student solution manual 8th edition

5/22/2018 Student Solution Manual 8th Edition

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STU DENT SOLUTIONS MAN UALJ. Richard Christman

Professor EmeritusU.,S. Coast Guard Academy

FUNDAMENTALS OFPHYSICSEighth Edition

David Halliday(Jnivers iQ of P itts burgh

Robert ResnickRens s elaer Polytechnic Institute

Jearl WalkerCleveland State Univers iQ

John Wiley Sons, Inc.


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Cover Image: @ Eric Heller/Photo ResearchersBicentennial Logo Design: Richard J. Pacifico

Copyright @ 2008 John Wiley Sons, Inc. All rights reserved.No part of this publication may be reproduced, stored in a retrieval system or transmittedin any form or by any means, electronic, mechantcal, photocopying, recording, scanning,or otherwise, except as perrnitted under Sections 107 or 108 of the I 97 6 lJnited StatesCopyright Act, without either the prior written permission of the Publisher, orauthorizationthrough payment of the appropriate per-copy fee to the CopyrightClearance Center, Inc ., 222 Rosewood Drive, Danvers, MA 01923, or on the web atwww.copyright.com. Requests to the Publisher for permission should be addressed to thePermissions Department, John Wiley Sons, Inc., 111 River Street, Hoboken, NJ07030-5714, (201) 148-6011, fax (201) 748-6008, or online athttp : //www. wilelz. c om/ go/p ermi s s ions.To order books or for customer service, please call 1-800-CALL-WILEY (225-5945).rsBN- 13 978- 0-47 r-779s8-2Printed in the United States of Amer'ca10 9 8 7 6 s 4 3 2 |Printed and bound by Bind-Rite Graphics.


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PREFACE

This solutions manual is designed for use with the textbook Fundamentals of Physics, eighthedition, by David Halliday, Robert Resnick, and Jearl Walker. Its primary puqpose is to showstudents by example how to solve various types of problems given at the ends of chapters in thetext.Most of the solutions start from definitions or fundamental relationships and the final equationis derived. This technique highlights the fundamentals and at the same time gives students theopportunity to review the mathematical steps required to obtain a solution. The mere pluggingof numbers into equations derived in the text is avoided for the most part. We hope students willlearn to examine any assumptions that are made in setting up and solving each problem.Problems in this manual were selected by Jearl Walker. Their solutions are the responsibility ofthe author alone.The author is extremely grateful to Geraldine Osnato, who oversaw this project, and to hercapable assistant Aly Rentrop. For their help and encouragement, special thanks go to the goodpeople of Wiley who saw this manual through production. The author is especially thankful forthe dedicated work of Karen Christman, who carefully read and coffected an earlier version ofthis manual. He is also grateful for the encouragement and strong support of his wife, MaryEllen Christman.J. Richard ChristmanProfessor EmeritusIJ.S. Coast Guard AcademyNew London, CT 06320


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TABLE OF CONTENTS

Chapterl .., . .. . .... . o. .. . IChapter?. .i . . . . . . . . . . . . . .4Chapter 3 . . . . . . . . . . . 0 . . . . . 10Chaptet 4 . .. . . . . . . . . . . . . . .I4Chaptgr5 . . . . . . . . . . . . . . . . .21Chapter6. . . . . . . . . . . . . . . . .28ChapterT o.......... o.... .37ChapterS . . . . . . . . . . . . . . . . .42Chapter9. ............... 50Chapter 10 . . . . . . . . . . . . . . . . 58Chapter 11 .......... r... ..63Chapter 12.. .. . .. . .. .. .. ..71Chapter 13 ... o... . .. . ... ..77Chapter 14 .. . . . . r . . . . . . . . . 84Chaptgr 15 . . . . . . . . . . . . . . . o 89Chapter 16.. .. ... . .. . . . .. .95Chaptgr 17 . . . . . . . . . . . . . . , . 101Chapter 18 . . . . . . . . . . . . . . r . 109Chaptgr 19 . . . , . . . . . . . . . . . . 115Chaptgr20. . . . . . . . . . . . . . . . 122Chapter2l ............... a I28Chaptet 22. . . . . . . . . . . . . . . . 134

Chapter 23Chapter 24Chapter 25Chapter 26Chapter 27Chapter 28Chapter 29Chapter 30Chapter 31Chapter 32Chapter 33Chapter 34Chapter 35Chapter 36Chapter 37Chapter 38Chapter 39Chapter 40Chapter 4lChapter 42Chapter 43Chapter 44

. . .. .. . .140o.o.o..146.......154. . . . . . . 159.......162...... .170. . . . . . .175. . . . . . . 193. . . . . . , l9l....,..199...... .205...... .213...... .221...... .229...... .235...... .239...... .243...... .247...... .251...... .254...... .260...... .264


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Chapter L

3use the given conversion factors.(a) The distance d, in rods is

d - 4.0 furlongs(b) The distance in chains is

1yd - (0.9144mX106 pmlm) - 9.144 x 105 pm.

- (4.0 furlongsX2Ol . 168 m/furlong)5.0292mf rod

d - 4.Lfurrongs - (4'0 furlongsX20l ' 168 m/furlong) : 4|chains .20.L7 mlchain-I(a) The circumference of a sphere of radius R is given by 2r R. Substitute R - (6.37 x106mX10-tk^lm)should obtain 4.00 x 104km.(b) The surface area of a sphere is given by 4trR2, so the surface area of Earth is 4n(6.37 x103 k*)'(c) The volume of a sphere is given by (4nlrR3, so the volume of Earth is G"13X6.37 x103 k*)3 _ 1.08 x 1012 km3.t7None of the clocks advance by exactly 24h in a 24-h period but this is not the most importantcriterion for judging their quality for measuring time intervals. What is important is that theclock advance by the same amount in each 24-h period. The clock reading can then easily beadjusted to give the correct interval. If the clock reading jumps around from one 24-h period toanother, it cannot be coffected since it would impossible to tell what the coffection should be.The followittg table gives the coffections (in seconds) that must be applied to the reading oneach clock for each 24-h period. The entries were determined by subtracting the clock readingat the end of the interval from the clock reading at the beginning.

Chapter I


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CLOCK Sun.-Mon.

Mon.-Tues.

Tues.-Wed.

Wed.-Thurs.

Thurs.-Fri.

Fri.-Sat

ABCDE

-16-3-58+67+70

-16+5-58+67+55

-15-10-s8+67+2

-17+5-s8+67+20

-15+6-58+67+10

-15-7-58+67+10

Clocks C and D are the most consistent. For each clock the same coffection must be appliedfor each period. The coffection for clock C is less than the coffection for clock D, so we judgeclock C to be the best and clock D to be the next best. The coffection that must be applied toclock A is in the range from 15 s to 17s. For clock B it is the range from -5 s to *10 s, for clockE it is in the range from -70 s to -2s. After C and D, A has the smallest range of correction,B has the next smallest range, and E has the greatest range. From best the worst, the ranking ofthe clocks is C, D, A, B, E.21(a) Convert grams to kilograms and cubic centimeters to cubic meters: 1 g _ I x 10-' kg and1cm3 - (1 x To-2 m)3

rs-(1 e)(#) (,%) :rx1o3kg\t(b) Divide the mass (in kilograms) of the water by the time (in seconds) taken to drain it. Themass is the product of the volume of water and its density: M - (5700m3X1 x 103 kg/nt')5.70 x 106kg. The time is t - (10.0hX3600s/h) - 3.60 x 104 s, so the mass flow rate n is

- M- 5.70 x 106kg-t 3.60 x 104 s 1 58 kg/s "3s(a) The amount of fuel she believes she needs is (750mi)l(0mif gal): 18.8gal. This is actuallythe number of IJ.K. gallons she needs although she believes it is the number of IJ.S. gallons.(b) The ratio of the U.K. gallon to the U.S. gallon is (4.545963 IL)l(3.785 3060L) : 1.201.The number of U.S. gallons she actually needs is(18.8 IJ.K. galx 1.201 U.S gallu.K. gal) : 22.5 IJ.S. gal .

R

39The volume of a cord of wood is V - (8 ftx 4 tt)(4 ft) : I28ft3.Appendix D) to obtain V - I28ft3X0 .3048 ^l ft)3 - 3.62m3. Thusto (l 13.62) cord - 0 .28 cord.2 Chapter I

IJse l ft1.0 m3 of wood coffesponds


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4T(a) The difference in the total amount between 73 freight tons and 73 displacement tons is

(8 barrel bulk/freight ton)(73 freight ton)- (7 barrel bulk/displacement ton)(73 displacementton): 73 barrel bulk .NowSO

l banel bulk - 0.141 5m3 - (0. 41 5m3)(28.378U.S. bushel): 4.01 5U.S. bushel ,

45

73barceI bulk - (73 barrel bulk)(4.01 5 U.S. bushellbarcel bulk) - 293 U.S. bushel .(b) The difference in the total amount between 73 register tons and 73 displacement tons is

(z}barrel bulk/register ton)(7 3 register ton)- (7 barrel bulk/displacement ton)(73 displacementton): 949 barrel bulk.

Thusg4:gbarrel bulk - (949barrel bulk)(4.01 5U.S. bushellbarcel bulk):3810U.S. bushel .

57(a) We want to convert parsecs to astronomical units. The distancebetween two points on a circle of radius r is d - 2r stn 0 12, where0 is the angle subtended by the radtal lines to the points. See thefigure to the right. Thus r - d,lz sin 0 12 and

l pc - t ^",., - 2.06 x 105 AU,2 sin( I" 12)where " - (I13600)" - (2.78 x 10-4)' was used. Finally

1 AU - (1 AU) 1Q.06 x 105 AU lpc) : 4.9 x 10-u p, .

m-1x10-3km,0.12 ATJ lmin.

(b) A light year is(1.86 x 10s mrlsxl.0 yX3 65.3 daly)Q|hld)(3600 s/h) - 5 .87 x 1012 mt

andl Au : 92'9 x },9^6 Tt,; - l.5g x 1o-' ly. 5.87xlot2milly L'J\)

/-"., " 012

Chapter I


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Chapter 21(a) The average velocity during any time interval is the displacement during that interval dividedby the duration of the interval: ?)av' - Lr I Lt, where Lr is the displacement and Lt is the timeinterval. In this case the interval is divided into two parts. During the first part the displacementis Lr 1 : 40 km and the time interval is

Ltr : (40km) -1.33h.(30 k*/h)During the second part the displacement is Lrz: 40km and the time intenral is

Ltz- (4gq)(60 l*ttr - o '67h 'Both displacements arc in the same direction, so the total displacement is Lr40km +40km:80km. The total time interval is Lt: Lt1 * Ltz- 1.33h+0.67h:2.00h.The average velocity is (80km)nl'avg : 40k*/h.(2.0 h)(b) The average speed is the total distance traveled divided by the time.distance is the magnitude of the total displacement, so the average speed(c) Assume the automobile passes the origin at r g0time t : 0. Then its coordinate as a function of Gm)time is as shown as the solid lines on the graph 60to the right. The average velocity is the slope ofthe dotted line. -- --Q- - --J -- -r - 40

200.5

(a)(b)(c)

In this case the totalis 40 km fh.

