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STU DENT SOLUTIONS MAN UALJ. Richard Christman
Professor EmeritusU.,S. Coast Guard Academy
FUNDAMENTALS OFPHYSICS
Eighth Edition
David Halliday(Jnivers iQ of P itts burgh
Robert ResnickRens s elaer Polytechnic Institute
Jearl WalkerCleveland State Univers iQ
John Wiley & Sons, Inc.
Cover Image: @ Eric Heller/Photo Researchers
Bicentennial Logo Design: Richard J. Pacifico
Copyright @ 2008 John Wiley & Sons, Inc. All rights reserved.
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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 the
text.
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 the
opportunity 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 her
capable assistant Aly Rentrop. For their help and encouragement, special thanks go to the good
people 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 Academy
New London, CT 06320
TABLE OF CONTENTS
Chapterl .., . .. . .... . o. .. . IChapter?. .i . . . . . . . . . . . . . .4Chapter 3 . . . . . . . . . . . 0 . . . . . 10
Chaptet 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 . . . . . . . . . . . . . . , . 101
Chapter 18 . . . . . . . . . . . . . . r . 109
Chaptgr 19 . . . , . . . . . . . . . . . . 115
Chaptgr20. . . . . . . . . . . . . . . . 122Chapter2l ............... a I28Chaptet 22. . . . . . . . . . . . . . . . 134
Chapter 23Chapter 24Chapter 25Chapter 26Chapter 27Chapter 28Chapter 29Chapter 30Chapter 31
Chapter 32Chapter 33
Chapter 34Chapter 35
Chapter 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
Chapter L
3
use 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.
t7
None 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 be
adjusted 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 on
each 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
CLOCK Sun.
-Mon.Mon.-Tues.
Tues.
-Wed.
Wed.
-Thurs.Thurs.-Fri.
Fri.-Sat
AB
C
DE
-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 judge
clock 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 and
1cm3 - (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
39
The 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. Thus
to (l 13.62) cord - 0 .28 cord.
2 Chapter I
IJse l ft1.0 m3 of wood coffesponds
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 .
Now
SO
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.
Thus
g4:gbarrel bulk - (949barrel bulk)(4.01 5U.S. bushellbarcel bulk):3810U.S. bushel .
57
(a) We want to convert parsecs to astronomical units. The distance
between two points on a circle of radius r is d - 2r stn 0 12, where
0 is the angle subtended by the radtal lines to the points. See the
figure 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
and
l Au : 92'9 x },9^6 Tt,; - l.5g x 1o-' ly.J 5.87xlot2milly L'J\)
/-"., " 012
Chapter I
Chapter 2
1(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 60
to the right. The average velocity is the slope ofthe dotted line.
-- --Q- - --J -- -r - 40
20
0.5
(a)
(b)
(c)
In this case the totalis 40 km fh.
1.0 1.5t (h)
2.0
-5
Substitute,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 - 0
r(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 is
Lr - r(4s) - tr(0) - I2m- 0 - +12m. The displacement is in the positive r direction.
4 Chapter 2
(0 The average velocity during an interval is defined as the
divided by the duration of the interval: uavs- L, I Lt. For the
the 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 average
velocitybetween 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)
9.0
6.0
3.0
0.0
-3.019
If ur is the velocity at the beginning of a time interval (at time t) and u2 is the velocity at the
end (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 t
The 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.
27
Solveu2 -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 is
2lzq.0m- (15.55m1sx2.00s)] 1 rr ^ r^Z&-
33
(a)
Ug
Chapter 2 5
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 the
ground 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.t-
9.8 ^l12
ul
Substitute uo : 0, 1) : -24mls, and g
h-
(b) Solve 1) : uo - gt for t:
(c)t (s)
2
t (s)2
0
a-10(-)-zo
-30The accelerationhits the ground:
as shown on the
0
u -10(m/s)' '_20
-30
47
(a) At the highestupward, set u
6 Chapter 2
point the velocityin u2
0
a-5(-is2)
-10
_15
of the ball is instantaneously
and solve for u$ us- \EA.zero. Take the A axis to be
Substitute g
is constant untila- -g.8 m/s2.right.
the wrenchIts graph is
t (s)
A:50mtoget
(b) It will be in the atr
solutions 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)
20
0
_20
-40
0
a-5(*/r')
-10
_15
68t (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+I
where 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.
51
The 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-c
9.8 ^lt' V (9.8 mls2)2 g.g m/s2
t (s)
Chapter 2
origin of the coordinate system at the point where the key is dropped and take the y axis to be
positive 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, the
coordinate of the key is given by A - -+g*. Thus
-3.03s.
This means12m
U6: - 4.0 m/s .
3.03 s
--I5
First 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 s857 ^lt2
The positive sign
89
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 velocity
9l
(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-att
f 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)
(c) Solve n for t. The result is
- 13 s.
97
The driving time before the change in speed limit was t6 - Lr luo, where Lr is the distance
and 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.
99
Let t be the time to reach the highest point and us be the initial velocity. The velocity at the
highest 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 H2
LgQt)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 the
acceleration 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 values
yield 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 the
second 3.0-s interval is Uz - Uz:72m - 27 m- 45m.
- *ot'
2(0.25 103 m)
Chapter 2
Chapter
1
(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 m
clr -25.0 m
The inverse tangent is -58.0o or -58.0" + 180' - I22". The first angle has a positive cosine
and 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.
3
The r component is given by ar - (7 .3 m) cos 250o -- -2.5 m and the A component is given
by a,a
the 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 distance
from one corner to the diametrically opposite corner:
d - \/ (3.00 m)2 + (3 .70 m)2 + (4.30 m)2
see this, look at the diagram of the room, with the
displacement 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 end
point 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 same
but the path length would be (. + ut * h" The path length is the same as the magnitude of the
displacement if the fly flies along the displacement vector.
10 Chapter 3
(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 the
dotted lines on the upper diagram. Pretend there is
a hinge where the front wall of the room joins the
floor and lay the wall down as shown on the lowerdiagram. The shortest walking distance between the
lower left back of the room and the upper rightfront corner is the dotted straight line shown on the
diagram. 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 and
a 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 135o
The 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
39
Since ab cos Q : a*b* * oab, r a"br,
The
b-
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 from
cos 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 30.0o
component 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 the
A 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 equations
simultaneously for p and q. The result is
b*ca - bac* (3.46 mX8 .66m) - (2.00 mX-5.00 m)P:ffi- --6'67
andaac* - arca -(3.00 mXS .66 m)q:ffi- -4'34'
The scalar product is
d . b - abcos d : (10X6.0) cos 60o : 30
(b) The magnitude of the vector product is
- ab sin d: (10X6.0) sin 60o : 52 .
51
Take the r axis to run west to east and the A axis to run south to north, with the origin at the
starting 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 displacement
12 Chapter j
47
(a)
ld*61
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.
7l
According to the problem statement i+ E - 6.0i+ 1.0j and A- E - -4.0i +7.0j. Add these
to 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
Chapter 3 13
Chapter 4
7
The 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. Its
direction 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/s
and the angle O it makes with the positive r axis satisfies
tanQ:'!:.^l,n - 2.43 .
2.9 k-/h
The 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) :8i
14 Chapter 4
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 - 4.50m
and I
a : -;(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 are
uyn - 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 distance
above 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 s
and 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 at
the time of release. Take the A axis to be upward and the r axis to be horizontal. Place the
origin at the point of release and take the time to be zero at release. Let r and y (: -730 m) be
the 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
(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 is
ua: -uscos 06 - gt: -(202mls) cos(53o) - (9.80 mls2x5.00 s) : -17Imf s .
39
Take 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)z
The 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.
47
You 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 s
Now 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 is
1.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 goes
over the fence.
gr2
(9.8 mls2xl07m)
l6 Chapter 4
51
Take the A axis to be upward and the r axis to be horizontal. Place the origin at the point where
the ball is kicked, or the ground, and take the time to be zero at the instant it is kicked. r and
A are the coordinates of ball at the goal post. You want to find the kicking angle 0s so that
A- 3.44m when rand Ainto the second the result is
y: rtan0sgn2
2r'ocos2 gs
You 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 -^^2
tft 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 0s
r tan?o + y + c : 0. It has the solution
tan 0or*
50m *2(19.6 m)
The two solutions are tan?sand 0s
on the goal post.
s3
Let h be the height of a step and u be the width. To hit step n, the ball must fall a distance
nh 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 be
positive in the upward direction and the r axis to be horizontal. The coordinates of the ball at
time 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
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 less
than n and greater than n - 1. The ball hits the third step.
67
To 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 be
horrzontal. 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 the
initial 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
d t : _rr")LT] I 1
""|;;:li
: -t? knots) sin 4s")r i + r (z'knots) cos 4s " r i
and the velocity of ship B is given by
d n : _ll; ;LT:] I _rrllnff
: -tQ8 knots) sin 40'r i - tQ8 knots) cos 40"r i
The velocity of ship A relative to ship B is
6tn:6A-6s- (1 .0 knots) i + 1f 8. knots) j .
The magnitude is
uln: 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 is
0 - tan-1UAB
A
UAB rThis direction is 1.5o east of north.
, 38.4 knots- tan-I -. - 88.5o
1.0 knots
_92y 2(-2.0m)
18 Chapter 4
(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
75
Relative 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.
77
Since 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 and
coordinate 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 s
the initial velocity of the bullet. It is hortzontal. If r is the honzontalr : uOt and
Ug -y- 30m -t 6.3 x 10-2 s4.8 x 102 mls
t07
(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 s
to 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
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 is
l-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
Chapter 5
-5
Label 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') j
Thus
- (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. The
angle 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.
13
In 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 case
the 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.
25
According 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
Chapter 5 2l
is u - (1600 korlhx1000m/km)1p600 s/h) : 444mls, so o :the magnitude of the force is Ir - (500 kgQaT ^lst) - I .2 x
29
The 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 the
direction of the initial velocity and the A axis to be in the direction of the electrical force. Place
the origin at the initial position of the electron. Since the force and acceleration are constant
theappropriateequations 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 is
t
4.5 x 1 30x9.11 x 10-3t kg 1.2 x 107 m/s
,a2
)10-3 mg-t6 y(#)':;( )(
35
The free-body diagram is shown at the right. F* is the normal force
of 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 the
normal 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 6
2 a, 2 gstn1
(-3.s0 mls)2
2 | 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 is
2(-3.50 m/s)
0.674s.
- 1.35 s., 2uo 2uot::: a g stn? (9.8 mlrt; ritt 32.0o
The velocity is
u - uol_ at: uo* gtsrnl - -3.50 mls + (9.8 mls2xl.35 s)sin 32o - 3.50 mls,
22 Chapter 5
45
The 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' IorI
msll F3 on+ msll F+ on smgy 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 be
upward. Then Newton's second law for this link is Fz on 1 - mg : ma. Thus
Fz 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 z
of 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 is
F+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
(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 is
applie,C only to block 1 and that block 2 exerts the force - i on
block 1. Newton's third law has thereby been taken into account.
Newton's second law for block 1 is F f - TrL1a, where o, is
the acceleration. The second law for block 2 is f -- TrL2e,. Sincethe blocks move together they have the same acceleration and the
same 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 : TTLI f 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 same
acceleration 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 the
monkey 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 and
e,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 package
is 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
(b) Newton's second law equations are F -mpg: ffipap for the package and F -rrLrng : TrL,n,arn
for the monkey. If the acceleration of the package is downward, then the acceleration of the
monkey 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) - 120N
61
The 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 gives
Fo: M(g a) and the second gives M(g a) (M m)g - (M m)a. Solve for m:m:ZMal(g + a).
73
Take 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 -^^ 1^2
m 310kg -u't+m/s '
In part (b) m: Wlg _- (3 10N) lQ.8^ls') - 31.6kg and
Fcos0-f (450N)cos38o 125N ,1 ..r r2nr m 36.1 kg
79
Let 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 are
rrLl: 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
(b) The acceleration of an object of mass TtLl 1 m2 is
F 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 10o
and
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 rn
Chapter 5
t02 N)2 r02 N)2
26
-6.44 x 102N.
99
The free-body diagrams for the two boxes are shown below.
Flrr
Fxz
Here T is the tension in the cord, FxTr is the norrnal force of the left incline on box 1, and FNz
is the norrnal force of the right incline on box 2. Different coordinate system are used for the
two boxes but the positive r direction are chosen so that the accelerations of the boxes have the
same sign. The r component of Newton's second law for box 1 gives T - TTLI7 stn01 : TTL1& and
the r component of the law for box 2 gives mzg stn 02 - T : TTL27. These equations are solved
simultaneously 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 N
3.0 kg + 2.0kg
101
Free-body diagrams for the two tins are shown on
the right. T is the tension in the cord and Fy is
the normal force of the incline on tin 1. The posi-
tive n direction for tin I is chosen to be down the
incline 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 second
law for tin 1 gives T + Tft1g stn {3 - ma, and for tin 2gives mzg T F - TrL2cL. The second equation is
solvedforT,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
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 normal
force 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 bureau
does 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. The
magnitude of the force of friction is given by f - l.rrcFw, where
ltr* 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 - 0
and 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 55kg
28 Chapter 6
0.5 6mf s2
13
(a) The free-body diagram for the crate is shown on the right. f is
the 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 force
of 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 at
rest, / must be less than lr"Fx, or T cos? <the tension force is sufficient to just start the crate moving T cos 0 -for T:
lrr(mg 7 sin d). Solve
T' : F'mgcos 0 + p,, stn0 cos 15o + 0.50 sin 15o
(b) The second law equations for the moving crate are 7 cos 0 f : ma and -Fl/ + T sin 0
mg:0. Now ff - p,n(mg 7 sin 0). This expression is substituted for f in the first equation to obtain
? cos 0 - pn(mg - T sin 0) : ffia, so the acceleration is
T(cos 0 + p,7" sin 0)a:
* -ltt 9
Its numerical value ls
e, = - (0.35X9 .8^ls2) - 1.3 ^lr'
23
The free-body diagrams for block B and for the knotjust above block ,4 arc shown on the right. T1 is the
magnitude of the tension force of the rope pullingon blo ck B , Tz is the magnitude of the tension force
of the other rope, f is the magnitude of the force offriction exerted by the horizontal surface on blockB, FAr is the magnitude of the norrnal force exerted
by the surface on block B, We is the weight ofblock A, and W n is the weight of block B . 0
(- 30") is the angle between the second rope and
the horizontal.
For each object take the r axis to be horizontal and the A axis to be vertical. The fr component
of Newton's second law for block B is then T1 f :0 and the A component is Fy Wn:0.The r aomponent of Newton's second law for the knot is T2 eos 0 - Tr - 0 and the A component
is Tzsin? Wt:0. Eliminate the tension forces and find expressions for f and F11r in terms
Chapter 6 29
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 T1
to obtain f - Waf tan?. For the blocks to remain stationary f must be less than F"Fx orWaltan? <Solve for We:
We: FrWntanT - (0.25)(711N) tan3Oo - 1.0 x I02 N.
27
(a) The free-body dragrams for the two blocks are
shown on the right. T is the magnitude of the ten-
sion force of the string , Fx t is the magnitude of the
normal force on block A, Fw s is the magnitude ofthe nonnal force on blo ck B , f a is the magnitude ofthe friction force on block A, f e is the magnitudeof the friction force on block B, TTLy is the mass ofblock A, and TTL s is the mass of block B . 0 is the
angle of the incline (30'). We have assumed that the
incline goes down from right to left and that blockA is leading. It is the 3.6-N block.
For each block take the r axis to be down the plane and the A axis to be in the direction of thenormal force. For block A the r component of Newton's second law is
mtgsin0- fa-T:TrLAa,g
and the A component is
Fnt-Tftagcos0:0.Here a,s is the acceleration of the block. The magnitude of the frictional force is
ft : FkAFxt : FkAmtgcos 0,
where Fxa- mtg cos9, from the second equation, is substituted. Fnt is the coefficient ofkinetic friction for block A. When the expression for f a is substituted into the first equation theresult is
Tftsg sin 0 - FntThsg cos 0 - T - ?rLga,n .
The same analysis applied to block B leads to
Trlpg sin 0 - FnsTrlyg cos 0 + T - rTL Ba, p .
We must first find out if the rope is taut or slack. Assume the blocks are not joined by a rope andcalcul ate the acceleration of each. If the acceleration of A is greater than the acceleration of B,then the rope is taut when it is attached. If the acceleration of B is greater than the accelerationof A, then even when the rope is attached B gains speed at a greater rate than A and the ropeis slack.
30 Chapter 6
Set T:0 in the equation you derived above and solve for as and a,p. The results arc
a,A- g(sin0 - ltp7cos 0): (9.8 mlrt;1ritt30o - 0.10cos30") : 4.05^lt'
and
a,s - g(sin0 - ltn6 coS 0): (9.8 mlr';1ritr30o - 0 .20cos30o): 3.20mlrt.
We have learned that when the blocks are joined, the rope is taut, the tension force is not zero,and the two blocks have the same acceleration.
Now go back to mAg sin 0 - Fntmeg cos 0 -T - TrL4a, and ThBgsrn) - FnsmBgcos 0 +T -TTL6a, where a has been substituted for both a,s and a,s. Solve the first expression for T,substitute the result into the second, and solve for a. The result is
a: gsin g - ll'naml* H"nnm6 ^
rYL4 + mB g cos a
- (9 .g^lrt; ,i,r 30o [(0'10X3'6 N) + (0'20X7'2 N)l (9.g mlrr; ,os 30o'\' L 3.6N+7.2N l\-/
- 3.5 ^lrt .
Strictly speakirg, values of the masses rather than weights should be substituted, but the factorg cancels from the numerator and denominator.
(b) IJse ffLsg sin 0 - Fneme7 cos 0 - T - rTLAa to find the tension force of the rope:
T - mAg sin 0 - FntTTLsgcos e - TtLAa,
- (3.6 N) sin 30o - (0. 10X3.6 N) cos 30o - (3.6 N/9.8 mls2x3 .49 mlst) - 0.21 N .
35
Let the magnitude of the frictional force be au, where e, : 70 N . s/m. Take the direction of the
boat's motion to be positive. Newton's second law is then -aa : m du ldt Thus
f du:Ju, u mJo /
where us is the velocity at time zero and u is the velocity at time t. The integrals can be
evaluated, with the result-uattn*- -A.
Take 1) : ,ol2 and solve for f :
t -m rn2 - loookg hz - g.9s.a 70N.s/m
49
(a) At the highest point the seat pushes up on the student with a force of magnitude F'^/ (: 556 N).Earth pulls down with a force of magnitude W (: 667 N). The seat is pushing up with a forcethat is smaller than the student's weight in magnitude. The student feels light at the highestpoint.
Chapter 6 31
(b) When the student is at the highest point, the net force toward the center of the circular orbitis W - Fy and, according to Newton's second law, this must be mu'lR, where u is the speed ofthe student and R is theradius of the orbit. Thus mu2f R-W - f^i:667N- 556N- 111].{.
The force of the seat when the student is at the lowest point is upward, so the net force towardthe center of the circle is F^i W and lr^/ W - mu2 I R. Solve for Fn:
,.,/,/- rytW:ltlN+ 667N- 77gN.
(c) At the highest point W F^/ : mu2 f R, so f^imu2 lR increases by a factor of 4, to 444N. Then F^/ :667 N - 444N - 223NI.
(d) At the lowest point W + Fl/ : m'u'I R, so F^/mu2 I R is still 444 N, F^r : 667 N + 444N - 1.11
53
-W-mu2fR,so-F1/x 103 N.
The free-body diagram for the plane is shown on the right" F is the
magnitude of the lift on the wings and m LS the mass of the plane. Since the
wings are tilted by 40" to the honzontal and the lift force is perpendicular
to the wings, the angle 0 is 50o. The center of the circular orbit is to the
right of the plane, the dashed line along r being a portion of the radius.
Take the r axis to be to the right and the A axis to be upward. Thenthe r component of Newton's second law is F cos g - mu2 I R and the A
component is F sin I mg - 0, where R is the radius of the orbit. Thefirst equation gives It - mu2 I Rcos 0 and when this is substituted into the
second, (*r'lR)tan? - mg results. Solve for R:
n12
R- tr
tanl.I
The speed of the plane is 'tr :480 kmfh - 133 mls, so
R _ 033 mls)2 tansoo : 2.2 x 103 m .
9.8 mls'
59
(a) The free-body dragram for the ball is shown on the right. f,is the tension force of the upper string , ip is the tension force
of the lower string, and m is the mass of the ball. Note that
the tension force of the upper string is greater than the tension
force of the lower string. It must balance the downward pullof gravity and the force of the lower string.
32 Chapter 6
Take the r axis to be to the left, toward the center of the circular orbit, and the A axis to be
upward. Since the magnitude of the acceleration is a -- ,2 I R, the r component of Newton'ssecond law is
Tucosg+ Tacos0-ry,where u is the speed of the ball and R is the radius of its orbit. The A component is
2,, sin 0 - Tzsrn? - mg - 0.
The second equation gives the tension force of the lower string: Tt : Tu - mg f stn 0 . Since the
triangle is equilateral 0 - 30o. Thus
Tr: 35 N - (1'3a kgX9"9 m/s2)
- 8.74N .
srn 30o
(b) The net force is radially inward and has magnitude Fnet, str - (7, + TD cos 0
8.74N) cos 30" - 37.9 N.
(c) [Jse Fn"t,str: nr,uz f R. The radius of the orbit is [(1.70m) l2)Jtan30" - 1.47 m. Thus
(1.a7 mX37.eN)1.3 4kg - 6.45 mls .
65
The first sentence of the problem statement tells us that the maximum force of static frictionbetween the two block is f ,,^u*:12N.When the force F is applied the only horizontal force on the upper block is the frictional force ofthe lower block, which has magnitude f and is in the forward direction. According to Newton'sthird law the upper block exerts a force of magnitude f on the lower block and this force is inthe rearward direction. The net force on the lower block is F - f .
Since the blocks move together their accelerations are the same. Newton's second law for the
upper block gives f - Tftt& and the second law for the lower block gives F - f : TTL6a, where cr
is the common acceleration. The first equation gives a - f l*t Use this to substitute for a, inthe second equation and obtain F - f : (mal^t)f .Thus
Ii-(r*e) f\ mt/
If f has its maximum value then F has its maximum value, So the maximum force that can be
applied with the block moving together is
F - (t.ffi) lzN):27N
f l2N ._ ,)LL: TrLl 4.0 N
RFn"t, str
The acceleration is then
Chapter 6 33
77
(a) The force of friction is the only horizontal force on the bicycle and provides the centripetalforce need for the bicycle to round the circle. The magnitude of this force is f : mu2 f r, wherem is the mass of the bicycle and rider together, u is the speed of the bicycle, and r is the radiusof the circle. Thus
f : (85.0 kgXg.oo ml s)2 - ,n
25.0m(b) In addition to the frictional force the road also pushes up with a nonnal force that is equalin magnitude to the weight of the bicycle and rider together. The magnitude of this force is.F1r/ - mg: (85.0kg)(9.8 ^lt')-833N. The frictional and normal forces are pe{pendicular to
each other, so the magnitude of the net force of the road on the bicycle is 4r"t - , E' * ** -vL Va'
- 877 N.
81
The free-body diagrams are shown on the
right. T is the tension in the cord, F* ^
is
the norrnal force of the incline on block A,F* u is the normal force of the platfonn on
block B, I is the angle that the incline makes
with the horizontal (which is also the angle
between the normal force and the vertic aI),
and ,f-'ir the frictional force of the platforrnon block B . The r axis for each block is also
shown.
The r component of Newton's second law for block A gives mgsinO T- Trlye,, the frcomponent of the second law for block B is T fFxn - TTLsg - 0.Note that the blocks have the same acceleration.
The magnitude of the frictional force LS ptrFxs: ltkTftsg,where ?rLpg was substituted for Fxn,and the r component for B becomes T - Fnmng - TTLsa,. The equations mgsin 0 - T - TLnaand T - Fnmng: TTLsa are solved simultaneously for T and a. The results are
rF _ TftlTrLB(sin 0 + Fil (4.0 kgX2.0 kg)(sin 30o + 0.50) 1 ' \rt *^+r", 10ro rJl\
Masin 0 - Fnrns (4.0kg) sin3Oo - (0.50x2.0kg) ., , r 2e,- g: :1.6m/sms+ mB r 4.0k9 + 2.0k9
8s
(a) If u is the speed of the car, m ts its mass, and r is the radius of the curve, then the magnitudeof the frictional force on the tires of the car must be f : muz f r or else the cat does not negotiatethe curve. Since m - W I g, where W is the weight of the car,
r. wu2 (10.7 x 103 N(13.4m1s)2 . ,_,1 1
JL-J./rLz\lWl\.a' gr (9.g mls2x61.0 m)
and
(275 N)2 + (833 N)2
34 Chapter 6
(b) The norrnal force of the road on the car is l7^/ - W and the maximum possible force of static
friction is .f", -u* - FrFxforce that is required is less than the maximum possible, the car successfully rounds the curve.
9t
Let F be the magnitude of the applied force and f be the magnitude of the frictional force.
Assume the cabinet does not move. Then its acceleration is zero and, according to Newton'ssecond law, F - f . The norrnal force is F1,'
maximum force of static friction is Fr,max
if F is less than 378 N the cabinet does not move and the frictional force is fgreater than 378 N, then the cabinet does move and the frictional force is f(0.56X556 N) - 311 N.
(a) The cabinet does not move and f : 222I\.
(b) The cabinet does not move and f : 334 N.
(c) The cabinet does move and f : 311 N.
(d) The cabinet does move and f : 311 N.
(e) The cabinet moves in attempts (c) and (d).
99
(a) The free-body diagram for the block is shown on the right. The
magnitude of the frictional force is denoted by f , the magnitude - .
of the normal force is denoted by lrl/ , and the angle between the
incline and the horizontal is denoted by 0. Since the block is sliditgdown the incline the frictional force is up the incline. The positive
r direction is taken to be down the incline. For the block whenit is sliding with constant velocity the n component of Newton'ssecond law gives mg sin 0 f - 0 and the 'A component gives
mg cos 0 -Fl/ - 0. The second equation gives F1/ - mg cos9, so the magnitude of the frictionalforce is fobtain mg sin 0 - Tftuamg cos 0 : 0. Thus the coefficient of kinetic friction is Fp - tan 0 .
When the block is sliditrg up the incline the frictional force has the same magnitude but is directed
down the plane. The r component of the second law equation becomes mg sin 0 + Fnmg cos e -rrLa, where a, is the acceleration of the block. Thus a - (sin 0 + [L7"cos0)g - 29 sind, where tan?was substituted for Fn and tan 0 - sin 0 f cos 0 was used.
If d is the displacement of an object with constant acceleration a, us is its initial speed and uis its final speed, then u2 ulr: Zad. Set u equal to zero and a, equal to 29 stn? and obtaind - -ufil2n -- -ufrlag sin9. The negative sign indicates that the displacement is up the plane.
(b) Since the coefficient of static friction is greater than the coefficient of kinetic friction the
maximum possible static frictional force is greater than the actual frictional force and the blockremains at rest once it stops.
Fl'/
fr/0
Chapter 6 3s
105
The box is subjected to two horizontal forces: the applied force of the worker, with magnitude
F, and the frictional force, with magnitude f .Newton's second law gives F - f : ma) where mis the mass of the box and a is the magnitude of its acceleration. The magnitude of the frictionalforce is fthe normal force of the floor. In this case F1y - mg and fbecomes F - Ltnmg - me' so ltt - (lr - ma) l*g.Let u be the final speed of the box and d, be the distance it moves. Then u2 - Zad, and
a,- u2 lza- (1.0 mlilz l2(1 .4m):0.3 57mf s2. The coefficient of kinetic friction is
ltt - F - ma
- (85 N - (a0kgX0.357 m/s2)
m9 (40 kgX9.8 m/s2)0.18
36 Chapter 6
Chapter 7
3
(a) Use Eq. 2-16: u2 - u|+Zar, where us is the initial velocity, u is the final velocity, r is the
displacement, and a is the acceleration. This equation yields
(b) The initial kinetic energy is
K,;,- i*rtThe final kinetic energy is
Ky - **r': iO.67 x 10-27 kg)(2.9 x r07 ^ls)2:6.9 x 10-13J.
The change in kinetic energy is LK :6.9 x 10-13 J - 4.8 x 10-13 J :2.I x 10-13 J.
t7(a) Let F be the magnitude of the force exerted by the cable on the astronaut. The force ofthe cable is upward and the force of gravity is mg is downward. Furtherrnore, the accelerationof the astronaut is glI0, upward. According to Newton's second law, F mg: mgl10, so
Fdone by F is
- ILmgd,- 11(72 kgX9.8 m/s2X15 m)
-Wp: Fd l.16 x 104J.10 10
(b) The force of gravity has magnitude mg and is opposite in direction to the displacement.
Since cos 180" - - 1, it does work
ws
(c) The net work done is W :1.16x 104J- 1.06x 10aJ - 1.1x 103 J. Since the astronaut started
from rest the work-kinetic energy theorem tells us that this must be her final kinetic energy.
(d) Since K - i*r'her final speed is
- 5.3 m/s .u-
t9(a) Let F be the magnitude of the force of the cord on the block. This force is upward,while the force of gravity, with magnitude M g, is downward. The acceleration is g 14, down.Take the downward direction to be positive. Then Newton's second law is Mg F- Mgl4,soFWp
2Km
2(lJ x 103 J)
Chapter 7 37
(b) The force of gravity is in the same
(c) The net work done on the block isfrom rest this is its kinetic energy K(d) Since K - LtW r', where u is the
According
where u6 is
direction as the displacement, so it does work Ws : M gd.
Wr: -3MSdl4+ Mgd - Il[gd,l4. Since the block stafts
after it is lowered a distance d.
speed,
t2K tsdu:V tw:Vzafter the block is lowered a distance d. The result found in (c) w-as used.
29
(a) As the
force is
l2w=u1: I * +'i:
(b) The velocityof the paftrcle is uy - 5.O^ls when it is at r -theorem for n y. The net work done on the particle is Wenergy theorem yields -3 (*'r - rl) : **(r? - ,?). Thus
- 4.0 m the work done by the
-3(*'r - *?)
in the kinetic energy:
(at r f). The theorem yields
,)2 - 6.6mf s.
n y. Solve the work-kinetic energy
-3(*tr *), so the work-kinetic
body moves along the r axis from n6
w : l_"n'
F*d,r: I_"n' -6rd,n - -3 *1"_',-
to the work-kinetic energy theorem, this is the change
W:A,K:L*@?-r?),
the initial velocity (at t) and u.s is the final velocity
35
(a) The graph shows F as a function of r tf rs is positive.
The work is negative as the object moves from rrSince the area of a triangle is ] (Uur.)(altitude), the workdone from r:.0 to n: n0 is -+(roXFo) and the work done
from rThe net work is the sum, which is zero.
38 Chapter 7
- , t-+[(s.0m/s)2 - (8 .O^ls)r] + (3.0m), - 4.7 m.V6N
F(*)Fs
0
-Fs
2(-21 J) + (g.o m/s)2
2.}ks
(b) The integral for the work is
w: Io'"o
ro(X t) dr:Fs(* \ P*o
")1, -0
45
(a) The power is given by P - F'u and the work done by f from
rtz rtzw' -
Jr, P dt:
Jr, ,, dt.
Since F is the net force, the magnitude of the acceleration is avelocity is us: 0, the velocity as a function of time is given by ,
nt^t
w : l .' "'
(tr' l *)t dt - )rr' l *)e3J tt
43
The power associated with force f is given by P : F . 6, where u* is the velocity of the objecton which the force acts. Let d (: 3 7") be the angle between the force and the honzontal. ThenP - F.6- Fucos d- 022NX5.O^ls)cos 37":4.9 x I02'W'.
time t1 to time t2 is given by
- F l* and, since the initial: uo * at - (F lm)t Thus
t?)
For f1 - 0 and t2
14/ : i [ts'o xl2 I (t.o s)2 - 0 83 JYv z L l5kg l
(b) For dr - 1.0 s and t2 - 2.0 s,
w :1 [(s oryfl lrz.o r), - (1.0 s)r] - 2.s r., L lsLs If(zu
(c) For C1 - 2.0 s and tz - 3.0 s,
w:1 fry] [(:0,), -(20,),] - 421z L lsk
(d) Substitute u- (Flm)t into P- Fu to obtain P- Ttztl* for the power at any time t. Atthe end of the third second P - (5.0 N)2(3.0 s) 115 kg - 5.0 'W'.
47
The net work Wnet is the sum of the work Wu done by gravrty on the elevator, the work W.done by gravity on the counterweight, and the work W, done by the motor on the system:
Wn t - W" + W. + W". Since the elevator moves at constant velocity, its kinetic energy doesnot change and according to the work-kinetic energy theorem the net work done is zere. Thismeans Wu + W"* W"
Chapter 7 39
gravity on it is W"- -TrLegd- -(1200kgX9.8*lr';1S+m): -6.35 x 105J. The counterweightmoves downward the same distance, so the work done by gravity on it is W" : rn"gd -(950kgX9.8 mlr';1S+m) - 5.03 x 105J. Since Wr: 0, the work done by the motor on thesystem is W, _ -W" - W. - 6.35 x 105 J - 5.03 x 105 J - 1.32 x 105 J. This work is done in a
time interval of At - 3.0 min - 180 s, so the power supplied by the motor to lift the elevator is
p -%- 1'32 x 1o5J -7.35x roz'w'.Lt 180 s
63
(a) Take the positive r direction to be in the direction of travel of the cart. In time Lt thecart moves a distance Lr - u Lt, where u is its speed. The work done is W - F* Lr(Fu cos 0) LL where f is the force of the horse and 0 is the angle it makes with the horizontal.Now 6.0 milh600s, so It - t(40 lbXS.Sftls)cos30ol(600s): 1.8 x 105 ft.lb.(b) The averagepower is P - F*u - Fucos 0 -(401bX8.}ftfs)cos30o:3.0 x 102ft.lb/s.Since 1ft-Ibls - 1.818x 10-'hp, thepoweris P - (3.0x 102ft.lblsxl.818x 10-thplft. lb/s) -0.55 hp.
69
(a) The applied force f is in the direction of the displacement i, so the work done by the forceis We : Fd -- (209 NX1.50m) - 314J.
(b) The crate rises is distance Ly - d stn 0, where 0 is the angle that the incline makes withthe horizontal. The work done by the gravitational force of Earth is Ws
-(25.0kg)(9.8 mls2xl.50m) sin 25.0"- -155 J.
(c) The nonnal force of the incline on
so this force does no work.
(d) The net work done on the qate is
7lLet Wr (: 1 10 N) be the first weight hung on the scale and r be the elongation of the springwith this weight on it. Let Wz (- 240 N) be the second weight hung on the scale and n2 be theelongation of the spring when this weight is hung on it. In each case the spring pulls upwardwith a force that is equal to the weight hung on rt, so according to Hooke's law Wr -- krt andWz: krz, where k is the spring constant. Now 11 and 12 are not the readings on the scales butnz- nr is the difference of the scale readings. Subtract the two Hooke's law equations to obtainWz Wr: k(*z - rr). Thus
Wz Wt 240N - 110N
the crate is perpendicular to the displacement of the crate,
Wnet: We "Ws - 3I4J - 155 J - 159J.
_ /v,/ f v L .<r-T\-rJ'\ IILTL\ . a
,u - Z - W.J n rV t\/Irr.n2-rr 60 x 10-3m -40 x 10-3m
When W1 is hung on the spring the elongation is n1 - Wrlk: (110N)/(6.5 x 103].1/m)1.7 x l0-2m: 17mm. The reading on the scale is 40mm- ITmm:23mm.
40 Chapter 7
(b) When the third weight is hung from the spring the elongation of the spring is r - 30 mm -23mm:7.0mm. The weight is W : kr : (6.5 x 103 N/r")(7.0 x 10-3 m) : 45 N.
73
The elevator is moving upward with constant velocity, so the force F that is moving it must be
equal in magnitude to the total weight of the elevator and load. That is, F _ Wrorur _ Mtotutg,
where Mrorut is the total mass. The power required is P - Fu, where u is the speed of the
elevator. Thus P - Mrorurgrr - (4500kg+ 1800kgX9.8^1s2X3.80 mls): 2.35 x 10sW.
77
(a)) Since the wind is steady the acceleration of the lunchbox is constant and x: - ust -r *.ot',where us is the velocity at time zero and a, is the acceleration. According to the graph the
coordinate is about 0.40m at time t-0.50s, so us- (0.40 m)l(0.50s) - 0.80m/s. The kineticenergy at t -- 0 is Ko : i*rfi(b) At t - 5.0 s the velocity is zero) so the kinetic energy is zero.
(c) According to the work-kinetic energy theorem the work done by the wind force is the change
in the kinetic energy, which is -0 .64 J.
Chapter 7 4l
Chapter 8
3
(a) The folce of gravity is constant, so the work it does is given by W : F . i, where F is the
force and d is the displacement. The force is vertically downward and has magnitude mg, where
m is the mass of the flake, so this reduces to W '_ mgh, where h is the height from which the
flake falls. This is equal to the radius r of the bowl. Thus
W : mgr - (2.00 x 10-'t gX9.8 mlt'1122.0 x r0-2m): 4.31 x 10-3 J.
(b) The force of gravity is conservative, so the change in gravitational potential energy of the
flake-Earth system is the negative of the work done: A(J: - -W : -4.31 x 10-3 J.
(c) The potential energy when the flake is at the top is greater than when it is at the bottom by
lAt/|. If U - 0 at the bottom, then [J: - +4.31 x 10-3 J at the top.
(d) If Lr - 0 at the top, then U: - -4.31 x 10-3 J at the bottom.
(e) All the answers are proportional to the mass of the flake. If the mass is doubled, all answers
are doubled.
-3
The potential energy stored by the spring is given by (Ji - Ln*', where k is the spring constant
and r is the displacement of the end of the spring from its position when the spring is inequilibrium. Thus
2(2s J)8.9 x 103 N/*., 2[-I
K- .,ro (0.075 m)2
2
(a) Neglect any work done by the force of friction and by air resistance. Then the only forcethat does work is the force of gravity, a consen'ative force. Let Ki be the kinetic energy ofthe truck at the bottom of the ramp and let K y be its kinetic energy at the top. Let [.Li be the
gravrtational potential energy of the truck-Earth system when the truck is at the bottom and let
t-I y be the gravitational potential energy when it is at the top. Then K 1 + U fpotential energy is taken to be zero when the truck is at the bottom, then LI I : ffigh, where h isthe final height of the truck above its initial position. Kithe truck, and Ky - 0 since the truck comes to rest. Thus mgh - *.*rt and h- u2 l2g.Substitute 1) - 130 korlh - 3 6.Im/s to obtain
It, : (36 .I mls)2 : 66.5 m.2(9.8 m/s2)
If L is the length of the ramp, then Isinl5o :66.5m or L-(66.5m)lsin15" -257m.
42 Chapter I
The truck is not a particle-like object since its wheels turn and the cylinders of its motor move.
However, if there is no frictional force between the tires and the roadway, these moving parts
have no influence on the rate with which the truck slows. If there is friction, then when the
driver takes his foot off the gas pedal the tires exert a forward frictional force on the road and
the road exerts a backward frictional force of the same magnitude on the truck. This, along withair resistance, helps slow the truck. The frictional force is greater if the driver shifts to a lowergear.
(b) The answers do not depend on the mass of the truck. They remain the same if the mass is
reduced.
(c) If the speed is decreased h and L both decrease. In fact, h is proportional to the square ofthe speed. If u is half its former value, then h is one-fourth its former value.
11
(a) The only force that does work as the flake falls is the force of gravity and it is a conservativeforce. If Ki is the kinetic energy of the flake at the edge of the bowl, K y is its kinetic energy at
the bottom, Ut is the gravitational potential energy of the flake-Earth system with the flake at the
top, and Uy is the gravitational potential energy with it at the bottom, then K1+Uf : Kt+U,i,.Take the potential energy to be zero at the bottom of the bowl. Then the potentral energy at the
top is Ut,: mgr, where r is the radius of the bowl and m is the mass of the flake. Ki - 0 since
the flake starts from rest. Since the problem asks for the speed at the bottom, write *,*r' forK y, The energy conservation equation becomes mgr - i*r', so
,tr: \B - lrfr.s^ls2xo.zzom): 2.08m/s.
(b) Note that the expression for the speed (u - \M) does not contain the mass of the flake.The speed would be the same, 2.08 mls, regardless of the mass of the flake.
(c) The final kinetic energy is given by Kf -- K +Ut-LIy. Since Ki is greater than before, K1is greater. This means the final speed of the flake is greater.
15
(a) Take the gravitational potenttal energy of the marble-Earth system to be zero at the positionof the marble when the spring is compressed. The gravitational potential energy when the marble
is at the top of its flight is then U s - mgh,, where h is the height of the highest point. This ish - 20 m. Thus
(.Is - (5.0 x 10-'t gX9.8 mlt'11ZOm) : 0.98 J.
(b) Before firing the marble is at rest and is again at rest at the top of its traJectory. Both the forceof the spring and the force of gravity, the only two forces acting, are conservative. Conservationof mechanical energy is expressed as LU n + At/" : 0, where [-I s is the gravitational potentialenergy and Lt" is the spring potential energy. This means L[]r- -LUn- -0.98J.(c) Take the spring potential energy to be zero when the spring has its equilibrium length. Thenits initial potential energy is (-1
" - 0.98 J. This must be *.tt*', where k is the spring constant and
Chapter 8 43
n is the initial compression. Solve for k:
2(0.e8 J) : 3.1 x L02 N/-.(0.080 m)2
31
Information given in the second sentence allows us to compute the spring constant. Solve F - krfor k: F 27ON
k -;- ffi- 1.35 x 101 N/-.(a) Now consider the block sliding down the incline. If it starts from rest at a height h abovethe point where it momentarily comes to rest, its initial kinetic energy is zero and the initialgravitational potential energy of the block-Earth system is rngh, where m is the mass of the
block. We have taken the zero of gravitational potenttal energy to be at the point where theblock comes to rest. We also take the initial potentral energy stored in the spring to be zero.Suppose the block compresses the spring a distance r before coming momentarily to rest. Thenthe final kinetic energy is zero, the final gravitational potential energy is zero, and the final springpotenttal energy is *tt*'. The incline is frictionless and the normal force it exerts on the blockdoes no work, so mechanical energy is conserved. This means mgh - Lrtt*', so
, kr2 (1 .35 x 104 N/-)(0.055 m)2 ,\ I Fh- 2*g- -0.174m.
If the block traveled down a length of incline equal to [., then (. srn3Oo
(0.174m)f sin3Oo - 0.35m.
(b) Just before it touches the spring it is 0.055 m away from the place where it comes to rest
and so is a vertical distance h' - (0.055 m) sin3Oo - 0.0275m above its final position. Thegravitational potential energy is then
mgh' - ( r2kg)(9.8 mls2xo .0275 m) - 3.23 J .
On the other hand, its initial potential energy is
msh - ( l2kg)(9.8 mls2x0 .174m) : 20.5 J.
The difference is its final kinetic energy: Ky :20.5 J - 3.23 J - 17.2 I. Its final speed is
- 1.7 m/s .
45
(a) The force exerted by the rope is constant, so the work it does is Wforce and i ir the displacement. Thus
W : Fdcos 0 - (7 .68 NX4.06 m) cos 15.0o - 30. 1 J .
(b) The increase in thermal energy is LE6 - f d - (7 .42NX4.06 m) : 30. I J.
, 2(J"K- .,
ro
2Krm
2(17 .2 J)
12kg
44 Chapter B
(c) We can use Newton's second law of motion to obtain the frictional and normal forces, then
use Fttand the A axis nofinal to the floor. The r component of Newton's second law is Fcos 0 - f :0and the u component is F^r + F sin g mg _ 0, where m is the mass of the block, Fw is the
nofinal force of the floor, F is the force exerted by the rope, and 0 is the angle between that
force and the horizontal. The first equation gives
f : F cos 0 - (7 .68 N) cos 15.0o - 7.42 N
and the second gives
F,^/ - mg - F sin 0 - (3.57kgX9.8 mls2) - Q.68N)sin 15.0o -33.0N.
Thus Fr,
47
(a) Take the initial glavitational potential energy to be LLi - 0. Then the final gravitational
potential energy is U y - -mg L, where L is the length of the tree. The change is U y Ut, _
-mgL - -(25kgX9 .8^ls2x I2m) - -2.9 x 103 J.
(b) The kinetic energy is K - L*r': itz5kgX5.6 mls)2:3.9 x 102J.
(c) The changes in the mechanical and thermal energies must sum to zero. Since the change in
thermal energy is LErn: f L, where f is the magnitude of the average frictional force,
L lZm
69
The change in the potential energy of the block-Earth system as the block goes from A to B isthe same for the two cases and, since mechanical energy is conserved, the change in the kinetic
energy of the block is the same. The change in the kinetic energy is
.1- ,2n) :
;*1Q.60 ^ls)2 - (2.00 mls)21 - (1.381lkg)m.
Ks : KA+ L,K - I*f+.00 mls)2 + (1.3s J/kg)m- (9.38 J lkg)m2\and the speed at B is
- 4.33 m/s .
7S
Since the blocks start from rest and the mechanical energy of the system consisting of the blocks
and Earth is conserved, the final kinetic energy is the negative of the change in potential energy.
LK:)*tLFor the second trial
2Kslm 2(e.38 J lke)
Chapter 8 45
If block B falls a distance d, block A moves a distance d, along the incline and rises a verticaldistance d srn 0 , where 0 is the angle of the incline. Thus
K - -A[_I - - t?*egd,) + (mtgd sin?)] - gdlmn - mssin 0l
- (9.8^1s2X0.25m)l2.0kg - (1.0kg)sin30"l - 3 .7 J.
83
(a) IJse conservation of mechanical energy. Let Ki be the initial kinetic energy, K y be the finalkinetic energy, and (. be the compression of the spring. Then the change in kinetic energy is
LK : Kf - Ki. The block travels the distance d+1. along the incline and the vertical component
of its displacement has magnitude (d + [) srn?, where 0 is the angle of the incline. Thus the
change in the potential energy is LLI - mg(d,+ l)sin 0 + *ttl', where k is the spring constant.
Since mechanical energy is conserved the final kinetic energy of the block is
K f : K,i, - mg(d,+ ,t) sinl - !,re2
t6 J - (1 .0 kgX9.8 m/s'X0.60 m + 0.20m) sin 40o ltroo N/*)(0.20 m)z - 7.0 J .
2'
(b) Now K y - 0 and K,; is the unknown. Conservation of mechanical energy gives
Kr: mg(d,+/) sin 0 + )*n- (1.0 kgXg .8^1s2X0.60 m * 0.40 m) sin 40o .;Q00N/*)(0.40 m)2 : zzJ .
87
Neither the kinetic energy or the potentral energy
the change in the total thermal energy is equal toIf F is the magnitude of the applied force and d
The thermal energies of the cube and the floorand
L4tn,floor : Fd - L4*r,cube - (15NX3.0m) - 20J:25J .
109
(a) Take the potential energy of the ball-E arth system to be zero when the ball is at the bottomof its swing. Then the initial potenttal energy is 2mg L, where m is the mass of the ball and
L is length of the rod. The initial kinetic energy is zero since the ball is at rest. Write *,*u',where u is the speed of the ball, for the final kinetic energy, at the bottom of the swing. Since
mechanical energy is conserved 2mg L : L*r' and
changes, so conservation of energy tells us thatthe work done by the applied force: LE.;,,: W .
is the distance the cube travels, then W : F d.
both change, so LEtn- L4rn,cube + A.Etrr, floor
46 Chapter 8
'u:z\EL - 2 (9.8 mls2xo .62m) : 4.9mls.
(b) At the bottom of the swing the force of gravity is downward and the tension force of the rod
is upward. If T is the magnitude of the tension force, Newton's second law is T -mg: m?r'lL,SO
T -mg*mu'lf -mg+4mg - 5mg - 5(0.0g}kgX9.8^lst) - 4.5N.
(c) The diagram on the right is the free-body diagram for the ballwhen the tension force of the rod has the same magnitude as the force
of gravity. We wish to solve for 0. The component of the force of .- //
gravity along the radial direction is mg cos 0 and is outward. The
net inward force is T - mg cos0 and, according to Newton's second
law this must equal muT f L, where u is the speed of the ball. Thus
T--muT lL*mgcosLWe now need to find the speed of the ball in terms of0. Take the potential energy to be zero when the rod ishorizontal. Since it starts from rest its kinetic energy is
also zeto. As can be seen on the diagram on the right,when the rod makes the angle 0 with the vertical, the
ball has dropped through a vertical distance L cos 0. Thepotential energy is then -mg L cos g. Write the kinetic energy as *.*r' and the conservation ofenergy equation as 0 - -mgLcos 0+**u'. Thus mu2 - 2mgLcos9. Substitute this expression
into the equation developed above for T: T - Zmg cos 0 + mg cos0- 3mg cosd. According to
the condition of the problem, this must be equal to ffig, so 3 mg cos 0 - mg, or cos 0 - I 13. Thismeans 0 :7Io.(d) Notice that the mass of the ball cancels from the equation for cos 0, so 0 does not depend on
the mass. The answer to (c) remains the same.
111
(a) At the top of its flight the velocity of the ball has only a horizontal component and this isthe same as the horizontal component of the initial velocity. Let ug be the initial speed and 0o
be the angle with which the ball is thrown. Then the kinetic energy at the top of the flight is
K -i*t 6coS 0oi)2_ ]fto x 10-'t g)t(s .0^ls)cos30"1': |.2J.
(b) [Jse conservation of mechanical energy. When the ball goes from the window to the point3.0m below, the potential energy changes by LU- -mgd, where d (: 3.0m) is the distance
from the window to the lower point. Since mechanical energy is conserved the change inthe kinetic energy is +mgd. If us is the initial speed and u is the speed at the lower point,
I*u'
- /rt .0^ls)2 + 2(9.8^ls2x3.0m - t1m/s.
(c) and (d) Notice that the mass cancels from the conservation of energy equation and none ofthe quantities in that equation depend on the initial angle, so the answer to part (b) does notdepend on the mass or the angle.
Lcos?
Chapter I 47
ttg(a) After the cue loses contact with the disk all the kinetic energy of the disk is converted tothermal energy. Thus the increase in thermal energy of the disk and court is LEn_ i*u'ito .42kgX4 .2mls)2 - 3.7 J.
(b) The change in thermal energy is LEtn : f d, where f is the force of friction of the court on
the disk and d is the distance the disk travels. Thus f : L0tnf d - (3.7 J)lIZm) - 0.31N. Overthe entire l4m the increase in thermal energy is LErn - (0.31NX14N) - 4.3J.
(c) A11 of the energy transferred from the cue to the disk ends up as thermal energy so the workdone by the cue is 4.3 J .
t2t(a) and (b) The force is the negative of the slope of the curve. Take the potential energy to be
-2.8J when the particle is at tr:1.0m and -17.5 J when the particle is at r:4.0m. Then
( -r7 .3 J) - (-2.9 J)F-- 40ffi --.4.8N.
The magnitude is 4.8N and it is in the positive r direction.
(c) and (d) When the particle is at r - 2.0m the potential energy is about tIthe kinetic energy is K - *.^r' - *f2.0 kg)(- I .5 mf s)z
E^"when the particle is at r- 1.5m and r- 13.5m. The particle moves between these twocoordinates.
(e) When the particle is at renergy is K - E*"" (Ji - (-5.5 J) - (-17.5 J) : lzJ. The speed of the particle is
- 3.5 m/s.
I23
(a) Let d, be the distance the ear travels. Its vertical position lowers by dstn?, where 0 is the
angle of the incline, so the potentral energy changes by - Ltl - mgd sin 9, where m is the
mass of the car. The kinetic energy changes by LKspeed and u y is the final speed. The change in the mech aiical.tt.tgy is LE^"" : LK + A (J: --nlgd,sin 0*+*(r?-'r?).Convert the given speeds to meters per second. They zte ui: 8.33 m/sand uy:11.1 mls, so
AEn,, - -(1500 kgXg .8^ls2x50 m) sin 5.0o .:(1500 kg) [{t t. t
- -2.4 x 104J.
The mechanical energy decreases by 2.4 x 104 J.
48 Chapter 8
^ls)'- (8 .33mlr)t]
2(r2 I)2.0 kg)
(b) The change in mechanical energy is given by - f d, where f is the magnitude of the frictionalforce. Thus f : -LE^""f d- -(-2.4 x 104 Dl(50m): 4.7 x 102N.
127
(a) The potential energy does not change, so the change in the mechanical energy is equal to thechange in the kinetic energy. Let m be the mass of the block, u0 be its speed at the beginningof the acceleration period and u be its speed at the end. Then LE^r" - lm lr,' - ,'oliCtske) ltlo ^ls)2 - (t0mlr)'] :6.0 x 103 J.
(b) The ayeruge rate with which energy is transferred to the block is the total energy transferred
divided by the time for the transferraL since the acceleration a, is constant the time is givenby LtAE-.
"l Lt - (6.0 x 103 Dl(10 s) : 6.0 x 102
'w.
