❧
차례
1장
서 론
5
2장
일차원 운동
9
3장
벡터와 이차원 운동
17
4장
운동의 법칙
27
5장
에 너 지
37
6장
운동량과 충돌
46
7장
회전 운동과 중력의 법칙
57
8장
회전 평형과 회전 동역학
66
9장
고체와 유체
75
10장
열물리학
84
11장
열과정에서의 에너지
90
12장
열역학 법칙
97
13장
진동과 파동
105
14장
소리(음파)
113
15장
전기력과 전기장
122
16장
전기 에너지와 전기용량
132
17장
전류와 전기 저항
142
18장
직류 회로
147
- i -
3
67352_ch01.indd 1
2/9/11 1:09:44 PM
19장
자 기
160
20장
유도 전압과 유도 계수
168
21장
교류 회로와 전자기파
176
22장
빛의 반사와 굴절
183
23장
거울과 렌즈
190
24장
파동 광학
200
25장
광학 기기
208
26장
상대성 이론
215
27장
양자 물리학
221
28장
원자 물리학
227
29장
핵물리학
233
30장
핵에너지와 소립자
239
- ii -
4
67352_ch01.indd 1
2/9/11 1:09:44 PM
1장
서
론
PROBLEM SOLUTIONS
1.1
(a)
The units of volume, area, and height are:
Introduction
5
[V ] = L , [ A] = L , and [h] = L
3
2
continued on next page
We then observe that L3 = L2 L or [V ] = [ A][h]
1.2
Substituting
dimensions
equation
correct .
Thus, the
equationinto
V =the
Ahgiven
is dimensionally
Vcylinder = π R 2 h = (π R 2 ) h = Ah , where A = π R 2
(b)
Vrectangular box = wh = ( w ) h = Ah, where A = w = length × width
1.2
1.3
x
B= 2
t
Substituting dimensions into the given equation T = 2π
sionless constant, we have
[T ] =
[ ]
[ g]
or
T=
L
=
L T2
g , and recognizing that 2π is a dimen-
T2 = T
1.4
Thus, the dimensions are consistent .
, which says that a constant
force F multiplied by a duration of time Δt equals the change in momentum, Δp.)
1.4
1.3
1.5
(a) Solving
From the universal gravitation law, the constant G is G = Fr 2 Mm. Its units are then
[G ] =
1.4
1.9
1.6
(a)
(a)
(b)
[ F ] ⎡⎣ r 2 ⎤⎦ ( kg ⋅ m s2 ) ( m 2 )
=
=
kg ⋅ kg
[ M ][ m ]
m3
kg ⋅ s 2
78.9 ± 0.2 has 3 significant figures with the uncertainty in the tenths position.
In the equation 12 m v 2 = 12 m v02 + mgh,
3.788 ×10 9 has 4 significant figures
(c)
2.46 ×10 −6 has 3 significant figures
(d)
0.003 2 = 3.2 × 10 −3 has 2 significant figures . The two zeros were originally included
only to position the decimal.
1.10
1.5
(a) Computing
Area = ( length ) × ( width ) = ( 9.72 m )( 5.3 m ) = 52 m 2
because 5.620 has only four significant figures.
1.13
1.6
The least accurate dimension of the box has two significant figures. Thus, the volume (product of
the three dimensions) will contain only two significant figures.
because rounding in part (b) was carried out too soon.
V = ⋅ w ⋅ h = ( 29 cm )(17.8 cm )(11.4 cm ) = 5.9 × 10 3 cm 3
1.11
51
67352_ch01.indd 1
2/9/11 1:09:44 PM
6
Chapter 1
1.7
1.16
1.8
1.10
1.15
1.9
1.17
1.19
1.21
1.10
1.11
1.23
(b) In m ⎛ 1 km ⎞ ⎛ 1 mi ⎞ ⎛ 3 600 s ⎞
v = 38.0
⎟⎜
⎟ = 85.0 mi h
⎜
⎟⎜
s ⎝ 10 3 m ⎠ ⎝ 1.609 km ⎠ ⎝ 1 h ⎠
Yes, the driver is exceeding the speed limit by 10.0 mi h .
1.609 km ⎞
2
= ( 348 mi ) ⎛⎜
⎟ = 5.60 × 10 km = 5.
⎝ 1.000 ⎛mi9⎠ gal ⎞ ⎛ 3.786 L ⎞ ⎛ 10 3 cm 3 ⎞ ⎛ 1 m 3 ⎞
= 0.204 m 3
6.00 firkins = 6.00 firkins ⎜⎜
⎟⎟ ⎜
⎟⎟ ⎜⎜
⎟⎜ 6
3 ⎟
1
firkin
1
gal
1
L
10
cm
⎝ the accuracy
⎠ of the original
⎠ ⎝ cant figures
⎠ ⎝ because⎠of
In (a), the answer is limited⎝ to three signifi
(b)
data value,
348
miles.
In
(b),
(c),
and
(d),
the
answers
are
limited
to
four
signifi
cant
⎞
186 furlongs ⎛ 1 fortnight ⎞ ⎛ 1 day ⎞ ⎛ 220 yds ⎞ ⎛ 3 ft ⎞ ⎛ 100
cmfigures
v = = of the accuracy
⎜⎜ to which the⎟⎟ kilometers-to-feet
⎜
⎟
because
conversion
factor
is
given.
⎜⎜
⎟
⎜
⎟
4
⎟8 ⎜ 1 furlong ⎟ ⎜ 1 yd ⎟ ⎜⎝ 3.281 ft ⎟⎠
5 28014ft days
1 fathom
t
1 fortnight
8
.
64
×
10
s
⎛
⎞
⎛
⎞
⎠ ⎝ ⎟ = 2 × 10⎠ ⎝fathoms ⎠ ⎝
⎠
d = ( 250 000 mi ) ⎜ ⎝
⎟⎜
⎝ 1.000 mi ⎠ ⎝ 6 ft ⎠
(a)
The answer is limited to one significant figure because of the accuracy to which the conversion
from fathoms to feet is given.
This means that the proteins are assembled at a rate of many layers of atoms each second!
diameter 5.36 in ⎛ 2.54 cm ⎞
=
⎜
⎟ = 6.81 cm
2
2 ⎝ 1 in ⎠
(a)
r=
(b)
A = 4π r 2 = 4π ( 6.81 cm ) = 5.83 × 10 2 cm 2
(c)
V=
2
4 3 4
3
π r = π ( 6.81 cm ) = 1.32 × 10 3 cm 3
3
3
Volume of cube = L 3 = 1 quart (Where L = length of one side of the cube.)
⎛ 1 gallon ⎞ ⎛ 3.786 liiter ⎞ ⎛ 1000 cm3 ⎞
= 947 cm3
Thus, L 3 = 1 quart ⎜
⎟
⎜⎝ 4 quarts ⎟⎠ ⎜⎝ 1 gallon ⎟⎠ ⎜⎝ 1 liter ⎟⎠
(
and
1.12
1.31
1.13
1.23
A reasonable guess for the diameter of a tire might be 3 ft, with a circumference (C = 2π r = π D =
distance travels per revolution) of about 9 ft. Thus, the total number of revolutions the tire might
make is
n=
67352_ch01.indd 6
total distance traveled ( 50 000 mi )( 5 280 ft mi )
= 3 × 10 7 rev, or ~10 7 rev
=
9 ft rev
distance per revolution
We assume an average respiration rate of about 10 breaths/minute and a typical life span of
70 years. Then, an estimate of the number of breaths an average person would take in a lifetime is
Volume of cube = L 3 = 1 quart (Where L = length of one side of the cube.)
⎛ 3.156 × 10 7 s ⎞ ⎛ 1 min ⎞
⎛
breaths ⎞
8
n = ⎜ 10
70
yr
⎟⎟ ⎜
⎜⎜
(
)
⎟ = 4 × 10 breaths
⎟
min
1
yr
60
s
⎝
⎠
⎝
⎠
Thus,The very large mass of prokaryotes
⎝
(c)
implies⎠ they are important to the biosphere. They are
or
1.35
1.15
L = 3 947 cm3 = 9.82 cm
⎛
m ⎞ ⎛ 3 600 s ⎞ ⎛ 1 km ⎞ ⎛ 1 mi ⎞
8
c = ⎜ 3.00 × 10 8
⎟⎜
⎟⎜
⎟⎜
⎟ = 6.71 × 10 mi h
s ⎠ ⎝ 1 h ⎠ ⎝ 10 3 m ⎠ ⎝ 1.609 km ⎠
⎝
1.27
1.33
1.14
)
responsible
for fixing carbon, producing oxygen, and breaking up pollutants, among many
n ∼ 108 breaths
other biological roles. Humans depend on them!
(
)
The x coordinate is found as
x = r cosθ = 2 .5 m cos 35° = 2.0 m
and the y coordinate
y = r sin θ = ( 2 .5 m ) sin 35° = 1.4 m
2/9/11 1:09:50 PM
Introduction
Introduction
Introduction
11
Introducti
11
7
1.39
ReferThe
to the
Figure given
in Problem
1.39
Refer to the Figure given in Problem
1.38 above.
Cartesian
coordinates
for the1.38
twoabove.
given The Cartesian coordinates for the two gi
1.16
fer to the Figure given in Problem 1.38
above. (a)
The
Cartesian
coordinates
for
the
two
given
points
are:
1.12
The
sum
is
rounded
to
points are:
nts are:
x1 = r1 cos θ1 = ( 2.00 m ) cos50.0° = 1.29 m
x2 = r2 cos θ2 = ( 5.00 m ) cos ( −50.0°) = 3.21 m
y1 = r1 siin θ1 = ( 2.00 m ) sin 50.0° = 1.53 m
y2 = r2 sin θ2 = ( 5.00 m ) sin ( −50.0°) = −3.83 m
The distance between the two points is then:
Δs = ( Δx ) + ( Δy ) = (1.29 m − 3.21 m ) + (1.53 m + 3.83 m ) = 5.69 m
2
2
1.41
1.17
(a)
2
With a = 6.00 m and b being two sides of this right
triangle having hypotenuse c = 9.00 m, the Pythagorean
theorem gives the unknown side as
( 9.00 m )2 − ( 6.00 m )2 = 6.71 m
b = c2 − a2 =
a 6.00 m
tan θ = adjacent
=
= 0.894
b
6.71 mside
cos θ =
hypotenuse
(b)
1.18
1.49
2
(c)
sin φ =
b 6.71 m
=
= 0.746
c 9.00 m
Using the sketch at the right:
w
= tan 35.0° , or
100 m
w
w = (100 m ) tan 35.0° = 70.0 m
14
1.19
1.53
Chapter 1
(a)
Given that a ∝ F m, we have F ∝ ma. Therefore, the units of force are those of ma,
[ F ] = [ ma] = [ m][a] = M ( L T 2 ) = M L T-2
(b)
1.54
1.20
1.55
L
M⋅L
[ F ] = M ⎛⎜ 2 ⎞⎟ = 2
T
⎝T ⎠
so
newton =
kg ⋅ m
s2
(a) The rate of filling in gallons per second is
The term s has dimensions of L, a has dimensions of LT−2, and t has dimensions of T. Therefore,
the equation, s = k a m t n with k being dimensionless, has dimensions of
L = ( LT −2 ) ( T )
m
n
or
L1T 0 = L m T n−2 m
The powers of L and T must be the same on each side of the equation. Therefore, L1 = Lm and
m =1
Likewise, equating powers of T, we see that n − 2 m = 0, or n = 2 m = 2
Dimensional analysis cannot determine the value of k , a dimensionless constant.
Assumes an average weight of 0.5 oz of aluminum per can.
1.56
1.57
1.21
Assume an average of 1 can per person each week and a population of 300 million.
⎛ 365.2 days ⎞ ⎛ 8.64 × 10 4 s ⎞
7
Introduction
(a)
⎟⎟ ⎜
⎟ = 3.16 × 10 s
(a) 1 yr = (1 yr ) ⎜⎜
1
yr
1
da
y
⎝
⎠
⎝
⎠
(b)
67352_ch01.indd 7
15
Consider a segment of the surface of the Moon which has an area of 1 m2 and a depth of
1 m. When filled with meteorites, each having a diameter 10−6 m, the number of meteorites
along each edge of this box is
2/9/11 1:09:51 PM
8
Chapter 1
(b)
1.20
Introduction
15
2
Consider alength
segment
of the
surface1ofmthe Moon
of an
edge
6 which has an area of 1 m and a depth of
n
=
=
=
10
−6
−6
1 m. When
filled with
meteorites,
meteorite
diameter
10each
mhaving a diameter 10 m, the number of meteorites
along each edge of this box is
The total number of meteorites in the filled box is then
( )
N = n 3 = 10 6 3 = 10 18
At the rate of 1 meteorite per second, the time to fill the box is
16
1.22
1.61
1y
⎞ = 3 × 10 10 yr, or
t = 1018 s = (1018 s ) ⎛⎜
⎟
7
⎝ 3.16 × 10 s ⎠
Chapter 1
The volume of paint used is given by V = Ah, where A is the area covered and h is the thickness
of the layer. Thus,
V 3.79 × 10 −3 m 3
=
= 1.52 × 10 −4 m = 152 × 10 −6 m = 152 μ m
25.0 m 2
A
h=
1.21
1.23
1.63
~1010 yr
The volume of the Milky Way galaxy is roughly
⎛ πd2 ⎞ π
21
VG = At = ⎜
⎟ t ≈ 4 10 m
4
⎝
⎠
(
) (10 m )
2
19
orr VG ∼10 61 m3
If, within the Milky Way galaxy, there is typically one neutron star in a spherical volume of radius
r = 3 × 1018 m, then the galactic volume per neutron star is
V1 =
3
4 3 4
π r = π ( 3 × 1018 m ) = 1 × 10 56 m 3
3
3
or
V1 ∼ 10 56 m 3
The order of magnitude of the number of neutron stars in the Milky Way is then
n=
1
67352_ch01.indd 8
VG 10 61 m 3
∼
V1 10 56 m 3
or
n ∼ 10 5 neutron stars
mi ⎛ mi ⎞ ⎛ 1.609 km ⎞
km
= ⎜1
⎟⎜
⎟ = 1.609
h ⎝ h ⎠ ⎝ 1 mi ⎠
h
2/9/11 1:09:52 PM
2장
일차원 운동
PROBLEM SOLUTIONS
2.1
2.2
2.3
2.2
(a)
Boat A requires 1.0 h to cross the lake and 1.0 h to return, total time 2.0 h.
Boat B requires 2.0 h to cross the lake at which time the race is over.
Boat A wins, being 60 km ahead of B when the race ends.
(b)
Average velocity is the net displacement of the boat divided by the total elapsed time. The
winning boat is back where it started, its displacement thus being zero, yielding an average
velocity of zero .
(a)
Distances traveled between pairs of cities are
Δx1 = v1 ( Δt1 ) = (80.0 km h ) ( 0.500 h ) = 40.0 km
Δx2 = v2 ( Δt 2 ) = (100 km h ) ( 0.200 h ) = 20.0 km
24
Chapter 2
Δx3 = v3 ( Δt3 ) = ( 40.0 km h ) ( 0.750 h ) = 30.0 km
Thus, the total distance traveled is Δx = ( 40.0 + 20.0 + 30.0 ) km = 90.0 km, and the elapsed time
is Δt = 0.500 h + 0.200 h + 0.750 h + 0.250 h = 1.70 h.
2.4
2.5
2.3
2.6
2.4
2.7
Δx 90.0 km
=
= 52.9 km h
Δt
1.70 h
(a)
v=
(b)
Δx = 90.0 km (see above)
(a)
(a)
Δx
The time for a car to make the trip is t =
v ) ⎛ 1 h ⎞ + 130 km = 180 km
Displacement = Δx = (85.0 km h ) ( 35.0 min
⎜
⎟
⎝ 60.0 min ⎠
(b)
1h ⎞
The total elapsed time is Δt = ( 35.0 min + 15.0 min ) ⎛⎜
⎟ + 2.00 h = 2.83 h
60.0
min ⎠
⎝
Δx 180 km
so,
v=
= 2
= 63.6 km h
Motion in One Dimension
Δx 2.Δ
000
t × 10
2.84mh
v=
=
= 10.04 m s
Δtvelocity
19over
.92 sany time interval is
The average
We assume that you are approximately 2 m tall and that the nerve impulse travels at uniform
speed. The elapsed time is then
Δt =
2.9
2.5
(a)
25
2m
Δx
=
= 2 × 10 −2 s = 0.02 s
v 100 m s
The total time for the trip is t total = t1 + 22 .0 min = t1 + 0.367 h , where t1 is the time spent
traveling at v1 = 89.5 km h. Thus, the distance traveled is Δx = v1 t1 = vt total, which gives
(89.5 km h ) t1 = ( 77.8 km h ) ( t1 + 0.367 h ) = ( 77.8 km h ) t1 + 28.5 km
or,
(89.5 km h − 77.8 km h ) t1 = 28.5 km
From which, t1 = 2 .44 h for a total time of t total = t1 + 0.367 h = 2.81 h
(b)
67352_ch02.indd 17
9 Δx = v1 t1 = vt total, giving
The distance traveled during the trip is
2/9/11 1:10:35 PM
24
10
Chapter 2
Thus, the
total
distance
is a total time of t total = t1 + 0.367 h = 2.81 h
From
which,
t1 = traveled
2 .44 h for
(b)
The distance traveled during the trip is Δx = v1 t1 = vt total, giving
Δx = v ttotal = ( 77.8 km h ) ( 2.81 h ) = 219 km
2.11
2.6
26
(a)
From v 2f = vi2 + 2a ( Δx ) , with vi = 0 , v f = 72 km h , and Δ x = 45 m, the acceleration of
the cheetah is found to be
⎡⎛
km ⎞ ⎛ 10 3 m ⎞ ⎛ 1 h
⎢⎜ 72
⎟⎜
⎟⎜
2
2
h ⎠ ⎝ 1 km ⎠ ⎝ 3 600
v − vi
⎝
a= f
=⎣
2 ( Δx )
2 ( 45 m )
Chapter 2
(b)
2
⎞⎤
⎟⎥ − 0
s ⎠⎦
= 4.4 m s 2
The cheetah’s displacement 3.5 s after starting from rest is
1
1
2
Δx = vi t + at 2 = 0 + ( 4.4 m s 2 ) ( 3.5 s ) = 27 m
2
2
2.7
2.13
The plane starts from rest ( v0 = 0 ) and maintains a constant acceleration of a = +1.3 m s 2 . Thus,
we find the distance it will travel before reaching the required takeoff speed ( v = 75 m s ) , from
v 2 = v02 + 2a ( Δx ) , as
v 2 − v02 ( 75 m s ) − 0
=
= 2.2 × 10 3 m = 2.2 km
2a
2 (1.3 m s 2 )
2
Δx =
Since this distance is less than the length of the runway, the plane takes off safely.
2.8
2.15
( Δx )1 + L
v1 =
=
=time
+ Ltot1complete the trip is
The maximum
allowed
( Δt )1 t1
total distance
1600 m ⎛ 1 km h ⎞
t total =
=
⎜
⎟ = 23.0 s
required average speed 250 km h ⎝ 0.278 m s ⎠
The time spent in the first half of the trip is
t1 =
half distance
800 m ⎛ 1 km h ⎞
=
⎜
⎟ = 12 .5 s
v1
230 km h ⎝ 0.278 m s ⎠
Thus, the maximum time that can be spent on the second half of the trip is
t 2 = t total − t1 = 23.0 s − 12 .5 s = 10.5 s
and the required average speed on the second half is
v2 =
2.9
2.17
⎛ 1 km h ⎞
half distance 800 m
=
= 76.2 m s ⎜
h
⎟ = 274atkm
than
the
instantaneous
t = 4.00 s.
t2
10.5 s
s⎠
⎝ 0.278 m velocity
The instantaneous velocity at any time is the slope of
the x vs. t graph at that time. We compute this slope
by using two points on a straight segment of the
curve, one point on each side of the point of interest.
(a)
(b)
(c)
(d)
67352_ch02.indd 24
( 5.00 − 10.0 ) m
= − 2 .50 m s
( 4.00 − 2 .00 ) s
( 5.00 − 5.00 ) m
vt=4.50 s =
= 0
( 5.00 − 4.00 ) s
0 − ( − 5.00 m )
vt=7.50 s =
= 5.00 m s
(8.00 − 7.00 ) s
vt=3.00 s =
2/9/11 1:10:42 PM
than the instantaneous velocity at t = 4.00 s.
2.17
2.7
The instantaneous velocity at any time is the slope of
the x vs. t graph at that time. We compute this slope
by using two points on a straight segment of the
curve, one point on each side of the point of interest.
Motion in One Dimension
25
11
We assume that10
you
are−approximately
2 m tall and that the nerve impulse travels at uniform
.0 m
0
(a) vt=1.00 s =
= 5.00 m s
speed. The elapsed
2 .00time
s − 0is then
Δx ( 5.200m− 10.0 ) m −2
m ss
0.02
Δvt t==3.00 s ==
= 2 × 10= −s 2=.50
4.00m− 2s .00 ) s
v (100
5.00 ) m
( 5.00 −over
The average
any time
vt=4.50 s velocity
=
= 0interval is
(c)
−
5
.
00
4.00 ) s
(
Δx 0x −
f − xi
( − 5.00 m )
= 5.00 m s
(d) vvt==7.50Δst == t − t
f
(8.00 −i 7.00 ) s
Δx 4.0 m − 0
(a) v =
=
= + 4.0 m s
Δt 1.0 s − 0
Δx − 2 .0 m − 0
=
= − 0.50 m s
(b) v =
Δt
4.0 s − 0
(b)
2.8
Δvx
0 − 4.0 m Δv ( 60 − 55) mi h ⎛ 0.447 m s ⎞
,=we have Δt = = −=1.0 m s
⎜
⎟ = 3.7 s
Δt 5.0 s − 1.0 s a
0.60 m s 2 ⎝ 1 mi h ⎠
2.21
2.10
From va =
2.11
2.23
We choose the positive direction to point away from the wall. Then, the initial velocity of the ball
is vi = −25.0 m s and the final velocity is v f = +22.0 m s. If this change in velocity occurs over
a time interval of Δt = 3.50 ms (i.e., the interval during which the ball is in contact with the wall),
the average acceleration is
a=
2.12
2.27
(a)
Δv v f − vi +22.0 m s − ( −25.0 m s )
=
=
= 1.34 × 10 4 m s 2
3.50 × 10 −3 s
Δt
Δt
With v = 120 km h , v 2 = v02 + 2a ( Δx ) yields
2
2
v 2 − v02 ⎡⎣(120 km h ) − 0 ⎤⎦ ⎛ 0.278 m s ⎞
2
a=
=
⎜
⎟ = 2.32 m s
2 ( Δx )
2 ( 240 m )
⎝ 1 km h ⎠
2.29
2.13
Δt =
The required time is
(a)
Δx = v ( Δt ) = ⎡⎣( v + v0 ) 2 ⎤⎦ Δt becomes
⎛ 2 .80 m s + v0
40.0 m = ⎜
2
⎝
which yields
(b)
2.31
2.14
v − v0 (120 km h − 0 ) ⎛ 0.278 m s ⎞
=
⎟ = 14.4 s
⎜
2 .32 m s 2 ⎝ 1 km h ⎠
a
(b)
a=
v0 =
⎞
⎟ (8.50 s )
⎠
2
( 40.0 m ) − 2 .80 m s = 6.61 m s
8.50 s
v − v0 2 .80 m s − 6.61 m s
=
= − 0.448 m s 2
Δt
8.50 s
Apply Δx = v0 + 12 at 2 to the 2.00-second time interval during which the object moves from
xi = 3.00 cm to x f = − 5.00 cm. With v0 = 12.0 cm s, this yields an acceleration of
2 ⎡( x f − xi ) − v0 t ⎤⎦ 2 ⎡⎣( −5.00 − 3.00 ) cm − (12.0 cm s ) ( 2.00 s ) ⎤⎦
=
a= ⎣
t2
( 2.00 s )2
or
67352_ch02.indd 25
a = −16.0 cm s 2
2/9/11 1:10:43 PM
32
26
12
2.15
2.33
Chapter
Chapter 2
2
(b)
cheetah’s displacement
3.5 s after
starting
is2 for the full 40 s interval yields
UsingThe
the uniformly
accelerated motion
equation
Δx from
= v0 t rest
+ 12 at
2
Δx = ( 20 m s ) ( 40 s ) + 12 ( −1.0 m s 2 ) ( 40 s ) = 0 , which is obviously wrong. The source of the
error is found by computing the time required for the train to come to rest. This time is
v − v0 0 − 20 m s
=
= 20 s
a
−1.0 m s 2
Thus, the train is slowing down for the first 20 s and is at rest for the last 20 s of the 40 s interval.
t=
The acceleration is not constant during the full 40 s. It is, however, constant during the first
20 s as the train slows to rest. Application of Δx = v0 t + 12 at 2 to this interval gives the stopping
distance as
Δx = ( 20 m s ) ( 20 s ) +
2.16
2.35
1
( −1.0 m s2 ) ( 20 s )2 = 200 m
2
We choose x = 0 and t = 0 at the location of Sue’s car when she first spots the van and applies
the brakes. Then, the initial conditions for Sue’s car are x0 S = 0 and v0 S = 30.0 m s. Her
constant acceleration for t ≥ 0 is aS = −2.00 m s 2. The initial conditions for the van
are x0V = 155 m, v0V = 5.00 m s , and its constant acceleration is aV = 0. We then use
Δx = x − x0 = v0 t + 12 at 2 to write an equation for the x-coordinate of each vehicle for t ≥ 0.
This gives
Sue’s Car:
Van:
1
( −2.00 m s2 ) t 2
2
1
xV − 155 m = ( 5.00 m s ) t + ( 0 ) t 2
2
xS − 0 = ( 30.0 m s ) t +
or
xS = ( 30.0 m s ) t − (1.00 m s 2 ) t 2
or
xV = 155 m + ( 5.00 m s ) t
In order for a collision to occur, the two vehicles must be at the same location ( i.e., xS = xV ) .
Thus, we test for a collision by equating the two equations for the x-coordinates and see if the
resulting equation has any real solutions.
