laser. In addition, with a radius o 290 nm, these lasers are much smaller than CD or DVD lasers, and the size o the pits is even smaller than in DVDs. All o the eatures o the Blu-ray disc combine to provide a much larger storage capacity than even the best DVD. Blu-ray discs can actually store up to 25 GB (gigabytes) o inormation. DVD (4.7 GB)

Blu-ray disc (25 GB)

Laser beam

Wavelength of laser (650 nm) Wavelength of laser (405 nm) 1.2 mm

1.2 mm

Metallic layer Recording surface

Figure 39 Features o a DVD and a Blu-ray disc

a) Pits in a CD

b) Pits in a DVD

Figure 38 The pits o a DVD are much smaller than those o a CD

With the advent o high defnition in audiovisual technology, the quantity o inormation required to guarantee image and sound quality has increased, and the available media are no longer suicient.

The storage capacities o these new media can be improved even urther. Scientists are currently developing a multi-layer Blu-ray disc that stores more than one TB (terabyte) o inormation.

1 Optoelectric device: Electronic apparatus that emits, detects or modulates light 2 Byte: Unit o measure o the quantity o inormation 1 megabyte (MB) 5 106 octets 1 gigabyte (GB) 5 109 octets 1 terabyte (TB) 5 1012 octets

CHAPTER 2 Refection o Light

71

mirrors Mirror, mirror on the wall. . . For thousands o years, scientists and ordinary people alike have been using mirrors on a daily basis. Ancient civilizations made mirrors by polishing metal suraces such as brass, bronze and silver. In the 14th century, manuacturing techniques improved signifcantly: a glass surace was covered with tin and lead (silvering). Later, these materials were replaced by aluminum, which was used or its reective properties and its resistance to oxidation. The aluminum was coated with a layer o copper or lead.

Mirrors are used in several scientifc instruments, including periscopes, microscopes and telescopes (see Figure 40). Mirrors can even be made rom liquids. Isaac Newton came up with this idea in the 17th century, but it was only technically possible by the end o the 20th century. Rotating mercury produces a perectly parabolic mirror (see Figure 41). One o the frst large-scale liquid mercury mirrors was built at Université Laval in Québec City. This type o mirror is used today in some o NASA’s telescopes. Each type o mirror has specifc uses. A concave mirror concentrates light; this is why it is used in ashlights and telescopes. Convex mirrors expand the feld o view and are used to manuacture rearview mirrors. Solar urnaces are one o the most recent and original applications o mirrors. The Sun’s rays are concentrated by several reective suraces, allowing solar energy to be ocused on a target. The giant solar urnace built in Odeillo, France in the 1960s, is a research centre or energy storage and production.

Figure 40 The telescope developed by Isaac Newton had two mirrors

Mirrors used in astronomy and optics are manuactured by a dierent process. A coat o metal (aluminum or gold) is applied to prevent light rom penetrating the glass, which could alter the curve by the phenomenon o reraction or dispersion. This way, the light rays come into direct contact with the reective element (in this case, aluminum) and do not pass through any distorting media.

72

UNIT 1 Geometric Optics

Figure 41 The liquid mirror of a telescope consists of mercury

chapter

2

rfion o Lig

2.1 Types o refection • The two types o reection are: – specular reection – diuse reection

2.2 Geometry o refection • The incident ray is the light ray that travels toward the reective surace. • The normal is an imaginary line that is perpendicular to the reective surace and originates rom the point o incidence. • The angle o incidence (θi) is the angle ormed by the incident ray and the normal. • The reected ray is the light ray that travels away rom the reective surace. • The angle o reection (θr) is the angle ormed by the reected ray and the normal.

2.3 Refection on a plane mirror: laws o refection • The frst law o reection states that the incident ray, the reected ray and the normal are all located in the same plane. • The second law o reection states that the angle o incidence is equal to the angle o reection (θi 5 θr ).

2.4 Refection on spherical mirrors • • • •

A spherical mirror is a spherical cap cut rom a reective hollow sphere. A mirror is concave i the hollow side o the spherical cap is used as the mirror. A mirror is convex i the domed side o the spherical cap is used as the mirror. A concave mirror, also called a converging mirror, converges the light rays that are parallel to its principal axis (P) toward a point located inside its curvature, reerred to as the ocal point (F). The principal axis (P) is the axis o symmetry o the spherical mirror. • A convex mirror, also called a diverging mirror, causes light rays that are parallel to its principal axis (P) to diverge. The principal points of a spherical mirror • The centre o curvature (C) is the point at the centre o the sphere rom which the spherical mirror originates. • The ocal point (F) is the point at which the light rays parallel to the principal axis (P) converge. • The vertex (V) is the geometric centre o the mirror’s surace. • The three principal points o a spherical mirror are located on the mirror’s principal axis (P).

CHAPTER 2 Refection o Light

73

(cont.) Characteristic lengths of a spherical mirror • The ocal length (f ) is the distance between the vertex (V) and the ocal point (F). • The radius o curvature (R) is the distance between the centre o curvature (C) and the vertex (V). Since the spherical mirror derives rom a sphere, the radius o curvature is also equal to the distance between the centre o curvature and any other point on the surace o the mirror. Principal rays of a concave mirror • Any incident ray that is parallel to the principal axis (P) o a concave mirror is reected toward the ocal point (F) o this mirror. • Any incident ray that passes through the ocal point (F) o a concave mirror is reected parallel to the mirror's principal axis (P) o the mirror. • Any incident ray passing through the centre o curvature (C) o a concave mirror is reected back on itsel. Principal rays of a convex mirror • Any incident ray that is parallel to the principal axis (P) o a convex mirror is reected in a direction originating rom the mirror’s ocal point (F). • Any incident ray directed toward the virtual ocal point o a convex mirror is reected parallel to the mirror's principal axis (P). • Any incident ray directed at the centre o curvature (C) o a convex mirror is reected back on itsel. Spherical aberration • For small angles o incidence (θi), that is, when the size o the mirror is smaller than its radius o curvature (R), the ocal point (F) is located halway between the vertex (V) and the centre o curvature (C). I this condition is not met, the light rays will not be ocused at a single point, and the spherical mirror will cause a spherical aberration.

2.5 Images Image characteristics • An image is a representation o an illuminated object produced by a series o points caused by the convergence o light rays originating rom various points on the object or rom the extension o these rays. • An image can be real or virtual. • An image is real i the optical system deviates (or directs) the light rays originating rom a source point by converging them toward an image point. • A real image can be projected onto a screen. • An image is virtual i its image points appear to originate rom a fctitious extension o light rays that are deviated by the optical system. • A virtual image cannot be projected onto a screen. • An image is either upright or inverted. An image is upright i it is oriented in the same direction as the object; it is inverted i it is rotated by 180° with respect to the object.

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(cont.) Images formed by a camera obscura • A camera obscura is an opaque box with a tiny hole in its ront side. The opposite interior wall is covered by a screen. • Images ormed by camera obscuras are real, inverted and are usually smaller than the object. • Magnifcation (M ) is the ratio between the image height (hi) and the object height (ho). In a camera obscura, the image height (hi) is inversely proportional to the distance between the illuminated object and the pinhole (do), but it is directly proportional to the depth o the camera obscura (di). M5

hi d 5 i ho do

Images formed by plane mirrors • Images ormed by plane mirrors are upright and virtual. • In a plane mirror, the distance between the object and the mirror (do) is equal to the distance between the image and the mirror (di). • The number o images ormed (N ) by two mirrors depends on the angle (θ) between the two mirrors and can be determined according to the ollowing rule:

N5

( 360θ ˚ ) 2 1

Images formed by spherical mirrors • When representing images ormed by spherical mirrors, we use the properties o their three principal rays. • Concave mirrors can orm fve kinds o images. They can be real or virtual, inverted or upright, enlarged or reduced, and with positions that vary according to the object’s position. Concave mirrors do not orm an image when the object is placed at the ocal point (F). • Convex mirrors can only orm one kind o image. These images are always virtual, upright, reduced and located behind the mirror. • The mirror equation establishes the relationship between ocal length (f ), image distance (di) and object distance (do). The mirror equation is: 1 1 1 5 1 f do di • To solve mathematical problems relating to the ormation o images by spherical mirrors, it is extremely important to use a sign convention (1 or - ). • According to the sign convention used in this textbook, magnifcation (M) is written as ollows:

M5

hi d 5- i ho do

• We use these signs so that the mathematical calculations will be consistent with the experimental results and geometric constructions o the images ormed by the mirrors. CHAPTER 2 Refection o Light

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CHAPTER 2

Refection o Light

1. Examine the ollowing diagram and explain the mathematical relationship between the angle o inci­ dence (θi) and the angle ormed by the reected ray and the plane mirror

θi

θr

V F

2. The distance between the vertex (V) and the ocal point (F) o a spherical mirror is called the ocal length (f ) a) What happens to the ocal length o a conver­ ging mirror when its centre o curvature (C) is increased, that is, when the mirror is less curved? b) At its maximum, what is the ocal length? c) What type o mirror would you have in this case? 3. The ollowing diagram illustrates the trajectory o a ray reected by two mirrors a) What types o mirrors are shown in the diagram? b) Will the ray that is reected by mirror M2 pass through point A, B, C or D? M1 M2

C

F

V

A B C D

4. The optical pathway shown in the fgure below contains an error Explain the error M3

V

F

C

M1

M2

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UNIT 1 Geometric Optics

5. In the ollowing diagram, which mirror (M1, M2 or M3) will the ray strike ater it is reected by mirrors M4 and M5?

M5 C M1 V M2 M3 F M4 C

6. Explain why interior decorators sometimes recommend hanging mirrors in small rooms 7. A person is standing ar away rom a concave mir­ ror Describe what happens to the image as the person walks closer to the mirror (type, orientation and size) 8. A person holds the concave side o a spoon on his nose and slowly brings the spoon away rom his ace He notices that his image disappears at a certain distance What is the signifcance o this distance? 9. A polished metal sphere is used as a diverging mirror (f 5 - 20 cm) above a birdbath A bird that is 25 cm in height and located at a distance o 50 cm rom the mirror looks directly at it What is the height and position o the bird’s image?

rcion o Lig

W

hen light passes from one transparent medium to another of a different nature, the direction in which it travels changes. This phenomenon, which has been known since antiquity, is called refraction. In this chapter, you will discover the properties of transparent media that affect the behaviour of light. You will study the geometry of refraction and the laws that govern it. This will help you to understand phe­ nomena involving refraction. Finally, you will learn how the refraction of light is put to use in a techno­ logical application.

rviw Refraction of light                                          9

3.1 3.2 3.3 3.4 3.5

pnomnon o cion               78 Indx o cion                       79 Gomy o cion                   82 Lws o cion                        84 tol innl fcion                  87 Chapter 3 Refraction of Light

77

3.1 phenomenon o eacion See refecion o ligh, p 41

A light ray usually travels in a straight line but can be deviated i the ray strikes a refective surace. This phenomenon occurs due to the refection o light. The trajectory o a light ray can also be deviated when the ray moves rom one transparent medium to another with dierent optical properties.

See Ligh oagaion in media, p 34

This phenomenon, which is called reraction, is responsible or many com­ mon optical illusions: a straw that looks broken in a glass o water (see Figure 1), a pool that appears shallower than it really is, or desert mirages in the scorching heat.

The surface that separates * Ineace two transparent, homogeneous media

Reraction reers to the change in direction in which light travels when it meets the boundary between two media that have dierent optical properties. The surace that separates the two media is called the interace*.

possessing different optical properties

Figure 1 A straw in a glass of water looks broken This optical illusion is caused by the refraction of light

Miages Mirages are not limited to deserts They can also occur above a hot, uniform surface, like these “puddles” of water on a highway in the summer that disappear as you approach them (see Figure 2 ) The increase in temperature near the ground alters the air’s index of refraction As the light from the sky approaches the ground’s sur­ face, it is refracted by a series of air layers of increasing tempe­ rature Each layer changes the direction in which the light travels This phenomenon creates an image that resembles water

Figure 2 A mirage on a road on a hot day

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3.2 Indx of fcion The speed o light in a vacuum (c) is a physical constant whose value has been established by the International Bureau of Weights and Measures. It is precisely equal to 299 792 458 m/s. However, or practical purposes, it is customary to use the rounded­o value o c ≈ 3 3 10 8 m/s. It is important to remember that this number applies only in a vacuum (it also represents the maximum speed by which a material particle can travel). When light pen­ etrates a transparent medium other than a vacuum, its speed decreases as a result o the medium’s optical properties. The property o a medium to reract light is called refractivity. Thereore, the more reractive a medium, the slower the speed o the light travelling through it. It is possible to quantify a medium’s refractivity by calculating the index of refraction (n). This is the ratio between the speed of light in a vacuum (c) and the speed of light in a transparent medium (v).

n5 c v

tbl 1 Approximate indices of refraction (n) for various media1

This defnition implies that:

Mdium

• the index o reraction (n) is always greater than 1, except in a vacuum, where it is exactly equal to 1 because c 5 1 c • the index o reraction (n) is a number without any units because the units o c and v are the same (m/s) and thereore cancel out in the ratio

Indx of fcion (n)

Vacuum

1000 000

Air

1000 29

Ice

131

Water

133

exml Calculate the index of refraction (n) for diamond if light travels through it at a speed of 124 3 108 m/s

Ethyl alcohol

136

Quartz (melted)

146

Turpentine

147

Data:

Calculation:

Glass

150

c 5 300 3 10 8 m/s v 5 124 3 10 8 m/s n 5?

c v 300 3 10 8 m/s n5 124 3 10 8 m/s n 5 242

Plexiglas

151

Crown glass

152

Polystyrene

159

Carbon disulphide (CS2)

163

Flint glass

166

Zircon

192

Diamond

242

Gallium phosphide

350

n5

The index o reraction characteristic o light passing rom a vacuum to an­ other transparent medium is reerred to as the absolute index o reraction (n) or that medium. The index o reraction can be considered a meas urement o the actor by which the speed o light (c) is reduced when it travels rom a vacuum to a reractive medium. The higher the index o reraction o a transparent medium, the more it will reduce the speed o the light travelling through it compared to its speed in a vacuum. Thereore, glass (n 5 1.50) slows light down more than water (n 5 1.33) does. Table 1 lists the indices o reraction or various transparent media.

1 These indices of refraction (n) are measured using a yellow light with a wavelength of 589 nm Values may vary based on the surrounding physical conditions

Chapter 3 Refraction of Light

79

Any reference to an index of * No refraction implies the “absolute index of refraction”

The values listed in Table 1 on the previous page indicate that the index o reraction* (n) o air is very close to that o a vacuum. For this reason, n 5 1 is oten used as the index o reraction o air. The error in this approximation is less than 0.03%, which is practically negligible. When the index o reraction reers to light passing between two transpar­ ent media other than a vacuum, the relative index o reraction (n 1 → 2) can be defned as the ratio between the index o reraction o the second medium (containing the reracted ray) and the index o reraction o the frst medium (con­ taining the incident ray):

n1 → 2 5

how a rainbow is formd You can see a rainbow if it is rain­ ing (or if there are suspended water droplets) in front of you and the Sun is behind you In addition, your eyes, the centre of the rain­ bow and the Sun’s rays must be roughly aligned The raindrops behave like minia­ ture prisms Light from the Sun enters the water droplets Each of the colours that make up white light exits and is refracted differently Red, which has the longest wavelength, is refracted the least, whereas violet, which has the shortest wavelength, is refracted the most (see Figure 4 on the following page) Since all droplets behave in the same way, a uniform spectrum of colours is formed (see Figure 3)

n2 n1

exampl A light ray passes from water to glass, then travels along the same trajectory in the opposite direction What are the relative indices of refraction? 1. From water to glass Data: n1 5 133 n2 5 150

Calculation: nglass n1 → 2 5 nwater → glass 5 nwater nwater → glass 5

150 133

nwater → glass 5 113 The light ray travels from a less refractive medium to a more refractive medium, therefore nwater → glass > 1 2. From water to glass Data: n1 5 150 n2 5 133

Calculation: nwater n1 → 2 5 n glass → water 5 nglass nglass → water 5

133 150

nglass → water 5 089 The light ray travels from a more refractive medium to a less refractive medium, therefore nglass → water < 1

Figure 3 A rainbow above the Grand Canyon in the United States

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It is important to bear in mind that the values provided in Table 1 are com­ monly used or all wavelengths o the visible spectrum. Since the variation o the indices o reraction (n) as a unction o wavelength is not very sig­ nifcant in this area (less than 2% or most o the commonly­used media), the error is negligible.

Furthering

your understanding

rlionsi bwn  indx of fcion (n) nd  wvlng (l) of lig The note below Table 1, which lists the indices of Furthering your understanding Furthering refraction (n) of various transparent media, explains that the measurements were taken with yellow light, whose wavelength is equal to 589 nm Measurements of indices of refraction are always taken with mono­ chromatic light 1 because the index of refraction depends on the wavelength (l) of the light used This dependence is described by Cauchy’s law, which is valid for the visible spectrum

white light, which is polychromatic2, into numerous rays your understanding of different colours; that is, the colours of the visible light spectrum (see Figure 4) This decomposition phe­ nomenon is called "dispersion" The same phenomenon explains how rainbows are formed

Scn Wi lig Visibl scum

n (l) 5 A 1

B l²

The positive constants A and B depend on the physical characteristics of the refractive medium

pism

rd Ong Yllow Gn Blu Viol

This law holds that n (l) increases as l decreases Therefore, a given medium will be more refractive for violet light (l ≈ 400 nm) than red light (l ≈ 700 nm), which means that violet light will deviate more than will red light This explains why a prism can break down 1 Composed of a single colour (and therefore having a single wavelength) 2 Composed of several monochromatic lights

SeCtION 3.2

Figure 4 White light is dispersed by the prism Since the index of refraction n (l) increases as l decreases, the colour red is deviated less than violet light

Indx of fcion

1. What is the index of refraction (n) of a liquid in which light travels at a speed of 250 3 108 m/s? 2. Zircon is often used in costume jewellery to imitate diamonds The index of refraction (n) of zircon is 192 At what speed does light travel in this transparent medium? 3. Calculate the decrease in the speed of light as it passes from air to zircon 4. What is the speed of light when it travels through glass?

5. Consider a light ray that moves from air to quartz How much time will it take to cross through a piece of quartz that is 100 m thick? 6. In which case can a relative index of refraction be less than 1? 7. A light ray travels from medium A to medium B The indices of refraction (n) of the two media are 150 and 136, respectively Which medium is less refractive? Explain your answer

Chapter 3 Refraction of Light

81

For the sake of clarity, * Convenion index 1 refers to the incident ray, and index 2 refers to the index of the refracted ray

3.3 Geomey of efcion Refection takes place entirely in a single medium. In the case o reraction, however, the incident ray and the reracted ray move in two optically dier­ ent media*.

3.3.1

Inciden y

The incident ray is the light ray that travels toward the interface (see Figure 5).

3.3.2

Noml

As in the case o refection, the normal is an imaginary line and is oten repre­ sented by a dotted line. It is perpendicular to the interace, crosses the point o incidence, and travels through the two media. The plane that contains the inci­ dent ray and the normal is called the plane o incidence (see Figure 6).

3.3.3

angle of incidence (θi)

The angle formed by the incident ray and the normal is called the angle of incidence and is represented by the symbol θi (see Figure 7).

Normal Plane of incidence

Incident ray

Angle of incidence (θi)

Interface Point of incidence

Mediu

Mediu m1 Mediu m2

m1

Medi um 2

n2

n1

Figure 5 The incident ray travels toward

n1 n2

Figure 6 The normal is an imaginary line that

the interface separating two media with different refractivities

is perpendicular to the interface and originates from the point of incidence The plane of incidence is the plane containing the incident ray and the normal

3.3.4

Mediu m1 Mediu m2

n1 n2

Figure 7 The angle of incidence (θi) is formed by the incident ray and the normal

refced y nd ngle of efcion (θr)

The refracted ray is the light ray that crosses the interface and enters medium 2. The angle of refraction (θR) is the angle formed by the refracted ray and the normal. Two separate situations may occur, depending on whether medium 2 is more or less reractive than medium 1, in other words, depending on whether n1 is greater or less than n2.

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UNIt 1 Geometric Optics

Situation 1: n1 < n2 In this situation, medium 2 is more reractive than medium 1. This means that light travels more slowly in medium 2 than in medium 1, and the reracted ray bends toward the normal (see Figure 8). The angle o reraction (θR) is thereore smaller than the angle o incidence (θi). Situation 2: n2 < n1 In this situation, medium 1 is more reractive than medium 2. This means that light travels more slowly in medium 1 than in medium 2. Here, con­ trary to the frst situation, the reracted ray bends away rom the normal (see Figure 9). The angle o reraction (θR ) is thereore greater than the angle o incidence (θi).

θi

θi

n1 n2

n2

Angle of refraction (θR)

Refracted ray

Figure 8 When n1 < n2, the angle of refrac­ tion (θR) is smaller than the angle of incidence (θi): the refracted ray bends toward the normal

SeCtION 3.3 1. Consider the following diagram:

n1

Refracted ray Angle of refraction (θR)

Figure 9 When n2 < n1, the angle of refraction (θR) is greater than the angle of incidence (θi): the refracted ray bends away from the normal

Gomy of fcion a) Which medium has the higher index of refraction (n) ? b) Which medium is more refractive? 2. A light ray travels from glass to water How does the refracted ray behave? 3. An interface has a relative index of refraction that is less than 1 How will a light ray crossing it behave? In other words, will it bend away from or toward the normal? Explain your answer

Chapter 3 Refraction of Light

83

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

3.4 Laws of rfracion Reraction is a phenomenon that has ascinated scholars throughout history. Although the identity o the person who discovered the laws o reraction has not been ofcially established, they are called “Snell’s laws” ater the Dutch scientist Willebrord Snell.

3.4.1

Firs law of rfracion

The frst law o reraction holds that the incident ray and the reracted ray are located on opposite sides o the normal originating rom the point o incidence, and that all three are in the same plane. SheN KUO Chinese scientist (1031–1095) Astronomer, mathematician, geologist, climatologist, inventor and poet, Shen Kuo was a Chinese scientist who lived during the Song HISTORY HIGHLIGHTS HISTORY HIGHLIGHTS Dynasty (960–1279) Among his many scientifc occupations, Shen Kuo was also interested in the reraction o light in the Earth’s atmosphere and ound a correct explanation or the appearance o rainbows He noticed that the posi­ tion o these luminous phenomena in the sky appear in the opposite HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY direction o the Sun Thus, ater­ noon rainbows are always visible in the East He also made the connection between rainbow or­ mation and the reraction o solar light through raindrops His under­ standing o the reraction o light in the Earth’s atmosphere allowed him to deduce that the Sun, as it is observed by humans on the Earth’s surace, is not seen in its real position From this deduction, he was able to iner the dierence between the apparent and the real positions o the Sun Besides his interest in reraction phenomena, Shen Kuo was renowned or having been the frst scientist to describe a magnetic needle compass, which he explains in his book Mengxi Bitan, published in 1088 It was more than a century later beore this discovery was brought to the West

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According to this law, the incident ray determines the plane in which rerac­ tion will occur. This means that i the incident ray is rotated by a certain angle in relation to the normal, the plane and the reracted ray will do the same (see Figure 10).

Figure 10 The incident ray, reracted ray and normal are always located in the same plane

3.4.2

Scond law of rfracion

The second law o reraction holds that the ratio o the sine o the angle o incidence (θi) and the sine o the angle o reraction (θR) is a constant. sin θi 5 constant sin θR It is possible to demonstrate that this constant is equal to the relative index o reraction o the two media through which the light ray travels. sin θi n2 5 n1 → 2 5 sin θR n1 Or, in its more common orm: n1 3 sin θi 5 n2 3 sin θR

One of the consequences of this law is that the light ray will not be refracted if the angle of incidence (θi) is zero; that is, perpendicular to the interface. Regardless of the values of the indices of refraction:

appeNDIX 5 rviw of tigonomy, p 404

I θi 5 0° → sin θi 5 0 Because sin θR 5 sin θi 3

n1 n2

( )

503

n1 n2

( )

50

Thereore: θR 5 0° exml a The fgure below shows the trajectory o a ray entering a transparent medium (n 1) with an un­ known index o reraction (n 2) Calculate this index Data: θi 5 30° θR 5 24° n1 5 1 n2 5 ?

Calculation: θi (30°)

n1 sin θi n2 5 sin θR n2 5 n2 5

1x sin 30° sin 24°

n1 n2

Air x

0500 0407

θR (24°)

n2 5 123 exml B A yellow light passes rom water (n 5 133) to crown glass (n 5 152) The rays penetrate the crown glass with an angle o incidence o 35° Calculate the angle o reraction (θR) at the point where the light enters the crown glass Data: n1 5 133 n2 5 152 θi 5 35° θR 5 ?

Calculation: n1 sin θi 5 n2 sin θR Thereore θR 5 sin­1 θR 5 sin­1

(

(

n1 sin θi n2

(133) (sin 35°) 152

)

)

θR 5 30°

an sunss By the time we see a sunset, the Sun is actually long gone Figure 11 illustrates that the Sun is actually below the horizon when we see it set To understand this phenomenon, it is important to keep in mind that the atmosphere is made up o dierent layers Some o the prop­ erties o these layers, such as pressure and temperature, vary with altitude, causing changes in the density o the layers The Sun’s rays are reracted in each layer and bend toward the ground The rays bend even more when the Sun is close to the horizon (early and late in the day) Thereore, the reason we see the Sun ater it has set is due to the reraction o its rays in the atmosphere

Sun’s apparent position Horizon

Sun’s actual position

Observer at point A Earth Atmosphere

Figure 11 The downward reraction o the Sun’s rays entering the atmosphere allows us to see the Sun even ater it has set

Chapter 3 Reraction o Light

85

SeCtION 3.4

Laws of rfracion

1. Determine the sine o the ollowing angles: a) 30° e) 744° b) 60° f) 0° c) 45° g) 90° d) 126° h) 20° 2. Determine the angle corresponding to each o the ollowing sines: a) 0342 d) 0333 b) 0643 e) 100 c) 0700 3. Light moves rom air to diamond with an angle o incidence (θi) o 60° What is the corresponding angle o reraction (θR)? 4. A transparent medium has an index o reraction o 130 What is the angle o incidence (θi) o a light ray in air i its angle o reraction (θR) in the medium is 45°? 5. A light ray moves rom air to a transparent medium with an angle o incidence (θi) o 50° and an angle o reraction (θR) o 40° What is the index o rerac­ tion (n) o the transparent medium? 6. Reproduce the ollowing gures on a sheet o paper In each case, draw the angle o reraction (θR) and the reracted ray a)

Air Water

c)

d) Air

Diamond Water

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8. Calculate the angle o incidence (θi) i the angle o reraction (θR) is 10° when light moves: a) rom diamond (n 5 242) to air b) rom air to diamond c) rom air to water (n 5 133) d) rom water to diamond 9. Calculate the index o reraction o medium 2 i medium 1 is air and the ray travelling through the two media has the ollowing angles o incidence (θi) and reraction (θR) a) 40° and 30° b) 30° and 12° c) 77° and 50° 10. A light ray travels through the air, passes through a water­lled aquarium, and then exits back into the air Determine the angle o reraction (θR) at each interace (nwater 5 133 ; nglass 5 150) Assume that the initial angle o incidence (θi) is 20° 11. A light ray penetrates an unknown material and slows to a speed o 267 3 108 m/s What is the index o reraction (n) o this unknown material? Compare this index o reraction to that o water Which o the two is higher?

b)

Air Glass

7. What is the angle o reraction (θR) o a light ray in each o the ollowing media i its angle o inci­ dence (θi) in air is 30°? a) water (n 5 133) b) diamond (n 5 242) c) ethyl alcohol (n 5 136) d) zircon (n 5 192)

Glass

12. A scuba diver looks upward, toward the surace o the water His line o vision orms a 25° angle with the normal at the surace What is the angle o inci­ dence (θi) in the air o the light rays that reach the diver’s eyes? 13. A scuba diver who is under water shines her fash­ light at an angle o 30° with respect to the vertical At what angle, in relation to the vertical, does the beam o light exit the water?

3.5 tol innl fcion Total internal refection is a phenomenon that occurs when a ray o light passing rom a highly reractive medium to a weakly reractive medium is not reracted but completely refected. Even i the phenomena o reraction and reection are studied separately, in reality they can occur simultaneously. For example, a body o water reects some light and reracts the rest. However, under the second law o reraction, reraction can disappear in certain geometric confgurations. I a light ray travels rom a highly reractive medium (medium 2) to a weak­ ly reractive medium (medium 1), the reracted ray will bend away rom the normal because the angle o reraction (θR) is larger than the angle o inci­ dence (θi). At the same time, part o the light will be reected toward medium 1 (see Figure 12a). I the angle o incidence (θi) increases progressively, an angle o reraction (θ R) o 90° will eventually be reached. At that point, the reracted ray runs along the interace, and the corresponding angle o inci­ dence (θi) is reerred to as the critical angle (θc) (see Figure 12b). I the angle o incidence (θi) is larger than the critical angle (θc), reraction is no longer observed. Rather, total internal reection o the light occurs (see fgures 12c and 13). Refracted rays

No refracted rays

Medium 1

n1 n2

Medium 2 a

c

b

Critical angle (θc)

Total internal reflection

Light source

Figure 12 Total internal refection o a ray travelling rom a highly reractive medium to a weakly reractive medium

The critical angle (θ c ) depends on the relative index o reraction o the two media. It is calculated as ollows: The second law o reraction holds that: n1 sin θi 5 n2 sin θR The critical angle is obtained when the angle o reraction is equal to 90°: n1 3 sin θc 5 n2 3 sin 90° But since sin 90° 5 1, we obtain: n2 n → θc 5 sin­1 2 sin θc 5 n1 n1

Figure 13 From a certain angle, the lower surace o the water­air interace behaves like a perect mirror (the object is refected) This phenomenon occurs as a result o the total internal refection o light on this interace

( )

Thereore, we observe that the critical angle (θc) is equal to the inverse sine o the relative index o reraction o the two media.

Chapter 3 Reraction o Light

87

examl Calculate the critical angle (θc) or the water­air interace Data: nwater 5 133 nair 5 100

Calculation: θc 5 sin­1 θc 5 sin­1

( ) ( ) n2 n1

100 133

θc 5 488° The phenomenon o total internal reection occurs only when two conditions are met:

*

Oical fbrs Very thin glass or plastic fbres that conduct light and transer data

Optical instrument consis­ * prisco ting o lenses and prisms that allows the viewer to see above an obstacle

1. The light ray travels rom a highly reractive medium to a weakly reractive medium (n2 < n1); 2. The angle o incidence is greater than the critical angle (θi > θc). Total internal relection is used in numerous technological applications such as the transer o inormation through optical fbres* and the ormation o images in periscopes*.

SeCtION 3.5

toal inrnal rcion b) What is the index o reraction (n) o the rst medium i the critical angle (θc) is 40°?

1. What is the critical angle (θc) o glass when light passes rom the glass to air? 2. The critical angle (θc) o a transparent medium is 405° when a light ray passes rom this medium into air What is the index o reraction (n) o this medium? 3. Does total internal refection in an aquarium occur at the separation o the water and the glass or at the boundary o the glass and the water? Explain your answer

6.

From the inside o an aquarium a light ray travels toward the glass to orm an angle o incidence (θi) o 30° in the water a) Determine the angle o reraction (θR) o the light ray emerging rom the glass into the air b) I the angle o incidence (θi) in the water is 52°, at what angle will the light ray emerge rom the glass?

4. I a scuba diver is underwater and looks up, he or she will see a round “hole” directly above him or her at the surace o the water The rest o the surace will look like a mirror Explain this phenomenon 5. Assume that air is the second medium: a) What is the critical angle (θc) i the index o reraction (n) o the rst medium is 168?

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UNIt 1 Geometric Optics

Glass (n  1.50) Air (n  1.00) Water (n  1.33)

θi (30°)

APPLICATIONS Oicl fbs In recent years, optical fbres have replaced electri­ city in data transmission. Optical fbres are covered by a thin transparent sheath with a low index o reraction which prevents light rom escaping. In the feld o medicine, optical fbres have numerous applications, which can be divided into two catego­ ries: observation and treatment. In the frst category, optical fbres are used to view things, as in the case o endoscopies. The optical fbre is inserted into the body and conducts light that is reected by the internal organs and allows images to be transmitted (see Figure 14). In the second category, optical fbres are used to treat patients. The same opti­ cal fbre cable can carry a laser beam which removes or destroys diseased tissue.

(specializing on the nose) use lasers to remove benign tumours (polyps), inammation, oreign objects or mal­ ormations causing an obstruction o the nasal pas­ sages. These are generally minor procedures with ew side eects. Optical fbres are also widely used in cancer treatment. For example, in a colonoscopy, the inner wall o the colon (large intestine) is observed. In addition to a cam­ era, the endoscope can carry tiny orceps to remove a tissue sample that will be studied by a gastroenterolo­ gist, who determines whether or not the tissue is cancer­ ous. However, once the diagnosis has been made, not all cancers can be treated with such a ocused laser beam: it depends on the size o the tumour and its location. More recently, doctors specializing in joints (knees, wrists, shoulders, etc.) have begun to use an arthro­ scope, which is an optical fbre that has been adapted to explore joints. Two small incisions must be made: one to insert the optical fbre and the other to perorm the surgery. In the past, surgeons have had to open up the entire joint to observe the problem (ligaments, cartilage, meniscus, etc.) and repair it. Video camera

Skin

Figure 14 A doctor performing an endoscopic examination Lens

It is also possible to remove a kidney using surgical tools that are controlled rom a distance while the procedure is viewed through an instrument called a laparoscope (see Figure 15). However, surgeons require training on how to manipulate these new instruments. Endoscopic surgery has become extreme­ ly specialized. For example, some rhinology surgeons

Ligaments Scissors

Grasping forceps Kidney

Figure 15 Removal of a kidney by laparoscopy

Chapter 3 Refraction of Light

89

the laws of refracion Scientifc laws and principles are generally named ater the individuals who discovered them, as in Newton’s laws, Ohm’s law and Archimedes’ principle. The scientifc authorship o some laws, however, is doubtul and not as clear­cut. This is the case or the laws o reraction, commonly known as Snell’s laws, but also as the Snell­Descartes law. René Descartes (1596–1650) (see Figure 16) and Willebrord Snell (1581–1626) (see Figure 17) were, respectively, French and Dutch scientists. Some works explain the laws’ name based on their separate and independent discovery by the two scien­ tists. However, this assertion has been called into ques­ tion by historians and scientists alike. Dutch scholar Christian Huygens (1629–1695) observed in one o his books that “Snell discovered the law o reraction in 1621, but the law was published by Descartes, who had seen Snell’s manuscripts.” The laws o reraction were published by Descartes in 1637, more than a decade ater Snell’s death. This accusation is based on two acts. First, Descartes lived in Holland or over 20 years. Furthermore, the

Figure 16 René Descartes (1596–1650)

90

UNIt 1 Geometric Optics

proo o the laws o reraction presented a number o weak points, such that French scientist Pierre de Fermat (1601–1665) completely rejected them. However, the history o the laws o reraction does not end here. In 1950, historians studied the manuscripts o a British scientist by the name o Thomas Harriot (1560– 1621) (see Figure 18) and realized that he had discovered these laws in their present orm in 1602, that is, 19 years beore Snell. He simply had not published them. Does the story o the laws o reraction end with Harriot? No, it actually goes much urther back in history. The recent discovery o Arabic manuscripts rom the end o the 10th century rekindled the debate. Roshdi Rashed, a specialist in the history o Arab science, studied the treatise by Ibn Sahl (940–1000) called On the Burning Instruments, which was written between 983 and 985. Rashed discovered that Sahl had clearly ormulated the laws o reraction in the orm o a geometric relationship, more than six centuries beore Descartes had published it. So, how then should we reer to these laws: “Snell’s laws,” “Harriot’s laws” or “Ibn Sahl’s laws”? To avoid any controversy, it is best to call them the “laws o reraction.”

Figure 17 Willebrord Snell (1581–1626)

Figure 18 A drawing rom the treatise On the Burning Instruments by Ibn Sahl, where the frst law o reraction is frst mentioned

Chapter

3

rfcion of Lig

3.1 pnomnon of fcion • Reraction reers to the change in direction in which light travels at the interace o two media that have dierent optical properties. • The interace is the surace that separates two transparent, homogeneous media possessing dierent optical properties.

3.2 Indx of fcion • The speed o light in a vacuum is approximately equal to c ≈ 3.00 3 108 m/s. • When light penetrates a transparent medium other than a vacuum, its speed decreases. • The property o a medium to reract light is called reractivity. • The index o reraction (n) is the ratio between the speed o light in a vacuum (c) and the speed o light in a transparent medium (v).

n5 c v • The index o reraction (n) is always greater than 1, except in a vacuum, where it is exactly equal to 1, c because c 5 1. • The higher the index o reraction (n) o a transparent medium, the more it will reduce the speed o the light (c) travelling through it, as compared to its speed in a vacuum. • Even i the index o reraction o air is not exactly equal to that o a vacuum, it is customary to use the same value (n 5 1) or the two media because the error is negligible. • The relative index o reraction (n1 → 2) is equal to the ratio between the index o reraction o the sec­ ond medium (containing the reracted ray) and the index o reraction o the frst medium (containing the incident ray).

n1 → 2 5

n2 n1

Chapter 3 Refraction of Light

91

3.3 Geomery o reracion • The incident ray is the light ray that travels toward the interace. • The normal is an imaginary line that is perpendicular to the interace, crosses the point o incidence and travels through the two media. • The angle o incidence (θi) is the angle ormed by the incident ray and the normal. • The reracted ray is the light ray that crosses the interace and enters the second medium. • The angle o reraction (θR) is the angle ormed by the reracted ray and the normal. • When a light ray passes rom a less reractive medium to a more reractive medium (n1 < n2), the angle o reraction (θR) is smaller than the angle o incidence (θi) and the reracted ray bends toward the normal. • When a light ray passes rom a more reractive medium to a less reractive medium (n2 < n1), the angle o reraction (θR) is greater than the angle o incidence (θi), and the reracted ray bends toward the normal.

3.4 Laws o reracion • The frst law o reraction states that the incident ray and the reracted ray are located on opposite sides o the normal originating rom the point o incidence, and that all three are in the same plane. • The second law o reraction states that the ratio o the sine o the angle o incidence (θi) and the sine o sin θi the angle o reraction (θR) is a constant 5 constant . Once the constant is defned, sin θR

(

)

the second law o reraction is written as ollows: n1 3 sin θi 5 n2 3 sin θR

3.5 toal inernal refecion • Total internal reection occurs when a light ray travels rom a highly reractive medium to a weakly reractive medium (n2 < n1) and the angle o incidence is greater than the critical angle (θi > θc). • The critical angle (θc) corresponds to the value o the angle o incidence when the value o the angle o reraction is equal to 90°. • The critical angle (θc) can be calculated using the ollowing ormula: θc 5 sin­1

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UNIt 1 Geometric Optics

( ) n2 n1

Chapter 3

rfcion of Lig

1. Match each o the ollowing terms to their corre­ sponding numbers on the gure shown below: angle o incidence (θi), normal, reracted ray, angle o refection (θr), incident ray, angle o reraction (θ R), refected ray, interace 2 1

3

6

8 11

12

9

10

4

2. A light ray hits a piece o polyethylene (n 5 150) at angles o incidence (θi) o: a)

0°

b)

30°

c)

60°

5. The index o reraction (n) o crown glass is 153 or violet light and 151 or red light I the speed o light in a vacuum is 300 3 108 m/s, what are the speeds o the violet light and the red light in the crown glass? 6. Light travels in water at three­quarters o its speed in air I the angle o incidence (θi) in air is 10°, what will be the angle in water?

7

5

4. Use geometry to prove that a light ray that strikes a sheet o glass always emerges in a direction parallel to the incident ray

Determine the angle o reraction (θ R) in the three cases 3. a) Two light rays, one red and one violet, encoun­ ter a glass surace I nred 5 152, nviolet 5 154 and the angle o incidence (θi) is 30°, calculate the angle o reraction (θR) o each colour o light ray (rom air to glass) b) Use the angle you obtained in part a) as a new angle o incidence (θi), and calculate the angle o reraction (θR) or each o these two wave­ lengths as they pass rom glass to air

7. Why does total internal refection never occur when a light ray passes rom a less dense medium to a more dense medium? Draw a diagram to explain your answer 8. Why do images appear clearer under water through a diving mask? 9. A coin sits at the bottom o a pool, under 12 m o water and 10 m rom the edge, as shown in the illustration below The beam rom a fashlight shines on the coin What angle must the fashlight orm with the pool wall in order or the light beam to shine on the coin?

θi Air Water θR 1.2 m

1.0 m

Chapter 3 Refraction of Light

93

10. Several transparent liquids are poured careully, one ater the other, into a glass container, as shown in the illustration below The liquids are immiscible, which means that they do not mix A light ray with an angle o incidence (θi) o 10° in glass is directed toward the top, through the liq­ uids What will be the angle o reraction (θR) in air?

12. Some periscopes use two prisms placed at right angles to each other as shown in the illustration below Look closely at the gure and answer the ollow­ ing questions: a) Why does the incident ray cross side AC o prism P1 without being reracted? b) Why is the incident ray completely refected on side AB o prism P1?

Air (n  1.00)

Oleic acid (n  1.43)

c) What is the angle o refection (θr) o the inci­ dent ray on this side? d) The index o reraction (n) o prisms P1 and P2 must be greater than a certain value in order or the incident light rays to be deviated at an angle o 90°, as shown in the gure What is this value? A

Water (n  1.33)

Air

Carbon disulphide (n  1.63) P1

Glass (n  1.50)

B C

θi (10°)

11. In which o the ollowing media does light travel aster: a medium with a critical angle (θc) o 27°or a medium with a critical angle (θc) o 32°? In both cases, the second medium is air

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P2

Lenses

A

lens is a transparent object with at least one curved surface. Lenses control the direction of light rays passing through them and are one of the essential components of many optical instruments. In this chapter, you will become familiar with the different types of lenses, learn how they bend light through the phenomenon of refraction, discover how images are formed with lenses, and study notions of optical power and optical aberrations.

Review Refraction of light                                          9 Converging lenses and diverging lenses                       10 The eye                                                 11

4.1 4.2 4.3 4.4 4.5

Different types of lenses                96 Refraction in lenses                     99 Optical power of lenses                106 Images formed by lenses               113 Optical aberrations of lenses           123

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95

4.1 Different types of lenses There are a number o types o lenses that vary according to shape and, conse­ quently, in the way they deviate light. This chapter will cover only spherical lenses, meaning lenses with at least one surace that is a portion o a sphere, the other surace being either spherical or at. These lenses will be consid­ ered thin—in other words, they are much thinner than the radius o curva­ ture (R) o the spheres they derive rom—in order to make the reasoning and mathematical ormulas behind them less complicated. See Phenomenon of refraction, p 78

Lenses are made o a transparent material, usually glass, which, because o the phenomenon o refraction and its laws, deviates the trajectory o light rays that pass through it. Despite their diversity, lenses can be classifed into two main categories: converging lenses and diverging lenses.

4.1.1

Converging lenses

There are three main converging spherical lenses, all characterized by at least one rounded surace (see Figure 1).

C2

V1

V2

C1

a) Biconvex lens

C2

V1

b) Planoconvex lens

V2

C1

C2

V1

V2

c) Positive meniscus lens

Figure 1 The most common converging lenses

Biconvex lenses are ormed rom two convex spherical suraces. Planoconvex lenses are defned by a at surace and a convex spherical one. Positive menis­ cus lenses are made up o two spherical suraces: one concave and the other convex. These are also called converging meniscus lenses. In Figure 1, C1 and C2 represent the centres o the curvature, while V1 and V2 indi­ cate their vertices. In the case o the planoconvex lens, V1 corresponds to the intersection o the at surace and the principal axis. The principal axis (P) is the line that links the centres and vertices o the spheres. This axis passes through the optical centre (O) o the lens. A lens is said to be converging when rays that are parallel to the principal axis strike the lens and converge to a single point on the opposite side o the lens (see Figure 2 on the following page). To simpliy diagrams o optical assemblies and paths, symbols are used. By convention, the symbol associated with converging lenses is a vertical double arrow (see Figure 3 on the following page).

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UNIT 1 Geometric Optics

C1

V2 O

C2

V1

P

Figure 3 Converging lenses are represented by a vertical double arrow

Figure 2 A biconvex spherical lens is converging

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY 4.1.2

Diverging lenses

There are also three main diverging spherical lenses, characterized by being thicker at the edges than at the centre. (see Figure 4).

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY C1

V1

V2

C2

a) Biconcave lens

C1

V1

V2

C2

b) Planoconcave lens

C1

V1

V2

c) Negative meniscus lens

Figure 4 The most common diverging lenses

Biconcave lenses are formed from two concave spherical surfaces. Plano­ concave lenses are defined by a flat surface and a concave spherical one. Negative meniscus lenses are made up of two spherical surfaces: one convex and the other concave. These are also known as diverging meniscus lenses. A lens is said to be diverging when rays that are parallel to its principal axis (P) and enter one of its surfaces and are deviated to different points that grow fur­ ther apart as they exit the lens. The rays that diverge after passing through the lens seem to come from a single point on the side of the incident rays (see Figure 5). Diverging lenses are represented by a double inverted arrow (see Figure 6).

C

F O

Figure 5 A biconcave spherical lens is diverging

SPECTACLES The frst glasses used expressly to improve vision appeared in the 13th century The invention’s authorship is unknown, but historians cite Roger Bacon (1214–1294), Salvino degli Armati (who died in 1317) and Alessandro di Spina (who died in 1313) as possible candidates What is known is that in 1267 Roger Bacon provided scientifc proo that small letters could be enlarged by viewing them through glass with a particular surace The frst eyeglasses, or spectacles consisted o two lenses ramed with wood or horn and topped with two rods joined with a nail These armree glasses were worn directly on the nose

Figure 6 Diverging lenses are represented by a double inverted arrow

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97

SECTION 4.1

Different types of lenses

1. Reproduce the following table on a sheet of paper and complete it

3. Copy and complete the following table on a sheet of paper Characteristics of converging spherical lenses

Lens

Type

Converging / diverging lens

Symbol

Converging lens

First surface

Biconvex Biconcave lens

Diverging lens

Positive meniscus

Second surface Convex spherical

Concave spherical Flat

Convex spherical

4. Copy and complete the following table on a sheet of paper Characteristics of diverging spherical lenses Diverging lens

First surface Flat

Biconcave Negative meniscus

Second surface Concave spherical Concave spherical

Convex spherical

5. Are negative meniscus lenses converging or diver­ ging? 6. A positive meniscus lens has two sides: one concave and the other convex On which of the two sides is the curvature more pronounced? 7. Copy and complete the following sentences: a) lenses are thicker at the centre than at the edges b) Diverging lenses are at the centre than at the edges 8. How can a converging lens be used to burn an object? Where should the lens be placed in respect to the object? 9. A drop of water on a smooth surface can act as a lens Using a diagram, illustrate how this effect is produced and determine the type of lens a drop of water represents 2. How does a diverging lens differ from a converging lens: a) in terms of shape? b) in terms of its effect on light?

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UNIT 1 Geometric Optics

4.2 Refraction in lenses When light travels through a lens, it undergoes two reractions. I the lens is glass, or instance, the frst reraction will occur at the air­glass interace, and the second at the glass­air interace. To illustrate this notion o double rerac­ tion, consider the way light travels through a glass prism (see Figure 7). Normal

Incident ray

Emerging ray

θ i1

Normal θ R 1 θ i2

θ R2

Prism Refracted ray

Figure 7 In a glass prism, double reraction deviates the incident ray to the base o the prism

The frst reraction occurs at the air­glass interace. Since the index o rerac­ tion (n) o glass is greater than that o air, the angle o reraction θR1 will be smal­ ler than the angle o incidence θi , in accordance with the laws of refraction. The 1 reracted ray will thereore tend to bend toward the normal and toward the base o the prism. The second reraction occurs at the glass­air interace, where the confg­ uration is reversed. The ray that emerges rom the prism will bend away rom the normal and even closer toward the base o the prism.

4.2.1

See Laws of refraction, p 84

Refraction in converging lenses

For simplicity’s sake, this section will cover only biconvex spherical lenses. Similar reasoning can be applied to the other types o converging lenses.

Convergence of rays A biconvex converging lens can be represented by joining two prisms to a glass rectangular parallelepiped (see Figure 8). This representation, though approximate, makes it easier to understand the ocusing properties o converging lenses. I light rays parallel to the bases o the prisms are directed at this lens rom the let, three things will happen. First, any incident ray that passes through the upper prism will undergo a double reraction that will deviate it downward. Next, any incident ray that passes through the lower prism will undergo a double reraction that will deviate it upward. Finally, any incident ray that passes through the rectangular parallelepiped will not undergo any deviation because it is parallel to the normal on its surace.

Prism Biconvex converging lens Rectangular parallelepiped

Prism

Figure 8 Simplifed representation o a biconvex converging lens

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99

Because of symmetry, the three rays converge to the same point, referred to as the principal focal point (F) (see Figure 9). In similar fashion, the curvature of a biconvex converging lens focuses a light ray parallel to the lens’s principal axis (P) on the principal focal point (F) (see Figure 10).

F′

O

F

P

f

Figure 9 The combination o two prisms and a glass rectangular parallelepiped ocuses light rays parallel to the bases o the prisms on the principal ocal point (F)

Figure 10 A biconvex converging lens ocuses a ray parallel to its principal axis (P) on a point called the principal ocal point (F) o the lens The enlarged portion shows how the frst reraction occurs at the air-glass interace, and the second at the glass-air interace

Converging lens terminology Figure 10 shows some important characteristics of lenses. 1. Optical centre (O) The optical centre (O) of a lens corresponds to its geometric centre. 2. Principal focal point (F) The principal focal point (F) is the point at which rays parallel to the prin­ cipal axis (P) converge. The principal focal point is located on the princi­ pal axis. This point is also called the image focal point. The principal focal point of a converging lens is said to be real because light rays are directed at this point. 3. Secondary focal point (F′) The secondary focal point (F′) is the point symmetrical to the principal focal point (F) in relation to the optical centre (O) of the lens. It is located on the principal axis on the side of the incident rays. The secondary focal point is also called the object focal point. 4. Focal length (f ) The focal length (f ) is the distance (OF) between the optical centre (O) and the principal focal point (F). 5. Focal plane The focal plane is the plane perpendicular to the principal axis (P) that passes through the principal focal point (F). This plane contains the points where all incident rays converge, whether they are parallel to the principal axis or not (see Figure 11 on the following page).

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UNIT 1 Geometric Optics

Focal plane

F

O

P

f

Figure 11 When the incident ray is not parallel to the lens’s principal axis (P), the rays do not converge on the principal focal point (F), but are focused instead on a plane called the focal plane

Principal rays of a converging lens Like spherical mirrors, converging lenses have three principal rays. 1. First principal ray This is any incident ray parallel to the lens’s principal axis (P). This ray is refracted toward the lens’s principal focal point (F) (see ray 1 in Figure 12). 2. Second principal ray This is any incident ray that passes through the secondary focal point (F′). This ray is refracted parallel to the lens’s principal axis (P) (see ray 2 in Figure 12). 3. Third principal ray This is any incident ray that passes through the lens’s optical centre (O). This ray does not undergo any deviation* (see ray 3 in Figure 12).

Ray 1

Ray 1

Ray 3 Ray 2

Ray 3 Ray 2

F′

O

a) Real representation of principal rays

F

P

F′

If the angle of incidence (θ ) is * Note not zero, this is valid only in the case i

of thin lenses

O

F

P

b) Diagram of principal rays

Figure 12 The principal rays of a converging lens

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101

4.2.2

Refraction in diverging lenses

For simplicity’s sake, this section will cover only biconcave spherical lenses. Similar reasoning can be applied to the other types of diverging lenses.

Divergence of rays A biconcave diverging lens can also be represented by combining two prisms and a glass rectangular parallelepiped, but the assembly is different from that of a biconvex lens (see Figure 13). Prism

Biconcave diverging lens

Rectangular parallelepiped

Prism

Figure 13 Simplifed representation o a biconcave diverging lens

This representation has the advantage of showing in simple fashion how a diverging lens makes light rays diverge. If a light ray parallel to the bases of the prisms is directed at the lens from the left, three things happen. First, any incident ray that passes through the upper prism will undergo a double refraction that will deviate it upward. Next, any incident ray that passes through the lower prism will undergo a double refraction that will deviate it downward. Finally, any incident ray that passes through the rectangular parallelepiped will not undergo any deviation because it is parallel to the nor­ mal on its surface. Because of symmetry, the three rays that diverge appear to come from a single point—the principal focal point (F) (see Figure 14).

Figure 14 The combination o two prisms and a glass rectangular parallelepiped makes light rays parallel to the bases o the prisms diverge

In similar fashion, the curvature of biconcave diverging lenses makes light rays parallel to the lens’s principal axis (P) diverge (see Figure 15 on the following page).

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UNIT 1 Geometric Optics

Focal plane

F

O

F′

P

f

Figure 15 A biconcave diverging lens makes light rays parallel to its principal axis (P) diverge. When extended, the emerging rays appear to come from the principal focal point of the lens (F).

Diverging lens terminology The terminology used for diverging lenses is practically identical to that used for converging lenses. However, there are some important geometric differences. 1. Principal focal point (F) The principal focal point (F) is the point from which rays emerging from the lens appear to originate. It is located on the principal axis (P) and, contrary to converging lenses, on the side of the incident rays. As with converging lenses, the principal focal point is also called the primary focal point or image focal point. However, the principal focal point of a diverging lens is said to be virtual because light rays do not actually pass through this point but only appear to come from it. 2. Secondary focal point (F′) The secondary focal point (F′) is the point symmetrical to the principal focal point in relation to the optical centre (O) of the lens. It is located on the principal axis (P) on the side of the emerging rays. The secondary focal point is also called the object focal point. 3. Focal length (f ) The focal length (f ) is the distance (OF) between the optical centre (O) and the principal focal point (F). 4. Focal plane The focal plane is the plane perpendicular to the principal axis (P) that passes through the principal focal point (F). Contrary to converging lenses, this plane is on the side of the incident rays.

Principal rays of a diverging lens Diverging lenses have three principal rays. They obey the following rules: 1. First principal ray This is any incident ray parallel to the lens’s principal axis (P). This ray is refracted by the diverging lens in such a way that it appears to come from the lens’s principal focal point (F) (see ray 1 in Figure 16 on the following page).

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103

2. Second principal ray This is any incident ray that points in the direction of the secondary focal point (F′). This ray is refracted parallel to the lens’s principal axis (P) (see ray 2 in Figure 16). 3. Third principal ray This is any incident ray that passes through the lens’s optical centre (O). This ray does not undergo any deviation (see ray 3 in Figure 16).

Ray 1 Ray 3

Ray 1 Ray 3 F

O

F′

P

F

F′

O

Ray 2

Ray 2

a) Real representation of principal rays

b) Diagram of principal rays

Figure 16 The principal rays of a diverging lens

Furthering

your understanding

Thick lenses and thin lenses A lens’s properties depend on its geometric characteristics and the material it is made out of. When a lens’s thickness (t) (equal to distance V 1 V2) is significant compared to the radii of curvature of the spherical surfaces that define it, the lens is said to be thick. In this case, any incident ray that passes through the lens’s optical centre (O) is refracted in such a way that the emerging ray undergoes a lateral displacement. However, the two rays remain parallel. When extended, the incident and emerging rays cross the principal axis (P) at the lens’s nodal points (N1 and N2) (see Figure 17a). When a lens’s thickness (t) is insignificant compared to the radii of curvature of the spherical surfaces that define it, the lens is said to be thin. In this case, the distance between the nodal points (N1 and N2) is very small, and the lateral displacement between the incident and emerging rays so slight as to be negligible (see Figure 17b). Thus, when lenses are thin, the two nodal points and the optical centre (O) are considered to be overlapping. This suggests that any incident ray passing through the optical centre of a thin lens does not undergo a deviation. This approximation has the advantage of greatly simplifying lensrelated mathematical formulas.

Lateral displacement

V1

N1

O

V2 N2

a) Thick lens Lateral displacement

V1 N1 O

V2 N2

b) Thin lens

Figure 17 Deviation of a light ray passing through the optical centre (O) of a converging lens

104

UNIT 1 Geometric Optics

P

SECTION 4.2

Refraction in lenses

1. Reproduce the following table on a sheet of paper and complete it Ray diagram

Type of ray

4. Copy the following diagram on a sheet of paper

Converging/ diverging lens

F’

O F′

0

F

a) Is the lens converging or diverging? b) What is point F' called? c) Draw the optical path of each ray after it passes through the lens Explain your answer

F′

0

5. A light ray is directed at a glass lens F′

0

F 30°

2. Though both are diverging, biconcave and plano­ concave lenses are different Biconcave lenses have two concave surfaces, while the plano­concave lenses second two have only one In your opinion, which lenses deviate light more if the radii of curvature (R ) of both types are identical? Explain your answer 3. On a sheet of paper, indicate the component corres­ ponding to each number in the following diagram:

3 1

2

5

a) What type of lens is it? b) What does the line dividing the lens in two equal parts represent? c) Use the second law of refraction to trace the light ray’s path when it emerges from the lens d) Based on the deviation of the refracted ray, state whether the lens is diverging or converging

4

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105

4.3 Optical power of lenses Optical power is a measure that quantifes the capacity o a lens or optical system to deviate light rays. Optical power also goes by such sel­explanatory names as ocusing power, dioptric power, reractive power and convergence power. The more an optical system is able to converge or diverge a light ray, the greater its optical power will be. This measure, which characterizes lenses and optical systems, is used by optometrists when prescribing eyeglasses and contact lenses.

4.3.1

Optical power of a lens

The optical power (P) o a lens depends on its radii o curvature (R) and the index o reraction (n) o the material it is made o. The optical power o a lens is inversely proportional to its radii o curvature (see Figure 18). This means that the larger the radii o curvature, the smaller the optical power o the lens will be. At the limit, i the radii o curvature are infnite, it is no longer a lens but a plate, at and parallel o surace, with an optical power o zero because the rays that cross it are not deviated. As a result, a lens with high optical power has a short ocal length (f) and, inversely, a lens with low optical power has a long ocal length.

O

F

P

O

F

P

f

f

a) Lenses with high optical power have small radii of curvature (R).

b) Lenses with low optical power have large radii of curvature (R).

Figure 18 The optical power o lenses with dierent radii o curvature

*

Note I the lens is placed in a medium other than air, the optical power (P) must take the medium’s index o reraction (n) into account: P 5 n  It should be f noted, however, that unless otherwise specifed, lenses are considered to be immersed in air

A lens’s optical power (P) is thereore inversely proportional to ocal length (f ) and, i the lens is immersed in air*, it is defned by the ollowing relationship:

P5

1 f

where P 5 Optical power, expressed in dioptres (d) f 5 Focal length, expressed in metres (m) Since the unit o ocal length (f ) is the metre (m), the unit o optical power is the inverse o the metre (m­1). In optics, the dioptre is commonly used as a unit o optical power. The symbol or dioptre is the Greek letter d (delta). Optometrists prescribe eyeglasses or contact lenses in dioptres. The dioptre is defned by the ollowing relationship: 1 d 5 1 m -1

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UNIT 1 Geometric Optics

4.3.2

Sign convention

In geometric optics, dierent algebraic signs are used; in other words, variables can be either positive or negative. The signs ( 1 or - ) assigned to these quantities depend on the convention adopted. In this textbook, the principal axis (P) o lenses is considered an axis that runs rom let to right, its origin being the optical centre (O). I the light rays also travel rom let to right, then the ollowing conditions apply: 1. The radii o curvature (R1 and R2) o the spheres generating the lenses do not have the same sign. They are positive when the surace hit by the incident ray is convex (bulged), and negative when the surace is concave (hollow). In other words, the radius o curvature is positive i the centre o curvature is on the side o the emerging rays and negative i the centre o curvature is on the side o the incident rays. When the surace is at, the radius o curvature is infnite. (See Table 1 for a summary of this convention for each of the six lenses introduced at the beginning of the chapter.) 2. Focal length (f ) is positive or a converging lens, and negative or a diverg­ ing lens (see Figures 19a and 19b). Table 1 The most common sign convention or lenses (light is presumed to come rom the let) Biconvex

Planoconvex

Positive meniscus | R1 |  | R2 |

Biconcave

Planoconcave

Negative meniscus | R2 |  | R1 |

R1

Positive (1)

Infnite

Negative (-)

Negative (-)

Infnite

Negative (-)

R2

Negative (-)

Negative (-)

Negative (-)

Positive (1)

Positive (1)

Negative (-)

Lens type

Shape

O

F

f

Figure 19a Focal length (f) is positive or a biconvex (converging) lens

P

F

P

f

Figure 19b Focal length (f) is negative or a biconcave (diverging) lens

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107

Because optical power (P) is inversely proportional to focal length (f ), they will both have the same sign. Thus, the optical power of a converging lens is positive, while that of a diverging lens is negative (see Figure 20). Focal plane

Convergence

Focal length (f )

2 1.5 Optical power (P ) 0.5 0.67

Divergence

-2

Metres (m)

1

0.5

0

-0.5

-1

-1.5

1

2

0

-2

-1

-0.67 -0.5

Dioptres (δ)

Figure 20 Relationship between optical power and focal length. If the focal plane is translated horizontally, the focal length is modified, which changes the optical power.

Example A A diverging lens has a focal length (f ) of 16 cm. What is its optical power (P ) in dioptres (δ) ? Data: f  -0.016 m P?

Solution : Because the lens is diverging, the focal length (f ) is negative: f  -16 cm  -0.016 m The optical power is therefore: P

1 1   -62.5 m -1  -62.5 δ f -0.016 m

Example B What is the focal length (f ) of a lens whose optical power (P ) is 10 δ ? Is the lens converging or diverging? Data: P  10 δ f?

Solution : Optical power (P ) is inversely proportional to focal length (f ) : P

1 1 1 1 ⇒f    0.1 m  10 cm f P 10 δ 10 m-1

The lens is converging because its optical power and focal length are positive.

4.3.3

Optical power of a system of lenses

When several lenses are placed side by side, the optical system created has a total optical power (PT) that depends on the individual optical power (Pi) of each lens. The relationship between total optical power and individual optical power is simple.

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UNIT 1 Geometric Optics

For an optical system made of n lenses, total optical power (PT) can be written as follows: P T 5 P1 1 P2 1 … 1 Pn where P T 5 Total optical power, expressed in dioptres (d) P1, P2, Pn 5 Individual optical powers, expressed in dioptres (d)

describes an optical system * Afocal that does not produce deviation in the light beam that passes through it

For example, in a system of two lenses with the optical power of P1 and P2, total optical power will be: PT 5 P1 1 P2. In fact, a system of lenses is equivalent to a single lens with the same optical centre (O) whose optical power (P) is equal to the algebraic sum of each lens’s individual power. When total optical power is zero, the system is said to be afocal*. Because of the simplicity of this relationship, optometrists and ophthal­ mologists work with lenses’ optical power rather than their focal length.

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

Example An optical system is created out o two lenses whose respective optical powers (P ) are 135 d and -400 d a) Calculate the system’s total optical power (PT ) b) What is the ocal length (fT) o this combination o lenses? Data: P1 5 135 d P2 5 -400 d PT 5 ? fT 5 ?

Solution : a) For an optical system created out o two lenses, total optical power (PT ) is shown as: P T 5 P1 1 P2 5 135 d 1 (-400 d) 5 -05 d b) The ocal length (f ) o this combination o lenses is calculated on the basis o its optical power (P ): PT 5

4.3.4

1 1 1 1 ⇒ fT 5 5 5 5 -2 m fT PT -05 d -05 m-1

Lens-maker’s equation

Optical power (P) and focal length (f ) are mathematically related. To discover a lens’s optical power, it is necessary to know how to calculate the value of a lens’s focal length on the basis of its fabrication parameters. But what does the focal length of a thin lens in air depend on? On the one hand, focal length (f ) varies with the refractivity of the material the lens is made of. The more refractive the material is, the more the light will be deviated and the shorter the focal length will be. But since the lens is in air, the focal length will be inversely proportional to the difference between the index of refraction (n) of the material and the air. The inverse expression of this rela­ tionship is more interesting from a mathematical point of view. 1 α (n 2 1) f where α signifes "is proportional to"

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY HANS LIPPERSHEy German-Dutch optician (1570–1619) A lens-maker by proession, Hans Lippershey requested a patent or the frst telescope in 1608 He called his invention kijker—literally “looker” It is said that while watching children play with lenses in his store, he had the idea o combining a converging and diverging lens in a tube His invention— the telescope—was well received, attracting the notice o the army Lippershey’s telescope was instrumental in the making o the astronomical telescope In homage to the amous optician’s discovery, a lunar crater and asteroid now bear his name

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109

On the other hand, as Figure 18 on page 106 shows, the larger the lenses’ radii o curvature (R1 and R2), the longer the ocal length (f ). At the limit, i the radius o curvature is infnite (at surace), the ocal point will be an infnite distance away. This relationship, expressed as a unction o the inverse o the distance, is shown as: 1 1 1 α 2 f R1 R2

(

)

By combining the previous two relationships o proportionality, we obtain: 1 1 1 5 (n 2 1) 2 f R1 R2

(

)

This equation is known as the lens­maker’s equation. It is very useul in calculating the optical power (P) o lenses. By taking into account the rela­ tionship between ocal length (f) and optical power, the ollowing result is obtained:

P5 where

1 1 1 5 (n 2 1) 2 R1 R2 f

(

)

P 5 Optical power, expressed in dioptres (d) f 5 Focal length, expressed in metres (m) R1 and R2 5 Radii of curvature of the lens, expressed in metres (m) n 5 Index of refraction of the lens material

Example A A biconvex lens’s radii of curvature are 20 cm If the index of refraction (n) of the lens is 166: a) What is the lens’s focal length (f )? b) What is the lens’s optical power (P )? Data: R1 5 20 cm R2 5 20 cm n 5 166 f 5? P 5?

Solution : a) Table 1 on page 107 gives the signs of the radii of curvature (R ) : R1 5 120 cm 5 102 m R2 5 -20 cm 5 -02 m The lens-maker’s equation is therefore: 1 1 1 1 1 5 (n 2 1) 2 5 (166 2 1) 2 R1 R2 (102 m) (-02 m) f

(

5 (066) ⇒f5

)

(

)

( (021 m) 1 (021 m) ) 5 66 m-

1

1 5 015 m 5 15 cm 66 m-1

b) The lens’s optical power (P ) is inversely proportional to its focal length (f ): P5

110

UNIT 1 Geometric Optics

1 1 5 5 66 d 015 m f

Example B If the exercise in the previous example was recreated with a biconcave lens, how would the results differ? Data: R1 5 20 cm R2 5 20 cm n 5 166 f 5? P 5?

Solution : a) In this case: R1 5 -20 cm 5 -02 m R2 5 120 cm 5 102 m The lens-maker’s equation is therefore: 1 1 1 5 (n21) 2 5 (166 2 1) R1 R2 f

(

)

( (-021 m) 2 (1021 m) )

( (021 m) 2 (021 m) ) 5 -66 m-

1

5 (066) -

1 5 -015 m 5 -15 cm -66 m-1

⇒f5

b) The lens’s optical power will be: P5

1 1 5 5 -66 d -015 m f

The two previous examples show that when biconcave and biconvex lenses with the same fabrication parameters are compared, the absolute values of the results are the same. However, in algebraic value, f and P are positive for the biconvex lens because it is converging, and negative for the biconcave lens because it is diverging.

Furthering

your understanding

The lens-maker’s equation when a lens is immersed in a medium other than air The lens-maker’s equation is valid only when the lens is in air 1 1 1 5 (n 2 1) 2 fair R1 R2

(

)

If the lens is immersed in a medium other than air, the medium’s index of refraction (n) must be taken into account in the formula In this case, the lens-maker’s equation is as follows: 1 (n 2 n′) 5 n′ fmedium

(R1 2 R1 ) 1

2

where n 5 Index of refraction of the lens n′ 5 Index of refraction of the medium The focal length (f ) of the lens will therefore be different when it is immersed in a medium other than air To evaluate this difference, calculate the relationship between focal lengths 1 fair 1 fmedium

1 1 2 ) ( R R f 5 ⇒ (n 2 n′) 1 1 (R 2 R ) f n′ (n 2 1)

1

2

medium air

1

2

5

Therefore: fmedium n′ 3 (n 2 1) 5 (n 2 n′) fair If, for example, a glass lens is immersed in water, the variation in focal length (f ) is calculated by making n’ 5 index of refraction of water 5 133 and n 5 index of refraction of glass 5 15 fwater n′ 3 (n 2 1) 133 3 (15 2 1) 5 5 5 391 fair (n 2 n′) (15 2 133) ⇒ fwater 5 391 3 fair Therefore, the focal length (f ) of the glass lens immersed in water is approximately four times longer than that of the lens in air

(n 2 1) (n 2 n′) n′

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111

SECTION 4.3

Optical power of lenses

1. Is the optical power o a diamond lens lower, higher or equal to the optical power o a glass lens that has the same shape? 2. What is the ocal length o two converging lenses placed side by side and whose optical powers are 25 d and 40 d, respectively? 3. A converging lens and a diverging lens have ocal lengths o 100 cm and ­150 cm, respectively What is the ocal length o the combination o these two lenses? 4. Reproduce the ollowing table on a sheet o paper and complete it Type of lens

Shape

R1

R2

9. What is the ocal length (f ) o a diverging lens whose optical power (P ) is -25 dioptres (d) ? 10. Two converging lenses are placed in combination One has a ocal length (f ) o 100 cm, and the other o 250 cm What is the ocal length o the two lenses combined? 11. Two lenses are placed in combination The frst has an optical power (P ) o 25 dioptres (d) The optical power o the two lenses combined is 40 dioptres What is the ocal length (f ) o the second lens? What type o lens is it? 12. A biconcave lens is made out o crown glass The radii o curvature (R ) o its suraces are 12 cm and 7 cm What is the optical power o the lens? 13. A biconvex glass lens has one radius o curvature (R ) that is double that o the other Knowing that its ocal length (f ) is equal to 20 cm, what are the lens’s radii o curvature? 14. A converging diamond lens and a crown glass lens have the same shape Which will have the longer ocal length (f )? Explain your answer

Planoconcave Negative (2)

Positive (1)

5. Determine the optical power (P) o a converging lens whose ocal length (f ) is 35 cm 6. The optometrist prescribes you a pair o 325­dioptre eyeglasses (d) What will the lenses’ ocal length (f) be? 7. A lens’s optical power (P ) is -55 dioptres (d) a) What is its ocal length (f ) ? b) What type o lens is it? 8. What is the optical power (P ) o a diverging lens whose ocal length (f ) is -200 cm?

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UNIT 1 Geometric Optics

15. An optical system is created using two lenses with ocal lengths (f ) o 12 cm and 20 cm What is their combined ocal length? 16. A 4­dioptre (d) biconvex lens, and a biconcave lens whose ocal length (f ) is 7 cm are combined a) What is the optical power (P) o the optical system? b) Is the system converging or diverging?

4.4 Images formed by lenses Like spherical mirrors, lenses form images. Though many similarities exist between the two optical systems, certain fundamental differences must be taken into account. Mirrors, for instance, use the phenomenon of reflection to deviate light rays and produce images. In lenses, it is refraction that causes light’s trajectory to alter and images to be formed. This difference is important. In the case of refraction, light rays must pass through the material of the optical system, which is not the case for reflection. One of the consequences of this difference is that the lenses’ focal length (f ) does not depend solely on the radius of curvature (R) as it does with spherical mirrors. It also depends, as shown by the lens-maker’s equation, on the index of refraction (n) of the material the lenses are made of. Thus, two mirrors with the same radius of curvature will have the same focal length, but two lenses with the same radius of curvature can have different focal lengths if they are made from different materials.

See Reflection on a plane mirror: the laws of reflection, p. 46. See Phenomenon of refraction, p. 78.

See Lens-maker’s equation, p. 109.

This section covers the geometric construction of images formed by lenses, and the mathematical formalism that describes them.

4.4.1

Geometric construction of images formed by lenses

Through their ability to refract light rays passing through them—either by making them converge or diverge—lenses form different kinds of images. Like the images formed by spherical mirrors, it is possible to geometrically construct the formation of these images. For simplicity’s sake, converging lenses and diverging lenses will be treated separately.

Converging lenses To construct the images formed by converging lenses, the properties of the converging lenses’ three principal rays are used. To illustrate this method, consider an object that rests on the lens’s principal axis (P) at a distance from the lens that is more than twice the focal length (f ) (see Figure 21). The examples on the following page will be examined subsequently.

F′

O f

2f

F

F′

P

O

F

P

F′

O

Ray 2

f

F

P Image

See Principal rays of a converging lens, p. 101.

O

F

F′

Ray 2

2f

P Image Ray 2 Ray 3

Ray 1

a) Draw the first principal ray.

Ray 1

Ray 1

b) Draw the second principal ray.

c) Draw the image between the principal axis (P) and the intersection of the two principal rays on the side of the refracted rays.

Ray 1

d) Draw the third principal ray to verify the image.

Figure 21 Formation of an image by a biconvex (converging) lens when the object is located at a distance of more than twice the focal length (f ) from the lens

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113

F′

O

F

P

a) Object located at an infinite distance away

Figure 21 on the previous page shows that the fi rst two principal rays are sufficient to obtain an image. The third ray merely verifies that the image is in the right place. The image obtained in this case is real, inverted, smaller than the object and located farther than the focal point (F), but closer than 2f. To study the different kinds of images it is possible to obtain with a converging lens, the object will be moved along the principal axis (P). Depending on the object’s position, five other cases are possible (see Figure 22). The characteristics of images formed by a converging lens are summarized in Table 2. Table 2 Characteristics of images formed by a converging lens

O

F

F′

P

b) Object located at a distance equal to 2f

O

F

F′

Type

Infinite

Real

Greater than 2f

Real

Inverted

Smaller than the object

Between f and 2f

Equal to 2f

Real

Inverted

Identical to the object

Equal to 2f

Between f and 2f

Real

Inverted

Larger than the object

Greater than 2f

F

P

d) Object located at a distance equal to f

O

F

F′

Size

Lens distance

Point image

Equal to f

c) Object located at a distance between f and 2f

O

Orientation

Equal to f

No image

P

Smaller than f

F′

Image characteristics

Object distance

P

Virtual

Upright

Larger than the object

On the object side, farther from the lens than the object is

Converging lenses therefore form five different kinds of images—real or virtual, inverted or upright, enlarged or reduced, and with positions that vary according to the object’s position with respect to the lens. Converging lenses do not form an image when the object is at the principal focal point (F).

Diverging lenses To construct the images formed by diverging lenses, the properties of the lenses’ three principal rays are again used. Here, again, the illustration of the geometric method for image construction deals with an object that rests on the principal axis (P) of the lens, at a distance that is greater than twice its focal length (f ) (see Figure 23 on the following page). As with converging lenses, the two principal rays are sufficient to obtain the image, while the third serves to verify it. Moreover, it is worth remembering that diverging lenses only form one kind of image. If the object’s position on the principal axis (P) changes, only the size of the image is likely to change, but will always be smaller than the object. Table 3 lists the properties of images formed by a diverging lens.

e) Object located at a distance smaller than f

Figure 22 Images formed by a biconvex (converging) lens. The object’s position determines the image’s characteristics. See Principal rays of a diverging lens, p. 103.

114

UNIT 1 Geometric Optics

Table 3 Characteristics of images formed by a diverging lens Image characteristics

Object position

Type

Orientation

Size

Position

Any position

Virtual

Upright

Smaller than the object

Between the lens and its principal focal point (F) (on the side of the object, between the object and the lens)

Ray 1

F

F′

O f

Ray 1

F

P

Image

f

2f

2f

a) Draw the first principal ray and extend the refracted ray to the lens’s focal point (F).

c) Draw the image between the principal axis (P) and the intersection of the two extended principal rays on the side of the incident rays.

Ray 1

F

Ray 2 P

F′

O

O

F′

Ray 1

Ray 2 P

Ray 2 P

F′

O

F Image

Ray 3

b) Draw the second principal ray and extend the refracted ray to the side the object is on.

d) Draw the third principal ray to verify the image.

Figure 23 Formation of an image by a biconcave (diverging) lens when the object is located at a distance of more than twice the focal length (f ) from the lens

In summary, diverging lenses produce images that are always virtual, upright, reduced and located on the side of the object, between the object and the lens.

4.4.2

Mathematical formalism of lenses

As with spherical mirrors, the parameters that characterize lenses are linked by mathematical equations. This section deals with the derivation of these equations, and their use in solving problems involving lenses.

See Mathematical formalism of spherical mirrors, p. 61.

Thin-lens equation To find the mathematical relationships linking the parameters that characterize lenses, consider the case of a biconvex lens with an object at a distance that lies between f and 2f (see Figure 24). Note that these mathematical relationships do not depend on lens type, nor on the object’s position. The thin-lens equation links the lens’s focal length (f ), the image distance (di) and the object distance (do). It reads as follows:

Object ho F′ f do

F

O

P

f

hi di

Image

Figure 24 The different parameters that character-

1 1 1   di f do

ize lenses: focal length (f ), image distance (di), object distance (do), object height (ho) and image height (hi).

Thin-lens equation resembles the spherical mirror equation.

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115

Note that when the object is placed an infnite distance away rom the lens, do is 1 1 infnite (do → ∞) and the raction d tends to zero ( d → 0). In this case, f 5 di, o o which means that the image is positioned on the ocal point. This result was obtained in the diagram in Figure 22a on page 114. Magnifcation (M) is always expressed the same way.

M5

hi d 5 i ho do

These two ormulas are derived mathematically as ollows: X

A ho Object F′

O

F

Y

B

hi Image

P

Z do

f di

Figure 25 Parameters involved in the ormation o images by a convex (converging) lens

Figure 25 shows that triangles ABO and ZYO are similar It ollows that: YZ YO 5 BA BO Since YZ 5 hi, BA 5 ho, YO 5 di and BO 5 do, we can say that: hi d 5 i ho do Two other triangles in Figure 25 are similar: XOF and ZYF From this, we can say that: OX YZ 5 OF YF Since OX 5 ho, OF 5 f, YZ 5 hi and YF 5 (di – f ) : hi ho h (d 2 f ) 5 ⇒ i 5 i (di 2 f ) f ho f Above, we saw that: hi d 5 i ho do This is the equation or magnifcation (M) It is identical to the equation derived or spherical mirrors I the two equations are combined: hi (d 2 f ) 5 i ho f d hi 5 i ho do

116

UNIT 1 Geometric Optics

⇒

di d d (d 2 f ) f 5 i 5 i 2 5 i 21 f do f f f

And i the two sides o the equation are divided by di : di di di di 1 1 1 1 ⇒ 5 5 2 2 do f di do f di By rearranging the terms, we obtain: 1 1 1 5 1 f do di

Sign convention To take the different characteristics of images into account, defining a sign convention is essential. You will remember that the presence of signs (1 and -) is intended to make mathematical calculations consistent with experimental results and with geometric constructions of images formed by lenses. Different sign conventions exist, but the one used in this textbook is as follows: Distances are measured rom the optical centre (O) o lenses Focal lengths (f ) are positive or converging lenses, and negative or diverging lenses The image distances (di) and object distances (do) are positive or real images and real objects The image distances (di) and object distances (do) are negative or virtual images and virtual objects The image heights (hi) and object heights (ho) are positive when they are upright and negative when they are inverted 6. To abide by this convention, magnifcation (M ) must be written:

1. 2. 3. 4. 5.

M5

hi d 5- i ho do

When applied to the thin­lens equation, this sign convention not only gives the images’ characteristics, but predicts their position as well. Example A Converging lens, object located at a distance greater than 2f An object 8 cm high is located 24 cm rom a converging lens whose ocal length (f ) is 10 cm Determine: a) the image position (di) b) the magnifcation (M ) c) the image height (hi) Data: Because the lens is converging, f is positive: f 5 10 cm • The object is real: d 5 130 cm o • The object is upright: h 5 18 cm o di 5 ? g 5? hi 5 ? •

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117

Example A (cont. ) Solution: a) Calculation o image position: 1 1 1 1 1 1 1 5 1 ⇒ 5 2 ⇒ di 5 f do di di f do 1 1 2 f do Numerical application: di 5

1 1 1 1 5 5 5 5 15 cm 1 1 1 1 3 1 2 2 2 2 f do 10 cm 30 cm 30 cm 30 cm 30 cm

The image distance is positive, which means that the image is real The numerical value indicates that the distance is between f and 2f b) Calculation o magnifcation (M): h d 15 cm 1 M5 i 5- i 55 - 5 - 05 ho do 30 cm 2 The magnifcation is negative; the image is inverted The absolute value o magnifcation is less than 1; the image is smaller than the object The numerical value shows that the image is twice as small as the object c) Calculation o image height: h M 5 i ⇒ hi 5 M 3 ho ho Numerical application: hi 5 M 3 ho 5 - 05 3 8 cm 5 - 4 cm The image height is negative; the image is inverted (this result was obtained in the previous question) These results are consistent with those obtained using the geometric construction in Figure 21 on page 113 and are summarized in Table 2 on page 114

Example B Converging lens, object located at a distance equal to 2f An object 8 cm high is located 20 cm rom a converging lens whose ocal length (f ) is 10 cm Determine: a) the image position (di) b) the magnifcation (M ) c) the image height (hi) Data: f 5 110 cm • d 5 120 cm o • h 5 18 cm o di 5 ? M 5? hi 5 ? •

Solution : a) Calculation o image position: di 5

118

UNIT 1 Geometric Optics

1 1 1 1 5 5 5 5 20 cm 1 1 1 1 2 1 1 2 2 2 f do 10 cm 20 cm 20 cm 20 cm 20 cm

Example B (cont.) The image distance is positive, which means that the image is real The numerical value indicates that the distance is exactly equal to 2f b) Calculation o magnifcation: h d 20 cm M5 i 5- i 55 -1 ho do 20 cm The magnifcation is negative; the image is inverted The absolute value o the magnifcation is equal to -1; the image is the same size as the object c) Calculation o image height: hi 5 M 3 ho 5 -1 3 8 cm 5 -8 cm The image height is negative; the image is inverted (this result was obtained in the previous question) These results are consistent with those obtained using the geometric construction in Figure 22b and are summarized in Table 2 on page 114 Example C Converging lens, object located at a distance between f and 2f An object 8 cm high is located 15 cm rom a converging lens whose ocal length (f ) is 10 cm Determine: a) the image position (di) b) the magnifcation (M ) c) the image height (hi) Data: f 5 110 cm • d 5 115 cm o • h 5 18 cm o di 5 ? M 5? hi 5 ? •

Solution : a) Calculation o image position: 1 1 1 1 5 5 5 5 30 cm di 5 1 1 1 1 3 2 1 2 2 2 f do 10 cm 15 cm 30 cm 30 cm 30 cm The image distance is positive, which means that the image is real The numerical value indicates that the distance rom the lens is greater than 2f b) Calculation o magnifcation: h d 30 cm M5 i 5- i 55 -2 ho do 15 cm The magnifcation is negative; the image is inverted The absolute value o magnifcation is greater than 1; the image is larger than the object The numerical value shows that the image is twice as large as the object c) Calculation o image height: hi 5 M 3 ho 5 - 2 3 8 cm 5 -16 cm The image height is negative; the image is inverted (this result was obtained in the previous question) These results are consistent with those obtained using the geometric construction in Figure 22c and are summarized in Table 2 on page 114

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119

Example D Converging lens, object located at a distance equal to f An object 8 cm high is located 10 cm rom a converging lens whose ocal length (f ) is 10 cm Determine: a) the image position (di) b) the magnifcation (M ) c) the image height (hi) Data: f 5 110 cm • d 5 110 cm o • h 5 18 cm o di 5 ? M 5? hi 5 ? •

Solution : a) Calculation o image position: di 5

1 1 1 2 f do

5

1 1 1 2 10 cm 10 cm

5

1 0

b) Does not apply because there is no image c) Does not apply because there is no image The equation is indeterminate No image is ormed, which is consistent with Figure 22d and as summarized in Table 2 on page 114

Example E Converging lens, object located at a distance smaller than f An object 8 cm high is located 4 cm rom a converging lens whose ocal length (f ) is 10 cm Determine: a) the image position (di) b) the magnifcation (M ) c) the image height (hi) Data: f 5 110 cm • d 5 15 cm o • h 5 18 cm o di 5 ? M 5? hi 5 ? •

Solution : a) Calculation o image position: di 5

1 1 1 2 f do

5

1 1 1 2 10 cm 5 cm

5

1 1 2 2 10 cm 10 cm

5

1 5 -10 cm -1 10 cm

The image distance is negative, which means that the image is now virtual The numerical value indicates that the image is arther rom the lens than the object

120

UNIT 1 Geometric Optics

Exmle E (cont. ) b) Calculation o magnifcation: h d -10 cm M5 i 5- i 55 12 5 cm ho do The magnifcation is positive; the image is upright. The absolute value o the magnifcation is greater than 1; the image is larger than the object. The numerical value shows that the image is twice as large as the object. c) Calculation o image height: hi 5 M 3 ho 5 2 3 8 cm 5 16 cm The image height is positive; the image is upright (this result was obtained in the previous question). These results are consistent with those obtained using the geometric construction in Figure 22e and are summarized in Table 2 on page 114. Exmle F Diverging lens An object 8 cm high is located 30 cm rom a diverging lens whose ocal length (f ) is 10 cm. Determine: a) the image position (di) b) the magnifcation (M ) c) the image height (hi) Data: Because the lens is diverging, f is negative: f 5 -10 cm • The object is real: d 5 130 cm o • The object is upright: h 5 18 cm o di 5 ? M 5? hi 5 ? •

Solution : a) Calculation o image position: 1 1 1 1 5 5 5 5 -7.5 cm di 5 -4 1 1 1 1 3 1 2 2 2 30 cm f do -10 cm 30 cm -30 cm 30 cm The image distance is negative, which means that the image is virtual. The numerical value indicates that the image is closer to the lens than the object is. b) Calculation o magnifcation: h d -7.5 cm 1 M5 i 5- i 55 1 5 10.25 30 cm ho do 4 The magnifcation is positive; the image is upright. The absolute value o the magnifcation is less than 1; the image is smaller than the object. The numerical value shows that the image is one-quarter the size o the object. c) Calculation o image height: hi 5 M 3 ho 5 0.25 3 8 cm 5 2 cm The image height is positive; the image is upright (this result was obtained in the previous question). These results are consistent with those obtained using the geometric construction in Figure 23 on page 115 and are summarized in Table 3 on page 114.