(c) and (d) If the accelerating force has magnitude F, then the instantaneous rate of energytransfer is given by Pfor F. For u- 10m/s, P- (15kgX2.0^ls2x10m/s) - 3.0 x 102W and for u- 30m/s,P - (15 kgX2 .0^lrtx:0*ls) : 9.0 x 102 W
131
The kinetic energy gained per unit time is equal to the potentral energy lost per unit time.mass L,m of water passes over the falls the gain in kinetic energy is LK - L,m gh, where hthe height of the falls. The rate of production of electrical energy is
D_3L^mgh 3,l4t:io200^'lSX1000kl^'X9.8^lS2X100m):8.8X108w.
133
(a) When the ball is at D the potenttal energy is mg L greater than when it is at A and the kineticenergy is **r3 less. Since mechanical energy is conserved, -+mu|+mgL - 0 andus- 1/ffi.(b) Let T be the tension in the rod when the ball is at B and Iet u be the speed of the ball then.Newton's second law gives T - mgu2 . When the ball is at B the potential energy is m,g L less than when it is at A and the kineticenergy is greater by **@' - r2i, so i*(r' - ufi) - mgL - 0 and u2 - u20+2gL - 4gL, whereZgL was substituted for uzr. Thus T - mgtmu'lL - mg+4mg - 5mg.
(c) When the ball is at C the potentral energy is the same as when it is at A and the kineticenergy is **r3 less. A11 of the kinetic energy is converted to thermal energy. The decrease inmechanical energy is i*r|: **zgL - mgL, where 2gL was substituted for u2o.
(d) When the ball has settled at B the potential energy is mg L less than when it started at A andthe kinetic energy is i*r?, - mg L less. The mechanical energy has decreased by 2mg L.
Ifis
Chapter I 49
Chapter 9
15
You need to find the coordinates of the point where the shell explodes and the velocity of thefragment that does not fall straight down. These become the initial conditions for a projectilemotion problem to determine where it lands.
Consider first the motion of the shell from firing to the time of the explosion. Place the
origin at the firing point, take the r axis to be horizontal, and take the A axis to be verticallyupward. The A component of the velocity is given by u - aya gt and this is zero at timet - uya I g - (ro I il sin gs, where us is the initial speed and 0s is the firing angle. The coordinates
of the highest point on the traJectory are
tr : uort - uotcos gs - 6 sin 9s cos 9s -Iand
1 ,t2: + 6 sin2 osA: ulat - 19, ,i stn- 9s
Since no horrzontal forces act, the horizontal component of the velocity of the center of mass is
constant. At the highest point the velocity of the shell is ?re coS 06, in the positive n direction.This is the velocity of the center of mass. Let M be the mass of the shell and let Vo be thevelocity of the fragment that does not fall straight down. Then the velocity of the center of mass
is given by MVolzM :Vof 2, since the masses of the fragments are the same. Since the velocityof the center of mass is constant, ue cos 0s : Vof 2. This means
Vo : 2u0 cos 0o : 2(20 mls) cos 60o : 20 m/s .
Now consider a projectile launched horrzontally at time t - 0 with a speed of 20 m/s from thepoint with coordinates frs : I7.7 m, Ao : 15.3m. Its A coordinate is given by A- Uo trgt',andwhenit1andsthisisZero.Thetimeoflandingist:\Mandthef,:coordinateofthelanding point is
r-ro*Vot:fr1+Vo - 17.7 m+ (20m/s) 53 m.
23
(a) Take the initial direction of motion to be positive and let J be the magnitude of the impulse,m be the mass of the ball, ui be the initial velocity of the ball, and u y be the final velocity ofthe ball. The impulse is in the negative r direction and the impulse-momentum theorem yields
-J : TrLUf - TrL't)i. Solve for uy to obtain
u y- *rt* t -
(0.40kgX14m/s) - 32.4N' s - _67m/s .m 0.40kg
The final speed of the ball is 67 mf s.
50 Chapter 9
(20 mls)2n sin 60o cos 60o
9.8 mls'
2ao
I
35
(a) Take the force to be in the positive direction, at least for earlier times. Then the
r- ft'oxlo-3 rf,-xL- ft'oxto-3Jt
ro, , -,,n . ,.,,6. ,? t
l ,^,\ . ^^ ^r 3'0x l0-3
- Li,u.o
x r061tz f cr.o x t0e;t3 1
""^ - g.oN . s.
lo
The impulse is in the positive direction.
(b) Since J - FavE Lt, where FavE is the average force and Lt is the duration of the
(b) The negative sign indicates that the direction of the velocity is opposite to the initial directionof travel. That is, it is in the negative r direction.
(c) The magnitude of the average force is Fave- Jf Lt - (32.4N. s)l(27 x l0-3 s - I.2x 103N.
(d) The impulse is in the negative r direction, the same as the force.
impulse is
kick,
Fa',n: Ivg Lt 3.oxlo-3s e'v
(c) To find time at which the maximum force occurs set the derivative of .F with respect to timeequal to zeto and solve for f. The result is t - 1.5 x 10-3 s. At that time the force is
4r,u,. - (6.0 x 106X1.5 x 10-') - (2.0 x 10eX1.5 x 10-3)2 : 4.5 x 103 N .
(d) During the kick the ball gains momentum equal to the impulse. Since it starts from rest, itsmomentum just after the player's foot loses contact is p - J . Let m be the mass of the ball and
u be its speed as it leaves the foot. Then, since n : p lm,
J 9.0N.sl): _ :20mlS.m 0.45 kg
39
No external forces with horizontal components act on the man-stone system and the verticalforces sum to zeto, so the total momentum of the system is conserved. Since the man and
the stone are initially at rest the total momentum is zero both before and after the stone iskicked. Let ms be the mass of the stone and l)s be its velocity after it is kicked; let mrn be
the mass of the man and I)Tn be his velocity after he kicks the stone. Then msl), * mrnlJnl: 0
and 'u,rrl
'L)m
man moves in the direction opposite to the direction of motion of the stone.
47
(a) Let m be the mass and uti be the velocity of the body before the explosion. Let TTL1, TrLz,
and ff\ be the masses of the fragments. (The mass of the third fragment is 6.00 kg.) Write
Chapter 9 51
uti for the velocity of fragment l, -uzi for the velocity of fragment 2, and u3ri + q, j for the
velocity of fragment 3. Since the original body and two of the fragments all move in the rAplane the third fragment must also move in that plane. Conservation of linear momentum leads
to muni: nllrtj - TTL2uzi+ m3rhri+ m3qa j, or (mut* TTL2v2 - Tr\?r3r) i - (mftr -t rfuu3a)i : 0.
The r component of this equation gives
,u3r- Tftu,t* mzur: (20.0kgX200m/s) + (4.00kgX500m/s) _ 1.00 x 103 m/s.Tr\ 6.0 kg
The A component gives
u3a : -TTLtur - - (10'0 kgX100 m/s)
- -r67m/s .Tr\ 6.0 kg
Thus dz - (1.00 x 103 mls)i- (167mls)j. The velocity has amagnitude of 1.01 x 103 mls and
is 9.48" below the r axis.
(b) The initial kinetic energy is
111Ki
The final kinetic energy is
1.1.,1Ky: ;*tr? +;rrL2u3+ r*tr3:1 t1ro.0 kg)(100 mls)2 + (4.00 kgXsoo m/s)2 + (6.00 kextor a^^)t]
2L- 3.63 x 106J.
The energy released in the explosion is 3.63 x 106 J - 4.00 x 10s J - 3.23 x 105 J.
6l(a) Letmr be the mass of the cartthat is originallymoving, uybe its velocitybefore the collision,and urf be its velocity after the collision. Let Tft2 be the mass of the cart that is originally at rest
and uzf be its velocity after the collision. Then, according to Eq. 9-67,
urf-- n" ',*'utt." mt t TTLz
Solve for rft2 to obtain
Tn2: L't; - ttf / l '2^ls - 016t"A)
(0'340kg) - 0'099 Iu, r ur f
TTLI - ( / (0.340 kg) - 0.099 kg .
(b) The velocity of the second caft is given by Eq . 9 -68:
uzr: Lrrn: t l (l .zmls):1.9 mls.Tft1 a'rrL2 ' 10.340 kg + 0.099 kgl
52 Chapter 9
(c) The speed of the center of mass is
Trlluu * Tft2uzt (0.340 kgX I.zm/s) _ n tucom : ffi - 0 30kgi.0 Jrrftg- - 0 .93 m/ s .
Values for the initial velocities were used but the same result is obtained if values for the finalvelocities are used. The acceleration of the center of mass is zero.
63
(a) Let rTrl be the mass of the body that is originally moving, uy be its velocity before thecollisior, and urf be its velocity after the collisioll. Let Tft2 be the mass of the body that isoriginally at rest and uzf be its velocity after the collisioll. Then, according to Eq. 9-67,
Solve for TtL2 to obtain
TTLI - TfL2utr - uyi, -" mt*mZ
uu - utflll2: lll,1
ut f + Ltlli
Substitute urf : ?,)rnl4 to obtain rrL2 - 3*tl5 - 3(2.0k9 15: I.2kg.(b) The speed of the center of mass is
ucom - TTt tuYi * mzuz,
- (2.0 kgX4.0 m/s)
- 2.5 m/s.TTII + m2 2.0 kg + I.2kg
77
(a) The thrust of the rocket is given by T - Rur"r, where R is the rate of fuel consumption and'urcr is the speed of the exhaust gas relative to the rocket. For this problem Rurct- 3.27 x 103 mls, so T - (480kelsx3.27 x 103 mls): I.57 x 106N.
(b) The mass of fuel ejected is given by Mna: RLt, where Lt is the time interval of thebum. Thus Mna: (480 kels)(250 s)
My: Mt - M6et: 2.55 x 105kg - 1.20 x 105kg - 1.35 x l05kg.(c) Since the initial speed is zero, the final speed is given by Eq.9-88:
u1:Urcl'"ffi-(3.27X103mls)/0nffi-2.08X103m/s.
79
(a) Take the r axis to be positive to the right in Fig. 9-72 of the text and take the A axis tobe perpendicular to that direction. Consider first the slow barge and suppose the mass of coalshoveled in time Lt is LM. If ?r-" is the velocity of the barge and 0 is the velocity of the coalas it leaves the barge, then the change in the momentum of the coal-barge system during thisinterval is Af - A M(t - u-"). The momentum of the coal changed from 6rLM to i tlt andthe momentum of the barge did not change. The force that must be exerted on the barge to keep
Chapter 9 53
its velocity constant is F,- Lf ILt- (LM ILD(0 d"). Now i"shoveledpetpendicularlytothelengthoftheboatthen 0 - uri*Urj. ThusF, -(LMlLt)Uoj.Ua is the slight transverse speed the coal must be given to get it from one barge to the other.It is not given in the problem statement, so we assume it is so small it may be neglected. Theforce that must be applied to the slower barge is essentially zero.
Now consider the faster barge, which receives coal with mass LM. Initially the coal has velocityt but after it comes to rest relative to the barge its velocity is d y , the same as the velocity ofthe barge. The momentum of the coal changes from LMi to LMdy and the momentum of the
barge does not change. The force that must be applied to the barge is Ff - (LMlLt)(df t).Now dy: ufi and 0 - uri*Urj, so the r component of the force is
Ff*
The rate with which coal is shoveled is converted from kg/min to kg/s in the first factor and thebarge speeds are converted from km/h to m/s by the last factor.
(b) The A component of the force that is applied to the faster barge is Ftr: -(LMlLt)fLa. IfUa is small, Ff , is essentially zero.
9t(a) If m is the mass of a pellet and u is its velocity as it hits the wall, then its momentum isp - n Lu - (2.0 x 10-' t gX500 mls) : 1.0 kg . ml s, toward the wall.
(b) The kinetic energy of a pellet is K - *.*r': *Q.O x 10-3 kgX500m/s)2 :2.5 x 102 J.
(c) The force on the wall is given by the rate at which momentum is transferred from thepellets to the wall. Since the pellets do not reboutrd, each pellet that hits transfers momentumptransferred is
p AltrEt avg - (1.0kg . mlsxl0 s-t) - 10N.A'The force on the wall is in the direction of the initial velocity of the pellets.
(d) If Af is the time interval for a pellet to be brought to rest by the wall, then the average forceexerted on the wall by a pellet is
Fn,*: P - l'okg'm/s :r.7 x 103N.vs a, 0.6 x 1o-3 s
The force is in the direction of the initial velocity of the pellet.
(e) In part (d) the force is averaged over the time a pellet is in contact with the wall, while inpart (c) it is averaged over the time for many pellets to hit the wall. Most of this time no pelletis in contact with the wall, so the average force in part (c) is much less than the average forcein (d).
93
(a) The initial momentum of the car is Fn: Tft6t - (1400kgX5 .3^ls)jfinal momentum is Ft - (7400kg.mls)i. The impulse on it equals the
J- - Ff - Fr, - (7400 kg . mlsxi - i).54 Chapter 9
- (7400 kg .mfs) j and the
change in its momentum:
(b) The initial momentum of the car it Fn - (7400kg .mfs)i and the final momentum rs Ft:0.The impulse acting on it is i - Ft - Ft,
(c) The average force on the car is
-AF_TA, A' 4.6 s
and its magnitude is Favs- (1600N/t - 2300N.
(d) The average force is
(1600NXi - j)Fuue
Favg
and the final kinetic energy is
Ki: !*rr',"2J
(-2.1 x lo4DiLt 350 x 10-3 s
and its magnitude is FavE : 2.1 x 104 N.
(e) The average force is given above in unit vector notation. Its r and A components have equalmagnitudes. The r component is positive and the A component is negative, so the force is 45"below the positive r axis.
97
Let TTL p be the mass of the freight car and u p be its initial velocity. Let rrls be the mass of thecaboose and u be the common final velocity of the two when they are coupled. Conservation of thetotal momentum of the two-car system leads to mp.'u7r: (mpimg)u, so u : ?rFTftpl(*r+mc).The initial kinetic energy of the system is
Ky ^?.-rT1 ffi2pr2p
(*e+mc)2 2 (me+mc)
Since 27% of the original kinetic energy is lost K y - 0.73K,i,, or
(:?rLp,")I
, (0.73)(me + mc)
Following some obvious cancellations this becomes rftpl(^e + mc)ms - (0.27 10.73)me - 0.37,mp - (0.37)(3. 18 x 104kg) - 1.18 x 104kg.
101
(a) Let 'ut,i be the speed of ball 1 before the collisions and ur f be its speed afterwards. Let uz.f
be the speed of ball 2 after the collision. Let m be the mass of each ball. Then conservationof momentum leads to the r component equation rrLU1,;- mutylcos?l * muzl cos02 and the Acomponent equation 0 _ -mury sin0l * muzy srn02. The masses cancel from these equations.The r component equation gives urf cosgl - uti-uzf cos02 and the y component equation gives
Chapter 9 55
ur f sin dlobtain
u?r - (uzf sin 0)2 + (urr - uzf cos 0r)'- [(1.1^ls)sin60o]'+12.2mls - (1.1 mls)cos60"lt
- 3 .62m2 lr' .
The speed is utf: @: r.gmls.(b) Divide ur1 sin01 - u2f stn02by ,ry cos91 :'t)ti,- uzf cos02 and use tan01- (sin 0)l@os01)to obtain
tan01: uzf srn?z -
urt - uzf cos 02 Q.zmls) - (1.1 mls) cos 60' - 0 '577 '
The angle is 0r - 30o.
(c) The initial kinetic energy is L*r?,i : )rn(z.2mf s)2 - 2.42m, itt joules tf m is in kilograms.The final kinetic energy is i*r? f + Lr'rr - L*(l .g mls)2 + i*(I .I mf s)2 - 2.4m, in joules tf mis in kilograms. The collision is elastic (at least to the number of significant digits given in theproblem).
107
(a) The acceleration of the center of mass of the two-particle system is the net external force onparticles of the system divided by the total mass of the system:
_ 1", - [(-4.00N)i + (5.00Nj] + [(2.00N)i + (-4.00$j]Tft1 1 m2 2.00 x 10-'kg + 4.00 x 10-3 kg
- (-3 .33 x r02 mlrt) i + ( r.67 x r02 mlr') j .
Since the acceleration is constant and the center of mass is initially at rest, the displacementduring the interval is
A4o-: |a1tt1': )X-3.33 x t02mlrt)i+( r.67 x 102 mlrt)jl(2.oo x t0-'r)t
- ( -6.67 x 10-4 m) i + (3 .33 x 10-4 m)j .
The magnitude of the displacement is
la'l (-6.67 x 10-4 m)2 + (3.33 x 10-4 m)2 - 7.45 x 10-4m.
(b) If 0 is the angle made by the displacement and the positive r direction then tan? - (3.33 xl0-4 $l(-6.67 x l0-4m) - -4.99 and 0- -26.5o or 153o. Since the displacement has anegative r component and a positive A component the correct answer is 153o.
(c) The velocity of the center of mass is
u-.o*:d"o Lt- t(-3.33 x r02mlr')i+(r.67 x r02mlst)il(2.O0 x 10-'s)
- (-0 .667 mls)i + (0 .333^ls)j
56 Chapter 9
--+
CLcom
and the kinetic energy of the center of mass is
l(,o*: lr*1 * rrl2)u?o :ir*1 * mz)@'.o **r?o^r)
- )rr.00 x 10-3 km + 4.00 x 10-'t g)t(0 .667 mls)2 + (0.3 33mls)21
'. .67 x 10-3 J .
113
Let M "
be the mass of the sled and us be its initial speed. Let M* be the mass of water scoopedup and u y be the final speed of the sled and the water it contains. Before the water is scoopedup the momentum of the sled-water system is Mruo and afterwards it is (Mr*M-)rf. The finalspeed is
M"uo Qgookgx25g*/9 - lgom/s.u7: A/Ir+M-: ,You should recognuze this as a completely inelastic collision between the sled and the water.
115
(a) Put the origin at the center of Earth. Then the distance rcom of the center of mass of ther,arth-Moon system is given by
TTL11qr11t1Tcom:
mM+mE,where TTL 11,1 is the mass of the Moon , TTL E is the rnass of Earth, and r y is their separation. Thesevalues are given in Appendix C. The numerical result is
(7 .36 x 1022 kgX3 .82 x 108 m)rcom: -4.64 x 106m.
(b) The radius of Earth is Rn -- 6.37 x 106m, so r"o*l Rn: 0.73 and the distance from thecenter of earth to the center of mass of the earth-Moon system is 73% of Earth's radius.
tt7(a) The thrust is T _ Rurer, where R is the mass rate of fuel consumption and nrct is thespeed of the fuel relative to the rocket. This should be equal to the gravitational force M g,where M is the mass of the rocket (including fuel). Thus Rurer - M g and R(6100 kgx9.8 m/ s\ l(1200 mls) : 50 kgls.(b) Now Rur"t - M g - M a, where a, is the acceleration. This means
R _ M(g + a) -
(6100 kgX9.8 m/s2 + 21 m/s2) - t.6x ..2kg/s .nrer 1200 m/s
129
Write Eq. 9-68 in the form uzf - 2mrtnl(m+ M), where m is the mass of the incident objectand M is the mass of the target. Solve for M:
M _m(Zuyi - uzf)- (3.0kgX2(8.0m/s) - (6.0m/s)l _ 5.0kg.uzf 6.0 mls
Chapter 9 57
Chapter L0
13
Take the time t to be zero at the start of the interval. Then at the end of the interval t - 4.0 s,
and the angle of rotation is 0 - clot + |at2. Solve for aoi
et_ *t, tzoraa - *e.0radls2)(4.0 s)20
Now use u) : ws * at to find the time when the wheel is at rest (ar - 0):
-L _ tis 24 radf s
a 3.0 ,uW: -8'os '
That is, the wheel started from rest 8.0 s before the start of the 4.0 s interval.
2t(a) Use I rev - 2r rad and I min - 60 s to obtain
200 rev (200 rev)(2n radfrev)u):ffi(b) The speed of a point on the rim is given by u- u)r, where r is the radius of the flywheeland w must be in radians per second. Thus 't): (20.9rad/sX0.60m: I2.5m/s.
(c) If a is the angular velocity at time t, us is the angular velocity at t: 0, and o is the angular
acceleration, then since the angular acceleration is constant a : Loy * at and
e.:a - ao : (1000rev/min) -(200rev/min) - g00 revf minzt 1.0 min
(d) The flywheel turns through the angle 0, which is
0 - aat.:a* - (200rev fmin)(1.0 min) .;(800 rev f mi,r'X1.0 min)2 : 600 rev.
29
(a) Earth makes one rotation per day and 1d is (24hx3600s/h) - 8.64 x 104 s, so the angular
speed of Earth is (2nrad)l(8.64 x 104 s) : 7.3 x 10-5 rad/s.
(b) Use u - ar, where r is the radius of its orbit. A point on Earth at a latitude of 40o goes
around a circle of radius r: Rcos 40", where R is the radius of Earth (6.37 x 106m). Its speed
is rtr - a(Rcos 40") : (7.27 x 10-s radfsx6.37 x 106 m)(cos40o) : 3.6 x 102 m/s.
(c) At the equator (and all other points on Earth) the value of r,,, is the same (7 .3 x 10-5 rad/s).
58 chapter I0
(d) The latitude is 0o and the speed is u - wR - (7.3 x 10-s rudfsx6.37 x 106 *) - 4.6x102 m/s.
33
The kinetic energy is given by K : itr', where I is the rotational inertia and w is the angularvelocity. [Jse
u): (602revf min)(Zrradf rcv) :63.0rad f s.60 s/min
Then
r-#: ffi:r23ke m2
35
Use the parallel axis theorem: J - 1ron'+Mh2, where /.o* is the rotational inertia about aparallelaxis through the center of mass, M is the mass, and h is the distance between the two axes. In thiscase the axis throughthe centerof mass is atthe 0.50mmark, so h- 0.50m-0.20m-0.30m.Now
/.o,n: iuOSO
r - 4.67 x LT-z kg.m2 +(0.56kgX0.30 m)2:9.7 x 10-'kg.mZ
37
Since the rotational inerlia of a cylinder of mass M and radius R is Ienergy of a cylinder when it rotates with angular vel ocrty w is
K -)tr':Iu^2o2.(a) For the first cylinder
K - lf t .z5kgX0 .25 m)2e35 radf s)2- 1.1 x 103 J .
4',
(b) For the second
K - ltt .zskgX0 .7 5 m)2e35 rud,f s)2 : 9.7 x 103 J .
4',
4l[Jse the parallel-axis theorem. According to Table l0-2, the rotational inertia of a uniform slababout an axis through the center and pe{pendicular to the large faces is given by
I, lv[ 'rom : O(o' + b2) .
Chapter I0 59
A parallel axis through a corner is a distance h -
J -/ro* + Mh2 + b2) + +(o,
+ b21 -+ (0.084 *)'] : 4.7 x
2)' from the center, so
+@'+b2)J
l0-4 kg . mZ- 0.r72kg
3 [(0.035 m)2
45
Two forces act on the ball, the force of the rod and the force
of gravity. No torque about the pivot point is associated withthe force of the rod since that force is along the line fromthe pivot point to the ball. As can be seen from ttre diagram,the component of the force of gravity that is pe{pendicular tothe rod is rng sin 0, so if (. is the length of the rod then the
torque associated with this force has magnitude r : rngt sin 0 -(0.75kgX9 .8^lr2x 1.25m) sin 30o - 4.6N . ffi. For the positionof the ball shown the torque is counterclockwise.
47
Take a torque that tends to cause a counterclockwise rotation from rest to be positive and atorque that tends to cause a clockwise rotation from rest to be negative. Thus a positive torque
of magnitude rrFrsin 91 is associated with Ft and a negative torque of magnitude rzFzsin d2 is
associated with F2. Both of these are about O. The net torque about O is
T : rtFt sin 91 - r2F2stn02
- (l .30 m)(a .20 N) sin 7 5.0" - (2. 15 mX4.90 N) sin 60.0o - -3.85 N ' ln .
49
(a) Use the kinematic equation a - clo t at, where es
final angular velocity, a is the angular acceleration, and
:a-eo: 6.20radfs -t 220 x 10-3 s
is the initial angular velocity, a is the
t is the time. This gives
z8.2radlrt.
(b) If I is the rotational inertia of the diver, then according to Newton's second law for rotation,
themagnitudeofthetorqueactingonheris r: Ia- (L2.0kg.m2)(28.2radls2) - 3.38x102N.m.
63
Let (. be the length of the stick. Since its center of mass is (, 12 from either end, its initialpotential energy is L*gL where m is its mass, and its initial kinetic energy is zero. Its finalpotential energy is zero and its final kinetic energy is it r', where I is its rotational inertiafor rotation about an axis through one end and a is its angular velocity just before it hits the
floor.Conservationofenergyyie1ds}'*gt-it,,)ofu-_\/@.Thefreeendofthestick
60 Chapter I0
(o l2)2 + (b l2)2
is a distance (. from the rotation axis, so its speed as it hits the floor is uAccording to Table I0-2, f - **t', So
u - \Fnt. - 3(9 .8^1s2Xtr.00 m) : 5.42m/s .
69
(a) Choose clockwise rotation of the pulley to be positive and take its angular position 0 to be
zero at time t - 0. Then the angular position at time t is 0 - **t', where a is its angular
acceleration. Thus (x:20 l* - 2(1 .30rad) lQl .0 x 10-'r)' - 3 .14 x I02 radf s2.
(b) The string does not slip on the pulley, so the acceleration of either block is a,
(0.0240 m)(3 .14 x 102 rudls') - 7 .54^lt' .
(c) The forces on the hanging block are the tension force of the string and the gravitational
force of Earth. Newton's second law for this block gives mg - T1
(6.20kex9 .8mf s2 - 7.S4mlst) - r4.0N.
(d) The net torque on the pulley is r(Tt Tz), so r(Tr T) - I a, where I is the rotationalinertia of the pulley. Thus
Tz: Tt Iar 0.024m
79
Use conservation of energy. Take the potential energy to be zero when the rod is horizontal.
If L is the length of the rod, the center of mass of the rod is initially a distance (I sin q 12above the pin and the initial potential energy is LL : mgl(sin 0) 12, where m is the mass of the
rod. The initial kinetic energy is zero since the rod starts from rest. The final kinetic energy isrotational and is given by K I - Lt r?, where I is the rotational inertia of the rod and a y is itsangular speed as it passes the horizontal. The conservation law gives mg(L sin 0) 12 - Lt r?, so
ut.s ms(L sin 0) I I .
We now need the rotational inertia for rotation about the pin. According to Table l0-2 the
rotational inertia of the rod about an axis through the center of mass is /ro-The parallel-axis theorem tells us that the rotational inertia for rotation about the pin is I -Oll})mLz +mQl2)' - (Il3)*L2. The angular speed as the rod passes the horizontal is
- 3.1rad f s.
87
Take the positive direction to be toward the right for the block and take clockwise to be thepositive direction of rotation for the wheel. Let T be the tension force of the cord. The horizontalcomponent of Newton's second law for the block gives P - T
msl3 I I.
3ms(I sin d) 39 srn? 3(9 .8^lr'; ritr 40"
Chapter I0 6l
the block and a is its acceleration. The torque on the wheel is Tr, where r is the radius of thewheel, so Newton's second law for rotation gives Tr: Ia, where I is the rotational inertia ofthe wheel and a is its angular acceleration. Since the cord does not slip on the wheel , e,: re,.When this substitution is made for a, is the equation for the block, the result is P - T - Tnre,?
so T - P mra. IJse this to substitute for T in the equation for the wheel. The result isPr - mr2a: Ia and the solution for a is
Pr (3.0NX0.20m)r r \J.vr\/\v.z.vLLL) A / 1t 2(L: *r- - 4.6radf S"mrT + I (2.0 kgXO .20 m)2 + 0.050 kg . m2
89
(a) For constant angular acceleration a : u)0+ at, so a - (a - ,il|t. Take w :0 and to obtainthe units requested use t - (30 s)/(60 s/min) : 0.50 min. Then
e, : - 33 '33 rev/min
- _-66.7 rev f mrn2 .
0.50 min
The negative sign indicates that the direction of the angular acceleration is opposite that of theangular velocity.
(b) The angle through which the turntable turns is
0 - aot.:a* - (3 3.33revfmin)(0.50min) + )e66.7 revfmin2x0.50min)2
9l(a) According to Table 1,0-2, the rotational inertia of a uniform solid cylinder about its centralaxis is given by Ic: *tW R', where M is its mass and R is its radius. For a hoop with mass Mand radius Rn Table I0-2 gives 111 - M RL for the rotational inertia. If the two bodies havethe same mass, then they will have the same rotational inertia if R2 f 2 - R'r, or Rn - Rf ,n.(b) You want the rotational inertia to be given by I - M k2 , where M ts the mass of the arbitrarybodyan.dkistheradiusoftheequivalenthoop.Thusk-ffi.
115
(a) The kinetic energy of the box is given by K6: *.*r', where nL is its mass andu is its speed.
Thus the speed of the box is u- \M and, since the
angular speed of the wheel is cr,' :'ulr: W. Theis
K-: )rr': :t'#: #: :loJ.
(b) IJse conservation of energy. As the box falls a distance h the potential energy of the box-wheel-Earth-mount system changes by LU - -mgh and the change in the kinetic energy isLK -6.0J+ 10J- 16J. Since LK + Lt-I-0, h- LKlmg
0.27 m.
cord does not slip on the wheel, the
rotational kinetic energy of the wheel
62 Chapter I0
Chapter LL
-Tifr. work required to stop the hoop is the negative of the initial kinetic energy of the hoop.The initial kinetic energy is given by K - )trz + i*r', where I is its rotational inertia, mis its mass, a is its angular speed about its center of mass, and u is the speed of its center ofmass. The rotational inertia of the hoop is given by I - mRz, where R is its radius. Since thehoop rolls without sliding the angular speed and the speed of the center of mass are related bya-ulR,Thus
1
K -;*R'L
The work required is W :
/ u2\ I
(trr) + ;*r2 - ml,2 - (r40keX0.150m/s)2 - 3. 15 J.
-3. 15 J.
t7(a) An expression for the acceleration is derived in the text and appears as Eq. I 1 - 13:
0com1 * /,o,,' lMRS)
where M is the mass of the yo-yo, 1ro* is its rotational inertia about the center, and Ro is theradius of its axle. The upward direction is taken to be positive. Substitute Iro - 950 g . cmz,M - 120 g, fto : 0.3 2 cm, and g -- 980 cmf s2 to obtain
acom: :13cm lr'.
(b) Solve the kinematic equation Ucom: i.o"o t2 for t and substitute !/com: I20 cm:
t-
(c) As it reaches the end of the
55 cm/s.
-4.4s.
string its linear speed is ucom
(d) The translational kinetic energy is K - **r?r*: }fo.I20kgX0.55 mls)z- 1.8 x 10-2J.
(e) The angular speed is given by u) : u"o l Ro and the rotational kinetic energy is K - *.Iro*r'*I"o u?r*l Ra
(0 The angular speed is u) - uro lRo - (0.55 mls)|Q.2 x 10-'*) - 1.7 x 102rcdls - 27rcvf s.
23
(a) Let F - F*i* Frj and i - ri+ aj.Then
2(120 cm)
13 cm lr'
-/\A.r\nAi - ix It - (ri+Ail x (F"i+ Frj) - (*F, - UF*)k.
Chapter I I 63
The last result can be obtained by multiplying out the quantities in parentheses and using i * i : t,/\AnnA'\A
(, i x i : 0, and j x j - 0. Numerically,Jxl--l
F - t(3.0m)(6.0N) - (4.0mX-8.0N)l t - (s0N . m)k.
(b) Use the definition of the vector product: li * f1T.andFwhentheyatedrawnwiththeirtailsattheSamepoint.NowT:\ffi:
(3.0 m)z + (4.0 m)2
rFsinQ: l and d:90o
- 5.0m and F -50 N . m, the same as the magnitude of the vector product. This means
29
(a) [Jse f - mr'x 6, where r- is the position vector of the object, u- is its velocity vector, and mis its mass. The position and velocity vectors have nonvanishing r and z components, so they
arc written r'_ ri+ rk and 6 - u*it urk. Evaluate the vector product term by term, making
sure to keep the order of the factors intact:
r' x,i - ("i + zk) x (u*i + u"t ; - *r*i x i + ru"i x tt + zu*kx i + zu"kx t .
Now use i x
Thus| - m(-nuz
- (0.2s ke)
(b) [Jse F - i x F,with Ii - Fj.
f - @?+ zk) x (Fi)
- (2.0mX4.0N)i -
xt- -j,kxi:+j,and 0 to obtain
Fxd-(-rr"+zu*)j
kxk-i:0,i
+ zu)j
l-fzo mXs.o mls) + (-2.0 mX-s.o m/s)]
:trFixj+zFkx(-2.0mX4.0N)k -
j-0.
i- rPk- zFi(8.0N .m)i + (8.0N .m)t.
33
(a) The angular momentum is given by the vector product ivector of the particle and t7 is its velocity. Since the positionplane we may write v-_ ri+ yj and 6 - u*i* rri.Thus
and velocity vectors are in the nA
r- x d - (ri+ yj) x (r*i+ urj) - *r*i x i+ ru,
x i - -k, and j x j - 0 to obtain
i x j + au*j x ?+ yurj x jA/\IJseixi- 0,?xj-k,i
(-8.0N)2 + (6.0N)2
64 Chapter I l
r'x6-(rua-UU*)k.
Thus
' :;fl'r, ll?il, --60^rs)-(8'mXs'm/s)l t- etTx r02r,g m2rlr)t
(b) The torque is given by i - ix f . Since the force has only an r component we may writeF - F*i and
i - (*i+ail x (F*i; - -uF*t- -(8.0mX-7.0N)t- (56N.m)t.
(c) According to Newton's second law for rotation, f - a[1at, so the time rate of change of the
angular momentum is 56 kg . nf lt', in the positive z direction.
37
(a) Sineer: dLldt,the average torque acting during anyinterval is givenby rav'- (Lf -Lt)lLt,where Li is the initial angular momentum, L y is the final angular momentum, and At is the
time interval. Thus
0.800 kg . m2 ls - 3.00 kg . # It - -1 .47 N .m.:,avg1.50 s
In this case the negative sign simply indicates that the direction of the torque is opposite the
direction of the initial angular momentum, which is taken to be positive.
(b) The angle turned is 0 : ast * i*t'. If the angular acceleration a is uniform, then so is the
torqueand e,:rlI. Furtherrnore, es:Ltf I, so
0_Lit+LrtzI 0.140kg.m2
(c) The work done on the wheel is
W : r0 : (-l .47 N . m)(20.3 rad) : -29.8 J.
(d) The average power is the work done by the flywheel (the negative of the work done on the
flywheel) divided by the time interval:
Po'n:-Y'avg Lt 1.50s Lt
43
(a) No external torques act on the system consisting of the man, bricks, and platforrll, so the
total angular momentum of that system is conserved. Let Ii be the initial rotational inertia of the
system and let I y be the final rotational inertia . If a,; is the initial angular velocity and w y is the
final angular velocity, then INt: Ifaf and
u) r- f !\ u);- ( 6'0 kg ' m2 \ ,v/s) - 3.6 rev/s .' \ r;)
ui: [z.Okg ,re )(r'ztevfs):
3'6rev/s '
Chapter I I 65
(b) The initial kinetic energy
ratio isKf
is Ki: LInr?, the final kinetic energy is Ky: ittr'f , and their
- t4: (2.okg-mz)(3.6rcvf s)2-
3.0 .K,; Itw? (6.0 kg . m2X1.Zrev f s)2
(c) The man did work in decreasing the rotational inertia by pulling the bricks closer to his body.This energy came from the man's store of internal energy.
45
(a) No external torques act on the system consisting of the two wheels, so its total angularmomentum is conserved. Let 11 be the rotational inertia of the wheel that is originally spinningand 12 be the rotational inertia of the wheel that is initially at rest. If w,i is the initial angularvelocity of the first wheel and w I is the common final angular velocity of each wheel, then
Itwt: (1r + Iz)of andI1
uY - IJ IrQi '
Substitute 12:2Ir and a,; - 800rev f mrnto obtain w.S :267 rev/min.
(b) The initial kinetic energy is Ki: LItr? and the final kinetic energy is Ky: i(tt + I2)wzy.
The fraction lost is
AK-
Kt-Kf -
fru?-(t,+tr)a?- wzn - 3'?Ki Ki Ip?
(800 rev f mrn)z
wl
0.667
49
No external torques act on the system consisting of the train and wheel. The total angularmomentum of the system is initially zero and remains zero. Let I (- M R') be the rotationalinertia of the wheel. Its final angular momentum is L*- Iw: MRzw, where M is the mass
of the wheel. The speed of the track is w R and the speed of the train is a R - u. The angularmomentum of the tratn is Lt - m(a R - u)R, where m is its mass. The direction of rotation ofthe track is taken to be positive. If the train is moving slowly relative to the track, its velocityand angular momentum are positive; if it is moving fast its velocity and angular momentum are
negative. Conservation of angular momentum yields 0 - M R2w + m(wR a)R. When thisequation is solved for u, the result is
muR MU
lttt + *1r*: $f + dRSubstitute M - I.Lm, R - 0.43 m, and n : 0. 15 m/s to obtain
66 Chapter I I
u): -0.l|radfs.(1.1mt m)(0.43 m)
67
(a) If we consider a short time interval from just before the wad hits to just after it hits and sticks,
we may use the principle of conservation of angular momentum. The initial angular momentumis the angular momentum of the falling putty wad. The wad initially moves along a line that isd,lZ distant from the axis of rotation, where d is the length of the rod. The angular momentum ofthe wad is mudf 2. After the wad sticks, the rod has angular velocrty w and angular momentumIw, where I is the rotational inertia of the system consisting of the rod with the two balls andthe wad at its end. Conservation of angular momentum yields mudf 2- Iw. If M is the mass
of one of the balls, J - (2M +m)(d,12)'. Wher:n'Lud,lz - (2M +m)(dl2)zw is solved for w, the
result is
Zmu 2(0.0500 kg)(3.00 m/s)u)-W:iU o,iimt -o'l48tadfs'
(b) The initial kinetic energy is Kiratio is KylKu- IwZ l*r'. When I -lZtW+m)d,'l4 and u,):2mulQM+m)d, ut. substituted,this becomes
K1 0.0s00 kga _ n n l^t- U.VLLJ
K,i 2M * m 2(2.00 kg) + 0.0500 kg
(c) As the rod rotates the sum of the kinetic and potential energies of the Earth-rod-wad systemis conserved. If one of the balls is lowered a distance h, the other is raised the same distanceand the sum of the potential energies of the balls does not change. We need consider only thepotential energy of the putty wad. It moves through a 90o arc to reach the lowest point on itspath, gaining kinetic energy and losing gravitational potential energy as it goes. It then swingsup through an angle 0, losing kinetic energy and gaining potentral energy, until it momentarilycomes to rest. Take the lowest point on the path to be the zero of potential energy. It starts adistance d,lz above this point, so its initial potential energy is U,i: mgd,lz. If it swings throughthe angle 0, measured from its lowest point, then its final position is (d l2)(I - cos 9) above thelowest point and its final potential energy is LI 1 - mg(dlz)(I - cos0). The initial kinetic energyis the sum of the kinetic energies of the balls and wad: KiAt its final position the rod is instantaneously stopped, so the final kinetic energy is K y : 0.
Conservation of energy yields mgd,l2+ *fZtW + m)(d,12)2w2: rng(d,|z)(I - cos0). When thisequation is solved for cos 0, the result is
cos o - -12 (ry)G)u2
[ ](ry) (0 1*"radtil,:-o0226
2
The result for 0 is 9l .3o. The total angle of the swing is 90" +91 .3o
Chapter I I 67
73
(a) and (b) The dtagram on the right shows the parti-cles and their lines of motion. The origin is markedO and may be anywhere. The angular momentum ofparticle t has magnrtude (.r: m'ur1 sin 01 : mu(d+h)and it is into the page. The angular momentum of par-
ticle 2 has magnitude (.2
is out of the page. The net angular momentum has
magnitude Lt0-4 kgX5 .46mlsx0 .0420m) : 6.65x t0-s kg .m2 lr'and is into the page. This result is independent of the
location of the origin.
(c) and (d) Suppose particle 2 is traveling to the right. Then L - mu(d,+h)+muh - mu(d+2h).This result depends on h, the distance from the origin to one of the lines of motion. If the originis midway between the lines of motion, then h - -d,12 and L:0.
77
Use conservation of energy. If the wheel moves a distance d along the incline its center of mass
drops a vertical distance h - d stn 0, where 0 is the angle of the incline. The potential energy
of the Earth-wheel system changes by LLI : -mgh - -mgd sin 0, where m is the rnass of the
wheel. The change in the kinetic energy is L^K : lm,u!"m + )to', where ucom is the final speed
of the center of mass, u is the final angular speed of the wheel, and I is the rotational inertiaof the wheel. Since the wheel rolls without sliding ucom : u)r, where r is the radius of the axle.
Thus A,K : Lr'(mr2 + /). Since energy is conserved mgd,sin 0 - io'(*r2 + /) and
2 2mgd,sin 0 2(10.0 kgXg .8^ls1(2.00 m) sin 30ou-: : :
m,r2 + I (l0.0kg)(0.200 )2 +(0.600kg .m2)
(a) The rotational kinetic energy is Krot: itr': ltO.600kg. m2X196 ndz lt'):58.8J.(b) The square of the speed of the center of mass is ,?o - a2r2: (196 rud2 lrtx0.200m)2 -7.84m2 lr' and the translational kinetic energy is Krrunr- **r'.r^: lCf 0.0kgX7.84 m2 lr')39.2 J.
85
(a) In terms of the radius of gyration k the rotational inertia of the meffy-go-round is I - M k2
and its value is (180kgX0.910*)t - l49kg . m2.
(b) Recall that an object moving along a straight line has angular momentum about any pointthat is not on the line. Its magnitude is mud, where m rs the mass of the object, u is the speed ofthe obj ect, and d is the distance from the origin to the line of motion. In particular, the angularmomentum of the child about the center of the meffy-go-round is L"- mu&, where R is the
radius of the meffy-go-round. Its value is (44.0kg)(3.00 mls)(l .20m): 158kg' m2 ls.
68 Chapter I I
td,
+
h
t
(c) No external torques act on the system consisting of the child and the merry-go-round, so
the total angular momentum of the system is conserved. The initial angular momentum is given
by muR; the final angular momentum is given by (/ + mRz)u, where a is the final commonangular velocity of the merry-go-round and child. Thus mu R - (/ + mR2)a and
muR 158 kg . m2 lsa- :w I+mRz r4gkg. m2+(44.0kg)(r.zlm)z -0'744radf s'
87
The car is moving along the n axis, going in the negative r direction. Let r- be the vector fromthe reference point to the particle and u* be the velocity of the car. Then the angular momentum
of the car is given by the vector product mi x 6, which is -mAu, t since d has only an rcomponent and, in all cases, r- has only r and A components. According to Newton's second
law in angular form the torque is the rate of change of the angular momentum.
(a) and (b) The reference point is at the origin, so A - 0 and f - 0. The angular momentum isconstant so the torque is also zero.
(c) The reference point is a distance lil - 5.0m from the r axis so the magnitude of the angularmomentum is ( - (3.0kgX2 .0^ls4)f3(5.0m: (30kg. m2 Ito)tt. Since y is positive the angular
momentum is in the negative z direction. Thus f - -(30 kg . m2 f sa1t3 k.
(d) The torque is
#L-(30kg .m,lsa)t3 t<l - -(90kg . nl f sa1t2k.
(e) and (f) The reference point is the same distance from the r axis as in parts (c) and (d), so the
Tagnitudes of the angular momentum and torque are the same. Now, howeveq A is negatiy., so
I ad, i are in the positive z direction. Thus [- +(30kg .m2 |to)tt k and i - +(90kg. m2 f sa\tzk.
95
Two horizontal forces act on the cylinder: the applied force Fupp in the positive r direction
and the frictional force / along the r axis. Newton's second law for the center of mass isFapp + f * : mc;catrt, nt where n'L is the mass of the cylinder and acom n is the r component of the
acceleration of its center of mass. 4oo acts the top of the cylinder and i acts at the bottom,both at the rim, so the magnitude of the net torque on the cylinder is R(F - f ") and Newton'ssecond law for rotation is R(F f ") - f a, where I is the rotational inertia of the cylinderand a is its angular acceleration. Since the cylinder rolls without sliding the acceleration of the
center of mass and the angular acceleration ate related by ocom
simultaneously for ocom, e,) and f*. The solutions are
Ocom :2R2 Fapp
mR2+I)2RFupp
Q.:mR2+I)
Chapter II 69
and
r 2mR2 Fupp nJn mR'z+I -r '
According to Table l0-2 the rotational inertia of the cylinder for rotation about its central axisis given by I : i^R'. Substitution of this expression leads to
4Fapp 4(12N) 1r- rZacom: G- 3(lokg) - r.om/s )
e,:4Fapp: 4(12N) -1r--^sr-Z3mR 3(10 kg)(O. 1o m) I o raol s )
and
f ,- Furo- + - 4.0N.33Sjnce f" is positive, the frictional force is in the positive r direction and can be written
f : (4.0 N) i.
70 Chapter I l
Chapter L2
-5
Three forces act on the sphere: the tension force f of the rope
(which is along the rope), the force of the wall Fr (which is hori-zontally away from the wall), and the force of gravity mj (which isdownward). Since the sphere is in equilibrium they sum to zero. Let0 be the angle between the rope and the vertical. Then the verticalcomponent of Newton's second is 7 cos 0 - mg - 0. The hoizontalcomponent is trl/ - T sin 0 :0.(a) Solve the first equation for T: T - mg f cos 0. Substitute cos 0 -Lltffi to obtain
rng (0.85 kgX9.8 m/s2y
TL
- 9.4N.
(b) Solve the
,l/t2+12to
0.080 m
second equation for 'Fli: F1r - T srn?. Use sin0obtain
F1u. -Tr 4.4N.L NTT+F L 0.080 m
7The board is in equilibrium, so the sum of the forces and the sum of the torques on it are each
zero. Place the r axis along the diving board. Take the upward direction to be positive. Take
the vertical component of the force of the left pedestal to be F1 and suppose this pedestal is at
n - 0. Take the vertical component of the force of the right pedestal to be F2 and suppose thispedestal is at r: d. Let W be the weight of the diver, applied at tr: L. Set the expression forthe sum of the forces equal to zero:
F1* Fz W :0.Set the expression for the torque about the right pedestal equal to zero:
&d+w(L - d) :0.(a) and (b) The second equation gives
F1: -+\\-: (H) (s8oN): -t2x ro3N
The result is negative, indicating that this force is downward.
+r2
\\
l
L2+12
Chapter 12 7l
(c) and (d) The first equation gives
F2: W - il -580N+ 1.2 x 103N- 1.8 x 103N.
The result is positive, indicating that this force is upward.
(e) and (0 The force of the diving board on the left pedestal is upward (opposite to the forceof the pedestal on the diving board), so this pedestal is being stretched. The force of the divingboard on the right pedestal is downward, so this pedestal is being compressed.
11
Place the r axis along the meter stick, with the originat the zero position on the scale. The forces on it are
shown on the diagram to the right. The coins are at
n : nt (: 0.120 m) and m is their total mass. The knifeedge is at rmass of the meter stick is M and the force of gravityacts at the center of the stick, fr : n3 (: 0.500m).
Since the meter stick is in equilibrium the sum of the
torques about rz must vanish: M g(U - r) - mg(*z -nr) - 0. Thus,
M-fr2-frtfr3-fr2
0.455 m - 0.120mm:(0.500m-0.455m
(10.0 g) : 74.4 g.
2tConsider the wheel as it leaves the lower floor. There is no
longer a force of the floor on the wheel , and the only forces
on it are the force F applied horizontally at the axle, the forceof gravity mg vertically downward at the center of the wheel,and the force of the step corner, shown as the two components
fn and fr. If the minimum force is applied the wheel does notaccelerate, so both the total force and the total torque on it are
zeto.
Calculate the torque around the step corner. Look at the second
diagram to see that the distance from the line of F to the
corner is r h, where r is the radius of the wheel and h isthe height of the step. The distance from the line of mg tothe corner r'2+? -hY - ffi. Thus F(r h)-mgffi - o. The solution for F is
mg
2(0.0600 mX0.0300 m) - (0.0300 m)2..\
rfS
o
is
F:
Ms
2rh - h2
r-h
72 Chapter I2
0.0600m - 0.0300m0.800kgX9 .8*ls') - 13.6N
33
(a) Examine the box when it is about to tip. Since it willrotate about the lower right edge, that is where the normalforce of the floor is applied. This force is labeled Fl,. on
the diagram to the right. The force of friction is denoted
by f , the applied force by F, and the force of gravity byW. Note that the force of gravity is applied at the center
of the box. When the minimum force is applied the boxdoes not accelerate, so the sum of the horizontal forcecomponents vanishes:
F-f:0,the sum of the vertical force components vanishes:
Fl,. -W:0,and the sum of the torques vanishes:
FL-W-0.2
Here L is the length of a side of the box and the origin was chosen to be at the lower right edge.
Solve the torque equation for F:
p -W - 89oN : 445N.22
(b) The coefficient of static friction must be large enough that the box does not slip. The box is onthe verge of slipping tf lt, - f I F*. According to the equations of equilibrium Fn : W - 890 NI
and f : It - 445 N, so F, - (445 N) l$90 N) - 0.50.
(c) The box can be rolled with a smaller applied force ifthe force points upward as well as to the right. Let 0 be
the angle the force makes with the horizontal. The torqueequation then becomes F Lcos 0 + F Lsin 0 -W L 12 : 0,
with the solution
It_2(cos 0 + sin0)
You want cos0+ sin0 to have the largest possible value. This occurs if 0_ 45o, a result youcan prove by setting the derivative of cos 0+ sin0 equal to zero and solvingfor 0. The minimumforce needed is
F- W 89ON- 315 N.
W
4 cos 45o 4 cos 45"
43
(a) The shear stress is given by F lA, where F is the magnitude of the force applie d parullel toone face of the aluminum rod and A is the cross-sectional area of the rod. In this case F is the
Chapter 12 73
weight of the object hung on the end: F - ffig, where m is the mass of the object. If r is theradius of the rod then A - nr2. Thus the shear stress is
F mg (1200 kgX9.8 m/s2) .
i- #- :6.5 x ro6N/*t.
(b) The shear modulus G is given by
FIAG- LrlL)where L is the protrusion of the rod and A,r is its vertical deflection at its end. Thus
L^r:(Fl!)L_: (6.5 x 106N/m2X0.053m)-1 1 x 10-5m.G 3.0 x 1010 N/m2
f-
55
(a) The forces acting on the bucket are the force of gravity, down, and the tension force of cableA, up. Since the bucket is in equilibrium and its weight is Wg8.01 x 103N, the tension force of cable A is Tt:8.01 x 103N.
(b) [Jse the coordinate axes defined in the diagram. Cable A makes an angle of 66" with thenegative A axis, cable B makes an angle of 27" with the positive U axis, and cable C is alongthe r axis. The A components of the forces must sum to zero since the knot is in equilibrium.This means Ts cos 27" - Ttcos 66" - 0 and
rs:ffiffi rt: (#) (8 or x lo3N)- 3 6sx ro3r{
(c) The r components must also sum to zero. This means Tg *Tn sin27" -Tasin 66o - 0 and
Ts:TAsin 66" - T6sin 27" - (8.01 x 103N)sin 66" - (3.65 x 103N)sin 27o - 5.66 x 103N.
61
(a) The volume of the slab is (43mxl2m)(2.5m)- 1.29 x 103m3 and,
3.2 x 103 kgl*t, its mass is mcomponent of the gravitational force parallel to the bedrock surface is
106 kgXg .8^lrt; ritr 26"
(b) The maximum possible force of static friction is f ^u*- F'Fw, whereof static friction and Fli is the norrnal force of the bedrock surface onsecond law (with the acceleration equal to zerc) gives the norrnal force as
F,mg cos 0 - (0.3 9)(4. 13 x 106 kgX9 .8 ^lrt; ro s26"
(c) The bolts must support a total shearing force of 1 .77 x I07 N - I.42 xEach bolt ean support a shearing force of (3.6 x 108 N/*t )((6.4 x 10-4 m2
74 Chapter I2
since the density is
4.13 x 106 kg. Themg sin d
F, is the coefficientthe slab. Newton'smgcosO, so f^* -
lo7 N - 3.5 x 106N.:2.3 x 105 N so the
number of bolts required is (3.5 x 106N/(2.3 x 10sN): 15.2. Round up to the nearest integer:
16 bolts are required.
63
Let Ts be the tension in the horizontal cord, f be the magnitude of the frictional force on blockA, .Fl/ be the magnitude of the normal force on that block, and ?rL4 be the mass of that block.Assume block A is stationary. Then Newton's second law for block A gives Ta f - 0 andFl/ - m.,q,g
ltrFx, whete F, is the coefficient of static friction between the block and the table surface. Thismeans Tn must be less than Frmeg.Now consider block B. Let Ts be the tension in the cord attached to it and M6 be its mass.
Newton's second gives T6 - wL6g:0, so Ts: TrLp!.
Next consider the knot where the three cords join and let T be the tension in the third cord.
Newton's second law gives 7 sin 0 - TaTequation gives Ta -T sin 0 - ffLpgsin 0f cos0 - TTLsgtan?.