( 30.0 m s ) t − (1.00 m s2 ) t 2 = 155 m + ( 5.00 m s ) t
⇒
xS = xV
(1.00
or
m s 2 ) t 2 − ( 25.00 m s ) + 155 m = 0
Using the quadratic formula yields
t=
− ( −25.00 m s ) ±
( −25.00 m s )2 − 4 (1.00 m s2 ) (155 m )
2 (1.00 m s 2 )
= 13.6 s or 11.4 s
The solutions are real, not imaginary, so a collision will occur . The smaller of the two solutions
is the collision time. (The larger solution tells when the van would pull ahead of the car again
if the vehicles could pass harmlessly through each other.) The x-coordinate where the collision
occurs is given by
xcollision = xS
2.17
2.37
2.38
67352_ch02.indd 26
t =11.4 s
= xV
t =11.4 s
= 155 m + ( 5.00 m s ) (11.4 s ) = 212 m
v − v0 24.0 m s 2 − 0
=
= 8.14 m s 2
Δt
2.95 s
(a)
a=
(b)
From a = Δv Δt , the required time is Δt =
(c)
Yes. For uniform acceleration, the change in velocity Δv generated in time Δt is given by
Δv = a ( Δt ). From this, it is seen that doubling the length of the time interval Δt will always
double the change in velocity Δv. A more precise way of stating this is: “When acceleration
is constant, velocity is a linear function of time.”
(a)
From
v f − vi
a
=
20.0 m s − 10.0 m s
= 1.23 s
8.14 m s 2
2/9/11 1:10:45 PM
27
13
Motion in One Dimension
2.18
2.39
At the end of the acceleration period, the velocity is
v = v0 + ataccel = 0 + (1.5 m s 2 ) ( 5.0 s ) = 7.5 m s
This is also the initial velocity for the braking period.
v f = v + at brake = 7.5 m s + ( −2.0 m s 2 ) ( 3.0 s ) = 1.5 m s
(a)
After braking,
(b)
The total distance traveled is
v + v0 ⎞
⎛ vf + v ⎞
Δxtotal = ( Δx )accel + (Δ x ) brake = ( v t )accel + (v t ) brake = ⎛⎜
⎟ taccel + ⎜
⎟ t brake
⎝ 2 ⎠
⎝ 2 ⎠
7.5 m s + 0 ⎞
⎛ 1.5 m s + 7.5 m s ⎞ 3.0 s = 32 m
Δxtotal = ⎛⎜
)
⎟ ( 5.0 s ) + ⎜
⎟(
2
2
⎝
⎠
⎝
⎠
2.41
2.19
(a)
Take t = 0 at the time when the player starts to chase his opponent. At this time, the
opponent is distance d = (12 m s ) ( 3.0 s ) = 36 m in front of the player. At time t > 0, the
displacements of the players from their initial positions are
1
1
Δxplayer = ( v0 )player t + aplayer t 2 = 0 + ( 4.0 m s 2 ) t 2
2
2
1
2
Δxopponent = ( v0 )opponent t + aopponent t = (12 m s ) t + 0
2
and
When the players are side-by-side, Δxplayer = Δxopponent + 36 m
[1]
[2]
[3]
Substituting Equations [1] and [2] into Equation [3] gives
1
( 4.0 m s2 ) t 2 = (12 m s ) t + 36 m
2
or
t 2 + ( − 6.0 s ) t + ( −18 s 2 ) = 0
Applying the quadratic formula to this result gives
t=
− ( − 6.0 s ) ±
( − 6.0 s )2 − 4 (1) ( −18 s2 )
2 (1)
which has solutions of t = −2.2 s and t = +8.2 s. Since the time must be greater than zero,
we must choose t = 8.2 s as the proper answer.
(b)
2.45
2.20
(a)
1
1
2
Δxplayer = ( v0 )player t + aplayer t 2 = 0 + ( 4.0 m s 2 ) (8..2 s ) = 1.3 × 10 2 m
2
2
From the instant the ball leaves the player’s hand until it is caught, the ball is a freely falling
body with an acceleration of
a = − g = −9.80 m s 2 = 9.80 m s 2 ( downward )
(b)
At its maximum height, the ball comes to rest momentarily and then begins to fall back
downward. Thus, vmax = 0 .
height
(c)
Consider the relation Δy = v0 t + 12 at 2 with a = − g . When the ball is at the thrower’s hand,
the displacement is Δy = 0, giving
0 = v0 t − 12 gt 2
67352_ch02.indd 27
2/9/11 1:10:46 PM
14
28
2.18
Chapter 2
(a)
(b)
(d)
2.21
2.47
(a)
In order
for thehas
trailing
athlete totbe
able
to catch
the leader,
his speed
(v1) was
mustthrown,
be greater
This
equation
two solutions,
= 0,
which
corresponds
to when
the ball
than tthat
leading athleteto(vwhen
),
and
the
distance
between
the
leading
athlete
and theat
= 2 vof0 the
g, corresponding
the
ball
is
caught.
Therefore,
if
the
ball
is
caught
and
2
line
be great
enough
to give
trailing athlete sufficient time to make up the
tfinish
= 2 .00
s, must
the initial
velocity
must
have the
been
deficient distance, d.
2
gt ( 9.80 m s ) ( 2 .00 s )
v
=
=
= 9.80 m s
0
During a time
2 t the leading
2 athlete will travel a distance d2 = v2 t and the trailing athlete
will travel a distance d1 = v1t . Only when d1 = d2 + d (where d is the initial distance the
trailing
thev leader)
willmaximum
the trailing
athlete have caught the leader.
From
v 2athlete
= v02 + was
2a ( Δbehind
y ), with
= 0 at the
height,
Requiring that this condition be satisfied gives the elapsed time required for the second
2
athlete to overtake
0 − ( 9.80 m s )
v 2 −the
v02 first:
=
= 4.90 m
( Δy )max =
2a
2 ( − 9.80 m s 2 )
The velocity of the object when it was 30.0 m above the ground can be determined by
applying Δy = v0 t + 12 at 2 to the last 1.50 s of the fall. This gives
1
m
2
−30.0 m = v0 (1.50 s ) + ⎛⎜ −9.80 2 ⎞⎟ (1.50 s )
2⎝
s ⎠
(b)
or
v0 = −12.7 m s
The displacement the object must have undergone, starting from rest, to achieve this velocity at a point 30.0 m above the ground is given by v 2 = v02 + 2a ( Δy ) as
( Δy )1 =
v 2 − v02 ( −12.7 m s ) − 0
=
= −8.23 m
2a
2 ( −9.80 m s 2 )
2
The total distance the object drops during the fall is then
The rocks have the same
acceleration,
but
the
rock
thrown
downward
has
a
higher
average
speed
between the two
( Δy )total = (−8.23 m ) + (− 30.0 m ) = 38.2 m
levels, and is accelerated over a smaller time interval.
2.22
2.51
(a)
From v 2 = v02 + 2a ( Δy ) with v = 0, we have
( Δy )max =
v 2 − v02 0 − ( 25.0 m s )
=
= 31.9 m
2a
2 ( −9.80 m s 2 )
2
(b)
The time to reach the highest point is
(c)
v − v0 0 − 25.0 m s
=
= 2.55 s
a
− 9.80 m s 2
The time required for the ball to fall 31.9 m, starting from rest, is found from
t up =
2 ( Δy )
2 ( − 31.9 m )
=
= 2.55 s
a
− 9.80 m s 2
The velocity of the ball when it returns to the original level (2.55 s after it starts to fall from
rest) is
1
Δy = ( 0 ) t + at 2 as t =
2
(d)
v = v0 + at = 0 + ( − 9.80 m s 2 ) ( 2 .55 s ) = − 25.0(the
m ssame as earlier!)
2.53
2.23
(a)
After its engines stop, the rocket is a freely falling body. It continues upward, slowing
under the influence of gravity until it comes to rest momentarily at its maximum altitude.
Then it falls back to Earth, gaining speed as it falls.
(b)
When it reaches a height of 150 m, the speed of the rocket is
v = v02 + 2a ( Δy ) =
67352_ch02.indd 28
( 50.0 m s )2 + 2 ( 2 .00 m s2 ) (150 m ) = 55.7 m s
2/9/11 1:10:47 PM
Motion in One Dimension
(b)
15
29
1
After the engines
continues
moving
an initial velocity of
Applying
Δx = vi (stop,
Δt ) +the
each of the
time upward
intervalswith
gives
( Δt ) to
2 arocket
v0 = 55.7 m s and acceleration a = − g = −9.80 m s 2 . When the rocket reaches maximum
height, v = 0. The displacement of the rocket above the point where the engines stopped
for
(that is, above the 150 m level) is
2
v 2 − v02 0 − ( 55.7 m s )
=
= 158 m
2a
2 ( − 9.80 m s 2 )
2
Δy =
The maximum height above ground that the rocket reaches is then given by
hmax = 150 m + 158 m = 308 m
(c)
The total time of the upward motion of the rocket is the sum of two intervals. The first is
the time for the rocket to go from v0 = 50.0 m s at the ground to a velocity of v = 55.7 m s
at an altitude of 150 m. This time is given by
t1 =
( Δy )1
v1
=
( Δy )1
( v + v0 )
2
=
2 (150 m )
= 2 .84 s
( 55.7 + 50.0 ) m s
Motion in One Dimension
41
The second interval is the time to rise 158 m starting with v0 = 55.7 m s and ending with v = 0.
This time is
t2 =
( Δy )2
v2
=
( Δy )2
( v + v0 )
2
=
2 (158 m )
= 5.67 s
0 + 55.7 m s
t up = t1 + t 2 = ( 2 .84 + 5.67 ) s = 8.51 s
The total time of the upward flight is then
2.58
(d)
(a)
(b)
2.55
2.24
The time for the rocket to fall 308 m back to the ground, with v0 = 0 and acceleration
1
a = − g = −9.80 m s 2 , is found from Δy = v0 t + at 2 as
2
t down =
2 ( Δy )
=
a
2 ( − 308 m )
= 7.93 s
− 9.80 m s 2
(c)
so the total time of the flight is
(a)
(d)
The acceleration of the bullet is
t flight = t up + t down = (8.51 + 7.93) s = 16.4 s
v 2 − v02 ( 300 m/s ) − ( 400 m/s )
5
2
=
= − 3.50 × 10 m s
2 ( Δx )
2 ( 0.100 m )
2
a=
(e)
(b)
(f)
2.25
2.59
67352_ch02.indd 29
2
The time of contact with the board is
s
v − vdid
( 300 − 400 ) amconstant
− 4 drawings
0 not
If thet speed
rate,
the
× 10
s
=
= change at
= 2.86
5
− 3.50 × 10than
mthose
s 2 given above.
would havea less regularity
Once the gymnast’s feet leave the ground, she is a freely falling body with constant acceleration a = − g = −9.80 m s 2 . Starting with an initial upward velocity of v0 = 2.80 m s, the vertical
displacement of the gymnast’s center of mass from its starting point is given as a function of time
by Δy = v0 t + 12 at 2.
(a)
At t = 0.100 s,
Δy = ( 2.80 m s ) ( 0.100 s ) − ( 4.90 m s 2 ) ( 0.100 s ) = 0.231 m
(b)
At t = 0.200 s,
Δy = ( 2.80 m s ) ( 0.200 s ) − ( 4.90 m s 2 ) ( 0.200 s ) = 0.364 m
(c)
At t = 0.300 s,
Δy = ( 2.80 m s ) ( 0.300 s ) − ( 4.90 m s 2 ) ( 0.300 s ) = 0.399 m
(d)
At t = 0.500 s,
Δy = ( 2.80 m s ) ( 0.500 s ) − ( 4.90 m s 2 ) ( 0.500 s ) = 0.175 m
2
2
2
2
2/9/11 1:10:48 PM
30
16
Chapter 2
(ii)
(d)
2.26
2.61
the sline
to the
At any
t = 0instant,
.500 s, the instantaneous
Δy = ( 2.80 macceleration
s ) ( 0.500 s )equals
− ( 4.90themslope
s 2 ) ( 0of
.500
0.175 m
)2 =tangent
v vs. t graph at that point in time.
The falling ball moves a distance of (15 m − h ) before they meet, where h is the height above the
At they
t = 2.0
s, the
slopeΔy
of =the
is
ground(a)where
meet.
Apply
v0tangent
t + 12 at 2 ,line
withtoathe
= −curve
g, to obtain
− (15 m − h ) = 0 −
1 2
gt
2
or
h = 15 m −
1 2
gt
2
[1]
1
Applying Δy = v0 t + at 2 to the rising ball gives
2
h = ( 25 m s ) t −
1 2
gt
2
[2]
Combining Equations [1] and [2] gives
1
2
1
2
( 25 m s ) t − gt 2 = 15 m − gt 2
15 m
or
t=
= 0.60 s
12 minutes.25 m s
2.65
2.27
(a)
of running 1.0 mile in
The keys have acceleration a = − g = −9.80 m s 2 from the release point until they are
caught 1.50 s later. Thus, Δy = v0 t + 12 at 2 gives
2
Δy − at 2 2 ( + 4.00 m ) − ( − 9.80 m s ) (1.50 s ) 2
=
= +10.0 m s
t
1.50 s
2
v0 =
or
(b)
v0 = 10.0 m s upward
The velocity of the keys just before the catch was
v = v0 + at = 10.0 m s + ( − 9.80 m s 2 ) (1.50 s ) = − 4.70 m s
or
2.28
2.73
v = 4.70 m s downward
.
The time required for the stuntman to fall 3.00 m, starting from rest, is found from
Δy = v0 t + 12 at 2 as
−3.00 m = 0 +
(a)
1
( −9.80 m s2 ) t 2
2
so
t=
2 ( 3.00 m )
= 0.782 s
9.80 m s 2
With the horse moving with constant velocity of 10.0 m s, the horizontal distance is
Δx = vhorse t = (10.0 m s ) ( 0.782 s ) = 7.82 m
(b)
2.74
67352_ch02.indd 30
The required time is t = 0.782 s as calculated above.
The distance the glider moves during the time
2/9/11 1:10:49 PM
장 ‫ث‬ਫ਼ࠪࢇ८ࡕࡋѰ
PROBLEM SOLUTIONS
3.1
LI
Your sketch should be drawn to scale, and be similar to that pictured below. The length of R and
the angle θ can be measured to find, with use of your scale factor, the magnitude and direction of
the resultant displacement. The result should be approximately 421 ft at 3° below the horizonntal .
→
Thus, the answer computed above is only approximately correct.
3.3
3.2
(a)
(b)
LI
In your vector
LI diagram, place theLItailLIof vector B at the
tip of vector A. The vector sum, A + B, is then found as
shown in the vector diagram and should be
LI LI
A + B = 5.0 units at –53°
LI LI LI
LI
To find the
LI vector difference A L−I B = A + (−B), form the
vector −B (same
magnitude as B, opposite direction) and add
LI
it to vector A as shown in the diagram. You should find that
LI LI
A − B = 5.0 units at +53°
PROBLEM SOLUTIONS
3.1
3.3
3.5
from the positive x-axis.
LI
LI LIto scale, and be similar to that pictured below. The length of R
LI
Your
sketch
should
be=drawn
and
We are
given
that R
A + B . When two vectors are added graphically,
θ
the
angle
can
be
measured
to
fi
nd,
with
use
of
your
scale
factor,
the
magnitude
and
direction
of
the second vector is positioned with its tail at the tip of the first vector.
n
tal
.
the
resultant
displacement.
The
result
should
be
approximately
421
ft
at
3°
below
the
horizon
The resultant then runs from the tail
LI of the first vector to the tip of the
second vector. In this case, vector A will be positioned with its tail at the
origin and its tip at the point (0, 29). The resultant is then drawn, starting
at the origin (tail of first vector) and going 14 units in
LI the negative
y-direction to the
point
(0,
−14).
The
second
vector,
B
,LImust then start
LI
from the tip of A at point (0, 29) and end on the tip of R at point (0, −14)
as shown in the sketch at the right. From→this, it is seen that
LI
B is 43 units in the negative y-direction
, illustrating that the sum of a set of vectors is not affected by
the order in which the vectors are added.
3.7
.
LI
The
LI displacement vectors A = 8.00 m westward and
B = 13.0LIm north can be drawn to scale as at the right. The
vector C represents the displacement that the man in the
maze must undergo to return to his starting point. LI
The
scale used to draw the sketch can be used to find C to be
15 m at 58° S of E
17
67352_ch03.indd 49
2/9/11 1:11:53 PM
18
50
1
3.5
3.9
Chapter 3
Using a vector diagram, drawn to scale,
like that shown at the right, the displacement
from Lake
LI B back to base camp is given by the
vector D. Measuring the length of this vector
and multiplying by the chosen scale factor should
give the magnitude of this displacement as 310 km.
Measuring the angle θ should yield a value of 57°.
Thus, the displacement from B to the base camp is
D
LI
D = 310 km at θ = 57° S of W
Vectors and Two-Dimensional Motion
3.6
3.11
59
LI
The x- and y-components of vector A are its projections on lines parallel to the x- and y-axes,
respectively, as shown in the sketch. The magnitude of these components can be computed using
the sine and cosine functions as shown below:
LI
LI
Ax = A cos 325° = + A cos 35° = ( 35.0 ) cos 35° = 28.7 units
and
LI
LI
Ay = A sin 325° = − A sin 35° = − ( 35.0 ) sin 35° = −20.1 units
3.7
3.13
(a)
LI
A = 3.10 km at 25.0° north of east
Her net x (east-west) displacement is − 3.00 + 0 + 6.00 = + 3.00 blocks, while her net y
(north-south) displacement is 0 + 4.00 + 0 = + 4.00 blocks. The magnitude of the resultant
displacement is
R = ( Σx ) + ( Σy ) = ( 3.00 ) + ( 4.00 ) = 5.00 blocks
2
2
2
2
and the angle the resultant makes with the x-axis (eastward direction) is
Σy
4.00 ⎞
−1
θ = tan −1 ⎛⎜ ⎞⎟ = tan −1 ⎛⎜
⎟ = tan (1.33) = 53.1°
⎝ Σx ⎠
⎝ 3.00 ⎠
The resultant displacement is then 5.00 blocks at 53.1° N of E .
(b)
3.14
3.8
3.15
Because of the commutative property of vector addition, the net displacement is the
The total distance traveled is 3.00 + 4.00 + 6.00 = 13.0 blocks .
same regardless of the order in which the individual displacements are executed.
(a)
a coordinate
system
the positive
eastward
positive
y-axis
After We
3.00choose
h moving
at 41.0 km/h,
thewith
hurricane
is 123 x-axis
km at 60.0°
N ofand
W the
from
the island.
In the
northward.
The hiker
thendue
undergoes
four
successiveof
displacements,
given beloware:
along
next 1.50
h, it travels
37.5 km
north. The
components
these two displacements
with their x- and y-components:
Displacement
x-component (eastward)
y-component (northward)
123 km
–61.5 km
+107 km
37.5 km
0
+37.5 km
Resultant
–61.5 km
144 km
Therefore, the eye of the hurricane is now
R = (− 61.5 km ) + (144 km ) = 157 km from the islaand
2
67352_ch03.indd 50
2
2/9/11 1:11:55 PM
19
51
Vectors and Two-Dimensional Motion
.
3.9
5.
3.17
The
traveled) by the athlete is
Ay (and
= 40.distance
0
−25.followed
0
Ax =path
shown in the sketch, along with the vectors for the initial
2
2
position,
A = Ax2 fi
+nal
Ay2 position,
= ( −25.and
0 ) +change
=position.
47.2 units
( 40.0 )in
The
speed we
for fithe
time interval
Fromaverage
the triangle,
nd elapsed
that
⎛ A ⎞
⎛ 40.0 ⎞
φ = tan −1 ⎜ y ⎟ = tan −1 ⎜
⎟ = 58.0°, so θ = 180° − φ = 122°
⎜ A ⎟
⎝ 25.0 ⎠
⎝ x ⎠
LI
Thus, A = 47.2 units at 122° counterclockwise from thee +x -axis .
3.19
3.10
I LI
I
The components of the displacements a, b, and c are
ax = a ⋅ cos 30.0° = + 152 km
(north)
bx = b ⋅ cos 110° = − 51.3 km
cx = c ⋅ cos180° = − 190 km
ay = a ⋅ sin 30.0° = + 87.5 km
and
by = b ⋅ sin110° = + 141 km
(east)
cy = c ⋅ sin180° = 0
Thus, Rx = ax + bx + cx = − 89.3 km , and Ry = ay + by + cy = + 229 km
(
)
R = Rx2 + Ry2 = 246 km, and θ = tan −1 Rx Ry = tan −1 ( 0.390 ) = 21.3°
so
City C is 246 km at 21.3° W of N from the starting point.
3.20
3.21
3.11
(a) single displacement required to sink the putt in one stroke is equal to the resultant of the
The
three actual putts used by the novice. Taking east as the positive x-direction and north as the positive y-direction, the components of the three individual putts and their resultant are
Ay = + 4.00 m
A =0
x
62
Bx = ( 2.00 m ) cos 45.0° = +1.41 m
By = ( 2.00 m ) sin 45.0° = +1.41 m
C x = − (1.00 m ) sin 30.0° = −0.500 m
C y = − (1.00 m ) cos 30.0° = −0.866 m
Rx = Ax + Bx + C x = + 0.910 m
Ry = Ay + By + C y = +4.55 m
Chapter 3
The magnitude and direction of the desired resultant is then
R = Rx2 + Ry2 = 4.64 m
Thus,
3.23
3.12
67352_ch03.indd 51
and
⎛ Ry ⎞
θ = tan −1 ⎜ ⎟ = +78.7°
⎝ Rx ⎠
LI
R = 4.64 m at 78.7° north of east
.
(a)
With the origin chosen at point O as shown in Figure P3.2, the coordinates of the original
position of the stone are x0 = 0 and y0 = +50.0 m .
(b)
The components of the initial velocity of the stone are v0 x = +18.0 m s and v0 y = 0 .
(c)
The components of the stone’s velocity during its flight are given as functions of time by
2/9/11 1:11:55 PM
..
52
20
Chapter 3
(c)
(c)
The
The components
components of
of the
the stone’s
stone’s velocity
velocity during
during its
its flflight
ight are
are given
given as
as functions
functions of
of time
time by
by
the vector difference
vx = v0 x + ax t = 18.0 m s + ( 0 ) t
v y = v0 y + a y t = 0 + ( − g ) t
and
(d)
or
or
vx = 18.0 m s
vy = − ( 9.80 m s 2 ) t
The coordinates of the stone during its flight are
x = x 0 + v0 x t +
1 2
1
ax t = 0 + (18.0 m s ) t + ( 0 ) t 2
2
2
or
x = (18.0 m s ) t
1 2
1
ay t = 50.0 m + ( 0 ) t + ( − g ) t 2 or
y = 50.0 m − ( 4.90 m s 2 ) t 2
2
2
1
We find the time of fall from Δy = v0 y t + ay t 2 with v0 y = 0 :
2
and y = y0 + v0 y t +
(e)
t=
(f)
2 ( Δy )
=
a
2 ( − 50.0 m )
= 3.19 s
− 9.80 m s 2
At impact, vx = v0 x = 18.0 m s, and the vertical component is
vy = v0 y + ay t = 0 + ( − 9.80 m s 2 ) ( 3.19 s ) = − 31.3 m s
v = vx2 + vy2 =
Thus,
and
or
3.13
3.25
(a)
(18.0 m s )2 + ( −31.3 m s )2 = 36.1 m s
⎛ vy ⎞
⎛ −31.3 ⎞
θ = tan −1 ⎜ ⎟ = tan −1 ⎜
= −60.1°
v
⎝ 18.0 ⎟⎠
⎝ x⎠
I
v = 36.1 m s at 60.1° below the horizontal
The time required for the ball to travel Δx = 36.0 m horizontally to the goal is found from
Δx = v0 x t = ( v0 cos θ ) t as t = Δx v0 cos θ .
At this time, the vertical displacement of the ball is given by Δy = v0 y t + 12 ay t 2. The vertical
distance by which the ball clears the bar is d = Δy − h , or
⎛ 36.0 m ⎞ 1
⎛
⎞
36.0 m
d = v0 sin 53.0° ⎜
⎟ + −9.80 m s2 ) ⎜
⎟ − 3.05 m
⎜ v0 cos 53.0° ⎟ 2 (
⎝ ( 20.0 m s ) cos 53.0° ⎠
⎝
⎠
2
yielding d = + 0.89 m. Thus, the ball clears the crossbar by 0.89 m .
(b)
64
The ball reaches the plane of the goal post at t = Δx v0 cos θ , or
Chapter 3
t=
36.0 m
= 2.99 s
( 20.0 m s ) cos 53.0°
At this time, its vertical velocity is given by v y = v0 y + ay t as
v y = ( 20.0 m s ) sin 53.0° + ( −9.80 m s 2 ) ( 2.99 s ) = −13.3 m s
Since v y < 0, the ball has passed the peak of its arc and is descending when it crosses the
crossbar.
3.26
67352_ch03.indd 52
(a)
At
2/9/11 1:11:57 PM
independent
the
local acceleration
of
Vectors of
and
Two-Dimensional
Motion
gravity.
3.14
3.27
21
53
(ii)
observerheight
on thevtruck
moves with the same horizontal motion as does the apple. This
At theAn
maximum
y = 0, and the time to reach this height is found from
observer does not detect any horizontal motion of the apple relative to him. However, this
vy − v0 y accelerated
0 − v0 y v0vertical
does detect
motion
of the apple. The
curve best65
and Two-Dimensional
Motion
vobserver
as tthe
= uniformly
=
= y Vectors
y = v0 y + a y t
ay as seen
− gby the gobserver on the truck is (b).
describing the path of the apple
continuedThe
on next pagedisplacement that has occurred during this time is
13.