ChaptER 4 Lenses

121

SECTION 4.4

Images formed by lenses

1. Use the principal rays to locate the image of the tree formed by the following lens.

7. Using a scaled diagram, locate the image of 5-cm object placed 15 cm from a diverging lens whose focal length (f ) is 25 cm. 8. A lamp 10 cm high is placed 60 cm from a diverging lens whose focal length (f ) is 20 cm. Using a scaled diagram, locate the image and determine its size. 9. Check your answers to the previous three questions by applying the thin-lens equation.

2. Use the principal rays to locate the image of the tree formed by the following lens.

10. A lens has a focal length (f ) of 20 cm and a magnification (M ) of 4. What is the distance between the object and the image? 11. Reproduce the following table on a piece of paper and complete it. Lens

F′

O

F

P

f (cm)

20

-20

do (cm)

25

25

20

15

di (cm)

20

-10

M

-1

Real/virtual image Orientation

3. Using a scaled diagram, find the image of a 5-cm candle located at the following distances from a converging lens whose focal length (f ) is 15 cm: a) 30 cm b) 25 cm c) 15 cm d) 10 cm In each case, describe the type, height and orientation of the image. 4. Repeat the previous question with a diverging lens. 5. An object 8 cm high is placed 80 cm from a converging lens whose focal length (f ) is 15 cm. Using a scale diagram, locate the image and determine its height. 6. Using a scaled diagram, locate the image of a 10-cm candle placed 20 cm from a converging lens whose focal length (f ) is 25 cm.

122

UNIT 1 Geometric Optics

Converging Diverging

Real

Virtual

10

-0.5 Real Inverted

12. A lens has a magnification (M ) of 2. If its focal length (f ) is 10 cm, determine: a) the image position b) the object position c) the type of lens 13. A lens has a magnification (M ) of 0.5. If its focal length (f ) is -20 cm, determine: a) the object position b) the image position c) the type of lens 14. A candle is placed 36 cm from a screen. At what point between the candle and the screen should a converging lens with a focal length (f ) of 8 cm be placed in order to obtain a clear image on the screen?

4.5 Optical aberrations of lenses Aberrations are deects in optical systems that prevent them rom ocusing the light rays passing through them on a single point. Spherical lenses can have numerous aberrations, which all into two categories: chromatic aberrations and geometric aberrations, which include spherical aberrations. Chromatic aberrations are produced only when the light used is polychromatic, while geometric aberra­ tions are produced when either monochromatic and polychromatic light are used.

4.5.1

Chromatic aberrations

Chromatic aberrations are produced only when the incident light is composed o a number o monochromatic lights, as is the case with white light. The lens-maker’s equation shows that a lens’s ocal length (f ) depends on the index o reraction (n) o the lens material. Because the index o rerac­ tion varies in relation to the wavelength (l) o the light passing through the lens, the ocal length will also vary as a unction o the wavelength. Thus, the lens—more light­reracting or violet light (l ≈ 400 nm) than or red (l ≈ 700 nm)—will ocus the dierent­coloured lights that compose white light on dierent points o the principal axis (P). As a result, the ocusing distance o violet light will be shorter than that o red light. This decomposition o light is an optical phenomenon known as dispersion. Chromatic aberration is thereore an optical deect caused by the phenomenon o dispersion, which makes the dierent­coloured lights crossing the lens ocus on dierent points (see Figure 26). When a converging lens is involved, chromatic aberration can be corrected by combining the lens with a diverging one o lower optical power (in absolute value) made rom a dierent material. The materials used—such as crown glass* and fint glass*—are chosen or the dierence in their dispersions. Such a two­ lens combination is called an achromatic doublet, or colour­corrected lens (see Figure 27).

See Lens-maker’s equation, p 109

glass Highly transparent and * Crown nearly nondispersive glass used to

*

make optical lenses Flint glass Highly dispersive leaded glass

P

P

Figure 26 Chromatic aberration of a biconvex lens The different monochromatic lights that compose white light are focused on different points of the principal axis (P) The shortest focusing distance corresponds to violet light, while the longest characterizes red light

Figure 27 An achromatic doublet, which combines a converging lens and a diverging one, focuses different-coloured lights on a same point

CHAPTER 4 Lenses

123

4.5.2 O

P

Figure 28 Spherical aberration of a biconvex lens Light rays passing close to the lens’s principal axis (P) do not converge on the same point as rays passing at the periphery

Spherical aberrations

Though a range o geometric aberrations exist, only spherical aberrations will be discussed here. As their name suggests, spherical aberrations are caused by the spherical shape o the suraces that deine the lens. Like spherical mirrors, spherical lenses do not converge all the rays hitting their surace on the same point. Light rays that pass close to a lens’s principal axis (P) do not converge on the same point as rays that pass at the pe­ riphery. Spherical aberration is thereore an optical deect caused by lenses’ spherical shape, which makes rays close to the principal axis and rays peripheral to it ocus on dierent points (see Figure 28). Spherical aberrations can be avoided by making lenses parabolic in shape. This method, however, is difcult and costly, and the preerred solution is to use a combination o lenses o dierent spherical curvatures.

Materials used in contact lenses The idea of placing a material directly in the eyes to correct vision problems is not new As early as 1508, Leonardo da Vinci was sketching out ideas for what would eventually become the contact lens It was not until 1887, however, that a glass prototype tolerated by the eye was developed By the 1930s, contact lenses made from polymethyl methacrylate—better known by its trade name, Plexiglas—had begun to be developed Yet the material—rigid and highly impermeable to oxygen—had disadvantages: if contact lenses do not allow the cornea to be oxygenated, their use can lead to corneal disease The first rigid contact lenses permeable to oxygen were made from silicone rubber The new material, however, also had its drawbacks as its physiochemical properties encouraged the development of fat and protein deposits Soft contact lenses (or soft lenses) appeared in the early 1960s

SECTION 4.5

2. A converging lens is hit with a beam of green light and then a beam of red

UNIT 1 Geometric Optics

Silicone hydrogel, developed in 1999, is the standard material in use today for contact lenses It enhances oxygen permeability, prevents ocular dryness and limits organic deposits Soft lenses made from silicone hydrogel can be worn for a month without having to be removed (see Figure 29) Figure 29 A monthly-use silicone hydrogelbased lens

Optical aberrations of lenses

1. What is a chromatic aberration? How can it be corrected?

124

To combine the advantages of Plexiglas and silicone rubber, a new material was synthesized: silicon acrylate It and its variants were permeable enough to provide proper corneal oxygenation, and minimized deposits of fats and proteins

a) Does the lens’s focal length (f ) change? b) If so, with which light is the shorter focal length obtained? 3. What causes the spherical aberration of a lens?

APPLICATIONS Unifocal and multifocal lenses Because they have only one principal focal point (F), single lenses are said to be unifocal and are charac­ terized by a single optical power. They are used to correct some of the visual disorders associated with far and near vision. However, when a person is simultaneously near­ and far­sighted, other solutions are available. He or she can use two pairs of unifocal eyeglasses, one for far vision and the other for near or, more practically, a pair of bifocal lenses. In this type of lens, one area (the bottom portion) is reserved for near vision, while the other (the top) is for far. Each area has a different optical power, and the transition between the two is abrupt (see Figure 30).

progressive lenses. These lenses have three areas— the two present in bifocal lenses, and an interme­ diate area designed to ease the transition between them. The optical power of this intermediate area shifts gradually between the first two areas’ ex­ tremes (see Figure 31), considerably improving the wearer’s comfort.

Far vision

This abrupt transition in optical power can disrupt vision. A more natural correction is obtained with

Near vision

Far vision

Near vision

Figure 30 Eyeglasses with bifocal lenses, also known as dual-focal-length lenses The bottom area is for near vision while the upper area is for far vision The transition between the two areas is abrupt

Intermediate area with progressive optical power

Figure 31 A progressive lens, which provides a gradual transition between the upper and lower areas

In general, the intermediate area is 15 to 20­mm high, and the optical power ranges between 1 and 2.5 dioptres (d). The more recently created progres­ sive lenses are described as digital, because they are manufactured by computer­controlled ma­ chines. These lenses have the advantage of correspond­ ing very closely to prescriptions and being better adapted to facial morphology. The result is made­ to­measure lenses that provide maximum comfort with the needed correction.

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125

Eyeglasses Vision problems are probably as old as humanity itsel, and eorts to correct them date to antiquity. The Greek philosopher Aristotle was describing vision problems such as presbyopia and myopia as early as the 4th century BCE, while the frst recorded use o lenses to modiy vision dates to the Roman empire. The philosopher Seneca (1st century CE) noticed that when he read a text through a glass globe flled with water, the letters appeared larger and clearer. At around the same time, the Latin writer Pliny the Elder reported that Roman emperor Nero watched the gladiatorial games through an emerald. The monastic scribes o 1000 CE used reading stones, glass hemispheres that could be slid over the pages o a manuscript, acting as a magniying glass (see Figure 32). The feld o optics, clearly, had progressed little—at least in the West. At around the same time, the 11th­century Egyptian Alhazen was publishing the frst scientifc treatise on optics based on his research. The work was translated into Latin in the 12th century and soon became a standard reerence throughout Europe. The frst glasses used to correct vision—spectacles— appeared in the 13th century in Italy. The invention’s

Figure 32 A reading stone

126

UNIT 1 Geometric Optics

authorship is unknown, though the names o Roger Bacon (1214–1294), Salvino degli Armati (who died in 1317) and Alessandro di Spina (who died in 1313) have been cited by historians. These eyeglasses were rudimentary, consisting o two quartz or beryl crystal biconvex lenses inserted in a wooden or horn rame (see Figure 33). Without the arms, or sidepieces, o today’s eyeglasses, they had to be held in ront o the eyes. The arms were not to appear on eyeglass rames until the 17th century. Initially, eyeglasses had only biconvex lenses, com­ pensating or presbyopia. Following Gutenberg’s in­ vention o the printing press in Germany in the 16th century, their use became more widespread. Growing interest in the invention led to the correction o my­ opia using biconcave lenses. In the early 16th century, the feld o geometric optics developed. European scientists o the time, among them Kepler, Descartes and Snell, dedicated themselves to the new discipline. Thanks to scien­ tifc and technical progress, the making o lenses and eyeglasses was improved, paving the way or two major inventions o the 17th century: the telescope and the microscope.

Figure 33 Spectacles consisted of two lenses framed with wood or horn and topped with two rods held together with a nail

chapter

4

Lenses

4.1 Different types of lenses • A lens consists o a transparent object that has at least one curved surace and controls the direction o the light rays passing through it. • A lens is said to be converging when it ocuses rays parallel to its principal axis (P) and entering one o its suraces on a point beyond its other surace. • Biconvex lenses are ormed rom two convex spherical suraces. • Planoconvex lenses are defned by a at surace and a convex spherical one. • Positive meniscus lenses are made up o two spherical suraces: one concave and the other convex. • Biconvex lenses, planoconvex lenses and positive meniscus lenses are converging lenses. • A lens is said to be diverging when it deviates rays parallel to its principal axis (P) and entering one o its suraces to dierent points that grow arther apart past its other surace. • Biconcave lenses are ormed rom two concave spherical suraces. • Planoconcave lenses are defned by a at surace and a concave spherical one. • Negative meniscus lenses are made up o two spherical suraces: one convex and the other concave. • Biconcave lenses, planoconcave lenses and negative meniscus lenses are diverging lenses.

4.2 Refraction in lenses • When light travels through a lens in air, it undergoes two reractions: the frst at the air­glass interace, and the second at the glass­air interace. • The optical centre (O) o a lens corresponds to its geometric centre. • The principal ocal point (F) o a converging lens is the point at which rays parallel to the principal axis (P) converge ater passing through the lens. It is located on the side where the light rays emerge. • The principal ocal point (F) o a diverging lens is the point rom which light rays emerging rom the lens appear to originate. It is located on the side o the incident rays. • The secondary ocal point (F′) o a lens is the point symmetrical to the principal ocal point (F) in relation to the optical centre (O). • The ocal length (f) o a lens is the distance between the optical centre (O) and the principal ocal point (F). • The ocal plane is the plane perpendicular to the principal axis (P) that passes through the principal ocal point (F). • Converging lenses and diverging lenses have three principal rays.

CHAPTER 4 Lenses

127

(cont.) • Any incident ray parallel to the principal axis (P) is reracted by a converging lens through its principal ocal point (F). • Any incident ray parallel to the principal axis (P) is reracted by a diverging lens in such a way that it appears to come rom its principal ocal point (F). • Any incident ray that passes through the secondary ocal point (F′) o a converging or diverging lens is reracted parallel to the principal axis (P). • Any incident ray that passes through the optical centre (O) o a converging or diverging lens does not undergo any deviation.

4.3 Optical power of lenses • Optical power (P) is a measure that quantifes the capacity o a lens or optical system to deviate light rays. • The optical power (P) o a lens depends on its radii o curvature (R) and the index o reraction (n) o the material it is made rom. • The unit o optical power is the dioptre (d). • A lens’s optical power is inversely proportional to its ocal length (f ).

P5

1 f

• According to the sign convention, the radii o curvature (R1 and R2) are positive when the surace hit by the incident ray is convex (bulged), and negative when the surace is concave (hollow). • The sign convention stipulates that ocal length (f ) and optical power (P) are positive or a converging lens and negative or a diverging lens. • The optical power (P) o a system o lenses is equal to the algebraic sum o each lens’s individual power. • When total optical power is zero, the optical system is said to be aocal. • The lens­maker’s equation calculates a lens’s ocal length (f ) rom its radii o curvature (R) and the index o reraction (n) o the material it is made rom. • The lens­maker’s equation reads as ollows: 1 1 1 5 (n 2 1) 2 f R1 R2

(

128

UNIT 1 Geometric Optics

)

4.4 Images formed by lenses • The phenomenon o reraction causes lenses to alter light’s path and orm images. • To construct the images ormed by lenses, the properties o their three principal rays are used. • Converging lenses orm fve dierent kinds o images. These images can be real or virtual, inverted or upright, enlarged or reduced, and in positions that vary according to the object’s position. • Converging lenses do not orm an image when the object is at the principal ocal point (F). • Diverging lenses only orm one kind o image. These images are always virtual, upright, reduced and located between the lens and its principal ocal point (F). • The thin­lens equation links the lens’s ocal length (f ), the image distance (di) and the object­lens distance (do). It reads as ollows: 1 1 1 5 1 f do di • According to the sign convention used in this textbook, magnifcation is expressed as ollows:

M5

hi d 5- i ho do

• The presence o the negative sign is intended to make mathematical calculations consistent with experimental results and with geometric constructions o images ormed by lenses.

4.5 Optical aberrations of lenses • Aberrations are deects in optical systems that prevent them rom ocusing the light rays passing through them on a single point on the principal axis (P). • Chromatic aberration is an optical deect caused by the phenomenon o dispersion, which makes the dierent­coloured lights crossing the lens ocus on dierent points on the principal axis (P). • Spherical aberration is an optical deect caused by lenses’ spherical shape, which makes rays close to the principal axis, and rays peripheral to it ocus on dierent points. • Chromatic and spherical aberrations can be corrected by combining lenses with dierent curvatures and optical powers.

CHAPTER 4 Lenses

129

CHAPTER 4

Lenses

1. Why do spherical lenses produce chromatic aberrations while spherical mirrors do not? 2. A planoconvex lens is to be manuactured with the ollowing characteristics: a) ocal length (f ): 10 cm b) material: glass

7. Consider an optical system made up o two lenses The rst is a planoconvex lens and the second is a planoconcave lens

Planoconvex lens

Planoconcave lens

What must the lens’s radius o curvature (R ) be? 3. A converging optical system whose ocal length is 40 cm consists o three lenses The characteristics o the rst two lenses are as ollows: Lens

Converging/ diverging

Focal length (f )

1

Converging

20 cm

2

Diverging

10 cm

What are the characteristics o the third? 4. What does the lens­maker’s equation become in the case o a planoconvex lens? Explain why such a lens is converging 5. A planoconvex lens consists o a fat surace, and a spherical surace whose radius o curvature (R ) is 25 cm The index o reraction (n) corresponding to blue light is nb 5 1523, while the index o rerac­ tion o red light is nr 5 1514 a) What is the distance between the two principal ocal points? b) What is the phenomenon called when dierent­ coloured lights are ocused on dierent points on the principal axis (P)? 6. A diverging meniscus with radii o curvature o 10 cm and 22 cm has a ocal length (f ) o 35 cm What material is the lens made rom?

130

UNIT 1 Geometric Optics

Both lenses have the same radius o curvature R 5 15 cm a) What is the system’s optical power (P ) i the two lenses are made rom the same material? b) Calculate the system’s optical power (P ) i the materials are dierent These materials are shown in the ollowing table: Lens

Type

Material

1

Planoconvex

Glass

2

Planoconcave

English crystal

8. An object 5 mm high is placed 10 cm rom a lens whose optical power is 5 dioptres (d) Determine: a) the image position b) the magnication (M ) c) the type o lens 9. A lens’s magnication (M ) is 05, and its optical power is -5 dioptres (d) Determine: a) the object position b) the image position c) the type o lens 10. A biconvex glass lens has a radius o curvature (R ) o 12 cm Knowing that the second radius o curva­ ture is double the rst, determine: a) its ocal length (f ) b) the image distance (di) i an object is 20 cm to the let o the lens c) the magnication (M )

Applied Geometric Optics

G

eometric optics is a science that enables the analysis and explanation o numerous natural phenomena, such as vision, the ormation o rainbows or the appearance o mirages. This science is also used in the production o technological devices that make it possible to take photographs, see the ininitely small or examine the infnitely large.

Review Refection o light                                          8 Reraction o light                                          9 Converging lenses and diverging lenses                       10 The eye                                                 11

In this chapter, you will be able to understand how a complex organ such as the eye unctions, and you will discover the most common visual disorders. You will also become amiliar with highly useul optical devices such as the camera, the microscope and the telescope.

5.1 5.2 5.3 5.4

The camera                             132 The human eye                          134 The light microscope                    139 The telescope                           141 CHAPTER 5 Applied Geometric Optics

131

5.1 The camera The word “photography” comes rom two Greek words: phôtos, meaning “light,” and graphein, meaning “to write.” As such, photography literally means “writing with light.” The camera is the perect instrument or capturing images. Although it has undergone numerous improvements over time, its operational principles have remained more or less the same. These consist in allowing light rays rom the object to be photographed to enter the opaque camera body, equipped with lenses and apertures allowing the object’s image to orm on a lightsensitive surace (see Figure 1). In the past, this surace was a light-sensitive flm, meaning it underwent a chemical reaction when it came into contact with light rays. Cameras today, which are digital, do not contain flm but are equipped with electronic sensors that are sensitive to the amount o light they receive. This light is then transormed into electric impulses that are coded in a binary electronic system and stored in a memory card. The term “digital” is based on this electronic orm o processing the inormation conveyed by light. Diaphragm Objective

Lens

Shutter O

F

Light-sensitive surace (flm or sensor)

See Images ormed by a camera obscura, p 56

Figure 1 The ormation o an image inside a camera The objectives in current cameras contain numerous lenses For reasons o clarity, a single lens is represented here and numerous other components have been omitted

The modern camera is a sophisticated camera obscura. The equivalent o the pinhole is an aperture secured by a diaphragm whose diameter is electronically controlled and varies according to the quantity o light one wants to let through to the light-sensitive surace (see Figure 2). Whereas the diaphragm is situated inside the objective, the shutter is located directly in ront o the light-sensitive surace. When a photograph is being taken, the shutter allows light to reach the light-sensitive surace. As such, the pairing o diaphragm and shutter make it possible to control the quantity o light required to orm the image. Figure 2 Consisting o fne overlapping blades, a camera’s diaphragm makes it possible to adjust the quantity o light passing through the objective

132

UNIT 1 Geometric Optics

Accurate focalization of the image is achieved using a system of lenses called the objective. The distance between the objective and the light-sensitive surface can be adjusted, making it possible to bring the image into focus to ensure that the image forms accurately on the light-sensitive surface. This way, the clearest image possible is obtained. The image formed by a camera is real and inverted. In addition to containing high-tech components, modern cameras feature many fully computerized functions, including autofocus, lens aperture and exposure time.

SECTION 5.1

The camera

1. Is the optical system o a camera’s objective converg­ ing or diverging? 2. List the characteristics o an image ormed on a camera’s light­sensitive surace 3. This is a diagram o a camera Explain why the object­ ive is mobile

Objective

4. What is the distance between the objective and the light­sensitive surace when an object being photo­ graphed is situated at an infnite distance? Explain your answer

5. A camera’s objective with a ocal length (f ) o 6 cm is situated 7 cm rom the light­sensitive surace At what distance rom the lens is the object located i a clear image orms on the light­sensitive surace? 6. An amateur photographer takes a series o photo­ graphs o a lunar eclipse using a camera with a ocal length (f ) o 6 cm I the average radius o the Moon is 174 3 106 m and the average radius o the Moon’s orbit is 384 3 108 m, what is the size o the image o the Moon on the light­sensitive surace? 7. A tourist wants to photograph a building that is 78 m tall and situated 400 m away rom him He uses a dis­ posable camera with a fxed objective, whose ocal length (f ) is 5 cm, and is equipped with a flm located on the lens’s ocal plane The base o the building can be considered to coincide with the lens’s princi­ pal axis (P) a) When looking through the viewfnder, at what angle does the tourist see the building? b) Where does the image orm? c) Calculate the size o the image on the flm

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5.2 The human eye The human eye is a highly complex organ shaped like a ball, with a diameter o about 2.5 cm in adults. It allows the continual ormation o images o a person’s environment. The ormation o images in the human eye ollows the same laws that lenses are subject to.

5.2.1 Crystalline lens Aqueous humor

Retina

Cornea

Pupil Iris Ciliary muscles Vitreous humor

Optic nerve

Figure 3 Anatomy of the human eye

Table 1 Average indexes of refraction of the eye’s transparent media Transparent medium

Index of refraction

Cornea

1376

Aqueous humor

1336

Crystalline lens

1386 to 1406

Vitreous humor

1337

Geometric optics of vision

A section o the eye (see Figure 3) shows that it contains our transparent media. From the outside in, one inds the cornea, the aqueous humor, the crystalline lens and the vitreous humor. Because these media are transparent, light can pass through them and reach the retina. The iris is a circular membrane situated in ront o the crystalline lens; its colouring gives eyes their colour. At its centre, there is an opening called the pupil. The pupil’s size is regulated by the iris according to the quantity o surrounding light, just like the diaphragm in a camera. Under normal conditions, the pupil’s diameter is approximately 4 to 5 mm. This diameter decreases to 2 mm when light is very bright and can reach 8 mm in extreme darkness. In accordance with their indexes o reraction and their shape, the dierent transparent media change the trajectory o light rays and orm images on the retina, which perorms the same unction as a camera’s lightsensitive surace. The average indexes o reraction o the eye’s transparent media are listed in Table 1. Table 1 shows that in contrast to the indexes o reraction o the other three transparent media, that o the crystalline lens is not uniorm. It increases rom the centre toward the outer edges.

Since the external surace o the cornea comes into contact with air (index o reraction o approximately 1), the largest reraction occurs at the air-cornea interace. Light rays that enter the eye continue to be reracted by the other media, however to a lesser degree since the dierence between their indexes o reraction is small. Since the position o an image depends on the position o the object, the eye must ensure that the image is always ormed on the retina rather than in ront o or behind it. In a camera, this ocalization is ensured by rotating the objective. Since the eye does not have any mobile components, this task is perormed by the crystalline lens. The crystalline lens has the shape o a biconvex lens, meaning it is converging, and its fexibility allows the ciliary muscles to change its shape. When an object is very distant, the reraction o the cornea is suicient to ocus the rays toward the back o the eye. The crystalline lens is not put to use in this case and takes on a thin, oblong shape (see Figure 4a on the following page). When the object being observed is close, the ciliary muscles

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contract, the crystalline lens bulges and its radii o curvature become smaller (see Figure 4b). This has the eect o increasing the optical power (P ) o the crystalline lens and thereby diminishes its ocal length (f ). As such, the crystalline lens is a converging lens with a variable ocal length.

See Optical power of lenses, p 106

When an image does not orm on the retina, one’s vision becomes blurred and requires correction. The ability o the crystalline lens to adjust its shape in order to allow or clear vision is called accommodation.

The maximum distance or clear vision is called ar point (FP). In a normal eye, it is infnite. In act, when the object is at ininity, the crystalline lens is at rest and the image orms at the eye’s ocal point (F). The minimum distance or clear vision is called near point (NP). This distance, corresponding to the maximum accommodation o the crystalline lens, is roughly equal to 25 cm in a normal eye in a young adult.

F

O

Figure 4 shows that the image is real and inverted. In a normal eye (said to be emmetropic), the position o this image is always the same—on the retina—regardless o the object’s position.

a) The object is far away

O

F

The ocal length (f ) o an emmetropic eye in the case o the minimum accommodation (ar point) is about 17 mm. Since optical power (P ) is inverse to the ocal length, we can write: b) The object is close

1 1 PFP 5 5 ≈ 59 δ f 17 3 10-3 m

Figure 4 Accommodation by the crystalline lens helps ensure that the image always forms on the retina

Since it is commonly acknowledged that the cornea’s optical power (P) represents approximately 43 δ, we may conclude that the optical power o the crystalline lens at rest is 16 δ.

See Geometric construction of images formed by lenses, p 113

The maximum optical power o the crystalline lens is reached when the object is at the near point, that is to say 25 cm rom the centre o the eye’s optical system. The eye’s optical power can be calculated using the thin-lens equation: do 5 25 cm 5 25 3 10-2 m di 5 17 cm 5 17 3 10-3 m PNP 5

1 1 1 1 1 5 1 5 1 ≈ 63 δ f do di 25 3 10-2 m 17 3 10-3 m

The amplitude o accommodation o the crystalline lens is equal to: PNP  PFP 5 63 δ  59 δ 5 4 δ As such, the optical power (P) o an emmetropic eye’s crystalline lens can vary by 4 δ, that is to say between 16 δ at rest and 20 δ when it swells.

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When light rays orming an image are correctly projected onto the retina, they are transormed into a nerve impulse with the help o specialized cells covering the retina’s surace. These light-sensitive cells called cones and rods, detect the intensity and wavelength o the light rays that reach them. The nerve impulse is then transported to the brain through the optic nerve. Finally, the brain reverses the images to give the environment its proper appearance.

5.2.2

Visual disorders and their correction

When the image o an object does not orm on the retina, the eye presents disorders. The main visual disorders are myopia, hyperopia, presbyopia and astigmatism.

Myopia

O

Myopia is a disorder characterized by images orming in ront o the retina (see Figure 5a). It is most oten caused by excessive lengthening o the eyeball or by excess curvature o the cornea or crystalline lens.

F

a) An image originating at infnity does not orm on the retina but in ront o it

O

In the frst case, the distance between the cornea and the retina is so great that the optical system constituted by the cornea and the crystalline lens is unable to orm images on the retina. In the second case, the curvature o the cornea or the crystalline lens is too pronounced, making the eye’s optical power (P) too large. As a result, the image can no longer reach the retina. Myopic persons have good vision at close range, but distant objects produce blurred images. In order to correct this visual deect, it is necessary to increase the eye’s ocal length (f ), thereby reducing its optical power (P). This is possible through the use o concave (diverging) glasses or contact lenses (see Figure 5b).

F

Concave (diverging) lens

b) A concave (diverging) lens can correct myopia

Figure 5 A myopic eye

A myopic eye’s optical power (P) can also be reduced by adjusting the shape o the cornea’s ront side. This is possible with laser technology, which slices o a thin layer o the cornea to reduce its curvature, thereby diminishing the optical power and moving the ocusing plane toward the retina.

Hyperopia In terms o optics, hyperopia is the opposite o myopia: the image orms behind the retina (see Figure 6a on the following page). This disorder is caused when the eyeball is too short or when the curvature o the cornea or crystalline lens is insufcient. In the frst case, the distance between the cornea and the retina is so small that the optical system composed o the cornea and the crystalline lens orms images

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beyond the retina. In the second case, the curvature o the cornea or the crystalline lens is not pronounced enough, leading to the eye’s optical power (P) being too small. As a result, the image orms behind the retina. Moderately hyperopic persons nevertheless have good vision at long range because the crystalline lens, normally at rest, can accommodate by bulging, and moving the ocal plane closer so that the image coincides with the retina. However, this continuous accommodation can cause headaches. On the other hand, when objects are close, the distance between the image and the crystalline lens grows, making it increasingly diicult or the crystalline lens to accommodate. Vision at close range becomes problematic and images are blurred.

O

F

a) An image originating at infnity orms not on the retina but behind it

With age, the crystalline lens accommodates less and less because it loses its lexibility. Hyperopic persons end up having difculty seeing at long range as well as at close range. Correcting this disorder requires an increase in the eye’s optical power (P). This is possible through the use o convex (converging) glasses or contact lenses that move the image toward the retina (see Figure 6b). A hyperopic eye’s optical power (P) can also be increased by adjusting the shape o the cornea’s ront side. To do this, a laser is used to sculpt the cornea’s outer surace, restoring its rounded appearance. The optical power increases and the image is moved toward the retina.

O

F

Convex (converging) lens

b) A convex (converging) lens can correct hyperopia

Figure 6 A hyperopic eye

Presbyopia In terms o optics, presbyopia is similar to hyperopia: the image orms behind the retina. However, presbyopia has a dierent cause. The ageing process o the crystalline lens results in the loss o its elasticity, which fnds expression in its reduced accommodation properties: its optical power (P) decreases. In general, presbyopia occurs ater the age o 40. In persons who were initially emmetropic, presbyopia is accompanied by increasingly blurred vision but does not aect vision at long range. This disorder can be corrected with converging contact lenses or glasses.

Astigmatism Astigmatism is a disorder caused by irregularities in the curvature o the cornea or the crystalline lens. This fnds expression in the presence o dierent ocal planes, leading to images that are not only blurred but also distorted in vision at close and at long range. Astigmatism can be corrected by using toric* (cylindrical) contact lenses or glasses, as well as with laser surgery.

See Geometric construction of images formed by lenses, p 113

glasses Glasses, one o the * Toric sides o which is shaped like a toroid A toroid is a surace o revolution, generated by the rotation o a circle about an axis external to the circle and situated in its plane It is the shape o a buoy, or example

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Latest developments in cataract surgery The ormation o cataracts is an aiction that renders the crystalline lens opaque and prevents light rom reaching the retina Generally caused by the normal ageing process o the crystalline lens, it is the primary cause o blindness in the developing world The disease aects 20% o the world’s population aged 65 years and older and 60% o persons aged 85 years and older For many years, the treatment o cataracts involved the use o surgery: an incision was made, the opaque crystalline lens destroyed, the ragments removed and an artifcial crystalline lens implanted This replacement lens was inlexible and incapable o eecting the slightest accommodation However, the implant’s optical power (P ) was based on a personalized calculation to provide maximum comort Moreover, these implants, known as monoocals, required glasses or correction Cataract surgery techniques have greatly evolved Surgeons today use emtosecond lasers, which allow or making very small incisions (measuring just millimetres), thus minimizing healing times and avoiding the need or stitches Moreover, new multiocal implants, meaning they have several ocal lengths, have been developed These implants make it possible not only to solve the problem o cataracts but also serve to correct visual disorders such as myopia, hypero-

SECTION 5.2

2. Why do distant objects appear blurry when seen by a swimmer with normal vision when she swims beneath the water’s surface? How do swimming goggles or masks correct this problem? 3. How can one tell whether a person is myopic or hyperopic by observing how the person’s glasses change the size of his or her eyes?

UNIT 1 Geometric Optics

Figure 7 An eye aected by cataracts: the crystalline lens has become opaque and prevents light rom reaching the retina

The human eye

1. An eye’s visual disorder is corrected with a lens with an optical power (P) of 52 δ Is the eye myopic or hyperopic?

138

pia or presbyopia Research is underway that aims to replace the opaque crystalline lens by introducing liquid silicone, which would harden on the spot and be modelled with a laser, thereby making it possible to correct several visual disorders simultaneously

4. The image distance of a person who can clearly see very distant objects is equal to 2 cm The per­ son can also clearly see objects situated at 04 m Calculate the amplitude of accommodation of this person’s crystalline lenses 5. A patient’s presbyopic eye has a near point of 45 cm a) What does this visual disorder consist of? b) What are the required characteristics of a lens that could correct this disorder and enable the patient to read a computer screen placed 25 cm from his or her eyes? The distance between the crystalline lens­ cornea system and the retina is equal to 17 mm

5.3 The light microscope

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

The light microscope is an instrument that makes it possible to see objects so small that they are invisible to the naked eye. This is achieved with an ingenious arrangement of lenses that deviate light rays and form an image that is much larger than the object. Although current microscopes are highly sophisticated and contain numerous lenses, they are all comprised of two essential optical systems: the objective and the eyepiece (see Figure 8).

ANTONIE VAN HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY LEEUwENHOEk Dutch scientist (1632–1723)

Eyepiece

Microscope body

Objective

Adjustment nob

Specimen

Mirror

Figure 8 A light microscope’s components

A light microscope is composed of at least two converging lenses that function as the objective and the eyepiece respectively. The objective is a lens with a very small focal length (f ) that is placed in close proximity to the object being observed.

Antonie Van Leeuwenhoek, a abric merchant by proession, had his curiosity to thank or becoming a scientist Although he did not receive any science education, he constructed no ewer than 400 microscopes in order to conduct countless observations that he regularly HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY transmitted to the Royal Society o London, o which he became a member in 1680 His basic microscopes enabled him to describe a universe that was completely unknown at the time: the infnitely small Even though at frst he was only interested in microscopy to check the purity o his abrics, his meticulous observations led him to make many discoveries Over the course o almost 50 years, he described protozoans, bacteria, algae, sperm cells, red blood cells, the anatomy o numerous insects, the structure o many plants and o many other organisms In tribute to his discoveries, his name was given to the medal awarded by the Royal Netherlands Academy o Arts and Sciences to deserving microbiology researchers

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L1

When an object being observed is situat ed at a distance located between the focal length and twice this length, the image is real, inverted and larger than the object (see Figure 9). This is the case with object AB. One can see that the objective (lens L 1 ) forms an image A 1B1 that is larger than AB. This image becomes the object of the eyepiece (lens L 2). The eyepiece is positioned in such a way that image A 1B1 is located in a position between lens L 2 and its secondary focal point (F' 2). This position is particular as it allows the eyepiece to form image A2B2, which is virtual but greatly enlarged. Finally, an image of A2B2 is formed on the eye’s retina.

L2

B A2

A1 F1

A F’1

F’2

F2 B1

Objective

Eyepiece

B2

Figure 9 The formation of images in a light microscope

It should be noted that objectives in high-performance microscopes are not comprised of a single lens but of a series of lenses that correct different optical aberrations. As a rule, microscopes have several objectives that are mounted on a turret. Each objective has a specific magnification and focal length (f ). To obtain a clear image, it is necessary to adjust the focus by changing the distance between the object and the objective.

SECTION 5.3

The light microscope

1. Light microscopes are comprised o two essential optical systems What are they? 2. Are the lenses used in light microscopes converging or diverging? 3. Is the image ormed by the eyepiece lens o a light microscope real or virtual? 4. Calculate the position and size o the fnal image ormed by a light microscope’s two lenses using the ollowing data: – the object measures 1 cm and is situated 25 cm rom the objective (f 5 2 cm) – the eyepiece is 12 cm rom the objective with a ocal length (f ) o 23 cm

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5. The objective in a light microscope has a ocal length (f ) o 15 cm and the eyepiece a ocal length o 2 cm The distance separating the two lenses is 14 cm I you are looking at a letter that is 1 mm tall and situated 17 cm rom the objective, determine: a) the position o the frst image that orms b) the frst image type c) the position o the second image that orms d) the second image type e) the height o the fnal image f) the microscope’s magnifcation g) the orientation o the fnal image in relation to the object

5.4 The telescope In contrast to the microscope, which is used to look at objects that are close, the telescope is an instrument speciically designed or observing distant objects. The many kinds o telescopes are grouped into two types that can be distinguished by the manner in which they ocus light rays. The rst type uses converging curved mirrors and is called a refecting telescope in reerence to the properties o its mirrors, which refect light. The second type uses converging lenses and is called a reracting telescope in reerence to the optical phenomenon produced by its lenses.

5.4.1

The refecting telescope

The relecting telescope uses mirrors to orm astronomical images. There are many dierent models which can be distinguished rom one another by the organization and type o mirrors used. In order to understand how a refecting telescope unctions, the Newtonian model will be studied in this section (see Figure 10).

Eyepiece

Lens Secondary mirror (plane mirror)

Light rays rom a

Primary mirror (parabolic mirror)

Principal axis (P)

This telescope is essentially comprised distant object o a primary mirror, a secondary mirror Figure 10 How a Newtonian telescope works and an eyepiece. The primary mirror is a parabolic mirror that makes it possible to ocus the light coming rom disSee Spherical aberrations, tant objects without incurring spherical aberration. The greater this mirror’s p 124 diameter, the more light it captures. The secondary mirror is a small plane mirror situated at 45° in relation to the primary mirror’s principal axis (P). It serves to send the ocal image toward the eyepiece, in a direction that is perpendicular to the principal axis. The eyepiece is equipped with a converging lens that unctions as a magniying glass. Its purpose is to enlarge the image ormed by the primary mirror, which can nally be seen through the outside end o the eyepiece. To bring the image into ocus, the eyepiece has to be moved in relation to the secondary mirror. Refecting telescopes have the main advantage o remaining unaected by chromatic aberration by virtue o the act that light does not pass through the primary mirror. Moreover, it is possible to construct mirrors the size o which greatly exceeds that o lenses and thereby allow or the observation o celestial objects that are less bright. This technology is used in the world’s biggest telescopes, whose parabolic mirrors have diameters approaching or even surpassing 10 m. In view o scientic advances and the perection o manuacturing processes, we can expect to see the records in this eld beaten once again.

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5.4.2

The refracting telescope

The reracting telescope is primarily equipped with two lenses, the objective and the eyepiece, which are placed at either end o a tube (see Figure 11). As a rule, both lenses are converging. The objective, which receives light rom distant objects, has a ocal length (fobjective) greatly exceeding that o the eyepiece (feyepiece). In terms o optics, the reracting telescope resembles a microscope. The undamental dierence stems rom the act that the observed object is extremely distant and the light rays that reach the objective are thereore parallel. The image ormed by the objective is inverted and real but smaller than the one ormed by a microscope. Due to the positioning o the lenses, the image is located within the eyepiece’s secondary ocal point (F'). The image becomes an object or the eyepiece lens, which in turn orms a second virtual image that is much larger than the rst.

Objective

Eyepiece

Light rays from a distant object

fobjective feyepiece

F’

Real image Objective

Eyepiece

Final virtual image

Figure 11 The formation of an image by a refracting telescope

See Chromatic aberration, p 123

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UNIT 1 Geometric Optics

Reracting telescopes today are comprised o several lenses that are designed and positioned in such a way as to minimize problems associated with spherical and chromatic aberration. However, the largest reracting telescope objectives barely exceed one metre in diameter. This inherent problem in the manuacture o “giant” lenses, when compared with what can be achieved with parabolic mirrors, gives refecting telescopes a clear advantage.

Furthering

your understanding

Space telescopes In 1946, US astrophysicist Lyman Spitzer proposed sending telescopes into space His idea was inspired by the act that the Earth’s atmosphere disrupts electromagnetic waves coming rom space The layers o gas surrounding the Earth distort light radiation and either entirely or partially absorb certain wavelengths, such as those o g (gamma) rays, x-rays and ultraviolet and inrared radiation Furthermore, the air’s natural turbulence and the light pollution produced by artiicial light sources considerably decrease the quality o images taken rom Earth A space telescope is 100 times more eective than an earthbased telescope in detecting visible light

The Hubble telescope will be replaced in 2014 by the James Webb Space Telescope (JWST) Expected to weigh just under 6 tonnes, this reecting telescope will be sent into orbit 15 million km above the Earth Like the Hubble, it will be equipped with two mirrors, the important dierence being that its primary mirror will have a diameter o 65 m—a surace area seven times larger than its predecessor This telescope o the uture will undoubtedly make it possible to penetrate new scientifc secrets and to push the limits o human knowledge even urther

The best-known space telescope is the Hubble Space Telescope (HST), named in memory o US astronomer Edwin Hubble Launched in 1999, it orbits around the Earth at an altitude o about 600 km The Hubble is a relecting telescope weighing 11 tonnes and is equipped with two hyperbolic mirrors: a primary mirror with a diameter o about 24 m and a secondary mirror with a diameter o 30 cm The inormation gathered by this telescope is priceless and has allowed or breakthroughs in the understanding o numerous phenomena such as the expansion o the universe, the ormation o planets and the presence o black holes at the centres o galaxies Figure 12 An artist’s conception o the James Webb Space Telescope

SECTION 5.4

The telescope

1. How do reracting telescopes work? 2. Explain the undamental dierence between a reracting telescope and a refecting telescope

rst image and the second image Then calculate the magnications Compare these magnications to those between the object and the nal image

3. Explain why the secondary mirror o a Newtonian telescope is positioned at 45° in relation to the primary mirror’s principal axis (P) F’

4. Are the world’s biggest telescopes refecting or reracting telescopes? Explain your answer 5. Copy and complete the ray diagram o a reracting telescope pictured opposite onto a piece o paper Measure the object and the rst image and then the

Objective

Eyepiece

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The microscope It can sometimes happen that an invention’s authorship is subject to controversy. In the case o the microscope, however, the context does not raise any questions. The individuals claiming to be the inventor o this instrument were both manuacturers o eyeglasses and each worked on this invention on his own, at the same time and in the same city. So who invented the frst microscope in the 1590s in Middelburg in the Netherlands? Was it Hans Lippershey or Hans Janssen, with the assistance o his son Zacharias? No one can confrm this. However, we do know that these men’s microscopes resembled a telescope more than a modern microscope (see Figure 13). They were essentially composed o a tube with an objective at one end and an eyepiece at the other. The word “microscope” itsel was frst mentioned in 1625. It reers to the instrument perected at the beginning o the 17th century by Galileo, who had taken inspiration rom Lippershey’s model.

In spite o Galileo’s improvements, the lenses used in these early instruments greatly distorted images. Nevertheless, it was with these basic microscopes that Antonie Van Leeuwenhoek turned microscopy into a scientifc discipline at the end o the 17th century. It wasn’t until the end o the 18th century that microscopists began to correct these distortions (or optical aberrations), in particular by using combinations o magniying lenses and corrective lenses in the objectives and eyepieces. Microscopes today oten contain more than 10 lenses in order to correct optical aberrations. In the 19th century, microscopy made another great advance. Microscopists now understood the degree to which the quality o the light illuminating the object and “carrying” the image to the eye through the lenses was important or the sharpness o the image. They developed new, oten very complex, ways o illuminating objects. Cellular details that had never beore been seen under ordinary light conditions could now be detected. The inventor o the lighting technique called phase contrast, the Dutchman Frits Zernike (see Figure 14), was awarded the Nobel Prize in 1953 or this invention. This technique is still used in laboratories today.

Figure 13 Representation o a microscope developed by Zacharias Janssen and his ather Hans

Figure 14 Frits Zernike, winner o the 1953 Nobel Prize in Physics or his discovery o the phase contrast lighting technique

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UNIT 1 Geometric Optics

Figure 15 The electron microscope uses a beam o electrons to illuminate a specimen and create a magnifed image o it Electron microscopes have a greater resolving power than light microscopes

chapter

5

alid Gomi Ois

5.1 The camera • The camera is an instrument used or capturing images. • The most important components o a camera are: the objective, the diaphragm, the shutter and the lightsensitive surace. • Accurate ocalization o the image on the light-sensitive surace is achieved using a system o lenses called the objective. • The image ormed by a camera is real and inverted.

5.2 The human eye • The human eye contains our transparent media: the cornea, the aqueous humor, the crystalline lens and the vitreous humor. • The iris is a circular membrane situated in ront o the crystalline lens and its colouring gives eyes their colour. • The opening at the centre o the iris is called the pupil, and its size is regulated by the iris according to the quantity o surrounding light. • The largest reraction o light entering the eye occurs at the air-cornea interace. • The crystalline lens is the shape o a biconvex (converging) lens. • The crystalline lens is a fexible lens whose curvature can be adjusted by the ciliary muscles. • The curvature adjustment o the crystalline lens ensures that the image is ormed on the retina at close range and at long range. • The ability o the crystalline lens to change its shape to allow or clear vision is called accommodation. • The image ormed on the retina is real and inverted. • The brain reverses images to give the environment its proper appearance. • A normal eye is said to be emmetropic. • The maximum distance or clear vision is called ar point. It is innite or an emmetropic eye. • The minimum distance or clear vision is called near point. It is roughly equal to 25 cm in young adults. • An emmetropic eye’s optical power (P) or distant vision is equal to about 59 δ. • The amplitude o accommodation o the crystalline lens in an emmetropic eye is roughly equal to 4 δ. • The image ormed on the retina is transormed into a nerve impulse with the help o specialized cells in the retina.

CHAPTER 5 Applied Geometric Optics

145

(cont.) Le phénomène • • •

de la réfraction

Myopia is a disorder characterized by images orming in ront o the retina. La réfraction correspond au changement de direction de propagation de la lumière à l’interface entre Myopia can bedont corrected using concave (diverging) glasses or contact lenses or with laser surgery. deux milieux les propriétés optiques sont différentes.

• a disorder imagesla orming behind the retina. • Hyperopia La propriétéisque possèdecharacterized un milieu deby réfracter lumière se nomme la réfringence. • canlabesurface corrected usingdeux convex (converging) glasses contact lenses orpossèdent with laserpas surgery. • Hyperopia Le dioptre est séparant milieux transparents et or homogènes qui ne mêmes propriétés optiques. • les Presbyopia is a disorder characterized by images orming behind the retina. It is caused by the ageing o the crystalline lens. • Astigmatism is a disorder caused by irregularities in the curvature o the cornea or the crystalline lens. • Astigmatism can be corrected using toric (cylindrical) contact lenses or glasses as well as with laser surgery.

5.3 The light microscope • The light microscope is an instrument that makes it possible to see objects so small they are invisible to the naked eye. • A light microscope is comprised o two optical systems: the objective and the eyepiece. • The lenses used in light microscopes are converging. • The image ormed by the eyepiece o a light microscope is virtual.

5.4 The telescope • The telescope is a specially designed instrument or observing distant objects. • There are two types o telescopes: the refecting telescope and the reracting telescope. • To capture light, refecting telescopes use converging mirrors whereas reracting telescopes use converging lenses. • The refecting telescope is not aected by chromatic aberration. • The reracting telescope has one objective and one eyepiece.

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UNIT 1 Geometric Optics

CHAPTER 5

Applied Geometric Optics

1. A camera is comprised o, among other things, a diaphragm, an objective and a light­sensitive surace What are the equivalent components in the human eye? 2. The human eye and the camera show many similari­ ties However, ocusing an image on the light­sensitive surace is perormed dierently in each case Explain this dierence 3. Which optical devices produce virtual images, and which ones produce real images? 4. The two gures below illustrate how images are ormed in two optical devices a) Which devices are shown in the gures? b) What is the role o each lens? c) Although each device has two converging lenses, there is a undamental dierence between the devices What is this dierence?

5 Explain why refecting telescopes are not aected by chromatic aberration, unlike reracting telescopes 6. Reproduce and complete the ollowing table on a sheet o paper Optical device Microscope Refecting telescope

Component

Optical phenomenon (Refection/Reraction)

Objective Eyepiece Primary mirror Secondary mirror

7. You have one or two converging lenses, an optical bench, and a candle or an object On a piece o paper, draw the lens or lenses and the candle so as to simulate the ollowing optical instruments: a) a camera b) a microscope c) a reracting telescope

1 F’

2

L1

L2

B A1

A2 A F’1

F1

F’2

F2 B1

B2

CHAPTER 5 Applied Geometric Optics

147

148

CONTENTS ChapTEr 6

Fmes o reeence              151 ChapTEr 7

Quntities nd Units              165 ChapTEr 8

Vectos                            183

Mecnics is te science o te motion nd equi­ libium o bodies Instumentl in umn tecno­ logicl develoment om te invention o te weel to te luncing o geosttiony stellites into sce, mecnics uses tools tt ccutely exlin mny ysicl enomen Fist, te descition o n object’s motion o os­ ition in sce t  given oint in time equies te defnition o  me o eeence, wic deends s muc on te tye o oblem to be solved s on te desied ccucy Fo exmle, it is ossible to solve numeous oblems in kinemtics nd dynmics using oely oiented Ctesin coodintes

Next, mecnics is concened wit te ysicl quntities tt e exessed quntittively toug numbes nd units o mesue In ode o scientists to undestnd ec ote, no mtte wee in te wold tey e conducting tei exeiments, it is imotnt to ve n intentionl system o units nd to defne nd stnddize te symbols eesenting quntities nd units Lstly, tee e mny ysicl quntities tt cn­ not be ully descibed by numbes, but o wic it is imotnt to indicte its diection in sce Vectos e essentil o undestnding enomen involving tese quntities, known s vecto quntities 149

61 The purpose of a frame of reference ChapTEr

6

FraMES OF rEFErENCE

62 Inertial frames of reference 63 Coordinate systems

71 Quantity, measurement and unit of measure

UNIT

2

prELIMINarY NOTIONS OF MEChaNICS

72 The International System of Units

ChapTEr

7

QUaNTITIES aND UNITS

73 Fundamental standards of the basic units of mechanics 74 Derived units in the International System 75 Multiples and submultiples of units 76 Scalar quantities and vector quantities

81 Properties of vectors ChapTEr

8

82 Adding vectors VECTOrS 83 Subtracting vectors 84 Multiplying a vector by a number

150

Fmes of refeence

S

tudying a body in motion requires constant knowledge of its position and of time. For this it is essential to choose a precise frame of reference against which this motion will be analyzed. The choice of the frame of reference will depend on the nature of the experiments and studies to be conducted. For exam­ ple, GPS is a geographic positioning system that uses the data transmitted by numerous satellites in station­ ary orbit around the Earth. Its operation is based on a frame of reference whose origin coincides with the centre of the Earth and whereby time is measured using very precise atomic clocks.

In this chapter, you will learn about the purpose of a frame of reference, become familiar with inertial frames of reference, use Cartesian coordinates and discover polar coordinates.

61 62 63

Te uose of  fme of efeence    152 Inetil fmes of efeence             155 Coodinte systems                     158 ChapTEr 6 Frames of Reference

151

61 The urose of  frme of reference The description of the motion of a travelling object is strongly con­ nected to the observer’s position. To better understand this notion, it is helpful to consider the following situation. In the middle of a long walkway in an airport, a moving sidewalk is moving at a constant velocity. Two people, P1 and P2, are standing stationary on this moving sidewalk. Two security guards, G1 and G2, are located on either side of the sidewalk (see Figure 1). If the security guards and person P2 were asked to describe the motion of person P1, each one would have a different answer. The two security guards would agree that person P1 is moving, but guard G1 would say that the person is moving toward the right whereas guard G2 would claim that P1 is moving toward the left. As for person P2, she would say that person P1 is station­ ary relative to her. In fact, the distance between person P1 and the guards varies, but the distance that separates him from person P2 does not change. Therefore, in a single situation, three observers give three different descriptions. G2

p1

p2

G1

Figure 1 The motion of person P1 will be described differently by the three observers (G1, G2 and P2).

This difference in interpretation does not only concern the position, but can also affect the trajectory followed by a travelling object. As an example, we can consider a situation where person P1 lets an apple fall from his hand while he is on the moving sidewalk. Person P2 will see the apple fall at person P1’s feet and, from her viewpoint, the fall will follow a vertical motion, perfectly perpendicular to the moving sidewalk’s plane (see Figure 2a). This will be so for any per­ son moving on the moving sidewalk (such as P1 and P2). Security guard G1 will not see the same trajectory. Because of the initial velo­ city given to the apple by the sidewalk moving toward the right, he will observe a trajectory that has the shape of a section of a parabola (see Figure 2b). This will be so for any stationary observer located off the moving sidewalk (such as G1 and G2).

a2 p1

p1

p2 G1

G2 p1

p1

p2

G1

a) The trajectory of the apple falling will appear vertical to persons P1 and P2.

Figure 2 The trajectory of an apple depending on the viewpoint of the observer

152

UNIT 2 Preliminary Notions of Mechanics

b) The trajectory of the apple falling will appear parabolic to security guards G1 and G2.

This difference in interpretation can also affect measurable quantities that characterize motion, such as velocity. If you assume that the moving sidewalk is moving at a constant velocity vAB toward the right and that person P2 has decided to walk, also toward the right at a certain constant velocity v B, observ ers will observe different phenomena. Person P1 will see person P2 move away from him at velocity vB. The security guard, who is stationary in the walkway, will see person P2 move away from them at velocity (v AB 1 vB), which is greater than velocity vB (see Figure 3). All the examples mentioned up to this point show that the perception of motion of a travelling object depends on the observer. In fact, any motion must be described relative to a frame of reference. Generally speaking, a frame of reference is associated with a spatial reference composed of an origin O and three axes (x, y and z) perpendicular to one another. This spatial refer­ ence enables one to precisely determine the position in space of a body in motion but also all of its mechanical characteristics*. For quantities that have a time component, a time reference is added to the frame of reference in order to evaluate time**. The previous examples involved two frames of reference. The first (RA) is stationary and connected to the walkway (or, more generally, to the ground) where the security guards (G1 and G2) are standing. The second (RB) is moving and connected to the moving sidewalk, which is moving relative to the walkway, and on which the two people (P1 and P2) are standing (see Figure 3). It is possible to analyze the above example concerning velocities from the perspective of frames of reference. Therefore, you can say that the velocity of frame of reference RB rela­ tive to that of frame of reference RA is equal to v AB and that the velocity of person P2 measured in frame of reference RB is vB. Given that person P1 is stationary relative to the moving sidewalk, he will see P2 moving away from him at velocity vB. However, guards G1 and G2 who are stationary in frame of refer­ ence RA, will see person P2 moving away at velocity vA  (vAB 1 vB). This relationship can be generalized to any body in motion whose velocity is measured relative to two frames of reference (RA and RB), RB being stationary relative to RA.

that to situate a reference * Note point in a plane, only two axes (x and y ) are required. is important to specify at this ** Itpoint that within the framework of classical mechanics, time is absolute, i.e. it elapses in the same way regardless of the frame of reference chosen.

G2

p1

p2

y x Fme of efeence rB

y

Fme of efeence ra

G1

x

Figure 3 Frame of reference RA is connected to the ground while frame of reference RB, connected to the moving sidewalk, is stationary relative to RA. The velocity of person P2 depends on the frame of reference against which it is measured.

vA  (vB 1vAB) where vA  Velocity of the body relative to frame of reference RA vB  Velocity of the body relative to frame of reference RB vAB  Velocity of frame of reference RB relative to frame of reference RA

ChapTEr 6 Frames of Reference

153

Example A person is on a train travelling at a speed of 20.0 m/s. This person is walking on the train at a speed of 1.2 m/s in the direction of the train’s motion (whose velocity is measured relative to the passengers sitting on the train). What is the velocity of this person relative to an observer who is stationary on the platform of a station the train is going through? Data : vAB  20.0 m/s (velocity of the train)

Solution : vA  (vB 1 vAB)  20.0 m/s 1 1.2 m/s  21.2 m/s

vB  1.2 m/s (velocity of the person on the train) The person has a velocity of 21.2 m/s relative to the observer on the platform. In physics, it is said that motion is relative to the frame of reference against which it is studied. This notion of relativity is not recent. It dates back to early works on mechanics, and was then developed in other disciplines of physics. The choice of a frame of reference can seem purely arbitrary. However, expe­ rience has shown that frames of some reference reveal themselves to be bet­ ter suited than others depending on the nature of the study being conducted. Consequently, a frame of reference, when it is carefully chosen, allows for a simpler description of motion.

SECTION 61

The purpose of a frame of reference

1 What is a rame o reerence made up o? 2 A person is stationary on a moving sidewalk that is moving at a given velocity At a specifc time, she drops a ball that she was holding in her hand What shape will the ball’s trajectory have or: a) an observer on the moving sidewalk? b) an observer on the ground, o the moving sidewalk? 3 A person is on a moving sidewalk that is moving at 13 m/s She decides to walk at a constant speed o 13 m/s in the direction opposite to the moving sidewalk At a given moment, she drops the ball she was holding in her hand a) What is the velocity o the person relative to an observer who is on the ground, o the moving sidewalk? b) What shape will the ball’s trajectory have or this observer?

154

UNIT 2 Preliminary Notions of Mechanics

4 A person is swimming against the current in a river The velocities o the person (v ) and the current (v ’) are constant The swimmer goes rom point A to point B, over a distance (d ) o 800 m At point B, he encounters a log that is heading down river He continues to swim or a time (tBC) o 18 minutes until he reaches point C, then turns around and returns to point A where he arrives at the same time as the log Calculate the velocity (v ’) o the river’s current Review The relationship between the constant velocity (v ), the distance travelled (d ) and the time interval (∆t) is: d  v  ∆t (see page 12)

62 Inetil fmes of efeence By defnition, an inertial rame o reerence is a rame o reerence in which an isolated body* is either at rest or in uniorm rectilinear motion*. In this type o rame o reerence, a body retains its state at rest or in uniorm rectilinear motion as long as the inuence o other bodies does not make it lapse rom this state.

body A body upon which no * Isolted net orce is exerted. ectiline motion * Unifom A motion that occurs in a straight line and at a constant velocity.

It is important to note that all rames o reerence in rectilinear motion rela­ tive to an inertial rame o reerence are themselves inertial. However, rames o reerence that have accelerated motion or rotational motion are considered non­inertial. For example, a vehicle that accelerates cannot be considered an inertial reerence. In act, i an isolated body in the car was initially at rest or in uniorm rectilinear motion, it can no longer maintain this state because it will be submitted to orces that are described as imaginary because they depend on the rame o reerence. The same occurs or a vehicle that brakes or turns. The rame o reerence then no longer meets the condition necessary to qualiy as an inertial rame o reerence. A rame o reerence fxed to the ground can be considered inertial or most experiments in mechanics that are done on Earth, provided they are o short duration and are not inuenced by the motion o the Earth’s rotation. This rame o reerence, called terrestrial rame o reerence (see Figure 4), is generally used in laboratories. By defnition, this rame o reerence ollows the same motion o rotation as that o the Earth. The experiment o the apple alling on the moving sidewalk described in the previous section is inter­ esting in many ways (see Figure 2 on page 152). It involves two rames o reerence that can be consid­ ered inertial given that the all o the apple is quick and that it is practically uninuenced by the Earth’s rotation. The frst rame o reerence is fxed to the ground and represents a terrestrial rame o reer­ ence. The second is linked to the moving sidewalk and, because it describes a uniorm rectilinear motion relative to the irst, it can also be consid­ ered inertial. This experiment demonstrates that a mechanical phenomenon taking place in an inertial rame o reerence moving at a constant velocity is the same as i the rame o reerence was stationary. This can be proven by examining Figure 2a, which shows the two people (P1 and P2) moving with the moving sidewalk. I person P1 drops an apple rom his hand, person P2 will see it all vertically. The situation would be the same i, instead o being on a moving sidewalk, both people were stationary on the walkway and the experiment was repeated: person P2 would see the apple all vertically.

z x

y

Figure 4 The terrestrial rame o reerence is xed to a point on the surace o the Earth and turns with it. It can be considered inertial or experiments conducted on Earth as long as they are o short duration and are not infuenced by the motion o the Earth’s rotation.

ChapTEr 6 Frames o Reerence

155

*

princile Statement o a general law that has not been proven, but whose consequences have been confrmed.

The Italian scientist Galileo Galilei (1564–1642), who conducted similar exper­ iments on a ship, ormulated a principle* that can today be paraphrased as ollows: “The physical laws that are valid in a given inertial rame o reerence are also valid in all other rames o reerence that move at constant speeds relative to the frst.” This principle, known as Galileo’s principle of relativity, is very important, because it allows or the notion that it is impossible to know, by conducting experiments in mechanics, i a laboratory is at rest or in motion at a constant speed. A person can experience this principle when he or she is sitting aboard an airplane ying at a constant speed relative to the ground, based on a recti­ linear motion when the blinds o the cabin windows are closed. Everything is happening as it would be on Earth and the person has no way o knowing i the airplane is moving or not without looking out the window, i.e. fnding their position with respect to a rame o reerence outside the airplane. To study the motion o satellites around the Earth or example, the terrestrial rame o reerence mentioned earlier is no longer adequate because o the rotation o the Earth. To overcome this problem, the geocentric frame of reference must be used. As its name indicates, this system situates the origin o its spatial reer­ ence at the centre o the Earth (see Figure 5). The operation o a GPS is based on the geocentric rame o reerence. This rame o reerence can be considered inertial with a very good approxima­ tion. However, because it revolves around the Sun, it can only be used i this motion has no eect on the experiments conducted or i their duration is very short, relative to the duration o the Earth’s rotation around the Sun. For certain studies, like those whose subject is the motion o the planets in the solar system or o interplanetary probes, these conditions are not all met. It is then better to use a rame o reerence based on the centre o the Sun: this is the heliocentric frame of reference (see Figure 6).

z Toward fxed star E3 (North Star)

z Toward fxed star E3 y

Toward fxed star E2

O Sun

x

Toward fxed star E1

y Toward fxed star E2

Figure 5 The geocentric rame o reerence situates the origin o its spatial reerence at the centre o the Earth. The three axes o the reerence are perpendicular to one another. One o the axes (z ) is coincident with the Earth’s rotational axis and is usually directed toward the North Star. The two other axes (x and y ) are situated on the equator’s plane and are directed toward distant stars, which are considered fxed. The geocentric rame o reerence does not ollow the Earth’s diurnal rotation.

156

UNIT 2 Preliminary Notions o Mechanics

O

Earth

x Toward fxed star E1

Figure 6 The heliocentric rame o reerence situates the origin o its spatial reerence at the centre o the Sun. The three axes o reer­ ence (x, y and z ) are perpendicular to one another and are directed toward distant stars that are considered fxed. In this rame o reerence, the Earth describes an elliptical orbit in its motion around the Sun.

The heliocentric rame o reerence, also known as Kepler’s rame o reerence, can be considered inertial with greater precision than the geocentric rame o reerence and even greater precision than the terrestrial rame o reerence. Finally, it can be concluded that the inertial character o a rame o reerence depends on the degree o precision with which it matches its defnition, i.e. a rame o reerence in which an isolated body is either at rest or in uniorm rectilinear motion. I, or example, the experiments are o relatively short duration in relation to the motion o the rame o reerence, then it can be considered that the rame o reerence is inertial. This is the reason that some rames o reerence are considered more or less inertial depending on the time scale considered.

Coenicn fme of efeence There is another system besides the terrestrial, geocentric and heliocentric rames o reer­ ence. Known as the Copernican rame o reerence, its centre is that o the solar system. This system is dierent rom the heliocentric system: not only is the Sun not exactly situated at the centre o the solar system but it is moving around this system. As in the case o the heliocentric system, the three axes (x, y and z) o the reerence o this system are directed toward distant stars that are considered fxed. This rame o reerence was named in homage to the Polish astronomer Nicolaus Copernicus (see Figure 7 ). It is considered inertial with excellent precision or all bodies in motion in the solar system.

Figure 7 Nicolaus Copernicus (1473–1543)

SECTION 62

Inetil fmes of efeence

1 Give the defnition o the heliocentric rame o reerence

3 What is an inertial rame o reerence? Give a ew examples

2 What is the Earth’s motion in the geocentric rame o reerence?

4 Give a ew examples o non-inertial rames o reerence

ChapTEr 6 Frames o Reerence

157

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

63 Coodinte systems A rame o reerence is made up o two reerences. The frst, the spatial reer­ ence, is necessary to identiy the position o a body in space. The second, the time reerence, makes it possible to measure the time elapsed between two events. A spatial reerence is made up o an origin (O) and three axes perpen­ dicular to one another (x, y and z). A coordinate system associates a group of three numbers to each point in space that make it possible to situate a body spatially.

rENé DESCarTES French philosopher, scientist and mathematician (1596–1650) René Descartes had a consider­ able infuence on all o his centu­ ry’s scientists. Even at a very young age, he stood out because o his precocious intelligence to HISTORY HIGHLIGHTS HISTORY the extent thatHIGHLIGHTS his ather called him “the little philosopher.” He was the rst to unite algebra and geometry (which had up until then been two distinct sciences) into the one discipline known today as analytic geometry. His works made it possible to solve geo­ metric problems algebraically. It was in his book, titled La Géométrie (Geometry), which was published in 1637, that the concept o coor­ dinates was rst mentioned. It is thereore in recognition o his work that the coordinates were named Cartesian. Besides mathematics, Descartes indelibly marked philosophy. For example, it is to him that we owe the amous saying I think therefore I am. To this day, a person who is logical, methodical and rational is described as being Cartesian.

158

In this textbook, bodies will be situated exclusively in the plane and not in space. In this case, the spatial reerence is made up o only two perpendicu­ lar axes (x and y), and only two real numbers are required to orm the coordi­ nate system. There are two distinct coordinate systems used to situate a body in a plane: the Cartesian coordinate system and the polar coordinate system.

631

The Ctesin coodinte system

In the Cartesian coordinate system, each point P o the plane can be situated with a pair o coordinates x and y, noted (x, y). The x and y coordinates are respectively called the abscissa and ordinate o point P. The value o each coordinate is determined by an orthographic projection issued rom point P on each o the axes (see Figure 8).

y (m) 5

P(3, 4)

4 3 2 1 -5 -4 -3 -2 -1

-1

1

2

3

4

5 x (m)

-2

This coordinate system, which is the most widely used, was named in homage to the French scientist René Descartes.

632

The pol coodinte system

-3 -4 -5

Figure 8 In this Cartesian coordinate system, point P has an abscissa x  3 and an ordinate y  4. This point is written P(3, 4). Values are in metres.

The Cartesian coordinate system is not the only way to situate a point in a plane. There is another, known as the polar coordinate system. In this system, point 0 is not called origin as it is in the Cartesian coordinate system, but pole, and the axis o abscissa is known as the polar axis. A point P is entirely deter­ mined by a pair o coordinates: a radial coordinate and an angular coordinate. The radial coordinate, which represents the distance connecting the pole to point P, is called the radius (r). The angular coordinate, which is the angle q between the polar axis and the radius, is called the polar angle (see Figure 9 on the following page). This angle has its origin on the polar axis, and the positive direction is the trigonometric direction, i.e. counter­clockwise. In this system, point P o the plane is written: P(r, q).

UNIT 2 Preliminary Notions o Mechanics

The polar coordinate system is a system particularly well­suited to problems that involve rotations or present rotational symmetries.

y (m) 5 4

P(5, 53.1°) r5m

3

633

reltionsi between Ctesin coodintes nd ol coodintes

It is possible to go rom a Cartesian coordinate system to a polar coordinate system and vice versa, using trigonometric relationships that connect the di­ erent coordinates (see Figure 10). According to Figure 10, applying the Pythagorean theorem to triangle POX makes it possible to write: r2  x2 1 y2 ⇒ r  x2 1 y2 On the other hand, angle q can be proven using the trigonometric relationship:

2 θ

1 -5 -4 -3 -2 -1 O -1 -2

1

2

3

4

5 x (m)

-3 -4 -5

Figure 9 In a polar coordinate system, O is the pole, r (radial coordinate) is the radius, q (angular coordinate) is the polar angle and the x­axis is called polar axis. In this example, point P has a ray r of 5 m and a polar angle q of 53.1°. This point is written P(5, 53.1°).

y y ⇒ q  tan­1 x x

()

tanq 

y (m)

These two equations make it possible to convert Cartesian coordinates to polar coordinates. Thereore, conversely, it is possible to fnd equations that enable you to go rom polar coordinates to Cartesian coordinates. Still using Figure 10, it can be written:

and

x cosq  ⇒ x  r  cosq r sinq 

y ⇒ y  r  sinq r

To sum up: 1. To go rom Cartesian coordinates to polar coordinates, the two ollowing equations are used:

5 4

Y

P

3 r

2

θ

1

X -5 -4 -3 -2 -1 O -1 -2

1

2

3

4

5 x (m)

-3 -4 -5

Figure 10 Trigonometric relationships connecting the coordinates are necessary to go from a Cartesian coordinate system to a polar coordinate system and vice versa.

r  x2 1 y2

( yx )

q  tan­1

2. To go rom polar coordinates to Cartesian coordinates, the two ollowing equations are used: x  r  cosq y  r  sinq

ChapTEr 6 Frames of Reference

159

Exmple a Using the data in Figure 8 on page 158, fnd the polar coordinates o point P.

Exmple B Using the data in Figure 9 on the previous page, fnd the Cartesian coordinates o point P.

Solution :

Data: x3m y4m

Data: r5m q  53.1°

r  x 2 1 y 2  32 1 42  9 1 16  25  5 m y 4 q  tan­1  tan­1  53.1° x 3 In a polar coordinate system, point P is thereore written P(5, 53.1°).

()

Furthering

()

Solution : x  r  cosq  5  cos53.1°  3 m y  r  sinq  5  sin53.1°  4 m In a Cartesian coordinate system, point P is written P(3, 4).

your understanding

a few three­dimensionl coordinte systems Locating a point in space can be done with several coordinate systems. In all cases, three distinct coordinates are required. Pour aller loin The most commonly used system is the Cartesian coordinate sys­ tem in which every point P is identifed with three coordinates: x, y and z. The value o each o the coordinates is determined by an orthographic projection issued rom point P or each o the three axes o the spatial reerence. The point is noted P(x, y, z ) [see Figure 11]. A point P in space can also be situated with a spherical coordinate system. As its name indicates, this system is very useul or re­ solving problems presenting a spherical symmetry. In this case, the z (m) z

three coordinates are a radial coordinate r and two angular coordi­ nates j and q , so that point P is noted P(r, j, q ) [see Figure 12 ]. When the symmetry o the problem is cylindrical, it is more advantageous to use the cylindrical coordinate system. Equations are generally less complex, and the mathematical ormalism is thereore greatly simpliied. The cylindrical coordinate system is a generalization o the polar coordinate system in a three­ dimensional space. It uses the two polar coordinates r and q and the third Cartesian coordinate, z. Point P is thereore noted P(r, q , z) [see Figure 13 ].

z (m) z

z (m) P (x, y, z) y

ϕ

r

P (r, ϕ, θ )

P (r, θ, z ) y (m)

y (m)

y (m)

r

θ

θ x x (m)

x (m)

x (m)

Figure 11 Three­dimensional Cartesian coordinate system

SECTION 63

Figure 12 Spherical coordinate system

Figure 13 Cylindrical coordinate system

Coordinte systems

1 In a Cartesian plane, situate the points whose polar coordinates are: P1(2, 45°); P2(1, -180°); P3(2, 135°) 2 What are the Cartesian coordinates of the three points in question 1? 3 Calculate the polar coordinates of the following points: P1(2, 1); P2(3, -3); P3(-2, 2)

y 15

4 For each of the three points, P1, P2 and P3, determine: a) their Cartesian coordinates b) their polar coordinates

P3 10 5

-15

-10

-5

5 -5 -10 -15

160

UNIT 2 Preliminary Notions o Mechanics

P1

P2

10

15 x

APPLICATIONS Te GpS The GPS, an acronym that stands or Global Positioning System, is a product o research initiated in 1960 by the US army and launched the frst experimental satellites in 1978. At the outset, their purpose was essentially military: using this system, they wanted to increase the efciency o cruise missile guidance and target localisation. However, starting in 1995, the GPS, at the time already operational worldwide, gradually became available or civil applications. Since the year 2000, the GPS has increasingly been used by ordinary citizens, to the point that you can now buy relatively inexpensive systems intended or identi­ ying locations and navigating during car trips or in the wilderness. The GPS is made up o three distinct segments. The “space” segment is composed o 41 satellites situated on stable orbits at approximately 20 000 km rom the surace o the Earth. The “control” segment consists o fve stations; their command centre, located in Colorado, ensures that the satellites unction properly and that the inormation transmitted (in the orm o ephemeris or clock parameters) is updated. Finally, the GPS has a “user” segment that includes all the users o the position­ ing system. There can be an infnite number o users at the same time because the satellite signal transmitted can be captured by any device situated in its range.

Because the signal travels very quickly, it is crucial that the receiver clock be perectly synchronized with those o the satellites. For example, one error o a millionth o a second could cause an error o 300 metres in the position. Using the signals o our satellites enables the receiver clock to synchronize itsel correctly. To know the latitude, longitude and altitude o its posi­ tion, a receiver calculates the distance that separates it rom at least our satellites. The intersection o the our imaginary spheres having a satellite as their centre corresponds to the receiver’s position (see Figure 14). To ensure the optimal precision o the system, the consequences o relativity must also be taken into account. The satellite’s great speed relative to the re­ ceiver’s reerence system results in time elapsing more slowly (by a ew microseconds per day). However, the gravity exerted on the satellite is less than that on the ground; this accelerates the satellite’s clock (by ap­ proximately 45 microseconds per day). The two eects do not compensate or one another, so temporal correc­ tions must thereore be made regularly.

The GPS uses a geocentric rame o reerence, whose origin coincides with the centre o the Earth. The time is measured using very precise atomic clocks installed onboard satellites. These clocks emit coded microwave signals that can be captured and analyzed by a receiver designed or this purpose. Each portion o the signal contains inormation about the position o the emitting satellite and the time at which it was emitted. The GPS receiver contains a quartz clock that enables it to measure the time at which the signal was received by each o the satellites. The receiver determines the time taken by a signal to travel to the receiver by comparing the time it was received to the time it was emitted. With the speed o microwaves being known (3  108 m/s), the receiver can then calculate the distance o the emit­ ting satellite. Distance calculations take into account the reraction o waves through the atmosphere.

Figure 14 Positioning using the GPS. The intersection of imaginary spheres corresponds to the receiver’s position.

ChapTEr 6 Frames of Reference

161

Marine positioning systems Human beings have always travelled. While they were limited to travelling on land or along coastal shorelines, obvious reerence points such as rivers or mountains would have enabled them to situate themselves geo­ graphically relatively easily and determine routes. When they began to travel the seas away rom the shoreline, where there were no visible reerence points to situate themselves, they had to develop the means to know their position. An understanding o astronomy was signifcant in helping them do so. It was frst observed that some stars remained fxed in the sky, e.g. the North Star, and that others ol­ lowed the same trajectory every night, one shaped like an arc o circle around the North Star. Much later, technical tools were conceived. Invented in China during antiquity and brought to the West by the Arabs during the Middle Ages, the compass was a signifcant invention. Its needle always indicates magnetic north, almost equivalent to true north, and makes it possible to orient onesel even i the Sun or stars are behind the clouds. Beginning in the 7th century, the astrolabe (see Figure 15), whose invention is generally attributed to the Greek scholar Hipparchus (190 to 120 BCE), was commonly used by Arabic astronomers. It was frst used or astronomical studies, astrology, and teaching astronomy. With it, the time can be calculated using the altitude o a given star and a graduated tray (called a tympan—see Figure 15) appropriate or the ob­ server’s latitude. Conversely, i the altitude and direction

Figure 15 An astrolabe’s set of tympans

162

UNIT 2 Preliminary Notions of Mechanics

o two stars are measured at a given time, latitude can also be determined by triangulation. In the 15th cen­ tury, the Portuguese developed a simplifed version o the astrolabe intended or navigation. This astrolabe enabled navigators to measure the angle ormed by a celestial object (the Sun, the Moon or the North Star) with respect to the horizon. Then, using the astronom­ ical tables, it was possible to precisely calculate their position in terms o latitude. It was in 1730 that the sextant (see Figure 16) frst appeared. Simultaneously invented by John Hadley (1682–1744), an English mathematician, and Thomas Godrey (1704–1749), an American inventor, it was also used to measure the angular altitude o celestial bodies above the horizon. Its mechanism is based on the laws o light reection with mirrors. Its main advantage over the astrolabe is that the two directions whose angle must be measured are observed at the same time, making the measurement somewhat inde­ pendent rom the ship’s motion. Thereore, the light rays rom a frst object (the celestial body observed) are reected by the sextant’s set o mirrors and enter into the observer’s lens. The light rays rom a second object (the horizon) enter the lens directly. By having the image o the celestial body and the horizon coincide in the lens, the angle between the two objects can be determined. Today, positioning systems using satellites have greatly acilitated navigation. However, the sextant is still used in the event o a ailure o a GPS or other systems.

Figure 16 A sextant

CHAPTER

R E V IE W

6

Frames of Reference

61 Te uose of  fme of efeence • The description o the motion o a travelling object is strongly dependent on the observer’s position. • In physics, it is said that motion is relative to the rame o reerence against which it is studied. • A rame o reerence is made up o a spatial reerence and a time reerence. • When rame o reerence RB is travelling relative to rame o reerence R A that is considered at rest, there is a relationship that connects the velocities o any bodies in motion measured relative to each rame o reerence. vA  (vB 1vAB) where vA  Velocity of the body relative to frame of reference RA vB  Velocity of the body relative to frame of reference RB vAB  Velocity of frame of reference RB relative to frame of reference RA

62 Inetil fmes of efeence • By defnition, an inertial rame o reerence is a rame o reerence in which an isolated body is either at rest or in uniorm rectilinear motion (motion in a straight line and at a constant velocity). • All rames o reerence in uniorm rectilinear motion relative to an inertial rame o reerence are themselves inertial. • Frames o reerence that have accelerated motion or rotational motion are considered non­inertial. • Galileo’s principle o relativity states that the physical laws that are valid in a given inertial rame o reerence are also valid in all other rames o reerence that move at constant velocities relative to the frst. • The terrestrial rame o reerence is linked to the surace o the Earth or to any body at rest placed on the ground. • The geocentric rame o reerence situates the origin o its spatial reerence at the centre o the Earth. • One o the three axes o the geocentric rame o reerence is generally coincident with the Earth’s rotation axis and is usually oriented toward the North Star, while the other two are directed toward two distant stars that are considered fxed. • The heliocentric rame o reerence situates the origin o its spatial reerence at the centre o the Sun. The three axes o the reerence (x, y and z) are perpendicular to one another and are directed toward distant stars that are considered fxed. • Frames o reerence are more or less inertial depending on the time scale considered and the degree o precision required.

ChapTEr 6 Frames of Reference

163

63 Coodinte systems • A coordinate system associates to each point in space a group of three numbers that makes it possible to situate a body spatially. • In the Cartesian coordinate system, each point P of the plane can be situated by a pair of coordinates x and y, noted (x, y). • The Cartesian coordinates x and y are respectively called the abscissa and ordinate of point P. • In the polar coordinate system, the axis of abscissa is known as the polar axis. • In the polar coordinate system, point P is entirely determined by a pair of coordinates (r, q ): a radial coordinate (r) and an angular coordinate (q). • The radial coordinate (r), which represents the distance connecting the origin of the reference to point P, is called the radius. • The angular coordinate (q), which is the angle between the polar axis and the radius, is called the polar angle. • To go from Cartesian coordinates to polar coordinates, the following two equations are used: r  x2 1 y2

( yx )

q  tan­1

• To go from polar coordinates to Cartesian coordinates, the following two equations are used: x  r  cosq y  r  sinq

ChapTEr 6 1 To study the motion o a travelling object on an inclined plane, you choose a rame o reerence corresponding to the foor o the laboratory The experiment in question lasts a ew seconds a) What type o rame o reerence is it? b) Is this rame o reerence inertial?

164

UNIT 2 Preliminary Notions of Mechanics

Fmes of refeence 2 You are aboard a rail car without windows This car is stationary or is moving at a constant velocity compared to the Earth Is there an experiment that you can do on the train to nd out i you are moving? Explain your answer

Quniis nd Unis

U

nits of measure are as old as the first human civilizations. The Nippur cubit (of Mesopotamia), the Greek stadion, the Roman foot, the aune of Paris, and the metre of today are examples of the variety of units used over time to measure length. The same could be said for other quantities in common use such as those for surface, volume, mass and time. As science evolved, new physical quantities were introduced, and the appropriate units were developed along with them. In this chapter, you will study the International System of Units (abbreviated to SI, for the French Système international d’unités) and the funda­ mental standards of the basic units of mechanics and you will also deduce units derived from the SI. You will use multiples and submultiples of the units used in mechanics and differentiate scalar quantities from vector quantities.

7.1

Quniy, msumn nd uni of msu                            166

7.2 7.3

t Innionl Sysm of Unis     167

7.4

Divd unis in  Innionl Sysm                                 172

7.5 7.6

Mulils nd submulils of unis   173

Fundmnl sndds of  bsic unis of mcnics                    169

Scl quniis nd vco quniis                              176 Chapter 7 Quantities and Units

165

7.1 Quaiy, masum ad ui of masu Agreement between a * Covio number o people, groups or countries.

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

Physics studies the properties o matter and establishes the laws that account or natural phenomena. This science relies only on what can be measured and reproduced with the aim o veriying scientists’ theories. Physics is thereore concerned with dierent quantities and creates new ones as scientifc disco­ veries are made. A quantity is the property o a phenomenon, body or substance that can be expressed quantitatively in the orm o a number and unit o measure. When we measure the dimensions o a table, or instance, and say that its length is equal to 2 m, we are stating:

51.85 cm

the nIppUr CUbIt The cubit is a unit o length that dates rom antiquity. It roughly HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY represents the distance rom the elbow to the longest fngertip. The Nippur cubit is probably the oldest known standard. A prototype dating rom 2650 BCE was discovered in Mesopotamia, a region corresponding to presentday Iraq. It is careully stored in the Istanbul Archaeology Museum in Turkey. This cubit, which measures approximately 51.85 cm, was subdivided into 30 equal parts called fngers. This standard is considered a reerence or all o antiquity’s measurement systems. The Egyptians used the cubit, but divided it into 28 fngers instead o 30. The Roman oot was also defned by this standard. It consisted o 16 Egyptian fngers, or 16 parts o the Nippur cubit divided by 28, making it roughly 29.6 cm long.

166

a quantity a number a unit o measure

→ → →

the length 2 the metre

A unit of measure is a real quantity, defned and adopted by convention*, that can be compared to any other quantity o the same kind to express the relationship between two quantities in the orm o a number. Measuring a quantity comes down to comparing this quantity, using a mea­ suring instrument, with another quantity o the same kind taken as a ree­ rence. Measuring the dimensions o the table above can be carried out, or instance, with a metre­long string. In this case, comparing the length o the table with that o the string gives the number 2. Table length String length

5 2

We can thereore establish that the table length is equal, in this case, to 2 m. Table length String length

5

Table length 1m

5 2 ⇒ Table length 5 2 3 1 m 5 2 m

Measurement can also be perormed by transorming one quantity into another according to a relationship expressed by an explicit ormula. This is the case, or instance, when a orce is transormed into extension by a spring, or lumi­ nous intensity is transormed into electric current by a light­sensitive diode. In the frst case, the quantity measured is length instead o orce, and in the second, it is the intensity o the electric current instead o luminous intensity. In general, instruments o measurement are calibrated to measure the quantity o origin, i.e. the quantity beore transormation. Each o these approaches nevertheless raises questions. Who deined the exact length o the string used to determine the table’s length? Who decided that the reerence would be one metre? What about other quantities’ units o measure? These are questions this section will attempt to answer.

UnIt 2 Preliminary Notions o Mechanics

7.2 t Innionl Sysm of Unis The development o units o measure has gone hand in hand with the devel­ opment o peoples and their socioeconomic needs. The history o humanity has been marked by the adoption o numerous systems o units that depended not only on geographic location, but also on the trade practised, the language used and the era. As an example, the English oot was equivalent to 30.48 cm, the Roman oot to about 29.6 cm, and the Sumerian oot to little more than 27.65 cm. In Europe, during the Middle Ages and the Renaissance, a range o units bearing the same name indicated dierent measurements rom one city to the next. Moreover, the names given to these units reerred to parts o the human body (oot, palm, fnger, thumb, etc.), everyday objects (rod, grain, etc.) or physi­ cal abilities, like the acre—Saxon or feld—which represented the area that could be ploughed by one person in a day. Because this anarchic development raised problems in administrative, com­ mercial and scientifc sectors, there were numerous attempts to standardize the units over the centuries. The most recent o these was the International System o Units, universally known by its abbreviation SI. This conven­ tion was established and defned in 1960 at the 11th General Conerence on Weights and Measures (CGPM). This conerence oversees all decisions on the SI. On January 1, 2010, it consis­ ted o delegates rom 54 member states, including Canada, and 27 associate states. Its frst meeting was in 1889 and, since 1960, it has met every our years in Paris, France. The SI has adopted seven base units that correspond to seven distinct quan­ tities (see Table 1). tbl 1 The SI’s seven base quantities and units Quniy

Quniy symbol

Uni

Uni symbol

Length

s, l, h, r, x

metre

m

Mass

m

kilogram

kg

Time

t

second

s

Electric current

I, i

ampere

A

Temperature

T

kelvin

K

Amount of substance

n

mole

mol

Luminous intensity

Iv

candela

cd

A clarifcation is required in regard to the quantities listed in Table 1. As indi­ cated in the frst column heading, length is included as a quantity. Yet here, the meaning o the word “length” is dierent rom the one used in everyday lie. According to the defnition given or the word “quantity” in Section 7.1 on the previous page, the quantity “length” is the property o a body that can be expressed in the orm o a number and unit o measure, in this case the metre (in the SI). This means, or instance, that the height o a building, the width o a feld, the length o a table, the depth o a well, or the distance travelled by a vehicle are all properties represented by the quantity “length.”