If block B does not slip Tn pgtan? must be less than FTTTLA7. Since TTL7 is the greatest it can be
without block A slippitrg, TtLs tan 0 - Fsmag or
D-,
rh,l\tlttI f, ___l-t <-!*g o
|f,"' v ITLy 10 kg
65
The force diagram for the rod is shown on
force of the rope and Fn is the force of1 80" 0r 0z: I 80" 60o - 0z: I20"hinge is TL sina - mg(Llz)sin01 and thisbe in equilibrium. Thus
the right. T is the tensionthe hinge. The angle u is
02. The net torque about the
must be zero if the rod is to
:. ms(L 12) sin 91 ms(L 12) sin 91srnu- rL _ sin 01 )
since T1 - mglz. This means a: 0t and 02: I20o - 60" : 60o
81
The force diagram for the cube is shown on the right. Fr is thenoffnal force of the floor on the cube, ,f ir the force of frictionof the flooq and m is the mass of the cube. Assume the cubeis stationary but it is about to tip. Only the lower right edge
of the cube exerts a force on the floor and the line of actionof the normal force is through the right side of the cube. Thehorizontal somponent of Newton's second law for the center of
T
I
I
I
I
I
I
Chapter 12 75
mass is P fP(. - mgL12, where (. is the distance between O and the point of application of P.
(a) The cube slides if f is greater than F'Fw, where F, is the coefficient of static frictionbetween the floor and the cube. According to the Newton's second law equations f : P and
,Fl,. : n'Lg. Thus sliding occurs if P >for the cube to slide but not tip as P increases Fsrng must be less than mgLl2l. ot Lr, must be
less 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.
85
The force diagram for the ladder is shown on the right. Fs is the force
of the ground on the ladder, F* is the force of the wall, and W is the
weight 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 lever
arrn of h, the vertical component of the force of the ground has a lever
affn 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 the
gravitational force has a lever ann of (. 12. Thus Fn*h - Fnrl+ Fh(l - d,lL) + WtlT - 0. The
r axis was taken to be horizontal with the positive r direction to the right and the A axis was
taken 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 means
75N-(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 negative
r direction. Its magnitude is (0.80)F 75N. If the ladder does not slip this must be less
than 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
Chapter L3
1
The magnitude
and Tfi,2 are the
for r:
of the forcemasses, T is
of one particle on the other is given by It - Gmflt Lz I ,', where TrLl
their separation, and G is the universal gravitational constant. Solve
^ 1 rc.67 x 1 0- 1r N . m2 lke\(s .2ke)e.4 kg)
7
At 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 center
of 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 find
M"-,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 kg
C.Values for A[., Mr, and d can be found in Appendix
17
_ d\M - (lso x toem)
-T1
The gravitational acceleration is given by as: GMf r2, where
is 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 mass
R 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)
itsThe density of aradius. The ratio
uniform sphere is givenof the density of Mars to
Pnrpn Mp Rtu
by p : 3M l4nR3, where Mthe density of Earth is
3
- 0.74.I
iIX
5
5
.6
A0j
km
k*g+
0,
GmtffLz
F
(6.67 x 10-ll m3 lr' . kgX5.98 x L024 kg)
Chapter 13 77
(b) The value of an at the surface of a planet is given by ag: GMIR', so the value for Mars is
Mxa R12
(9.8 mlr2) - 3.8 ^lt'lVL r t'F.
A7I/I- ,r- t' aI
M
(c) If u is the escape speed, then, for a particle of mass m
WHagE:0'11km
k*
Gry
g+
0,rX
5
5
.6
70
1
**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 is
i*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 yields
GMm I-;+;*,2:
Replace GM with onR2 and cancel m in the energy
-asR+)*:
GMmR+h
equation to obtain
onR2
(R+ h)
The solution for h is
h- 2anR2
2onR - u2 -R
- z(g.o m/s2)(soo x to3 m)2
2(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)
(c) Initially the particle is aU, - -GMml@+ h) and
the potentral energy is U y
of energy yields
The solution for u is
u-
distance h above the surface and
the initial kinetic energy is K,; --GMmlR.Write *.*r'r for the
GMm GMm
is at rest. The potenttal energy is
0. Just before it hits the asteroid
final kinetic energy. Conservation
1.+ - mL)"
2R+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 T6
The 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 radius
of either of the stars. The final potentral energy is given by LI 1 - -GM'lrf and the energy
equation 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)2
500 x 103m+ 1000 x 103m
(6.67 x 10-rr m3 lr'.kgX103o kg)
10lo m
Chapter 13 79
45
Let 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-GTzM
The 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) andRp
Ro - 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 m
6t
(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 kinetic
80 Chapter 13
4r213
4n2(300 x 10e m)3
(6.67 x 10- 11 m3 I ,' . kgX r .gg x 1030 kg)
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/ 4.20X107m
where 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.
79
Use 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
83
(a) The centripetal acceleration of either star is given by o, - u2r, where w is the angular speed
and 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 is
the 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) is
1o-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) is
IJora
The 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
so
x 10ll m - 3 .46 x 10-12 m-1
Ke: GMnMs [;
Ke - (6.67 x
- 2.74 x
and the speed is u,p
82 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,t
Tt
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.
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.
93
Each star is a distance r from the central star and a distance 2r from the other orbiting star, so
it is attracted toward the center of its orbit with a force of magnitude
rt-GM: *ry12 (2r)2 4r2 \r-r
According 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 ? is4rr3 /2
Chapter 13 83
Chapter 14
1
The air inside pushes outward with a force given by piA, where Pi is the pressure inside the room and A is the area of the window. Similarly, the air on the outside pushes inward with a force given by PoA, where Po is the pressure outside. The magnitude of the net force is F = (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 .
J. The change in the pressure is the force applied by the nurse divided by the cross-sectional area of the syringe:
F F 42N 5 D.p = A = 7fR2 = 7f(1.1 X 1O-2m)2 = 1.1 x 10 Pa.
11 The pressure P at the depth d of the hatch cover is Po + pgd, where p is the density of ocean water and Po is atmospheric pressure. The downward force of the water on the hatch cover is (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 be applied 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.
19
When the levels are the same the height of the liquid is h = (hI + h2)/2, where hl and h2 are the original heights. Suppose hI is greater than h2. The final situation can then be achieved by taking liquid with volume A(hl - h) and mass pA(hl - h), in the first vessel, and lowering it a distance h - h2. The work done by the force of gravity is W = pA(hl - h)g(h - h2)' Substitute h = (hI + h2)/2 to obtain
27
W = ~pgA(hl - h2)2
= ~(1.30 X 103 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,
84 Chapter 14
and Uz to be at the top of the atmosphere, where the pressure ts pz: 0. Take the density to be
1.3 kg/*'. Then,
Uz-Atv 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. Nowtheintegral
Pz : pr - lr^ Poe(r I) da : pr * pogh .
Since p2 : 0, this means
2pr 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 because
the 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. Substitute
V"- 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:
Thus msphere, or V"
4nllJ?, : Ol'-
6 r'o (?) (8oo kel^'xo oeom)3 - r zkg
Air 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 m3V-
The density ism L.22kg
n- :r v 9.09 x lo-a m31.3 x 103 kgl^t
49
Use the equation of continuity. Let u1 be the speed of the water in the hose and u2 be its speed
as it leaves one of the holes . Let A1 be the cross-sectional area of the hose. If there are l/ holes
you 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 the
equation of continuity becomes urAtlN: uzAz. Thus u2: (AtllVAr)ur - (R2 llYr')ut, where
R is the radius of the hose and r is the radius of a hole. Thus
fi (0.9s cm)2u2: Tfur,
s3
Suppose 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)r N Ar \'"'," )Now the rate of mass flow is L* lAf - pAu, where p is the density of water and A is the
atea of the hose. The area of the hose is ApAuobtained from Table I4-I of the text. Thus
/ 1rt)P - 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
--33
(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 area
Thus u2
where 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-l
59
(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 the
altitude 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 the
speed 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 speed
where 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 has
fallen 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
(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.
75
Let p (: 998 kg/-') be the density of water and pt (: 800 kg/-') be the density of the other
liquid. Let d*r, be the length of the water column on the left side, d*n (: 10.9cm) be the
length of the water column on the right side, and ds (: 8.0 cm) be the length of the column
of 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, so
po * 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^t
Before the other liquid is poured into the tube the length of the water column on the left side is
the 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
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 arn
5.7 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 Frn
2
The 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-r2n
For frequencies greater than
9.8 ^lt' :498H2.
acceleration exceeds g for some part of the motion.
I l-s: 2"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- mw2rTn
acceleration ) u) is the angular frequency, and f is the frequency. The relationship a : 2n f wasused to obtain the last form.
Chapter 15 89
Substitute It - m(2nf)'r,n and F^r - mg into F <largest amplitude for which the block does not slip is
rnl- F'9 . : (o'soX?'8 T/s1) - o.o3 1 m.(2n f)2 (2n x 2.0Ht)2
A larger amplitude requires a trargw force at the end points of the motion. The surface cannot
supply the larger force and the block slips.
t9(a) Let
A / 2trt\n1 -zcos ( , /
be the coordinate as a function of time for particle 1 and
12: Iror(+.;)be the coordinate as a function of time for particle 2. Here T is the period. Note that since the
range of the motion is A, the amplitudes are both A 12. The arguments of the cosine functionsare in radians.
Particle 1 is at one end of its path (*t - Al2) when tZntlT+116 - 0 or t- -f lI2. That is, particle 1 lags particle 2 by one-twelfth a period. We
want the coordinates of the particles at t - 0.50 s. They are
nt :4 ro, (2zr x o'50 s)I 2v\'\)\ l.5s )
and A lznxo.5os zr\fr2:^cost -+= l--0.4334.2 \ 1.5s 6/
Their separation at that time is n1 - n2: -0.2504+ 0.433A - 0.1834.
(b) The velocities of the particles are given by
?,1 -drt:4 (Znt\
dt T-sm\ t /and
27
When the block is at the end of its path and is momentarily stopped, its displacement is equal tothe amplitude and all the energy is potential in nature. If the spring potential energy is taken tobe zero when the block is at its equilibrium position, then
E - lrrl: )r.t x t02N/,o) (0.024m)2 : 3.7 x t0- 2 I .
90 Chapter 15
Lt't- d"
-z- dt T \T 6/Evaluate these expressions for t - 0.50 s. You will find they are both negative, indicating that
the particles are moving in the same direction.
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 is
itt*',"-itt*',"
The fraction of the energy that is kinetic is
E-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 the
amplitude 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 s39.5 radf s .
(b) When 0 - n12, then 010,"- l12, cos(ZntlT): I12, and
1
4v-E
-1 Y-1 1- 3-
844EKE
a_o - -+ o*sin
T (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. Thus
sin(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 is
0,n cos(2rt lT) -
When 0-n14,
Again the negative
- 1 24 radlr'
Chapter 15 9l
43
(a) A uniform disk pivoted at its center has a rotational inertia of + M R2, where M is its mass
and R is its radius. See Table I0-2. The disk of this problem rotates about a point that
is displaced from its center by R + L, where L is the length of the rod, so, according to the
parallel-axis theorem, its rotational inertia is Lm n2 + M (L + R)2. The rod is pivoted at one end
and has a rotational inertia of mL'13, where m is its mass. The total rotational inertia of the
disk 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, on
the 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.
51
If the torque exerted by the spring on the rod is proportional to the angle of rotation of the rod and
if the torque tends to pull the rod toward its equilibrium orientation, then the rod will oscillate
in simple harmonic motion. If r -- -C0, where T is the torque, 0 is the angle of rotation, and
e 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. This
immediately 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 the
spring. 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 the
length of the spring is given by (,0+ L0 12 and its elongation is Lr: L0 12. The force it exerts
on the rod has magnitude It - k Lr- kL?f 2, where k is the spring constant. Since 0 is small
we may approximate the torque exerted by the spring on the rod by r: -FLf 2, where the pivot
point was taken as the origin. Thus r - -(kL' lqg. The constant of proportionality C that
relates 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. See
Table L0-2. Thus the period of oscillation is
t'r1 -^ trt--t"l ,:2rChapter 15
+ m)sd
0.205 kg . m2
(0.500 kg + 0 .270kgX9 .8^ls2x0 .447 m)
mL2 l12kL2 l4
0.600 kg
3(18s0 N/*)
92
- 0.0653 s
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 was
its reciprocal. Thus
(b) The angular frequency is
t_ 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 of
the 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 the
same 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
, mw - V * 4rrr':
- 0.39H2.
(c) Now the acceleration due to gravity and the acceleration of the elevator are in the same
direction 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 'uTL
stroke, 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/-.
^Tfr2n
8.00 N/* (0.230 kg/s)2
1.50 kg 4(1.50 kg)'
9.8 mlrt + 2.0^lt'
Chapter 15 93
(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 so
the amplitude is 5.0 cm.
(e) The block has maximum speed as it passes the equilibrium point. At the initial position, the
block 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 the
potential energy is
1, ^ I I(,1 y- -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 the
same 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
Chapter L5
15
The 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- tFrtp
and the angular wave number is
l4lm-r
(c) The frequency is f :100H2, so the angular frequency is a - 2nf :2r(100H2):628rad/s.
(d) The positive sign is used since the wave is traveling in the negative r direction.
2t(a) Read the amplitude from the graph. It is the displacement at the peak and is about 5.0 cm.
(b) Read the wavelength from the graph. The curye crosses A - 0 at about r : 15 cm and agatnwith the same slope at about tr- 55cm, so )- 55cm- 15cm:40cm-0.40m.(c) The wave speed is
where T is the tension in the string and pr, is the linear mass density of the string. Thus
u-c.t
0.0300 kg lm
10-3kglmu- - L2mf s.
Chapter 16 95
(d) The frequency is
and the period is
( e) The maximum string speed is
v 12m/s f = - = = 30Hz A 0.40m
1 1 T = - = -- = 0.033 s .
f 30Hz
U m = WYm = 27i}Ym = 2IT(30Hz)(5.0cm) = 940cm/s = 9.4m/s.
(f) The angular wave number is
27i 27i k = - = = 16m- I .
A 0.40m
(g) The angular frequency is w = 2IT f = 2IT(30 Hz) = 1.9 X 102 rad/s.
(h) According to the graph, the displacement at x = 0 and t = 0 is 4.0 X 10-2 m. The formula for the displacement gives y(O,O) = Ym sin ¢. We wish to select ¢ so that 5.0 x 10-2 sin ¢ =
4.0 X 10-2. The solution is either 0.93 rad or 2.21 rad. In the first case the function has a positive slope at x = 0 and matches the graph. In the second case it has negative slope and does not match the graph. We select ¢ = 0.93 rad.
(i) A positive sign appears in front of w because the wave is moving in the negative ;]; direction.
31
The displacement of the string is given by
Y = Ym sin(kx - wt) + Ym sin(kx - wt + ¢) = 2Ym cos(~¢) sin(kx - wt + ~d»,
where ¢ = IT /2. The amplitude is
A = 2Ym cos(~¢) = 2Ym cos (IT /4) = 1.41Ym·
35
The phasor diagram is shown to the right: Yl m and Y2m represent the original waves and Ym represents the resultant wave. The phasors corresponding to the two constituent waves make an angle of 90° with each other, so the triangle is a right triangle. The Pythagorean theorem gives
y~ = Y?m + Yim = (3.0 cm)2 + (4.0 cm)2 = 25 cm2 .
Thus Ym = 5.0 cm.
41
Possible wavelengths are given by A = 2L / n, where L is the length of the wire and n is an integer. The corresponding frequencies are given by f = v / A = nv /2L, where v is the wave
96 Chapter 16
T
LM
speed. The wave speed is given by u -tt is the linear mass density of the wireobtain the last form. Thus
where T is the tension in the wire,of the wire. p, : M I L was used to
tm,-and M is
llrLlM,the mass
- n(7.glHr)
(a) For rL : 1, f - 7 .9I Hz.
(b) For ft:2, f : 15.8 Hz.
(c) For fr:3, f :23.7H2.
43
(a) The wave speed is given by u - \mt, where r is the tension in the string and p is thelinear mass density of the string. Since the mass density is the mass per unit length, F : M I L,where M is the mass of the string and L is its length. Thus
- 82.0 m/s .
(b) The longest possible wavelength .\ for a standing wave is related to the length of the stringby L - ,\f2, so )- 2L-2(8.40m): 16.8m.
(c) The frequency is f - ul^ - (82.0^ls)/(16.8m) - 4.88Hz,.
47
(a) Thc resonant wavelengths are given by ^
- zLf n, where L is the length of the string and nis an integer, and the resonant frequencies are given by f - ul^ - nnlzL, where u is the wavespeed. Suppose the lower frequency is associated with the integer n,. Then since there are noresonant frequencies betweeo, the higher frequency is associated with n*1. That is, fr: TL't)l2L
is the lower frequency and fz: (n * l)u lzL is the higher. The ratio of the frequencies is
fz: n+rfin'
The solution for n' is rL : - f ,fz h 420H2-3I5Hz r'
The lowest possible resonant frequency is f - uf 2L - hl, - (3 l5Hz)13
(b) The longest possible wavelength is ) _ 2L. If f is the lowest possible frequency thenu - xf - 2Lf - 2(0.7s mX105 Hr) - 158 m/s.
53
The waves have the same amplitude, the same angular frequency, and the same angular wavenumber, but they travel in opposite directions.
^TL+r2L(10.0 mX0.100 kg)
(96.0 NXS.40 m)
0.l20kg
Chapter 16 97
(a) The amplitude of each of the constituent waves is half the amplitude of the standing wave or0.50 cm.
(b) Since the standing wave has three loops the string is three half-wavelengths long. If L is thelength of the string and
^ is the wavelength, then L -3^12, or ,\ - 2L13:2(3.0m)13 -2.0m.
The angular wave number is k - 2nl) - 2nl(2.0m):3.1m-l.(c) If u is the wave speed, then the frequency is
u 3u 3(100 m/s)I: )- n- ,(3.0"r)
The angular frequency is u):2nf :2r(50H2) - 3.1 x I02rudf s.
(d) Since the first wave travels in the negative r direction, the second wave must travel in thepositive r direction and the sign in front of u must be a negative sign.
6t(a) The phasor dtagram is shown to the right: Ut, Uz, and Us
represent the original waves and Urn represents the resultantwave. The honzontal component of the resultant is arnh
Ut Az - Ur arl3 - 2yrf 3. The vertical component is
U*, : U2 :
Ar f 2. The amplitude of the resultant is
Urn:m:
I
5: ,al : 0.83ato
-1 3-tan-r:-0.644rad:37"
4
5A : 6At
stn(kr - wt + 0 .644 rad) .
The graph below shows the wave at trme / - 0. As time goes on it moves to the right with speed't) : a lk.
(b) The phase constant for the result ant is
6 _tan-r u*, _ tan-r ( ^M^\arnh \ 2Yr 13 /(c) The resultant wave is
-U'n-Ut
Chapter 1698
69
(a) Take the form of the displacement to be A@,t): arnstn(kr - at). The speed of a point on
the cord is u(r, t) - 0y l 0twave speed, on the other hand, is given by, -
^lf - alk. The ratio is
urn :u
(b) The ratio of the speeds depends
Different waves on different cords have
and wavelength, regardless of the wavethe cords.
77
only on the ratio of the amplitudethe same ratio of speeds if they have
speeds , linear densities of the cords,
aArn 7
n - Ngrn:alK
2narn
to the wavelength.the same amplitudeand the tensions in
(a) If r is the tension in the wire and LL
a
length. Thus
is its linear mass density,
mlL, where m is the mass
then the wave speed is
of the wire and L is its
(b) A one-loop standing wave has two nodes, one a each
apart, so the wavelength is ^
- 2L - 2(1 .50m) - 3.00m.
: I44mls .
end, and these are half a wavelength
A two-loop standing wave has three nodes, one at each end and one at the midpoint. Since the
nodes are half a wavelength apaft, the wavelength is ,\ - L - 1.50 m.
(c) and (d) The frequency is f - ulx. For the one-loop wave f : (I44mls)l(3.00m):48.0H2and for the two-loop wave f : (I44mls)/(1.50m) :96.6H2.
87
(a) The transverse rope velocity is given by , : aUrn, where a is the angular frequency and Arnis the amplitude. The angular frequency is a - 2n f, where f is the frequency. Thus
uuoa : :ant a Znf 2r(5.0 Hz)
5.0 m/s- 0.I6m.
(b) The wave speed is u - \trn where r is the tension in the rope and LL is the linear mass
density of the rope. The linear mass density is l.L- mlL, where m is the mass of the rope and
L is its length. The wave speed is ^f
, where ^
is the wavelength. Since the rope is vibrating inits fundamental mode ) - 2L Thus
r-puz-irxrf(c) Thc general forrn for the displacement at coordrnate r and time t for a standing wave that has
nodes atn -0and r- Iis A:Amstn(2rrlDsrn(2nft. Here Un isthemaximumdisplacementof any of the points along the rope. Since, in this case, the rope is vibrating in its fundamental
(r20NX1.s0m)8.70 x 10-3 kg
Chapter 16 99
mode Urn is the maximum displacement of the point at its center and thus has the value calculatedin part (a). The displacement at any coordinate r is
a@,t):(0.16m)sin|#")sin|2rr(5.0Hz)t]-(0.I6m)sin[(1.6m-')"]sin[(31s-,)'].
89
(a) The wave speed is given by u- \mu where r is the tension in the rubber band and lr isthe linear mass density of the rubber band. According to Hooke's law the tension is r - k L(..The length of the stretched rubber band is (.+ L(,, so the linear mass density is l.L: ml!+ Lt).The wave speed is
(b) The time for a pulse to travel the length of the rubber band is
(.+ alt_T_((,+Lt)
[f L(. is much less than (, we may neglect the LI in the numerator. Then
which is proportional to L l161-.(c) It L(. is much greater than (, we may
u:
which is independent of L(..
in the numerator. Then
k L(,(t + Lt)
m((+ Lt)k L(,((,+ L!)
m[.
KN
100 Chapter 16
Chapter L7
-5
Let t y be the time for the stone to fall to the water and t, be the time for the sound of the splash
to travel from the water to the top of the well. Then the total time elapsed from dropping the
stone to hearing the splash is t_ ty *tr. It d is the depth of the well, then the kinematics offree fall gives d : *gt?, or t y
or f" : d/ur.Thus the total time is
t-
This equation is to be solved for d. Rewrite it as
dt-us
and square both sides to obtain
dI
us
2d, .-lL L
I -z-t d*\a'as u'"
NIow multiply by gu? and realrange to get
- Zu,(gt I rr s)d + gu?* : 0 .gd2
This is a quadratic equation for d. Its solutions are
d- 2u r(gt | 'u s) *
The physical solution must yieldin front of the square root. Once
2s
d : 0 for t : 0, so we take the
values are substituted the resultthe negative signobtained.
solution withd-40.7mis
1_
If d, is the distance from the location of the earthquake to the seismograph and u" is the speed ofthe S waves, then the time for these waves to reach the seismograph is ts: d/ur. Similarly, thetime for P waves to reach the seismograph is tp: d/up.The time delay is
Af - d d
--us up
d(up - u ") )
u sup
2d
g
ar?@t + ur)2 4g2u?t2
Chapter 17 101
SO
.l _ lrsup Lt (4.5kn'lsx8.0 km/sX3.0 minx60 s/min) 1 ^u-
Notice that values for the speeds were substituted as given, in km/s, buttime delay was converted from minutes to seconds.
2(a) Use
^ - ulf, where u
x 103 km.
that the value for the
(b) Now ^
-- ulf , whereand tissue. Thus
is the speed of sound in air and f is the frequency. Thus
\ _ 343mls/t
- 4.5 x 106 H;- 7 '62 x 10-5 m '
u is the speed of sound in tissue. The frequency is the same for air
\ _ 1500 m/slt
- 4.5 x 106 Hr- 3 '33 x 1o-4 m '
t9Let Lr be the distance from the closer speaker to the listener. The distance from the other
speaker to the listener is L2: lW d is the distance between the speakers. The
phase difference at the listener ls, 2n(Lz - Lt)
where ^
is the wavelength.
(a) For a minimum in intensity at the listener, 0 - (2n + l)n, where n is an integer. Thus
"\ - 2(Lz L) lQn + 1). The frequency is
t _ u (2n + l), (2n + 1X3 a3 mls)J r - (2n+ 1X343 Ht).) 2lrry-1,1 2l -37sm]
I
To obtain the lowest frequency for which a minimum occurs set n equal to 0. The frequency is
f^in,l : 343H2.
(b) To obtain the second lowest frequency set n equal to 1. This means multiply F*i* I by 3.
(c) To obtain the third lowest frequency set n equal to 2. This means multiply lr",in, I by 5.
For a maximum in intensrty at the listener, d : 2nn, where n is any positive integer. Thus
)-:lEa-r']and
u nu n(343 m/s):::- n(686 Hz) .
lW-11 (3.7 5 m)2 + (2.00 m)2 - 3.7 5 m
(d) To obtain the lowest frequency for which a maximum occurs set n equal to 1. The frequencyis f^u*,I : 686 Hz.
102 Chapter 17
(e) To obtain the second lowest frequency set n equal to 2. This means multiply f-in, 1 by 2.
(0 To obtain the third lowest frequency set n equal to 3. This means multiply .F*6, t by 3.
25
The intensity is the rate of energy flow per unit area pe{pendicular to the flow. The rate at whichenergy flows across every sphere centered at the source is the same, regardless of the sphere
radius, and is the same as the power output of the source . If P is the power output and I is the
intensity adistance r fromthesource,then P- IA:4rr2 I,where A(:4rr2) isthesurfacearea of a sphere of radius r. Thus P - 4r(2.50 m)2(1.91 x 10-4W/ttr') - 1.50 x I0-2'W'.
29
(a) Let 11 be the original intensity and 12 be the final intensity. The original sound level is
hreference intensity. Since Cz : Cr + 30 dB,
or
(10 dB) Iog(I2l Iil - (10 dB) log(/t I IO + 30 dB ,
(10 dB) tog(I2l Iil - (10 dB) log(/r I Iil - 30 dB .
Divide by 10dB and use log(I2lIil - log(hlIo): Iog(I2lI) to obtain log(I2lIr):3. Now use
each side as an exponent of 10 and recogntze that
1glog(/2 /I) : Izl I,
The result is IzlIt: 103. The intensity is multiplied by a factor of 1.0 x 103.
(b) The pressure amplitude is proportional to the square root of the intensity so it is multipliedby a factor of .,4000 :32.
43
(a) The string is fixed at both ends and, when vibratittg at its lowest resonant frequency, exactlyhalf a wavelength fits between the ends . If L is the length of the string and l is the wavelength,then
^ - 2L The frequency is f - ulx- ul2L, where u is the speed of waves on the string.
Thus 'u : 2L f - 2(0.220 m)(920 Hz) : 405 m/s.
(b) The wave speed is given by u - \mt where r is the tension in the string and pt is thelinear mass density of the string. If M is the mass of the string, then p : M I L since the stringis uniform. Thus
r: truz: f*:W(4os ^ls)z-5e6N.
(c) The wavelength is ) - 2L - 2(A.220m) : 0.440m.
Chopter 17 103
(d) The frequency of the sound wave in arc is the same as the frequency of oscillation of thestring. The wavelength is different because the wave speed is different. If u o is the speed ofsound in air the wavelength in atr is
343 m/s)- 0 .373 m.
92AHz
45
(a) Since the pipe is open at both ends there are displacement antinodes at both ends and an
integer number of half-wavelengths fit into the length of the pipe. If L is the pipe length and A isthe wavelength then
^ - 2Lln, where n is an integer. If u is the speed of sound then the resonant
frequencies are givenbV f :nf ^-nuf2L.
Now L-0.457m, so f :n(344mls)12Q.457m)-376.4nH2. To find the resonant frequencies that lie between 1000H2 and 2000H2, first set
f - 1000Hz and solve for rL, then set f :2000H2 and agaun solve for rL. You should get 2.66and 5.32. This means ft:3,4, and 5 are the appropriate values of n There are three resonance
frequencies in the given range.
(b) For fr:3, f :3(376.4H27 - Il29Hz.(c) For fr: 4, f : 4(376.4H2) - 1506 Hz.
47
The string is fixedatboth ends so the resonantwavelengths are given by,\ - 2Lln,where L is the
length of the string and n is an integer. The resonant frequencies are given bV f - n l A: nu f 2L,where t' is the wave speed on the string. Now 'tr : tmt where T is the tension in the string andp is the linear mass density of the string. Thus f : (nlzL),,mr. Suppose the lower frequencyis associated with TL : TL1 and the higher frequency is associated with TL : rL1 + I . There are no
resonant frequencies between so you know that the integers associated with the given frequencies
differ by 1. Thus fi - (nrlzL)tmt and
, ual-t\a
^,l
TLr*l F TLt tr 1 n 1
fz
This means fz h - (llzL)\mt and
r : 4L2 ttffz f )t- 4(0.300 m)2(0.650 x 10-t kgl*Xl3 20Hz - 880 Ht)':45.3N.
53
Each wire is vibrating in its fundamental mode so the wavelength is twice the length of the wire() : 2L) and the frequency is
f: ullT: l-) 2Ly p)
the wire, T is the tension in the wire, and p is thewhere u (: \mD is the wave speed forlinear mass density of the wire.
104 Chapter 17
!l-t,
Suppose the tension in one
in the other wire is r * Lrand fz: 606 Hz.
Now
and
SO
This means
wire is r and the oscillationand its frequency is fz. You
frequency of that wire is f r. The tension
want to calculate Lr lr for fi: 600 Hz
f.-1 Ert n\ LL
1 - (g6H'\t\600 Hrl I - o'o2o '
Lr-T
f'- f -(5oo.oHz)
fzft
G),--05
(a) The expression for the Doppler shifted frequency is
.f'-frya * us
where f is the unshifted frequency, u is the speed of sound, u p is the speed of the detector (the
uncle), and us is the speed of the source (the locomotive). All speeds are relative to the air. Theuncle is at rest with respect to the air, so ?/p : 0. The speed of the source is u g
the locomotive is moving away from the uncle the frequency decreases and we use the positive
sign in the denominator. Thus
f'U f Ug \ 343mls + 10.00m/s
(b) The girl is now the detector. Relative to the atr she is moving with speed u ptoward the source. This tends to increase the frequency and we use the positive sign in the
numerator. The source is moving at us: 10.00m/s away from the girl. This tends to decrease
the frequency and we use the positive sign in the denominator. Thus (u + u n) - (u + u il and
f' - f : 5oo.oHz.
(c) Relative to the arr the locomotive is m s away from the uncle. I-Jse
the positive sign in the denominator. Relat s moving at up: 10.00m/stoward the locomotive. Use the positive si S
: 486.2H2.u * upu * ug
(d) Relative to the air the locomotive is m s away from the girl and thegirl is moving at u p he positive signs in both the
f - 500 .0H2.
343mls
) : 48s.8 Hz.
LO
ir\atr thr
the
\mliml
,atoco
=(uc
('
ob,o
1
3
5
obk
m:tin
\4
A
ini
e
)
re
rir,3,
,tvir
he
n)
10v
tive
gn
//,(:'I
lovth
Ur,
=20the
erat<
0.0(
n^0(
= 20.ve.
)anV
r)
rr1m(
f1-Y-t
rtirlJq
,,S
?1
u1
+
+
slol
u,
dav<
nl
lst,
U1
)mrf
tl:rirhu
:)
lsth
'h
S
;
ml1e
TT
t:ls
m/
e
0
IC
mlml
)0n
Jse
f'
).0(
unor.
10.0r
eunltor.
00n
00"
:0.0(
.IJlnd
tne
3^l3^lgatlocor
- (o,
LrT
numerator and the denominator. Thus (u +
Chapter 17 105
69
(a) The half angle 0 of the Mach cone is given by sin 0 - u lu s, where uandus is the speed of the plane. Since us: I.5u, sin 0 -ull.S'u: If L.5.
(b) Let h be the altitude of the plane and suppose the Mach cone
intersects Earth's surface a distance d behind the plane. The situ-ation is shown on the diagram to the right, with P indicating the
plane and O indicating the observer. The cone angle is related to
h and d,by tan?- hld' so d - hf tan?. The shock wave reaches
O in the time the plane takes to fly the distance d:
t - dl, - hlutan? - (5000 n)11.5(331^ls)tan 42o :11s.
is the speed of sound
Thismeans 0-42o.
77
(a) Use the Doppler shift equation, which gives the detected frequency J .
f'- fU LUg
where f is the emitted frequency, u is the speed of sound, Vn is the speed of the detector, and
ug is the speed of the source. You are the detector and are station?ry, so upthe source and is moving away from you, which lowers the frequency, so the positive sign isused in the denominator. Thus
f'- (1000H2)330 m/s :9.7 x I02Hz.
330 mls + 10m/s
(b) The detector is the cliff, which is stationary. The source is the siren, which is moving towardthe cliff. This increases the frequency and the minus sign is used in the denominator. Thus
f'- (1000H2)330 mls
330 mls - 10m/s
(c) The beat frequency is fb"ut:1.0 x 103 Hz - 9.7 x l02Hz:60Hz,.
81
(a) The rate with which sound energy is passing through the surface of a sphere with radius ris P - 4nrz I, where I is the intensity. Since energy is conserved this must be the power of the
source. Thus P - 4r(0.0080 W/*'Xt0 -)' _ 10 W.
(b) The sound inrensity is f - Pf 4rr2 - (10w/4r(5.0*)2 - 0.032Wfm2.
(c) The sound level is 0 : (10d8 log(Illo), where Is is the standard reference intensity, whichis given as 10-tzWl^'in the text. At a point 10m from the source it is
0.0080W lmz
106 Chapter 17
C : (10 dR)Ios10- 12
:99 dB
85
(a) The intensity is given by I : *prr'sl, where p is the density of the medium, u is the speed
of sound, a is the angular frequency, and sTn is the displacement amplifude. The displacementand pressure amplitudes are related by Lprn - pl)usrn, so srn: Lprnlpw and f - (Lprn)z lLpr.For waves of the same frequency the ratio of the intensity for propagation in water to the intensityfor propagation in air is
I-- (Lp,,-\' pouo
h \lo-," I p*u- '
where the subscript a, denotes arc and the subscript ?t) denotes water.
Since Io : I-,
/t0.998 x 103 kel^'Xr 4v2mls)^tV e.2rr.g7ffir; se'7
LPrn* -LPr,,o
The speeds of sound are given in Table 17 -I and the densities are given in Table l4-1.(b) Now Lprn* : Lpnlo, so
- parra : Q.2tkglm3)(343mls) _ z.g1 x 10-4puuu (0.998 x 103 kglm'Xt Yzm/s) -'v
87
(a) When the right side of the instrument is pulled out a distance d the path length for soundwaves increases by 2d. Since the interference pattern changes from a minimum to the nextmaximum, this distance must be half a wavelength of the sound. So 2d:
^12, where .\ is the
wavelength. Thus ,\(343mls)/a(0.O165 m) : 5.2 x 103 Hz.(b) The displacement amplitude is proportional to the square root of the intensity (see Eq. 17 -27).Write t/Tconstant of proportionality. At the minimum, interference is destructive and the displacementamplitude is the difference in the amplitudes of the individual waves 1 srn
where the subscripts indicate the paths of the waves. At the maximum, the waves interfereconstructively and the displacement amplitude is the sum of the amplitudes of the individualwaves i srn
for ssAn and ssBD. Add the equations to obtain ss7.n- (1f00 + \m)lzcsubtract them to obtain ssso: (vm- vC00)l2C: I}f C. The ratio of the amplitudes issstnl ts"r:2.(c) Any energy losses, such as might be caused by frictional forces of the walls on the ak inthe tubes, result in a decrease in the displacement amplitude. Those losses are greater on path Bsince it is longer than path A.
101
(a) The frequency is increased by reflection from the flowing blood, so the blood must be flowingto the right, with a positive velocity component in the direction of the original source of theultrasound.
I*h
Chapter t 7 107
(b) Use the Doppler shift equation twice. It is
f' - fU f.Ug
where f is the emitted frequency, u is the speed of soutrd, Vn is the speed of the detector, and
u s is the speed of the source. First, take the source to be the ultrasound generator, which is
stationdry, and the detector to be the blood. Thus us - 0 and up _ u6eos0, where u6 is the
speed of the blood. Since the detected frequency is greater than the generator frequency, use the
plus sign in the numerator. Thus
f ' - fu * uacos?
u
In the next step the blood is the source and f is the emitted frequency. The generator is the
detector and is stationary. Take up:0 and us:lrbcos0. The detected frequency is higher, so
we use the minus sign in the denominator of the Doppler shift equation. The detected frequency
is r,,-r,#-r(ry) (#) :rffi,,where f ' was replaced by the expression developed previously. The solution for u6 is
ub:(ffi) (#) :( )(#) -oeom/s
(c) If 0 increases, cos g decreases. This means the numerator of the expression for f " decreases
and the denominator increases. Both changes result in a decrease in f " .
108 Chapter 17
Chapter L8
ISince a volume is the product of three lengths, the change in volume due to a temperature change
LT is given by LVlinear expansion. See Eq. 18-11. Since V- (4nlrB3, where R is the original radius of the
sphere,
AV
The value for
15
If V. is the original volume of the cup, e.a is the coefficient of linear expansion of aluminum, and
LT is the temperature increase, then the change in the volume of the cup is LV.See Eq. 18- 11. If P is the coefficient of volume expansion for glycerin then the change in the
volume of glycerin is LVnthe original volume of the cup. The volume of glycerin that spills is
LVn LV" - (p - 3ao)V"LT
2tConsider half the bar. Its original length rs (.s : L0 f 2 and its length after the temperature increase
is ( : to + qh LT . The old position of the half-bar, its new position, and the distan ee n that one
end is displaced form a right triangle, with a hypotenuse of length (., one side of length (0, andthe other side of length r. The Pythagorean theorem yields 12 : ,(.2 - 1'z0 - tl\ * a LT)z - [3.
Since the change in length is small we may approximate (1 * crLT)2 by I +2uLT, where the
small term (a LT)z was neglected. Then
and
12 - tA + 2lla LT - 4 - 2[?ra LT
n:hrffi:ry :7.5x to-2m.
25
The melting point of silver is 1235 K, so the temperature of the silver must first be raised from15.0oC (- 288 K) to 1235 K. If m is the mass of the silver and c is its specific heat, this requires
energy
Q - cm(Ty -7,;): (236Jlkg.KX0.130kgXI235oC -288oC) - 2.gI x 104J
- 3u (+R') ^r -
(23 x to-6\3 )
the coefficient of linear expansion
lc")Gil(lO cm)3(100 Co) : 29 cm3
was obtained from Table l8-2.
to-u lc')) (1oo cm3x6 c") - o .2|cm3
Chapter I B 109
Now the silver at its melting point must be melted. If L e is the heat of fusion for silver thisrequires
Q:mLp -(0.130kgX105 x 103J/kg)- 1,36 x 104J.
Thetotal energyrequiredas heat is 2.9I x 104J+ 1.36 x 104J- 4.27 x 104J. The specificheatof silver can be found in Table 18-3 and its heat of fusion can be found in Table 18-4.
27
Mass n'L (: 0.100kg) of water, with specific heat c (: 4I90Jlkg.K), is raised from an initialtemperature T,i (: 23"C) to its boiling point Ty (: 100'C). The heat input is given by A_cm(71 - Tt). This must be the power output of the heater P multiplied by the time t; A : Pt.Thus
r__a cm(Ty -T) (4190 llkg.K)(0.100kgx100'C - 23"C)t-;
4t(a) There are three possibilities:
1. None of the ice melts and the water-ice system reaches thermal equilibrium at a temperature
that is at or below the melting point of ice.
2. The system reaches thermal equilibrium at the melting point of ice, with some of the ice
melted.
3. All of the ice melts and the system reaches thermal equilibrium at a temperature at or above
the melting point of ice.
First suppose that no ice melts. The temperature of the water decreases from Twt (: 25"C) to
some final temperature Ty and the temperature of the ice increases from Tp (: -15"C) to Ty
If m,yy is the mass of the water and cys is its specific heat then the water loses energy
Q:cyrTft'yy(Twt,-Tf)
If m7 is the mass of the ice and c7 is its specific heat then the ice absorbs energy
Q:crTrrtlf-Ttt),Since no energy is lost these two energies must be the same and
cyrTTl,14r(Twt, - Ty) : ct'trLtQf - Ttr) .
The solution for the final temperature is
rn CyrTTLryyTWt t ctTTLTTtf
,WW
- (4190 J/kg . KX0.200 kgX25"C) + (2220 J/kg . KX0.100 kgX- 15"C)
(4190 I lks . K)(0.200 ke) + (2220 I lkg. KXO.100 ke)
This is above the melting point of ice, so at least some of the ice must have melted. The
calculation just completed does not take into account the melting of the ice and is in elror.
110 Chapter 18
Now assume the water and ice reach thermal equilibrium at Ty : OoC, with mass m (< *r) ofthe ice melted. The magnitude of the energy lost by the water is
Q : cyyITLltgTwt ,
and the energy absorbed by the ice is
Q : crTTLr(O - Ty) + mLp ,
where L p is the heat of fusion for water. The first term is the energy required to warrn all the
ice from its initial temperature to OoC and the second term is the energy required to melt mass
m of the ice. The energy lost by the water equals the energy garned by the ice, so
CyrrTLyrf Wt: -CTtTLlTtt + mL p .
This equation can be solved for the mass m of ice melted:
C14r Tft1,yTW t a C I Tft 1T t tm-Le
(4t90 I lks . K)(0.200 kgx2s"C) + (2220 I lkg. KXO.100 kg)(- 15"C)
333 x 103Jlke: 5.3 x I0-2 kg - 53 g.
Since the total mass of ice present initially was 100 g, there is enough ice to bring the watertemperature down to 0oC. This is the solution: the ice and water reach thermal equilibrium at atemperature of OoC with 53 g of ice melted.
(b) Now there is less than 53 g of ice present initially. All the ice melts and the final temperature
is above the melting point of ice. The energy lost by the water is
Q-cyrTTlyr(Twt,-Tf)
and the energy absorbed by the ice and the water it becomes when it melts is
Q : cr?rLr(0 - Ty) + cyrTTLlQf - 0) * mrL7, .
The first term is the energy required to raise the temperature of the ice to 0o C, the second termis the energy required to raise the temperature of the melted ice from 0"C to Ty, and the thirdterm is the energy required to melt all the ice. Since the two energies are equal,
cyrTTlryy(Twt, - Ty) : ctTTLr?Tp,) + cyrwLTTf + TrLTLp .
The solution for T1 is
T1: cwmwTwt* ctmtTt't - mtLPcyr(mw + *i
Substitute given values to obtain T1 - 2.5oC.
Chapter t I 111
43
The internal energy is the same at the beginning and end of a cycle, so the energy Q absorbed as heat equals the work done: Q = W. Over the portion of the cycle from A to B the pressure p is a linear function of the volume V and we may write p = a + bV, where a = (10/3) Pa and b = (20/3 Pa/m3. The coefficients a and b were chosen so that p = 10 Pa when V = 1.0 m3 and p = 30 Pa when V = 4.0 m3 . The work done by the gas during this portion of the cycle is
The BC portion of the cycle is at constant pressure and the work done by the gas is W BC =
P ~ V = (30 Pa)(1.0 m3 - 4.0 m3) = -90 J. The CA portion of the cycle is at constant volume, so no work is done. The total work done by the gas is W = W AB + W BC + W CA = 60 J - 90 J + 0 =
-30J and the total energy absorbed as heat is Q = W = -30J. This means the gas loses 30J of energy in the form of heat.
49
(a) The change in internal energy ~Eint is the same for path iaf and path ibf. According to the first law of thermodynamics, ~Eint = Q - W, where Q is the energy absorbed as heat and W is the work done by the system. Along iaf ~Eint = Q - W = 50cal- 20 cal = 30 cal. Along ibf W = Q - ~Eint = 36cal- 30 cal = 6 cal.
(b) Since the curved path is traversed from f to i the change in internal energy is - 30 cal and Q = ~Eint + W = -30 cal- 13 cal = -43 cal.
(c) Let ~Eint = Eint, J - E int, i. Then E int, J = ~Eint + E int, i = 30 cal + 10 cal = 40 cal.
(d) and (e) The work WbJ for the path bf is zero, so QbJ = E int, J - E int, b = 40 cal - 22 cal =
18 cal. For the path ibf Q = 36 cal so Qib = Q - QbJ = 36 cal- 18 cal = 18 cal.
51
The rate of heat flow is given by
TH-Tc Pcond = kA L '
where k is the thermal conductivity of copper (401 W /m· K), A is the cross-sectional area (in a plane perpendicular to the flow), L is the distance along the direction of flow between the points where the temperature is T Hand Tc. Thus
Pcond = (40 I W /m . K)(90.0 x 10-4 m2)(l25°C - 10.0°C) = 1.66 X 103 J / s . 0.250m
The thermal conductivity can be found in Table 18-6 of the text.
112 Chapter 18
--65
Let h be the thickness of the
is
slab and A be its area. Then the rate of heat flow through the slab
p i - 4A(TH-Tc)
rcono h
)
where k is the thermal conductivity of ice, Tu is the temperature of the water (0'C), and Tc isthe temperature of the air above the ice (-10"C). The energy leaving the water freezes tt, the
energy required to freeze mass m of water beitrg A : L pm, where L p is the heat of fusion forwater. Differentiate with respect to time and recogntze that dQ ldt: Pco'd to obtain
Pco,ro - L dmFE.
Now the mass of the ice is given by m - pAh, where p is the density of ice and h is the
thickness of the ice slab, so dm I dt - pA(dh I dt) and
Prono : LppA#
Equate the two expressions for Pro,,o and solve for dh I dt:
dh k(Tn - Tc)dt LFph '
Since l cal
k- (0.0040caIls.cm.KX4.186Jlcal)10 x l0-2mf cm)- I.674W/-.K. The SIvalue forthedensity of ice is p:O.gzglt t - 0 .g2 x 103 kgl^t.Thus
dh1.1 x 10-u ^ls - 0.40 cmfh.dt (333 x 103 Ilkg(o.92 x 103 kel^'Xo.osom)
73
(a) The work done in process 1 is Wt - .\ptVt, so according to the first law of thermod5mamics
the change in the internal energy is AEnt : A-Wr: l\prV,-4.\ptV - 6.}p,iV. The work donein process 2 canbe computed as the sum of the areas of a triangle (with altitude 3pnl2-pn: p,;f 2
andbase 4.0V) andrectangle (with sidespi and 4.0V,): Wz: *Onl2)(4.0V)+4.\prV - 5.\p,iV.The change in the internal energy is the same so the energy transferred to the gas a heat is
Q: LBint+Wz - 6.oprV + 5.op,iV
(b) The change in internal energy is the same for all processes that start at state a and end at
state b, namely 6.}ptV.
75
At the colder temperature the volume of the disk is V - n R2 (., where R is its radius and (. is itsthickness. After it is heated its volume is n(ft+Aft)r({+ Ll): rR'(,+rR2 L(,+2rR(, LR+. . .,
where terms that are proportional to the products of small quantities are neglected. Thus the
Chapter 18 113
change in the volume is LV - nR2 L(. + 2rR(. LR. Now AR - Ra LT and L!,where LT is the change in temperature and a (: 3.2 x 10-6 x 10-u lC" from Table 18-2) isthe coefficient of lin ear expansion for Pyrex. Thus
Lv - 3r R2 ta LT - 37r(0.0800 m)',(0.00500 m) (3 .2 x 10- 6 c"x60.0oc - 10.0"c)
- 4.83 x 10-8 m3 .
77
Let pt be
work done
t::
the inby th
#
itial pressure, V be the initial volume, and Vy be the final volume. Then the
e gas is
pdv:
81
The magnitude of the energy transferred as heat from the aluminum is given by Q : ?rL4ct(Ta -D and the magnitude of the energy transferred as heat into the water is given by Q - mu)c*(T -T-), wher:- cs is the specific heat of aluminum (900 Ilkg.K from table 18-3), cu) is the specificheat of water (4190 J lkg. K from the same table) , Tt is the initial temperature of the aluminum,T* is the initial temperature of the water, and T is the common final temperature. The solutionfor T is
T- Trl,gcaTA I TTL,C-T-
rTLgCl * ffiusCu
(2.50 kgX900 J lke. KX92.0'C) + (8.00 kgXa190 J lke. KX5.00.C)(2.s0 kgX900 I lke. K) + (8.00 kgX4190 J lke. K)
- 10.5oC .
Note that it is not necessary to convert the temperatures to the Kelvin scale.
83
Let m be the mass of the ice cube and crce be its specific heat. Then the energy required to bringthe ice cube temperature to 0o C is
Q : TTLCice LT - (0.700 kg)(2220 J lkg . K)(150 K) - 2.331 x 10s J.
This is less than the 6.993 x 105 J that are supplied, so the ice cube is brought to OoC and all orpaft of it melts. The specific heat for ice can be found in Table 18-3.
If L e is the neat of fusion for water, then the energy required to melt all the ice is
mLp -(0.700kgx333 x 103 Ilkg-2.331 x 10sJ.
This less than the 6.993 x 105 J 2.33I x 10s J -- 4.662 x 10s J that arc avatlable, so allthe ice melts. The heat of fusion for water can be found in Table l8-4. Now energy E -4.662x10sJ-2.33Ix10sJ-2.33IxlOsJisusedtoraisethetemperatureofthewaterfromOoC. The final temperature is
ry1 _ E 2.33Ix105Jt : '/9-5oC .
TTLC*a1sy (0.700 kgX4190 J lkg. K) ' ' "
ll4 Chapter I B
avz dv -- \rri vn')- :23J.
Chapter 19
7(a) Solve the ideal gas law pV : nRT for n. First, convert the temperature to the Kelvin scale:
T - 4A.0 + 273. 15 - 313.15K. Also convert the volume to m3: 1000cm3 - 1000 x 10-6m3.Then
TL: pv
- (1.01 x 10spaX1000 x 10-6m3) _ 3.gg x l'-zmol .RT (8.3lJlmol .KX3 13.15K)
(b) Solve the ideal gas law pV : nRT for T:
rn _ pv (1.06 x 10s paX1500 x 10-u rn') ^(\., r.r _ .ono rr1 -
"R- :493 K - 220" L.
13
Suppose the gas expands from volume V to volume \ during the isothermal portion of theprocess. The work it does is
w: ['' pdv -nR, l:: Y -nRrtn\,Jvn
where the ideal gas law pV - nRT was used to replace p with nRT f V. Now Vand Vy - nRT lpt, so Vr lV: pi,lpr. Also replace nRT with ptV to obtain
W:ptVlnU.pf
Since the initial gauge pressure is 1 .03 x 10s Pa, pt: 1 .03 x 105 Pa+ 1 .01 3 x 105 Pa - 2.04 x 10s Pa.
The final pressure is atmospheric pressure: pf - 1.013 x 105 Pa. Thus
w : (2.04x 105 pa)(O .r40m3)rr, 2'04 x 105-Pa
- 2.00 x 104J.1.013 x lOsPa L'
During the constant pressure portion of the process the work done by the gas is W : p ilV -Vf).Notice that the gas starts in a state with pressure p f, so this is the pressure throughout this portionof the process. Also note that the volume decreases from Vy to V. Now Vy : pV lp f, so
14/-Pf(u '#)-(pf-Pt)v(1.013 x 10spa - 2.04 x 105pa)(O.r40m3) - -l .44 x 104J.
The total work done by the gas over the entire process is W5.60 x 103 J.
Chapter 19 115
t9
According to kinetic theory the rrns speed is
where T is the temperature onAccording to Table 19-I, the
10-3 kg/mol, so
the Kelvin scale and M is the molar mass. See Eq. 19-34.molar mass of molecular hydrogen is 2.02 gf mol _ 2.02 x
29
(a)
n] :UrmS
According to Eq. 19-25, the mean free path for molecules in a gas is given by
3(9.31 J/mol : KX2._15) - 1.8 x r02m/s .
2.02 x 10-3 kg/mol ^ '\
1
35
(a)
tDrd,'lYlV )
where d is the diameter of a molecule and l/ is the number of molecules in volume I/. Substituted - 2.0 x 10-10 m and IY lV - 1 x 106 molecules/m3 to obtain
)-fr"(2.0 x 10-to tn;z(l x 106 m-3)
-6 x 1012m.
(b) At this altitude most of the gas particles are in orbit around Earth and do not suffer random-rzrng collisions. The mean free path has little physical significance.
The average speed isfuT:-v lr)
where the sum is over the speeds of the particles and tf is the number of particles. Thus
u _ (2.0 + 3.0 +4.0+ 5.0+ 6.0 + 7.0 + 8.0+ 9.0 + 10.0 + 11.0)km/s _ 6.5km ls.10
(b) The rrns speed is given by
olvrms
Now
Du'+ (11.0)2]
116 Chapter 19
+ (7.0)' + (g .0)'+ (9.0)2 + (10.0)', km2 lr' - 505 km' lr' ,
505km2 f s2
10
SO
- 7.1 km fs.
4l(a) The distribution function gives the fraction of particles with speeds between u and u * du, so
its integral over all speeds is unity: f p(r)du:1. Evaluate the integral by calculating the areaunder the curve in Fig . 19 -24. The area of the triangular portion is half the product of the base
and altitude, or *oro. The area of the rectangular portion is the product of the sides, or &u0.Thus f p(r)du: *ouo+auy: ].orr, So lroro- l and auo-2f3.(b) The average speed is given by
oluavg
For the triangular portion of the distribution P(u) : au lro and the contribution of this portion is
* lr"o " du:
^'a:au3
3
I'P@)du
l*P@)du
2:0,,,
where 2l3us was substituted for a. P(r): a, in the rectangular portion and the contribution ofthis portion is
Thus 1)av' ir, 1 uo
(c) The mean-square
,kr-
The contribution of the triangular section is
* 1," ,'du:
The contribution of the rectangular portion is
|er', - rfi): lrA:,uo: 1.2,
A^1.,,J
4uo uo
b
o ,, du: I rsrd - rl) : !ra: l ra
a, ['"0 Lu dn :Juo
_ I.Zug and uuueluo
speed is given by
n I,'n"
Thus urms - 1.3luo and ?rrms lro - 1.3.