The vertical
boat moves with a constant horizontal velocity (or its velocity relative to Earth has
2
components of
⎛ v + v ⎞ ⎛ 0 + v0 y ⎞ ⎛ v0 y ⎞ v0 y
( Δy )max = ( vy )av t = ⎜ y 0 y ⎟ t = ⎜
⎜
⎟=
⎟
2
⎝
⎠ ⎝ 2 ⎠ ⎝ g ⎠ 2g
If ( Δy )max = 3.7 m , we find
v0 y = 2 g ( Δy )max = 2 ( 9.80 m s 2 ) ( 3.7 m ) = 8.5 m s
and if the angle of projection is θ = 45°, the launch speed is
v0 =
3.15
3.29
(a)
v0 y
sin θ
=
8.5 m s
= 12 m s
sin 45°
At the highest point of the trajectory, the projectile is moving horizontally with velocity
components of vy = 0 and
vx = v0 x = v0 cos θ = ( 60.0 m/s ) cos 30.0° = 52.0 m s
(b)
The horizontal displacement is Δx = v0 x t = (52.0 m s)(4.00 s) = 208 m and, from
Δy = ( v0 sin θ )t + 12 ay t 2 , the vertical displacement is
1
2
Δy = ( 60.0 m s ) ( sin 30.0°)( 4.00 s ) + ( −9.80 m s 2 ) ( 4.00 s ) = 41.6 m
2
The straight line distance is
d = ( Δx ) + ( Δy ) = ( 208 m ) + ( 41.6 m ) = 212 m
2
2
3.16
3.31
2
2
The speed of the car when it reaches the edge of the cliff is
v = v02 + 2a ( Δx ) = 0 + 2 ( 4.00 m s 2 ) ( 50.0 m ) = 20.0 m s
Now, consider the projectile phase of the car’s motion. The vertical velocity of the car as
it reaches the water is
vy = − v02 y + 2ay ( Δy ) = − ⎡⎣ − ( 20.0 m s ) sin 24.0°⎤⎦ + 2 ( − 9.80 m s 2 ) ( − 30.0 m )
2
or
vy = − 25.6 m s
(b)
The time of flight is
t=
(a)
v y − v0 y
ay
=
− 25.6 m s − ⎡⎣ − ( 20.0 m s ) sin 24.0°⎤⎦
= 1.78 s
− 9.80 m s 2
The horizontal displacement of the car during this time is
Δx = v0 x t = ⎡⎣( 20.0 m s ) cos 24.0°⎤⎦ (1.78 s ) = 32 .5 m
67352_ch03.indd 53
2/9/11 1:11:57 PM
54
22
3.17
3.33
Chapter 3
We choose our origin at the initial position of the projectile. After 3.0 s, it is at ground level, so
the vertical displacement is Δy = − H .
To find H, we use Δy = v0 y t +
1 2
ay t , which becomes
2
1
2
− H = ⎡⎣(15 m s ) sin 25°⎤⎦ ( 3.0 s ) + ( −9.80 m s 2 ) ( 3.00 s ) , or H = 25 m
2
3.35
3.18
(a)
The jet moves at 3.00 × 10 2 mi h due east relative to the air. Choosing a coordinate system
with the positive x-direction eastward and the positive y-direction northward, the components of this velocity are
I
( v JA ) x = 3.00 ×10 2
(b)
and
(d)
I
( v JA ) y = 0
and
The velocity of the air relative to Earth is 1.00 × 10 2 mi h at 30.0° north of east. Using the
coordinate system adopted in (a) above, the components of this velocity are
I
I
I
I
( v AE ) x = v AE
(c)
mi h
cos θ = (1.00 × 10 2 mi h ) cos30.0° = 86.6 mi h
( v AE ) y = v AE sin θ = (1.00 × 10 2
mi h ) sin30.0° = 50.0 mi h
Carefully observe the pattern of the subscripts in Equation 3.16 of the textbook. There,
two objects (cars A and B) both move relative to a third object (Earth, E). The velocity of
object A relative to object B is given in terms of the velocities of these objects relative to E
I
I
I
as v AB = v AE − v BE . In the present case, we have two objects, a jet (J) and the air (A), both
moving relative to a third object, Earth (E). Using the same pattern of subscripts as that in
Equation 3.16, the velocity of the jet relative to the air is given by
I
I
I
v JA = v JE − v AE
I
From the expression for v JA found in (c) above, the velocity of the jet relative to the ground
I
I
I
is v JE = v JA + v AE . Its components are then
I
I
I
( v JE ) x = ( v JA ) x + ( v AE ) x = 3.00 × 10 2
and
I
I
I
mi h + 86.6 mi h = 3.87 × 10 2 mi h
( v JE ) y = ( v JA ) y + ( v AE ) y = 0 + 50.0
mi h = 50.0 mi h
This gives the magnitude and direction of the jet’s motion relative to Earth as
I
v JE =
and
I 2 I
v JE x + v JE
2
y
I
⎛ ( v JE )
θ = tan −1 ⎜ I y
⎜ ( v JE )
x
⎝
Therefore,
and
=
(3.87 × 10
2
mi h ) + ( 50.0 mi h ) = 3.90 × 10 2 mi h
2
2
⎞
⎛ 50.0 mi h ⎞
⎟ = tan −1 ⎜
⎟ = 7.36°
2
⎟
⎝ 3.87 × 10 mi h ⎠
⎠
I
v JE = 3.90 × 10 2 mi h at 7.36° north of east
(North)
3.19
3.37
During the trip of duration t, Ithe displacement of the
plane relative to the ground, d PG , is to have a magnitude
of 750 km and be directed due north. We choose the
positive y-axis to be directed northward and the positive
x-axis directed eastward. During Ithe trip, the plane’s
displacement
relative to the air d PA has magnitude
I
d PA = ( 630 km h ) t and is directed at some angle α
relative to the y-axis.
The displacement
I
I of the air relative
to the ground, d AG , has magnitude d AG = ( 35.0 km h ) t and
is assumed to be at angle β from the y-axis.
I
I
I
Since these relative displacements are related by d PA = d PG − d AG
67352_ch03.indd 54
dAG
␤
dPG
dPA
␣
x (East)
I
I
I
d PG = d PA + d AG
2/9/11 1:11:58 PM
Vectors and Two-Dimensional Motion
40.
− ( 630 km h ) ⋅ t ⋅ sin α + ( 35.0 km h ) ⋅ t ⋅ sin β = 0
I
Equating y-components in the vector triangle gives d PA
35.0 ⎞
sin α = ⎛⎜
⎟ sin β
⎝ 630 ⎠
I
I
cos α + d AG cos β = d PG , or
and
⎡⎣( 630 km h ) cos α + ( 35.0 km h ) cos β ⎤⎦ t = 750 km h
(a)
(b)
[2]
km h − 35.0 km h ] t = 750 km h
or
t=
750 km h
= 1.26 h
595 km h
The wind blows northward as a tailwind (α = β = 0°) , and Equation [2] yields
[ 630
(c)
[1]
The wind blows toward the south ( β = 180°) and is a headwind for the plane (α = 0°) .
Then, Equation [2] gives
[ 630
70
23
55
I
d PA = ( 630 km h ) t and is directed at some angle α
relative
to the y-axis.
The displacement
(a)
I of the air relative
I
to the ground, d AG , has magnitude d AG = ( 35.0 km h ) t and
is assumed to be at angle β from the y-axis.
I
I
I
I
I
I
Since these relative displacements are related by d PA = d PG − d AG , or d PG = d PA + d AG , they
form
as shown above. Equating the x-components in the vector triangle gives
I a vector triangle
I
− d PA sin α + d AG sin β = 0, or
km h + 35.0 km h ] t = 750 km h
or
t=
750 km h
= 1.13 h
665 km h
The wind blows due East, so β = 90.0°. Then Equation [1] requires that
35.0 ⎞
sin α = ⎛⎜
⎟ sin 90.0° = 0.056
⎝ 630 ⎠
Chapter 3
and
α = 3.18°
or the plane must fly 3.18° W of N relative to the air to maintain a due north heading
relative to the ground.
Finally, Equation [2] for this case gives
⎡⎣( 630 km h ) cos 3.18° + ( 35.0 km h ) cos 90.0°⎤⎦ t = 750 km h
or
3.41
3.20
72
(a)
t=
750 km h
( 629 km h + 0 )
= 1.19 h
Both the student (S) and the water (W) move relative to Earth (E). The velocity of the stuI
I
I
I
I
dent relative to the water is given by v SW = v SE − v WE , where v SE and v WE are the velocities
of the student relative to Earth and the water relative to Earth, respectively. If we choose
I
I
downstream as the positive direction, then v WE = + 0.500 m s , v SW = −1.20 m s when the
I
student is going up stream, and v SW = +1.20 m s when the student moves downstream.
The velocity of the student relative to Earth for each leg of the trip is
I
I
I
( vSE )upstream = v WE + ( vSW )upstream = 0.500 m s + ( −1.20 m s) = −0.700 m s
Chapter 3
and
I
(v )
SE downstream
I
I
= v WE + ( v SW )downstream = 0.500 m s + ( +1.20 m s ) = +1.70 m s
The distance (measured relative to Earth) for each leg of the trip is
d = 1.00 km = 1.00 × 10 3 m. The times required for each of the two legs are
t upstream = I
v SE
67352_ch03.indd 55
d
upstream
=
1.00 × 10 3 m
= 1.43 × 10 3 s
0.700 m s
2/9/11 1:11:59 PM
56
24
Chapter 3
. The times required for each of the two legs are
and
t upstream = I
v SE
d
t downstream = I
v SE
d
=
upstream
1.00 × 10 3 m
= 1.43 × 10 3 s
0.700 m s
=
downstream
1.00 × 10 3 m
= 5.88 × 10 2 s
1.70 m s
so the time for the total trip is
t total = t upstream + t downstream = 1.43 × 10 3 s + 5.88 × 10 2 s = 2.02 × 10 3 s
(b)
I
If the water had been still ( v WE = 0 ), the speed of the student relative to Earth would have
I
I
I
been the same for each leg of the trip, v SE = v SE upstream = v SE downstream = 1.20 m s. In this
case, the time for each leg and the total time would have been
d
1.00 × 10 3 m
t leg = I =
= 8.33 × 10 2 s, and
v SE
1.20 m s
(c)
3.42
3.45
3.21
(a)
(a)
t total = 2t leg = 1.67 × 10 3 s
The time savings going downstream with the curreent is always less
than the extra time required to go the same distance against the current.
The time
speedofofflthe
relative
to =
shore
The
ightstudent
is found
from Δy
v0 y tis+ 12 ay t 2 with Δy = 0, as t = 2 v0 y g.
2 v0 x v0 y
This gives the range as
R = v0 x t =
On Earth this becomes
REarth = 2 v0 x v0 y gEarth
and on the Moon,
RMoon = 2 v0 x v0 y gMoon
g
Dividing RMoon by REarth , we find RMoon = ( gEarth gMoon ) REarth. With gMoon = gEarth 6, this gives
RMoon = 6 REarth = 6 ( 3.0 m ) = 18 m
(b)
3.47
3.22
(a)
Similarly, RMars = ( gEarth gMars ) REarth = 3.0 m 0.38 = 7.9 m
The known parameters for this jump are: θ0 = −10.0°, Δx = 108 m, Δy = −55.0 m, ax = 0,
and ay = − g = −9.80 m s.
Since ax = 0, the horizontal displacement is Δx = v0 x t = ( v0 cos θ0 ) t , where t is the total
time of the flight. Thus, t = Δx ( v0 cos θ0 ).
The vertical displacement during the flight is given by
Δy = v0 y t +
1 2
gt 2
ay t = ( v0 sin θ0 ) t −
2
2
or
⎛
⎞ g ⎛ Δx ⎞
⎡ g ( Δx )2 ⎤ 1
Δx
Δy = v0 sin θ0 ⎜
= ( Δx ) tan θ0 − ⎢
⎟− ⎜
⎟
⎥ 2
2
⎜ v0 cos θ0 ⎟ 2 ⎝ v0 cos θ0 ⎠
⎣ 2 cos θ0 ⎦ v0
⎝
⎠
Thus,
( )
[ Δy − ( Δx ) tan θ0 ] = − ⎢ 2 cos
2
θ
(
2
)
⎡ g Δx
⎣
− g ( Δx )
=
2 [ Δy − ( Δx ) tan θ0 ] cos 2 θ0
2
or
v0 =
⎤ 1
⎥ 2
0 ⎦ v0
2
− ( 9.80 m s 2 ) (108 m )
2
2 [ −55.0 m − (108 m ) tan ( −10.0°)] cos 2 ( −10.0°)
yielding
the skier glides through the air much like a bird, prolonging
the jump.
3.48
67352_ch03.indd 56
The vertical displacement from the launch
point (top of the building) to the top of
the arc may be found from
2/9/11 1:12:00 PM
Vectors and Two-Dimensional Motion
3.48
3.23
57
25
(b)
The vector difference−1.143 × 10 5 m 3 s 2
= 40.5 m s
v0 =
yielding
−69.75 m
(b)
Rather than falling like a rock, the skier glides through the air much like a bird, prolonging
the jump.
The vertical displacement from the launch
AL = (top
v1t =of( 90
.0building)
km h ) ( 2to
.50the
h )top
= 225
point
the
of km
the arc may be found from
BD = AD − AB = AL cos 40.0° − 80.0 km = 92.4 km
From the triangle BLD,
or
3.49
BL =
( BD ) + ( DL )
BL =
( 92.4 km )2 + ( AL sin 40.0°) = 172 km
3.24
3.51
2
2
Since Car 2 travels this distance in 2.50 h, its constant speed is
v2 =
76
2
Chapter 3
172 km
= 68.8 km h
2.50 h
The distance, s, moved in the first 3.00 seconds is given by
s = v0 t +
1 2
1
2
at = (100 m s ) ( 3.00 s ) + ( 30.0 m s 2 ) ( 3.000 s ) = 435 m
2
2
Choosing the origin at the point where the rocket was launched, the coordinates of the rocket at
the end of powered flight are
x1 = s ( cos 53.0°) = 262 m
y1 = s ( sin 53.0°) = 347 m
and
The speed of the rocket at the end of powered flight is
v1 = v0 + at = 100 m s + ( 30.0 m s 2 ) ( 3.00 s ) = 190 m s
so the initial velocity components for the free-fall phase of the flight are
v0 x = v1 cos 53.0° = 114 m s
(a)
v0 y = v1 sin 53.0° = 152 m s
and
When the rocket is at maximum altitude, vy = 0. The rise time during the free-fall phase can
be found from vy = v0 y + ay t as
t rise =
0 − v0 y
ay
=
0 − 152 m
= 15.5 s
− 9.80 m s 2
The vertical displacement occurring during this time is
0 + 152 m
⎛ v + v0 y ⎞
Δy = ⎜ y
t rise = ⎛⎜
⎟
2
2
⎝
⎝
⎠
The maximum altitude reached is then
s⎞
3
⎟ (15.5 s ) = 1.18 × 10 m
⎠
H = y1 + Δy = 347 m + 1.18 × 10 3 m = 1.53 × 10 3 m
(b)
After reaching the top of the arc, the rocket falls 1.53 × 10 3 m to the ground, starting with
zero vertical velocity ( v0 y = 0 ). The time for this fall is found from Δy = v0 y t + 12 ay t 2 as
t fall =
67352_ch03.indd 57
2 ( Δy )
=
ay
2 ( −1.53 × 10 3 m )
−9.80 m s 2
= 17.7 s
2/9/11 1:12:02 PM
58
26
Chapter 3
as
3.8
I I I
3
Find the resultant R
I2 (=ΔyF)1 + F2 2graphically
I × 10by
m)
( −of1.53
F
F
placing the ttail
of
at
the
head
=
= 17.7 s
=
2
1
fall
2
−9.80 m s
ay
The total time of flight is
t = t powered + t rise + t fall = ( 3.00 + 15.5 + 17.7 ) s = 36.2 s
(c)
The free-fall phase of the flight lasts for
t 2 = t rise + t fall = (15.5 + 17.7 ) s = 33.2 s
The horizontal displacement occurring during this time is
Δx = v0 x t 2 = (114 m s ) ( 33.2 s ) = 3.78 × 10 3 m
and the full horizontal range is
R = x1 + Δx = 262 m + 3.78 × 10 3 m = 4.04 × 10 3 m
3.25
3.53
Choose the positive direction to be the direction of each car’s motion
to
I relative
I
I Earth.
The
velocity
of
the
faster
car
relative
to
the
slower
car
is
given
by
v
=
v
−
v
SE , where
FS
I
IFE
v FE = + 60.0 km h is the velocity of the faster car relative to Earth, and v SE = 40.0 km h is
the velocity of the slower car relative to Earth.
I
Thus, v FS = + 60.0 km h − 40.0 km h = + 20.0 km h and the time required for the faster car to
move 100 m (0.100 km) closer to the slower car is
t=
67352_ch03.indd 58
d
0.100 km
3600 s ⎞
=
= 5.00 × 10 −3 h ⎛⎜
⎟ = 18.0 s
vFS 20.0 km h
⎝ 1h ⎠
2/9/11 1:12:03 PM
장 ࡋѰࢂ‫ئ‬৕
PROBLEM SOLUTIONS
4.1
4.2
4.3
4.2
⎛ 2000 lbs ⎞ ⎛ 4.448 N ⎞⎡
4 2
w = 2 tons ⎜
⎜
⎟ =( 2 × 10 )N⎤⎥ = 3.1 × 10 2 N
⎝ 1 ton ⎟⎠ ⎝ 1 lb ⎠⎢
⎢⎣ 2 ( 0.82 m ) ⎥⎦
(320 m s )2 − 0
The acceleration of the bullet is given2 by
=
(a) ΣFx = max = ( 6.0 kg ) 2.0 m s = 12 N
2 ( 0.82 m )
(
)
(
Then, ax =
(b)
4.4
4.5
4.3
98
(
)
Chapter 4
f 1 800 N
x
(d)
F 2 000 N
ΣF 2 000 N − 1 800 N
a=
=
= 0.200 m s 2
1 000 kg force to the right on m. Reaction: m exerts an equal magniAction: Mmexerts a gravitational
(b)
tude gravitational force to the left on M.
The distance moved is
(f)
(c)
→
F
→
Action: The charge +Q exerts an electrostatic force to the right on the charge
–q. Reaction:
f
1 2 an equal
1 magnitude2electrostatic
2
The charge
–
q
exerts
force
to
the
left
on
the
charge + Q.
Δx = v0 t + at = 0 + 0.200 m s (10.0 s ) = 10.0 m
2
2
Action: The magnet exerts a force to the right on the iron.
Reaction: The iron exerts an
The final velocity is
v = v0 + at = 0 + 0.200 m s 2 (10.0 s ) = 2.00 m s
equal magnitude force to the left
on the magnet.
(
)
(
)
(a)
a larger
mass
than the
sphere
a larger graviIf F =The
825 sphere
N is thehas
upward
force
exerted
on feather.
the manHence,
by the the
scales,
the experiences
upward acceleration
of the
tational
force
=
mg
than
does
the
feather.
man (and
hence,
theFacceleration
of
the
elevator)
is
g
(b)
4.5
4.13
ΣFx
12 N
=
= 3.0 m s 2
4.0 kg
m
= 25 N
From From
v = v0 the
+ at,second
the acceleration
given to theoffootball
is
(a)
law, the acceleration
the
y
boat is
(e)
4.6
4.11
4.4
)
ΣFy of fall
825isNless
− mforgthe sphere
825 N than for
825theNfeather. This
The time
is because air resistance
ay =
= = 798 N man= 0.266
=
g=
− 9.8 m s 2 = 1.2 m s 2
m −
s 2than
affects
the motion
of
the
feather
more
that
of
the
sphere.
m
75
kg
mman
m
mankg
man
3 000
.
(c) In a vacuum, the time of fall is the same for the sphere and the feather.
In the absence of air
Taking eastward as the positive x-direction, the average horizontal acceleration of the car is
resistance, both objects have the free-fall acceleration
g.
I
(b) The vacceleration
is inmthes −same
25.0
0 direction as2 FR and has magnitude
x − v0 x
=
= +5.00 m s
x =
(d) aIn
a vacuum,
force
5.00
s on the sphere is greater than that on the feather. In the absence
Δt FR the total
a
=
of air resistance, the total force is just the gravitational force, and the sphere weighs more
m
the feather.
Thus, than
the average
horizontal force acting on the car during this 5.00-s period is
N of E
ΣFx = max = ( 970 kg ) ( + 5.00 m s2 ) = + 4.85 × 103 N = 4.85 atkN65.2°
eastward
4.14
4.15
4.6
Since
thewith
twovforces
are perpendicular to each other, their resultant is
Starting
0 y = 0 and falling 30 m to the ground, the velocity of the ball just before it hits is
(
)
v1 = − v02 y + 2a y Δy = − 0 + 2 −9.80 m s 2 ( −30 m ) = −24 m s
The Laws of Motion
99
the rebound,
continued On
on next
page the ball has v y = 0 after a displacement Δy = +20 m. Its velocity as it left the
ground must have been
(
)
v2 = + v 2y − 2a y Δy = + 0 − 2 −9.80 m s 2 ( 20 m ) = +20 m s
Thus, the average acceleration of the ball during
the 2.0-ms contact with the ground was
27
67352_ch04a.indd 89
2/9/11 1:14:53 PM
96
28
(
)
= + 0 − 2 −9.80 m s 2 ( 20 m ) = +20 m s
Chapter 4
Thus, the average acceleration of the ball during the 2.0-ms contact with the ground was
aav =
v2 − v1 +20 m s − ( −24 m s )
=
= + 2.2 × 10 4 m s2
Δt
2.0 × 10 −3 s
upward
The average resultant force acting on the ball during this time interval must have been
(
)
Fnet = maav = ( 0.50 kg ) +2.2 × 10 4 m s2 = +1.1 × 10 4 N
LI
F net = 1.1 × 10 4 N=upward
( F m )2 + ( −g )2 =
or
4.7
4.17
(a)
( F m )2 + g 2
Since the burglar is held in equilibrium, the tension in the
vertical cable equals the burglar’s weight of 600 N .
y
→
T2
Now, consider the junction in the three cables:
→
T1
ΣFy = 0, giving T2 sin 37.0° − 600 N = 0
or
T2 =
37.0°
x
w 600 N
600 N
= 997 N in the inclined cable
sin 37.0°
Also, ΣFx = 0, which yields T2 cos 37.0° − T1 = 0
100
Chapter 4
or
(b)
T1 = ( 997 N ) cos 37.0° = 796 N in the horizontal cable
continued on next page
If the left end of the originally horizontal cable was attached to a point higher up the wall,
the tension in this cable would then have an upward component. This upward component
would support part of the weight of the cat burglar, thus
decreasing the tension in the cable on the right .
4.18
4.19
4.8
Using
thex reference
shown
in the
sketch
From ΣF
= 0, T1 cosaxis
30.0°
− T2 cos
60.0°
= 0 at the right,
we see that
[1]
or
T2 = (1.73) T1
The tension in the vertical cable is the full weight
of the feeder, or Tvertical = 150 N
y
→
T2
→
T1
60.0°
30.0°
x
150 N
Then ΣFy = 0 becomes
T1 sin 30.0° + (1.73 T1 ) sin 60.0° − 150 N = 0
which gives T1 = 75.1 N in the right side cable .
Finally, Equation [1] above gives T2 = 130 N in the left side cable .
4.20
4.21
4.9
The Laws of Motion
If the Force
hip exerts
no force
on two
the leg,
the are
system
must
(a)
diagrams
of the
blocks
shown
at be
the
in equilibrium
with
the
three
forces
shown
in
the
freeright. Note that each block experiences a downward
body diagram.
gravitational force
(
T1
101
T2
)
= ( 3.50 kg ) 9.80 m s 2 = 34.3 N
Thus ΣFx =F0g becomes
Also, each has the same upward acceleration as the
elevator, in this case a y = +1.60 m s 2 .
T2
Fg
Fg
Applying Newton’s second law to the lower block:
67352_ch04a.indd 96
2/9/11 1:15:09 PM
97
29
The Laws of Motion
.
4.7
The weight
of the
bag of sugar
onlaw
Earth
is lower block:
Applying
Newton’s
second
to the
ΣFy = ma y
or
⇒
T2 − Fg = ma y
(
)
T2 = Fg + ma y = 34.3 N + ( 3.50 kg ) 1.60 m s 2 = 39.9 N
Next, applying Newton’s second law to the upper block:
ΣFy = ma y
or
(b)
⇒
T1 − T2 − Fg = ma y
(
)
T1 = T2 + Fg + ma y = 39.9 N + 34.3 N + ( 3.50 kg ) 1.60 m s 2 = 79.8 N
As the acceleration of the system increases, we wish to find the value of a y when the upper
string reaches its breaking point (i.e., when T1 = 85.0 N). Making use of the general relationships derived in (a) above gives
(
)
T1 = T2 + Fg + ma y = Fg + ma y + Fg + ma y = 2Fg + 2ma y
102
4.23
4.22
4.10
Chapter 4or
ay =
T1 − 2Fg
2m
=
85.0 N − 2 ( 34.3 N )
= 2.34 m s 2
2 ( 3.50 kg )
(a)
Force
m = 1.00
kgdiagrams
and mgof= the
9.80two
N blocks are shown at
the right. Note that each block experiences a
downward
gravitational
force
m⎞
⎛ 0.200
a = tan −1 ⎜
= 0.458°
⎟
⎝ 25.0 m ⎠
25.0 m
a
→
25.0 m
a
0.200 m
T
→
T
→
mg
Since a y = 0, this requires that ΣFy = T sina + T sina − mgg = 00,
giving
4.24
4.11
4.25
2T sina = mg
T=
or
9.80 N
= 613 N
2sina
In
each
case,
is measuring
the tension
in theand
cord
The
forces
onthe
thescale
bucket
are the tension
in the rope
theconnecting
weight
2
to
tension
can
be kg
determined
by
applying
Newton’s
second
law
of it.
theThis
bucket,
mg
=
5.0
9.80
m
s
=
49
N.
Choose
the
positive
(
)
I
(with
a = upward
0 for equilibrium)
to the object
attached
direction
and use Newton’s
second
law: to the end of this cord.
(
(a)
ΣFy = ma y
ΣFy = 0 ⇒
)
→
T
T − mg = 0
(
→
)
T − 49 N = ( 5.0 kg ) 3.0 m s 2
or
T = mg = ( 5.00 kg ) 9.80 m s 2
T = 64 N
a
→
mg
The Laws of Motion
4.12
4.27
We choose reference axes that are parallel to and
perpendicular to the incline as shown in the force
diagrams at the right. Since both blocks are in
equilibrium, ax = a y = 0 for each block. Then,
applying Newton’s second law to each block gives
n1
T1
or
⇒
y
q
n
− T1 + T2 + mg sinq = 0
[1]
T2
For Block 2 (mass 2m):
⇒
4.28
mg
x
mg
2m
n2
T +y
+x
− T2 + 2mg sinq = 0
or T2 = 2mg sinq
67352_ch04a.indd 97
+x
T2
m
T1 = T2 + mg sinq
ΣFx = max
+y
m
For Block 1 (mass m):
ΣFx = max
103
[2]
(a)
Substituting Equation [2] into Equation [1] gives
(b)
From Equation [2] above, we have
q
2 mg
T1 = 3mg sinq
T2 = 2mg sinq
Let m1 = 10.0 kg, m2 = 5.00 kg, and q = 40.0°.