Chapter 7 Quantities and Units

167

It should be noted too that the suggested symbols for quantities, in the second column of Table 1 on the previous page, are those recommended by the International Bureau of Weights and Measures. An s has been added to the letters symbolizing the quantity “length” to represent, in this textbook, displace­ ment. This letter was chosen over the letter d to avoid confusion between displacement and distance travelled. The use of SI units and their symbols is subject to certain important rules. It is essential that these rules be obeyed to ensure data is understood and conveyed appropriately. th pricipal ruls for wriig SI uis • The names o units are considered to be common names. They begin with a lowercase letter (even i the unit symbol is an uppercase letter or begins with an uppercase letter), except when they are at the beginning o a sentence or in a title in uppercase letters. Examples: 10 amperes, 2 metres, etc. •

In general, unit symbols are written in lowercase letters, but i the name o the unit derives rom a proper name, the frst letter o the symbol is uppercase. The litre is an exception to this rule. Examples: m (metre), A (ampere), Ω (ohm), J (joule), l or L (litre), etc.

•

Unit symbols are mathematical entities, not abbreviations. They must thereore not be ollowed with a period, except i they are at the end o a sentence. They are not pluralized. Examples: 10 m high and not 10 m. high 20 kg and not 20 kgs

•

Unit symbols are preceded by a space. Examples: 7.5 m and not 7.5m

SeCtIOn 7.2

th Iraioal Sysm of Uis

1. A truck with a length o 10 m and a mass o 10 000 kg drives down a road at a constant speed o 100 km/h What quantities and units o measure are cited in this example? 2. Spot the typographical errors and rewrite the ollowing sentences correctly a) A worker applies a orce o 100 Newton to a crate b) The table’s mass is equal to 30 kgs c) The game lasted 45 sec longer

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UnIt 2 Preliminary Notions o Mechanics

3. A can label bears the inormation “Net weight: 500 g” Is there an error? I so, indicate the error and correct it 4. The Sumerian fnger and the Egyptian fnger are defned by the Nippur cubit, whose length is equal to 05185 m Knowing that the Sumerian and Egyptian fngers are respectively 1/30 and 1/28 o the Nippur cubit, calculate the number o Sumerian and Egyptian fngers in 1 metre

7.3 Fundmnl sndds o  bsic unis o mcnics A standard is the realization o a given quantity’s defnition, with a deter­ mined value and an associated uncertainty o measurement, and is used as a reerence. A standard can be a material object or it can be based on a physical pheno­ menon. Until 1960, two standards based on a material object or artiact were in use: the metre and the kilogram. Today, the international prototype o the kilogram is the only material standard still used to defne an SI base unit (see Table 2). A standard does not have to be a material object. It can be based on a physical phenomenon, in which case it is known as a natural standard. The example o the metre standard’s modifcations shows that both it and the SI have evolved along with science and the improvements made to measuring instruments. Many other defnitions, like that o the metre, have evolved over time as well (see Table 2). tbl 2 The denitions o the SI’s seven base units, according to the 8th edition o the ocial brochure o the International Bureau o Weights and Measures

Uni

Dfniion

metre

The metre is the length o the path travelled by light in vacuum during a time interval o 1/299 792 458 o a second.

kilogram

The kilogram is the unit o mass; it is equal to the mass o the international prototype o the kilogram.

second

The second is the duration o 9 192 631 770 periods o the radiation corresponding to the transition between the two hyperne levels* o the ground state o the caesium 133 atom.

ampere

The ampere is that constant current which, i maintained in two straight parallel conductors o innite length, o negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a orce equal to 2 3 10-7 newton per metre o length.

kelvin

The kelvin, unit o thermodynamic temperature, is the raction 1/273.16 o the thermodynamic temperature o the triple point* o water.

mole

The mole is the amount o substance o a system which contains as many elementary entities as there are atoms in 0.012 kilogram o carbon 12.

candela

The candela is the luminous intensity, in a given direction, o a source that emits monochromatic radiation o requency 540 3 1012 hertz and that has a radiant intensity in that direction o 1/683 watts per steradian*.

Source : BIPM, 2006.

It is important to mention that the seven base quantities—length, mass, time, electric current, thermodynamic temperature, amount o substance, and lumi­ nous intensity—are independent quantities. This is not the case with their basic units. The defnition o the metre, or instance, requires the second, and that o the mole requires the kilogram.

lvl Separation * hyfn o an energy level in an atom into several sublevels whose energies are very similar.

oin Point obtained in * til a pressure-temperature diagram corresponding to where three states o a pure body coexist: liquid, solid and gas. For water, this point corresponds to T 5 273.16 K and P ≈ 611 Pa. SI unit o measure or * Sdin solid angles that, in mathematics, are the three-dimensional equivalent o the fat or bi-dimensional angle.

The basic units o mechanics are the metre, the kilogram and the second. In this textbook, only these units will be used.

Chapter 7 Quantities and Units

169

7.3.1

the mere

The metre standard, which has been modifed our times since 1889, is used to deine the unit o length. It has gone rom being a material standard to a natural standard (see Figure 1). The current deinition was adopted in 1983 at the 17th General Conerence on Weights and Measures (CGPM). Today, the SI unit o the metre is “the length o the path travelled by light in vacuum during a time interval o 1/299 792 458 o a second,” making the speed o light exactly 299 792 458 m/s. This value o the speed o light is thereore now considered constant, by convention.

Figure 1 The metre standard was once an X-shaped bar o platinum-iridium alloy (90% platinum and 10% iridium) with two notches indicating the ends o the unit. Since 1960, this bar has no longer been used to defne the unit o length.

* mg Microgram.

7.3.2

the kilogram

The kilogram standard, commonly known as the international prototype kilogram, is used to defne the unit o mass. The only material standard still in use in the SI, it was specially manu­ actured rom platinum­iridium (90% platinum and 10% iri­ dium) in 1889. Cylindrical in shape, with a height and diameter both equal to 39 mm, it weighs exactly, by defnition, one kilo­ gram. The prototype is stored in a vacuum under a triple bell jar (see Figure 2). Despite this, it is subject to surace contam­ ination that leads to an increase in mass evaluated at 1 mg* (10 ­6 g) per year. For this reason, the reerence mass o the stan­ dard is the mass it has immediately ater cleaning—a job per­ ormed according to a specifc method most recently established in 1990.

7.3.3

the secod

The second standard is used to deine the unit o time. Initially, the second was defned as a raction o the duration o an average solar day, but the old standard was abandoned because o the Earth’s rotational irregularities. The current defnition, which dates rom 1967, is no longer based on astro­ nomical data, but on a property o matter—the energy tran­ sition between two hyperfne levels o the caesium 133 atom. This deinition has a much greater precision than previous reerences; in act, the second is considered to be the most pre­ cisely known o the SI units. Caesium 133 atoms are used in atomic clocks and are precise to the order o 10 ­15 s—an error o approximately one second every 30 million years. Figure 2 The international prototype kilogram has been kept in the International Bureau o Weights and Measures located in Sèvres, France since 1889, the year it was made. It is the only material standard still used today to defne an SI base unit.

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Furthering

your understanding

a w sdd fo  kilogm? The international prototype kilogram is the only material standard still in use. It is the standard that is the least precisely deined because o the variations in mass it has undergone, which have been evaluated at a ew dozen micrograms (mg) since its manuacture.

The goal o the undertaking was to develop a new and stable standard whose mass did not vary with time. Moreover, the standard’s crystalline perection, isotopic purity and perect sphericity make it possible to defne the kilogram in terms o a precise number o silicon 28 atoms.

To remedy these problems o imprecision and instability in the value o the kilogram standard, scientists rom a number o countries have joined orces to manuacture a new standard, this time made rom ultra-pure silicon and in the shape o a perect sphere whose mass exactly equals one kilogram.

The next meetings o the General Conerence on Weights and Measures (CGPM) will have the mandate o deciding whether or not to proceed with this historical change.

To make it, many constraints had to be overcome. Firstly, the silicon sphere had to be o extreme purity. The material used was purifed in such a way that the silicon ingots obtained contained the highest concentration o the silicon 28 isotope ever produced. The techniques used in preparing it, moreover, produced a crystal o exceptional perection. Scientists then proceeded to shape and polish the crystal using techniques borrowed rom telescope mirror manuacturing. The standard obtained has a diameter o 93.75 mm and is considered the most perect sphere ever manuactured by humans (see Figure 3 ). Its surace imperections are in the order o 0.3 nm, and the variations in its radius are no more than 70 nm.

SeCtIOn 7.3

Figure 3 A pure silicon sphere, which may replace the international prototype kilogram that has been in use since 1889

Fudml sdds of  bsic uis of mcics

1. Is the material standard o the metre still in use? 2. What is the only material standard o an SI base unit still used today?

3. The most recent defnition o the metre standard is based on fxing an important quantity in physics Which one? 4. Why is the second no longer defned by the duration o an average solar day?

Chapter 7 Quantities and Units

171

7.4 Drivd uis i h Iraioal Sysm tabl 3 The units for area and volume, expressed in terms of metres Quaiy

Symbol

Ui

Symbol

s, l, h, r, x

metre

m

Area

A

square metre

m2

Volume

V

cubic metre

m3

Length

The SI’s base units are the oundation or defning all SI units o measure. As ar as mechanics is concerned, the metre (m), the kilogram (kg) and the second (s) are its base units. Note that powers such as “square” and “cubic” used in unit names are placed beore the unit name (except in the case o seconds, where powers are placed ater the unit name). For example: square metre, cubic metre, etc. (see Table 3).

tabl 4 The units for velocity and acceleration, expressed in terms of metres and seconds

Quaiy

Symbol

Ui

Symbol

s, l, h, r, x

metre

m

Time

t

second

s

Velocity

v

metre per second

m/s

Acceleration

a

metre per second squared

m/s2

Length

tabl 5 The units for mechanical quantities with proper names, expressed in terms of SI base units

Symbol

Ui

Symbol

exprssio i SI bas uis

Force

F

newton

N

kg m s-2

Energy

E

joule

J

kg m2 s-2

Power

P

watt

W

kg m2 s-3

Quaiy

By combining units o length and time, it is possible to defne units o velocity and acceleration (see Table 4). Some mechanical quantities have units named ater indivi­ duals who have marked scientifc history. This is the case with orce, energy and power. However, these units are also expressed in terms o SI base units (see Table 5). Since the units o these quantities are complex combina­ tions o base units, the special names and specifc symbols make the units simpler to use. They also make it easier to remember the nature o the quantities involved. The classical rules o algebraic multiplication and divi­ sion apply to orm the products and quotients o unit symbols. Multiplication must be indicated by a space or middle dot.

exampl Kilogram metre per second is written kg m s-1 or kg  m  s-1 m • Metre per second is written or m s-1 s •

SeCtIOn 7.4

Drivd uis i h Iraioal Sysm

1. The unit used in electrical energy sales is the kilowatt hour (kWh) What is the value of this unit in the SI? 2. The work (W ) of a force (F ) is equal to the product of this force and the displacement (∆s) Express work in terms of SI base units

172

UnIt 2 Preliminary Notions of Mechanics

3. Process the units to show that weight is a force 4. Determine the unit of pressure in the SI Remember that p 5 F/A (p 5 pressure; F 5 force ; A 5 area)

7.5 Mulils d sumulils of uis The names and symbols o the multiples and submultiples o SI units are determined according to a decimal base that uses specifc prefxes (see tables 6 and 7). It should be noted that each o the symbols or multiple preixes is written in uppercase, except or da (deca), h (hecto) and k (kilo).

tl 6 Prexes used to orm the

tl 7 Prexes used to orm the

names o multiples o SI units

names o submultiples o SI units

Fco 101

In the case o the symbols or submultiple prefxes, how­ ev er, all are written in lower case. Moreover, all prefx names (o both multiples and submultiples) are written in lowercase, except at the beginning o a sentence.

deca

nm

Symol

da

10-1

deci

d

centi

c

hecto

h

103

kilo

k

10-3

milli

m

M

10-6

micro

µ

G

10-9

nano

n

T

10-12

pico

p

P

10-15

emto



E

10-18

atto

a

zepto

z

yocto

y

mega giga

1012

tera

1015

peta

1018

In the case o unit names ormed with the prefxes o a mul­ tiple or submultiple, there is no space or hyphen between the name o the prefx and that o the unit.

Fco

102

109

m 5 10­6 kg 5 1 mg and not m 5 10­6 kg 5 1 µkg

Symol

10-2

106

For historical reasons, the kilogram is the only basic unit that contains a prefx. For this particular unit, multiples and submultiples are ormed on the basis o the word “gram” whose symbol is g. Thus, we write:

nm

exa

1021

zetta

Z

10-21

1024

yotta

Y

10-24

Example : milligram and not milli­gram exml a Using multiples or submultiples o SI units, write the value o a period o time equal to 5 millionths o a second. Data: t 5 5 3 10-6 s

Solution: t 5 5 3 10-6 s 5 5 micoseconds 5 5 µs

exml b Using multiples or submultiples o SI units, write the value o an energy equal to 17 3 103 J. Data: E 5 17 3 103 J

Solution: E 5 17 3 103 J 5 17 kilojoules 5 17 kJ

a o i naSa’s sysms of uis The Mars Climate Orbiter was a probe sent by NASA to the planet Mars to study its meteorology (see Figure 4 ). The $125 million spacecrat should have begun orbiting the red planet on September 23, 1999, but it was lost because o a navigation error. The probe few over the planet at an altitude much lower than scientists had planned and programmed it to do, causing it to destruct. NASA launched an investigation into the cause o the malunction, examining all hypotheses and conducting a number o investigations. Finally, the inquiry determined the problem’s cause. Two dierent teams working on the probe’s navigation device had not used the same system o units; one used the imperial system, while the other used the SI. The error on the approach orbit was thereore due to a simple error o unit conversion—one that cost $125 million.

Figure 4 The Mars Climate Orbiter

Chapter 7 Quantities and Units

173

APPEnDiX 5 Scefc oao, p. 402.

Example C Using multiples or submultiples o SI units, write the value o a mass m 5 3.2 3 10-3 kg. Data: m 5 3.2 3 10-3 kg

Solution: m 5 3.2 3 10-3 kg 5 3.2 3 10-3 3 (103 g) 5 3.2 g

The grouping ormed by the prefx symbol and the unit symbol constitutes a new and inseparable unit symbol that can be considered a mathematical entity in itsel. This means that it can be multiplied, divided or raised to a power. Example D Express in cubic metres (m3) the value o the ollowing quantity: V 5 4 3 106 cm3 Data: V 5 4 3 106 cm3 1 cm 5 10-2 m

A ormula or an * Homogeey equation is homogeneous when its two sides have the same units.

Solution: V 5 4 3 106 cm3 5 4 3 106 3 (10-2 m)3 5 4 3 106 3 (10-6 m3) 5 4 3 106 3 10-6 m3 5 4 3 100 m3 5 4 m3

Knowledge o units o measurement, mastery o their multiples and submultiples, and the use o mathematical analysis o units are useul in veriying the homogeneity* o ormulas. Example E Veriy the homogeneity o the ormula that defnes gravitational potential energy: Egp 5 mgh Data: Quay

Symbol

Correspodg quay  he Si

U

Symbol

Expresso  Si base us

Gravitational potential energy

Egp

Energy

joule

J

kg m2 s-2

Mass

m

Mass

kilogram

kg

kg

Gravitational acceleration

g

Acceleration

metre per second squared

m s-2

m s -2

Height

h

Length

metre

m

m

Solution: The unit or potential energy is: kg m2 s-2 For the ormula to be homogeneous, the unit o the product (mgh) must have the same unit as that o the energy. Unit o (mgh) 5 unit o (m) 3 unit o (g) 3 unit o (h) 5 kg 3 (m s-2) 3 m 5 kg m2 s-2 The ormula is homogeneous because the unit o the product (mgh) is the same as that o the energy. When solving problems, it is oten a good idea to veriy the homogeneity o the ormulas to identiy possible mathematical errors made by manipulating the variables in the calculations. It should be noted that when a ormula is homogeneous, it does not mean the calculations are accurate. However, i a ormula is not homogeneous, the result is necessarily alse.

174

Unit 2 Preliminary Notions o Mechanics

SeCtIOn 7.5

Mulils d submulils of uis

1. Express the duration “250 seconds” using each o the ollowing prefxes Use decimal notation and scientifc notation a) micro b) milli c) kilo d) mega 2. Express an energy value E o 150 MJ in millijoules (mJ)

9. The period o a simple pendulum is given by the ollowing ormula: T 5 2π

l g

where T 5 Period l 5 Length g 5 Gravitational acceleration

3. Express a mass m o 102 mg in kilograms (kg) 4. The area A o a rectangle is equal to 235 cm2 Express this area in square metres (m2)

θ

5. Perorm the ollowing conversions Show all units and cancellations a) 25 km/h in metres per second b) 150 km/h in metres per second c) 20 m/s in kilometres per hour d) 50 m/s in kilometres per hour 6. Given that 1 J 5 1 kg m2 s-2, veriy the homogeneity o the ormula or kinetic energy 1 Ek 5 3 m 3 v2 2 7. Write the names o the ollowing units using the submultiple prefxes defned by the SI: a) l 5 10-6 m b) t 5 10-15 s c) m 5 10-9 kg 8. Write the symbols or the names o the ollowing units using the symbols or the multiple prefxes defned by the SI: a) l 5 103 m b) E 5 106 J c) P 5 109 W

I

Fg

Veriy the homogeneity o the ormula given that the period’s SI unit is the second 10. Hooke’s law links applied orce (F ) to an extended spring (∆x ) using the ollowing ormula:

Fr5 k 3 ∆x The proportionality actor between orce and extension is called the spring constant (k) Determine the unit o this constant

Chapter 7 Quantities and Units

175

7.6 Scalar quaiies ad vecor quaiies In physics, there are two kinds o quantities: scalar quantities and vector quantities.

7.6.1 table 8 Some examples of scalar quantities used in mechanics

Scalar quaiies

Scalar quaiies

Scalar quantities are completely defned by a real number and a unit o mea­ sure. An example is the measuring o the table’s length at the beginning o the chapter, which gave the result:

Length

l52m

Area Volume Time Energy

The “length” quantity o the table is, in this case, defned by the number 2 and the “metre” unit o measure, represented by its symbol (m). The same goes or many other physical quantities used in mechanics (see Table 8).

Power

7.6.2

Vecor quaiies

Vector quantities are not completely defned by a real number and a unit o measure. A simple example makes this easier to understand. I we say that a orce F 5 100 N is applied to a door, the situation is not completely defned. Even though the intensity o the orce is known, the way it is applied is not. For example, orce can be applied to open the door or to close it—two dier­ ent situations (see Figure 5).

100 n

100 n

Figure 5 In each situation a force of 100 N is applied to the door. The two situations differ, however, in the direction of the applied force in space.

176

UnIt 2 Preliminary Notions of Mechanics

For the orce represented in Figure 5 to be completely deined, it requires additional inormation, relating to its direction in space, which can be pro­ vided by a vector.

tbl 9 Some examples of vector quantities used in mechanics

Vco quiis

A number o physical vector quantities are used in mechanics (see Table 9).

Displacement

A vector is a mathematical entity represented by an arrow characterized by: • its magnitude (also called norm), which represents the intensity o the quantity • its direction, which is that o the line and arrowhead carrying the vector (see Figure 6)

Velocity Acceleration Force

Vector quantities are represented by letters with an arrow on top. A vector v is written v. The magnitude o a vector v corresponds to the size o the quan­ tity it represents. It can be symbolized by ||v|| or simply v. The latter notation will be used in this textbook. Note that when the term “magnitude” is used, it is generally symbolized by |v|.

See pois of vcos, p. 184.

y

Figure 6 represents a vector quantity represented by the vector F . The magnitude o vector F corresponds to the length o the line segment AB. (This can be calculated mathemat ically using the Pythagorean theorem or the right­angled triangle ABC. This method will be covered in detail in Chapter 8.) The unit o the quantity represented by vector F must be added to the numerical value calculated. Also in Figure 6, the vector’s direction is defned by the direction o the line d that carries the vector F , i.e. rom A to B.

50 B

40

d

30 20

F

10 A 50





40

30





20

10





10



20



30



40



50

10

20

30

40

50

x

Figure 6 The vector F represents a vector quantity; it is characterized by its magnitude (length AB) and its direction (line d from A to B).

SeCtIOn 7.6

Scl quiis d vco quiis

1. What is the difference between scalar quantities and vector quantities?

3. Name three scalar quantities and three vector quantities

2. Why are quantities directed in space called vector quantities?

4. What characterizes a vector?

Chapter 7 Quantities and Units

177

APPLICATIONS aomic clocks The development o the quantum theory led to the construction o the frst unctional atomic clock in 1955. According to the quantum theory, an atom that is subjected to certain types o radiation emits deter­ mined quantities o energy, or quanta o energy. The amount o energy emitted by the atoms is directly pro­ portional to the requency o the electromagnetic waves. The energy dierences between the states o an atom correspond to perectly defned values. The same is true or the requency o an electromagnetic wave that can change the state o an atom or that is generated by the change o state o an atom The prin­ ciple o the atomic clock is based on the requency o the electromagnetic wave and on the number o pe­ riods, much like a grandather clock keeps track o the number o oscillations o its pendulum and ad­ vances the hands o the dial once every period.

o space probes to ar­o places also requires the accu­ rate measurement o very long time spans. In scientifc research, time must be measured as reliably and as accurately as possible. For example, some o the eects o the theory o relativity proposed by Einstein (1879–1955) can be detected or infnitely small objects moving at very high speeds. As a result, researchers who wish to study these eects must measure time very accurately and use cesium atomic clocks or this purpose.

The most stable and accurate atomic clock is currently the cesium clock (see Figure 7). The SI unit o measure or time, which is the second, is defned in relation to this clock. In order to obtain the greatest possible accu­ racy, the atomic second is calculated based on an aver­ age taken rom 250 cesium atomic clocks around the world. The “true” second is called the atomic second, not to be conused with the solar second, which is an astronomical second. Today, in many felds, it is extremely important to measure time with great accuracy. The development o much state­o­the­art technology depends on it. For example, or GPS signals, which travel at the speed o light, an accuracy o three billionths o a second is required to determine a location to within one metre. Satellite transmissions and several modern electronic systems require even greater precision. In telecommu­ nications, the more accurately time is measured, the more bits (inormation units) can be transmitted each second without becoming scrambled. The launching

178

UnIt 2 Preliminary Notions of Mechanics

Figure 7

A cesium atomic clock

t kilogm The balance was created or trade purposes, enabling people to evaluate the quantities involved in order to acilitate trade. The Egyptians and Babylonians appear to have been the frst to establish a system o weights and measures that allowed merchants to weigh the products they were buying and selling. For centuries, balances worked with weights (the spring balance was only invented in the 18th century). It was important or these weights to match the established standards as closely as possible. At the beginning o the Middle Ages in France, the units o weight were relatively uniorm and defned according to the Paris pound. Magistrates were responsible or determining the accuracy o the measurement standards used in various cities by comparing them with the standards kept in the king’s palace. However, beginning in the 9th century, some o the lords saw no problem with using dierent measures than the ofcial standard to serve their business interests. This led to such a multiplication o units o measure that rom one province or village to another, the units were not the same and did not even have the same value. Over the centuries, several attempts were made to standardize the measures, but they were unsuccessul because the lords reused to give up their privilege to determine the standard. During the French Revolution (1789), the people demanded “one weight, one measure, one custom” so that they would no longer have to put up with arbi­ trary measurements. In 1793, Antoine de Lavoisier (1743–1794), the ounder o modern chemistry, sug­ gested using the mass o a cubic decimetre (one litre) o pure water at 4°C as a unit o mass, which was called a grave in French. Instead, the decision was made to defne a “gram” as the mass o a cubic cen­ timetre o water at 0°C. However, because the gram was very small, use o this as a standard was difcult. This led authorities to suggest, in 1795, the kilogram as the basic unit o the metric system, which was the precursor to the SI system. At that time, the kilogram was defned as the mass o a cubic decimetre o water at 4°C. This is why the kilogram is the only base unit in the SI system that has a prefx.

In 1889, at the First General Conerence o Weights and Measures, the kilogram was ofcially adopted as a unit o mass. The standard was made rom a plati­ num­iridium alloy (90% platinum and 10% iridium), a dense material that is practically unalterable. The kilogram is the SI’s only material standard still in use, and it is also the least accurate standard because o variations in its mass in the order o tens o micro­ grams (μg) since it was frst made (see Figure 8). Scientists are currently developing a new stable stand­ ard whose mass will not change over time (see p. 171).

Figure 8 The beam balance dates from antiquity. With this type of balance, an object is suspended from a hook, then a counterweight is moved along a graduated beam. When the object weighed and the counterweight are in a position of equilibrium, the weight can be read on the graduated beam. This type of balance is still used in many countries.

Chapter 7 Quantities and Units

179

7

chapter

Quaiies ad Uis

7.1 Quaiy, measureme ad ui o measure • A quantity is the property o a phenomenon, body or substance that can be expressed quantitatively in the orm o a number and unit o measure. • A unit o measure is a real scalar quantity, defned and adopted by convention, that can be compared to any other quantity o the same kind to express the relationship between two quantities in the orm o a number.

7.2 the Ieraioal Sysem o Uis • The International System o Units (SI) was established in 1960 at the 11th General Conerence on Weights and Measures (CGPM). • The SI has adopted seven base units that correspond to seven distinct quantities. The ollowing table shows the quantities used in mechanics. Quaiy

Symbol

Ui

Symbol

Length

s, l, h, r, x

metre

m

Mass

m

kilogram

kg

Time

t

second

s

• The writing o SI units and their symbols is subject to important rules.

7.3 Fudameal sadards o he basic uis o mechaics • A standard is the realization o a given quantity’s defnition, with a determined value and an associated uncertainty o measurement, and is used as a reerence. • A standard does not have to be a material object. It can be based on a physical phenomenon, in which case it is known as a natural standard. • The kilogram standard, commonly known as the international prototype kilogram, is used to defne the unit o mass. It is the only material standard still in use in the SI. • The ollowing table shows the basic units o mechanics as defned by the SI. Ui

180

Defiio

metre

The metre is the length o the path travelled by light in vacuum during a time interval o 1/299 792 458 o a second.

kilogram

The kilogram is the unit o mass; it is equal to the mass o the international prototype o the kilogram.

second

The second is the duration o 9 192 631 770 periods o the radiation corresponding to the transition between the two hyperfne levels o the ground state o the caesium 133 atom.

UnIt 2 Preliminary Notions o Mechanics

7.4 Deved uns n e inenonl Sysem • The SI’s base units are the oundation or defning all SI units o measure. • From the base units it is possible to defne the units o other quantities in mechanics. • Some mechanical quantities have units named ater individuals who have marked scientifc history. These units are also expressed in terms o SI base units.

7.5 Mulles nd submulles of uns • The names and symbols o the multiples and submultiples o SI units are determined according to a decimal base that uses specifc prefxes. • The symbols or multiple prefxes are written in uppercase, except or da (deca), h (hecto) and k (kilo). • The symbols or submultiple prefxes are written in lowercase. • All prefx names (o both multiples and submultiples) are written in lowercase, except at the beginning o a sentence. • For historical reasons, the kilogram is the only basic unit that contains a prefx.

7.6 Scl qunes nd veco qunes • Scalar quantities are quantities that are completely defned by a real number and a unit o measure. • Length, area, volume, time, energy and power are examples o scalar quantities. • Vector quantities are not completely defned by a real number and a unit o measure. • Vector quantities require inormation relating to their direction in space. • Vector quantities are represented by vectors. • A vector is a mathematical entity represented by an arrow characterized by: – its magnitude (also called norm), which represents the intensity o the quantity – its direction, which is that o the line carrying the vector • Displacement, velocity, acceleration and orce are examples o vector quantities.

ChaptEr 7 Quantities and Units

181

Chapter 7

Quiis d Uis

1. Reproduce the ollowing table on a sheet o paper and complete it Fco

nm

Symbol

deca

Explain why momentum is a vector quantity

milli

5. Express the unit or momentum (p) in SI base units

giga

6. In the law o universal gravitation:

10-1

2. Reproduce the ollowing table on a sheet o paper and complete it SI bs ui

Unit based on a property o the cesium 133 atom Unit defned in relation to a distance covered by light

Unit that has been modifed our times since 1889 Most precisely known SI unit

3. Reproduce the ollowing table on a sheet o paper and complete it Scl/vco

Force Mass Velocity Acceleration Energy

182

F5G

m1 m2

r2 F is a orce, m1 and m2 are masses, r 2 is a length (the distance between the masses), and G is the gravitational constant What is the unit o this constant in the SI ? 7. Knowing the exact dimensions o the international prototype kilogram (a cylinder whose height and diameter are equal to 39 3 10-3 m), calculate the relative density o the alloy used to manuacture it Give the result in kilograms per cubic metre (kg/m3) and in grams per cubic centimetre (g/cm3)

Only unit defned in relation to a material standard

Quiy

p 5 mv where: p 5 momentum m 5 mass v 5 velocity

10-15

Ifomio

4. Momentum is a physical quantity whose intensity is calculated by the ollowing ormula:

UnIt 2 Preliminary Notions o Mechanics

8. Knowing that the SI unit o energy (E ) is (kg m2 s-2) and that o power (P ) is (kg m2 s-3), fnd the relationship that links energy to power

Vcos

M

athematics has played a major role in the devel­ opment of physics. In order to study motion from a physics standpoint, you must use the notion of vectors as developed in the field of mathematics. Generally speaking, vectors stem from a combination of algebra and geometry and are an essential tool to describe and understand many physical phenomena such as velo­ city and force. In this chapter, you will be introduced to the notion of vectors. You will discover their properties, and you will become famil iar with the operations that can be done with vectors.

8.1 8.2 8.3 8.4

pois of vcos                    184 adding vcos                          190 Subcing vcos                     193 Mulilying  vco by  numb       196 Chapter 8 Vectors

183

8.1 proeres of vecors To describe certain physical phenomena, it is important to know the direction o quantities such as displacement, velocity and orce. A vector is a mathe­ matical tool that is used to express such quantities.

8.1.1 See Scalar quaes ad vecor quaes, p. 176.

Scalars ad vecors

A physical quantity can be described as a scalar or a vector. A scalar, or scalar quantity, is completely defned by a real number and a unit o mea­ surement. Time, mass and volume are examples o physical quantities that are expressed with scalars. For example, a 25 kg mass is perectly defned by the number 25, and the “kilogram” unit o measurement is represented by its symbol (kg). A vector, or vector quantity, is not completely defned by a real number and a unit o measurement. For this quantity to be completely deined, additional inormation related to its direction in space is required. Displacement, vel­ ocity, acceleration and orce are examples o physical quantities that are expressed with vectors. For example, a 3 m southward displacement is com­ pletely defned by the number 3, the “metre” (m) unit o measurement and its direction in space (toward the south).

8.1.2

a) Two vectors with different directions

Magude ad dreco of a vecor

A vector is represented by an arrow. The beginning o the arrow is called the tail (origin) and the tip is called the head (end point). The length o the arrow is proportional to the vector’s magnitude, and the direction is indicated by the line segment and arrowhead (see fgures 1 and 2).

b) Two vectors with opposite directions

Vector

Magnitude

→

6m

s

Direction

Figure 1 A vector has a magnitude and a direction. The properties above show the displacement → s of 6 m c) Two vectors with the same direction

on the horizontal toward the right.

Figure 2 The direction of vectors

An arrow drawn over the symbol or magnitude is used to represent a vector, or example, s symbolizes the displacement vector. Two letters with an arrow on top may also be used. In that case, the letters represent the two points loca­ → ted at the origin and at the end o the arrow, or example: AB (see Figure 3).

→

→

s

A AB

B

Figure 3 There are two ways of desig→ nating a vector. Here, the displacement s is from point A to point B.

184

→

For any v vector, the || v || symbol or simply the letter v is used to express its magnitude. For example, or a velocity o 5 km/h eastward, we write: → ||v || 5 5 km/h or v 5 5 km/h. Magnitude will be expressed as v in this textbook. In a Cartesian plane, the tail o a vector does not necessarily correspond to the origin o the plane, nor is a vector represented by its position. In act, it is possible to move the arrow representing a vector without changing its value or direction. There are dierent ways o describing the direction o a vector in a plane. The most common methods are the cardinal points method and the trigonometric angle method.

Unit 2 Preliminary Notions of Mechanics

SW

SE S

Cdinl oins mod

N

N NW

The cardinal points method indicates the direction o a vector using one or two cardinal points and an angle. The compass rose that appears on most geographical maps makes reerence to this method. The most pre­ cise way o using this method is to combine two car­ dinal points and an angle (see Figure 4). Start rom the second cardinal point, indicated in brackets, then move rom the angle indicated in the direction o the frst cardinal point.

NE

W

W

E

SW

SE

E

60°

8 m [60° S of W]

S

S

b) The combination of two cardinal points and an angle. The 8 m [60° S of W] displacement vector indicates that a displacement of 8 m occurred at 60° south of west.

a) The compass rose N

Figure 4 The direction of a vector according to the cardinal points method

tigonomic ngl mod

W

In physics, the direction o a vector is generally described using the trigonometric angle method. This method uses the Cartesian plane: the positive x-axis corresponds to 0° and the angles are measured counter­ clockwise. A complete rotation corresponds to 360° (see Figure 5).

E

60°

90° y 8 m [60° S of W]

II

I

S

240°

180°

x III 8 m [240°]

0° [360°]

IV 270°

Figure 5 The direction of a vector according to the trigonometric angle method. The 8 m at 240° displacement vector is the result of a 240° rotation counter-clockwise starting at the positive x-axis.

exml a Graphically represent a billiard ball that moves by →s 4 dm at 60°. Data: s 5 4 dm at 60°

exml B A force applied to a rope attached to a post is represented as follows. Using the two methods, describe the vector representing this force.

Solution:

→

F  500 N

135°

d  4 dm 60°

exml C Using the cardinal points method, how can we indicate that a bicycle is travelling at a velocity of 30 m/s directed at 112.5°?

exml D Using the trigonometric angle method, how can we indicate the displacement →s of a hiker who has just walked 200 m [25° E of S] ?

Solution: → The vector is v 5 30 m/s [22.5° W of N]. 112.5° v  30 m/s

Solution: Angles are calculated in the trigonometric sense. The vector is → F→ 5 500 N at 225° or F 5 500 N [45° S of W]

Solution: → The vector is s 5 200 m at 295°. 25°

d  200 m

S

It should be noted that these are only two­dimensional vectors (in a plane).

Chapter 8 Vectors

185

See addg vcs, p. 190. See Subcg vcs, p. 193.

8.1.3

Cms f  vc

The vectors located in a Cartesian plane can be decomposed into an x­com­ → ponent and a y­component. For any vector v located in a Cartesian plane whose direction is described by an angle, the components are described as follows:

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

→

v

vy θ vx

vx 5 v cos θ

vy 5 v sin θ

Decomposing a vector into its components makes it possible to add and subtract vectors. This will be addressed in the following sections. The components of a vector can be positive or negative. In general, an x­com­ ponent is positive when it is directed toward the right and negative when it is directed toward the left; a y­vertical component is positive when it is directed upward and negative when it is directed downward. This convention is in line with that of the Cartesian plane (see Figure 6). pierre-SiMon LapLaCe

 y

French mathematician, astronomer and physicist (1749–1827)

HISTORYLaplace HIGHLIGHTS HISTORY HIGHLIGHTS Pierre-Simon was the rst to have used the term “vecteur” in French, a word that comes rom the Latin vector and reers to a chariot rider (this word had already appeared in the works o Newton). Laplace ormulated audacious hypotheses, many o which are still accepted today. He theorized, among other things, that the solar system resulted rom the condensation o a nebula; that some visible stars were not part o the Milky Way but were actually galaxies; that black holes might exist. Undeniably, his most important contribution is that he systematically applied mathematical analysis to physics and astronomy. His theories had a proound infuence on the development o science in the 18th and 19th centuries.

186

vx 



x

→

s

α

α

vy

sy

→

v

sx



a) The conventional representation o the Cartesian plane

b) The x-component is positive and the y-component is negative.

c) Both components are negative.

Figure 6 The components o a vector can be positive or negative

exml A velocity vector v has a magnitude o 10.0 m/s and a direction o 30° below the x-axis, as illustrated in the ollowing gure. What are its components?

vx α vy →

v

Data: v 5 10.0 m/s α 5 30° Calculation: θ 5 360° 2 30° 5 330° vx 5 v cos θ 5 10.0 m/s cos 330° 5 8.66 m/s vy 5 v sin θ 5 10.0 m/s sin 330° 5 -5.00 m/s The vector is written as ollows based on its components: v 5 (8.66, -5.00) m/s

Unit 2 Preliminary Notions o Mechanics

When the vector’s components are known, it is possible to determine the magni­ tude of the corresponding vector by applying the Pythagorean relationship. In the case of any right­angled triangle for which c is the hypotenuse and the other sides are a and b the Pythagorean theorem states that, c2 5 a2 1 b2. Consequently, we can verify the following relationship for any vector located in a plane:

appenDix 5 rvw of goomy, p. 404.

→

v

vy θ vx

v 5 vx2 1 vy2

Using trigonometry*, it is therefore possible to determine the direction of a vector from its components. To make it easier to calculate the angle, it is use­ ful to draw a diagram of the components and form a right­angled triangle. We → can then verify the following relationship for any vector v located in a plane: tan α 5

vy vx

⇒ α 5 tan-1

Part of mathematics * tgoomy that aims to establish, through calculations, using circular functions, the relationships between the sides and angles of a triangle.

vy

(v ) x

The previous formula always results in an angle between 0° and 90° if the values of vx and vy are positive. However, when one or two components are negative, the formula does not directly provide θ, which is obtained through trigonometry. Therefore, to determine the value of θ in these cases, begin by calculating α using the above formula. When the components have opposite signs, use their absolute values. Next, use a diagram to determine angle θ by adding or subtracting α from the angle being sought (see Figure 7). eml An airplane’s displacement is a vector →s whose components are sx 5 -24 km and sy 5 15 km. What are the magnitude and direction of vector →s ?

→

v

θα

a) The two components are positive, so θ 5 α.

→

v

θ  180°  α

α

b) The x component is negative and the y component is positive, so θ 5 180° – α. →

s

sy  15 km

θ  270°  α

θ

α sx  24 km →

Solution: s 5 sx2 1 sy2 5 (-24 km)2 1 (15 km)2 5 576 km2 1 225 km2 5 801 km2 5 28 km v 15 km α 5 tan-1 y 5 tan-1 5 tan-1(0.625) 5 32° vx |-24 km|

()

(

θ 5 180° 2 32° 5 148° → The vector is d 5 28 km at 148°.

)

v α

c) The x component is positive and the y component is negative, so θ 5 270° 1 α.

Figure 7 Example of calculation of angle θ from angle α Chapter 8 Vectors

187

Vecors ha have more ha wo compoes

z

This chapter deals with two-dimensional vectors. But what about the other dimensions? In three dimensions, vectors have three components (vx , vy and vz ). To recompose a vector, the Pythagorean relationship must be applied twice. In a three-dimensional axis system, three axes must be used: the abscissa (x ), the ordinate (y ) and the applicate (z ) (see Figure 8).

→

v

Albert Einstein’s (1879-1955) theory o general relativity describes the universe as an infnite space in three dimensions that is curved around a ourth dimension. It is impossible or a human being to perceive this ourth dimension; nonetheless, several experiments have proven the accuracy o this theory. Mathematics that describe the general theory o relativity use vectors that can have more than three components.

8.1.4

vz vy y vx

x

Figure 8 In three dimensions, vector v has three components (vx, vy and vz ).

Vecors ad he Caresa plae

In the Cartesian plane, vectors are represented using their components accord­ ing to the ollowing general notation: →

v 5 (vx , vy ) →

In Figure 10, or example, vector v 5 (5, ­3) has a x-component o fve units to the right and a y-component o three units downward. The components o such a vector must not be conused with the coordinates o a point despite the act that their notation is similar. In act, it is possible to place the vector anywhere in the plane. y

→

v

-3 →

v

-3

5

5 x →

-3

v 5

Figure 9 Vector → v 5 (5, -3) can be placed anywhere in the plane. It is important not to conuse the components (5, -3) o such a vector with the coordinates o a point despite the act that their notation is similar.

However, it is possible to fnd the components o a vector using the coordinates o the points that represent its tail and its head. To fnd out the x­component, calculate the variation o abscissas (vx 5 ∆x 5 x2 – x1). To determine the y­com­ ponent, calculate the variation o ordinates (vy 5 ∆y 5 y2 – y1). The ollowing example illustrates how to proceed.

188

Unit 2 Preliminary Notions o Mechanics

exml A man lives at point A (-2, 3) and works at point B (4, 5). What are the components → of the vector AB representing the trip the man takes to get to work? (Each square represents 1 kilometre.)

y

B(4, 5) A(-2, 3)

Solution: → Calculate the variation of the abscissas and the variation of the ordinates of vector AB in order to determine its components:

2 6 x

(AB)x 5 ∆x 5 x B 2 x A 5 4 km 2 (-2 km) 5 6 km (AB)y 5 ∆y 5 y B 2 y A 5 5 km – 3 km 5 2 km

SeCtion 8.1

t s f vcs

1. Determine if the following examples are scalars or vectors Explain your answers a) the price of a car b) the time a car takes to accelerate to 100 km/h c) the speed of a car that travels at 90 km/h toward the east d) a quantity of gas of 38 L e) the displacement between two cities

5. Points A, B and C are the positions of soccer players Vectors AB , BC and CA represent the ball’s displacements Write the components of these vectors (Each square represents 1 metre) y A B

2. The following vectors represent the wind’s direction as indicated by a weathervane Express the direction of these vectors using a trigonometric angle → u

a)

→

u →

u

30°

30° b)

20°

c)

v

40°

20°

20° v→ v →

C

40° w

→

30°

x

w 40° w

3. Calculate the components of the following force vectors →

6. Find the components, magnitude and direction of the following displacement vectors (Each square represents 1 metre) y

a) F1 5 166 N at 125° →

b) F2 5 235 N of SW →

c) F3 5 575 N [15° E of S] 4. Determine the magnitude and direction of each of the following velocity vectors a) v1x 5 84 m/s and v1y 5 -59 m/s b) v2x 5 -881 m/s and v2y 5 302 m/s c) v3x 5 -551 m/s and v3y 5 -875 m/s

s2

s1

x s3

Chapter 8 Vectors

189

8.2 addg vecors In the same way that it is possible to add scalar quantities, it is possible to add vector quantities. However, adding vectors is a more complex operation because one must take the direction o the vectors into account. There are several methods o adding vectors: the graphical method, the com­ ponent method and addition in the Cartesian plane.

8.2.1

Grphcl mehod

The basic principle o the graphical method consists o graphically arran­ ging the vectors one ater the other. In other words, the tail o the second vec­ tor must correspond to the head o the frst. Then draw the resultant vector (resultant) that corresponds to the arrow that starts rom the tail o the frst vector and ends at the head o the second vector. When this method is used to add vectors, it is oten called the triangle method (see Figure 10). s1

s2

s1

s2

sR

Figure 10 Adding vectors with the graphical triangle method. The tail of vector s2 corresponds to the head of vector s1 . The resultant vector sR begins at the tail of vector s1 and ends at the head of vector s2 . In this example, the addition of two displacements results in another displacement.

I more than two vectors must be added, the graphical method is still valid. and then is oten called the polygon method (see Figure 11). sR s1

s2

s3

s3 s1 s2

Figure 11 Adding vectors with the polygon method. The tail of vector s2 corresponds to the head of vector s1 and the tail of vector s3 corresponds to the head of vector s2 . The resultant vector sR begins at the tail of vector s1 and ends at the head of vector s3 . In this example, adding three displacements results in another displacement.

In this textbook, the graphical method is used only to represent the addition o vectors in a diagram. To obtain a numerical result, represent the vectors accor­ ding to a determined scale. Then, draw the vectors using a protractor, and mea­ sure the quantity and direction o the resulting vector.

8.2.2

Compoe mehod

The component method uses trigonometric calculations rather than graphical tools. This method requires decomposing the vectors to be added, then adding the x­components to each other on the one hand, and the y­components to each other on the other hand. To obtain the magnitude and direction o the resultant vector, we must proceed as indicated on page 187. The ollowing example illus­ trates the addition o vectors with the component method.

190

Unit 2 Preliminary Notions of Mechanics

exml → A mail carrier works in the countryside. He must deliver two letters. The vectors u 5 15.0 km → at 30° and v 5 25 km at 320° correspond to his successive displacements. What is his total → displacement s ? Solution: → → → s 5u 1v ux 5 u cos θ 5 15 km cos 30° 5 12.99 km vx 5 v cos θ 5 25 km cos 320° 5 19.15 km

θ

uy 5 u sin θ 5 15 km sin 30° 5 7.50 km

sx  32.14 km

vy 5 v sin θ 5 25 km sin 320° 5 -16.07 km

α

sy  8.57 km

→

s

sx 5 ux 1 vx 5 12.99 km 1 19.15 km 5 32.14 km sy 5 uy 1 vy 5 7.50 km 2 16.07 km 5 -8.57 km

s 5 sx2 1 sy2 5 (32.14 km)2 1 (-8.57 km)2 5 1032.98 km2 1 73.44 km2 5 1106.42 km 5 33.26 km -8.57 km| 5 tan (0.27) 5 14.9° ( vv ) 5 tan (|32.14 km )

α 5 tan-1

y

-1

-1

x

θ 5 360° – 14.9° 5 345.1° →

The resultant vector is s 5 33.26 km at 345.1°.

8.2.3

adding in  Csin ln

Adding vectors using the Cartesian plane consists of adding the vectors’ x­ components to each other and the y­components to each other according to the following rule: If then

→

→

u 5 (a, b) and v 5 (c, d) → → → w 5 u 1 v 5 (a, b) 1 (c, d) 5 (a 1 c, b 1 d)

Figure 12 illustrates how to add vectors in the Cartesian plane by applying this rule. y

→

u

b a

→

v

d c

x ac →

w

bd

Figure 12 The addition of two vectors in a Cartesian plane. The resultant vector w is equal to u 1 v .

Chapter 8 Vectors

191

exmpl A plane is fying nonstop. The rst part o the fight is expressed by the vector s1 5 (-400, 900) m. The second part o the fight is expressed by the vector s2 5 (700, 100) m. What is the plane’s total displacement sR ? Solution: → → sR 5 s1 1 s2 5 (-400, 900) 1 (700, 100) 5 (-400 1 700, 900 1 100) 5 (300, 1000) We obtain the ollowing plane. (Each square represents 100 km). y

s2

x sR

s1

The total displacement o the plane is sR 5 (300, 1000) km.

SeCtion 8.2

addg vcrs

1. Determine the resultant vector of the following vector additions → a) v →

→

a

u

b)

→

→

u

v

→

or

b

y →

c

→

a

→

c

or

3. Three forces are applied to an object located at point (2, 3) Determine the components of the resultant force vector FR  (Each square represents 1 newton)

or

→

b

or

F1

2. Using the component method, add the following vectors a) forces F1 5 45 N at 135° and F2 5 20 N at 200° b) speeds v1 5 150 km/h SW and v2 5 20 km/h W c) displacements s1 5 54 m [15° S of W] and s2 5 182 m [5° N of W]

192

Unit 2 Preliminary Notions o Mechanics

x F2

F3

8.3 Subcing vcos In physics, there are situations where it is necessary to subtract vector quan­ tities. Since the operation opposite to addition is subtraction, there is a con­ nection between the methods or adding and subtracting vectors. Beore addressing the dierence o vectors, it is essential to defne the notion o opposite vectors. The opposite o a vector is a vector o identical magni­ tude, but o opposite direction (see Figure 13). For example, velocity vectors v1 5 4 m/s at 45° and v2 5 4 m/s at 225° are opposite, because they have the same magnitude and there is an angle o 180° between them.

Figure 13 Vectors v and -v are opposite because they have the same magnitude but are in opposite directions.

y

In the Cartesian plane, opposite vectors have components o the same magnitude but opposite signs. For example, displacement vectors s 5 (­8, 6) and ­s 5 (8, ­6) are opposite (see Figure 14).

8.3.1

-v

v

Gicl mod

s

Subtracting two vectors actually consists o adding the frst vector to the opposite o the second u 2 v 5 u 1 (­v ). With the graphical method, place the tail o the second vector at the head o the frst, but only ater having drawn the arrowhead on the opposite side o the line segment used to orm the arrow representing the vector (see Figure 15).

x -s

-v u

v

Figure 14 Displacement vectors → s 5 (-8, 6) m

w

and -s 5 (8, -6) m are opposite in the Cartesian plane. →

u

Figure 15 The subtraction o vectors by the graphical triangle method. The tail o vector -v corresponds to the head o vector u . The resultant vector w starts at the tail o vector u and ends at the head o vector -v .

8.3.2

Comonn mod

This method requires decomposing the vectors that must be subtracted, then subtracting the x­components rom one another on the one hand, and the y­ components rom one another on the other hand. To obtain the magnitude and direction o the resultant vector, we must proceed as indicated on page 187. The ollowing example illustrates the subtraction o vectors using the component method. exml Three children want the same toy. The frst child pulls on it, exerting a orce F1 o 8.00 N at 45.0°. The second child exerts a orce F2 o 6.00 N at 280°. The third child exerts an unknown orce F3 . The resultant orce FR is 3.03 N at -58.5°. Determine the magnitude and direction o the unknown orce F3 . Solution: Since FR 5 F1 1 F2 1 F3 , then F3 5 FR 2 F1 1 F2 . F1x 5 F1cos θ 5 8.00 N cos 45.0° 5 5.66 N F2x 5 F2cos θ 5 6.00 N cos 280° 5 -1.04 N FRx 5 FR cos θ 5 3.03 N cos (-58.5°) 5 1.58 N

Chapter 8 Vectors

193

F3x 5 FRx 2 F1x 2 F2x 5 1.58 N 2 5.66 N 2 1.04 N 5 -5.12 N

θ

F3x  5.12 N

F1y 5 F1 sin θ 5 8.00 N sin 45° 5 5.66 N F2y 5 F2 sin θ 5 6.00 N sin 280° 5 -5.91 N

α F3y  2.33 N

FRy 5 FR sin θ 5 3.03 N sin (-58,5°) 5 -2.58 N

F3

F3y 5 FRy 2 F1y 2 F2y 5 -2.58 N 2 5.66 N 2 (-5.91) N 5 -2.33 N F3 5 F3x2 1 F3y2 5 (-5.12 N)2 1 (-2.33 N)2 5 5.63 N F3y |2.33| α 5 tan-1 F 5 tan-1 5 24.5° |5.12| 3x

( )

( )

θ 5 α 1 180° 5 24.5° 1 180° 5 205° The unknown force F3 has a magnitude of 5.63 N and a direction of 205° above the positive x-axis.

8.3.3

Subracg  h Carsa pla

In the Cartesian plane, subtracting vectors consists of subtracting x­com­ ponents from each other and subtracting y­components from each other according to the following rule: →

→

u 5 (a, b) and v 5 (c, d) → → → w 5u 2 v 5 (a, b) 2 (c, d) 5 (a 2 c, b 2 d)

Symbolically, we have: If then

Figure 16 illustrates how to subtract vectors in the Cartesian plane by apply­ ing this rule. y

u

b

v

d

a c x w

bd

ac

Figure 16 Subtracting two vectors in the Cartesian plane. The resultant vector w is equal to u 1 (-v ).

The following example illustrates the subtraction of vectors in the Cartesian plane: exampl Two forces are exerted on an object, F1 5 (5, -7) N and F2 5 (4, -1) N. Calculate F1 2 F2 . Solution: F1 2 F2 5 (5, -7) 2 (4, -1) 5 (5 2 4, -7 2 -1) 5 (1, -6)

194

Unit 2 Preliminary Notions of Mechanics

From that, we obtain the following plane. (Each square represents 1 newton.)

y

F1

F1  F 2

x -F2

F1 2 F2 5 (1, -6) N

SeCtion 8.3

Subcg vcs

1. Determine which notation applies to the vector that is opposite to 25 m/s southwest A 25 m/s at 225° B 25 m/s [45° N of E] C 25 m/s at 45° D 25 m/s [45° S of W] 2. Determine the pairs of opposite vectors y

4. Using the component method, calculate the differences of the following vectors: a) displacement s , if s1 5 45 m at 135° and s2 5 20 m at 200° b) velocity v , if v1 5 60 km/h SE and v2 5 40 km/h NW c) force F if F1 5 243 N [10° E of S] and F2 5 137 N [35° N of W] 5. Calculate the following velocity differences (Each square represents 1 km/h):

a

b

y

c d x f

e g

h

x

3. Determine the resultant vector of the following v vector differences a or b or  u a) v



u

b)

u

v

 u

v2

v1

v



b

v3

c c

a

or

a

or

b

or

c

a

or

b

or

c

or

a) v1 – v2 5 b) v3 – v1 5 c) v2 – v3 5

Chapter 8 Vectors

195

8.4 Mulplyg a vcor by a umbr Vectors are a broad feld o study. Beyond the addition and subtraction o vec­ tors, there are other mathematical operations that are possible. This section deals with multiplying a vector by a number (scalar).

8.4.1

Mulplyg a vcor by a scalar

The multiplication o a vector by a number is the equivalent o a repeated addi­ tion. For example, multiplying a vector by 3 is equivalent to adding the same vector three times (see Figure 17). The resultant vector has a greater magnitude but its direction is the same as the initial vector. The vector’s direction is the same i the number is positive but it is opposite i the number is negative. 3

→

v

→

3v

→

v

→

v

→

v

Figure 17 The multiplication o vector v by the number 3 is the equivalent o the addition o three vectors. For example, i we triple the velocity, the size o the velocity triples but its direction does not change.

Thereore, or any vector v multiplied by k, the vector’s magnitude becomes: I u 5 kv then u 5 kv

8.4.2

Mulplyg  h Carsa pla

In the Cartesian plane, the multiplication o a vector by a number becomes the multiplication o the vector’s components by this number. I v 5 (a, b), then kv 5 (ka, kb)

exampl A orce F 5 (-2, 3) N is applied to an object at a given point in time. I the direction o the orce is reversed and its quantity is quadrupled, what is the vector o the fnal orce F ? Solution: F 5 -4F 5 (-4 3 -2, -4 3 3) 5 (8, -12)

y

We obtain the graph shown here. (Each square represents 1 newton.) →

-4F

→

F

x

F 5 -4F 5 (-4 3 -2, -4 3 3) 5 (8, -12) N.

196

Unit 2 Preliminary Notions o Mechanics

Furthering

your understanding

Mullyg vcs The addition, subtraction and multiplication o a vector by a number are not the only mathematical operations that are possible with vectors. It is also possible to multiply a vector by another vector. To do so, two cases are possible: frst, the scalar product o two vectors gives an answer as a scalar; second, the vector product gives an answer as a vector. The ollowing examples o operations help distinguish between these two cases. The concept o work is an example o a scalar product. In act, work is the product o efcient orce by displacement (W 5 F  ∆s or W 5 Feff 3 ∆s 5 F 3 cos θ 3 ∆s ). Because work does not have a direction, it is a scalar (see Figure 19 ).

The rectilinear portion o a conductor with a length o L, crossed by a current with an intensity I and placed in a uniorm magnetic ield B, is subject to an electromagnetic orce (F 5I L × B or F 5 I 3 B 3 sin θ ). Force is a vector and its direction is determined by the right-hand rule (see Figure 20 ).

F

B

→

θ θ L

→

∆s

Figure 18 The multiplication o efcient orce by displacement results in a scalar: work.

SeCtion 8.4

Figure 19 The resultant vector F is perpendicular to the plane ormed by vectors L and B .

Mullyg  vc by  umb

1. Choose the vector that corresponds to the operation described a) If velocity v 5 15 m/s at 100°, what is the value of -25v ? 1 a 5 375 m/s [10° W of S] 2 b 5 375 m/s at 280° 3 c 5 125 m/s SE

2. Calculate the components of the resultant vector if the velocity is v 5 (-4, 7) (The velocity is in km/h) a) 15v b) -08v 3 v c) 5

b) If a force F 5 1127 N NW, what is the value of 8F ? 1 a 5 9016 N [45° W of N] 2 b 5 1927 N at 135° 3 c 5 9016 N SE

Chapter 8 Vectors

197

CHAPTER

8

Vectors

8.1 Properties of vectors • In physics, a scalar quantity is completely defined by a real number and a unit of measurement. A vector quantity is defined by a real number, a unit of measurement and information related to its direction in space. • A vector is represented by an arrow whose length is proportional to the vector’s magnitude. • A vector can be decomposed into a x-component and a y-component. The following equations make it possible to calculate the value of the components.

vx  v cos θ

vy  v sin θ

• The Pythagorean relationship enables us to determine a vector’s magnitude from its components.

v  vx2  vy2

• Calculating a vector’s direction from its components is done using trigonometry and a diagram. The following equation is used to calculate the angle formed by the components.

v α  tan-1 vy x

• In the Cartesian plane, a vector is expressed by its components.

( )

→

v  (vx , vy )

8.2 Adding vectors • The graphical method consists of arranging vectors one after the other. The resultant vector starts from the tail of the first vector and ends at the head of the last vector. • The component method consists of adding the x-components to each other, then the y-components to each other. Recomposing these sums using the formulas makes it possible to obtain the resultant vector. • In the Cartesian plane, vector addition consists of adding x-components to each other, then the y-components to each other.

8.3 Subtracting vectors • The subtraction of two vectors consists of adding the first vector to the opposite of the second. An opposite vector is a vector whose magnitude is identical, but whose direction is opposite.

8.4 Multiplying a vector by a number • The multiplication of a vector by a number is the equivalent of a repeated addition. The magnitude of the resultant vector is equal to the magnitude of the vector multiplied by the number. The resultant vector will be in the opposite direction if the number is negative. • In the Cartesian plane, the multiplication of a vector by a number corresponds to the multiplication of the vector’s components by this number: if v  (a, b), then kv  (ka, kb).

198

UNIT 2 Preliminary Notions of Mechanics

Chapter 8

Vcos

1. Calculate the components o the ollowing vectors (In d, each graduation equals 5 N) a) 228 m/s at 160° b) 79 N [35° E o N] y d) c)

4. Participants in an orienteering race must use a map and compass or GPS to fnd all the reerence points I someone wants to cheat and go directly to the fnish line, he or she must do a vector calculation

40˚

55 m x

2. Calculate the magnitude and direction o the ollowing vectors (In d, each graduation equals 20 m) a) Fx 5 -1607 N and Fy 5 1915 N b) v 5 (45, -62) c) d) y

5. Complete the ollowing operations on velocity vectors: v1 5 20 km/h at 50°, v25 40 km/h at 200° and v3 5 30 km/h at 310° a) 2v1 1 3v2 5 b) 5v3 – v1 5 c) v2 1 4(v1 – v3) 5

→

F

750 N

x

600 N

3. Calculate the magnitude and direction o the displacement vector AB 1 CD  (Each square represents 1 metre) y

B A

The course is as ollows: 1 Walk 150 m eastward 2 Turn 90° to the right and walk 300 m 3 Travel 450 m at 30° south o west 4 Go 100 m [225° W o N] a) Draw the course to scale on a sheet o paper b) Perorm the calculations to fnd the distance and direction o the fnish point with respect to the departure point c) Find the dierence in distance between the real course and the cheater’s course

6. Complete the ollowing operations on orce vectors F1 5 (5, -3), F2 5 (-8, 12) and F3 5 (-6, -7) (Forces are expressed in newtons) a) -4F3 1 3F2 – F1 5 b) 75F1 – 2F2 5 c) 4(F1 – F3) – (6F2 – F1 ) 5 7. A kayaker is able to paddle at a speed o 12 km/h She wants to head due north, but she must cross a lake that is highly exposed to a westerly wind This wind makes her drit to the east by 4 km/h In what direction must the kayaker direct her kayak in order to reach her destination? What will the resultant velocity be?

x

C

D

Chapter 8 Vectors

199

200

ConTenTs ChApTer 9

Uimly rctilia Mti. . . . .203 ChApTer 10

Uim Acclatd rctilia Mti . . . . . . . . . . . . . . . . . . . . . . . . . . . . .221 ChApTer 11

T Mti  pjctil. . . . . . . . .243

T

 bjctiv  kimatic i t dcib t mti  bdi but d t accut  t cau  t mti. T tudy  mti dat back t atiquity ad Gk cla w cidd autiti  t ubjct  ctui. T wd “kimatic” div m t Gk kinêma, wic ma "mti." Kimatic cct vid cmiv dciti  w bdi mv. Kimatic df t yical quatiti quid t dcib a mti, uc a tim, iti, vlcity ad acclati, ad it tudi t latii btw t vaiabl. T latii a xd i matmatical tm ad

a d t a kimatic quati. Kimatic quati ca b ud t dcib mti i a taigt li ad at ctat vlcity, acclatd mti a wll a t mti  jctil, i.. t mti  bjct laucd it t ai ad tat llw a cuvd tajcty. Aid m ti ttical act, kimatic quati av may actical ad tclgical alicati. F xaml, kimatic quati l t dtmi t cifc caactitic  a giv mti, uc a t bakig ditac i a mgcy ituati. Vid gam tat ivlv fig  vitual jctil al ly  kimatic quati. 201

9.1 Position, displacement and distance

ChApTer

9

UniForM reCTiLineAr MoTion

9.2 Graphical representation of position as a function of time 9.3 Velocity 9.4 Graphical representation of velocity as a function of time

UniT

3 ChApTer

KineMATiCs

10

UniForMLY ACCeLerATeD reCTiLineAr MoTion

10.1 Characteristics of uniformly accelerated rectilinear motion 10.2 Equations for uniformly accelerated rectilinear motion 10.3 Free fall

11.1 Describing the motion of projectiles ChApTer

11

The MoTion oF proJeCTiLes

11.2 The motion of objects launched horizontally 11.3 The motion of objects launched at an angle

202

Unifom rctilina Motion

A

train generally moves at a constant velocity and in a straight line, except when it pulls out of a station, turns or comes to a stop. A car equipped with electronic cruise control can travel many kilometres on a highway in a straight line and at a constant velocity. In physics, this motion is called uniform rectilinear motion.

your understanding of velocity, which was covered in previous years in your science and technology classes.

In this chapter, you will learn about the concepts and terminology used in analyzing uniform rectilinear motion. You will draw and interpret graphs illustrat­ ing uniform rectilinear motion. You also will build on

rviw Relationship between constant velocity, distance and time       12

9.1 9.2

position, dislacmnt and distanc   204

9.3 9.4

Vlocity                                 211

Gaical sntation of osition as a function of tim        208

Gaical sntation of vlocity as a function of tim        214 ChApTer 9 Uniform Rectilinear Motion

203

9.1 poto, dlacemet ad dtace When describing the motion o an object, it is important to speciy the position o the object at all times. Since the object is moving, this involves deter­ mining displacement and distance. A scientifc description o motion requires an accurate defnition o each o these variables.

9.1.1

poto

Position is the location of a point or object. The word “object” in physics has a broad meaning: it can reer to a celestial body, an electron, a vehicle, a person, etc. The term “body” is common as well. The position o an object is always given in relation to a frame of reference. There are many everyday examples o rames o reerence. For instance, a per­ son talking on the telephone can describe his or her position to the person on the other end o the line in a number o ways: “I am 50 metres south o the school,” or “I am at the corner o Boucher and Boulanger streets.” In the latter example, the rame o reerence is made up o street names.

See Frame of referece, p 151

Since the scope o this chapter is limited to an analysis o rectilinear motion, a one­dimensional rame o reerence will be used here. The coordinate sys­ tem is usually an x­axis, whose direction and point o origin must be selected and specifed. In this context, the symbol or position is simply x. The position o an object can be identifed by the distance between the point o origin and the point at which this object is located at a given time, on either the negative or positive side o the x­axis. Position thereore involves direction: it is a vector. However, since this chapter is concerned solely with motion along a line (that is, in one dimension), it is not necessary to provide an angle to speciy the direction o the vector. This can be achieved using a plus or minus sign: the position is positive i the object is located on the posi­ tive side o the x­axis and negative i the object is located on the negative side o the x­axis.

See Vector, p 183

When describing motion, the point o origin is oten placed at the object’s orig­ inal location, and the x­axis is oriented in the direction o the initial motion. For example, two students are walking to school along the same rectilinear xB 5 -0.5 km

xf 5 1.0 km

xA 5 0 km

school B

-0.5

A

0

0.5

Figure 1 The position of various objects is represented using a one-dimensional coordinate system

204

UniT 3 Kinematics

1.0

1.5

x (km)

street (see Figure 1). Student A travels 1.0 km rom home to school. When indi­ cating Student A’s position, we assume that the x­axis is oriented in the direc­ tion in which the student is walking and that the point o origin is located at the student’s home. Thereore, Student A’s initial position is xiA 5 0 km, and her fnal position is xA 5 1.0 km. Student B lives on the same street but 500 m ar­ ther rom the school. His initial position is xiB 5 -0.5 km and his fnal position is xB 5 1.0 km. When we say that a person’s position is x 5 1.0 km, we do not speciy which part o the body is located precisely at point x 5 1.0 km. We assume that the object is reduced to a particle, which is a very small body, and that the entire mass o the object is concentrated at its centre. The position given is the posi­ tion o the object’s centre. This method avoids the need to describe the motion o each point on the object.

9.1.2

Dislacmnt

Displacement, which is represented by ∆x, is defned as the change in position. The symbol ∆ is the uppercase Greek letter delta. Thereore, ∆x is pronounced “delta x.” I the initial position is xi and the fnal position is x, the displacement is: ∆x 5 x 2 xi Since position is a vector, then displacement, which is the change in position, is also a vector. In the case o rectilinear motion, plus and minus signs are also used to indicate the direction o the displacement vector: a plus sign (1) i the vector is directed toward the positive side o the x­axis and a minus sign ( ­) i it is directed toward the negative side o the x­axis. In this context, the symbols ∆x, x and x do not have an arrow above them  i because these quantities are components (see Figure 2). y

→

∆s

∆y θ ∆x

x

Figure 2 Rectilinear motion is considered to be a specifc two-dimensional motion, in which displacement is represented by ∆s  Displacement ∆x is components x o displacement ∆s  This is why ∆x does not have an arrow above it

Displacement depends solely on the initial and inal positions, and as a result, very dierent motions can result in identical displacements. For example, i Student A leaves her house and walks straight to school, her dis­ placement is ∆x 5 1.0 2 0 km 5 1.0 km (see Figure 1 on the previous page). I Student A leaves home but walks to Student B’s house beore going to school, Student A’s displacement remains the same: ∆x 5 1.0 km because xiA 5 0 km and xA 5 1.0 km in this case as well. It is important to realize, however, that ChApTer 9 Uniorm Rectilinear Motion

205

Student A walked more and travelled a longer distance: she travelled 0.5 km to go to Student B’s house, 0.5 km to walk back to her house, and 1.0 km to get to school, for a total of 2.0 km. This example emphasizes that displacement ∆x is a different physical quan­ tity from distance (L) [see Figure 3]. x 5 -0.5 km

xf 5 1.0 km

x 5 0 km

school

A

-0.5

Dplacmt ∆x

0

Dtac (L)

0.5

1.0

1.5

x (km)

Figure 3 Distinction between displacement (∆x) and distance (L)

9.1.3

Dtac

Distance, represented by the symbol L, is the length of the trajectory, without taking direction into account. A trajectory is the series of successive points through which an object passes as it moves. Although displacement may be negative, distance is always positive because it is a length. Distance is not a vector. In the case of rectilinear motion, distance is equal to the sum of the absolute value of each displacement between two direction changes. As a result, dis­ placement is always less than or equal to distance. The following examples based on Figure 1 on page 204 help to visualize this concept. exampl A On Saturday morning, Student A visits Student B, who lives 500 m farther from the school than Student A does What is Student A’s displacement? What distance has she travelled? x 5 -0.5 km

B

school

A

-0.5

0

Data: With the x-axis pointed toward the school and the point of origin of the axis located at Student A’s house: xi 5 0 km ∆x 5 ? xf 5 -05 km L5?

206

UniT 3 Kinematics

xf 5 1.0 km

x 5 0 km

0.5

1.0

1.5

x (km)

Solution: ∆x 5 xf 2 xi 5 -05 km 2 0 km 5 -05 km The motion in this example always occurs in the same direction L 5 |∆x| 5 05 km

examl B Beore going to school, which is located 1 km rom her house, Student A walks to Student B’s house What is Student A’s displacement? What distance does she travel? x 5 -0.5 km

B

-0.5

xf 5 1.0 km

x 5 0 km

scl

A

0

0.5

Data: With the x-axis still pointed toward the school and the point o origin o the axis located at Student A’s house: ∆x 5 ? xi 5 0 km x 5 10 km L5?

1.0

1.5

x (km)

Solution: ∆x 5 x 2 xi 5 10 km 2 0 km 5 10 km In this example, the motion occurs in two dierent directions The frst displacement is 05 km on the negative side o the x-axis, and the second displacement is 15 km on the positive side o the x-axis (toward the school) L 5 |∆x1| 1 |∆x2| 5 |-05| km 1 |15| km 5 20 km

seCTion 9.1

pt, dlacmt ad dtac

1. Read the ollowing statements Which physical quantity (position, displacement or distance) do they provide inormation about? a) Jean travels 200 km by train rom Gatineau to Montréal b) Myriam is at La Courgette restaurant c) Marie-Christine pedalled or 30 minutes on her stationary bicycle 2. A bus leaves Montréal, travels to Québec City and returns to Montréal What is the displacement? 3. A cyclist travels 10 km along a rectilinear street a) What was her initial position? b) What is her fnal position? c) What is her displacement? 4. A cyclist travels 10 km in hal an hour along a rectilinear street During the second hal-hour, she travels back along the same road or 8 km a) What was her position ater hal an hour? b) What was her displacement during the frst hal-hour?

c) What was her fnal position? d) What was her displacement during the second hal-hour? e) What distance did the cyclist travel in an hour? 5. A basketball player drops a ball rom a height o 1 m The ball bounces on the ground, rises 06 m, alls again, bounces and rises 03 m a) What is the magnitude o the displacement? b) What distance has the ball travelled? 6. A bus leaves rom Montréal and travels to Québec City, which is 250 km away On the same day, the bus returns to Montréal then goes back to Québec City What is the ratio between the magnitude o its displacement and the distance? (Assume that the highway between Montréal and Québec City orms a straight line)

ChApTer 9 Uniorm Rectilinear Motion

207

9.2 Graphcal represetato of posto as a fucto of tme Uniform rectilinear motion is motion in a straight line at a constant velocity. For equal time intervals, the distance covered will be equal as well. When describing a motion, in addition to indicating the object’s successive positions, it is important to speciy when the object is located at these posi­ tions. This can be done in various ways. 1. The various times and positions can be listed in a table. 2. The time­position pairs can be plotted on a graph o position as a unction o time. 3. The relationship between time and position can be illustrated through an equation. t (m)

x (km)

0

0

5

-05

10

0

20

10

This section covers the frst two methods. Uniorm rectilinear motion can be broken down into very short time intervals. Based on the type o motion, the time intervals can be hours, seconds, millionths o a second, etc. For example, it would not be very practical or even useul to provide the position o a car travelling between two cities at every millionth o a second. In most situations, it is acceptable to give the position at relatively long time intervals.

Position x (km)

a) Time and position values for Student A when she walks to Student B’s house before going to school

1.0

The example o Student A and Student B (see Figure 1 on page 204) provides an interesting situation to analyze. I Student A walks to Student B’s house beore going to school, her motion can be described by listing specifc values o time and position in a table (see Figure 4a). These values can be plotted on a graph o position as a unc­ tion o time, which allows us to visualize the overall motion (see Figure 4b).

0.5

0 5

10

15

20 Time t (min)

-0.5

b) Graph of Student A’s position as a function of time when she walks to Student B’s house before going to school

Figure 4 Motion of a person walking at a constant velocity, beginning in one direction (toward the negative side of the x-axis), then in the opposite direction (toward the positive side of the x-axis)

Student A travels 0.5 km in 5 minutes, and there­ ore 1 km in 10 minutes and 6 km in 60 minutes (see Section 9.3 on page 211 for the formal definition of velocity). Ater returning to her house with Student B, she needs an additional 10 minutes to travel 1.0 km rom her house to school. Five minutes ater she leaves her house, Student A arrives at Student B’s house (see Figure 4b). Ten minutes ater she leaves, Student A walks past her house, at x 5 0 km. Fiteen minutes ater she leaves, Student A is 0.5 km away rom her house. Twenty minutes ater she leaves, she is 1.0 km away rom her house and has thereore reached her school.

For uniorm rectilinear motion, a graph o position as a unction o time consists o many line segments i there are several sections o dierent motion. I all o the motion occurs at the same velocity and in the same direc­ tion, then the graph will consist o a single straight line.

208

UniT 3 Kinematics

b) What is Student A’s position at time t 5 25 min? c) What is Student A’s displacement between times t 5 0 min and t 5 20 min? d) What distance did Student A travel between times t 5 0 min and t 5 20 min?

Position x (km)

examl The ollowing day, the motion o Student A described on page 208 is illustrated in the graph shown here a) What was dierent about the motion illustrated in the fgure on the previous page?

1.0

0.5

0 5

Solution: -0.5 a) The student waited fve minutes at the same place (at Student B’s house) beore walking to school b) x25 5 -025 km

10

15

20

25 Time t (min)

c) ∆x 5 x 2 xi 5 x20 2 x0 5 05 km 2 0 km 5 05 km d) L 5 |∆x0-5| 1 |∆x5-10| 1 |∆x10-20| 5 |-05| km 1 0 km 1 |10| km 5 15 km

Klmt t  gway Many highways and some bicycle paths have kilometre posts Highway exits are usually numbered according to their distance rom a reerence point The highway is thereore a rame o reerence or the people who use it In most Canadian provinces, the exit number o a highway represents the distance in kilometres between the beginning o the highway and the exit (see Figure 5) As a result, exit numbers are not consecutive This numbering method allows users to easily determine the distance travelled as well as the distance remaining The exit numbers on some highways in the United States are consecutive whole numbers and provide no inormation about distance This is the case in Nova Scotia as well In Québec, Highway 40 begins at Pointe-Fortune at the Ontario border, passes through Montréal, Trois-Rivières (kilometre 199), Québec City, and ends at Beauport (kilometre 325) In Ontario, Highway 401 begins in Windsor, passes through London (kilometre 189), Toronto (kilometre 371), runs along the St Lawrence River to Québec (kilometre 828), where Highway 20 begins and the exit numbers start over at zero

1. Which of the following statements describes the motion of the person illustrated in this graph?

Position x (m)

seCTion 9.2

a) A person walks in one direction, pauses for 30 seconds, and returns to the starting point, walking twice as fast

10

-10

the exit number on a highway represents the distance in kilometres between the beginning o the highway and the exit

Gacal tat f t a a fuct f tm

20

0

Figure 5 In Québec, as in most Canadian provinces,

10

20

30

40 Time t (s)

(continued on the next page)

ChApTer 9 Uniorm Rectilinear Motion

209

b) A person walks in one direction, pauses for 10 seconds, and returns to the starting point, walking half as fast c) A person walks in one direction, pauses for 10 seconds, and returns to the starting point, walking twice as fast d) A person walks in one direction, pauses for 10 seconds, and walks an equal distance in the same direction

8 6 4

6 4 0

20

-2 -4

40

60 80 Time t (min)

-2 -4

D 8 6 4

Position x (km)

0

Position x (km)

8

2

2

C

20

40

60 80 Time t (min)

Time t 8 6 4 2

2

0

0

20

-2 -4

40

60 80 Time t (min)

-2 -4

20

40

60 80 Time t (min)

Position x (km)

3. The following graph shows a cyclist’s position as a function of time while riding a bike along a rectilinear path

15

10

5

0

10

20

30

40

50

60

70

80 Time t (min)

a) How long did it take the cyclist to travel 5 km? b) What was the cyclist’s position at t 5 40 min?

210

4. A child runs to a store located two blocks from her house After buying what she needs, she runs back home Refer to the graph of position as a function of time, shown below, to answer the following questions Position x

B

Position x (m)

A

Position x (m)

2. On a rectilinear street, a person walks east 2 km for 20 minutes, sits on a bench for 10 minutes, walks east another 2 km for 20 minutes, then runs back to the starting point in 30 minutes Which of the following graphs accurately illustrates this person’s motion?

c) What distance did the cyclist travel after 60 minutes? d) After cycling for 60 minutes, at what distance from the starting point was the cyclist located? e) What total distance did the cyclist travel? f) What was the cyclist’s displacement between t 5 0 and t 5 60 min? between t 5 0 and t 5 80 min? g) How long did the cyclist stop before heading back to the starting point? h) How long did the cyclist ride?

UniT 3 Kinematics

a) Did the child take less time going to or coming from the store? Explain your answer b) Assume that the difference in travelling time occurred because the child had to run up a hill in one case and down the same hill in the other case Is the store or the house located at a higher altitude? Explain your answer 5. During a soccer game, Player A, who is stationary, kicks a ball on the ground at 10 m/s toward Player B, who is 30 m away The ball takes three seconds to travel to Player B Player B, who is also stationary, immediately kicks the ball back on the ground at 15 m/s The ball returns to Player A in two seconds Show the motion of the ball in a graph of position as a function of time

9.3 Vlocty Velocity describes the rate at which a position changes. Velocity v is equal to displacement ∆x divided by time interval ∆t. The equation or velocity is:

v5

x 2 xi ∆x 5  t 2 ti ∆t

where t i 5 initial time, expressed in seconds (s) t  5 fnal time, expressed in seconds (s) xi 5 initial position, expressed in metres (m) x 5 fnal position, expressed in metres (m) The deinition o velocity corresponds to the equation relating velocity, dis­ placement and time. I we know two o these variables, we can determine the third. For example, i we are looking or the fnal position, we can isolate it in the equation as ollows: x 2 xi ⇒ x  5 x i 1 v (t  2 t i) t 2 ti

The units o velocity are obtained by dividing the units o distance by the units o time. The SI unit o velocity is metres per second (m/s). Depending on the type o motion, the physical quantity mea sured is usually expressed in kilometres per hour (km/h) or metres per second (m/s). For example, i you say that a car is travelling at 27.8 m/s, ew people will know what you are talking about. However, the equivalent in kilometres per hour, that is, 100 km/h in this case, makes more sense. In contrast, the speed o light is usually expressed in the orm c ≈ 3 3 108 m/s, rather than in kilometres per hour. As discussed in Section 9.1, displacement is a vector and can be positive or negative or a one­dimensional motion. It ollows rom the defnition o veloc­ ity that velocity is also a vector. In physics, or a one­dimensional motion, the direction o the velocity is represented by a positive or negative sign, though in everyday situations we never reer to negative velocity. The sign or v is the same as the sign or ∆x. (Note that ∆t is always positive.) For an object moving toward the positive side o the x­axis, v > 0; or an object moving toward the negative side o the x­axis, v < 0. In a graph o position x as a unction o time t, the slope (also reerred to as rate o change) o a straight line or line segment corresponds to the change in posi­ ∆x tion ∆x divided by the change in time. The ratio corresponds to the defnition ∆t o velocity.

Slope 5

Position x

v5

xf

xi

0

ti

tf

Time t

Figure 6 For a uniorm motion, the graph o position as a unction o time is represented by a straight line whereby the slope corresponds to the velocity between (ti, xi) and (t, x)

∆x 5v ∆t AppenDiX 3

Thereore, in a graph o position as a unction o time, velocity corresponds to the slope o the line segment connecting points (ti, xi) and (t, x) (see Figure 6).

Gacal dtmato of t slo of a stagt l, p 392

ChApTer 9 Uniorm Rectilinear Motion

211

For uniform rectilinear motion, a graph of position as a function of time consists of a straight line or of several line segments if there are several sec­ tions of different motion (see Figure 4b on page 208). The higher the magnitude of the velocity, the greater the absolute value of the slope of the line. In other words, the line will have a greater inclination. exampl A A car travels at 800 km/h along a straight road between two towns that are 100 km apart How many minutes will it take the car to travel between the two towns? Solution: ∆x ∆x ⇒ ∆t 5 ∆t v 100 km ∆x ∆t 5 5 5 0125 h 800 km/h v

v5

The last step is to convert this time into minutes: ∆t 5 0125 h 3

exampl B The graph o position as a unction o time shown here represents the trajectory o Student A (see Section 9.1 on page 204 ) It consists o two successive uniorm rectilinear motions, each represented by a line segment in the graph What is the velocity, in km/h, o Student A: a) during the frst 5 minutes? b) during the last 15 minutes?

Position x (km)

Data: v 5 800 km/h ∆x 5 100 km ∆t 5 ?

60 min 5 750 min 1h

1.0

0.5

0

5

10

15

20 Time t (min)

-0.5

Data: a) t i 5 0 xi 5 0 t  5 5 min x  5 -05 km v5?

Solution:

b) t i 5 5 min x i5 -05 km t  5 20 min x 5 10 km v5?

b) v 5

Furthering

x 2 xi -05 km 2 0 km km ∆x 5 5 -010 5  min 5 min 2 0 min t 2 ti ∆t km km 60 min In kilometres per hour, the velocity is: v 5 -010 3 5 -600 min h 1h

a) v 5

x 2 xi km 10 km 2 (-05) km ∆x 5  5 5 1010 min 20 min 2 5 min t 2 ti ∆t km km 60 min In kilometres per hour, the velocity is: v 5 1010 3 5 1600 min h 1h

your understanding

Moto at vry hgh vlocty When an object moves at a very high velocity, that is, at a velocity that approaches the speed o light, certain eects occur These eects do not coincide entirely with the relationship between velocity, distance and time established by the classical theory o mechanics discussed in this chapter The theory o special relativity, proposed by Albert Einstein in 1905, analyzes this issue The theory o relativity is based on the ollowing postulate: the speed o light in a vacuum (c ≈ 3 3 108 m/s) is the same or all observers, regardless o the motion o the light source or o the observer For

example, i a light source moving at a velocity o 05c emits a beam o light toward a stationary observer, the observer will perceive the light to have a velocity o c, and not 15c, contrary to the classical model o velocity addition The speed o light c is the maximum possible velocity or all interactions and all bodies in the universe Einstein deduced rom his postulate that the motion o an observer aects measurements o time and length For example, time would pass more slowly or an astronaut travelling in a rocket at a speed o 05c than it would or a person back on Earth! While one year (continued on the next page)

212

UniT 3 Kinematics

Furthering Furthering

your understanding your understanding Furthering

your understanding

Mt at vy g vlcty (cont.) elapsed on Earth, only 087 years would have elapsed in the rocket: this is called time dilation For example, a clock would be slower in the rocket than in would be on Earth, and the passengers in the rocket would age more slowly than people on Earth, as well! Another implication o the theory o relativity is length contraction According to classical mechanics, an object’s length remains the same whether it is in motion or at rest According to the theory o relativity, a moving object contracts in the direction o the motion, and the contraction increases as velocity increases (see Figure 7 ) The theory o relativity also states that as an object approaches the speed o light, it becomes heavier and has more diiculty accelerating Although we cannot get a rocket to fy at a speed o 05c, other experiments have veried Einstein’s theory Relativity, however, does not have an impact in everyday lie because at low velocities, the eects o relativity cannot be detected Figure 7 According to the theory o relativity, the length o a moving object that reaches about 90% o the speed o light is reduced by about hal

Vlcty

1. A cyclist who is in good shape rides or 50 minutes at 30 km/h What distance does he cover? 2. The distance between Earth and the star Sirius is 824 3 1013 km Light travels at a speed o 300 3 108 m/s How much time, in years, does it take or the light emitted rom Sirius to reach Earth? 3. A car travels on a highway at 100 km/h The driver sees a deer on the road There is a 10 second elapse beore the driver brakes How many metres has the car travelled during this time? 4. Water rom a river fows at an average velocity o 50 cm/s How long does it take a water molecule to travel 100 km?

5. The graph below illustrates a pedestrian’s position as a unction o time What is the pedestrian’s velocity, in kilometres per hour (km/h), on his way to his destination? on the return trip? Position x (km)

seCTion 9.3

6

4

2

0

0.5

1.0

1.5

2.0

2.5 Time t (h)

6. A jogger runs at a velocity o 4 m/s or 20 seconds, then 2 m/s or 10 seconds Draw a graph o the jogger’s position as a unction o time ChApTer 9 Uniorm Rectilinear Motion

213

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

9.4 Gaphcal ptat f vlcty a a fuct f tm

niCoLe oresMe French philosopher, economist and scholar (c. 1320–1382) Nicole Oresme played a major role in the evolution o medieval science He developed a basis or graphical representation that is still in use today: a diagram with two axes, each representing a variable, where one variable is dependent on the other He graphically represented velocity as a unction o time and concluded that the surace area under the curve corresponds to displacement Oresme was also interested in uniorm accelerated motion and discovered rom his graphs that i the initial velocity is zero, the distance is proportional to the square o the time However, he did not conduct any experimental verifcations o this concept

Draw a graph o velocity as a unction o time

1.0

0.5

0

5

10

15

20 Time t (min)

-0.5

Data: The velocities o the two motion sections were calculated earlier (see Example B on page 212): rom 0 to 5 min, v 5 -60

Solution: We obtain the ollowing graph o velocity as a unction o time

km h

10

0 5

km rom 5 to 20 min, v 5 160 h

10

-10

15

20 Time t (min)

Note that this graph is a valid representation but is not entirely accurate in reality For instance, Student A cannot switch rom a velocity o -6 km/h to a velocity o 16 km/h in an infnitely small period o time She must slow down, stop, then accelerate in the other direction; rather than the abrupt change shown at t 5 5 min, the velocity varies in a continuous manner At this scale, the graph cannot show this type o variation

Velocity v (m/s)

A graph of velocity as a function of time allows us to determine displacement for a given time interval. Since velocity is equal to displacement ∆x divided by the time interval ∆t, displacement is equal to velocity multiplied by time.