1.0 1.5t (h) 2.0-5Substitute,inturn, t- I,2,3, and4sintotheexpression r(t):3t-4*+#,where r isinmeters and t is in seconds:r(Is): (3^lsxl s) - (mls2xl r)2 + (l^11311 s;3 - 0r(zs) : (3mlsx2 s) - (4mls\(2s)2 + Qmls';12 s)' - -2mr(3s): (3^lsx3 s) - (4mls2x3 s)2 + Omlr311: r13 - 0(d) r(4s): (3mlsx4s) - (4mls2x4s)2 + Q^lr3;1+s;3(e) The displacement during an interval is the coordinate at the end of the interval minusthe coordinate at the beginning. For the interval from t -- 0 to t - 4 s, the displacement isLr - r(4s) - tr(0) - I2m- 0 - +12m. The displacement is in the positive r direction.4 Chapter 2


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(0 The average velocity during an interval is defined as thedivided by the duration of the interval: uavs- L, I Lt. For thethe displacement is Lr - r(4 s) r(2 s) - I2m (-2m) -Lt: 4 s - 2 s : 2s. Thus Lr l4m'uavg:E-

^ -7m/s.(d) The solid curye on the graph to the rightshows the coordrnate r as a function of time.The slope of the dotted line is the averagevelocitybetween t- 2.0s and t- 4.0s.

displacement over the intervalinterval from t- 2s to t-4sI4m and the time intenral is

r r2.o(m)9.06.03.00.0

-3.019If ur is the velocity at the beginning of a time interval (at time t) and u2 is the velocity at theend (at tz), then the average acceleration in the interval is given by eav'Take h : 0, u1 : 18 m/s, t2 - 2.4 s, and u2

auus -30 m/s - 18 m/s - -zo^lr' .2.4s tThe negative sign indicates that the acceleration is opposite to the original direction of travel.25(a) Solve u- us* at for t: t - (u - uo) f a. Substitute u-0.1(3.0 x 108 mls):3.0 x I07 ml s,u0 : 0, and e,:9.8 ^lr'. The result is t - 3.06 x 106 s. This is ,zmonths.(b) Evaluate r4.6 x 1013 m.27Solveu2 -u3+2a(r-r0 fora. TakeffO:0. Then a-(r'-ril|2tr. Use'u0:1.50x 105 mls,u - 5.70 x 106m/s, and r - 1.0cm - 0.010m. The result is

a_ (5.70 x 106m/s)2 - (1.50 x 105m/s)2 - 1.62x l0r, ^/r, .2(0.010m)Take frs- 0, and solve r- ust * *ot' for a: a_ 2(r ust)f t2. Substitute r- 24.0m,- 56.0k*/h - 15.55 m/s, and t - 2.00 s. The result is2lzq.0m- (15.55m1sx2.00s)] 1 rr ^ r^Z&-

33(a)Ug

Chapter 2 5


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The negative sign indicates that the acceleration is opposite the direction of motion of the car.The car is slowing down.(b) Evaluate u(30.3 km lh).45(a) Take the A axis to be positive in the upward direction and take t - 0 and A - 0 at the pointfrom which the wrench was dropped. If h is the height from which it was dropped, then theground is at A : -h. Solve u2 - uzo + 2gh for h:

u2h- 2s- g.8 m/s2:

(24mls)2 :29.4m.2(9.8 m/s2)

(to -u) 24mls -2.45s.- 9.8 ^l12

ulSubstitute uo : 0, 1) : -24mls, and g

h-(b) Solve 1) : uo - gt for t:

(c) t (s)2

t (s)20

a-10(-)-zo-30

The accelerationhits the ground:as shown on the

0

u -10(m/s)' '_20-30

47(a) At the highestupward, set u6 Chapter 2

point the velocityin u2

0

a-5(-is2)-10_15

of the ball is instantaneouslyand solve for u$ us- \EA. zero. Take the A axis to beSubstitute g

is constant untila- -g.8 m/s2.right.the wrenchIts graph is t (s)


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A:50mtoget(b) It will be in the atrsolutions aret:0 and t

2(9.8^ls2x50m) - 31 m/s.until A-0 again. Solve A: 2uo I g . Rej ect the first and- uot t7t' for t. Since Aaccept the second:

-6.4s.the two

.L _ 2uo 2(31 m/s)t- -I 9.8 ^lt'(c)a60(m) 40

20

u40(n/s) 200

_20-40

0

a-5(*/r')-10_15

68 t (s)The acceleration is constant while the ball is inflight: a,on the right.

49(a) Take the A axis to be upward and place the origin on the ground, under the balloon. Sincethe package is dropped, its initial velocity is the same as the velocity of the balloon, +lTm/s.The initial coordinate of the package is ao: 80 m; when it hits the ground its coordinate is zero.Solve a : Uo + uot - *gt' for t:

t-uo+Iwhere the positive solution was used. A negative value for t coffesponds to a time before thepackage was dropped.(b) Use 't):7)0 - gt: l}mls - (9.8 mls2x5.4s) - -.41 m/s. Its speed is 4Imf s.51The speed of the boat is given by u6 _ d lt, where d is the distance of the boat from the bridgewhen the key is dropped (IZm) andt is the time the key takes in falling. To calculate t, put the

;-i- ll L r :J.-TD-c9.8 ^lt' V (9.8 mls2)2 g.g m/s2

t (s)

Chapter 2


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origin of the coordinate system at the point where the key is dropped and take the y axis to bepositive in the upward direction. Take the time to be zero at the instant the key is dropped. Youwant to compute the time t when A - -45 m. Since the initial velocity of the key is zera, thecoordinate of the key is given by A - -+g*. Thus-3.03s.

This means 12mU6: - 4.0 m/s .3.03 s--5First find the velocity of the ball just before it hits the ground. During contact with the groundits average acceleration is given by Lua'avg N ,where Lu is the change in its velocity durittg contact and Lt is the time of contact.To find the velocity just before contact take the A axis to be positive in the upward directionand put the origin at the point where the ball is dropped. Take the time t to be zero when it isdropped. The ball hits the ground when A - -15.0m. Its velocity then is found fromu2- -\ga,SO - -l -ze.g^ls2x-15.0m) : -17.rmls.The negative sign is used since the ball is traveling downward at the time of contact.The average acceleration during contact with the ground is

0 - (- 17 .I m/s)rt:*avs zo.o x 1o-3 s 857 ^lt2The positive sign89

indicates it is upward.

The velocity at time t is given byintegration. use the condition that2.5(2.0 s)2 - 7.0 m/s. The velocity9l(a) First convert the final velocity to meters per second: 1) : (60 k*/hx1000 m/km)1Q600 s/h) -16.7mf s. The average acceleration is ult - (16.7^ls)/(5.4s):3.1 ^lr'.(b) Since the initial velocity is zero,the distance traveled is r - Lot' - Ltl .L ^ls2x5 .4 r)2 - 45 m.8 Chapter 2

u:u-attf adt: f S.\tdt:2.5t2+C, where e is aconstantof+ITmlsatt- 2.0stoobtain C:u-2.5t2 - ITmls-- 4.0 s * 7.0^lsr2.5t2 - 7.0^ls*2.5(4.0 r)2 : 47 m/s.

2(-45 m)


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(c) Solve n for t. The result is- 13 s.

97The driving time before the change in speed limit was t6 - Lr luo, where Lr is the distanceand u6 is the original speed limit. The driving time after the change is to - A^r lro, where I)q isthe new speed limit. The time saved is

t6- te- Lr(*;) -(700km)(0.62ramilkm)( | I )-r2h.\ss milh 65 mi lh) L'L )This is about t h and TZmrn.99Let t be the time to reach the highest point and us be the initial velocity. The velocity at thehighest point is zero, so 0 - us - gt and us- gt. Thus H- uot i7t': gtz ig*where the substitution was made for ns. Let H2 be the second height. It is given by H2LgQt)2 - 2gt2 : 4H. The balls must be thrown to four times the original height.I07(a) Suppose the iceboat has coordinate Ut at time h and coordinate Az at time t2. If a is theacceleration of the iceboat and tre is its velocity att - 0, thenAr: ?rltt+Lot? andy2: u0t2**t7.Solve these simultaneously for a and us. The results are

2(azh - aftz)t1t2(t2 - tt)and ug:ffiTake h - 2.0s and tz: 3.0s. The graph indicates that at: 16m and Uz - 27 m. These valuesyield a, - 2.0^lt2 and us - 6.0m/s.(b) The velocity of the iceboat at t - 3.0s is ,u: uo* at:6.0 mls * (2.0m1s2x3.0s) - L}mf s.(c) The coordinate at the end of 3.0 s is az - 27 m. The coordinate at the end of 6.0 s isUz: uotz+ |"*3- (6.0^lsx6.0s) + l(2.0m1s2x6.0r)' - 72m. The distance traveled during thesecond 3.0-s interval is Uz - Uz:72m - 27 m- 45m.

- *ot'2(0.25 103 m)

Chapter 2


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Chapter1(a)use a: lmt(b) The tangent of the

3

to obtain e, : (-25.0 m)2 + (+40.0 m)2 : 47 .2m.angle between the vector and the positive r axis is

tano - aa - 4o'o mclr -25.0 mThe inverse tangent is -58.0o or -58.0" + 180' - I22". The first angle has a positive cosineand a negative sine. It is not correct. The second angle has a negative cosine and a positive sine.It is correct for a vector with a negative r component and a positive A component.3The r component is given by ar - (7 .3 m) cos 250o -- -2.5 m and the A component is givenby a,athe components can also be computed using ar - -Q.3 m) cos 70" and e,a - -Q,3 m) sin 70o .It is also 20" from the negative A axis, so you might also use a,n - -Q.3 ) sin 20" and a,a :-(7.3 m) cos 20". These expressions give the same results.7(a) The magnitude of the displacement is the distancefrom one corner to the diametrically opposite corner:d - \/ (3.00 m)2 + (3 .70 m)2 + (4.30 m)2see this, look at the diagram of the room, with thedisplacement vector shown. The length of the diag-onal across the floor, under the displacement vector,is given by the Pythagorean theorem: L- tM,where (. is the length and u is the width of the room.Now this diagonal and the room height form a right tri-angle with the displacement vector as the hypotenuse,so the length of the displacement vector is given by

d-{L2+h2-(b), (c), and (d) The displacement vector is along the straight line from the beginning to the endpoint of the trip. Since a straight line is the shortest distance between two points the length ofthe path cannot be less than the magnitude of the displacement. It can be greater, however. Thefly might, for example, crawl along the edges of the room. Its displacement would be the samebut the path length would be (. + ut * h" The path length is the same as the magnitude of thedisplacement if the fly flies along the displacement vector.10 Chapter 3


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(e) Take the r axis to be out of the page, the y axis to be to the right, and the z axis to be upward.Then the r component of the displacement is 'u) : 3.70 m, the A component of the displacementis 4.30m, and the z component is 3.00m. Thus i- (3.70m)i +(4.30m)j+(3.00m)t. You maywrite an equally correct answer by interchanging the length, width, and height.(f) Suppose the path of the fly is as shown by thedotted lines on the upper diagram. Pretend there isa hinge where the front wall of the room joins thefloor and lay the wall down as shown on the lowerdiagram. The shortest walking distance between thelower left back of the room and the upper rightfront corner is the dotted straight line shown on thediagram. Its length is

T,umtn(3.70m * 3.00 m)2 + (4.30 m)2 - 7 .96m .

2(a) Let i- d,+6. Then r, : an*b* - 4.0m- 13m - -g.0m andrr: &y*b, - 3.0m +7.0m -10m. Thus r-: (-9.0m)i*(10m)j.

w

h

(b) The magnitude of the resultant is r : ^ lr2 + ,2 -Y'*''a (-9.0m)2 + (10m)2 _ 13 m.

- I2.2m.

(c) The angle 0 between the resultant and the positive r axis is given by tan? - ,a lr* -(10 m)l(-9.0m) - -1.1. 0 is either -48o or I32". The first angle has a positive cosine anda negative sine while the second angle has a negative cosine and positive sine. Since the ncomponent of the resultant is negative and the A component is positive, 0 - I32o .t7(a) and (b) The vector d has a magnitude 10.0m and makes the angle 30o with the positive naxis, so its components are a,n- (10.0m)cos30o - 8.67m and aa: (10.0m)sin30o - 5.00m.The vector 6'hur a magnitude of 10.0m and makes an angle of 135o with the positive n axis,so its components are b*- (10.0m)cos 135o - -7.07 m and ba _ (10.0m) sin 135oThe components of the sum are rn: a* * b*- 8.67 m - 7.07 m- 1.60m and ra: ea * ba:5.0m*7.07 m- 12.1 m.(c) The magnitude of iis r - ym:(d) The tangent of the angle 0 between i and the positive r axis is given by tan? _ rr lr* -(lz.Lm)l(l.60m) : 7.56. e is either 82.5o or 262.5o. The first angle has a positive cosine anda positive sine and so is the correct answer.