2/9/11 1:15:11 PM
→
a
+y
n1
T1
98
m
q
Chapter 4
T1 = T2 + mg sinq
[1]
+x
T2
mg
(d)
Action:
M exerts
For Block
2 (mass
2m):a gravitational force to the right on m. Reaction: m exerts an equal magnin2 +y
tude gravitational force to the left on M.
ΣFx = max ⇒ − T2 + 2mg sinq = 0
2m
T
4.11
2
= maThe
⇒
−T
T2 + mg
=0
(e) ΣF
Action:
charge
+Q
ansinq
electrostatic
force to the right on the charge –q. Reaction:
x
1 + exerts
or T2 =x 2mg sinq
[2]
The charge – q exerts an equal magnitude electrostatic force to the left on the charge + Q. +x
or
(a) Substituting Equation [2] into Equation [1] gives
T1 = 3mg sinq
q
(f) Action: The magnet exerts a force to the right on the iron.
Reaction: The iron exerts
an
2 mg
equal magnitude force to the left on the magnet.
(b) From Equation [2] above, we have
T2 = 2mg sinq
⎡⎣
⎤⎦ = 53.7 N
If
4.29
4.13
(a)
The resultant external force acting on this system, consisting of all three blocks having a
total mass of 6.0 kg, is 42 N directed horizontally toward the right. Thus, the acceleration
produced is
a=
4.28
ΣF
42 N
=
= 7.0 m s2 horizontally to the right
m
6.0 kg
→
(b) mDraw
a free body diagram of the 3.0-kg block and apply Newton’s second law to the
Let
a
1 = 10.0 kg, m2 = 5.00 kg, and q = 40.0°.
horizontal forces acting on this block:
(a) Applying the second law to each object gives
→
ΣFx = max ⇒ 42 N − T = ( 3.0 kg ) 7.0 m s 2 , and therefore T = 21T N
(
(c)
(a)
or
(b)
(a)
8.00 kg
F
(
T
70.0°
T
mg
(a)
(b)
)
T = mg = (8.00 kg ) 9.80 m s 2 =
= 78.4
78.4 N
N
ΣFx = T + T cos 70.0° − F = max = 0
= 78.4 N
F = T (1 + cos 70.0° ) = ( 78.4 N ) (1 + cos 70.0° ) = 105 N
(
)
.
Part (b) of the sketch above gives a force diagram of the pulley near the foot. Here, F is the
First, we consider the glider, plane, and connecting
n1
magnitude of the force the foot exerts on the pulley. By Newton’s third law, this
is the same
rope to be a single unit having mass
as the magnitude of the force the pulley exerts on the foot. Applying the second law gives
mtotal = 276 kg + 1 950 kg = 2 226 kg
For this system, the tension in the rope is an internal
force and is not included in an application of Newton’s
second law. Applying the second law to the horizontal
motion of this combined system gives
ΣFx = mtotal ax
67352_ch04a.indd 98
T
Part
Part (b)
(b) of
of the
the sketch
sketch above
above gives
gives aa force
force diagram
diagram of
of the
the pulley
pulley near
near the
the foot.
foot. Here,
Here, F
F is
is the
the
magnitude
of
the
force
the
foot
exerts
on
the
pulley.
By
Newton’s
third
law,
this
is
the
magnitude of the force the foot exerts on the pulley. By Newton’s third law, this is the same
same
as
as the
the magnitude
magnitude of
of the
the force
force the
the pulley
pulley exerts
exerts on
on the
the foot.
foot. Applying
Applying the
the second
second law
law gives
gives
or
4.33
.
m2
))
Assuming frictionless pulleys, the tension is uniform
through the entire length of the rope. Thus, the tension
at the point where the rope attaches to the leg is the
same as that at the 8.00 kg block. Part (a) of the
sketch at the right gives a force diagram of the
suspended block. Recognizing that the block has
zero acceleration, Newton’s second law gives
ΣFy = T − mg = 0
(b)
(b)
→
n
The force accelerating the 2.0-kg block is the force exerted on it by the 1.0-kg block.
q
Therefore, this force is given by
I
→
9.8mms 2s 2, [(or
0.10F
sin 25° ] = 0.49
m right
s2
) cos
m2g
F = ma = ( 2.0 kg )−7.0
= 25°
14 N+ horizontally
to the
((
4.31
4.14
)
⇒
mtotal
glider + plane
+ rope
F(thrust)
mtotal g
F = ( 2 226 kg ) 2.20 m s2
2/9/11 1:15:14 PM
mtotal = 276 kg + 1 950 kg = 2 226 kg
For this system, the tension in the rope is an internal
force and is not included in an application of Newton’s
second law. Applying the second law to the horizontal
motion of this combined system gives
(
The Laws of Motion
99
31
)
On the rebound,
ball
a displacement
Its3velocity
as kN
it left the
ΣFx =the
mtotal
ax has⇒v y =F0=after
kg ) 2.20 m Δy
s2 ==+20
4.90m.
× 10
N = 4.90
( 2 226
ground must have been
(b)
n2
To determine the tension in the rope connecting the
glider and the plane, we consider a system consisting
of the glider alone. For this system, the rope is a external
agent and the tension force it exerts on our system
(glider) is included in a second law calculation.
ΣFx = mglider ax
⇒
( (
mglider
) )
T = ( 276 kg ) 2.20 m s 2 = 607
= N
32.7 N
T
mglider g
.
4.37
Trailer
300 kg
→
T
→
wT
→
nT
Car
1 000 kg
→
→
→
nc
nc
q
→
→
→
Rcar
F
T
wc
→
F
Choosing the +x-direction to be horizontal and forward, the +y-direction vertical and upward, the
common acceleration of the car and trailer has components of ax = +2.15 m s 2 and a y = 0 .
(a)
The net force on the car is horizontal and given by
( ΣFx )car = F − T = mcar ax = (1 000 kg) ( 2.15 m s2 ) =
(b)
2.15 × 10 3 N forward
The net force on the trailer is also horizontal and given by
( ΣFx )trailer = +T = mtrailer ax = (300 kg) ( 2.15
)
m s2 = 645 N forward
(c)
Consider the free-body diagrams of the car and trailer. The only horizontal force acting on
the trailer is T = 645 N forward, and this is exerted on the trailer by the car. Newton’s third
law then states that the force the trailer exerts on the car is 645 N toward the rear .
(d)
The road exerts two forces on the car. These are F and nc shown in the free-body diagram
of the car.
From part (a),
( )
Also, ΣFy
car
F = T + 2.15 × 10 3 N = 645 N + 2.15 × 10 3 N = + 2.80 × 10 3 N
= nc − wc = mcar a y = 0, so nc = wc = mcar g = 9.80 × 10 3 N
The resultant force exerted on the car by the road is then
Rcar = F 2 + nc2 =
3
2
3
2
= 1.02 × 10 4 N
at q = tan −1 ( nc F ) = tan −1 ( 3.50 ) = 74.1° above the horizontal and forward. Newton’s third
law then states that the resultant force exerted on the road by the car is
1
2
a y t 2 below
= 0 + the
m s 2 )(and
1.00rearward
s ) = 1.22
1.02 × 10 4 N at 74.1°
horizontal
. m
( 2.44
2
4.17
.
4.39
( 2.80 × 10 N ) + (9.80 × 10 N )
1
1
2 ( Δx ) 2 ( 2.00 m )
ax t 2 = 0 + ax t 2 gives ax =
=
= 1.78 m s 2
2
2
t2
(1.50 s )2
(a)
Δx = v0 t +
(b)
Considering forces parallel to the incline, Newton’s
second law yields
ΣFx = ( 29.4 N ) sin 30.0° − fk = ( 3.00 kg ) 1.78 m s 2
67352_ch04a.indd 99
2/9/11 1:15:17 PM
=
100
32
2 ( 2.00 m )
= 1.78 m s 2
(1.50 s )2
(b) 4Considering
Chapter
forces parallel to the incline, Newton’s
(
)
second law yields
(b)
If
the=left
endNof
the
originally
was
to a point higher
or
N 30.0°
y up the wall,
k = 9.36
ΣF
− fk = horizontal
1.78 m
s 2 attached
(f29.4
) sin
(3.00 kg) (cable
)
x
the tension in this cable would then have an upward component. This upward component
→
kn
would
support
part
the weight
of the
cat burglar,
Perpendicular
to
theofplane,
we have
equilibrium,
sothus
n
or
fk = 9.36
N
m
(
4.18
decreasing the tension in the cable on the right .
ΣFy = n − ( 29.4toNthe
= 0haveorequilibrium,
n = 25.5 N
) cos30.0°
Perpendicular
plane, we
so
or
f cos30.0°
9.36inNthe
ΣFy reference
= n − ( 29.4
==0sketch
or atn the
= 25.5
N
Using Then,
the
right,
m kaxis
=N )kshown
=
0.367
25.5 N
n
we see that
f
9.36 N
Then,
mk = k =
= 0.367
(c) From part (b) above,
n 25.5fkN= 9.36 N
)
fk
x
30.0°
w mg 29.4 N
30.0°
fk = 9.36 N
(c)
From part (b) above,
(d)
Finally, v 2 = v02 + 2ax ( Δx ) gives
the incline
v = v02 + 2ax ( Δx ) = 0 + 2 (1.78 m s 2 )( 2.00 m ) = 2.67 mdown
s
.
4.51
When the block is on the verge of moving, the static friction force has a
magnitude fs = ( fs )m ax = m s n.
Since equilibrium still exists and the applied force is 75 N, we have
ΣFx = 75 N − fs = 0
or
(f )
s
m ax
= 75 N
In this case, the normal force is just the weight of the crate, or n = mg. Thus, the coefficient of
static friction is
ms =
(f )
s
m ax
n
(f )
=
s
m ax
mg
=
75 N
( 20 kg) (9.80
m s2 )
= 0.38
After motion exists, the friction force is that of kinetic friction, fk = m k n.
4.19
116
4.52
4.57
.
Since the crate moves with constant velocity when the applied force is 60 N, we find that
ΣFx = 60 N − fk = 0 or fk = 60 N. Therefore, the coefficient of kinetic friction is
(
)
f
f
60 N
a = 0 on a horizontal floor with only two vertimk = k = k =
= 0.31y
2
cal
forces
acting
on
it,
the
upward
normal
force
exerted on the block by the floor must equal
n
mg
9.80
m
s
20
kg
(
)
(
)
Chapter 4
the downward gravitational force. That is,
(a)
(a)
When the block
is resting in equilibrium
→
→
T1
a2
→
T2
→
n
→
a1
→
m1
T1
4.00 kg
→
T2
m2
m3
→
a3
2.00 kg
1.00 kg
→
→
→
m2g
fk
→
m3g
m1g
(b)
Note that the suspended block on the left, m1, is heavier than that on the right, m3. Thus, if
the system overcomes friction and moves, the center block will move right to left with each
block’s acceleration being in the directions shown above .
fk = m k n = ( 0.350 )( 9.80 N ) = 3.43 N
First, consider the center block, m2, which
has no vertical acceleration. Then,
Assuming the cords do not stretch, the speeds of the three blocks must always be equal.
Thus, the magnitudes of the blocks’ accelerations must have a common value, a.
I
I
I
a1 = a 2 = a 3 = a
67352_ch04a.indd 100
Taking the indicated direction of the acceleration as the positive direction of motion for
each block, we apply Newton’s second law to each block as follows:
2/9/11 1:15:21 PM
.
4.21
(a)
First, consider
, which
hasatno
Force
diagramsthe
of center
the twoblock,
blocksm2are
shown
thevertical acceleration. Then,
T1
right. Note that each block experiences a downward
2
ΣFy = n force
− m2 g = 0
or
n = m2 g = (1.00 kg ) ( 9.80 m s ) = 9.80 N
gravitational
(
101
33
The Laws of Motion
T2
)
Fg = ( 3.50 kg ) 9.80 m s 2 = 34.3 N
This means the friction force is:
fk = m k n = ( 0.350 )( 9.80 N ) = 3.43
N F
T2
Fg
g
Also,
each the
hascords
the same
upward
acceleration
asofthe
Assuming
do not
stretch,
the
speeds
the
three
blocks
must
always
be
equal.
2
elevator,
this case aof
+1.60
m saccelerations
.
y =the
Thus, theinmagnitudes
blocks’
must have a common value, a.
I
I
I
a 2 = a 3 second
=a
a1 =Newton’s
Applying
law to the lower block:
Taking the indicated direction of the acceleration as the positive direction of motion for
each block, we apply Newton’s second law to each block as follows:
For m1:
m1 g − T1 = m1 a
or
T1 = m1 ( g − a ) = ( 4.00 kg ) ( g − a )
[1]
For m2:
T1 − T2 − fk = m2 a
or
T1 − T2 = (1.00 kg ) a + 3.43 N
[2]
For m3:
T2 − m3 g = m3 a
or
T2 = m3 ( g + a ) = ( 2.00 kg ) ( g + a )
[3]
Substituting Equations [1] and [3] into Equation [2], and solving for a yields
( 4.00 kg) ( g − a ) − ( 2.00 kg) ( g + a ) = (1.00 kg) a + 3.43 N
a=
( 4.00 kg − 2.00 kg) (9.80
m s 2 ) − 3.43 N
4.00 kg + 2.00 kg + 1.00 kg
= 2.31 m s 2
I
I
2
a1 = 2.31 m s 2 downward , a 2 = 2.31 m s to the left ,
so
I
and a 3 = 2.31 m s 2 upward
(c)
Using this result in Equations [1] and [3] gives the tensions in the two cords as
T1 = ( 4.00 kg ) ( g − a ) = ( 4.00 kg )( 9.80 − 2.31) m s 2 = 30.0 N
The Laws of Motion
and
T2 = ( 2.00 kg ) ( g + a ) = ( 2.00 kg )( 9.80 + 2.31) m s 2 = 24.2 N
117
(d)
4.20
4.58
4.59
.2
The resultant y-component is
45.0°
N
.0
N
Rx = ΣFx = ( 60.0 N ) cos 45.0° − ( 60.0 N ) cos 45.0° = 0
east
60
.0
60
From the final calculation in part (b), observe that if the frictionnorth
force had a value of
→
zero (rather than 3.53 N), the acceleration of the system would increase in magnitude.
Fsail
Then, observe from Equations [1] and [3] that this would mean T1 would decrease while
3
( 4.2 × 10 N ) sin 30° = 2.6 m s2
T2 would
northincrease .
=
30°
30°
800 kg
m
→
The sketch at the right gives an edge view of the sail (heavy
vwind
(a)
The
horizontal
component
of the of
resultant
force exerted
y
line) as
seen
from above.
The velocity
the wind,
on the light by the cables is
45.0°
x
Ry = ΣFy = ( 60.0 N ) sin 45.0° + ( 60.0 N ) sin 45.0° = 84.9 N
Hence, the resultant force is 84.9 N vertically upward .
(b)
67352_ch04a.indd 101
The forces on the traffic light are the weight, directed downward, and the 84.9 N vertically
upward force exerted by the cables. Since the light is in equilibrium, the resultant of these
forces must be zero. Thus,
2/9/11 1:15:24 PM
102
34
Chapter 4
.
4.23
m = 1.00
and
9.80c N
(b)
The kg
forces
onmg
the=traffi
light are the weight, directed downward,
vertically
25.0 m and the 84.9 N
25.0
m
a
upward force exerted by the cables. Since the light is in equilibrium,
the resultant of a
these
0.200 m
⎞
−1 ⎛ 0.200 m
.
→
→
aforces
= tanmust
= 0.458°
w = 84.9 N downward .
⎜⎝ be zero.⎟⎠ Thus,
T
T
25.0 m
→
4.60
.2
4.61
(a)
For the
suspended
block, ΣF
T −is50.0
N = 0, soforce
the tension in the rope is T = 50.0 N.
(a)
accelerates
they ==
box
the
0, thisthat
requires
that ΣF
Since The
a y =force
T sina
+ Tfriction
sina − mgg = 00,between the box and truck.
y
Then,
considering the horizontal
forces on the 100-N block, we find
9.80Then,
N the normal force exerted on the box by the
(b)
givingWe assume the truck is on level Tground.
=
2sina
truck equals the weight of the box, n = mg. The maximum acceleration the truck can have
before the box slides is found by considering the maximum static friction force the truck
bed can exert on the box:
mg
(f )
s
m ax
= m s n = m s ( mg )
Thus, from Newton’s second law,
am ax =
4.71
.2
4.62
(f )
s
m ax
m
=
m s ( mg )
2
of
m sLaws
= m s g = ( 0.300 )( 9.80 m s 2 ) = 2.94 The
m
Motion
123
When
an object of mass m is on this frictionless incline, the only force acting parallel to the
(a)
incline is the parallel component of weight, mg sinq , directed down the incline. The acceleration
is then
a=
(a)
(b)
mg sinq
= g sinq = ( 9.80 m s 2 ) sin 35.0° = 5.62 m s 2 (directed down the incline)
m
Taking up the incline as positive, the time for the sled projected up the incline to come to
rest is given by
v − v0 0 − 5.00 m s
= suitcase
= moves with2 constant
= 0.890 velocity,
s
Sincetthe
− 5.62 m s
a
The distance the sled travels up the incline in this time is
⎛ v + v0 ⎞
⎛ 0 + 5.00 m s ⎞
t=⎜
Δs = vav t = ⎜
⎟⎠ ( 0.890 s ) = 2.23 m
⎝
⎝ 2 ⎟⎠
2
(b)
The time required for the first sled to return to the bottom of the incline is the same as the
time needed to go up, that is, t = 0.890 s. In this time, the second sled must travel down the
entire 10.0 m length of the incline. The needed initial velocity is found from
1
Δs = v0 t + at 2 as
2
v0 =
4.23
124
2
Δs at −10.0 m ( −5.62 m s )( 0.890 s )
− =
−
= −8.74 m s
t
2
0.890
s
2
→
n
Chapter 4or
8.74 m s down the incline
→
mb
.2
4.72
4.73
T
→
T
mr
→
F
y
(a)
Choose the positive x-axis to be down the incline and
the y-axis perpendicular to this→as shown in the
mg
free-body diagram of the toy. bThe acceleration of the
toy then has components of
Δv x + 30.0 m s
a y = 0, and ax =
=
= + 5.00 m s 2
6.00 s
Δt
(b)
Applying
Newton’s
second
law
to the rope yields
Applying
the second
law to
the toy
gives
(a)
→
T
→
a
x
q
m
→
mg q
ΣFy = T − mg cosq = ma y = 0 , or
T = mg cosq = ( 0.100 kg ) ( 9.80 m s 2 ) cos 30.7° = 0.843 N
4.74
67352_ch04a.indd 102
The friction force exerted on the mug by the moving tablecloth is the only horizontal force the
mug experiences during this process. Thus, the horizontal acceleration of the mug will be
2/9/11 1:15:27 PM
=
+ 30.0 m s
= + 5.00 m s 2
6.00 s
4.27
103
35
The Laws of Motion
Applying the second law to the toy gives
ΣFx =reference
mg sinq axes
= mathat
⇒
We choose
are parallel
tosinq
and = max mg = ax g,
(a)
x
perpendicular to the incline as shown in the force 2
⎛ 5.00 m s ⎞
−1 ⎛ a x ⎞
diagrams
Since
and at the right.
=blocks
sin −1 ⎜are in
= 30.7°
q = sin
⎟
⎜⎝ both
g ⎠ block.⎝Then,
9.80 m s 2 ⎟⎠
equilibrium, ax = a y = 0 for each
(b)
applying
ΣFyNewton’s
= T − mgsecond
cosq =law
ma yto=each
0 , orblock gives
+y
n1
T1
m
+x
T2
q
.2
4.75
4.74
2
−2
mg
For Block
T = 1mg(mass
= ( 0.100
kg ) ( 9.80
s 2 )×cos
0.843cm
N
s ) =m6.00
1030.7°
m == 6.00
(cosq m):
) ( 0.490
ΣFx = max ⇒ − T1 + T2 + mg sinq = 0
First,
we willforce
compute
theon
needed
accelerations:
The friction
exerted
the mug
by the moving tablecloth is the only horizontal force
n2 the
or
mug experiences during this process. Thus, the horizontal acceleration of the mug will be
ay = 0
2m
(1) Before it starts to move:
a
T2
q = sin −1 x
g
v y − v0 y 1.2 m s − 0
=
= 1.5 m s 2
ay =
(2) During the first 0.80 s:
0.80 s
t
(3)
While moving at constant velocity: a y = 0
(4)
During the last 1.5 s:
ay =
q
v y − v0 y
t
=
0 −1.2 m s
The
= − 0.80
mLaws
s 2 of
1.5 s
2 mg
Motion
125
The spring scale reads the normal force the scale exerts on the man. Applying Newton’s second
law to the vertical motion of the man gives
ΣFy = n − mg = ma y
4.25
.2
4.79
4.76
(
n = m g + ay
or
)
(a)
When a y = 0,
n = ( 72 kg ) ( 9.80 m s 2 + 0 ) = 7.1 × 10 2 N
(b)
When a y = 1.5 m s 2 ,
n = ( 72 kg ) ( 9.80 m s 2 + 1.5 m s 2 ) = 8.1 × 10 2 N
(c)
When a y = 0,
n = ( 72 kg ) ( 9.80 m s 2 + 0 ) = 7.1× 10 2 N
(d)
When a y = −0.80 m s 2 ,
⎞
⎛
104 N
m s 2 ) = 6.5 × 10 2 N
(9.80mms2s)2 =− 0.80
⎟⎠ ()6.86
⎜⎝ n = ( 72 kg
The acceleration the car has as it is coming to a stop is
Let
2
v 2 − v02 0 − ( 35 m s )
a=
=
= − 0.61 m s 2
2 ( Δx )
2 (1 000 m )
Thus, the magnitude of the total retarding force acting on the car is
4.26
4.81
.2
4.80
⎛ w⎞
⎛ 8 820 N ⎞
F=ma =⎜ ⎟ a =⎜
( 0.61 m s2 ) = 5.5 × 102 N
⎝ g⎠
⎝ 9.80 m s 2 ⎟⎠
(a)
Consider
theright
firstgives
free-body
T 250 N
The sketch
at the
the force diagram of the person.
diagram
in
which
the magnitude
child and of the normal force
The scale simply reads the
theonchair
are treated
as aseat. From Newton’s second law,
exerted
the student
by the
combined
system.
The
weight
T 250 N
we obtain
of this system is wtotal = 480 N,
and its mass is
a
wtotal
mtotal =
= 49.0 kg
g
Taking upward as positive, the
acceleration of this system is
found from Newton’s second
law as
67352_ch04a.indd 103
wtotal 320 N 160 N
T 250 N
n
a
wchild 320 N
2/9/11 1:15:46 PM
36
104
4.30
Taking upward as positive, the
acceleration of this system is
found from Newton’s second
Chapter 4
law as
The figure atΣF
the
forces
=right
2T −shows
wtotal =the
mtotal
a y acting on the
y
block. The incline is tilted at q = 25°, the mass of
the block is m = 5.8 kg, 2while
N ) applied
− 480 Nforce
( 250the
Thus
a =
= + 0.408 m s 2
pulling the block up ythe incline
is Fkg= 32 N.
49.0
Since a y = 0 for this block,
0.408 m s 2 upward
ΣFy = n − mg cosq = 0
⫹y
n
F
⫹x
The
fk Laws of Motion
q
127
or
m
mg
q magnitude
and
normal
force isforce that
n = the
mg cosq
(b) theThe
downward
child exerts on the chair has the same
as the
upward normal force exerted on the child by the chair. This is found from the free-body
(a) Since
theofincline
is considered
diagram
the child
alone as frictionless for this part, we take the friction force to be
fk = 0 and find
n = mchild a y + wchild − T
so
ΣF = T + n − wchild = mchild a y
ΣFxy = F − mg sinq
= max
or
Hence,
4.82
67352_ch04a.indd 104
(a)
⎛ 320 N ⎞
n=⎜
( 0.408 m s2 ) + 320 N − 250 N = 83.3 N
⎝ 9.80 m s 2 ⎟⎠
In the vertical direction, we have
2/9/11 1:15:50 PM
5ࢠ 에 너 지
5.1
5.3
If the weights are to move at constant velocity, the net force on them must be zero. Thus, the
force exerted on the weights is upward, parallel to the displacement, with magnitude 350 N. The
work done by this force is
W = ( F cos θ ) s = [ ( 350 N ) cos 0°] ( 2.00 m ) = 700 J
5.2
5.5
(a)
The gravitational force acting on the object is
w = mg = ( 5.00 kg ) ( 9.80 m s 2 ) = 49.0 N
and the work done by this force is
(
)
(
Wg = − ΔPEg = − mg y f − yi = + w yi − y f
or
(b)
)
Wg = w ( L sin 30.0° ) = ( 49.0 N )( 2.50 m ) sin 30.0° = 61.3 J
The normal force exerted on the block by the incline is n = mg cos 30.0°, so the friction
force is
fk = μ k n = ( 0.436 )( 49.0 N ) cos 30.0° = 18.5 N
This force is directed opposite to the displacement (that is q = 180°), and the work it does is
W f = ( fk cos θ ) L = [(18.5 N ) cos 180° ]( 2.50 m ) = − 46.3 J
(c)
Since the normal force is perpendicular to the displacement, so the work done by the normal force is Wn = ( n cos 90.0° ) L = 0 .