Since v 5

ti

t f Time t (s)

Figure 8 In a graph o velocity as a unction o time, the surace area under the shaded curve is equal to displacement

214

exampl The graph shown here illustrates the position o Student A (see Section 9.1 on page 204)

Velocity v (km/h)

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

Position x (km)

One way to represent motion is to draw a graph of velocity as a function of time. In the case of uniform rectilinear motion, since velocity is constant, a graph of the velocity as a function of time will be a straight horizontal line. If there are several sections of different motions with different velocities, the graph will include several horizontal line segments, as illustrated by the fol­ lowing example.

UniT 3 Kinematics

∆x , we have ∆x 5 v 3 ∆t  ∆t

The value ∆x 5 v 3 ∆t 5 v 3 (tf 2 ti) corresponds to the surface area under the curve between times ti and tf, from a graph of velocity as a function of time (see Figure 8). The units of the area under the curve are units of velocity (m/s) multiplied by units of time (s), resulting in units of length (m), which in turn correspond to units of displacement.

40 30 20 10

Solution: a) Between 0 s and 20 s, the surface area under the curve is equal to: ∆x0-20 5 base 3 height 5 20 s 3 30 m/s 5 600 m Since ∆x0-20 5 xf – xi 5 x20 – x0, then x20 5 x0 1 ∆x0-20 5 200 m 1 600 m 5 800 m

0

20

40

60 Time t (s)

b) Between 20 s and 40 s, the surface area under the curve is equal to: ∆x20-40 5 base 3 height 5 20 s 3 30 m/s 5 600 m Since ∆x20-40 5 xf – xi 5 x40 – x20, then x40 5 x20 1 ∆x20-40 5 800 m 1 600 m 5 1400 m

6

15 10 5

Velocity v

Velocity v Position x

20

0

5

10

15 Time t (s)

Position x

3. The initial position of the object whose motion is shown in question 2 is 20 metres on the reference Time a t graph Time t of position as a function of time x-axis Draw Time t Time t for this motion, from t 5 0 s to t 5 15 s 4. The graph below illustrates the position of a drop of water in a horizontal pipe whose diameter begins to decrease at a certain point Draw a graph of velocity as a function of time Time t Time t Time t

Position x (m)

Time t Time t Time t Time t Time t Position x

Time t Time t Time t Time t Time t

Time t Time t Time t

Velocity v v Velocity

Position x Velocity v

Position Velocity Velocity v v x

4

Position x

Position x

5

Velocity Velocity Velocity v v v

3

Position Position x x

Velocity v Velocity v

Time t Time t

2. The motion of an object is illustrated in the graph shown here What is the object’s displacement: a) after 5 seconds? b) after 15 seconds?

Position x x Position

Velocity v

2

Velocity Position x v

1

Position x

Position x

1. Which of the graphs below describe: a) a stationary object? b) an object travelling in a uniform rectilinear motion? c) an object that returns to its starting point?

Velocity v (m/s)

Gacal tat f vlcty a a fuct f tm

seCTion 9.4

Time t t Time t

Time t t Time t

Velocity v (m/s)

examl A car’s motion is illustrated by the graph shown here The car’s initial position is located at 200 m a) What is the car’s position at t 5 20 s? b) What is the car’s position at t 5 40 s?

Time t

120

80

40

Time t Time t Time t

Time t Time t

0

20

40

60

80 Time t (s)

ChApTer 9 Uniform Rectilinear Motion

215

APPLICATIONS spd gu Police ocers use two types o kinemometers to meas­ ure a car’s velocity: a radar gun and a laser gun. The radar gun emits a beam o microwaves that refect against the vehicle and return to the radar. The refection on the moving car alters the requency o the waves, a phenomenon called the Doppler eect. The radar gun measures requency change and calculates velocity. This method is also used to measure the velocity o baseballs, tennis balls and hockey pucks. In contrast, the laser gun measures the time lapse between the emission and reception o laser pulses (see Figure 9). It sends an inrared laser beam toward the vehicle. The beam consists o about a hundred very short pulses emitted at intervals o a ew thousandths o a second. With a wavelength (λ) o approximately 900 nanometres (nm), the beam is invisible and does not interere with driving. Each laser pulse is refected against the target vehicle and returns to the gun, which measures the time it takes to travel to and rom the vehicle. Since the speed o inrared light is known (it is equal to the speed o light, or 3 3 108 m/s), the detection system can calculate the distance o the vehicle at each pulse. By analyzing the change in the vehicle’s distance rom one pulse to another, the laser gun calculates the vehicle’s velocity (see Figure 10). Since the person operating the laser gun is usually on the side o the road, the laser beam strikes the vehicle at an angle. The change in the distance measured is less

rcvd pul

than the distance the vehicle travels along the road. As a result, the velocity measured is slightly lower than the vehicle’s true velocity. Laser gun manuacturers claim that their devices have an accuracy o approximately ± 2 km/h. Like all measuring instruments, laser guns must be cali­ brated regularly and used properly. They are usually programmed to detect incorrect measurements and intererence. For example, i the inrared laser beam sweeps dierent parts o a vehicle, the velocity measured will not be constant or all o the impulses. In this situation, the device shows an error message, and the measurement must be repeated.

Figure 9 A laser gun

emttd pul

Δx

Figure 10 An everyday situation that can be interpreted using the relationship v 5 Δt

216

UniT 3 Kinematics

Masuing tim The measurement o time involves two components. First, to determine the chronology o certain events, it is necessary to associate these events with specifc points in time. Second, it is important to speciy the duration o an event. When did such­and­such an event occur? How long did it last? These two undamental questions have impelled humans to measure time. When measuring time, it is important to consider a periodic or regular phenomenon. Throughout the ages and all over the world, people have measured time by observing the Sun’s apparent motion. The Sun’s changing positions in the sky indicated the part o the day. Solar time can be measured with greater accuracy using a sundial. The principle o the sundial is an ancient one: about 4000 years ago, the Egyptians and the Chinese told time by observing the shadow cast by a stick on the ground. The frst known sundial is Egyptian and dates back to the 14th century BCE. The needs o society led to the design o other instru­ ments or measuring time. For example, as early as the 15th century BCE, the Egyptians used the clepsydra, which is a water clock made o a water­flled container with markings on it. The container is pierced with a hole, and the gradual lowering o the water level shows how much time has elapsed. The hourglass is based on the same principle. The Romans were already using it beore the Common Era. In order or the level o sand to lower at a constant rate, the bulbs must have a spe­ cifc shape (see Figure 11). Both the clepsydra and the hourglass are based on gravity. Over the centuries, technical and scientifc advances oered urther possibilities or measuring time. The frst mechanical clocks date back to the 14th century. They relied on gravity as well. The downward motion o a weight moved the gears, which transmitted the motion to the needles o a dial. The downward motion was control­ led by an oscillating piece that ensured a constant speed. Toward the end o the 16th century, Galileo (1564–1642) improved on the design o clocks, applying the principle

o a pendulum’s regular oscillation. The frst pendulum clock dates back to 1657. In 1658, Christian Huygens, a Dutch mathematician, astronomer and physicist (1629–1695), described a new type o clock that was truly revolutionary: the balance spring clock, also called a spring­wound clock. Smaller and smaller clocks and watches began to appear rom this point orward (see Figure 12). In the 18th century, navigational needs, such as accu­ rately determining longitude, required increasingly pre­ cise chronometers. In Europe, the race was on to invent new timepieces. In 1759, Englishman John Harrison (1693–1776) built a chronometer that lost less than one second per day, which was a tremendous advance­ ment. Today, most clocks and watches rely on the vibration o a small quartz crystal. The crystal generates an alter­ nating current, which is used by a microprocessor to measure time and to control the display. In addition to being compact, these watches are extremely accu­ rate. High­quality quartz watch es lose less than one second per month.

houglass bulb Conncting tub

Figure 11 The bulbs o an hourglass have a specifc shape, which ensures that the level o sand alls at a constant rate

Figure 12 In a pocket watch, the energy stored in a wound-up spring eeds the motion o the balance wheel

ChApTer 9 Uniorm Rectilinear Motion

217

chapter

9

Unifom rilin Moion

9.1 posto, dslacemet ad dstace • Position is the location o a point or object. The position o an object is always given in relation to a rame o reerence. For rectilinear motion, the coordinate system is usually an x­axis whose direction and point o origin must be selected and specifed. • The magnitude o the position is the distance between the point o origin and the point at which the object is located at a given time, on either the negative or positive side o the x­axis. • When describing a motion, the point o origin is oten placed at the object’s original location, and the x­axis is oriented in the direction o the initial motion. • Displacement, which is represented by ∆x, is defned as the change in position. I the initial position is xi and the fnal position is x, the displacement is: ∆x 5 xf 2 xi • Displacement depends solely on the initial and fnal positions, and as a result, very dierent motions can result in identical displacements. I an object’s motion ends at its original location, ∆x 5 0. • Distance, represented by the symbol L, is the length o the trajectory, without taking direction into account. Distance is always positive and greater than or equal to displacement. A trajectory is the series o successive points through which an object passes as it moves.

9.2 Grahcal reresetato of osto as a fucto of tme • Uniorm rectilinear motion is motion in a straight line at a constant velocity. For equal time intervals, the distance covered will be equal as well. • The values o time and position, which are needed to describe motion, can be listed in a table. They can also be plotted on a graph o position as a unction o time, which allows us to visualize the overall motion. • For uniorm rectilinear motion, a graph o position as a unction o time consists o a straight line or several line segments i there are several sections o dierent motion.

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UniT 3 Kinematics

9.3 Vlocity • Velocity describes the rate at which a position changes. Velocity v during a time interval ∆t is equal to displacement ∆x divided by the time interval ∆t.

v5

∆x ∆t

• The units of velocity are obtained by dividing the units of distance by the units of time. Depending on the type of motion, the physical quantity measured is usually expressed in kilometres per hour (km/h) or metres per second (m/s). • Velocity can be positive or negative. For an object moving toward the positive side of the x­axis, v > 0; for an object moving toward the negative side of the x­axis, v < 0. • In a graph of position x as a function of time t, the slope of a straight line or line segment corresponds to the change in position ∆x divided by the change in time.

Slope5

∆x 5v ∆t

9.4 Gaical sntation of vlocity as a function of tim • In the case of uniform rectilinear motion, since velocity is constant, a graph of velocity as a function of time will consist of a straight horizontal line. If there are several sections of different motion with different velocities, the graph will include several horizontal line segments. • In a graph of velocity as a function of time, the surface area under the curve between times ti and tf is equal to displacement.

ChApTer 9 Uniform Rectilinear Motion

219

ChApTer 9

Ufom rctla Moto

1. Which o the ollowing motion(s) is/are rectilinear motions? Which is/are uniorm rectilinear motions? a) An apple alls rom a tree b) A cyclist brakes in order to stop at a trac light c) A truck exits a highway without slowing down d) A baseball player hits the ball over the ence e) A car travels along a straight highway with its cruise control on f) A person uses an escalator to go up one foor

A

7. The ollowing graph describes the motion o a car What is the car’s displacement during the second hour o the drive?

B

Time t

Position x (m)

20

0 60

80 Time t (s)

-20

Determine: a) his position at t 5 25 s b) his displacement between t 5 0 s and t 5 30 s c) his displacement between t 5 0 s and t 5 80 s d) the distance between t 5 0 s and t 5 30 s e) the distance between t 5 0 s and t 5 80 s f) his velocity at t 5 25 s g) his velocity at t 5 40 s

220

UniT 3 Kinematics

80

40

40

40

120

C

4. The graph shown below illustrates the motion o a person who is not quite sure where he is going

20

6. A laser gun emits inrared pulses at 0003 second intervals toward a car travelling in the direction o the gun I the car is 400 m away when the rst pulse strikes it, how long will it take the rst pulse to reach the car and return to the gun?

Velocity v (km/h)

3. The graph shown here illustrates the motion o three runners Which runner is the astest? Which runner is moving in the opposite direction o the other two?

Position x

2. The average distance between the Earth and the Moon is 384 400 km Light travels at a speed o 300 3 108 m/s How long does it take light refected by the Moon to reach the Earth?

5. Two cars leave rom the same place at the same time and travel 200 km on the same road Car A travels at 80 km/h, and car B travels at 100 km/h How much earlier will car B arrive than car A?

0

0.5

1.0

1.5

2.0 Time t (h)

8. Farouk gives Gabriella a 30 m head-start during a 100 m race Farouk can run at 10 m/s, but Gabriella cannot run aster that 6 m/s a) On a sheet o paper, draw a graph o position as a unction o time or the two runners b) Does Farouk win the race? I so, when and at which position does he catch up to Gabriella? 9. Québec City is located approximately 250 km rom Montréal A reight train leaves Montréal or Québec City travelling at 50 km/h At the same time, a passenger train leaves Québec City or Montréal travelling at 75 km/h How much time will elapse (in minutes) beore the trains cross paths?

Unifomly accld rcilin Moion

A

car travelling along a highway can move in a straight line at a constant velocity for a fairly long time interval. However, this uniform rectilinear motion does not constitute the only type of motion it can perform. In fact, before the car can reach its cruising speed it must accelerate. Furthermore, if a slower car appears in front of it, it has to brake or, in other words, decelerate.

rviw Relationship between constant velocity, distance and time       12 Gravitational force                                        13

In this chapter, you will learn how to describe and analyze uniformly accelerated rectilinear motions (UARM), in particular those occurring near the Earth’s surface. As such, you will see that parachut­ ists in free fall have, at the very beginning of their jump, have uniformly accelerated rectilinear motion.

10.1

Ccisics of unifomly ccld cilin moion       222

10.2

equions fo unifomly ccld cilin moion                     230

10.3

F fll                               235

Chapter 10 Uniformly Accelerated Rectilinear Motion

221

10.1 Characerisics of uniformly acceleraed recilinear moion The velocity o uniorm rectilinear motion (URM) is constant, whereas the velocity o uniormly accelerated rectilinear motion (UARM) varies rom one moment to the next. This variation in velocity constitutes the acceleration. For example, in the course o a car journey, the velocity constantly changes according to what occurs on the road: the presence o other vehicles, driving through towns, going up– or downhill, the type o road, etc. The speedometer provides the car’s speed at a given point in time. Changes in velocity are con­ sidered as acceleration. The units are SI units. In order to describe a UARM, it is necessary to introduce and defne new con­ cepts, such as instantaneous velocity and acceleration. The ollowing variables will be systematically used in equations related to instantaneous velocity and acceleration. The units are SI units. ti 5 Initial time, expressed in seconds (s) tf 5 Final time, expressed in seconds (s) xi 5 Initial position, expressed in metres (m) xf 5 Final position, expressed in metres (m) vi 5 Initial velocity, expressed in metres per second (m/s) vf 5 Final velocity, expressed in metres per second (m/s) a 5 Acceleration, expressed in metres per second squared (m/s2)

10.1.1

Insananeous velociy and average velociy

Instantaneous velocity is the velocity at a given moment. It is equal to the displacement (∆x) divided by time interval (∆t) on condition that this time interval, is very short. See Velociy, p 211

In Chapter 9, velocity was defned or relatively long time intervals; rom this chapter onward, such velocity will be described as average velocity. Since in­ stantaneous velocity concerns a precise point in time, the time interval involved is very brie. The defnitions o average velocity and instantaneous velocity thereore can be expressed in the ollowing ways: Average velocity x 2 xi ∆t being ∆x v5 5 f ∆t tf 2 ti fairly long

Instantaneous velocity x 2 xi ∆t being ∆x v5 5 f very short ∆t tf 2 ti

The notion o instantaneous velocity is essential to describing a motion per­ ormed at a non­constant velocity. In everyday lie, the notion o instantaneous velocity is requently used. For example, a cyclist can, at a given point in time, travel at 20 km/h, even though he was travelling at a velocity o 12 km/h just moments earlier, while he was cycling up a hill, or instance. These two cases concern the cyclist’s velocity at a specifc point in time. Average velocity, on the other hand, is determined over a long time interval. For example, the aver­ age velocity o the same cyclist is 20 km/h i he travelled 50 km in 2.5 hours. When an accelerated motion is being analyzed, average velocity is not a very useul concept, as it is determined only by the initial and fnal time and thereore does not provide inormation about what occurred between these two points in time.

222

UNIt 3 Kinematics

A graph o position as a unction o time or a uniorm rectilinear motion comprises a straight line (or several straight line segments), since the position varies regularly as a unction o time. For all equal time intervals, the distances are also equal. For example, an object travels 10 m between t 5 0 s and t 5 2 s and also 10 m between t 5 4 s and t 5 6 s (see Figure 1). I the motion is accelerated at equal time intervals, the distances vary. As such, an object has an accelerated motion i it travels 4 m between t 5 0 s and t 5 2 s, whereas between t 5 2 s and t 5 4 s, it advances rom 4 to 16 m and thereore travels 12 m (see Figure 2). A graph o position as a unction o time or an accelerated motion always produces a curve. In the case o uniorm rectilinear motion, the slope o the straight line in a graph o position as a unction o time corresponds to velocity. In order to fnd out the velocity at a given point in time during an accelerated motion, we have to determine the curve’s slope at that precise point in time. To do this, we frst have to draw a straight line with the same slope as the curve at the time under consideration: this straight line is the tangent to the curve. In Figure 2, the slope o the tangent to the curve at t 5 2 s is smaller than the slope o the tangent drawn at t 5 4 s. From this we can deduce that the velocity increases with the passing o time. exml Calculate the instantaneous velocity at t 5 3 s for the object whose motion is represented in Figure 2

25 20 15 10 5 0

1

2

3

4

5

6 Time t (s)

Figure 1 An example of uniform rectilinear motion The line’s slope corresponds to the velocity

30

20

10

0

tngn  4 s

1

2

3

4

5 Time t (s)

Figure 2

Position x (m)

Any two points on the tangent can be chosen to calculate the slope but it is more precise to choose the furthest points possible, in this case the points at the edges of the grid The points’ coordinates and their units are as follows:

30

tngn  2 s

Data: t53s v 5? Solution: 1 Draw the tangent to the curve at t 5 3 s

Position x (m)

Dmining insnnous vlociy using  g

Position x (m)

10.1.2

An example of uniformly accelerated rectilinear motion The curve’s slope (velocity) varies from one moment to the next

30

20

B

appeNDIX 3 t gicl dminion of  slo of  ngn o  cuv, p 392

10

Point A: (16 s, 0 m) Point B: (50 s, 20 m) A 0

1

2

3

4

5 Time t (s)

2 Calculation of the velocity at t 5 3 s: x 2 xi x 2 xA ∆x 20 2 0 m v5 5 f 5 B 5 5 59 m/s t 2 t t 2 t ∆t 50 2 16 s f i B A

Chapter 10 Uniformly Accelerated Rectilinear Motion

223

Position x (m)

In order to determine the velocity at a point in time t based on the curve in a graph o position as a unction o time, it is also possible to use the intervals method. This procedure consists o drawing a secant that intersects the curve at two points situated symmetrically on either side o the time t under consideration; as such, the two points’ abscissas are t – ∆t and t 1 ∆t. The slope o the secant is roughly equal to that o the tangent and corresponds to the velocity at time t (see Figure 3).

30

20

10

It is important not to choose points that are too ar apart: the smaller the quantity o ∆t, the closer the slope o the secant is to that o the tangent. The intervals method becomes less precise when the curve bends irregularly.

Secant

10.1.3 Tangent 0

1

2

3

4

5 Time t (s)

Figure 3 Intervals method: the slope of the secant is almost the same as that of the tangent

accelerion

Acceleration is a variation in the velocity’s magnitude (it can increase or decrease), its direction, or both. In everyday lie, a simple defnition o acceleration sufces: accel­ eration corresponds to an increase in the velocity’s magnitude. In physics, however, in order to describe all types o motion, a more extensive defnition o acceleration is required.

Acceleration corresponds to the change in velocity (∆v) divided by the time interval (∆t) during which this change in velocity occurs.

a5

v 2 vi ∆v 5 f tf 2 ti ∆t

Velocity v

Units o acceleration are units o velocity divided by units o time. The most commonly used units are metres per second divided by seconds, in other words, metres per seconds squared.

vf

m/s 5 m/s2 s

vi

0

ti

tf

Figure 4

Time t

For uniformly accelerated motion, the graph of velocity as a function of time produces a straight line

224

UNIt 3 Kinematics

In a graph o velocity as a unction o time, acceleration corresponds to the slope o the straight line that connects the initial point (ti, vi) and the fnal point (t  , v  ). For a uniormly accelerated rectilinear motion, the veloc­ ity increases regularly, meaning that or equal time intervals the changes in velocity are also equal. The curve in a graph o velocity as a unction o time has the same slope or equal time intervals: it is a straight line (see Figure 4).

exml The fgure shown opposite illustrates a runner’s velocity as a unction o time a) What is the acceleration between: t 5 0 s and t 5 2 s ? b) What is the acceleration between: t 5 2 s and t 5 6 s ? c) What is the acceleration between: t 5 6 s and t 5 10 s ?

Velocity v (m/s)

8

Acceleration a

Since the acceleration is constant, a graph of acceleration as a function of time produces a horizontal straight line (see Figure 5). The straight line is located above or below the time axis, depending on whether the acceleration is positive or negative.

0

Time t

Figure 5 Uniormly accelerated motion This graph represents acceleration as a unction o time or the motion described in Figure 4

6 4 2

Solution: 0 a) At ti 5 0 s, vi 5 0 m/s ; at t 5 2,0 s, v 5 60 m/s v 2 vi ∆v 60 2 0 m/s a5 5  5 5 30 m/s2 t 2 ti ∆t 20 2 0 s

2

4

6

8

10 Time t (s)

b) The velocity remains equal to 6 m/s, the acceleration is zero c) At ti 5 60 s, vi 5 60 m/s ; at t 5 100 s, v 5 0 m/s v 2 vi ∆v 0 2 60 m/s a5 5  5 5 -15 m/s2 t 2 ti ∆t 100 2 60 s In the third section o the motion, the velocity is positive but the acceleration is negative, which indicates that the velocity is decreasing The velocity vector is directed toward the positive x-values (v > 0), but the acceleration vector is directed toward the negative x-values (a < 0)

X-y ubs Motions that occur at truly constant acceleration are uncommon In act, numerous actors, such as air resistance, may prevent a body’s acceleration rom being constant The motion o electrons in an x-ray tube, however, oers an example o constant acceleration The tube, in which a vacuum has been created (see Figure 6 ) contains two electrodes: the cathode (negative charge) and the anode (positive charge) Electrons exit the cathode and are accelerated toward the anode, which is situated a ew centimetres away Their acceleration is in the region o 10 17 m/s2 Since nothing opposes their motion inside the tube, their acceleration is constant The electrons strike the anode at a very high velocity and lose their energy, leading to the emission o x-rays

cod nod

Figure 6 Since there is a vacuum inside an x-ray tube, the electrons’ acceleration rom the cathode to the anode is constant

Chapter 10 Uniormly Accelerated Rectilinear Motion

225

10.1.4

Disinguishing beween velociy and acceleraion

Velocity can be positive or negative, since the fnal position may be greater or smaller than the initial position. The sign o the velocity indicates the direc­ tion o movement in relation to the axis o reerence: • positive velocity: toward positive x­values • negative velocity: toward negative x­values Due to the vector characteristics o velocity, the defnition o acceleration in ∆v physics not only takes into account the change in the velocity’s magni­ ∆t tudes but also the velocity’s direction in relation to the axis o reerence. To determine whether the acceleration is positive or negative, it is necessary to take into account both the magnitude and sign o the initial and fnal velocities. Velocity v

Considering the dierent situations possible, it can be shown that the sign o the acceleration compared to the sign o the velocity indicates how the velocity varies: • i the acceleration and the velocity have the same sign, it means that the object is travelling increasingly ast (the velocity’s mag­ nitude increases) • i the acceleration and the velocity have opposite signs, it means that the object is travelling increasingly slowly (the velocity’s magnitude decreases)

Time t

For example, i the motion o a car is analyzed by orienting the x­axis in the same direction as the motion, the velocity is positive regardless o whether the car moves increasingly ast or slowly. However, the acceleration is positive when the car starts o and trav­ els increasingly ast, and negative when it slows down and brakes.

Figure 7

Graph of velocity as a function of time for a car that starts off and travels increasingly fast

x

Figure 8

Motion of an accelerating car, illustrated with arrows whose length represents the velocity The acceleration is in the direction of the positive x-values Velocity v

The defnition o acceleration in physics has other unusual conse­ quences. For example, an object can accelerate, even i its instan­ taneous velocity is temporarily zero: this is the case with an object that reverses its direction. For one point in time its velocity, but not its acceleration, is zero. Immediately beore becoming tempo­ rarily immobilized, the object had a velocity vi. Immediately ater beginning to move again, the object has a velocity v: its accelera­ tion thereore is not zero. I the acceleration were zero at the same time as the velocity, the object would not begin moving again. Time t

Figure 9

Graph of velocity as a function of time for a

braking car

x

Figure 10 Motion of a decelerating car, illustrated with arrows representing successive velocity vectors The acceleration is in the direction of the negative x-values

226

I the car travels toward the negative x-values and its velocity changes rom vi 5 ­5 m/s to v 5 ­2 m/s, the velocity increases (­2 > ­5), and thereore ∆v > 0, and the acceleration is positive. However, the velocity’s magnitude is smaller (|­2| < |­5|), and it is commonly said that the car decelerates.

UNIt 3 Kinematics

The motion o a car that starts and accelerates with a constant acceleration is a good example. The axis o reerence is oriented in the same direction as the car’s movement. The velocity is posi­ tive and its magnitude increases, which can be represented in a graph o velocity as a unction o time (see Figure 7). To repre­ sent this situation more clearly, it is oten useul to illustrate the motion using arrows that represent the velocity vector in succes­ sive points in time (see Figure 8). I the car brakes ater reaching its maximum speed, the magni­ tude o its velocity decreases and its acceleration becomes nega­ tive (see fgures 9 and 10).

Velocity v (m/s)

Sins’ cclion The 100 m is a spectacular race, the leading event in major athletic competitions, such as the Olympic Games Usain Bolt, a Jamaican sprinter, set a record o 958 seconds or it in 2009

14

The starting acceleration was exceptional: Bolt’s velocity increased by roughly 5 m/s during each o the frst two seconds (see Figure 11) This acceleration is equivalent to that o a sports car that goes rom 0 to 100 km/h in 55 seconds

12 10 8

Ater the irst two seconds, the velocity increased at a slower rate or another 4 seconds, beore decreasing slightly In act, ater 6 seconds o highly intensive eort, the reserves o uel in the muscle cells that can be rapidly mobilized (ATP and phosphocreatine) begin to run out

6 4 2 0

2

Figure 11

10.1.5

4

6

8 10

Time t (s)

The change in velocity as a unction o time during the 100 m Olympic event

Sufc  und  cuv in  g of vlociy s  funcion of im

In the case o uniormly accelerated rectilinear motion, it is possible to deter­ mine acceleration based on a graph of velocity as a function of time by cal­ culating the slope o the straight line (see Figure 4 on page 224). Using such a graph also makes it possible to deduce the displacement, as in the case o uniorm rectilinear motion.

See Gicl snion of vlociy s  funcion of im, p 214

By isolating ∆x in v 5

Velocity v

Since velocity (v) is defned as displacement (∆x) divided by time interval (∆t), displacement is equal to the product o velocity and time, or in other words: ∆x ∆t

we obtain ∆x 5 v 3 ∆t

v

The value ∆x 5 v 3 ∆t corresponds to the surace area under the curve in a graph o velocity as a unction o time or uniorm rectilinear motion (see Figure 12). In act, since the units o the surace area under the curve are units o velocity (m/s) multiplied by units o time (s), which produce units o length (m), the surace area under the curve corresponds to a displacement.

0

∆x 5 Total area under the curve 5 Area o the triangle 1 area o the rectangle 1 5 (v  2 v i) (t  2 t i) 1 v i (t  2 t i) 2 1 1 5 v  (t  2 t i) 2 v i (t  2 t i) 1 v i (t  2 t i) 2 2 1 ∆x 5 (v i 1 v ) (t  2 t i) 2

Time t

Figure 12 For uniorm rectilinear motion, the displacement corresponds to the surace area under the curve Velocity v

For uniormly accelerated rectilinear motion, the surace area under the curve is trapezoidal (see Figure 13). It corresponds to the sum o the areas o a rectangle with a height vi and a triangle with a height (v − vi), both o which have a width (t − ti). The displacement is calculated as ollows:

tf

ti

vf vi

0

ti

tf

Time t

Figure 13

For uniormly accelerated rectilinear motion, the displacement corresponds to the surace area under the curve

Chapter 10 Uniormly Accelerated Rectilinear Motion

227

Vvelocity v (m/s)

exampl A car moves from rest to a velocity of 25 m/s in 8 seconds with a constant accelerationWhat is the car’s displacement between t 5 0 s and t 5 8 s? Data: ti 5 0 s vi 5 0 m/s tf 5 8 s vf 5 25 m/s

25 20 15 10 5 0

1 2 3 4 5 6 7 8 Time t (s)

Since the acceleration is constant, the graph of velocity as a function of time produces a straight line Solution: 1 The displacement is produced by the area of the triangle under the curve: ∆x 5 (250 m/s) 3 8 s 5 100 m 2 At t 5 8 s, the car is located 100 m from its starting point

10.1.6

Surfac ara undr h curv in a graph of acclraion as a funcion of im

Acceleration a

With the help o a graph o acceleration as a unction o time, it is possible to deduce the change in velocity. The process is similar to the one or determin­ ing the displacement with a graph o velocity as a unction o time. Acceleration is defned as the change in velocity (∆v) divided by the time inter­ val (∆t), which is a 5 ∆v . Thereore, the change in velocity is equal to the ∆t product o acceleration and time.

a

∆v ∆t we obtain ∆v 5 a 3 ∆t

By isolating ∆v in a 5 0

ti

tf

Time t

Figure 14

For uniformly accelerated rectilinear motion, the change in velocity corresponds to the surface area under the curve

SeCtION 10.1 1.

228

The product a 3 ∆t corresponds to the surace area under the curve in a graph o acceleration as a unction o time (see Figure 14). The units o the surace area under the curve are units o acceleration (generally m/s2) multiplied by units o time (s), which produce units o velocity (m/s).

Graphical rprsnaion of vlociy as a funcion of im

Give an example for each situation described below a) an object with zero velocity and zero accel­ eration b) an object with non­zero velocity and zero accel­ eration c) an object with non­zero velocity and non­zero acceleration d) an object with zero velocity and non­zero accel­ eration UNIt 3 Kinematics

2. A car’s velocity is directed southward whereas its acceleration is directed northward Is the car accel­ erating or decelerating? 3. Which of the following statements are true? Explain your answer A if ∆x 5 0, then vave 5 0 B if a 5 0, then v 5 0 C if v 5 0, then a 5 0 D if a 5 0, then v  0

Time t

Time t

2

b)

Velocity v

6

Position x

Acceleration a

5

Time t

Time t

Time t

Time t

3

c) x

Time t

4

d) x

Time t

40 30 20 10

0

1

2

3

4

5 Time t (s)

6. The graph below shows an object’s velocity as a function of time Determine: a) the instantaneous acceleration at t 5 2 s b) the instantaneous acceleration at t 5 35 s c) the average acceleration between t 5 2 s and t56s

8. For a rectilinear motion, the graph below describes an object’s velocity as a function of time Determine: a) the displacement between t 5 1 s and t 5 6 s b) the average velocity between t 5 1 s and t 5 6 s c) the acceleration at time t 5 25 s d) the velocity at t 5 4 s Velocity v (m/s)

Position x (m)

5. The graph below shows a cyclist’s position as a function of time a) What is the cyclist’s velocity at t 5 2 s? b) What is the average velocity between t 5 1 s and t 5 5 s?

Velocity v (m/s)

Velocity v

Time t

x

4

Velocity v

x

Velocity v

Time t

1

a)

Velocity v

3

Velocity v

2

Velocity v

1

7. Match each set of velocity vectors with the corre­ sponding graph

Position x

4. Which of the following graphs without a doubt apply to a: a) uniform rectilinear motion? b) uniformly accelerated rectilinear motion? c) a motion or object reversing its direction?

20 10 0

-10

1

2

3

4

5

6 Time t (s)

-20

10 5 0 1 -5

2

3

4

5

6 Time t (s)

Chapter 10 Uniformly Accelerated Rectilinear Motion

229

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

10.2 equions for uniformy ccrd rciinr moion Using graphs to describe and analyze motion oers certain advantages, including the possibility o visualizing the motion. Along with graphic lan­ guage, it is necessary to establish equations that relate the motion’s variables, namely position, velocity, acceleration and time, to one another.

10.2.1

GalIleO GalIleI Italian mathematician, physicist HISTORY HIGHLIGHTS HISTORY HIGHLIGHTS and astronomer (1564–1642) Galileo is known or his discoveries in hydrostatics and pendulum motion, as well as or his studies o uniormly accelerated motion based on experiments conducted using an inclined HISTORY HIGHLIGHTS HISTORY plane He alsoHIGHLIGHTS made major discoveries in astronomy While he did not develop the reracting telescope, he was the frst to use it to observe the sky He discovered the mountains on the Moon and observed that the Milky Way is composed o stars His discovery that our satellites orbit Jupiter showed that not all celestial bodies revolve around the Earth His observation and analysis o the phases o Venus also contributed to the end o the geocentric model, in use since antiquity, and to the acceptance o the heliocentric model, according to which celestial bodies revolve around the Sun Since this idea contradicted dominant thought at the time, Galileo was orced to renounce his theory under pressure rom the Catholic Church

230

UNIt 3 Kinematics

a body’s moion on n incind pn

The relationship between position, velocity and time in an accelerated motion has preoccupied scientists or centuries. Today it is known that veloc­ ity increases linearly with time i the acceleration is constant (see Figure 3 on page 224). In the Middle Ages, however, it was not easy to establish such a relationship. The concept o acceleration was not clearly defned. Moreover, it was not easy to measure velocities accurately due to a lack o instruments or measuring time that were sufciently precise or short time intervals. At the time, people wondered whether velocity increased in proportion to time or in proportion to distance in the case o a alling object subject to the eects o gravitational orce. Likewise, the relationship between distance and time was unknown. In 1604, Italian scientist Galileo, while experimenting with marbles rolling on an inclined plane (and thereore with a constant acceleration), succeeded in determining that distance was proportional to time squared. Today, this result is written as ollows: ∆x α t 2

A ew years later, Galileo managed to deduce that velocity increased linearly with time, rather than with distance.

10.2.2

equions for uniformy ccrd rciinr moion

The equations relating distance (or position), velocity, acceleration and time can be deduced rom the defnitions presented in the previous section. The frst equation ollows rom the defnition o acceleration I acceleration is constant, instantaneous acceleration is equal to average acceleration and thereore:

a 5 aave 5

v 2 vi ∆v 5  t 2 ti ∆t

The usual way to present this frst equation, ater having isolated the fnal velocity, is as ollows: equion 1 v  5 v i 1 a ∆t A second equation is obtained by considering the surace area under the curve in the graph o velocity as a unction o time 1 ∆x 5 x  2 x i 5 (v i 1 v ) (t  2 t i) 2 This equation is usually rewritten, ater isolating xf , as ollows: equion 2 1 x  5 x i 1 (v i 1 v ) ∆t 2 By replacing Equation 1 in Equation 2, we obtain: 1 1 1 x  5 x i 1 (v i 1 v i 1 a ∆t)∆t 5x i 1 (2v i)∆t 1 (a ∆t)∆t 2 2 2 which, once transormed, becomes a third equation: equion 3 1 x  5 x i 1 v i ∆t 1 a (∆t ) 2 2 Finally, rom Equation 1, one can extract ∆t 5

v 2 vi a 1 (v 1 v ) ∆t, we obtain: 2 i 1 v 2 2 v i2 5 xi 1 a 2

and by replacing this expression in Equation 2, x  5 x i 1 x 5 xi 1

v 2 vi 1 (v 1 v )  a 2 i

(

)

(

)

which can be rewritten as: equion 4

It is important to remember that Equations 1 to 4 are only valid for uniform­ ly accelerated motion. They relate the values ti, tf, xi, xf, vi, vf and a to one another. If one of these values is unknown, it can be determined by using an equation containing this unknown value, and in which all other parameters are known. In general, when studying a motion, it is considered that ti 5 0 s, as if start­ ing the chronometer at the beginning of the motion. In this context, Equation 1 becomes vf 5 vi 1 at f or, more simply put, vf 5 at 1 vi, where t is the time elapsed since the beginning of the motion. This equation has the same form as the straight line equation (y 5 ax 1 b). This is why a graph of velocity vf as a function of time t for uniformly accelerated rectilinear motion produces a straight line with a slope a and y­intercept vi (see Figure 15).

Final velocity v f

v 2 5 v i2 1 2 a ∆x

Slope a

vi 0

Time t

Figure 15

For uniormly accelerated motion, the graph o velocity as a unction o time produces a straight line The line’s slope is equal to the acceleration

Chapter 10 Uniormly Accelerated Rectilinear Motion

231

Final position xf

1 In the same way, Equation 3 becomes x 5 xi 1 vit 1 2 at 2. This is the equa­ tion or a parabola. A graph o fnal position x as a unction o time t there­ ore produces a parabola with an y­intercept xi (see Figure 16). I the initial velocity vi is zero, the equation can be reduced to: 1 x  2 x i 5 ∆x 5 at 2 2 which matches Galileo’s discovery, according to which distance is propor­ tional to time squared or an object with an initial velocity o zero.

xi 0

Time t

Figure 16 For uniormly accelerated motion, a graph o fnal position x as a unction o time t produces a parabola

10.2.3

anlyzing uniformly cclrd rcilinr moions

The structure o the equations or uniormly accelerated rectilinear motion indicates how to approach kinematic problems related to constant acceleration. Since these equations relate the parameters o motion at an initial point in time to the parameters o motion at a fnal point in time, it is important to clearly identiy these time points rom the start. Moreover, the presence o numerous variables, ti, t, xi, x, vi, v and a, compels us to proceed methodically. Gnrl mhod of solving kinmic problms 1. Draw a diagram o the situation 2. Choose the direction and the origin o the x-axis, and indicate them in your diagram In general, the x-axis is oriented in the direction o the initial motion, and the origin is chosen at the initial position; as such, xi 5 0, which eliminates one term rom most equations 3. Identiy the initial and fnal time points 4. Identiy the known parameters and indicate their values by adding the appropriate signs For example, i the object initially moves in the direction o the positive values o the x-axis, then vi > 0 Likewise, identiy the unknown parameters It may be useul to list all the parameters, known as well as unknown, in a small table that can be reerred to at a glance 5. Find the equation in which the quantity sought is the only unknown, and isolate this unknown and its units Veriy whether the result obtained is probable Oten, several unknowns need to be ound, and there are several dierent ways to accomplish this It is important to remember that the equations or uniormly accelerated recti­ linear motion are only valid or the motion segment during which acceleration is constant. A car travelling between two stoplights will generally accelerate, drive at a constant velocity and then slow down. Over this distance, there are three motion segments that need to be dealt with in succession, each one in turn, rather than all o them together. exmpl a A cyclist accelerates rom rest to a velocity o 10 m/s in 12 seconds, with a constant acceleration a) What is his acceleration? b) What is the distance travelled during this phase o acceleration? Solution: 1 Draw a diagram o the situation 2 Orient the x-axis in the direction o the displacement In the diagram at the top o the next page, the arrows represent dierent velocity vectors in succession

232

UNIt 3 Kinematics

O

x

3 The initial time is obviously the starting point, and the fnal time is when the cyclist reaches 10 m/s 4 The initial position xi is zero since we established the origin o the x-axis at the starting point The chronometer is started at the starting time point: ti 5 0 s The motion segment under consideration lasts 12 seconds, thereore t 5 12 s As the cyclist sets o rom rest, his initial velocity is zero: vi 5 0 m/s The fnal velocity is known: v 5 10 m/s The unknowns are the acceleration a and fnal position x ti 5 0 s

xi 5 0 m

v i 5 0 m/s

t  5 12 s

x  5?

v  5 10 m/s

a 5?

5 a) Calculation o the acceleration: Since ti and t as well as vi and v are known, one can use equation 1, v  5 v i 1 a ∆t and isolate a a5

v 2 vi ∆t

5

10 2 0 m/s 5 083 m/s2 12 2 0 s

The acceleration is positive and thereore directed toward the positive x-values, just as expected b) Calculation o the distance: Since a is known, one can determine x with equation 3 x  5 x i 1 v i ∆t 1 x 5 0 1 0 1

1 a (∆t ) 2 2

1 3 (083 m/s2) 3 (12 s)2 5 60 m 2

The distance is equal to x  2 x i, which is 60 m exml B A car is travelling at a constant velocity when the trafc signal at the intersection situated 40 m urther ahead changes to red The driver instantly steps on the brakes (we can disregard the reaction time) and the car comes to a stop in 30 seconds, just in ront o the intersection (the acceleration is constant) a) What was the car’s initial velocity? b) What was the acceleration during braking? Solution: 1 Draw a diagram o the situation 2 Direct the x-axis toward the displacement. In the diagram below, the arrows represent dierent velocity vectors in succession

O

x

3 The initial time corresponds to the start o braking, and the fnal time is when the car comes to a stop: v 5 0 m/s 4 The initial position xi is zero since we established the origin o the x-axis at the start o braking The chronometer is started at the beginning o the motion being considered: ti 5 0 s The unknowns are the acceleration and initial velocity vi ti 5 0 s

xi 5 0 m

vi 5 ?

t 5 3 s

x  5 40 m

v  5 0 m/s

a 5?

Chapter 10 Uniormly Accelerated Rectilinear Motion

233

5 a) Calculation o the car’s initial velocity: The three equations comprising acceleration a also contain initial velocity vi ; thereore, these equations cannot be directly solved However, we can use the ollowing equation to fnd: 1 x  5 x i 1 (v i 1 v ) ∆t to fnd vi: 2 2x  2 3 40 m vi 5 5 5 27 m/s t 30 s b) Calculation o the acceleration during braking: The acceleration a can be calculated with any one o the three equations, or example with v 5 vi 1 a ∆t a5

v 2 vi ∆t

0 2 27 m/s 30 2 0 s

5

5

-27 m/s 5 -9 m/s2 30 s

The acceleration obtained is negative, meaning it is directed toward the negative x-values, in the opposite direction o the velocity: this is the case when there is braking

SeCtION 10.2

equaions for uniformly acclrad rcilinar moion

All the exercises in this section consider motion to be uniformly accelerated 1. During a drag race, a dragster starts o and drives 402 m in 60 seconds Determine the acceleration, in metres per second squared (m/s2), and the fnal veloc­ ity, in kilometres per hour (km/h) 2. Starting rom rest, a coin slides down an inclined plane and accelerates, travelling 10 m in 10 second a) What is the coin’s acceleration? b) I the coin slides down the same inclined plane once again but or 20 seconds instead o one, what distance would it travel? 3. A cyclist travelling at 200 km/h begins to accelerate while passing a telephone pole and continues to the next pole, located 250 m urther At this instant, her velocity reaches 300 km/h a) What is her acceleration? b) How much time did the cyclist take to travel rom the frst to the second telephone pole? 4. The moment a car passes in ront o you, it begins to brake in order to stop at a red light situated 250 m ahead The car decelerates at a rate o 400 m/s2 Determine the car’s initial velocity, in kilometres per hour (km/h), and the duration o the braking

234

UNIt 3 Kinematics

5. A car enters a tunnel at 240 m/s and accelerates at a rate o 20 m/s2 What will its velocity be at the tun­ nel’s exit 80 seconds later? 6. At a point on a sloping plane, a ball rolls down at 10 m/s It then travels 60 m in 40 seconds a) What is its fnal velocity? b) What is its acceleration? 7. A car initially at rest accelerates at a rate o 60 m/s2 What distance does it travel between t 5 100 s and t 5 150 s? 8. A skier descending a slope accelerates rom 350 m/s to 1140 m/s in 420 seconds a) What is her displacement? b) What is her average velocity? 9. A woman is driving her car Ater noticing an obsta­ cle, her reaction time is 070 seconds When she brakes at the maximum, her car decelerates at a rate o ­750 m/s2 I her initial velocity was 500 km/h: a) what is the minimum time she’ll need in order to stop ater noticing the obstacle? b) what distance does she travel during this time?

10.3 F ll Free fall is motion that occurs when an object is subject only to gravitational force. Since this chapter deals with uniformly accelerated rectilinear motions, this section will be limited to an object’s vertical motion, either up or down. An object travelling vertically upward evidently does not fall, but this motion also constitutes a free fall, as the object is only subject to gravitational force. When an object is only subject to the Earth’s gravitational force, the magni­ tude of its acceleration, symbolized by the letter g, is roughly constant near the Earth’s surface, and its average value is: g 5 980 m/s2

Gravitational acceleration has a downward direction and varies slightly depend­ ing on the latitude and altitude; however, these variations are considered negli­ gible in the situations analyzed in this section. If air resistance is taken into account, objects generally do not fall with an acceleration equal to g. Air resistance opposes motion and causes real accel­ eration to decrease, especially at a high velocity, as air resistance is propor­ tional to velocity squared. However, considering the acceleration of an object in free fall as equal to g allows us to analyze motion more easily. This approx­ imation adequately corresponds to reality for relatively low velocities. This acceleration is the same for objects of different masses. If a small metal marble and a bowling ball are dropped from the same height, they will both land on the ground at the same time, since they are not greatly affected by air resistance. Other objects, however, cannot reach the ground at the same time if they are subject to very different air resistance due to their respective sizes.

t s o  w column fowing om   We can observe the eect o acceleration due to gravity on a water column gently lowing rom a tap (see Figure 17 ) The water column’s width narrows during its all because as it alls, the water accelerates and its velocity increases Since the discharge is the same no matter at which point in the water column, it is necessary that the water column’s section, ie its diameter, decrease in the course o the all The water column retains its cohesion due to the orce o attraction created by the hydrogen (H) bonds between the water molecules

Figure 17

The water column gently fowing rom a tap narrows in the course o its all

Chapter 10 Uniormly Accelerated Rectilinear Motion

235

10.3.1

The analysis o motions connected with ree all requires that we adjust some o the conventions used so ar.

y v

a

axis of rfrnc for nlyzing fr fll

a

Rise

To analyze vertical motion, the upward y­axis is commonly used as a rame o reerence. Since the x­axis generally has a horizontal orientation, it is preer­ able to use the y­axis or vertical motions. The kinematic equations or con­ stant acceleration remain valid, except that x is replaced with y.

v Fall

Ground

Figure 18 The direction of velocity and acceleration for a ball that rises and falls

The higher an object, the greater the value o y. Because heights and altitudes are measured rom the ground up, the origin o the y­axis is generally estab­ lished at ground level. Gravitational acceleration always has a downward direction, regardless o whether the object rises or descends (see Figure 18). Since the y­axis has an upward orientation, it is written: ay 5 -g 5 -980 m/s2

10.3.2

Fr fll nlysis

The analysis o ree all is perormed in the same way as the analysis o uni­ ormly accelerated rectilinear motion. A parameter, such as position or velocity, must oten be given a value not explicitly mentioned in the statement, and so the statement has to be trans­ lated into algebraic language. For example, i an object leaves the ground, we write yi 5 0. I an object rises, its velocity is positive; i it alls, its velocity is negative, and we have to remember to add the minus sign to the velocity’s value. Certain situations concern an object that reaches its maximum height. In order to translate this into algebraic language, one has to consider that at this point in time, the object reverses direction: it was rising and it will begin to all. At its maximum height, the object’s velocity is thereore zero. In other cases, the motion o an object returning to the ground is analyzed. One might be tempted to write that the object comes to a stop and that v 5 0. This would be incorrect: it is important to remember that the equations only allow or the analysis o a ree all and not a collision with the ground. As soon as the ball begins to touch the ground, it is aected by another orce in addition to gravitational orce and is no longer in ree all. The fnal velocity o an object returning to the ground is its velocity just beore it comes into contact with the ground. This velocity is negative and o a maximum magni­ tude, but it is not zero. To characterize an object returning to the ground, we write that its fnal posi­ tion is approximately zero: y 5 0. exmpl A ball is thrown upward from the ground at a velocity of 250 m/s a) What is the maximum height reached by the ball? b) How long does the ascent take? c) How long does the ball remain in the air?

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Data: To answer the frst two questions, the fnal time corresponds to the moment when the ball reaches its maximum height ti 5 0 s

yi 5 0 m

v i 5 250 m/s

t 5 ?

y 5 ?

v  5 0 m/s

a 5 -980 m/s2

Solution: v 2 2 v i2 (0 m/s)2 2 (250 m/s)2 a) v2 5 vi2 1 2a ∆y thereore ∆y 5  5 y 5 5 319 m 2a 2(-980 m/s2) v 2 vi 0 m/s 2 250 m/s 5 255 s 5 t 5 -980 m/s2 a c) To answer the third question, the fnal time corresponds to the moment the ball arrives very near the ground, thereore y 5 0 m b) v 5 vi 1 a ∆t thereore ∆t 5

ti 5 0 s

yi 5 0 m

v i 5 250 m/s

t 5 ?

y 5 0 m

v 5 ?

a 5 -980 m/s2

v 2 5 v i2 1 2a ∆y 5 (250 m/s)2 1 2(-980 m/s2) 3 (0 m 2 0 m) 5 (250 m/s)2 as such, v  5 ±25 m/s Since the ball is alling at the time it touches the ground, it is directed toward the negative y-values and the correct sign is the minus sign: v 5 -250 m/s -250 2 250 m/s v 2 vi v  5 v i 1 a ∆t thereore t  5  5 510 s 5 -980 m/s2 a As the previous example shows, the fight time (the time the ball spends in the air) is twice the rise time. As such, the all takes as much time as the rise. The ball returns to the ground at a velocity that is o the same magnitude as the starting velocity, but in the opposite direction. In reality, on account o air resistance, things do not happen in this way. The ball would rise less high and the return velocity would be lower than the starting velocity: the all would thereore takes longer than the rise.

Furthering

your understanding

tminl vlociy nd skydiving A parachutist jumps rom an altitude o 4000 m I she accelerated at a constant rate o 980 m/s each second, her velocity would reach 98 m/s (353 km/h) ater 10 seconds

Once the parachute has opened, air resistance abruptly increases and the parachutist decelerates until reaching a new terminal velocity o about 5 m/s (18 km/h)

Air resistance ensures that the velocity does not reach such a high rate During the very frst seconds o the all, the velocity remains low and resistance is limited because it is proportional to the velocity squared The parachutist can be considered to be in ree all: her velocity increases in proportion to time Ater a ew seconds, air resistance becomes substantial and the parachutist is no longer in ree all Her velocity continues to increase but at an increasingly slower pace The velocity reaches a limit when air resistance becomes so great that it osets gravitational orce exactly As such, the parachutist reaches a terminal velocity o approximately 200 km/h The terminal velocity varies depending on dierent actors, primarily the parachutist’s size and the surace area he or she presents to the air (cross-sectional area) [see Figure 19]

Figure 19 In the spread-eagle position, the parachutist encounters more air resistance than she would in a vertical position Her terminal velocity is lower and her dive lasts longer Chapter 10 Uniormly Accelerated Rectilinear Motion

237

SeCtION 10.3

Fr fall

Position y

Position y

1. A ball is dropped Which o the ollowing graphs represents the ball’s position as a unction o time? B A

D

Position y

C

Time t

Time t

Position y

Time t

Time t

2. From a building with a height o 15 m, a man throws a ball upward (along a vertical situated beyond the building) at a velocity o 10 m/s What is the ball’s fight time, and at what velocity does it hit the ground? 3. From a building with a height o 15 m, a man throws a ball downward (along a vertical situated beyond the building) at a velocity o 10 m/s What is the ball’s fight time, and at what velocity does it hit the ground? 4. A child dives o a swimming pool’s 30 m spring­ board What is the child’s velocity when he hits the water? 5. Parachutists land on the ground at a velocity o 50 m/s During training, rom what height do they need to jump in order to simulate a landing? 6. A diver jumps rom a diving board without pushing o vertically and reaches the water 143 seconds later What is the height o the diving board?

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7. A pitching machine throws a ball vertically upward The ball leaves the machine at a height o 12 m and reaches a height o 50 m What was the ball’s initial velocity? 8. A fying sh makes a vertical jump out o the water at a velocity o 10 m/s Determine: a) the maximum height reached by the sh b) the total time the sh spends in the air c) the acceleration o the sh 1 second beore alling back into the water 9. A student wants to throw a textbook to a classmate leaning out o a school window directly above him He throws the book upward at a velocity o 80 m/s His classmate catches the textbook while it is travel­ ling upward at a velocity o 30 m/s a) What upward distance did the textbook travel? b) How long did the textbook remain in the air? 10. A girl throws a pebble into a deep well at a velocity o 40 m/s (downward) The pebble strikes the water 20 seconds later a) At what depth is the well’s water surace located? b) What is the pebble’s velocity when it strikes the water? 11. A person standing on a balcony uses his right hand to throw ball A vertically upward at an initial velocity o 10 m/s (along a vertical situated beyond the balco­ ny) At the same time, he uses his let hand to throw an identical ball B downward The two balls leave his hands at a height o 5 m above the ground How long ater ball B will ball A reach the ground? 12. A diver standing on a diving board drops her nose clip rom a height o 11 m rom the water’s surace Water resistance slows the nose clip at a constant rate o 30 m/s2 What depth will the nose clip reach?

APPLICATIONS accloms There are various means of detecting accelerations. For example, an object suspended from the rear­view mirror of a car accelerating forward appears to be trans­ ported further back. The greater the acceleration, the wider the angle of inclination.The suspended object thus makes it possible to evaluate the acceleration. To detect accelerations, accelerometers are integrated into various objects, such as a car’s air bags. Air bags need to deploy when the car experiences a collision, i.e. when a very abrupt deceleration occurs (see Figure 20). To detect such an acceleration, a tiny accel­ erometer is integrated into the air bag: this detection device measures less than 1 mm2. It contains a struc­ ture made of silicon that acts like an electric capacitator: when the car decelerates forcefully, the capacitator deforms and the attached electronic circuit detects the change in capacitance. This device is an example of microelectromechanical systems (MEMS).

Some video game controllers detect their own motion, speed and orientation with the help of accelerometers. Accelerometers are also integrated into laptop com­ puters. If the laptop drops, the accelerometer detects the fall and commands the hard drive’s read head to retract in order to prevent damage upon impact with the ground. Many cell phones also contain acceler­ ometers that detect the phone’s orientation in order to correct its image display. Accelerometers can also be found in some navigation systems (GPS), in antilock braking systems (ABS) of vehicles, and in some washing machines in order to detect vibrations caused by an unevenly spread load. Accelerometers can also be used to count the number of steps a person makes or to determine how a box was handled.

Figure 20 An air bag, a video game controller and a cell phone are some examples of the many objects that contain an accelerometer

Chapter 10 Uniformly Accelerated Rectilinear Motion

239

Free fall For Aristotle, in the 4th century BCE, there existed two kinds o motion: violent motion, produced by an external orce (such as that o a ball being thrown), and natural motion. He described the latter as the motion o objects toward their “natural” position: re rises into the air, a stone alls to the ground. Shaped identically, a heavier stone alls aster than a lighter stone. Since it wasn’t easy to test these conceptions, Aristotle’s views endured or almost 2000 years. There weren’t any means to accurately measure velocity or eliminate air resistance. Although Oresme, in the 14th century, predicted that i the initial velocity was zero, the dis­ tance was proportional to time squared, his analysis remained without consequence. Galileo (1564–1642) turned his attention to the pro­ blem. Because it was dicult to take measurements o time, position and velocity or a rapidly alling object, he experimented with marbles rolling on a slightly inclined plane tted with a groove (see Figure 21). Galileo considered the motion o a marble rolling on an inclined plane to be similar to that o a marble

alling vertically. In both cases, the marbles acceler­ ated under the infuence o gravitational orce, but the marble rolling on the inclined plane accelerated less than the one in ree all; it happened as i gravitational orce had been “diluted.” Galileo’s reasoning was cor­ rect, or i the slope o the inclined plane is increased, the marble’s acceleration increases; ultimately, a marble “rolling” on an inclined plane at 90° drops in ree all. To measure a all’s duration, Galileo used a water clock. Since the marbles did not roll ast, air resist­ ance was negligible. Galileo could thus prove that dis­ tance was proportional to time squared. Moreover, he noted that dierent marbles descended the inclined plane with the same acceleration. About 450 years later, during the Apollo 15 space mis­ sion, astronaut David Scott dropped a eather and a hammer on the Moon. Since there was no air, the eath­ er and the hammer touched the ground at the same time, a result that supports the modern conception o ree all.

Figure 21 Galileo established that an object reely descending on an inclined plane describes uniormly accelerated rectilinear motion This resco, painted by Giuseppe Bezzuoli (1784–1855) or the Institute and Museum o History and Science, in Florence, Italy, shows Galileo demonstrating his fndings to Don Giovanni de’ Medici

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chapter

10

Unifomly accld rcilin Moion

10.1 Ccisics of unifomly ccld cilin moion • In uniormly accelerated rectilinear motion (UARM), velocity varies regularly rom one moment to the next. • Instantaneous velocity is the velocity at a precise point in time. • In a graph o position as a unction o time, the instantaneous velocity corresponds to the slope o the tangent to the curve at a precise point in time. • Average acceleration corresponds to the change in velocity ∆v divided by the time interval ∆t during which this change in velocity occurs. • In physics, the notion o acceleration includes situations in which the velocity’s magnitude decreases. The acceleration is then in the opposite direction o the velocity’s. • In a graph o velocity as a unction o time, average acceleration corresponds to the slope o the straight line connecting the initial point (ti, vi) and the nal point (t, v). • A graph o velocity as a unction o time or uniormly accelerated rectilinear motion presents a straight line in which the slope is equal to the acceleration. Displacement corresponds to the surace area under the curve. • In graph o acceleration as a unction o time, the surace area under the curve corresponds to the change in velocity.

10.2 equions fo unifomly ccld cilin moion • Four equations establish the relationship between acceleration, v f 5 v i 1 a ∆t position, velocity and time, at the initial and nal time. 1 x f 5 x i 1 (v i 1 v f) ∆t • Generally speaking, when studying a motion, it is considered that ti 5 0, 2 1 as i starting the chronometer at the beginning o the motion. The origin o x f 5 x i 1 v i ∆t 1 a ∆t 2 2 the x­axis is oten established as the starting point, and thereore xi 5 0. 2 5 v 2 1 2a ∆x v • To solve a kinematic problem, one must draw a diagram o the situation; f i choose the direction and origin o the x­axis; identiy the initial and nal time points; identiy the known and unknown parameters; nd the equation in which the quantity sought is the only unknown; and solve the equation.

10.3 F fll • Free all is a motion occurring under the sole infuence o gravitational orce, regardless o whether the object travels upward or downward. • In the absence o air resistance, the acceleration o an object in ree all, close to the Earth’s surace, is g 5 9.80 m/s2 downward. • Air resistance slows down an object in ree all and so causes its real acceleration to decrease. The consideration that all objects experience the same acceleration in ree all is an approximation enabling us to analyze motion more easily. • To analyze vertical motion, the upward y­axis is generally used as a rame o reerence, with the origin at ground level. In this context, ay 5 ­g 5 ­9.80 m/s2. Chapter 10 Uniformly Accelerated Rectilinear Motion

241

Unifomly accld rcilin Moion

1. Using the variables yi , y , vi and v , how can we write: a) that a moving object comes to a stop? b) that an object reaches its maximum height? c) that an object is set in motion rom rest? d) that an object thrown vertically leaves rom the ground? e) that an object at rest is dropped rom above the ground? f) that an object reaches the ground?

6. An ant travels along a straight line with its posi­ tion varying as a unction o time, as indicated in the graph below Classiy its velocities in increas­ ing order at the points indicated Position x

Chapter 10

B

E C

A

D

Time t

Velocity v

Velocity v

2. A ball is dropped in ree all Which o the ollowing graphs represents the magnitude o the ball’s veloc­ ity as a unction o time? A B

D

Velocity v

C

Time t

Time t

Velocity v

Time t

7. A motorcyclist travelling at 12 m/s accelerates at a rate o 60 m/s2 in the same direction How long does it take her to travel 63 m?

Time t

3. The acceleration due to the gravitational orce on the Moon is approximately 16 m/s2 How much time does it take a hammer to reach the Moon’s surace i it is dropped rom a height o 18 m? What happens i the hammer is replaced with a eather? 4. On an airport runway, an airplane starts o and constantly accelerates or 500 seconds over a distance o 1500 m beore taking o a) What is the airplane’s acceleration? b) At the time o takeo, what is its velocity in kilometres per hour (km/h)? 5. Using a graph o velocity as a unction o time represent the motion o a bus that, ater it starts o, accelerates at 10 m/s2 or 60 seconds, travels at a constant velocity or 60 seconds, and then decelerates at a rate o 20 m/s2 or 30 seconds

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8. In a classic superhero movie, the bad guy drops his hostage rom the top o a tower on the Queensboro Bridge in New York City The tower measures 106 m in height The superhero hesitates or 12 seconds beore deciding to jump in order to catch the hostage How much time does it take an object dropped rom a height o 106 m to reach the ground? Is this movie scene realistic? 9. A cyclist coasting along slows down rom 100 m/s to 60 m/s over a distance o 50 m Over what dis­ tance does he continue to coast beore coming to a stop? 10. A hot­air balloon leaves the ground and rises at a velocity o 80 m/s When it reaches an altitude o 60 m, its passenger drops a sandbag a) How much time does it take the sandbag to reach the ground? b) What distance does the sandbag travel in the fnal second? 11. In the fnal second o its ree all, an object covers hal the height o its total all From what height did it all?

te Moion of projeciles

A

projectile is an object that is launched into the air and, under the sole infuence o terrestrial gravita­ tional orce, describes a curved trajectory in the proc­ ess. According to this deinition, the weight thrown by shot putters, or long jumpers themselves, can be considered projectiles, as can bullets rom guns, and shells. When we say that gravitational orce alone acts on an object, we implicitly consider that air resistance is negligible. Since air resistance increases with veloc­ ity, this means that the description o the motion o projectiles in the chapter is only valid or relatively slow­moving objects, which are less aected by air resistance.

Review Relationship between constant velocity, distance and time       12 Gravitational force                                        13

In this chapter, you will study the characteristics o a projectile’s motion. To do so, you will distinguish projectiles launched horizontally rom projectiles launched at an angle. As you will notice, this process is an extension o the analyses made in the previous two chapters.

11.1 11.2

Describing e moion of rojeciles   244

11.3

te moion of objecs lunced  n ngle                            252

te moion of objecs lunced orizonlly                 249

ChaptER 11 The Motion of Projectiles

243

11.1 Describing he moion of projeciles In chapters 9 and 10, position, displacement, velocity and acceleration were defned in the context o rectilinear motion. The same variables can be used to describe a two­dimensional curved motion. The defnitions are similar, yet here, the vectorial nature o each variable is more evident. y

∆s

si sf

x

Figure 1 The position and displacement vectors or two points on a two-dimensional trajectory The displacement vector ∆s is rom the initial position to the fnal position See Componens of a vecor, p 186

11.1.1

In a two­dimensional curved motion, posi­ tion, symbolized by s , corresponds to a vec­ tor going rom the origin to the point where an object is at a given time. The vector si corre­ sponds to the initial position at time ti and the vector s corresponds to the fnal position at time t. The displacement is again a change in position: ∆s 5 s 2 si (see Figure 1). Each o the position, displacement, velocity and acceleration vectors can be decomposed, or de­ scribed by its x­ and y­components. This is the most practical way to analyze projectile motion.

Independence of horizonal and verical moions

When a person throws a ball, his or her hand exerts a orce on the ball as long as the hand pushes the ball. What happens when the ball leaves the hand? The ball continues to move along a curved trajectory and then ends up alling on the ground.

Figure 2 The trajectory o a ball thrown horizontally according to the theory o impetus

Why does the ball continue to move? Scientists struggled to answer this ques­ tion or years, with little success. In the Middle Ages, they believed that the hand conerred to the ball an “impetus,” a kind o internal orce that kept the ball moving ater it let the hand. According to this theory, the impetus would peter out because o air resistance, and the ball would then all vertically. According to the theory o impetus, the trajectory o a ball thrown hori­ zontally consists o a horizontal segment, a curved segment and a vertical downward segment (see Figure 2). At the beginning o the motion, when the impetus is high, gravitational orce will have no eect and the ball will conti­ nue to move horizontally. As the impetus runs out, the trajectory will curve and the ball will all vertically under the eect o gravitational orce. Today’s technological tools make it possible to analyze a projectile’s trajec­ tory in detail. An example o this is the motion o balls in ree all released simultaneously rom the same height (see Figure 3).

Figure 3 The motion o two balls in ree all released simultaneously One o the balls is launched horizontally

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In Figure 3, one ball is dropped vertically at the same time as another is launched horizontally. The positions o the balls are photographed at dierent times separated by equal intervals. This is done with a stroboscope, a device that emits short luminous ashes at regular intervals while the camera shutter remains open.

Here, we can see that at each instant, the vertical position o the two balls is the same, as shown by the horizontal lines superimposed on the photo. The two balls reach the lowest point at the same time. The ball launched horizon­ tally begins to all rom the very start. This observation contradicts the tra­ jectory predicted by the theory o impetus. Clearly, this ancient theory is not valid because it does not accurately reect reality.

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

The vertical lines superimposed on the photo, separated by equal time inter­ vals, illustrate the displacements o the ball launched horizontally. Here, we can see that or each time interval, the horizontal displacement o the ball is the same. In other words, the horizontal velocity o the ball does not change over the course o the motion, even i the ball is accelerated downward because o gravitational orce. Vertical motion and horizontal motion are independent rom each other. This observation has undamental implications. Horizontal motion, along the x­axis, is uniorm rectilinear motion. Horizontal acceleration is zero (ax 5 0). In the vertical direction, along the y­axis, an object is subject to constant accel­ eration, directed downward and equal in magnitude to g (ay 5 ­g). The vertical motion is uniormly accelerated rectilinear motion. It was Galileo who frst hypothesized that, in a context where air resistance is negligible, vertical and horizontal motions are independent. His hypothesis has since been confrmed by experiments on projectile motion. Chapter 14 explains why vertical and horizontal motions are considered to be independent. It all stems rom orces. Because no horizontal orce is present i air resistance is disregarded, there is no horizontal acceleration and hori­ zontal velocity is constant. In the vertical direction, only gravitational orce is present, which explains the projectile’s downward acceleration.

11.1.2

Equins f rjecie min

To describe the motion o projectiles, we must decompose the motion and separately describe what happens horizontally and vertically. To do this, we use a system o coordinates with a horizontal x­axis and a vertical y­axis. Though the points o origin o the axes of reference are quite arbitrary, the ori­ gin o the y­axis generally is generally fxed at ground level. The origin o the x­axis is fxed vertically on the projectile’s departure point; or example, i the projectile is released rom the top o a cli, the origin o the x­axis is at its oot (see Figure 4). y

NICColo FoNtaNa taRtaglIa Italian mathematician (1499–1557) Tartaglia was severely wounded in the head by a French soldier when his native city, Brescia, was HISTORY HIGHLIGHTS HISTORY HIGHLIGHTS invaded in 1512 He recovered, but stammered for the rest of his life (earning him the name Tartaglia from the Italian word for stammer, tartagliare) He went on to earn a living teaching mathematics He is principally know for having found the general solution of the thirddegree equations On the subject of projectiles, Tartaglia reckoned that a trajectory was made up of three parts: a diagonal rectilinear section at the beginning, a curved section in the middle, and a vertical section at the end He experimented with real cannons to determine the departure angle that produced maximum range He concluded that no part of the trajectory was rectilinear His book Nova Scientia (A New Science) founded the science of ballistics

x

Figure 4 Locating the origin in a two-dimensional coordinate system used to describe the motion of a projectile

See axis f reference fr nyzin free f, p 236

ChaptER 11 The Motion of Projectiles

245

To describe the motion o a projectile, the variables that must be taken into account are time (t), horizontal position (x), vertical position (y), horizontal velocity (vx), vertical velocity (vy) and vertical acceleration (ay 5 ­ g). Note that there are no arrows over the symbols o the components; there are only arrows over the symbols o the vectors. The x­ and y­positions and the velocity component vy vary as a unction o time. The velocities and accelerations in x and in y are not the same; to prop­ erly dierentiate them, it is very important to place the x and y subscripts beside the velocity and acceleration components. Moreover, as in previous chapters, the “i" subscript is placed beside variables describing the initial time, and the “” subscript beside those describing the fnal time. See Equaions for uniformly acceleraed recilinear moion, p 230

The equations o projectile motion are shown in the box below beside the equations of uniformly accelerated rectilinear motion. Because acceleration is constant vertically (ay 5 ­ g), it is possible to use equations o uniormly accelerated rectilinear motion. Along the x­axis, since acceleration is zero (ax 5 0), velocity is constant and the motion is uniorm. The units are SI units. Moion of a projecile General equations or uniormly accelerated rectilinear motion (UARM) v 5 vi 1 a ∆t 1 x  5 xi 1 (vi 1 v) ∆t 2 1 x  5 xi 1 vi∆t 1 a (∆t ) 2 2 v2 5 vi2 1 2a∆x where ∆t 5 ∆x 5 ∆y 5 vx 5 vy 5 ay 5

Equations describing horizontal motion (along the x-axis)

Equations describing vertical motion (along the y-axis) vy 5 viy 1 ay ∆t

vx 5 vix

1 y  5 yi 1 (viy 1 vy) ∆t 2 1 y  5 yi 1 viy ∆t 1 ay (∆t ) 2 2

x  5 xi 1 vix ∆t x  5 xi 1 vix ∆t

vy2 5 viy2 1 2ay ∆y

vx 5 vix

Time interval, expressed in seconds (s) Change in horizontal position, expressed in metres (m) Change in vertical position, expressed in metres (m) Horizontal velocity, expressed in metres per second (m/s) Vertical velocity, expressed in metres per second (m/s) Vertical acceleration (ay 5 -g), expressed in metres per second squared (m/s2)

The example below shows how the equations o projectile motion can be used to interpret a particular situation. Example A soccer ball is kicked at a 45° angle over a feld (see fgure opposite) Draw the graphs or acceleration, velocity and position, horizontally and vertically See Uniform Recilinear Moion, p 203

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y

45˚

Solution: 1 Because the ball leaves at a 45° angle, its initial horizontal velocity is equal to its initial vertical velocity With the coordinate system used, xi 5 yi 5 0

x

See Crcerisics of uniformly ccelered reciliner moion, p 222

Acceleration ax

Position x

Velocity vx

2 Horizontally, acceleration is zero, and velocity is constant Position increases proportionally over time, as shown by the equation x f 5 vixt (assuming that ti 5 0 and t f 5 t)

Time t

Time t

Time t

Time t

Acceleration ay

Position y

Velocity vy

3 Vertically, acceleration is constant and equal to -g The graph for velocity as a function of time therefore has a negative slope Vertical velocity is initially positive, turns temporarily to zero at the peak of the trajectory, then becomes negative thereafter The y-position increases, reaches a maximum, then decreases, according to the equation 1 y f 5 viyt 1 ayt 2 (assuming that ti 5 0 and t f 5 t) 2

Time t

Time t -g

11.1.3

te wo-dimensionl velociy vecor

The average velocity over the time interval ∆t is equal to the displacement ∆s ∆s divided by the time interval ∆t: vave 5 while the instantaneous velocity ∆t corresponds to the average velocity or a very small time interval Δt. For a very small time interval, the displacement vector ∆t becomes tangent to the curve. The instantaneous velocity vector ∆s becomes tangent to the curve. The instantaneous velocity vector v, proportional to ∆s, is thereore tangent to the trajectory. Velocity vectors can be drawn at dierent points o a trajectory, like those shown opposite (see Figure 5). The horizontal component vx is constant. From the beginning o the motion to its end, com­ ponent vy, positive at frst, decreases in magnitude as the object rises. It becomes zero at the peak o the trajectory, and then becomes negative as it drops and increases in magnitude.

y

See Insnneous velociy nd verge velociy, p 222

v v vx v

vy

vx vy v

vy vx

vx

x

θ v

vy

Figure 5 The velocity vectors of a projectile at different points in its trajectory The component vx is constant, and the component vy varies over time

ChaptER 11 The Motion of Projectiles

247

See Cmpnens f  vecr, p 186

If components vx and vy of the velocity vector at a given instant are known, the magnitude of the velocity is given by the Pythagorean relationship. v 5 vx2 1 vy2 and the angle θ between the velocity vector and the x­axis is given by the fol­ lowing expression: tanθ 5

SECtIoN 11.1

vy vx

Describing he min f prjeciles

1. O the ollowing situations, which correspond to a projectile motion as it is defned in physics? A A alcon soars through the air B A child throws a snowball C A rocket leaves its launcher to go into orbit D The Manneken Pis in Brussels urinates a stream o water into a basin

3. You are driving in a car with the windows open while eating an apple Suddenly, you drop the apple and it alls out the window The apple will hit the road: A directly below the place you dropped it B behind the place you dropped it C in ront o the place you dropped it 4. Associate each o the variables to its characteristic in terms o projectile motion The x-axis is in the same direction as the projectile’s horizontal motion, and the y-axis is directed upward a) x 1 zero b) vx 2 decreases, then increases c) vy 3 increases d) ax 4 constant e) ay 5 equal to -g 5. Indicate which o the ollowing equations are written correctly A vy2 5 viy2 1 ay ∆t

2. A marble rolls o a table What is its trajectory rom the moment it leaves the table? (Air resistance is negligible)

B y 5 yi 1 viy ∆t 1 ay ∆t C vy2 5 viy2 1 2ay(y 2 yi) D x 5 xi 1 vix∆t E v 5 vx1 vy F v 5 vix 2 1 vfy 2

a

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B

C

D

E

11.2 te moion of objecs lunced orizonlly The equations in the previous section make it possible to quantitatively analyze projectile motion. In this section, we will consider only projectiles launched horizontally. Projectiles launched at an angle will be covered in the next sec­ tion). In the case of projectiles launched horizontally, the initial velocity vector is horizontal, and viy 5 0.This situation occurs, for instance, when an object moves over a horizontal surface and then falls off the edge (see Figure 6). Five basic equations, from the equations cited in the previous section, are used to numerically analyze projectile motion. x  5 xi 1 vix ∆t (1)

Figure 6 From the moment it alls o the edge, the kayak behaves like a projectile launched horizontally (i the water fow’s direction is perfectly horizontal)

vy 5 viy 1 ay ∆t (2) 1 y  5 yi 1 (viy 1 vy) ∆t (3) 2 1 y  5 yi 1 viy∆t 1 ay (∆t ) 2 (4) 2 vy2 5 viy2 1 2ay ∆y (5) The following examples show how these equations are used. Emle a From the roo o a building 500 m high, a ball is thrown horizontally with a velocity o 500 m/s What is the ball’s position at t 5 100 s, t 5 200 s and t 5 300 s? Solution: 1 We set the origin o the coordinate system at the oot o the building The initial position is thereore si 5 (xi, yi) 5 (0 50) m

appENDIx 5 problem solving, p 400

y vi

The initial velocity vector is horizontal Thereore, vix 5 500 m/s and viy 5 0 m/s v fx x

θ vf

v fy

2 Calculation o the position at t 5 100 s Data: ti 5 0 s

xi 5 0 m

yi 5 50 m

vix 5 500 m/s

viy 5 0 m/s

ax 5 0 m/s2

t  5 100 s

x 5 ?

y  5?

vx 5 500 m/s

vy 5 ?

ay 5 -980 m/s2

Calculation: From the ve equations, we choose those that provide x  and y  as a unction o known values: equations (1) and (4) x  5 xi 1 vix ∆t 5 0 1 500 m/s 3 100 s 5 500 m 1 1 y  5 yi 1 viy ∆t 1 ay (∆t ) 2 5 500 m 1 0 1 (-980 m/s2) (100 s)2 5 4510 m 2 2 ChaptER 11 The Motion o Projectiles

249

For the position at t 5 2 s and at t 5 3 s, only the value o t  changes; the initial values remain the same In this way, we can fnd the positions indicated in the table below t (s)

x (m)

y (m)

0

00

500

100

50

451

200

100

304

300

150

59

Example B From the roo o a building 500 m high, a ball is thrown horizontally with a velocity o 500 m/s a) How long does it take or the ball to touch the ground? b) How ar rom the building will the ball be when it touches the ground? c) What is the magnitude o the ball’s velocity at the moment it touches the ground? What is the direction o the velocity vector at this time point? Solution: The fnal time corresponds to the instant when the ball is about to come into contact with the ground Thus, the fnal velocity is not zero, but we can write that y  5 0 According to the results in Example A, we can predict that t  will be a little more than 3 s Data: ti 5 0 s

xi 5 0 m

y i 5 50 m

vix 5 500 m/s

viy 5 0 m/s

ax 5 0 m/s2

t 5 ?

x 5 ?

y 5 0 m

vx 5 500 m/s

vy 5 ?

ay 5 -980 m/s2

Calculation: a) From the fve equations, choose the one where t  is the only unknown: equation (4) 1 1 y  5 yi 1 viy ∆t 1 ay (∆t ) 2 thereore 0m 5 500 m 1 (0 m/s x ∆t) 1 (-980 m/s2 3 ∆t 2) 2 2 1 5 500 m 1 2 (-980 m/s2 3 ∆t 2)

-2 3 (500 m) -2 3 (500 m) 5 319 s thereore tf 5 -980 m/s2 -980 m/s2 b) The only equation providing x  is equation (1): x  5 xi 1 vix ∆t 5 0 1 (500 m/s 3 319 s) 5 160 m ∆t 2 5

c) The y-component o velocity is given by equation (2): vy 5 viy 1 ay ∆t 5 0 1 (-980 m/s2) 3 (319 s) 5 -313 m/s This component is negative, because at the moment it touches the ground, the ball is travelling downward We can now calculate the magnitude o the velocity vector using the Pythagorean relationship: v 5 vx2 1 vy2 5 (500 m/s)2 1 (-313 m/s)2 5 317 m/s and the direction o v relative to the x-axis: v y -313 m/s tan θ 5 v 5 5 -626 thereore θ 5 tan-1(-626) 5 -809° 500 m/s x Note: When doing the last calculation, do not orget to select degree as the angle unit on your calculator

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SECtIoN 11.2

te min f bjecs rwn riznlly

Time tTime t Time tTime t Time tTime t

Acceleration Acceleration Acceleration ay ay a y

F

Acceleration Acceleration Acceleration ax ax ax

Time tTime t Time tTime t Time tTime t

Position Position Position y y y

D

Acceleration Acceleration Acceleration ay ay a y

Velocity Velocity Velocity vx vx vx

E

Time tTime t Time tTime t Time tTime t

Acceleration Acceleration Acceleration ax ax ax

C

B

Position Position Position y y y

Velocity Velocity Velocity vy vy vy Position Position Position x x x

A

Velocity Velocity Velocity vx vx vx

Velocity Velocity Velocity vy vy vy Position Position Position x x x

1. A marble rolls along a table top and then alls through a vacuum until it hits the ground Which o the ollowing graphs accurately illustrates the motion o the marble?

Time tTime t Time tTime t Time tTime t

Time tTime t Time tTime t Time tTime t

Time tTime t Time tTime t Time tTime t

2. A UN airplane carrying emergency supplies fies horizontally at 920 m/s and drops a supply package rom the hold at a height o 1950 m a) What will be the duration o the package’s all? b) What distance will the package cover horizontally? c) What will the velocity’s horizontal and vertical components be when the package hits the ground? 3. A pebble thrown horizontally rom the top o a building takes 756 s to reach the street How high is the building?

5. Serving, a tennis player hits the ball at a height o 250 m The ball leaves the racket horizontally at 160 km/h a) At what distance rom the player will the ball touch the ground? b) Given that the serve ball must touch the ground at a distance o 1189 to 1829 m ater passing over a net approximately 1 m high, what can you conclude about the direction o the ball’s initial velocity in reality? 6. A metal ball is thrown horizontally at 30 m/s rom a 15-m height What distance does it travel beore hitting the ground? 7. A diver dives horizontally rom a cli top Her initial velocity is 425 m/s The top o the cli is 200 m above the ocean surace Calculate: a) the time it takes or the diver to reach the water b) the horizontal displacement o the diver c) the magnitude o the driver’s velocity just beore she touches the water 8. A diver runs along a diving board and jumps into the pool without pushing o vertically He travels a horizontal distance o 25 m beore reaching the water I the diving board is 3 m above the water, how ast was the diver running along the diving board? 9. An eagle fying horizontally at a velocity o 30 km/h and at a height o 100 m above the ground is gripping a dead mouse with its talons The eagle is attacked by another eagle and drops the mouse a) How long will it take the mouse to hit the ground? b) What is the direction o the velocity vector o the mouse when it hits the ground?

4. A ball thrown horizontally rom a 30-m height hits the ground ater covering a 40-m horizontal distance Calculate: a) the duration o the all b) the initial velocity

ChaptER 11 The Motion of Projectiles

251

11.3 the moion of objecs lunched  n ngle An object launched at an angle begins its motion with a velocity vector that is above or below the horizontal. This is the case, or instance, when a ball is kicked in the air in a soccer match, or a stream o water pours rom a down­ ward­pointing pipe (see Figure 7). Analyzing the motion o objects launched at an angle is similar to ana­ lyzing the motion o objects launched horizontally. The only dierence is that the vertical component o the initial velocity is not zero: v iy ≠ 0. The object can be launched with a vertical velocity vector directed upward (viy > 0) or downward (viy < 0). Inormation on the initial velocity vector is oten provided in polar coordinates: vi and θi. For example, a soccer ball is kicked with a velocity o 20 m/s at a 25° angle above the horizontal. To integrate these coordinates into the equations, components vix and viy must be calculated frst (see Figure 8). I θi is expressed relative to the horizontal x­axis, these components are as ollows:

Figure 7 In a shower jet, each o the water molecules is a projectile whose initial velocity vector is directed below the horizontal See the polr coordine sysem, p 158

vix 5 vi cos θi and viy 5 vi sin θi Questions on the motion o objects launched at an angle are similar to the ones asked in the previous section: How much time does an object spend in the air (ight time)? Where will it touch the ground? At what velocity? I the object is launched upward, however, new questions arise: What is the maxi­ mum height attained? How long does it take to get there?

y

vi v iy θi v ix

Figure 8 The components o an initial velocity vector

x

Exmple a A soccer player kicks the ball with a velocity o 200 m/s at 250° above the horizontal a) How long does it take or the ball to reach its maximum height? b) What is the ball’s fight time? c) What distance does the ball travel horizontally? d) What is the value o the ball’s velocity at the moment it touches the ground? What is the direction o the velocity vector at the moment it touches the ground? Solution: From vi 5 200 m/s and θi 5 250°, we can determine that: vix 5 vi cos θi 5 200 m/s 3 cos 250° 5 181 m/s viy 5 vi sin θi 5 200 m/s 3 sin 250° 5 845 m/s 1 For question a), the nal time corresponds to the moment the ball passes the peak o its trajectory At this moment, the velocity vector is horizontal (see Figure 5 on page 247 ) Algebraically, we translate this by writing vy 5 0 Data: ti 5 0 s

xi 5 0 m

yi 5 0 m

vix 5 181 m/s

viy 5 845 m/s

ax 5 0 m/s2

t 5 ?

x 5 ?

y 5 ?

vx 5 181 m/s

vy 5 0 m/s

ay 5 -980 m/s2

Calculation: v 2 v iy -845 m/s 0 m/s 2 845 m/s a) vy 5 viy 1 ay ∆t thereore ∆t 5 y 5 5 5 0862 s 2 ay -980 m/s2 -980 m/s

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2 For questions b), c) and d) , the fnal time corresponds to the moment the ball is about to come into contact with the ground: y  5 0 Data: ti 5 0 s

xi 5 0 m

yi 5 0 m

vix 5 181 m/s

viy 5 845 m/s

ax 5 0 m/s2

t 5 ?

x 5 ?

y 5 0 m

vx 5 181 m/s

vy 5 ?

ay 5 -980 m/s2

Calculation: 1 1 1 b) y 5 yi 1 viy ∆t 1 ay (∆t) 2 thereore 0 5 0 1 viy ∆t 1 ay (∆t) 2 rom which -viy ∆t 5 ay (∆t) 2 2 2 2 One o the solutions is ∆t 5 0: it is the frst moment or which yi 5 yf -2v iy -2 3 845 m/s The other solution is ∆t 5 and t  5 a 5 5 1724 s -980 m/s2 y Note: The descent thereore takes 172 2 086 5 086 s c) x  5 xi 1 vix ∆t 5 0 1 (181 m/s 3 172 s) 5 312 m d) vy 5 viy 1 ay ∆t 5 845 m/s 1 (-980 m/s2 3 172 s) 5 -845 m/s v 5 vx2 1 vy2 5 (181 m/s)2 1 (-845 m/s)2 5 200 m/s tan α 5

|v y| 845 m/s 5 5 0467 thereore α 5 250° and θ 5 360° 2 250° 5 335° 181 m/s |v x|

The results o Example A demonstrate the trajectory’s symmetry. The dura­ tion o the rise (0.86 s) is equal to the duration o the descent (0.86 s). The magnitude o the ball’s velocity as it returns to the ground is the same as at the beginning, except that the velocity vector is directed 25° below the horizontal instead o 25° above. This symmetry is due to the act that the trajectory is shaped like a parabola. The peak o the parabola corresponds to the peak o the trajectory. Galileo was the frst to demonstrate that with negligible air resistance the trajectory o a projectile is parabolic. Exmle B From a roo 500 m high, a ball is thrown with a velocity o 500 m/s at 250° above the horizontal a) How long does it take or the ball to reach the ground? b) What is the magnitude o the ball’s velocity at the moment it touches the ground? What is the direction o the velocity vector at this moment? c) What is the maximum height reached by the ball? Solution: From vi 5 500 m/s and θi 5 250°, we can determine that: vix 5 vi cos θi 5 500 m/s 3 cos 250° 5 453 m/s viy 5 vi sin θi 5 500 m/s 3 sin 250° 5 211 m/s 1 For questions a) and b), the fnal time corresponds to the instant the ball is about to come into contact with the ground: y  5 0 Data: ti 5 0 s

xi 5 0 m

yi 5 500 m

vix 5 453 m/s

viy 5 211 m/s

ax 5 0 m/s2

t 5 ?

x 5 ?

y 5 0 m

vx 5 453 m/s

vy 5 ?

ay 5 -980 m/s2

ChaptER 11 The Motion o Projectiles

253

appENDIx 5 Solving second-degree equions, p 401

Calculation: 1 a) y  5 yi 1 viy ∆t 1 ay ∆t 2 thereore 0 m 5 500 m 1 211 m/s (t ) 2 490 m/s (t 2) 2 This is a second-degree equation o the orm ax2 1 bx 1 c 5 0, except that the variable is t  instead o x By comparing ax2 1 bx 1 c 5 0 à -49t 2 1 2,11t  1 50 5 0, we see that a 5 -4,90, b 5 211 and c 5 500 The solution is thereore: -b ± b 2 2 4ac -211 ± 2112 2 4(-49)50 5 5 -299 or 342 s 2a 2(-49) The elapsed time is necessarily positive, thereore t  5 342 s t 5

b) vy 5 viy 1 ay ∆t 5 211 m/s 1 (-98 m/s2 3 342 s) 5 -314 m/s v 5 vx2 1 vy2 5 (453 m/s)2 1 (-314 m/s)2 5 317 m/s |v y| 314 m/s 5 5 6,93 thereore α 5 818° et θ 5 360° 2 818° 5 278° |v x| 453 m/s 2 For question c), the fnal time corresponds to the instant the ball reaches its highest point tan α 5

ti 5 0 s

xi 5 0 m

yi 5 500 m

vix 5 453 m/s

viy 5 211 m/s

ax 5 0 m/s2

t 5 ?

x 5 ?

y 5 ?

vx 5 453 m/s

vy 5 0 m/s

ay 5 -980 m/s2

Calculation: v 2 v iy c) vy2 5 viy2 1 2ay∆y thereore y 2 y i 5 y 2a y and y  5

(0 m/s)2 2 (211 m/s)2 1 50 m 5 502 m 2 3 (-980 m/s2)

The results obtained above can be compared to the ones in Example B in Section 11.2 (see page 250). The fight time, at 3.42 s instead o 3.19 s, is longer because the ball rises beore alling back down. The velocity o the ball when it reaches the ground, 31.7 m/s, is the same as when it was launched horizon­ tally. This result can be inerred or any launch angle.