11

l- /-..t t. /./ \/(

(1.60m)2 + (12.I

Chapter 3


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39Since ab cos Q : a*b* * oab, r a"br,

Theb-cos d: a*b*+a,abr+&"b,ab

magnitudes of the vectors given in the problem are e, : (3.0)2+(3.0)2+(3.0)2(2.0)2 + (1.0)2 + (3.0)2 :3.7. The angle between them is found fromcos Q: (3.0X2.0) + (3.0X1.0) + (3.0X3.0) _ 0 .926(5.2)(3.7)

and the angle is d:22o .43(a) and (b) The vector d, is along the r axis, so its r component is arcomponent is zero.(c) and (d) The r component of 6 is b* - b cos 0 -- (4.00 m) cos 30.0ocomponent is ba - b sin 0 - (4.00 m) sin 30.0o - 2.00 m.(e) and (f) The r component of c*is cr -- ccos(0 + 90")- (10.0m)cos I20" - -5.00m and theA component is ca: csin(9 + 90o): (10.0m)sin I20o - 8.66m.(g) and (h) In terms of components cr : pa* * qb* and ca : paa + eba. Solve these equationssimultaneously for p and q. The result is

b*ca - bac* (3.46 mX8 .66m) - (2.00 mX-5.00 m)P:ffi- --6'67and aac* - arca -(3.00 mXS .66 m)q:ffi- -4'34'

The scalar product isd . b - abcos d : (10X6.0) cos 60o : 30

(b) The magnitude of the vector product is- ab sin d: (10X6.0) sin 60o : 52 .

51Take the r axis to run west to east and the A axis to run south to north, with the origin at thestarting point. Let f,dest - (90.0 k-) j be the position of the destination and r'rthe position of the sailor after the first leg of his journey and iz be the remaining displacement12 Chapter j

47(a)

ld*61


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required to complete the journey. The total journey is the vector sum?'dest : r'I + iz and n: ri,,, - iz - (90 km)i - (50.0 km) i.The magnitude of the remaining trip is

of the two parts, so

T2 (50.0 km)2 + (50.0 km)2 _ 103 km .The tangent of the angle with the positiye r direction is tan d : r2a lrr* - (90.0 km) 160.0 km) -1.80. The angle is either 60.9' or 180'+ 60.9o :24Io. Since the sailor must sail northwest toreach his destination the correct angle is 24Io . This is equivalent to 60.9" north of west.7lAccording to the problem statement i+ E - 6.0i+ 1.0j and A- E - -4.0i +7.0j. Add theseto obtain 2A-2.0i+ 8.0j and then A- 1.0i++.0j. The magnitude of A'is

'?.* + r?a

(1.0)2 + (4.0)2

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Chapter 47The average velocity is the total displacement divided by the time interval. The total displacementi is the sum of three displacements, each calcul ated as the product of a velocity and a timeinterval. The first has a magnitude of (60.0 km lh)(40.0 min) l$0.0 min lh)direction is east. If we take the r axis to be toward the east and the A axis to be toward thenorth, then this displacement is i1 - (40.0 km) i.The second displacement has a magnitude of (60.0 km lh)(20.0 min) 1rc0.0 min lh) : 20.0 km. Itsdirection is 50.0o east of north, so it may be written

iz - (20.0km) sin50.0'i + (20.0km)cos 50.0"i - (15.3 km)i* (12.9k*)i.The third displacement has a magnitude of (60.0 km lh)(50.0 min) l(60.0 min lh) _ 50.0 km. Itsdirection is west, so the displacement may be written fi - (-50 km) i. The total displacement is

r-: r-r+ i2+ i, - (40.0km)i+ (15.3km)i+ (I2.9km)j - (50k*)i- (5.3 km) i + (12.9 km) j .The total time for the trip is 40 min * 20 mtn + 50 mininterval to obtain an average velocity of du,,e - (2.9k^lh) i + (7 .05k*/h)i. The magnitude ofthe average velocity is

lr-uu*l : -7.6km/sand the angle O it makes with the positive r axis satisfies

tanQ:' :.^l,n - 2.43 .2.9 k-/hThe angle is d : 68o.11(a) The velocity is the derivative of the position vector with respect to time:

in meters per second for t in seconds.(b) The acceleration is the derivative of the velocity with respect to time:

6 - *(i. 4*i +,t) : 8,i + r

d-* (r,i.o) :8i14 Chapter 4


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in meters per second squared.17(a) The velocity of the particle at any time t is given by d - 6o + dt, where do is the initialvelocity and d is the acceleration. The r component is 'trr:'uyrl a,nt - 3.00 mls - (1.00 mls2)tand the g component rs uy - 'tr1a* aat - -(0.500 mlsz)t When the particle reaches its maximumr coordinate 'trr:0. This means 3.00 mls - (1.00 mls\t - 0 or t - 3.00s. The A component ofthe velocity atthis time is rra - (-0.500 mls2x3.00s): -1.50m/s. Thus 6 - (-1.50 mls)j.(b) The coordinates of the particle at any time t are r- nyrt + *.o*t' and A - uyat + *,oot' . Att - 3.00 s their values are

r:(3.00 mlsx3.00s) - )ft00 mls2x3.00s)2 - 4.50mand Ia : -;(o.5oo m/s2x3.oo s)2 _ -2.25m .Thus r': (4.50 m) i - (2.25m)i.29(a) Take the y axis to be upward and the r axis to be horizontal. Place the origin at thepoint where the diver leaves the platforrn. The cornponents of the diver's initial velocity areuyn - 3.00m/s and uya - 0. At t - 0.800 s the horizontal distance of the diver from the platforrnis :x : 'uy*t - (2.00 mlsx0.800 s) : 1.60 m.(b) The driver's a coordinate is a- -+g*- -+(9.8 mls2x0.800s)2 - -3. 13m. The distanceabove the water surface is 10.0m - 3.13 m - 6.86m.(c) The driver strikes the water when A: -10.0m. The time he strikes is- 1.43 sand the horizontal distance from the platform is r:1)0rt - (2.00 mlsxl .43 s): 2.86m.31(a) Since the projectile is released its initial velocity is the same as the velocity of the plane atthe time of release. Take the A axis to be upward and the r axis to be horizontal. Place theorigin at the point of release and take the time to be zero at release. Let r and y (: -730 m) bethe coordinates of the point on the ground where the projectile hits and let t be the time when ithits. Then 1.A : -uot cos 96 1gt' ,,where 0o - 53.0o. This equation gives

-730 m + |(g.go mls2x5.oo s)2 :202m/s.(5.00 s) cos(53.0o)(b) The horizontal distance traveled is r: uotsin0s - (202*1sX5.00 s) sin(53.0") : 806m.

y + *gt,ug: -T-'/ f cos 0g

2(- 10.0 m)

Chapter 4 15


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(c) and (d) The n component of the velocity is't)r : 't)0 sin 06 - (202^ls) sin(53.0o) : 161 m/s

and the A component isua: -uscos 06 - gt: -(202mls) cos(53o) - (9.80 mls2x5.00 s) : -17Imf s .

39Take the A axis to be upward and the r axis to the horizontal. Place the origin at the firingpoint, let the time be zero at firing, and let 0o be the firing angle. If the target is a distance d,away, then its coordinates are r- dandA:0. The kinematic equations are d:uotcos0s and0 - uotsin 0o - Lgt'. Elimrnate t and solve for de. The first equation gives t - dlurcosgs. Thisexpression is substituted into the second equation to obtain 2rB sin 0s cos 9s - gd,- 0. [Jse thetrigonometric identity sin 9s cos ds _ | sin(zlil to obtain ufr sin (200 - gd,or

sin(200): 4- Q'Smfl2)(45'] m) - 2.12x 10-3 .' ufi (460 mls)zThe firing angle is 0o: 0.0606o. If the gun is aimed at a point a distance (. above the target,then tan2o: (, 1d, or (, - dtan?s - (45.7 m)tan0.0606' - 0.0484m - 4.84cm.47You want to know how high the ball is from the ground when its horizontal distance from homeplate is 97.5 m. To calculate this quantity you need to know the components of the initial velocityof the ball. [Jse the range information. Put the origin at the point where the ball is hit, take the yaxis to be upward and the r axis to be horizontal . If r (: 107 m) and y (- 0) are the coordinatesof the ball when it lands, then tr : uyrt and 0 - uyat - *St', where t is the time of flight of theball. The second equation gives t - 2uoa I g and this is substituted into the first equation. Use1)0n - uya, which is true since the initial angle is 0o : 45o . The result is r : 2u\a I g. Thus

u,a - :22.9mf sNow take r and A to be the coordinates when the ball is at the fence. Again ra: uyat-Lgt'. The time to reach the fence is given by t - rluo* - (97.5m)lQ2,9mf s) : 4.26s.When this is substituted into the second equation the result is1.a: utat - ,gt' - (22.9^lsx4 .26s) - )e.8^ls2x4 .26s)2:8.63m.Since the ball started I.22m above the ground, it is 8.63 m + 1,.22m : 9.85 m above the groundwhen it gets to the fence and it is 9.85 m - 7 .32m - 2.53 m above the top of the fence. It goesover the fence.

gr2

(9.8 mls2xl07m)

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51Take the A axis to be upward and the r axis to be horizontal. Place the origin at the point wherethe ball is kicked, or the ground, and take the time to be zero at the instant it is kicked. r andA are the coordinates of ball at the goal post. You want to find the kicking angle 0s so thatA- 3.44m when rand Ainto the second the result is y: rtan0s gn22r'ocos2 gsYou may solve this by trial and effor: systematically try values of 0g until you find the two thatsatisfy the equation. A little manipulation, however, will give you an algebraic solutioll.Use the trigonometric identity 1/ cos' 0o: 1 + tan2 0s to obtain

1 ar2 . I -^^2tft tan2 os - ntanoo + Y + ;T- o'This is a quadratic equation for tanfls. To simplify writing the solutior, let c - ig*'lr| -ifq.B0 mls2x50*)t lQ5mls)z - 19.6m. Then the quadratic equation becomes ctanz 0sr tan?o + y + c : 0. It has the solution

tan 0o r*50m *

2(19.6 m)The two solutions are tan?sand 0son the goal post.s3Let h be the height of a step and u be the width. To hit step n, the ball must fall a distancenh and travel horizontally a distance between (n - l)u and nu. Take the origin of a coordinatesystem to be at the point where the ball leaves the top of the stairway. Take the U axis to bepositive in the upward direction and the r axis to be horizontal. The coordinates of the ball attime t are given by nthe level of step n:

The r coordinate then is

+ 4(A + c)c(50 m)2 - 4(3.44 m + 19.6 mX 19.6 m)

2n(0.203 m)n-uor (r.szmls) - (0.30e n)\fr,.Chapter 4 t7


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Try values of n until you find one for which r lu is less than n but greater than n 1. Fornnlw - 2. 15. This is also greater than rL. For n- 3, r_0.535m and rlu - 2.64. This is lessthan n and greater than n - 1. The ball hits the third step.67To calcul ate the centripetal acceleration of the stone you need to know its speed while it is beingwhirled around. This the same as its initial speed when it flies off. [Jse the kinematic equationsof projectile motion to find that speed. Take the A axis to be upward and the r axis to behorrzontal. Place the origin at the point where the stone leaves its circular orbit and take the timeto be zeto when this occurs. Then the coordinates of the stone when it is a projectile are givenby r- uot and A_ -+g*. It hits the ground when r- 10m and A- -2.0m. Note that theinitial velocity is horizontal. Solve the second equation for the time: t : {:di. Substitutethis expression into the first equation and solve for uoi

ug:r - (10m) 15.7 m/s .The magnitude of the centripetal acceleration is a,: u2 lr - (15.7*1il2 l(1 .5m): 160 ^/ ,'.73(a) Take the positive r direction to be to the east and the positive A direction to be to the north.The velocity of ship A is given by

dt : _rr")LT] I 1""|;;:li

: -t? knots) sin 4s")r i + r (z'knots) cos 4s " r iand the velocity of ship B is given by

dn : _ll; ;LT:] I _rrllnff : -tQ8 knots) sin 40'r i - tQ8 knots) cos 40"r iThe velocity of ship A relative to ship B is

6tn:6A-6s- (1 .0 knots) i + 1f 8. knots) j .