(d)
If a shorter ramp is used to increase the angle of inclination while maintaining the
same vertical displacement y f − yi , the work done by gravity will not change , the
work done by the friction force will decrease (because the normal force, and hence
the friction force, will decrease and also because the ramp length L decreases), and the
work done by the normal force remains zero (because the normal force remains
.
perpendicular to the displacement).
s 20.0 m
5.7
5.3
5.6
(a)
(a)
ΣF = F sin θ + n − mg = 0
→
They 35 N force applied by the shopper makes a 25° angle with the
n displacement of the cart
(horizontal). The work done on the cart by the shopper is then
n = mg − F sin θ
→
ΣFx = F cos θ − μ k n = 0
n=
F
Fk mkn
18.0 kg
q 20.0°
F cos θ
μk
→
37
67352_ch05a.indd 129
mg
2/9/11 1:18:54 PM
38
136
Chapter 5
5.4
(a)
∴
mg − F sin θ =
F cos θ
μk
( 0.500 )(18.0 kg ) ( 9.80 m s2 )
μ k mg
F=
=
= 79.4 N
μ k sin θ + cos θ
( 0.5000 ) sin 20.0° + cos 20.0°
(b)
WF = ( F cos θ ) s = [( 79.4 N ) cos 20.0° ]( 20.0 m ) = 1.49 × 10 3 J = 1.49 kJ
(c)
fk = F cos θ = (79.4 N) cos 20.0° = 74.6 N
Wf = ( fk cos θ ) s = [( 74.6 N ) cos 180° ]( 20.0 m ) = −1.49 × 10 3 J = − 1.49 kJ
5.9
5.4
(a)
The work-energy theorem, Wnet = KE f − KEi , gives
5 000 J =
5.11
5.5
(b)
W = ( F cos θ ) s = ( F cos 0° )( 25.0 m ) = 5 000 J, so F = 200 N
(a)
KE =
(b)
Since the kinetic energy of an object varies as the square of the speed, doubling the speed
will increase the kinetic energy by a factor of 4 as shown below:
1
1
2
m v 2 = ( 65.0 kg )( 5.20 m s ) = 879 J
2
2
KE f
KEi
5.6
5.13
2
⎛v ⎞
2
= ⎜ f ⎟ = ( 2) = 4
2
1
v
v
m
2
i
⎝ i ⎠
1
2
m v 2f
1 2 1
2
mvi = ( 6.50 × 10 7 kg ) (12.0 m s ) = 4.68 × 10 9 J
2
2
KEi =
(b)
Wnet = KE f − KEi = 0 − 4.68 × 10 9 J = −4.68 × 10 9 J
(c)
Wnet = ( FR cos θ ) Δx , and θ = 180° since the resultant force acting on the ship is a retarding
frictional force (note that the normal force the water exerts on the ship simply cancels
out the weight of the ship). Thus, if the ship comes to rest after a displacement of
Δx = 2.50 km, the resultant force acting on the ship is
(a)
(a)
(b)
Wnet
− 4.68 × 10 9 J
=
= 1.87 × 10 6 N
( Δx ) cos θ ( 2.50 × 103 m ) cos 180°
As the
the bullet
bullet penetrates
penetrates the
the tree
tree trunk,
trunk, the
the only
only force
force doing
doing work
work on
on it
it is
is the
the force
force of
of
As
resistance
exerted
by
the
trunk.
This
force
is
directed
opposite
to
the
displacement,
so
resistance exerted by the trunk. This force is directed opposite to the displacement, so
,
and
the
magnitude
of
the
average
the
work
done
is
=
f
cos180
°
Δ
x
=
KE
−
KE
W
(
)
net = ( f av cos180°) Δx = KE f − KE i , and the magnitude of the average
the work done is Wnet
av
f
i
resistance force
force is
is
resistance
1
2
0 − ( 7.80 × 10 −3 kg ) ( 575 m s )
KE f − KEi
2
fav =
=
= 2.34 × 10 4 N Energy
( Δx ) cos 180°
− ( 5.50 × 10 −2 m )
139
If the friction force is constant, the bullet will have a constant acceleration and its average
velocity while stopping is v = ( v f + vi ) 2. The time required to stop is then
Δt =
67352_ch05a.indd 136
=
(a)
FR =
5.15
5.7
5.15
1
( 2 .50 × 103 kg) v2 − 0 , or v = 2.00 m s
2
−2
Δx 2 ( Δx ) 2 ( 5.50 × 10 m )
=
=
= 1.91 × 10 −4 s
0 + 575 m s
v
v f + vi
2/9/11 1:19:02 PM
Energy
2 ( KEmax 2 )
KEmax
3.2 × 10 4 J
=
=
= 23 m s
m
m
62 kg
Since
continues to
move attoconstant
speed, the net work done on the cart in the
We usethe
thecart
work-energy
theorem
find the work.
second aisle is again zero. With both the net work and the work done by friction
1 2 1 2
1
(Wkg
unchanged,
the =
workmvdone
W = ΔKE
mvthe
− 5).6is×also
10 2 J
0 − ( 70
)( 4.=0Wmnet s−)2W=friction
shopper
f − by
i =shopper
2
2
2
unchanged . However, the shopper now pushes horizontally on the cart, making
W = ( F cos θ ) s = ( fk cos 180°) s = ( −μ k n ) s = ( −μ k mg ) s,
F ′ = Wshopper ( Δx ⋅ cos 0° ) = Wshopper Δx smaller than before when the force was
( −5.6 × 10 2 J)
W
F = Wshopper
s = −( Δx ⋅ cos35
= − ° ).
= 1.2 m
so
μ k mg
( 0.70 ) ( 70 kg ) ( 9.80 m s2 )
v=
5.17
5.8
(c)
(a)
(b)
5.7
5.9
5.19
137
39
(a)
The work the beam does on the pile driver is given by
Wnc = ( F cos 180°) Δx = − F ( 0.120 m )
Here, the factor cos180° is included because the force F exerted on the driver by the beam is
directed upward, but the Δx = 12.0 cm = 0.120 m displacement undergone by the driver while in
contact with the beam is directed downward.
From the work-energy theorem, this work can also be expressed as
140
Chapter 5
(
) (
)
Wnc = KE f − KEi + PE f − PEi =
(
)
(
1
m v 2f − vi2 + mg y f − yi
2
)
Choosing y = 0 at the level where the pile driver first contacts the top of the beam, the driver
starts from rest ( vi = 0 ) at yi = +5.00 m and comes to rest again ( v f = 0 ) at y f = −0.120 m.
Therefore, we have
− F ( 0.120 m ) =
yielding
5.23
5.10
5.25
5.11
F = 8.78 × 10 5 N directed upward
(a)
PEi = mgyi = ( 0.20 kg ) ( 9.80 m s 2 )(1.3 m ) = 2.5 J
(b)
PE f = mgy f = ( 0.20 kg ) ( 9.80 m s 2 )( −5.0 m ) = −9.8 J
(c)
ΔPE = PE f − PEi = −9.8 J − 2.5 J = −12 J
While the motorcycle is in the air, only the conservative gravitational force acts on cycle and
rider. Thus, 12 m v 2f + mgy f = 12 m vi2 + mgyi , which gives
h = y f − yi =
5.12
5.27
1
m ( 0 − 0 ) + ( 2 100 kg ) ( 9.80 m s 2 ) ( −0.120 m − 5.00 m )
2
vi2 − v 2f
2g
=
( 35.0 m s )2 − ( 33.0 m s )2
2 ( 9.80 m s 2 )
= 6.94 m
The magnitude of the force a spring must exert on an object of mass m to give it an acceleration
of a = 0.800 g is F = ma = 0.800 mg.
Then, by Newton’s third law, this object exerts an oppositely directed force of equal magnitude
on the spring. If this reaction force is to stretch the spring 0.500 cm, the required force constant of
the spring is
−3
2
F 0.800 mg 0.800 ( 4.70 × 10 kg )( 9.80 m s )
= 7.37 N m
k=
=
=
Δx
Δx
0.500 × 10 −2 m
67352_ch05a.indd 137
2/9/11 1:19:03 PM
138
40
Chapter 5
5.29
5.13
(a)
(c)
(b)
None of the calculations in parts (a), (b), or (c) involve the initial angle.
increased,
magnitude
friction
force.
crate a energy
PEg =greater
mgy =than
0, atthe
Taking y =it0,has
anda hence
ground
level,
the This
initialgives
totalthe
mechanical
resultant
force (and
of the projectile
is hence, an acceleration) in the direction of motion, meaning the
speed of the crate will1increase with time .
( Etotal )i = KEi + PEi = mvi2 + mgyi
2
F0, the magnitude of 2the friction force will 5be
If the applied
than
2
1 force is made smaller
50.0ofkgthe
m s )This
+ ( 50
.0 kgthe
80 mhas
s )a(142
m ) = force,
4.30 ×and
10 J
= (that
.20 × 10 2force.
) (1applied
) ( 9.crate
greater than
means
resultant
2
acceleration, in the direction of the friction force (opposite to the direction of motion).
The work done on the projectile is equal to the change in its total mechanical energy.
The crate will now slow down and come to rest.
1
(Wnc )rise = ( KE f + PE f ) − ( KEi + PEi ) = m ( v 2f − vi2 ) + mg ( y f − yi )
2
1
2
2
= ( 50.0 kg ) ⎡⎣(85.0 m s ) − (120 m s ) ⎤⎦ + ( 50.0 kg ) ( 9.80 m s 2 ) ( 427 m − 142 m )
2
= −3.97 × 10 4 J
(c)
If, during the descent from the maximum height to the ground, air resistance does one and a
half times as much work on the projectile as it did while the projectile was rising to the top
of the arc, the total work done for the entire trip will be
(W )
nc total
= (Wnc )rise + (Wnc )descent = (Wnc )rise + 1.50 (Wnc )rise
= 2.50 ( −3.97 × 10 4 J ) = −9.93 × 10 4 J
Then, applying the work-energy theorem to the entire flight of the projectile gives
(W )
nc total
= ( KE + PE ) just before
hitting ground
− ( KE + PE )at launch =
1 2
1
mv f + mgy f − mvi2 Energy
− mgyi
2
2
143
and the speed of the projectile just before hitting the ground is
vf =
=
5.14
5.33
2 (Wnc )total
m
(
+ vi2 + 2 g yi − y f
2 ( −9.93 × 10 4 J )
50.0 kg
)
+ (120 m s ) + 2 ( 9.80 m s 2 ) (142 m − 0 ) = 115 m s
2
Since no nonconservative forces do work, we use conservation of mechanical energy, with the
zero of potential energy selected at the level of the base of the hill. Then,
1
1
m v 2f + mgy f = m vi2 + mgyi with y f = 0 yields
2
2
yi =
v 2f − vi2
2g
=
( 3.00 m s )2 − 0
2 ( 9.80 m s 2 )
= 0.459 m
Note that this result is independent of the mass of the child and sled.
5.15
5.35
.
5.34
(a) On
a frictionless
track, no external
do work on the system consisting of the block
Using
conservation
of mechanical
energy,forces
we have
and the spring as the spring is being compressed. Thus, the total mechanical energy of the
system is constant, or KE f + ( PEg ) f + ( PEs ) f = KEi + ( PEg )i + ( PEs )i. Because the track is
horizontal, the gravitational potential energy when the mass comes to rest is the same as
just before it made contact with the spring, or ( PEg ) f = ( PEg )i . This gives
1
1
1
1
m v 2f + kx 2f = m vi2 + kxi2
2
2
2
2
Since v f = 0 (the block comes to rest) and xi = 0 (the spring is initially undistorted),
x f = vi
67352_ch05a.indd 138
m
0.250 kg
= (1.50 m s )
= 0.350 m
k
4.60 N M
2/9/11 1:19:05 PM
Energy
(b)
5.37
5.16
5.36
(a)
145
139
41
If the track
was
not is
frictionless,
some
of the
would be
overfriction
force
constant, the
bullet
willoriginal
have a kinetic
constantenergy
acceleration
andspent
its average
coming friction
betweenisthe
mean
energy would be
velocity
while stopping
v =block
vi ) track.
2. TheThis
timewould
required
to that
stopless
is then
( v f +and
stored as elastic potential energy in the
−2 spring when the block came to rest. Therefore, the
5=.50
m
×(10
(
) ofg 2)the
Wg ) AΔ→tC==Δ
mg
y
−
yC)) ==be
.0×N
3.s00case.
m
(149
(maximum
(xPE=g )2A( −Δx( PE
compression
spring
would
less
A
1in
.91
0)(−4this
= C
0 + 575 m s
v
v f + vi
We
choose
the zeroof
ofmechanical
potential energy
at the level of the bottom
From
conservation
energy,
of the arc. The initial height of Tarzan above this level is
yi = ( 30.0 m ) (1 − cos 37.0°) = 6.04 m
艎⫽30.0 m
q yi yf
Then, using conservation of mechanical energy, we find
1
1
m v 2f + 0 = m vi2 + mgyi
2
2
or
(b)
v f = vi2 + 2 gyi = 0 + 2 ( 9.80 m s 2 ) ( 6.04 m ) = 10.9 m s
In this case, conservation of mechanical energy yields
v f = vi2 + 2 gyi =
5.41
5.17
( 4.00 m s )2 + 2 ( 9.80 m s2 ) ( 6.04 m ) = 11.6 m s
(a)
When the child slides down a frictionless surface, the only nonconservative force acting on
the child is the normal force. At each instant, this force is perpendicular to the motion and,
hence, does no work. Thus, conservation of mechanical energy can be used in this case.
(b)
The equation for conservation of mechanical energy, ( KE + PE ) f = ( KE + PE )i , for this
situation is 12 m v 2f + m gy f = 12 m vi2 + m gyi . Notice that the mass of the child cancels out
of the equation, so the mass of the child is not a factor in the frictionless case.
(c)
Observe that solving the energy conservation equation from above for the final speed gives
v f = vi2 + 2 g ( yi − y f ). Since the child starts with the same initial speed (vi = 0) and has
the same change in altitude in both cases, v f is the same in the two cases.
(d)
Work done by a nonconservative force must be acccounted for when friction is present.
This is done by using the work-energy theorem rather than conservation of mechanical energy.
(e)
From part (b), conservation of mechanical energy gives the final speed as
v f = vi2 + 2 g ( yi − y f ) = 0 + 2 ( 9.80 m s 2 ) (12.0 m ) = 15.3 m s
5.18
5.49
Choose PEg = 0 at the level of the base of the hill and let x represent the distance theEnergy
skier moves
151
along the horizontal portion before coming to rest. The normal force exerted on the skier by the
snow while on the hill is n1 = mg cos 10.5° and, while on the horizontal portion, n2 = mg.
Consider the entire trip, starting from rest at the top of the hill until the skier comes to rest on the
horizontal portion. The work done by friction forces is
Wnc = ⎡⎣( fk )1 cos 180° ⎤⎦ ( 200 m ) + ⎡⎣( fk )2 cos 180° ⎤⎦ x
continued on next page
= − μ k ( mg cos 10.5° )( 200 m ) − μ k ( mg ) x
(
Applying Wnc = KE + PEg
) − ( KE + PE )
f
g i
to this complete trip gives
− μ k ( mg cos 10.5° )( 200 m ) − μ k ( mg ) x = [ 0 + 0 ] − [ 0 + mg ( 200 m ) sin 10.5° ]
or
67352_ch05a.indd 139
⎛ sin 10.5°
⎞
x=⎜
− cos 10.5°⎟ ( 200 m ). If μ k = 0.0750 , then x = 289 m .
⎝ μk
⎠
2/9/11 1:19:06 PM
140
42
5.51
5.19
Chapter 5
As the piano
constant
up to
the apartment,
thethe
total
mustthe
bedriver
done on
Choosing
at theatlevel
wherespeed
the pile
driver
first contacts
topwork
of thethat
beam,
y =is0 lifted
it is from rest ( vi = 0 ) at yi = +5.00 m and comes to rest again ( v f = 0 ) at y f = −0.120 m.
starts
Therefore, we have
3
4
Wnc = ΔKE + ΔPE
1 g = 0 + mg y f − yi = ( 3.50 × 10 2N )( 25.0 m ) = 8.75 × 10 J
− F ( 0.120 m ) = m ( 0 − 0 ) + ( 2 100 kg ) ( 9.80 m s ) ( −0.120 m − 5.00 m )
2
The three workmen (using a pulley system with an efficiency of 0.750) do work on the piano at a
rate of
yielding
F = 8.78 × 10 5 N directed upward
(
5.20
(a)
)
⎛
⎞
PThe
0.750stretching
= 0spring
.750 [ 3 (is165
371 J s object. Therefore, the force
)] = 371
net =force
⎜⎝ 3Psingle
⎟ the
theW
weight
of W
the=suspended
worker ⎠
constant of the spring is
so the time required to do the necessary work on the piano is
Δt =
5.53
5.20
Wnc 8.75 × 10 4 J
⎛ 1 min ⎞
=
= 236 s = ( 236 s ) ⎜
= 3.93 min
⎝ 60 s ⎟⎠
Pnet
371 J s
(
The work done on the older car is (Wnet )old = KE f − KEi
)
old
=
The work done on the newer car is
152
(W )
Chapter 5
net new
(
= KE f − KEi
)
new
=
1
1
mv 2 − 0 = mv 2
2
2
1
2
⎛1
⎞
m ( 2 v ) − 0 = 4 ⎜ m v 2 ⎟ = 4 (Wnet )old
⎝2
⎠
2
and the power input to this car is Pnew =
(W )
net new
Δt
=
4 (Wnet )old
= 4 Pold
Δt
or the power of the newer car is 4 times that of thhe older car .
5.54
5.21
5.59
(a)
The work done on the particle by the force F as
the particle moves from x = xi to x = x f is the area
under the curve from xi to x f.
Fx (N)
B
6
4
(a)
For x = 0 to x = 8.00 m,
2
1
W = area of triangle ABC = AC × altitude
2
W0→8
(b)
1
= (8.00 m )( 6.00 N ) = 24.0 J
2
=
67352_ch05a.indd 140
C
E
⫺2
⫺4
2
4
6
8
10 12
x (m)
D
For x = 8.00 m to x = 10.0 m,
W8→10 = area of triangle CDE =
(c)
A
0
1
CE × altitude
2
1
( 2.00 m )( −3.00 N ) = − 3.00 J
2
W0→10 = W0→8 + W8→10 = 24.0 J + ( − 3.00 J ) = 21.0 J
2/9/11 1:19:07 PM
Energy
5.22
5.65
43
141
w athlete’s
700 total
N energy is conserved is summarized by the equation
(d) The statement that the
The person’s mass is m = =
= 71.4 kg. The net upward force acting on the body
ΔKE + ΔPE = 0 or KEg2 + PE
=
KE
9.80
m
s 21 + PE1. In terms of mass, speed, and height, this
2
2
is Fnet becomes
= 2 ( 355 N12)m
− v700
N = 10.01 N. 2The final upward velocity can then be calculated from the
2 + mgy2 = 2 m v1 + mgy1 . Solving for the final height gives
work-energy theorem as
1
1
mgy1 + m v12 − m v22
( v12 − v22 )
1
2 m v 2 −2 1 m v 2
or
y
y
=
+
Wnet = yKE
2 =f − KEi =
2
1
i
2g
2mg
2
or
The given numeric values for this case are y1 = 0, v1 = 9.20 m s (at the trampoline level),
1
F cos
10.0 N ) cos
0° ]( 0)..250
4 kg ) vattained
− 0 is then
) = ( 71.height
(and
v =θ 0) s(=at[(maximum
height
The m
maximum
2
net
2
( v12 − v22 ) + (9.0 m s )2 − 0 = 4.1 m
which givesy2 = y1v+= 0.265 m= s0 upward
2g
2 ( 9.80 m s 2 )
5.67
5.23
(a)
The equivalent spring constant of the bow is given by F = kx as
k=
(b)
Ff
xf
(a)
(a)
230 N
= 575 N m
0.400 m
The work done pulling the bow is equal to the elastic potential energy stored in the bow in
its final configuration, or
W=
5.71
5.71
5.24
=
1 2 1
2
kx f = ( 575 N m )( 0.400 m ) = 46.0 J
2
2
The
The two
two masses
masses will
will pass
pass when
when both
both are
are at
at yy ff =
= 22..00
00 m
m above
above the
the table.
table. From
From conservation
conservation
of
energy,
(
KE
+
PE
+
PE
)
=
(
KE
+
PE
+
PE
)
of energy, ( KE + PEg + PEs ) f = ( KE + PEg + PEs )i
g
s f
g
s i
1
( m1 + m2 ) v 2f + ( m1 + m2 ) gy f + 0 = 0 + m1gy1i + 0, or
2
2 m1 g y1i
vf =
− 2 g yf
2mm
gy
1 1+ m21i
vf =
− 2 g yf
m1 + m2
2 ( 5.00 kg ) ( 9.80 m s 2 )( 4.00 m )
− 2 ( 9.80 m s 2 ) ( 2 .00 m )
m s 2 )( 4.00 m )
2 ( 5.00 kg ) ( 98..80
00 kg
− 2 ( 9.80 m s 2 ) ( 2 .00 m )
8.00 kg
This yields the passing speed as v f = 3.13 m s .
This yields the passing speed as
When m1 = 5.00 kg reaches the table, m2 = 3.00 kg is y2 f = 4.00 m above the table.
Thus, ( KE + PEg + PEs ) f = ( KE + PEg + PEs )i becomes
=
=
(b)
1
( m1 + m2 ) v 2f + m2 g y2 f + 0 = 0 + m1gy1i + 0, or v f =
2
(
2 g m1 y1i − m2 y2 f
m1 + m2
)
Thus,
158
Chapter 5
(c)
67352_ch05b.indd 141
vf =
2 ( 9.80 m s 2 ) ⎡⎣( 5.00 kg )( 4.00 m ) − ( 3.00 kg )( 4.00 m )⎤⎦
= 4.43 m s
8.00 kg
When the 5.00-kg object hits the table, the string goes slack and the 3.00-kg object becomes
a projectile launched straight upward with initial speed v0 y = 4.43 m s. At the top of its arc,
vy = 0 and vy2 = v02 y + 2ay ( Δy) gives
2/9/11 1:19:46 PM
142
44
gives
Chapter 5
or
2
2
5.00 × 10 5 2N + ( 3.60 × 10 5 N m ) ( 0.500 m )
vw
y − v0+y k y0 − ( 4 43 m s )
= 0.768 m
ΔyxC== load =h 0 =
= 1.00 m
2
.25s×
2 kaCy + kh 2 ( −9.80 5m
)105 N m + 3.60 × 105 N m
(b)
5.73
5.25
The work done compressing the springs equals the total elastic potential energy at equilib2
rium. Thus, W = 12 kC xC2 + 12 kh ( xC − 0.500 m ) , or
Since the bowl is smooth (that is, frictionless),
1
mechanicalWenergy
conserved
or ( 0.768 m )2 + 1 3.60 × 10 5 N m ( 0.268 m )2 = 1.68 × 10 5 J
= ( 5is.25
× 10 5 N m
)
)
2(
2 + PE )i . Also, if we choose
( KE + PE ) f = ( KE
y = 0 (and hence, PEg = 0) at the lowest point
in the bowl, then y A = + R, yB = 0, and yC = 2 R 3.
(a)
( PE ) = mgy = mgR
or ( PE ) = ( 0.200 kg ) ( 9.80 m s )( 0.300 m ) =
g A
2R/3
A
2
0.588 J
g A
5.26
5.87
(b)
KEB = KEA + PEA − PEB = 0 + mgyA − mgyB = 0.588 J − 0 = 0.588 J
(c)
KEB = 12 m vB2 ⇒ vB =
(d)
( PE )
(e)
KEC = KE B + PE B − PEC = 0.588 J + 0 − 0.392 J = 0.196 J
g C
⎡ 2 ( 0.300 m ) ⎤
= mgyC = ( 0.200 kg ) ( 9.80 m s 2 ) ⎢
⎥ = 0.392 J
3
⎣
⎦
1
1
2
KEB = m vB2 = ( 0.200 kg )(1.50 m s ) = 0.225 J
Choose PEg =2 0 at the2 level of the river. Then yi = 36.0 m, y f = 4.00, the jumper falls 32.0 m,
and the cord stretches 7.00 m. Between the balloon and the level where the diver stops momentarily, ( KE + PEg + PEs ) f = ( KE + PEg + PEs )i gives
0 + ( 700 N )( 4.00 m ) +
or
5.91
.
5.27
2 ( 0.588 J )
= 2.42 m s
0.200 kg
2 KEB
=
m
1
2
k ( 7.00 m ) = 0 + ( 700 N )( 36.0 m ) + 0
2
k = 914 N m
When the cyclist travels at constant speed, the magnitude of the forward static friction force on
the drive wheel equals that of the retarding air resistance force. Hence, the friction force is proportional to the square of the speed, and her power output may be written as
P = fs v = ( k v 2 ) v = k v 3
where k is a proportionality constant.
If the heart rate R is proportional to the power output, then R = k ′P = k ′( k v 3 ) = k ′k v 3, where k ′ is
also a proportionality constant.
The ratio of the heart rate R2 at speed v2 to the rate R1 at speed v1 is then
R2 k ′k v23 ⎛ v2 ⎞
=
=
R1 k ′k v13 ⎜⎝ v1 ⎟⎠
3
⎛R ⎞
v2 = v1 ⎜ 2 ⎟
⎝ R1 ⎠
giving
13
Thus,
Thus, if
if R = 90.0 beats min at v = 22.0 km h , the speed at which the rate would be
136 beatsR min
= 90.is0 beats min at v = 22.0 km h , the speed at which the rate would be
136 beats min is
⎛ 136 beats min ⎞
v = ( 22.0 km h ) ⎜
⎝ 90.0 beats min ⎟⎠
13
= 255.2 km h
and the speed at which the rate would be 166 beats min is
67352_ch05b.indd 142
⎛ 166 beats min ⎞
v = ( 22.0 km h ) ⎜
⎝ 90.0 beats min ⎟⎠
13
= 277.0 km h
2/9/11 1:19:48 PM
45
Energy
143
1 3 h , hitting
speed
of the
projectile
just
theatground
Thus, and
if Rthe
the speed
whichisthe rate would be
= 90
.0 beats
v = 22
.0 ⎞before
km
⎛ min
136 atbeats
min
=
v = ( 22.0 km h ) ⎜
25
5
.2
km
h
⎝ 90.0 beats min ⎟⎠
and the speed at which the rate would be 166 beats min is
⎛ 166 beats min ⎞
v = ( 22.0 km h ) ⎜
⎝ 90.0 beats min ⎟⎠
67352_ch05b.indd 143
13
= 277.0 km h
2/9/11 1:19:49 PM
ࢠ ࡋѰԛ˕঻Ѫ
6.1
6.3
(a)
If pball = pbullet , then
(
(b)
)(
)
3.00 × 10 −3 kg 1.50 × 10 3 m s
mbullet vbullet
=
= 31.0 m s
mball
0.145 kg
vball =
The kinetic energy of the bullet is
KEbullet
(
)(
3.00 × 10 –3 kg 1.50 × 10 3 m s
1
2
= mbullet vbullet
=
2
2
)
2
= 3.38 × 10 3 J
while that of the baseball is
1
( 0.145 kg) (31.0 m s ) = 69.7 J
2
mball vball
=
2
2
2
KEball =
The bullet has the larger kinetic energy by a factor of 48.5.