Ski juming Ski jumping is a spectacular Olympic sport Beore jumping, skiers accelerate down an inrun angled at some 35° beore reaching an approximate speed o 25 m/s, or nearly 90 km/h The take-o area o the ramp is angled at approximately 8° below the horizontal At the ramp exit, the skiers propel themselves upward This creates a situation in which the athletes’ initial velocity vector is horizontal, or even directed slightly upward To achieve the longest jump possible, skiers nearly lie on their skis, arms glued to their bodies, and position their skis in a V This position reduces air resistance and increases lit, a orce exerted by air that carries the skiers during their jumps By the time they touch ground, the skiers have travelled 100 m in the air, a distance measured rom the end o the ramp to the landing zone

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Figure 9 In this position, skiers reduce air resistance and increase lit, a orce exerted by air that carries the skiers during their jumps

11.3.1

Rnge

The range o a projectile is defned as the horizontal distance it travels between its launching point and its landing point. Calculating range can be done case by case, as in the example o the soccer player (see the example on page 252), with the kinematic equations or x and y. For launching and landing points situated at the same height, we can also use the general range equation:

Range 5 where

v i2 sin 2θi g

Range 5 Range expressed in metres (m) vi 5 Initial velocity, expressed in metres per second (m/s) θi 5 Departure angle relative to the horizontal, expressed in degrees g 5 Gravitational acceleration, expressed in metres per second squared (m/s2)

The general range equation derives rom the kinematic equations, as shown in the box below. For departure and arrival points of the same height, range corresponds to the horizontal position x f when y f 5 yi 1 According to equation (4), y f 5 yi 1 viy ∆t 1 ay (∆t ) 2, for yf 5 yi we obtain: 2 1 1 2 y f 2 yi 5 0 5 viy ∆t 1 ay (∆t ) and ay (∆t ) 2 5 -viy ∆t 2 2 -2v iy Therefore, ∆t 5 ay When we substitute this expression for ∆t in equation (1), x f 5 xi 1 vix ∆t, with xi 5 0 and ay 5 -g, we obtain: -2v iy 2v v Range 5 x f 5 0 1 vix 5 ix iy -g g

( )

Since vix 5 vi cos θi and viy 5 vi sin θi, we can write Range 5 x f 5

2v i2 cos θi sin θi g

According to a known trigonometric identity, sin 2θ 5 2sin θ cos θ, which means we can write: v 2 sin 2θi Range 5 i g

This equation gives the range or an initial velocity vi and a departure angle θi, on condition that the arrival and departure points are at the same height. We can see that or θi 5 0 or θi 5 90°, the range is zero. In the frst case, which cor­ responds to launching a projectile horizontally, the projectile never rises into the air but hits the ground immediately; in the second, the projectile rises and alls along the same vertical axis.

ChaptER 11 The Motion of Projectiles

255

11.3.2

table 1 Range as a function of angle, for vi 5 10 m/s

θ i (°)

Range (m)

0

0

10

35

20

66

30

88

40

100

45

102

50

100

60

88

70

66

80

35

90

0

Maximum range

The initial speed o a projectile oten depends on the characteristics o the thrower, be it a human being, a bow, a gun, or something else. A person will throw a ball at a maximum speed that depends on their muscles and throwing technique. A bullet leaves the barrel o a irearm at a speed that depends on the amount o powder in the cartridge, but the speed will be the same whether the weapon is directed 10° rom the horizontal, or 60°. However, it is worthwhile determining the angle that ensures that the projectile goes as ar as possible, i the arrival point is considered to be at the same height as the departure point. To do this, we use the range equation. We choose an initial velocity—10 m/s, or instance—and then calculate the range o any projectile or various values o θi: 10°, 20°, 30°, etc. (see Table 1). Table 1 shows that maximum range is obtained with a 45° angle (in the absence o air resistance). This is due to the presence o sin 2θi in the equation. The sine unction’s maximum value, equal to 1, is obtained with a 90° angle. The range is thereore at a maximum when 2θi 5 90°, in other words, when θi is at 45°. Galileo, once again, was the frst to demonstrate this result. Table 1 also indicates that or a given range, there are two possible values or θi (see Figure 10). One o these angles is smaller than 45°, and the other is larger than 45°, equidistant on either side o 45°.

θ  60°

θ  45°

θ  30°

Maximum range

Figure 10 At equal initial velocity, range varies as a function of launching angle The maximum range is obtained for an angle θi of 45° For two angles equidistant from 45° (here, 30° and 60°), the range is the same

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Furthering

your understanding

te effec f ir resisnce n rjeciles Because of air resistance, a projectile’s trajectory is not parabolic Air resistance reduces velocity’s horizontal and vertical components As a result, a projectile does not go as far or as high as when there is no resistance Because velocity’s horizontal component is constantly diminishing because of air resistance, the trajectory is no longer symmetrical The descent occurs over a shorter horizontal distance than the rise (see Figure 11) The projectile accelerates less during the descent, and the impact velocity is lower than the launch velocity Air resistance creates a situation in which maximum range no longer corresponds to a 45° inclination of the initial velocity vector with respect to the horizontal The inclination that produces the maximum range depends on the initial velocity as well as

SECtIoN 11.3

the frontal surface and shape of the projectile For example, for a baseball to go as far as possible, it must be hit at an angle of approximately 35° above the horizontal y

Trajectory in a vacuum Trajectory in air x

Figure 11 Air resistance alters the shape of the projectile’s trajectory

te min f bjecs lunced  n ngle

1. In a flm scene, a motorcycle travels at 94 km/h through a feld with an ascending slope, arrives at the edge o a cli and alls o The motorcycle takes 55 seconds to hit the ground at the bottom o the cli The slope o the feld is 20° a) What is the height o the cli? b) I the cli is vertical, at what distance rom its oot will the motorcycle crash? 2. From the roo o a building 500 m high, a ball is thrown with a velocity o 500 m/s at 250° below the horizontal a) How long does the ball take to hit the ground? b) What is the magnitude o the ball’s velocity at the moment it touches the ground? At that moment, what is the direction o the velocity vector? c) At what distance rom the building will the ball hit the ground? 3. Ater a snow storm, Juliet steps out onto a narrow balcony and, rom a point 70 m above the ground,

throws a snowball at Romeo at an angle o 20° above the horizontal Down below, Romeo catches the ball 13 seconds later when it is 10 m above the ground a) What was the magnitude o the snowball’s initial velocity? b) At what distance horizontally is Romeo rom the balcony? 4. A projectile thrown at a 40° angle reaches a certain point located at the same height as the departure point At what other angle could the same projectile be thrown, with the same initial velocity, to end up at the same point? 5. What is the maximum range o a baseball thrown at 350 m/s? 6. A soccer player kicks a ball at 180 m/s The ball lands 260 m away on a horizontal feld At what angles relative to the horizontal could the ball have been kicked?

ChaptER 11 The Motion of Projectiles

257

APPLICATIONS gof In gol, good players can send a ball fying at over 200 km/h. The launching angle depends on where the ball is in relation to the player’s eet, but above all, on the direction o the ront ace o the gol club head. The clubs known as drivers, which are longer than their counterparts, irons, are used or drives. At one time, club heads were made o wood; today, drivers are made rom composite materials. The lot, which is the angle between the vertical and the ront ace o the club head (or a club held vertically), ranges rom 8° to 30° or drivers and 16° to 50° or irons (see Figure 12). The longest shots are made with clubs with a small lot. Clubs with larger lots are used or shorter, but higher, shots—when the ball must be hit out o a sand trap or instance, or over an obstacle.

concavities altered airfow around the ball and re­ duced air resistance. Maximum range, which exceeds 200 m or the best players, is obtained with departure angles o approxi­ mately 10° (see Figure 13). Without rotation, the depar­ ture angle producing the maximum range would be close to 25° (this value is not equal to 45° because o air resistance).

Because o its angle, the club head rotates the ball at the moment o contact at a requency o up to 6000 turns per minute. In gol, a backspin makes the top o the ball turn backwards, which creates a situation in which air exerts an upward orce that lits the ball. The ball stays in the air longer and goes arther than it would without rotating. In the 19th century, balls that were not smooth were discovered to travel arther, and gol balls covered with tiny dimples began to be manuactured. These

lof

lof

90°

Driver

Figure 12 Loft of a driver and an iron

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90°

Iron

90°

Figure 13 Maximum range is obtained with departure angles of approximately 10°

te science of rojeciles It has taken several centuries to understand what ac­ tors determine the path o projectiles, despite the act that they have been used since the dawn o time. As in other areas o physics, it took time to move beyond impressions and build real scientifc theory. The area o physics that studies projectiles is sometimes called ballistics. The weapons known as ranged weapons (bows, cata­ pults, trebuchets), already used in antiquity, multi­ plied in the Middle Ages. Aiming was rarely precise and was based on experience. In the 14th century, Jean Buridan (1292–1363), a French scientist, pos­ ited that a projectile set in motion by a orce gained a momentum (impetus) that caused the motion to continue. Air resistance gradually reduced the impe­ tus. Buridan also believed that a projectile’s motion travelled in a straight line at the start. In Europe, the frst cannons appeared in the 14th cen­ tury. For many years, gunners used the straight shot, in other words, fring in a nearly straight line because they did not know how to point the cannon to reach a tar­ get otherwise. In the 16th century, the Italian scientist Tartaglia (1499–1557) confrmed that a projectile’s tra­ jectory is curved at any point and determined, without proving it, that the angle producing the largest range was approximately 45°. This was a relatively accurate

assertion at the time because cannonballs were signif­ cantly slower than today’s projectiles, which consider­ ably reduced the eect o air resistance. In his book, Discourses and Mathematical Demon­ strations Relating to Two New Sciences, published in 1638, Galileo demonstrated that in the absence o air resistance, a projectile’s trajectory was parabolic and the maximum range was obtained with a 45° angle. It was not until the 18th century that air resistance was taken into account, by English scientist Benjamin Robins (1707–1751). Little by little, the theory o pro­ jectiles was developed and the practice o artillery improved. Wind velocity, the Earth’s rotation and the dierence in altitude between departure points and arrival points began to be taken into account. By the end o the 19th century, gunners had learned to use indirect fre; in other words, they could point a cannon toward a target without seeing it, using inormation conveyed by observers or terrestrial or aerial devices. The progress o ballistics is taking more peaceul turns today; the science’s application in video games is an example. In some video games or ball sports (gol, soccer, etc.), players control velocity and departure angle and are required to take wind into account. The sotware integrated into these games uses the kine­ matic equations to determine projectile trajectories.

poo hR à venir

Figure 14 A catapult in action in the Middle Ages

Figure 15 An up-to-date example of applications for the kinematic equations: video games

ChaptER 11 The Motion of Projectiles

259

chapter

11

t Moion of pojils

11.1 Describing he moion of projeciles • The motion o projectiles reers to a situation in which an object launched into the air alls to the ground under the sole inuence o terrestrial gravitational orce, describing a curved trajectory in the process. • To describe the motion o a projectile, a coordinate system with a horizontal x­axis and a vertical y­axis is used. Generally, the origin o the x­axis is fxed vertically on the projectile’s departure point. The origin o the y­axis is the ground. • Vertical and horizontal motions are independent o each other. Horizontal motion, in the x­direction, is uniorm rectilinear motion. Horizontal acceleration is zero (ax 5 0). In the vertical direction, y, an object is subject to constant acceleration, one directed downward and equal in magnitude to g (ay 5 ­g). The vertical motion is uniormly accelerated rectilinear motion. • The variables that must be taken into account are time (t), horizontal position (x), vertical position (y), horizontal velocity (vx), vertical velocity (vy) and vertical acceleration (ay). • The basic equations describing projectile motion are shown in the table opposite.

along he x-xis

x f 5 xi 1 vix ∆t

• In a diagram o trajectory, the instantaneous velocity vector v is tangent to the trajectory at any point.

along he y-xis vfy 5 viy 1 ay ∆t 1 y f 5 yi 1 (viy 1 vfy) ∆t 2 1 y f 5 yi 1 viy ∆t 1 ay (∆t) 2 2 vfy2 5 viy2 1 2ay ∆t

11.2 the moion of objecs lunched horizonlly • For a projectile launched horizontally, the initial velocity vector by horizontal and viy 5 0.

11.3 the moion of objecs lunched  n ngle • Inormation on the initial velocity vector is oten provided by polar coordinates: vi and θi. The components must thereore be calculated: vix 5 vi cos θi and viy 5 vi sin θi. • A trajectory is symmetrical and shaped like a parabola (i air resistance is negligible). The duration o the rise is equal to the duration o the all. A projectile alls to the ground at the same magnitude o velocity it had when it let i the departure and arrival points are at the same height. • The range o a projectile is defned as the horizontal distance it travels between the departure point and the arrival point. For departure and arrival points that are at the same altitude, range is given by v i2 sin 2θi the equation: Range 5 g • Maximum range is obtained with a 45° angle (in the absence o air resistance).

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ChaptER 11

te Moion of projeciles

1. Using the variables dened in this chapter, how can we write: a) that a projectile reaches its maximum height? b) that a projectile is launched horizontally? c) that a projectile reaches the ground? d) that a projectile is travelling downward? e) that a projectile’s departure point is at the same height as its arrival point? f) that a projectile’s range is at the maximum? 2. Why must archers aim above the centre o the target? 3. a) I a projectile is thrown rom a point lower than its arrival point, at which point o its trajectory is the velocity’s magnitude at a maximum? At a minimum? b) I a projectile is thrown rom a point higher than its arrival point, at which point o its trajectory is the velocity’s magnitude at a maximum? At a minimum? 4. A stuntman must cross a 500-m wide canyon on a motorcycle On the canyon edge, he sets up a ramp at an incline o 120° Both sides o the canyon are the same height At what speed, in kilometres per hour, must the stuntman drive to just make it to the other side o the canyon? 5. A child throws a ball onto the roo o a house, then catches it 33° with a baseball glove 10 m above the ground, as shown in the gure 6.2 m opposite The ball leaves the roo at a velocity o 32 m/s a) How long will the ball stay in the air ater leaving the roo? b) What is the horizontal distance between the glove and the edge o the roo? c) What are the magnitude and direction o the ball’s velocity vector just beore touching the glove?

6. In pétanque, a shooter removes the opposing team’s ball by throwing her own at an initial velocity o 80 m/s, 200° above the horizontal The ball leaves her hand at a height o 10 m At what distance does it touch the ground? 7. The mouth o a hose is held horizontally above the ground, as shown opposite The stream o water makes a projectile motion Describe how, with a ruler and a calculator, you can determine the initial velocity o the water fowing rom the pipe 8. In a tennis serve, a ball leaves the racket at 140 km/h at an angle o 40° below the horizontal The racket hits the ball at a height o 24 m At what distance above the net will the ball pass i the net is 09 m high and 12 m away rom the person serving? 9. A ball is released rom the top o a peaked roo It rolls down 400 m, accelerating at a rate o 500 m/s2, and then alls o the edge The roo is sloped at 400° rom the horizontal, and the edge o the roo is 300 m rom the ground The ground consists o concrete fagstones 50 cm wide, except in one area, 150 to 200 m rom the wall, where there is a fower bed Determine where the ball will all to gauge the risks o it alling directly on the bed and damaging the fowers

40.0°

3.00 m

ChaptER 11 The Motion of Projectiles

261

262

CONTENTS ChapTER 12

Dierent Tyes o Forces . . . . . . . . .265 ChapTER 13

Bodies Subject to  Number o Forces . . . . . . . . . . . .287 ChapTER 14

Newton’s Lws . . . . . . . . . . . . . . . . . . .303

Dynmics is te study o te cuses o motion. Te questions rised by dynmics originte rom te descrition o motion introduced in kinemtics. Wy does  body move? Wy does it continue to move? Wy does it sto? Tese questions uzzled ilosoers nd scientists or centuries beore Glileo nd Newton fnlly rovided stisctory nswers to tem t te end o te 17t century. Tese investigtions re te ocus o dynmics,  word tt origintes rom te Greek dunamis, wic mens “orce.” Forces determine motion. Dynmics tereore studies te dierent tyes o orces, suc s grvittionl

orce nd riction, nd ten nlyzes teir eects on bodies. One o its most remrkble lictions is Newton’s lws, wic, mong oter tings, llows us to determine te ccelertion o  body subjected to vrious orces. Te concets o dynmics ve mny rcticl nd scientifc lictions. For exmle, tey el us to understnd vrious enomen, suc s te motion o te tides, te roulsion o veicles, nd te motion o edestrins. Tey lso el us to understnd ow  cr turns nd ow red blood cells cn be serted rom lsm. 263

12.1 The notion o orce 12.2 Gravitational orce ChapTER

12

DiFFERENT TYpES OF FORCES

12.3 Normal orce 12.4 Force o riction 12.5 Tension 12.6 Centripetal orce

uNiT

4

ChapTER

DYNamiCS

13

BODiES SuBjECT TO a NumBER OF FORCES

13.1 Free-body diagram 13.2 Resultant o several orces 13.3 Equilibrium

14.1 Newton’s frst law and inertia ChapTER

14

NEWTON’S LaWS

14.2 The notion o orce and Newton’s second law 14.3 Newton’s third law

264

Different Tyes of Forces

T

he chapters in the previous unit were about kinemat­ ics, the study o the various aspects o the motion o bodies, without taking into account the causes o motion. This is the frst chapter devoted to dynamics, the study o the causes o motion, or in other words, orces. Forces are implicated in many natural phenomena, such as the planets’ rotation around the sun or the tides. They also play a role in dierent aspects o everyday lie. A pedestrian walking on a sidewalk, an accelerat­ ing car, or a person scaling a rock ace are examples o situations in which various types o orces play a part.

Review Characteristics of force                                    13 Gravitational force                                        13 Relationship between mass and weight                       13 Force of friction                                          14

In this chapter, you will study fve examples o orces: gravitational orce, normal orce, the orce o riction, tension, and centripetal orce. You will also have the opportunity to become amiliar with the eects o these orces.

12.1 12.2 12.3 12.4 12.5 12.6

Te notion of force                    266 Grvittionl force                    267 Norml force                          273 Force of friction                       275 Tension                                278 Centrietl force                      279 ChapTER 12 Different Types of Forces

265

12.1 The noton of force A force is a vector quantity that acts on an object, causing it to experience a deformation or even change in motion. A orce may change the shape o an object. Deormation can be temporary, allowing the object to return to its initial shape. This is the case with a ball that bounces o the ground, or example. The ball deorms with each bounce and returns to its shape ater leaving the ground. Deormation can also be permanent. An object that breaks is an example o permanent deormation caused by a orce. See accelerton, p 224

A orce may also change the motion o an object. A change in motion may be an acceleration, a deceleration (negative acceleration), a starting motion i the object is initially at rest, a standstill i the fnal velocity is zero, or a change in the motion’s direction. The example o a car making on­road manoeuvres illustrates the various changes in motion. In order or the initially stationary car to take­o at a green light, a orce (that o the engine) must be exerted on it to initiate its motion. This orce must also continue to be exerted in order or the car to accelerate. The action o certain orces is required as well in order or the car to change direction (tire riction on the road). When there is a descending slope, the car will accelerate due to a orce (gravitational orce). Furthermore, a orce must intervene to slow down and eventually bring the car to a stop (the orce o braking riction). Acceleration implies that a orce is being exerted in the same direction as the motion. In the case o deceleration, the orce opposes the motion. I the orce is not applied in the same direction as the motion, it will change the motion’s direction. The unit o orce is the newton (N). We can express the newton using SI units.

a) A balance used in a grocery store

b) An X-ray of a bathroom scale

Figure 1 Two examples of dynamometers

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1N51

kg 3 m s2

Forces are usually measured with a dynamometer which is graduated in newtons. Some grocery stores make balances available to their customers to allow them to weigh the ruits and vegetables they are buying (see Figure 1a). This type o balance is a dynamometer, graduated in mass units and is used to measure the mass o objects (see A Brief History of... The spring scale p. 375). Another example o a dynamometer is the bathroom scale ound in many homes (see Figure 1b). Although both o these devices are reerred to by either the term “balance” or “scale,” in actual act they are dynamometers. Most o these devices unction by stretching or compressing one or many springs. The deormation o the springs allows or measuring the mass or weight o an object, as opposed to scales or balances which use masses that are compared with that o an object.

12.2 Grvittionl force The gravitational force (Fg ) is the mutual attraction between two bodies on account of their mass. Isaac Newton (1642–1727) ormulated the law o universal gravitation. This law states that bodies exert a orce o attraction on each other that depends on their mass. Newton was the frst to provide an explanation or the cause o the motion o planets. He could demonstrate that it was due to the orce o gravi­ tational attraction that the planets revolved around the sun in elliptical orbits.

12.2.1

Lw of universl grvittion

One o the most widely known scientiic anecdotes must surely be the one about Isaac Newton observing an apple all rom an apple tree. It is said that this event led him to establish the law o universal gravitation. This law states that two bodies are attracted to each other by a orce called gravitational orce. According to the law o universal gravitation, the apple alls toward the Earth because it is attracted to it, in the same way that the Earth alls toward the apple, since the Earth is attracted to the apple as well (even though its motion is imperceptible). The gravitational orce o attraction is proportional to the masses o the two bod­ ies. Its intensity is small and not really noticeable in objects with a low mass. However, this orce is large when exerted by objects o extreme mass, such as planets or stars. The gravitational orce o attraction is also inversely proportional to the distance (separating the two interacting bodies) squared. This means that the smaller the distance between two masses, the greater the gravitational orce o attraction, and that the greater the distance, the lesser the interaction. In mathematical terms, the law o universal gravitation is ormulated as ollows: Gm1m2 r2 N 3 m2 where G 5 Gravitational constant 5 667 3 10-11 kg2 m1 5 A body’s mass, expressed in kilograms (kg) m2 5 The other body’s mass, expressed in kilograms (kg) r 5 Distance between the two bodies, expressed in metres (m) Fg 5

The distance r is always measured rom the bodies’ centres (see Figure 2a). For example, or an object situated on the Earth’s surace, the distance would be approximately equal to the Earth’s radius (see Figure 2b).

r

r

a) The distance between the Earth and the Moon

b) A person situated on the Earth’s surface

Figure 2 The distance (r ) is measured from the bodies’ centres

ChapTER 12 Different Types of Forces

267

The ollowing examples show how the law o universal gravitation is used. They demonstrate why orce o gravity is considered negligible, except or objects o an extremely large mass. Exmple a What is the gravitational force exerted between a stationary, 700-kg bowling ball and a 16-kg bowling pin if they are 18 m apart? Data: m1 5 mball 5 7 00 kg m2 5 mpin 5 16 kg r 5 18 m Fg 5 ?

Solution: Gm1m2 Fg 5 r2

( 5

667 3 10-11

N 3 m2 kg2

) 3 (7 00 kg) 3 (16 kg)

(18 m)2

5 23 3 10-12 N Note: A force of 23 3 10-12 N is approximately equivalent to the weight of half a billionth of a water drop

Exmple B

What is the gravitational force between a 700-kg bowling ball and the Earth? The Earth’s mass is 598 3 1024 kg and its mean radius is 637 3 106 m Data: m1 5 mball 5 700 kg m2 5 mEarth 5 598 3 1024 kg r 5 rEarth 5 637 3 106 m Fg 5 ?

Solution: Gm1m2 Fg 5 r2 5

(

667 3 10-11

N 3 m2 kg2

) 3 (700kg) 3 (598 3 10

24 kg)

(637 3 106 m)2 5 688 N

12.2.2

See Free fll, p 235

Grvttonl ccelerton

Gravitational acceleration (g) is the constant acceleration experienced by an object in free fall close to the surface of the Earth or of another celestial body. In the 16th century, Galileo demonstrated in his experiments that objects in free fall close to the surace o the Earth had a constant accelera­ tion, known as gravitational acceleration. One o these experiments consisted in veriying whether two objects o dierent masses that all rom the same height reach the ground at the same time, in the absence o air resistance on the object. This result conficts with common sense, which suggests instead that the object with the greater mass should reach the ground rst. This is why Galileo’s experiment was important: it established that an object’s mass bears no infuence on the duration o its all. In other words, gravitational acceleration is the same or objects o dierent masses in the absence o air resistance. g 5 980 m/s2

The demonstration on the top o the ollowing page accounts or this phenomenon.

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Any body of mass m2 that falls to the surface of the Earth (m1) is subject to an acceleration of 980 m/s2 A demonstration of this follows If we apply the formula expressing the law of universal gravitation and use the Earth’s mass m1 and the Earth’s radius as the distance between m1 and m2, we obtain: Fg 5

Gm1m2 r2

(

667 3 10-11

N 3 m2 kg2

) 3 (598 3 10

24 kg)

3 m2

5 (637 3 106 m)2 5 980 N/kg 3 m25 980 m/s2 3 m2 In the above calculation, it is possible to replace the newtons per kilogram (N/kg) by metres per second squared (m/s2) since: kg 3 m s2 N 1 m kg 3 m Therefore: 5 3 5 2 kg kg s s2 Since metres per second squared (m/s2) are the units of acceleration, 980 m/s2 is the acceleration that will affect any body that falls to the surface of the Earth 1N51

When two bodies are attracted to one another by gravitational force, both are subject to a relative acceleration. This acceleration is the gravitational accel­ eration (g). Each of the two bodies experiences an acceleration that is directed toward the other body. An observer used to feeling the Earth’s gravity may also get the impression that just one of the two bodies is attracted by the other. And yet, this is not the case. The Earth’s mass is immense in relation to that of objects that can be attracted toward it; however, the Earth also moves, however imperceptibly, toward the object. In other words, we can say that the Earth and the object “fall” toward one another. Gravitational acceleration is not uniform everywhere on the planet. In fact, since the Earth isn’t perfectly spherical, its radius can vary according to its position on the globe and affect the value of g. Moreover, a body situated at a higher altitude has less gravitational acceleration. In this textbook we use 9.80 m/s2 as the value for all calculations involving the acceleration of an object in free fall toward Earth. The following formula allows for the calculation of the gravitational force exerted by the Earth on an object:

Fg 5 mg where Fg 5 Gravitational force, expressed in newtons (N) m 5 Mass of the object experiencing gravitational acceleration, expressed in kilograms (kg) g 5 Gravitational acceleration, expressed in m/s2

ChapTER 12 Different Types of Forces

269

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

The ollowing example demonstrates how to use the ormula to calculate the gravitational orce (Fg) exerted by the Earth. Exle Calculate the gravitational orce exerted by the Earth on a 75-kg person Data: g 5 980 m/s2 m 5 75 kg Fg 5 ?

jOhaNNES KEpLER

HISTORY HIGHLIGHTS HISTORY HIGHLIGHTS German astronomer, physicist and mathematician (1571–1630) Kepler is primarily known or having demonstrated three laws that describe the motion o plan ets, known as Kepler’s laws The frstHIGHLIGHTS law states that HISTORY HIGHLIGHTS HISTORY the planets ollow elliptical orbits and that the sun is situated at one ocus o the ellipse According to the second law, a planet travels aster close to the sun and more slowly when it is urther away The third law is a mathematical relationship between the revolving time and the dimension o the orbit Oten considered to be a orerunner o Newton, Kepler deduced rom these laws that a orce must exist that attracts the planets to the sun However, he never succeeded in discovering the origin o this orce

12.2.3

Solution: Fg 5 mg 5 75 kg 3 980 m/s2 5 735 N Fg 5 735 N downward

mss nd wegt

In everyday language, we sometimes have a tendency to conuse mass and weight. These two concepts are nevertheless very dierent rom the viewpoint o physics. The mass (m) o a body is the measure o the quantity o matter it comprises, that is to say the total mass it is composed o. The SI unit o mass is the kilo­ gram (kg). The instrument or measuring mass is the scale. Weight is the measure o the gravitational orce (Fg ) exerted on a body by a celestial body. Since this is a orce, weight is expressed in newtons (N). The instrument or measuring weight is the dynamometer. In contrast to the mass o an object, which does not vary according to the gravitational feld, the weight o an object can vary according to the celestial body the object is closest to. As such, a body does not have the same weight on Earth as on the Moon because the gravitational acceleration o attraction is not the same. The ollowing example illustrates the dierence between mass and weight. Exle I 1 mL o water has a mass o 1 g, what are the mass and weight values o a 675-mL bottle o water on Earth and on the Moon? (The mass o the container can be considered as negligible) The mass and radius o the Earth are 598 x 1024 kg and 637 x 106 m, respectively, while the mass and radius o the Moon are 735 x 1022 kg and 174 x 106 m Data: V 5 675 mL m5? mEarth 5 598 3 1024 kg rEarth 5 637 3 106 m mMoon 5 735 3 1022 kg rMoon 5 174 3 106 m Fg Earth 5 ? Fg Moon 5?

Solution: 1 Since 1 mL o water has a mass o 1 g, 675 mL o water has a mass (m ) 5 675 g 5 0675 kg on Earth as well as on the Moon 2 Fg Earth5 mg 5 0675 kg 3 980 m/s2 5 66 N G 3 mMoon 3 mwater 3 FgMoon5 r Moon2

( 5

667 3 10-11

5 11 N

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uNiT 4 Dynamics

N 3 m2 kg2

) 3 (735 3 10

(174 3 106 m)2

22 kg) 3

(0675 kg)

The example at the bottom of the previous page shows that the weight of an object can vary, whereas its mass remains constant.

Grvity ssist For almost fty years now, humans have been trying to conquer space In addition to orbital lights, space walks and the irst steps on the Moon, probes have been sent out to the other planets in the solar system in order to explore them In order to solve the problem o the consumption o limited uel reserves posed by the great travelling distances, astrophysicists made use o the orce o gravity in order to accelerate space probes This principle is called gravity assist It consists in precisely calculating the probe’s trajectory so it passes near a planet with a great mass—Jupiter, or example The planet’s very strong gravity attracts the probe, resulting in a changed trajectory and an increased velocity in relation to the Sun This technique was used during the missions o the Voyager 1 (see Figure 3 ) and Voyager 2 probes During the 1970s and 1980s, these space probes took the irst photographs o the

Furthering Furthering

your understanding Furthering your understanding

planets Jupiter, Saturn, Uranus and Neptune With the help o these probes, rings similar to those o Saturn were discovered circling other giant planets Furthermore, the probes made it possible to observe several moons in Figure 3 Space probe Voyager 1, orbit around these planets NASA scientists de - launched in 1977 velo ped this program at this particular time because they knew the planets would have a avourable alignment in the early 1980s These probes are still active and are gradually moving toward interstellar space These objects are the urthest man-made objects rom the Earth

your understanding

Tidl force nd te Roce limit Gravitational orce has various eects, some o which can be impressive The tidal orce and the Roche limit are two such examples

in shape generate a great quantity o heat And so Io ’s core is in usion and causes intense volcanic activity in spite o the very cold surace temperature (≈ -120 °C) at that distance rom the Sun

Gravitational orce does not act uniormly on the entire body on which it is exerted As such, the side o a body that is urthest The Roche limit is the minimal distance away rom a celestial body is less inlupossible between a planet o a given size enced by its gravitational orce than the and its natural satellite I an asteroid Figure 4 Io, volcanic moon o Jupiter closer side This change in the gravitationpasses too close to a planet, meaning al orce between two points o a body within the Roche limit, the tidal orce is known as tidal orce Tidal orce causes a deormation in the exerted by the planet on the asteroid will be greater than the body under its inluence This explains how the Moon subjects gravitational orce, which allows the asteroid to remain whole: the Earth to a deormation o its crust and magma, and above all and so the asteroid will break apart Furthermore, it may hapto a transormation o the shape o its oceans, commonly known pen that a large quantity o matter is situated too close to a as the tides planet and that the tidal orce is greater than the gravitational The tidal orce phenomenon occurs elsewhere in the solar system as well One o the moons o Jupiter, called Io, is subject to the tidal orces o Jupiter and its biggest satellites in such a way as to signifcantly deorm it (see Figure 4 ) These changes

orce, which would cause the matter to compress and orm a celestial body This explains the existence o the rings o Saturn, or example, which are made up o scattered matter revolving around the planet

ChapTER 12 Dierent Types o Forces

271

SECTiON 12.2

Gravtatonal force

1. Use the law of universal gravitation to calculate gravitational force in the following situations a) An apple weighing 150 g on the surface of the planet Mars (the mass of Mars is 637 3 1023 kg and its radius is 34 3 106 m) b) A 70 000-kg space shuttle and a 100-kg astronaut on a space walk 15 m from the centre of the shuttle c) A 77-kg man and a 56-kg woman standing 15 m from each other 2. In your own words, explain the difference between: a) weight and mass b) a balance and a dynamometer 3. What is the gravitational force acting on a 79-kg Olympic diver who leaps into the air? 4. A carabiner used for rock-climbing is capable of withstanding different forces according to its position (see fgure below ) Would the carabiner resist, regardless of its position, if a team of three climbers, with masses of 72 kg, 84 kg and 81 kg, were suspended from it?

9 kN

24 kN

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5. What is the gravitational acceleration on the surface of planet Venus (the mass of Venus is 483 3 1024 kg and its radius is 605 3 106 m)? 6. What is the acceleration of the following objects when they are in free fall, if air resistance is negligible? a) a parachutist weighing 68 kg b) a 50 mg rain drop c) a 45 tonne rock 7. A space probe has a weight of 650 N on the surface of the planet Mercury (the mass of Mercury is 328 x1023 kg and its radius is 244 x106 m) How much would the probe weigh on Pluto (the mass of Pluto is 13 X 1023 kg and its radius is 115 X 106)? 8. The European rocket Ariane has a mass of 750 metric tonnes What force must its engines produce in order for the rocket to lift off?

12.3 Norml force The normal force (FN ) is the resistant reaction o a body’s surace to a orce exerted by another body in contact. In mathematics and in physics, we say that a straight line is normal to a plane when it is perpendicular to this plane. A vector can also be normal to a plane (see Figure 5). I a surace is not plane, it is still possible or a vector to be nor­ mal to it. The vector will be perpendicular to the plane that is tangent to this surace at the point to which the vector is applied (see Figure 6).

Figure 5 A vector (blue) normal to plane A

Figure 6 A normal vector to a curved surface

The normal orce is a reaction orce. It is exerted on a body by a surace with which it is in contact. The normal orce is always per­ pendicular to the contact surace. I a body is in contact with another body without a orce being applied, the normal orce will be zero. I the body exerts a orce on the other body, the normal orce is o the same magnitude as this rst orce but in the opposite direction. As will be explained in Chapter 14, the nor­ mal orce ollows rom Newton’s third law. The normal orce prevents a person rom sink­ ing into the foor. The person’s weight (Fg) is a downward orce exerted on the foor. The nor­ mal orce is an upward orce that the loor exerts on the person. Given that the person does not have vertical motion, the two orce vectors are o the same magnitude but oppose each other: their eects cancel each other out (see Figure 7). The example on the ollowing page illustrates this phenomenon.

See Newton’s tird lw, p 315 Fg

See Subtrcting vectors p 193

FN

Figure 7 The normal force (FN ) is of the same magnitude as the person’s weight (Fg ) but it is exerted in the opposite direction Its direction is perpendicular to the contact surface of the two bodies

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273

Examle What is the normal force exerted by a table supporting a 130-kg box? Data: m 5 130 kg FN 5 ? Solution: Fg 5 mg 5 130 kg 3 980 m/s2 5 127 3 102 N Fg 5 127 3 102 N downward Since FN 5 -Fg FN 5 127 3 102 N upward The normal force is the result of a body’s cohesion. A body’s internal cohesion is all of the chemical bonds linking its atoms and molecules. The normal force is a body’s resistance to deformation.

SECTiON 12.3

Normal force

1. What is the magnitude and direction of the normal force: a) exerted by the ground on the unit formed by a 60-kg woman and her 8600-g bicycle? b) exerted by a telephone pole on a car that hit the pole with a force of 4500 N toward the front? c) exerted by the side of a mini-golf course on a ball that hits the side with a force of 4 N (see fgure below)

2. What is the normal force exerted by a pickup truck’s cargo container if it contains a mass of 1750 kg? 3. If a force of 38 N were applied at point P of a sphere with a centre C (see fgure below ), what would the normal force be?

38 N p

25° 150° 4N

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uNiT 4 Dynamics

C

12.4 Force o riction The orce o riction (Ff ) is a orce that opposes motion. The direction o this orce is usually opposite to the motion. I a orce is applied to a stationary object in order to set it in motion, the orce o riction is in the opposite direction to the motion one tries to convey to the object.

12.4.1

Te nture o riction

Friction is a contact orce. I one body is in con­ tact with another solid body, there necessarily will be riction (see Figure 8). The particles (molecules, atoms, electrons) at the surace o the two bodies interact through electromagnetic orces. Particles are attracted to each other, which opposes motion. The attraction can cause certain particles to be torn out rom one solid by the other. Friction also occurs in luids*and is caused by the same electromagnetic orces. The riction attributed to a gas is o a lesser magnitude than that exerted by a liquid, as ewer particles by unit area are involved.

12.4.2

Ff

Figure 8 The rictional orce (Ff ) opposes the motion o the box and at the same time acts in a direction parallel contact surace A substance lacking a denite * Fluid shape Gases and liquids are fuids

Te coefcient o riction

When there is riction between two solid bodies, two situations can occur. First, the two bodies can be stationary relative to each other. In this case, we speak o starting riction or static riction. Static riction is a orce that pre­ vents sliding. The riction between the foor and the soles o a person’s shoes illustrated in Figure 8 is an example o static riction. Second, a body can slide alongside another. This is known as kinetic riction. The riction between the foor and the box illustrated in Figure 8 is an example o kinetic riction. Friction depends on the nature o the contact suraces (materials), the roughness o the contact suraces and the state o the contact suraces (dry, lubricated). It is independent o the contact pressure, the shape o the con­ tact suraces, the area o the contact suraces and the velocity o sliding. Friction depends on two actors. First, it is proportional to the normal orce (F N), which is exerted perpendicular to the suraces o the bodies in contact and ensues rom the bodies’ resistance to deormation. The greater this orce, the greater the riction. Conversely, the smaller this orce, the smaller the riction. Second, riction depends on the nature o the substances in contact. The abil­ ity o a substance to create riction depends on the substance with which it is in contact. The orce o riction between two given substances is proportional to a coefcient o riction (μ). This coecient is a ratio between the orce o ric­ tion and the normal orce established experimentally, and it depends solely on the two substances in contact. Since it is a ratio, the coecient o riction does not have any units. There are two types o this coecient: the coecient o kinetic riction and the coecient o static riction. The coecient o kinetic riction makes it possible to calculate the value o the orce o riction between two bodies sliding along one another. The coecient o static riction, however, allows one to evaluate the maximum value that riction can attain beore two

ChapTER 12 Dierent Types o Forces

275

bodies begin to slide along one another (see Table 1). Thus, if we try to displace an object at rest on a surface, we have to increase the applied force until it sets into motion. The static friction is at its maximum the instant just before the object sets into motion. The coefficient of static friction is always greater than the coefficient of kinetic friction (µs > µk). This is because electromagnetic links form more readily between objects that adhere to one another without sliding. Finally, it should be noted that friction does not depend on the contacting surface areas at all. The following formulas allow for the calculation of the forces of friction: Ff (static) ≤ µsFN Ff (kinetic)  µkFN where Ff  Frictional force, expressed in newtons (N) FN  Normal force, expressed in newtons (N) µs  Coefficient of static friction µk  Coefficient of kinetic friction

Table 1 Approximate coefficient of friction of different substances Coefficient of static friction (µs)

Coefficient of kinetic friction (µk)

Copper on copper

1.6

1.0

Steel on dry steel

0.41

0.38

Steel on lubricated steel

0.15

0.09

Oak on dry oak

0.5

0.3

Rubber on dry asphalt

1.2

0.8

Rubber on wet asphalt

0.6

0.5

Rubber on dry concrete

1.0

0.7

Rubber on wet concrete

0.7

0.5

0.006

0.005

Substance in contact

Rubber on ice

The following example shows how to calculate the force of friction. Example What is the force of friction Ff exerted when a 1500-kg car skids on a wet asphalt road? Data: m  1500 kg k  0.5 Ff  ?

Solution: Fg  mg  1500 kg  9.80 m/s2  1.47  104 N Since FN  -Fg FN  1.47  104 N Ff  µkFN  0.5  1.47  104 N  7.35  103 N Ff  7.35  103 N

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Elementry interctions In foundational physics, forces are called interactions. An inter­ action is defined as the mutual and reciprocal action between two objects. In the 20th century, physicists determined four ele­ mentary interactions that make it possible to explain all physi­ cal phenomena. They are: gravitational force, electromagnetic force, weak nuclear force, and strong nuclear force. Gravitational force is the mutual attraction of all objects that have mass. It explains the interactions between objects of great mass, such as planets, stars and galaxies. Electromagnetic force is the attrac­ tion or repulsion of charged particles. In addition to electricity and magnetism, this interaction accounts for the links between atoms, chemical phenomena and even the existence of light. Weak nuclear force is involved in nuclear reactions (see Figure 9 ). Strong nuclear force is the interaction between heavy subatom­ ic particles, such as protons and neutrons; it explains the cohe­ sion of atomic nuclei.

SECTION 12.4

Figure 9 Particles interact and disintegrate during nuclear reactions.

Force of friction

1. Determine the force of friction described in each situation. a) A man applies a force of 400 N to push a concrete block to the left. He does not succeed in moving it. b) A teenager drops a 2.35-g penny from the top of the CN Tower in Toronto. The penny falls at a constant velocity. 2. Calculate the maximum force that can be applied to the following objects without causing them to move. a) a 345-g wooden (oak) sculpture resting on a piece of oak furniture b) a 125-kg concrete block on the rubber mat of a conveyor belt c) a 0.8-g ball in a steel ball bearing

3. Explain why the force of friction depends on the normal force. 4. A rubber hockey puck weighs 160 g. a) What is the value of the force of friction when the puck slides across the ice? b) When the puck is stopped, what is the minimum force required to put the puck back in motion? 5. What is the minimum force required to move a 1200kg car on an ice-covered surface? 6. In a strong-man competition, contestants pull tractor tires across a concrete surface. The largest tire ever pulled had a mass of 415 kg. However, this record was set in the rain. What would the maximum mass of the tire have been in good weather?

ChapTER 12 Different Types of Forces

277

12.5 Tenson A force that acts by pulling * Tracton a body

Tension (FT ) is the traction* that a cable (or any other long and thin object) exerts on a body. The orce exerted in a cable when we pull at both o its extremities in oppo­ site directions is an example o tension (see Figure 10). Many dynamics problems involve systems o masses suspended by cables. In order to clearly understand these situations, an understanding o the notion o tension is essential. To simpliy its analysis, it is necessary to establish some approximations. As such, we can generally assume that: • the mass o the cable is negligible • the width o the cable is negligible • the cable does not stretch It should be noted that when a cable passes through a pulley, the only eect produced is the change in the tension’s direction and not in its magnitude, as long as the pulleys are fxed (see Figure 11).

Figure 10 The force present in the cable is tension

The ollowing example shows that in order to calculate the tension, the opposing orce must be known. FT FT

FT

Figure 11 Tension is applied to a cable

Example A 750-g mass is lifted using a cable suspended on a frictionless pulley What is the tension in the cable? Data: m 5 0750 kg FT 5 ?

in order to lift a mass Pulleys only change the tension’s (FT ) direction and not its magnitude

SECTiON 12.5

Solution: Fg 5 mg 5 0750 kg 3 980 m/s2 5 735 N Since FT 5 Fg FT 5 735 N FT 5 735 N at 230˚

Tenson

1. An 8-kg mass rests on the foor A rope is attached to this mass and an upward tension o 50 N is applied What is the normal orce applied by the foor?

tension in the rope

2. In situations in which tensions are analyzed, it is generally assumed that a rope’s mass is negligible I we take into account the mass o a rope placed on a table, as in the opposite image, what happens to the

a way that the hang-

278

uNiT 4 Dynamics

when the rope is gradually moved in such ing part becomes increasingly long? Explain your answer

FT 5 ?

50°

12.6 Centrietl orce For an object moving in a circular pattern, a lateral force directed toward the centre of the circle must impact its motion. This force can be any of the forces discussed earlier in this chapter. In the context of circular motion, the force is centripetal.

12.6.1

Secifc circlr motion terminology

All bodies have a tendency to preserve their state of motion. For a body to turn, a force has to be exerted on it and change the direction of its motion. This force is centripetal force. The centripetal force (Fc ) maintains a body in a circular motion. To get a body to move in a circular fashion, the centripetal force has to be directed toward the centre of rotation (see Figure 12). The word “centripetal” comes from a combination of two Latin words: centrum, meaning “centre,” and petere, meaning “tend toward.” The word centripetal thus means “tending toward the centre.” The example of a car travelling at a constant velocity and making a turn illustrates the action of centripetal force. The car has a tendency to preserve its state of uniform rectilinear motion. The reason it can turn right and maintain a circular motion is due to the action of a force directed toward the centre of rotation. This force is the centripetal force. While the car is turning, the passengers also have a tendency to continue their rectilinear trajectory. They therefore have the impression of being thrown to the left (see Figure 13).

v a Fc C

Figure 12 A person spinning a slingshot above his head What we call the centripetal force (Fc ) is the force directed toward the centre of rotation C It allows body A to have a circular trajectory by constantly changing the direction of velocity (v )

Fc

Fc

Figure 13 Due to inertia, any body has a tendency to preserve its state of motion The car turns to the right and maintains a circular motion due to the action of the force directed toward the centre of rotation: centripetal force

12.6.2

uniorm circlr motion

A body that follows a circular trajectory and turns with a constant magnitude of velocity performs a uniform circular motion. Here, we cannot speak of con­ stant velocity, since velocity is a vector. The magnitude of velocity is constant in a uniform circular motion, but its direction constantly changes. The velocity vector is always tangent to the circle traced by the trajectory and perpendicu­ lar to the radius of curvature (see Figure 14). As such, it constantly changes direction. Therefore, to change the direction of a motion, it is necessary to apply a force: centripetal force. If the centripetal force disappears, the body’s motion becomes rectilinear.

ChapTER 12 Different Types of Forces

279

v

The centripetal orce can take on dierent orms according to the situation at hand. A ew examples will sufce to show this. A slingshot composed o a leather strap is used to make a pebble turn in order to hurl it toward a target. As the pebble turns, its velocity vector constantly changes direction. The moment the strap is released, the pebble preserves its velocity and continues its motion in the direction o the tangent to the circular trajectory. In this situation, the cen­ tripetal orce is the tension that the leather strap applies to the pebble (see Figure 15a). This tension is directed toward the centre o rotation.

v

In the case o the Moon revolving around the Earth, the centripetal orce is the gravitational orce (see Figure 15b). Furthermore, i a person turns a bucket o water with his arm out straight in a horizontal circle, the normal orce exert­ ed by the bottom o the bucket on the water acts as the centripetal orce (see Figure 15c). Finally, when a bicycle turns, the riction between the tires and the ground constitutes the centripetal orce (see Figure 15d).

r v

r r

Figure 14 The velocity vector is always tangent to the circle and perpendicular to the radius of curvature

FT

12.6.3

Centrpetal force eqaton

The mathematical relationship allowing or the calculation o centripetal orce is: a) The tension in the strap of a slingshot

mv 2 r where m 5 Mass, expressed in kilograms (kg) v 5 Velocity, expressed in metres per second (m/s) r 5 Radius of the trajectory, expressed in metres (m) Fc 5

Fg

b) The gravitational force between the Earth and the Moon FN

c) The normal force exerted by the bottom of a bucket on the water contained in the bucket

Ff

d) The frictional force of a bicycle’s tires against the ground

Figure 15 The centripetal force can take on different form according to the situation at hand See Resltant of several forces, p 290

280

uNiT 4 Dynamics

According to this ormula, centripetal orce is proportional to the mass. This means that the greater the mass o a rotating body, the greater the orce required to maintain it in rotation. According to this ormula, the centripetal orce is proportional to the velocity squared, thus or a velocity that is twice as great, a centripetal orce our times as great is required. Finally, the cen­ tripetal orce is inversely proportional to the radius o the circle described by the body in motion. In act, or a small radius o curvature, the trajectory must be greatly deviated. The ollowing example shows how to use this or­ mula. Example A father spins his daughter around him while holding her by the hands The centre of gravity of the 350-kg girl is located 115 m from her father’s body At what velocity does the little girl spin if her father exerts a tension of 700 N? Data: m 5 350 kg r 5 115 m Fc 5 FT 5 700 N v5?

Solution: F 3r mv 2 Fc 5 ⇒ v2 5 c r m F 3r v5 c m v5

700 N 3 115 m 5 480 m/s 350 kg

Dierent orces may combine to assume the role o a centripetal orce. In some cases, the centripetal orce may be the resultant of several forces. For example, a roller coaster car that travels in a vertical loop undergoes a centripetal orce

that depends on two forces. The car is subjected to a gravitational force (Fg) and a normal force (FN) (see Figure 16). When the car is at its highest point, the direction of the two forces is the same, that is, downward: Fc 5 FN 1 Fg. The gravitational force is added to the normal force to curve the trajectory. The normal force is weaker, and there is less strain on the rails. When the car is at its lowest point, the normal force is directed upward while the gravitational force is directed downward; therefore: Fc 5 FN 2 Fg. At this point on the tra­ jectory, the normal force must compensate the gravitational force in addition to curving the trajectory. The normal force must be higher than at the top, and there is more strain on the rails.

FN

Fg

a) The gravitational orce has the same direction as the normal orce

Te Grnd unfed Teory Ater James Clerk Maxwell (1831–1879) succeeded in uniying the analysis o magnetic, electric and light phenomena, physicists tried to uniy all elementary interactions The Grand Unifed Theory is the search or a single interaction that would explain all physical phenomena in the universe So ar, this research has led to the concept o electroweak interaction, which uniies electromagnetic and weak nuclear orces

best hope is the Large Hadron Collider (see Figure 17) It is the most powerul particle accelerator ever built Spanning the border between France and Switzerland, it measures about 27 km in circumerence and cost several billion dollars It is considered the largest experimental device or validating scientifc theories ever built

SECTiON 12.6

Fg

b) The gravitational orce is opposite to the normal orce

Figure 16 A roller coaster car’s vertical circular loop

Today, research on uniication with strong nuclear orce continues and looks promising However, the problem lies in uniying all these orces with gravitational orce Even Albert Einstein (1879–1955), who worked on this problem or many years, ailed to solve it The intensity o the energy required or the hypotheses in question makes experiments complicated and costly At present, the

FN

Figure 17 One o the Large Hadron Collider’s many detectors

Centretl orce

1. On a horizontal section of track, an 81-kg skier makes a turn at a velocity of 50 km/h What is the magnitude of the centripetal force if the radius of curvature of his trajectory is 22 m? 2. A 40-kg child is swinging from a swing that has a chain measuring 32 m Her velocity is 80 m/s when she reaches the lowest point What is the value of the centripetal force at this point?

3. One of Saturn’s satellites travels in a circular orbit at a velocity of 19 481 m/s Saturn’s mass is 569 3 1026 kg What is the radius of the satellite’s orbit?

ChapTER 12 Dierent Types o Forces

281

APPLICATIONS Centrfgaton Centriugation is a technique that allows or the sepa­ ration o a liquid’s particles or o two fuids with dierent densities. In a way, it is an accelerated sedi­ mentation. The separation o red blood cells rom blood plasma and the extraction o solid matter rom wastewater are some o its applications. In biology, centriugation is useul or separating cell components (nuclei, mitochondria, etc.). A centriuge is a device that turns and subjects a mixture to rotational motion. The centripetal orce exerted on the particles comes rom the surrounding mixture. Beyond a certain rotational velocity, the available centripetal orce is insucient to maintain the particles in rotation: the particles thereore tend to move away rom the rotational axis. The mixture is generally placed in a perorated container that retains larger particles and lets smaller ones, such as the molecules o a liquid, pass through. A salad spinner is a centriuge: the sides o the turn­ ing perorated basket block the lettuce leaves and let water pass through. Following the same princi­ ple, a juice extractor lets you extract the juice rom ruits and vegetables. In the nuclear industry, centriugation makes it pos­ sible to obtain the enriched uranium required to op­ erate most nuclear reactors. In nature, the isotope 235U represents only 0.7% o all uranium atoms. However,

reactors require uranium containing approximately 4% o uranium­235. To obtain this grade, it is rst neces­ sary to produce gaseous uranium hexafuoride (UF6) rom the ore, using a series o chemical reactions. The gas is injected into very ast centriuges: the molecules containing uranium­238 tend to travel outward aster, and the gas becoming gradually enriched with ura­ nium­235 accumulates close to the rotational axis, rom where it can be removed with a pump. Centriugation takes its name rom the centriugal orce that appears to be exerted outward on a rotating object, such as on passengers in a vehicle making a turn. Although this orce is mentioned in common descriptions, it does not really exist. Centriugal orce is a way o describing the eects o inertia. For example, in the rotating drum o a washing machine the water rom the clothes appear to be subjected to a centriugal orce moving them toward its outer walls, but in act, it is the centripetal orce directed toward the centre o the drum that is not large enough to keep the water rotating (see Figure 18). Centripetal acceleration is expressed as a unction o g, i.e. the gravitational acceleration on Earth. Centripetal acceleration can reach 3 times the magni­ tude o g in the airground rides o an amusement park (see Figure 19), 30 times g in a salad spinner, and between 1000 times g and 300 000 times g in laboratory centriuges.

poto hR à venr

Figure 18 The rotating drum of a washing machine is a centrifuge

282

uNiT 4 Dynamics

Figure 19 Acceleration in the fairground rides of an amusement park can reach 3 times the magnitude of g

Figting friction Friction is present in many situations; or example, the riction between the cross hairs responsible or the cohesion o abrics. The riction between nails and wood ensures the solidity o houses. The operation o a vehicle’s brakes and the clutch o a car with manual transmission are based on riction. Nevertheless, we primarily note the harmul eects o riction on motion. Moving a stone block by pushing it along the ground requires enormous eort. To make this task easier, humans understood millennia ago that they could slide cylindrical logs underneath the block. This was an advance, but the necessity to constantly place logs in ront o the block made it impossible to move it quickly. It took the invention o the wheel and axle to eliminate this inconvenience and allow or the develop­ ment o carts and wagons in the 4th millennium BCE. Even though the riction between the axle and the hub o a wheel can be great, the wheel­axle system oers a clear advantage: while the wheel makes a turn and advances a distance that is equal to its circumerence, the riction

between the axle and the hub is only exerted on the hub’s circumerence, which is much smaller than that o the wheel. The energy required to fght riction is much less than i the object supported by the wheel were pushed directly along the ground (see Figure 20). It is possible to decrease the riction between the axle and the hub by lubricating the contact surace. Tallow was ound on the hubs o an Egyptian wagon used in 1400 BCE. Later, the number o lubricants increased, especially rom the industrial revolution onward. First animal and vegetable oils were used; or example, in the 1860s, salad oil was used in the bearings o bicycles. Subsequently, mineral and synthetic oils as well as solid lubricants (such as graphite) or others containing nanoparticles, appeared. The invention o ball bearings urther reduced the ric­ tion in rotating systems. Leonardo da Vinci (1452–1519), who sketched the frst ball bearings fve hundred year ago, did not put his plan into practise. (see Figure 21). The development o the bicycle in the latter decades o the 19th century led to the development o eective ball bearings.

l

a) Maximum friction

l

b) Decreased friction The load travels the same distance (l ), but friction is limited to the axle and hub, whose circumference is much smaller than that of the wheel The friction at the points of contact between the wheel and the ground is negligible

Figure 21 Ball bearings made of wood produced according to Leonardo da Vinci’s sketches

Figure 20 The wheel is a simple system that has proved to be effective in decreasing friction when moving heavy loads

ChapTER 12 Different Types of Forces

283

ChapTER

RE V i E W

12

Dfferent Tyes of Forces

12.1 Te noton of force • A force is a vector quantity that acts on an object, causing it to experience a deformation or even change in motion. • A change in motion can be an acceleration, a deceleration, a starting motion if the object is initially at rest, a standstill, or a change in the motion’s direction. • The unit of force is the newton. 1N51

• The instrument for measuring forces is the dynamometer.

kg 3 m s2

12.2 Grvttonl force • The gravitational force ( Fg ) is the mutual attraction between two bodies on account of their mass. • In mathematical terms, the law of universal gravitation is formulated as follows: Gm1m2 r2 N 3 m2 where G 5 Gravitational constant 5 667 3 10-11 kg2 Fg 5

• The distance (r) is always measured from the bodies’ centres. • The gravitational acceleration is the constant acceleration experienced by an object in free fall close to the surface of the Earth or to that of another celestial body. • We use g 5 9.80 m/s2 as the value for all calculations involving the acceleration of an object in free fall on Earth. • We calculate the gravitational force exerted by the Earth on an object using the formula opposite:

Fg 5 mg

• The mass (m) of an object is the measure of the quantity of matter it comprises. • The weight is the measure of the gravitational force (Fg) exerted on a body by a celestial body. It can vary according to the celestial body it is closest to.

12.3 Norml force • The normal force (FN ) is the resistant reaction of a body’s surface to a force exerted by another body through contact. • The normal force is always perpendicular to the contact surface between two bodies.

284

uNiT 4 Dynamics

12.4 Force of friction • Friction (Ff ) is a force that opposes motion. This force is usually exerted opposite to the motion or to the tendency of the motion. • Friction is a contact force. The particles (molecules, atoms, electrons) at the surface of the two bodies interact through electromagnetic forces, which oppose motion. • We refer to starting friction or static friction when two bodies do not slide along one another. • We say that friction is kinetic when one body slides alongside another body. • Kinetic friction and the maximum value of static friction are related to the normal force by the coefficient of friction (µ), a number with no units.. • The following formulas express the forces of friction: Ff (static) ≤ µsFN

Ff (kinetic)  µkFN

• The coefficient of static friction is always greater than the coefficient of kinetic friction (µs > µk).

12.5 Tension • Tension (FT ) is the traction that a cable (or any other long and thin object) exerts on a body. • It is generally assumed that the cable’s mass and the cable’s width are negligible and that the cable does not stretch. • When a cable passes through one or more fixed pulleys, the effect produced is a change in the tension’s direction, not its magnitude.

12.6 Centripetal force • The centripetal force (Fc) maintains a body in a circular motion. It is always directed toward the centre of rotation. • In a uniform circular motion, the magnitude of velocity is constant but its direction constantly changes. The velocity vector is always tangent to the circle traced by the trajectory and perpendicular to the radius of curvature. • The centripetal force can take on different forms according to the situation at hand. For example, it can be a tension, a gravitational force, a normal force or a frictional force. • The equation for centripetal force is: Fc 

mv 2 r

CHAPTER 12 Different Types of Forces

285

CHAPTER 12

Different Types of Forces

1. What is the minimum mass that could break a cara­ biner used or rock­climbing?

9 kN

24 kN

2. A 629­kg space probe passes 1500 km rom the surace o the planet Uranus. The mass o Uranus is 8.8 3 1025 kg and its radius is 2.71 3 107 m. What is the orce o attraction exerted by the planet on the probe? 3. While moving, a man tries to push a 230­kg piano by himsel along a horizontal oor. I he applies a horizontal orce o 845 N, will he be able to move it? (The piano has no wheels and both the piano and the oor are made o oak). 4. A horizontal orce o 150 N is applied to a copper block resting on a copper horizontal surace. What must the maximum mass be so that the orce is sufcient to move it? 5. What is the maximum mass o a rubber block that can be pulled by a rope whose maximum tension is 3500 N? The block is sitting on dry asphalt. 6. The 1968 Olympic Games were held in Mexico City, the capital o Mexico. Several records were set at those games. Some o them can be explained by the lower air density at high altitudes, given that Mexico is located at an altitude o 2200 m. What about the eect o gravity? Calculate the dierence in gravitational acceleration between sea level and Mexico City. Can this dierence explain the records that were set? 7. For the past 20 years, cars have been equipped with ABS brakes. This system prevents the wheels rom locking when the driver brakes. Explain why

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it is important to prevent the wheels rom lock­ ing, taking static riction and kinetic riction into account. 8. Roads are sometimes constructed rom concrete rather than asphalt so that they will last longer. What impact does this have on saety? For these two suraces, compare the orces o riction on a 1250­kg car when it brakes: a) i the road is wet b) i the road is dry c) i the road is dry but the wheels lock, which causes the car to slide 9. NASA wants to build a centriuge to train its astro­ nauts to handle the strong accelerations they will experience when the space shuttle lits o. A composite material capable o withstanding a nor­ mal orce o 120 000 N without rupturing is used to manuacture the base o the cabin on which the astronaut’s seat rests. What is the maximum mass that this base can support without rupturing when the apparatus turns at a velocity o 250 km/h and the astronaut’s cabin is located at a distance o 12.5 m rom the centre o the rotation? 10. A 1250­kg car makes a turn with a radius o 50 m on a horizontal wet asphalt road. What is the maxi­ mum velocity the vehicle can reach without skid­ ding? 11. At what velocity should a circular space station turn to create artifcial gravity similar to gravity on Earth? The circular corridor in which this fctitious orce would be created is located at a distance o 175 m rom the centre o the rotation.

Bodies Subjec o  Number of Forces

B

odies are rarely subject to a single force. Even sta­ tionary bodies are generally acted on by several forces at a time. A number of forces can be exerted on struc­ tures such as bridges and skyscrapers, such as gravi­ tational force, normal force and tension, for example. To keep them stationary (or able to move within cer­ tain defined limits) and maintain their equilibrium, engineers and architects must determine all of the for­ ces that will be acting on the different points of these structures.

Review Gravitational force                                        13 Force of friction                                          14 Equilibrium between two forces                             14

In this chapter, you will learn more about dynamics by studying the effects of several forces on a single body. You will use the concept of vector addition to discover the concepts of resultant force and equilib­ rant force. As well, you will look at the concepts of static and dynamic equilibrium, and apply your knowledge to the case of the inclined plane.

13.1 13.2 13.3

Free-body digrm                    288 Resuln of severl forces           290 Equilibrium                            293 ChaptER 13 Bodies Subject to a Number of Forces

287

13.1 Free-body diagram A free-body diagram is a diagram that shows, with arrows, all of the forces exerted on a body. A free­body diagram is a graphic tool that can be useful in describing cer­ tain situations from a physical point of view. The word “body” here is syn­ onymous with “object”, and the term “free body” expresses the idea that the object is being represented without its environment. The following explains how a free­body diagram is made. Mehod for drawing a free-body diagram 1. Determine the body whose situation is to be analyzed 2. Represent the body with a point that corresponds to the object’s centre (whenever possible) 3. Show, with arrows, all o the orces acting on the body, making the origin o each arrow coincide with the point drawn in the previous step; the length o the arrows must be proportional to the magnitude o the orces illustrated 4. Generally, make the origin o the rame o reerence coincide with the point that symbolizes the body See Differen types of Forces, p 265

The action of the different types of forces studied up to now, such as gravita­ tional force (Fg ), normal force (FN ), friction (Ff ) and tension (FT ) can be repre­ sented in a free body diagram (see Figure 1). Figure 1 presents three examples of free­body diagrams. y

y →

→

FN

FT2 →

FT1 x

x →

Fg

→

Fg

a) A crate on the ground

b) A crate suspended rom two ropes y →

Flift

→

→ Fdrive forcex

Ff

→

Fg

c) An airplane in fight

Figure 1 Some examples o ree-body diagrams

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SECtION 13.1

Free-body digrm

1. On a piece of paper, draw a diagram showing the forces acting on each of the following bodies The length of the arrows must be proportional to the magnitude of the forces Indicate the symbol for each force a) a book lying on a table

2. For each of the following diagrams, name the force or forces not shown a) a marble rolling on the ground v

FN

Fg

b) a weight held by two ropes b) a raindrop falling to the ground FT

c) a stone block being pulled by a person

c) a weight sliding down an inclined plane

Fg

d) the carriage of an amusement ride in rotation

d) a body pushed on a horizontal surface FN

F

Fg

ChaptER 13 Bodies Subject to a Number of Forces

289

13.2 Resuln of severl forces The resultant force (FR ) is the vector sum of all forces exerted on a body. In general, a body is subject to a number of forces oriented in different direc­ tions. When several forces are exerted on a body, it is possible to determine a single force that can represent them all while producing the same effect. This is the resultant force. See Vecor quniies, p 176

Because forces are vector quantities, we must decompose them to add the components. For a given free body, the resultant force FR of n forces is writ­ ten: FR 5 F1 1 F2 1 … 1 Fn where FR 5 Resultant orce, expressed in newtons (N) F1, F2, Fn 5 Individual orces, expressed in newtons (N)

See adding vecors, p 190

See Grphicl mehod, pp 190 and 193

The methods of vector addition outlined in Chapter 8 are used to add up forces. The following examples illustrate how each of these methods is used to deter­ mine the resultant force in a situation where a body is subject to several forces. Exmple a A ootball player is pushed simultaneously by two opponents The situation is illustrated in the fgure opposite a) Draw a diagram representing the orces exerted on the player b) Using the grphicl mehod, determine the vector that represents the resultant orce Solution: a) Free-body diagram

→

F2

→

F1

b) Resultant orce →

F2

→

F1

→

FR →

F2

→

F1

FR 5 F1 1 F2

See Componen mehod, pp 190 and 193

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Exmple B A boat’s engine exerts a drive orce (Fdrive ) o 8500 N at 45°, while the orce exerted by the wind (Fwind ) is 2000 N at 210°, and the orce o the current (Fcurrent ) is 1500 N at 110° a) Draw a ree-body diagram representing these orces b) Determine the magnitude and direction o the resultant orce using the componen mehod.

Solution: a) Free-body diagram

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

b) Resultant force Data: Fdrive 5 8500 N at 45°

y →

Fwind 5 2000 N at 210°

Fdrive →

Fcurrent 5 1500 N at 110°

Fcurrent

FR 5 ? x →

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

Fwind

SIMON StEVIN FR 5 Fdrive 1 Fwind 1 Fcurrent

Flemish engineer, physicist and mathematician (1548–1620)

Along the x-axis: Fdrive 5 Fdrive cos θ 5 8500 N  cos 45° 5 6010 N

Stevin was an early proponent of the decimal system He declared that the universal use of the decimal system was only a matter of timeHIGHLIGHTS HISTORY HIGHLIGHTS HISTORY

Fwind 5 Fwind cos θ 5 2000 N  cos 210° 5 -1732 N Fcurrent 5 Fcurrent cos θ 5 1500 N  cos 110° 5 -513 N Along the y-axis : Fdrive 5 Fdrive sin θ 5 8500 N  sin 45° 5 6010 N Fwind 5 Fwind sin θ 5 2000 N  sin 210° 5 -1000 N

→

FR

Fcurrent 5 Fcurrent sin θ 5 1500 N  sin 110° 5 1409 N FRx 5 6010 N  2 1732 N 2 513 N 5 3765 N FRy 5 6010 N 2 1000 N 1 1409 N 5 6419 N

FRy = 6419 N θ FRx = 3765 N

FR 5 FRx2 1 FRy2 5 (3765 N)2 1 (6419 N)2 5 7442 N FRy 6419 N θ 5 tan-1 5 tan-1 5 596° 3765 N FRx

( )

(

)

Thus, the resultant force is FR 5 7442 N at 596° Exmle C In the Cartesian plane opposite, the point represents a mechanical part attached to three springs Each square represents 1 N Determine the magnitude and direction of the resultant force using the Cresin coordine meod.

Stevin’s work in physics was quite diverse He showed experimentally that two bodies of different masses fell at the same speed—three years before Galileo predicted they would Like Galileo, he defended the hypothesis that the Earth revolved around the Sun (heliocentric theory) One of his contributions to dynamics was to establish a method for decomposing and adding forces

y

See adding in e Cresin lne, p 191, and Subrcing in e Cresin lne, p 194

→

F1

→

F2

x →

F3

ChaptER 13 Bodies Subject to a Number of Forces

291

Data: F1 5 (-5, 7)

appENDIx 5 Review of rigonomery, p 404

Solution: FR 5 F1 1 F2 1 F3 5 (-5, 7) 1 (6, 4) 1 (-6, -7) 5 (-5 , 4)

F2 5 (6, 4) F3 5 (-6, -7)

FR 5 FRx2 1 FRy2 5 (-5 N)2 1 (4 N)2 5 64 N

FR 5 ?

FRy

( F ) 5 tan (-54 ) 5 39°

α 5 tan-1

-1

Rx

θ 5 180° 2 39° 5 141° FR 5 64 N at 141°

SECtION 13.2

Resuln of severl forces

1. The diagrams below show orces acting on various bodies In each case, determine the magnitude and direction o the resultant orce using the method o your choice a)

6.0 N

b)

8.0 N

6.0 N

c)

3.0 N

4.0 N

4.0 N

d)

6.0 N

10.0 N 53°

6.0 N 3.0 N

4.0 N

e) f)

15.0 N

15.0 N

45°

37° 2.0 N

2. For each o the ollowing situations, determine the magnitude and direction o the resultant orce a) A child pulls a sleigh At a particular moment, he exerts a 10-N horizontal orce sideways on the sleigh The resistance due to riction is increased to 8 N First, draw a diagram showing the orces exerted on the sleigh b) Two kids play tug-o-war with an old inner tube taken rom a tire One exerts a orce o 45 N upward and the other, a orce o 60 N at 45° downward c) A car is stuck in a ditch Three people push it out with the ollowing orces: F1 5 300 N at 180°, F2 5 500 N at 200° and F3 5 400 N at 170° d) Four children are fghting over a toy by pulling on it, as represented in the Cartesian plane below Each square represents 1 N

9.0 N

45°

y

9.0 N 15.0 N

g)

12.0 N

→

8.0 N

F1

12.0 N

F2

h) 65°

→

5.0 N

5.0 N

10.0 N 45°

45° 3.0 N

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UNIt 4 Dynamics

x

→

30°

15.0 N

→

F4

F3

13.3 Equilibrium Equilibrium is a body’s state when the resultant force acting on it is zero. When a body is subject to several forces that, given their magnitude and direc­ tion, cancel each other out, it is said to be in a state of equilibrium. A body in equilibrium is not subject to any variation in motion (see Figure 2). The result­ ant force FR acting on a body in a state of equilibrium is written: FR 5 F1 1 F2 1 … 1 Fn 5 0 where FR 5 Resultant orce, expressed in newtons (N) F1, F2, Fn 5 Individual orces, expressed in newtons (N)

13.3.1

Sic equilibrium or dynmic equilibrium

When considering a body’s state of equilibrium, a distinction is made between static equilibrium and dynamic equilibrium. A body is in a state of static equilibrium when the resultant force exerted on the object is zero and the object is stationary. A body is in a state of dynamic equilibrium when the resultant force is zero and the object is moving at a constant velocity.

Figure 2 When the resultant orce acting on a body is zero, it is in a state o equilibrium: it thereore does not undergo any variation in movement

There are, therefore, two complementary ways of looking at equilibrium: on the one hand, if the sum of the forces exerted on a body is zero, it implies that the object does not undergo any change in motion; on the other hand, if a body maintains a constant velocity, it means that the resultant force exerted on the object is zero. The following example shows a case of static equilibrium. Exmle a The fgure opposite shows a crane moving a concrete block weighing 12 000 N I the tension in cable 1 (FT 1 ) is 3117 N and the system is in equilibrium, what is the tension in cable 2 (FT 2 )? Data: Fg 5 12 000 N at 270°

FR 5 0

FT 1 5 3117 N at 160°

FT 2 5 ? N at 75°

Solution: We can consider the y orces: FRy 5 Fgy 1 FT 1y 1 FT 2y 0 N 5 -12 000 N 1 3117 N sin 160° 1 FT 2y 0 N 5 -12 000 N 1 1066 N 1 FT 2y FT 2y 5 10 934 N 5 FT 2 sin θ 5 FT 2 sin 75°

y →

FT2 →

75°

FT1 20°

x

→

Fg

10 934 N 5 11 319 N sin 75° FT 2 5 11 319 N

FT 2 5

We can veriy this last value by also analyzing the equilibrium along the x-axis

ChaptER 13 Bodies Subject to a Number o Forces

293

The ollowing example illustrates a case o dynamic equilibrium. Example B A child is sitting in a wagon pulled by a horizontal rope The tension (FT ) on the rope is 600 N The air resistance (Ff air ) is 200 N, and the riction on each o the wheels (Ff ) is 100 N Is the wagon in a state o equilibrium?

y

→

FN →

→

Ffair

FT x

→

Ffground →

Fg

Data : FT 5 600 N Ff ground 5 4  100 N 5 400 N Ff air 5 200 N Fg 5 ? FN 5 ?

Solution: Along the y-axis: There is no change in the motion (FNy 5 Fgy ) Along the x-axis: FR 5 FT 1 Ff ground 1 Ff air FRx 5 FTx 2 Ff groundx 2 Ff airx 5 600 N 2 400 N 2 200 N 5 0 N Yes, this represents a state o dynamic equilibium thereore the child moves at a constant velocity Example B illustrates the defnition o the state o dynamic equilibrium. The child on the wagon is travelling at a velocity that is not zero, but is constant; he is in a state o equilibrium in the same way he would be i he were stationary. A resultant orce o zero implies that its two components (x and y) are zero. In other words, when the vector addition is made, the sum o the x­components must be zero, and the sum o the y­components must also be zero.

Sable, unsable and neural equilibrium The notion o equilibrium is a complex one Any in-depth study o the subject frst requires a knowledge o the moment o a orce The moment o a orce is the capacity o a orce to make a body turn around a point called a pivot True equilibrium is achieved when the sum o the orces is zero, and the sum o the moments o orce is zero A body that is in a state o equilibrium does not undergo any change in its linear motion, nor in its rotational motion It is also possible to distinguish several kinds o equilibrium: stable, unstable and indierent (see Figure 3 ) Stable equilibrium is a situation where a slight additional orce (called perturbation) can shit a body, but the object returns to its starting position as soon as the orce’s action stops Unstable equilibrium is a situation where a slight

294

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additional orce shits a body and the movement continues and intensifes even when the orce stops A body balanced on the end o a circus perormer’s nose is in unstable equilibrium Indierent equilibrium is also possible In this case, the addition and subsequent withdrawal o a slight orce creates a new state o equilibrium in a new position

a) Stable equilibrium

b) Unstable equilibrium

c) Indierent equilibrium

Figure 3 A marble is in stable, unstable or indierent equilibrium, depending on the surace it is on

13.3.2

Equilibrn force

In a case where the resultant force (FR ) is not zero, it is possible to determine a single force that will cancel out the effect of the resultant force—that is, the effect of all the other forces. This force is called the equilibrant force, written as Feq. The equilibrant force and the resultant force are opposite vectors, meaning they have the same magnitude and are parallel, but act in opposite directions. We can therefore say that:

See Subrcing vecors, p 193

Feq 5 -FR The following examples illustrate situations that bring equilibrant force into play. Exmle a A race car (the red car) collides side-on with another car (the blue car) and is subject to a orce F1 o 20 000 N at 300° At the same time, it hits a third vehicle (the yellow car) and is subject to a orce F2 o 9000 N at 2350° The situation is shown in the fgure opposite →

What is the equilibrant orce that must be exerted on the red car or it to maintain the same motion?

F1

→

F2

Data: F1 5 20 000 N at 300° F2 5 9000 N at 2350° FR 5 ? Free-body diagram:

Solution: For the red car to keep moving at a constant velocity, Feq 5 -FR where FR 5 F1 1 F2

y

→

F1 235º

30º x →

F2

Along the x-axis: F1x 5 F1 cos θ 5 20 000 N  cos 300° 5 17 321 N F2x 5 F2 cos θ 5 9000 N  cos 2350° 5 -5162 N Along the y-axis: F1y 5 F1 sin θ 5 20 000 N  sin 300° 5 10 000 N F2y5 F2 sin θ 5 9000 N  sin 2350° 5 -7372 N

FRx  12 158 N FRy  2628 N θ

FRx 5 F1x 1 F2x 5 17 321 N 2 5162 N 5 12 159 N FRy 5 F1y 1 F2y 5 10 000 N 2 7372 N 5 2628 N FR 5 FRx2 1 FRy2 5 (12 159 N)2 1 (2628 N)2 5 12 440 N F 2628 N 5 122° θ 5 tan-1 Ry 5 tan-1 FRx 12 159 N

( )

(

)

The resultant orce is thereore FR 5 12 440 N at 122°, and the equilibrant orce is Feq 5 12 440 N at 1922°

ChaptER 13 Bodies Subject to a Number o Forces

295

Example B A very small asteroid is subject to the gravitational attraction o a larger asteroid, Jupiter and the Sun The situation is shown in the fgure opposite Calculate the orce required to keep the small asteroid in a state o equilibrium Each square represents 1 N

y

→

Fa

→

FJ

Data: Fa 5 (-3, 6) FJ 5 (6, 7) FS 5 (-7, -10) Feq 5 ?

x

→

FS

Solution: FR 5 Fa 1 FJ 1 FS 5 (-3, 6) 1 (6, 7) 1 (-7, -10) 5 (-4, 3) Feq 5 -FR 5 (4, -3)

θ

FRx  4 N α

FR 5 FRx2 1 FRy2 5 (4 N)2 1 (-3 N)2 5 5 N F |-3| α 5 tan-1 Ry 5 tan-1 5 37° FRx 4 θ 5 360° 2 37° 5 323°

( )

FRy  3 N

( )

Feq 5 5 N at 323°

Furthering

your understanding

Spagheifcaion A orce can cause a change in an object’s motion It can also cause an object to deorm In practice, however, the only consequence o a orce is a change in motion I we consider each particle o an object separately, what we reer to as deormation appears as changes in the positions o the particles In its most extreme orm, this phenomenon occurs near a black hole and is called spaghettifcation A black hole is an extremely massive dead star (generally 10 times more massive that the Sun) In theory, the gravitational orce exerted by a black hole is such that the particles o a body that approaches it are subjected to dierent levels o gravitational attraction Since the particles located closer to the black hole would experience a greater acceleration, the body would become elongated like a strand o spaghetti (see Figure 4)

Figure 4 The « spaghettifcation » o a spacecrat The curved surace represents the black hole’s gravitational orce

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Inclined lne

13.3.3

All of the notions studied in this chapter can be synthesized in an examina­ tion of the inclined plane. An incline plane is a plane surface set at an angle to the horizontal. Because the object is no longer on a horizontal surface, the analysis of forces acting on a body on an inclined plane is changed. The forces applied to a body on an inclined plane are easier to analyze when they are drawn on a Cartesian coordinate system that is also inclined. The axes of the ordinates (y) and abscissas (x) are no longer vertical and hori­ zontal. To simplify things, the axes are rotated at an angle of θ, which corre­ sponds to the elevation angle of the inclined plane, and the x­axis is pointed toward the bottom of the inclined plane. The gravitational force must then be decomposed parallel and perpendicular to the plane in order to determine its x and y components (see Figure 5). In Figure 5, the magnitudes of the two components are determined using the following equations: Fgx 5 Fg sin θ

and Fgy 5 -Fg cos θ

See te Cresin coordine sysem, p 158

appENDIx 5 Review of rigonomery, p 404 y

y →

FN

→

FN

y Fgx

Fgx

→

FN

→

θ

θ

x

Fg

Fgy

Fgy

→

Fg

x θ

θ

Fgy

Figure 5 On an inclined plane, the gravitational force decomposes parallel to the plane (along the x-axis) and perpendicular to the plane (along the y-axis) The component parallel to the plane is written as Fgx , and the component perpendicular to the plane is written as Fgy 

When the forces perpendicular to the inclined plane are analyzed, we see that the normal force (FN ) is opposite to the y­component of gravitational force (Fgy ) (see Figure 6). Since FN is in the opposite direction to Fgy and the magnitudes of FN and Fgy are equal, there is equilibrium along the y ­ axis. It is written: FRy 5 FNy 1 Fgy 5 0

→

Fg

x θ

Figure 6 Normal force (FN ) and the y-component of gravitational force (Fgy ) are in opposite directions y

or FNy 5 -Fgy →

What about forces exerted on the object if we consider that it is immobile or that it descends at a constant velocity? There is equilibrium along the x­axis if the force of friction (Ff ) is opposite to the x­component of the gravitational force (Fgx). In other words, if the force of friction and the component of the gravi­ tational force parallel to the inclined plane have the same magnitude and are parallel to each other, but point in opposite directions, the x­component of the resultant force (FRx) is zero (see Figure 7). It is written: FRx 5 Ffx 1 Fgx 5 0

or

Ffx 5 -Fgx

Fƒ Fgx →

x

Fg θ

Figure 7 In a situation of equilibrium, the force of friction (Ff ) and the x-component of gravitational force (Fgx ) are opposite

ChaptER 13 Bodies Subject to a Number of Forces

297

It is important to remember that when an object is stationary, it is in a state of equilibrium known as static, and that when the object is descending at a con­ stant velocity, it is in a state of equilibrium known as dynamic. The following example shows how the forces exerted on an object on an inclined plane are analyzed. Example A 100-kg object slides down a plane inclined at 40° What must the force of friction be for it to descend at a constant velocity?

y →

FN →

Fƒ

→

Data: m 5 100 kg θ 5 40° g 5 980 m/s2 Ff 5 ?

Fgx

→

θ

Fg x θ

Solution: Fg 5 mg 5 100 kg  980 m/s2 5 980 N Fgx 5 Fg sin θ 5 980 N  sin 40° 5 629 N toward the bottom of the inclined plane Ff 5 -Fgx 5 -629 N Ff 5 629 N toward the top of the inclined plane

SECtION 13.3

Equilibrium

1. In each o the ollowing situations, at what moment is there equilibrium between orces? a) A cyclist accelerates until he reaches maximum speed and then stops pedalling and coasts b) A parachutist jumps rom a plane and ree-alls toward the ground Ater a minute, she deploys her parachute and alls at a constant speed or 15 minutes 2. A crate weighs 50 N It is attached to two ropes, as shown in the fgure opposite What is the tension FT 2 in the right-hand rope i 35.5° the tension in the horiFT 2 zontal rope (FT 1) is 70 N? F

Cartesian plane shows the action o the orces Each square represents 1 N What is the toy’s equilibrant orce? y

→

F1

→

F2

x →

F3

T1

3. A child plays with a toy containing three magnets exerting orces on each other The ollowing

50 N

4. A 25-kg weight is stationary on an inclined plane What is the magnitude o the orce o riction i the elevation angle is 30°? 30°

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APPLICATIONS Skyscrers Prior to the 19th century, elevated buildings had to have bearing walls that were very thick at the base to support their weight. The walls of cathedrals of the Middle Ages, for instance, were routinely two to three metres thick at their lowest part. The Pulitzer Building, a 14­storey structure completed in 1890 in New York, had walls that were three metres thick. Constructing taller buildings with even thicker bearing walls would have been costly and risky, because the ground and foundations would not have been able to support the weight of the walls.

reduce swaying and maintain the building’s vertical equilibrium. In short, the development of materials and construction techniques has enabled builders to erect increasingly taller buildings, and everything suggests the trend is likely to continue (see Figure 8). Using their knowledge of forces and materials, the engi­ neers that design these skyscrapers are able to insure their solidity and stability. There are many forces that need to be taken into account: gravitational force and force due to winds are among the most important.