The magnitude isuln: u2tnr*'Io, (1 .0 knots)z + (38.4 knots)2 - 38.4 knots

(b)The angle 0 that 6te makes with the positive r axis is0 - tan-1 UAB AUAB r

This direction is 1.5o east of north., 38.4 knots- tan-I -. - 88.5o1.0 knots

_92y 2(-2.0m)

18 Chapter 4


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(c) The time t for the separation to become d is given bymile, t - (l60nautical miles) lQS. knots): 4.2h.(d) Ship B will be 1.5o west of south, relative to ship A.t d lr tu. Since a knot is a nautical

75Relative to the cur the velocity of the snowflakes has a vertical component of 8.0 m/s anda horizontal component of 50 km/htan 0 - unlu, - (13.9^ls)/(8.0 mls): I.74. The angle is 60o.77Since the raindrops fall vertic aIIy relative to the train, the honzontal component of the velocity ofa raindrop LS u1" - 30^ls, the same as the speed of the train. If u, is the vertical component of thevelocity and 0 is the angle between the direction of motion and the vertical, then tan 0 - uh/ur.Thus 't)u:'uhf tan? - (30 mls)ltan70" - 10.9mf s. The speed of a raindrop is u - G*,?:91(a) Take the positive A axis to be downward andcoordinate of the bullet is given by A - i7t'. Ifbullet hits below the target, then

place the origin at the firing point. Then the At is the time of flight and A is the distance the

(b) The mvzzle velocity isdistance to the target, then

: 6.3 x 10-2 sthe initial velocity of the bullet. It is hortzontal. If r is the honzontalr : uOt and

Ug -y- 30m -t 6.3 x 10-2 s 4.8 x 102 mlst07(a) Use A : uyat - *gt' and 'na - 't)ga - gt, where the origin is at the point where the ball is hit,the positive y direction is upward, and uya is the vertical component of the initial velocity. Atthe highest point ua:0, so uya - gt and a: *gt': ifg.8^ls2x3.0r)t - 44m.(b) Set the time to zero when the ball is at its highest point. The vertical component of the initialvelocity is then zero and the ball's initral y coordinate ts 44m. The ball reaches the fence at timet- 2.5 s. Then its height above the ground is a: Uo_ Lst' :44m- itg.8^ls\(2.5 s)t(c) Since the ball takes 5.5 s to travel the horizontal distance of 97 .5 m to the fence, the horizontalcomponent of the initial velocity is 1)0r: Q7.5m)l(5.5 s) - 17.7 mf s. Since the ball took 3.0 sto rise from the ground to its highest point it must take the same time, 3.0s to fall from thehighest point to the ground. Thus it hits the ground 0.50 s after clearing the fence. The pointwhere it hits is (17 .7 ^1sX0.50 s) : 8.9 m.

(30 mls)2 + (10.9m1s)z - 32mf s.

2(0.019 m)

Chapter 4 T9


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111(a) The position vector of the particle is given by i : 6ot + |dtz,t - 0 and d is the acceleration. The r component of this equation

(15 .2^ls)2 + (15 .6mls)2 : 22m/s.

where ds is the velocity at timeis r : ?Jont* +e*tT. Sinc a uo*

0 this becomes r - *.o*t' . The solution for t is t- - \l zesm)l(4.0 ^ls')The a coordinate then is A : uyat + *ort' - (8 .0 ^1sX3.81 s) + *tz.o^1s2x3.81 s)2 : 45 m.(b) The r component of the velocity is 'ur: rr0* + e*t - (4.0 mls2x3.81s) - 15.2mls andthe a component is rra: uya + Lort: 8.0 mls * (2.0m1s2x3.81 s) - 15.6mf s. The speed isl-j?r: lr'"*r; --12t(a) and (b) Take the r axis to be from west to east and the A axis to be from south to north.Sum the two displacements from A to the resting place. The first is Lfij sin 37o) : (60 km) i+1+S km) j and the second is Liz- -(65 k*)i. The sum is Ar - (60 km) i-(20k*)i. The magnitude of the total displacement is Lrthe tangent of the angle it makes with the east is tan? - (-20 km)/(60 km) - -0 .33. The angleis 18" south of east.(c) and (d) The total time for the trip and rest is 50h+ 35h+ 5.0h:90h, so the magnitude ofthe ayeruge velocity is (63 km) lQ0 h) - 0.70 kmlh. The average velocity is in the same directionas the displacement, l8o south of east.(e) The average speed is the distance traveled divided by the elapsed time. The distance is75 km + 65 km - 140 km, so the average speed is (140 k*) lQ0 h) : 1.5 km/h.(0 and (g) The camel has I20h 90 h - 30 h to get from the resting place to B. If Lin isthe displacement of B from A and A4.r, is the displacement of the resting place from A, thedisplacement of the camel during this time is Lr'B- Lr-rest: (90km)i-(60km)i- (-20k*)j:(30k-)i+1zok-)i.Themagnitudeofthedisp1acementis-36kmandthe magnitude of the average velocity must be (36 km) lQO h) _ I.zk*/h. The angle 0 that theaverage velocity must make with the east is given by tan Qangle is 34" north of east.

2rfa*

20 Chapter 4


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Chapter 5-5Label the two forces Ft and Fr. According to Newton's second law, F, + Fr: md,F, - rnd, - Fr. In unit vector notation Ft - (20.0 N) i and

d - -(1 2^lrt;1ritr 30o) i - (I2mlr';1.os 30o)j - -6.0*lrt) i - (10. a^lr') jThus

- (z.0ks)(-6.o^lrt)i +(2.0ks)(-1l. mlrt)i - e0.0N)i - (-32N)i - (2rr.Di(b) and (c) The magnitude of F2 is F2: @- (-32N)2 + (-21 X;z - 3gN. Theangle that F2makes with the positive r axis is given by tan? - FzalFr* - (21 N) lQ2N) - 0.656.The angle is either 33" or 33o + 180' - 213o. Since both the r and A components are negativethe coffect result is 2I3". You could also take the angle to be 180o - 213o - -I47o.13In all three cases the scale is not accelerating, which means that the two cords exert forces ofequal magnitude on it. The scale reads the magnitude of either of these forces. In each casethe magnitude of the tension force of the cord attached to the salami must be the same as themagnitude of the weight of the salami. You know this because the salami is not accelerating.Thus the scale reading is ffig,where mtsthe mass of the salami. Its value is (11.0kgX9 .8*lr2) -108N.t9(a) The free-body diagram is shown in Fig. 5 - 16 of the text. Since the acceleration of theblock is zero, the components of the Newton's second law equation yield T - mg sin 0:0 andl7/r,' mg cos 0(8.5 kexg .8^lrtl rin 30" : 42 N.(b) Solve the second equation for ,F^/: Fry - mgcos 0 - (8.5kg)(9.8^lrt;ros30o :72N.(c) When the string is cut it no longer exerts a force on the block and the block accelerates . The rcomponent of the second law becomes -mgsin e - me,, so a,: - gsin 0 - -(9.8^lr';ritt30o-4.9mf s2. The negative sign indicates the acceleration is down the plane.25According to Newton's second law F - ma, where F is the magnitude of the force, a" is themagnitude of the acceleration, and m is the mass. The acceleration can be found using theequations for constant-acceleration motion. Solve u : lJ1 * at for a: e, : u lt The final velocity

Fz

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is u - (1600 korlhx1000m/km)1p600 s/h) : 444mls, so o :the magnitude of the force is Ir - (500 kgQaT ^lst) - I .2 x29The acceleration of the electron is vertical and for all practical puqposes the only force actingon it is the electric force. The force of gravity is much smaller. Take the fr axis to be in thedirection of the initial velocity and the A axis to be in the direction of the electrical force. Placethe origin at the initial position of the electron. Since the force and acceleration are constanttheappropriateequations arer_ uot and a- Lot': ifgl*)*, where F- ma, wasusedtosubstitute for the acceleration a. The time taken by the electron to travel a distance n (: 30 mm)horizontally is t - r luo and its deflection in the direction of the force is

(444mls)/(1.8 s) : 247 ^lr' and105 N.

1Fa: ;-/.m

(b) The time ist

4.5 x 1 30x9.11 x 10-3t kg 1.2 x 107 m/s

,a2

)10-3 m-t6 y(#)':;( )(

35The free-body diagram is shown at the right. F* is the normal forceof the plane on the block and mj ts the force of gravity on the block.Take the positive r axis to be down the plane, in the direction ofthe acceleration, and the positive A axis to be in the direction of thenormal force. The r component of Newton's second law is thenmg sin 0 - ma) so the acceleration is a : g stn 0 .(a) Place the origin at the bottom of the plane. The equations formotion alongthe r axis are r: uot+)atz and 'u: n0*at. The blockstops when 'u : 0.According to the second equation, this is at the time t - -us f a. The coordinate when it stops is

fr : 'tJ.( -u0A't_Ltn

'

3) *)"(+)2--1 62 a, 2 gstn1(-3.s0 mls)22 | tq .B^lrt; ritt 32.0" ] : -t.t8m

Ug Ug -3.50 m/s(c) Now set r : 0 and solve tr : uot + \atz for

g srn? (9.g mlrt;,in 32.0"t. The result is2(-3.50 m/s)

0.674s.

- 1.35 s. 2uo 2uot::: a g stn? (9.8 mlrt; ritt 32.0oThe velocity is

u - uol_ at: uo* gtsrnl - -3.50 mls + (9.8 mls2xl.35 s)sin 32o - 3.50 mls,22 Chapter 5


31/274

45The free-body dtagrams for the links are drawn below. The force affrows are not to scale.

4onr{ F:onz{ fLon:{ fton+{ FA'll-\rrrJl-\-'rr-rlllltrlt ltl ltl 'o' IorImsll F3 on+ msll F+ on sgy mgililr' on'Link I Link 2 Link 3 Link 4 Link 5

(a) The links are numbered from bottom to top. The forces on the bottom link are the force ofgravity md, downward, and the force F, onr of hnk2, upward. Take the positive direction to beupward. Then Newton's second law for this link is Fz on 1 - mg : ma. ThusFz onl : m(a+ g) - (0.100kgX2.50 mlr'+ 9.8 mlr') - I.23N.(b) The forces on the second link arc the force of gravity md, downward, the force Fr on zof tink I, downward, and the force Fz onz of link 3, upward. According to Newton's thirdlaw Ft onz has the same magnitude as Fzonr. Newton's second law for the second link isF3 onz - fl on2 - mg : mQ,, SO

as expected since there is no friction. The velocity is down the plane.