6.2
6.5
6.4
6.6
6.3
177
Use
p =the
mvimpulse-momentum theorem, F (Δt) = Δp = mv − mv , we find the average force to be
From
av
f
i
(
)(
)
(
)(
)
(a)
p = 1.67 × 10 −27 kg 5.00 × 10 6 m s = 8.35 × 10 −21 kg ⋅ m s
(b)
p = 1.50 × 10 −2 kg 3.00 × 10 2 m s = 4.50 kg ⋅ m s
(c)
p = ( 75.0 kg ) (10.0 m s ) = 750 kg ⋅ m s
(d)
p = 5.98 × 10 24 kg 2.98 × 10 4 m s = 1.78 × 10 29 kg ⋅ m s
(a)
(b)
6.9
6.4
6.7
Momentum and Collisions
(a)
(
)(
)
2 ( KE )
2 ( 275 J )
KE 12 mv 2 v
so
v=
=
= 22.0 m s
=
=
p
25.0 kg ⋅ m s
p
mv
2
2 ⋅s
2
N
m 2 v 2 13.5
( mv
kg )⋅ m=–3s p= 9.00 × 10 3 N = 9.00 kN
= p = =25.0
=
m=
=
1.50 × 10 s 1.14 kg
v2m 22.02mm s 2m
We choose the positive direction to be the direction of the final velocity of the ball.
ANSWERS TO MULTIPLE CHOICE QUESTIONS
(
)
I = Δp = m v f − vi = ( 0.280 kg ) ⎡⎣ +22.0 m s − ( −15.0 m s ) ⎤⎦
1.
6.8
6.10
Assuming that the collision was head-on so that, after impact, the wreckage moves in the original
or
I = +10.4 kg ⋅ m s = 10.4 kg ⋅ m s in the direction of the final velocity
direction of the car’s motion, conservation of momentum during the impact gives
(b) The
Theimpulse
averagedelivered
force the by
player
exerts
on the ball is
(a)
a force
is equal
to the area under
the
Force
versus
Time
I
10.4 kg ⋅ m s
= 173
curve. F
From
this N
av = the=figure at the right,
Δt
0.060 0 s
is seen to be a triangular area having a
By Newton’s
law,
the−3ball
exerts a force of equal magnitude back on the player’s fist.
base
of 1.50 msthird
= 1.50
× 10
s and
altitude of 18 000 N. Thus,
Take the direction of the ball’s final velocity
46(toward the net) to be the + x-direction.
(a)
67352_ch06.indd 169
(
)
I = Δp = m v f − vi = ( 0.060 0 kg ) 40.0 m s
2/9/11 1:25:36 PM
(
m v 2f − vi2
178
6.5
6.11
56.
58.
60.
62.
6.13
6.6
6.12
Chapter 6
=
)
( 0.060 0 kg) ⎡⎣( 40.0
2
2
m s ) − ( 50.0 m s ) ⎤⎦
Momentum and Collisions
= − 27.0 J
2
175
47
(a)
Taking
as the positive
direction,
v = 4M
gC forward
m
8 lb ⎞
⎛ 0.224
(b)
= 6.4 × 10 3 N ⎜
= 1.4 × 10 3 lb
⎝ 1 N ⎟⎠
0.961 m I = m ( Δv ) = ( 70.0 kg ) ( 5.20 m s − 0 ) = + 364 kg ⋅ m s = 364 N ⋅ s forward
(b) It is unlikely that the man has sufficient arm strength to guarantee the safety of the child
during aI collision.
collision would tear the child from his arms.
+ 364 The
kg ⋅ violent
m s forces during the
40.5
=
= + 438 kg ⋅ m s2 = 438 N forward
(b) g Fav =
Δt are soundly
0.832 based
s
(c) The laws
on physical principles: always wear a seat belt when in a car.
(a)
2
2
The
(a) velocity of the ball just before impact is found from v y = v0 y + 2a y Δy as
(
)
(
)
v1 = − v02 y + 2a y Δy = − 0 + 2 −9.80 m s 2 ( −1.25 m ) = − 4.95 m s
and the rebound velocity with which it leaves the floor is
(
)
v2 = + v 2f − 2a y Δy = + 0 − 2 −9.80 m s 2 ( +0.960 m ) = + 4.34 m s
The impulse given the ball by the floor is then
I LI
I
I
I
I = FΔt = Δ mv = m v 2 − v1
( )
(
)
= ( 0.150 kg ) ⎡⎣ + 4.34 m s − ( −4.95 m s ) ⎤⎦ = +1.39 N ⋅ s = 1.39 N s upward
6.14
6.15
6.7
(a)
(a)
Choose upward as the positive direction:
The impulse equals the area under the F versus t graph. This area is the sum of the area of
the rectangle
triangle.
I = m v f plus
− vi the
= (area
65.0ofkgthe
m Thus,
s
) +1.80
(
)
I = ( 2.0 N ) ( 3.0 s ) +
(b)
(
1
( 2.0 N ) ( 2.0 s ) = 8.0 N ⋅ s
2
I = Fav ( Δt ) = Δp = m v f − vi
)
8.0 N ⋅ s = (1.5 kg ) v f − 0, giving v f = 5.3 m s
(c)
(
)
I = Fav ( Δt ) = Δp = m v f − vi , so v f = vi +
I
m
8.0 N12.0
⋅s N ⋅s
v f = −2.0
m s +m s +
= 3.3 =m 4.00
s
= −2.00
ms
1.5 kg2.00 kg
6.8
6.16
6.17
(a)
Impulse
= area
under
curve
= (two
triangular
areas
of altitude
The impulse
is the
area
under
the curve
between
0 and
3.0 s. 4.00 N and base 2.00 s) +
(one rectangular area of width 1.00 s and height of 4.00 N). Thus,
This is:
I = (4.0 N)(3.0 s) = 12 N ⋅ s
(b)
The area under the curve between 0 and 5.0 s is
I = (4.0 N)(3.0 s) + (−2.0 N)(2.0 s) = 8.0 N ⋅ s
For parts (c) and (d), we use I = Fav (Δt) = Δp = m(v f − vi ), giving v f = vi + I m
67352_ch06.indd 175
(c)
At 3.0 s:
v f = vi +
I
12 N ⋅ s
= 0+
= 8.0 m s
m
1.50 kg
(d)
At 5.0 s:
v f = vi +
I
8.0 N ⋅ s
= 0+
= 5.3 m s
m
1.50 kg
2/9/11 1:25:54 PM
=
176
48
6.19
6.9
6.20
6.10
6.21
−2 ( 3.00 kg ) (10.0 m s ) sin 60.0°
= −260 N
0.200 s
Chapter 6
Thus,
(a)
FromΔx
Newton’s
the m
average
force the −2rain exerts on the roof is
2 ( Δxthird
2 (1.20
) = law,
) = 9.60
Δt =
=
× 10 s
vav v f + vi 0 + 25.0 m s
(b)
Fav =
Δp m ( Δv ) (1 400 kg ) ( 25.0 m s )
=
=
= 3.65 × 10 5 N
Δt
Δt
9.60 × 10 −2 s
(c)
aav =
⎞
⎛
Δv
25.0 m s
1g
= 26.5 g
=
= 260 m s 2 = 260 m s 2 ⎜
−2
2⎟
Δt 9.60 × 10 s
⎝ 9.80 m s. ⎠
(
)
We shall take toward the pitcher as the positive direction. Then, the velocity of the ball just before
v
Consider
thethe
thrower
contact
with
bat isfirst, with velocity after the throw of thrower . Applying conservation of
momentum yields
( 65.0 kg ) vthrower + ( 0.045 0 kg ) ( 30.0
or
m s ) = ( 65.0 kg + 0.045 0 kg ) ( 2.50 m s )
vthrower = 2.48 m s
Now, consider the (catcher + ball), with velocity of vcatcher after
⎛ the catch.
⎞ I From momentum
conservation,
⎜⎝
⎟⎠ v gloves
(b)
or
6.11
6.23
0 kg ) ( 30.0 m s ) + ( 60.0 kg ) ( 0 )
( 60.0 kg + 0.045 0 kg ) v she =exerts
( 0.045
As she throws the gloves,catcher
a force on them. As described by Newton’s third law,
the gloves exert a force
of equal magnitude in the opposite direction on the girl. This force
I
vcatcher = 2.25 × 10 −2 m s
causes
her to accelerate from rest to reach the velocity v girl.
The velocity of the girl relative to the ice, vGI, is vGI = vGP + vPI , where vGP = velocity of girl relative to plank, and vPI = velocity of plank relative to ice. Since we are given that vGP = 1.50 m s,
this becomes
vGI = 1.50 m s + vPI
(a)
[1]
Conservation of momentum gives mG vGI + mP vPI = 0, or vPI = − ( mG mP ) vGI
[2]
Then, Equation [1] becomes
⎛m ⎞
vGI = 1.50 m s − ⎜ G ⎟ vGI
⎝ mP ⎠
or
⎛
mG ⎞
⎜⎝ 1 + m ⎟⎠ vGI = 1.50 m s
P
giving
vGI =
(b)
1.50 m s
= 1.15 m s
⎛ 45.0 kg ⎞
1+ ⎜
⎝ 150 kg ⎟⎠
Then, using Equation [2] above,
⎛ 45.0 kg ⎞
vPI = − ⎜
(1.15 m s ) = − 0.345 m s
⎝ 150 kg ⎟⎠
or
6.24
vPI = 0.345 m s directed opposite to the girl’s motion
Originally, with both girl and plank at rest, the total momentum of the girl-plank system is zero.
With negligible friction between the plank and the ice, the total momentum of the girl-plank system is conserved.
(a)
I
I
I
I
The velocity of the girl relative to the ice is given by v GI = v GP + v PI , where v GP is the velocI
ity of the girl relative to the plank and v PI is the velocity of the plank relative to the ice.
Conservation of momentum of the girl-plank system then gives
67352_ch06.indd 176
2/9/11 1:25:56 PM
⎛
⎜⎝
6.12
6.5
6.25
⎞I
⎟⎠ v GP
Momentum and Collisions
177
49
Use
= mv a subscript a for the astronaut and t for the tank, conservation of momentum gives
(a) p Using
(a)
(b)
(c)
ma va f + mt vt f = ma vai + mt vti . Since both astronaut and tank were initially at rest, this
becomes
p = 1.67 × 10 −27 kg 5.00 × 10 6 m s = 8.35 × 10 −21 kg ⋅ m s
(
)(
)
(
)(
)
⎛m ⎞
ma va×f +
+ 0 × 10 2orm s =va f4.50
= − ⎜kg ⋅t m
vt f
tf = 0
p = 1.50
10m−2t vkg
3.00
⎝ ma ⎟⎠ s
The mass of the astronaut alone (after the oxygen tank has been discarded)
is ma = 75.0 kg. Taking toward the spacecraft as the positive direction, the velocity
imparted to the astronaut is
⎛ 12.0 kg ⎞
vaf = − ⎜
( −8.00 m s ) = +1.28 m s
⎝ 75.0 kg ⎟⎠
and the distance she will move in 2.00 min is
d = vaf t = (1.28 m s ) (120 s ) = 154 m
(b)
6.26
6.27
6.13
By Newton’s third law, when the astronaut exerts a force on the tank, the tank exerts a
force back on the astronaut. This reaction force accelerates the astronaut towards the
spacecraft.
toward the left
The boat and
sherman
move asbe
a single
unit gives
having mass
Requiring
thatfitotal
momentum
conserved
( mclub vclub + mball vball ) f = ( mclub vclub + mball vball )i
or
( 200 g ) ( 40
m s ) + ( 46 g ) vball = ( 200 g ) ( 55 m s ) + 0
⎛
and vball = 65 m s⎜
⎝
6.29
6.14
(a)
⎞
−2
⎟⎠ ( 300 m s ) = 2.0 × 10 m s
man
m1
v1
wife
m2
v2
Before impact
(b)
6.30
6.15
6.31
couple
m1 m2
vf
After impact
The collision is best described as perfectly inelastic , because the skaters remain in contact
after the collision.
(c)
m1 v1 + m2 v2 = ( m1 + m2 ) v f
(d)
vf =
(e)
vf =
m1 v1 + m2 v2
m1 + m2
( 70.0 kg ) ( 8.00
m s ) + ( 50.0 kg ) ( 4.00 m s )
= 6.33 m s
70.0 kg + 50.0 kg
.
For each skater, the impulse-momentum theorem gives
When Gayle jumps on the sled, conservation of momentum gives
( 50.0 kg + 5.00 kg ) v2 = ( 50.0 kg ) ( 4.00
m s) + 0
or the speed of Gayle and the sled as they start down the hill is v2 = 3.64 m s.
67352_ch06.indd 177
2/9/11 1:25:59 PM
178
50
184
Chapter 6
After Gayle and the sled glide down 5.00 m, conservation of mechanical energy (taking y = 0 at
(b)
the level of the top of the hill) gives
(
)
(
1
55.0 kg v32 + 55.0 kg
2
) ( 9.80
)
m s 2 ( −5.00 m ) =
(
1
55.0 kg
2
) ( 3.64 m s)
2
+0
so Gayle’s speed just before the brother hops on is v3 = 111 m s.
After her brother jumps on, conservation of momentum yields
( 55.0 kg + 30.0 kg ) v4 = ( 55.0 kg ) (
)
111 m s + 0
and the speed of Gayle, brother, and sled just after her brother hops on is v4 = 6.82 m s.
After all slide an additional 10.0 m down (to a level 15.0 m below the level of the hilltop),
conservation of mechanical energy from just after her brother hops on to the end gives the final
speed as
(
)
1
( 85.0 kg ) v52 + ( 85.0 kg ) 9.80 m s2 ( −15.0 m )
2
1
2
= ( 85.0 kg ) ( 6.82 m s ) + ( 85.0 kg ) 9.80 m s 2 ( −5.00 m )
2
(
6.33
6.16
)
or
v5 = 15.6 m (s22.5 g ) ( +35.0 m s ) + ( 300 g ) ( −2.50 m s ) − 0
=
= 1.67 m s
22.5 g
(a)
If M is the mass of a single car, conservation of momentum gives
( 3M ) v f
or
(b)
= M ( 3.00 m s ) + ( 2M ) (1.20 m s )
v f = 1.80 m s
The kinetic energy lost is KElost = KEi − KE f , or
KElost =
1
1
1
2
2
2
M ( 3.00 m s ) + ( 2M ) (1.20 m s ) − ( 3M ) (1.80 m s )
2
2
2
(
)
M
2
2
With M = 2.00 × 10 4 kg, vthis
yields
KE
( v1×−10v24)2J .
1 + v2 − 2v
1 vlost
2 = 2.16
3
6.17
6.35
Let M = mass of ball, m = mass of bullet, v = velocity of bullet, and V = the initial velocity of the ballbullet combination. Then, using conservation of momentum from just before to just after collision gives
⎛ m ⎞
v
or V = ⎜
⎝ M + m ⎟⎠
( M + m ) V = mv + 0
Now, we use conservation of mechanical energy from just after the collision until the ball reaches
maximum height to find
0 + ( M + m ) g hmax =
2
1
V2
1
m ⎞ 2
( M + m ) V 2 + 0 or hmax = = ⎛⎜⎝
⎟ v
2
2g 2g M + m⎠
With the data values provided, this becomes
hmax =
6.36
67352_ch06.indd 178
⎞
⎛
1
0.030 kg
⎟
2 ⎜
2 9.80 m s ⎝ 0.15 kg + 0.030 kg ⎠
(
)
2
( 200
m s ) = 57 m
2
(a)
2/9/11 1:26:02 PM
( M + m ) V 2 + 0 or hmax =
2
V2
1 ⎛ m ⎞ 2
=
⎜
⎟ v
2g 2g ⎝ M + m⎠
179
51
Momentum and Collisions
With the data values provided, this becomes
But Fnet = F2 −1 mg = F⎛2 − F1, 0.030
wherekg
F2 is the⎞ 2upward force
the floor exerts on the player
2
hmax = the jump and2 F⎜1 is the force exerted⎟by( 200
= 57the
m jump. Thus,
during
the flm
oors )before
2 9.80 m s ⎝ 0.15 kg + 0.030 kg ⎠ .
(
)
F2 = Fnet + F1 = + 260 N + 637 N = + 897 N = 897 N upward
6.36
6.18
6.37
6.15
(a)
The leftmost part of the
m1
(a)
impulse
sketchThe
depicts
the equals the area under the F versus t graph. This area is the sum of the area of
the from
rectangle
RThus,
3.75 m
i 0triangle.
situation
whenplus
the the area ofvthe
actor starts from rest
1
until just before
he N ) ( 3.0 s ) +
I = ( 2.0
2
makes contact with
R
his costar. Using
conservation of energy
over this period gives
v2 0
( KE + PE )1 = ( KE + PE )i
or
vf 0
m2
y0
Start
m1 m2
h
v1
v0 and Collisions
Momentum
Just Before
Just After
1
m1 v12 + 0 = 0 + mgR
2
End
187
so his speed just before impact is
(
v1 = 2gR = 2 9.80 m s 2
) ( 3.75 m ) = 8.57 m s
Now, employing conservation of momentum from just before to just after impact gives
( m1 + m2 ) v0 = m1v1 + m2 ( 0 )
v0 =
or
( 80.0 kg ) ( 8.57 m s )
m1 v1
=
= 5.08 m s
m1 + m2
80.0 kg + 55.0 kg
Finally, using conservation of energy from just after impact to the end yields
( KE + PE ) f = ( KE + PE )0
0 + ( m1 + m2 ) gh =
or
1
( m1 + m2 ) v02
2
( 5.08 m s ) = 1.32 m
v02
=
2g 2 9.80 m s 2
2
and
h=
(
)
188
Chapter 6
6.38
6.39
6.19
(a)
Because momentum is conserved even in a perfectly inelastic collision such as this, the
ratio is p f pi = 1 .
(b)
p f = pi
KEi =
⇒
( m1 + m2 ) v f
= m1 v1i + m2 ( 0 )
1
1
1
m1 v1i2 + m2 ( 0 ) = m1 v1i2
2
2
2
KE f
and
( m1 + m2 ) v 2f = ( m1 + m2 )
=
or
KE f =
vf =
m1 v1i
m1 + m2
1
( m1 + m2 ) v 2f
2
m12 v1i2
m1
=
so
2
2
2
Consider the sketches
above,
situation
before
KEi
m1 v1iwhich show
m1 +and
m2 just after collision.
m1 vthe
( m1 + mjust
1i
2)
(about 320 mph)
6.40
6.20
6.41
Conserving
momentum
in y-direction:
First,
find the horizontal
speed, v0energy
andspeed
embedded
impact.
x , of the
First, we
we will
use conservation
of mechanical
toblock
find the
of thebullet
block just
and after
embedded
bulAfter
this
instant,
the
block-bullet
combination
is
a
projectile,
and
we
fi
nd
the
time
to
reach
the
let just after impact:
floor by use of Δy = v0 y t + 12
( KE + PEs ) f = ( KE + PEs )i becomes 12 ( m + M ) V 2 + 0 = 0 + 12 kx 2
and yields
V=
kx 2
=
m+M
(150
N m ) ( 0.800 m )
= 29.3 m s
( 0.0120 + 0.100 ) kg
2
Now, employ conservation of momentum to find the speed of the bullet just before impact:
mv + M 0
67352_ch06.indd 179
2/9/11 1:26:04 PM
180
52
6.18
(m + M ) V
Chapter 6
and yields
I
LI
Δp
F av =
Δt
V=
(150
2
+0= 0+
1 2
kx
2
N m ) ( 0.800 m )
= 29.3 m s
( 0.0120 + 0.100 ) kg
kx 2
=
m+M
2
Now, employ conservation of momentum to find the speed of the bullet just before impact:
mv + M ( 0 ) = ( m + M ) V, or
⎛ 0.112 kg ⎞
⎛m+M⎞
v=⎜
V =⎜
( 29.3 m s ) = 273 m s
⎝ m ⎟⎠
⎝ 0.0120 kg ⎟⎠
6.42
6.43
6.21
(a)
(a)
Conservation of momentum gives
Over a the short time interval of the collision, external forces have no time to impart significant impulse to the players. The two players move together after the tackle , so the collision
is completely inelastic.
y (north)
(b)
vf
m1 m2
m1
v1i
x (east)
q
v2i
m1 90.0 kg
m2 95.0 kg
v1i 5.00 m/s
v2i 3.00 m/s
m2
px f = Σpxi
⇒
or v f cosq =
( m1 + m2 ) v f cosq = m1v1i + 0
( 90.0 kg ) ( 5.00 m s )
m1 v1i
=
m
+
m
( 1 2 ) 90.0 kg + 95.0 kg
v f cosq = 2.43 m s
and
( m1 + m2 ) v f sinq = 0 + m2 v2i
( 95.0 kg ) ( 3.00 m s )
m2 v2i
v f sinq =
=
m
+
m
( 1 2 ) 90.0 kg + 95.0 kg
py f = Σpyi
giving
⇒
(
)
and
v f sinq = 1.54 m s
Therefore, v 2f sin 2 q + cos 2 q = v 2f = (1.54 m s ) + ( 2.43 m s ) , and
2
2
v f = 8.28 m 2 s 2 = 2.88 m s
Also,
tanq =
190
Chapter 6
Thus,
(c)
6.44
v f cosq
=
1.54 m s
= 0.634
2.43 m s
and
q = tan −1 ( 0.634 ) = 32.4°
I
v f = 2.88 m s at 32.4° north of east
KElost = KEi − KE f =
=
67352_ch06.indd 180
v f sinq
1
1
1
m1 v1i2 + m2 v2i2 − ( m1 + m2 ) v 2f
2
2
2
1
1⎡
2
continued
m s ) on= next
785page
J
( 90.0 kg ) ( 5.00 m s )2 + ( 95.0 kg ) ( 3.00 m s )2 ⎤⎦ − (185 kg ) ( 2.88
⎣
2
2
(d)
The lost kinetic energy is transformed into other forms of energy, such as thermal energy
and sound.
(a)
Momentum is conserved, even in a perfectly inelastic collision. Thus, p f = pi or
2/9/11 1:26:07 PM
190
6.22
m1 v1i2 +
Chapter 6
6.23
Momentum and Collisions
181
53
1 before the gloves are
2
The1 total
momentum of
the2 system (girl plus gloves)2 is zero
thrown.
=
kg ) ( 5.00= 1m s ) + ( 95.0 kg ) ( 3.00 m s ) ⎤ − (185 kg ) ( 2.88 m s ) = 785 J
KElost⎡⎣(=90.0
KEfriction
i − KE f between
⎦
Neglecting
the
girl
and
the
ice,
the
total
momentum
is
also
zero
after
2
2
⎡
⎤
m
2
4
4
4
2the2 ⎤
1
⎛
⎞ 2
⎛
⎞ 2
⎡
2
2
gloves are thrown, giving ⎢⎜⎝ 3 − 1⎟⎠ v1 + 3 v1 v2 + ⎜⎝ 3 − 2 ⎟⎠ v2 ⎥ = 2 ⎢ − 3 v1 + 3 v1 v2 − 3 v2 ⎥
⎣
⎦
⎣
⎦
(d) The lost kinetic energy is transformed into other forms of energy, such as thermal energy
I
m ⎞I
m I
2 ⎛
and sound. I
or ( M − m ) v girl + vm12v−gloves
2v1=v20+ v22 and
= − ( v1v−girlv=2 )− ⎜⎝
⎟ v gloves
M
− m⎠
3
(a) Momentum
is conserved,
even in a perfectly inelastic collision. Thus, p f = pi or
Conservation
of momentum
gives
(b) As she throws the gloves, she exerts a force on them. As described by Newton’s third law,
the gloves exert a force of equal magnitude in the opposite direction on the girl. This force
I
25.0tog reach
s ) + (10.0
( 25.0 gher
) v1 to
(10.0 g ) vfrom
) ( 20.0thecmvelocity
f +accelerate
2 f = (rest
causes
v girl.g ) (15.0 cm s )
(a)
(c)
(
6.44
6.45
6.22
1
1
m2 v2i2 − ( m1 + m2 ) v 2f
2
2
)
or
2.50v1 f + v2 f = 65.0 cm s
[1]
The velocity of the girl relative to the ice, vGI, is vGI = vGP + vPI , where vGP = velocity of girl relaFor
know relative
that v1i −
− v1 f we
− vare
2 i =Since
2 f .given that v
tive head-on,
to plank, elastic
and vPIcollisions,
= velocitywe
of plank
tovice.
GP = 1.50 m s
(
)
Thus,
20.0 cm s − 15.0 cm s = −v1 f + v2 f or v2 f = v1 f + 5.00 cm s
[2]
Substituting Equation [2] into [1] yields 3.50v1 f = 60.0 cm s, or v1 f = 17.1 cm s .
Equation [2] then gives v2 f = 17.1 cm s + 5.00 cm s =. 22.1 cm s .
6.46
6.23
6.49
First, consider conservation of momentum and write
Choose the +x-axis to be eastward and the +y-axis northward.
If vi is the initial northward speed of the 3 000-kg car, conservation of momentum in the
y-direction gives
0 + ( 3 000 kg ) vi = ( 3 000 kg + 2 000 kg ) ⎡⎣( 5.22 m s ) sin 40.0° ⎤⎦
or
vi = 5.59 m s
(12.0 kg ) ( +3.00 m s )
Observe that knowledge=of the
initial
of the=2 0.537
000-kgmcars was unnecessary for this solution.