But by the end of the 19th century, a solution to the prob­ lem had been found: advances in metallurgy made it possible to produce long pieces of steel. By assembling these pieces into cage­shaped frames comprising riv­ eted or soldered columns and beams, taller and lighter structures could be built. In the 20th century, advan­ ces in the manufacturing of reinforced concrete, i.e. concrete reinforced with a steel framework, put a new and improved material for forming skyscraper struc­ tures into builders’ hands. A well­designed frame must support the weight above it at any point, ensuring the equilibrium of forces throughout the building. Steel frames support, without breaking, a heavier weight than brick or stone walls, and weigh much less. Furthermore, because the exte­ rior walls of a steel­frame skyscraper do not have to support the weight of the building, the glassed surface can be increased considerably. In a skyscraper, the weight of the storeys is transferred via columns to the footing of the foundation, which sits on the ground, distributing the weight over a larg­ er surface. When necessary (for very tall buildings, for instance), the footing is supported by concrete pillars that extend in to the bedrock. Skyscrapers are not perfectly stationary: they are de­ signed to sway in the wind by as much as a few dozen centimetres if they are very high. To prevent the dis­ comfort of occupants when the swaying gets too much, buildings are stiffened with reinforced concrete, and the support structure on the outside of the building is enhanced. Some skyscrapers even have dampers to

Figure 8 When it was inaugurated in January 2010, the Burj Khalifa tower, a 739-metre structure in Dubai, United Arab Emirates, was the tallest skyscraper in the world

ChaptER 13 Bodies Subject to a Number of Forces

299

Bridges The frst bridges were probably made rom tree trunks thrown over small bodies o water and astened to each bank. These “beam bridges” had a drawback i they were only supported at either end, they would bend under the eects o a load. Obviously, any bridge must be stable and in equilibrium. At any point, the normal orces, tensions and compressions must com­ pensate or the weight o the structure and load. In the case o a beam bridge, beams may rest on pillars (called bridge piers) made o rock, masonry or wood and erected in the middle o a body o water. Using this method, the ancient Romans built wooden bridges several dozens o metres long. Masonry arch bridges are much more durable than wooden bridges. An arch is made rom stones cut into wedges. The weight applied to the top o the arch press­ es the stones against each other and prevents them rom moving ; eort is transerred to the ends o the arch, which rest on piers or the riverbank. In roughly 600 BCE, the Babylonians used this technique to build a bridge 120 metres long over the Euphrates River.

south o France, it is 275 metres long and 48 metres high. It once supported an aqueduct supplying water to the city o Nîmes. The amous Pont d’Avignon, built over the Rhône in the 12th century, is another masonry bridge. O its 22 original arches, our still stand today. In the 19th century, advances in metallurgy then made construction o metal truss bridges possible, which even­ tually took the place o masonry arch bridges. The Pont de Québec (1917) in Québec City, and the Jacques­Cartier Bridge (1930) in Montréal are cantilever bridges whose main “beams” (trusses) extend as cantilevers over the gap and hold up their central sections. For many years, suspension bridges were made only o vines or ropes supporting a wooden platorm. The frst modern suspension bridges date rom the early 19th century. Metal suspension cables are attached to tall bridge towers at the ends o the bridge. The plat­ orm is suspended rom the suspension cables with vertical cables called suspenders. This technology made it possible to build long bridges without inter­ mediate piers (see Figure 10).

Later, the Romans also constructed a number o mason­ ry bridges. Many still stand, like the amous Pont du Gard (see Figure 9). Built in the 1st century BCE in the

Figure 9 The Pont du Gard (1st century BCE) in France is a masonry arch bridge

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Figure 10 In the foreground: the Pierre-Laporte Bridge (1970) in Québec City is a suspension bridge In the background: the Pont de Québec (1971) is a cantilever bridge

chapter

re V I e W

13

Objs Subj o  Numb of Fos

13.1 Free-body digrm • A ree­body diagram is a diagram that shows, with arrows, all o the orces exerted on an object. A ree­body diagram is a representation o the object that simplifes a situation or problem.

13.2 Resuln of severl forces • The resultant orce is the vector sum o all orces exerted on an object.

FR 5 F1 1 F2 1 … 1 Fn

• The methods o adding vectors discussed in Chapter 8 (graphical method, component method and Cartesian coordinate method) can be used to add orces.

13.3 Equilibrium • Equilibrium is an object’s state when the resultant orce acting on it is zero. The object is thereore not subject to any change in motion.

FR 5 F1 1 F2 1 … 1 Fn 5 0

• I a resultant orce is zero, its two components (FRx and FRy ) are zero. • The equilibrant orce is the single orce that cancels out the eect o all the other orces exerted on a body. It has the same magnitude as the resultant orce, but acts in the opposite direction. • To analyze the orces exerted on an object on an inclined plane, the gravitational orce must be decomposed parallel to the inclined plane (along the x­axis), and perpendicular to the inclined plane (along the y­axis). The magnitudes o the two components are determined with trigonometry.

Feq 5 -FR

Fgx 5 Fg sin θ and

Fgy 5 -Fg cos θ

ChaptER 13 Bodies Subject to a Number of Forces

301

ChaptER 13

Objecs Subjec o  Number of Forces

1. Using a ree-body diagram, represent the orces coming into play in the ollowing situations: a) a skier making a jump

2. In a piece o metal, an electron is attracted by the nucleii o a number o atoms The ollowing Cartesian plane represents the action o the orces at work Each square represents 10-6 N Calculate the resultant orce on this particle y

→

b) a skater competing in a race

F1

→

F2

x →

F3

3. Calculate the coefcient o static riction i a 6-kg weight is stationary on a plane inclined at 35° c) a yo-yo at the end o its string

4. In a laboratory, three dynamometers are attached to a weight I the equilibrant orce is Feq 5 08 N at 110°, what orce is applied to the third dynamometer i the two others are F1 5 05 N at 50° and F2 5 1 N at 310° ? 5. A 15-kg weight slides down an inclined plane What is the magnitude o the orce o riction i there is equilibrium and the normal orce is 905 N?

d) a weight pulled up an inclined plane

6. A 75-kg weight is suspended rom two ropes What angle does the second rope orm i the frst is at 25°, and its tension FT1 is 500 N? The system is in a state o equilibrium 25° FT 2

FT 1

75 kg

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Newon’s Lws

T

he three laws o physics discussed in this chap­ ter were ormulated by Isaac Newton in his 1688 book Philosophiae naturalis principia mathematica (Mathematical Principles of Natural Philosophy). It is one o the major books in the history o science, compa­ rable with Lavoisier’s Elementary Treatise on Chemistry (1789) or Darwin’s Origin of the Species (1859). Newton’s three laws are the oundation o the mechan­ ics known as classical mechanics: that which describes the world at a macroscopic scale.

Review Characteristics of force                                    13 Gravitational force                                        13 Relationship between mass and weight                       13 Equilibrium between two forces                             14

Over the course o the 20th century, the limits o New­ tonian theory have become evident. But three centuries ater its development, it is still classical, or Newtonian, mechanics that makes it possible to predict the motion o the planets, artifcial satellites or objects on Earth. In this chapter, you will discover Newton’s laws and you will learn how to use them to determine the accelera­ tion o a body or the magnitude o a orce exerted on this body.

14.1 14.2

Newon’s frs lw nd ineri        304

14.3

Newon’s ird lw                    315

te noion o orce nd Newon’s second lw                            309

ChaptER 14 Newton’s Laws

303

14.1 Newon’s frs law and ineria Newton’s frst law states that, in the absence o a resultant external orce acting upon it, a body will remain at rest or in uniorm rectilinear motion (its acceleration will be zero). Logical implication of a * Corollary stated principle See Saic equilibrium or dynamic equilibrium, p 293 See Graviaional orce, p 267 See Normal orce, p 273

This frst law had a corollary*, namely that to change a body’s motion, i.e. to put it in motion, accelerate it, slow it down, stop it or change its direction, a non­zero external orce must be applied to it. Consequently, Newton’s frst law means that i no external orce is exerted on an object, the object’s acceleration will be zero. This implies that i the object is at rest, it will remain at rest, and that i the object is moving, it will continue to do so at the same velocity and in the same direction. The frst law thereore concerns static equilibrium and dynamic equilibrium, which was discussed in the previous chapter. By external orce, we mean a orce exerted on an object by another object, in contrast to an internal orce, which is exerted by a part o an object on anoth­ er part o the same object (see Figure 1). A person contracting their biceps is an example o internal orce. The arm bends, but the person as a whole is not put in motion, nor do they accelerate. However, a hockey player checked by an opponent is subject to an external orce exerted by another player: in this case, he accelerates in a certain direction. An object can be subject to a number o external orces. The vector sum o these orces is called the “resultant orce” (written FR ).

a) Internal force: a person contracts their biceps and their arm bends

According to Newton’s frst law, i the resultant orce exerted on an object is zero, the acceleration will also be zero. This is the case with an apple placed on a table: the apple is subject to two external orces, gravitational orce (Fg), directed downward, and the normal orce (FN ) exerted by the table which pre­ vents the apple rom alling (see Figure 2). The two orces act in opposite direc­ tions. The apple’s acceleration is zero. According to Newton’s frst law, the resultant external orce is zero, implying that the magnitudes o gravitational orce and normal orce are equal. →

FN

→

Fg

b) External force: a hockey player is checked by an opponent

Figure 1 Internal force and external force

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Figure 2 An apple on a table is subject to two external forces of the same magnitude, gravitational force (Fg ) and normal force (FN ) The resultant force (FR ) is zero

14.1.1

te frs lw nd e rcice o everydy lie

At frst glance, the frst law appears to go against what we experience in every­ day lie. Firstly, i a crate is pushed along the ground with a constant drive orce, the box moves at a constant speed instead o accelerating (see Figure 3a). Secondly, when the pushing stops, the crate stands still (see Figure 3b). And yet the frst law states that i the pushing stops, i.e. i the resultant external orce is zero, the acceleration is zero: thereore the crate should continue at a constant speed with no pushing required! →

v

→

→

→

FN

→

v0

FN

→

Fdrive

→

→

Fg

Fg

a) A person pushes a crate along the ground, the crate moves at a constant speed

b) I the person stops pushing, the crate comes to a stop

Figure 3 Two situations that appear to contradict the frst law

This apparent contradiction stems rom the act that a nearly imperceptible orce has been let out o the equation: riction. The force of friction (Ff ) exerted by the ground points in the direction opposite to the motion and opposes the drive orce (F drive) pushing the crate. In the situation described here, the riction is as great as the orce moving the crate (see Figure 4). The resultant orce is thereore zero in the horizontal direction and, according to Newton’s frst law, the acceler­ ation is zero. This is why the crate moves orward at a constant speed when pushed. I the pushing stops, riction slows the crate down, bringing it to a stop. →

→

See Force o ricion, p 275

→

a0

FN

→

→

Ff

Fdrive

→

Fg

Figure 4 Friction counteracts the drive orce: the crate moves at a constant speed when it is pushed with a constant orce

ChaptER 14 Newton’s Laws

305

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

To ully understand the frst law, consider a situation where riction is very low: a person skating, or instance. When the person stops pushing with her skates and simply lets hersel glide, the resultant orce is nearly zero, and the person continues to move with a deceleration so low that her speed is nearly constant. From a scientiic perspective, the discovery o the irst law was a veritable revolution. Since the time o the ancient Greeks, within the feld o physics it was thought that a orce had to be applied to an object or it to be in motion. An object not subject to orce was stationary. The only exception was the spe­ cifc case o the celestial bodies in the sky, whose unchanging circular motion was considered “natural.” The frst law contradicted these intuitive conceptions and made it possible to clariy the notion o orce. It meant that i a body was already in motion, this motion would continue in the absence o orce. To stop this already moving body or curve its trajectory, a orce had to be exerted.

The title page o Philosophiae naturalis principia mathematica.

ISaaC NEwtoN

HISTORY HIGHLIGHTS HISTORY HIGHLIGHTS British mathematician, physicist and astronomer (1642–1727) Newton’s interests extended to optics and beyond At around the age o 25, he developed the “method o fuxions,” today known as dierential and inte­ HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY gral calculus, and used this mathematical tool to analyze curved trajectories with precision In mechanics, Newton used Galileo’s results to establish his three undamental laws With these laws and the method o fuxions, he showed that an orbit caused by a orce o attraction inversely proportional to the square o the distance can take the orm o an ellipse, as the planets’ orbits do Newton proved that his theory o gravitational attraction was consistent with observations time and time again, and used it to explain such phenomena as the tides, the Earth’s equatorial bulge, and the motion o comets Newton is considered to be one o the greatest geniuses in human history His mechanics, com­ plete and coherent, had a major infuence on scientic thought

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For example, when an object tied to the end o a string is swung in a hori­ zontal plane, the centripetal orce curving its trajectory is the tension in the string. I the string breaks, the resultant orce on the object becomes zero; acceleration also becomes zero, and the object continues in a direction tan­ gent to the circle (see Figure 5). Similarly, in the track and feld event known as the hammer throw, the head o the hammer leaves in a straight line when the athlete lets go o the handle.

v

Figure 5 An object in circular motion at the end o a string When the string breaks, the object continues in a direc­ tion tangent to the circle as it begins its reeall

The case o a planet rotating around the Sun is particularly interesting. Because its motion is not rectilinear, this means that the planet necessarily undergoes an acceleration. According to the frst law, the planet is subject to a orce. Newton described this orce as the gravitational attraction exerted by another celestial body, the Sun. I the gravitational orce exerted on a planet suddenly disappeared, the planet’s acceleration would become zero, and the planet would leave its orbit to continue in a direction tangent to the orbit.

14.1.2

Ineri nd is rcicl consequences

Inertia is an object’s tendency to retain its state of motion. An object’s inertia is quantifed by its mass. The larger the mass, the more the body tends to stay at rest i it is at rest, or continue its motion in a straight line i it is already in motion. The notion o inertia and Newton’s irst law explain countless everyday occurrences. The ollowing are some examples: • A car takes a curve and, because o its inertia, tends to continue in a straight line and go o the road. As surprising as it may seem, it is riction (an exter­ nal orce) that changes the car’s direction as it negotiates the curve. I the riction is low, or instance on an icy road, the car is at a greater risk o continuing in a straight line and going o the road. • In winter, people stamp their boots on the ground to rid them o snow and slush. The ground abruptly stops the boots’ motion, but the snow contin­ ues its motion because o its inertia. Because it is not stuck to the boots, it comes o. • In a salad spinner, the rotation o the basket spins the wet lettuce. The water’s inertia makes it tend to continue in a straight line: it lies o the lettuce tangentially, passes through the holes in the basket, hits the sides o the spinner, then runs to the bottom o the bowl. The lettuce behaves the same way, but cannot pass through the basket. • A magician pulling on a tablecloth set with objects is a classic trick (see Figure 6). I the tablecloth is pulled slowly, the riction between the table­ cloth and the objects acts long enough to put the objects in motion. I the tablecloth is pulled quickly, the riction is very small and the objects remain in roughly the same place because o their inertia.

a) A woman prepares to pull quickly on a tablecloth

b) The objects remain in place because of their inertia

Figure 6 Nothing magic about it: this classic trick works because of the objects’ inertia

ChaptER 14 Newton’s Laws

307

Furthering

your understanding

Ficinal rces When a bus brakes, the passengers eel themselves projected orward They have the impression o being subjected to a orce   but this orce does not exist! The eel­ ing comes rom the act that, because o their inertia people tend to continue in a straight line within a rame o reerence (the bus) that is accelerating A “orce” appearing thus in an accelerated rame o reerence is said to be a fctional orce This type o orce has no physical origin because it is impossible to determine an object exerting it It simply does not exist The best­known case o fctional orce is what is known as “centriugal orce” We know, or instance, that a person driving a car who negotiates a right­hand curve is subject to a centripetal orce directed to the centre o the curve This orce is the ric­ tion between the seat and the person I the centripetal orce is insufcient, the per­ son, because o their inertia, will ollow a trajectory that is less curved than that o the car: he thereore eels himsel thrown against the let door (Once he touches the door, a horizontal normal orce directed to the centre o the curve, also plays a part in curving his motion) This analysis corresponds to the one made when we describe a situation by placing ourselves in an inertial rame o reerence (the ground, or instance) The person driving analyzes his motion in relation to the car, which is an accelerated rame o reerence He interprets the situation by saying that a “centriugal orce” threw him against the let door This orce does not exist, any more than the orce exerted on the passengers o the braking bus

Figure 7 In a car taking a right curve, the driver eels himsel pushed letward by a fctional orce known as centriugal orce In reality, the phenomenon can be explained by inertia

It is oten useul to use an accelerated rame o reerence and to reer to fctional orces However, in this textbook we will only study inertial rames o reerence where no fctional orces come into play

SECtIoN 14.1

Newn’s frs law and ineria

1. When you shake the water o your hands ater washing them, you are unconsciously applying Newton’s frst law Explain what happens 2. You swing a ball at the end o a string in a horizontal plane above your head Which o these our bird’seye-view trajectories will the ball ollow i you let go o the string when the ball passes point P? Explain your answer A

B

C

D

3. A screenwriter is asking you about her script or a science fction flm She wants you to veriy the accuracy o the ollowing scene: “On board the vessel, the hero detects an asteroid and veers sharply to avoid collision Throughout the manoeuvre, he remains standing, arms resolutely crossed without holding onto anything” What is wrong with this scene?

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14.2 te noion of force nd Newon’s second lw Newton’s irst law implies a connection between orce and acceleration. Newton’s second law denes this connection.

14.2.1

te noion of force

The arrangement in Figure 8 can be used to experi­ mentally determine the connection between orce and acceleration. On a table covered with a substance oering very little riction, Telon, or instance, a body o mass m is pulled with a dynamometer mea­ suring the applied orce. By noting the object’s posi­ tion over time, we can calculate its acceleration. I the orce exerted on the object is doubled, the experiment shows that the acceleration doubles. The acceleration is thereore proportional to the orce.

m

Figure 8 The force exerted using the dynamometer causes a horizontally directed acceleration on block m

aαF I the orce is kept constant, but the mass m is doubled, we discover that the acceleration is halved. This result refects the act that mass is the measure­ ment o a body’s inertia: the greater the mass, the more dicult it is to change the body’s motion. Acceleration is thereore inversely proportional to mass. aα

1 m

To take the proportionalities shown above into account, we write: F 5 ma The equation F 5 ma corresponds to the denition o orce. It means that i a certain orce is applied to an object, this orce is equal to the mass multiplied by the acceleration experienced by the object. In the International System o Units (SI), mass is expressed in kilograms (kg), acceleration in metres per sec­ ond squared (m/s2), and orce in newtons (N); we can thereore write: 1 N 5 1 kg 3 1 m/s2 5 1 kg 3 m/s2 1 newton (N) 5 1 kilogram metre per second squared (kg m/s2) I a single orce is applied to a body, the body’s acceleration is in the direction o the orce. The complete relationship between orce, mass and acceleration includes the spatial directions o orce and acceleration as well as their mag­ nitudes. Vector notation is thereore required, giving us: F 5 ma

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309

14.2.2 See Resuln of severl forces, p 290

Newon’s second lw

A body is rarely subject to a single orce. In such cases, a body’s accelera­ tion depends on all o the orces applied. Because these orces are vectors, the resultant force is the vector sum o the orces. Newton’s second law, also known as the fundamental principle of dynamics, states that the resultant force ( FR ) exerted on a body is equal to the product of the mass (m) of this body and its acceleration (a ). FR 5 ma where FR 5 Resultant orce, expressed in newtons (N) m 5 Mass, expressed in kilograms (kg) a 5 Acceleration, expressed in metres per second squared (m/s2)

The direction o the acceleration (a) is the same as the direction o the result­ ant orce. Thereore, Newton’s frst law is a particular case o the second law: i the resultant orce is zero, acceleration is zero, and vice versa. Newton’s second law is undamental to mechanics and should be approached systematically to ensure it is used correctly. Keep in mind that the orc­ es are vectors. Because the sum o vector orces requires adding the x­ and y­ components, applying the second law oten consists in writing the equations: FRx 5 max and

FRy 5 may

absorbing collisions In a collision, the orce exerted depends on the distance over which the deceleration occurs Take the example o a baseball player who, with arms sti, catches a ball arriving at 100 km/h and eels its ull impact Yet i he “cushions the ball” by drawing back his hand as he catches it, the shock is easier to absorb According to the equa­ tion v2 5 vi2 1 2a∆x, the greater the distance ∆x travelled by the ball during the collision, the lower the ball’s acceleration in absolute value Moreover, according to Newton’s second law, i acceleration is low, the orce that the hand is subject to is also reduced The same principle can be applied to a number o situations A per­ son alling rom a height, or instance, such as a parachutist touch­ ing down, inds it to his advantage to bend his knees instead o keeping his legs sti The fexing o the knees reduces his average deceleration and thereore the orces on his body In the same way, cars now are designed in such a way as to allow the parts surround­ ing the passenger compartment to deorm considerably in case o collision (see Figure 9 ) Because o these deormations, the

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passengers are subject to orces considerably lower than the ones they would experience in a totally rigid car

Figure 9 In a car accident, deormation o the ront o the vehicle reduces the orces on its passengers

The orces in question in the second law are those applied to a specifc body. To determine a body’s acceleration, there is no need to take the orces applied to other objects in the environment into account. The ollowing method explains how to proceed when applying Newton’s second law. Generl meod for solving dynmic roblems (wi Newon’s second lw) 1. Identiy the body whose motion you wish to analyze 2. Identiy all the orces acting on this body and draw them in a free-body digrm 3. Choose an appropriate coordinate system, generally with the positive x­axis pointing in the direction o the body’s presumed acceleration (so that ax > 0 and ay 5 0, which simplifes the analysis) 4. Write the equations derived rom Newton’s second law: FRx 5 max FRy 5 may 5. Solve the equations (or example, determine the acceleration)

See Free-body digrm, p 288

The ollowing examples show how this problem­solving method is applied. Exmle a A 40­kg object is pushed over a horizontal surace with a orce o 20 N Assume that riction is negligible a) What is the object’s acceleration? b) What is the normal orce?

Data: m 5 40 kg Three orces are exerted on the object: – A horizontal orce: an applied orce F whose magnitude is F 5 20 N – Two orces in the vertical direction: gravitational orce Fg and the normal orce FN ax 5 ?

y →

FN

→

F

x

→

Fg

FN 5 ? Solution: a) Along the x­axis: FRx 5 F 5 max F 20 N 5 5 50 m/s2 m 40 kg b) Along the y­axis: FRy 5 FNy 1 Fgy 5 FN 2 Fg 5 may thereore ax 5

But in the vertical direction, the object does not accelerate upward or downward, thereore: ay 5 0 FN 2 Fg 5 0 thereore FN 5 Fg 5 mg 5 40 kg 3 980 m/s2 5 39 N

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Example B Using a rope with a 30° incline above the horizontal, a 40­kg object, initially at rest, is pulled over a horizontal surface with a force of 20 N The friction is equivalent to 12 N a) What is the object’s acceleration? b) What distance is covered by the object in 200 seconds if it is initially at rest? c) What is the magnitude of the normal force?

30°

Data: m 5 40 kg ax 5 ?

FT 5 20 N xf 5 ?

θ 5 30°

Ff 5 12 N

vix 5 0 m/s

t50s

FN 5 ?

Four forces are exerted on the object: tension (FT ), friction (Ff ), gravitational force (Fg ) and normal force (FN )

Solution: a) Along the x­axis: FRx 5 FTx 1 Ffx 5 FT cos θ 2 Ff 5 max F cos θ 2 Ff (20 N 3 cos 30°) 2 12 N therefore ax 5 T 5 5 13 m/s2 m 40 kg b) ∆t 5 200 s

See Equaions for uniformly acceleraed recilinear moion, p 230

xi 5 0

vix 5 0

xf 5 ?

vfx 5 ?

ax 5 133 m/s2

We use Equation 2 for uniformly acceleraed recilinear moion: 1 1 xf 5 xi 1 vix∆t 1 ax∆t 2 5 0 1 0 1 3 133 m/s2 3 (200 s) 5 266 m 2 2 c) Along the y­axis: FRy 5 FTy 1 FNy 1 Fgy 5 FT sin θ 1 FN 2 mg 5 may But in the vertical direction, the object does not accelerate upward or downward, therefore: ay 5 0 and therefore FN 5 mg 2 FT sin θ 5 (40 kg 3 980 m/s2) 2 (20 N 3 sin 30°) 5 29 N Because tension has a non­zero vertical component, the normal force (FN) is not equal to the gravitational force (Fg)

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Exmle C A block slides down a plane with a 30° incline Assuming there is no riction, what is its accele­ ration? Data: θ 5 30° ax 5 ? To simpliy the analysis, we choose an x­axis aligned parallel to the inclined plane and pointing in the direction o the block’s acceleration; thereore ay 5 0 y

→

FN

x

30° →

Fg

30°

Solution: FRx 5 Fgx 1 FNx 5 Fg sin θ 1 0 N 5 max thereore ax 5

Fg sin θ mg sin θ 5 5 g sin θ 5 980 m/s2 3 sin 30° 5 49 m/s2 m m

te bounce of  bll When a ball bounces diagonally on the ground, interesting things happen which can be observed in a tennis ball or baseball Because the ball ar­ rives with a non­zero horizontal velocity component, the ball tends to slide along the ground (see Figure 10 ) The ground then exerts riction in the opposite direction to the sliding tendency In the diagram, the ball seeks to slide to the right at the same time as it is subject to a orce o riction directed to the let

Direcion fer e bounce

Direcion of roion

fer e bounce This riction has two eects First, according to Newton’s second law, it induces an acceleration directed toward the let and because the accel­ Ff eration is in the opposite direction to that o the horizontal motion, the velocity’s horizontal component diminishes The ball then takes o again at an angle closer to the vertical than it had when it arrived As it leaves, Figure 10 At the moment o contact with the ground, riction the riction gives the ball a rotation, as any orce will do when it is not modifes the direction o the ball relative to the vertical and gives applied to the centre o an object and the ball begins to rotate in the the ball a rotation direction indicated in the diagram

ChaptER 14 Newton’s Laws

313

SECtIoN 14.2

the nin f frce and Newn’s secnd law

1. A car carrying ve people accelerates less than the same car carrying just one person Explain 2. A 1100-kg car accelerates at a rate o 3 m/s2 on a horizontal road a) What is the resultant orce acting on the car? b) I the orce acting on the car rom the engine is equal to 5000 N, what is the magnitude o the orce opposing the car’s motion? 3. A 102-g spider is suspended rom an elevator ceiling by a thread a) What is the tension in the thread i the elevator rises at a constant velocity? b) What is the tension in the thread i the elevator accelerates at a rate o 20 m/s2 as it rises? c) What is the tension in the thread i the elevator decelerates at a rate o 20 m/s2 as it rises? 4. A 60-kg girl takes the elevator to the foor above a) At the start, the elevator accelerates at 10 m/s2 What is the normal orce exerted on the girl? b) The elevator then rises at a constant velocity What is the normal orce exerted on the girl? c) As the elevator decelerates beore coming to a stop, the normal orce is equal to 500 N What is the magnitude o the deceleration? 5. A 170-g hockey puck slides over the ice and is subject to a riction orce o 05 N a) What is the puck’s acceleration? b) I its initial speed is 15 m/s, what distance will the puck cover beore coming to a stop? 6. Starting rom a rest position, how long will it take a 50-kg cyclist riding a 10-kg bicycle to reach a speed o 40 m/s i the resultant orce acting on the bicycle is 48 N? 7. A man pushes a 10-kg grocery cart containing 30 kg o ood with a orce o 50 N directed 30° below the horizontal The cart moves at a constant velocity a) What is the magnitude o the orce o riction? b) What is the magnitude o the normal orce?

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8. A 75 000-kg airplane is tted with two jet engines At the moment o takeo, the thrust o each jet engine is raised to 60 000 N The airplane must reach a speed o 220 km/h to take o a) What is the minimum length o the runway required or takeo i air resistance and riction are not taken into account and i acceleration is considered constant? b) I air resistance and riction are taken into account, will the length o the runway required be longer or smaller than that ound in a)? 9. A 50-kg block slides, without riction, down an inclined plane, accelerating at a rate o 30 m/s2 What is the plane’s incline relative to the horizontal? 10. A 50-kg block slides down an inclined plane at a constant speed and the orce o riction is 20 N What is the plane’s incline relative to the horizontal? 11. Braking abruptly, a 1000-kg car driving at 100 km/h is subject to a orce o riction rom the ground o 7000 N I the deceleration is constant, what distance will the car travel beore stopping: a) on a horizontal road? b) on a descending road with a 100° incline? 12. A plumb line suspended rom a car ceiling orms a 10° angle with the vertical when the car accelerates on a horizontal road Determine the car’s acceleration

10°

14.3 Newon’s ird lw Newton’s third law states that if body A exerts a certain force FA→B on body B, body B exerts a force FB→A on body A that is equal in magnitude, but opposite in direction. This law is oten ormulated more succinctly (though less precisely) as: to every action there is an equal and opposite reaction. This is why the third law is also known as the action­reaction law. This law also helps to clariy the notion o orce: it states that any orce is exerted by an object on another. In other words, orces only exist i there is an interaction between two bodies. I object A exerts a orce on object B, object B necessarily exerts a orce on object A. When determining which orces are present in a given situation, it is essential to take this into account. The algebraic statement o Newton’s third law is written: FA→B 5 ­FB→A where FA→B 5 Force exerted by A on B, expressed in newtons (N) FB→A 5 Force exerted by B on A, expressed in newtons (N)

The negative sign highlights the act that the two orces are acting in opposite directions. The two orces have the same magnitude, thereore FA→B 5 FB→A. The choice o FA→B or FB→A to designate the action is arbitrary; either o the two orces can be considered the action or the reaction because the two orces exist simultaneously. The symbols used refect the act that one o the orces is exerted on one o the objects, and the other orce is exerted on the other object. The action and the reaction do not act on the same object. I a orce exists, it must always be pos­ sible to identiy the other member o the action­reaction pair. When a soccer player kicks a ball with his oot, or instance, the ball is subject to a orce F→b and acceler­ ates. At the same time, the player’s oot is subject to another orce in the oppo­ site direction that slows it. This orce Fb→ is exerted by the ball on the oot (see Figure 11). Here is another example o Newton’s third law. A boy and a girl are ace to ace on a skating rink. The girl shoves the boy: he accelerates under the eect o the orce Fg→b exerted by the girl. However, she then accelerates backward because o the reaction Fb→g exerted on her by the boy. The girl can never push the boy without being pushed hersel.

Ff→b

Fb→f

Figure 11 An action­reaction pair: the foot exerts a force on the ball and, at the same time, the ball exerts a force on the foot

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See Inerial frames of reference, p 155

→

FE → b →

Fb → E

Figure 12 An action­reaction pair: the Earth exerts a orce o attraction on the ball, and the ball exerts a orce o attraction on the Earth

The case o a ball in reeall is another example o Newton’s third law. I the ball alls, it is because o gravitational attraction: the ball is subject to a orce FE→b exerted by the Earth. The reaction to this orce is a orce Fb→E exerted by the ball on the Earth (see Figure 12). Because o this reaction, the Earth accelerates toward the ball (this acceleration can be measured in a heliocentric coordinate system). Because the Earth’s mass is 6 3 10 24 kg, the Earth’s acceleration is imperceptible, but it is there all the same. When the orces present in a given situation must be determined and a reac­ tion cannot be identifed, it is because the orce does not exist. The case o a person in a moving bus clearly illustrates this act. I the bus abruptly brakes, the person has the impression that a “orce” is pushing her orward yet there is no reaction to this orce, and no agent exerting such a orce can be estab­ lished. Though seemingly apparent, this orce does not really exist. The per­ son is pushed to the ront o the bus because o her inertia. When she was in motion and tending to conserve this motion, the bus slowed down. Example On a windless day, a sailor sets up a battery­powered an at the back o his sailboat and directs it so that it blows air into the sail The sail is perpendicular to the sailboat’s axis, as shown in the image below Will the sailboat move orward? Why or why not? y

→

F(air → sail)

→

F(air → fan) x

Identiy the orces present and determine which orces are acting on the sailboat, as well as on the an and the sail, which are part o the sailboat The an pushes the air orward (orce Fan→air) According to Newton’s third law, the an and the sailboat are pushed backward (Fair→an) as shown in the diagram This is the frst horizontal orce acting on the sailboat The moving air will push the sail (orce Fair→sail ) and the sailboat orward This is the second horizontal orce acting on the sailboat However, the air pushed by the an is slowed due to air resistance In addition, some o the air slips past the sides o the sail As a result, the orce exerted by the air on the sail (Fair→sail ) is smaller than the orce exerted by the air on the an (Fair→an) Solution: The sailboat will not move orward Because Fair→sail  Fair→an, the resultant orce exerted on the sailboat is directed backward As a result, the sailboat cannot move orward and may even move backward Note: Forces exerted on the air are not shown in the diagram Because the question involves the motion o the sailboat, we consider only the orces exerted on the sail­an combination

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Grund rulsin A pedestrian beginning to walk accelerates forward According to Newton’s second law, an external force is therefore being exerted in the same direc­ tion What kind of force is this? It is not the force exerted by the leg muscles because that is a force from within the body The only agent that can propel the pedestrian forward is the ground When the pedestrian begins her stride, the internal force exerted by the muscles causes the rear foot to tend to slide backward, which generates a friction exerted by the ground on the foot This friction (an external force) is directed forward: this is the force propelling the pedestrian The reaction to this force is the friction exerted by the foot on the ground (this force subjects the ground to an acceleration backward) [see Figure 13 ] In the same way, a car accelerates because the engine turns the wheels, which tend to slide on the ground The friction of the ground against the tires, which opposes this sliding tendency, is directed forward and causes the vehi­ cle to accelerate No friction, no propulsion For proof, look at the case of a car on the ice as its driver attempts to accelerate

Ff (g→f)

Ff (f→g)

Figure 13 The friction exerted by the ground on the rear foot

(Ff (g→f)) results in an acceleration of the pedestrian forward

SECtIoN 14.3 1. An apple is on a table a) Is the normal force the force of reaction to the gravitational force exerted on the apple? Why or why not? b) If you answered no to question a), what is the reaction to the gravitational force exerted on the apple?

Newn’s ird lw

FN

Fg

2. An apple is on a table What is the reaction to the normal force exerted on the apple? 3. A small car and an imposing sport utility vehicle (SUV) collide Which of the two vehicles is subject to the greater force?

4. Two objects with respective masses of m and 2m are placed next to each other The arrows in the following diagrams illustrate the gravitational forces exerted on the objects Choose the accurate representation Explain your answer A

2m

B

m

C

m

2m

m

2m

D.

2m

m

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APPLICATIONS Je rulsin A jet­powered airplane and a rocket move orward because they discharge gases backward. A child’s bal­ loon, when it is infated and released, moves accord­ ing to the same principle. An airplane’s jet engine exerts on the gases a orce directed backward, while the gases exert on the air­ plane a orce o equal magnitude directed orward.

through a large slit and accumulates it. With the coor­ dinated contraction o its muscles, the octopus expels the water in the pocket through another opening. The octopus is then pushed orward in the direction oppo­ site to that o the expelled water.

The air that fows into the airplane’s jet engine is com­ pressed beore being mixed with kerosene. The sub­ sequent combustion dilates the gases and increases the pressure. As a result, the gases leave the jet nozzle at high speed. These gases are thereore all subjected to a orce exerted by the jet engine (see Figure 14). Today’s large airplanes are equipped with jet engines that can exert (and be subject to) thrusts o over 300 000 N. Jet propulsion is also used in rockets, though the gases ejected contain no air because the rocket does not stay long in the atmosphere. The ejected gases derive rom the chemical reaction o uel and other substances stored in the rocket’s tanks (see Figure 15). Jet propulsion also exists in nature. Octopi can crawl and swim with their tentacles, but their best means o locomotion is jet propulsion. An octopus has a kind o pocket, called a mantle cavity, that draws in water

aDMISSIoN

air inke

CoMpRESSIoN

Figure 15 The combustion of fuel and other substances stored in rocket’s tank produce gases that are ejected by powerful motors The gases travel toward the ground, and the motors, which are attached to the rocket, are pushed upward This is how the rocket lifts off

CoMBUStIoN

Cmbusin cmber

EXhaUSt

turbine Gs Je engine (irlne)

Figure 14 Cross section of an airplane’s jet engine The blue arrows represent the movement of the air and combustion gases The gases are pushed to the right, while the jet engine, which is attached to the airplane, is pushed to the left

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Je nzzle

Grviionl rcion In his book The Physics, the Greek philosopher Aristotle (384–322 BCE) stated that objects ell to the Earth to return to their “natural place,” but he ailed to establish any link between what happened on Earth and what happened in the sky. According to Aristotle, the Moon orbits around our planet because it was the nature o cosmic movements to be circular, and not because o any kind o attraction. Many scientists would later speculate on a orce o attraction between celestial bodies, but ailed to make convincing arguments or its existence. The German astronomer Johannes Kepler (1571–1630), noted that a high tide occurred in the region o the Earth that aced the Moon, and considered that it was due to an attraction (his theory was incomplete because a high tide was occurring at the same time on the other side o the Earth). The British scientist Robert Hooke (1635–1703) hypothesized that the Sun exerted a magnetic orce o attraction on the planets. Furthermore, although he proposed that attraction was inversely proportional to the square o the distance 1 (F α 2 ), he was unable to prove it. r It was Newton, thanks to his innitesimal calculations, who supplied the proo, building on the work o Tycho Brahe and Kepler. Tycho Brahe (1546–1601), a Danish astronomer, developed highly precise instruments to determine the position o the planets and the stars. With the help o his sister Sophie, he accumu lated

observation data taken over several decades. Kepler used Brahe’s data to discover that the planetary orbits were elliptical. Newton showed theoretically that the existence o a orce o attraction inversely proportional to the square o the distance meant that orbits were elliptical. His theory also showed that the same orce that made objects drop to the Earth’s surace kept the Moon in orbit around the Earth (see Figure 16). Newton or­ mulated his theory o gravitational attraction in an equation that would become amous, known as the law o universal gravitation (see Law of universal gravitation on page 267): Fg 5

Gm1m2 r2

General relativity, hypothesized by Albert Einstein in 1915, abandoned the concept o orce and mapped out a new vision o gravitation. According to his theory, mass curves the space around it, and this curvature infuences the motion o any other body nearby. The Moon thereore orbits around the Earth because it is ollowing the shortest path in a curved space (see Figure 17). On a human scale, relativity provides the same results as Newtonian mechanics. Yet it also describes hither­ to unknown phenomena, such as black holes and the curvature o light near a very massive body.

a B

E

C D

Figure 16 According to Newton’s theory, the force that brings a cannonball to the ground is the same that keeps the Moon orbiting around the Earth If the cannonball is projected at a fast enough speed, it can be put into orbit around the Earth

Figure 17 According to the theory of general relativity, gravitation is not a force of attraction, but an effect of the deformation of space­time due to the mass of the objects that occupy it ChaptER 14 Newton’s Laws

319

chapter

re V I e W

14

Nwon’s Lws

14.1 Newon’s frs law and ineria • In the absence o a resultant external orce acting upon it, the body’s acceleration will be zero. • Newton’s frst law states that to change a body’s motion, i.e. to put it in motion, accelerate it, slow it down, stop it or change its direction, a non­zero external orce is required. I no external orce is exerted on an object, the object’s acceleration will be zero, and the object will remain at rest or continue its motion in a straight line at a constant velocity. • An object’s tendency to retain its state o motion is called inertia. An object’s mass quantifes its inertia. The larger the mass, the more the body tends to stay at rest i it is at rest, or continue its motion in a straight line i it is already in motion.

14.2 the noion o orce and Newon’s second law • The resultant orce exerted on a body is equal to the product o the mass o this body and its acceleration.

FR 5 ma

• In the International System o Units (SI), because mass is expressed in kilograms (kg), and acceleration in metres per second squared (m/s2), the unit o orce, called the newton (N), is defned as ollows: 1 N 5 1 kg m/s2. • Newton’s frst law is a particular case o the second law: i the resultant orce is zero, acceleration is zero, and vice versa. • To solve a dynamic problem with Newton’s second law: 1. Identiy the body in question. 2. Draw a diagram o all the orces acting on the body. 3. Choose an appropriate coordinate system, generally with the positive x­axis pointing in the direction o the body’s presumed acceleration. 4. Write the equations derived rom Newton’s second law (FRx 5 max and FRy 5 may). 5. Solve the equations.

14.3 Newon’s third Law • I a body A exerts a certain orce FA→B on a body B, body B exerts a orce FB→A on body A that is equal in magnitude, but opposite in direction.

FA→B 5 ­FB→A

The negative sign in the statement highlights the act that the two orces are acting in opposite directions, but have the same magnitude: FA→B 5 FB→A. • Newton’s third law states that a orce never exists alone. Forces only exist i there is an interaction between two bodies. • Either o the two orces FA→B and FB→A can be considered the action or the reaction. The action and the reaction do not act on the same object.

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ChaptER 14

Newon’s Lws

1. Indicate which o Newton’s laws applies to each o these situations a) At a bend in an icy road, a car continues in a straight line and goes o the road b) You kick a ball and your oot decelerates c) While you push a car stuck in the snow, your eet slide and you move backward d) As you get your bike going, you push twice as hard on the pedals and accelerate twice as ast 2. A small car and a sport utility vehicle (SUV) collide Which o the two vehicles is subject to the greater acceleration? 3. Leaving an object on the rear-door windowsill o a car is not recommended Why? 4. The Earth exerts a orce o 1000 N on a satellite in orbit at an altitude o 8000 km What orce does the satellite exert on the Earth? 5. A 1200-kg car is parked on a hill inclined at an angle o 120° relative to the horizontal Determine the magnitudes o the orce o riction and the normal orce acting on the car 6. A rocket carrying a ull load has a mass o 292 3 106 kg Its engines exert a thrust o 334 3 107 N a) What is the gravitational orce acting on the rocket at launch time? b) What is the resultant orce acting on the rocket at launch time? c) What is the rocket’s acceleration when it leaves the launcher? d) As the rocket gains in altitude, the engines’ thrust remains constant, but the rocket’s mass decreases Explain why e) While the engines maintain their thrust, will the rocket’s acceleration increase, decrease or remain constant? Explain your answer

7. A screenwriter consults you about her script or a science ction lm She wants you to veriy the accuracy o the ollowing scene: “The main space vessel is transporting several smaller vessels that look like war planes Each o these vessels has only one thruster in the back During a space battle, one o the vessels takes several sharp turns (with a small radius o curvature) as the gas umes leaving its thruster trace the same sharp turns behind” Is this a realistic scene? 8. A 200-kg crate is on a horizontal foor The crate is pulled with a rope held 300° above the horizontal I the orce o riction is 200 N, what tension must there be in the rope to put the crate in motion? What is the normal orce at this moment? 9. On a horizontal surace, a 200-kg sled is pulled at a constant speed with a horizontal rope whose tension is 300 N a) What is the coecient o riction? Reminder: Ff (kinetic) 5 µk 3 FN b) I two people weighing 600 kg each sit on the sled, what orce will be required to pull the sled at a constant speed? 10. A 1500-kg car driving at 900 km/h brakes; the ground exerts a orce o riction o 5000 N I the deceleration is constant, what distance does the car cover beore coming to a stop: a) on a horizontal road? b) on a descending road inclined 50°? 11. A dog pulls the rope o a 50-kg sled up a hill with an incline o 20° above the horizontal The rope is directed 30° rom the slope and the dog exerts a orce o 200 N on the rope The orce o riction is 50 N What is the sled’s acceleration?

ChaptER 14 Newton’s Laws

321

322

CONTENTS ChApTER 15

Work and Mecanical power . . . .325 ChApTER 16

Mecanical Energy . . . . . . . . . . . . . . .339 ChApTER 17

Elastic potential Energy . . . . . . . . . .363

Energy is a ysical concet tat reresents te caacity o a system or body to roduce work or to cange te state o oter systems. Tis concet, wic is central to all uman activity, as been a key element in scientifc, economic, industrial and social develoment. In mecanics, energy is very closely connected to te notion o work, wose defnition, in tis context, diers rom te oular meaning o te word. Te same is true or te defnition o ower, wic is te rate at wic work is erormed. Mecanics involves several tyes o energy: kinetic energy, otential energy and mecanical energy.

Kinetic energy is te energy o bodies in motion, otential energy is te energy stored in a body, and mecanical energy is te sum o te frst two. Among te various tyes o otential energy, it is imortant to distinguis gravitational otential energy, wic is due to te osition o a body in relation to a gravitational feld, and elastic otential energy. Te reason we study elical srings is to analyze and construct models o elastic objects and, esecially, to reveal te caacity o tese objects to store and release elastic otential energy according to te tecnical alications in wic tey are used. 323

ChApTER

UNIT

15.1 Work 15.2 Mechanical power

16.1 Kinetic energy

5

ENERGY AND ITS TRANSFORMATIONS

15

WORK AND MEChANICAl pOWER

16.2 Work-energy theorem ChApTER

16

MEChANICAl ENERGY

16.3 Gravitational potential energy 16.4 Conservation of mechanical energy 16.5 Conservation of total energy

ChApTER

324

17

ElASTIC pOTENTIAl ENERGY

17.1 Behaviour of constrained helical springs 17.2 The energy stored in a spring

Work and Mecanical power

W

hile they are commonly used in everyday language, the notions o work and power have very specifc meanings in physics. Both are closely connected to the application o orces on bodies. For instance, dogs pulling a sled on a snow-covered path are doing work, because they are exerting a orce that causes the displacement o their load. Furthermore, i this team o dogs does this work quickly, it is said that it is powerul.

In this chapter, you will learn about mechanical work by using the notion o scalar product, and you will become amiliar with the concept o mechanical power. In certain cases, the concepts o work and power do not have the usual meanings we tend to associate with them. For example, a weightliter supporting a heavy load while motionless is not perorming any work rom a physics perspective.

Review Characteristics of force                                    13 Effective force                                            15 Relationship between work, force and displacement            16 Relationship between power and electrical energy              19

15.1 15.2

Work                                  326 Mecanical ower                    331 ChApTER 15 Work and Mechanical Power

325

15.1 Work In everyday language, work is defned as various human activities coordinated to produce something or accomplish a task. It is somewhat connected to the notion o eort. However, a task can be exhausting without corresponding to the defnition o work in the physics sense o the term.

15.1.1

Work, force and displacement

Work (W) is equal to the scalar product of the force (F ) acting on the body through the displacement of this body (∆s ). (The notion o scalar product is defned in the Furthering your understanding section on the ollowing page.) In physics, we speak o work when we apply a orce to a body to displace it (see Figure 1). The relationship between work, orce and displacement is written as ollows (all units are SI units): W5F

•

∆s 5 F 3 ∆s 3 cos θ

where W 5 Work, expressed in joules (J) F 5 Force acting on the body, expressed in newtons (N) ∆s 5 Displacement of the body, expressed in metres (m) θ 5 Angle between the force and the displacement

→

F θ B

→

∆s

A

Figure 1 By pulling on the rope, the skier exerts a force (F ) that causes the sled to move forward from point A to point B, which corresponds to its displacement (∆s ) The work provided is equal to the scalar product of the force and the displacement

It is important to understand that a orce only does work i its point o application is in motion. In Figure 1, the point o application o F is in motion rom point A to point B. See Displacement, p 205

326

This notion o displacement is undamentally linked to work, in the sense that a orce can be exerted without work being done. Thereore, a person who is carrying a load on his back while remaining stationary is not doing any work in the physics sense o the term, even i he is making an eort and getting tired. It should also be noted that work (W ) is a scalar quantity while orce (F ) and displacement (∆s) are vector quantities.

UNIT 5 Energy and its Transformations

The unit o work being the joule (J), we can write, in accordance with the defnition o work. 1J51N31m51N3m 1 joule (J) 5 1 newton metre (N m) Based on the defnition o work, we can also write: W 5 F 3 ∆s 3 cos θ 5 (F 3 cos θ ) 3 ∆s The quantity (F 3 cos θ)is the magnitude o the orthogonal projection o orce (F ) on vector ∆s (see Figure 2). This projection represents the vector component o the orce in the direction o the displacement. Because this orce is the only one at work, it is called eective orce (Feff ). These geometric considerations enable us to deine work using eective orce. We can thereore write: W5 F

•

∆s 5 Feff

Furthering

•

→

F θ →

B

∆s

→

A

Feff

Figure 2 Eective orce (Feff ) is the vector component o the orce in the direction o the displacement

∆s 5 Feff 3 ∆s 5 F 3 cos θ 3 ∆s

your understanding

Scalar roduct The scalar product o two vectors u and v , orming an angle θ (see Figure 3), is defned as ollows: u • v 5 u 3 v 3 cos θ

It should be noted that, as indicated by its name, the result o the scalar product o two vectors is a scalar This is a very important property because it stipulates that the scalar product o two vector quantities results in a scalar quantity

In this equation, the sign • represents the scalar product o vectors u and v (arrows on top o frst u and v) , or which u and v are the magnitudes

Examle Calculate the scalar product o two vectors u and v knowing that u 5 50, v 5 60 and the angle between u and v 5 60°

B

B

→

→

v

v

θ

θ →

u

A

Figure 3a

Data: u 5 50 v 5 60 θ 5 60° u •v 5?

C

The scalar product o vectors u and v is equal to the product o the two vectors’ magnitudes multiplied by the cosine o the angle between the two vectors

A

Solution: u • v 5 u 3 v 3 cos θ 5 50 3 60 3 cos 60° 5 30 3 05 5 15

v  cos θ →

u

H

C

Figure 3b

The magnitude o the orthogonal projection o vector v on vector u is equal to v 3 cos θ

Figure 3b demonstrates that the quantity v 3 cos θ corresponds to segment AH on the right, which is in act the orthogonal projection o vector v on vector u  We can thereore say that the scalar product o two vectors u and v , orming an angle θ, is equal to the product o the magnitude o vector u multiplied by the magnitude o the orthogonal projection o vector v on vector u 

Ortogonal vectors The scalar product depends on the angle between the two vectors Thereore, when two vectors are perpendicular to one another, the scalar product is zero, because the cosine o the right angle is zero Thereore, the scalar product o two vectors can be zero even i the vectors are not zero I u  v ⇒ θ 5 90°, then u • v 5 u 3 v 3 cos 90° 5 u 3 v 3 0 5 0

ChApTER 15 Work and Mechanical Power

327

particular values of work

15.1.2

Given that the values of the cosine function vary between -1 and 11 and that work depends on the cosine of the angle between force (F ) and displacement (∆s ), some angular values are particularly interesting.

Maximum work: q 5 0° When force (F ) and displacement (∆s ) have the same direction, the angle θ between them is zero and its cosine is equal to 1. This maximum value of an angle’s cosine indicates the work’s maximum value. θ 5 0° ⇒ cos 0° 5 1 W5F

•

∆s 5 F 3 ∆s 3 cos θ 5 F 3 ∆s 3 1 5 F 3 ∆s Wmax 5 F 3 ∆s

When the angle θ between force (F ) and displacement (∆s ) is zero, there is maximum work (see Figure 4a). In this case, F 5 Feff .

Zero work: q 5 90° A force (F ) that is perpendicular to displacement (∆s) does no work. Because the cosine of the right angle is zero, we have: θ 5 90° ⇒ cos 90° 5 0 W5F

•

∆s 5 F 3 ∆s 3 cos θ 5 F 3 ∆s 3 0 5 0

When the force (F ) is perpendicular to displacement (∆s ), there is zero work done (see Figure 4b). In this case, Feff 5 0.

Effective work: 0° ≤ q < 90° The work is said to be effective when force (F ) provokes or maintains the motion. This is possible if effective force (Feff ) has the same direction as the displacement (∆s ). In this case, the angle θ between force (F ) and displacement (∆s ) is between 0° and 90°, and the work’s value is positive (see Figure 4c). Therefore, there is effective work when W > 0 (0° < θ < 90°).

→

F →

F

→

→

∆s

∆s

a) Maximum work (F 5 Feff )

b) Zero work (Feff 5 0 )

Resistive work: 90° < q ≤ 180°

→

F

→

F

θ θ

→

Feff →

∆s

c) Effective work (Feff and ∆s are in the same direction)

→

Feff

d) Resistive work (Feff and ∆s are in opposite directions)

Figure 4 Particular values of work

328

→

∆s

UNIT 5 Energy and its Transformations

Work is said to be resistive when force (F ) is opposed to motion. This happens if the effective force (Feff ) is in the opposite direction of displacement (∆s ). In this case, the angle θ between force (F ) and displacement (∆s ) is between 90° and 180°, and the work’s value is negative (see Figure 4d). Therefore, work is resistive when W < 0 (90° < θ ≤ 180°).

Examle A A crane is liting, at a constant speed, a mass equal to 5000 kg to a height o 20 m rom the ground a) Calculate the work done by the crane b) What type o work is this? Data: m 5 5000 kg h 5 20 m W5?

Solution: a) Work done by the crane Calculation o the orce exerted by the crane: Since the motion is at a constant speed, the magnitude o the orce (F ) exerted by the crane is equal to the magnitude o the weight (Fg ) o the load: F 5 Fg 5 mg 5 5000 kg 3 980 m/s2 5 49 3 104 N

→

F

Calculation o the work done by the crane: The angle between the displacement vector and the orce vector being zero, we can write: W 5 F • ∆s 5 F 3 ∆s 5 49 5 104 N 3 20 m 5 980 3 105 J

• →

Fg

The work done by the crane is equal to 98 3 105 J

→

∆s

b) Type o work Force (F ) and displacement (∆s ) have the same direction; thereore it is eective work The angle between the two vectors (F and ∆s ) is zero; thereore it is maximum work

Examle B A person is pulling a crate with a drawbar pull F 5 600 N at 25° The displacement o the crate is defned by vector ∆s 5 500 m at 0° a) Draw a diagram o the problem Represent and identiy the orces that are exerted on the crate b) Determine the orces that do work and those that do not c) Calculate the magnitude o the eective orce d) Calculate the work required to move the crate Data: F 5 600 N at 25° ∆s 5 500 m at 0° θ 5 25° 2 0° 5 25°

Solution: a)

y →

FN →

F

c) Calculation o the magnitude o the eective orce: Feff 5 F 3 cos θ 5 600 N 3 cos 25° 5 544 N

θ x

→

∆s

→

Fg

b) The normal orce (FN ) and the weight (Fg ) are perpendicular to the displacement and do no work The drawbar pull (F ) is the only orce that does work

d) Calculation o the work required to move the crate: W 5 Feff 3 ∆s 5 544 N 3 500 m 5 272 3 103 J

FN 5 Normal orce Fg 5 Weight

F 5 Drawbar pull

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Mechanical work and biological work In the mechanical sense o the term, the notion o work involves displacement o a orce’s point o application A person who is carrying a load on her back while remaining stationary thereore does not do any mechanical work Nonetheless, she is making an eort and is becoming tired in the same way as she would be i she were pulling a sled or pushing a cart Biomechanics and exercise physiology specialists make a distinction between mechanical work and biological work Studying how muscles unction shows that muscular work can be divided into two distinct categories: dynamic work and static work As its name indicates, dynamic work is akin to mechanical work For example, when a weightliter lits weights o the ground and raises them over his head, some o his muscles contract because o the shortening o the muscular fbres, which enables him to do mechanical work In act, the athlete has applied an overall orce that has caused the displacement o the weights rom the ground to a cer-

SECTION 15.1

tain height But what happens when he wants to keep the weights above his head and remain stationary in ront o the judges? He will not be doing any mechanical work because the height o the weights does not change and he, himsel, is remaining stationary However, this is not the case with muscular work To remain standing with a load over his head, the athlete uses many muscles that will work statically This work is necessary as much to maintain the shortening o the fbres o the contracted muscles as to ensure the increase o muscle tone or others The weightliter’s muscles consume chemical energy to maintain the static position Visible maniestations o static work are atigue, shortness o breath, increase in body temperature and perspiration Considered by specialists to be more tedious than dynamic work, static work requires more signiicant muscular eort It is or this reason, among others, that ergonomists recommend eliminating static positions held or too long

Work

1. A sled is being pulled on a horizontal surace with a 620-N orce at a 42° angle above the horizontal What quantity o work is required to pull the sled 160 m? 2. A orce acts at an angle o 30° with respect to the displacement’s direction What orce must be applied to do 9600 J o work over a 25-m displacement? 3. A 640-N orce does 12 500 J o work over a 24-m displacement What is the angle between the orce and the displacement?

6. A 71-kg window washer is standing on a 179-kg platorm suspended along the side o a building I an electric motor raises the platorm by 58 m along the side o the building, what is the quantity o work done? 7. A person is pushing a box on a foor with a 250-N orce The box is moving at a constant speed over 1275 m What quantity o work does the person do on the box?

4. A woman with a mass o 60 kg bungee jumps rom a bridge She reealls over a distance o 20 m beore the elastic starts stretching What quantity o work does the gravitational orce do during this 20-m all? 5. To move a large rock, a bulldozer exerts a orce o 5000 N at a velocity o 2 m/s or 20 s What is the work done by the bulldozer?

330

UNIT 5 Energy and its Transormations

Ff

F 250 N

15.2 Mecanical ower

HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

In everyday lie, it is difcult to make a clear distinction between work and power. The dierence between these two quantities is more obvious i you consider the ollowing example. Using a rope, a person is capable o liting a load rom the ground to the roo o a house. This same task can be carried out by a piece o equipment, the electric winch. The person and the winch thereore do the same work; but the electric apparatus accomplishes this task much aster. It is said that the electric winch’s mechanical power is greater than that o the person who did the work. We thereore notice that mechanical power involves the time required to get the work done. This quantity then characterizes the speed at which the work is done. Mechanical power (P) is the quantity of work (W) done per unit of time. This relationship is written as ollows: JAMES WATT P5

Scottish engineer (1736–1819)

W ∆t

where P 5 Mechanical power, expressed in watts (W) W 5 Work, expressed in joules (J) ∆t 5 Time interval, measured in seconds (s) Like work, power is a scalar quantity. According to the International System o Units (SI), the unit o power is the watt and its symbol is W. 1W5

1J 5 1 J/s 1s

1 watt 5 1 joule per second The ollowing examples show how this equation can be used. Examle A What is the power generated by a power shovel that does 60 kJ o work in 15 s? Data: W 5 60 kJ 5 60 3 103 J ∆t 5 15 s

Solution: P5

W 60 3 103 J 5 5 40 3 103 J/s 5 40 kW ∆t 1,5 s

Examle B A horse pulling a 100-kg sleigh or 500 minutes covers a distance o 500 m on horizontal ground The rope used to pull the sleigh orms an angle o 236° with the ground The orce exerted by the horse on the rope is assumed to be constant and equal to 300 N What is the power generated by the horse? Data: m 5 100 kg ∆t 5 500 min 5 300 s ∆s 5 500 m F 5 300 N θ 5 236° W5? P5?

Solution: 1 Calculation o the horse’s work: W 5 F • ∆s 5 F 3 ∆s 3 cos θ 5 300 N 3 500 m 3 cos 236° 5 137 3 105 J

From a very young age, James Watt demonstrated a scientifc curiosity and an aptitude or learning His health was ragile and he thereore did not attend HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY much school; he was mostly home-schooled by his mother Ater a period o study in a mathematical instrument-making workshop in London, where he learned the principles o mechanics, James Watt returned to Scotland He ound work at the University o Glasgow doing maintenance and repairs o scientifc equipment It is while he was in this position that he succeeded in improving one o the original steam engines He subsequently submitted many patents in this feld James Watt built steam engines that remained unsurpassed or nearly 50 years His inventions had a signifcant impact on the industrial development o Great Britain in the 19th century The watt, the International System’s unit o power, was named in his honour

2 Calculation o the power generated by the horse: W 137 3 105 J P5 5 5 458 W ∆t 300 s

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Example C A car is travelling up a slope with a 60° incline with respect to the horizontal, at a constant velocity of 360 km/h (friction is negligible) If the vehicle’s mass is equal to 1500 kg, calculate: a) the work done by the vehicle’s engine in 600 min b) the power generated by the vehicle’s engine Data: θ 5 60° 1 000 m 1h v 5 360 km/h 3 3 5 100 m/s 1 km 3600 s m 5 1500 kg ∆t 5 600 min 5 360 s Solution: a) Since friction is negligible, the forces acting on the car are the engine’s force (F ), the vehicle’s weight (Fg ) and the normal force (FN ) They are represented in the following diagram, along with the system of coordinates chosen to solve this problem y

→

FN x →

F Fgx

θ →

Fg

θ

Fgy

As the vehicle is moving along the x-axis, the only forces or force components that contribute to the motion and which should be taken into account are those that are parallel to the x-axis We can therefore write: FRx 5 F 2 Fgx 5 max therefore F 2 Fg sin θ 5 max Because velocity is constant the acceleration a x is zero By taking this into account and by projecting the force vectors on to the x-axis, we have: F 2 Fg sin θ 5 0 Knowing that Fg 5 m g, we obtain: F 5 m g sin θ The distance travelled in 6 min is: ∆s 5 v 3 ∆t 5 100 m/s 3 360 s 5 36 3 103 m The drive force being parallel to the displacement, the work done by the car will therefore be: W 5 F 3 ∆s 5 m 3 g 3 sin θ 3 ∆s 5 1500 kg 3 980 m/s2 3 sin 60° 3 36 3 103 m 5 553 3 106 J 5 553 MJ b) The power generated by the engine is calculated by taking into account the work done and the time required to do it: P5

332

UNIT 5 Energy and its Transformations

W 553 3 106 J 5 154 3 104 J/s 5 154 kW 5 ∆t 360 s

horseower In order to quantify the power generated by a steam engine, the Scottish engineer James Watt (1736–1819) compared it to the power of a horse Before the advent of the Industrial Revolution, horses were used for jobs where a certain degree of power was required This comparison is at the source of the term “horsepower,” a unit of power that is not part of the SI, but which is still used today, particularly in the automotive industry By observing horses at work, James Watt estimated that the strongest could, per minute, lift a 33 000-pound mass to a height of 1 foot If you consider that a pound is equivalent to 04536 kg and that a foot measures 03048 m, we can calculate the power of a horse in SI units: P5

5

W ∆t Fg

•

∆t m 3 g 3 ∆s ∆t

5

m 3 ∆s 3g ∆t kg m 3 1 foot 3 03048 lb ft 3g 60 s

33 000 pounds 3 04536 5 5 7604

a) In the imperial system: kg 3 m 1 hp 5 7604 3g s m kg 3 m 5 7604 3 980 2 s s kg 3 m2 5 745 5 745 W s3 b) In the metric system: kg 3 m 1 hp(M) 5 75 3g s m kg 3 m 5 75 3 980 2 s s kg 3 m2 5 735 5 735 W s3

∆s

5

The British unit of power is known as horsepower and its symbol is “hp” The metric unit is the metric horsepower and its symbol is “hp(M)” There is a slight difference between the British unit of power, as based on the observations of James Watt, and the metric unit of power By incorporating the value of the gravitational acceleration g 5 980 m/s2, we obtain the following results:

kg 3 m 3g s

The Imperial system has retained this value,while this number kg 3 m has been rounded down to 75 3 g in the metric system s (which is not the SI)

Horsepower and hp(M) are not part of the SI, because the mass estimated by James Watt is purely arbitrary But the watt (W), named in his honour, is the SI unit of power that is expressed as follows, according to the three base units of mechanics in the SI: kg 3 m2 1W5 1 s3 Nearly a century after James Watt developed high performance steam engines, horses were still used for transportation Created in 1861, public transit in Montréal consisted of horse-drawn cars on wheels (see Figure 5) in the summer In winter, sleds replaced the cars

Figure 5 Horse-drawn tramway on rue Saint-Denis, in Montréal, 1887

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333

Furthering

your understanding

Relationship between power, force and velocity As shown in the beginning o Section 151, a orce only does work i its point o application is in motion The displacement o the point o application is thereore done at a certain velocity (v ) It is possible to express power based on this velocity (v ) i this velocity is constant To do so, you only need to take into account the relationship between displacement (∆s ) and velocity (v )

on page 332 by using the relationship between power (P ), orce (F ) and velocity (v ) In this example, velocity was constant and we ound that the power generated by the engine was equal to:

∆s 5 v 3 ∆t

By using the constant velocity (v ) o the displacement o the point o application o orce (F ), ie the velocity o the vehicle, and by taking into account the act that the angle between these two vectors is zero, we can calculate the power generated by the engine

Power will thereore be written: P5

F • ∆s F • v 3 ∆t W 5 5 5F •v ∆t ∆t ∆t

Power (P ) is thereore equal to the scalar product o orce (F ) and velocity (v ) P5F •v To veriy i this relationship can be used to calculate power in the same way as the ormula defned at the beginning o Section 152, it is possible, or example, to solve problem b) in example C

SECTION 15.2

2. How long would it take an electric engine of 25 kW to do 75 × 104 J of work? 3. A crane’s engine lifts a mass of 150 metric tonnes to a height of 65 m in 350 min How much power is generated by the crane? 4. What is the average power of an engine that lifts a 250-kg mass to a height of 30 m in 20 s? 5. A crane has a power of 40 kW How long will it take the crane to perform 70 kJ of work?

UNIT 5 Energy and its Transormations

W 553 3 106 J 5 154 3 104 J/s 5 154 kW 5 ∆t 360 s

P5F •v 5F3v F 5m 3 g 3 sin a

⇒ P 5 m 3 g 3 sin a 3 v

P 5 1500 kg 3 980 m/s-2 3 sin 60° 3 100 m/s kg 3 m2 5 154 3 104 5 154 kW s3 This result is identical to the one obtained previously

Mechanical power

1. What power is provided by a crane that does 6 3 104 J of work in 5 min?

334

P5

6. What power does the crane in question 5 generate if it lifts a concrete container weighing 125 metric tonnes over a vertical distance of 574 m in 35 s? 7. Water is pumped to the top of an elevated tank that is 92 m high The discharge toward the top of the elevated tank is 75 L/s Every litre of water has a mass of 1 kg What power is required to maintain the discharge? 8. A 613-kg mass is placed on a lift truck that generates up to 950 W of power What is the constant velocity of the load while the truck is lifting it?

APPLICATIONS Te ower of wind turbines When the wind blows, the air masses in motion exert a orce on the obstacles that are in their path. The orce exerted by the wind on wind turbines’ blades thereore causes them to turn, thereby producing work (see Figure 6). The greater the quantity o work done per unit o time, the greater the power o the wind turbine. Rotor rotation direction

its maximum level when the wind speed is nearly 14 m/s (approximately 50 km/h). Beyond this value, the wind turbine is slowed to avoid submitting it to signifcant constraints, but it retains its optimal power. When wind speed surpasses 25 m/s (90 km/h), the wind turbine is completely stopped or saety reasons. This corresponds to the wind turbine’s “cut-out speed.

Blade

Ø 125 m 5 MW

Rotor diameter

Ø 70 m 1.5 MW Ø 46 m 0.6 MW Ø 30 m 0.30 MW

Wind turbines are oten classifed according to the electric power they are able to generate (see Table 1). Table 1 Classifcation o wind turbines based on power Tye of wind turbine Low power Average power High power

power generated P < 300 kW 300 kW < P < 1,5 MW P > 15 MW

Generally speaking, when a wind turbine’s power is high, it is very tall and has a large rotor diameter (see Figure 7). The power o a wind turbine obviously depends on the speed o the wind that causes its blades to turn. However, this relationship is not linear. The variation o a wind turbine’s power based on wind speed can be represented by a typical curve (see Figure 8). The graph in Figure 8 shows that a minimum wind speed o 3 m/s (approximately 10 km/h) is required to start the wind turbine. This speed corresponds to the wind turbine’s “cut-in speed.” The power then reaches

Figure 7 Generally speaking, the greater a wind turbine’s height and rotor diameter (Ø), the greater the wind turbine’s power is

The popularity o “green energy” has led to a remarkable increase in the number o wind turbines around the world. According to the Global Wind Energy Council (GWEC), worldwide, wind power increased by 31% in 2009 to reach a total o 157.9 GW. In Canada, there has been a 40% growth, which corresponds to 3.3 GW o power. As a means o comparison, the Robert-Bourassa hydroelectric generating station, the most impressive one in Hydro-Québec’s network, can generate 5.6 GW o power. Power of a wind turbine based on wind speed

Power (kW)

Figure 6 Front view o a three-bladed horizontal-axis wind turbine

3

14 Wind speed (m/s)

25

Figure 8 Typical curve o the variation o power o a wind turbine based on wind speed

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335

The steam engine The idea o using the power o steam goes back to Antiquity. Since the 1st century AD, the Greek engineer and mathematician Hero o Alexandria developed a rudimentary steam engine, the aeolipile, made up o a vessel upon which was xed a sphere with two small exhaust nozzles (see Figure 9). When the vessel was placed near a source o heat, the water it contained started boiling and produced steam that caused the sphere to rotate. However, this machine was used only or entertainment purposes and could not create enough energy to accomplish work. It was not until the 17th century that there was renewed interest in steam power. The French inventor Denis Papin (1647–1714) established principles or the operation o the steam engine and constructed a piston engine that could convert the thermal energy o steam into mechanical energy. The objective o this invention was to lit heavy loads or activate a ountain’s water jets. Subsequently, the English mechanic Thomas Savery (1650–1715) created a new steam engine to pump the water out o coal mines. This machine had little success, because it did not resist well to the pressures required to make it work and risked exploding. Savery then started working with the English mechanic Thomas Newcomen (1663–1729), and together they

invented the atmospheric engine in which outside air pressure was the source o the drive orce. This steam engine was used to pump water out o the mines but its low output made its use inecient or other work. In 1763, the Scottish engineer James Watt (1736–1819) was put in charge o repairing one o Newcomen’s atmospheric engines. He then proposed improvements that revolutionized the design o steam engines (see Figure 10). To optimize the atmospheric engine’s output and power, Watt added a condenser and closed the extremity o the cylinder to prevent heat loss. In 1769, he patented his rst steam engine and continued to perect it. He invented the valve that made it possible to apply pressure when the piston was raised or lowered. He also created the centriugal governor, the fywheel and the beam, innovations that increased the power and eectiveness o the engine. Watt’s steam engine had signicant importance during the advent o industrialization in the 18th and 19th centuries. It led to the acceleration o industrial production and contributed to the rapid development o new means o transportation such as the steamboat and steam locomotive. Watt’s engine dominated industrial production or over a century beore being replaced in the 20th century with the internal combustion engine and the electric motor.

Beam

piston rod Cylinder Valve

Centrifugal governor Cistern containing the condenser

Figure 9 Hero of Alexandria’s aeolipile

336

UNIT 5 Energy and its Transformations

Figure 10 James Watt’s steam engine

Flywheel

ChApTER

15

Work and Mecanical power

15.1 Work • Work (W) is equal to the scalar product of the force (F ) acting on the body by its displacement (∆s ). W5F

•

∆s 5 F 3 ∆s 3 cos θ

• A force only does work if its point of application is in motion. • The unit of work is the joule (J). 1J51N3m 1 joule (J) 5 1 newton metre (N m) • The orthogonal projection of force (F ) on vector ∆s is called effective force (Feff ). • There is maximum work when the angle (θ ) between force (F ) and displacement (∆s ) is zero. In this case, F 5 Feff . • There is zero work when force (F ) is perpendicular to displacement (∆s ). In this case, Feff 5 0. • There is effective work when the angle (θ ) between force (F ) and displacement (∆s ) is between 0° and 90°. In this case, W > 0. • There is resistive work when the angle (θ ) between force (F ) and displacement (∆s ) is between 90° and 180°. In this case, W < 0.

15.2 Mecanical ower • Mechanical power (P) is the quantity of work (W) done per unit of time. P5

W ∆t

• According to the SI, the unit of power is the watt and its symbol is W. 1W5

1J 51 J/s 1s

1 watt 5 1 joule per second

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337

ChApTER 15

Work and Mecanical power

1. A tractor pulling a plow exerts a drawbar pull o 750 3 103 N over a distance o 320 km I it has a power o 25 kW, how much time does it need to do this work? 2. It takes an elevator 250 minutes to climb rom the lobby to the 10th foor o an apartment building The tenants have asked the landlord to increase the elevator’s speed The landlord responded that i the elevator went aster, it would perorm more work and consume more energy Is the landlord right? Explain your answer 3. A 50-N orce is causing a 60-kg object to move on a rough horizontal foor at a constant velocity o 25 m/s The orce is parallel to the foor a) Calculate the work done in 25 s b) What is the power generated? c) What is the magnitude o the orce o riction that opposes the motion?

F

l

0m 1.5

Direction of te motion

40 m

1300 kg

338

6. A student is pulling a sled at a constant velocity and with a 60-N orce The sled’s rope is not parallel to the ground; there is a height dierence rom one end o the rope to the other equal to 90 cm What is the quantity o work done by the student i he must travel a distance equal to 400 m?

h  90 cm

4. An elevator and its passengers have a mass o 1300-kg  The elevator’s engine causes it to rise to the 12th foor in 75 s, over a distance o 40 m a) Calculate the work done by the elevator’s engine b) What is the power generated by the elevator?

5. A chairlit is transporting skiers to the summit o a 300-m mountain The average mass o a skier, including equipment, is 80 kg The chairlit can transport three skiers to the summit every 30 s a) Determine the power required to accomplish this task (assuming that the chairlit does not slow down when skiers are embarking on it) b) I riction increases the power required by 25%, what power must the chairlit’s engines provide?

UNIT 5 Energy and its Transformations

7. A car with a 2000-kg mass is travelling up a slope at a constant velocity o 90 km/h The air resistance opposed to this motion is 450 N The slope orms an angle o 6º with respect to the horizontal a) On a diagram, represent all the orces acting on the car when it is travelling on the slope b) Calculate the orce exerted by the engine to maintain the car’s velocity c) What work is done by the car’s engine in 5 min? d) What power is generated by the engine?

Mcnicl engy

A

waterall, an airplane’s fight, a pendulum’s motion and a stretched spring are some examples among many others o situations that involve energy. Energy is a undamental concept in physics. Generally speaking, energy is considered to be the capacity to generate motion and transorm objects. In mechan­ ics, the main orms o energy are kinetic energy, that depends on the motion o an object, and potential energy, which depends on the position o this object in its environment.

rviw Relationship between work and energy                       16 Kinetic energy                                            17 Potential energy                                          17 Law of conservation of energy                              18 Solar energy                                             19

In this chapter, you will learn how to distinguish and calculate kinetic energy and gravitational potential energy, and you will use the law o conservation o mechanical energy to solve dynamics problems. You will also discover that the concept o energy acilitates the analysis o certain situations that you have already studied using Newton’s laws and kinematic equations.

16.1 16.2 16.3 16.4

Kinic ngy                         340

16.5

Consvion of ol ngy          354

Wok-ngy om                 342 Gviionl onil ngy       347 Consvion of mcnicl ngy                                 350

Chapter 16 Mechanical Energy

339

16.1 Kinic ngy HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

Kinetic energy is the energy of an object in motion. Kinetic energy depends on an object’s mass and velocity. The heavier an object is and the faster its motion, the greater its kinetic energy will be (see Figure 1). For example, a large bowling ball travelling at the same speed as a small ball has greater kinetic energy. And a ball travelling at 6 m/s has a greater kinetic energy than if it were travelling at 4 m/s.

JaMes presCott JoUle British physicist (1818–1889) The son o a British brewer, James Prescott Joule worked in the amily business until the age o 35, at which time his interest in science shited rom being HISTORY HIGHLIGHTS HISTORY merely a hobbyHIGHLIGHTS to his main ocus Starting in the 1840s, he began to investigate the equivalence o work and heat, two energytranser processes He discovered that the electric current that owed through coils released heat and quantitively studied the phenomenon, now known as the Joule eect He conducted a historic experiment demonstrating that mechanical work can also heat water Based on his results, Joule predicted that the temperature o Niagara Falls would warm by 01 °C ater alling 52 metres Joule was the frst scientist to clearly describe the concept o energy His experiments contributed to establishing the law o conservation o energy The unit o work and energy in the International System was named in his honour, in recognition o his contributions in many felds (including electromagnetism, electricity, and thermal and mechanical energy)

340

Figure 1 An object in motion, like a bowling ball, possesses kinetic energy When the ball hits the pins, the eect produced depends on the quantity o kinetic energy that the ball possessed beore impact

The relationship between the kinetic energy, mass and velocity of an object in motion is represented by the following equation (whose form will be ex­ plained in Section 16.2). 1 mv 2 2 where Ek 5 Kinetic energy, expressed in joules (J) m 5 Mass, expressed in kilograms (kg) v 5 Velocity, expressed in metres per second (m/s) Ek 5

From this equation, we can deduce that kinetic energy increases in a way that is directly proportional to the mass of an object. Therefore, if object A has a mass twice as large as that of object B and both are travelling at the same vel­ ocity, the kinetic energy of object A is double that of object B. Furthermore,

UNIt 5 Energy and its Transormations

kinetic energy increases in a way that is directly proportional to velocity squared. This means that if, for example, the velocity of an object doubles, the kinetic energy of this object quadruples. However, it should be noted that kinetic energy is a scalar quantity: it does not depend on the direction of the velocity but on its magnitude. An object travelling at 10 m/s upward has the same kinetic energy as if it were travel­ ling at 10 m/s horizontally. Finally, the unit of kinetic energy is the joule (J) (as it is for work) named in honour of James Prescott Joule, a British scientist and pioneer of heat and energy research. According to the definition of kinetic energy, the joule is equivalent to:

m s

( )

1 J 5 1 kg 3 1

2

51

kg 3 m2 s2

exml What is the kinetic energy of a 1200-kg car travelling at 100 km/h? Data: m 5 1200 kg v 5 100 km/h Ek 5 ?

Solution: 1 The units of velocity v must be converted into metres per second v 5 100

1h km 1000 m 3 5 278 m/s 3 1 km 3600 s h

2 We can then calculate Ek  1 1 Ek 5 mv 2 5 3 1200 kg 3 (278 m/s2) 5 464 3 105 J 2 2

seCtIoN 16.1

Kinic ngy

1. A 15-tonne airplane is fying at 850 km/h What is its kinetic energy?

5. A soccer ball moving at 150 m/s has a kinetic energy o 484 J What is the ball’s mass?

2. At a given moment in an X-ray tube, an electron (o a mass equal to 911 3 10-31 kg) is travelling at a velocity o 20 3 106 m/s What is its kinetic energy at this precise moment?

6. A 750-kg skier skies down a hill at a velocity o 200 m/s At what velocity must he ski or his kinetic energy to double?

3. Which car has a greater kinetic energy: a 900-kg car travelling at 100 km/h or a 1800-kg car travelling at 50 km/h?

7. A spherical meteorite that is 100 km in diameter is travelling at 200 km/s Its average density is 300 g per cubic centimetre What is its kinetic energy?

4. At what velocity must a child throw a 200-g stone or the kinetic energy o the stone to equal 100 J? Chapter 16 Mechanical Energy

341

16.2 Work-energy heorem The work-energy theorem states that the total work done upon an object, between an initial instant and a fnal instant, is equal to the change in kinetic energy between these two instances. See Work, p 326

When only one force F acts upon an object, the work done by this force is equal to W 5 F 3 ∆s 3 cos q, where ∆s is the magnitude of the displacement, and q is the angle between F and ∆s . However, generally speaking, there are several forces that can act upon an object: gravitational force, normal force, and occasionally the force of friction, for example. In this case, the total work (Wtot) done upon the object is equal to the sum of the work done by each of the forces acting upon this object: Wtot 5 WFg 1 WFN 1 WFf 1 … The total work is also expressed based on the resultant orce FR acting upon the object. We can therefore write: Wtot 5 FR 3 ∆s 3 cos q where q 5 Angle between FR and ∆s There is a close connection between the total work done by a resultant force upon an object and kinetic energy. In fact, any work includes a transfer of energy. This is expressed in the work­energy theorem, which is formulated as follows: Wtot 5 ∆Ek 5 Ek f 2 Ek i Wtot 5 Total work, expressed in joules (J) ∆Ek 5 Change in kinetic energy, expressed in joules (J)

Ek f 5 Final kinetic energy, expressed in joules (J) Ek i 5 Initial kinetic energy, expressed in joules (J) This theorem states that the total work done by forces on an object produces a change in kinetic energy. For example, when an object falls, the gravitational force does positive work that will result in an increase in kinetic energy, so that Ekf > Eki, because as the object falls its velocity increases. Even if the resultant force is * Noe oriented in the direction of the motion, individual forces can be directed differently

See Muliples and submuliples of unis, p 173

342

The proof of the work­energy theorem on the following page illustrates how physicists work with the formulas they establish. It naturally introduces the expression of kinetic energy. The proof concerns a situation where the resultant force is constant and oriented in the direction of the motion*. The situation therefore occurs in only one dimension. This proof also enables us to recognize the effectiveness of the homogeneity o unctions method.

UNIt 5 Energy and its Transformations

1. A shopping cart is travelling along the x-axis under the influence of a resultant force FR oriented in the direction of the displacement ∆x (see Figure 2).

APPENDIX 5

y

Problem-solving, p. 400.

→

FR x

→

∆x θ  0°

Figure 2 A supermarket cart is moving in the direction of x-positive under the influence of resultant force FR .

The total work is equal to:

Wtot  FR cos 0° ∆x

(equation 1)

However, according Newton’s second law: FR  ma

(equation 2)

For a constant resultant force FR , acceleration ax is constant and therefore: vf2  vi2  2a ∆x vf2  vi2 or a  2 ∆x

(equation 3)

See Equations for uniformly accelerated rectilinear motion, p.230.

By combining equations 1, 2 and 3, vf2  vi2 we obtain: Wtot  ma ∆x  m ∆x 2 ∆x 1 1 Wtot  mvf2  mvi2 2 2 2. Because these two physical quantities are part of the same equation, the unit of work and kinetic energy are the same: the joule (J). We can also ensure that the unit of kinetic energy is the same as the unit of work by comparing the expressions of work and that of kinetic energy. According to the equations W  F ∆s and F  ma, 1J1N1m1

kg  m kg  m2  1m1 2 s s2

1 According to the equation Ek  mv 2, 2 m 2 kg  m2 1 J  1 kg  1 1 s s2

( )

The following examples show how the work-energy theorem is used. Example A Under the influence of a force, a 10-kg object goes from being at rest to a velocity of 4.0 m/s. a) What is the object’s initial kinetic energy? b) What is the object’s final kinetic energy? c) What is the total work done on the object? Data: m  10 kg vi  0 vf  4.0 m/s Eki  ? Ekf  ? Wtot  ?

Solution: 1 1 a) Eki  mvi 2  10 kg  (0 m/s)2  0 J 2 2 1 1 b) Ekf  mvf 2  10 kg  (4.0 m/s)2  80 J 2 2 c) Wtot Ekf  Eki  80 J  0 J  80 J

CHAPTER 16 Mechanical Energy

343

exampl B A dog is pulling a 250-kg sled over a horizontal surace o 400 m with a orce o 500 N using a horizontal rope The sled was initially at rest and we assume that riction is negligible a) What is the total work done? b) What is the sled’s fnal velocity? Data: m 5 250 kg F 5 500 N q 5 0° vi 5 0 ∆s 5 40 m Wtot 5 ? v 5 ?

Solution : a)

y

→

FN

→

Ft

x

→

Fg

There is zero work done by the normal orce and gravity because these orces are at a 90° angle to the displacement: WF 5 WF 5 0 N

g

Wtot 5 WF 1 WF 1 WF 5 FT 3 ∆s 3 cos q 1 0 1 0 T

N

g

5 500 N 3 400 m 3 cos 0° 5 200 J 1 b) Eki 5 mvi2 5 0 2 Ek 5 Eki 1 Wtot 5 0 J 1 200 J 5 200 J 2Ec 1 2 3 200 J Ek 5 mv2 ⇒ v 5 5 5 400 m/s 2 m 25 kg 2

J 5 kg

The units o v are metres per second:

( kg 3s m ) 5 2

kg

m2 m 5 s s2

We can answer question b) from example B using Newton’s second law, accord­ ing to which FR 5 ma. The following example demonstrates this method. exampl C A dog is pulling a 250-kg sled over a horizontal surace o 400 m with a orce o 500 N using a horizontal rope The sled was initially at rest and we assume that riction is negligible What is the sled’s fnal velocity? Data: m 5 250 kg ∆x 5 400 m F 5 500 N vi 5 0 m/s v 5 ?

y →

FN →

Ft

x →

Fg

344

UNIt 5 Energy and its Transormations

Solution: 1 In the horizontal direction, Nwon’s scond lw indicates: FT 500 N FR 5 FT 5 ma ⇒ a 5 5 5 200 m/s2 m 250 kg

See Nwon's scond lw, p 310

2 By using the 4 equations rom the quions fo unifomly ccld cilin moion, we have: v2 5 vi2 1 2a ∆x

See equions fo unifomly ccld cilin moion, p 230

v2 5 (0 m/s)2 1 2 3 200 m/s2 3 400 m 5 160 m2/s 2 v 5 160 m2/s 2 v 5 400 m/s Example C demonstrates that the result obtained using the notions o work and energy is the same as the one obtained using Newton’s second law. This highlights an interesting act. Historically, physicists spent many years defning the concepts o work and kinetic energy and ater much trial and error, they suc­ ceeded in developing a theory that corroborates Newton’s laws, thereby consti­ tuting an important achievement in the history o physics.