Fz onz- m(&+ g)+ Fr onz- (0.100kg)(2.50 mlr'+9.8 mls2)+ 1.23N - 2.46N,where Newton's third law was used to substitute the value of F2on I for Ft onz.(c) The forces on the third link are the force of gravity mj, downward, the force Fz on z of link2, downward, and the force F+ on z of link 4, upward. Newton's second law for this link isF+onZ- F2on3 -mg:TnA, SO

F+on3 - m(a+ g)+ Fzon3 - (0.100NX2.50 mlr'+9.8 mls2)+ 2.46N - 3.69N,where Newton's third law was used to substitute the value of F3 on2 for F2 on3.(d) The forces on the fourth link are the force of gravity mj, downward, the force Ft on + oflink 3, downward, and the force Fs on + of link 5, upward. Newton's second law for this link isFS on + - F3 on 4 - mg - mA, SO

Fs on+ - m(a+ g)+ Ft on4 - (0.100kgX2.50 mlr'+ 9.8 mls2) + 3.6gN - 4.92N,where Newton's third law was used to substitute the value of Faon 3 for F3 on4.(e) The forces on the top link arc the force of gravrty mrt, downward, the force Fq on s oflink 4, downward, and the applied force F, upward. Newton's second law for the top link isF - F+ on S - mg - nLa, SO

F -m(a+ g)+ Fqon5 - (0.100k9(2.50 mlrt+9.8 mls2)+ 4.92N- 6.15N,where Newton's third law as used to substitute the value of Fs on 4 for F+ on s.

Chapter 5 23


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(0 Each link has the same mass and the same acceleration, so the same net force acts on eachof them: Fnet - ma - (0.100 kg(z.50 mls2) - 0.25 N.s3(a) The free-body dragrams are shown to the right. F is the appliedforce and f is the force of block 1 on block 2. Note that F isapplie,C only to block 1 and that block 2 exerts the force - i onblock 1. Newton's third law has thereby been taken into account.Newton's second law for block 1 is F f - TrL1a, where o, isthe acceleration. The second law for block 2 is f -- TrL2e,. Sincethe blocks move together they have the same acceleration and thesame symbol is used in both equations. Use the second equationto obtain an expression for a: a - f l*r. Substitute into the firstequation to get F - f : TTLIf l*r. Solve for f :

n Fmz Q.2NXl.2kg)f - - m1 * TTL2 z.3kg + r.zk|- 1' 1 N 'If F is applied to blo ck2 instead of block 1, the force of contact

llrr(b)is t _ Fml _ Q.2NX2.3 kg) .ti-

---.1N.

a' TrLl I mz 2.3 kg + l.2kg(c) The acceleration of the blocks is the same in the two cases. Since the contact force f isthe only horizontal force on one of the blocks it must be just right to give that block the sameacceleration as the block to which F is applied. In the second case the contact force acceleratesa more massive block than in the first, so it must be larger.57(a) Take the positive direction to be upward for both the monkey and the package. Suppose themonkey pulls downward on the rope with a force of magnitude F. According to Newton's thirdlaw, the rope pulls upward on the monkey with a force of the same magnitude, so Newton'ssecond law for the monkey is F - mrng - mrnarn, where mrn is the mass of the monkey ande,rn is its acceleration. Since the rope is massless F is the tension in the rope. The rope pullsupward on the package with a force of magnitude F, so Newton's second law for the packageis F + Flr - mpg : mpap, where mp is the mass of the package, a,p is its acceleration, and Fxis the normal force of the ground on it.Now suppose F is the minimum force required to lift the package. Then F'lr - 0 and e,,p - 0.According to the second law equation for the package, this means F - mpg. Substitute wpg forF in the second law equation for the monkey, then solve for am. You should obtain

F - rrlrng (mo - m,n)g (15 kg - 10kg)(9.8 ^lr\ An - _ r 2aTn : ,rr-, : *-, lo kg - +'Y mf s '24 Chapter 5


33/274

(b) Newton's second law equations are F -mpg: ffipap for the package and F -rrLrng : TrL,n,arnfor the monkey. If the acceleration of the package is downward, then the acceleration of themonkey is upward, so o??" - -ap. Solve the first equation for -F: p - mp(g + op) : ffip(g - arn).Substitute the result into the second equation and solve for arni

anL- (TY - *'')gmp*Tftm. 15kg+10kg /(c) The result is positive, indicating that the acceleration of the monkey is upward.(d) Solve the second law equation for the package to obtain

p - Trl,p(g - a,nr): (15kgX9.8mf s2 -2.0*ls2) - 120N61The forces on the balloon arc the force of gravity mj, down , and the force of the atr Fo, up.Take the positive direction to be up. When the mass is M (before the ballast is thrown out)the acceleration is downward and Newton's second law is Fo M g - - M a. After the ballastis thrown out the mass is lVI wL, where m is the mass of the ballast, and the accelerationis upward. Newton's second law is Fo (M m)g - (M m)a. The first equation givesFo: M(g a) and the second gives M(g a) (M m)g - (M m)a. Solve for m:m:ZMal(g + a).73Take the r axis to be horizontal and positive in the direction that the crate slides. ThenFcosd facceleration (the only nonvanishing component). In part (a) the acceleration is

e,r : F cos? - f - (450N)cos38" - 125N n n A -^^ 1^2m 310kg -u't+m/s 'In part (b) m: Wlg _- (3 10N) lQ.8^ls') - 31.6kg andFcos0-f (450N)cos38o 125N ,1 ..r r2nr m 36.1 kg79Let F be the maglitude of the force, a4 (: L2.0mls2) be the acceleration of object I, and a,2(:3.30 mlt'; be the acceleration of obj ect2. According to Newton's second law the masses arerrLl: Flol and TrL2- Ff e,2.(a) The acceleration of an object of mass rTL2 - mr is

(L- F -TrL2 - Tft1 - or u, I2.0 - +'O m/ S 'Chapter 5 25


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(b) The acceleration of an object of mass TtLl 1 m2 isF F eta,zA'- _Tft2 * mr (F lor) + (F lo) a1 a aZ (12.0 ^ls2x3.3o *lt') - 2.6mf s212.0 mlr' + 3.3 0 mf s2

9l(a) Both pieces arc station dry, so you know that the net force on each of them is zero. The forceson the bottom piece are the downward force of gravity, with magnitude Tftzg, and the upwardtension force of the bottom cord, with magnitude T6. Since the net force is zero,

T6: TTL27 : (4.5kgX9 .8^lr2) - 44N.(b) The forces on the top piece are the downward force of gravity, with magnitude r(Ltg, thedownward tension force of the bottom cord, with magnitude Ta, and the upward force tension ofthe top cord, with magnitude ?7. Since the net force is zero,

T1 : Tb t mrg :44 N + (3.5 kg)(9.8 mlst) - 78 N .(c) The forces on the bottom piece are the downward force of gravity, with magnitude rrLsg, andthe upward tension force of the middle cord, with magnitude Trn Since the net force is zeto,

T,n : TTLsg : (5.5 kgXg .8 ^ls') - 54 N .(d) The forces on the top piece are the downward force of gravity, with magnitude w4g, theupward tension force of the top cord, with magnitude T7 (: I99 N), and the downward tensionforce of the middle cord, with magnitude Trn Since the net force is zero)T,n: Tt - Tft3g : IggN - (4.8 kg)(9 .8^ls2) - 152N.

9s(a) According to Newton's second law the magnitude of the net force on the rider is It - ma, :(60.0kg)(3.0 mls') - 1.80 x 102N.(9) Take the t t force to__b. thl vector sum of the force of the motorcycle and the force of Earth:Fnet: Fr, + Fn. Thus Frn: 4r.t - Fs. Now the net force is parallel to the ramp and thereforemakes the angle 0 (: 10') with the horrzontal, so Fnt_ (F cos 0)i+(lrsin 0)j, where the r axisis taken to be horrzonta| and,the A axLS is taken to be vefitcal. The force of Earth is F" - -mgj,so F,- - (F cos il?+ (F sin o + milj.Thus

F,n* - It cos 0 - (1.80 x I02 N) cos 10oand

Frna - (1.80 x 102) sin 10o + (60.0kg)(g.g mls2) - 6.19 x 102 N.The magnitude of the force of the motorcycle is

TA-r rnChapter 5

t02 N)2 r02 N)226

-6.44 x 102N.


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99The free-body diagrams for the two boxes are shown below.

FlrrFxz

Here T is the tension in the cord, FxTr is the norrnal force of the left incline on box 1, and FNzis the norrnal force of the right incline on box 2. Different coordinate system are used for thetwo boxes but the positive r direction are chosen so that the accelerations of the boxes have thesame sign. The r component of Newton's second law for box 1 gives T - TTLI7 stn01 : TTL1& andthe r component of the law for box 2 gives mzg stn 02 - T : TTL27. These equations are solvedsimultaneously for T. The result is

T : mtmzg (sin 91 * sin oz)?TL1 * mZ \ ^ (3.0 kgx2.0 kgX9.8 m/s2) (sin 30o + sin 60.) : 16 N3.0 kg + 2.0kg101Free-body diagrams for the two tins are shown onthe right. T is the tension in the cord and Fy isthe normal force of the incline on tin 1. The posi-tive n direction for tin I is chosen to be down theincline and the positive r direction for tin 2 is cho-sen to be downward. The sign of the accelerations ofthe two tins arc both then positive. Newton's secondlaw for tin 1 gives T + Tft1g stn {3 - ma, and for tin 2gives mzg T F - TrL2cL. The second equation issolvedforT,withtheresultT- TrL2@_ a)-F-(2.0kgX9.8^lrt-5.5m1s2)-6.0N-2.6N.The first Newton's law equation is solved for sin p, with the result

sin C:mra-Tmts (1.0kgX9.8 m/s)The angle is I7o

T

\r

Chapter 5 27


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Chapter 5

1(a) The free-body diagram for the bureau is shown on the right.F is the applied force, "f- i, the force of friction, fi" is the normalforce of the flooq and mj is the force of gravity. Take the r axisto be honzontal and the A axis to be vertical. Assume the bureaudoes not move and write the Newton's second law equations. Ther componentis F- f :0 andthe A componentis Fl/ -rng:0.The force of friction is then equal in magnitude to the appliedforce: fforce of gravity: ,Fl/ - mg. As F increases, f increases untilf : ltrrFx. Then the bureau starts to move. The minimum forcethat must be applied to start the bureau moving is

It - FrFx : ltr,mg - (0 .45)(45 kgX9.8 mls2) - 2.0 x I02 N .(b) The equation for F is the same but the mass is now 45 kg - 1,7 kg - 28 kg. Thus

F - F,mg - (0 .45)(28 kgX9.8 mlst) - 1.2 x T02 N .

(a) The free-body dragram for the crate is shown on the right. F isthe force of the person on the crate,,f-ir the force of friction, Fr itthe noffnal force of the floor, and mj ts the force of gravity. Themagnitude of the force of friction is given by f - l.rrcFw, whereltr* is the coefficient of kinetic friction. The vertical componentof Newton's second law is used to find the normal force. Sincethe vertical component of the acceleration is zero, F'l/ - mg - 0and Fl/ - mg. Thus

f : F;Fw : ltkmg - (0.35X55 kgX9.8 mls2) - l.g x I02 N .(b) IJse the horizontal component of Newton's second law to find the acceleration. SinceF-f:ffia,

3

(F - f) (220N - 18eN)a: - -m 55kg28 Chapter 6

0.5 6mf s2


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13(a) The free-body diagram for the crate is shown on the right. f isthe tension force of the rope on the crate, Fn is the norrnal force ofthe floor on the crate, rnfi rs the force of gravity, and ,i ir the forceof friction. Take the r axis to be horizontal on the right and the Aaxis to be vertically upward. Assume the crate is motionless. The rcomponent of Newton's second law is then 7 cos e - f : 0 and the Acomponent is Tsin 0+F^/ *mg:0, where 0 (: 15") is the angle be-tween the rope and the horizontal. The first equation gives / - T cos Iand the second gives Fx - mg - T sin 9. If the crate is to remain atrest, / must be less than lr"Fx, or T cos?