(d) The number of particles with speeds between 1.5us and 2uo is given by N [1."19,0 f g) du.
The integral is easy to evaluate since P(u): a, throughout the range of integrattoll. Thus thenumber of particles with speeds in the given range is IYa(2.0u0 1.510) - 0.5lYaus- IVl3,where 2l3us was substituted for a. The fraction of particles in the given range is 0.33.
ffi
Chapter 19 Il7
45
When the temperature changes by LT the internal energy of the first gas changes by nrCr LT,the internal energy of the second gas changes by nzCz LT, and the internal energy of thethird gas changes by ntCz LT. The change in the internal energy of the composite gas isLEin,specific heat of the mixture. Thus
e- ntet*nzez+nzCzT1,1 tnZ4n3
(2.40mo1)(12.0 J/mol K) + (1.50mo1)(12.8 J/mol . K) + (3.20mo1)(20.0J lmol ' K)2.40mol + 1.50 mol + 3 .20mol
- 15.8 J/mol . K .
53
(a) Since the process is at constant pressure energy transferred as heat to the gas is given by
a - nCp LT, where n is the number of moles in the g&s, Cp is the molar specific heat at
constant pressure, and LT is the increase in temperature. For a diatomic ideal gas with rotatingmolecules Cp
77Q : ,nR
LT : ,(4.00
mol)(8.3 IaJ lmol . KX60.0 K) - 6.98 x 103 J .
See Table I9-3 for the expression for Cp.
(b) The change in the internal energy is given by LEin - nCv LT, where Cv is the specificheat at constant volume. For a diatomic ideal gas with rotating molecules Cv - lrn, so
LE,nt:1nn, LT : ]f^.00mo1)(8.3
vJ lm,ol . KX60.0K) - 4.gg x 103 J.
2
See Table l9-3 for the expression for Cy.
(c) According to the first law of thermodynamics, L4int: Q W , so
W : A - AEi,,, - 6.98 x 103 J - 4.gg x 103 J - l.gg x 103 J .
(d) The change in the total translational kinetic energy is
LK :3=rn LT :t=fo.00mol)(8.3 14J lmol . KX60.0K) - 2.gg x 103 J.22',
67
Let po be the density of air suffounding the balloon and p n be the density of hot air in the
balloon. The buoyant force on the balloon has magnitude pogV, where V is the volume of the
envelope, and it is upward. The magnitude of the force of gravity is W + pngV , where W is the
weight of the basket and envelope combined, and it is downward. The second term is the weight
118 Chapter 19
of the hot air. If ft., (: 2.67 kN) is the net force on the balloon then Fnet: pagv - W - pngV ,
SO
Pn9 - pogV -W - Fnt- (tt.qX/m3XZ.tS x 103 m3) - 2.45 x 103 N - 2.67 x lo3 N
V
- 9.55 N/*t
69
The molar specific heat of a monatomic ideal gas is e vconstant. Let n be the number of moles of gas and LTchange in the internal energy is
- (3 l2)R, where R is the universal gas
be the change in temperature. Then the
2.18 x 103 m3
The ideal gas law tell us that the number of moles of hot air per unit volume is nlv - plRT,where p is the pressurc and T is the temperature. Multiply by the molar mass M of air to obt arnpn - pM lRT. The pressure is atmospheric pressure (1.01 x 105 Pa) so the temperature shouldbe
n1 _ pM (1.01 x 10s Pa)(0 .028 kg/mol) . tI - - -2<ALRpn (8.3 r J lmol . KX9.55 N/m')
AEnt - nCv LT - (3 lz)nRLT - (3 l2)(2.00mo1)(8.31Jlmol .KXl5.0K) - 374J .
Since the process is adrabatic the energy transferred as heat is Q : 0.
According to the first law of thermodynamics the work done by the gas is W --374 L
Since the gas is monatomic the internal energy is translational kinetic energy.
of atoms is the product of the number of moles and the Avogadro constant:(2.00 moI)( 6.02 x 1023 atoms lmol-
1)
peratomis (374J)l(I .20 x l}z4atoms):3.11 x 1022Jfatom.
A LE'n-
The numberiV - nlYekinetic energy
71
The mean free path is given by I - Iltnnd,2(lVlv),where d, is the diameter of a molecule andl/ is the number of molecules in volume V . According to the ideal gas law IY lV - p lkf, wherep is the pressure, T is the temperature on the Kelvin scale, and k is the Boltzmann constant.Thus
'\- -y-J1nd2p J2"(290 x 10- 12 m)2(2.00 atm)(1.01 x 10s Palatm) "J
The average time between collisions is T - \lruue, where uavs is the average speed of themo1ecu1es.Thisisgivenby,Uavg-@,whereRistheuniversa1gaSconstantandMis the molar mass (see Eq. 19-30). The molar mass of oxygen is 32.0 x 10-3 kg/mol (see table19- 1), so
8(8.31 I lmI. KX400 K)r(32.0 x 10-3 kg/mol)
ol :uavg -514mls.
Chapter 19 ll9
The average time between collisions is (7.31 x 10-8-)/(5 l amls) - L.42 x 10-e s and thefrequency of collision is the reciprocal of this or 7 .04 x 10e collisions/s.
77
(a) Let p,i be the initial pressure, V be the initial volume, p f be the final pressure, and Vy be thefinal volume. According to the ideal gas law pV : p fVf since the initial and final temperatures
are the same. Thus
pr: fiu,- #f Qzatm) - 8'oatm.
(b) The final temperature is the same as the initial temperature: 300 K.
(c) Since p : nRT f V , the work done by the gas is
r,\' : ["' p d,v : I:,' ry dv - nRr tn (+) : pivtn (+)J vn,
(32atm)(1.01 x 105 pafatmxl.0LXl.00 x 10-3m3 lL)tn(#i) :4.5x 103J.
(d) Since the process is now adtabatic, ptv: : pfvi, wherc l is the ratio of the heat capacity at
constant pressure to the heat capacrty at constant volume. Since the gas is monatomtc ^l - 11667 .
The final pressure is
rY\ - (v\t = (r,L)tuu',r2atm):3.zatm.Pr: \r, ) Pt -- \oor,
(e) Let T,; be the initial temperature and Ty be the final temperature. Then according to the idealgas law nR: pVlft,: pfVr lTt and
rr ^: 4r,- (3'2 atmxl'g-? (3oo K) : r2oK .rf
ffi" (32-t*)(l oL) \JVr
(0 The first law of thermodynamics tells us that the change in the internal energy Afint is equal
to the negative of the work done by the gas during an adiabatic process. The change in the
internal energy is L4rnt_ nCvlf T) -- (CvlR)(pf - p), where Cv is the molar specificheat for constant volume processes. The ideal gas law was used to write the equation in the
last form. For a monatomic ideal gas Cv: Ql2)R, So W- -L4int- -(3 lz)(p{t -ptv).Now pfvf - (3.2atm)(4.0LXl.01 x 10sPafatmxl.O x 10-3m3 lL)- 1.29 x 103J and ptV(32arm)(1.0LXl.01 x 10s Pulatm)(l.0 x 10-3m3 lL) - 3.23 x 103J, so
w:-:(r.2gx l03J- 3.23x 103D:z.gx 103J.
(g) Now 1 : 1.4, so
120 Chapter 19
pf
(h) The final temperature is now
rr-1 ^: 'fi7n: (a'6atmxa.O-L(300K): r70K.tf
ffi" @\JwL(i) The molar specific heat is Cv - 5 f2, so the work done by the gas is W: -]@fVf - ptV).Now prVr: (4.6atm)(4.0LXl.01 x l05Pulatm)(l.0 x 10-3m3 lL)-1.86 x 103J and pV is
still 3.23 x 103 J, so
w : -:(1.86 x l03J - 3.23 x 103D: 3.4x 103J.
81
(a) Since a molecule must have
Thus
and C - 3ro'.(b) The average
(c) The average
speed, the integral of P(u)
Cuz d,u : (I l3)Crl: 1
over all value of u must be L
speed of the particles ls
Io"o P(u)u d,u : 3rit
lo"o u3 d,u - (3 l4)ro .
of the square of the speed is
some
lo"o
;' Ir'o 'o o'
0.77 5us.
(rr)uus : lo"o
P(u)u2 d, : 3u - (3 l s)r"
so the nns speed is urms- l@r:
83
(a) According to the ideal gas law p : nRT f V , so the work done by the gas as the volume goes
from Vt to Vy is
tv: I:,'' pd,v: I:: rydv'-nRrtn(V)
- (3.s0mol)(8.3 rllmol .KX283K)ln f #)-) LLL
\+.oo # )
The temperature was converted from degrees Celsius to kelvins.
(b) The internal energy of an ideal gas does not change unless the temperature changes, so
according to the first law of thermodynamics the energy transferred as heat is a-2.37 x 103 J.
(3 ls)r"
Chapter 19 l2I
Chapter 20
-3
(a) Since the gas is ideal, its pressurc p is given inV, and the temperature T by p : nRT fV . Theexpansion is
terms of the number of moles TL, the volumework done by the gas during the isothermal
rvzw: I pdv:Jv,
Substitute Vz - 2.0014 to obtain
rvznRr
Jn +:nRrI"#
W : nRT In2 - (4.00 mol)(8.3 14J lmol . KX400 K) ln} - 9.22 x 103 J .
(b) Since the expansion is isothermal, the change in entropy is given by A,S - /tt lT) dQ -Q lT , where A is the energy absorbed as heat. According to the first law of thermodyramics,L4rnt- A W . Now the internal energy of an ideal gas depends only on the temperature and
not on the pressure and volume. Since the expansion is isothermal, LEin - 0 and Q : W . Thus
9.22 x 103J
400 K:23.IJfK.
(c) A,S - 0 for all reversible adiabatic processes.
1
(a) The energy that leaves the aluminum as heat has magnitude Q : Thaco(Toi, - Tf), where mais the mass of the aluminum, ca is the specific heat of aluminum, Toi is the initial temperature
of the aluminum, and Ty is the final temperature of the aluminum-water system. The energy thatenters the water as heat has magnitude Q: mu)c*(Tf -T*a), where mu) is the mass of the wateqcu) is the specific heat of water, and T-t is the initial temperature of the water. The two energies
are the same in magnitude since no energy is lost. Thus maco(Toi - Tf) : ffiusc-(Tf - T*t) and
T1macoToi, * TfLuc-T-,i,
maco I TTLucu
The specific heat of aluminum is 900 J lk1. K and the specific heat of water is 4190 J lkg. K.Thus
(0.200 kg)(900 J/kg . KX100' C) + (0.0500 kgx4190 J lke. KX20" C)
(0.200 kg)(900 J/kg . K) + (0.0s00 kg)(4190 I lke. K)
- 5J.00 C .
This is equivalent to 330 K.
122 Chapter 20
AS: Y-
Ty
(b) Now temperatures must be given in kelvinsi Tot, - 393K, T*r:293K, andTy -330K. For
the aluminum, dQ : maco d,T and the change in entropy is
ASo : IE:ffio"" l::,r:TTlacotn+- (0 .200kgxe0 0Ilkg.K)h:# - -22.rJfK.' 373K
(c) The entropy change for the water is
AS- : IE:m1D"- [:' #:TTL,uc-rn+,J T*i
- (0.0500 kg)(41 90 I lke. K) 1,r 33K : *24.g J fK.' 2g3K
(d) The change in the total entropy of the aluminum-water system is A,S - AS" + AS- --22.1 J fK+ 24.9 I fK - *2.8 J/K.
25
(a) The efficiency is
t-Tn-TcTs (235 + 273) K
Note that a temperature difference has the same value on the Kelvin and Celsius scales. Since
the temperatures in the equation must be in kelvins, the temperature in the denominator was
converted to the Kelvin scale.
(b) Since the efficiency is given by t0.236(6.30 x 104D:1.49 x 104J.
29
(a) Energy is added as heat during the portion of the process from a to b. This portion occurs
at constant volume (V), so Qrn - nCv LT, where Cv is the molar specific heat for constant
volume processes. The gas is a monatomic ideal g?s, so Cv - )n and the ideal gas law gives
ATgiven. We need to find po. Now po is the same as pc and points c and b are connected by an
adiabatic process. Thus p"Vl : pbvl and
po: pc: (#)' ,r- (*)''' (1 013 x 106pa) - 3 .167 x 104pa
The energy added as heat is
3
Qrn: =f
t.013 x 106Pa -- 3.167 x 104PaXl.00 x 10-3 m3): r.47 x 103 J.z
Chapter 20 123
(b) Energy leaves the gas as heat during the portion of the process from c to a. This is a constant pressure process, so
5 5 Qout = nOp I::lT = 2(Pa Va - Pc Vc) = "2Pa(Va - Vc)
= ~(3.167 X 104 Pa)(-7.00)(1.00 X 1O-3 m3) = -5.54 X 102 J,
where Op is the molar specific heat for constant pressure processes. The substitutions Va - Vc =
Va - 8.00Va = -7.00Va and Op = ~R were made.
(c) For a complete cycle, the change in the internal energy is zero and W = Q = 1.47 X 103 J -5.54 X 102 J = 9.18 X 102 1.
(d) The efficiency is E = W/Qin = (9.18 X 102 J)/(1.47 X 103 J) = 0.624.
37
An ideal refrigerator working between a hot reservoir at temperature TH and a cold reservoir at temperature Tc has a coefficient of performance K that is given by K = Tc/(TH - Tc). For the refrigerator of this problem, TH = 96° F = 309K and Tc = 70° F = 294K, so K =
(294 K)/(309 K - 294 K) = 19.6. The coefficient of performance is the energy Qc drawn from the cold reservoir as heat divided by the work done: K = IQcl/IWI. Thus IQcl = KIWI =
(19.6)(1.0 J) = 201.
39
The coefficient of performance for a refrigerator is given by K = IQcl/IWI, where Qc is the energy absorbed from the cold reservoir as heat and W is the work done by the refrigerator, a negative value. The first law of thermodynamics yields Q H + Qc - W = 0 for an integer number of cycles. Here Q H is the energy ejected to the hot reservoir as heat. Thus Qc = W - Q H· Q H is negative and greater in magnitude than W, so IQcl = IQHI -IWI. Thus
K = IQHI-IWI IWI .
The solution for IWI is IWI = IQHI/(K + 1). In one hour,
IWI = 7.54MJ = 1.57MJ. 3.8 + 1
The rate at which work is done is (1.57 x 106 J)/(3600 s) = 440 W.
(a) Suppose there are nL molecules in the left third of the box, nc molecules in the center third, and nR molecules in the right third. There are N! arrangements of the N molecules, but nL! are simply rearrangements of the n L molecules in the right third, nc! are rearrangements of the nc molecules in the center third, and nR! are rearrangements of the nR molecules in the right third. These rearrangements do not produce a new configuration. Thus the multiplicity is
124 Chapter 20
N! W=----
nL! nc! nR!
(b) If half the molecules are in the right half of the box and the other half are in the left half of the box, then the multiplicity is
N! W B = -( N-/-;-2-)! -( N-/-:-2-)! .
If one-third of the molecules are in each third of the box, then the multiplicity is
N! WA = (N/3)! (N/3)! (N/3)! .
The ratio is
(c) For N = 100,
49
WA (N/2)!(N/2)!
W B (N/3)! (N/3)! (N/3)! .
SO! SO! = 4.16 x 1016. 33!33!34!
( a) and (b) The most probable speed is given by v p = J 2RT / M and the rms speed is given by Vrms = J3RT / 1\11, where T is the temperature on the Kelvin scale, M is the molar mass, and R
is the universal gas constant. See Eqs. 19-34 and 19-3S. Thus L'lv = (J3 - J2) JRT/M.
According to Table 19-1 the molar mass of nitrogen is 0.028 kg/mol. For T = 2S0 K,
and for T = SOO K,
(8.31 J /mol . K)(2S0 K) 8 / ----'-----,-----,---- = 7 m s
0.028 kg/mol
(8.31 J/mol· K)(SOOK) = l.2 x 102m/s 0.028 kg/mol
(c) Thc energy transferred as heat when the temperature changes by the infinitesimal dT at constant volume is dQ = nCv dT, where n is the number of moles and Cv is the molar specific heat for constant volume processes. Thus the entropy change is
J dQ iTf nCv (Tf) L'lS= - = --dT=nCvln - .. T T. T T~
2
Here Ti is the initial temperature and T f is the final temperature. Since nitrogen is diatomic with rotating molecules Cv = SR/2 (see Table 19-3),
( Tf) (SOOK) L'lS = (S/2)nRln Ti = (S/2)(l.Smol)(8.31 J/mol· K)ln 2S0K = 22J/K.
55
The temperature of the ice is raised to O°C, then the ice melts and the temperature of the resulting water is raised to 40°C. As the temperature of the ice is raised the infinitesimal dT the energy
Chapter 20 125
transferred to it as heat is dQ = mCice dT, where Cice is the specific heat of ice and Tn is the mass of the ice. The entropy change is
J dQ iTf mCice (Tf) tJ.S1 = - = -- dT = me' In -T T ·lce T Ti ~
Table 18 - 3 gives the specific heat of ice as 2220 J /kg . K. The initial temperature on the Kelvin scale is Ti = -20° + 273 ° = 253 K and the final temperature is Tf = 273 K, so
( 273 K) tJ.S1 = (0.600kg)(2220J/kg· K)ln 253K = 101 J/K.
The heat of fusion of water is L f = 333 x 103 J /kg, so the entropy change on melting is
tJ.S2 = mL f = (0.600kg)(333 x 103 J/kg) = 732J/K. T 273K
The initial temperature of the water is Ti = 273 K and its final temperature is T f = 40° + 273° =
313 K. The specific heat of water is Cwater = 4190 J /kg . K, so the change in the entropy of the water as its temperature is raised is
tJ.S3 = mCwater ln (i) = (0.600kg)(4190J/kg· K)ln (~~~~) = 344J/K.
The change in entropy for the complete process is tJ.S = tJ.S1 + tJ.S2 + tJ.S3 = 101 J /K + 732 J /K + 344J/K = 1.18 x 103 J/K.
63
(a) The coefficient of performance of a refrigerator is given by
where I Q L I is the energy extracted as heat from the low temperature reservoir and I Q H I is the energy transferred as heat to the high temperature reservoir. In this case the low temperature reservoir is the interior of the refrigerator and the high temperature reservoir is the room. The solution for IQHI is
K + 1 4.60 + 1 IQHI = IQLI-K = (35.0kJ) = 42.6kJ. 4.60
(b) Over a cycle the change in the internal energy of the system is zero, so according to the first law of thermodynamics the work done per cycle is IWI = IQHI - IQLI. Thus K = IQLI/IWI and
IWI = IQLI = 35.0kJ = 7.61 kJ. K 4.60
126 Chapter 20
67
(a) and (b) If there are l/ particles in all, with n in one box and lfmultiplicity is W - tf ! lnl(.Af n)t. The least multiplicity occurs for nW:.^f!/ff!O! - 1. The greatest multiplicity occurs for rL: (ff - Dlz or
n
-0n-
in the other, the
or n - .Af and is(l/ + l) 12 and is
W:." ' tt
''
[(r lz)(n - 1)] !10 l2)(lr + 1)l ! '
For l[: 3 this is W :3!lIl2! : 3 and for l/: 5 this is W :5!12!3!: 10.
(c) and (d) The entropy is given by ,S - klnW , where k is the Boltzmann constant. The greatest
entropy occurs if the multiplicity is the greatest. For lf1.5 x T0-23 JIK and for l'r:5 it is Sr - (1.38 x 10-23 JllfJln 10 - 3.2 x 10-23 JlK.
Chapter 20 127
Chapter 2L
1
The magnitude of the force that
where r is the distance between
either charge exerts on the other is given by
n- 1 lqtllqrl'r +r.., T'2' '
them. Thus
t-. /(s.gg x 10eN.m2 lc')(26.0 x 10-6c)(47.0 x 10-6c) 1 .
- t/v 5.70 N
5
The magnitude of the force of either of the charges on the other is given by
n 1 q(Q-q)r - 4"r, rz
where r is the distance between the charges. You want the value of g that maximizes the function
f (q) - q(Q - il. Set the derivative df ldq equal to zero. This yields A - 2q:0, or q: Q12,
7
Assume the spheres are far apart. Then the charge distribution on each of them is sphericallysymmetric and Coulomb's law can be used. Let q1 and q2 be the original charges and choose the
coordinate system so the force on ez is positive if it is repelled by qt Take the distance between
the charges to be r. Then the force on ez is
Fo --1 $Qz
4nes 12
The negative sign indicates that the spheres attract each other.
After the wire is connected, the spheres, being identical, have the same charge. Since charge
is conserved, the total charge is the same as it was originally. This means the charge on each
sphere is (q + q) f 2. The force is now one of repulsion and is given by
r;1 I (q+q)2'r b 4" r, 4r'z
lqtllqrl4nesF
128 Chapter 2l
Solve the two force equations simultaneously for h and e2. The first gives
-4treor| Fow g.gg x 10eN. nf lc'
et+qz-2r : 2(0.500 m) - 2.00 x 10-6 C.
QtQz :-
and the second gives
Thus
and substitution into the second
Qr+
-(3.00 x 10-tz gz'S
Qz:Qt
equation gives
-3.00 x 10-12 g2- 2.00 x 10-6 C.
Qr
Multiply by qt to obtain the quadratic equation
q? - e.oo x lo-u c)q, - 3.oo x lo-12 g2:0.
The solutions are
Qr:2.00 x 10-6C+
If the positive sign is used, $:3.00 x 10-e g and if the negative sign is used, $_ -1.00 x10-o g. IJse Qz
-1.00 x 10-o g and rf qt
identical, the solutions are essentially the same: one sphere originally had charge -1.00 x 10-e g
and the other had charge +3.00 x 10 -6 C.
19
If the system of three particles is to be in equilibrium, the forceon each parlicle must be zero. Let the charge on the third particleb. qo. The third particle must lie on the r axis since otherwise the
two forces on it would not be along the same line and could notsum to zero. Thus the A coordinate of the particle must be zero.
The third particle must lie between the other two since otherwisethe forces acting on it would be in the same direction and wouldnot sum to zero. Suppose the third particle is a distance r fromthe particle with charge e, as shown on the diagram to the right.The force acting on it is then given by
F^- 1 lqqo a.ooqqof_nro 4"rrl7 @)-u'where the positive direction was taken to be toward the right. Solve this equation for r. Cancelingconlmonfactorsyields Ll*'- 4.00 lQ-n)2 andtakingthe squarerootyields Il* -2.00 l(L- r).The solution is r :0.333L.
<_ f __+<_ L_f ___+
eo 4.00q
4nesF60.0360 N
8.99 x 10eN . m2 lC'
(-2.00 x 10-e C)2 + 4(3.00 x 10-12 C2)
Chapter 21 129
The force on
Solve for eo:
The force on
qis
F-: I lry* looq'1 - nq- 4"r,LA* L'z ):u'Q0: -4.00qr2 1fz: -0.444q, where r -- 0.333L was used.
the particle with charge 4.00q is
n I 14.00q2 4.00qqol 1 14.00q2D:'4q 4resL L2 e-02) 4resL L2
4res L L2 L2 I \'r '
4.00(0.444)q 1l(0.444)L2
With eo: -0.444q and r :0.333L, all three charges are in equilibrium.
25
(a) The magnitude of the force between the ions is given by
r-., q2-r @,where g is the charge on either of them and r is the distance between them. Solve for the charge:
q : r {4?t€oF - (5.0 x 1o-to -) :3.2 x 10-1eC.
(b) Let lf be the number of electrons missing from each ion. Then IY e : q and
17 q 3.2x10-leC./\/ -
tl-rv
; -L.
3s
(a) Every cesium ion at a corner of the cube exerts a force of the same magnitude on the chlorineion at the cube center. Each force is a force of attraction and is directed toward the cesium ionthat exerts rt, along the body diagonal of the cube. We can pair every cesium ion with another,
diametrically positioned at the opposite corner of the cube. Since the two ions in such a pairexert forces that have the same magnitude but are oppositely directed, the two forces sum to zero
and, since every cesium ion can be paired in this way, the total force on the chlorine ion is zeto.
(b) Rather than remove a cesium ion, superpose charge -e at the position of one cesium ion.
This neutralizes the ion and, as far as the electrical force on the chlorine ion is concerned, it isequivalent to removing the ion. The forces of the eight cesium ions at the cube corners sum tozeto, so the only force on the chlorine ion is the force of the added charge.
The length of a body diagonal of a cube is l8o, where a is the length of a cube edge. Thus the
distance from the center of the cube to a corner is d - (tftlD".The force has magnitude
D_ I e2 1 e2r' 4"r, d' "q@-
(g.gg x 10eN.m2/c2x1.6-0 x 10-lec)2 _ L.gx 10_eN.(3 l4)(0.40 x 10-g m)2
130 Chapter 2l
3.7 x 10-e 58.99 x 10eN . m2 lC'
Since both the added charge and the chlorine ion are negative, the force 1s one of repulsion. Thechlorine ion is pulled away from the site of the missing cesium ion.
37
None of the reactions given include a beta decay, so the number of protons, the number ofneutrons, and the number of electrons ate each conserved. Atomic numbers (numbers of protonsand numbers of electrons) and molar masses (combined numbers of protons and neutrons) can
be found in Appendix F of the text.
(a) tg has 1 proton, I electron, and 0 neutrons and eBe has 4 protons, 4 electronse and
94neutrorls. One of the neutrons is freed in the reactioll. X must be boron with a molar mass of5 elmol + 4 elmol - 9 gf mol: eB.
(b) tzc has 6 protons, 6 electrons, and 12 - 6:6 neutrons and lH has 1 proton, 1 electron, and
0 neutrons, so X has 6 + 1
nitrogen with a molar mass of 7 glffiol + 6 glmol - 13 gf mol: 13N.
(c) 15N has 7 protons, 7 electrons, and 15 7-8 neutrons; tg has 1 proton, I electron, and 0
neutrons; and aHe has 2 protons, 2 electrons, and 4 2
protons, 6 electrons, and 8+0 2-6 neutrons. Itmustbe carbonwith a molarmass of6elmol+ 6el^ol - 12glmoI: t2C.
39
The magnitude of the force of particle I on particle 2 is
- 1 lqtllqrl4nes a?, + a?r
'
The signs of the charges are the same, so the particles repel each other along the line that runs
throughthem.Thislinemakesaf|angle0withtheraxissuchthatcoS0-d,zl|m,SothelL component of the force is
F*
- (g .gg x 10e c'/N .m2) 24(1 .60 x 10-ro 97z(6.00 x 1o-'-)-1.31 x 10-22N
[(2.00 x 10-t *)' + (6.00 x 10-z m)2]3/z
50
The magnitude of the gravitational force on a proton near Earth's surface is mg, where m isthe mass of the proton (I .67 x 10-27 kg from Appendix B). The electrostatic force between twoprotons is F - Q lanro)("' I d,'), where d is their separation. Equate these forces to each otherand solve for d. The result is
F
- 0.119m.( 1 .60 x 10- te C)2
(I .67 x 10- 27 kg)(9.8 ^lt')(8.99 x 10-eY .m2 lC')(t--,1 t ez
V 4Tes mg
Chapter 2 I 131
60
The magnitude of the force of particle 1 on particle 4is
lo-tn cX3.2o x lo-tq c)17 I lq'llqolr1 4"rrT - (8 .99 x roe N .m2 lc'r(3'20 x
The charges have opposite signs, so the particles attract each other and the vector force is
(0.0300 m)2
Fr: -(1 .02-24 N)(cos 35.0o) i - (r.02 x 10-24 N)(sin 35.0") j- -(g .36 x 1o-" N) i - (5.g5 x 1o-" N) j.
3 repel each other. The force of particle 2 on particle 4 is
lqrllqol ?
-l
zn) Jd;
x loeN . m2 lcrr(3.20x lo-le cX3.2o x 1o-tn c)
"
Particles 2 and
IF2
F3
4neg
- -(2.30 x l0-to $ j
Particles 3
(0.0200 m)2
force of particle 3 on particle 4 isand 4
1
4Tes
repel each other and the
l2qtllqol ,
-l
di
x t0eN. m2 lcr)(6.40 x 10-1e-qx3.20-x 10-1e c)
(0.0200 m)2_ -(4.60 x 10-24 C) ? .
The net force is the vector sum of the three forces. The r component is F"4.60N - -5.44 x 10-24N and the A component is Fa: -5.85 x 10-2sN - 2.30 x I0-24NI:-2.89 x 10-24N. The magnitude of the force is
(-5 .44 x 10-24 N)2 + ( -2.89 x L0-24N)2 - 6.T6 x 1024 N.
The tangent of the angle 0 between the net force and the positive r axis is tan? : F, I F" :(-2.89 x 10-24 N/(-5 .44 x 10-z+N) - 0.531 and the angle is either 28" or 208". The laterangle is associated with a vector that has negative r and A components and so is the correctangle.
69
The net force on particle 3 is the vector sum of the forces of particles 1 and 2 and for this tobe zero the two forces must be along the same line. Since electrostatic forces are along the linesthat join the particles, particle 3 must be on the r axis. Its A coordinate is zero.
Particle 3 is repelled by one of the other charges and attracted by the other. As a result, particle3 cannot be between the other two particles and must be either to the left of particle 1 or to the
right of particle 2. Since the magnitude of gr is greater than the magnitude of qz, partrcle 3 must
132 Chapter 2l
F3 + F?a
be closer to particle 2 than to particle I and so must be to the rightcoordinate of particle 3. The the r component of the force on it is
of particle 2. Let r be the
r-, I I %et ezez Ia-r:l-r-lLn
4Tes L 12 @_ D'lIf F" - 0 the solution for :r is
L - 2.72L .-erlq,
-Qr I q,-(-5.00q) lQ.00q)
-(-5.00q) lQ.00q) -
Chapter 21 133
Chapter 22
3
Since the magnitude of the electric field produced by a point particle with charge q is given
by E- lqlla"eor2, where r is the distance from the particle to the point where the field has
magnitude E, the magnitude of the charge is
lql -4neor2|- :5.6x10-llc.
5
Since the charge is uniformly distributed throughout a sphere, the electric field at the surface isexactly the same as it would be if the charge were all at the center. That is, the magnitude ofthe field is
7,-.,- q'n 4Tr€oH'
where q is the magnitude of the total charge and R is the sphere radius. The magnitude of the
total charge is Z e, so
E _ z" -
(g.gg x 10eN .T2/c2x94x1.q0 x 10-1e c) _ 3.07 x t02'].{/c.4TesRz (6.64 x 10-15 m)2
The field is normal to the surface and since the charge is positive it points outward from the
surface.
7
At points between the partrcles, the individual electric fields are
in the same direction and do not cancel. Charge ez has a greater dPmagnitude than charge er, so a point of zero field must be closer Qz Qr
to % than to e2. It must be to the right of qr on the diagram.
Put the origin at the particle with charge ez and let r be the coordinate of P, the point wherethe field vanishes. Then the total electric field at P is given by
E-*lp #lwhere % and Qz are the magnitudes of the charges. If the field is to vanish,
Qz: Qt
rz (r - d)2
L34 Chapter 22
Take the square root of both sides to obtain Viii/x = vfiil / (x - d). The solution for x is
x-( Viii )d-( yI4.Oq1 )d Viii - vfiil V4.Oql - vfiil
( 2.0 ) = d = 2.0d = (2.0)(50 cm) = 100 cm. 2.0 - 1.0
The point is 50 cm to the right of ql .
2 Choose the coordinate axes as shown on the diagram to the right. At the center of the square, the electric fields produced by the particles at the lower left and upper right comers are both along the x axis and each points away from the center and toward the particle that produces it. Since each particle is a distance d = V2a/2 = a/ Vi away from the center, the net field due to these two particles is
,Y , q ,
a
/
-g /
/
, , , d , ,
/ /
/ /
/
a
/ /
/ ,/ /, , , ,
= _1 _ _ q_ = (8.99 x 109 N· m2/C2)(1.0 x 10-8 C) = 7.19 x 104 N/C. 41rEo a2/2 (0.050 m)2/2
/
/ /
,d , ,
/ :r: /
~2q
2q , , ,
At the center of the square, the field produced by the particles at the upper left and lower right comers are both along the y axis and each points away from the particle that produces it. The net field produced at the center by these particles is
1 [2q q 1 1 q 4 Ey = 4- 2/2 - ~/ = -4 - 2/2 = 7.19 x 10 N/C. KEO a a 2 KEO a
The magnitude of the net field is
and the angle it makes with the x axis is
E () = tan- I ~ = tan- l (l) = 45° .
Ex
It is upward in the diagram, from the center of the square toward the center of the upper side.
21
Think of the quadrupole as composed of two dipoles, each with dipole moment of magnitude p = qd. The moments point in opposite directions and produce fields in opposite directions at
Chapter 22 135
points on the quadrupole axis. Consider the point P on the axis, a distance z to the right of the
quadrupole center and take a rightward pointing field to be positive. Then the field produced
by the right dipole of the pa.r is qdlLnro(z d,l2)t and the field produced by the left dipoleis -qdlTTes(z * d,lD3. Use the binomial expansions (z dlZ\-z r z-3 3z-a(-dlz) and
(z + dlz)-t N z-3 - 32-4@12) to obtain
E- qd
2nes
E,: hlr":^ t1
4"r, lA
3d1I
2r^ 7
dr(L + a, - r)2 4Tes
lltL+"-*loL
l-1
le*3d1
2ro l6qdz
Let Q :2qd2. Then
E- 3Q
4Tesza
27
(a) The line ar charge density ) is the charge per unit length of rod. Since the charge is uniformlydistributed on the rod, l - -qlL - -(4.23 x 10-tsC) 1Q.0815m) - -5.19 x 10-14Cfm.(b) and (c) Position the origin at the left end
of the rod, as shown in the dtagtarct Let drbe an infinitesimal length of rod at r. The
charge in this segment is dq - ), dr. Since the
segment may be taken to be a point partrcle, the
electric field it produces at point P has only an
n component and this component is given by
), drdE"- 4Tes(L+a-r)zThe total electric field produced at P by the whole rod is the integral
4n esz4
drPo
tlll0 n L L+a
ll )l:
L + a) 4nes a,(L + a)
When - q I L is substituted for ^
the result is
E*1 q (8.99 x 10eN. m2 lC'X4 .23 x 10-ts C)
1.57 x 1o-' N/C .
4nes a(L + a) (0.120 m)(0.0815 m + 0.120 m)
The negative sign indicates that the field is toward the rod and makes an angle of 180" with the
positive r direction.
(d) Now
E*-_ 1 q -_(S.qqx
tOeX.lt2/C]X+.ZJ_xtO-t5C)__l .szx 10-*N/C.4res a(L + a) (50 mX0.0815 m + 50 m)
136 Chapter 22
E* :4Tesa2 (50 m)2
35
At a point on the axis of a uniformly charged disk a distance
magnitude of the electric field isz above the center of the disk, the
(e) The field of a point particle at
q (8.99 x
or
the origin is
loeN . m2 lC'X4 .23 x lo-ts C)- -1.52 x 10-tN/C.
where R is the radius of the disk and a is the surface charge density on the disk. See Eq. 22-26.The magnitude of the field at the center of the disk (zfor the value of z such that ElE.- Il2. This means
EE"
,21Jz2+R2 2
Square both sides, then multiply them by z2 + R2 toand z- Rlrn -(0.600m)lrn -0.346m.
39
2'obtatn z2 - (22 l4)+(R2 lq. Thus z2 - R'13
The magnitude of the force acting on the electron is F - €8, where E is the magnitude of the
electric field at its locatiorl. The acceleration of the electron is given by Newton's second law:
a _ ! : eE -
(1.60 x 10-1e CX2.0O x l04N/C) _ 3.51 x 101' *^, .m m 9.11 x 10-gt kg
43
(a) The magnitude of the force on the particle is given by F - qE, where q is the magnitude ofthe charge carried by the particle and E is the magnitude of the electric field at the location ofthe particle. Thus
103 N/C.
the field points upward.
F": eE - (1.60 x 10-tn cxl.5 x 103 N/c) : 2.4 x l0-16 N .
(c) A proton is positively charged, so the force is in the same direction as the field, upward.
Chapter 22 137
E-!_3.ox1o-l N 1Fq 2.0 x 10-eC - l'5 x
The force points downward and the charge is negative, so
(b) The magnitude of the electrostatic force on a proton is
z2+R2
(d) The magnitude of the gravitational
Fs : mg - (1.67 x
force on the proton is
1o-" kg)(9.8 mlst) - 1,.64 x ro-26N.
The force is downward.
(e) The ratio of the force magnitudes is
F"- ''0, "
t,o t:,T - 1.5 x roroFs I.64x10-26N
L'v
45
(a) The magnitude of the force acting on the proton is F - e E, where E is the magnitudeof the electric field. According to Newton's second laW the acceleration of the proton isa, - F l* - eE lm) where m is the mass of the proton. Thus
(1.60 x 10-te CX2.00 x 104 N/C) _ l.gZ x 101' ^l ,'a*L67 x 10-zt kE
(b) Assume the proton starts from rest and use the kinematic equation u2
r, - iot' and n : at) to show that,E + Zar (or else
?): \r - l@grx 1or, : r.g6x losm/s.
57
(a) If q is the positive charge in the dipole and d, is the separation of the charged particles, themagnitude of the dipole moment is p: qd: (1.50 x 10-n CX6.20x 10-u*) - 9.30 x 10-15 C.ril.
(b) If the initial angle between the dipole rnoment and the electric field is 0s and the final angleis 0, then the change in the potentral energy as the dipole swings from 0 : 0 to e - 1 80o is
L(J : -pU(cos 0 - cos 0o): -(9.30 x 10-15 C.mX1100Nic)(cos l80o - cos0)
- 2.05 x 10-11 J.
79
(a) and (b) Since the field at the point on the r axis with coordinate tr :2.0 cm is in the positiver direction you know that the charged particle is on the r axis. The line through the pointwith coordinates r - 3.0cm and y - 3.0cm and parallel to the field at that point must pass
through the position of the particle. Such a line has slope (3.0)l(4.0) - 0.75 and its equation isy:0.57 + (0.75)r. The solution for A - 0 is tr: -1.0cm, so the particle is located at the pointwith coordinates r: -1.0cm and A - 0.
138 Chapter 22
(c) The magnitude of the field at the point on the r axis with coordinate tr : 2.0 cm is given byE - Qlanrilql(2.0cm - r)2, So
q:4nesr2E -(0.020m+0.010m)2(100-N/c) _1.0 x 10-n c.8.99x10eN.nflC'
81
(a) The potential energy of an electric dipole with dipole moment f in an electric field E is
' : _L lo:-'l::,; Tffi ;]f;:T;l :11 :, :::1 ;1;T
Here we used d,.6: (trnb** eabai arb, to evaluate the scalar product.
(b) The torque is
F-FxE-(p*i*prilx (tr*i) - -patr*k- -(4.00Xr.24 x 10-30C.mX4000N/C) - -(1.98 x 10-26N.m)t.
(c) The work done by the agent is equal to the change in the potential energy of the dipole. Theinitial potential energy is Ut, - -1.49 x 10-26 J, as computed in part (a). The final potential
energy is
u7:: ,1o,ll1,1f .*h1;Hil:1,:IT :l'-,0-26 I
Theworkdonebythe agent is W: (1.98 x L0-26D-(-1.49 x I0-26D:3.47 x 10-26J.
Chapter 22 139
Chapter 23
1
The vector area A and the electric field E are shown on the dia-gram to the right. The angle 0 between them is 180o 35o : L45" ,so the electric flux through the area is O -- E.A- EAcosd -(1800N/C)(3.2 x 10-t ^)' cos l45o- -1.5 x l0-2 N. m2 f C.
2
Let A be the area of one face of the cube, Eu be the magnitude of the electric field at the upperface, and E2 be the magnitude of the field at the lower face. Since the field is downward, theflux through the upper face is negative and the flux through the lower face is positive. The fluxthrough the other faces is zero, so the total flux through the cube surface is OThe net charge inside the cube is given by Gauss' law:
n:ll .'\f:L:?:^l5
rr x 10-t2c2lN m2x100m)2(I00N/c - 60 oN/c)
t9(a) The charge on the surface of the sphere is the product of the surface charge density o and
the surface area of the sphere (4nr2, where r is the radius). Thus
(b) Choose a Gaussian surface in the form a sphere, concentric with the conducting sphere andwith a slightly larger radius. The flux through the surface is given by Gauss' law:
q:4nr2o:4n e)t,r., x 10-6 c l^') : 3.7 x r0-5 c.
o- q- :4.rx1o6N.mzfc.€e 8.85 x 10-t2 Cz/N .m/
#
23
The magnitude of the electric field produced by a uniformly charged infinite line is E : ^127r€sr,where l is the linear charge density and r is the distance from the line to the point where the
field is measured. See Eq. 23-12. Thus
\ _FA - )TegEr - 2n(8.85 x l0-12 gz/N .mzx4.5 x l04N/c)(2.0m):5.0 x 10-u cl*.140 Chapter 23
27
Assume the charge density of both the conducting rod and the shell are uniform. Neglect fringing.Symmetry can be used to show that the electric field is radial, both between the rod and the
shell and outside the shell. It is zero, of course, inside the rod and inside the shell since they are
conductors.
(a) and (b) Take the Gaussian surface to be a cylinder of length L and radius r, concentricwith the conducting rod and shell and with its curved surface outside the shell. The area of the
curved surface is 2rr L The field is normal to the curyed portion of the surface and has uniformmagnitude over it, so the flux through this portion of the surface is O - 2rrLE, where E is the
magnitude of the field at the Gaussian surface. The flux through the ends is zero. The charge
enclosed by the Gaussian surface is Qr - 2.00Q1 : -Qr. Gauss' law yields 2TresLE: -Qr,so
E _ _g- (8.99 x 10e_+r .ml/q2x3.10_x 10-12 c) - _0 .2r4N/c.ZresLr (11.00 m)(26.0 x 10-3 *)
The magnitude of the field is 0 .2L4N/C. The negative sign indicates that the field points inward.
(c) and (d) Take the Gaussian surface to be a cylinder of length L and radius r, concentric withthe conducting rod and shell and with its curved surface between the conducting rod and the
shell. As in (a), the flux through the curved portion of the surface is O : 2nr LE, where E is
the magnitude of the field at the Gaussian surface, and the flux through the ends is zero. The
charge enclosed by the Gaussian surface is only the charge Qr on the conducting rod. Gauss'
law yields ZTesrLE : Qr, so
Q r 2(8.99 x 10e NI . nf lC'X3 .40 x l0- rz C)E- ffi- -+0.SssN/C.
The positive sign indicates that the field points outward.
(e) Consider a Gaussian surface in the form of a cylinder of length L with the curved portionof its surface completely within the shell. The electric field is zero at all points on the curvedsurface and is parallel to the ends, so the total electric flux through the Gaussian surface is zero
and the net charge within it is zero. Since the conducting rod, which is inside,,the Gaussian
cylinder, has charge Q1 , the inner surface of the shell must have charge -At : -3.40 x 10-12 C.
(0 Since the shell has total charge -2.00Q1 and has charge -Ar on its inner surface, it musthave charge -At : -3.40 x 10-12 C on its outer surface.
35
(a) To calculate the electric field at a point very close to the center of a large, uniformly charged
conducting plate, we may replace the finite plate with an infinite plate with the same area charge
density and take the magnitude of the field to be E - o leo, where o is the area charge densityfor the surface just under the point. The charge is distributed uniformly over both sides of the
original plate, with half being on the side near the field point. Thus
q 6.0x10-og a.o:il-ffi -_4.69 x lo-4cl^'
Chapter 2 3 l4l
The magnitude of the field is
_ (j _ 4.69 X 10-4 C/m2 _ 7
E- EO - 8.85 x 1O-12C2/N.m2 -5.3 x 10 N/C.
The field is normal to the plate and since the charge on the plate is positive, it points away from the plate.
(b) At a point far away from the plate, the electric field is nearly that of a point particle with charge equal to the total charge on the plate. The magnitude of the field is E = q I 411"Eor2, where r is the distance from the plate. Thus
E = (8.99 X 109 N . m2IC2)(6.0 x 10-6 C) = I (30m)2 60N C.
41
The forces on the ball are shown in the diagram to the right. The gravitational force has magnitude mg, where m is the mass of the ball; the electrical force has magnitude qE, where q is the charge on the ball and E is the electric field at the position of the ball; and the tension in the thread is denoted by T. The electric field produced by the plate is normal to the plate and points to the right. Since the ball is positively charged, the electric force on it also points to the right. The tension in the thread makes the angle () (= 30°) with the vertical.
T
qE
mg
Since the ball is in equilibrium the net force on it vanishes. The sum of the horizontal components yields qE - T sin () = 0 and the sum of the vertical components yields T cos () - mg = O. The expression T = qE I sin (), from the first equation, is substituted into the second to obtain qE = mgtan().
The electric field produced by a large uniform plane of charge is given by E = (j 12Eo, where (j
is the surface charge density. Thus
and
q(j - = mgtan() 2Eo
2Eomgtan () (j = -----='-----
45
q
2(8.85 X 10-12 C2 IN· m2)(1.0 x 10-6 kg)(9.8 m/s2) tan 30°
2.0 x 10-8 C
= 5.0 X 10-9 C/m2 .
Charge is distributed uniformly over the surface of the sphere and the electric field it produces at points outside the sphere is like the field of a point particle with charge equal to the net charge on the sphere. That is, the magnitude of the field is given by E = qI411"Eor2, where q is the
142 Chapter 23
magnitude of the charge on the sphere and r is the distance from the center of the sphere to the
point where the field is measured. Thus
q : 4nesr2 E _(0.15 m)2(3.0 x 103 N/c) : 7 .5x l0- e c .
8.99 x 10eN. m2 lC'The field points inward, toward the sphere center, so the charge is negative: - 7 .5 x 10-e C.
49
To find an expression for the electric field inside the shell in terms of A and the distance fromthe center of the shell, select A so the field does not depend on the distance.
I-Jse a Gaussian surface in the form of a sphere with radius r s, concentric with the sphericalshell and within it (a <distance r s from the shell center.
The charge that is both in the shell and within the Gaussian sphere is given by the integral
Qens
distribution has spherical symmetry, we may take dV to be the volume of a spherical shell withradius r and infinitesimal thickness dr: dV - 4rr2 dr. Thus
eenc - 4tr I_*
prz dr - on I_* | * dr:4nA
I_' r dr:2nArr? - a2).
The total charge inside the Gaussian surface is q + eenc : q + 2r A(r? - a2).
The electric field is radial, so the flux through the Gaussian surface is O - 4rrln, where tr isthe magnitude of the field. Gauss' law yields
4r esUr? : q + 2n Afr? - a2) .
Solve for E:
E-+l++2nA-ry\4nes l,ri rzn I
For the field to be uniform, the first and last terms in the brackets must cancel. They do ifq-2nAa2 - 0 or A- qfLra2: (45.0 x 10-ts C)l2r(2.00 x I0-2m)2: l.7g x 10-tt Cl^2.
s9
(a) The magnitude tr1 of the electric field produced by the charge q on the spheri cal shell isEt: ql4T€oRZ, where Ro is the radius of the outer surface of the shell. Thus
q - reohR\
(b) Since the field at P is outward and is reduced in magnitude the field of Q must be inward. Ais a negative charge and the magnitude of its field at P is E2:450N/C - 180N/C :270N/C.The value of Q is
- (450 N/CXo.2o m)2_ : 2.0 x 10- s C
8.99 x 10eN . nP lC'
e : 4nesEzRZ:- . (?70 N/cX0'20 m)2 - - - 1.2x 10-e c .
8.99 x 10eN. m2 lqz L'/
Chapter 23 143
(c) Gauss' law tells us that since the electric field is zero inside a conductor the net charge insidea spheri cal surface with a radtus that is slightly larger than the inside radius of the shell must bezero. Thus the charge on the inside surface of the shell is +L2 x 10-e C.
(d) The remaining charge on the shell must be on its outer surface and this is 2.0 x lQ-e g1.2 x 10-rg:*0.8 x 10-eC.
69
(a) Draw a spheri cal Gaussian surface with radius r, concentric with the shells. The electricfield, if it exists, is radtal and so is norrnal to the surface. The integral in Gauss' law isf E . d,A - 4nr'E, where E is the radial component of the field. For r <is zero. Gauss' law gives 4nr2E:0, so E:0.(b) For a <4nrz E - qo I eo and E - qo l4lreor2 .
(c) For r >and E - (qo + qd l4nesr2 .
(d) Consider first a spherical Gaussian with radius just slightly greater than a. The electric fieldis zero everywhere on this surface, so according to Gauss' law it encloses zero net charge. Sincethere is no charge in the cavity the charge on the inner surface of the smaller shell is zero. Thetotal charge on the smaller shell is eo and this must reside on the outer surface. Now considera spherical Gaussian surface with radius slight larger than the inner radius of the larger shell.This surface also encloses zero net charge, which is the sum of the charge on the outer surfaceof the smaller shell and the charge on the inner surface of the larger shell. Thus the charge onthe inner surface of the larger shell is -eo. The net charge on the larger shell is Qa, with -Qoon its inner surface, so the charge on its outer surface must be eu - ( -eo) - eu + ea.
76
(a) The magnitude of the electric field due to a large uniformly charged plate is given by o l2ro,where o is the surface charge density. In the region between the oppositely charged plates thefields of the plates are in the same direction, so the net field has magnitude Eelectrical force on an electron has magnitude eE - eo f es and the gravitational force on it isffig, where m is it mass. If these forces arc to balance, they must have the same magnitude, so
mg:eofesand
(9.11 x 10-" kg)(9.8 mls2x8.85 x 10-12 gz/N.m2) : 4.9 x Io-22 C lrrl1.60 x 10-le C
m9€oo:e
(b) The gravitational force is downward, so
is negatively charged the electrical force on
must be downward.
the electrical force must be upward. Since an electronit is opposite to the electric field, so the electric field
79
(a) Let A be the net charge on the shell, % be the
charge on its outer surface. Then Q - Qt * eo and et
144 Chapter 2 3
charge on its inner surface and eo be the
- Q - eo - (-10lrC) - (-1a pC) - *4 p,C.
(b) Let q be the charge on the particle. Gauss' law tells us that since the electric field is zero inside the conducting shell the net charge inside any spherical surface that entirely within the shell is zero. Thus the sum of the charge on the particle and on the inner surface of the shell is zero, so q + qi = 0 and q = -qi = -4 fLC.
Chapter 23 145
Chapter 24
3(a) An ampere is a coulomb per second, so
84A.h- (roc'h\ /
\ T) (roooi) :30x 1o5c
(b) The change in potential energy is Ltl : qLV - (3.0 x 10sq(12V):3.6 x 106J.
-5ifr. electric field produced by an infinite sheet of charge has magnitude E - of Zes, where o isthe surface charge density. The field is normal to the sheet and is uniform. Place the origin of acoordinate system at the sheet and take the r axis to be parallel to the field and positive in the
direction of the field. Then the electric potentral is
v:v, [" rd,r:vr-Er,Jo
t-J v
where V, is the potentral at the sheet. The equipotential surfaces are surfaces of constant r; thatis, they are planes that are parallel to the plane of charge. If two surfaces are separated by Lrthen their potentials differ in magnitude by LV : ELr - (ol2eo)A". Thus
Lr:2eoLv- 2(8.85 x 10-12c2/N.m:X50v) - g.g x 10_3m.o 0.10 x 10-6clm2
The electric potential V at the surface of the drop, the charge q on the drop, and the radiusof the drop are related by V - qf 4res.R. Thus
R-#,(b) After the drops combine the total volume is twice the volume of an original drop, so the
radius R' of the combined drop is given by (n')3 - 2R3 and R' - 21 /3 R. The charge is twice,the charge of original drop: q' : 2q. Thus
V,:1 q: -_L
2q _22/tV_22/t(5oov):1gov.4"r, R' "qffi-
t
29
The disk is uniformly charged. This means that when the full disk is present each quadrant
contributes equally to the electric potentral at P, so the potential at P due to a single quadrant
is one-fourth the potentral due to the entire disk. First find an expression for the potential at Pdue to the entire disk.
146 Chapter 24
t9(a)
R
Consider a ring of charge with radius r and width dr. Its area is Znr dr and it contains charge
d,q:2nord,r.Allthechargeinitisadistance!mEfromP,SothepotentialitproduceSatPis
or dr
2eo
The total potential at P is
o
6The potential Vsq at P due to a single quadrant is
a,7 I 2rrordrd,v: :4nes lmE
%q:I:*1ffi-Dl7 .73 x 10-rs C l^' - 0 .25q *l
I8(8.85 x 10-12 C2/N.m2)
- 4.71 x 10-5 V.
39
Take the negatives of the partial derivatives of the electric potential with respect to the coordinates
and evaluate the results for r - 3.00 m, A : -2.00 m, and z - 4.00 m. This yields
E*
Ea: -XE": -y - -2(2.00 Y l^o)raz : -2(2.00 Y l^ox3.00mX -2.00mX4.00m)
: 96.0v/m.0z
Ll
The magnitude of the electric field is
E- : :1.50x 102Y1*.