Furthering

your understanding

roionl kinic ngy Objects rotating around an axis, such as a grindstone or the blades o a an, also have kinetic energy, because they are masses in motion Each particle o a rotating object moves at a given velocity (at a lower velocity close to the axis and a higher velocity ar rom the axis) and has kinetic energy By adding the kinetic energies o all the particles that constitute the object, we obtain the rotational kinetic energy o the object as a whole

to the cart is then lower than when the heavy items are at the ront o the cart (see Figure 3)

This energy depends on the breakdown o mass Two objects o the same mass and with the same radius turning at the same velocity do not necessarily have the same rotational kinetic energy For example, a bicycle wheel whose rim is light and hub is heavy has less kinetic energy that a wheel whose rim is heavy and hub is light Initiating rotation in the frst wheel requires less energy rom the cyclist, which makes accelerations easier In the same way, making a shopping cart turn requires less eort when the heavy items are located in the back, near the person who is pushing it, rather than in the ront o the cart Since the cart pivots on the back wheels, the rotational kinetic energy that must be provided

Figure 3 This shopping cart requires more kinetic energy to turn with the load in the ront o the cart than i the load was in the back

Chapter 16 Mechanical Energy

345

seCtIoN 16.2

Wrk-nrgy hrm

1. A child pulls the rope o a sled sliding on a horizontal surace For each o the orces listed, indicate i there is negative work, zero work or positive work a) gravitational orce b) normal orce c) riction d) tension in the rope 2. A locomotive exerts a 20 000-N orce upon a train It propels our cars, each with a 15-tonne mass The locomotive’s mass is 40 tonnes and riction is considered negligible Initially at rest, the train accelerates over a distance o one kilometre a) What is the work done by the locomotive on the train? b) What is the total work? c) What is the velocity o the train at the end o the kilometre? 3. A child lits a bucket ull o water rom the bottom o a 5 m-deep well by pulling on a rope The 100-kg bucket accelerates at a rate o 0250 m/s2 Determine: a) the work done on the bucket by the child b) the work done on the bucket by gravity c) the total work d) the velocity o the bucket when it reaches the top o the well

6. A young girl pulls the rope o a 200-kg sled that was initially at rest She exerts a 100-N orce on the rope at a 370° angle rom the horizontal The surace is horizontal and the riction is equivalent to 750 N What is the velocity o the sled ater 500 m? 7. A 250-g ball is thrown vertically rom the ground with a velocity o 100 m/s The ball reaches a height o 510 m beore alling back down a) What is the change in the ball’s kinetic energy, between the initial point and the peak o the trajectory? b) What is the work done by gravity? 8. An object is subjected to a non-zero resultant orce oriented in the direction o its displacement Which o the ollowing statements are true? A The object accelerates in the direction o its displacement B The fnal velocity o the object is lower than its initial velocity C All the orces acting on the object are oriented in the direction o its displacement D The fnal kinetic energy is greater than the initial kinetic energy E The total work is negative

4. A 170-g hockey puck propelled on the ice at 150 m/s travels 383 m beore stopping What is the magnitude o the orce o riction?

9. A 600-kg skier starts o at rest and begins the descent o a hill with a 300˚ incline What distance must he travel to reach a velocity o 100 km/h, assuming that there is zero riction and considering that he is not pushing with his arms or legs?

5. A 200-g bullet hits a wooden target with a velocity o 500 m/s It comes to a stop over a distance o 100 cm a) How much kinetic energy did the bullet lose in the target? b) What average orce did the target exert on the bullet?

10. A ship with a mass o 100 000 tonnes is moving orward at 500 km/h Once the engines are shut down, the ship moves orward 400 km beore coming to a stop I you assume that the resistance o the water and the air is constant over these 400 km, what is the magnitude o the total resistance?

346

UNIt 5 Energy and its Transformations

16.3 Gviionl onil ngy The gravitational potential energy of an object is associated with its mass and position above a reference level, usually the Earth’s surface. In mechanics, the important forms of energy are kinetic (see Section 16.1) and potential energy. Gravitational potential energy is connected to the position of an object in its environment. The term “potential” means that the energy is stored in an object or system and can be transformed into another kind of energy, e.g. kinetic energy. Based on the physical situation, there can be different types of potential ener­ gy (see Figure 4). Elastic potential energy is the energy of a stretched or com­ pressed object or physical system. A stretched elastic or bow contains elastic potential energy. If the bow’s string is released, the elastic potential energy will then be transformed into the kinetic energy of the arrow and the bow. Elastic potential energy will be the subject of the next chapter. Gravitational potential energy is the energy of a body raised above the ground. It results from the existence of gravity. If an object raised above the ground is released, it will accelerate downwards and its potential energy will be trans­ formed into kinetic energy.

a) The stretched bow contains elastic potential energy that will be transformed into the kinetic energy of the arrow and the bow if the string is released

To gain further understanding of this concept, we may consider the case of an object held at a height h above the ground (see Figure 5). If we let the object fall, it will drop from distance h. The gravitational force does work noted as: WF 5 Fg 3 h 3 cos 0° 5 mgh g

According to the work­energy theorem, this work is transformed into kinet­ ic energy. The object landing on the ground therefore has a kinetic energy equivalent to mgh. One way of describing where this energy came from is to consider that when the object was at a height h above the ground, it had an energy called gravitational potential energy. This potential energy gradually transformed itself into kinetic energy during the fall.

b) At the top of the hill, this stationary skier has gravitational potential energy that will be transformed into kinetic energy if the skier skis down

m

h

m →

Fg Initial instant

Mid-point

Figure 4 Two situations where an object has potential energy

m Final instant

Figure 5 The gravitational force does work WF 5 mgh during the fall of an object from height h g

Strictly speaking, gravitational potential energy does not only belong to the object that is above the ground: it belongs to the system made up of the object and the Earth. In fact, if there weren’t any gravitational force, there would be no work done by this force and, therefore, no gravitational poten­ tial energy. To simplify communication we say that the object located above the ground has a gravitational potential energy, without specifying each time that the Earth is part of the system. Chapter 16 Mechanical Energy

347

In the situation described in Figure 5 on the previous page, variable h is measured with respect to the ground because, most o the time, it is considered that there is zero gravitational poten­ tial energy at ground level. This choice acilitates calculations i the object starts on the ground, or ends up there, or i all o an object’s positions are indicated with respect to the ground. h Epg 5 0

Ground

Figure 6 The potential energy is measured using the appropriate reerence level at which Epg 5 0

Depending on the context, it can be preerable to choose a di­ erent reerence level or the gravitational potential energy. For example, there is a person inside a house holding an apple in her hand 1 m rom the foor (see Figure 6). I the apple alls, it will not drop urther than the foor. It is then valid to choose the foor rather than the ground as the reerence level to calculate the gra­ vitational potential energy o the apple. The change in potential energy thereore takes place over a distance o 1 m. The distance between the foor and the ground is not taken into account. The ollowing equation enables us to determine the gravitational potential energy o an object: Epg 5 mgh

where Epg 5 Gravitational potential energy o an object, expressed in joules (J) m 5 Mass o an object, expressed in kilograms (kg) g 5 Gravitational acceleration, expressed in metres per second squared (m/s2) h 5 Height above the reerence level (usually the ground), expressed in metres (m) The unit o potential energy is the joule (J), as it is or kinetic energy and work. The product mgh has the ollowing units: 1 kg 3 m/s2 3 m 5

1 kg 3 m2 51J s2

exmpl a A 100-kg object is initially located on the ground and is then raised to a height o 500 m above the ground a) What is the initial potential energy? b) What is the fnal potential energy? Data: m 5 100 kg With respect to the ground: hi 5 0 h 5 500 m Epgi 5 ? Epg 5 ?

348

UNIt 5 Energy and its Transormations

Solution: a) Epgi 5 mghi 5 100 kg 3 980 m/s2 3 0 m 5 0 b) Epg 5 mgh 5 100 kg 3 980 m/s2 3 500 m 5 490 J

exml B A pendulum of length L has a 500-g ball suspended from a cord attached to a hook The pendulum is raised until the cord forms an angle q with the vertical If L equals 10 m and q equals 300 °, what is the potential energy of the mass in this position? Data: L 5 10 m m 5 0500 kg q 5 300° h5? Epg 5 ?

Solution: The reference level chosen for the gravitational potential energy is the lowest point at which the pendulum’s mass passes if it is released (see fgure opposite) h 5 L 2 L’ 5 L 2 cos q 5 10 m 2 10 m 3 cos 300° 5 0134 m Epg 5 mgh 5 0500 kg 3 980 m/s2 3 0134 m 5 0657 J

L θ

L’

Initial instant: Epg is at a maximum Ek  0

h Epg  0

The gravitational potential energy of an object at a given height is constant as long as the object remains at the same height. During a freefall, the work done by the gravitational force gradually transforms the gravita tional potential energy that the object had before falling into kinetic energy (see Figure 7). If the air resistance is not considered, any initial potential energy is thereby transfor­ med into kinetic energy. This transformation is the consequence of the conserva­ tion of mechanical energy, which will be the subject of the next section.

During the descent: Epg decreases Ek increases

At the end of the descent: Epg  0 Ek is at a maximum

Figure 7 During a fall, gravitational potential energy is gradually transformed into kinetic energy

seCtIoN 16.3

Gviinl nil ngy

1. A 600-kg woman climbs a staircase o 10 stairs, each one measuring 200 cm in height What is her potential energy gain?

4. An 80-kg man bungee jumps rom a bridge and alls 30 m What is his initial potential energy? What is his fnal potential energy?

2. A 100-kg ball is at the top o a slope that is 200 m long and has a 120° incline What is its potential energy?

5. While a person is walking 500 m on a slope with a 300˚ incline with respect to the horizontal, his potential energy increases by 19 600 J What is the person’s mass?

3. A pendulum has a 200-kg mass suspended to a 100-m cord that is attached to a hook The pendulum is raised until the cord orms a 600° angle with the vertical What is the potential energy o the mass in this position?

6. A cirrocumulus cloud is a thin cloud located at a high altitude I a 1000-tonne cloud has a potential energy o 780 gigajoules (GJ), what is its average altitude with respect to the ground?

Chapter 16 Mechanical Energy

349

16.4 Conservaion of mechanical energy Mechanical energy (Em) is equal to the sum of kinetic energy Ek and gravitational potential energy (Epg) of an object or system. The ollowing equation demonstrates the relationship between these two orms o energy: Em 5 Ek 1 Epg where Em 5 Mechanical energy, expressed in joules (J) Ek 5 Kinetic energy, expressed in joules (J) Epg 5 Gravitational potential energy, expressed in joules (J)

In the absence o riction or air resistance, and i there is no orce adding ener­ gy to or removing energy rom an object or system, the mechanical energy o this object or system remains constant during motion. It is said that mechan­ ical energy is conserved. In other words, mechanical energy E m at the fnal instant is equal to mechanical energy Emi at the initial instant: this is the law of conservation of energy. While in reality, there is almost always riction, the law o conservation o mechanical energy is a valid approximation i there is low riction. The ollowing equations both demonstrate the law o conservation o energy: Emi 5 Em where Emi 5 Initial mechanical energy, expressed in joules (J) Em 5 Final mechanical energy, expressed in joules (J) or Eki 1 Epgi 5 Ek 1 Epg Eki 5 Kinetic energy in the initial instant, expressed in joules (J) Epgi 5 Gravitational potential energy in the initial instant, expressed in joules (J) Ek 5 Kinetic energy in the fnal instant, expressed in joules (J) Epg 5 Gravitational potential energy in the fnal instant, expressed in joules (J) The result o this law is that any reduction in an object’s gravitational poten­ tial energy must be compensated or by an increase in its kinetic energy, so that its total mechanical energy will be conserved. Mechanical energy is actually only transormed. For example, the gravitational potential energy o an object is transormed into kinetic energy when the object alls. Inversely, when an object rises ater having been thrown, its kinetic energy is transorm­ ed into gravitational potential energy. The situation o a skateboarder who is skateboarding on a semi­cylindrical (hal­ pipe) ramp is a good illustration o the conservation o mechanical energy (see Figure 8 on the following page). Beore letting himsel drop rom the top o the ramp, the skateboarder is stationary. He has maximum gravitational potential energy and zero kinetic energy. When the skateboarder goes down the slope, his gravitational potential energy is transormed into kinetic energy. At the bottom o the ramp, he has zero gravitational potential energy i you consider the bot­ tom o the ramp as the reerence level. At this point, the skateboarder has

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maximum kinetic energy and, if friction is negligible, this kinetic energy is equal to the skateboarder’s gravitational poten­ tial energy at the top of the ramp.

Ep

The skateboarder’s momentum makes him go up the length of the opposite wall and his kinetic energy is then transformed into gravitational poten­ tial energy. Again at the top of the ramp, he has zero kinetic energy and maximum potential energy. In order to apply the law of conserva­ tion of mechanical energy, the ener­ gies of an object or system must be compared at two different instants.

Ep

Ep

Ep

Ek

Ek

Ek

Figure 8 The relative ratios o gravitational potential energy and kinetic energy o a skateboarder in motion

Gnl mod fo solving oblms of consvion of mcnicl ngy 1. Choose the initial instant and the fnal instant as a unction o the data and unknowns in the problem 2. Defne the reerence level or the gravitational potential energy 3. Calculate the value o kinetic and potential energies in the initial instant and fnal instant Usually, one o the energies remains unknown: you must thereore write its algebraic expression 4. Integrate all the terms into the equation o the conservation o mechanical energy and solve it to determine the unknown element The following examples demonstrate how the equations for the law of conser­ vation of energy are used. exml a A 200-kg object initially at rest at a height o 100 m is released At what velocity will it hit the ground i air resistance is negligible? Data: m 5 200 kg vi 5 0 m/s hi 5 100 m h 5 0 m v 5 ? Solution: 1 Initial instant: when the object is at its maximum height; fnal instant: when the object hits the ground (right beore touching it) 2 Reerence level or Epg : the ground 1 3 Eki 5 mvi2 5 0 J 2 Epgi 5 mghi 5 200 kg 3 980 m/s2 3 100 m 5 196 J Epg 5 mgh 5 0 J Ek 5 ? 4 Eki 1 Epgi 5 Ek 1 Epg so 0 J 1 196 J 5 Ek 1 0 J and Ek 5196 J 2Ek 2 3 196 J 1 Ek 5 196 J 5 mv2 so v 5 5 5 140 m/s 2 m 200 kg Chapter 16 Mechanical Energy

351

A situation similar to the previous example may also be analyzed, but without specifying the object’s mass. Example B An object located at a height of 50.0 m has a velocity of 5.00 m/s. What velocity will it have when it hits the ground if air resistance is negligible? Data: vi  5.00 m/s hi  50.0 m hf  0 m vf  ? Solution: 1. Initial instant: when the object is at its maximum height; final instant: when the object hits the ground (right before touching it). 2. Reference level for Eg : the ground. 3. Given that we do not know the mass, we cannot calculate the energies, except for Epgf. We can only write their algebraic expressions. 1 Eki  mvi2 2 Epgi  mghi 1 Ekf  mvf2 2 Epgf  mghf  0 J 4. Eki  Epgi  Ekf  Epgf 1 2 1 mv  mghi  mvf2  0 J 2 i 2 By multiplying the two sides of the equation by 2, we obtain: mvi2  2mghi  mvf2 By dividing the two sides of the equation by m, we obtain: vf2  vi2  2ghi  (5.00 m/s)2  2(9.80 m/s2) 50.0 m 25.0  980 m2/s2  1.00  103 therefore vf  31.7 m/s

See The motion of objects launched horizontally, p. 249.

Note: vf2  vi2  2ghi corresponds to the equation for vertical freefall (for hf  0) covered in Chapter 11 (see Equation 5 on page 253 ). This equation states that, regardless of its mass, the object will have the same final velocity. The notions of energy are therefore once again consistent with those of kinematics, according to which objects of different masses have the same acceleration during their fall. Example B illustrates how powerful the law of conservation of mechanical energy is. The direction of the initial velocity is not specified in the problem and is not needed to express kinetic energy, which is a scalar quantity. This means that, regardless of the direction, the magnitude of the velocity when hitting the ground is the same (the direction of the final velocity vector varies according to the initial angle). In Chapter 11, numerous examples had to be analyzed before reaching the same conclusion (see Section 11.2, Example B, on page 250; Section 11.3, Example B on page 253 and Exercise 2 on page 257).

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UNIT 5 Energy and its Transformations

seCtIoN 16.4

Cnvin f mcnicl ngy

Note: For the following exercises, air resistance is consid­ ered negligible. 1. A ball is dropped vertically The reerence level or the gravitational potential energy is the ground Which o the ollowing statements is alse? A During its all, the ball loses potential energy B At mid-height, the ball’s kinetic energy and its potential energy are equal C The ball’s mechanical energy is constant D At mid-height, the ball’s velocity is equivalent to hal its maximum value 2. A person holds a 100-kg stone at a height o 200 m above the ground and she suddenly lets it go a) What is the potential energy o the stone beore she lets it go? b) What is the kinetic energy o the stone beore she lets it go? c) What is the potential energy o the stone at the moment it hits the ground? d) What is the kinetic energy o the stone at the moment it hits the ground? e) What is the magnitude o the stone’s velocity at the moment it hits the ground? 3. A diver dives o a diving board at a velocity o 400 m/s What is the magnitude o his velocity at the moment he hits the water, i the diving board is 100 m above the water? 4. During a hockey game, a player misfres a slap shot The puck, starting with a velocity o 250 m/s, goes above the glass and reaches a 120-m height beore it is stopped by the saety net used to protect spectators What was the velocity o the puck at the moment it hit the saety net? 5. Children are sledding down an icy hill and they begin their descent at the top The sled and the children have a combined mass o 90 kg I riction is negligible, determine: a) the total energy o the sleigh at point A

b) the velocity o the sled at point B c) the velocity o the sled at point C

a 10 m

C B

3m

6. A shot putter throws her shot At the moment the 400-kg weight leaves her hand, it has a velocity o 140 m/s and is 200 m above the ground The weight reaches a maximum height o 650 m above the ground beore alling back down a) What is the magnitude o the weighted ball’s velocity at the peak o the trajectory? b) What is the magnitude o the weighted ball’s velocity when it reaches the ground? 7. A basketball player attempts a shot at the basket The ball, with a velocity o 8 m/s, reaches the basket, which is 305 m high, with a velocity o 800 m/s I the ball (treated as a single point) let the hands o the player at a height o 220 m, what was its initial velocity? 8. A skier starts o at rest and then skis down a hill with a 350º incline What velocity, in kilometres per hour, will she reach ater skiing 200 km, assuming that there is zero riction and considering that she is not pushing with her arms or legs? 9. From the ground, a 200-g ball is thrown vertically upwards with a velocity o 200 m/s At what height will its kinetic energy be equal to: a) its gravitational potential energy? b) a quarter o its gravitational potential energy? 10. A 2 m-long pendulum passes its lowest point with a velocity o 400 m/s, then swings back up beore temporarily coming to a stop What angle does it then orm with the vertical, i the air resistance is negligible?

Chapter 16 Mechanical Energy

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16.5 Conservaion of oal energy The law of conservation of total energy states that the total quantity of energy is constant in an isolated system. In other words, the initial total energy is equal to the fnal total energy. An isolated system is a system that does not exchange matter or energy with its environment.

Figure 9 With each contact with the ground, the ball loses mechanical energy Lost energy is transformed into thermal energy, which is dissipated into the ground and the ball

Besides kinetic energy and gravitational potential energy, there are other kinds o energy that must be considered in order to thoroughly analyze a pro­ cess concerning the transer o energies. The situation o a ball alling to the ground is a prime example (see Figure 9). When the ball alls, its gravitational potential energy is transormed into kinetic energy. But when the ball ends up stationary on the ground ater a ew bounces, there is no longer any potential energy or kinetic energy, because: hf 5 0 ⇒ Epgf 5 mghf 5 mg 3 0 5 0 1 1 vf 5 0 ⇒ Ekf 5 mv f2 5 3 m 3 02 5 0 2 2 What happened to the mechanical energy? The ball’s mechanical energy has been transormed into mechanical energy on a microscopic scale: the shocks increased the agitation o the atoms constitu­ ting the ball and the ground, and modifed their positions. This atomic energy is called thermal energy. The motion o the ball in the air also increased the agitation o the molecules in the air. The outcome o these shocks between the ball and the ground was thereore an increase in the temperature o the materials involved in the process, namely, the ball, the ground and the air. Taking into consideration thermal energy also leads us to consider motion subject to riction or uid resistance. Thereore, when an object slides on a horizontal surace and comes to a stop, the mechanical energy is not maintai­ ned because o the presence o riction. In this case, the energy lost is transor­ med into thermal energy distributed between the object and the surace. In such situations, we can write the law o conservation o total energy in the ollowing orm: Eki 1 Epi 5 Ekf 1 Epf 1 ∆Eth where ∆Eth, the change in thermal energy, is equal to the absolute value of the (negative) work done by friction or another resistive force The law o the conservation o total energy takes into account kinetic energy, potential energy and thermal energy, but it also includes all other orms o energy. The law states that in any transormation o energy, the sum o all orms o energy is constant in an isolated system, i.e. a system that does not exchange any energy with an outside environment. This law is undamental to all scientifc felds: physics, chemistry, biology, astronomy, geology, etc. Table 1 identifes the main orms o energy.

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UNIt 5 Energy and its Transformations

tbl 1 Main forms of energy engy

Dfniion

Kinetic energy

Energy associated with the motion of an object

Gravitational potential energy

Energy associated with the position of an object above the ground

Elastic energy

Energy present in stretched or compressed objects (special case of electric energy)

Electric energy

Energy present in the interactions of electric charges

Nuclear energy

Energy present in atomic nuclei (eg in uranium nuclei)

Chemical energy

Energy stored in chemical bonds (special case of electric energy)

Thermal energy

Combination of kinetic and potential energy on a microscopic scale

Radiant energy

Energy transported by an electromagnetic wave (eg visible light, microwave)

The law o conservation o total energy enables us to characterize energy using its essential property; the energy can pass rom one orm to another, but the total energy is constant in an isolated system. It also makes it possible to analyze any process by defning the energy trans­ ers. The case o a student who toasts some bread beore leaving or school is an interesting example. The bread is heated by the warm air (thermal energy) and inrared radiation (radiant energy) produced by the toaster’s elements. These ele­ ments are heated by the electric energy supplied by a hydroelectric power plant. The electric energy was produced rom the kinetic energy o alling water. At the top o this waterall, this water had gravitational potential energy. Prior to this, the water lowed in a river (kinetic energy) and came rom clouds (gravitational potential energy). Clouds were ormed as a result o the grouping o evaporated water molecules (kinetic energy) because o the Sun’s rays (radiant energy). And the Sun’s radiation is produced by the nuclear energy released at the centre o the Sun. Regardless o which process is analyzed, we discover that, ultimately, the energy used on Earth originates rom two sources: the Sun’s nuclear energy (mostly) and the nuclear energy released by the radioactive elements pre­ sent within the Earth. Human activities do not “produce” energy, they only transorm it. Thereore, a thermal power plant transorms into electric energy the chemical energy contained in the ossil uels extracted rom the ground. Furthermore, the chemical energy contained in ossil uels comes rom sub­ stances that originate in the ossilization o plants ormed by the Sun’s radiant energy tens or hundreds o million years ago.

Chapter 16 Mechanical Energy

355

Mri clliin Meteorite impacts occur requently on Earth It is estimated that a meteorite o 10 m in diameter hits the planet every year, and a meteorite o 1 km in diameter, approximately every million years

potential energy, leading to the ormation o crater walls and the dispersion o dust into the atmosphere (see Figure 10 )

Meteorites hit the Earth with an average speed o 20 km/s A meteorite o 1 km in diameter, with a density o approximately 3 g/cm3, has an approximate mass o 15 billion tonnes and kinetic energy o around 1020 J This energy is equivalent to the energy released by the explosion o approximately 70 000 megatonnes o trinitrotoluene (TNT), a chemical explosive The energy o explosives is generally expressed in equivalents o TNT The nuclear bomb dropped on Hiroshima, Japan, on August 6, 1945, corresponded to 15 kilotonnes o TNT When a meteorite impacts the Earth, kinetic energy is mainly transormed into thermal energy The ensuing pressures and temperatures are sufcient to melt or vaporize the meteorite and the impacted rocks Part o the total energy is transormed into

seCtIoN 16.5

Cnrvain f al nrgy

1. Name the orm o energy involved in each case a) the energy associated with the wind b) the energy rom the Sun c) the energy present in a tree branch bent by the wind d) the energy present in a glass o orange juice 2. Julie starts riding her bicycle, reaches a velocity o 20 km/h in our seconds on a at road, and then coasts until she comes to a complete stop Name the orms o energy present in this situation and the transormation rom one orm o energy to another 3. Gary is on the diving board at the swimming pool He dives, enters the pool, and reaches the lowest point o his dive Name the orms o energy present and the transormation rom one orm o energy to another

356

UNIt 5 Energy and its Transormations

Figure 10 A crater, 1300 m in diameter, ormed approximately 50 000 years ago in Arizona, USA by the impact o a meteorite, 50 m in diameter

4. An electric current ows through the flament o an incandescent light bulb What orms o energy is the electric energy transormed into? 5. A 170-g hockey puck propelled on the ice at 10 m/s slows and then stops How much thermal energy have the ice, the puck and the air received? 6. A 600-kg skier starts rom the top o a 500-m hill with a 250° incline to the horizontal At the bottom o the hill, he moves at 120 m/s How much mechanical energy is transormed into thermal energy because o riction and air resistance?

APPLICATIONS hydolcic ow lns Hydroelectricity is a clean and renewable source o energy used to fll a signifcant portion o energy needs. Approximately one quarter o the electric energy con­ sumed in the world is hydroelectric in origin. In 2008, nearly 94% o the electricity produced in Québec was rom a hydroelectric source, while in Canada, hydro­ electricity made up approximately 60% o the electric energy produced. Hydroelectric power plants are generating stations where the orce o water is used to turn turbines. These gener­ ators, relying on turbines, produce an electric current. Storage plants and run­o­the­river plants are the two main types o plants producing hydroelectricity. Run­o­the­river plants directly exploit the current o a river without having to create very large reservoirs or high wateralls. The power o these plants is thereore determined mainly by the ow o water. Storage plants require the construction o dams to retain the water o a river in large water reservoirs that increase the hydrau­ lic head (see Figure 11).

W svoi

Dm

This hydraulic head determines the quantity o energy produced. In act, the higher the water level is, the greater the water pressure near the bottom o the reser­ voir will be. The water entering through the penstock then ows more quickly. Turbines are installed at a lower height than the entry o the penstock. When the water alls toward the tur­ bines, its gravitational potential energy is transormed into additional kinetic energy. The water then does work upon the blades o the turbine, which is a type o water wheel, and transers rotational kinetic energy to it (see Figure 12). The axis o the turbine turns the rotor o a generator, in which rotational kinetic energy is converted into elec­ tric energy. The current produced generates magnetic orces that act upon the rotor. These orces would slow down and quickly stop the rotor and the turbine i the turbine was not lead by the water. When the current consumed in the electric network connected to the generator increases, magnetic orces also increase. To maintain the rotation o the generator (which is essential to keep the requency o the alterna­ ting current at 60 Hz), the power plant operators must increase the mechanical energy provided to the turbine by increasing the ow o water.

Gno hydulic d

pnsock tubin

Figure 11 Cross-sectional view of a dam The hydraulic head corresponds to the difference between the water level upstream of the dam and the turbine

Figure 12 Turbine of a hydroelectric power plant

Chapter 16 Mechanical Energy

357

the noions of work and energy The notions o work and energy are abstract and it has taken centuries to develop them. Simon Stevin (1548– 1620) and Galileo (1564–1642) came close to an unders­ tanding o the notion o work, without calling it as such. They understood that the product o orce and displace­ ment is important in the study o simple machines. For example, or a lever, the product o orce and displace­ ment is the same on either end o the lever (see Figure 13). Galileo also studied the pendulum and believed that the velocity it acquired enabled it to return to its ini­ tial level. Also in the 17th century, the Dutch physi­ cist Christiaan Huygens (1629–1695) discovered that when two hard spheres collide, the sum o the quan­ tities mv2 are the same beore and ater the collision. Gottried Leibniz (1646–1716) gave the name “live orce” to the product mv2 and ormulated the princi­ ple o conservation o live orce. Huygens also showed that or a body in reeall, the live orce is proportional to the height o the all. However, it is not only energy that is connected to motion. During the 18th century, Daniel Bernoulli (1700–1782), Leonhard Euler (1707–1783) and Pierre­Simon Laplace

(1749–1827) believed that heat came rom the vibration o molecules. Other scientists thought dierently: that heat was ormed rom a uid. Antoine Lavoisier (1743–1794), in his Traité élémentaire de chimie (Elementary Treatise on Chemistry) (1789), avoured this idea and named this uid “caloric.” According to this dated theory, the calo­ ric, made up o minuscule heat particles, owed rom a warm body toward a cool body until the temperatures o both bodies became equal. However, the caloric theory did not adequately explain the production o heat by riction, because experiments showed that the amount o heat available appeared inexhaustible. In 1807, the British scientist Thomas Young (1773–1829) was the frst to use the word “energy” in its modern sense. In the 1840s, Julius Robert von Mayer (1814–1878), a German physicist and chemist, showed that work and heat corresponded to energy transers and ormulated the law o conservation o energy. With thermometers becoming more precise, Mayer and James Prescott Joule (1818–1889) established that heat could be produced by doing work, which voided the caloric theory. The modern theory o energy was born.

F1

∆s1 ∆s2

F2

Figure 13 The work done by a person at one end of the lever (F1∆s1) is equal in absolute value to the work done by the weight at the other end of the lever (F2∆s2)

358

UNIt 5 Energy and its Transformations

chapter

16

Mnil engy

16.1 Kinic ngy • Kinetic energy is the energy o an object in motion. It is a scalar quantity: it does not depend on the direction o the velocity but only on its magnitude. Ek 5

1 mv 2 2

• The unit o work and kinetic energy is called the joule: 1 J 5 1

kg 3 m 2 s2

16.2 Wok-ngy om • The work­energy theorem states that the total work done between an initial instant and a fnal instant is equal to the change in kinetic energy between these two points in time. Wtot 5 ∆Ek 5 Ekf 2 Eki • The total work done on an object is equal to the sum o the work done by each o the orces acting on the object.

16.3 Gviionl onil ngy • Potential energy is associated with the position o an object in its environment. • The gravitational potential energy o an object is related to its mass and its position above a reerence level, usually the Earth’s surace. For an object o mass m, it is equivalent to: Epg 5 mgh • The unit o potential energy is the joule (J), because the units o the product mgh is the product kg 3 m 2 , which corresponds to a joule. s2 • During a reeall, the work done by the gravitational orce gradually transorms the gravitational potential energy that the object had beore alling into kinetic energy.

Chapter 16 Mechanical Energy

359

chapter

16

Mnil engy

16.4 Conservaion of mechanical energy • Mechanical energy, noted Em, is equal to the sum o kinetic energy Ek and potential energy Ep o an object or system. Em 5 Ek 1 Ep • In the absence o riction or air resistance, and i there is no orce adding energy to, or removing energy rom an object or system, the mechanical energy o this object or system remains constant during motion: this is the conservation o mechanical energy. Eki 1 Epi 5 Ekf 1 Epf • To solve problems o conservation o mechanical energy, you must: 1. Accurately determine the initial instant and fnal instant. 2. Defne the reerence level or the gravitational potential energy. 3. Calculate the value o kinetic and potential energies at the initial instant and fnal instant. Usually, one o the energies remains unknown; you must write its algebraic expression. 4. Integrate all the terms in the equation o the conservation o mechanical energy and solve the equation to determine the unknown element.

16.5 Conservaion of oal energy • The law o conservation o total energy states that the total quantity o energy is constant in an isolated system. In other words, the total initial energy is equal to the total fnal energy. An isolated system is a system that does not exchange matter or energy with its environment. • There are other orms o energy besides kinetic energy and gravitational potential energy. • Thermal energy corresponds to mechanical energy on a microscopic scale: it depends on the velocity and position o atoms. • I there is riction, part o the mechanical energy is transormed into thermal energy. Thereore, Eki 1 Epi 5 Ek 1 Ep 1 ∆Eth where ∆Eth, the fnal thermal energy, is equal to the absolute value o the (negative) work done by riction or another delaying orce.

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Chapter 16

Mcnicl ngy

1. Two children are riding in a roller coaster car Friction is negligible

a

4. A 500-g ball is dropped rom a height o 200 m, hits the ground and bounces to a 150-m height Determine the heat energy dissipated in the ball, the ground and the air during the process

C B D

a) At what point is the gravitational potential energy greatest? At what point is it smallest? b) At what point is the kinetic energy greatest? At what point is it smallest? c) Is the mechanical energy at point A greater, equal to or inerior to the mechanical energy at point D?

5. A 100 m-long pendulum is held in a horizontal position What will be the velocity o the mass attached to the end o the cord when it passes the lowest point o its motion?

2. The ollowing fgure illustrates two slides, one straight, the other, concave From the top, a marble is simultaneously released on both slides

slid X

slid Y

a) In which slide will the marble have the greater velocity when it reaches the bottom? b) In which slide will the marble reach the bottom frst? 3. An eagle ying horizontally 100 m above the ground at 300 km/h holds a dead mouse in its talons It releases its prey and lets it drop At what velocity does the mouse land on the ground i the air resistance is negligible?

6. A 70-kg man and a 65-kg woman who are the same height are sitting on identical swings The amplitudes o their oscillation (the angle obtained relative to the vertical) are also equal What can be deduced about their respective maximum velocities? 7. By pushing with its tail, a 10-kg ying fsh jumps out o the water vertically with a velocity o 10 m/s What height does it reach? 8. A 700-kg hiker is climbing a mountain with an elevation o 2000 m a) What is the gain in the hiker’s gravitational potential energy? b) The potential energy gained results rom the muscular eort rendered possible by the conversion o chemical energy into mechanical energy The conversion efciency is only 25% (ie the mechanical energy produced is only equivalent to 25% o the chemical energy) Knowing that a gram o atty tissue provides approximately 30 kJ o chemical energy, estimate the hiker’s weight loss attributable to the hike, assuming that all the chemical energy comes rom atty tissue

Chapter 16 Mechanical Energy

361

9. At the Chute-Montmorency near Québec City, water alls rom a height o 830 m The average ow rate o the Montmorency River is 35 000 litres per second Estimate the maximum power that a hydroelectric plant located at the bottom o the alls would provide W reminde: P 5 ∆t where P 5 Power W 5 Work ∆t 5 Time interval 10. As a 20-kg block slides on a horizontal surace, its velocity goes rom 50 m/s to 30 m/s over a distance o 10 m What additional distance will it travel beore coming to a stop? 11. A skateboarder is going down a plane with a 450º incline without riction At the moment when she is at a height o 300 m, her velocity is 400 m/s Ater having skated a horizontal section, she goes back up a plane with a 300º incline, still without riction What distance does she skate on the ascending plane?

3.00 m 45.0°

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12. From being at rest, a 300-kg child goes down a waterslide 200 m long and with a 250º incline, on which riction is negligible He then arrives on a horizontal section that has a drier surace to slow down and stop the people rom sliding The child slides a distance o 800 m beore stopping What is the magnitude o the orce o riction? 13. At Niagara Falls, the water alls rom a height o 52 m At the bottom o the alls, the water’s kinetic energy is almost completely transormed into thermal energy What is the increase in water temperature at the bottom o the alls? The specifc heat capacity o water is equivalent to 4186 J/(kg 3 °C) reminde: The quantity o heat Q provided to the matter and the increase in temperature ∆T are related by the equation Q 5 mc ∆T, where Q is the quantity o heat, m is the mass, c is the specifc heat capacity and ∆T is the change in temperature 14. A pendulum consists o a 600-g mass attached to a 120-m cord When the cord orms a 300° angle with the vertical, the mass has a velocity o 300 m/s a) What is the maximum angle that the cord orms with the vertical? b) When the cord is vertical, what is the tension in the cord?

elsic ponil engy

M

any lexible materials and objects used today possess a property that enables them to return to their initial shape and volume when the orce applied to them is no longer exerted. This property, known as elasticity, allows them to store energy when they undergo deormation. Given the act that this energy only maniests itsel i it is transormed into another orm o energy, it is said to be potential. As such, elastic bodies can store energy in the orm o elastic potential energy. This is the case with rubber bands, springs, bows and certain plastics and organic mate­ rials, or example. It is due to this type o energy that

trampolines make it possible to perorm acrobatic moves, and have the impression even i only or a very short period o time, o escaping the Earth’s gravita­ tional feld. In this chapter, you will become amiliar with helical springs, you will calculate their spring constant, and you will explore Hooke’s law. Finally, you will cal­ culate the energy stored in a spring.

rviw Characteristics of force                                    13 Equilibrium between two forces                             14 Relationship between work and energy                       16 Potential energy                                         17 Law of conservation of energy                              18

17.1

Bviou of consind licl sings                        364

17.2

t ngy sod in  sing         370 Chapter 17 Elastic Potential Energy

363

17.1 Beaviour of consrained elical springs There are many kinds o springs, but the ones with a helical shape are prob­ ably the most common. When these springs are in operation, they are constrained, meaning they come under the inluence o orces that tend to deorm them. The behaviour o helical springs can be modelled using mathe­ matical equations.

17.1.1

helical springs

A spring is a mechanical device, which, owing to its elastic properties, can store energy or subsequent release, absorb shocks, or maintain a contact orce between suraces. There are many types o springs which dier in their shapes, their constituent materials and in the way they unction. Springs are said to be helical when they are in the shape o a helix. A helix can be defned as a rotation around an axis, combined with a translation along the same axis (see Figure 1a). We distinguish three types o helical springs according to the way each one unc­ tions: compression springs, tension springs and torsion springs (see Figure 1). As their name indicates, compression springs are designed to be compressed. As such, when a compression spring is in operation, the external orce is directed toward the spring, causing its length to decrease. Conversely, the ten­ sion spring is designed to be stretched. Thereore, in the course o the spring’s operation, the external orce is directed so as to increase its length. The tor­ sion spring requires a pair o orces in order to operate. The two orces, called “couple o orces” in physics, have the same magnitude but opposite direc­ tions. The torsion spring will not be studied in this textbook.

*

Spiral A complete turn in a helical structure

The physical properties o helical springs depend on the diameter o the wire used, the diameter and length o the spring, the number o spirals,* and the type o material used in the production o the wire. These materials vary widely and are closely connected with the spring’s intended use. Dierent types o steel are used in the production o springs as well as various specia­ lized metal alloys.

Spiral F

Spiral helix

Spiral

F

F

F

a) Compression spring

Figure 1 Different helical spring types

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UNIt 5 Energy and its Transformations

b) Tension spring

c) Torsion spring

17.1.2

hook’s lw

Hooke’s law stipulates that an elastic object’s deformation is proportional to the applied deforming forces. When a orce F is applied to a helical com­ pression or tension spring along its axis, the spring’s length decreases or increases. Within a certain deormation limit, the spring re­ covers its initial length as soon as the orce F is removed. As long as this is the case, we fnd ourselves within the elastic range o the spring. Figure 2a shows a tension spring lying on a sur ace with one o its extremities ixed to a vertical support (surace riction can be dis­ regarded). When no orce is exerted on its other extremity (A), the spring’s length is li. Point O, which corresponds to the spring’s ixed extremity, is considered to be the ori­ gin o the one­dimensional horizontal rame o reerence. When a rightward orce F is applied to the spring (see Figure 2b), its length increases and reaches the value l, which cor­ responds to an elongation o the spring that is equal to ∆x. At the same time, the spring’s ree extremity (A) experiences a displacement that can be represented by the vector ∆l . The spring, in turn, develops a letward orce Fr , that is equal and opposite to F .

li a

x xi

O

a) When there is no orce F applied to the spring, its length is equal to li lf li

∆l

Fr

a F

O

∆x

xi

xf

x

b) When the spring is stretched by a rightward orce F , its length reaches the value l At the same time, a restoring orce Fr , directed letward, that is equal and opposite to F , is exerted by the spring This orce, which is always opposite to the displacement ∆l o the ree extremity (A), tends to restore the spring to its initial position

Figure 2 Deormation o a helical compression spring under the infuence o a orce

Fr 5 -F This orce, which is exerted along the spring’s axis, is called the restoring force. Always opposite to the displacement ∆l o the ree extremity (A), the restoring orce tends to restore the spring to its initial length li. The relationship between the restoring orce Fr and the displacement ∆l o the spring’s ree extremity (A) was studied by English scientist Robert Hooke. This relationship, known as Hooke’s law, stipulates that a spring’s restoring orce Fr is proportional and opposite to the displacement ∆l o the spring’s ree extre­ mity. Mathematically, this law is written: Fr 5 -k 3 ∆l In this law, the negative sign (­) indicates that the restoring orce Fr is oppo­ site to the displacement ∆l . The proportionality constant k is called the spring constant o a spring. The value o the constant k depends on the physical char­ acteristics o the spring. The greater the value is, the more diicult it be­ comes to change the length o the spring, and vice versa. Chapter 17 Elastic Potential Energy

365

Since this is a one­dimensional problem, it is possible to dispense with the vector notation by projecting the force and displacement vectors onto the x­axis. Hooke’s law therefore becomes: Fr 5 k ∆x where Fr 5 Restoring orce o the spring, expressed in newtons (N) k 5 Spring constant o the spring, expressed in newtons per metre (N/m) ∆x 5 Elongation o the spring, expressed in metres (m)

dformion A permanent * plsic deormation produced by an applied orce, which remains when the orce causing the deormation is removed

As such, according to Hooke’s law, the magnitude of the restoring force Fr is proportional to the elongation ∆x of the spring. This proportionality implies, for example, that if the elongation ∆x doubles or triples, the magnitude of the restoring force Fr is also multiplied by two or three, respectively. Hooke’s law is valid as long as the experiments are conducted within the elastic range of the spring. In the opposite case, the spring can suffer irreversible plastic deformations* and not return to its initial length. Furthermore, the unit of the spring constant k is the newton per metre (N/m). This is how it can be determined:

Fr 5 k 3 ∆x ⇒ k 5 where Unit o k 5

Fr ∆x

Unit of Fr Unit of ∆x

5 N/m

In short, the restoring force of a spring is characterized by: The magnitude → k 3 ∆x The direction → ollowing the spring’s axis and opposite to the displacement o the ree extremity o the spring

exml a A tension spring with a spring constant k equal to 10 N/m is fxed to a horizontal surace A person stretches this spring so that its length increases by 20 mm What is the restoring orce Fr developed by this spring? (The spring’s surace riction can be disregarded) Data: k 5 10 N/m ∆x 5 20 mm 5 20 3 10-2 m Fr 5 ? Solution: Fr 5 k 3 ∆x 5 10 N/m 3 20 3 10-2 m 5 020 N The restoring orce Fr o the spring is directed along the spring’s axis and is opposite to the displacement o the pulled extremity The magnitude o the spring’s restoring orce is equal to 020 N

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HISTORY HIGHLIGHTS HIGHLIGHTS HISTORY

exml B A 50-cm-long tension spring with a negligible mass is fxed to the ceiling in a vertical position When a metal sphere with a mass equal to 100 g is attached to it, its length reaches 65 cm What is the spring constant k o this spring? Data: li 5 50 cm l 5 65 cm ∆x 5 65 cm 2 50 cm 5 15 cm 5 015 m m 5 100 g 5 0100 kg g 5 980 m/s2 k 5? Solution: The orces to be considered in this problem are the weight Fg o the sphere and the restoring orce Fr o the spring

rOBert hOOke English scientist (1635–1703) Robert Hooke, a contemporary o Isaac Newton, at a very young age demonstrated scientifc curiosity and a remarkable aptitude or inventiveness and engineering While a student at Oxord, he met Robert Boyle (1627–1691) and became his assistant In 1678, HISTORY HIGHLIGHTS HISTORY HIGHLIGHTS he published the law bearing his name, according to which the extension o an elastic body is proportional to the applied orce He expressed this law in its Latin orm, ut tensio sic vis, (which is also the ofcial motto o the École Polytechnique de Montréal), meaning "as the extension, so the orce"

Given that the spring is in equilibrium, we can write: FR 5 0 ⇒ Fr 1 Fg 5 0 By projecting the two orces onto the x-axis (see fgure opposite), we obtain: FRx 5 -Fr 1 Fg 5 0 ⇒ Fr 5 Fg Since we have: Fg 5 m 3 g Fr 5 k 3 ∆x ⇒ k 3 ∆x 5 m 3 g

}

The spring constant k is: m3g k5 ∆x 0100 kg 3 980 m/s2 5 015 m

li

∆x

lf

Fr

5 65 N/m Fg x

a) No orce is applied to b) When the metal sphere the spring: its length is attached to the spring, is equal to li its length reaches l The equilibrium condition o the spring is written: Fg 5 -Fr 

It is important to note that the whole reasoning followed in this section for a tension spring would apply equally to a compression spring. Hooke’s law remains valid and all results obtained are similar. However, we must take into account that in the case of compression, the forces F and Fr , as well as the displacement ∆l , are inversed when compared to the case of tension (see Figure 2 on page 365). This is why the formulation of Hooke’s law at the beginning of this section is stated in a general context.

In addition to his work on elasticity, Hooke’s research let its mark in many other scientifc areas In biology, or example, Hooke is considered to be the inventor o the compound microscope and the creator o comparative plant anatomy We also owe him the concept o the cell as well as the frst description o a biological cell His book Micrographia, published in 1665, in which he collected his microscopic observations, became an immediate success

Finally, it must be pointed out that Hooke’s law is a very important law in physics. It forms the basis of the study of the mechanics of the strength of materials and continuum mechanics, which pertain, among other things, to the formation of solids. Chapter 17 Elastic Potential Energy

367

Furthering

your understanding

th sring consan of a combinaion of srings When several springs are joined together, it is possible to replace them with a single spring that is equivalent to all of them combined, which is characterized by an equivalent spring constant Furthering yourk understanding paralll combinaion of wo srings

k1

If two springs are combined parallel to each other (see Figure 3a), the equivalent spring constant keq is equal to the sum of the spring constants of each spring

k2

k1

keq 5 k1 1 k2 Sris combinaion of wo srings The equivalent spring constant of a series combination of springs (see Figure 3b) is more complex It is written: 1 1 1 5 1 keq k1 k2

k2

In other words, the inverse of the equivalent spring constant of a series combination of springs is equal to the sum of the inverse spring constants of each spring It should be noted that these two results (series and parallel combinations) can be generalized to any number of springs

a) Parallel combination of two springs

b) Series combination of two springs

Figure 3 The spring constant of a combination of springs

SeCtION 17.1

Bhaviour of consraind hlical srings

1. What is a helical spring? 2. What are the dierent types o helical springs? 3. What is the dierence between a compression spring and a tension spring? 4. What do the physical properties o helical springs depend on? 5. What is the direction o the restoring orce o a spring? 6. Determine the magnitude o the restoring orce o a spring when the extremity is displaced by 55 cm The spring constant is 48 N/m

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7. A orce o 100 N is required to compress a spring over a displacement o 4 cm What is the spring constant o the spring? 8. A spring is suspended rom a hook in the ceiling A person attaches a mass o 510 g to the spring and the spring stretches over a distance o 0500 m What is the spring constant? 9. Compression springs A and B have spring constants that are equal to 68 N/m and 48 N/m, respectively Which spring is more difcult to compress?

10. The spring o a mountain bike’s suspension experi­ ences a compression o 185 cm when a orce o 855 N is applied to it Determine the orce required in order to compress the spring by 495 cm

spring constant o 24 3 103 N/m each I the student’s weight is evenly distributed over the six springs, what is the compression applied to each o the springs? 13. A person using springs in a workout routine stret­ ches a 03 m spring with a orce o 365 N What would the spring’s extension be i this person applied a orce o: a) 400 N? b) 223 N? c) 20 N? 14. A student stretches a spring horizontally over a length o 15 mm and applies a orce o 018 N a) Determine the spring constant b) What orce does the spring exert on the student?

11. A spring hangs rom the ceiling o a physics labo­ ratory The lower extremity o the spring is located 180 m rom the foor The teacher suspends a mass o 100 g rom the lower extremity o the spring, and the spring stretches by 50 cm a) What is the spring constant? b) What orce would a person have to apply to pull the mass o 100 g downward over a distance o 20 cm? c) The teacher replaces the mass o 100 g with a mass o 300 g What is the displacement o the mass in relation to the foor? 12. A student with a mass o 62 kg is standing on an upholstered chair containing six springs with a

15. A tension spring that is hung vertically has a spring constant o 48 N/m What is the spring’s elongation when subjected to a gravitational orce o 24 N? 16. A sh scale is equipped with a spring whose spring constant is 600 N/m What is the mass o a sh that deorms the spring by 75 cm? 17. Express the unit o the spring constant as a unction o SI base units used in mechanics

Chapter 17 Elastic Potential Energy

369

17.2 the energy sored in  sring It is possible to see the effects of the energy stored in a spring by observing the acrobatic moves performed on a trampoline. When an athlete lets herself drop on a trampoline, her mechanical energy is stored in the form of elastic potential energy (Epe) in the tension springs surrounding the trampoline bed (see Figure 4). The restoring forces of the springs work and, in turn, return this energy to the athlete, enabling her to rise up into the air again and again. The mathematical expression of the elastic potential energy Epe of a spring is as follows: 1 k (∆x)2 2 where Epe 5 Elastic potential energy of the spring, expressed in joules (J) k 5 Spring constant of the spring, expressed in newtons per metre (N/m) ∆x 5 Elongation of the spring, expressed in metres (m) Epe 5

Figure 4 Tension springs surrounding a trampoline bed See Work, force nd dislcemen, p 326

Where does this equation come from? The mathematical expression of the elastic potential energy Epe of a spring can be deduced from the one relating to the work of the force F that is applied to change its length. However, cal­ culating this work is not as direct as the calculations presented in Chapter 15. In that chapter, the force considered in the calculation of the work was constant and independent of the displacement. In this case, work was written: W 5 F  ∆s 5 F 3 ∆s 3 cos θ It is possible to represent the work of a constant force using a graph of force as a function of position (see Figure 5). Force ∆s Feff

F 3 cos θ

a

B

posiion

Figure 5 Graph of force as a function of position in the case of a constant force The area of the shaded rectangle represents the work of the force for displacement ∆s from point A to point B

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Figure 5 on the previous page shows that the force preserves its value through­ out its displacement from point A to point B. The area of the shaded rectangle is equal to the product of its length and width, which amounts to:

Area of the rectangle 5 ∆s 3 (F 3 cos θ ) 5 F 3 ∆s 3 cos θ 5 W The area of the rectangle is equal to the value of the work done by the constant force. This result can be generalized with the assertion that the work of a force is equal to the surface area under the curve in a graph representing force as a function of position. This statement can be used to calculate the work of force F which is applied to a spring to change its length. This force, whose norm is equal to that of the restoring force Fr, is not constant. By the same measure that a spring is stretch­ ed (or compressed), the magnitude of force F increases, making it a variable force that depends on the displacement of the free extremity of the spring (see Figure 6). Foc ∆x F

k 3 ∆x

xi

xf posiion

Figure 6 Graph of the force F applied to a spring as a function of the position of its free extremity The shaded triangular area represents the work of this force when the free extremity of the spring is displaced from position xi to position xf

The surface area under the curve is a measure of the work of the force F applied to a spring. It is equal to:

Area of the triangle 5 W 5

1 1 3 [∆x 3 (k 3 ∆x )] 5 3 k 3 (∆x )2 2 2

As such, the work of the force F applied to a tension (or compression) spring to stretch (or compress) it is equal to half of the product of the spring constant and the elongation (or compression) squared. 1 k (∆x )2 2 where W 5 Work of the restoring force of the spring, expressed in joules (J) k 5 Spring constant of the spring, expressed in newtons per metre (N/m) ∆x 5 Elongation of the spring, expressed in metres (m) W5

Chapter 17 Elastic Potential Energy

371

See Work-nrgy horm, p 342

Therefore, work W corresponds to a process of transfer of energy. It is related to energy by the equation: ∆E 5 W In the case of a spring, the work of the force F which is applied to it to change its length allows it to store elastic potential energy Epe. ∆Epe 5 Epef 2 Epei 5 W where Epef 5 Final elastic potential energy Epei 5 Initial elastic potential energy If we consider that a the spring does not store any elastic potential energy before being stretched or compressed (Epei 5 0), we can deduce that: ∆Epe 5 Epef 2 Epei 5 Epef 2 0 5 Epef Assuming that in this case Epef 5 Epe, we can conclude that the elastic poten­ tial Epe of a spring is equal to the work W of the force F that is applied to it change its length. Epe 5 W From this, we get the equation expressing the elastic potential energy of a spring.

Epe 5

1 3 k 3 (∆x)2 2

exampl The length of a compression spring initially unextended is reduced by 500 cm when a force of 200 N is applied to it Calculate the elastic potential energy stored in this spring (the friction can be disregarded) Data: Solution: F 5 200 N 1 First, calculate the spring constant of the spring: ∆x 5 500 cm 5 00500 m Fr 5 F 5 k 3 ∆x, Epe 5 ? 200 N F therefore k 5 5 5 4 3 102 N/m 00500 m ∆x 2 The elastic potential energy stored in the compressed spring is: 1 3 k 3 (∆x)2 2 1 5 3 (4 3 102 N/m) 3 (00500 m)2 5 05 J 2 The elastic potential energy stored in the spring is equal to 05 J Epe 5

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UNIt 5 Energy and its Transformations

Wy do fs l so ig? The fea is a parasitic insect that lives o the blood o its hosts (see Figure 7 ) This tiny animal, whose length varies rom 2 to 8 mm, is a true leaping champion In act, a fea can leap more than 30 cm high, representing up to 150 times its own size I humans were capable o such eats, they would be able to leap to heights comparable to that o the Eiel Tower! The secret o this exceptional champion is a substance called resiline, which supports the insect’s third pair o legs Resiline is a protein with phenomenal elastic power Whereas the best synthetic elastic substances can only release 80% o absorbed energy, resiline can manage almost 97% Moreover, this surprising substance can release its elastic potential energy in an extremely short time and allows or repetitive motions without aecting its elasticity Australian researchers succeeded in making synthetic resiline by transerring the gene responsible or the protein into a bacterium The uture is cerFigure 7 Engraving o a fea eatured in the book Micrographia tain to hold surprises in connection with the applications o this substance by Robert Hooke, published in 1665 Why not use this synthetic substance to achieve new sports records?

SeCtION 17.2

t ngy sod in  sing

1. An apple with a mass of 0100 kg is fastened to a verti­ cal tension spring with a spring constant of 960 N/m a) Calculate the spring’s elongation b) Calculate the elastic potential energy stored in the spring 2. A force of 125 N stretches a spring to 0250 m from its equilibrium position a) What is the elastic potential energy stored in the spring? b) If the spring contracts to 0150 m from its equi­ librium position, what is the change in elastic potential energy? 3. A spring has a spring constant of 440 x 10 4 N/m What is the change in elastic potential energy stored in the spring when it stretches from 125 cm to 15 cm?

4. The spring constant of a spring is 750 N/m a) What distance from its equilibrium position would the spring have to be stretched in order to store an elastic potential energy of 45 J? b) In order to double the elastic potential energy stored in the spring, what further distance would it have to be stretched? 5. A spring stretched to 0400 m from its equilibrium position has an elastic potential energy of 5 3 102 J a) What is the magnitude of the force required to produce this extension? b) If the force producing the extension increases to 1000 N, by how much does the elastic potential energy change?

Chapter 17 Elastic Potential Energy

373

APPLICATIONS Suspension in moor veicles The suspension in motor vehicles consists o a set o mechanical parts that ensure an elastic link between the wheels and the vehicle’s body (see Figure 8). Its main role is to minimize the transer o shocks to the body, produced by the wheels’ jolts, which are caused by the unevenness o roads. I roads were perectly level, motor vehicles would not require suspension. The suspension used in most mass­produced cars inclu­ des, among other things, two important devices: a heli­ cal spring and a shock absorber. When the wheels o a car encounter irregularities on the road, they experience vertical orces that displace the vehicle upward. The helical spring fxed to each o the wheels makes it pos­ sible to reduce the vertical displacements caused by road imperections. This reduction o vertical displacement occurs on account o the elastic properties o the spring. When the wheel experiences an upward orce, the spring is compressed, absorbing a certain amount o compres­ sion energy, which it transorms into elastic potential energy. At the same time, the spring develops a resto­ ring orce, which is directed downward. The restoring orce ensures that the wheel maintains a better grip on the road. Thanks to the suspension, the vertical motion becomes soter and is elt over a longer period o time. However, when the wheel lands back on the road, the spring is still compressed, and the restoring orce, now directed upward, creates a new vertical displacement. As such, even though the spring absorbs energy each time, it creates a sometimes uncomortable periodic oscillatory movement that is difcult to control. This behaviour is due to the act that while springs absorb energy well enough, they don’t dissipate it very eectively.

The device responsible or dissipating the energy is the shock absorber. There are several types, but the most widely used are hydraulic shock absorbers. They trans­ orm the kinetic energy generated by the deormation o the spring, which alternates between being compressed and being stretched, into thermal energy. Hydraulic shock absorbers can mitigate the oscillations when the spiral is compressed or stretched. Two­tube models are composed o two cylinders that can slide into each other and one o which contains viscous oil. With each oscillation o the suspension, a piston moves inside the oil­containing cylinder and is slowed down by the liquid that escapes through the calibrated openings. This has the eect o absorbing the movement and increasing the liquid’s temperature. The thermal energy stored in the oil is subsequently dissipated by the shock absorber’s body. As shown in Figure 8, the shock absorb­ er is inserted into the helical spring, allowing both to work in unison: the shock absorber absorbs the energy while the helical spring dissipates it. As such, suspen­ sion allows or the eective absorption o a road’s une­ venness, ensuring a comortable and sae ride.

Sock absorber

Weel

helical spring Body suppor

Spring moion

Figure 8 A car’s suspension connects the wheels to the body while absorbing the wheels’ sudden movements caused by imperfections or unevenness on the road as much as possible

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UNIt 5 Energy and its Transformations

t sing scl The spring scale is a device that employs the elastic properties o certain metallic materials. It usually has a helical spring with a ring attached to one end and a hook attached to the other and is used to measure the mass o objects. When an object is attached to the hook, the stretching o the spring causes a pointer to move, indicating the value o the mass by pointing to gradua­ tions on a scale marked in units o mass (See Figure 9). Since the restoring orce Fr o the spring balances the weight Fg o the object, we have (see Example B on page 367): Fr 5 Fg 5 k 3 ∆x 5 m 3 g We can then deduce: g 3m k The stretch ∆x o the spring is proportional to the mass m o the object being weighed. The proportionality constant is equal to the quotient o the two magnitudes: the accele­ ration o the gravitational feld g and the spring constant o the spring k. The scale can thereore be directly grad­ uated in units o mass ater calibration, and is considered accurate so long as the values o g and k do not vary.

the beginning o the 19th century. There are two reasons or this ban. First, these devices can provide inaccurate measurements o mass i the spring becomes permanent­ ly stretched. Second, the measured value is correct only in a given gravitational feld. Since the earth’s gravita­ tional feld varies according to geography and topogra­ phy, measurements o the same mass can vary rom one place to another. Even i the dierence is minor, spring scales are not reliable. Dynamometers, which are based on the same principle as spring scales, are still used. Graduated in newtons (N), they are no longer used to measure mass but the orces o traction and compression to which springs are subjected.

∆x 5

The frst spring scales likely appeared in the late 17th century in Germany. They were a great success all across Europe because they were small and gave a rapid reading. Unlike other scales at the time, they were the only weighing devices that did not compare the mass being measured with calibrated masses, which made them very practical. Spring scales are actually dyna­ mometers that measure the orce o attraction exerted by the Earth’s gravitational feld on bodies by balan­ cing it with the deormation o the spring. The springs or these devices are usually made o steel or iron and the graduated scales are made o brass. Spring scales can measure masses ranging rom a ew grams to a ew hundred kilograms. Despite their ease o use and popularity with the public, spring scales have been banned rom commercial transactions in France since

ring

poin Gdud scl

hook

Figure 9 A helical spring scale with fat ace and straight pointer

Chapter 17 Elastic Potential Energy

375

chapter

17

elsi ponil engy

17.1 Behaviour of consrained helical springs • A spring is a mechanical device, which, owing to its elastic properties can store energy for subsequent release, absorb shocks, or maintain a contact force between surfaces. • Springs are said to be helical when they are in the shape of a helix. • We distinguish three types of helical springs according to the way each one functions: compression, tension, and torsion. • The physical properties of helical springs depend on the diameter of the wire used, the diameter and length of the spring, the number of spirals, and the type of material used in the production of the wire. • Hooke’s law stipulates that an elastic object’s deformation is proportional to the applied deforming forces. Mathematically, Hooke’s law is written:

Fr 5 -k ∆l

• Always opposite to the displacement ∆l of the free extremity of the spring, the restoring force Fr tends to restore the spring to its initial length. • In cases where the problem is one­dimensional, it is possible to dispense with the vector notation by projecting the force and displacement vectors onto the x­axis. Hooke’s law therefore becomes:

Fr 5 k ∆x

• The magnitude of the restoring force Fr is proportional to the elongation (or compression) ∆x of the spring. • The unit of the spring constant k is the newton per metre (N/m).

17.2 the energy sored in a spring • The work of a force is equal to the surface area under the curve in a graph representing force as a function of position. • The restoring force Fr of a spring as well as the force F that is applied to it in order to change its length are variable forces that depend on the displacement of the free extremity of the spring. • The work of the restoring force F applied to a tension (or compression) spring to stretch (or compress) it is equal to half of the product of the spring constant and the elongation (or compression) squared. • The work W of force F applied to a spring is related to the change in the elastic potential energy ∆Epe according to the following equation: • If the initial elastic potential energy of a spring is zero, the elastic potential energy stored in a spring Epe is written:

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UNIt 5 Energy and its Transformations

W5

1 k (∆x) 2 2

∆Epe 5 W

Epe 5

1 k (∆x) 2 2

elsic ponil engy

1. a) An ideal spring with a spring constant of 2000 N/m is compressed over a distance of 0400 m What is the elastic potential energy stored during this spring compression? b) If this spring transfers all of its elastic potential energy into kinetic energy onto a mass of 2 kg, what velocity would this mass have? We can assume that the initial velocity of the mass is zero 2. A spring has a spring constant of 650 N/m Initially, the spring is compressed over a length of 0100 m in relation to its equilibrium position a) What is the elastic potential energy stored in the spring? b) What additional length would the spring have to be compressed in order to triple its potential energy? 3. With gravitational potential energy, if the height doubles so does the potential energy However, with elastic potential energy, if the extension of a spring doubles, the energy does not double a) How does the stored elastic potential energy change if the extension doubles? b) Explain this phenomenon using graphs of force as a function of position for the gravitational poten­ tial energy and the elastic potential energy 4. A person with a mass of 650 kg goes bungee jump­ ing At the lowest point, he is located 30 m below his starting point If, at the equilibrium point, the bungee cord measures 15 m, what is its spring constant? hin: Assume this is an isolated system At its lowest point, the bungee cord must convert all the gravitational potential energy that the per­ son lost into elastic potential energy

5. For the following graph: a) calculate the spring constant of the spring b) name the unit of the surface area under the curve Force F (N)

Chapter 17

50 40 30 20 10 0

0.1

0.2

0.3.

0.4

0.5

0.6

0.7

0.8 x (m)

6. A spring is stretched 350 cm from its equilibrium position A force of 105 N is exerted on the spring a) How much elastic potential energy is stored in the spring? b) If the spring’s extension is reduced to 200 cm, what is the change in the elastic potential energy? 7. A 020­kg mass is hanging from a vertically sus­ pended spring that has a spring constant of 55 N/m When the spring is released from its equilibrium position (not stretched), the mass falls Use the law of conservation of energy to determine: a) the velocity of the mass after it has fallen 15 cm b) the distance travelled by the falling mass before a change in direction occurs 8. A horizontal spring with a spring constant of 12 N/m is fastened to the edge of a laboratory table and is used to launch marbles toward targets on the ground, 93 cm below A marble with a mass of 83 3 10­3 kg is launched by a spring that under­ goes an initial compression of 4 cm What distance does the marble travel horizontally before hitting the ground?

Chapter 17 Elastic Potential Energy

377

378

CON T E N T S 1

2

Laboratory Saety                            380

4

Interpreting Measurement Results            396

1.1 Danger symbols                          380

4.1 Uncertainty                              396

1.2 Saety symbols used in the Quantum collection                        383

4.2 Signifcant fgures                         397

1.3 Saety rules                              383

4.4 Signifcant fgures in results o mathematical operations                               398

Methods Used in Physics                    385

4.5 How to round o a number                 399

4.3 Measurements with unknown uncertainty     398

2.1 Summary table o methods                 385 2.2 The our steps o problem-solving            386

5

Mathematics in Physics                      400

2.3 Experimental method                      387

5.1 Problem-solving                           400

2.4 Analysis                                 388

5.2 Solving second-degree equations            401 5.3 Scientifc notation                         402

3

5.4 Review o trigonometry                    404

Presenting and Analyzing Scientifc Results   389

5.5 Review o geometry                       405

3.1 Table                                   389

5.6 Review o some ormulas used in this textbook                             406

3.2 Cartesian graph                           390 3.3 Graphical determination o the slope o a straight line                            392 3.4 Graphical determination o the slope o a tangent to the curve                     392

6

Reerence Tables                            413 6.1 Physical quantities and units                413

3.5 Linear regression on a scatter diagram        393

6.2 Physical constants                        414

3.6 Laboratory report                         394

6.3 Solar system                             414

7

Nobel Laureates in Physics                  415

379

1 Laboratory Safety

APPENDIX

1

It is extremely important to pay attention to health and saety in the labora­ tory. You must thoroughly understand the meaning o all danger symbols. You must also be amiliar with saety rules and ollow them to avoid accidents.

1.1

Danger symbols

Various danger symbols and saety symbols are commonly used in the labo­ ratory. These symbols always provide a warning. You will thereore need to take special precautions whenever you handle substances or materials label­ led with any o these symbols. Anyone handling dangerous substances must know what these symbols mean and which precautions to take when using these substances.

Danger symbols on household products Many products ound in homes, at school and in the workplace contain dan­ gerous substances. These products must be used with caution. Manuacturers warn o the dangers associated with their products by labelling the containers with symbols. Each symbol ound on a household product has a specifc mean­ ing (see Table 1).

Table 1 Meanings o danger symbols ound on household products Symbol

Meaning This product may burn the skin or eyes I swallowed, it causes injury to the throat and stomach

Figure 1 Oven cleaner is a corrosive

Corrosive

product

This product may explode i its container is heated or punctured

Explosive This product and the umes it releases are highly fammable i they are placed near a heat source, fames or sparks

Flammable This product is toxic and may cause serious health complications It may be atal i swallowed and, in certain cases, i inhaled

Poison

380

APPENDIX 1 Laboratory Saety

WHMIS danger symbols

APPENDIX

1

The Workplace Hazardous Material Information System (WHMIS) is the Canadian standard for all information and communication concerning dan­ gerous materials. The WHMIS danger symbols identify substances that pose health risks. These substances include household products and solvents found in schools and homes. Each WHMIS symbol indicates which precau­ tions should be taken (see Table 2).

Table 2 Saety precautions or products identied with WHMIS danger symbols Symbol

Precaution Handle this product with care; keep it away rom heat and fames

Compressed gas Keep this product away rom heat, fames and sparks

Flammable and combustible material Keep this product away rom heat, fames, sparks and any combustible materials

Oxidizing material Avoid all contact with this product

Material causing immediate and toxic eects Avoid all contact with this product

Material causing other toxic eects Avoid all contact with this product

Inectious material Avoid all contact with skin and eyes Wear saety glasses, gloves, a lab coat or an apron

Corrosive material Use only in a well-ventilated area

Dangerously reactive material

APPENDIX 1 Laboratory Saety

381

GHS danger symbols

APPENDIX

1

In 2003, the member countries o the United Nations (UN) adopted the Globally Harmonized System o Classiication and Labelling o Chemicals (GHS). The GHS danger symbols are gradually replacing the various systems, such as the WHMIS, used in dierent countries. The new system was desi­ gned to increase human and environmental saety. Each GHS danger symbol has a specifc meaning. The symbols also indicate which precautions should be taken when handling certain products (see Table 3).

Table 3 Risks and precautions identied by GHS danger symbols Symbol

382

Risks

Precautions

Unstable, explosive material or object

Risk o mass explosion, blast, projection or re

Handle the container with caution; keep away rom heat and fames; do not puncture

Flammable gas, liquid or solid

Risk o re or explosion

Keep the container and its contents away rom heat, fames and sparks

Oxidizing gas, liquid or solid

Risk o re, intensication o re or explosion

Keep the container and its contents away rom heat, fames, sparks and any other combustible material, such as wood, gas or solvents

Gas under pressure

Risk o explosion

Handle the container with caution; keep away rom heat and fames; do not puncture

Corrosive material

Risk o burns, irritation o the skin or eyes, and eye injury

Avoid all contact with skin and eyes; in case o contact, rinse with water Wear saety glasses, gloves and a lab coat or apron when handling this product

Highly toxic material

Risk o poison i swallowed, i in contact with skin or i inhaled

Avoid all contact with this product Let an experienced person handle it or you

Harmul material

Risk o skin allergies, irritation o the skin, eyes and respiratory tracts, dizziness or drowsiness

Avoid all contact with this product Let an experienced person handle it or you

Material that presents various hazards

Risk o allergic reactions, asthma or breathing diculties i inhaled Risk o genetic deects, cancer or organ damage Risk o inertility or harm to etus Risk o harm to breasted babies

Avoid all contact with this product Let an experienced person handle it or you

Material that is hazardous to the aquatic environment

Risk o harmul long-term eects

Avoid all contact with this product Let an experienced person handle it or you

APPENDIX 1 Laboratory Saety

1.2

Safety symbols used in the Quantum collection

APPENDIX

1

It is extremely important that you understand saety symbols when working in the laboratory. Each saety symbol used in the Quantum collection has a specifc meaning and indicates the precautions that must be taken (see Table 4). Table 4 Saety symbols used in the Quantum collection Symbol

1.3

Precautions

Protect eyes

Wear saety glasses

Protect hair

Tie hair back

Protect skin

Wear gloves

Protect clothing

Wear a lab coat or an apron

Caution: may cause burns

Exercise caution when using hot substances or objects

Caution: hot

Wear heat-resistant mitts

Caution: dangerous umes

Work under a ume hood or in a well-ventilated area

Caution: sharp

Exercise caution when using objects with sharp edges

Safety rules

Saety depends upon responsible behaviour and attitude: this includes obeying certain rules. It is important to be thoroughly amiliar with these saety rules and, above all, to ollow them at all times. Laboratory safety rules 1 Inorm the person in charge about any health problems such as an allergy or illness that might interere with the activity 2 Identiy the closest frst-aid kit, fre extinguisher, fre blanket, emergency shower, laboratory eyewash and fre alarm Learn how to adequately use each o these

APPENDIX 1 Laboratory Saety

383

APPENDIX

1

3 Handle the equipment with care, act calmly and pay attention to what you are doing 4 Wear saety glasses Wear gloves and a lab coat or an apron when handling materials that are dirty or corrosive 5 Do not wear jewellery, clothing that might get in the way, or untied shoes Tie back your hair 6 Never eat or drink during an activity 7 Keep your work surace clean and neat 8 Follow directions concerning the use o dangerous products 9 Beore beginning an experiment, make sure that you ully understand every aspect o the procedure, such as how to handle the equipment and materials, waste disposal and saety symbols (see Table 4 on page 375) 10 Obtain the approval o the person in charge beore beginning the experiment i you have developed your own procedure or i you have modied an approved procedure 11 Never leave an experiment unattended 12 Never touch or taste a substance Never smell a substance directly 13 Wash your hands ater each experiment 14 Immediately inorm the person in charge o any accident, even i it does not appear serious Examples include injury, spills and equipment breakage

Handling electrical devices and heating appliances 1 Beore using an electrical device or heating appliance, make sure that it works properly Inorm the person in charge o any problems 2 Never distract someone who is working with an electrical device or heating appliance Figure 2 In the laboratory, tie back your hair and wear saety glasses, gloves and a lab coat

3 Never use a burner to heat a fammable substance; use a hot plate instead 4 Never leave an electrical device or heating appliance that is in use unattended 5 Never touch an electrical device or electrical outlet with wet hands 6 Never touch an electrical cord that appears to be damaged 7 Never leave electrical cords out on the foor 8 When you are nished with an electrical device, unplug it by pulling on the plug (not the cord), and then put it away

384

APPENDIX 1 Laboratory Saety

2 Methods Used in Physics

APPENDIX

There are dierent ways o inding an eective solution to a problem. Each method involves a logical process o trial and error.

2.1

2

Summary table of methods

To solve a problem in physics, you may use a general problem­solving method or more specifc methods depending on the nature o the problem. The choice o a specifc method is based on the problem to be solved. Two or more methods may be combined to arrive at a result. Five o these methods are explored in the Third Year o Secondary Cycle Two (see Table 5).

Table 5 Methods used in physics Method Modelling Modelling provides a concrete representation o an element that is difcult to imagine This representation can take on several orms, such as a text, a drawing, a mathematical ormula or a chemical equation Over time, the model becomes more refned and complex It may even be modifed or rejected A model must take into account the characteristics o the element that is being studied It must also help people understand a given reality, explain certain properties o that reality and predict new observable phenomena Observation method This method consists o observing a phenomenon and interpreting acts while taking into account various criteria Through observation, you can learn new acts that result in a dierent understanding o the phenomenon Analysis This method involves analyzing an object or a system with the aim o identiying the elements rom which it is constructed, as well as the interactions between these elements This analysis helps in understanding the purpose o the object or system, as well as how it works and how it is made In some cases, this method involves using a broader understanding o a system to determine the unction o its parts and the relationships between them (see Section 2.4 on page 388)This aspect o the analytical method is particularly useul in studying phenomena and applications Experimental method This method seeks an answer to a problem through an experiment Students can begin looking or an answer and defning the ramework o the experiment The person conducting the experiment must then develop an experimental procedure in order to identiy a certain number o variables to be manipulated The aim o the procedure is to identiy and compare observable or quantifable elements and check them against the initial hypotheses Analysis at the end o the experiment raises new questions, helps to ormulate new hypotheses, adjust the experimental procedure and take the limitations o the experiment into account (see Section 2.3 on page 387 ) Empirical method This method involves feld research without any manipulation o variables This makes it possible to explore the elements o a problem or to obtain a new representation o the problem This method is oten based on intuition It can lead to the development o hypotheses and possible avenues or other research projects Surveys are an example o the empirical method

APPENDIX 2 Methods Used in Physics

385

APPENDIX

2

2.2

The our steps o problem-solving

Regardless o the method used to solve a problem, the same our steps are always ollowed. Figure 3 illustrates the steps used to ind a solution to a scientifc problem. You can always return to an earlier step i you need to modiy a hypothesis, action plan or procedure. I this is the case, continue with the procedure rom the point at which the modiication was made. Adjustments will probably have to be made to the rest o the procedure. Defne the problem Begin by describing the problem Next, determine your goal, taking into account the context o the problem Then, you can ormulate your working hypothesis by reerring to the relevant scientifc concepts

Develop an action plan You must explore all problem-solving methods to determine which one is the most appropriate to the situation Then, develop an action plan that can work with the available resources and applicable constraints You must also plan out each step o the action plan

Carry out an action plan Follow the steps o your action plan Make a note o any inormation that may be useul during the problem-solving process

Analyze the results You must establish connections between the inormation you obtained and scientifc concepts You may also draw conclusions and provide an explanation or a solution Figure 3 The our steps o problem-solving

Attention! You may, at any time, return to an earlier step to make modifcations Continue with the procedure rom this step It is very important to keep a record o all modifcations

386

APPENDIX 2 Methods Used in Physics

2.3

Experimental method

APPENDIX

The problem­solving process applied to the experimental method includes four steps (see Table 6).

2

Table 6 The our steps o problem-solving applied to the experimental method Step

Example: Determining the melting point o ice

Defne the problem a) Defne the problem to be solved

a) Ice begins to melt at a certain temperature I must accurately determine this temperature

b) Identiy the goal

b) Determine the melting point o ice

c) Use relevant scientifc concepts (ormulas, theories) or solving the problem, i necessary

c) Every pure substance possesses a unique melting point

d) Formulate a hypothesis

d) I believe that the melting point o ice is 1°C

Develop an action plan a) Speciy the variables you need to observe and measure

a) Variables: state o water based on temperature, and water temperature (as a unction o time)

b) Make a list o the equipment required and make a drawing o the set-up, i necessary

b) Equipment and materials: 250 mL o crushed ice, a 500-mL beaker, a thermometer, a stopwatch, etc

c) Write out the steps o the procedure

c) Procedure 1 Place the ice in the beaker 2 Place the thermometer in the beaker Etc

Carry out an action plan a) Perorm the experiment in a sae manner

a) I ollow saety rules while conducting the experiment

b) Gather the data and make a note o any useul observations

b) I record the temperature every 30 seconds and when the ice begins to melt

c) Process the data (by making diagrams, calculations, etc) gathered rom observations made during the experiment

c) I write this inormation in a table I prepare a broken-line graph representing water temperature as a unction o time I use a colour code to indicate the state o the water at each temperature

Analyze the results a) Analyze your data and observations

a) I observe that ice begins to melt at 0°C and that this temperature remains stable until all o the ice has melted

b) Draw conclusions that either confrm or disprove the initial hypothesis

b) I conclude that the melting point o ice is 0°C This conclusion does not correspond to my initial hypothesis

c) Formulate explanations and, i necessary, suggest new hypotheses or improvements that could be made to the experiment

c) The period during which the temperature remains steady is a plateau The experiment could be repeated to determine whether a plateau occurs when water transorms to vapour

APPENDIX 2 Methods Used in Physics

387

APPENDIX

2

2.4

Analysis

In physics, you can use analysis to recognize interactions among the various elements that make up a system, a phenomenon or an object. For example, by analyzing an object such as a camera (see Figure 4), you can identify its structural components: the objective (which contains the lens or system of lenses), the diaphragm, the shutter and the light-sensitive surface. You can also discover the interconnections among the various components. Determining the function of each part and the relationship among these parts helps to illustrate the dynamics of the system, the purpose of which, in the case of the camera, is to take photographs.

Diaphragm Objective

Lens

Shutter

O

F

Light-sensitive surace (flm or sensor) f

Figure 4 Analysis of a camera provides an understanding of how it works

Analysis of the components of a camera helps us to understand the underlying physical phenomena. For example, the objective is mobile so that the distance between the lens and the light-sensitive surface may be altered. By moving the lens according to the distance of the object to be photographed, the photographer ensures the proper geometric arrangement of the components. Light rays from the object can then form a clear image on the light-sensitive surface.

388

APPENDIX 2 Methods Used in Physics

3 Presenting and Analyzing Scientifc Results

APPENDIX

3

In physics, data is often presented in a table, graph or diagram, which helps in understanding and interpreting the data. The laboratory report is another essential tool for presenting results obtained from an experimental method.

3.1

Table

Tables organize data (numbers or words) into columns and rows, which makes it easier to analyze information. Tables are also used to record data that will be presented subsequently in the form of a graph or diagram (see Section 3.2). The units of measure can be provided in parentheses in the table under the column title; this avoids having to repeat them on each line. Data may be recorded for an independent variable and a single dependent variable. Table 7 Average monthly precipitation

Table title

or Montréal Independent variable

Value o the independent variable

Month

Precipitation (mm)

January

87

February

66

March

91

April

81

May

91

June

96

July

97

August

100

September

97

October

91

November

98

December

93

Dependent variable Unit o measure

Value o the dependent variable

We can also record data for several dependent and independent variables in tables with double entries.

Table 8 Temperature dierence observed between two pieces o cardboard ater 30 minutes Under white cardboard Under black cardboard Independent variables

Initial temperature (°C)

22

22

Final temperature (°C)

27

33

15

1 11

Variation (°C)

Dependent variables

Calculations

APPENDIX 3 Presenting and Analyzing Scientifc Results

389

3.2

APPENDIX

3

Cartesian graph

When data is organized into a table, it is not always easy to determine the relationship between the quantities measured. However, when these values are plotted on a graph, the relationship is more obvious. There are many types o graphs and diagrams. This appendix describes the Cartesian graph. The axes o a Cartesian graph cross at the coordinates’ point o origin. In science, it is customary to use perpendicular axes. The data in the table are plotted on the two axes: the axis o abscissas, the horizontal axis usually carrying the independent variable, and the axis o ordinates, the vertical axis usually carrying the dependent variable.

Broken-line graph A broken­line graph is used to illustrate continuous data. It provides a graphical representation o the relationships between the independent vari­ able and the dependent variable. To create a broken­line graph, begin by drawing the axes. Make them long enough so that you can clearly indicate all values o the variables. Represent each value with a point. Connect each point to the next. This creates the broken line (see Figure 5, which was created based on Table 7 on the previous page).

Average monthly precipitation or Montréal

Title

Unit o measure

Dependent variable

Precipitation (mm)

Axis o ordinates 105 100 95 90 85 80 75 70 65 60 55 50 0

Point representing a value Broken line connecting each point

J

F

Independent variable

M

A

M

J

Month

J

A

S

O

N

D Axis o abscissas

Figure 5 Example o a broken-line graph

Line o best ft In a diagram, you may draw a broken­line graph or a line o best ft, depending on the situation. I the variation in the dependent variable is discontinuous, as in the previous example, draw a broken­line graph. However, i the variation o the dependent variable is continuous, draw a line o best ft (see Figure 6). This may be either a straight line or a curve.

390

APPENDIX 3 Presenting and Analyzing Scientifc Results

Cooling time o 250 mL o hot water

APPENDIX

Title

3

Unit o measure Dependent variable

Temperature (°C)

Axis o ordinates 60 55 50 45 40 35 30 25 20 15

Point representing a value Curve o best ft in a scatter plot Axis o abscissas 10 20 30 40 50 60 70 80 90 100

0

Time (min) Unit o measure y Independent variable

Figure 6 Example o a line o best ft

yαx

Variations o some mathematical unctions Mathematical relationships among physical quantities may be represented by various mathematical functions (see Table 9). The graphical representation of these functions depends on how the variables are chosen. For example, the quadradic function y  ax 2 forms a parabola on a Cartesian graph when the yy values of y are plotted on the axis of ordinates and the values of x are plotted 2 y yααare xx plotted on the axis of abscissas. However, if the values of x (instead of x) on the axis of abscissas, this same function forms a straight line. This is refer­ red to as a change of variable.

x

y

y y α x2

The graphical representations of a few typical functions, with and without a xx change of variable, are shown below.

x

Table 9 Variations o some unctions y

yyy

yy

y yαx yαx

y

2

y yααxx2 y α x1

Function x

x that a) The unction y  ax is a straight line passes through the origin The slope o the line is equal to a y

xx

b) Variations o the unction y  ax 2

xxx2 2

c) Variations o the unction y  a 3

y y yy

y

1 x

yy

2

yαx y α x2 y yααx1 1 x

How to obtain a straight line

x

y

x 2 xx x2

x

To obtain a straight line, plot y as a unction o x 2 The slope o the liney is equal to a

y

y

11 xx

To obtain a straight line, plot y as a unction o 1  The slope o the line is equal to a x

()

y α x1 1 yαx x

x

1 x 1 x

APPENDIX 3 Presenting and Analyzing Scientifc Results

391

3.3

APPENDIX

3

Graphical determination of the slope of a straight line

The graph in Figure 7 represents a quantity y (measured in arbitrary q units) as a unction o x (in p units). The equation or the straight line is: y  ax 1 b where a  slope b  ordinate at the origin The slope o the straight line is: a  slope 

vertical displacement (y 2 – y 1) 20 q – 6 q 14 q  2    47 q/p horizontal displacement (x – x 1) 35 p – 05 p 30 p

The ordinate at the origin is 3 q. The equation o the straight line is:

( qp )x 1 3 q

y (q)

y  47

Vertical displacement  14 q

25

20

15

10

5

0

Horizontal displacement  3.0 p

1.0

2.0

3.0

4.0

5.0

x (p)

Figure 7 Graphical determination o the slope o a straight line

3.4 x (m)

In a graph o position as a unction o time, the velocity at a particular moment is determined by calculating the slope o a tangent to the curve drawn at a specifc point. To draw the tangent o the curve at point A, you must draw a line:

A

Dx

• touching the curve at point A only • with the same slope as the curve at this point

Dt 0

t (s)

Figure 8 Tangent line to the curve at point A

392

Graphical determination of the slope of a tangent to the curve

This line is a tangent to the curve. The tangent is properly oriented i the points on the curve move away rom the tangent more or less symmetrically on each side o point A (see Figure 8). To calculate the slope, use the method previously explained (see Section 3.3).

APPENDIX 3 Presenting and Analyzing Scientifc Results

Linear regression on a scatter diagram

3.5

APPENDIX

3

The plotting o experimental results may produce an elongated scatter o points roughly orming a straight line (see Figure 9). I you want to extract the parameters o a linear law in the orm y  ax 1 b, draw a line that passes through the maximum number o points. I the points are not per­ ectly aligned (which is usually the case), the task will be difcult and the result ar rom perect.

Variation in the position as a function of time Position (m)

Position (m)

Some spreadsheet programs use sophisticated mathematical calculations or determining the line that passes closest to all o the points on the graph (see Figure 10). The purpose o this linear regression is to determine the coef­ cients a and b o the unction y  ax 1 b.

30 25

Variation in the position as a function of time 30 25

20

20

15

15

10

10

5

5

0

0 0

2

4

6

8

10

12

14

16

18

20 Time (s)

Figure 9 A scatter diagram representing the variation in the position o a moving object as a unction o time

0

2

4

6

8

10

12

14

16

18

20 Time (s)

Figure 10 The straight line represents the linear regression o a scatter diagram

In the example shown in Figure 10, the line represents the linear relationship between time (t) and position (x) o a moving object. Since the line pas­ ses through the origin, we can conclude that b  0. The graphical determi­ nation o the slope o the line (see Section 3.3, on the previous page) gives a  1.5 m/s which is the velocity o the moving object. The linearity o the relationship between time and position illustrates uniorm motion characteri­ zed by the typical equation: xv3t where x  position (m) v  velocity o the moving object  15 m/s t  time (s)

APPENDIX 3 Presenting and Analyzing Scientifc Results

393

APPENDIX

3

3.6

Laboratory report

A laboratory report is a summary o the experimental method used. It de­ scribes each step o the method and presents the results. A laboratory report must always have a title page. The ollowing example shows the dierent parts o a laboratory report.

Purpose of the experiment The goal o the experiment is explained at the beginning o the laboratory report. The goal is to answer one o the questions regarding a phenomenon that has been observed. For example, the question might be, “How does this phenomenon occur?” or “What conditions must be present in order or the phenomenon to occur?” The goal is expressed in a statement using scientifc words or expressions. This statement must begin with an action verb in the infnitive, such as “to measure” or “to determine.”

Hypothesis The hypothesis is a proposition that must be validated through experimentation. It must be based on knowledge or observations. The results o the experiment will either confrm or disprove the hypothesis. The hypothesis is a statement that always begins with “I assume that . . .” or “I believe that . . .” The wording used should, i possible, identiy two variables: the independent variable and the dependent variable.

