COMBINED EDITION Solutions Manual

COMBINED EDITIONSolutions ManualIgor NowikowBrian HeimbeckerChristopher T. HowesJacques ManthaBrian P. SmithHenri M. van Bemmel

Physics: Concepts and ConnectionsCombined Edition Solutions ManualAuthorsIgor NowikowBrian HeimbeckerChristopher T. HowesJacques ManthaBrian P. SmithHenri M. van BemmelNELSONDirector of PublishingDavid SteeleFirst Folio Resource GroupProject ManagementRobert TempletonPublisherKevin MartindaleCompositionTom DartProject DeveloperDoug PanasisProofreading and Copy EditingChristine SzentgyorgiPatricia TrudellMatt SheehanProject EditorLina Mockus-O’BrienEditorsShirley TessierLisa KafunMark PhilpottKevin LinderIllustrationsGreg DuhaneyClaire MilneDesign and ArtworkArtPlus Design and CommunicationsCOPYRIGHT 2003 by Nelson, adivision of Thomson Canada Limited.Printed and bound in Canada.1 2 3 4 05 04 03 02For more information contact Nelson,1120 Birchmount Road Toronto,Ontario, M1K 5G4. Or you can visit ourInternet site at http://www.nelson.comALL RIGHTS RESERVED. No part of thiswork covered by the copyright hereonmay be reproduced, transcribed, or usedin any form or by any means—graphic,electronic, or mechanical, includingphotocopying, recording, taping, Webdistribution, or information storage andretrieval systems—without the writtenpermission of the publisher.For permission to use material from thistext or product, contact us byTel 1-800-730-2214Fax 1-800-730-2215www.thomsonrights.comEvery effort has been made to traceownership of all copyrighted material andto secure permission from copyrightholders. In the event of any questionarising as to the use of any material, wewill be pleased to make the necessarycorrections in future printings.

Table of ContentsISolutions to Applying theConcepts QuestionsChapter 1Section 1.21.31.4Chapter 2Section 2.12.4Chapter 3Section 3.13.33.43.5Chapter 4Section 4.34.4Chapter 5Section 5.35.45.55.6Chapter 6Section 6.16.26.36.46.5Chapter 7Section 7.37.47.57.67.7Chapter 8Section 7Chapter 9Section 9.29.39.49.59.69.89.99.10Chapter 10Section 10.110.210.610.710.910.1110.1210.13Chapter 11Section 11.111.211.311.411.5Chapter 12Section 12.312.412.612.8Chapter 13Section 2323232323242424Chapter 14Section 14.214.314.514.614.714.814.9Chapter 15Section 15.5Chapter 16Section 16.4Chapter 17Section 17.217.317.417.517.617.8Chapter 18Section 18.218.318.418.5II Answers to End-ofchapter 535353524242424252526272828Table of ContentsChapter 1Chapter 2Chapter 3Chapter 4Chapter 5Chapter 6Chapter 7Chapter 8Chapter 9Chapter 10Chapter 11Chapter 12Chapter 13Chapter 14Chapter 15Chapter 16Chapter 17Chapter 18373839404345464750525556576162636566III Solutions to End-ofchapter ProblemsChapter 1Chapter 2Chapter 3Chapter 4Chapter 5Chapter 6Chapter 7Chapter 8Chapter 9Chapter 10Chapter 11Chapter 12Chapter 13Chapter 14Chapter 15Chapter 16Chapter 17Chapter 211iii