4t
The work required is equal to the potential energy of the system, rclative to a potential energy ofzero for infinite separation. Number the particles l, 2, 3, and 4, in clockwise order starting withthe particle in the upper left corner of the affangement. The potential energy of the interaction
of particles I and 2 is
(g.gg x 10eN. m2 lc')(2.30 x 10-t, cx -2.30 x 10-rz c)0.640 m
- -7.43 x 10-14 J .
The distance between particles I and 3 is ,Do and both these particles are positively charged,
so the potential energy of the interaction between particles I and 3 is Un: -UrrlfrChapter 24 147
r r QtQzL./a)-- -L- a
4Tega
r2+D2
R2+Dz
+5.25 x 10- 14 I. The potentral energy of the interaction between particles 1 and 4 is (Ju(Jn : -7.43 x 10-14 J. The potentral energy of the interaction between particles 2 and 3 is(Jzz _ (Jn: -7 .43 x 10- t4 J. The potential energy of the interaction between particles 2 and 4is Uz+
is Ut+
The total potentral energy of the system is
Ut - [Jn + Ur3 + Uru + [Jzz * Uz+ i Uz+
-7.43 x lo-LaJ+5.2s x 10-147- -1.92 x 10-13J.
This is equal to the work that must be done to assemble the system from infinite separation.
59
(a) Use conservation of mechanical energy. The potential energy when the moving particle is at
any coordinate A is qV , where V is the electric potential produced at that place by the two fixedparticles. That is,
(l:q 2Q
4Tes
where r is the coordinate and a is the charge of either one of the fixed particles. The factor2 appears since the two fixed particles produce the same potential at points on the A axis.Conservation of mechanical energy yields
Ky-K +q2Q 2Q
4nes 4nes
where K is the kinetic energy of the movin g particle, the subscrrpt 'i refers to the initial positionof the movin g particle, and the subscript f refers to the final positioll. Numerically
Ky2(-r5 x 10-6 C;1SO x 10-u C)
- 3.0J.4n(8.85 x 10-12 Cz/N .m2) (3.0m)2 + (4.0m)'
(b) Now K1- 0 and we solve the energy conservation equation for Uf .Conservation of energyfirst yields tJ y : K,i, + Ut The initial potenttal energy is
: K,+2qQ" 4reg
rT - zqQ
- 2(-15 x 10-6 CX50 x 10-6 C)
\-/L /---._:4nesffi 4r$.85 x 10-12cz/N .m2)2.7 I.
Thus Ky: l.zJ- 2.7 J- -1.5J.Now
*a2
+A?
1l@l
+y?
148 Chapter 24
Uy4nes
2qQ
63
If the electric potential is zero at infinity, then the electric potential at the surface of the sphere
is given by V - qf 4Tesr, where q is the charge on the sphere and r is its radius. Thus
q : 4nesrv - (0' 15 mX1500 V)
- : 2.5 x 10-8 C .
8 .gg x 10eN . nP lC' /
65
(a) The electric potential is the sum of the contributions of the individual spheres. Let h be
the charge on one, ez be the charge on the other, and d be their separation. The point halfwaybetween them is the same distance d,l2 (: 1.0m) from the center of each sphere, So the potentialat the halfway point is
V: $*Qz' 4nesd,f 2 1.0 m
(b) The distance from the center of one sphere to the surface of the other is d, - R, where R isthe radius of either sphere. The potential of either one of the spheres is due to the charge on thatsphere and the charge on the other sphere. The potentral at the surface of sphere 1 is
,i:*[*.h]=(g.ggx loeN.nf lc')
[:2.9 x 103 V.
(c) The potential at the surface of sphere 2 is
Vz I n, *gZ1ld,- R RJ
- (g .gg x loeN. m2 lc') L
- -8.9 x 103 V.
1.0 x 10-a g 3.0 x 10-s g
0.030m 2.0m - 0.030m
1.0 x 10-8 g
2.0m - 0.030m 0.030m
3.0 x 10-s g
75
The initial potential energy of the three-particle system is Ur:2(q2 f 4reoL) * flrr*"d, where q isthe charge on each particle, L is the length of a triangle side, and Urr*"d is the potenttal energyassociated with the interaction of the two fixed particles. The factor 2 appears since the potential
Chapter 24 149
energy is the same for the interaction of the movable particle and each of the fixed particles. Thefinal potential energy is uy - 2lq'f 4nes(Ll2)l + Un*ed, and the change in the potential energy is
2q2
4nesL
This is the work that is done by the external agent. If P is the rate with energy is supplied bythe agent and t is the time for the move, then Pt : LU, and
, LIJ 2q' z(B.gg x 10e N . m2 lc'Xo .rzc)2 1 o', 1 .t- p-ffi- :1.83x10's.
This is 2.1 d.
77
(a) {Jse Gauss' law to find an expression for the electric field. The Gaussian surface is acylindrical surface that is concentric with the cylinder and has a radius r that is greater thanthe radius of the cylinder. The electric field is normal to the Gaussian surface and has uniformmagnitude on tt, so the integral in Gauss' law is f E . dA:2rrEL, where L is the length ofthe Gaussian surface. The charge enclosed is
^L, where
^ is the charge per unit length on the
cylinder. Thus ZrrRLE : ),Lf es and E - ),f 2nesr.
Let E n be the magnitude of the field at B and r p be the distance from the central axis to B. LetEs be the magnitude of the field at C and rs be the distance from the central axis to C. SinceE is inversely proportional to the distance from the central axis,
trs: ryEp: '#(160N/c) :64N/c.Tg u
5.0 cm'
(b) The magnitude of the field a distance r from the central axis is E - (, e liU n so thepotenttal difference of points B and C is
ve vs: l,': TEp d,r: -rBEptn(P\\rc /= -(o 'a2omx16oN/c)ln (ffi) -z.ev.
(c) The cylinder is conducting, so alI points inside have the same potential, namely VB, so
Vt Vn: 0.
8s
Considerfrom thepotentralthe point
a point on the z axis that has coordinate z. Allpoint. The distance is r- lW, where Ris taken to be zero at points that are infinitelyis
points on the ring are the same distanceis the radius of the ring. If the electricfar from the ring, then the potential at
a
150 Chapter 24
V-4Tes
where A is the charge on the ring. Thus
- (8 .gg x loeN. m2 lc'X16.0 x lo-u c)
Forr>
r7 1av - 4"r"; '
where the zero of potential was taken to be at infinity.
(b) To find the potentral in the region 11 <for the electric field, then integrate along a radial path from 12 to r. The Gaussian surface is
a sphere of radius r ) concentric with the shell. The field is radial and therefore normal to the
surface. Its magnitude is uniform over the surface, so the flux through the surface is O - 4nr2 E.The volume of the shell is (nl3)(rt, - r?), so the charge density is
3Qp:ar@) - r?)
and the charge enclosed by the Gaussian surface is
1l_lnl
-o3oo{
93
(a)
Gauss' law yields/n3-rJ\
4TesrzU-QI' 'r\#)and the magnitude of the electric field is
D- A r3-'3,t) 4"rrffiIf V, is the electric potential at the outer surface of the shell (r : r) then the potential a distance
r from the center is given by
/ 4r\q: (+) (,'
(0.0300 m)2 1(0.0400 *)2
Chapter 24 151
The potential at the outer surface is found by placing r : 12 in the expression found in part (a).
It is V, - Q f 4Tesr2. Make this substitution and collect like terms to find
r/_ a 1 (v3 t "i \4"ofir\z T 7)'Since p : 3Q lan@] - ,3r) this can also be written
v-*(+ + +)(c) The electric field vanishes in the cavrty, so the potential is everywhere the same inside and
has the same value as at a point on the inside surface of the shell. Put r - 11 in the result ofpart (b). After collecting terms the result is
rr_ A 3(r3-r?)4?t€o zez - f)'
or in terms of the charge density
v - +?3-'?)zeo
(d) The solutions agree at r : rr and at r : 12.
95
The electric potential of a dipole at a point a distance r away is given by Eq. 24-30:
rr_ I pcos?v - 4"r, 12 '
where p is the magnitude of the dipole moment and e is the angle between the dipole momentand the position vector of the point. The potential at infinity was taken to be zero. Take the zaxis to be the dipole axis and consider a point with z positive (on the positive side of the dipole).For this point r : z and 0 : 0. The z component of the electric field is
E=--ry:-a ( o \- p- A" - 0, \a"rrrz ) 27reo*'
This is the only nonvanishing component at a point on the dipole axis.
For a point with a negative value of z, r: -z and cos 0 - -1, So
E=- -L r--p^\ -- - P =
d z \ +ere orz ) 2T esz3 '
103-
(a) The electric potentral at the surface of the sphere is given by V : qf 4TeoR, where q is the
charge on the sphere and R is the sphere radius. The charge on the sphere when the potential
reaches 1000 V is
q : 4nesrV : - (0'010 mX1000 v)
- :1.11 x 10-e C .
8.99 x 10eN. m2 lC'I52 Chapter 24
The number of electrons that enter the sphere is jV - qle - (1.11 x 10-e ql(l.60 x 10-1e g; -6.95 x 10e. Let.R be the decay rate and tbethe time for the potential to reach it final value. Sincehalf the resulting electrons enter the sphere jV - (Pl2)t and t - zl.VlP - 2(6.95 x 10e) 1Q.70 x108 s-r) : 38 s.
(b) The increase in temperature is LT : l/A E lC , where E is the energy deposited by a singleelectron and C is the heat capacity of the sphere. Since iV - (Plz)t, this is LT - (Plz)tLElCand
t -'zg Y- 2(14J/KX5"0K) :2.4x 107 s.P LE (3.70 x 108 s-rx100 x 103 eVXl.60 x 10- re J leY)
H'
This is about 280 d.
Chapter 24 153
Chapter 25
-5
ftl The capacitance of a parallel-plate capaeitor is given by e - esA I d", where A is the areaofeach plate and d is the plate separation. Since the plates are circulag the plate area is A - T R2 ,
where R is the radius of a plate. Thus
e_4d 1.30 x 10-3 m
(b) The charge on the positive plate is given by q - CV, where V is the potential differenceacross the plates. Thus q: (I .44 x 10-tof'XI20V): 1.73 x 10-8 C - 17.3nC.
15
The charge initially on the charged capacitor is given by q- CrVo, where Cr (: 100pF) is the
capacitance and Vo (: 50 V) is the initial potential difference. After the battery is disconnectedand the second capaeitor wired in parallel to the first, the charge on the first capacitor is
er : CrV, where u (: 35 V) is the new potential difference. Since charge is conserved in theprocess, the charge on the second capacitor is Qz: q er, where ez is the capacitance of the
second capacitor. Substitute CtVo for q and CrV for gt to obtain q2: Cr(Vo-V). The potentialdifference across the second capacitor is also V , so the capacitance is
(1 : ez: vo v n 5ov -#(roopF) :43pF.L,Z-V- , t't: g5V \
t9(a) After the switches are closed, the potentral differences across the capacitors are the same andthe two capacitors are in parallel. The potenttal difference from a to b is given by Vou - Q lC"r,where A is the net charge on the combination and Ceq is the equivalent eapacitance.
The equivalent capacitance is Ceq: Ct+Cz - 4.0 x 10-6F. The total charge on the combinationis the net charge on either pair of connected plates. The charge on capacitor I is
er: CtV - (1.0 x 10-u pX100V) - 1.0 x 10-4 C
and the charge on capacrtor 2 is
Qz : CzV - (3.0 x 10-u pX100 v) : 3.0 x 10-4 C ,
so the net charge on the combination is 3.0 x 10-4 C 1.0 x 10-4 C - 2.0 x l0-4 C. Thepotential difference is
Voa
(b) The charge on capacitor 1 is now
154 Chapter 2 5
-2.0 x 10-4C-
4.0 x 10-6 F
Qt:CrVob-(1.0
50v.
x 10-u FXSO V) : 5.0 x 10-5 C.
(c) The charge on capacitor 2 is now ez: CzVou - (3.0 x 10-upXS0V; - 1.5 x 10-4C.
29
The total energy is the sum of the energies stored in the individual capacitors. Since they areconnected in parallel, the potenttal difference V across the capacitors is the same and the totalenergy is (J: : *(ct+C)172: *Q.o x 10-6F+ 4.0 x lO-upX:00D2 - 0.27 J.
35
(a) Let q be the charge on the positive plate. Since the capacitance of a parallel-plate capacitoris given by esAld. the charge is qseparation is dt and the potential difference is V' . Then q : esAV' ld,' and
v, dt d' eoA - - d' -- 8.oo mr- ,rAq: rrA d v-ftv:t6b#(6.oov): T6.ov.
(b) The initial energy stored in the eapacitor is
(Ji2- 2d 2(3.00 x 10-3 mm)
and the final energy stored is
uy: lc,(v,)r: ;# (v,)r: : r.zox ro-ro J.
(c) The work done to pull the plates apart is the difference in the energy: W - tI y Ut -1.20 x 10-10J -4.51 x 10-11 J -7.49 x 10-11 J.
capacitance of a cylindrical capacitor is given by
c - K(1^- 2" *oL
._,e - In(W,
where Co is the capacitance without the dielectric, K is the dielectric constant, L is the length, ais the inner radius, and b is the outer radius. See Eq. 25-14. The capacitance per unit length ofthe cable is
43
The
C 2n nes zn(2.6x8.85 x 10- t' p l^) n 4 , 1 1::L tn(bla) ln [(o.60mm)le. 10**]]--8'1 x 10-"Ffm- 81pFfm'
45
The capacitance is given by Cdielectric, K is the dielectric constant, A is the plate area, and d, is the plate separation. The
Chapter 2 5 155
electric field between the plates istheplates. Thus d-Vltr and C
given by E : V ld, where V is the potential difference between: K€yAE lV. Solve for A:
CVA-K€gE
For the area to be a minimum, the electric field must be the greatest it can be without breakdownoccurritr.g. That is,
0.63 nl
51
(a) The electric field in the region between the plates is given by E - V I d, where V is thepotenttal difference between the plates and d is the plate separation. The capacitance is given byC - neyAf d,, where A is the plate area and K is the dielectric constant, so d - nesAf e and
VC (50 VX100 x 10-tz F)E -
=: -1.o x 1o4Yl^.KesA 5.4(8.85 x 10-t2F lmxl00 x 10-4 m2)
(b) The free charge on the plates is qf : CV - (100 x 10-t'fXS0V) : 5.0 x 10-e C.
(c) The electric field is produced by both the free and induced charge. Since the field of a large
uniform layer of charge is q f 2esA, the field between the plates is
,;r_ qf , Qf Qt Qt'rJ 2r"A= vr4 zrrA zroA'
where the first term is due to the positive free charge on one plate, the second is due to the
negative free charge on the other plate, the third is due to the positive induced charge on one
dielectric surface, and the fourth is due to the negative induced charge on the other dielectric
surface. Note that the field due to the induced charge is opposite the field due to the free charge,
so the fields tend to cancel. The induced charge is therefore
Qt: Qf - €e AE: 5.0 x 10-e c- (8.85 x 10-12 F lm)(100 x 10-4 m2xl.0 x 104 v l^):4.I x l0-eC-4. 1nC.
6r
Capacitors 3 and 4 are in parallel and may be replaced by a capacitor with capacitance Ctq :e3 + e+- 30pF. Capacitors l, 2, and the equivalent capaeitor that replaced 3 and 4 are allin series, so the sum of their potential differences must equal the potential difference across the
battery. Since all of these capacitors have the same capacitance the potential difference across
each of them is one-third the battery potential difference or 3.0V. The potential difference across
capac rtor 4 is the same as the potential difference across the equivalent capacitor that replaced 3
and,4, so the charge on capacrtor 4 is e+: CqVq - (15 x 10-upX:.0V) :45 x 10-6 C.
156 Chapter 25
(7.0 x 10-8 F114.0 x 103 V)2.8(8.85 x 10-t2F lmxlS x 106 V lm)
69
(a) and (b) The capacitors have the same plate separation d and the same potenttal difference Vacross their plates, so the electric field are the same within them. The magnitude of the field ineither one is E -Vld, - (600V)l(3.00 x 10-'-) -2.00 x 105 Yl^.(c) Let A be the area of a plate. Then the surface charge density on the positive plate is
oA105 Vlm)- 1,.77 x 10-u Cl^', where CV was substituted for q and the expression esAld, forthe capacitance of a parallel-plate capacitor was substituted for C.
(d) Now the capacitance is rces Ald, where rc is the dielectric constant. The surface charge densityon the positive plate is oB: K€08 - KoA- (2.60XI.77 x 10-uCl^') - 4.60 x 10-u Cl^'.(e) The electric field in B is produced by the charge on the plates and the induced charge
together while the field in A is produced by the charge on the plates alone. since the fields ate
the same og * oindu_ced : oA, so oinduced: 04 - op: I.77 x 10-o gl^' - 4.60 x 10-u Cl^'--2.g3 x 10- u Cl^' .
73
The electric field in the lower region is due to the charge on both plates and the charge inducedon the upper and lower surfaces of the dielectric in the region. The charge induced on the
dielectric surfaces of the upper region has the same magnitude but opposite sign on the twosurfaces and so produces a net field of zerc in the lower region. Simtlarly, the electric field inthe upper region is due to the charge on the plates and the charge induced on the upper and lowersurfaces of dielectric in that region. Thus the electric field in the upper region has magnitudeE oo.,
: QK,,ryper€sA and the potential difference across that region is l/uppr, - f,upp"rd, where d is
the thickness of the region. The electric field in the lower region is -E1o*", : eKto*er€oA and thepotential difference across that region is I4o*., - Eu,*"rd. The sum of the potential differencesmust equal the potentral difference V across the entire eapacitor, so
v -Eupp., d,+Ero*., d- #L#. *]The solution for q is
KrpperKlower eOA x 10-12N .m2 lc')(z.oo x ro-2m2)Vq:
ffiupp.. * Klo*.. d
1.06 x 10-e C.
3.00 + 4.00 2.00 x l0-3 m(7.00 v)
Chapter 2 5 157
Chapter 25
7_
(a) The magnitude of the current density is given by Jparticles per unit volume, q is the charge on each particle, and u4 is the drift speed of theparticles. The particle concentration is n- 2.0 x 108 cm-3q:2e:2(L60 x 10-tsC) -3.20 x 10-" C, and the drift speed is 1.0 x 105m/s. Thus
J: (2 x 10r4m-t)(3.2 x 10-tnCXl.0 x 105 mls): 6.4A1^' .
(b) Since the particles are positively charged, the cuffent density is in the same direction as theirmotion, to the north.
(c) The current cannot be calculated unless the cross-sectional area of the beam is known, Then'i - JA can be used.
t7The resistance of the wire is given by R- pL lA, where p is the resistivity of the material, Lis the length of the wire, and A is the cross-sectional area of the wire. The cross-sectional area
is A- nr2radius of the wire. Thus
RA (50 x 10-'OXz.85 x 10-t m2) A ,\p: t: 2n^ :2.0 x lo-t g 'rr.
t9The resistance of the coil is given by R - pL lA, where L is the length of the wire, p is theresistivity of copper, and A is the cross-sectional area of the wire. Since each turn o1 wire has
length 2nr, where r is the radius of the coil, L - (250)2nr - (250)(2n)(0.12m) - 188.5m. lf r*is the radius of the wire, its cross-sectional areais A - Tr2* - ?T(0.65x 10-'*)tAccording to Table 26-I, the resistivity of copper is I.69 x 10-a 9.ril. Thus
n _ pL (l .69 x 10-s 9.mX188.5m) _ ^ A t_,IL- -
- L.+trr.A 1.33 x 10-6 m2
2tSince the mass and density of the material do not change, the volume remains the same. If Lsis the original length, L is the new length, Ao is the original cross-sectional are1 and A is the
new cross-sectional area, then LoAo: LA and A- LoAolL: LoAol3fo: Aol3. The newresistance is
P3 LoR-+: -s+: epo.AgAo13
158 Chapter 26
where Ro is the original resistance. Thus R - 9(6.0 Q) : 54 O.
23
The resistance of conductor A is given by
RA:#,
where ra is the radius of the conductor. If ro is the outside radius of conductor B and ri is itsinside radius, then its cross-sectional area is r(r2" - ,?) and its resistance is
Rn- .lL =".u'D n@2" - ,?) '
The ratio isRa:rrr_r?R" T
J'v'
39
(a) Electrical energy is transferred to internal energy at a rate given by
y2D-t. E'
where V is the potential difference across the heater and R is the resistance of the heater. Thus
P -(1?9p'r4r)
(b) The cost is given by
c - (1.0kwx5.0hx$0.050 /kw .h) - $0.25 .
43
(a) Let P be the rate of energy dissipation, i be the current in the heater, and V be the potential
difference across the heater. They are related by P - iV. Solve for i:
' P 1250w-10.9A.?'- v- rrsv
(b) According to the definition of resistance V - dR, where R is the resistance of the heater.
Solve for R:
R-Y:1ry- r0.6E2.i 10.9 A(c) The thermal energy t produced by the heater in time t (: 1.0 h - 3600 s) is
E - Pt : (1250 W)(3600 s) : 4.5 x 106 J .
Chapter 26 159
s3
(a) and (b) Calculate the electrical resistances of the wires. Let pc be the resistivity of wire C,
rs be its radius, and Lc be its length. Then the resistance of this wire is
Rc:pc
Let p n be the resistivityof this wire is
Rn: pD
If i is the current in the
Ln+-(2.ox1o-og.m)Trb
4- (t.o x 10-6e.m)Tr'D
1.0m
7r(0.50 x 10-t ^)'-2.540.
of wire D, r p be its radius, and L n be its length. Then the resistance
1.0m
7r(0 .25 x 10-'*)twire, the potentral difference between points I
LVz - iRs - (2.0 AX2.54Q) : 5. 1 V
and the potential difference between points 2 and 3 is
LVzt:'iRp - (2.0AX5.09 O) - 10V.
(c) and (d) The rate of enerry dissipation between points 1 and 2 is
Prz : i2 Rg - (2.0 A)2 (2.54o) : 10 w
and the rate of energy dissipation between points 2 and 3 is
Pzz : i2 Rn - (2.0 A)2(5.09 o) : 20 w .
--55
(a) The charge that strikes the surface in time Lt is given by Lq - 'i Lt, where i is the current.
Since each particle carries charge 2e, the number of particles that strike the surface is
iv - L.q- ry- (0.25 x 10-: A)-(1.0 s) :2.3 x 1012 .2e 2e 2(1 .6 x 10- tq C)
(b) Now let t/ be the number of particles in a length L of the beam. They will all pass throughthe beam cross section at one end in time t - L lr, where u is the particle speed. The current isthe charge that moves through the cross section per unit time. That is, i- 2elt{lt - 2el{uf L.Thus, lV - iLf 2eu.
Now find the particle speed. The kinetic energy of a particle is
K - 20MeV - QA x 106eD(l.60 x 10-t'llev) :3.2 x 10-t2 J.
SinceK-}'^u,,U:vw.ThemaSSofanalphaparticleisfourtimesthemaSSofaprotonor m:4(1 .67 x 10-27 kg) - 6.68 x 10-" kg, so
-5.09f).
and 2 is
2(3 .2 x 10- tz I)6.68 x 1O-zt kE
160 Chapter 26
:3.1 x 107 ml s
and
iv - ?L - (O'zs x 19,6=t)(?o
x t9-,im) : 5.0 x 103 .Zeu 2(I.60 x 10- le CX3. I x 107 m/s) v '
(c) Use conservation of energy. The initial kinetic energy is zero, the final kinetic energy is
20 MeV - 3.2 x 10- " J, the initial potential energy is qV - 2eV, and the final potential energy
is zero. Here I/ is the electric potential through which the partrcles are accelerated. Conservation
of energy leads to K y - LLi, -- 2eV , so
v-#- _10x1o6v.
59
Let Rn be the resistance at the higher temperature (800" C) and let Rr be the resistance
at the lower temperature (200" C). Since the potential difference is the same for the twotemperatures, the rate of energy dissipation at the lower temperature is Pr _ yz I R6 and the
rate of energy dissipation at the higher temperature is Ps - 172 I Rn, so P7 - (Rn I Rr)Pn.Now Rr : Ru + aRn LT, where LT is the temperature difference Ty - Ts
PrRs
PsPs s00w
Rn * aRn LT:
1*c-LT 1+(4.0 x 10-4 l.C)(-600"C)
75
If the resistivity ir po at temperatureTo, then the resistivity at temperatureT is p: po+apy! -70), where a is the temperature coefficient of resistivity. The solution for T is
T_ p - po* apoTo
aPo
Substitute p : Zpo to obtain
IT - To +
a:20.0"C +
4.3 x 10-3 K-l:250oC .
The value of a was obtained from Table 26-I.
Chapter 26 16l
Chapter 27
7
(a) Let i be the current in the circuit and take it to be positive if it is to the left in R1. I-Jse
Kirchhoffs loop rule: & - iRz - iRr - tz:0. Solve for i:
i _ !, - ? - t2y - 6.oy _ o.5oA.Rr+Rz 4.0O+8.0O
A positive value was obtained, so the current is counterclockwise around the circuit.
(b) and (c) If i is the current in a resistor with resistance R, then the power dissipated by thatresistor is given by P - z2 R. For Rr the power dissipated is
P1 - (0.50 A)'(4.0 O) : 1.0w
and for Rz the power dissipated is
P2 - (0.50 A)'(8.0 O) : 2.0'W .
(d) and (e) If i is the cuffent in a battery with emf t, then the battery supplies energy at the rate
P - it provided the current and emf are in the same direction. The battery absorbs energy at
the rate P : i,t if the current and emf are in opposite directions. For battery 1 the power is
P1 -(0.50AXLZV)-6.0W
and for battery 2 it isP2 - (0.50 AX6.0 V) - 3.0'W' .
(0 and (g) In battery l, the current is in the same direction as the emf so this battery suppliesenergy to the circuit. The battery is discharging. The cuffent in battery 2 is opposite the directionof the emf, so this battery absorbs energy from the circuit. It is charging.
13
(a) If i is the current and LV is the potential difference, then the power absorbed is given byP - i, LV. Thus
Lv: I- 5owi 1.0A
:50v'
Since energy is absorbed, point A is at a higher potential than point B; that is, Ve - Vn : 50 V.
(b) The end-to-end potentral difference is given by Va Vn - +'iR + t, where t is the emf ofelement C and is taken to be positive if it is to the left in the diagram. Thus t - Ve - Vs - oR :50 v - (1.0 A)(2.0 o) - 48 V.
162 Chapter 27
(c) A positive value was obtained for t, so it is toward the left. The negative terminal is at B.
2t(a) and (b) The circuit is shown in the diagram tothe right. The current is taken to be positive if itis clockwise. The potential difference across battery1 is given by Vthe current must be igives 2E irr irz iR - 0. Substitute i - t lrtand solve for R. You should get R _ 11 12 _
0.0160 - 0.012Q - 0.004Q.
Now assume that the potential difference across bat-tery 2 is zero and carry out the same analysis. Youshould find R: rz - 11. Since 11 >be positive, this situation is not possible. Only the potenttal difference across the battery withthe larger internal resistance can be made to vanish with the proper choice of .R.
29
Let r be the resistance of each of the thin wires. Since they are in parallel, the resistance R ofthe composite can be determined from
or R-r19. Now
and
R- oll=.rDz )
where p is the resistivity of copper. Here rd2 l4 was used for the cross-sectional area of any one
of the original wires and n D2 14 was used for the cross-sectional arca of the replacement wire.Here d and D are diameters. Since the replacement wire is to have the same resistance as thecomposite,
4p( -nD2
4p(9rd2
Solve for D and obtain D - 3d.
33
Replace the two resistors on the left with their equivalent resistor. They are in parallel, so theequivalent resistance is ftee - 1.0 O. The circuit now consists of the two emf devices and fourresistors. Take the current to be upward in the right-hand emf device. Then the loop rule gives
tz - 'iR"q - 3iR - ty where R - 2.0 O. The current is
. tz-trL: &q+3u
JL
19R r)
rr, : 4P(' rd,2
rzv - 5.0vl.0o+3(2.0o) -1.0A.
Chapter 27 163
To find the potential at point I take a path from ground, through the equivalent resistor andtz, to the point. The result is Vthe potential at point 2 continue the path through the lowest resistor on the digram. It isVz: V + iR: -11 V + (1.0 A)(2.0 O) : -9.0V.
47
(a) and (b) The copper wire and the aluminum jacket are connected in parallel, so the potentialdifference is the same for them. Since the potenttal difference is the product of the current andthe resistance, ic Rc - i,tRt, where i,g is the current in the copper, i, t is the current in thealuminum, Rc is the resistance of the copper, and Rt is the resistance of the aluminum. Theresistance of either component is given by ft - pL lA, where p is the resistivity, L is the length,and A is the cross-sectional area. The resistance of the copper wire is
Rg
and the resistance of the aluminum jacket is
P PNLn,g
Substitute these expressions into icRc: itRt and cancel the common factors L and 7r toobtain
'ic pc -
iePaA2 b2_a2 '
Solve this equation simultaneously with 'L - 'ic + t A, where i is the total current. You should get
is: o2 Pc'i(b2 - a2)pc * a2 pe
and
,i e - -(b2 : a2) Pci-(b2 - a2)pc * a2 pt '
The denominators are the same and each has the value
Thus
?,9 -(0.250 x I0-t
^)'(2.75 x 10-sg . m)(2.00A)
3.10 X 10-1s O.m3and
xg:
- 0.893 A.
164 Chapter 27
(b2 - a2)pc + a2pt: [(0.380 x 10-"tt)'- (0.250 x 10-'*)'] (1.69 x 10-a g.m)
+ (0 .250 x l0-'*)t (2.75 x 10-8 O .m)
-3. 10x 10-15o.m3.
[(0.380 x 10-' rr)t - (0 .250 x l0-' *)t] tt .69 x ro-a g . mX2.00 A)3.10x l0-l5e.m3
(c) Consider the copper wire. If V is the potential difference, then the current is given byV - icRc:'icpcLf ra2, so
r razV zr(0 .250 x 10-t ^)'(12.0
V) 1AL- -- -126m.icpc (1.11AXL.69 x 10-8 O .m)
s7
During charging the charge on the positive plate of the capacitor is given by Eq. 27 -33, withRC : r. That is,
q: Ct fr - e-'/'1 ,L J/where e is the capacitance, t is applied emf, and r is the time constant. You want the time forwhich q:0.990Ct, so
0.990:1-e-tl'Thus
e-t /' o.o 10 .
Take the natural logarithm of both sides to obtain tlr: - ln0.010: 4.61 and t- 4.61r.
65
(a), (b), and (c) At t : 0, the capaaitor is completely uncharged and the current in the capaeitor
branch is as it would be if the capacitor were replaced by a wire . Let 'i1 be the current in ^Rr and
take it to be positive if it is to the right. Let i,2 be the current in Rz and take it to be positiveif it is downward. Let i,z be the current in Rt and take it to be positive if it is downward. Thejunction rule produces 'ir : 'iz * 'iz, the loop rule applied to the left-hand loop produces
t - irRr - izRz: 0,
and the loop rule applied to the right-hand loop produces
izRz - hRt:0.
Since the resistances are all the same, you ean simplify the mathematics by replacing Rr, Rz,
and R3 with R. The solution to the three simultaneous equations is
'it- 2.5-: - 1.1 X 10-3 A3R 3(0.73 x 106 Q)
and
,i2: ,i3- !-- t'' * t?1y=. : 5.5 x ro-4 A.' 3R 3(0.73 x 106 O) v
(d), (e), and (0 At t - oo, the capacitor is fully charged and the current in the capaeitor branchis zero. Then \ : 'iz and the loop rule yields
t - hRt - irRz: 0.
Chapter 27 165
The solution is
,it: ,i2: L_ r'2 " 103 Y2R 2(0.73x tT"O- 8'2 x lo-4A'
(g) and (h) The potentral difference across resistor 2 is Vz: i2R2. At f : 0 it is
V2 - (5.5 x 10-4 AXO .73 x 106 O) : 4.0 x 102 V
and at f : m it isV2 - (8.2 x 10-4 AXO .73 x 106 0) : 6.0
(i) The graph of Vz versus t is shown to the right. Vz
sl2
x I02 V.
tl3
e l6
73
As the capacitor discharges the potential difference across its plates at time t is given byV - Voe-tl', where Vo is the potenttal difference at time t- 0 and r is the capacitive timeconstant. This equation is solved for the time constant, with result
tr: rn(vlw'Since the time constant is r : RC, where RR is the resistance and C is the capacitance,
n- trL- ctn(vlw'For the smaller time interval
R -- to'o * to-1t - : )48 o.(o.zzox to-6F)ln fg)
L"'\)
\ 5.00v /and for the larger time interval
R -- 6'oo x 1o-3 s - r.4g x ro4 e .
(0.220x 10-6F)ln fg)' \5.00v )75
(a) Let i be the cuffent, which is the same in both wires, and t be the applied potential difference.Then the loop equation gives t - iR,q - iRn - 0 and the current is
i- t - 6l'oY = :70. 1A.
Ra+ Rn 0.127 Q + 0.729n
166 Chapter 27
The current density in wire A is
70.1 A1.32 x I07 Al^':::
Trr2o 7r(1.30 x 10-t ^)'J-t
(b) The potentral difference across wire A is Vt : 'iRt - (70. 1 AX0.127 A) - 8.90 V.
(c) The resistance is Rt : p AL I A, where p is the resistivity, A is the cross-sectional area, and
L is the length. The resistivity of wire A is
Pa- 44L 40.0m
According to Table 26-l the material is copper.
(d) Since wire B has the same diameter and length as wire A and carries the same current, the
current density in it is the same, I.32 x 107 Al^' .
(e) The potential difference across wire B is Vs:'iRs - (70. 1AX0.729 O):51.1V.(f) The resistivity of wire B is
- ReAL 40.0m
26-I the material is iron.
- 9.68 x 10-8 O .fii.
The three circuit elements are in series, so the current is the same in all of them. Since the
battery is discharging, the potential difference across its terminals is Vaut _ t - ir, where t is
its emf and r is its internal resistance. Thus
iJt
Pe
According to Table
77
- t -v - lzv - ll .4v _ t- 0.012Qi sOA
This is less than 0.0200 Q, so the battery is not defective.
The resistance of the cable is Rcable
0.040 O. The cable is defective.
The potential difference across the motor is Tmotor- II .4V - 3.0V-8.4V and its resistance
is R*oto. - Vmotorl'i- (8.4V)l(50A) - 0.17 {1, which is less than 0.200O. The motor is notdefective.
85
Let ftso be the resistance of the silicon resistor at 20" and Rrc be the resistance of the iron resistorat that temperature. At some other temperature T the resistance of the silicon resistor is ftsftso + o.s Rso(T - 20"C) and the resistance of the iron resistor is Rr : Rra * ar Rrc(T - 20" C).Here a s and a 1 are the temperature coefficients of resistivity, The resistors are series so theresistance of the combination is
R - ftso + Rrc+ (asRso a arRroXT - 20oC) .
We want Rso * Bro to be 10000 and asRso * arRrc to be zero. Then the resistance of thecombination will be independent of the temperature.
Chapter 27 167
The second equation gives Rn- -(rs lrt)Rso and when this is used to substitute for Rn inthe first equation the result is Rso - (as l*r)Rso:1000O. The solution for Bso is
1000 f) 1000 (-,
as , _70 x lo-3 K-lIag )
a1 1
where values for the temperature coefficients of resistivity were obtained from Table 26-1. Theresistance of the iron resistor is Rro: 1000 O 85 O - 91 5 O.
9s
When the capacitor is fully charged the potential difference across its plates is t and the energy
stored in it is (l - +C t' .
(a) The current is given as a function of time by i _ (t lB)e-t/', where r (- RC) is thecapacitive time constant. The rate with which the emf device supplies energy is P6 : it and the
energy supplied in fully charging the capacitor is
ftSO- T_ -85O
T2RC : Ct2
This is twice the energy stored in the capacitor.
(b) The rate with which energy is dissipated in the resistor is Pp : iZR and the energy dissipatedas the capacitor is fully charged is
Es: lr*
,rdt -f Ir* "-tt, d,t:+:
R
Ep: lo*
Ppd,t:"-2t/r
d,t: S'! : tzRC _2R 2R+1,* ice'
97
(a) Immediately after the switch is closed the capacitor is uncharged and since the charge onthe capaaitor is given by q - CVc, the potential difference across its plates is zero. Applythe loop rule to the right-hand loop to find that the potential difference across Rz must also be
zero. Now apply the loop rule to the left-hand loop to find that t - irRr: 0 and 'h - t I Rt-(30 v)lQ0 x 103 o) - 1.5 x 10-3 A.
(b) Since the potential difference across Rz is zero and this potential difference is given byVnz : 'i2Rz, 'i2: 0.
(c) A long time Iater, when the eapacitor is fully charged, the current is zero in the capacitorbranch and the cuffent is the same in the two resistors. The loop rule applied to the left-hand loopgives €-iRr-iRz:0, soi -tl@t+R) -(30Y)lQ0x 103O+10x 1030):1.0x 10-3A.
99
(a) Rz and Rz are in parallel, with an equivalent resistance of RzRt l@, + Rz), and thiscombination is in series with Rt, so the circuit can be reduced to a single loop with an emf tand a resistance Req: Rr+ Rzful@2* R:) : (ftr Rz+ RrRz+ RzR:,)l(Rr + Rz).The curent is
t (Rz+ h)t
168 Chapter 27
Req RrRz + Br Rz * RzRt
The rate with which the battery supplies energy is
P-,it:
The derivative with respect to Rt is
(Rz + Rz)t'RrRz+ ftr Rz + RzRz
'
d,P gz (Rza RtXRr * Rz)tz srRTdRt RrRz+ RrR: + RzRt (ftr Rz+ RrRz+ RzR)z (ftr Rz+ RrRz+ RzRz)z )
where the last form was obtained with a little algebra. The derivative is negative for all (positive)values of the resistances, so P has its maximum value for Rz: 0.
(b) Substitute Rt - 0 in the expression for P to obtain
p - \rt : t: gJl: r4.4w.- RrRz Rr 10.0 O
101
If the batteries are connected in series the total emf in the circuit is IV t and the equivalentresistance is B+ frr, so the current is ,i - Ntl@+l[r). If R: r, then ,i - Ntl(N + 1)r.
If the batteries are connected in parallel then the emf in the circuit is t and the equivalentresistanceisE+rlIV, sothecurrentis i- Sl@*rlIV)- IVtl(Xn*r). If RIV E l(lv + l)r, the same as when they are connected in series.
Chapter 27 169
Chapter 28
3
(a) The magnitude of the magnetic force on the proton is given by Fs - eu B sin @, where u isthe speed of the proton, B is the magnitude of the magnetic field, and 0 is the angle betweenthe particle velocity and the field when they are drawn with their tails at the same point. Thus
t): {I , - 6'50x10-17N-- - 4.00x105m/s.
eB sin / (1.60 x 10-te CX2.60 x 10-3 T) sin 23,0"
(b) The kinetic energy of the proton is
K - **r': *O.67 x 10-" kg)(4.00 x 105 mf s;2- r.34 x 10-16J.
This is (r.34 x l0-16 Dl(l.60 x 10-t'Jlrv): 835 ev.
t7
(a) Sinse the kinetic energy is given by K -is its speed,
where m is the mass of the electron and u**r',
- 2.05 x 107 ml s.
(b) The magnitude of the magnetic force is given by eu B and the acceleration of the electron isgiven by ,'f r, where r is the radius of the orbit. Newton's second law is euB - mu'lro so
B -mu: (q.tt x to- 31-kgx2.05 x 197T/s) - 4.6g x 10-4T- 46gtrtT.er (1.60 x 10-te CX25.0 x t0-2 m)
(c) The frequency f is the number of times the electron goes around per unit time, so
f : u
- 2'05x107.m/s
-1.31 x107Hz:13.1MH2.2nr 2r(25.0 x t0-2 m)
(d) The period is the reciprocal of the frequency:
'r_ 1 - I _ 7.63x10-8s-76.3ns.r'
7:13ffi-29
(a) If u is the speed of the positrotr, then u stn S is the component of its velocity in the plane thatis peqpendicular to the magnetic field. Here 0 is the angle between the velocity and the field(89o). Newton's second law yields eBu sin d: m(u sin ilz f r, where r is the radius of the orbit.Thus r : (*, l"B) sin /. The period is given by
2rr 2nm 2r(9.11 x 10-" kg)
u stnf eB (1.60 x 10-te CX0.100 T) v 'v
The expression for r was substituted to obtain the second expression for T.
170 Chapter 28
2Km
(b) The pitch p is the distance traveled along the line of the magnetic field in a time interval ofone period, Thus p : uT cos /. Use the kinetic energy to find the speed: K - i*r' yields
- ^ - Z.65Ix L07m/s .
V
Thusp: (2.651 x 107 mlsx3.58 x 10-to r) cos 89.0o - I.66 x 10-4 m.
(c) The orbit radius is
ma smpT:
-.=--:
rnd (9.11 x 10-" kg)(2.651 x 107 mls)sin89.0o1.51 x 10-3m.
u-
eB (1.60 x 10-te CX0.l00 T)
4l(a) The magnitude of the magnetic force on the wire is given by Fn : 'iLB sin /, where i is the
current in the wire, L is the length of the wire, B is the magnitude of the magnetic field, and 0is the angle between the current and the field. In this case Q : 70o. Thus
Fn - (5000 A)(100 m)(60.0 x 10-u t) sin 70o - 28.2N .
(b) Apply the right-hand rule to the vector product FB - 'iE x E to show that the force is to the
wgst.
47
UtI
z,'The situation is shown in the left diagram above. The g axis isfield is in the positive r direction. A torque around the hinge isthe hinge and not with the other wires. The force on this wirehas magnitude I7 - IV ibB, where ,Af is the number of turns.
The right diagram shows the view from above. The magnitude
along the hinge and the magnetic
associated with the wire opposite
is in the positive z direction and
of the torque is given by
r - Facos 0 - IVibBa cos I: 20(0.10 AX0.10 mX0.50 x l0-' fXO.050 m) cos 30o
:4.3 x l0-3N.rn.
Chapter 28 l7l
IJse the right-hand rule to show that the torque is directed downward, in the negatrve A direction.Thus i--(4.3 x 10-3 11 .m)-i.
--I)(t) The magnitude of the magnetic dipole moment is given by p : IYI,A, where ^Af is the numberof turns, i is the current in each turn, and A is the area of a loop. In this case the loops are
circular, so A - nr2, where r is the radius of a turn. Thus
i- F =- 2'3oA'm2
- r. F.A
lhrr'z rL'tA'
(b) The maximum torque occurs when the dipole moment is pe{pendicular to the field (or theplane of the loop is parallel to the field). It is given by , - pB - (2.30A.m2X35.0 x 10-'T) -8.05 x l0-2 N.m.
s9
The magnitude of a magnetic dipole moment of a current loop is given by LL
the current in the loop and A is the area of the loop. Each of these loops is a circle and itsarea is given by A : T R2 , where R is the radius. Thus the dipole moment of the inner loop has
a magnitude of ptouter loop has a magnitude of po : 'irr2, - (7.00 Azr(0.300 m)z - I.979 A . m2.
(a) Both currents are clockwise in Fig. 28-51 so, according to the right-hand rule, both dipolemoments are directed into the page. The magnitude of the net dipole moment is the sum of the
magnitudes of the individual momentsi Fr"t: lti* Ito - 0.880A.m2+ 1 .9794.m2 -2.86A'rrf .
The net dipole moment is directed into the page.
(b) Now the dipole moment of the inner loop is directed out of the page. The moments are
in opposite directions, so the magnitude of the net moment is Fnet- Fo - ltr - 1.979 A . m2
0.880A .r#- 1.10A.m2. The net dipole moment is again into the page.
63
If .,Arr closed loops are formed from the wire of length L, the circumference of each loop is LIIV,the radius of each loop is R - L l2rltr, and the area of each loop is A : r Rz
L2 l4r 7gz. For maximum torgue, orient the plane of the loops parallel to the magnetic field, so
the dipole moment is pe{pendicular to the field. The magnitude of the torque is then
iLz B4rlY
To maximize the torque, take .,Af to have the smallest possible value, 1. Then
iL2 B (4.51 x l0-' exo .250 -)t (s .7 | x l0-' r) 1T:T
r: rviAl-(rtr?) (#) B -
65
(a) the magnetic potential energy is given
of the coil and B is the magnetic field.
172 Chapter 28
by U - -1,.8, where E is the magnetic dipole momentThe magnitude of the magnetic moment is lr: IV|A,
where i is the current in the coil, A is the area of the coil, and .A/ is the number of turns. The
moment is in the negative A direction, as you can tell by wrapping the fingers of your right hand
around the coil in the direction of the current. Your thumb is then in the negative A direction.Thus F: -(3.00X2.00AX4.00 x 10-'*t)i - -Q.40 x l0-2A.*2)i. The magnetic potential
energy is
' : _ly)},': .,'olf
it ::f:-l i?,r) : _7 2ox I o _, r,
where j . i : 0, j .i - 1, and j . k - 0 were used.
(b) The magnetic torque on the coil is
i-fi,x8-Qtrilx (B*i+Brj+ B,k;- FaB,i- ttaB*k
- (-2.40 x l0-2A. m2X-4.00 x 10-'r)i - (-2.40 x t0-2 A.m2)(2.00 x l0-'r)t- (9.6 x 10-5 N . m) i + 1+.80 x lo-s N . m)t,
where j x i - -k, j x j : 0, and j x t - i were used.
73
The net force on the electron is given by F - -e(E + 6 x 87, where E is the electric field, Eis the magnetic field, and 6 is the electron's velocity. Since the electron moves with constant
velocity you know that the net force must vanish. Thus
E-- -ix E - -(ri) x (Bk)- -uBj- -(100m/sxs.00r)j-(500vlm)j
75
(a) and (b) Suppose the particles are accelerated from rest through an electric potential differenceV. Since energy is conserved the kinetic energy of a particle is Kparticle's charge. The ratio of the proton's kinetic energy to the alpha particle's kinetic energy isKellf * - "l\e - 0.50. The ratio of the deuteron's kinetic energy to the alpha particle's kineticenergy is KalKr- el2e- 0.50.
(c) The magnitude of the magnetic force on a particle is quB and, according to Newton's second
law, this must equal mu2 f R, where u is its speed and R is the radius of its orbit. Sincert)-\m-\Nn,
R-
The ratio of the radius of the deuteron's path to the radius of the proton's path is
Ra:Re
Since the radius of the proton's path is 10cm, the radius of the deuteron's path is (1.4X10cm):14cm.
mu mtu{ mlW:l,
-
tlqB qB V m qB V mt2*tlc
Chapter 28 173
(d) The ratio of the radius of the alpha
Ro
R"
Since the radius of the proton's path is14cm.
77
Take the velocity of the particle to be 6 _ u*i * ,rjmagnetic force on the particle is then
the radius of the proton's path is
r.4 .
of the deuteron's path is (l.4Xl0 cm) -
and the magnetic field to be Bi. The
particle's path to
l4nlr E-Vr^or\1'--
10 cm, the radius
F - q6 x E - e(u*i+urj1 x
where q is the charge of the particle. We used i x i *
F 0.48N
(B i) - -QUaB k,
oandjxi--t. The charge is
10-2 c .Lr.: ^- rr:-uab -(4.0 x 103 mls)(sin 37"X5.0 x 10-r T)
81
(a) If K is the kinetic energy of the electron and m is its mass,
u-
then its speed is
- 6.49 x 107 m/s.
Since the electron is traveling along a line that is parallel to the horizontal component of Earth'smagnetic field, that component does not enter into the calculation of the magnetic force on the
electrorl. The magnitude of the force on the electron is eu B and since F - ma) where a, is the
magnitude of its acceleration , eu B - ma, and
& _ euB: (t.oo x to-te cX6.49 x 101T/sX55.0 x 10-6 T) :6.3 x l0r4 mlrr.m 9.11 x 10-3t kg
(b) If the electron does not get far from the r axis we may neglect the influence of the horizontalcomponent of Earth's field and assume the electron follows a circular path. Its acceleration is
given by o : 'u2 f R, where R is the radius of the path. Thus) /r At\ l-n7 t->,)
R - t_ (6.49 x ro7 mf sY - 6.72m.a 6.27 x 101 a ml s2
The solid curve on the dragram is the path. Suppose it subtends
the angle 0 at its center. d (- 0 .200 m) is the distance traveledalong the r axis and (. is the deflection. The right triangle
yields d : .R sin g, so sin 0 - d,l R and cos g
@. The triangle also gives (,- R Rcos9, so
t- R- R@. Substitute R- 6.i2m and d-0.2mto obtain L - 0.0030 m.
17 4 Chapter 28
2(12 x 103 eV)(l.60 x 10-to I leVl9.11 x 10-31
Chapter 29
1(a) The field due to the wire, at a point 8.0 cm from the wire, must be 3 9 pT and must be directedtoward due south. Since B - Foi,f Znr,
- 2rrB _ 2zr(0.080mX39 x 10-6T)
- 16A.lto 4n x 10-7 7 .ml A
(b) The cuffent must be from west to east to produce a field to the south at points above it.
7
(a) If the currents are parallel, the two magnetic fields arc in opposite directions in the regionbetween the wires. Since the currents are the same, the net field is zero along the line that runshalfway between the wires. There is no possible current for which the field does not vanish. Ifthere is to be a field on the bisecting line the currents must be in opposite directions. Then thefields are in the same direction in the region between the wires.
(b) At a point halfway between the wires, the fields have the same magnitude, Foi f 2rr. Thusthe net field at the midpoint has magnitude B - Foi, ln, and
. nrB 7r(0.040 m)(300 x 10-6 T)?,- : -
lto 4; x tO- f . ,r, lA = 30A
15
Sum the fields of the two straight wires and the circular arc. Look at the derivation of theexpression for the field of a long straight wire, leading to Eq. 29-6. Since the wires we arcconsidering are infinite in only one direction, the field of either of them is half the field of an
infinite wire. That is, the magnitude is Foi lar R, where R is the distance from the end of thewire to the center of the arc. It is the radius of the are. The fields of both wires are out of thepage at the center of the arc.
Now find an expression for the field of the ara at its center. Divide the arc into infinitesimalsegments. Each segment produces a field in the same direction. If ds is the length of a segment,
the magnitude of the field it produces at the arc center is (ltoi, lar R') dt If 0 is the anglesubtended by the arc in radians, then R0 is the length of the arc and the net field of the arc is
ttoi? lan R. For the ara of the diagram, the field is into the page. The net field at the center, dueto the wires and arc together, is
B-H+ l\ t:rcT: ry(z- r7).4nR 4nR 4rR 4rR \-
Chapter 29 175
For this to vanish, 0 must be exactly 2 radians.
19
Each wire produces a field with magnitude given by B : pyi f 2nr, where r is the distance fromthe corner of the square to the center. According to the Pythagorean theorem, the diagonal ofthe square has length .,,0.o, so r: "ltn and B - Foi,f tD,na. The fields due to the wires at the
upper left and lower right corners both point toward the upper right corner of the square. Thefields due to the wires at the upper right and lower left corners both point toward the upper leftcorner. The horizontal components cancel and the vertical components sum to
cos 45o- 2Poi
7I CL
Bnet - 4 ry'-ttvL 'lDno
- 2(47r x l0-7 T . ml A)(20 A)
- g.0 x 10_5 T .
zr(0.20 m)
2tFollow the same steps as in the solution of Problem 17 above but change the lower limit ofintegration to -L, and the upper limit to 0. The magnitude of the net field is
In the calculation cos 45" was replaced with Lltn In unit vector notation E - (8.0 x 10-t t)j.
B_rtoi,R fo = 4*==,=rr' 4n l-t(*2+R2)3/zI r lo ttoi,
HWI-, 4*R
(42- x 10-7 T .m/AX4.61 x 10-6 A) ,n
0.0216m + 0.0491 m
2r(0.0491 m) 0.0216 m
2.23 x 10-11 T.
FoiR4n JT'TW
4r x 10-t 7 .ml AX0.693 A) 0.136m1.32 x 10-7 T.
4r(0.251 m) (0. 136 m)2 a (0.251 m)2
31
The current per unit width of the strip is i l* and the current through a width d"r is (i, lu\ dr.Treat this as a long straight wire. The magnitude of the field it produces at a point that is adistance d from the edge of the strip is dB : (pa 12")(i I @ dr f r and the net field is
B - ltoi [o*- d,r
2nwJa r 2nw d
35
The magnitude of the force of wire 1 on wire 2 is given
by Foili2f 2rr, where h is the current in wire I, 'i2 is the
current in wire 2, and r is the separation of the wires.
The distance between the wires is r: lffir. Since the
cuffents are in opposite directions the wires repel each otherso the force on wire 2 is along the line that j oins the wiresand is away from wire 1.
17 6 Chapter 29
L
To find the r component
the angle 0 that the force
component.of the force is
of the force, multiply the magnitude of
makes with the r axis. This is cos 0 -the
dzl
force by
tw,the cosine of
Thus the r
n lloiti,zJ'n A
/,rr
(2n x
dz
d?+ d7
l0-7 7 .ml AX4.00 x 10-'axe .80 x l0-'R) 0.0500 m
2n (0.024m)z + (5.00 m)2
: 8.84 x 10-ll T .