55.0
kg +speed
12.0 kg
6.50
6.51
6.24
Consider conservation of momentum in the first event (twin A tossing the pack), taking the direcChoose
thevelocity
x-axis togiven
be along
the original
line of motion.
tion
of the
the backpack
as positive.
This yields
(a)
From conservation of momentum in the x-direction,
m ( 5.00 m s ) + 0 = m ( 4.33 m s ) cos 30.0° + m v2 f cosq
or
v2 f cosq = 1.25 m s
[1]
Conservation of momentum in the y-direction gives
0 = m ( 4.33 m s ) sin 30.0° + m v2 f sinq , or v2 f sinq = − 2 .16 m s
Dividing Equation [2] by [1] gives tanq =
[2]
− 2 .16
= −1.73 and q = −60.0°
1.25
Then, either Equation [1] or [2] gives v2 f = 2 .50 m s, so the final velocity of the second
I
ball is v 2 f = 2.50 m s at − 60.0° .
continued on next page
67352_ch06.indd 181
2/9/11 1:26:11 PM
54
182
Chapter 6
(b)
(b)
Momentum and Collisions
The velocity
of the girl
1
1 relative to the
2 ice is
KEi = m v1i2 + 0 = m ( 5.00 m s ) = m 12 .5 m 2 s 2
2
2
⎛ mG ⎞ I
⎛ mG + mP − mG ⎞ I
I
I
I
I
=
v
+
v
v
⎟⎠ v GP
1 GI 2 GP 1 PI 2= v GP − ⎜⎝ m + m ⎟⎠ v GP = ⎜⎝ m + m
G
P
G
P
KE f = m v1 f + m v2 f
2
2
⎛
⎞I
or
1
⎜⎝ m s )2 ⎟⎠+ v1GPm ( 2 .50 m s )2 = m 12 .5 m 2 s 2
= m ( 4.33
2
2
(
)
(
6.25
6.52
6.25
6.53
193
)
Using a subscript a for the astronaut and t for the tank, conservation of momentum gives
Since KE f = KEi, this is an elastic collision .
≤ θinitially
≤ +90°at
( −90°
) that
ma va f + mt vt f = ma vai + mt vti . Since both astronaut and tank
were
rest, have
this the
given becomes
value for the tangent function.
The mass of the third fragment must be
We shall choose southward as the positive direction. The mass of the man is
(a)
m=
w
730 N
=
= 74.5 kg
g 9.80 m s 2
Then, from conservation of momentum, we find
(m
m an
vm an + mbook vbook ) f = ( mm an vm an + mbook vbook )i
or
( 74.5 kg ) v
m an
+ (1.2 kg ) ( −5.0 m s ) = 0 + 0
vm an = 8.1 × 10 −2 m s
and
Therefore, the time required to travel the 5.0 m to shore is
(KE + PEg ) f = (KE + PEg )i
Δx
5.0 m
t=
=
= 62 s
2
vm an 8.11× 10 −22 m s
v12 f ( − 3.30 m s )
or
0 + m1 ghm ax = m1 v1 f + 0
and
hm ax =
=
= 0.556 m
2
2 g 2 9.80 m s 2
6.26
6.55
(
The sketch at the right gives before and after
views of the collision between these two objects.
Since the collision is elastic, both kinetic
energy and momentum must be conserved.
m1
)
v1i v0
v2i v0
m2
Before Impact
v1f 0
Conservation of Momentum:
m1
m2
v2f v
After Impact
m1 v1 f + m2 v2 f = m1 v1i + m2 v2i
m1 ( 0 ) + m2 v = m1 v0 + m2 ( −v0 )
or
⎛m
⎞
v = ⎜ 1 − 1⎟ v0
⎝ m2
⎠
[1]
(
)
Since this is a perfectly elastic collision, v1i − v2i = − v1 f − v2 f , and with the given velocities
this becomes
v0 − ( −v0 ) = − ( 0 − v )
67352_ch06.indd 182
or
v = 2v0
(a)
⎛m
⎞
Substituting Equation [2] into [1] gives 2 v0 = ⎜ 1 − 1⎟ v0
⎝ m2
⎠
(b)
From Equation [2] above, we have
[2]
or
m1 m2 = 3
v v0 = 2
2/9/11 1:26:14 PM
gC
6.28
6.27
6.57
Momentum and Collisions
55
183
(a) The
mass of the rifl
e is to have negligible mass and ignore any energy or momentum it may
Note:
We consider
spring
possess after being released. Also, we take toward the right as the positive direction.
w
30 N
⎛ 30 ⎞
m= =
I = ⎜I ⎟ kg
⎠
Conservation of gmomentum,
9.80 m p
s2f = ⎝p9.8
i , gives
⎛ mmotion
⎞
We choose the direction
to be negative. Then, conservation of
or of thev2bullet’s
m1 v1 + m2 v2 = 0
= − ⎜ 1 ⎟ v1
momentum gives
⎝ m2 ⎠
[1]
Since the surface is frictionless, conservation of energy,
KE1 f + KE2 f + PEs, f = KE1i + KE2i + PEs,i
with KE1i = KE2i = PEs, f = 0 gives
1
1
1
m1 v12 + m2 v22 = kd 2
2
2
2
[2]
where d = 9.8 cm is the initial compression of the spring. Substituting Equation [1] into
Equation [2] gives:
1
1 ⎛ m2 ⎞ 1
m1 v12 + m2 ⎜ 12 v12 ⎟ = kd 2
2
2 ⎝ m2 ⎠ 2
⎛
m ⎞
m1 ⎜ 1 + 1 ⎟ v12 = kd 2
m
⎝
2 ⎠
or
Choosing the negative sign (since m1will recoil to the left), this result yields
v1 = −d
and
(
k
= − 9.8 × 10 −2 m
m1 (1 + m1 m2 )
v1 = −1.7 m s
or
)
280 N m
( 0.56 kg ) (1 + 0.56 0.88 )
1.7 m s to the left
Then, Equation [1] gives the velocity of the second object as
⎛ 0.56 ⎞
v2 = − ⎜
( −1.7 m s ) = +1.1 m s
⎝ 0.88 ⎟⎠
6.58
6.75
6.28
or
(
7.1 m s )
1.1 m s to the(right
=
= 2.6 m
2 9.8 m s 2
) (
2
)
(
)
Use
mechanical
energy,
, to find the
speed
of the
blue
+ PEg bullet
= KEfrom
+ PEimmediately
g
Firstconservation
consider the of
motion
of the block
andKE
embedded
after
impact
until
B
A
the block
comes
to rest
after
acrossbead.
the horizontal
bead
at point
B just
before
it sliding
collidesdistance
with thedgreen
This givestable. During this time, a
kinetic friction force fk = m k n = m k ( M + m ) g, directed opposite to the motion, acts on the block.
The net work done on the block and embedded bullet during this time is
Wnet = ( fk cos180° ) d = KE f − KEi = 0 −
1
( M + m) V 2
2
so the speed, V, of the block and embedded bullet immediately after impact is
206
Chapter 6V
=
−2 fk d
=
− ( M + m)
(
)
2m k M + m gd
M+m
= 2m k gd
Now, make use of conservation of momentum from just before to just after impact to obtain
continued on next page
pxi = px f
67352_ch06.indd 183
⇒
mv0 = ( M + m )V = ( M + m ) 2m k gd
2/9/11 1:26:16 PM
56
184
Chapter 6
After
Gayle
and
the sled
downwas
5.00 m, conservation of mechanical energy (taking y = 0 at
and the
initial
velocity
ofglide
the bullet
the level of the top of the hill) gives
⎛ M + m⎞
1 ( 6.26 m s ) sin 30.0°2
1v0 = ⎜⎝ m 2⎟⎠ 2m k gd ( 90.0 kg ) ( 8.00
s) + m
( 55.0
55.0 kg v3 + 55.0 =kg 9.80 m s 2 (m
−5.00
kg ( 3.64 m s ) =+ 06.15 m s
) = kg )55.0
2 kg
2
90.0 kg + 55.0
The woman starts from rest (v0y = 0) and drops freely with a y = −g for 2.00 m before the impact
so Gayle’s
speed just before the brother hops on is v3 = 111 m s.
with
(a) the toboggan. Then,
y
→
(
6.76
6.77
6.29
)
(
)(
(
)
)
v1f
m1
After her brother jumps on, conservation of momentum yields
→
m1
v1i
x
m2
x
53.0°
q
At rest
m2
→
v2f
Just Before Collision
Just After Collision
The situations just before and just after the collision are shown above. Conserving momentum in both the x- and y-directions gives
( ) ( )
py
(p )
x
f
f
= py
i
Momentum and Collisions
⇒ m1 v1 f sin 53° − m2 v2 f sinf = 0
or
207
m2 v2 f sinf = m1 v1 f sin 53° [1]
= ( px )i ⇒ m1 v1 f cos 53° + m2 v2 f cosf = m1 v1i + 0
m2 v2 f cosf = m1 v1i - m1 v1 f cos 53°
or
[2]
Dividing Equation [1] by [2] yields
tanf =
v1 f sin 53°
v1i − v1 f cos 53°
Equation [1] then gives
(b)
=
( 2.0
v2 f =
(1.0 m s ) sin 53°
= 0.57
m s ) − (1.0 m s ) cos 53°
m1 v1 f sin 53°
m2 sinf
=
or
( 0.20 kg ) (1.0 m s ) sin 53°
=
( 0.30 kg ) sin 30°
f = 30°
1.1 m s
The fraction of the incident kinetic energy lost in this collision is
KE f
ΔKE KEi − KE f
=
= 1−
= 1−
KEi
KEi
KEi
1
2
( 0.20 kg ) (1.0 m s ) + ( 0.30 kg ) (1.1 m s )
( 0.20 kg ) ( 2.0 m s )
2
1
2
1
2
2
2
ΔKE
= 0.30 or 30%
KEi
67352_ch06.indd 184
2/9/11 1:26:19 PM
7
ࢠ ୣࢷࡋѰ˕ࣸԯࢂ‫ئ‬৕
PROBLEM SOLUTIONS
PROBLEM SOLUTIONS
Rotational Motion and the
Law of Gravity
7.1
(a)
Earth rotates 2p radians (360°) on its axis in 1 day. Thus,
7.1
(a)
Earth rotates
radians
on its ⎞axis in 1 day.
Thus, Motion and the Law of Gravity
Rotational
215
1 day
rad ⎛(360°)
Δq 2p 2p
−5
=
rad s
=
7.27
×
10
w=
⎜
⎟
4
Δt
1 day ⎝ 8.64 × 10 s ⎠
Δq
w
=
proportional
Δt to both the angular acceleration and the time interval. If the time interval is
Because
of its rotation
its axis,acceleration
Earth bulgeswill
at the
equator.
held constant,
doublingabout
the angular
double
the angular speed attained
during the interval.
(b)
7.3
7.2
s 60 000 mi ⎛ 5 280 ft ⎞
8
=
⎜
⎟ = 3.2 × 10 rad
r
1.0 ft ⎝ 1 mi ⎠
(a)
q=
(b)
The car travels a distance equal to the circumference of the tire for every revolution the tire
makes if there is no slipping of the tire on the roadway. Thus, the number of revolutions
made during the warranty period is
n=
7.4
7.5
7.3
The distance traveled is s = rq , where
revin⎞radians.
⎛ q is
⎛ 2 p rad ⎞ ⎛ 1 min ⎞
Main Rotor: v = rw = ( 3.80 m ) ⎜ 450
⎟⎜
⎟⎜
⎟ = 179 m s
⎝
min ⎠ ⎝ 1 rev ⎠ ⎝ 60 s ⎠
p rad
(a) For 30°,
s = rq = ( 4.1 m ) 30°
⎛
⎞
180°
vsound
m⎞
⎛
v = ⎜ 179
= 0.522 vsound
⎟
⎝
s ⎠ ⎜⎝ 343 m s ⎟⎠
Tail Rotor:
216
7.7
7.4
S
60 000 miles ⎛ 5 280 ft ⎞
7
=
⎜
⎟ ⎡= 5.0 ×⎛ 10 rev ⎞ ⎤ = 7.7 × 10 2 m
2p r
2p (1.0 ft ) ⎝ 1 mile ⎠ ⎢
⎜⎝
⎟⎠ ⎥
⎣
⎦
rev ⎞ ⎛ 2p rad ⎞ ⎛ 1 min ⎞
⎛
v = rw = ( 0.510 m ) ⎜ 4138
⎟⎜
⎟⎜
⎟ = 221 m s
⎝
min ⎠ ⎝ 1 rev ⎠ ⎝ 60 s ⎠
⎞
m⎞ ⎛ v
⎛
v = ⎜ 221 ⎟ ⎜ sound ⎟ = 0.644 vsound
Chapter 7
⎝
s ⎠ ⎝ 343 m s ⎠
(a)
From w 2 = w 02 + 2a ( Δq ), the angular displacement is
w 2 − w 02 ( 2.2 rad s ) − ( 0.06 rad s )
=
= 3.5 rad
2a
2 0.70 rad s 2
2
Δq =
7.8
2
(
)
(b)
From the equation given above for Δq , observe that when the angular acceleration is constant, the displacement is proportional to the difference in the squares of the final and initial
angular speeds. Thus, the angular displacement would increase by a factor of 4 if both of
these speeds were doubled.
(a)
The maximum height h depends on the drop’s vertical speed at the instant it leaves the tire
and becomes a projectile. The vertical speed at this instant is the same as the tangential
speed, vt = rw , of points on the tire. Since the second drop rose to a lesser height,
the tangential speed decreased
57
67352_ch07.indd 209
2/9/11 1:27:53 PM
58
214
54.
7.9
7.5
)
( 0.381 m ) ( 2p
rad )
2
(
( 0.510 m − 0.540 m ) =
− 0.322 rad s 2
)
(a) 56.5 rad
s × 10 4 rev min − 0
2.51
⎛ 2 p rad ⎞ ⎛ 1 min ⎞
2
(a) a =
⎜⎝
⎟⎠ ⎜⎝
⎟⎠ = 821 rad s
3.20
s
1
rev
60.0
s
(d)
(b)
7.10
7.11
7.6
(
Chapter 7
(
)
⎡ 2.51 × 10 4 rev min ( 2p rad 1 rev ) (1 min 60.0 s ) + 0 ⎤
⎛ w f + w0 ⎞
⎢
⎥ ( 3.20 s )
q =w t = ⎜
t
=
⎝
2
2 ⎟⎠
⎢⎣
⎥⎦
⎞
⎛
⎟⎠ = 51 rev
⎜⎝
= 4.21 × 10 3 rad
We will break the motion into two stages: (1) an acceleration period and (2) a deceleration period.
From Δq = w av t = ⎡⎣(w f + w i ) 2 ⎤⎦ t, we find the initial angular speed to be
The angular displacement during the acceleration period is
2 ( 37.0 rev) ( 2p rad 1 rev)
2 Δq
wi =
−w f =
− 98.0 rad s = 57.0 rad s
t
3.00 s
The angular acceleration is then
a =
7.13
7.7
7.12
(a)
w f − wi
t
2
98.0
0 − (rad
377s −
rad57.0
s ) rad s
2
==
= 13.7
= − 226
rad rad
s2 s
3.00
s
2 ( 314 rad )
The linear distance the car travels in coming to rest is given by v 2f = v02 + 2a ( Δx ) as
v 2f − v02
Δx =
2a
=
0 − ( 29.0 m s )
(
2 −1.75 m s2
2
)
= 240 m
Since the car does not skid, the linear displacement of the car and the angular displacement
of the tires are related by Δx = r ( Δq ). Thus, the angular displacement of the tires is
Δq =
(b)
Δx
240 m
⎛ 1 rev ⎞ 116 rev
=
= ( 727 rad ) ⎜
=
⎝ 2p rad ⎟⎠
r
0.330 m
When the car has traveled 120 m (one half of the total distance), the linear speed of the car is
v = v02 + 2a ( Δx ) =
( 29.0
(
)
m s ) + 2 −1.75 m s2 (120 m ) = 20.5 m s
2
and the angular speed of the tires is
w=
7.14
7.15
7.8
(a)
(a)
20.5 m⎞s ⎛ 0 + 10.5 rad s ⎞
=⎛
= 62.1 rad s
⎜⎝ 0.330 m
⎟⎠ t = ⎜⎝
⎟⎠ ( 5.25 s ) = 27.6 rad
2
The
angular
is
The initial
car travels
235speed
m at constant
speed in an elapsed time of 36.0 s. Its constant speed is
therefore
v=
(b)
v
r
Δs 235 m
=
= 6.53 m s
Δt 36.0 s
The angular displacement of the car during the 36.0 s time interval is one-fourth of a full
circle or p 2 radians. Thus, the radius of the circular path is
r=
Δs
235 m
470
=
=
m
Δq p 2 rad
p
During the 36.0 s interval, the car has zero tangential acceleration, but does have a centripetal acceleration of constant magnitude
p ( 6.53 m s )
v 2 ( 6.53 m s )
ac =
=
=
= 0.285 m s2
r
470 m
( 470 p ) m
2
2
This acceleration is always directed toward the center of the circle
67352_ch07.indd 214
2/9/11 1:28:03 PM
=
235 m
470
=
m
p 2 rad
p
59
Rotational
Motion andbut
thedoes
Law of
Gravity
215
During the 36.0 s interval, the car has zero tangential
acceleration,
have
a centripetal acceleration of constant magnitude
2 and the time interval. If the time interval is
proportional2 to both the angular
m s)
( 6.53 m s )2 = p acceleration
( 6.53
v
held constant,
acceleration
will double
=doubling the angular
ac =
= 0.285
m s2 the angular speed attained
r
470
p
m
470
m
(
)
during the interval.
7.3
(a)
This acceleration is always directed toward the center of the circle . Therefore, when the car
is at point B, the vector expression for the car’s acceleration is
I
a c = 0.285 m s2 at 35.0° north of west
7.16
7.17
7.9
In
fortangential
the archeologist
to make
it safely
thespeeds
river, the
(a)order
The
acceleration
of the
bug asacross
the disk
up vine
is must be capable of giving
2
him a net acceleration of ac = vmax
r upward as he passes through the lowest point on the swing
⎛ w f − w0 ⎞
with a speeda of= vra
max==r8.00 m s
t
⎜⎝ Δt ⎟⎠
⎛ 78.0 rev min − 0 ⎞ ⎛ 1 m
= 5.00 in ⎜
⎟⎠ ⎜⎝ 39.37 in
3.00 s
⎝
(
(b)
)
⎞ ⎛ 1 min
⎟⎠ ⎜⎝ 60 s
⎞ ⎛ 2p rad ⎞
= 0.346 m s 2
⎟⎠ ⎜
⎝ 1 rev ⎟⎠
The final tangential speed of the bug is
⎛
rev ⎞ ⎛ 1 m ⎞ ⎛ 1 min ⎞ ⎛ 2p rad ⎞
ms
vt = rw f = 5.0 0 in ⎜ 78.0
Rotational Motion⎜ and the Law
Gravity
⎟ = of1.04
⎝
min ⎟⎠ ⎜⎝ 39.37 in ⎟⎠ ⎜⎝ 60 s ⎟⎠ ⎝ 1 rev ⎠
(
)
219
(c)
Since the bug has constant angular acceleration, and hence constant tangential acceleration
(at = ra ), the tangential acceleration at t = 1.00 is at = 0.346 m s 2 as above.
(d)
At t = 1.00 s, the tangential velocity of the bug is
(
)
vt = v0 + at t = 0 + 0.346 m s 2 (1.00 s ) = 0.346 m s
and the radial or centripetal acceleration is
vt2
( 0.346 m s )
=
= 0.943 m s2
r ( 5.00 in )(1 m 39.37 in )
2
ac =
(e)
The total acceleration is a = ac2 + at2 = 1.00 m s 2 , and the angle this acceleration makes
I
with the direction of a c is
⎛ at ⎞m s2 −1 ⎛ 0.346 ⎞ −2
q = =tan −19.80
⎟ = 20.1°
rad s
× 10
⎜⎝ a ⎟⎠ =3 tan = ⎜⎝ 4.9
0.943 ⎠
4.0
×c 10 m
7.19
7.10
7.18
(a) The tension in the string must counteract
1609 mthe radial
The radius
of the of
cylinder
is r = weight,
2.5 mi and also supply
component
the object’s
1 mi
the needed centripetal acceleration.
ΣFc = T − mg cosq = mac =
or
(
T = m v 2 r + g cosq
)
T
mv 2
r
q
q
mg
⎡ ( 8.00 m s )
⎤
= ( 0.500 kg ) ⎢
+ 9.80 m s2 cos 20.0° ⎥
⎢⎣ 2.00 m
⎥⎦
2
(
)
= 20.6 N
(b)
67352_ch07.indd 215
The net tangential force acting on the object is Ft = mg sinq , so the tangential acceleration
has magnitude
2/9/11 1:28:05 PM
(
+ g cosq
)
⎡ ( 8.00 m s )2
⎤
= ( 0.500 kg ) ⎢
+ 9.80 m s2 cos 20.0° ⎥
⎢⎣ 2.00 m
⎥⎦
(
216
60
7.7
Chapter 7
(a)
(b)
= 20.6 N
From
w 2tangential
= w 02 + 2aforce
displacement
is sinq , so the tangential acceleration
( Δqacting
), the angular
The net
on the object
is Ft = mg
has magnitude
2
2
w 2 − w 02 ( 2.2 rad s ) − ( 0.06 rad s )
F
2
2
t
= = g sinq= = 9.80 m s sin 20.0°
= 3.5 rad
aΔq
t =
2 0.70 rad s 2 = 3.35 m s
m 2a
(
(b)
)
( )
)
From
equation
given above
for Δqto
, observe
that path.
when the angular acceleration is conand is the
directed
downward,
tangential
the circular
stant, the displacement is proportional to the difference in the squares of the final and initial
angular
speeds.
Thus, the
angular
displacement
would increase by a factor of 4
The radial
component
of the
acceleration
is
v 2 ( 8.00 m s )
=
= 32.0 m s2 toward the center of the path
r
2.00 m
2
220
ac =
Chapter 7
(c)
The total acceleration has magnitude
atotal = at2 + ac2 =
( 3.35 m s ) + ( 32.0 m s )
2 2
2 2
⎛a ⎞
⎛ 3.35 ⎞
q = tan −1 ⎜ t ⎟ = tan −1 ⎜
= 5.98°
⎝ 32.0 ⎟⎠
⎝ ac ⎠
at
Thus,
→
atotal
atotal = 32.2 m s2
or
→
q
ac
→
at
I
a total = 32.2 m s 2 at 5.98° to the cord and pointing below
the center of the circular path
(d)
No change in answers if the object is swinging toward the equilibrium point instead of
away from it.
(e)
If the object is swinging toward the equilibrium position, it is gaining speed, whereas it is
losing speed if it is swinging away from the equilibrium position. In both cases, when the
⎞ tangential,
1 rev the
cord is 20.0° from the ⎛vertical,
centripetal,
and total accelerations have the
= 1.5 × 10 2 rev
s
⎜⎝ 2p calculated
magnitudes and directions
in parts (a) through (c).
rad ⎟⎠
(
7.21
7.11
7.20
)
(a) From ΣFvrt2 = mac, we have
Since Fc = m
r
2
⎛ v 2 ⎞ ( 55.0 kg ) ( 4.00 m s )
T = m⎜ t ⎟ =
= 1.10 × 10 3 N = 1.10 kN
0.800 m
⎝ r ⎠
(b)
The tension is larger than her weight by a factor of
3
T
1.104.78
× 10rad
Ns − 0
=
2.04 stimes
=
= 6.55
mg ( 55.0 kg0.730
m s22
) 9.80 rad
(
7.22
7.25
7.12
)
(a)
(a)
If
the tension
in each of is
the two support chains, the net force acting on the child at the
T is
The
centripetal
acceleration
lowest point on the circular path is
2
⎡⎛
rev ⎞ ⎛ 2 p rad ⎞ ⎛ 1 min ⎞ ⎤
2
2
ac = rw = ( 9.00 m ) ⎢⎜ 4.00
⎥ = 1.58 m s
min ⎟⎠ ⎜⎝ 1 rev ⎟⎠ ⎜⎝ 60 s ⎟⎠ ⎥⎦
⎢⎣⎝
(b)
At the bottom of the circular path, we take upward as positive and apply Newton’s second
law. This yields ΣFy = n − mg = m(+ac ), or
n = m ( g + ac ) = ( 40.0 kg ) ⎡⎣( 9.80 + 1.58 ) m s2 ⎤⎦ = 455 N upward
(c)
67352_ch07.indd 216
At the top of the path, we again take upward as positive and apply Newton’s second law to
find ΣFy = n − mg = m(−ac ), or
2/9/11 1:28:07 PM
61
217
Rotational Motion and the Law of Gravity
⎡⎣
⎤⎦ = 455 N upward
and while
(c)
At thedecelerating,
top of the path, we again take upward as positive and apply Newton’s second law to
find ΣFy = n − mg = m(−ac ), or
⎡ 0 + ( 5.0 rev s ) ( 2 p rad 1 rev) ⎤
⎛ w f + wi ⎞
2
t=⎢
q2 = ⎜
⎥ (12 s ) = 1.9 × 10 rad
⎟
2 − 1.58 ) m s2⎦⎤⎦ = 329 N upward
⎝n = m2 ( g −⎠ ac ) =⎣ ( 40.0 kg ) ⎡⎣( 9.80
⎞
⎛ 1 rev equal
(d)
At a displacement
point halfwayisup, theq seat
upward
component
to the
⎡⎣(1.3
⎤⎦ ⎜
The total
= q1exerts
+q 2 =an
+ 1.9 )vertical
× 10 2 rad
⎟⎠ = 51 rev
⎝ 2p rad
2p and a component toward the center having
child’s weight (392 N)
magnitude
=
= 8.98
s
2
kg ) 1.58
Fc = mac = ( 40.00.700
rad sm s = 63.2 N. The total force exerted by the seat is
(
7.11
7.27
7.13
)
From Upon
Δq = w
+ astronaut’s
w2 i ) 2
(g)
moving
slower than
his feet
his head is
av t = 392
f N
Fstanding,
N directed
inward
and because
at
( (wthe
) + ( 63.2 Nhead
)2 =is 397
R =
closer to the axis of rotation. When standing, the radius of the circular path followed by
the head is rhead⎛=392
10.0Nm⎞ − 1.80 m = 8.20 m, and the tangential speed of the head is
= 80.8° above the horizontal
q = tan −1 ⎜
⎝ 63.2 N ⎟⎠
(
) = 5.74 m s
(a)
Since the 1.0-kg mass is in equilibrium, the tension in the string is
(
)
T = mg = (1.0 kg ) 9.8 m s 2 = 9.8 N
7.28
7.29
7.14
(b)
The tension in the string must produce the centripetal acceleration of the puck. Hence,
Fc = T = 9.8 N .