PART 1 Solutions to Applying the ConceptsIn this section, solutions have been provided only for problems requiring calculation.Section 1.24. a) c 3.00 108 m/s1 second 9 192 631 770 vibrationsTherefore, in 3632 s, there are 3.34 1013vibrations.b) 1 m 1 650 763.73 d (.150 m)(1 m)d 2.48 105 Section 1.32. a)b)c)d)e)f)3. a)b)c)d)e)4. a)b)c)d)e)5. a)b)c)d)e)4571463.1 m3.2 m3.4 m3.6 m3.4 m3.745 m309.6 m120 s671.6 s461.7 s4.0 m3.3 m3.33330.330.333i) 60 monthsii) 2.6 106 miniii) 1.8 103 div) 1.6 108 sSection 2.11. At t 2.0 s, v 10 m/s,d 12 (10 m/s 20 m/s)2.0 s 30 mAt t 7.0 s, v 15 m/s, d 12 (4.0 s)(20 m/s) 12 (7.0 s 4.0 s)(15 m/s) 40 m 27.5 m 67.5 mSection 2.4Section 1.41. a) 389 s 6.4833 min 0.10805 h 4.502 10 3 d 1.50 10 4 months 1.25 10 5 ai) 1.50 10 4 monthsii) 6.48 miniii) 1.25 10 5 aiv) 3.89 108 sb) 5.0 a 60 months 1825 d 43 800 h 2 628 000 min 157 680 000 s1. a) a 4.0 m/s2t 40.0 sv1 0 m/sv2 (4.0 m/s2)(40.0 s)v2 160 m/s v1 v2 b) i)d t2d 3200 m v1 t 12 ii) da t21d 2 (4.0 m/s2)(40.0 s)2d 3200 m2. a) d 152 mv1 66.7 m/sv2 0v 66.7 m/s v1 v 2 d t22 dt v1 v2t 4.5577 s v ata 14.6 m/s2b) t 4.56 s v2 v 1 tc) i)d2d 152 m v1 t 12 ii) da t2d (66.7 m/s)(4.56 s) 12 ( 14.6 m/s2)(4.56 s)2d 152 mSolutions to Applying the Concepts1

3395 km/h3. a) v 943.1 m/s3.6b) t 0.5 sv 943.1 m/sd 471.5 m v avg c) ataavg 78.59 m/s2d) t 8.7 sv2 943.1 m/sv2 v1aavg t943.1 m/s v178.59 m/s2 8.7 s2v1 2.6 10 m/s4. a) 31 km 0.12/km 3.72Total cost 3.72 1.50 2.00 7.22b) 19 min 0.32 h31 km 97 km/h0. 32 h20 kmd) 0.20 h100 km/h10 km 0.08 h125 km/h10 km 83 km/h0.12 h31 kme) 3.4 L 0.77/L9.1 km/L 2.62Section 3.11. a) dx 20 sin 30 dx 10 kmdy 20 cos 30 dy 17.32 kmdy 17 kmb) dx 40 cos 60 dx 20 kmdy 40 sin 60 dy 34.64 kmdy 35 kmc) dx 10 sin 10 dx 1.736 kmdx 1.7 km2dy 10 cos 10 dy 9.848 kmdy 9.8 kmd) dx 5 sin 24 dx 2.03 kmdx 2.0 kmdy 5 cos 24 dy 4.5677 kmdy 4.6 kme) dx 12 sin 45 dx 8.5 kmdy 12 cos 45 dy 8.5 kmf) dx 10 kmdy 0 km2.dx 20 sin 20 120 sin 50 150 30 sin 75 206.1 mdy 20 cos 20 120 cos 50 30 cos 75 88 md ( 206 .1 m)2 (88 m)2d 230 m88 tan 1 206.1 23 d 230 m [W23 N]Section 3.31. a) g 9.81 m/s2v 1 0d 100 m v 1t 12 gt2d 100 m 12 (9.81 m/s2) t2t2 20.387 s2t 4.52 sb) v1 10 m/sd 100 mg 9.81 m/s2 d v 1t 12 g t2 100 10t 12 ( 9.81) t2 4.905t2 10t 100 0 10 100 4( 4.905)(100) t 2( 4.905)t 5.6 sSolutions to Applying the Concepts