43
(a) Two of the currents are out of the page and one is into the page, so the net current enclosed
by the path is 2.0 A, out of the page. Since the path is traversed in the clockwise sense, a cuffentinto the page is positive and a cuffent out of the page is negative, as indicated by the right-hand
rule associated with Ampere's law. Thus 'i"n": -i and
f --+
6 E.d,i:- Fa,i:-(4n x 10-7T.ml A)(2.0A): -2.5 x 10-6T.rn.J
(b) The net current enclosed by the path is zero (two currents are out of the page and two are
into the page), so f E . d,g: Fo'irn, : 0.
s3
(a) Assume that the point is inside the solenoid. The field of the solenoid at the point is parallelto the solenoid axis and the field of the wire is perpendicular to the solenoid axis. The net fieldmakes an angle of 45" with the axis if these two fields have equal magnitudes.
The magnitude of the magnetic field produced by a solenoid at a point inside is given byBsor - Foisotn, where n is the number of turns per unit length and isol is the current in the
solenoid. The magnitude of the magnetic field produced by a long straight wire at a point a
distance r away is given by B*ire : l.L1iwire f 2nr, where 'i*ir" is the current in the wire. We want
Fonisol : Foiwir"f 2nr. The solution for r is
i*ir" 6.00 A4.77 x l0-2 m: 4.77 cm.
This distance is less than the radius of the solenoid, so the point is indeed inside as we assumed.
(b) The magnitude of the either field at the point is
Bror: Bwire: Fonirol: (n x 10-7 T .ml AX10.0 x I02 m-1;120.0 x 10-',,t) -2.51 x 10-5T.
Each of the two fields is a vector component of the net field, so the magnitude of the net fieldis the square root of the sum of the squares of the individual fields: B - JZ(2.51 x 10-s T)2 -3.55 x 10-5 T.
Chapter 29 177
57.
The magnitude of the dipole moment is given by p- IVdA, where t'r is the number of turns, iis the current, and A is the area. Use A - n R2, where R is the radius. Thus
l.L: ItinR2 - (200X0.30A)zr(0.050 m)2 - 0.47 A. m2 .
59
(a) The magnitude of the dipole moment is given by p : IY'|A, where ,Af is the number of turns,i is the current, and A is the area. Use A - r R2, where R is the radius. Thus
p : IYirR2 - (300X4.0 A)n(0.025 m)2 - 2.4 A. nf
(b) The magnetic field on the axis of a magnetic dipole, a distance z awa,y, is given by Eq . 29 -27:
B_lto P_
2r 23
Solve for zi
ry:llto lt1r/t- tP LznBl L
4r x 10-77.m/ A 2
21 5.0 x 10-6 1.36,4.. fif l/t
I : 46cm.
7lUse the Biot-Savart law in the form
- lto iLi x r'4r 13
Take Adto be Arj,and r-tobe ri*yj+zk. Then As*xi- Asjx @i+yj*tt<; - Ls(zi-rt<;,wherejXi:-t,ixj--0,andjXk:iwereuSed.Inad.ditiofl,r:ffi.TheBiot-Savart equation becomes
B
tto iLs(zi- zk)4n (*2 * y2 1 ,213 /2
E _ 4r x r0-7 T .mf A (2.0AX0.030mX5.0m)i _ (2.4x 10-ro T) i.4n (5.0m)3
(b) For fr:0, A:6.0m, and z:0, E:0.(c) For r - 7.0m, A :7.0m, and z :0,
B-
(a) For fr:0, A:0, and z - 5.0m,
E _ 4r x l0-7 T .mf A (2.0AX0.030mX-7.0m)k : (4.3 x 10_,, T)t.4r lQ.0 m)2 + (7.0 m)2l3 /z \
Chapter 29178
(d) For r - -3.0m, A: -4.0m, and z:0,
4r x 10-t7-ml A (2.0AX0.030m)(3.0m)k4rr t(-3.0 m)2 + ( -4.0 m)213/z
E (1 .4 x 1o-to T) t .
77
First consider the finite wire segment shown on the right.It extends from A - -d to A : a, d, where a is the
length of the segment, and it carries current i in the
positive A direction. Let dy be an infinitesimal lengthof wire at coordinate A. According to the Biot-Savartlaw the magnitude of the magnetic field at P due to thisinfinitesimal length is dB : (po laz')(e sin 0 lr') da. Nowr2 - a2 + R2 and sin 0 - Rl, - Rf \ffi, so
cJB:lto-iR4n (A2 + dA
and the field of the entire segment is
aI
I
,1
ds
r
ffida-pola"h
3l_l,ml
a-dR2+(a-d,)2
tto 4i ,--
= - (1.66).57r a
To calcul ate the field of the right side of
po 4i- -
_ (0.341).3r a,
Bbrt:
The field of the upper side of the square is
the square put d - a,l4 and R - 3a la. The
B-#iRI::'
where integral 19 of Appendix E was used.
All four sides of the square produce magnetic fields that are into the page at P, so we sum theirmagnitudes. To calculate the field of the left side of the square put d - 3a la and R - a,14. Theresult is
tro 4i l- 1rIrT r4r a l{Z
the same.
result is
Bright :
The field of the bottom side is the same. The total field at P is
tto4i t3r-
4n 3a l.fr1l
IIl:l
Jro]r
B - Brcn* Bupp., * Brieht+ Bro*er : #! fI .66 + I.66+ 0 .341+ 0 .341)
4r x 10-7 7 .ml A
d1ffi)
4n4(10 A)ffi(4.00):2.0 x 1o-4T.
Chapter 29 179
79
(a) Suppose the field is not parallel to the sheet, as shownin the upper dragram. Reverse the direction of the current.According to the Biot-Savart law, the field reverses, so itwill be as in the second diagram. Now rotate the sheet
by 180" about a line that is perpendicular to the sheet. Thefield, of course, will rotate with it and end up in the directionshown in the third diagram. The cuffent distribution is nowexactly as it was originally, so the field must also be as itwas originally. But it is not. Only if the field is parallelto the sheet will be final direction of the field be the same
as the original direction. If the current is out of the page,
any infinitesimal portion of the sheet in the form of a longstraight wire produces a field that is to the left above the
sheet and to the right below the sheet. The field must be as
drawn in Fig. 29-85.(b) Integrate the tangenttal component of the magnetic fieldaround the rectan galar loop shown with dotted lines. Theupper and lower edges are the same distance from the currentsheet and each has length L. This means the field has the
same magnitude along these edges. It points to the left alongthe upper edge and to the right along the lower.
If the integration is carried out in the counterclockwise sense, the contribution of the upper edge
is B L, the contribution of the lower edge is also B L, and the contribution of each of the sides
is zero because the field is pe{pendicular to the sides. Thus f A . dg:28L. The total currentthrough the loop is
^L Ampere's law yields 2BL: p1^L, so B - po\12.
81
(a) IJse a circular Amperian path that has radius r and is concentric with the cylindrical shellas shown by the dotted circle on Fig. 29-86. The magnetic field is tangent to the path and
has uniform magnitude on tt, so the integral on the left side of the Ampere's law equation is
f E . d,d : 2rr B. The current through the Amperian path is the current through the region
outside the circle of radius b and inside the circle of radius r. Since the current is uniformlydistributed through a cross section of the sheltr, the enclosed current is i(r' - b') l(o' - b'). Thus
I
\.]I
I
I
\
I
I
t/,I
rz 6z2nrB: , -riCL' - OL
ltai, r2 - b2
and
B-
(b) When rfield of a long straight wire. When r :
180 Chapter 29
tt
L :=-E- l+_L
correct expression for the
is correct since there is no
2n(a2 - b2) rto B- Foi,f 2nr, which is the
b it reduces to B : 0, which
field inside the shell. When b -inside a cylindrical conductor.
(c) The graph is shown below.
B (T)
10 x 10-4
8 x 10-4
6 x 10-4
4 x 10-4
2 x 10-4
0.01 0.02 0.03 0.04r (m)
89
The result of Problem I 1 is used four times, once for each ofthe sides of the square loop. A point on the axis of the loopis also on a pe{pendicular bisector of each of the loop sides.
The diagram shows the field due to one of the loop sides,
the one on the left. In the expression found in Problem 11,
replace L with a, and R with @- +m.The field due to the side is therefore
0 it reduces to B- poirl2naz, which is correct for the field
0.05 0.06
P
,'D\=*E
rR//
//
B_ Foza
The field is in the plane of the dotted triangle shown and \ i
is perpendicular to the line from the midpoint of the loopside to the point P. Therefore it makes the angle 0 with thevertical.
When the fields of the four sides are summed vectorially the horizontal components add to zero.The vertical components are all the same, so the total field is given by
Bn ur: 48 cos I _ 4Ba: 4Ba2Rffi
4 P'g'ia2
Thus
Brorut-
For fr : 0, the expression reduces tor(4r2 + a2)
4 p,s,ia2 ZtD poi
4rz * a2 + 2a2
+ 2a2
Brorut-na2tQo 7fa
Chapter 29 181
in agreement with the result of Problem 12.
9l
[Jse Ampere's law: f E.d,g: pyi"n, where the integral is around aclosed loop and i"n" is thenet current through the loop. For the dashed loop shown on the diagram iintegral t U' dd is zero along the bottom, right, and top sides of the loop as it would be if thefield lines are as shown on the diagram. Along the right side the field is zero and along thetop and bottom sides the field is pe{pendicular to dg. If (, is the length of the left edge, thendirect integration yields f E .d,i: Bl, where B is the magnitude of the field at the left sideof the loop. Since neither B nor (. is zero, Ampere's law is contradicted. We conclude that thegeometry shown for the magnetic field lines is in error. The lines actually bulge outward andtheir density decreases gradu ally, not precipitously as shown.
182 Chapter 29
Chapter 30
-5ifr. magnitude of the magnetic field inside the solenoid is B- Fon'ir, where nis the numberof turns per unit length and 'i, is the current. The field is parallel to the solenoid axis, so the
flux through a cross section of the solenoid is Q6 : ArB: pgrr2"n'ir, where A, (: nr?) is the
cross-sectional area of the solenoid. Since the magnetic field is zero outside the solenoid, this isalso the flux through the coil. The emf in the coil has magnitude
IYdQsc-L-.
dt
- po\Tr?lvn d'i,
R dtlturns in the coil and R is the resistance of the coil. The current50ms, so d,i,ldt - (3.0A)l(50 x 10-3 r):60A/r. Thus
0C
(4n' x t0-7 t . m/n)zr(0.016 m)2(120X220 x 102 m-t)(60 A^) : 3.0 x
5.3 0r0-2 A.
2t(a) In the region of the smaller loop, the magnetic field produced by the larger loop may be taken
to be uniform and equal to its value at the center of the smaller loop, on the axis. Eq. 29-26,with z - r and much greater than R, gives
B- Foi,R2
213
: ponrlxn#
and the current in the coil is
where lf is the number ofchanges linearly by 3.0A in
for the magnitude. The field is upward in the diagram.loop is the product of this field and the area (nr') of the
Os- n p,sir2 R2
The magnetic flux through the smallersmaller loop:
3T p,si,rz Rzu
2ra
.t,,0cR
213
(b) The emf is given by Faraday's law:
t:-#: (ry)#,(*)(c) The field of the larger loop is upward and decreases with distance away from the loop. Asthe smaller loop moves away, the flux through it decreases. The induced current is directed so
as to produce a magnetic field that is upward through the smaller loop, in the same direction as
Chapter 30 183
the field of the larger loop. It is counterclockwise as viewed from above, in the same directionas the current in the larger loop.
29
Thermal energy is generated at the rute €' I R, where t is the emf in the wire and R is theresistance of the wire. The resistance is given by E - pL I A, where p is the resistivity of copper,L is the length of the wire, and A is the cross-sectional area of the wire. The resistivity can be
found in Table 26-1. Thus
R - pr: (1.69 x 10-8 o 'px0.500m) _ T.076x r0-2 o.' A n-(0.500 x 10-t ^)'
Faraday's liw is used to find the emf. If B is the magnitude of the magnetic field through the
loop, then EE AdBldt, where A is the area of the loop. The radius r of the loop is r - LlLnand its area is rr2 : n L2 l4n' : L2 lan Thus
e : 1z dB (0.500q)'(10.0 x 10- rr ls): l.ggg x 10-4v.(' c dt 4?l
The rate of thermal energy generation is
D _ gz (1.989 x lo-4 v)2 A -r-E -3.68x10-6'W.
37
(a) The field point is inside the solenoid, so Eq. 30-25 applies. The magnitude of the inducedelectric field is
E_ I d,B I
; E r : ){e.s
x 10-t r lsx0 .0220m) : 7 .15 x 10-t v l*.(b) Now the field point is outside the solenoid and Eq. 30-27 applies. The magnitude of the
induced field is
E -i dF Rz:1(u., x 10-trlrr(0'o6oomf - r.43x t0-4 yl^.2 dt r 2' - ' -" (0.0820 m)
51
Starting with zero current when the switch is closed, &t time t - 0, the current in an RL series
circuit at a later time t is given by
where T7 is the inductive time constant, t is the emf, and R is the resistance. You want tocalcul ate the time t for which 'i - 0.999At f R. This means
,i - 1(r - e-t/'L) ,
o.sssr*: 1(r - e-t/"I) ,
184 Chapter 30
SO
0.ggg0-1-_e-t/'r-
or
rake the narural logarithm or both ,io.l r'::*: T: r;: ln(0.0010) - -6.e1. rhat is, 6.srinductive time constants must elapse.
55
(a) If the battery is switched into the circuit at time t * 0, then the current at a later time t is
given by
'i-L(, -e-t/'n\ qE\'-- / )
where r7 - L I R. You want to find the time for which 'i - 0.800 t I R. This means
0.800 : 1 - e-tl'r
ore-tl"r-o.2ao.
Take the natural logarithm of both sides to obtain -(t lrr) - ln(0.200) : - I .609. Thus
t - l.609rr- 1"6019L- 1'609(6'30 x-10:6H) - 8 .45x l0-e s.R 1.20 x 103 O
(b) At t - 1.012 the current in the circuit is
'i-*(r -e-'o) : (#) (r - e-'o) -7.37x r0-3A.
59
(a) Assume i is from left to right through the closed switch. Let 'fu be the current in the resistor
and take it to be downward. Let i,z be the cuffent in the inductor and also take it to be downward.The junction rule gives i - fi+i2 and the loop rule gives hR- L(dizldt):0. Since dildt:0,the junction rule yields (di,t I dt) - -(diz I dt). Substitute into the loop equation to obtain
L+ + irR:0.dt
This equation is similar to Eq. 30-44, and its solution is the function given as Eq. 30-45:)
i,t : 'ioe-Rt/L ,
where 'io is the current through the resistor at t - 0, just after the switch is closed. Now, justafter the switch is closed, the inductor prevents the rapid build-up of current in its branch, so at
that time, i,z - 0 and it - 'i. Thus 'io : 'i, so
'i1 : 'ie- Rt / L
Chapter 30 185
and
(b) When i2 = ii,
so
. . . . [1 -Rt/L] 1,2 = 1, - 1,1 = 1, - e
e- Rt / L = 1 _ e- Rt / L
e-Rt / L = ~ 2'
,
Take the natural logarithm of both sides and use In(1/2) = -In 2 to obtain (Rt / L) = In 2 or
L t = R In2.
63
(a) If the battery is applied at time t = 0, the current is given by
i = ~ (1 - e-t / TL ) ,
where £ is the emf of the battery, R is the resistance, and TL is the inductive time constant. In terms of R and the inductance L, TL = L / R. Solve the current equation for the time constant. First obtain iR
e- t / TL = 1 - £ '
then take the natural logarithm of both sides to obtain
t [ iR] -7£ =In 1-£
Since
In [1 _ iR] = In [1 _ (2.00 x 10-3 A)(10.0 X 103 Sl)] = -0.5108 £ 50.0V '
the inductive time constant is TL = t/0.5108 = (5.00 x 10-3 s)/(0.5108) = 9.79 x 10-3 s and the inductance is
L = TLR = (9.79 x 10-3 s)(10.0 X 103 0) = 97.9 H.
(b) The energy stored in the coil is
U B = ~ Li2 = ~(97.9 H)(2.00 x 10-3 Ai = 1.96 x 10-4 J. 2 2
69
(a) At any point, the magnetic energy density is given by UB = B2/2f-LO, where B is the magnitude of the magnetic field at that point. Inside a solenoid, B = f-Loni, where n is the number of turns
186 Chapter 30
73
(a)
per unit length and i is the current. For the solenoid of this problem, n- (950)l(0.850m) -1.118 x 103 m-1. The magnetic energy density is
us: )uon'r': )fo" x 10-7 T .ml AXl.1t8 x 103m \'(0.60 A)'_ 34.zJl^t.
(b) Since the magnetic field is uniform inside an ideal solenoid, the total energy stored in the
field is LI a - u nV , where V is the volume of the solenoid. V is calculated as the product ofthe cross-sectional area and the length. Thus
(-In - (34.2Jfm'Xt7.0 x 10-4m2X0.850m) - 4.94 x 10-2J.
The mutual inductance M is given by
tt- M#,where & is the emf in coil 1 due to the changing current 'iz in coil 2. Thus
M_ffi:ffi(b) The flux linkage in coil 2 is
Iy2pzr: M,il : (1 .67 x 10-'UX:.60A) : 6.01 x 10-3 Wb.
75
(a) Assume the cuffent is changittg at the rate di I dt and calcul ate the total emf across both coils.First consider the left-hand coil. The magnetic field due to the current in that coil points to the
left. So does the magnetic field due to the current in coil 2. When the current increases, bothfields increase and both changes in flux contribute emf s in the same direction. Thus the emf incoil 1 is
tr: -Qr + I\D +af,
The magnetic field in coil 2 due to the current in that coil points to the left, as does the field incoil 2 due to the current in coil 1. The two sources of emf are agarn in the same direction and
the emf in coil 2 is
tz: - (Lz+ M) + .
af,
The total emf across both coils is
t:tr+tz: -Q1 * Lz+2M\di.vr / dt '
This is exactly the emf that would be produced if the coils were replaced by a single coil withinductance Leq - Ll + Lz * 2M .
Chapter 30 187
(b) Reverse the leads of coil 2 so the current enters at the back of the coil rather than the frontas pictured in the diagram. Then the field produced by coil 2 at the site of coil 1 is opposite the
field produced by coil 1 itself. The fluxes have opposite signs. An increasing current in coil 1
tends to increase the flux in that coil but an increasing current in coil 2 tends to decrease it. The
emf across coil 1 is
tr0,f,
Similarly the emf across coil 2 is
tz: - (Lz M)
The total emf across both coils is
t--(Lr*Lz-2M)
This the same as the emf that would be produced by a single coil with inductance Leq
LraLz-2M.
79
(a) The electric field lines are circles that are concentric with the cylindrical region and the
magnitude of the field is uniform around any circle. Thus the emf around a circle of radius 'r
is€According to Faraday's law 2nrE - -rr2(d,B ldt) and
E _ -1 *dB - -+(0.050mX-10 x 10-3 T/s) :2.5 x 10-4 y l^.2' dt
Since the normal used to compute the flux was taken to be into the paBe, in the direction of the
magnetic field, the positive direction for the electric is clockwise. The calculated value of E is
positive, so the electric field at point ais toward the left and E -- -(2.5 x 10-4V/m)i.
The force on the electron is 7i - -"8 and, according to Newton's second law, its acceleration is
A'I eD \I.UU2(IU \-,/\-L.J 2(I\/ V/LIL)L /^A tn7-^_t^2.,?i- _ -_ _(4.4xl0,mls')iF eE (1.60 x 1o-tn cX -2.5 x 1o-4 v/m)i
9.Il x 10-:t kg
didt
didt
mmThe mass and charge of an
(b) The electric field at r -b arc zeto.
(c) The electric field at point c has the same magnitude as the field at pqinl a but now the fieldis to the right. That is E - (2.5 x 10-4V/m)i and d- -(4.4 x 107 mlrt)i.
81
(a) The magnetic flux through the loop is Os : B A, where B is the magnitude of the magnetic
field and A is the area of the loop. The magnitude of the average emf is given by Faraday's law; tave - B LAI Lt, where LA is the change in the area in time Lt Since the final area is zero,
the change in area is the initial area and tave: BAILt - (2.0TXO.20m)2 lQ.20s):0.40V.
188 Chapter 30
electron can be found in Appendix B.
0 is zero, so the force and acceleration of an election placed at point
(b) The average current in the loop is the emf divided by the resistance of the loop f iav' -tuunl l? - (0.40 v) lQ0 x 10-' o) : 20 A.
85
(a), (b), (c), (d), and (e) Just after the switch is closed the cuffent iz through the inductor is zero.
The loop rule applied to the left loop gives t - IrRr:0, so er : tlRt - (10V) 16.0Q) -2.0A.The junction rule gives 'i, -- 'i1 - 2.0 A. Since i,z - 0, the potential difference across Rz isVz : 'izRz - 0. The potential differences across the inductor and resistor must sum to t and, since
Vz: 0, Vt: t :10V. The rate of change of i,2 is dizldt - VrlL - (10V)l(5.0H) : 2.0A/*.(g), (h), (i), 0), (k), and (1) After the switch has been closed for a long time the current i2
reaches a constant value. Since its derivative is zero the potential difference across the inductoris Vr : 0. The potenttal differences across both Rr and Rz are equal to the emf of the batteryso z1 - tlRt: (10Y)l(5.0O) - 2.0A and i2: tlRr: (10V)l(10O) - 1.0A. The junctionrule gives 'i, - 'it * 'iz - 3.0 A.
9s
(a) Because the inductor is in series with the battery the current in the circuit builds slowly andjust after the switch is closed it is zero.
(b) Since all currents are zerojust after the switch is closed the emf of the inductor must match the
emf of the battery in magnitude. Thus L(diburldt) - t and diaur- tlL - (40V)|60 x 10-'H) -8.0 x 102 A/r.(c) Replace the two resistors in parallel with their equivalent resistor. The equivalent resistance
is
R"n- RrRz -
(2ok[)X2okq) - loko.1 Rt.1 Rz 20ko + 2oko
The current as a function of time is given by
where 4 is the inductive time constant. Its value rs r7 - LfReq - (50 x 10-3H)l(10 x 103 O) -5.0 x 10-6 s. At t: 3.0 x 10-u r, tlr" - (3 .0)l(5.0) - 0.60 and
ibut: 40 v-10 x 103 o [t - e-o'60] - l'8 x 1o-3 A'
(d) Differentiate the expression foliaut to obtain
'ibut[r - e-'/'"f ,
e-tlrr: !"-t/rrqL")
where 17 - LfReq was used to obtain the last form. At t - 3.0 x 10-6 s
diaut 40 V -ldt ffie-o'60 - 4'4 x I02A/t'
(e) A long time after the switch is closed the currents are constant and the emf of the inductoris zero. The current in the battery is 'iaut - t lRrq- (40V)/(10 x 103 O) : 4.0 x 10-3 A.
Chapter 30 189
d'iaut
dt
(0 The currents are constant and diaurld,t: 0.
97
(a) and (b) Take clockwise current to be positive and counterclockwise current to be negative.Then according to the right-hand rule we must take the nonnal to the loop to be into the page,
so the flux is negative if the magnetic field is out of the page and positive if it is into the page.
Assume the field in region 1 is out of the page. We will obtain a negative result for the fieldif the assumption is incorrect. Let r be the distance that the front edge of the loop is intoregion 1. Then while the loop is entering this region flux is -BtHr and, according to Faraday'slaw, the emf induced around the loop is t- BrH(drldt): B1Hu. The current in the loop is'i-tlR-BrHufR,so
Bt: g- (3 0 x 10-6AX0'020o) - r.0 x 10-5T.Hu (0.0150mXO.40m/s)
The field is positive and therefore out of the page.
(c) and (d) Assume that the field 82 of region 2 is out of the page. Let r now be the
distance the front end of the loop is into region 2 as the loop enters that region. The flux is
-B1H(D r) B2Hr, the emf is t'i - (Bz - Br)Hu f R. The field of region 2 is
Bz: Br. #- 1.0 x 10-t T +(-2.0 x 10-u e(0.020 Q) :3.3 x 10-6T.
The field is positive, indicating that it is out
(0.015 m)(0.40 m/s)
of the page.
190 Chapter 30
Chapter 3L
7
(a) The mass nl, cotresponds to the inductance, so m : I.25 kg.
(b) The spring constant k coffesponds to the reciprocal of the capacitance. Since the totalenergy is given by (Ji' - Q'lzC, where A is the maximum charge on the capacitor and C is the
capacitance,
r-,-Q' (tlsxlo-uc) t A,u- W -2.69x10-3F
and
k- _
(c) The maximum displacement nrn coffesponds to the
372N/-.
maximum charge, so
r,n: 1, .7 5 x 10-4 m .
(d) The maximum speed 'u?,,.1 corresponds to the maximum cuffent. The maximum current is
J-ew-&- -3.ozxlo-3A.\rc v'v-
Thus utn-3.02 x 10-'*/r.
Since the frequency of oscillation f is related to the inductance L and capacitance C byI l2n\/ LC, the smaller value of C gives the larger value of f. Hence, f^u*- IlZn\/TC* o and
15
(a)
f:IJmm
-cJ max
f.:Jmm
(b) You want to choose the additional
,m:JC^in
Since the additional capacitor isof the tuning capacitor. If C is
r:0.54MHz
in parallel with the tuning capacitor,
in picofarads, then
_ Il2n\m,
capacitance C so the ratio of the frequencies is
1.60MHz- 2.96 .
its capacitance adds to that
re - 2.96 .
Chapter 3 t l9l
The solution for e is
C- (36s pF) - (2.eq2(10 pF)- 36pF.(2.96)2 1
(c) Solve ff : 0.54 MHz. Thus
11 1L- @Ctr :2.2 x 10-4H.
27
Let t be a time at which the eapaeitor is fully charged in some cycle and let q*ax I be the charge
on the eapacitor then. The energy in the eapaeitor at that time is
u(t) - @: A'- e-RttL2C 2e" )
where
llmaxl : Q e-Rt/zL
was used. Here A is the charge at t : 0. One cycle lateg the maximum charge is
Qmaxl - Q e-R(t+r)l2L
and the energy is
u(t+T) -@- q e-R(t+TlL2C 2C" )
where T is the period of oscillation. The fractional loss in energy is
LLr U(t) - U(t + T) e-RtlL - e-R(t+r)/L
tf U@ e-HwAssume that Rf I L is small compared to 1 (the resistance is small) and use the Maclaurin series
to expand the exponential. The first two terms are:
,-RT/LE1 y.L
Replace T with 2n f u, where u is the angular frequency of oscillation. Thus
LLr =1
/ 87\ RT 2rR,u ^ [t L)- L- oL"
33
(a) The generator emf is amaximumwhen sin(oat--Tl4) - I or a6t-nl4 ^ (nl2)*2nn, where
n is an integer, including zero. The first time this occurs after f : 0 is wherT a4t - n14 - T12 or
t-3r: 3T . - 6.73x10-3s.4ua 4(350 s- t )
(b) The cuffent is amaximum when sin(a,'at-3nf 4) - I, or a6t - 3nl4 - n12*2nn. The firsttime this occurs after t : 0 is when
t- 5n -
tn .4ua 4(350 s- 1)
(c) The current lags the inductor by n 12 rad, so the circuit element must be an inductor.
192 Chapter 3l
39
(a)
(d) The current amplitude I is related to the voltage amplitude Vr by Vn
the inductive rcactance, given by X p - uaL. Furthermore, since there is only one element inthe circuit, the amplitude of the potential difference across the element must be the same as the
amplitude of the generator emf: Vr : trn Thus trn - IwyL and
r_ trn 30.0 vL_+- _0.138H.Iua rc20 x 10-3 AX350 rad ls)
The capacitive reactance is
Xs aaC 2r f 6C 2n(60.0 Hz)(70.0 x 10-o F) - 37 '9 {l '
The inductive reactance is
Xr: u)d,L - Lnf aL:2r(60.0H2)(230 x 10-'H) - 86.7 Q.
The impedance is
Z- R2+(Xr-Xd2 (200 o)2 + (37.9 o - 96.7 Q)2 :206Q .
(b) The phase angle is
Q_tan-' (ry)_tan-'((c) The current amplitude is
cT L'yy1,,-I
Z
36.0 V206Q
(d) The voltage amplitudes are
vp: IR - (0.t75AX200O) - 35.0V,
v7: IXr - (0.i75 AX86.7 O) - 15.2V,
and
Vc : I Xc - (0 .17 5AX3 7 .g O) : 6,63V .
Note that Xy >diagram is drawn to scale on the right.
45
(a) For a given amplitude trn of the generator
1,3.7"200 r)
-0.r75A.
Vr
emf, the current amplitude is given by
cc'yyy
):86.7Q-37.9Q
CT Qyy1,t-I
Z
Chapter 31 193
where R is the resistance, L is the inductance, C is the capacitance, and Wd is the angular frequency. To find the maximum, set the derivative with respect to Wd equal to zero and solve for Wd. The derivative is
dI = -Em [R2 + (Wd L - l/wdciJ -3/2 [Wd L - _1_] [L + _1_]. dWd WdC w~C
The only factor that can equal zero is WdL - (l/WdC) and it does for Wd = l/VLC. For the given circuit,
1 1 Wd = V LC = J (1.00 H)(20.0 X 10-6 F) = 224 rad/ s .
(b) For this value of the angular frequency, the impedance is Z = R and the current amplitude is
1= Em = 30.0V = 6.00A. R s.oon
(c) and (d) You want to find the values of Wd for which 1= Em /2R. This means
Em _ Em JR2 + (WdL - 1/wdC )2 2R
Cancel the factors Em that appear on both sides, square both sides, and set the reciprocals of the two sides equal to each other to obtain
Thus
R2 + (Wd L __ 1_)2 = 4R2. WdC
( Wd L __ 1_) 2 = 3R2 •
WdC
Now take the square root of both sides and multiply by wdC to obtain
w~(LC)±Wd(v'3CR) -1=0,
where the symbol ± indicates the two possible signs for the square root. The last equation is a quadratic equation for Wd. Its solutions are
±V3CR ± V3C2 R2 + 4LC Wd = 2LC
You want the two positive solutions. The smaller of these is
-V3CR+ V3C2R2 +4LC W2 =
2LC -V3(20.0 x 10-6 F)(S.OO n)
2(1.00 H)(20.0 x 10-6 F)
J3(20.0 X 10-6 F)2(S.00 n)2 + 4(1.00 H)(20.0 x 10-6 F) +~~--------------~------~-----------
2(1.00 H)(20.0 x 10-6 F)
= 219 rad/s
194 Chapter 31
+\/1(zo.o x lo-u pxs.oo o)2(1.00 HX20.0 x 10-6 F)
3(20.0 x 10- e p)2(5.00 Q)2 + 4(1 .00 Hx20.0 x 10-6 F)
2(1.00 HX20.0 x 10-o F)
49
[Jse the expressions found in Problem 31- 45:
+t6cR+A1
2LC
and
-fjcR+2LC
ThusLra
- et - Loz: ZtECntR:
R2LC
The impedance is given by
z-{Rr+(xr-xs)z,where R is the resistance, X 1 is the inductive reactance, and Xs is the capacitive reactance.Thus
z - {(12.0o)2+(1.30o -0)2 - l2.lQ.
conditioner is given byrate at which energy is supplied to the aff
and the larger is
a1+t/icft +
2LC
: 228 rad/s .
(e) The fractional width is
at-azag -
228radls - 2lgradls _ 0 .04 .
224 radf s
t,J2
Also use
tFLC'
55
(a)
s2Puus: t cos @,
3C
T
(b) The average
Chapter 31 195
where cos / is the power factor. Now
cos Q:2: ffi:o.ssz,SO
puun: [(t'O,?tl (0.gg2):1.1g x 103'w'.g Ln.to j\v'l
57
(a) The power factor is cos /, where 0 is the phase angle when the current is written i/sin(aat - il. Thus Q: -42.0" and cos d: cos(-42.0o) - 0.743.
(b) Since 0 <(c) The phase angle is given by tan$- (Xr XdlR, where Xr is the inductive reactance,
Xs is the capacitive reactance, and R is the resistance. Now tan Q : tan( -42.0o)negative number. This means X 7 Xs is negative, or Xs >predominantly capacitive.
(d) If the circuit is in resonance, X7 is the same as Xs, tanS is zero, and 0 would be zero.
Since d is not zero, we conclude the circuit is not in resonance.
(e), (0, and (g) Since tan $ is negative and finite, neither the capacitive reactance nor the
resistance is zero. This means the box must contain a capacitor and a resistor. The inductivercactance may be zero, so there need not be an inductor. If there is an inductor, its reactance
must be less than that of the capacitor at the operating frequency.
(h) The average power is
T) 1 c Tnnna- 1
Purs: tE^-I
cos d : tQ 5.0 VXI.20 AXO .7 43) - 33.4 W .
(i) The answers above depend on the frequency only through the phase angle O:,which is given.
If values are given for R, L, and C , then the value of the frequency would also be needed tocompute the power factor.
63
(a) If I're is the number of primary turns and lf" is the number of secondary turns, then
(b) and (c) The current in the secondary is given by Ohm's law:
I, R, 15O--o'164'
The current in the primary is
v,: ftu,: (#) lzov)- z4v
I,p- ffi I - (#) (0.16A) :3.2x 1o-3 A.
196 Chapter 3I
67
use the trigonometric identity, found in AppendiX E,
sin a-sin p--2sin (+)cos w),where a and B are any two angles. Thus
W -Vz- Asin(c..'at)-Asin(aat- I20"): zAsin(I20")cos(aat- 60o): ,trlcos(c,;6t-_60o),
where sin( 120") : ,n p was used. Similarly,
V-Vt - Asin(c,'at)-A sin(o,at_240o) : 2Asin(240') cos(a.'at-120") : -ttrlcos(cu 6t-120o) ,
where sin(240o) : -\E 12 was used, and
Vz Vz - Asin(a,'at - I20") - Asin(aat - 240o): zAsin(I20")cos(r,,'at - 180')
- \frA cos (wat - l Soo) .
All of these are sinusoidal functions of aa and all have amplitudes of tfrA.
7l(a) Let Vc be the maximum potential difference across the capacitor, Vr be the maximumpotential difference across the inductor, and Vp be the maximum potential difference across the
resistor. Then the phase constant O is
,(Vt V"\:tan-, (ry Vn\tan- \ vp / \ ,= )-
tan-t(1.00) :45.0o.
(b) Since the maximum emf is related to the current amplitude by t* _ I Z, where Z is theimpedanceand R-Zcos$,
R-ry: :70.,1 {1 .
73
(a) The frequency of oscillation of an LC circuit is f - I l2n\re, where L is the inductanceand C is the capacitance. Thus
1 -6 H.L-#r*
(b) The total energy is tl : ifP, where I is the current amplitude. Thus U - LtO.89 x10-' tt)(l .20 x lo-s A)2 - l.7g x 10-1r J.
Chapter 3l I97
(c) The total
Q-tffi-energy is also given by U - Q' l2C, where A is the charge amplitude. Thus
2(I .79 x 10-11 JX340 x 10-aF): 1.10 x 10-7 C.
83
(a) The total energy Lf of the circuit is the sum of the energy IJ n stored in the capacitor andthe energy LIa stored in the inductor at the same time. Since LIs - 2.00[Jp, the total energy istl:3.00t-In Now U- Q'lze and Un: q2 lzc, where a is the maximum charge, q is thecharge when the magnetic energy is twice the electrical energy, and C is the cryaeitance. Thus
Q'lzc - 3.ooq'lzC and q: A Itm - o .577e.
(b) If the capacitor has maximum charge at time t - 0, then qangular frequency of oscillation. This means at - cos- t (0 .577)where T is the period,
t-ryr -0.1 s3T2r
8s
(a) The energy stored in a eapacitor is given by Un - q2 lzc, where q is the charge and e is thecapaeitance. Now q2 is periodic with a period of T12, where T is the period of the driving emf,so LI n has the same value at the beginning and end of each cycle. Actually IJ n has the same
value at the beginning and end of each half cycle.
(b) The energy stored in an inductor is given by Liz f 2, where i is the current and L is theinductance. The square of the current is periodic with a period of T 12, So it has the same valueat the beginning and end of each cycle.
(c) The rate with which the driving emf device supplies energy is
where ris the currenr ampli tl;:;I o'#.Xgglitl:Ho,t, ,!I. angular rrsquency, and a,sa phase constant. The energy supplied over a cycle is
trs : Io'
ps d,t: It,n Ir'sin(r,,,af)
sin( aat - ild,t
: I trn [' sin(c,,ar) tsin(a,,a t)cos(il -cos(cr., at)sin(/)] dt ,Jo
where the trigonometric identity sin(a - P) - sinocos P-cosasinp was used. Now the integralof sin'(rot) over a cycle is T12 and the integral of sin(u,'at)cos(wat) over a cycle is zero, so
Es -- *t t,"cos @.
(d) The rate of energy dissipation in a resistor is given by
Pp: i2 R - /2 sin'(rot - d)and the energy dissipated over a cycle is
Ep: 12 ['sin2 (rat - il d,t : ]t'RT .
Jo
(e) Now trn: IZ, where Z is the impedance, and R- Z cos$, so Es - *f'TZ cos@ -*t'RT: ER.
198 Chapter 3l
Chapter 32
3
(a) [Jse Gauss' law for magnetism: f E.d,A- 0. Write f E.dA- Or + Qz+Qs, where Qr isthe magnetic flux through the first end mentioned, Q2 is the magnetic flux through the second
end mentioned, and Q g is the magnetic flux through the curved surface. Over the first end, themagnetic field is inward, so the flux is 01
field is uniform, normal to the surface, and outward, so the flux is Q2- AB - 7Tr2 B, where Ais the area of the end and r is the radius of the cylinder. Its value is
Q2 - z-(0 .I20m)2(1.60 x 10-'r) - +7.24 x 10-5 Wb : *72.4 trtWb .
Since the three fluxes must sum to zero,
Qs
(b) The minus sign indicates that the flux is inward through the curved surface.
-3Consider a circle of radius r (- 6.0 mm), between the plates and with its center on the axis ofthe capacitor. The cuffent through this circle is zero, so the Ampere-Maxwell law becomes
f '-+ dQn
f B.dg: lroroT,
where E is the magnetic field at points on the circle and Q p is the electric flux through thecircle. The magnetic field is tangent to the circle at all points on Lt, so f U . d,i : 2rr B. Theelectric flux through the circle is Q6 - r R'E, where R (: 3.0 mm) is the radius of a capacitorplate. When these substitutions are made, the Ampere-Maxwell law becomes
2nrB: ltoesTRz#
Thus
dE 2rB 2(6.A x 10-3 m)(2.0 x 10-t f)- - :2.4x10lrvl^.s.dt FoeoRz (4n x 10-, Hl^X8.85 x 10-12 FmX3.0 x 10-t nr)-
13
The displacement current is given by
. d,tria: eoA dt ,
Chapter 32 199
where A is the area of a plate and tr is the magnitude of the electric field between the plates.
The field between the plates is uniform, So E - V ld, where V is the potential difference across
the plates and d is the plate separation. Thus
. esA dVLd: d dt
Now esAld is the capacitance C of a parallel-plate capacitor without a dielectric, so
'ia-C#
2t(a) For a parallel-plate capacitor, the charge q on the positive plate is given by qwhere A is the plate areU d is the plate separation, and V is the potential difference between the
plates. In terms of the electric field E between the plates, V : Ed, So q: eoAtr : €6Oe, where
Q n is the total electric flux through the region between the plates. The true current into the
positive plate is i - dqldt - es dQ t ldt - 'id" totut, where 'id, torut is the total displacement current
between the plates. Thus id, totur - 2.0 A.
(b) Since id, totar: 60 dA t I dt : eyA dE I dt,
dE i d" ,otul 2.0 A2.3 x 101'V l^. s.:::
eoA (8.85 x 10- t2 F lmX1.0 m)2
(c) The displacement current is uniformly distributed over the area. If a is the area enclosed bythe dashed lines and A is the area of a plate, then the displacement current through the dashed
path is
id""n"- frrdbtar:ffi e.oA): o.5oA.
(d) According to Maxwell's law of induction,
f
6 B .di: ltyid,"n, - (4n x 10-'tll^X0.50A): 6.3 x 10-7 T.ril.J
Notice that the integral is around the dashed path and the displacement current on the rightside of the Maxwell's law equation is the displacement current through that path, not the total
displacement current.
3s
(a) The z component of the orbital angular momentum is given by Lorb,z: m(hf 2r, where h isthe Planck constant. Since Tftp: 0, Lorb,, : 0.
(b) The z component of the orbital contribution to the magnetic dipole moment is given by
Forb, z
(c) The potentral energy associated with the orbital contribution to the magnetic dipole moment
is given by [J[ - -Forb,"Bext, where Be*t is the z component of the externalmagnetic field. Since
lJorb,z:0, fJ :0.
200 Chapter 3 2
dt
(d) The z component of the spin magnetic dipole moment is either +lrn or -pe, so the potential
energy is either
[Ji - -paBe*t: -e.27 x 10-24 JIT)(35 x 10-'T) - _3.2 x 10-25 J.
or Ui - +3.2 x 10-25 J.
(e) Substitute m4 into the equations given above. The z component of the orbital angularmomentum is
Lorb,r: ry- (-3X6'626 x 10-34J's) - -3.2x r0- 3a J.s.zr 2n
(0 'Ihe z component of the orbital contribution to the magnetic dipole moment is
Forb, z
(g) The potential energy associated with the orbital contribution to the magnetic dipole momentis
[] - -Fora, "8"*t: -QJ8 x 10-23 JIT)(35 x l0-'T) - -9.7 x 10-25 J.
(h) The potential energy associated with spin does not depend on Tftp. It is +3.2 x 10-25 J.
39
The magnetization is the dipole moment per unit volume, so the dipole moment is given by
l.L - MV, where M is the magnetization and V is the volume of the cylinder. Use V: nrzL,where r is the radius of the cylinder and L is its length. Thus
Ltr: MrrzL - (5.30 x 103 AIN?r(0.500 x I0-2m)2(5.00 x L0-2 m):2.08 x I0-2 JIT .
45
(a) The number of atoms per unit volume in states with the dipole moment aligned with themagnetic field is N. - AepB /kr and the number per unit volume in states with the dipolemoment antialigned is lf- - Ae- 1t'B lkr, where A is a constant of proportionality. The totalnumber of atoms per unit volume is jV - ff. + l/- - A ("ru /kr * e- p"B /*') . Thus
A- =,.= t
=,.=et"B /kr * e- p,B /kr
The magn ettzatton is the net dipole moment per unit volume. Subtract the magnitude of the totaldipole moment per unit volume of the antialigned moments from the total dipole moment perunit volume of the aligned moments. The result is
a E IV peuB /kr - IY pe- FB /kr IY 1t ("ru /kr - e- p"B lkr) a iM: - -lYptanh(p,BlkT).
Chapter 32 201
(b) If p,B <the power series expansion of the exponential function.) Thebecomes
- p,B lkT.expression
(See Appendix E forfor the magnetization
47
(a)
M= IY p, [f t + ptB lkT) - (1 - p,B lkT)] IV p'B(1 + pB lkT) + (l - p,B lkT) KT
(c) If p,B >denominator of the expression for M. Thus
M = IV u"::=t
mvffi-tVtr.(d) The expression for M predicts that it is linear in B lkT for p,B lkf small and independent
of B lkf for p,B lkT large. The figure agrees with these predictions.
The field of a dipole along its axis is given by Eq . 29-27:
ts_lto p,L) G7,where LL is the dipole moment and z is the distance from the dipole. Thus the magnitude of the
magnetic field is
B2r(10 x 10-n -)'
(b) The energy of a magnetic dipole with dipole moment I, in a magnetic field E is given byU- -rt. E - -p,Bcos/, where O is the angle between the dipole moment and the field. Theenergy required to turn it end for end (from Q : 0o to 0 - 180") is
L(J : -pB(cos 180o - cos0") - 2LtB - 2(1 .5 x L0-23 JIT)(3.0 x 10-uf): 9.0 x ro-ze J - 5.6 x 1o-lo eV.
The mean kinetic energy of translation at room temperature is about 0.04 eV (see Eq. 19-24 orSample Problem 32-3). Thus if dipole-dipole interactions were responsible for aligning dipoles,collisions would easily randomtze the directions of the moments and they would not remainaligned.
53
(a) If the magn etization of the sphere is saturated, the total dipole moment is Ftotat - IY Lr, wherel/ is the number of iron atoms in the sphere and LL is the dipole moment of an iron atom. Wewish to find the radius of an iron sphere with ,,Af iron atoms. The mass of such a sphere is IVm,where m is the mass of an iron atom. It is also given by 4n pRt 13, where p is the density ofiron and R is the radius of the sphere. Thus IYm:4rpRt 13 and
4n pR3
202 Chapter 3 2
iv-3m
Solve for R and obtain
R - lt*u'o'u'1t't .
L 4rpp IThe mass of an iron atom is
m :56 u - (56 uX 1.66 x 10-27 kg/u) - 9.30 x 10-26 kg .
So
Substitute this into Ftotat - Ir,' Lt to obtain
4r pR3 p,Ftotal
:3m
3(9.30 x 10-26 kg)(8.0 x 1
4r(14 x 103 kelmt)(z] x 10-23 J lT)(b) The volume of the sphere rs
v,- !nt- +(1.82 x 1o5m)3 -2.53 x 1016m3
and the volume of Earth is
V"- +(6.37
x lo6m)3- 1.08 x Lo2'*',
so the fraction of Earth's volume that is occupied by the sphere is
2.53 x 1016 m3
022 I lT)R-L]t":1.8x105m.
1.08 x
The radius of Earth was obtained from
r' rr-r' : 2.3 x 10-5 .
1021 m3
Appendix C.
--))(r) The horizontal and vertical directions arepe{pendicular to each other, so the magnitude ofthe field is
B-ffi:#fLUfL
'l+3sin2 \*,,
where the trigonometric identity cos' \rn: I - sin2 ^rn
was used.
(b) The tangent of the inclination angle is
tan Q6- ";^: (ffi) (ffi :#) - 2 tan \*,where tan
^*: (sin
^*) l(cos ,\-) was used.
61
lto lt
(a) The z component of the orbital angular momentumwhere m4 can take on any integer value from -3 to +3,
(-3, -2, -1, 0, *1, *2, and +3).
can have the values Lorb," : TrL2h f 2r,inclusive. There ate seven such values
Chapter 32 203
(b) The z component of the orbital magnetic moment is given by plorb,z _ -Trl,peh f 4rffi, where
m is the electron mass. Since there is a different value for each possible value of m6 there are
seven different values in all.
(c) The greatest possible value of Lorb,z occurs rf m,p - +3 is 3hf 2n.
(d) The greatest value of L;orb, " is 3eh lanm.(e) Add the orbital and spin angular momentai Lnet,z: Lorb,rt Lr,r: (mthl2r)*(mrhl2n\ Toobtainthemaximumvalue, setm2equalto+3 and TTLs equalto+j. Theresult LS L1sqs - 3.5h12n.
(0 Write Lnesr: Mhl2n, where M is half an odd integer. M can take on all such values from
-3.5 to +3.5. There are eight of these: -3.5, -2.5, -1.5, -0.5, +0.5, +1.5, +2.5, and +3.5.
204 Chapter 32
Chapter 33
fIIf f is the frequency and I is the wavelength of an electromagnetic wave, then f
^_ c. The
frequency is the same as the frequency of oscillation of the current in the LC circuit of thegenerator. That is, f : llLr{Lc, where e is the capaeitance and L is the inductance. Thus
ZnJLC:=11
Lt.
The solution for L is1z (550 x 10-'^)'
5.00 x 10-21 H.r, -u 4rzCc2 412(17 x 10-12 FX3.00 x 108 mls)z
This is exceedingly small.
2T
The plasma completely reflects all the energy incident on Lt, so the rcdtation pressure is givenbypr:Llfc,where I istheintensity.Theintensityis I-PlA,where P isthepowerand Ais the area intercepted by the radration. Thus
2PP,
23
Let f be the fraction of the incident beam intensity that is reflected. The fraction absorbed isI- f . The reflected portion exerts a radiation pressure of p, - (2f Idlc and the absorbed portionexerts a radration pressure of po -- (1 f)Iaf c, where Is is the incident intensity. The factor 2
enters the first expression because the momentum of the reflected portion is reversed. The totalrudration pressure is the sum of the two contributions:
,_ 2flo+(1 f)IoPtotal:PrrPo:
"
J
To relate the intensity and energy density, consider a tube with length (. and cross-sectionalarea A, lying with its axis along the propagation direction of an electromagnetic wave. Theelectromagnetic energy inside is LI - uA(., where u is the energy density. A11 this energy willpass through the end in time t - ( I , so the intensity is
r-YAt
Thus upropagation direction.
(l + f)Io
: uc.
are inherently positive, regardless of the
uA(.c
A(density
Chapter 33 205
For the partial|y reflected and parttally absorbed wave, the intensity just outside the surface isI - /o+ f Io - (1 + f)Io, where the first term is associated with the incident beam and the secondis associated with the reflected beam. The energy density is, therefore,
,t.t, : I- (1 + /)/o .cc)
the same as radtation pressure.
25
(a) Since c: ^f,
where ) is the wavelength and f is the frequency of the wave,
f : 9- 3'oo x -108
m/s - 1.0 x 108 Hz .
^ 3.0m
(b) The angular frequency is
u) :2nf :2r(1 .0 x 108 H4: 6.3 x 108 rudf s.
(c) The angular wave number is
-2n2rk-T-rr_-2. lradlm.
(d) The magnetic field amplitude is
Brr: E'n: ?ooY13 ,c 3.00 x 108 m/s
(e) E -,rst be in the positi ve z direction when d ir in the positi ve A direction in order for E "
Eto be in the positive r direction (the direction of propagation).
(0 The time-averaged rate of energy flow or intensity of the wave is
r _ E, (3oo Y l^),I-ffi :I.2xI02Wl^'.
(g) Since the sheet is perfectly absorbitg, the rate per unit area with which momentum is deliveredto it is Il", so
d,p- IA: (l|gWlm\(2.0m\- , -
it c 3^0 _8.0x10-'N.
(h) The rcdration pressure is
p,- ry-#:4.ox1o-7pa.27
If the beam carries energy U away from the spaceship, then it also carries momentum p - U l"away. Since the total momentum of the spaceship and light is conserved, this is the magnitude of
206 Chapter 3 3
the momentum acquired by the spaceship . If P is the power of the laser, then the energy carried
awayintime t is (Ji - Pt Thusp - Ptf c and, rf m is mass of the spaceship, its speed is
'J- p- Pt: (10 I103yxldx8'64 x 104s/d)- r.gx 10-r^l'- r.gmm/s.m mc (1.5 x 103 kgX3.00 x 108 m/s)
3s
Let Is be in the intensity of the unpol artzed light that is incident on the first polari zing sheet.
Then the transmitted intensity is Ir - Lto and the direction of polanzation of the transmittedlight is 0r (: 40") counterclockwise from the A axis in the diagram.
The polarizrng direction of the second sheet is 0z (:20") clockwise from the A axis so the angle
between the direction of polarrzation of the light that is incident on that sheet and the polarizingdirection of the of the sheet is 40" + 20" : 60o. The transmitted intensity is
12: Ircos2 60o - )trcos2 60o
and the direction of pol arrzation of the transmitted light is 20o clockwise from the A axis.
The polariztng direction of the third sheet is fu (:40') counterclockwise from the A axis so the
angle between the direction of polanzation of the light incident on that sheet and the polarizrngdirection of the sheet is 20" + 40o : 60o. The transmitted intensity is
\ : Izcos2 60o _ lrrcos4
60o : 3. I x I0_2 .
3.Ioh of the light's initial intensity is transmitted.
43
(a) The rotation cannot be done with a single sheet. If a sheet is placed with its polarizrngdirection at an angle of 90" to the direction of pol arrzation of the incident radiation, oo radrationis transmitted.
It can be done with two sheets. Place the first sheet with its polarizrng direction at some angle
0 , between 0 and 90o, to the direction of pol arrzation of the incident radiation. Place the second
sheet with its polari zing direction at 90o to the polari zation direction of the incident radration.
The transmitted radiation is then polarized at 90o to the incident polarization direction. Theintensity is /s cos? 0 cost(gO" - 0): /0 cos2 0 srnz 0, where I0 is the incident radiation. If 0 is not0 or 90", the transmitted intensity is not zero.
(b) Consider n sheets, with the polariztngdirection of the first sheet making an angle of 0 - 90" lnwith the direction of polarrzation of the incident radration and with the polariztng direction ofeach successive sheet rotated 90" f n in the same direction from the polari zing direction of theprevious sheet. The transmitted rcdration is polari zed with its direction of polanzation makingan angle of 90o with the direction of polarrzation of the incident radration. The intensity isI - .Is cos'n (90' ln). You want the smallest integer value of n for which this is greater than
0.60,Is.
Chapter i 3 207
Start with n : 2 and calcul ate cos2n(90" l.r,). If the result is greater than 0.60, you have obtainedthe solutiotl. If it is less, increase n by 1 and try again. Repeat this process, increasing n by Ieach time, until you have a value for which cos2'(90" ld is greater than 0.60. The first one willbe n: 5.