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of Wt and Wn, then select Wt so f : F"Fx. The second Newton's law equation givesFyr - Ws immediately. The third gives Tz - Tr I cos 9. Substitute this expression into thefourth equation to obtain T1to obtain f - Waf tan?. For the blocks to remain stationary f must be less than F"Fx orWaltan? for the cube to slide but not tip as P increases Fsrng must be less than mgLl2l. ot Lr, must beless than Ll2(, - (8.0 cm) 12(7.0 cm) - 0.57.(b) The cube tips before it slides if P(. >greater than mgLl2(. ot Lr, must be greater than Llz(. - 0.57.85The force diagram for the ladder is shown on the right. Fs is the forceof the ground on the ladder, F* is the force of the wall, and W is theweight of the ladder. The horizontal component of Newton's second lawgives F + Fn* - F*- 0 and the vertical component gives Fno W - 0.Set the net torque about the point where the ladder touches the wall equal Fn'to zero. The honzontal component of the force of the ground has a leverarrn of h, the vertical component of the force of the ground has a leveraffn of (., where (. is the distance from the foot of the ladder to the wall,the applied force has a lever arrn of (1 dlL)h, where L is the length of the ladder, and thegravitational force has a lever ann of (. 12. Thus Fn*h - Fnrl+ Fh(l - d,lL) + WtlT - 0. Ther axis was taken to be horizontal with the positive r direction to the right and the A axis wastaken to be vertical with the positive direction upward.The vertical component of the second law gives F ga - W : 200 N. The torque equation gives

Fn*Now (.d,lL -Dtgn

_ J(1om)2 - (8.0m)2 :1 (2.0m)110m)-0.80,150N-(0.80)F-75N:6.0m, so ((. lh)Fno - (6.0 m)l(8.0m)(200N) : 150N, Iand (dlzh)W_ (6.0m)(200N/2(8"0m) - 75N. This means75N-(0.80),F.

(a) If .F' - 50N, Fn*:75N - (0.80)(50N) - 35N and Fn - (35N)i+1ZO0Ni.(b) If F:150N, Fn*:75N- (0.80)(150N;- -45N and Fn - (-45N)i+(200Dj.(c) When the ladder is on the verge of slipping the frictional force is to the left, in the negativer direction. Its magnitude is (0.80)F 75N. If the ladder does not slip this must be lessthan prFga: (0.38X200N - 76N. The applied force must be less than (75N+ 76N) 1Q.80)1.9 x 102 N. Thus the applied force that will just start the ladder moving is I.9 x 102 NI.

76 Chapter 12


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Chapter L3

1The magnitudeand Tfi,2 are thefor r:

of the forcemasses, T is

of one particle on the other is given by It - Gmflt Lz I ,', where TrLltheir separation, and G is the universal gravitational constant. Solve^ 1 rc.67 x 1 0- 1r N . m2 lke\(s .2ke)e.4 kg)

7At the point where the forces balance GM"*lr?- GMr*|r3, where M" is the mass of Earth,M, is the mass of the Sun, m is the mass of the space probe, rr is the distance from the centerof Earth to the probe, and 12 is the distance from the center of the Sun to the probe. Substitute12 - d - rt, where d is the distance from the center of Earth to the center of the Sun, to findM"-

,2,M"

(d - ,t)'Take the positive square root of both sides, then solve for 11 . A little algebra yields

\M+tM I .99 x 1030 kg + 5.98 x 1024 kgC.alues for A[., Mr, and d can be found in Appendix

17

_ d\M - (lso x toem) -1

The gravitational acceleration is given by as: GMf r2, whereis the distance from Earth's center. Substitute r- R+ h, whereis the altitude, to obtain as: GMI(ft+ h)'. Solve for h. YouAccording to Appendix C of the text, R- 6.37 x 106m and M

2.60 x 108 m.

M is the massR is the radiusshould get h-

of Earth and rof Earth and h@-R.

is its mass and R is

h:- 5.98 x 1024 kg, so

-6.37 x 106m- 2.6 x 106m.29(a)its

The density of aradius. The ratio uniform sphere is givenof the density of Mars toPnrpn Mp Rtu

by p : 3M l4nR3, where Mthe density of Earth is3 - 0.74.iX55.6A0j kmk*g+0,

GmtffLzF

(6.67 x 10-ll m3 lr' . kgX5.98 x L024 kg)

Chapter 13 77


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(b) The value of an at the surface of a planet is given by ag: GMIR', so the value for Mars isMxa R1 2 (9.8 mlr2) - 3.8 ^lt'lVL r t'F.A7I/I- ,r- t' a IM

(c) If u is the escape speed, then, for a particle of mass mWHagE:0'11 kmk*

Gryg+0,rX55.6701

**r':and

For Mars, Fu.67 x 10- F ,1u-V :5.0x10'm/s.

37(a) Use the principle of conservation of energy. Initially the particle is at the surface of theasteroid and has potentral energy [-Li: -GMmlR, where M is the mass of the asteroid, R isits radius, and m is the mass of the particle being fired upward. The initial kinetic energy isi*r'. The parttcle just escapes if its kinetic energy is zeto when it is infinitely far from theasteroid. The final potential and kinetic energies are both zero. Consenration of energy yields-GMmlR+ i*r':0. Replace GMIR with agR, where a,e is the gravitational accelerationat the surface. Then the energy equation becomes -a,sB+ *r'- 0. Solve for u:

x 103m):1.7 x 103m1 s(b) Initially the particle is at the surface; the potential energy is [Ii: -GMrnlR and the kineticenergy is Kicomes to rest. The final potential energy is tJy: -GMml(R+ h) and the final kinetic energyis K,; - 0. Conservation of energy yieldsGMm I-;+;*,2:Replace GM with onR2 and cancel m in the energy

-asR+)*:

GMmR+hequation to obtainonR2(R+ h)

The solution for h ish- 2anR22onR - u2 -R- z(g.o m/s2)(soo x to3 m)22(3.0^lrt;1soo x 103 m) - (1000 mls)2:2.5 x 105 m.

2anR 2(3.0 m/s';1soo

78 Chapter 13

(500 x 103 m)


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(c) Initially the particle is aU, - -GMml@+ h) andthe potentral energy is U yof energy yields

The solution for u isu-

distance h above the surface andthe initial kinetic energy is K,; --GMmlR.Write *.*r'r for the

GMm GMm

is at rest. The potenttal energy is0. Just before it hits the asteroidfinal kinetic energy. Conservation1.+ - mL)"2+h RReplace GM with onR2 and cancel m to obtain

onR' esR*lr'

-- 1..4 x 103 m/s .39(a) The momentum of the two-star system is conserved, and since the stars have the same mass,their speeds and kinetic energies are the same. use the principle of conservation of energy. Theinitial potential energy is Ut,- -GM'lrn where M is the mass of either star and ri is theirinitial center-to-center separation. The initial kinetic energy is zero since the stars are at rest.The final potential energy is U1: -zGM'lrn since the final separation is ,n12. Write Muz forthe final kinetic energy of the system. This is the sum of two terms, each of which is *M r' .Consenration of energy yields

_GMz __ _2GIVI2 + Muz

R+h

Ti T6The solution for u is

- 8.2 x 104 m/s .(b) Now the final separation of the centers is ry - 2R: 2 x 105 m, where R is the radiusof either of the stars. The final potentral energy is given by LI 1 - -GM'lrf and the energyequation becomes -GM'lrn: -GM'lrt + Mu2. The solution for u is

tlrt- /,u.67 x 10-1rm3 lr' 'keXl030ke) (^iffi mt*)-1.8 x 107m/s.

2anR2

2(3.0^lrt;1soo x 103 m) - 2(3.0^lr';1soo x 103 m)2500 x 103m+ 1000 x 103m

(6.67 x 10-rr m3 lr'.kgX103o kg)10lo m

Chapter 13 79


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45Let l/ be the number of stars in the galaxy, M be the mass of the Sun, and r be the radius of thegalaxy. The total mass in the galaxy is IY M and the magnitude of the gravitational force actingon the Sun is F - G M' lr'. The force points toward the galactrc center. The magnitude ofthe Sun's acceleration is a: 12 lR, where u is its speed. If T is the period of the Sun's motionaround the galacttc center then u- 2nRlT and a, -- 4n2RlT'. Newton's second law yieldsGIY M'I R' : 4r2 M RlT2. The solution for tf is

4r2 R3iv- GTzMThe period is 2.5 x 10t y, which is 7.88 x 101t r, so

rv - 4n2(2'2 x 7020 m)3 : 5.1 x 1010(6.67 )(7.88 x I01t r)t (2.0 x 1030 kg)4t(a) The greatest distance between the satellite and E,arth's center (the apogee distance) is Ro:6.37 x 106m+360 x 103 m - 6.73 x 106m. The least distance (perigee distance) is Rp - 6.37 x106m+180x 103m - 6.55 x 106m. Here 6.37 x 106m is the radius of Earth. Look at Fig. 13-13to see thatthe semimajoraxis is o : (Ro+R|)f 2 - (6.73x 106m+6.55x 106 lrri;12 - 6.64x 106m.(b) The apogee and perigee distances ate related to the eccentricity e by Ro - o( I + e) andRpRo - Rp - 2ae. Thus

Ro- Rp Ro- Rp 6,73x 106m- 6.55 x 106me: -0.0136.2a Ro * Rn 6.73 x 106 m * 6.55 x 106 m6t(a) IJse the law of periods: T2 : (4n'IGM)r3, where M is the mass of the Sun (1.99 x 1030kg)and r is the radius of the orbit. The radius of the orbit is twice the radius of Earth's orbit:r - 2r. - 2(1 50 x 10em):300 x 10em. Thus

T_

Divide by (365 dlfiQ{hldX60 minlhx60 s/min) to obtain T - 2.8y.(b) The kinetic energy of any asteroid or planet in a circular orbit of radius r is given byK - G M m l2r, where m is the mass of the asteroid or planet. Notice that it is proportional tom and inversely proportional to r. The ratio of the kinetic energy of the asteroid to the kinetic80 Chapter 13

4r2134n2(300 x 10e m)3

(6.67 x 10- 11 m3 I ,' . kgX r .gg x 1030 kg)


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energy of Earth is K I K. - (m ln'Le)(r.l i. SubstituteKIK.7s(a) Kepler's law of periods gives

m - 2.0 x lT-am" and r :2r" to obtain

-2.15 x 104s.(b) The craft goes a distance2nr tnaperiod, so its speed is us :2r(4.20x 107 m)l(2.L5x 104 s):1.23 x 104 mls.(c) The new speed is ,u:0.98u0 - (0.98XI.23 x 104 mls): l.2I x 104 mls.(d) The kinetic energy of the crafi is K - **r': *tz000kgXr.2\ x 104 mls)z -2.20 x 1011 J.(e) The gravitational potential energy of the planet-craft system is

u--GWr \_-_ t^_o/ 4.20X107mwhere the potential energy was taken to be zero when the craft is far from the planet.(0 The mechanical energy of the planet-craft system is E - K+U - 2.20x 1011 J -4.53 x 1911 1--2.33 x 10ll J.(g) The mechanical energy of a satellite is given by E : -GmM l2a, where a is the semimajoraxis. Thus

GmMo"- 2E

(6.67 x l0-11N . m2 lke\(3000kgX9.50 x t02s kg) - 4.08 x I07 m.2(-2.33 x 10rr J)(h) and (i) The new period is

-2.06 x 104s.Thechangeintheperiodis 2.06 x 104s- 2.I5 x 104s- -9 x 103s. Theperiodforthesecondorbit is smaller by 9 x 102(4.20 x 107 m)3 s.79Use 7;t - GmrTTlrnlr', where ms is the mass of the satellite, mrn is the mass of the meteor, andr is the distance between their centers. The distance between centers is r: R+d - 15m+3m -18 m. Here R is the radius of the satellite and d is the distance from its surface to the center ofthe meteor. Thus

7;1 _ 6.67 x 10-rr N .m2 lke')Q0kgX7.0 kg) ^b'- : Z.9 x l0-ll N.