c) v1 10 m/sd 100 mg 9.81 m/s2 d v 1t 12 g t2 100 10t 12 ( 9.81) t2 4( 4.905)(100) 10 100 t 2( 4.905)t 3.6 sd) vx1 5.0 m/sax 0vy1 0ay 9.81 m/s2v 100 m/sdx vx1t 12 axt2dy vy1t 12 ayt2i) dx 45 mii) dx 28 miii) dx 18 m2. a) At maximum height in trajectory, v2 0.g 9.81 m/s2 v v o g t vo v t g t 1.08 sb) v 2 v 12 2g dv12 v22dy 2gdy 0.539 mc) v1y 0v1x 25 cos 25 v1x 22.658 m/sdy v1y t 12 ayt2v2y v1ye) gtv2y g t v1yv2y 3.25 m/sv2y 3.25 m/s [down]v2 vy2 vx2v 22.9 m/s 22.658tan 3.25 81.8 v 22.9 m/s [S81.8 E] v t3. a) dd (18.5 m/s)(cos 18 )(10.9 s)d 191.84 mTherefore, the ball travels 192 m.b) v2y v1y2 2ay dyAt maximum height, v2y 0. (v1y2)dy 2ay (18.5 sin 18 )2dy 2ay (32.68 m2/s2)dy 2( 9.8 m/s2)dy 1.7 m v tc) i) dd (18.5 cos 8 )(10.9 s)d 200 mii) v2y v1y2 2ay dy (v1y2)dy 2ay (18.5 sin 8 )2dy 2aydy 0.34 m0.539 12 ( 9.81) t2t 0.331 sd) dx vx tdx 25 cos 25 (1.08 0.331)dx 31.97dx 32 m away from the soccer player.4. d v t31 m (18.5 m/s)(cos )(3.66 s)1.676 cos (3.66 s) 62.7 Therefore, the loft angle of the club is 63 .Section 3.42. vwind 80 km/hv wind 22.22 m/sv plane 200 km/hv plane 55.55 m/sSolutions to Applying the Concepts3

a) vog2 (55.55)2 (22.22)2vog 59.84 m/s22.22tan 55.55 21.8 v og 59.84 m/s [N21.8 E]b) vog 50.86 m/s22.22cos 55.55 66.42 vog 50.86 m/s [N23.6 W]c) vx 22.22 55.55 cos 70 vx 3.22 m/svy 55.55 sin 70 vy 52.20 m/sv2 (3.222)2 (52.202)2v 52.30 m/s 86.5 vog2 (22.22)2 (55.55)2 2(22.22)(55.55)cos 70 vog 52.30 m/ssin 70 sin 52.3055.55 86.45 86.5 vog 52.30 m/s [E86.5 N]Section 3.52. a)t 0.5 sv2 (120 km/h)2 (120 km/h)2v 169.7 km/h169.7 km/h47.1 m/sa 0.5 s0.5 sa 94.3 m/s2 94 m/s2120 km/h tan 1 120 km/h 45 94 m/s2 [W45 N]ab) v 1 120 km/h [E]v 2 120 km/h [N25 W]t 0.5 sTrigonometric Methodv2 (120 km/h)2 (120 km/h)2 2(120 km/h)(120 km/h)cos 115 v 202 km/h4sin 115 sin 202 km/h120 km/h 32.6 202 km/ha 0.5 s56.1 m/sa 0.5 s a 112 m/s2 [W33 N]Component Methodvx ( 120 cos 65 ) km/h 120 km/h 170.7 km/hvy (120 sin 65 ) km/h 0 108.8 km/hv2 ( 170.7 km/h)2 (108.8 km/h)2v 202 km/h202 km/ha 0.5 s 112 m/s2108.8 km/h tan 1 170.7 km/h 33 a 112 m/s2 [W33 N]c) v 1 120 km/h [E]v 2 100 km/h [N25 W] or [W65 N]v2 (120 km/h)2 (100 km/h)2 2(120 km/h)(100 km/h)cos 115 v 185.9 km/hv 186 km/hsin 115 sin 100 km/h186 k m/h 29.2 29 vx ( 100 cos 65 ) km/h 120 km/h 162.3 km/hvy (100 km/h sin 65 ) 0 90.6 km/hv2 (162 km/h)2 (91 km/h)2v 186 km/h90.6 km/h tan 1 162.3 km/h 29 186 km/ha 0.5 s51.6 m/sa 0.5 s a 103 m/s2 [W29 N]Solutions to Applying the Concepts