51
Consider a ruy that grazes the top of the pole, as shown inthe diagram to the right. Here 0r:35o, lt: 0.50m, and(,2- 1.50m. The length of the shadow is tr* L. r is given
byrthe law of refraction , TL2 stn 02 : rL1 sin 91 . Take T\ _ 1 and
TL2
0z: sin-, ( sin gr
) : sin- t ( sin 35g)
- 25.55o .rr \- )-Dr' \ l33 /-/-'
L is given by
L - (.ztat0z - (1 .50 m) tan 25.55o - 0 .72 m .
The length of the shadow is 0.35 m + 0 .72m - 1.07 m.
--33
Look at the diagram on the right. The two angles
labeled a have the same value. 0z is the angle ofrefraction. Because the dotted lines are pe{pendicularto the prism surface 0z * a - 90o and aBecause the interior angles of a triangle sum to 1 80o,
180o - 202* d: 180' and 0z - 012.
Now look at the next dragram and consider the triangleformed by the two normals and the ruy in the interior.
The two equal interior angles each have the value 0 -02.Because the exterior angle of a triangle is equal to the
sum of the two opposite interior angles, ,lt - 2(e - 0z)
and 0 - ?z+rb 12. Upon substitution for 02thrs becomes
0 - (d + ,1,) 12.
'Lr
According to the law of refraction the index of refraction of the prism material is
sin 0::stn 02
sin(/ + rD12
water
shador,V
-t' j/\.
208 Chapter 3 3
sin 012
05
(a) No refraction occurs at the surface eb, so the angle of incidence at surface o,c is 90" O.
For total internal reflection at the second surface, ne stn(90" il must be greater than TL,.
Here ne is the index of refraction for the glass and TLa is the index of refraction for air. Since
sin(90" il - cos @, you want the largest value of O for which nn cos $ >cos / decreases as 0 increases from zero. When 6 has the largest value for which total internalreflection occurs, then rre cos 0 - tra, or
, -1Q: cos r
The index of refraction for air was taken to be unity.
(b) Replace the air with water. If n* (: 1.33) is the index of refraction for water, then the largest
value of 0 for which total internal reflection occurs is
e) _ cos-r f _L\ : t
\ 152l - 48 e"
_ cos-l f ry) _ 2e.oo\ 1.52 /
*3) : 12 srnz o1rL' /
, -lQ: cos r
69
The angle of incidence 0 s for which reflected light is fully polarized is given by Eq. 33-49of the text. If nr is the index of refraction for the medium of incidence and TL2 is the index ofrefraction for the second medium, then 0e - tan-t (n, l"t) - tarrt (t.53 f I.33) :63.8o.
73
Let 01 (- 45") be the angle of incidence at the first surface and 0z be the angle of refractionthere. Let fu be the angle of incidence at the second surface. The condition for total internalreflection at the second surface is n srn 03 >of refraction n for which this inequality holds.
The law of refraction, applied to the first surface, yields nstn02- sin01 . Consideration of the
triangle formed by the surface of the slab and the ray in the slab tells us that 0z
Thus the condition for total internal reflection becomes 1 <this equation and use srnz e 2 + cos2 02 - 1 to obtain 1 <sin d2 - ( | I ")
sin 01 to obtain
e)
1<
The largest value of n for which this equation is true is the value for which 1
Solve for n:
sin2 45" : 1.22
Chapter 3 3 209
75
Let 0 be the angle of incidence and 0z be the angle
of refraction at the left face of the plate. Let nbe the index of refraction of the glass. Then, the
law of refraction yields sind - nsrn?2. The angle
of incidence at the right face is also 02. If fu isthe angle of emergence there, then n srn 02 - sin 93.
Thus sin 93 - sin 0 and fu - 0. The emerging ray isparallel to the incident ray.
You wish to derive an expression for r in terms
of 0. If D is the length of the ray in the glass,
then Dcos?z:t and D:tf cos02. Theangle ain the diagram equals 0 0z and r - lJ sin a -D sin(O - 0). Thus
M
-Jr
If all the angles e,
srn 02 = 02, sin(0applied to the point
The distances are given
(c) Take d to be 2(l .3 x
r= t(r- t:
t sin(0 - 0z)
cos 02
02, 02, and 0 0z are small and measured in radians, then sin 0 N 0,
0) = 0 02, and cos 02 = 1. Thus r E t(0 0). The law of refractionof incidence at the left face of the plate is now 0 E n?z, So 02 E 0 l, and
(n - r)t0
77
The time for light to travel a distance d in free space is t- dlr, where c is the speed of light(3.00 x 108 m/s).
(a) Take d to be 150 km : 150 x 103 m. Then,
+_d 150x103mf,:;- -
(b) At full moon, the Moon and Sun are on opposite
the light is d,- (1.5 x 108km) + 2(3.8 x 105km) -taken by light to travel this distance is
d 1.51 x 1011 mJ__l'- c 3.00 x 108 m/s
in the problem.
10e km) : 2.6 x 1012 m. Then,
d 2.6 x 1012 mJ__I-u c 3.00 x 108 m/s
5.00 x 10-4s.
sides of Earth, so the distance traveled by1.51 x 108km- 1.51 x 10llm. The time
500 s - 8.4 min .
210 Chapter 3 3
8.7x103s-2.4h
(d) Take d to be 6500 ly and the speed of light to be 1.00 IV lV. Then,
t-l:#ffi:65ooyThe explosion took place in the year 1054 - 6500: -5446 or B.C.5446.
79
(a) The amplitude of the magnetic field is B10-B 1. According to the argument of the trigonometric function in the expression for the electricfield, the wave is moving in the negative z direction and the electric field is parallel to the Aaxis. In order for E x E to b. in the negative z direction, E must be in the positive r directionwhen E is in the positive A direction. Thus
B* - ( 1,.67 x 10-* r) sinf(l .00 x 106 m- t), + wt]
is the only nonvanishing component of the magnetic field.
The angular wave number is k-1.00 x 106m-l so the angular frequency is u- kc: (1.00 x106 m-txf .00 x 108 mls) : 3.00 x 1014 s-l and
B* - (r.67 x 10-t r) sin[(1.00 x 106m-'), + (3.00 x 10r4 s-t)t] .
(b) The wavelength is ) - 2nlk -2nl(1.00 x 106m-t) - 6.28 x 10-6m.
(c) The period is T :2nf u:2nl(3.00 x 1014 s-1) : 2.09 x 10-14 s.
(d)Theintensityofthiswaveis/ - Ez,.l2poc- (5.00V1rr'1l'lZ(+rxl0-'ttl^X3.00x108 mls-0.0332W l*'. (0 A wavelength of 6.28 x 10-6m places this wave in the infrared portion of the
electromagnetic spectrum. See Fig. 33-1.
83
(a) The power is the same through any hemisphere centered at the source. The arca of ahemisphere of radius r is A - 2nr2. In this case r is the distance from the source to the aircraft.Thus the intensity at the atrcraft is I - PIA: Pf 2nr2 - (180 x 103 Wl2r(90 x 103 m)2 _
3.5 x 10-uW l^'.(b) The power of the reflection is the product of the intenstty at the arcc.r:aft
of the akqaft: P, - (3.5 x l0-uw l^'Xo .22m2) : 7.8 x 10-7'w.
(c) The intensity at the detector is P, f 2nr2 -- (7.8 x 10-t W) l2r(9010-17 w l^'.(d) Since the intensity is given by I - E2,.l2por,
and the cross section
x 103 m)2 - 1.5 x
Ern
(e) The rrns value of the
108 mls) : 2.5 x 10-16 T.magnetic field is Brrn, : Ernlfr"- (1.1 x 10-'Vl^)l(t[Dp.O0 x
2 p,scl
Chapter 33 2ll
9t
The critical angle for total internal reflection is given by 0"_ sin-t(tln). For n- 1.456 thisangle is 0":43.38o and for TL: I.470 it is 0.:42.86o.(a) An incidence angle of 42.00o is less than the critical angle for both red and blue light. Therefracted light is white.
(b) An incidence angle of 43. 10o is less than the critical angle for red light and greater than thecritical angle for blue light. Red light is refracted but blue light is not. The refracted light isreddish.
(c) An incidence angle of 44.00o is greater than the critical angle for both red and blue light.Neither is refracted.
103
(a) Take the derivative of the functions given for tr and B, then substitute them into
a2E ,o2E . ozg ,o2BM - c- A., ancl
0t, - c- Arz. '
The derivatives of tr are A'E l0t2 - -a2 E,n srn(kr - .,lt) and A'E l0r2 - -pz En sin(kr - olt),
so the wave equation for the electric field yields u2 - c2 k2. Since e - ck the function satisfiesthe wave equatiorl. Similarly, the derivatives of B are A2 B l\t' - -a2 Brnsin(kr .,lt) and
A2Bl0*'- -TtzBrnsrn(kr - u)t) and the wave equation for the magnetic field yields w2 : c2k2.
Since w - ck the function satisfies the wave equation.
(b) Let LL : kr * wt and consider f to be a function of u, which in turn is a function of n andt. Then the chain rule of the calculus gives
a2tr d,2f (ou\' d,2f )
at, - d,r, \e/ :
a,u\*-
anda2 tr d,2 f ( ou\' d,' f:#l;) :#k2012
Substitullgn into the wave equation agarn yields w2 - c2 k2, So the function obeys the waveequatioll. A similar analysis shows that the function for B also satisfies the wave equation.
212 Chapter 3 3
Chapter 34
5
The light bulb is labeled O and its image islabeled I on the digram to the right. Considerthe two rays shown on the dragram to the right.One enters the water at A and is reflected fromthe mirror at B. This ray is petpendicular to the
water line and mirror. The second ruy leaves
the lightbulb at the angle 0, enters the water at
C, where it is refracted. It is reflected from the
mirror at D and leaves the water at E. At C the
angle of incidence is I and the angle of refractionts 0'. At D the angles of incidence and reflectionare both 0'. At E the angle of incidence is g/ and
the angle of refraction is 0. The dotted lines thatmeet at I represent extensions of the emergingrays. Light appears to come from I. We want tocompute d3.
Consideration of the triangle OBE tells us that the distance d2+ fu is Ltan(90o - 0) - Lf tan?,where L is the distance between A and E. Consideration of the triangle OBC tells us thatthe distance between A and C is d1 tan 0 and consideration of the triangle CDE tells us thatthe distance between C and E is 2dztan?t, so L - drtan? + 2dztan?t, d2 + dz
Zdztan 0') I tan 0, and
, d1 tan 0 + 2dztan 0ltu
Apply the law of refraction at point C: sin d
water. Since the angles 0 and 0' are small we may approximate their sines by their tangents and
write tan 0 - ntan?t. Us this to substitute for tan? in the expression for fu to obtain
d3
I *'I
I
, ndt + 2dzrr3l
n, (1'33)(250 cm) + 2(20@
- 200cm - 350 cm,-42
where the index of refraction of water was taken to be 1.33.
9
(a) The radius of curvature r and focal length f are positive for a conca\/e mirror and are related
by f :rl2,sor-2(+1 8cm):+36cm.
mlffor
Chapter 34 213
(b) Since (tlD+(lli) - llf , where i is the image distance,
,i - 4- (18cm)(12cm) - -36cm.p- f 12cm- 18cm
(c) The magnification is m : -i lp - -(-36 cm) l(12 cm: 3.0.
(d) The value obtained for i is negative, so the image is virtual.
(e) The value obtained for the magnification is positive, so the image is not inverted.
(0 Real images are formed by mirrors on the same side as the object and virtual images are
formed on the opposite side. Since the image here is virtual it is on the opposite side of themir:ror from the object.
11
(a) The radius of curvature r and focal length f are positive for a concave mirror and are relatedby f -r12, so r:2(+l2cm)- +24cm.
(b) Since (tlD + (Ili) - Ilf , where i is the image distance,
i - &- (12cmX18cm) - 36cm.p- f 18cm- I}cm
(c) The magnification is m : -i lp - -(36 cm) 108 cm - -2.0.(d) The value obtained for i, is positive, so the image is real.
(e) The value obtained for the magnification is negative, so the image is inverted.
(0 Real images are formed by mirrors on the same side as the object. Since the image here isreal it is on the same side of the mirror as the object.
15
(a) The radius of curvature r and focal length f are negative for a convex mirror and are relatedby f : r 12, so r : 2(-10 cm) : -20cm.(b) Since (tld+(Lli,) - Ilf, where i is the image distance,
,i- h- - -4.44cm.
(c) The magnification is m - -ilp - -(-4.44cm)l(8 cm - +0.56.
(d) The value obtained for i is negative, so the image is virtual.
(e) The value obtained for the magnification is positive, so the image is not inverted.
(0 Real images are formed by mirrors on the same side as the object and virtual images are
formed on the opposite side. Since the image here is virtual it is on the opposite side of themirror from the object
27
Since the mirror is convex the radius of curvature is negative. The focal length is f - r 12 -(-40cm)fT--20cm.
214 Chapter 34
Since (r lil+ (I li)- (1 I il,
This yields p- +5.0cm if ipositive we select iThe magnification is m - -i lpnegative the image is virtual and on the opposite side of the mirror from the object. Since the
magnification is positive the image is not inverted.
29
Since the magnification m is mdistance, i - -mp. lJse this to substitute for i in (t lD + (lli)length. The solve for p. The result is
p-r(' :)-(+3ocm)(' #) :+1 2ocm
Since p must be positive we must use the lower sign. Thus the focal length is -30 cm andthe radius of curvature is rnegative the mirror is convex.
The object distance is 1,.2m and the image distance is 'i - -mp: -(0.20)(l20cm): -24cm.Since the image distance is negative the image is virtual and on the opposite side of the mirrorfrom the object. Since the magnification is positive the image is not inverted.
3s
Solve
'r + ?- TLz - TLr
p?,7for r. the result is
r _ i,p(nz - nt) -
(-13 cmX+10 cm) - 43 cm.
nti * nzp ((1.0X- 13 cm) + (1.5)(+10 cm)
Since the image distance is negative the image is virtual and appears on the same side of the
surface as the obj ect.
37
SolveTL r TL'trrL
-t-
piTL2 - TL1
for r. the result is
,i _ , nrrp -
(1.0X+30 cmX+70 cm)
(nz - nt)p - wrr (1.0 - 1.5)(+70 cm) - (1.5X+30 cm)
Chapter 34 215
ifY: . "'?,T
Since the image distance is negative the
surface as the obj ect.
image is virtual and appears on the same side of the
4t
IJse
where f is the focal length, n is the index of refraction, rr is the radius of curvature of the firstsurface encountered by the light and 12 is the radius of curvature of the second surface. Since
one surface has twice the radius of the other and since one surface is convex to the incominglight while the other is concave, set 12: -2rt to obtain
the lens maker's equation, Eq. 34-10:
i:(n-1) (* ;) ,
i-(n-1)(*.+):3(n - 1)
Solve for 11 :
11- 3(n - r)f -
3(1'5 - lX6omm) aF
2 - 2 :zl)mm.
The radii are 45 mm and 90 mm.
47
The object distance p and image distance i obey (tld + (Lli,)- (1 1il, where f is the focallength. In addition, p + i - L, where L (: 44 cm) is the distance from the slide to the screen.
Use 'i-L-pto substitute fori inthefirstequationand obtatnp2-pL+Lf :0. Thesolutionis
p:L+@- (4cm)+ :22cm.22
51
The lens is divergirg, so the focal length is negative. Solve (Ild + (Ili,)- Olfi for i. Theresult is
i- *- - -4.gcm.
The magnification is m- -ilp: -(-4.8 cm)l(+8.0cm)-0.60. Since the image distance is
negative the image is virtual and appears on the same side of the lens as the object. Since the
magnification is positive the image is not inverted.
2rr
--55
The lens is converging, so the focal length is positive.
result isSolve (I lil + (I li,)- (T I fi for i. The
pf (+45 cm)(+20 cm) , n,?'- p-f -(45c@:tJocm'
216 Chapter 34
The magnification is m - -i lp _ -(36 cm) l@S cm) : -0.80. Since the image distance ispositive the image is real and appears on the opposite side of the lens from the object. Since the
magnification is negative the image is inverted.
61
The focal length is
t _ rtrz (+30 cm)(-42cm)f :,e_ r';a,-d: -*3 1'8cm'
Solve (t lil + (l li) - (l I fl for i. the result is
,i- *- -s5cm.
The magnification is m: -ilp - -(55 cm) IQS cm) : -0.73.Since the image distance is positive the image is real and on the opposite side of the lends fromthe object. Since the magnification is negative the image is inverted.
75
Since m - -i lp, i - -mp: -(+1.25X+16 cm)The result is
p (+16 cmx _ zocm)f -
p?' -p + i (+16 cm) + ( _ zocm)
: *8o cm '
Since f is positive the lens is a converging lens. Since the image distance is negative the image
is virtual and appears on the same side of the lens as the object. Since the magnification ispositive the image is not inverted.
79
The image is on the same side of the lens as the object. This means that the image is virtual and
the image distance is negative. Solve (tlD + (Ili) - (llfl for i. The result is
.pf?': p- fand the magnification is
i fm:-p-- p-f '
Since the magnification is less than 1.0, f must be negative and the lens must be a diverginglens. The image distance is
: _ (+5.0 cm)(- 10 cm)i
and the magnification is m : -i lp - -(- 3.3 cm) l(5.0 cm) - 0 .66 cm.
Chapter 34 217
Since the magnification is positive the image is not inverted.
81
Lens I is converging and so has a positive focal length. Solve (Ilpr)+ Qli)_ (Ilil for theimage distance 'h associated with the image produced by this lens. The result is
i1 : hh.- (2ocmX+9'ocm) - r6.4cm.pt h QA cm) - (9.0 cm)
This image is the object for lens 2. The object distance is d-p, - (8.0cm)-(16.4cm) : -8.4effi.The negative sign indicates that the image is behind the second lens. The lens equation is stillvalid. The second lens has a positive focal length and the image distance for the image it formsis
i2: #r- -+3.rcm.
The overall magnification is the product of the individual magnifications:
/ it\ / ir\ / l6.4cm\ / 3.1cm\n7, - t71,1t7l2
\Since the final image distance is positive the final image is real and on the opposite side of lens2 from the object. Since the magnification is negative the image is inverted.
89
(a) If L is the distance between the lenses, then according to Fig. 34-20, the tube length iss: L- foa f"r:25.0cm -4.00cm- 8.00crrr:13.0cm.(b) Solve (tlD + (Ili) - (Ilf.,) for p. The image distance is 'L - foa* s - 4.00cm* 13.0cm -17.0 cffi, so
p : 4- (17'o cmX4'oo cm) - 5 .z3cm .i f oa 1.7 .0 cm - 4.00 cm
(c) The magnification of the objective is
m:-;- -m--3 2s
(d) The angul ar magnification of the eyepiece is
25 cm 25 cm,Ls: f"-g^oor*-3.13.
(.) The overall magnification of the microscope is
M : TTLrrLo - (--3.25X3.13) - -10.2.
93
(a) When the eye is relaxed, its lens focuses far-away obj ects on the retrna, a distance i behindthe lens. Set p - cc in the thin lens equation to obtain lli - llf , where f is the focal length of
218 Chapter 34
the relaxed effective lens. Thus iimage distance i remains the same but the object distance and focal length change. If p is thenew object distance and f is the new focal length, then
1
-+p
Substitute 'i - f and solve for f ' . You should obtain
f':4f + p 40.0 cm + 2.50 cm
(b) Consider the lensmaker's equation
1 /1 1\j:(n- 1) (e d ,
where 11 and 12 are the radii of curvature of the two surfaces of the lens and n is the index ofrefraction of the lens matertal. For the lens pictured in Fig. 34-46, 11 and 12 have about the samemagnitude, 11 is positive, and 12 is negative. Since the focal length decreases, the combination
0 lrr ) - 0 ld must increase" This can be accomplished by decreasing the magnitudes of eitheror both radii.
103
For a thin lens, (t ld+ (1 ltl - ( I I il,f is the focal length. Solve for z:
where p is the object distance, i is the image distance, and
fpp- f
Let p- f +*, where:r isinside. Then
positive if the object is outside the focal point and negative if it is
f(f + r)r
Now let i - f + n' , where r' is positive if the image is outside the focal point and negative if itis inside. Then
r,-i f:f(f+r) f:f'rrand m' - f2.
105
Place an object far away from the composite lens and find the image distance ,i. Since the imageis at a focal point, i : f , the effective focal length of the composite. The final image is producedby two lenses, with the image of the first lens being the object for the second. For the first lens,
Olpt)+ Olir): (1 1il, where h is the focal length of this lens and,fi is the image distance forthe image it forms. Sinc e pl : cc, i1 : f r.
11:x f'
Chapter 34 219
The thin lens equation, applied to the second lens, is (l lpz) + 0 li) _ (I I f), where pz is theobject distance, 'i2 is the image distance, and fz is the focal length. If the thicknesses of thelenses can be ignored, the object distance for the second lens is pz - -'ir. The negative signmust be used since the image formed by the first lens is beyond the second lens if ir is positive.This means the object for the second lens is virtual and the object distance is negative. If it isnegative, the image formed by the first lens is in front of the second lens and p2 is positive. Inthe thin lens equation, replac e pz with - fi and 'iz with f to obtain
11l-
hfThe solution for f is
1
f,
P hfz,-a' fi+ fz
t07
(a) and (b) Since the height of the image is twice the height of the fly and since the fly and itsimage have the same orientation the magnification of the lens is n'L: +2.0. Since TrL - -ilp,wherep is the object distance andi is the image distance, 'i - -2p. Now lp+il: d, so l-pl: d,
and p : d - 20 cm. The image distance is -40 cm.
Solve (t ld + (I li) - (l I f) for f . the result is
p?, (20 cmx-40 cm)I- :\:---: -/\ :*40Cm.J p.C:(20 :
(c) and (d) Now m: +0.5 and 'i- -0.5y). Since lp+il: d,0.5p - d and p:2d:40cm. Theimage distance is -20 cm and the focal length is
,. pi (40 cmx -20 cm) A,\I: ,.a' p+i (40cm)+(-20cm)
220 Chapter 34
Chapter 35
-5
(a) Take the phases of both waves to be zero at the front surfaces of the layers. The phase ofthe first wave at the back surface of the glass is given bV dl : hL - at, where fu (: 2n I
^t) is
the angular wave number and )1 is the wavelength in glass. Simrlarly, the phase of the second
wave at the back surface of the plastic is given by dz - kzL - at, where kz (: 2nl )z) is the
angular wave number and ),2 is the wavelength in plastic. The angular frequencies are the same
since the waves have the same wavelength in at and the frequency of a wave does not change
when the wave enters another medium. The phase difference is
/ 1- 1) Lfu-dz-(kl-k)L-'tn-t -/- tt
\,1 t \, ) ''-r
Now,\1glass. Similarly, ),2: Iui. lrr, where rL2 is the index of refraction of the plastic. This means that
the phase difference is h- Qz - (2nl\u)(u-rr)L.The value of L that makes this 5.65rad is
r -(il - d)\^*'l - 2n4tt - rd 2"(1.60 - 1i0) - 3'60 x 10-6m'
(b) 5.65 rad is less than Zrrad (: 6.28rad), the phase difference for completely constructive
interference, and greater than r rad (: 3. 1 4 rad), the phase difference for completely destructive
interference. The interference is therefore intermediate, neither completely constructive nor
completely destructive. It is, however, closer to completely constructive than to completelydestructive.
15
Interference maxima occur at angles 0 such that d stn 0
sources, ,\ is the wavelength, and m is an integer. Since d - 2.0m and ^that sin g :. 0.25m. You want all values of m (positive and negative) for which l0.25ml <
These are -4, -3, -2, -1,0, *1, *2, *3, and +4. For each of these except -4 and a4, there
are two different values for 0. A single value of 0 (-90') is associated with msingle value (-90") is associated with m- +4. There are sixteen different angles in all and
therefore sixteen maxima.
t7The angular positions of the maxima of a two-slit interference pattern are given by d sin 0 - ffi\,where d is the slit separation, ,\ is the wavelength, and m ts an integer. If 0 is small, sin 0 may be
approximated by 0 in radians. Then d0 : m),. The angular separation of two adlacent maxima
Chapter 3 5 221
is L0 - ) f d,. Let )'be the wavelength for which the angular separation is 10.0% greater. Then1.10^ f d- A'ld, or
^/- 1.10):1.10(589nm): 648nm.
t9The condition for a maximum in the two-slit interference pattern is d sin 0 - mA, where d is theslit separation, .\ is the wavelength, m is an integer, and 0 is the angle made by the interferingrays with the forward direction. If 0 is small, sin 0 may be approximated by g in radians. Thend0: m), and the angular separation of adjacent maxima, one associated with the integer m andthe other associated with the integer m+ 1, is given by L0 -
^ld.The separation on a screen a
distance D away is given by Ly - D L0 : ),D ld. Thus
29
The phasor diagram is shown to the right. Here E1 - 1.00, Ez - 2.00,and d:60o. The resultant amplitude E* is given by the trigonometriclaw of cosines:
fi-, - tr? + $ - z4r4zcos(l80o - O),
Ern: (1.00)2 1 (2.00)2 - 2(1 .00X2.00) cos 120" - 2.65
39
For complete destructive interference, you want the waves reflected from the front and back ofthe coating to differ in phase by an odd multiple of n rad. Each wave is incident on a mediumof higher index of refraction from a medium of lower index, so both suffer phase changes ofr rud on reflection . If L is the thickness of the coating, the wave reflected from the back surfacetravels a distance 2L farther than the wave reflected from the front. The phase difference is2L(2n l\), where )" is the wavelength in the coating. If n is the index of refraction of thecoating, I": \ln, where ) is the wavelength in vacuum, and the phase difference rs2nL(2rl\).Solve
2nL
for L. Here m is an integer. The result i
: (2m + I)n
L- (2m+ 1))u4n'
To find the least thickness for which destructive interference occurs, take m : 0. Then
^ (500 x 10-e *X5.40 m) A ^ - .,LA: -2.25 x 10-'m-2.25mm.
L _ I - 600 x 10-em
- 1 Zx 10_7 m.r-/ 4n 4U, r"
(T)S
222 Chapter 35
4t
Since nr is greater than n2 there is no change in phase on reflection from the first surface. Since
rL2 is less than T\ there is a change in phase of n rad on reflection from the second surface.
One wave travels a distance 2L fuither than the other, so the difference in the phases of the two
waves is anLl\z* n, where ),2 is the wavelength in medium 2. Since interference produces a
minimum the phase difference must be an odd multiple of r'. Thus 4n L I \, * n - (2m + I)n,where nL is an integer or zero. Replace ),2 with \lry where ) is the wavelength in arc, and
solve for l. The result is
4Lnz 2(380 nmxl. 1.34) 1018 nm
2mmFor nt: I,,\ -- 1018nm and for m- 2,,\ - (1018nm)12: 509 llm. Other wavelengths arcshorter. Only l - 509 nm is in the visible range.
47
There is a phase shift on reflection of r for both waves and one wave travels a distance 2Lfunher than the otheE so the phase difference of the reflected waves is ar L I \2, where ),2 is the
wavelength in medium 2. Since the result of the interference is a minimum of intensity the phase
difference must be an odd multiple of n. Thus 4r L I ),2 - (2m + l)n, where m is an integer or
zero. Replace )2 with \lny where ) is the wavelength in arc, and solve for.\. The result is
\ 4Lnz -
4(210nmXl.46)2m+1 2m+1 2m*1
For massociated with wavelengths that are not in the visible range.
53
(a) Oil has a greater index of refraction than at and water has a still greater index of refraction.
There is a change of phase of n rud at each reflection. One wave travels a distance 2L fuitherthan the other, where L is the thickness of the oil. The phase difference of the two reflected
waves is anLllo, where ,\ is the wavelength in oil, and this must be equal to a multiple of 2rfor a bright reflection. Thus 4nLf )."-Zmn, where m is an integer. Use
^- TLo\o, where no
is the index of refraction for oil, to find the wavelength in air. The result is
For m- I,,\ : 1l04nm; for m- 2, ^
: (1104nm)12: 552nm; and for m- 3, I -(1104nm)13 - 368nm. Other wavelengths ate shorter. Only ) - 552nm is in the visible range.
(b) A maximum in transmission occurs for wavelengths for which the reflection is a minimlrm.The phases of the two reflected waves then differ by an odd multiple of n rad. This means
4nL l^" : (2m + I)n and
, 4noL 4(I .20)(460 nm)A- : :
2m* I 2m* I
2208nm
h+l
ZnoL 2(I .20)(460 nm) 1104 nm
Chapter 3 5 223
For m(2208 nm)/5 _ 442nm. Other wavelengths are shorter. Only l 442nm falls in the visiblerange.
63
One wave travels a distance 2L further than the other. This wave is reflected twice, once fromthe back surface and once from the front surface. Since TL2 is greater than T\ there is no changein phase at the back-surface reflection. Since TL1 is greater than rL2 there is a phase change of nat the front-surface reflection. Thus the phase difference of the two waves as they exit material 2is anLl),2*n, where ),2 is the wavelength in material 2. For a maximum in intensity the phase
difference is a multiple of 2n. Thus an L I ),2 * nfor ),2 is
, 4L 4(41 5 nm)/\., : :L 2m-1 2m-1
1660 nm
2m-lThe wavelength in aff ls
^ - nz\z -
(1.59X1660nm) -
2639nm2m-1 2m-1
For m- 1, ) - 2639nm; for m- 2, ^
-- 880nm; for m- 3, ^
- 528nm; and for ml - 377llm. Other wavelengths are shorter. Only l - 528 nm is in the visible range.
7lConsider the interference of waves reflected from the top and bottom surfaces of the air film.The wave reflected from the upper surface does not change phase on reflection but the wavereflected from the bottom surface changes phase by n rad. At a place where the thickness ofthe aff film is L, the condition for fully constructive interference is 2L - (m + *).f, where ,\(- 683 nm) is the wavelength and ?rL is an integer. This is satisfied for m - I40:
r. (m+ +)) (140.5X693 x lo-em) . - cI - - -4.80x10-'m-0.048mm.rr- 2 2
At the thin end of the aLr film, there is a bright fringe. It is associated with mtherefore, 140 bright fringes in all.
75
Consider the interference pattern formed by waves reflected from the upper and lower surfacesof the air wedge. The wave reflected from the lower surface undergoes a r-rad phase changewhile the wave reflected from the upper surface does not. At a place where the thickness ofthe wedge is d, the condition for a maximum in intensity is 2dwavelength in aff and m is an integer. Thus d - (2m+ 1)) f 4. As the geometry of Fig. 35-47shows, d: R- IF4, where R is the radius of curvature of the lens and r is the radius ofaNewton,Sring.Thus(2m+l)^l4-R-|w.So1veforT,FirstfeaffangethetermsSothe equation becomes
224 Chapter 3 5
-R-(2m+ 1))
Now square both sides and
If /? is much larger than a
solve for 12. When you take the square
lQm + l)B) (2m + I)zSzr: V z 16 '
wavelength, the first term dominates the
Qr_dz:2Ll+ +l:
root, you should get
second and
81
Let Ql be the phase difference of the waves in the two arrns when the tube has air in it and Iet $2be the phase difference when the tube is evacuated. These ate different because the wavelengthin arr is different from the wavelength in vacuum. If ,\ is the wavelength in vacuum, then the
wavelength in air is \lr, where n is the index of refraction of air. This means
4r(n - I)L
where L is the length of the tube. The factor 2 arises because the light traverses the tube twice,once on the way to a mirror and once after reflection from the mirror.
Each shift by one fringe coffesponds to a change in phase of 2r rad, so if the interference patternshifts by t/ fringes as the tube is evacuated,
4n(n - l)L :2lYn
and60(500 x 10-e -) - 1.00030 .
2(5.0 x 10-2 ^)87
Suppose the wave that goes directly to the receiver travels a distance L1 and the reflected wavetravels a distance L2. Since the index of reftaction of water is greater than that of aff thislast wave suffers a phase change on reflection of half a wavelength. To obtain constructiveinterference at the receiver the difference L2 L2 in the distances traveled must be an oddmultiple of a half wavelength.Look at the diagram on the right. The right triangleon the left, formed by the vertical line from the waterto the transmitter T, the ray incident on the water,and the water line, gives D otriangle on the right, formed by the vertical line fromthe water to the receiver R, the reflected r?y, and thewater line gives D6: rf tan?. Since Do+ Du: D,
al rD6 --+
Chapter 3 5 225
l/,\tL- 1+ n
Ir
I
a
Itan? - D (_ Do
[Jse the
means
and
Lzo
Lzu
identity stn20- tanz0l0+lrrnz0) to show that sing_ (a+ r)l Dz + (a + r)2. This
rsin 0 a* r
SO
L2: L2o* Lzaa* r
Use the binomial theorem, with pz large and a2 * 12 small,LzB D+(a+r)2 l2D.The distance traveled by the directapproximate this expression: L1 E
L2-L1ED+
sin 0 a* r
D2+(a**)'.
to approximate this expression:
wave is L1 : J D'+ (a - r)2. IJse the binomial theorem toD + (a - r)2 lzD. Thus
a2+2ar+12 r\ a2 -2ar*12tt :-2ar
2D 2DDor a positive integer. Solve for r. The result isSet this equal to (m + *)X, where m is zero
r-(m+))(Df\a)),.
89
Bright fringes occur rtan angle 0 suchthat dsin0 -ffi\, where d is the slit separation, ^
is the
wavelength in the medium of propagation, and m $ an integer. Near the center of the pattern the
angles are small and sin 0 can be approximated by 0 in radians. Thus 0 - m^ld and the angularseparation of two adjacent bright fringes is L0 -
^1d,. When the arrangement is immersed in
water the angular sep aration of the fringes becomes L0' : )- f d, where .\- is the wavelength inwater. Since,\- -
^ ln-, where TLra is the index of refraction of water, L?t - \ln*d - @qlTL,y1.
Since the units of the angles cancel from this equation we may substitute the angles in degrees
and obtain L?', - 0.30o f L.33 - 0 .23o .
93
(a) For wavelength A dark bands occur where the path difference is an odd multiple of ^12.That is, where the path difference is (2m+ I)^12, where m is an integer. The fourth dark band
from the central bright fringe is associated with m - 3 and is 7^12:7(500nm)12 - 1750nm.
(b) The angular position 0 of the first bright band on either side of the central band is given bysin 0 - xld, where d is the slit separation. The distance on the screen is given by LA - Dtan?,where D is the distance from the slits to the screen. Because 0 is small its sine and tangent are
very nearly equal and Ly - D sin 0 - D^f d.
Dark bands have angular positions that are given by sin 0 - (m + |)A I d, and, for the fourth darkband, m - 3 and srn2a- Q l2)^f d. Its distance on the screen from the central fringe is Lyo-Dtan7a: Dsrn2a- 7D^fzd. This means that D^ld: 2Lyal7 - 2(L68 cm)17 :0.48cm.Note that this is Ly.
226 Chapter 3 5
D2 + @+ r)z
D2+(a+r)2
97
(a) If 1 is the incident intensity then the radration pressure for total absorption is
I 1.4 x 103 W l^'rnrrr c 3.00 x 10s mls
(b) The ratio is
ratio - 4'57 x lo-6Pa :4.7 x ro-ll
1.0 x 105 Pa
99
IVlinima occur at angles 0 for which sin 0 - (m+ ))^1d,, where ,\ is the wavelength, d, is the slitseparation, and m is an integer. For the first minimum, m:0 and sin01 :
^12d. For the tenth
minimum, m:9 and sin Orc
The distance on the screen from the central fringe to a minimum is A: Dtan?, where D is the
distance from the slits to the screen. Since the angle is small we may approximate its tangent
with its sine and write A
minima is
4.67 x 10- 6 Pa .
Ar: D (19^a'' d, \ 2
and
\ d La (0. 150 x 10-3 *X 18.0 x 10-' m) r nn 1 n .7A-6: -6.oox1o-'m.
103
The difference in the path lengths of the two beams is 2r, so their difference in phase whenthey reach the detector is d : 4nr f
^, where l is the wavelength. Assume their amplitudes arc
the same. According to Eq. 35 -22 the intensity associated with the addition of two waves isproportional to the square of the cosine of half their phase difference. Thus the intensity ofthe light observed in the interferometer is proportional to cos2 (2rr I D. Since the intensity ismaximum when rmaximum intensity I,n and I - I,ncos2 (2nr l^).
Chapter 3 5 227
Chapter 3 5
2
The condition for a minimum of intensity in a single-slit diffraction pattern is a sin 0 : ffiA,where a is the slit width,
^ is the wavelength, and m is an integer. To find the angular position
of the first minimum to one side of the central maximum, set m : 1:
If D is the distance from the slit to the screen, the distance on the screen from the center of thepattern to the minimum is Ar - Dtan01- (3.00m)tan(5.89 x 10-4rad): 1.J67 x 10-3m.
To find the second minimum, set m - 2:
The distance from the pattern center to the minimum is az: Dtan?2: (3.00m)tan(1.178 x10-3rad)- 3.534 x 10-3m. The separation of the two minima is Ly: A2 - hI.7 67 nun - 1,.77 mm.
T7
(a) The intensity for a single-slit diffraction pattern is given by
r_r sin2ot-tr_
AZ, )
where a- (nol,\)sin9, a, is the slit width and ^
is the wavelength. The angle 0 is measuredfrom the forward direction. You want J - Irnf 2, so
sin2a-1 1
1a'
(b) Evaluate srn2 * and *'12 for a- 1.39rad, and compare the results. To be sure that 1.39radis closer to the coffect value for a than any other value with three significant digits, you shouldalso try I .385 rad and 1 .3 95 rud.
(c) Since e : (nol )) sin 0,
Now alr:1.391n -0.442, so
o -sin- ' (0'442'\ )r'r \ a )'
228 Chapter 36
,,r- sin-' (]) - sin-' (m
m
1 0-e10-
IX
89
JO
m
m
g-r0-
)l
o_sin-'(*)
5
i ) - 5.89 x 1o-4 rad./
IX
89
00a1.
20z: ,in-t
I
1
i
(d) For al^ - 1.0,
(e) For a,lx: 5.0,
(0 For al\: 10,
The angular separation of the two points of half intensity, one on either side of the center of the
diffiaction pattern, is
Lo - 20 -2 sin- ' (0'442^ )''' \ a )'
L0 - zsin- t (O .442 f I.0) : 0.9 16 rad: 52.5o
L0 - 2sin- t (O .442 15.0) - 0 .177 rad : 10. 1"
L0 - zsin- t (o .442 I I0) - 0.08 84 rad - 5.06o
2t
(a) Use the Rayleigh criteria. To resolve two point sources, the central maximum of the diffractionpattern of one must lie at or beyond the first minimum of the diffraction pattern of the other.
This means the angular separation of the sources must be at least 0 a : I .22^ f d, where ,\ is the
wavelength and d is the diameter of the aperture. For the headlights of this problem,
(b) If D is the distance from the headlights to the eye when the headlights arc just resolvable
and !, is the separation of the headlights, then (.- Dtan?p = D?n where the small angle
approximation tan? p = 0 p was made. This is valid if 0 n is measured in radians. ThusD - ll0n- (1.4m)l(I .34 x 10-4rad): 1.0 x 104m - 10km.
25
(a) [Jse the Rayleigh criteria: two objects can be resolved if their angular separation at the
observer is greater than 0 pdiameter of the aperture (the eye or mirror) . If D is the distance from the observer to the objects,
then the smallest separation (. they can have and still be resolvable is (. - D tan? n t D0 a,where 0p is measured in radians. The small angle approximation tan?p E 0p was made. Thus
o n:1'2?(550 x lo-e m)
- 1.3 x 1o-4 rad .- LtJ 5.0 x 10-3 m
( _T.22Dx: 1.22(8.0 x 1010mX550 x 10-em) - 1.1 x I07m: 1.1 x 104km.d 5.0 x 10-3 m
This distance is gteater than the diameter of Mars. One part of the planet's surface cannot be
resolved from another part.
(b) Now d - 5.1m and
n _ 1.22(8.0 x l01o m)(550 x 10-n *)(.- 5lr" -1.1 x104m-l1km.
Chapter 36 229
29
(a) The first minimum in the diffraction pattern is at an angular position 0, measured
center of the pattern, such that sin 0 - 1 .22^ I d,, where ^
is the wavelength and d is the
of the antenna. If f is the frequency, then the wavelength is
^ - 9- 3'oo x 191 m/s
- r.36x 1o-3 m.f 220 x 10e Hz
Thus
from the
diameter
10-2 m3 .02 x 10-3 rad .
The angular width of the central maximum is twice this, or 6.04 x 10-3 rad (0.346").
(b) Now ) - I.6 cm and d - 2.3 m, so
o _sin-' (
):x 10-'-)
1.22(I .6 x 10-2 m)- 8.5 x 10-3 rad .
2.3m
The angular width of the central maximum is 1.7 x I0-2 rad (0.97").
39
(a) The angular positions 0 of the bright interference fringes are given by d sin 0 - ffi\, where d isthe slit separation, .\ is the wavelength , and m ts an integer. The first diffraction minimum occurs
at the angle 0t given by asrn?1 - ^,
where a is the slit width. The diffraction peak extends
from -0t to +0t, So you want to count the number of values of m for which -0r ( 0 <what is the same, the number of values of m for which - sin d1 <
-lIo<5.00, so the values of m are m- -4, -3, -2, -1, 0, *1, *2, *3, and +4. There are ninefringes.
(b) The intens rty at the screen is given by
I - I,n (cos2 P) f lry)' ,
'\ a )where aFor the third bright interference fringe , dsin g - 3), so P - 3r rad and cosz pe, - 3naf d - 3n 15.00 - 0.600n rad and
fry)' - ( sin o'6oozr\ 2
- o .2ss .\ a ) \o.6oon )The intensity ratio is I I Ir. - 0 .255.
45
The ruling separation is d - I lg00 mm- t) : 2.5 xsuch that d sin 0 - ffi\, where
^ is the wavelength
230 Chapter 36
10-3 mm. Diffraction lines occur at angles e
and m is afi integer. Notice that for a given
order, the line associated with a long wavelength is produced at a greater angle than the lineassociated with a shorter wavelength. Take ) to be the longest wavelength in the visible spectrum(700 nm) and find the greatest integer value of m such that 0 is less than 90o. That is, find thegreatest integer value of m for which m), <3.57 , that value is m : 3. There are three complete orders on each side of the m : 0 order. Thesecond and third orders overlap.
51
(a) Maxima of a diffraction gratingpattern occur at angles d given by dsin 0 - ffi\, where d is the
slit separation, ,\ is the wavelength , and m rs an integer. The two lines are adj acent, so their order
numbers differ by unity. Let m be the order number for the line with sin e - 0.2 and mr I be the
order number for the line with sin 0 - 0.3. Then 0.2d: m), and 0.3 d - (m+1)^. Subtractthe firstequation fromthe secondto obtain0.ld-
^, or d- ) l0.l - (600x 10-e^)10.1 -6.0x 10-6m.
(b) Minima of the single-slit diffraction pattern occur at angles 0 given by o sin e - m\, where &
is the slit width. Since the fourth-order interference maximum is missing, it must fall at one ofthese angles. If a is the smallest slit width for which this order is missing, the angle must be givenbyosin0:),. Itisalsogivenbydsin0_ 4^,soa:d,14-(6.0x10-u*)f4-1.5x10-6m.(c) First, set 0 - 90" and find the largest value of m for which m),, <order that is diffracted toward the screen. The condition is the same as m <d,l^ - (6.0 x 10-u*) 1rc00 x 10-n-) - 10.0, the highest order seen is the m:9 order.
(d) and (e) The fourth and eighth orders are missing so the observable orders are m - 0, 1,2,3,5, 6, 7, and 9. The second highest order is the m- 7 order and the third highest order is the
m - 6 order.
6t
If a grating just resolves two wavelengths whose average is lave and whose separation is A^,then its resolving power is defined by R - )ave I L^. The text shows this is IY m, where .A\i is
the number of rulings in the grating and rn is the order of the lines. Thus )ave lL^ - IYm and
r 7 )uun 656.3 nm/\/ -- o
-tvm L), (1X0.18nm)
3 .65 x 103 rulings .
73
We want the reflections to obey the Bragg condition 2d srn 0 - ffi\, where I is the angle betweenthe incoming rays and the reflecting planes,
^ is the wavelength, and m is an integer. Solve for
0:
10-e*)
For m: The crystal should be turned 45o 14.4o - 30.6o clockwise.
For m:2 it gives 0 - 29.7o. The crystal should be turned 45" - 29.7" - 15.3" clockwise.
For m:3 it gives 0 - 48.1o. The crystal should be turned 48.1o - 45" - 3. 1o counterclockwise.
o_sin_,1#]
1 this gives 0 - 14.4o
: ,irr-t I
25
25(0.1
2(0lX
X
,g-r m)m
] - sin-t(o .24tom)
Chapter 36 231
For m- 4rtgives e -82.8o. Thecrystalshouldbeturned 82.80-45o -37.8o counterclockwise.
There are no intensity maxima for m >1 for m greater than 4. For clockwise turns the smaller value is 15.3" and the larger value is30.6o. For counterclockwise turns the smaller value is 3.1o and the larger value is 37.8o.
77
Intensity maxima occur at angles 0 such that d sin 0 - ffi\,rulings and ) is the wavelength. Here the ruling separation
5.00 x 10-6 m. Thus
where d is the separation of adjacent
is IlQ00mm-t) - 5.00 x 16-r mnl -
2.50 x 10-6 m
For m- 1., l - 2500nm; for m- 2,,\ - l250nm; for m 3r,\I : 625 nm; for m- 5, ) - 500offi, and for n'L: 6,
^ - 4I7 rlm. Only the last three are in
the visible range, So the longest wavelength in the visible range is 625ilil, the next longest is500 trffi, and the third longest is 4L7 nm.
79
Suppose mo is the order of the minimum for orange light, with wavelength Ao, and mbe isthe order of the minimum for blue-green light, with wavelength ),as. Then a sin 0 _ rno).o and
astn? : mbe),ug. Thus ffio\oThe smallest two integers with this ratio are mbe - 6 and TrLo - 5. The slit width is
TTto\o 5(600 x 10-9 m ^) n ", ,.-10,:#- :3.0X10-'m.
Other values for TrLo and mbe are possible but these are associated with a wider slit.
81
(a) Since the first minimum of the diffraction pattern occurs at the angle 0 such that sin e - \lo,where
^ is the wavelength and a, is the slit width, the central maximum extends from 0r -
- sin-t(Xlo) to 0z: *sin-t(^lo). Maxima of the two-slit interference pattern are at angles 0such that sin d - mAf d, where d is the slit separation and m is an integer. We wish to knowthe number of values of m such that sin-t1*X1a7 hes between - sin-t(Xlq and +sin(^lo) or,
what is the same, the number of values of m such that rn I d, lies between -l I o and +l f a. Thegreatest m can be is the greatest integer that is smaller than dlo: (I4 prn)l(z.0 pm):7. (Them : 7 maximum does not appear since it coincides with a minimum of the diffraction pattern.)There are 13 such valuesi 0, + 1 , +2, *3; *4; +5, and *6. Thus 13 interference maximaappear in the central diffraction envelope.
(b) The first diffraction envelope extends from 0t: sin-t(Xlo) to 0z: sin-t(ZLlil. Thus wewish to know the number of values of m such that m I d, is greater than L I o and less than 2l o.
Since d - 7.0a, m can be 8, 9, 10, 11, 12, or 13. That is, there are 6 interference maxima in the
first diffraction envelope.
232 Chapter 36
d stn0 (5.00 x 10-u rn) sin 30.0o
93
If you divide the original slit into l/ strips and represent the light from each strip, when itreaches the screen, by a phasor, then at the central maximum in the diffraction pattern you add
l/ phasors, all in the same direction and each with the same amplitude. The intensity there
is proportional to 1Jz. If you double the slit width, you need 2lV phasors if they are each to
have the amplitude of the phasors you used for the naffow slit. The intensity at the central
maximum is proportional to (2LD2 and is, therefore, four times the intensity for the naffow slit.The energy reaching the screen per unit time, however, is only twice the energy reaching it per
unit time when the naffow slit is in place. The energy is simply redistributed. For example, the
central peak is now half as wide and the integral of the intensity over the peak is only twice the
analogous integral for the naffow slit.
95
(a) Since the resolving power of a grating is given by R - )/A^ and by IY nt , the range ofwavelengths that can just be resolved in order m is A) : ),lIVm. Here l/ is the number
of rulings in the grating and ) is the average wavelength. The frequency f is related to the
wavelengthby f^:c,where cisthespeedof light. Thismeans f L,\+^AJ:0, so
A.\: -+ Lf : -^2 Lf .
JC
where f - cl ^
was used. The negative sign means that an increase in frequency coffesponds to
a decrease in wavelength. We may interpret Lf as the range of frequencies that can be resolved
and take it to be positive. Thent2^ a.f :c
,\
l{*C
and
Lf:l{m),
(b) The difference in travel time for waves traveling along the two extreme rays is Lt : LL f c,
where LL is the difference in path length. The waves originate at slits that are separated by(,Af - I)d,, where d is the slit separation and tf is the number of slits, so the path difference isLL - (ltr - l)d, sin 0 and the time difference is
A,- (ff - Dd, sin 0
If t/ is large, this may be approximated by Lt - (lV d lc) sin 9. The lens does not affect the
travel time.
(c) Substitute the expressions you derived for Lt and Lf to obtain
^f Lt:(#) (ry) :
The condition d sin 0 - m), for a diffraction line was used
dsrn?-1m\
to obtain the last result.
Chapter 3 6 233
101
The dispersion of a grating is given by D = dO / dA, where 0 is the angular position of a line associated with wavelength A. The angular position and wavelength are related by R. sin 0 = mA, where R. is the slit separation and m is an integer. Differentiate this with respect to 0 to obtain (dO / dA) R. cos 0 = m or
D = R.O = m R.A R. cos 0
Now m = (R./ A) sin 0, so D = R. sin 0 __ tan_O
R. A cos 0 A The trigonometric identity tan 0 = sin 0/ cos 0 was used.
234 Chapter 36
Chapter 37
l1
(a) The rest length Ls (: l30m) of the spaceship and its length L as measured by the timingstationarerelatedbyL:Loll:L0\m,where1:Il\Fryandp:ulc.ThusL- (130m)@ -87.4m.(b) The time interval for the passage of the spaceship is
. L 87.4mLt_ r-J- -3.94 x 1o-7s.u (0140)Q.sg7s x 108 mA)
19
The proper time is not measured by clocks in either frame ,S or frame S' since a single clockat rest in either frame cannot be present at the origin and at the event. The full Lorcntztransformation must be used:
r,':1lr-utlt': jlt- grlrl
,
where p - ulc- 0.950 and 1: tl\m - rl@ - 3.2026. Thus
tr'- (3.2026) [tOO x l03m - (0.950)(2.9979 x 108 mls)(200 x 10-u r]
- 1.38 x 10s m - 138 km
and
t' - (3 .2026) 10-6 s -(0.950x 100 )
-3 .74 x 10-4 s - -3 74 p,s .
2.9979 x 108 m/s
29
IJse Eq. 37 -29 with It' - 0 .40c and ?J :0.60c. Then
0.40c + 0 .60c?t: - -0.81c.1 + (0 .a0Q(0.60c) I "'
33
Calculate the speed of the micrometeorite relative to the spaceship. Let ,S' be the reference framefor which the data is given and attach frame ,S to the spaceship. Suppose the micrometeorite isgoing in the positive n direction and the spaceship is going in the negative r direction, both as
viewed from S' . Then, in Eq. 37 -29, ?.,0' - A.82c and u
froo x t:
x 103m
Chapter 37 235
is the velocity of S' relative to S. Thus the velocity of the micrometeorite in the frame of the spaceship is
u' + v 0.82e + 0.82e 09 u = = = . 806e.
1 + u'v I e2 1 + (0.82e)(0.82e)1 e2
The time for the micrometeorite to pass the spaceship is
L 350m tlt = - = = 1.19 X 1O-6 s.
u (0.9806)(2.9979 X 108 m/s)
37
The spaceship is moving away from Earth, so the frequency received is given by
H f = fay 1+73'
where fa is the frequency in the frame of the spaceship, {3 = v I e, and v is the speed of the spaceship relative to Earth. See Eq. 37 - 31. Thus
f = (100 MHz)
39
1 - 0.9000 = 22.9 MHz. 1 + 0.9000
The spaceship is moving away from Earth, so the frequency received is given by
H f = fay 1+73'
where fa is the frequency in the frame of the spaceship, {3 = vie, and v is the speed of the spaceship relative to Earth. See Eq. 37-31. The frequency f and wavelength A are related by fA = e, so if AO is the wavelength of the light as seen on the spaceship and A is the wavelength detected on Earth, then
~+{3 A = AO -- = (450nm)
1-{3
1 + 0.20 0 = 550nm.
1 - 0.2
This is in the yellow-green portion of the visible spectrum.
43
Use the two expressions for the total energy: E = me2 + K and E = ,me2, where m is the mass of an electron, K is the kinetic energy, and, = 1 I J 1 - {32. Thus me2 + K = ,m2 and
K (100.000 x 106 eV)(1.602176462J/eV) 9 69 ,=1+-=1+ =16 5
me2 (9.10938188 x 10-31 kg)(2.99792458 x 108m/s)2 ..
236 Chapter 37
Now 'y2 - p'), so
required is the increased in
@ ldz} Let u1 be the initial
mc2 mc2
where mcT : 938 MeV was used.