4r213 4n2(4.20 x 107 m)3(6.67 x 10-11N .nP lke'X9.50 x l02s kg)

4r2 a3 4n2(4.08 x 107 m)3(6.67 x 10-11 N .m2 lkg'X9.50 x l02s kg)

Chapter 13 8l


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83(a) The centripetal acceleration of either star is given by o, - u2r, where w is the angular speedand r is the radius of the orbit. Since the distance between the stars is 2r the gravitational forceof one on the other is Gmz IQD', where m is the mass of either star. Newton's second lawgives Gmz IQD' : murLr. Thus

I ^Fu, x 1o-rrN .mlkgl3o. lorkgf-;l -2'2 x 10-7rcdfs'(b) As the meteoroid goes from the center of the two-star system to far away the kinetic energychanges by LK- -+mu2 and the potentral energy changes by Ltf - ZGmMlr, where M isthe mass of the meteoroid and u is its speed when it is at the center of the two-star system. Sinceenergy is conserved LK + A(J : 0 and

: 8.9 x 104 m/s.

Ia- 2

87(a) Since energy is conserved it is the(b) The potential energy at the closest

(Je: -GMnMs :-(6 .67 xT'p- -5 .40 x 1033 J

same throughout the motion and there is no variation.distance (perihelion) is1o-11 N .m2 lkg')(5.98 x lo24 kgX I .99 x 1030 kg)1.47 x 101r m

and at the furthest distance (aphelion) isIJo

raThe difference is 1 .8 x 1032 J.

(5.98 x T024 kgX | .99 x 1030 kg)L.52 x 10rr m

(c) Since energy is conserved the vanation in the kinetic energy must be the same as the variationin the potenttal energy, 1.8 x 1032 J.(d) The semimajor axis is a: (ro{ra)12 - (1.47 x 108km+1 .495 x 108 k^)12 - 1.50 x 108km.The kinetic energy at perihelion is 1l_l2")Now

sox 10ll m - 3 .46 x 10-12 m-1

Ke: GMnMs [;

Ke - (6.67 x- 2.74 xand the speed is u,p82 Chapter 13

1o-rr N .m2 lkg')(s.991033 J

x 1024 kgX r .gg x 1030 kgX3 .46 x l0- 12 m- I )

Gm,tTt

4(6.67 x 10-rr N .m2 lke\(3.0 x 1030kg)1.0 x 10ll m

2KIME 2(2.7 4 x 1033 J)l (5.98 x 1024 kg) - 3 .02 x 104 m/s.


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Since angular momentum is conserved uprp: nara and the speed at aphelion is 't)a: uprolro(3.A2 x 104 mlsxl .47 x 10rr rri;l .sz x 10rt *) - 2.93 x 104m/s. The varration is 3.02 x104 mls - 2.93 x 104 mls - 9.0 x I02 m/s.93Each star is a distance r from the central star and a distance 2r from the other orbiting star, soit is attracted toward the center of its orbit with a force of magnitude

rt-GM: *ry12 (2r)2 4r2 \r-rAccording to Newton's second law this must equal the product of the mass and centripetalacceleration ,2 I,. Each star travels a distance 2rr in a time equal to the period ?, so ?r :2rrf T,and the centripetal acceleration is 4n2 r lT2. Thus

The solution for ? is 4rr3 /2

Chapter 13 83


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Chapter 4

1The air inside pushes outward with a force given by piA, where Pi is the pressure inside theroom and A is the area of the window. Similarly, the air on the outside pushes inward witha force given by PoA, where Po is the pressure outside. The magnitude of the net force isF = Pi - Po)A. Since 1 atm = 1.013 x 105 Pa,

F = 1.0 atm - 0.96 atm) 1.013 x 105 Pa/atm) 3.4 m) 2.1 m) = 2.9 x 104 N .

JThe change in the pressure is the force applied by the nurse divided by the cross-sectional areaof the syringe: F F 42N 5D p = A = 7fR2 = 7f 1.1 X 1O-2m)2 = 1.1 x 10 Pa.

The pressure P at the depth d of the hatch cover is Po pgd, where p is the density of oceanwater and Po is atmospheric pressure. The downward force of the water on the hatch coveris Po pgd)A, where A is the area of the cover. If the air in the submarine is at atmospheric pressure then it exerts an upward force of PoA. The minimum force that must beapplied by the crew to open the cover has magnitude F = Po pgd)A - PoA = pgdA =l024kg/m3) 9.8 m/s2) 100m) 1.2 m) 0.60m) = 7.2 x 105 N.9

When the levels are the same the height of the liquid is h = (hI h2)/2, where hl and h2 arethe original heights. Suppose hI is greater than h2. The final situation can then be achieved bytaking liquid with volume A h l - h) and mass pA h l - h), in the first vessel, and lowering it adistance h - h2. The work done by the force of gravity is W = pA h l - h)g(h - h2) Substituteh = (hI h2)/2 to obtain

27

W = ~ p g A h l - h2)2= ~ 1 . 3 0 X 10 3 kg/m3) 9.8 m/s2) 4.00 x 10-4 m2) 1.56 m - 0.854 m)2= 0.635 J.

a) Use the expression for the variation of pressure with height in an incompressible fluid: P2 =PI - pg Y2 - YI). Take YI to be at the surface of Earth, where the pressure is PI = 1.01 X 105 Pa,8 Chapter 4


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and Uz to be at the top of the atmosphere, where the pressure ts pz: 0. Take the density to be1.3 kg/*'. Then,

Uz-At p9 (1.3 kel^'Xq.gr;7s) :7'9(b) Let h be the height of the atmosphere. Since the densitythe integral

Pz: Pt Ir^ ps da .where po is the density at Earth's surface. This expression predicts that0 atA-h. Assume g isuniformfrom A-0 to A-h. NowtheintegralPz : pr - lr^ Poe(r I) da : pr * pogh .

Since p2 : 0, this means2pr 2(1.01 x 105 Pa)h_Lyr: a\r.vr I r'-r r*/ = - 16 X 103m:16km.Pog (1.3 kgl^'Xq.8 m/s2) ^ v '

31(a) The anchor is completely submerged. It appears to be lighter than its actual weight becausethe water is pushing up on it with a buoyant force of p*gV , where p* is the density of waterand V is the volume of the anchor. Its effective weight (in water) is Weff: W - p*gV, whereW is its actual weight (the force of gravity). Thus

v -w -w"rc- 2o=oN

= - 2.045 x lo-2 m3 .P*9 (gg8 kg/m'Xq.8 m/s2)The density of water was obtained from Table I4-I of the text.(b) The mass of the anchor is m - pV, where p is the density of iron. Its weight in arr isW : mg: pgv - (7870kelm'Xg .8^ls\(2.045 x r0-2m3) - 1.58 x 103 N.35(a) Let V be the volume of the block. Then, the submerged volume is Vr:2Vf3. Accordingto Archimedes' principle the weight of the displaced water is equal to the weight of the block,so p-V, : pbV , where pu is the density of water, and pa is the density of the block. SubstituteV"- 2Vl3 to obtain pa- 2p*13- 2(gg8kg/mt)lZ: 6.7 x \}zkgl^t. The density of waterwas obtained from Table l4-l of the text.(b) If po is the density of the oil, then Archimedes' principle yields poV, - paV. SubstituteV, - 0.90V to obtain po: pbf 0.90: 7.4 x I02kgl^t.37(a) The force of gravity mg is balanced by the buoyant force of the liquid pgV": mgHere rra is the mass of the sphere, p is the density of the liquid, and V, is the submerged volume.

Chapter 14 85

x 103m- 7.gkm.varies with altitude, you must use

Take p: po(l -Alh),p:poatA-0 and p:can be evaluated:


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Thus msphere, or V"

4nllJ?, : Ol'-6 r'o (?) (8oo kel^'xo oeom)3 - r zkgAir in the hollow sphere, if ufry, has been neglected.(b) The density prn of the materral, assumed to be uniform, is given by p,* : m lV , where m isthe mass of the sphere and V is its volume. If ri is the inner radius, the volume is

-r?)- 4rT [(0.090 m)3 - (0.080 m)3] - 9.09 x 10-a m3-The density is m L.22kgn- :r v 9.09 x lo-a m3 1.3 x 103 kgl^t49Use the equation of continuity. Let u1 be the speed of the water in the hose and u2 be its speedas it leaves one of the holes . Let A1 be the cross-sectional area of the hose. If there are l/ holesyou may think of the water in the hose as l/ tubes of flow, each of which goes through a singlehole. The cross-sectional area of each tube of flow is At lN . If A2 is the area of a hole theequation of continuity becomes urAtlN: uzAz. Thus u2: (AtllVAr)ur - (R2 llYr')ut, whereR is the radius of the hose and r is the radius of a hole. Thusfi (0.9s cm)2u2: Tfur,s3Suppose that a mass A^m of water is pumped in time Lt The pump increases the potentialenergy of the water by A^mgh, where h is the vertical distance through which it is lifted,and increases its kinetic energy by i/r*u2, where u is its final speed. The work it does isLW : L*gh+ ).Lmuz and its power is

P - Lw - Lnz ( on* 1.,t) N Ar \'"'," )Now the rate of mass flow is L* lAf - pAu, where p is the density of water and A is theatea of the hose. The area of the hose is ApAuobtained from Table I4-I of the text. Thus/ 1rt)- pAu I gh+\" 2/l-.- (1.s 7 kgls) I (9 .B^1s2X3.0 m) * (5'0 m/s)2 IL /\J'\t LLL) ' 2 I:66W.86 Chapter I4


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--3(t) Use the equation of continuity: Apr : Azuz. Here A1 is ththe pipe at the top and u1 is the speed of the water there; A2 isof the pipe at the bottom and u2 is the speed of the water there.lg.0 cm2)/(8.0 cm2)] (5.0 mls) : z.Sm/s.(b) Use the Bernoulli equation: h + Lpu?+ pghr : pz+ Lpr'r* pghz,water, fu is its initial altitude, and hz is its final altitude. Thus

e cross-sectional arca ofthe cross-sectional areaThus u2where p is the density of

pz: p1 * )ofr? - ,l) + ps(hr - hz)2

+ (998 ke: 2.6 x 10s Pa.The density of water was obtained from Table I4-l59(a) Use the Bernoulli equation: h+ |pu?+ pgfu: pz+ lpu2z* pghz, where fu is the height ofthe water in the tank, h is the pressure there, and u1 is the speed of the water there; h2 is thealtitude of the hole, pz is the pressure there, and u2 is the speed of the water there. p is thedensity of water. The pressure at the top of the tank and at the hole is atmospheric, So pr : p2.Since the tank is large we may neglect the water speed at the top; it is much smaller than thespeed at the hole. The Bernoulli equation then becomes pghl : ipr? + pghz and

U2: 2(g.8*ls2xo.3o m) : 2.42m/s .The flow rate is Azuz- (6.5 x 10-4 m2)(2.42m1s): 1.6 x 10-'*'/r.(b) Use the equation of continuity: A2u2: Azut, where A3: Azl2 and rh is the water speedwhere the cross-sectional area of the stream is half its cross-sectional area at the hole. Thusu3 : (Az I At)uz : Zuz - 4.84 ml s. The water is in free fall and we wish to know how far it hasfallen when its speed is doubled to 4.84 m/s. Since the pressure is the same throughout the fall,Lputr + pghz

h2 h3: 1 : 4- (4.84mf s)2 - (2.42mf s)2 - 0.90m.2s 2e.8 m/s2)67(a) The continuity equation yields Au - aV and Bernoulli's equation yields ipr' - Lp + )pvz,where Lp: pz - pr. The first equation gives V- (Ala)u. Use this to substitute for V in thesecond equation. You should obtain *pr' : Lp + )p(Ala)2rr2 . Solve for u. The result is

2Lp

[(s. o mls)2 - (2.5 m/s)2]l^tx9 .8^ls2xlo m)of the text.

#)

2g(hr

Chapter 14 87


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(b) Substitute values to obtain't):V -3.06m/s.