Section 4.31. a)b)c)d)e)00 y: Fnet 0ay 0x: ax 60 N2. a) ax 60 kga 1.0 m/s2b) F 20 N20 Na 60 kga 0.33 m/s2c) a 0.33 m/s230 Nd) a 60 kga 0.50 m/s2Section 4.41.2 105Fnet1. a) sin 45 sin 90 Fnet 1.7 105 N [N45 E]b) Fnet (1.2 105 cos 30 ) 2 Fnet 2.1 105 N [E]c) Fx 1.2 105(cos 20 cos 10 )Fx 2.3 105 NFy 1.2 105(sin 20 sin 10 )Fy 2.0 104 NF2 Fx2 Fy2F 2.3 105 N2.0 10 4 tan 1 52.3 10 5.0 2.3 105 N [N85 E]Fd. i) Fx 1.2 105 5.0 104Fx 7.0 104 NFy 2.1 105 NF 2 (1.2 105)2 (7.0 104)2F 1.4 105 N1.2 10 5 tan 1 47.0 10 59.7 F 1.4 105 N [N30 E]ii) Fx (1.2 105 cos 30 )2 5 104Fx 1.578 105 NFy 0 1.6 105 N [E]Fiii) Fx 1.2 105(cos 20 cos 10 ) 5 104Fx 1.81 105 NFy 1.2 105(sin 20 sin 10 )Fy 2.02 104 N2.02 104 tan 1 1.8 10 5 6.4 F 1.8 105 N [N83.6 E]Section 5.31. a) F (12 000 kg)(9.81 m/s2)F 1.18 105 NGm1m2b) F r2F (6.67 10 11 N·m2/kg2)(12 000 kg)(5.98 1024 kg) (6.98 106 m)2F 9.82 104 Nc) distance from the surface 6.00 105 Nd) On the Moon,F (6.67 10 11 N·m2/kg2)(12 000 kg)(7.34 1022 kg) (1.74 106 m)2F 1.94 104 NSection 5.4 01. a) aTherefore, Fn Fg.Fn (9.81)(70)Fn 686.7 NFn 6.9 102 Nb) a 0Therefore, Fn Fg.Fn 6.9 102 Nc) ma Fn Fg(70 kg)( 2 m/s2) Fn (70 kg)( 9.81 m/s2)Fn 546.7 NFn 5.5 102 Nd) m( 9.81) Fn m( 9.81)Fn 0 NSolutions to Applying the Concepts5

Section 5.53. a) There is a constant velocity; therefore,Fk F and Fk kFnF (0.5)(30 kg)(9.81 m/s2)F 147.15 NF 1.5 102 Nb) Since there is no motion, Fs F.F 100 Nc) Fs sFn100 N s (30 kg)(9.81 m/s2) s 0.34d) Fn 20 (30 kg)(9.81 m/s2)Fn 314 Ni) F (0.5)(314.3 N)F 157.15 NF 1.6 102 Nii) F 100 N100 Niii) s 314.3 N s 0.32e) Fn 274.3 Ni) F (0.5)(274.3 N)F 137.15 NF 1.4 102 Nii) F 100 N100 Niii) s 274.3 N s 0.36Section 5.62. a) F 10 Nx 1.2 cmx 0.012 m10 Nk 0.012 mk 8.3 102 N/mb) k 3.0 N/mx 550 mmx 0.55 mF (3.0 N/m)(0.55 m)F 1.65 NF 1.7 Nc) F 20 Nk 3.0 N/m620 Nx 3.0 N/mx 6.7 md) F (2 kg)(9.81 m/s2)F 19.62 Nx 0.04 mk 4.9 102 N/mSection 6.11. The only two unbalanced forces are F and Ff.Fnet F Ff(eq. 1)F Fg sin 25 (eq. 2)Ff FnFf Fg cos 25 (eq. 3)Substituting equations 2 and 3 into equation 1,Fnet Fg sin 25 Fg cos 25 Fnet Fg(sin 25 cos 25 )Fnet (2.0 kg)(9.8 m/s2)(sin 25 cos 25 )Fnet (19.6 N)(sin 25 cos 25 )Fnet 6.51 NFnet ma6.51 N (2.0 kg)aa 3.26 m/s21d vi t a t2214.0 m (3.26 m/s2) t228.0 mt 23.26 m/st 1.6 s2. Since there is no friction, the only force thatprevents the CD case from going upward isthe deceleration due to gravity, F .Fnet F Fnet Fg sin 20 Since Fnet ma,ma mg sin 20 a g sin 20 a 3.35 m/s2v2 v1a tv2 v1t a4.0 m/st 23.35 m/st 1.2 sSolutions to Applying the Concepts