77
(a) Let u be the speed of either satellite,transformation equation the relative speed
(2.00 MeV)z 1 2(2.00 MeV)(0.5 1 1 MeV) : 2.46MeV .
the energy of the proton. The energy is
speed and u2 be the final speed. Then the
1
1
@ - 0 .999 987 .
7l
The
53
The energy equivalent of one tablet rs mcT - (320x 10-ut gX2.gg79 x 108 mls)2 : 2.88 x 10r3 J.
This provides the same energy as (2.88 x 1013 Dl(3.65 x L07 JIL)_ 7.89 x 105 L of gasoline.
The distance the car can go is d - (7.89 x 105 LXI2.75km lL) - 1.01 x 107km.
energyofthee1ectronisgivenbyE_mc2lre,whichyields
c: -0.gggg9994cxc
for the speed u of the electron. In the rest frame of Earth the trip took time t_ 26y. A clocktraveling with the electron records the proper time of the trip, so the trip in the rest frame of the
electron took time tt : t l^y. Now
.y : r_- 1533 MevXl .602 x 10_ 13 J/x,fev) : 3.0 x 103mc2 (9.11 x 10-rt kg)(2.9979 x 108 m/s)
and t' - (26y)/(3.0 x 103) : 8.7 x 10-'y. The distance traveled is 8.7 x 10-'ly.
73
Start with (pr)z : K2 +2Kmc2, where p is the momentum of the particle, /( is its kinetic energy,
and m is its mass. For an electron mc2 - 0.511MeV, so
pc: -Thus p:2.46MeV/c.
75
The workmcT llt
938 MeV
given by trwork is
189 MeV,938 MeV
W1 - (o.gg5o)2
relative to Earth. According to the Galilean velocityis urer : 2u : 2(2.7 x 104 km/h : 5 .4 x 104 krnf h.
lH'
(, l r)' (o.gg60)2
Chapter 37 237
(b) The correct relativistic transformation equation is
The fractional error is
2v Vre1 = 2 .
1+~ c2
2v - Vre1 1 fract err = = 1 - ---0-2 .
2v 1 + ~ c2
The speed of light is 1.08 x 109 km/h, so
1 fract err = = 6.3 X 10-10 .
1 + (2.7 X 104 km/hi (1.08 x 109km/h)2
238 Chapter 37
Chapter 3I
7
(a) Let ^R be the rate of photon emission (number of photons emitted per unit time) and let tr be
the energy of a single photon. Then the power output of a lamp is given by P - Rtr if all the
power goes into photon production. Now E - hf - hcf ),, where h is the Planck constant, f is
the frequency of the light emitted, and ^
is the wavelength. Thus P - Rhcl ), and R - ^P
lhc.The lamp emitting light with the longer wavelength (the 700-nm lamp) emits more photons per
unit time. The energy of each photon is less so it must emit photons at a greater rate.
(b) Let ,R be the rate of photon production for the 700 nm lamp Then
za )P (700 x 10-e m)(400J/s)H-Lw hc (6.626x 10- 34 J.sX2 .ggigx 108 -/t - I'41 x L021 photon/s '
t7The energy of an incident photon is tr - h ffrequency of the electromagnetic rudration, andmost energetic electron emitted is Krn - E - O
sodium. The stopping potential Vo is related toeVo-(hclD- O and
hc f ),, where h is the Planck constant, f is the
,\ is its wavelength. The kinetic energy of the
- (hcl D - O, where O is the work function forthe maximum kinetic energy by eVo: K,n so
, h, (6.626 x 19-34 J. sX2 .9979 x 108 m/s) 1 n r n_jI - "Vr. a:
Here eVo: 5.0 eV was used.
2l(a) The kinetic energy Krn of the fastest electron emitted is given by Krn: hf -O - (hclD-O,where O is the work function of aluminum, f is the frequency of the incident radtation, and ,\is its wavelength. The relationship f : cl
^ was used to obtain the second form. Thus
K,n(200 x 10-e mX | .602 x 10- ts J lev)
(b) The slowest electron just breaks free of the surface and so has zero kinetic energy.
(c) The stopping potential Vo is given by Krn: €Vo, so Vo: Krnl": (2.00eV) l" - 2.00V.
(d) The value of the cutoff wavelength is such that Krn - 0. Thus h"l,\ - O or
I-UO (4.2 eVX | .602 x 10- re J I eY)
Chapter 38 239
If the wavelength is longer, the photon energy is less and a photon does not have sufficientenergy to knock even the most energetic electron out of the aluminum sample.
29
(a) When a photon scatters from an electron initially at rest, the change in wavelength is givenby A,\ - (hl*cxl - cos 0), where m is the mass of an electron and O is the scattering angle.Now hf mc-2.43 x 10-12m-2.43pffi, so Al - (2.43pmx1 - cos30o):0.326pm. The finalwavelength is )' : ) + A,\ - 2.4pm+ 0.326pm - 2.73pm.
(b) Now A.\ - (2.43pmxl - cos 120"): 3.645pm and X -2.4pm+3.645pm - 6.05pm.
43
Since the kinetic energy K and momentum p are related by K - p2 lzm, the momentum of theelectronisp-'MandthewavelengthofitsmatterwaVeis^-hlp-hf'ffi.ReplaceK with €V, where V is the accelerating potential and e is the fundamental charge, to obtain
6.626 x 10-34J.s
2(9. 109 x 10-:tkg)(1 "602 x 10-teCX25.0 x 103V)
-7.75 x 10-12m- 7.7Spm.
47
(a) The kinetic energy acquired is K - qV, where q is the charge on an ion and V is theaccelerating potential. Thus K - (I .602 x 10-1e CX300V) - 4.80 x l0-r7 J. The mass ofa single sodium atom is, from Appendix F, m- (22.9898g1mo1)l(6.02 x 1023 afimfmol)3.819 x 10-23 g - 3.819 x 10-26 kg. Thus the momentum of an ion is
p:17: 2(3.819 x 10-20 kg)(4.80 x 10-17 J) - 1.91 x 10-2t kg.m/s.
(b) The de Broglie wavelength is
\h{2meV
K- I
2m
- 6.92
10ru\6.626 x 10-34 J .
590 x 10-e m)
, h 6.63x10-34J.s)- 'u: t ="r""'-'^^"-,. ' ", -3.47 x 10-13m.p 1.91 x I0-2t kg .m/s
49
Since the kinetic energy K and momentum p ate related by K - p2 lzm, the momentum of theelectronisp-\ffiandthewavelengthofitsmatterwaVeis^-hlp-hf,ffi.Thus
(*)
x l0)'
s)
2(9.11 x-2s J - 4.33
59
The angular wave number k is relatedrelated to the particle momentum p by
240 Chapter 38
x 10-6 eV.
to the wavelength l by k,\ - hlp, so k:2rpf h.
:2rfAandtheNow the kinetic
wavelength isenergy K and
61
For
themomentumarere1atedbyK-p2l2m,wheremisthemaSSofthepartic1e.Thusp:\ffiand
k- ZnrffiR
Ut - (Jo, Schrodinger's equation becomes
d,zr! , 8r2m r1.,+ ,", ltr- Lb]qr'-0.o,tr- n'
second derivative is d,2rb ld*' - -\tz{o"ik" - -kzr!. The result is
-k2q)+ryW -uoltb:0.-/ ' 7^r2
Substitute * - rho"ik* . The
Solve for k and obtain
67
(a) If rn is the mass of the particle and
barrier of height U and width L is given
where
If the change LU in U is small (as it is), the change in the transmission coefficient is given by
Lr:# L(-r: -zLr # L(.r .
Now
tr is its energy, then the transmission coefficient for aby
T _ e-2kL
dk :d(I 2yffi 2(u - E) 2(u - tr)
Thus
LT: -LTK AU\(J-E'
For the data of Sample Problem 38-7,ZkL:10.0, so kL:5.0 and
LT kL*- -(5.g; (o'o1oX6'8ev)
- -o .zo.T - Lr - E- -(l'u)6.gev-5.1ev
There is a 20% decrease in the transmission coefficient.
8n2m(U
Chapter 38 241
(b) The change in the transmission coefficient is given by
LT : # LL: -2ke-zkL LL - -zkr LL
andLT : -2k LL: - 2(6.67 x 10e m-tX0.010X750 x l0-" *) : -0.10.T
There is a I0% decrease in the transmission coefficient.
(c) The change in the transmission coefficient is given by
LT: #LE: -ZLe-2'L#LE: -2LT#LE.Now d,kf dtr - -dkld,U - -klz(U - E), so
+: kL#- (5 s;(o'oloxs'ltv) - 0.r5.
There is a 15% increase in the transmission coefficient.
79
The uncertaintyin the momentum is Lp: mLu - (0.50kg)(1.0mls):0.50kg.ffi/s, where Auis the uncertainty in the velocity. Solve the uncertainty relationship Lr Lp > n for the minimumuncertainty in the coordinate r: Lr -nf Lp - (0.60J.s) l2n(0.50kg.mls):0.19m.
242 Chapter 38
Chapter 3 9
13
The probability that the electron is found in any interval is given by P _ f lVl' d*, where the
integral is over the intenral. If the interval width Lr is small, the probability can be approximated
by P : lrfl' L*, where the wave function is evaluated for the center of the interval, say. For an
electron trapped in an infinite well of width L, the ground state probability density is
so P-ff)sin2 (T)
(a) Take L:100pffi, r - 25pm, and Lr - 5.0pm. Then
P-[ffi] sin2L*#] -ooso
(b) Take L - 100pffi, r - 50pm, and Lr - 5.0pm. Then
p lzfs'0 Pm)1 ' fzr(so Pm)l - o. to .
(c) Take L:100pffi, r:90pm, and Ln - 5.0pm. Then
P-rry] sin2[#ffi] -oooe5L loopn
lrhl': lrrn'(T) ,
En*nv: #ln.frl
25
The energy levels are given by
where the substitutions L*- L and La _ 2L were made. In units of h'l8mLz, the energy
levels are given by nI* "1 f
4. The lowest five levels are 81,1 : I.25, Et,2 - 2.00, EtJ - 3.25,E2; : 4.25, and Ez,z: Er,4 - 5.00. A little thought should convince you that there ate no otherpossible values for the energy less than 5.
The frequency of the light emitted or absorbed when the electron goes from an initial state z to afinal state f is f - (Et - Ei,)lh and in units of hl8mLz is simply the difference in the values ofn?.+nll+ for the two states. The possible frequencies are 0.75 (L,2--+ 1,1), 2.00 (1,3 -+ 1,1),
3.00 (2,1 ------+ 1,1), 3.75 (2,2 ----+ 1,1), 1.25 (1,3 + 1,2),2.25 (2,1 ----J I,2),3.00 (2,2 ----J 1,2),
1.00 (2,I ->
1,3), I.75 (2,2---+ 1,3), 0.75 (2,2-----+ 2,1), all in units of hl8mL2.
Chapter 39 243
There are 8 different frequencies in all. In units of h l8mL2 the lowest is 0.7 5, the second lowestis 1 .00, and the third lowest is I .25. The highest is 3.7 5, the second highest is 3.00, and thethird highest is 2.25.
33
If kinetic energy is not conserved some of the neutron's initial kinetic energy is used to excitethe hydrogen atom. The least energy that the hydrogen atom can accept is the differencebetween the first excited state (n : 2) and the ground state (na state with principal quantum number n is -(13.6eY)f n2, the smallest excitation energy is13.6eV-(13.6eV)lQ)2- l0.2eV. The neutron does not have sufficient kinetic energy to excitethe hydrogen atom, so the hydrogen atom is left in its ground state and all the initial kineticenergy of the neutron ends up as the final kinetic energies of the neutron and atom. The collisionmust be elastic.
37
The energy E of the photon emitted when a hydrogen atom jumps from a state with principalquantum numb er u to a state with principal quantum numb er (, is given by
E:A(; #)where A - 13.6eV. The frequency f of the electromagnetic wave is given by f : Elh and thewavelength is given by
^ - cl f. Thus
the Balmer series,
1 and the shortest
proposed wave function
1- T- E- A (L 1\
^ c hc hc\Zt ur)
The shortest wavelength occurs at the series limit, for which u - co. For(. : 2 and the shortest wavelength is .\s : 4lr,c I A. For the L5rman series , t :wavelength is ),r- hclA. The ratio is ),n I \t, : 4.
43
The
where a, is the Bohr radius.
show that the result is zero.
SO
is
,r!,:+ e_ rlot/ 'ftot
/2" 1
Substitute this into the right side of Schrodinger's equation andThe derivative is
drh:- I o-r/ad, - fiTas/2e
)
*, drb -
r2 ^-, / or- dr:
244 Chapter 39
and1
7Now the energy ofby o: h2rof nme2,
#(*#): #ll.;] e-"::ll.;] ,b
the ground state is given by tr - -me4 l8elhz and the Bohr radius is given
so E - -e2 f 8nesa. The potenttal energy is given by (Ji - -e2 f 4nesr, so
ut,h - ry |h. hll, : ry h[+ .?],t,
- trmeT t_1 z1 I l- 1 z1
n+o l-an;l 'Yrs - ; L-;
*;i 'b
ryw-
The two terms in Schrodinger's equation obviously cancel and the proposed function ,lt satisfies
that equation.
47
The radial probability function for the ground state of hydrogen is P(r) - 7412 I ot)e-2r /o" , where
a, is the Bohr radius. (See Eq. 39-44.) You want to evaluate the integral /r"" P(r)d,r. Eq. 15
in the integral table of Appendix E is an integral of this form. Set n- 2 and replace a in the
given formula with 2lo and r with r. Then
P(r) dr T2 e-2, /o d,r4
a3
4f*: *J,l,*49
(a)
(2 I a)3
th21o is real. Simply square it to obtain the probability density:
It^ol'_ Le-'/o cos2 o.32naj
(b) Each of the other functions is multiplied by its complex conjugate, obtained by replacing 'i
with -i in the function. Since eif e-if : e0 - I, the result is the square of the function withoutthe exponential factor:
lrhr,*, lt - #e-'/o srn2 o
lrhr,-,lt _ #e-'/o srn2 o .
The last two functions lead to the same probability density.
(c) For IrL4
is greatest along the z axis, and for a given distance from the nucleus decreases in proportion tocos2 d for points away from the z axis. This is consistent with the dot plot of Fig. 39-24 (a).
For TrLp : + 1 the radial probability density decreases strongly with distance from the nucleus, is
greatest in the fr,, a plane, and for a given distance from the nucleus decreases in proportion tosrn2 0 for points away from that plane. Thus it is consistent with the dot plot of Fig . 3g-24(b).
Chapter 39 245
(d) The total probability density for the three states is the sum:
l,hz,ol,+l,bzr*r|2+l,b,I-|2:#e_,/o[,o,,0+|,,n,0+
r2:
-e-r/a.
32na)
The trigonometric identity cos2 0 + stn2 0
depend on 0 or 0. It is spherically symmetric.
57
1,,"'rl
The wave function is { : tE "-kr. Substitute this function into Schrodinger's equation,
Since d,2rh ld*'- tFCtt'e-k* - k2th, the result is
h2 k2
ffirb+Uorh - E1h.
The solution for k is
Thus the function given for th is a solution to Schrodinger's equation provided k has the valuecalculated from the expression given above.
ry(to_ E).
246 Chapter 39
Chapter 40
2
Since L2 = L; -+ L; -+ L;, J Li + L~ = JL2 - L;. Replace L2 with IV, + 1)n2 and Lz with'mjln
to obtain
JLi+L~= JR(R+l)-'m~n.
For a given value of R, the greatest that 'mjl can be is R, so the smallest that J L~ + L~ can
be is jR(R + 1) - R2n = V£n. The smallest possible magnitude of'mjl is zero, so the largest
J Li + L~ can be is JR(!! + l)n. Thus
11
(a) For R = 3, the magnitude of the orbital angular momentum is L = JR(R -+ l)n = J3(3 + l)n = JI2Ti, = 3.46Ti,.
(b) The magnitude of the orbital dipole moment is /--Lorb = JR(R + 1)/--LB = VU/--LB = 3.46 /--LB·
(c) The largest possible value of'mjl is R, which is +3.
Cd) The corresponding value of the z component of the angular momentum is Lz = fn = +3n.
(e) The direction of the orbital magnetic dipole moment is opposite that of the orbital angular momentum, so the corresponding value of the z component of the orbital dipole moment is
/--Lorb, z = -3/--LB.
(f) The angle () between i and the z axis is
1 Lz 1 3n 0 () = cos- - = cos- -- = 30.0 . L 3.46n
(g) The second largest value of'mjl is 'mp = R - 1 = 2 and the angle is
1 Lz 1 2n 0 e = cos- - = cos- -- = 54.7 . L 3.46n
(h) The most negative value of 'mp is -3 and the angle is
L -3n e = cos- 1 ~ = cos- l -- = 1500 •
L 3.46n
Chapter 40 247
15
The acceleration 1s
0.t cos 0) (dB ldr)
where M is the mass of a silver atom, p is its magnetic dipole moment, B is the magnetic field,and 0 is the angle between the dipole moment and the magnetic field. Take the moment and thefield to be parallel (cos 0 - 1) and use the data given in Sample Problem 40-1 to obtain
(9.27 x 10-'ollrxl.4x 103T1*)a- -7.21 x 104*lrt
terms of the quantum nurnbers frr, fra, and fr2, the single-particle energy levels are given by
En*,rlsrlz: # (n'* * r?r ', ,2").
The lowest single-particle level coffesponds to nr: 1, na: 1, and nz: 1 and is ErJ1 -3(h2 f 8mL2). There are two electrons with this energy, one with spin up and one with spindown.
The next lowest single-particle level is three-fold degenerate in the three integer quantum numbers.The energy is 8t1,2: Er,2,t: E2J,r - 6(h'l8*L\. Each of these states can be occupied by a
spin up and a spin down electron, so six electrons in all can occupy the states. This completesthe assignment of the eight electrons to single-particle states. The ground state energy of thesystem is Es - (2)(r(h2 lSmL') * (6X6) (h' lSmL') - (42)(h, lB*L,).
31
(a) All states with principal quantum numb er n - I are filled. The next lowest states have rL : 2.
The orbital quantum number can have the values (. - 0 or 1 and of these, the t - 0 states have thelowest energy. The magnetic quantum number must be m4 - 0 since this is the only possibilityrf ( : 0. The spin quantum number can have either of the values TTLs: -+ or +|. Since thereis no external magnetic field, the energies of these two states are the same. Thus, in the groundstate, the quantum numbers of the third electron are either n- 2, (,- 0, wlp
tu : 2, | : 0, TfL1: 0, TTLs - +t.(b)Thenextloweststateinenergyis ann-2,(: l state. A11 n:3 statesarehigherinenergy.The magnetic quantum number can be TTLy: -1, 0, or +1; the spin quanfum number can be
TTLs: -+ or +|. If both external and internal magnetic fields can be neglected, all these states
have the same energy. The possible states are (2, I, l, +Il2), (2, t, 1, -1 l2), (2, 1, 0, +ll2),(2, 1, 0, -1 l2), (2, I, -1 , +ll2), and (2, l, -l , Il2).
37
(a) The cut-off wavelength )*in is charucteristic of the incident electrons, not of the targetmaterial. This wavelength is the wavelength of a photon with energy equal to the kinetic energy
248 Chapter 40
MF
a,- :M
25
In
of an incident electron. Thus
) - Y- (6;926 x 1=0 _11 l,' :l(3'00
x, l98 T1s) - 3.ss x 10-r m - 3s.s pm.' LE (35 x 103 eVX1.60 x 10-1eJ/eV)
(b) A K* photon results when an electron in atarget atom jumps fromthe l-shell to the K-shell.The energy of this photon is 25.5 1 keV - 3 .56 keV - 21 .95 keV and its wavelength is
\ _ hc (6.626 x 10-34 J. sX3.00 x 108 m/s)A: Lt
(c) A K p photon results when an electron in a target atom jumps from the M-shell toK-shell. The energy of this photon is 25.5 1 keV - 0.53 keV : 24.98 keV and its wavelength
, hc (6.626 x 10-34 J . sX3.00 x 108 m/s))- +- -4.96 x 10-rrm:49.6pm.LE (24.98 x 103 eVXl.60 x 10-te JleY)
4t
Since the frequency of an x-ray emission is proportional to (Z I)2, where Z is the atomicnumber of the target atom, the ratio of the wavelength Axu for the K o line of niobium to the
wavelength lcu for the Ko line of gallium is given by )Nal\cu: (Zca- D'lGr*- I)', where
ZNa is the atomic number of niobium (41) and the Zcu is the atomic number of gallium (3 1).
Thus )Nb I \cu - (3 0)' (4q2 - 9 f 16.
49
The number of atoms in a state with energy tr is proportional to e- E lr' , where T is the
temperature on the Kelvin scale and k is the Boltzmann constant. Thus the ratio of the number
of atoms in the thirteenth excited state to the nurnber in the eleventh excited state is
TLtz - n_ LE /kr
ut-e )
where LE is the difference in the energies: LE- En En: 2(L.2eY): 2.4eV. For the
given temperature , kT - (8 .62 x l0-2 eV/KX2000 K) : 0.1 724eV. Hence,
TLtz -1
Tt tt
6s
(a) The intensity at the target is given by I - P lA, where P is the power output of the source
and A is the area of the beam at the target You want to compute I and comp are the result with108 wl^'.The beam spreads because diffraction occurs at the aperture of the laser. Consider the part ofthe beam that is within the central diffiaction maximum. The angular position of the edge isgiven by sin d - 1.22^ld, where l is the wavelength and d is the diameter of the aperture (see
the
is
Chapter 40 249
Problem 50). At the target, a distance D away, the radius of the beam is r - D tan?. Since e issmall, we may approximate both sin0 and tan? by 0, in radians. Then r: D0:1.22D^ld, and
P PdzI-I,, rr' ?r(I .22D^)2
:2.1 x 105 W l^' ,
not great enough to destroy the missile.
(b) Solve for the wavelength in terms of the intensity and substitute f - 1.0 x 108 W f m2:
d,E\_ It\- nzD\ "I
4.0m
- L.4 x 10-7 m-1 .22(3000 x 103 m)
I40 nm.
7l(a) The length of the pulse is I : c Lt, where Lt is its duratioll. Thus L - (3.00 x 10* -/sxl0 x10-15 s - 3.0 x l0-6m. The number of wavelengths in the pulse is ,Af
l0- 6
^) l(500 x 10-e -) : 6.0.
(b) Solve for X:
x-#-r.ox 1or4s.
Since I year contains 356 days, each day contains 24hours, and each hour contains 3600 seconds,
the value of X in years is
(5.0 x 106W(4.0ttt)2
Tr 1t.2213000 x 103 m)(3.0 x 10-u *)] '
1.0 x 10la s :3.2 x 106y.(365 .2 d)(24h1dX3600 s/h)
5.0 x 106 W
zr(1.0x108W1*')
250 Chapter 40
Chapter 4L
1(a) At absolute temperature T - 0, the probability is zero that any state with energy above theFermi energy is occupied.
(b) The probability that a state with energy E is occupied at temperature T is given by
P(E) - ,
where k istheBoltzmannconstantand trp istheFermienergy. Now, tr-Ep -0.062eVand(E - Ep) lkf - (0 .062 eV)/(8.62 x 10-t ,Y lKX320 K) - 2.248, so
P(E)-+ -0.0956."2.248
I I
See Appendix B for the value of k.
11
The Fermi-Dirac occupation probability is given by Pro - I I ("^u lkr + 1) and the Boltzmannoccupation probability is given by Ps : e- Ltr /kr . Let f be the fractional difference. Then
^- LE /kr 1
$_Ps-Pro e t -W
J P" e- LE /k'r- '
[Jsing a common denominator and a little algebra yields
t - e-LE/kr
e-LElkr + 1 '
The solution for e- LE lkr is
e- r.E/kr: L1fTake the natural logarithm of both sides and solve for T. The result is
ry1 - Ltrrm.
,UIII \I f )
(a) Put f equal to 0.01 and evaluate the expression for T:
q-1 _ (1.00eVXl.60x 10-relleVlf
(1.38x lo-23 _ 2'5ox1o3K'
Chapter 41 251
(b) Put f equal to 0.10 and evaluate the expression for ?:
T- (1 .00 eVXl .60 x 10- te I 1eV|
(0. 1 2I)(6.626 x 10-34 J. s)2
- 5.30 x 103 K.(r.38 x LI-L3J/K) ln (#ffi)
t7(a) According to Appendix F the molar mass of silver is 107.870 glmol and the density isp : 10. a9 glc*3. The mass of a silver atom is
M _ 107.870 x 10-3 kg/mol
6.022 x 1023 mol- I
The number of atoms per unit volume is
p 10.49 x 103 kgl*t F (1 z. - no 1n-M
Since silver is monovalent this is the same as the number density of conduction electrons.
(b) The Fermi energy is
Eprn 9.109 x 10-3t kg
5.49 eV .- 8.80 x 10-le ; -
(c) Since Ee: *,*u'r,
Mup: tl -Vm
(d) The de Broglie wavelength is
hh tr^a: : : -\ /1p p TnLU p (9. 109 x 10-31 kg)( I .39 x 106 m/s) e 'H-'
(5.86 x 1028m r)2/3
x 10-lo m.
31
(a) Since the electron jumps from the conduction band to the valence band, the energy ofthe photon equals the energy gap between those two bands. The photon energy is given byhf : hcf A, where f is the frequency of the electromagnetic wave and.\ is its wavelength. ThusEs - hcf ), and
, hc (6.63 x 10-34 J.sX3.00 x 108m/s) _ ^ ^r 11 1A_ - __
Es (5.50 ev)(l.60 x lo-re J/ev) 2'26 x 10-' m : 226 nm '
Photons from other transitions have a greater energy, so their waves have shorter wavelengths.
252 Chapter 4I
2(8.80 x 10-1r J)
9.109 x 10-:t kg
6.626 x 10-34J.s
(b) These photons are in the ultraviolet portion of the electromagnetic spectrum.
37
Sample Problem 4I -6 gives the fraction of silicon atoms that must be replaced by phosphorus
atoms. Find the number the silicon atoms in 1.0 g, then the number that must be replacod, andfinally the mass of the replacement phosphorus atoms. The molar mass of silicon ts 28.08 6 gf mol,so the mass of one silicon atom is (28.086g1^01')l(6.022 x 1023 mol-l) - 4.66 x 10-23 g
and the nurnber of atoms in 1.0g is (1.0 g)l(4.66 x t0-23 g) - 2.I4 x 1022. According toSample Problem 41 -6 one of every 5 x 106 silicon atoms is replaced with a phosphorus atom.This means there will be (2.I4 x 1022)16 x 106) - 4.29 x 1015 phosphorus atoms in 1.0g ofsilicon. The molar mass of phosphorus is 30.97 58 g/mol so the mass of a phosphorus atom is(30.9758g/-ol)l$.022 x 10-23 mol-t) - 5.14 x 10-" g.The mass of phosphorus that must be
added to 1.0g of silicon is (4.29 x l0ltxS .14 x IA-23 g): 2.2 x 10-7 o
39
The energy received by each electron is exactly the
the conduction band and the top of the valence band
be excited across the gap by a single 662-keV photonSince each electron that jumps the gap leaves a holehole pairs that can be created.
difference in energy between the bottom of(1.1 eV). The number of electrons that can
is l/ - (662x 103 eYl(1.1eV) - 6.0 x 105.
behind, this is also the number of electron-
4e
(a)
Its derivative is
According to Eq. 4I-6P(E) - e@-Ep'')lkr + 1
.
-1 e@-EP)/kTdE l"(t-EF)/kr + Il' kT '
For E - Ee, e(tr-trF)/kr : e0: L, so the derivative at E + Ep is -1 f 4kT.
(b) Represent the tangent line by P - A+ BE, where A and B are constants. We want P - ll2and dPldU: -1l4kf for E- Ep. This means A+ BEe: Il2 and B- -1l4kf. Thesolution for A is A- (Il2)+ (EelLkT). Thus P- (Il2) (E Ee)l\kT. The intercept isfound by setting P equal to zero and solving for E. The result is E - E p + 2kT .
d,P(e)
Chapter 41 253
Chapter 42
13
(a) The de BroglieThe kinetic energy
wavelength is given byK and momentum ate ^
- h lp, where p is the magnitude of the momentum.related by Eq. 37 *54, which yields
pc- - (200MeV)2 a2(200MeV)(0.511MeV) - 200.5 MeV.
Thushc 1240 eV.Ilm\ - 1!--" " ' ^^^^- - 6. 18 x 10-6 nm - 6. 18 fm .t\- W 200J.106.V-(
(b) The diameter of a copper nucleus, for example, is about 8.6 fm, just a little larger than the de
Broglie wavelength of a 200-MeV electroll. To resolve detail, the wavelength should be smallerthan the target, ideally a tenth of the diameter or less. 200-MeV electrons are perhaps at the
lower limit in energy for useful probes.
t7
The binding energy is given by AEt.number (number of protons), A is the mass number (number of nucleons), TL17 is the mass
of a hydrogen atom, mn is the mass of a neutron, and Mpu is the mass of a 'llp" atom. Inprincipal, nuclear masses should have been used, but the mass of the Z electrons includedin Z M ru is canceled by the mass of the Z electrons included in Mpu, so the result is the
same. First, calcul ate the mass difference in atomic mass units: Lm(239 94)(1 .00867 u) (239.A5216 u) - L.94101 u. Since I u is equivalent to 93I.5 MeV,LEa" - (l .94101 u)(93 1.5 MeV/u) - 1808 MeV. Since there are 239 nucleons, the bindingenergy per nucleon is LEa"n: EIA- (1808MeV) lQ39): 7.56MeV.
t9If a nucleus contains Z protons and tf neutroos, its binding energy is Atr. - (ZmH + IVmnm)c2, where TTL7 is the mass of a hydrogen atom, mn is the mass of a neutron, and m isthe mass of the atom containing the nucleus of interest. If the masses are given in atomicmass units, then mass excesses are defined by L11 - (*n l)r2, Ln _ (mn l)r2, and
AE -(ZLs* NA", A) +(Z +t/ - A)"'- ZLs*l[A,r, - A, where A- Z+ IV was used.
For t?!X", Z : 79 and iV - I97 - 79 : 118. Hence
AEu. -- (79)(1 .29 MeV) + (118X8.07 MeV) - (-3l.zMeV) : 1560MeV.
This means the binding energy per nucleon is LEu"n: (1550MeV)1Q97) -7.92MeV.
254 Chapter 42
27
(a) The half-life Tr /z and the disintegration constant ,\ are related by Tr lz - (ln 2) I ^,
so
Tt/z - (ln 2)1Q.0108 h-t) - 64.2h.
(b) At time t, the number of undecayed nuclei remaining is given by
jV - ffo e-\t - l/o s-(ln2)t/Trf2 .
Substitute t - 3Tr /z to obtain,'A\r --3tn2_0.I25.l/r -e
In each half-life, the number of undecayed nuclei is reduced by half. At the end of one half-life, Niv-AIo/g-0.125Ah.(c) Use
iV - Afo e-\t
10.0d is 240h., so )t - (0.01086-t)Q40h):2.592 and
lr _2.ss2
lrr:e 2'J'- -0'0749'
35
(a) Assume that the chlorine in the sample had the naturally occurring isotopic mixture, so the
average mass number was 35.453, as given in Appendix F. Then the mass o1 226Ra was
m: 226
226 + zlzts lsY(o'10 g) : 76'l x 1o-' s '
The mass of a 226pu nucleus is (226uX 1.66I x 10-24 glu)226pu nuclei present was iV - (76.I x 10-3 gl(3.75 x 10-22 g): 2.03 x 1020.
(b) The decay rate is given by R- N^ - (lf h\lTr/2, where ^
is the disintegration constant,
Tr/z is the half-life, and tf is the number of nuclei. The relationship l - (ln2)lTrp was used.
Thus
R- - 2.7gx 10es-l .
43
If ,Af is the number of undecayed nuclei present at time t, then
ry - R_)N,dt
where R is the rate of production by the cyclotron and ^
is the disintegration constant. The
second term gives the rate of decay. Rearrange the equation slightly and integrate:
[* dtY - [' n,,J*o R-'\A/' l,
uui
Chapter 42 255
rv-R / R\ ^ti* (.^o x) e
After many half-lives, the exponential is small and the second term can be neglected. Then,,Af - Rl^, regardless of the initial value fig. At times that are long compared to the half-life,the rate of production equals the rate of dec ay and .,Af is a constant.
where Afg is the number of undecayed nuclei present at time f, : 0. This yields
r R - ^rr) tn*-xvo:t'
Solve for l/:
49
The fraction of undecayed nuclei remaining after time t is given by
,,Ar
tf, - e-'\t : e-(lnDtl\ 12
where ^
is the disintegratton constant and Trlz (- (ln 2)lD is the half-life. The time for half theoriginal 238IJ nuclei to decay is 4.5 x 10ey. For 2aapu atthat time
ry- (h2X4'5 x loey) -3g.0Tt /z 8.2 x 107 y
andtrF;
: e-38'o : 3'1 x 10-17 '
For za9g^ at that time(In2)t
- (ln 2X4.5 x 10e y) _ 9n0
Tt /z 3.4 x 105 y
and
{ : e-st7o - 3.31 x 10-3e83 .
lro
For any reasonably sized sample this is less than one nucleus and may be taken to be zero. Yourcalculator probably cannot evaluate e-er70 directly. Treat it as (e-el'70;100.
--35
Let Mc, be the mass of one atom of t33Cr and Msu be the mass of one atom of t3Znu. Toobtain the nuclear masses we must subtract the mass of 55 electrons from Mc" and the mass
of 56 electrons from Msu. The energy released is A- l(Mc, - 55*) - (Msa- 56m) -mf c2,
where m is the mass of an electron. Once cancellations have been made, Q : (Mc, - Msu)r' isobtained. Thus
Q - [136.9071 u - 136.9058u] c2 - (0.00 13u)c2 - (0.0013u)(g3zMeV/u): 1.21 MeV.
256 Chapter 42
59
Since the electron has the maximum possible kinetic energy no neutrino is emitted. Since
momentum is conserved, the momentum of the electron and the momentum of the residual sulfurnucleus arc equal in magnitude and opposite in direction. If p" is the momentum of the electronand ps is the momentum of the sulfur nucleus, then ps: -pu. The kinetic energy Ks ofthe sulfur nucleus is Ks - pLlLtWt - p?l2Ms, where Ms is the mass of the sulfur nucleus.
Now the electron's kinetic energy K" is related to its momentum by the relativistic equation(prc)z - K?+2K"fficT, where m is the mass of an electron. See Eq. 37-54. Thus
Kg - (P"c)z _ KZ + 2K"mc2
-(1 .71 MeV)' + 2(1 .71MeV)(0.511 MeV)
- p?*12*" : (m*l*")K*, so
TTLyc2 + mnc2 + K*: TTLyc2 + (m*l*t)K, + Ev
K*TTLy - ffir
2Mscz 2Msc2
- 7.83 x 10-s MeV - 78.3 eV
2(32u)(93 1.5 MeV l")
where mcz - 0.511 MeV was used.
67
The decay rute R is related to the number of nuclei l[ by R - ,LAf , where I is the disintegrationconstant. The disintegration constant is related to the half-life Tt/z by
^ - (ln 2)lTt/2, so
.A\r - RIA : RTr/zlln2. Since 1 Ci :3.7 x 1010 disintegrations/s,
Ar _ Q50 CiX3 .7 x 1010 s-' lct)(2.7 dX8 .64 x 104 s/d)]V
The mass of a le84n atom is M _ (198uXI.661 x 10-24 glu)- 3.29 x 10-22 g so the mass
required is M: (3.11 x lOltxf .2g x 10-t'g) - 1.02 x 10-'g- l.02mg.
73
A generaltzed formation reaction can be written X + x -+ Y, where X is the target nucleus, x isthe incident light particle, and Y is the excited compound nucleus ('oN.). Assume X is initiallyat rest. Then conservation of energy yields
ITLyc2 + mrr2 + K*: TTLyr2 + Ky i Ev ,
where ffLy, ffir, and rft,y are masses, K, and Ky are kinetic energies, and Ey is the excitationenergy of Y. Conservation of momentum yields
P*: PY
Now Ky - p+12^t
and
Chapter 42 257
(a) Let r rcpresent the alpha particle and X represent the 160 nucleus. Then (mv -wLy -mn)r' :(19.99244u - 15.99491 u - 4.00260u)(931.5 MeV/n) - -4.722MeV and
19.99244uKo:
19.99244u - 4.00260u(-4.722MeV + 25.0MeV) : 25.35 MeV.
(b) Let n represent the proton and X represent the leF nucleus. Then (mv TLy TTL')"' -(19 .99244u - 18.99841 u - 1.00783 u)(931.5 MeV/,r) - - 12.85 MeV and
19.99244uKo
19.99244u- 1.00783u(-12.85 MeV + 25.0MeV) - I2.80MeV.
(c) Let r represent the photon and X represent the 20Ne nucleus. Since the mass of the photonis zero, we must rewrite the conservation of energy equation: if E.y is the energy of the photon,then E^,t *mxc2 : TTLyc2 + Ky + Ev. Since ?TLy: TTLr, this equation becomes E^t: Ky + Ev.Since the momentum and energy of a photon are related by pt - E^t f c, the conservation ofmomentum equation becomes E^, l" - pv. The kinetic energy of the compound nucleus is
Kv - pTl2^t- 8'?1 f2mycz. Substitute this result into the conservation of energy equation toobtain
E^-3^Y zmva* EY '
This quadratic equation has the solutions
E^r : TTl,y c2 +
If the problem is solved using the relativistic relationship between the energy and momentumof the compound nucleus, only one solution would be obtained, the one corresponding to the
negative sign above. Since my c2 - ( L9 .99244u)(93 1 .5 MeV/u) : I .862 x 104 MeV,
E^t - (I.862 x 104 MeV) -:25.0MeV.
(I .862 x 104 MeV)2 - 2(I .862 x 104 MeV) (25.0 MeV)
The kinetic energy of the compound nucleus is very small; essentially all of the photon energygoes to excite the nucleus.
75
Let A be the area over which fallout occurs and a be the area that produces a count rate ofR-74000counts/s. The count rate is R-
^ltr, where,\ is the disintegration constant and l/ is
the number of radioactive nuclei in area a. The number of atoms in the entire fallout is M l*,where M is the mass of e0Sr produced and m is the mass of a single nucleus of e0Sr. Thus the
count rate for the area a is R- ^(Mlm)(alA).
The half-lifeT1 p is related to the disintegrationconstant by
^ - (Inz) lTr /2, so R - (ln 2lT, p)(Att ld@lA). Solve for a,:
(mvr2)2 2mvc2Ev
258 Chapter 42
a-AR(#)(#)
The molar mass of e0 Sr is 90 glmol, so the mass of a single e0 Sr nucleus is (90-l kgl^ol)/(6 .02 x1023 mol-t): 1.50 x l0-2skg. The half-life is (z9yx365dlfle4hld)(3600s/h) -9. l5 x 108s.
Therefore
&-(2000X106m\Q4000counts/s)()(#):7.3X10-2m2.
85
The number of undecayed nuclei at time t is given by iV - l[oe-'\t, where Ah is the number at
f : 0 and A is the disintegration constant. The rate of decay is R - -d,IV ld,t :,\Ah e-\t - ,\Arand the rate att:) is fto: Ah. Thus RlRo: Nllr{o- e-lt. The solution for t is
1Rt- -,\tr*0.
The disintegration constant is related to the half-life Tr/z by ) - (ln2)lTt/2, so
r - '+r,r*: -{+h(o .ozo):3.2x looy.L--lt
87
Let fiX represent the unknown nuclide. The reaction equation is
fx+fn*_!e+2tRr.Conservation of charge yields Z+0- -1 +4 or Z-3. Conservation of mass number yieldsA+ I :0+8 or A: 7. According to the periodic table in Appendix E, lithium has atomicnumber 3, so the nuclide must be lti.
Chapter 42 259
13
(a)
Chapter 43
If X represents the unknown fragment, then the reaction can be written
'ltu+fn-!ic.+fX,where A is the mass number and Z is the atomic number of the fragment. Conservation ofchargeyields 92+0- 32+2, so Z - 60. Conservationof massnumberyields 235+1_83+,4,so A_ 153. Look in Appendix F or G for nuclides with Zunknown fragment is t;ANd.
(b) and (c) Ignore the small kinetic energy and momentum carried by the neutron that triggers the
fission event. Then A - Kc" * KNo, where Kc" is the kinetic energy of the gefinanium nucleus
and /fxa is the kinetic energy of the neodymium nucleus. Conservation of momentum yieldspce + prva : 0, where pce is the momentum of the gennanium nucleus and ft.'-ro is the momentum
of the neodymium nucleus. Since pNd : -pce, the kinetic energy of the neodymium nucleus is
lfxd :
Thus the energy equation becomes
Piro PL" Mc"Kc"2MNd 2Mxd Mxo
and
e:Kc".mKc"-WKc"
Kc.: # Q:#(l7oMev): lloMev
Similarly,
lrNd : ffie -#(170 Mev) : 6o Mev .
The mass conversion factor can be found in Appendix C.
(d) The initial speed of the gennanium nucleus is
-, -1.60x l07m/s.V
(e) The initial speed of the neodymium nucleus is
uNd :
260 Chapter 43
2(60 x 106 eVX1.60 x 10-le J/eV)(153 uX 1.661 x 10-27 kg/,r)
:8.69 x 106 m/s.
15
(a) The energy yield of the bomb is E - (66 x 10-'-rgaton)(2.6 x l028MeV/megaton)I.72 x 1027 MeV. (The energy conversion factor is given in Problem 16.) At 200MeV per
fission event, (l .72 x 1027 MeV) lQ}AMeV) - 8.58 x L024 fission events take place. Since
only 4.0oh of the 23sg nuclei originally present undergo fission, there must have been (8.58 xrc2\l(0.040) : 2.L4 x 1026 nuclei originally present. The mass og 23s1; originally present was(2.I4 x 1026Xn5uXI.66l x 10-27 kg/.r) - 83.7kg. The mass conversion factor can be found
in Appendix C.
(b) Two fragments are produced in each fission event, so the total number of fragments is
2(8.58 x 10241 - !.72 x 102s.
(c) One neutron produced in a fission event is used to trigger the next fission event, so the
average number of neutrons released to the environment in each event is 1.5. The total number
released is (8.58 x 1024X1.5): L.29 x 1025.
23
(a) Let 'un,i be the initial velocity of the neutron, unf be its final velocity, and u y be the finalvelocity of the target nucleus. Then, since the target nucleus is initially at rest, conservation
of momentum yields n'Lnuni, - TTLnunf + rnu f and conservation of energy yields **rult -*,**ult + i*r'r. Solve these two equations simultaneously for r) y. This can be done, forexample, by using the conservation of momentum equation to obtain an expression for unf interms of 'u y and substituting the expression into the conservation of energy equatioll. Solve the
resultirg equation for ?)y. You should obtain uy: 2mrnni,l(*+ mn). The energy lost by the
neutron is the same as the energy gained by the target nucleus, so
The initial kinetic energy of the neutron is K -AKK
*.*r'zrt, so
4mrm(m+ mn)2
(b) The mass of a neutron is 1.0 u and the mass of a hydrogen atom is also 1.0 u. (Atomicmasses can be found in Appendix G.) Thus (A/O lK - 4(1 .0uxl.0u)l(1.0u+ 1.0u)2
(c) The mass of a deuterium atom is 2.0v, so (A/O lK - 4(1 .0uX2.0u)/(2.0u* 1.0 u)2 - 0.89.
(d) The mass of a carbon atom is l2u, so (A/{) lK - 4(1 .0uXI2u)lQzu* 1.0u)2 - 0.28.
(e) The mass of a lead atom is 207u, so (A/O lK - 4(1 .0u)(207u)l(207 u* 1.0u)2 - 0.019.
(0 During each collisior, the energy of the neutron is reduced by the factor I - 0.89 - 0.11. IfEi is the initial energy, then the energy after n collisions is given by E - (0.1L)n Ei. Take thenatural logarithm of both sides and solve for n. The result is
tn(E I trr,)tt:
ln 0.1I
ln(0 .025 eY l1 .00 eV)
ln 0.11-7.9.
Chapter 43 261
The energy first falls below 0.025 eV on the eighth collision.
25
Let Po be the initial power output, P be the final power output, k be the multiplication factor, t
be the time for the power reduction, and tgen be the neutron generation time. Then according to the result of Problem 18,
P = Po kt/tgen .
Divide by Po, then take the natural logarithm of both sides of the equation and solve for In k. You should obtain
In k = tgen In P . t Po
Hence
where _ tgen P _ 1.3 X 10-3 s 350.00MW _ -4
a - -t- In Po - 2.6000s In 1200.0MW - -6.161 x 10 .
This yields k = .99938.
29
Let t be the present time and t = 0 be the time when the ratio of 235U to 238U was 3.0%. Let N235 be the number of 235U nuclei present in a sample now and N 235 , 0 be the number present at t = O. Let N 238 be the number of 238U nuclei present in the sample now and N238, 0 be the number present at t = O. The law of radioactive decay holds for each specie, so
N N -.>..t 235 = 235,0 e
and N N -.>..t
238 = 238 0 e . ,
Divide the first equation by the second to obtain
r = ro e-('>"-'>")t ,
where r = N235/N238 (= 0.0072) and ro = N 235 , O/N238 , 0 (= 0.030). Solve for t:
1 r t=- In- .
..\235 - ..\238 ro
Now use ..\235 = (In 2)/T235 and ..\238 = (In 2)/T238, where T 235 and T 238 are the half-lives, to obtain
_ T235T238 I r_ (7.0 x 108 y)(4.5x 109 y) 10.0072_17 109 t-- n--- n -.Ix y. (T238 - T 235 ) In 2 ro (4.5 x 109 y - 7.0 X 108 y) In 2 0.030
262 Chapter 43
31
The height of the Coulomb barrier is taken to be the value of the kinetic energy K each
deuteron must initially have if they are to come to rest when their surfaces touch (see Sample
Problem 43-4). If r is the radius of a deuteror, conservation of energy yields
SO
K - -]- 4: (8 ,ggx4nes 4r \
This is l7}keV.
43
2K: 1 t.-LL 4rreg 2r )
10e m/F) - 2.74 x 10-14 J .
(a) The mass of a carbon atom is (I2.0 uX I.66I x 10-27 kg/,r)number of carbon atoms in 1.00kg of carbon is (1.00kg) 1Q.99 x 10-'ukg)(The mass conversion factor can be found in Appendix C.) The heat of combustion per atom is(3.3 x 107 JlkglO.Oz x 1025 alrrmfkg) - 6.58 x 10-t'Jlutom. This is 4.lIeYlatom.(b) In each combustion event, two oxygen atoms combine with one carbon atom, so the totalmass involved is 2(16.0u)+(12.0u) - 44v. This is (44uXL.661 x 10-27 kglu) -7.31x 10-26 kg.
Each combustion event produces 6.58 x 10-le J so the energy produced per unit mass of reactants
is (6.58 x 10-te DlQ.3r x L0-26 kg) -9.00 x 106 Jlkg.
(c) If the Sun were composed of the appropriate mixture of carbon and oxygen, the number ofcombustion events that could occur before the Sun burns out would be (2.0 x 1030 kgl(7.31 xL0-26 kg) - 2.74x 1055. The total energy released would be E - (2.74x 10ssx6.58 x 10-tnJ) -1 .80 x 1037 J. If P is the power output of the Sun, the burn time would be t(1.80 x 1037 DlQ9 x 1026 w) - 4.62 x 1010 s. This is 1460y.
Chapter 43 263
Chapter 44
11
(a) The conservation laws considered so far are associated with energy, momentum, angularmomentum, charge, baryon number, and the three lepton numbers. The rest energy of the muonis 105.7MeV, the rest energy of the electron is 0.511MeV, and the rest energy of the neutrino iszero. Thus the total rest energy before the decay is greater than the total rest energy after. Theexcess energy can be carried away as the kinetic energies of the decay products and energy canbe conserved. Momentum is conserved if the electron and neutrino move away from the decayin opposite directions with equal magnitudes of momenta. Since the orbital angular momentumis zero, we consider only spin angular momentum. All the particles have spin nlz. The totalangular momentum after the decay must be either h, (if the spins are aligned) or zero (if the spinsare antialigned). Since the spin before the decay ish, 12, angular momentum cannot be conseryed.The muon has charge -e) the electron has charge -e) and the neutrino has charge zeto, so thetotal charge before the decay is -e and the total charge after is -e. Charge is conserved. All theparticles have baryon number zero, so baryon number is conserved. The muon lepton numberof the muon is 11, the muon lepton number of the muon neutrino is *1, and the muon leptonnumber of the electron is 0. Muon lepton number is conseryed. The electron lepton numbers ofthe muon and muon neutrino are 0 and the electron lepton number of the electron is +1. Electronlepton number is not conserved. The laws of consenration of angular momentum and electronlepton number are not obeyed and this decay does not occur..
(b) Analyze the decay in the same way. You should find that only charge is not conserved.
(c) Here you should find that energy and muon lepton number cannot be conserved.
29
(a) Look at Table 44-5. Since the particle is abaryon, it must consist of three quarks. To obtaina strangeness of -2, two of them must be s quarks. Each of these has a charge of -e13, so thesum of their charges is -2e 13. To obtain a total charge of e, the charge on the third quark mustbe 5ef3. There is no quark with this charge, so the particle cannot be constructed. In fact, sucha particle has never been observed.
(b) Again the particle consists of three quarks (and no antiquarks). To obtain a strangeness ofzero, none of them may be s quarks. We must find a combination of three u and d quarks witha total charge of 2e. The only such combination consists of three u quarks.
4t
(a) The mass M wtthin Earth's orbit is used to calculate the gravitational force on Earth. If r is
264 Chapter 44
the radius of the orbit, R is the radius of the new Sun, and M s is the mass of the Sun, then
M - (L)'rtr*: (t ?9 .
19ll *)'t r.ssx 1030ke) -3.27 x rr2skg.tvr
\ n/ /Yrr \ s.lo x lol'
^ )
The gravitational force on Earth is given by GMmlr', where m is the mass of Earth and G isthe universal gravitational constant. Since the centripetal acceleration is given by ,' lr, where u
is the speed of Earth, GMmlr' -- mr)'l, and
IGMt)-l; - l.2l x 102 m/s .
(b) The period of revolution
T* 2n(1 .50 x 101t ttt)- 7.82 x 109 s.I.2l x I02 m/s
This is 248y.
45
The energy released would be twice the rest energy of Earth, or E - 2mc2 - 2(5.98 x1024 kgX3.00 x 108 ml s)z
C.
47
(a) Since ,S
-1 f3. To obtain a meson with charge quantum number *1, the s quark must be combined withan antiquark with strangeness 0 and charge quantum number *4f3. There is no such antiquark.
(b) Now S- +1, so the meson contains an S quark, which has a charge quantum number of+l 13. To obtain a charge quantum number of - 1 it must also contain a quark with charge
quantum number -4 12. There is no such quark.
51
(a) After time At the distance between the galaxy and Earth is r * ra Lt, where r is the
distance when the light is emiffed. The distance when the light reaches Earth must be cLt, so
cLt - r + raLt and Lt - r I @ - rcr).
(b) The detected waveleneth is longer than I by )c Lt, so L^l^ - a Lt - ar l@ - crr).
(c) Since c >
is
2nr :u
T.ff) '*ff) '*AA ar r ar1-r ar I- / czr\ r art2,\ :;[r
T] -':
T['+ ff)+ (T)'+ ...] -
(6.67 x 10-ll m3 lt'.kgX3 .27 x 102s kg)
1.50 x l0ll m
(d) If only the first term is retained L^l,\ - ar f c.
Chapter 44 265
(e) If u - Hr, where H is the Hubble constant, then A^/^_ ulc: Hrf c. Comparison withL^l,\ - ar f c gives e - H - 0.0218 m/s . ly.
(f) Solve A) f ^
- ra l@ - rc-) for r. The result is
- :(411)) - (3'00 x 108 m/sX0 91% : 6.6x 108 ly.
a(L + A ^lD
(0.02 lSmls . ly)(l + 0.050)
(g) According to the result of part (a)
Lt- r - (6;6 110tlI),(?'16*,
10ltT11I) ,=, :z.zx1016s.c- ra 3.00 x 108 mls - (0.02 l8m/s.lyX6.6 x 108 ly) -'-
This is 6.9 x 108 y.
(h) The time is Lt - rlc: (:6.6 x 108 l9l(1.00 lVlg:6.6 x 108y.
(i) The distance is cLt - (1.00 IVlg(6.9 x 108 y) : 6.9 x 108 ly.
0) [Jse the equation developed in part (0:
r:ffi:(k) The result of part (a) gives
At- r(c-rc-) 3.00 x 108 mls - (I .02 x 10e1VX0.021 Sm/s.ly) v"
This is 1.1 x 10ey.
(1) Galaxy B emits light Lt- 1.1 x 10ey - 6.9 x 108yDuring that time the universe expands, so that the distance of galaxy B from Earth at the time Aemits is
The separation of the galaxies at the time A emits is 1.05 x 10ely- 6.6 x 108 ly: 3.9 x 108 ly.
(4.1x 108 yX3. 16 x 107 r/vl]0.0218mls . ly
9.46 x 101t *fiy,n(L+ oA t): (1,.02x tOe ly) [t
+
-1.05 x 10ely.
266 Chapter 44