The density of water was obtained from Table l4-l of the text. The flow rate is Au(64 x 10-4 m2X3.06mls):2.0 x I0-2m3/s.75Let p (: 998 kg/-') be the density of water and pt (: 800 kg/-') be the density of the otherliquid. Let d*r, be the length of the water column on the left side, d*n (: 10.9cm) be thelength of the water column on the right side, and ds (: 8.0 cm) be the length of the columnof the other liquid. The pressure at the bottom of the tube is given by po + pd,t * p*d*L,where po is atmospheric pressure, and by po * p*d- n These expressions must be equal, sopo * pdt, * p-d*r : p0 * p*d*n The solution for d*r is

d-r- P*d-n-Pt'dt': -3.5gcm.P* 998 kgl^tBefore the other liquid is poured into the tube the length of the water column on the left side isthe same as the water column on the right side, namely 10.0cm. After the liquid is poured it is3.59cm. The length decreases by 10cm - 3.59cm - 6.41 cm. The volume of water that flowsout of the right arrn is T(L 50 cm)2 (6.41 cm) : 45.3 cm3.

88 Chapter 14


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Chapter L5

3(a) The amplitude is half the range of the displacement, or trrn: 1 .0 rrlln.(b) The maximum speedu* is relatedto the amplitude trmby n*: u)frrn, where a is the angularfrequency. Since ur - 2nf ,where / is the frequency, 'trrn:2rf rrn:2r(l20Hz)(l .0x 1g-: m):0.7 5 m/s.(c) The maximum acceleration is arn5.7 x 102 ^/r2 .7(a) The motion repeats every 0.500 s so the period must be T * 0.500 s.(b) The frequency is the reciprocal of the period: f : IIT - IlQ.500s):2.00H2.(c) The angular frequency is a - 2nf :2r(2.00H2) - l2.6radf s.(d)Theangu1arfrequencyisre1atedtothespringconstantkandthemaSSmbya_\ffi,Sok : mw2 - (0.500 kgX 12.57 radls)' : 79.0 N/m.(e) If trtn is the amplitude, the maximum speed is 't)rn _ urfrrn - 02.57 rud1sx0.350 m) _4.40 m/s.(0 The maximum force is exerted when the displacement is a maximum and its magnitude isgiven by Frn2The magnitude of the maximum acceleration is given by arn: tt2trrn, where a is the angularfrequency and nnl is the amplifude. The angular frequency for which the maximum accelerationis9isgivenby,-\MandthecoffeSpondingfrequencyisgivenby

Aat-r2nFor frequencies greater than

9.8 ^lt' :498H2.acceleration exceeds g for some part of the motion.

I l-s2"y .-,498zHz the

t7The maximum force that can be exerted by the surface must be less than F'FN or else the blockwill not follow the surface in its motion. Here, F" is the coefficient of static friction and F71/is the normal force exerted by the surface on the block. Since the block does not acceleratevertic ally, you know that Flr : mg, where m is the mass of the block. If the block follows thetable and moves in simple harmonic motion, the magnitude of the maximum force exerted on itis given by tr - ma,m- mw2rTnacceleration ) u) is the angular frequency, and f is the frequency. The relationship a : 2n f wasused to obtain the last form.

Chapter 15 89


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Substitute It - m(2nf)'r,n and F^r - mg into F


99/274

29(a) and (b) The totalthe amplitude. When

energy is given by E1r : i*r- the potential where k is the spring constant and rrn isf-Jt - *tt*' - ttt*?". The ratio is

-3 4.2radf s .cycle its angular speed is

- itt*',-,energy isitt*',"-itt*',"

The fraction of the energy that is kinetic isE-U(c) Since E - *tr*h, and fJ: - *tt*', UIE: 12 l*',". Solve ,'l*?*- Llz for tr. You should get:L - :rrnlt/r.39(a) Take the angular displacement of the wheel to be 0 - ?rn cos(2nt lT), where 0,n is theamplitude and T is the period. Differentiate with respect to time to find the angular velocity:O - -(2n lT)0,- sin(2zrt lD. The symbol O is used for the angular velocity of the wheel so itis not confused with the angular frequency. The maximum angular velocity is

Qrn T 0.500 s 39.5 radf s .(b) When 0 - n12, then 010,"- l12, cos(ZntlT): I12, and

1

4v-E

-1 Y-1 1- 3-844KE

a_o - -+ o*sinT (f) :

d2 0 / 2n\2dtr: I t ,)a- (J* \2(osoo '/sign is not significant. (;):

where the trigonometric identity cosz 4+ sin2 A - I was used. Thussin(2zrtlT)- :@:tZlz,(#) Qrrad)r):

The negative sign is not significant. During another portion of the+34.2radf s when its angular displacement is n l2rad.(c) The angular acceleration is0,n cos(2rt lT) -

When 0-n14,

Again the negative- 1 24 radlr'

Chapter 15 9l


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43(a) A uniform disk pivoted at its center has a rotational inertia of + M R2, where M is its massand R is its radius. See Table I0-2. The disk of this problem rotates about a point thatis displaced from its center by R + L, where L is the length of the rod, so, according to theparallel-axis theorem, its rotational inertia is Lm n2 + M (L + R)2. The rod is pivoted at one endand has a rotational inertia of mL'13, where m is its mass. The total rotational inertia of thedisk and rod is I - *mn2 + M(L+ R)2 + |mL2: 1f0.500kgX0.100 m)2 +(0.500kgX0.500m+0.100m)z + 1to .2J0kgX0.500 m)2 - 0 .205kg . m2.(b) Put the origin at the pivot. The center of mass of the disk rs,(.a - L+R - 0.500m+0.100m -0.600m away and the center of mass of the rod is (.r- L12: (0.500m)12:0.250m away, onthe same line. The distance from the pivot point to the center of mass of the disk-rod system is

t _ M la t m(., (0.500 kgx0.600 m) + (0.270 kgX0 .250 m) _ n Ad_-ffi: _u.+Til'r.(c) The period of oscillation is

T :2r -1.50s.51If the torque exerted by the spring on the rod is proportional to the angle of rotation of the rod andif the torque tends to pull the rod toward its equilibrium orientation, then the rod will oscillatein simple harmonic motion. If r -- -C0, where T is the torque, 0 is the angle of rotation, ande is a constant of proportionality, then the angular frequency of oscillation is atheperiodisT-2nfa-2rr\re,whereIistherotationa1inertiaoftherod.Thep1anistofind the torque as a function of 0 and identify the constant e in terms of given quantities. Thisimmediately gives the period in terms of given quantities.Let (.s be the distance from the pivot point to the wall. This is also the equilibrium length of thespring. Suppose the rod turns through the angle 0, with the left end moving away from the wall.If L is the length of the rod, this end is now (Llz)sind fuither from the wall and has moved(Ll2)(1 -cos 0)to the right. The spring length is now {ff lDz(I _- cos q2 +Vo+ (LlT)sin 0f',If the angle e is small we may approximate cos I with 1 and sin 0 with 0 in radians. Then thelength of the spring is given by (,0+ L0 12 and its elongation is Lr: L0 12. The force it exertson the rod has magnitude It - k Lr- kL?f 2, where k is the spring constant. Since 0 is smallwe may approximate the torque exerted by the spring on the rod by r: -FLf 2, where the pivotpoint was taken as the origin. Thus r - -(kL' lqg. The constant of proportionality C thatrelates the torque and angle of rotation is e - kL2 f 4.The rotational inertia for a rod pivoted at its center is I - mLz ll2, where m is its mass. SeeTable L0-2. Thus the period of oscillation is

t'r1-^ trt--t"l ,:2rChapter 15

+ m)sd0.205 kg . m2

(0.500 kg + 0 .270kgX9 .8^ls2x0 .447 m)

mL2 l12kL2 l4

0.600 kg3(18s0 N/*)

92- 0.0653 s


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57(a) You want to solve e-bt/2*-btlZmreversed when the argument of the logarithm was replaced by

logarithm of both sides to obtain(2* lb)1n3, where the sign wasits reciprocal. Thus

(b) The angular frequency ist_ ffih3-14.3s

mlVL

2.31 radf s .The period is T(14.3 s)/(2.72 s) : 5.27 .75(a) The frequency for small amplitude oscillations is f : (l lzr)\ffi,, where L is the length ofthe pendulum. This gives f - (Ilzr)tl (9.80^ls\l(2.0m):0.3 5Hz.(b) The forces acing on the pendulum are the tension force 7 of the rod and the force of gravitymj. Newton's second law yields f + mfi : md, where m is the mass and d is the accelerationof the pendulum. Let d _ d,. + d, where d,. is the acceleration of the elevator and u" is theacceleration of the pendulum relative to the elevator. Newton's second law can then be writtenm(d - d") + f : md,'. Relative to the elevator the motion is exactly the same as it would be inan inertial frame where the acceleration due to gravity is d - d,". Since j and d," are along thesame line and in opposite directions we can find the frequency for small amplitude oscillationsbyreplacing g with g*a" in the expression f -(Ilzr)\FgIL Thus

, m- V * 4rrr':

- 0.39H2.(c) Now the acceleration due to gravity and the acceleration of the elevator are in the samedirection and have the same magnitude. That is, d - d," - 0. To find the frequency for smallamplitude oscillations, replace g with zero in f - (llzr)\ffi,. The result is zero. Thependulum does not oscillate.83[Jse 'uTLstroke, or 0.38 m. Thus unl: 2n(3.0 Hz)(0.38 m) : 7 "zm/s.89(a) The spring stretches until the magnitude of its upward force on the block equals themagnitude of the downward force of gravity: ky - mg, where A is the elongation of thespring at equilibrium, k is the spring constant, and m is the mass of the block. Thusk - mg la - (1 .3 kgX9.8 */ sz)l (0.096 m) : 133 N/-.

^Tr2n

8.00 N/* (0.230 kg/s)21.50 kg 4(1.50 kg)'

9.8 mlrt + 2.0^lt'

Chapter 15 93


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(b)TheperiodisgivenbyT:Ilf:2nlu):2r1ffik_2rr-0.62S.(c) The frequency is f : IIT - Il0.62s - L.6Hz.(d) The block oscillates in simple harmonic motion about the equilibrium point determined bythe forces of the spring and gravity. It is started from rest 5.0cm below the equilibrium point sothe amplitude is 5.0 cm.(e) The block has maximum speed as it passes the equilibrium point. At the initial position, theblock is not moving but it has potenttal energy

U,i,

When the block is atthe equilibrium point, the elong atioirof the spring is A- g.6cm and thepotential energy is1, ^ I I(,1y- -msa. ;ka'

Write the equation for conservation of energy as [-Li: ,rl **r'and solve for u:- 0.51 m f s.

The frequency of oscillation is

- 3.2H2.(b) Because mechanical energy is conserved the maxlmum kinetic energy of the block has thesame value as the maximum potential energy stored in the spring, so i*u?^

9t(a)

l-mrrn -- N Zurn: l.2kg ,Affi(5.2mls) : 0.26 [m.(c) The position of the block is given by *: nmcos(c,,,t+0\ where n?n - 0.26m and u) - 2nf -2r(3.2H2):2\rudf s. Since r-0 at time t-0, the phase constant O must be either +rl2 or-r, 12. The velocity at t-0 is given by -urfrrn sin@ and this is positive, so 6 must be -TT12,The function is tr : (0.26m) cos[(20 radls)f - Tr l2].

2(Lh - tI r)m 2(-0.44 J + 0.61

94 Chapter 15


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Chapter L5

15The wave speed u is given by , - \mu where r is the tension in the rope and LL is the linearmass density of the rope. The linear mass density is the mass per unit length of rope:m 0.0600 kgp: i- ,Jo* - o.o3ookg f m.Thus

- I29 m/s.t7(a) In the expression given for y, the quantrty A* is the amplitude and so is 0.1Zmm.(b) The wave speed is given by u - tFn, where r is the tension in the string and Lr is thelinear mass density of the string, so the wavelength is

t- tFrtpand the angu

student solution manual 8th edition

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