3. To find the distance the skateboarder travelsup the ramp, we need to find the velocity ofthe skateboarder entering the second ramp atv1. Since there is no change in velocity on thehorizontal floor, v1 v2.For the acceleration on ramp 1,Fnet F ma mg sin 30 a g sin 30 a 4.9 m/s2v22 v12 2adv22 0 m/s 2(4.9 m/s2)(10 m)v2 9.9 m/sFor the deceleration on ramp 2,Fnet F Fnma mg sin 25 (0.1)mg cos 25 a 5.02 m/s2For d,v32 v22 2a d(0 m/s)2 (9.9 m/s)2 2( 5.02 m/s2) dd 9.8 m4. Fnet m(0.60g)Fnet also equals the sum of all forces in theramp surface direction: F f F engineFnet Fm(0.60g) mg sin 30 Fn Fenginem(0.60g) mg sin 30 (0.28)mg cos 30 FengineFengine (0.60)mg mg sin 30 (0.28)mg cos 30 Fengine mg(0.60 sin 30 (0.28) cos 30 )Fengine 3.36m NSection 6.21. a) For m1,Fnet m1aT m1 g m1aFor m2,Fnet m2am2 g T m2a(eq. 1)(eq. 2)Adding equations 1 and 2,m2 g m1 g a(m1 m2)m2 g m1 ga m1 m2a (15 kg)(9.8 m/s2) 0.20(10 kg)(9.8 m/s2) 25 kg 5.1 m/s2 [right]aSubstitute a into equation 2:T m2 g m2aT 71 Nb) For m1,Fnet m1aT m1 g sin 35 m1 g cos 35 m1a(eq. 1)For m2,Fnet m2am2 g T m2a (eq. 2)Adding equations 1 and 2,m2 g m1 g sin 35 m1 g cos 35 a(m1 m2)g(m2 m1 sin 35 m1 cos 35 )a m1 m2a (9.8 m/s2)[5.0 kg (3.0 kg) sin 35 0.18(3.0 kg) cos 35 ] 8.0 kg 3.5 m/s2 [right]aSubstitute a into equation 2:T m2 g m2aT (5.0 kg)(9.8 m/s2) (5.0 kg)(3.5 m/s2)T 32 Nc) For m1,Fnet m1aT m1 g sin 40 1m1 g cos 40 m1a(eq. 1)For m2,Fnet m2am2 g sin 60 T 2m2 g cos 60 m2a(eq. 2)Adding equations 1 and 2,m2 g sin 60 2m2 g cos 60 m1 g sin 40 1m1 g cos 40 a(m1 m2)a g(m2 sin 60 2m2 cos 60 m1 sin 40 1m1 cos 40 ) m1 m2a (9.8 m/s2)[(30 kg) sin 60 0.30(30 kg) cos 60 (20 kg) sin 40 0.20(20 kg) cos 40 ] 50 kg a 1.1 m/s2 [right]Solutions to Applying the Concepts7