Experimental procedure The experimental procedure is a plan that describes exactly how the experiment will be conducted. It includes clear, concise descriptions o each step. The procedure should be easily understood by anyone else who wishes to repeat the experiment.

394

APPENDIX 3 Presenting and Analyzing Scientifc Results

An experimental procedure usually includes these elements:

APPENDIX

• a list o equipment and materials with quantities specifed • a diagram o the set­up, such as a plan, drawing or photo • the steps to ollow when conducting the experiment (they must be numbered and described in short sentences)

3

Results The experiment results are an essential part o the laboratory report. The data obtained during the experiment may be presented in a table or diagram (see Sections 3.1 to 3.5, pages 389 to 393). Whatever orm you choose, it is important to indicate the units o measure used in the experiment. The observations recorded during the experiment must also be mentioned in this part o the report. It is important to provide an example o each type o calculation perormed with the data. For each calculation, show the data, equation, calculations and answer.

Analysis of the results In this part o the report, the meaning o the results obtained is discussed. You must also answer any questions your teacher asked at the beginning o the experiment. I the results are very dierent rom what was expected, you must indicate the source o any possible errors. For example, you could question whether the measuring instruments were accurate enough or whether handling or calculation errors occurred.

Conclusion In the last part o the report, you must state whether or not the hypothesis was confrmed and summarize what you observed by conducting the experiment. For example, you should use the results o the experiment to explain the relationship between the independent variable and the dependent variable. Improvements that could be made to the experiment are also discussed in this section. Finally, suggest possible avenues or new experiments that could lead to a better understanding o the phenomenon studied. You may also suggest experiments to explore similar phenomena.

APPENDIX 3 Presenting and Analyzing Scientifc Results

395

4 Interpreting Measurement Results

APPENDIX

4

4.1

Uncertainty

All measurements taken with instruments or devices involve uncertainty. One source of uncertainty is the measuring instrument itself. Another is the skill of the person using the instrument at taking and interpreting the read­ ings. No measurement can be taken with absolute certainty. Errors relating to measuring instruments depend on three factors: repeat­ ability, sensitivity and accuracy.

Absolute uncertainty and relative uncertainty The uncertainty of a reading associated with a measuring instrument is gene­ rally equal to half of the instrument’s smallest graduation. In the case of an electronic device, absolute uncertainty is the unit of the smallest graduation dis­ played. Absolute uncertainty is usually recorded in a table of results with a «  » symbol in front of the uncertainty. For example, the absolute uncertainty of a measurement taken with a 50­mL graduated cylinder whose smallest graduation is 1 mL, is 0.5 mL (see Figure 11). This is written as follows: V  (300  05) mL This means that the volume of the measured water is not exactly equal to 30.0 mL, but that it is between 29.5 mL and 30.5 mL. The last digit (the one on the far right) is always an estimated value. V  (300  05) mL ⇔ 295 mL  V  305 mL

Figure 11 The volume of the measured water is 300 mL  05 mL

To know how accurate a measurement is, uncertainty may be expressed in the form of a percentage. This is referred to as relative uncertainty. Relative uncertainty is the ratio between the absolute uncertainty and the measured value and is calculated as follows:

Relative uncertainty 

Absolute uncertainty 3 100% Value of measurement

The following example shows how to determine the relative uncertainty of the volume of a graduated cylinder: Example What is the relative uncertainty of the volume of solution in of the cylinder shown in Figure 11?

396

Data :

Calculation :

V  300 mL Absolute uncertainty   05 mL Relative uncertainty  ?

Absolute uncertainty 3 100% Value of measurement 05 mL  3 100% 300 mL  17% The volume of the water measured is 300 mL  2%

APPENDIX 4 Interpreting Measurement Results

Relative uncertainty 

Repeatability, sensitivity and accuracy

APPENDIX

The repeatability o a measuring instrument is its ability to repeat the same result or the same measurement taken under the same conditions. To deter­ mine the repeatability o an instrument or device, each measurement must be taken more than once.

4

The sensitivity o a device reers to its ability to detect small variations in measurement. I several measuring instruments or several calibres o the same instruments are available, it is important to choose the one that oers the greatest sensitivity. The accuracy o a measuring instrument reers to its ability to take measure­ ments with very ew errors. Calibrating a device beore using it improves its accuracy. For example, beore a scale is used, we must adjust the needle to zero using the thumbwheel.

4.2

Signifcant fgures

The digits recorded when taking a measurement are called signifcant fgures. Signiicant igures include digits that are certain and a inal, uncertain digit that is estimated when the measurement is taken. The degree o certainty o a measurement is the number o signifcant fgures it contains. Use the ollowing rules to determine the number o signiicant igures o a measurement: Rule 1 All digits other than zero are signifcant. 7886 has our signifcant fgures 194 has three signifcant fgures • 527266 992 has nine signifcant fgures • •

Rule 2 All zeros located between fgures other than zero are signifcant. • •

408 has three signifcant fgures 25 074 has fve signifcant fgures

Rule 3 Zeros located at the beginning o a number are not signifcant. •

00927 has three signifcant fgures: 9, 2 and 7

Rule 4 Zeros located at the end o a number are signifcant. • •

22 700 has fve signifcant fgures 0002 10 has three signifcant fgures: 2, 1 and the 0 on the ar right

Rule 5 When counting the number o signifcant fgures, do not take into account decimals, multiples, submultiples or powers o 10. •

Ek  45 786 J; Ek  45786 kJ; Ek  45786 3 104 J all have fve signifcant fgures (4, 5, 7, 8 and 6)

Note: In scientifc notation, all signifcant fgures are shown For example, the numbers shown in Rule 4 would be written as ollows: •

22 700  22700 3 104

•

0002 10  210 3 10-3

APPENDIX 4 Interpreting Measurement Results

397

APPENDIX

4

4.3

Measurements with unknown uncertainty

When it is not specifically indicated, uncertainty relates to the last significant figure. By convention, the uncertainty is equal to one unit of the least significant figure. •

c  4.176 J • °C-1 • g-1 means c  (4.176  0.001) J • °C-1 • g-1

•

P  0.22 W means P  (0.22  0.01) W

•

λ  0.500  10-6 means λ  (0.500  0.001)  10-6 m

•

T  1000 °C meansT  (1000  1) °C

•

F  100.2 N means F  (100.2  0.1) N

•

V  0.078 cm3 means V  (0.078  0.001) cm3

It is important to maintain the same number of significant figures when transforming units or when converting from units to their multiples or submultiples. •

l  22.4 km  22.4  103 m (three significant figures in both notations) and not l  22 400 m (five significant fi gures)

•

V  5.75 L  5.75  106 mL (three signifi cant figures in both notations) and not V  5750 mL (four significant figures)

4.4

Significant figures in results of mathematical operations

Significant figures are mainly used to determine the degree of certainty of a result obtained from the calculation of several measurements. For example, on a calculator, the result of the following calculation does not consist solely of significant figures. 27.53 N  74.5 m  2050.985 J The following rules help to express the results of calculations with the correct number of significant figures: Rule 1 Addition and subtraction In a calculation, the value with the smallest number of decimal places determines the number of decimal places in the answer. • 1.2 g  1.22 g  1.222 g  3.6 g, because 1.2 has only one decimal place. Rule 2 Multiplication and division In a calculation, the value with the smallest number of signifi cant fi gures determines the number of signifi cant fi gures in the answer. • 27.53 N  74.5 m  2.05  103 J, because 74.5 has only three signifi cant figures. • 1.2 m  1.55 s  0.77 m/s, because 1.2 has only two signifi cant figures.

398

APPENDIX 4 Interpreting Measurement Results

Rule 3 Complex calculations When calculations include additions (and subtractions) as well as multiplications (and divisions), these calculations must be perormed separately Data: F  2355 N ∆s  125 m ∆t  0021 s P1  523 3 104 W

Calculation: P2  P1 1

APPENDIX

4

(F 3 ∆s) ∆t

P2  523 3 104 W 1

(2355 N 3 125 m) 0021 s

 523 3 104 W 1 14 3 104 W, because 0021 has only two signifcant fgures  66 3 104 W, because 14 3 104 has only one decimal place

Note : When successive calculations are required, keep the maximum number o fgures in the calculator to do the intermediate calculations Only round o the fnal result, according to the rules explained below

4.5

How to round off a number

There are several methods for rounding off a number while taking into account the rules mentioned above. The method used in this textbook is called arithmetic rounding and is explained below. Rule 1 Select the last digit you need to retain as well as the following one. Consider the number 5074 68 To round o to the nearest hundredth, truncate the number at 5074 (last digit to retain  7; the ollowing digit  4) Rule 2 If the following digit is less than 5, do not change the last digit (rounding down). In the truncated number 5074, the digit 4, which ollows the last digit to be retained (7), is less than 5 Rounded o to the nearest hundredth, the result is 507 Rule 3 If the following digit is 5 or more, increase the last digit you need to retain by one unit (rounding up). Consider the number 5074 68 To round o to the nearest thousandth, truncate the number at 50746 (last number to retain  4; the ollowing digit  6) In the truncated number 50746, the digit 6, which ollows the last digit to be retained (4), is greater than 5 Rounded o to the nearest thousandth, the result is 5075 Note : This method amounts to separating the 0 decimal digits into two groups: The frst group is 0, 1, 2, 3, 4: rounded down The second group is 5, 6, 7, 8, 9: rounded up

APPENDIX 4 Interpreting Measurement Results

399

5 Mathematics in Physics

APPENDIX

5

Mathematics provides physics with the tools or better representing and better understanding the universe. It also helps to solve scientifc problems using algebraic expressions.

5.1

Problem solving

In science, physical phenomena are described by algebraic expressions that deine the relationship between the dierent variables involved in these phenomena. Because o its complexity, a problem must oten be solved in several steps, including literal calculation, the handling o units and numer­ ical calculation.

Literal calculation Literal calculation involves using letters only in mathematical expressions that describe physical phenomena. Literal calculation, which ollows the same rules as numerical calculation, allows you to isolate the variable whose value is required to calculate in a given problem. For example, the restoring orce (Fr) o a tension spring is related to the spring constant (k) and the elon­ gation o the spring (∆x). Fr  k 3 ∆x This algebraic expression may be used to calculate the restoring orce (F r). However, i you need to know the elongation o the spring (∆x), you must per­ orm a literal calculation, that is, isolate the variable ∆x. To do so, gradually isolate ∆x by perorming the necessary inverse mathematical operations on each side o the equation. 1 Isolate ∆x by dividing the terms of each side of the equation by k Fr Fr k 3 ∆x  ⇒  ∆x k k k 2 Express the equation in the usual manner by placing the isolated variable on the left - hand side of the equation and simplifying the right - hand side F ∆x  r k

Handling of units and numerical calculation Unlike purely mathematical calculations, scientiic calculations involve units that must be taken into account (and not just digits and numbers). Let us consider the previous example, using the ollowing data: Fr  25 N and k  15 N/m

400

APPENDIX 5 Mathematics in Physics

We obtain:

APPENDIX ∆x 

Fr 25 N ⇒ ∆x  N k 15



5

25 N 3 m  017 m 15 N

m

The handling o units allows you to eliminate certain units through simplifcation and retain only the unit you need in the answer.

5.2

Solving second-degree equations

Sometimes the variable you need to isolate will be in a second­degree equa­ tion in the orm y  ax2 1 bx 1 c. In this situation, ollow the steps or solv­ ing a second­degree equation and ind the value o x using the ollowing equation.

x

-b  b 2 2 4ac 2a

Example Determine the value of x in the following expression: 6x 2 1 4 1 8x  3x 1 4x 2 1 7 Solution : 1 Gather all of the terms on the same side of the equation by performing the appropriate mathematical operations among similar terms, and rearrange them so that they are in the form ax 2 1 bx 1 c  0 6x 2 2 4x 2 1 4

6x 2 1 4 1 8x  3x 1 4x 2 1 7 – 7 1 8x 2 3x  0 2x 2 1 5x 2 3  0

2 Determine the values of a, of b and of c a2 b5 c  -3 3 Calculate the possible values of x x

-b 1 b 2 2 4ac 2a - 5 1 52 2 4(2 -3) - 5 1 25 1 24 - 5 1 49   2(2) 4 4 •



 05 or -b 2 b 2 2 4ac x 2a - 5 2 49 - 5 2 52 2 4(2 -3) - 5 2 25 1 24   4 2(2) 4 •



 -3 4 Determine which of the values of x makes more sense based on the nature of the problem Keep only the value that is relevant Answer: The value of x is equal to 05 or -3

APPENDIX 5 Mathematics in Physics

401

APPENDIX

5

5.3

Scientifc notation

The order o magnitude o numbers used in science is oten very small or very large. These types o numbers are difcult to handle when perorming calculations. Scientifc notation helps to simpliy these numbers. The ollowing two examples show how to write numbers using scientifc notation. This notation is the product o a number called the mantissa whose absolute value is between 1 and 10 (excluding 10) and a power o 10.

Example A The number o particles contained in a mole o a substance (Avogadro’s number) is usually rounded o to 602 000 000 000 000 000 000 000. To simpliy this number, it is expressed in the orm o a number between 1 and 10 that is multiplied by a power o 10. To do this, move the decimal to the non-zero number on the ar let, and count how many places you have moved the decimal. This number becomes the exponent o the base 10 (see Figure 12 ).

6.02 000 000 000 000 000 000 000. 23 21 6.02  10

18

15

12

9

6

3

23

Figure 12 To express a large number using a power o 10, move the decimal to the let.

Example B The mass o a molecule o water is 0.000 000 000 000 000 000 000 029 9 g. To express this number using scientifc notation, that is, in the orm o a number between 1 and 10 that is multiplied by a power o 10, move the decimal toward the right to the frst non-zero number you meet, and count how many places you have moved the decimal. This number becomes the negative exponent o base 10 (see Figure 13).

Figure 13 To express a small number using a power o 10, move the decimal to the right.

402

APPENDIX 5 Mathematics in Physics

Calculation rules or scientifc notation

APPENDIX

Every measurement o a quantity must include the correct number o signif­ cant fgures. This number can be provided using scientifc notation. Use the ollowing rules when perorming calculations with numbers expressed in scientifc notation.

5

Rule 1 To add or subtract numbers expressed in scientifc notation, begin by converting them so that the numbers have the same exponents. The result is then multiplied by a power of 10 to which is attributed the sum of the exponents of the initial powers of 10 (a 3 10m) 3 (b 3 10n)  (a 3 b) 3 10(m 1 n) Example : (732 3 103) 3 (891 3 102)  (732 3 891) 3 10 (3 1 2)  652212 3 105  652 3 105 Rule 2 Each number should have the same exponent as the number with the largest exponent o 10. Once all of the numbers are expressed with the same exponent of 10, the exponent of 10 is not added or subtracted in the calculation m

( ab 33 1010 ) ( ab ) 3 10

(m 2 n)

n

Example : 1842 1842 3 106  3 10 (622) 10787 10787 3 102  1707 611 3 104  1708 3 104

(

) (

)

Rule 3 To add or subtract numbers in scientifc notation, frst convert the numbers so that they have the same exponents. Each number must have the same exponent as the number with the greatest exponent of 10 Once all the numbers are expressed with the same exponent of 10, the mantissas are added or subtracted and the exponent of 10 remains as is (342 3 106) 1 (853 3 103)  (342 3 106) 1 (0008 53 3 106)  (342 1 0008 53) 3 106  3428 53 3 106  343 3 106 (993 3 101) 2 (786 3 10-1)  (993 3 101) 2 (00786 3 101)  (993 2 00786) 3 101  98514 3 101  985 3 101

APPENDIX 5 Mathematics in Physics

403

5.4

APPENDIX

5

Review of trigonometry

Trigonometry is a branch o mathematics that deals with calculating the dimensions and angles o a triangle. Trigonometry, which uses trigonometric unctions, is essential to problem­solving in optics and mechanics.

Trigonometric functions The trigonometric unctions or the angle θ are as ollows (see Figure 14): y

y r x cos θ  r y tan θ  x sin θ 

r

y

θ x

x

Figure 14 Defnition o trigonometric

Pythagorean theorem

ratios

In the right­angled triangle (see Figure 15a), the Pythagorean theorem is as ollows: c 2  a 2 1 b 2, where c is the hypotenuse and a and b are the other sides

c a

For the unspecifed triangle (see Figure 15b), with angles A, B and C and the opposite sides a, b and c, we use the law o sines and the law o cosines.

b

Figure 15a Right-angled triangle

Law of sines B

The law o sines is as ollows: a

c

A

b

Figure 15b An unspecifed triangle

sin A sin B sin C   a b c C

To use the law o sines, you need to know two sides and an opposite angle or two angles and a side.

Law of cosines The law o cosines is as ollows: c 2  a 2 1 b 2 – 2ab cos C To use the law o cosines, you need to know three sides or two sides and the angle between them them. I C  90°, the equation corresponds to the Pythagorean theorem.

404

APPENDIX 5 Mathematics in Physics

Trigonometric identities

APPENDIX

5

The ollowing trigonometric identities are oten useul: cos θ  sin (90° 2 θ ) sin θ  cos (90° 2 θ ) sin θ cos θ sin2 θ 1 cos2 θ  1 tan θ 

sin 2θ  2sin θ cos θ cos 2θ  cos2 θ 2 sin2 θ

5.5

Review of geometry

The study o many physical phenomena, particularly in geometric optics, requires knowledge o the relationships between the angles that orm certain geometric fgures. The most important are set out below. Figure 16a The alternate interior angles formed by two parallel lines and a secant line are equal

Angles formed by two lines cut by a secant I two parallel lines are cut by a secant, then: • the alternate interior angles are equal (see Figure 16a) • the alternate exterior angles are equal (see Figure 16b) • the corresponding angles are equal (see Figure 16c)

Vertically opposite angles Vertically opposite angles are equal (see Figure 17).

Sum of the angles of a triangle In all triangles, the sum o the three interior angles is equal to 180° (see Figure 18). Figure 16b The alternate exterior angles formed by two parallel lines and a secant line are equal

θ1

θ2 θ3

Figure 17 The angles shaded in blue are equal because they are vertically opposite angles

Figure 18 In all triangles the following relationship exists between the interior angles: θ1 1 θ2 1 θ3  180°

Figure 16c The corresponding angles formed by two parallel lines and a secant line are equal

APPENDIX 5 Mathematics in Physics

405

APPENDIX 5.6

5

Review of some formulas used in this textbook

Appendix 5.6 gathers the most important formulas used in the Quantum collection.

Acceleration a  aave 

v 2 vi ∆v  f ∆t tf 2 ti

Average velocity and instantaneous velocity Average velocity:

v

x 2 xi ∆x  f tf 2 ti ∆t

Instantaneous velocity:

∆t being fairly long

x 2 xi ∆x  f tf 2 ti ∆t

v

∆t being very short

Calculation of the components of a vector →

v

vy θ vx

vx  v cos θ

vy  v sin θ

v  vx2 1 vy2

Centripetal force Fc 

mv 2 r

where

m  Mass, expressed in kilograms (kg) v  Rectilinear velocity, expressed in metres per second (m/s) r  Radius of curvature, expressed in metres (m)

Critical angle (θ c) θc  sin-1

( ) n2 n1

Determining the magnitude of a vector from its components using the Pythagorean relationship If then

406

→

→

u  (a, b) and v  (c, d) → → w  u 1 v  (a, b) 1 (c, d)  (a 1 c, b 1 d)

APPENDIX 5 Mathematics in Physics

Displacement (∆x) (when the problem is one - dimensional)

APPENDIX

5

∆x  x 2 xi

Elastic potential energy Epe of a spring Epe 

1 k (∆x)2 2

where

Epe  Elastic potential energy o the spring, expressed in joules (J) k  Spring constant o the spring, expressed in newtons per metre (N/m) ∆x  Elongation o the spring, expressed in metres (m)

Equation for velocity (v) v

x 2 xi ∆x   t 2 ti ∆t

where

t i  initial time, usually expressed in minutes (min) or seconds (s) t   fnal time, usually expressed in minutes (min) or seconds (s) xi  initial position, usually expressed in kilometres (km) or metres (m) x  fnal position, usually expressed in kilometres (km) or metres (m)

Equations of the motion of a projectile General equations for uniformly accelerated rectilinear motion (UARM) v  vi 1 a ∆t

Equations describing horizontal motion (along the x-axis) vx  vix

1 x   xi 1 (vi 1 v) ∆t 2 1 x   xi 1 vi∆t 1 a (∆t )2 2

x   xi 1 vix ∆t x   xi 1 vix ∆t

v2  vi2 1 2a∆x

vx  vix where

∆t  ∆x  ∆y  vx  vy  ay 

Equations describing vertical motion (along the y-axis) vy  viy 1 ay ∆t 1 y   yi 1 (viy 1 vy) ∆t 2 1 y   yi 1 viy ∆t 1 ay ∆t2 2 vy2  viy2 1 2ay ∆y

Time interval, expressed in seconds (s) Change in horizontal position, expressed in metres (m) Change in vertical position, expressed in metres (m) Horizontal velocity, expressed in metres per second (m/s) Vertical velocity, expressed in metres per second (m/s) Vertical acceleration (ay  -g), expressed in metres per second squared (m/s2)

APPENDIX 5 Mathematics in Physics

407

APPENDIX

5

Equations for uniformly accelerated rectilinear motion Equation 1 : v   v i 1 a ∆t 1 Equation 2 : x   x i 1 (v i 1 v ) ∆t 2 1 Equation 3 : x   x i 1 v i ∆t 1 a (∆t ) 2 2 Equation 4 : v 2  v i2 1 2 a ∆x

Equation to go from Cartesian coordinates to polar coordinates r  x2 1 y2

( yx )

θ  tan-1

Equation to go from polar coordinates to Cartesian coordinates x  r 3 cosθ y  r 3 sinθ

Equilibrant force Feq  -FR

Forces of friction Ff (static) ≤ µsFN Ff (kinetic)  µkFN

where

Ff  Force o riction, expressed in newtons (N) FN  Normal orce, expressed in newtons (N) µs  Coefcient o static riction µk  Coefcient o kinetic riction

Gravitational force exerted on an object Fg  mg

where

Fg  Gravitational orce, expressed in newtons (N)

m  Mass o the object experiencing gravitational acceleration, expressed in kilograms (kg) g  Gravitational acceleration, expressed in metres per second squared (m/s2)

Gravitational potential energy Epg  mgh

408

where

APPENDIX 5 Mathematics in Physics

Epg  Gravitational potential energy o an object, expressed in joules (J) m  Mass o an object, expressed in kilograms (kg) g  Gravitational acceleration, expressed in metres per second squared (m/s2) h  Height above the reerence level (usually the ground), expressed in metres (m)

Hooke’s law

APPENDIX

Fr  -k 3 ∆l

5

Hooke’s law (when the problem is one - dimensional and when the orce and displacement vectors are projected onto the x - axis) where

Fr  k ∆x

Fr  Restoring orce o the spring, expressed in newtons (N) k  Spring constant o the spring, expressed in newtons per metre (N/m) ∆x  Elongation o the spring, expressed in metres (m)

Index o reraction (n) n c v

Law o conservation o mechanical energy where

Emi  Em

or Eki 1 Epgi  Ek  1 Epg  where

Emi  Initial mechanical energy, expressed in joules (J) Em  Final mechanical energy, expressed in joules (J) Eki  Kinetic energy in the initial instant, expressed in joules (J) Epgi  Gravitational potential energy in the initial instant, expressed in joules (J) Ek  Kinetic energy in the fnal instant, expressed in joules (J) Epg  Gravitational potential energy in the fnal instant, expressed in joules (J)

Law o universal gravitation Fg 

Gm1m2 r2

where

N 3 m2 kg2 m1  A body’s mass, expressed in kilograms (kg) m2  The other body’s mass, expressed in kilograms (kg) r  Distance between the two bodies, expressed in metres (m)

G  Gravitational constant  667 3 10-11

Lens-maker’s equation 1 1 1  (n 2 1) 2 f R1 R2

(

)

where

f  Focal length, expressed in metres (m) R1 and R2  Radii o curvature o the lens, expressed in metres (m) n  Index o reraction o the lens material

Magnifcation (M ) in spherical mirrors (according to the sign convention used in this textbook) M

d hi - i ho do

APPENDIX 5 Mathematics in Physics

409

APPENDIX

5

Mathematical relationship between the radius of curvature (R ) and the focal length (f ) f

R 2

Mechanical power (P ) P

W ∆t

where

P  Mechanical power, expressed in watts (W) W  Work, expressed in joules (J) ∆t  Time interval, measured in seconds (s)

Mirror equation 1 1 1  1 f do di

Multiplication of a vector by a number If v  (a, b), then kv  (k a, k b)

Newton’s second law FR  ma

where

FR  Resultant force, expressed in newtons (N) m  Mass, expressed in kilograms (kg) a  Acceleration, expressed in metres per second squared (m/s2)

Newton’s third law FA→B  -FB→A

where

FA→B  Force exerted by A on B, expressed in newtons (N) FB→A  Force exerted by B on A, expressed in newtons (N)

Number of images formed (N) by two mirrors N

( 360°θ ) 2 1

where

N  Number of images formed θ  Angle between the mirrors, expressed in degrees (°)

Optical power of a lens (C ) P

410

1 f

where

APPENDIX 5 Mathematics in Physics

P  Optical power, expressed in dioptres (d) f  Focal length, expressed in metres (m)

Range of a projectile v 2 sin 2θ i Range  i g

APPENDIX

where Range  Range expressed in metres (m) vi  Initial velocity, expressed in metres per second (m/s) θ i  Departure angle relative to the horizontal, expressed in degrees g  Gravitational acceleration, expressed in metres per second squared (m/s2)

5

Relationship between kinetic energy, mass and velocity of an object in motion Ek 

1 mv 2 2

where

Ek  Kinetic energy, expressed in joules (J) m  Mass, expressed in kilograms (kg) v  Velocity, expressed in metres per second (m/s)

Relationship between mechanical energy (Em), kinetic energy (Ek) and gravitational potential energy (Epg) Em  Ek 1 Epg

where

Em  Mechanical energy, expressed in joules (J) Ek  Kinetic energy, expressed in joules (J) Epg  Gravitational potential energy, expressed in joules (J)

Relationship between the frequency (f ) and the period (T ) of a wave f (Hz) 

1 1 or T (s)  T (s) f (Hz)

Relationship between the period (T ) and the wavelength (λ) lv3T

where

l  Wavelength, expressed in metres (m) v  Wave velocity, expressed in metres per second (m/s) T  Wave period, expressed in seconds (s)

Relative index of refraction (n1 → 2) n1 → 2 

n2 n1

Resultant force FR  F1 1 F2 1 … 1 Fn

where

FR  Resultant force, expressed in newtons (N) F1, F2, Fn  Individual force, expressed in newtons (N)

APPENDIX 5 Mathematics in Physics

411

APPENDIX

5

Second law of refraction n1 3 sin θi  n2 3 sin θR

Thin-lens equation 1 1 1  1 f do di

Total optical power (P T) (for an optical system made of n lenses) P T   P1 1 P2 1 … 1 Pn

where

P T  Total optical power, expressed in dioptres (d) P1, P2, Pn  Individual optical power, expressed in dioptres (d)

Universal wave equation vl3f

where

v  Wave velocity, expressed in metres per second (m/s) l  Wavelength, expressed in metres (m) f  Wave frequency, expressed in hertz (Hz)

Velocity of a body measured relative to two frames of reference (RA and RB), RB being stationary relative to RA vA  (vB 1vAB)

where

vA  Velocity of the body relative to frame of reference RA vB  Velocity of the body relative to frame of reference RB vAB  Velocity of frame of reference RB relative to frame of reference RA

Work (W ) W  F • ∆s  F 3 ∆s 3 cos θ

where

W  Work, expressed in joules (J) F  Force acting on the body, expressed in newtons (N) ∆s  Displacement of the body, expressed in metres (m) θ  Angle between the force and the displacement

where

Wtot  Total work, expressed in joules (J) ∆Ek  Change in kinetic energy, expressed in joules (J)

Work-energy theorem Wtot  ∆Ek  Ekf 2 Eki

Ekf  Final kinetic energy, expressed in joules (J)

Eki  Initial kinetic energy, expressed in joules (J)

412

APPENDIX 5 Mathematics in Physics

6 Reference Tables 6.1

APPENDIX

6

Physical quantities and units Quantity/notation

Quantity symbol

Unit

Unit symbol

SI base unit

m/s²

m s-2

Acceleration

a

metre per second squared

Area

A

square metre

m2

m2

Centripetal force

Fc

newton

N

kg m2 s-2

Displacement

Δx

metre

m

m

Effective force

Feff

newton

N

kg m2 s-2

Elastic potential energy

Epe

joule

J

kg m2 s-2

Force of friction

Ff

newton

N

kg m2 s-2

Frequency

f

hertz

Hz

s-1

Gravitational acceleration

g

newton per kilogram or metre per second squared

N/kg or m/s2

m s-2

Gravitational potential energy

Epg

joule

J

kg m2 s-2

Heat

Q

joule

J

kg m2 s-2

Kinetic energy

Ek

joule

J

kg m2 s-2

Length

l, h, r, x

metre

m

m

Mass

m

kilogram

kg

kg

Mechanical energy

Em

joule

J

kg m2 s-2

Normal force

FN

newton

N

kg m2 s-2

Period

T

second

s

s

Power

P

watt

W

kg m2 s-3

Restoring force

Fr

newton

N

kg m2 s-2

Resultant force

FR

newton

N

kg m2 s-2

Spring constant

k

newton per metre

N/m

kg s-2

Temperature

T

• •

degree Celsius kelvin

• •

°C K

K

Tension

FT

newton

N

kg m s-2

Thermal energy

Eth

joule

J

kg m2 s-2

second

s

s

Time (or time interval)

t (or Δt)

Velocity

v

metre per second

m/s

m s-1

Volume

V

cubic metre

m3

m3

Wavelength

λ

metre

m

m

Weight (or gravitational force)

Fg

newton

N

kg m2 s-2

Work

W

joule

J

kg m2 s-2

APPENDIX 6 Reference Tables

413

APPENDIX

6.2

Physical constants

6

Physical constant Name

Symbol

Value in SI units

Atomic mass unit

U

1660 538 782 3 10-27 kg

Avogadro's number

N

6022 141 79 3 1023 mole-1

Elementary charge

E

1602 176 487 3 10-19 C

Electron mass

me

9109 382 15 3 10-31 kg

Gravitational constant

G

6674 28 3 10-11 m3 kg-1 s-2

Molar gas constant

R

8314 471 J mol-1 K-1

Neutron mass

mn

1674 927 211 3 10-27 kg

Planck’s constant

H

6626 074 506 896 3 10-34 J s

Proton mass

mp

1672 621 637 3 10-27 kg

Speed of light

C

2997 924 58 3108 ms-1

6.3

Solar system Mass (kg)

Radius (m)

Period of rotation on its axis (s)

Average orbital radius (m)

Period of revolution around its orbit (s)

Earth

598 3 1024

638 3 106

864 3 104

149 3 1011

316 3 107

Jupiter

190 3 1027

715 3 107

358 3 104

778 3 1011

375 3 108

Mars

637 3 1023

340 3 106

886 3 104

228 3 1011

594 3 107

Mercury

328 3 1023

244 3 106

505 3 106

579 3 1010

760 3 106

Moon

735 3 1022

174 3 106

236 3 106

384 3 108

236 3 106

Neptune

103 3 1026

248 3 107

580 3 106

450 3 1012

520 3 109

Pluto

13 3 1023

115 3 106

551 3 105

591 3 1012

782 3 109

Saturn

567 3 1026

603 3 107

384 3 104

143 3 1012

930 3 108

Sun

199 3 1030

696 3 108

214 3 106

-

-

Uranus

880 3 1025

256 3 107

620 3 104

287 3 1012

265 3 109

Venus

483 3 1024

605 3 106

21 3 107

108 3 1011

194 3 107

Object

414

APPENDIX 6 Reference Tables

7 Nobel Laureates in Physics 2009 2008 2007 2006 2005 2004 2003 2002 2001 2000 1999 1998 1997 1996 1995 1994 1993 1992 1991 1990 1989 1988 1987 1986 1985 1984 1983 1982 1981 1980 1979 1978 1977 1976 1975 1974 1973 1972 1971 1970 1969 1968 1967 1966 1965 1964 1963 1962 1961 1960 1959 1958

APPENDIX

Charles K Kao (1933–), Willard S Boyle (1924–), George E Smith (1930–) Yoichiro Nambu (1921–), Makoto Kobayashi (1944–), Toshihide Maskawa (1940–) Albert Fert (1938–), Peter Grünberg (1939–) John C Mather (1946–), George F Smoot (1945–) Roy J Glauber (1925–), John L Hall (1934–), Theodor W Hänsch (1941–) David J Gross (1941–), H David Politzer (1949–), Frank Wilczek (1951–) Alexei A Abrikosov (1928–), Vitaly L Ginzburg (1916–2009), Anthony J Leggett (1938–) Raymond Davis Jr (1914–2006), Masatoshi Koshiba (1926–), Riccardo Giacconi (1931–) Eric A Cornell (1961–), Wolfgang Ketterle (1957–), Carl E Wieman (1951–) Zhores I Alferov (1930–), Herbert Krœmer (1928–), Jack S Kilby (1923–2005) Gerardus’t Hooft (1946–), Martinus JG Veltman (1931–) Robert B Laughlin (1950–), Horst L Störmer (1949–), Daniel C Tsui (1939–) Steven Chu (1948–), Claude Cohen-Tannoudji (1933-), William D Phillips (1948–) David M Lee (1931–), Douglas D Osheroff (1945–), Robert C Richardson (1937–) Martin L Perl (1927–), Frederick Reines (1918–1998) Bertram N Brockhouse (1918–2003), Clifford G Shull (1915–2001) Russell A Hulse (1950–), Joseph H Taylor Jr (1941–) Georges Charpak (1924–) Pierre-Gilles de Gennes (1932–2007) Jerome I Friedman (1930–), Henry W Kendall (1926–1999), Richard E Taylor (1929–) Norman F Ramsey (1915–), Hans G Dehmelt (1922–), Wolfgang Paul (1913–1993) Leon M Lederman (1922–), Melvin Schwartz (1932–2006), Jack Steinberger (1921–) J Georg Bednorz (1950–), K Alexander Müller (1927–) Ernst Ruska (1906–1988), Gerd Binnig (1947–), Heinrich Rohrer (1933–) Klaus von Klitzing (1943–) Carlo Rubbia (1934–), Simon van der Meer (1925–) Subramanyan Chandrasekhar (1910–1995), William Alfred Fowler (1911–1995) Kenneth G Wilson (1936–) Nicolaas Bloembergen (1920–), Arthur Leonard Schawlow (1921–1999), Kai M Siegbahn (1918–2007) James Watson Cronin (1931–), Val Logsdon Fitch (1923–) Sheldon Lee Glashow (1932–), Abdus Salam (1926–1996), Steven Weinberg (1933–) Pyotr Leonidovich Kapitsa (1894–1984), Arno Allan Penzias (1933–), Robert Woodrow Wilson (1936–) Philip Warren Anderson (1923–), Sir Nevill Francis Mott (1905–1996), John Hasbrouck van Vleck (1899–1980) Burton Richter (1931–), Samuel Chao Chung Ting (1936–) Aage Niels Bohr (1922–2009), Ben Roy Mottelson (1926–), Leo James Rainwater (1917–1986) Sir Martin Ryle (1918–1984), Antony Hewish (1924–) Leo Esaki (1925–), Ivar Giaever (1929–), Brian David Josephson (1940–) John Bardeen (1908–1991), Leon Neil Cooper (1930–), John Robert Schrieffer (1931–) Dennis Gabor (1900–1979) Hannes Olof Gösta Alfvén (1908–1995), Louis Eugène Félix Néel (1904–2000) Murray Gell-Mann (1929–) Luis Walter Alvarez (1911–1988) Hans Albrecht Bethe (1906–2005) Alfred Kastler (1902–1984) Sin-Itiro Tomonaga (1906–1979), Julian Schwinger (1918–1994), Richard P Feynman (1918–1988) Charles Hard Townes (1915–), Nicolay Gennadiyevich Basov (1922–2001), Aleksandr Mikhailovich Prokhorov (1916–2002) Eugene Paul Wigner (1902–1995), Maria Gœppert-Mayer (1906–1972), J Hans D Jensen (1907–1973) Lev Davidovich Landau (1908–1968) Robert Hofstadter (1915–1990), Rudolf Ludwig Mössbauer (1929–), Donald Arthur Glaser (1926–) Emilio Gino Segrè (1905–1989), Owen Chamberlain (1920–2006) Pavel Alekseyevich Cherenkow (1904–1990), Il’ja Mikhailovich Frank (1908–1990), Igor Yevgenyevich Tamm (1895–1971) APPENDIX 7 Nobel Laureates in Physics

7

415

APPENDIX

7

416

7 Nobel Laureates in Physics (cont.) 1957 Chen Ning Yang (1922–), Tsung-Dao Lee (1926–) 1956 William Bradford Shockley (1910–1989), John Bardeen (1908–1991), Walter Houser Brattain (1902–1987) 1955 Willis Eugene Lamb (1913–2008), Polykarp Kusch (1911–1993) 1954 Max Born (1882–1970), Walther Bothe (1891–1957) 1953 Frits (Frederik) Zernike (1888–1966) 1952 Felix Bloch (1905–1983), Edward Mills Purcell (1912–1997) 1951 Sir John Douglas Cockcroft (1897–1967), Ernest Thomas Sinton Walton (1903–1995) 1950 Cecil Frank Powell (1903–1969) 1949 Hideki Yukawa (1907–1981) 1948 Patrick Maynard Stuart Blackett (1897–1974) 1947 Sir Edward Victor Appleton (1892–1965) 1946 Percy Williams Bridgman (1882–1961) 1945 Wolfgang Pauli (1900–1958) 1944 Isidor Isaac Rabi (1898–1988) 1943 Otto Stern (1888–1969) 1939 Ernest Orlando Lawrence (1901–1958) 1938 Enrico Fermi (1901–1954) 1937 Clinton Joseph Davisson (1881–1958), George Paget Thomson (1892–1975) 1936 Victor Franz Hess (1883–1964), Carl David Anderson (1905–1991) 1935 James Chadwick (1891–1974) 1933 Erwin Schrödinger (1887–1961), Paul Adrien Maurice Dirac (1902–1984) 1932 Werner Karl Heisenberg (1901–1976) 1930 Sir Chandrasekhara Venkata Raman (1888–1970) 1929 Prince Louis-Victor Pierre Raymond de Broglie (1892–1987) 1928 Owen Willans Richardson (1879–1959) 1927 Arthur Holly Compton (1892–1962), Charles Thomson Rees Wilson (1869–1959) 1926 Jean-Baptiste Perrin (1870–1942) 1925 James Franck (1882–1964), Gustav Ludwig Hertz (1887–1975) 1924 Karl Manne Georg Siegbahn (1886–1978) 1923 Robert Andrews Millikan (1868–1953) 1922 Niels Henrik David Bohr (1885–1962) 1921 Albert Einstein (1879–1955) 1920 Charles-Edouard Guillaume (1861–1938) 1919 Johannes Stark (1874–1957) 1918 Max Karl Ernst Ludwig Planck (1858–1947) 1917 Charles Glover Barkla (1877–1944) 1915 Sir William Henry Bragg (1862–1942), William Lawrence Bragg (1890–1971) 1914 Max von Laue (1879–1960) 1913 Heike Kamerlingh Onnes (1853–1926) 1912 Nils Gustaf Dalén (1869–1937) 1911 Wilhelm Wien (1864–1928) 1910 Johannes Diderik van der Waals (1837–1923) 1909 Guglielmo Marconi (1874–1937), Carl Ferdinand Braun (1850–1918) 1908 Gabriel Lippmann (1845–1921) 1907 Albert Abraham Michelson (1852–1931) 1906 Sir Joseph John Thomson (1856–1940) 1905 Philipp Eduard Anton von Lenard (1862–1947) 1904 Lord (John William Strutt) Rayleigh (1842–1919) 1903 Antoine-Henri Becquerel (1852–1908), Pierre Curie (1859–1906), Marie Curie (1867–1934) 1902 Hendrik Antoon Lorentz (1853–1928), Pieter Zeeman (1865–1943) 1901 Wilhelm Conrad Röntgen (1845–1923) No prizes were awarded in 1916, 1931, 1934 and from 1940 to 1942

APPENDIX 7 Nobel Laureates in Physics

GLOSSARY A Acceleration (a) Variation in the velocity’s magnitude (it can increase or decrease), its direction, or both. Amplitude (A) (o a wave) Maximum displacement o a point rom its equilibrium position. Angle o incidence (θi) Angle ormed by the incident ray and the normal. Angle o reection (θr) Angle ormed by the reected ray and the normal. Angle o reraction (θR) Angle ormed by the reracted ray and the normal.

C Cartesian coordinate system, Coordinate system whereby each point P o the plane can be situated with a pair o coordinates x and y, noted (x, y). Centripetal orce (Fc) Force that maintains a body in a circular motion. Coefcient o riction (µ) The experimentally established ratio between the orce o riction and normal orce depending on the substances in contact. Because it is a ratio, the coefcient o riction is a unitless number. Concave mirror Mirror whose surace converges the light rays that are parallel to its principal axis (which is the mirror’s axis o symmetry and is labelled P) toward a point located inside its curvature and reerred to as the ocal point (F). Also called a converging mirror. Converging spherical lenses Lenses characterized by at least one rounded surace. The most common are biconvex lenses, planoconvex lenses and positive meniscus lenses. Convex mirror Mirror whose surace diverges the light rays that are parallel to its principal axis (P). Also called a diverging mirror. Coordinate system A system which associates a group o three numbers to each point in space that make it possible to situate a body spatially.

D Displacement (Δx) Defned as the change in position. Distance (L ) The length o the trajectory, without taking direction into account. Diverging spherical lenses Lenses characterized by being thicker at the edges than at the centre. The most common are biconcave lenses, planoconcave lenses and negative meniscus lenses.

Dynamic equilibrium The state o an object moving at a constant velocity whose resultant orce is zero.

E Equilibrant orce (Feq) A single orce that cancels out the eect o the resultant orce exerted on a body. Equilibrium A body’s state when the resultant orce acting on it is zero.

F First law o reection Law stating that the incident ray, the reected ray and the normal are all located in the same plane. First law o reraction Law stating that the incident ray and the reracted ray are located on opposite sides o the normal originating rom the point o incidence, and that all three are in the same plane. Focal length (f ) Distance (OF) between the optical centre (O) and the principal ocal point (F). Focal plane (converging lens) Plane perpendicular to the principal axis (P) that passes through the principal ocal point (F). This plane contains the points where all incident rays converge, whether they are parallel to the principal axis or not. Focal plane (diverging lens) Plane perpendicular to the principal axis (P) that passes through the principal ocal point (F). Contrary to converging lenses, this plane is on the side o the incident rays. Force Vector quantity that acts on an object, causing it to experience a deormation or even change in motion. Force o riction (Ff ) Force that opposes motion. Frame o reerence A spatial reerence composed o an origin O and three axes (x, y and z) perpendicular to one another. Free-body diagram Diagram that shows, with arrows, all o the orces exerted on a body. Free all Motion that occurs when an object is subject only to gravitational orce. Frequency (f) (o a wave) Number o cycles completed by a wave in one second.

G Galileo’s principle o relativity A principle according to which the physical laws that are valid in a given inertial rame o reerence are also valid in all other rames o reerence that move at constant velocities relative to the frst.

GLOSSARY

417

Geocentric frame of reference Frame of reference situating the origin of its spatial reference at the centre of the Earth.

Law of conservation of total energy A law that states that the total quantity of energy is constant in an isolated system.

Gravitational acceleration (g) Constant acceleration experienced by an object in free fall close to the surface of the Earth or that of another celestial body.

Light wave Electromagnetic disturbance whose propagation allows for the transportation of light energy.

Gravitational force (Fg) Mutual attraction between two bodies on account of their mass.

Longitudinal wave Wave whose direction of propagation is parallel to the direction of the disturbance.

Gravitational potential energy (Epg) Energy associated with an object’s mass and position above a reference level, usually the Earth’s surface.

M

Gravity (See Gravitational force) Mass (m) Measure of the quantity of matter a body comprises, that is to say the total mass of the atoms of which is composed.

H Heliocentric frame of reference Frame of reference situating the origin of its spatial reference at the centre of the Sun.

Mechanical energy (Em) Energy equal to the sum of kinetic energy Ek and gravitational potential energy Epq of an object or system.

Hooke’s law Law that stipulates that an elastic object’s deformation is proportional to the applied deforming forces.

Mechanical power (P) Quantity of work done per unit of time.

N

I Image In optics, an image is a representation of an illuminated object produced by a series of points caused by the convergence of light rays that originate from various points on the object or from the extension of these rays. Incident ray (in the case of reflection) Light ray that travels toward the reflective surface. Incident ray (in the case of refraction) Light ray that travels toward the interface. Index of refraction Ratio between the speed of light in a vacuum (c) and the speed of light in a transparent medium (v). Inertia An object’s tendency to retain its state of motion. Inertial frame of reference A frame of reference in which an isolated body is either at rest or in uniform rectilinear motion. Isolated system System that does not exchange matter or energy with its environment.

K Kinetic energy (Ek) The energy of an object in motion.

Newton’s second law Law that states that the resultant force exerted on a body is equal to the product of the mass of this body and its acceleration. Newton’s third law Law that states that if body A exerts a certain force on body (FA→B ), body B exerts a force on body A (FB→A ) that is equal in magnitude, but opposite in direction. Normal (in the case of reflection) Imaginary line that is perpendicular to the reflective surface and originates from the point of incidence, that is, the intersection of the incident ray and the reflective surface. Normal (in the case of refraction) Imaginary line often represented by a dotted line. It is perpendicular to the interface, crosses the point of incidence, and travels through the two media. Normal force (FN) The resistant reaction of a body’s surface to a force exerted by another body.

O Optical centre (O) Corresponds to the geometric centre of the lens.

L Law of conservation of energy A law that states that mechanical energy at the final instant is equal to mechanical energy at the initial instant.

418

Newton’s first law Law that states that, in the absence of a resultant external force acting upon it, a body will remain at rest or in uniform rectilinear motion (its acceleration will be zero).

GLOSSARY

Optical power (P) Measure that quantifies the capacity of a lens or optical system to deviate light rays.

P Period (T) (o a wave) Time required or a wave to complete a cycle. Polar coordinate system Coordinate system in which point 0 is not called origin, but pole, and in which the abscissa axis is known as the polar axis. In this system, a point P o the plane is written: P (r, θ) in which r is the radius (distance connecting the pole to point P) and θ is the polar angle (angle between the polar axis and the radius). Position The location o a point or object. Principal ocal point (F) (converging lens) Point where rays parallel to the principal axis (P) converge. Principal ocal point (F) (diverging lens) Point that rays emerging rom the lens appear to originate rom.

Q Quantity The property o a phenomenon, body or substance that can be expressed quantitatively in the orm o a number and unit o measure.

R Range The horizontal distance travelled by a projectile between its launching point and its landing point. Real image Image produced when the optical system deviates (or directs) the light rays originating rom a source point by converging them toward an image point. Refected ray Light ray that travels away rom the reective surace. Refection Change in the direction o light ater it meets a surace that returns it to its original medium.

Resultant orce (FR) Vector sum o all orces exerted on a body.

S Scalar quantity A quantity that is completely defned by a real number and a unit o measurement. Secondary ocal point (F’)(converging lens) Point symmetrical to the principal ocal point (F) in relation to the optical centre (O) o the lens. Secondary ocal point (F’) (diverging lens) Point symmetrical to the principal ocal point in relation to the optical centre (O) o the lens. Second law o refection Law stating that the angle o incidence is equal to the angle o reection: θi 5 θr. Second law o reraction Law stating that the ratio o the sine o the angle o incidence (θi) and the sine o the angle o reraction (θR) is a constant. Spherical aberration Optical deect that occurs when the rays that are parallel to the principal axis (P) and that strike a concave spherical mirror at large angles o incidence (θi) are not ocused at one common ocal point (F). Spherical mirror Spherical cap cut rom a reective hollow sphere. Standard The realization o a given quantity’s defnition, with a determined value and an associated uncertainty o measurement, and is used as a reerence. Static equilibrium The state o a stationary object when the resultant orce exerted on the object is zero.

T Tension (FT) The traction that a cable (or any other long and thin object) exerts on a body.

Reracted ray Light ray that crosses the interace and enters medium 2.

Terrestrial rame o reerence Frame o reerence linked to the ground.

Reraction Change in direction in which light travels at the boundary o two media that have dierent optical properties. The surace that separates the two media is called the interace.

Total internal refection Phenomenon that occurs when a ray o light passing rom a highly reractive medium to a weakly reractive medium is not reracted but completely reected.

Reractivity The property o a medium to reract light.

Total work (Wtot) Sum o the work done by each o the orces acting on an object.

Relativity Notion according to which the description o a motion can change in relation to the rame o reerence used.

Trajectory The series o successive points through which an object passes as it moves.

Restoring orce (Fr) The orce exerted along the spring’s axis, which tends to restore the spring to its initial length.

Transverse wave Wave whose direction o propagation is perpendicular to the direction o the disturbance.

GLOSSARY

419

U

W

Uniform circular motion Movement o a body that ollows a circular trajectory and turns with a constant velocity.

Wave Travelling disturbance that carries energy rom one point to another.

Uniform rectilinear motion Motion in a straight line at a constant velocity.

Wavefront Imaginary line that connects all the points touched by the wave at the same moment in time.

Unit of measure A real quantity, defned and adopted by convention, that can be compared to any other quantity o the same kind to express the relationship between two quantities in the orm o a number.

Wave train Consecutive series o waves that travel in the same direction.

V Vector quantity A mathematical entity characterized by its magnitude and direction. For the analysis o situations in physics today, a vector quantity is defned by a real number, a unit o measurement and inormation related to its direction in space. Velocity (v) Describes the rate at which a position changes and is equal to the displacement per interval o time. Virtual image Image produced when its image points appear to originate rom a fctitious extension o light rays that are deviated by the optical system.

420

Wavelength (λ) Distance between two points o a wave separated by a complete cycle.

GLOSSARY

Weight Measure o the gravitational orce exerted on a body by a celestial body. Work (W) Equal to the scalar product o the orce acting on the body through the displacement o this body. Work-energy theorem Theorem that states that the total work done upon an object, between an initial instant and a fnal instant, is equal to the change in kinetic energy between these two instances.

INDEX A Aberrations optical, 140, 144 chromatic, 48, 123, 141-142 geometric, 123-4 spherical 42, 53, 124, 141-142 Absorption, 34, 143 Acceleration (a), 176, 184, 222, 224-226, 228, 230-234, 235, 266, 296, 304-306, 309-313, 343-345, 413 average, 224-225, 230, 406 centripetal, 282 constant, 221-240, 244-246, 268 gravitational (g), 13, 17, 174, 235-236, 244-246, 268-270, 282, 333, 348, 352, 375 horizontal, 246 sprinters’, 227 units of, 172, 174, 224 vertical, 246 Accelerometer(s), 239 Accomodation, 11, 135, 137 Accuracy of a measuring instrument, 397 Action-reaction pair, 315-316 Air bags, 239 Air resistance, 225, 235, 237, 240, 244, 245, 253, 254, 256257, 258, 259, 268, 350 Al-Haytham, Ibn (dit Alhazen), 42, 126 Amplitude (A), 5, 25-26, 37 Angle critical, 87-88, 406 of incidence (θi), 8-9, 44, 46, 51, 52, 58, 82-85, 87-88, 99, 101 of reflection (θr), 8, 44, 46, 51, 58 of refraction (θR), 9, 82-85, 87, 99 Area below the curve graph of acceleration as a function of time, 228 graph of the restoring force of a spring as a function of the displacement of its free extremity, 369 graph of velocity as a function of time, 214-215, 227 Aristotle, 126, 240, 319 Armati, Salvino degli, 97, 126 Astigmatism, 137 Astrolabe, 162

B Bacon, Roger, 97, 126 Ballistics, 243-259 Beam of light converging, 33 diverging, 33 parallel, 33 Blu-ray optical disc, 71 Bolt, Usain, 227 Buridan, Jean, 259

C Calculation(s) literal, 400 numerical, 400-401 Calculus differential and integral, 306, 319 Camera, 132-133, 134, 244 obscura, 55-56, 57, 62, 132 pinhole of, 55-56, 57, 58, 132 Cannon, 245, 256, 259 Cataracts, 138 Cartesian plane, 186, 188, 191, 194, 196 Centrifugation, 282 Chronometer, 217 Clepsydra, 217 Chrystalline 11, 134-135, 136-137, 138 Clock atomic, 161, 170, 178 mechanical, 217 pendulum, 217 quartz, 161, 217 spring-wound, 217 water, 217, 240 Coefficient of kinetic friction, 275-276 of static friction, 275-276 Collision(s) car, 310, 339, 340 meteorite, 356 Compass, 162 Compression(s), 17, 24, 300, 376, 371, 374, 375 spring, 364, 365, 367 Conservation law of conservation of total energy, 18, 340, 354-356, 358, 410 of mechanical energy, 349, 350-352, 407 of total energy, 354-356 Constant(s) Coulomb’s, 7 of restoring forces of a combination of springs, 368 physical (table), 414 spring, 365-367, 370, 371, 375, 413 Contact lenses, 106, 124, 136, 137 Coordinate system Cartesian, 158-160, 204, 236, 245, 297 polar, 158-160, 252 three-dimensional, 160 Copernicus, Nicolaus, 157

D Daguerreotype, 58 Deformation, 266, 271, 273-275, 293, 296, 310, 365, 374, 375 of space-time, 319 INDEX

421

plastic, 366 Descartes, René, 84, 90, 126, 158 Diagram ree-body, 288 Dioptre (δ), 106, 108-110, 125 Dispersion, 72, 81, 123 Displacement (Δx), 14-15, 17, 152, 156, 168, 176, 205-206, 211, 214-215, 222, 244, 247, 334, 365, 370-371, 373 as expressed by vectors, 184-185, 247 calculations or a graph o velocity as a unction o time, 227-228 lateral, o the emerging ray, 104 relationship between work, orce and, 16, 326-328, 374 Distance (L), 12, 167-168, 206-207, 208-209, 214, 216, 223, 230, 232, 240 Doppler eect, 216 Dynamometer, 266, 270, 309, 375

E Einstein, Albert, 29, 32, 35, 188, 212-213, 281, 319, 416 Elasticity, 365-368, 373 Elementary interactions, 277, 281 Energy, 16-19 absorbed, 34, 372, 373, 374 chemical, 330, 355, 413 conservation o, 18, 349, 350-352, 354-356, 407 dissipation o, 372, 374 electrical, 19, 355, 357 electrical power and, 19 green, 335 kinetic (Ec), 17, 18, 340-345, 347, 349, 350-352, 354355, 356, 357, 374, 411, 413 mechanical (Em), 18, 336, 339-358, 413 nuclear, 355 potential (Ep), 17, 18, 347, 349, 350-351, 356, 413 potential, elastic, 17, 347, 354-355, 363-375, 413 potential, gravitational (Epg), 17, 18, 347-349, 350-352, 354-355, 357, 368, 413 radiant, 355 relationship between work and, 16 rotational kinetic energy, 345, 357 solar, 19, 50, 72 stored in a spring, 17, 217, 370-372 table o main orms o, 355 thermal, 336, 354, 355, 356, 374, 413 transer o, 16, 340, 342, 355, 372 transormation o, 17-18, 325-377 Engines, steam, 331, 333, 336 Equation(s) mirror, 61-62, 64-68, 407 solving second-degree, 401 thin-lens, 115-121, 135, 407 universal wave, 27, 30-31

422

index

Equilibrium dynamic, 293-294, 298, 304 neutral, 294 o several orces, 14, 293-298, 299 stable, 294 static, 293, 298, 304 unstable, 294 Eye human, 11, 124, 134-138 indexes o reraction o, 134 Eyeglasses, 97, 106, 125, 126, 138

F Ferromagnetic substances, 7 Field(s) electric, 6-7, 29-30 magnetic, 7, 29-30, 197 Figures, signifcant, 397-398 in results o mathematical operations, 398-399 Focal length (f) o lenses, 100-101, 103, 106-111, 113-115, 117-121, 123, 135-136, 139-140, 142 o spherical mirrors, 50-51, 53, 61, 68 Focal plane, 108 o converging lenses, 100-101, 137 o diverging lenses, 103 Focal point (F) o converging lenses, 10, 100-101, 113-114, 125 o diverging lenses, 11, 102-104, 114-115, 125 o spherical mirrors, 49, 50-53, 60, 62 Focal point, secondary (F′) o converging lenses, 100-101, 140, 142 o diverging lenses, 102-104 Force(s), 12-15, 266 centriugal, 282, 308 centripetal (Fc ), 266, 279-281, 282, 308, 409, 413 characteristics o, 13 couple o, 364 defnition o, 308 dierent types o, 266-283, 288 drive, 305, 336 eective (Feff ), 15, 197, 327-329, 413 electric, 6-7, 197 electromagnetic, 275, 277, 281 equilibrant (Feq), 295-296, 409 equilibrium between two, 14, 293-298 external, 240, 304-305, 317 fctional, 155, 282, 308, 316 gravitational (Fg), 13, 235-236, 266, 267-271, 280-281, 296, 297, 304, 306, 342, 347-349, 408, 409, 413 inertial, 308, 316 internal, 304, 317 magnetic, 7, 319, 357 moment o a, 294

normal (FN ), 273-274, 275-276, 280-281, 297, 300, 304, 308, 311-313, 342, 413 notion o, 266, 306, 309, 315 o riction (Ff ), 14, 266, 271, 275-277, 280, 283, 297298, 305-308, 309, 313, 317, 342, 350-351, 354, 358, 409, 413 relationship between power, velocity and, 334 relationship between work, displacement and, 16, 326327 restoring (Fr), 365-367, 370-372, 374, 375, 412, 413 resultant (FR), 14, 281, 290-292, 293-294, 295, 297, 304306, 310, 342-343, 409, 413 sum o, 304 tension (FT), 278, 280, 288 tidal, 271 total, 304 Formula(s) lens-maker’s, 109-111, 113, 123, 409 used in this textbook, 406-412 Frame(s) o reerence, 152-162, 204, 209, 236, 308, 412 Copernican, 157 geocentric, 156, 161 heliocentric, 156-157, 316 inertial, 155-157 terrestrial, 155 Freeall, 235-237, 240, 244, 268-269, 306, 316, 349, 358 Frequency o a wave (f), 5, 6, 25-26, 27, 30-31, 32, 35, 37, 216, 411, 413 uniorm circular motion, 279 Friction, 14, 266, 271, 275-277, 280, 297-298, 305-308, 309, 313, 317, 342, 350-351, 354, 358, 409, 413 coefcient o, 275-276 fghting, 283 kinetic, 275-276 static, 275-276

G Galileo, Gailei, 29, 144, 156, 217, 230, 232, 240, 245, 253, 256, 259, 268, 291, 306, 358 General Conerence on Weights and Measures (CGPM), 167, 170, 171 Gol, 258,259 GPS, 37, 156, 161, 162, 239 Graph broken-line, 390-91 Cartesian, 390 curve o best ft, 390-391 determination o the slope o a straight line, 392 determination o the slope o a tangent to the curve, 392 o acceleration as a unction o time, 224, 228 o orce as a unction o displacement, 370-371

o position as a unction o time, 208, 211-212, 214, 223-224, 232 o velocity as a unction o time, 214-215, 224, 226, 227, 228, 231 variations o some unctions, 391 Gravitational attraction, 13, 267, 270, 277, 296, 306, 316, 319, 375 Gravity, 161, 217, 230, 235, 237, 240, 244, 245, 267-271, 288, 304, 347 Gravity assist, 271 Gun laser, 216 radar, 216

H Harrison, John, 217 Heat, 271, 336, 340, 358 Hertz, Heinrich Rudolph, 31, 37 Hooke, law o, 365-368, 410 Robert, 319, 363, 367, 368 Horsepower, 333 Hourglass, 217 Hubble Space Telescope (HST), 143 Huygens, Christiaan, 90, 217, 358 Hydroelectricity, 357 hydroelectric power plants, 357 Hyperopia, 136-137, 138

I Image(s) ormed by lenses, 113-122 ormed by a plane mirror, 34, 42-43, 46, 57-59 ormed by a camera obscura, 56-57, 132 ormed by a concave mirror, 59-60 ormed by a converging lens, 113-114, 116, 117-121, 134-137, 139-140, 142, 144 ormed by a diverging lens, 114-115, 121 ormed by a spherical mirror, 48, 59-69, 141 orientation, 55, 57, 58, 60, 61, 63, 114, 117 position, 55-56, 57, 58, 60, 114-115, 117, 134-135 real 55, 57, 60, 63, 64-66, 68, 114, 118-121, 132, 135, 140, 142 size, 55, 57, 58, 60, 61, 63, 69, 114, 116, 117 virtual, 8-9, 55, 57-58, 60-61, 63-69, 114-115, 117, 120121, 140, 142 Impetus, 244-245, 259 Inclined plane, 230, 240, 297-298, 408 Index o reraction, 78, 79-81, 84-85, 87, 99, 106, 109, 113, 123, 133 absolute, 79-80, 409 relative(n1→2), 80, 84, 87, 410 index

423

Inertia, 279, 282, 307-308, 309, 316 Interace, 78, 82-83, 85, 87-88, 99-100, 124, 134 Intererometer, 29 Interpreting measurement results, 396-399 signifcant fgures, 397-399 uncertainty, 396-397, 398

J James Webb Space Telescope (JWST), 143 Janssen, Hans, 144 Jet propulsion, 318 Joule, James Prescott, 340, 341, 358

K Kepler, Johannes, 126, 270, 319 Kilometre posts on highways, 209 Kilogram, 167, 169, 170, 171-174, 179

L Laplace, Pierre-Simon, 186, 358 Large Hadron Collider, 281 Laser, 29, 35, 71, 89, 136-138, 216 Law(s) action-reaction, 315-317 Cauchy’s, 81 Hooke’s, 365-368, 410 Newton’s (frst), 303-319 Newton’s (second), 309-313, 342-343, 345 Newton’s (third), 273, 315-317 o conservation o energy, 18, 349, 350-352, 354-356, 407 o conservation o mechanical energy, 340, 350-352 o conservation o total energy, 354-356 o reection (frst), 8, 46, 73 o reection (second), 8, 46, 73 o reraction (frst), 84 o reraction (second), 84-85, 87, 92 o universal gravitation, 267-268, 319, 425 Snell-Descartes, 84-86, 90 Length o a spring, 364-366, 367 o a wave (λ), 5, 6, 19, 25-27, 30-33, 35, 42, 43, 71, 80-81, 123, 216, 411, 413 Lenses, 95-130 Lenses, converging spherical, 10, 96-97, 104, 107-109, 111, 115, 135, 137, 139, 141-142 biconvex, 96 images ormed by, 113-114, 117-121 planoconvex, 96 positive meniscus, 96 principal rays o, 101, 113-114 reraction in, 99-101, 123

424

index

Lenses, diverging spherical, 10-11, 97, 107-109, 111, 136 biconcave, 97 images ormed by, 114-115, 117, 121 negative meniscus, 97 planoconcave, 97 principal rays o, 103-104, 114-115 reraction in, 102-104, 123 Lippershey, Hans, 109, 144 Lit, 254, 258, 288 Light beams, 33, 35, 89, 109, 212 extended light source, 33 inrared, 71, 216 laser, 29, 35, 71, 89, 136-138, 216 point source, 33 propagation o, 32-34, 42, 72, 78 properties o, 8-11 reection o, 8-9, 34, 41-76, 78, 82, 87-88, 141 162, 216 reraction o, 9-10, 34, 42, 48, 72, 78-90, 99-104, 113, 134, 142, 161 speed o, 29-30, 37, 79, 83, 170, 211, 212-213, 216, 414 visible, 6, 19, 30-31, 32, 43, 45, 48, 81, 143 waves, 29-40 Linear regression on a scatter diagram, 393

M Machines, simple, 358 Magnets, 7 Magnifcation, in the case o a spherical mirror, 56-57, 62-63, 64-69, 409 in the case o a thin-lens, 116, 117-121 Marconi, Guglielmo, 37, 416 Mars Climate Orbiter, 173 Mass breakdown o, 345 centre o, 205, 288 gravitational potential energy, 348, 359 orce o two bodies, 267-270, 277, 331 kinetic energy, and, 340 inertia as quantifed by, 279, 280, 307, 309-310 measure o, 167, 169, 375, 413 relationship between weight and, 13, 270-271, 375 Mathematics in physics ormulas used in this textbook, 406-412 problem-solving, 400-401 review o geometry, 405 review o trigonometry, 404-405 scientifc notation, 402-403 solving second-degree equations, 401 Maxwell, James Clerk, 29, 31, 281 Media opaque, 34

other than air, 111 reractive, 9, 79-80 translucent, 34 transparent, 34, 78-80, 81, 134 Measurement(s) length, characteristic, o a spherical mirror, 50 physical quantities and units, 166-177, 413 power, optical, o lenses, 106-108, 109, 111 quantities, scalar, 176, 184, 326-327, 331, 341 quantities, vector, 153, 176-177, 184-199, 204, 205, 226, 266, 290, 326-327 size o images ormed by a converging lens, 114 size o images ormed by a mirror, 55-56, 57-61, 63 Metallic paint, 42 Method(s) cardinal points (direction o a vector), 185 Cartesian plane coordinate (operations that can be done with vectors), 191-192, 194-195, 291-292 component (operations that can be done with vectors), 190, 193-194, 290-291 or drawing a ree-body diagram, 288 graphical (operations that can be done with vectors), 190, 193, 290 intervals (to determine instantaneous velocity), 224 o homogeneity o unctions (work-energy theorem), 342-343 trigonometric angle (direction o a vector), 185 Methods o problem-solving used in physics analysis, 388 by literal calculation, 400 by numerical calculation, 400 by the handling o units, 400-401 experimental, 387 our steps in, 386 general, to solve dynamic problems, 311-313 general, to solve problems o conservation o mechanical energy, 351-352 general, to solve kinematic problems, 231-234 summary table o, 385 Metre defnition o, 166-168, 169, 170, 172 Michelson, Albert Abraham, 29, 416 Microwaves, 6, 30, 161, 216 Microscope light 43, 139-140, 144 Mirage, 78, 85 Mirror(s), 46-69, 72 cylindrical, 48 elliptical, 48 hyperbolic, 48 manuacturing, 72 parabolic 48, 53, 141 plane, 8-9, 46, 57-59 Mirror(s), concave, 49-52, 53, 59-60, 62-67, 69, 73-74

principal rays o a, 52 Mirror(s), convex, 49, 52-53, 61, 67-68, 73-74 principal rays o a, 53 Mirror, spherical, 48-53, 59-74, 115 centre o curvature 50 characteristic lengths o a, 50 principal points o a, 50 sign convention, 63, 68-69 vertex, 50 Model geocentric, 230 heliocentric, 230 Motion apparent, 85, 217 at very high velocity, 161, 212-213 change in, 13-15, 266, 275-276, 293, 296, 297, 304, 326, 307, 324 curved, 244, 279-281, 306, 319 description o, 152-154 diurnal rotation, 155, 156 horizontal, 245 natural, 240, 306 o projectiles, 243-259, 408 uniorm circular, 279-280 uniorm rectilinear, 155-157, 203-217, 222, 223, 227, 245, 279, 304 uniorm rectilinear, accelerated, 155, 214, 221-241, 245, 246, 408 vertical, 235-236, 245 violent, 240 Myopia, 126, 136, 138

N Newton Isaac, 29, 48, 72, 186, 267, 270, 306, 319 law (frst), 304-308 law (second), 309-313, 317, 342-343, 344-345, 406 law (third), 273, 315-317, 412 Newtonian telescope 141 Nippur Cubit, the, 166 Nobel laureates in physics, 415-416 Normal orce, 273-274, 275-276, 280-281, 297, 300, 304, 308, 311-313, 342, 413 in the case o reection, 8, 44, 46, 49, 51-52, 73 in the case o reraction, 10, 82-84, 87, 99, 102

O Optical fbres, 88, 89 Optical centre (O), 100-101, 103, 104, 107, 109, 117 Oresme, Nicole, 214, 240

index

425

P Pendulum, 217, 230, 358 Period (T) (o a wave), 5, 25-27, 29 Photon, 32, 35, 36 Position equilibrium, 25-26 graphical representation o, 208-209 horizontal 246 o a body in space, 152-153, 158, 161, 162, 204-205, 236, 244, 246 o images ormed by lenses, 114 o images ormed by plane mirrors, 56-61 vertical, 246 Power, 106-111, 123, 125 electrical (P), 19, 335 mechanical, 331-334, 336, 411 optical (P), 135-137, 138 relationship between power, orce and velocity, 334 wind turbines, o, 335 Presbyopia, 126, 137, 138 Presenting and analyzing scientifc results broken-line graph, 390 Cartesian graph, 390 curve o best ft graph, 390-92 graphic determination o the slope o a straight line, 392 graphic determination o the slope o a tangent to the curve, 392 laboratory report, 394-395 linear regression on a scatter diagram, 393 table, 389 Principle undamental, o dynamics, 310 Galileo’s, o relativity, 156 Projectile(s) equations o the motion o a, 407 launched horizontally, 249-250 launched at an angle, 252-254 motion o, 243-259 range o, 245, 255-257, 258, 259, 410

R Radiation rom the sun, 19 inrared, 5, 19, 30, 143, 216, 355 ultraviolet, 5, 19, 30, 143 Rainbow(s), 30, 80, 81, 84 Range, 255-257, 410 maximum, 245, 256-257, 258, 259 Ray(s) gamma, 6, 30, 143 incident (in the case o reection), 8, 44, 46, 52-53

426

index

incident (in the case o reraction), 9, 80, 82, 84, 99, 101-104, 107 inrared 19, 355 light, 33 reected, 8, 44, 46, 51-52 reracted, 8-9, 80, 82-83, 84, 87, 99, 113, 115 ultraviolet 19 wave, 32 X-, 6, 30, 36, 143, 225 Reection, 8-9, 34, 41, 42-76, 78, 82, 113, 162, 216 diuse, 42 law (frst), 8, 46, 73 law (second), 8, 46, 73 specular, 42 total internal, 87-88, 92 Reraction, 9-10, 34, 42, 48, 72, 78-90, 96, 99-104, 113, 134, 142, 161 law (frst), 84 law (second), 84-85, 87, 92 Reractivity, 11, 79-81, 82-83, 87-88, 109, 123 Relativity, 154, 161, 212-213 general, 188, 319 special, 212-213 Repeatability o a measuring instrument, 397 Review o geometry, 405 o some ormulas used in this textbook, 406-412 o trigonometry, 404-405 Robins, Benjamin, 259 Roche limit, 271 Röntgen, Wilhelm, 36, 416 Rounding o a number, how to, 399 Rules or perorming calculations with numbers expressed in scientifc notation, 403 saety, 383-384

S Scalar multiplying a vector by a, 196 product, 197, 326-327, 334 quantity, 176, 184, 326-327, 331, 341 Scientifc notation, 402 calculation rules or, 403 Scott, Dave, 240 Second (s), 161, 167, 169, 170, 172 Sensitivity o a measuring instrument, 397 Sextant, 162 Shen, Kuo, 84 Ski jumping, 254 Skydiving, 237 Slope determination o the slope o a straight line, 392 determination o the slope o a tangent to the curve, 392

Snell, Willerbrord, 84, 90, 126 Solar urnace(s), 50, 72 Solving second-degree equations, 401 Spaghettifcation, 296 Spectacles, 97, 126 Spectrum electromagnetic, 6-7, 19, 30-31, 80-81 visible light, 6, 19, 31, 48, 80-81 Speed guns, 216 Speed o light, 29-30, 37, 79, 83, 170, 211, 212-213, 216, 414 Spina, Alessandro di, 97, 126 Spiral(s), 24, 364 Solar system, 146, 157, 186, 414 Spring(s), 266 compression, 364, 367 deormation, 372, 373 elongation o a, 363, 365-366, 371,373, 375 helical, 364-369, 374 length o a, 365 parallel combination o two, 368 scale, 375 series combination o two, 368 tension, 364-367, 370-371, 412 torsion, 364 Standard material, 166, 169-171, 375 natural, 169-170 Stevin, Simon, 291, 358 Stroboscope, 244 Sundial, 217 Symbol(s), danger, 380-382 saety, 383 Systems o coordinates, 245 Cartesian, 158-160, 204, 236, 245, 297 heliocentric, 156-157, 316 polar, 158-160, 252 three-dimensional, 160 System o Units (SI), 13, 31, 167-174, 309, 331, 340 System, isolated, 18, 354-355

T Tangent, 223 Tartaglia, Niccolo Fontana known as, 245, 259 Telephone, cellular 31, 37, 239 Telescope(s) reecting, 48, 72, 141 reracting, 109, 142, 230 X-ray, 36 Time measuring, 217 units o, 169, 170, 413 Tension, 17, 278, 280, 300, 306

spring, 364-367, 370-371, 412 Theorem o work-energy, 342-345, 347, 411 Pythagorean, 159, 177, 404 Theory caloric, 358 Grand Unifed, 281 heliocentric, 291 quantum, 29, 35 Trajectory, 152, 162, 206 circular or curved, 244-245, 247, 279-281, 306 o light rays, 46, 78, 96, 113, 133 parabolic, 253, 257, 259 Travelling object, 152-153

U Umbra ormation, 33 Uncertainty absolute, 396 unknown, 398 relative, 396-397 Unit(s) handling o, 400-401 measure o, 166-167, 413 physical quantities and their, 413 SI base, 167-174 submultiples o SI, 173

V Van Leeuwenhoek, Antonie, 139, 144 Vector(s), 176-177, 183-199, 204, 205, 211, 244, 246, 273, 310, 406 addition,186, 190-192, 290, 294 multiplication, 196-197, 327, 410 orthogonal, 327 subtraction, 186, 193-195 velocity (v ), 247-248, 249, 252, 280 Velocity (v), 184, 211-213, 226, 230-231, 235-236, 244, 246-247, 252, 255-257, 340-341, 407, 412 average 12, 222, 247, 412 calculating changes in, 228 components, horizontal, 244, 248, 257, 313 components, vertical, 244, 248, 252, 257 constant 12, 14, 152-153, 155-156, 208, 279, 293-294, 298, 305-306, 334 determining velocity (based on the curve in a graph o position as a unction o time), 223-224 graphical representation o velocity as a unction o time, 213-214 instantaneous, 222, 223-224, 226, 247, 412 relationship between power, orce and, 334 speed guns to measure, 216 index

427

terminal, 79, 237 units of, 172, 413 Von Mayer, Julius Robert, 358

W Water column, 235 Watt, James, 331, 333, 336 Wave(s) characteristics of, 5, 24-27 electromagnetic, 4, 6, 19, 24, 29, 30-33, 37, 143 light, 29-40, 32 longitudinal, 4, 24 mechanical, 4, 24, 34, 37 period, 5, 25-27, 29 radio, 30, 31, 37 ray, 32 sound, 24 train, 32 transverse 4, 24 types of, 4-7 universal wave equation, 27, 30, 407 wavefront, 32-33 wavelength, 5, 6, 19, 25-27, 30-33, 35, 42, 43, 71, 80-81, 123, 136, 143, 216, 411, 413 Weight, 13, 266, 273, 299, 300, 373 relationship between mass and, 13, 270-271, 413 Work (W), 326-330, 331, 335, 340, 342-343, 345, 357, 358,

428

index

412, 413 biological, 330 effective, 328 maximum, 328 mechanical, 330 muscular, 330 of gravitational force, 347-349 particular values of, 328-330 relationship between energy and, 16 relationship between force, displacement and, 16, 197, 326-327, 370-372 resistive, 328 zero, 328

X X-ray tubes, 225

Z Zernike, Frederik, 144, 416

SOURCES Legend t: top b: bottom c: centre l: left r: right Cover Martin Kawalski/Istockphoto (snowboarding jump), Olaru Radian-Alexandru/Istockphoto (chronograph), Mustafa Deliormanli/Istockphoto (Newton’s pendulum), Michele Galli/ Istockphoto (goggles), Miguel Salmeron/Getty Images (pogo stick) Review p. 2: Steve Rabin/Istockphoto (lightening), Peeter Viisimaa/ Istockphoto (sunset), VolkerMoehrke/Zefa/Corbis (Newton’s pendulum), Ekaterina Starshaya/Shutterstock (water drop), Charlotte Erpenbeck/Shutterstock (magnetic field) Istockphoto (merry-go-round) • p. 3: NASA Unit 1 – Geometric Optics p. 20 and 21: Ioana Drutu/Shutterstock • p. 23: Mustafa Deliormanli/Istockphoto • p. 29t: Sheila Terry/Science Photo Library • p. 29b: Bettmann/Corbis • p. 31t: Science Source/ Photo Researchers, Inc. • p. 31b: Tim Arbaev/Shutterstock • p. 34l: Mark Jensen/Istockphoto • p. 34c: Stacey Newman/ Istockphoto • p. 34r: Istockphoto • p. 35: Firefly Productions/ Corbis • p. 36l: Photo Researchers • p. 36r: Javier Larrea/maXx image • p. 37l: Hulton-Deutsch Collection/Corbis • p. 37r: Bettmann/Corbis • p. 41: Antonio Jorge Nunes/Shutterstock • p. 42: Sheila Terry/Science Photo Library • p. 43: Ahmed Benssaada • p. 45: Istockphoto • p. 48: Sheila Terry/Science Photo Library • p. 50: José Fuste Raga/maXx images • p. 57: Anastasia Pelikh/Istockphoto • p. 58: Michael Freeman/Corbis • p. 70: Maurice van der Velden/Istockphoto • p. 72l: Adam Hart-Davis/Science Photo Library • p. 72r: NASA/Photo of Paul Hickson (University of British Columbia) • p. 77: GIPhotostock/ Photo Researchers, Inc. • p. 78t: Angelika Schwarz/Istockphoto • p. 78b: Ed Darack/Science Faction/Corbis • p. 80: Daniel Grill/Istockphoto • p. 84: Photo of Hans/Creative Commons • p. 87: Noe Lecocq/Creative Commons Attribution Share Alike 3.0 • p. 89: Phillippe Psaila/Science Photo Library • p. 90l: Sheila Terry/Science Photo Library • p. 90c: Science Photo Library • p. 90r: Wikipedia Commons • p. 95: doc-stock/ Corbis • p. 97: Vassil/Wikipedia Commons • p. 109: Bettmann/ Corbis • p. 124: Blaz Kure/IStockphoto • p. 125l: fotohunter/ Shutterstock. Modification: Pierre Rousseau • p. 125l: STILLFX/ Shutterstock. Modification: Pierre Rousseau • p. 126l: Carl Zeiss AG. • p. 126r: Picture Library/Art Resource, NY • p. 131: Rémi Boucher • p. 132: Luca di Filippo/Istockphoto • p. 138: Western Ophthalmic Hospital/Science Photo Library • p. 139: Bettmann/ Corbis • p. 143: NASA • p. 144 t: akg-images • p. 144l: W. F. Meggers Collection /Emilio Segre Visual Archives/American Institute of Physics/Science Photo Library • p. 144r: Steve Allen/ Science Photo Library Unit 2 – Preliminary Notions of Mechanics p. 148, 149 and 150: Shutterstock • p. 151: Detlev Van Ravenswaay/Science Photo Library • p. 157: Paul Almasy/ Corbis • p. 158: Wikipedia Commons • p. 162l: Erich Lessing Art Resource, NY • p. 162r: Private Collection/The Bridgeman Art Library • p. 165: Olaru Radian-Alexandru/Istockphoto • p. 170t: NIST • p. 170b: BIPM • p. 171: CSIRO • p. 173: NASA • p. 178: Andrew Brookes, National Physical Laboratory/Science Photo Library • p. 179: Sheila Terry/Science Photo Library • p. 183: GustoImages/Science Photo Library. Modification: Pierre Rousseau • p. 186: Maria Platt-Evans/Science Photo Library

Unit 3 – Kinematics p. 200, 201 and 202: cultura/Corbis • p. 203: Paul Hill/ Istockphoto • p. 209: Courtesy of Transports Québec • p. 214: Bibliothèque nationale de France • p. 216: dst/maXx images • p. 217l: Istockphoto • p. 217r: Slobo Mitic/Istockphoto • p. 221: 2happy/Shutterstock • p. 225: Museum Boerhaave Leiden • p. 230: akg-images/Rabatti – Domingie • p. 235: yungshu chao/ Istockphoto • p. 237: Brian Erler/Getty Images • p. 239tl: Patti McConville/Getty Images • p. 239bl: Steve White/CPI/The Canadian Press • p. 239r: AP Photo/Paul Sakuma • p. 240: Scala/ Art Resource, NY • p. 243: GustoImages/Science Photo Library • p. 244: Loren Winters/Visuals Unlimited, Inc. • p. 245: Science Photo Library • p. 248: Istockphoto • p. 249: Ty Milford/Aurora Photos/Corbis • p. 252: Istockphoto • p. 254: Grzegorz Momot/epa/Corbis • p. 258: Ben Welsh/maXx Images • p. 259l: Leonard de Selva/Corbis • p. 259r: Getty Images • p. 261: Ryasick photography/Shutterstock Unit 4 – Dynamics p. 262, 263 and 264: Mustafa Deliormanli/Istockphoto • p. 265: Greg Epperson/maXx image • p. 266t: Michael Krinke/ Istockphoto• p. 266b: Gusto images/Science Photo Library • p. 270: SSPL/Science Museum/Art Resource, NY • p. 271t: NASA/JPL/Science Photo Library • p. 271b: NASA/Science Photo Library• p. 272l: Tom Grundy/Shutterstock • p. 272r: ESA S. Corvaja 2008 • p. 277: Science Source/Photo Researchers • p. 278: Alan Crawford Istockphoto • p. 281: Christophe Vander Eecken/Reporters/Science Photo Library • p. 282l: lukaszfus/ Shutterstock • p. 282r: Serp/Shutterstock • p. 283: Alessandra Benedetti/Corbis • p. 286: Tom Grundy/Shutterstock • p. 287: Alistair Forrester Shankie/Istockphoto • p. 291: akg-images • p. 293: Istockphoto • p. 299: Bloomberg/Getty Images • p. 300l: Elena Elisseeva/maXx images • p. 300r: Wikipedia Commons • p. 303: NASA • p. 304l: Duomo/Corbis • p. 304r: Greg Fiume/ Corbis • p. 306: Bettmann/Corbis • p. 307r and l: Limagier.com • p. 310: Dedyukhin Dmitry/Shutterstock • p. 318: Gary Rothstein/ epa/Corbis Unit 5 – Energy and its Transformations p. 322, 323 and 324: Herbert Hopfensperger/Age Fotostock • p. 325: Chris Harris/Photolibrary • p. 331: Sheila Terry/Science Photo Library • p. 333: André Vigneau/Archives photographiques de la STM • p. 336l: Dr Jeremy Burgess/Science Photo Library • p. 336r: Getty Images • p. 339: Arvind Balaraman/Shutterstock • p. 340l: Science Photo Library • p. 340r: Jonas Velin/ Istockphoto • p. 345: Stéphanie Colvey • p.347t: Slobodan Vasi /Istockphoto • p. 347b: Istockphoto • p. 354: Adam HartDavis/Science Photo Library • p. 356: David Parker/Science Photo Library • p. 357: Courtesy of Alstom, Denis Félix / Interlinks Image 2007 p. 363: Martin Harvey/Photolibrary • p. 367: Rita Greer, 2009/Department of Engineering Science, Oxford University • p. 369l: Anthia Cumming/Istockphoto • p. 369r: Judy Ledbetter/Istockphoto • p. 370: Michael Jaeger/ Photolibrary • p. 373: Private Collection/The Bridgeman Art Library • p. 375: Scott Williams/Istockphoto Appendices p. 378: Shutterstock (measuring instruments), Erich Schrempp/ Photo Researchers, Inc. (refraction), NASA (solar system), Chepko Danil VitalevichShutterstock (masses), Ariadna de Raadt/ Istockphoto (calculator) • p. 380: Jason Smith/Shutterstock • p. 384: Image source/Corbis • p. 396: Janicke Morrissette/Le bureau officiel

SOURCES

429

Illustrations Arto Dokouzian: p. 380, 381, 382, 383 Peter Gudella/Shutterstock: p. 14 Jacques Perreault: p. 4, 5, 6, 8, 9, 10, 11 Late Night Studio: p. 6 Marc Tellier: p, 13, 14, 15 Michel Rouleau: Units 1 to 5 and p. 338, 391, 392, 393, 404, 405, 406

430

SOURCES

PHYSICS The QUANTUM collection offers a flexible and concrete approach to the Physics Program for the Third Year of Secondary Cycle Two. It is a major asset that helps students develop their competencies, build their understanding of the scientific concepts in the Program and have a better grasp of the issues pertaining to their daily environment. The QUANTUM collection is: • an encyclopaedic textbook that offers substantial treatment of all concepts in the Program • an impressive number of exercises with varied levels of difficulty • a wide variety of laboratories, some previously tested and others new and original • a clear presentation of scientific formulas, mathematical equations, procedures and examples of calculations • numerous highly precise diagrams and illustrations to help students understand the concepts • a Review section for Physics concepts addressed in the first and second years of Secondary Cycle Two • appendices that address certain scientific procedures, strategies and techniques and that delve deeper into various aspects of problem-solving

• indispensable reference tables • animation clips that take into account different learning styles and facilitate the comprehension of certain phenomena • a choice of new learning and evaluation situations related to students’ interests and designed to help develop competencies • suggestions for the use of ICTs in the classroom • varied evaluation tools adapted to the reality of the classroom • a clear structure that encourages the development of student autonomy • flexible lesson plans

QUANTUM SUBSTANTIAL, SOLID AND RIGOROUS CONTENT!

The components of the QUANTUM collection – PHYSICS, Third Year of Secondary Cycle Two

• Student Textbook

• Teaching Guide

ISBN 978-2-7652-1444-1