(c)
From Fc = m puck vt2 R , we find
(d)
R Fc
(1.0 m ) ( 9.8 N ) = 6.3 m s
vt = of the
=numeric data from Problem
Substitution
7.27 into the results for (a) and (c) shown
mpuck
0.25 kg
above will yield the answers given for that problem.
(
)
Rotational Motion and the Law of Gravity
223
(a) Since the mass m2 hangs in equilibrium on the end of the string,
Friction between the tires and the roadway is capable of giving the truck a maximum centripetal
acceleration of
v2
( 32.0 m s ) = 6.83 m s2
= t, max =
r
150 m
2
ac, max
If the radius of the curve changes to 75.0 m, the maximum safe speed will be
vt, max = r ac, max =
7.30
7.31
7.15
( 75.0 m ) ( 6.83
)
m s2 = 22.6 m s
, v0 y = 0.
(a)
(a)
The external forces acting on the 2water are the gravitational force
ac = rw 2 = ( 2.00 m ) ( 3.00 rad s ) = 18.0 m s 2
(b)
Fc = mac = ( 50.0 kg ) 18.0 m s2 = 900 N
(c)
We know the centripetal acceleration is produced by the force of friction. Therefore, the
needed static friction force is fs = 900 N. Also, the normal force is n = mg = 490 N. Thus,
the minimum coefficient of friction required is
(
ms =
( fs )max
n
=
)
900 N
= 1.84
490 N
Such a large coefficient of friction is unrealistic, and she will not be able to stay on the
merry-go-round.
7.32
67352_ch07.indd 217
(a)
At A, taking upward as positive, Newton’s second law gives ΣFy = n − mg = m(+ac ). Thus,
2/9/11 1:28:10 PM
218
62
7.16
7.33
Chapter 7
The time
to stop
(i.e.,
reachby
a speed
of wbodies
= 0) with a = −2.00 rad s 2 is
The forces
exerted
on the
2.0-kg
the other
are Fx and Fy as shown in the diagram at the right.
3.0 kg
→
The magnitudes of these forces are
(6.67 × 10
Fx =
−11
)
N ⋅ m 2 kg2 ( 2.0 kg ) ( 4.0 kg )
2.0 m
( 4.0 m )2
2.0 kg
= 3.3 × 10 −11 N
and
Fy
( 6.67 × 10
=
−11
)
N ⋅ m 2 kg 2 ( 2.0 kg ) ( 3.0 kg )
( 2.0 m )
→
Fy
2
F
4.0 kg
␪
→
Fx
4.0 m
= 1.0 × 10 −10 N
The resultant force exerted on the 2.0-kg is F = Fx2 + Fy2 = 1.1 × 10 −10 N
⎛ Fy ⎞
directed at q = tan −1 ⎜ ⎟ = tan −1 ( 3.0 ) = 72°. The
above
the +x-axis
answer
m1 = 2 .00 kg and m2 = 3.00 kg is
F
physically equivalent.⎝ x ⎠
7.34
7.35
7.17
8
Wethe
know
that m1point
+ m2the
= 5.00
kg, orism1.92
m− m
from
both bodies. The force exerted on the
At
half-way
spaceship
× 10kg
2 = 5.00
1.
ship by the Earth is directed toward the Earth and has magnitude
FE =
=
GmE ms
r2
6.67 × 10 −11 N ⋅ m 2 kg 2 5.98 × 10 24 kg 3.00 × 10 4 kg
(
)(
(1.92 × 10 m )
8
)(
2
) = 325 N
The force exerted on the ship by the Moon is directed toward the Moon and has a magnitude of
FM =
=
GmM ms
r2
6.67 × 10 −11 N ⋅ m 2 kg 2 7.36 × 10 22 kg 3.00 × 10 4 kg
(
)(
(1.92 × 10 m )
8
(
)(
2
)(
) = 4.00 N
)
6.67 × 10 N ⋅ m 2 kg2 1.991 × 10 30 kg (1.00 kg )
The resultant force
directed toward Earth =. −2.08 × 1013 J
= − is ( 325 N − 4.00 N ) = 321 N
6.38 × 10 6 m
7.37
7.18
(a)
−11
If air resistance is ignored, the only force acting on the projectile during its flight is the
gravitational force. Since the gravitational force is a conservative force, the total energy of
the projectile remains constant. At r = RE , the projectile has speed
v = vesc 3 = 2GM E RE 3, and its total energy is
E = KE + PEg =
or
E=−
1 2 ⎛ GM E m ⎞ 1 ⎛ 1 2GM E ⎞ GM E m
mv + ⎜ −
= m
⋅
−
2
RE ⎟⎠ 2 ⎜⎝ 9
RE ⎟⎠
RE
⎝
8 GM E m
⋅
9
RE
When the projectile reaches maximum height at r = rmax, and is momentarily at rest, the
kinetic energy is zero and we have
E = KE + PEg = 0 −
or
(b)
rmax =
7.38
(
)
9
9
RE = 6.38 × 10 6 m = 7.18 × 10 6 m
8
8
The altitude of the projectile when at r = rmax is
h = rmax − RE =
67352_ch07.indd 218
GM E m
8 GM E m
=− ⋅
9
rmax
RE
9
R
6.38 × 10 6 m
RE − RE = E =
= 7.98 × 10 5 m
8
8
8
The radius of the satellite’s orbit is
2/9/11 1:28:12 PM
8 GM E m
Rotational Motion and the Law of
⋅
9
RE
=−
(
63
219
)
(c)
Since the bug 9has constant
angular acceleration, and hence constant tangential acceleration
9
or
rmax = RE = 6.38 × 10 6 m = 7.18 × 10 6 m
(at = r
as above.
8
8
(d)
(b)
At
1.00 s,ofthe
bug
Thet =
altitude
thetangential
projectilevelocity
when atofr the
is is
= rmax
h = rmax − RE =
7.38
7.39
7.19
Gravity
9
R
6.38 × 10 6 m
RE − RE = E =
= 7.98 × 10 5 m
8
8
8
The radius of the satellite’s orbit is
(a) At the midpoint between the two masses, the forces exerted by the 200-kg and 500-kg
masses are oppositely directed, so from F = GMm r 2 and r1 = r2 = r, we have
ΣF =
or
GM1m GM 2 m Gm
−
= 2 ( M1 − M 2 )
r
r12
r22
(6.67 × 10
ΣF =
−11
N ⋅ m 2 kg2 ) ( 50.0 kg ) ( 500 kg − 200 kg )
( 0.200 m )2
= 2.50 × 10 −5 N toward the 500-kg
(b)
At a point between the two masses and distance d from the 500-kg mass, the net force will
be zero when
G ( 50.0 kg ) ( 200 kg )
( 0.400 m − d )
2
=
G ( 50.0 kg ) ( 500 kg )
d2
or
d = 0.245 m
Note that the above equation yields a second solution d = 1.09 m. At that point, the two
. gravitational forces do have equal magnitudes, but are in the same direction and cannot add to zero.
7.40
7.41
7.20
The equilibrium position lies between the Earth and the Sun on the line connecting their centers. At
The
radii of
orbits of the
two satellites
this point,
thethe
gravitational
forces
exerted onare
the object by the Earth and Sun have equal magnitudes
and opposite directions. Let this point be located distance r from the center of the Earth. Then, its
rA =from
rB =may
hA +the
RE Sun
= Ris
RE = ×2R
hB +determine
RE = 2REthe
+R
E +1.496
E = 3R
distance
10E 11 m −and
r , and we
value
of Er by requiring that
(
)
From Kepler’s third law, the ratio of the squares of the periods of the two satellites is
3
27
TB2 ⎛ 4p 2 rB3 ⎞ ⎛ GME ⎞ rB3 ⎛ 3RE ⎞
=
=
=
⋅
=
8
TA2 ⎜⎝ GME ⎟⎠ ⎜⎝ 4p 2 rA3 ⎟⎠ rA3 ⎜⎝ 2RE ⎟⎠
thepage
ratio of their periods is
continued Thus,
on next
TB
=
TA
7.42
7.43
7.21
(a)
(
TB2
=
TA2
) ( 5.98 × 10
)
27
= 1.84
7.61 × 10 6 m
8
(
2
24
Rotational Motion and the Law of Gravity
kg
) = 6.89 m s
227
2
satellite’s
period
is T = orbit
110 min
The radius
of the
satellite’s
is 60.0 s 1.00 min
r = h + RE = 2.80 × 10 6 m + 6.38 × 10 6 m = 9.18 × 10 6 m
Then, modifying Equation 7.23 (Kepler’s third law) for orbital motion about the Earth
rather than the Sun, we have
(
)
3
4p 2 9.18 × 10 6 m
⎛ 4p 2 ⎞ 3
T =⎜
=
r
⎝ GM E ⎟⎠
6.67 × 10 −11 N ⋅ m 2 kg 2 5.98 × 10 24 kg
2
(
)(
)
yielding
67352_ch07.indd 219
2/9/11 1:28:15 PM
64
220
(
)(
The total acceleration has magnitude
yielding
T 2 = 7.66 × 10 7 s2
and
(b)
The constant tangential speed of the satellite is
Chapter 7
3
)
(
6
circumference of orbit 2p r 2p 9.18 × 10 m
=
=
period
T
8.75 × 10 3 s
)
vt = 6.59 × 10 3 m s = 6.59 km s
or
(c)
)
⎛ 1h ⎞
T = 8.75 × 10 3 s ⎜
= 2.43 h
⎝ 3 600 s ⎟⎠
(c)
vt =
228
(
4p 2 9.18 × 10 6 m
⎞ 3
⎟⎠ r = 6.67 × 10 −11 N ⋅ m 2 kg 2 5.98 × 10 24 kg
⎛
⎜⎝
Chapter 7
The satellite’s only acceleration is centripetal acceleration, so
(
)
5.98s ×2 10 24 kg ( 600 kg )
⎛
2
3
on next page
6.59 × 10 3⎞ km
v
= 1.47 × 10continued
N
t
⎟⎠ 6
a = a⎜⎝c =
=
= 4.736 m s⎤22 toward center of Earth
⎡
r
9.18 × 10 m
⎣ 2 6.38 × 10 m ⎦
(
7.45
7.22
7.44
(
)
(
((
)
)(
(
)
)
3
= 6.3 × 10 23 kg
vt2
2
= rw 2 = ( 0.800 m ) (p rad s ) = 7.90 m s2
r
(b)
ac =
(c)
We imagine that the weight of the ball is supported by a frictionless platform. Then, the
rope tension need only produce the centripetal acceleration. The force required to produce
the needed centripetal acceleration is F = m ( vt2 r ). Thus, if the maximum force the rope
can exert is 100 N, the maximum tangential speed of the ball is
rFmax
=
m
( 0.800 m ) (100 N )
5.00 kg
= 4.00 m s
From Kepler’s third law, the mass of Jupiter can be expressed in terms of one of its satellite’s
orbital radius and period as M Jupiter = 4p 2 GT 2 r 3.
(
giving
(b)
)
For Io,
r = 4.22 × 10 8 m
M Jupiter =
T = 1.77 days ⎡⎣(8.64 × 10 4 s) 1 day ⎤⎦ = 1.53 × 10 5 s,
and
(
4p 2 4.22 × 108 m
(6.67 × 10
−11
N ⋅ m kg
2
2
)
3
) (1.53 × 10 s)
5
2
= 1.90 × 10 27 kg
For Ganymede,
r = 1.07 × 10 9 m
giving
67352_ch07.indd 220
) (
)
gravitational
force
small
of material
w = 0.500
rev s =atp the
radstar’s
s. equator supplies the centripetal
The angular
velocity
of on
theaball
is parcel
acceleration, or
(a) vt = rw = ( 0.800 m ) (p rad s ) = 2.51 m s
(a)
7.52
)
30
⎡ 2 1.99 ×
⎤
⎞
⎛ 4p 2 ⎞ 3 ⎛
⎣
4p 2 10 kg ⎦ = 1.63 × 10 4 rad6 s
⎜
⎟ 9.4 × 10 m
=⎜
r =
3
2 ⎟
2
3
−11
⎝ GT ⎠
⎜⎝ 6.67
10.0××1010
m
N ⋅ m 2 kg 2 2.8 × 10 4 s ⎟⎠
( vt )max =
7.51
7.24
(
From The
Kepler’s
thirdmoves
law (Equation
written
the
form suitable for bodies orbiting Mars, we
(a)
satellite
in an orbit7.23),
of radius
r =in
2R
E and the gravitational force supplies the
2
3
, so the massHence,
of Mars, computed from the given data, must be
have Trequired
= 4p 2centripetal
GM Mars racceleration.
M Mars
7.46
7.47
7.23
)
M Jupiter =
and
T = 7.16 days ⎡⎣(8.64 × 10 4 s) 1 day ⎤⎦ = 6.19 × 10 5 s,
(
4p 2 1.07 × 10 9 m
(
)(
)
3
6.67 × 10 −11 N ⋅ m 2 kg2 6.19 × 10 5 s
)
2
= 1.89 × 10 27 kg
(c)
Yes. The results of parts (a) and (b) are consistent. They predict the same mass within the
limits of uncertainty of the data used to compute these results.
(a)
When the passenger is at the top, the radial forces producing the
centripetal acceleration are the upward force of the seat and
the downward force of gravity. The downward force must exceed the
upward force to yield a net force toward the center of the circular
path.
2/9/11 1:28:17 PM
(
)(
Rotational Motion
Law of
1.89the
× 10
kgGravity
2 = and
, or × 10 5 s
6.19
27
)
221
65
2
⎛
s ) ⎞mass within the
⎛
The
of staticoffriction
( 4.00themsame
v 2 ⎞ are consistent.
Yes. force
The results
parts
=
m ⎜ (a)
g + and⎟(b)
m s2predict
+
= ( 70.0 kg ) ⎜ 9.80They
⎟ = 826 N
⎜⎝ these results. 8.00 m ⎟⎠
limits of uncertainty of the
⎝ datar used
⎠ to compute
7.23
(a)
(c)
7.52
7.53
7.25
(a)
Whenofthe
at the
The radius
thepassenger
satellite’sisorbit
is top, the radial forces producing the
centripetal acceleration are the upward force of the seat and
downward
force of6 gravity. The downward
force must exceed the 6
2
rthe
=R
E + h = 6.38 × 10 m + 1.50 × 10 mi (1 609 m 1 mi ) = 6.62 × 10 m
upward
force to yield a net force toward the center of the circular
(a) path.
The required centripetal acceleration is produced by the gravitational force, so
(
(
)
)
m vt2 r = GM E m r 2, which gives vt = GM E r .
(
232
7.55
7.26
(
(a)
(c)
(b)
)
When the car is at the top of the arc, the normal force is upward and the weight downward.
The net
seatforce
mustdirected
exert thedownward,
greatest force
on the
The
toward
thepassenger
center of the circular path and hence supplying the centripetal acceleration, is ΣFdown = mg − n = m vt2 r . Thus, the normal force is
n = m g − vt2 r .
(
(
)
)
If r = 30.0 m and n → 0, then g − vt2 r → 0. For this to be true, the speed of the car must
be
vt = rg =
7.27
7.57
)
24
⎛
N ⋅ m 2 ⎞ 5.98 × 10 kg
vt = ⎜ 6.67 × 10 −11
= 7.76 × 10 3 m s
⎟⎠ radial
(b) At the lowest⎝ point on the path,
forces
contributing to
kg 2 the
6.62 ×
10 6 m
the centripetal acceleration are again the upward force of the seat and
(b) the
Thedownward
time for one
complete
revolution
is the upward force must
force
of gravity.
However,
now exceed the downward force to yield a net force directed toward
the center2ofp the
path.
2 p 6.62
× 10 6 m
r circular
T
=
=
= 5.36 × 10 3 s = 89.3 min
Chapter 7
vt
7.76 × 10 3 m s
2
3
((30.0 m ) ( 9.80
) =× 1017.1
Nm= s2.1 kN
) (1.0m) =s2.1
The speed the person has due to the rotation of the Earth is vt = rw , where r is the distance from
the rotation axis and w is the angular velocity of rotation.
The person’s apparent weight, Fg, apparent , equals the magnitude of the upward normal force exerted
on him by the scales. The true weight, Fg, true = mg, is directed downward. The net downward
force produces the needed centripetal acceleration, or
⎛ v2 ⎞
ΣFdown = − n + Fg,true = −Fg,apparent + Fg,true = m ⎜ t ⎟ = mrw 2
⎝ r ⎠
(a)
At the equator, r = RE , so Fg,true = Fg,apparent + mRE w 2 > Fg,apparent
(b)
At the equator, it is given that ac = RE w 2 = 0.0340 m s 2, so the apparent weight is
Fg,apparent = Fg,true − mRE w 2 = ( 75.0 kg ) ⎡⎣( 9.80 − 0.0340 ) m s2 ⎤⎦ = 732 N
At either pole, r = 0 (the person is on the rotation axis) and
(
)
Fg,apparent = Fg,true = mg = ( 75.0 kg ) 9.80 m s2 = 735 N
67352_ch07.indd 221
2/9/11 1:28:19 PM
ࢠ ୣࢷૡ୉˕ୣࢷѰࠇଝ
PROBLEM SOLUTIONS
PROBLEM SOLUTIONS
.
8.1
Resolve the 100-N force into components parallel
to and perpendicular to the rod, as
2.00
F( = F cos ( 20.0° + 37.0° ) = F cos57.0°
F
20.0
37.0
m
F = 100 N
F⬜
and
Pivot
F⊥ = F sin ( 20.0° + 37.0° ) = F sin 57.0°
The lever arm of F⊥ about the indicated pivot is 2.00 m, while that of F( is zero. The torque due
to the 100-N force may be computed as the sum of the torques of its components, giving
t = F( ( 0 ) − F⊥ ( 2.00 m ) = 0 − [(100 N ) sin 57.0° ]( 2.00 m ) = −168 N ⋅ m
or
.
8.3
8.3
t = 168 N ⋅ m clockwise
First resolve all of the forces shown in Figure P8. into components parallel to and perpendicular
to the beam as shown in the sketch below.
(25 N) cos 30°
(25 N) sin 30°
(30 N) cos 45°
O
(30 N) sin 45°
(10 N) cos 20°
C
2.0 m
(10 N) sin 20°
4.0 m
(a)
t O = + [( 25 N ) cos30° ]( 2.0 m ) − [(10 N ) sin 20° ]( 4.0 m ) = + 30 N ⋅ m
or
(b)
t C = + [( 30 N ) sin 45° ]( 2.0 m ) − [(10 N ) sin 20° ]( 2.0 m ) = + 36 N ⋅ m
or
8.5
.
(a)
8.5
t O = 30 N ⋅ m counterclockwise
t C = 36 N ⋅ m counterclockwise
Rotational Equilibrium and Rotational Dynamics
251
Rotational Equilibrium and Rotational Dynamics
251
t = Fg ⋅ ( lever arm ) = ( mg−2) ⋅ [ Csinq ]
d = (1.20 × 10 m ) cos 48.0° = 8.03 × 10 −3 m, and the torque is
= ( 3.0 kg ) ( 9.8 m s 2 ) ⋅ [( 2.0 m ) sin 5.0° ] = 5.1 N ⋅ m
(b)
The magnitude of the torque is proportional to sinq , where
q is the angle between the direction of the force and the
line from the pivot to the point where the force acts.
Note from the sketch that this is the same as the angle
the pendulum string makes with the vertical.
ᐉ
q
q
Pivot
Fg mg
the pendulum string makes with the vertical.
67352_ch08.indd 243
2/9/11 1:29:50 PM
Rotational Equilibrium and Rotational Dynamics
251
67
q is the angle between the direction of the force and the
line from the pivot to the point where the force acts.
t =F
arm ) =that
Note
from
the sketch
is the]same as the angle
( mgthis
) ⋅ [ Csinq
g ⋅ ( lever
the pendulum string makes with the vertical.
= ( 3.0 kg ) 9.8 m s 2
Since sinq increases as q increases, the torque also increases with the angle.
8.5
(a)
8.9
8.4
Require that Σt = 0 about an axis through the elbow and perpendicular to the page. This gives
(
)
Σt = + ⎡⎣( 2.00 kg ) 9.80 m s 2 ⎤⎦ ( 25.0 cm + 8.00 cm ) − ( FB cos 75.0° ) (8.00 cm ) = 0
or
8.10
8.13
8.5
FB =
(19.6 N ) ( 33.0 cm ) =
( 8.00 cm ) cos 75.0°
312 N
.
Consider the torques about an axis perpendicular to the page through the left end of the scaffold.
Σmi xi
Requiring that x cg =
= 0 gives
Σmi
( 5.0 kg ) ( 0 ) + ( 3.0 kg ) ( 0 ) + ( 4.0 kg ) ( 3.0 m ) + ( 8.0 kg ) x = 0
( 5.0 + 3.0 + 4.0 + 8.0 ) kg
or 8.0 x + 12 m = 0 which yields x = −1.5 m
Also, requiring that ycg = Σmi yi Σmi = 0 gives
( 5.0 kg ) ( 0 ) + ( 3.0 kg ) ( 4.0 m ) + ( 4.0 kg ) ( 0 ) + ( 8.0 kg ) y = 0
( 5.0 + 3.0 + 4.0 + 8.0 ) kg
or 8.0 y + 12 m = 0 yielding y = −1.5 m
Thus, the 8.0-kg object should be placed at coordinates
8.19
8.6
m) .
( −1.5 m,− 1.5
= 2.33 × 10 3 N
Consider the torques about an axis perpendicular to the page through the left end of the rod.
Ty T cos30.0
6.00 m
Ry
Tx T sin30.0
Rx
3.00 m
100 N
4.00 m
(100 N ) ( 3.00 m ) + ( 500 N ) ( 4.00 m )
Σt = 0 ⇒ T =
( 6.00 m ) cos 30.0°
500 N
continued on next page
T = 443 N
ΣFx = 0 ⇒ Rx = T sin 30.0° = ( 443 N ) sin 30.0°
67352_ch08.indd 251
2/9/11 1:30:08 PM
68
252
Chapter 8
(b)
force
theT column
is N
found
from
ΣFx =The
0 ⇒
Rx =
sin 30.0°exerts
= ( 443
) sin 30.0°
ΣFN
0
⇒
F + T − mg = 0
Rx = 222
the right
y =toward
− T30.0°
= 343
N−N
226
N = N717
ΣFy =or
0 ⇒ RFy =+ mg
T cos
− 100
− 500
= 0N upward
⎛ N ) cos 30.0° = 216
⎞ N upward
Ry = 600 N − ( 443
⎜⎝
⎟⎠ = 139 g
8.21
8.7
Consider the torques about an axis perpendicular to the page and through the left end of the plank.
Σt = 0 gives
− ( 700 N ) ( 0.500 m ) − ( 294 N ) (1.00 m ) + ( T1 sin 40.0° ) ( 2 .00 m ) = 0
or
→
T2
T1 = 501 N
0.500 m
Then, ΣFx = 0 gives − T3 + T1 cos 40.0° = 0,
or
→
T1
700 N
→
T3
T3 = ( 501 N ) cos 40.0° = 384 N
40.0°
1.00 m
1.00 m
mg 294 N
Σt )
= − ( 700 N ) x max − ( 200 N ) ( 3.00 m ) −
− 994 N
+ T1 sin 40.0° = 0,
From ΣFy = 0, Tleft2 end
(80.0 N )( 6.00 m ) + [( 900 N ) sin 60.0°]( 6.00 m ) = 0
or
T2 = 994 N − ( 501 N ) sin 40.0° = 672 N
which gives x max = 5.14 m
(a)
Considering a pivot at the lower end of the beam, we get
Σt )lower = 0
end
⇒
⎛C
⎞
+ FS C sinq − mg ⎜ cosq ⎟ = 0
⎝2
⎠
Fs
8.8
8.23
and the spring force is
mg ( C cosq 2 )
mg
=
C sinq
2 tanq
/
2
FS =
d=
(b)
or
d=
q
Rx
mg
2k tanq
From the first condition of equilibrium,
and
8.31
8.9
FS
k
mg
Ry
From Hooke’s law, FS = kd , the distance the spring is
stretched is then
ΣFx = 0
⇒
Rx − FS = 0
or
Rx = FS =
ΣFy = 0
⇒
Ry − mg = 0
or
Ry = mg
mg
2 tanq
.
The moment of inertia for rotations about an axis is I = Σmi ri2 , where ri is the distance mass mi is
from that axis.
(a)
For rotation about the x-axis,
I x = ( 3.00 kg ) ( 3.00 m ) + ( 2 .00 kg ) ( 3.00 m ) +
2
2
( 2.00 kg ) ( 3.00 m )2 + ( 4.00 kg ) ( 3.00 m )2 =
(b)
67352_ch08.indd 252
99.0 kg ⋅ m 2
When rotating about the y-axis,
2/9/11 1:30:11 PM
253
69
Rotational Equilibrium and Rotational Dynamics
(b)
When rotating about the y-axis,
I y = ( 3.00 kg ) ( 2 .00 m ) + ( 2 .00 kg ) ( 2 .00 m ) +
2
2
( 2.00 kg ) ( 2.00 m )2 + ( 4.00 kg ) ( 2.00 m )2 =
(c)
44.0 kg ⋅ m 2
For rotations about an axis perpendicular to the page through point O, the distance
ri for each mass is ri =
( 2 .00 m )2 + ( 3.00 m )2
= 13.0 m. Thus,
2 ( Δy )
2 ( −1.00 m )
⇒ +t 2.00
= + 4.00=
0.857
IO = ⎡⎣( 3.00 + 2.00
kg ⎤⎦ 13.0 m 2 2 == 143
kg s⋅ m 2
a ) −2.72
m s
(
8.32
8.10
8.35
)
The required torque in each case is t = I a . Thus,
(a) Consider the force diagrams of the cylinder and man given at
the right. Note that we shall adopt a sign convention with
clockwise and downward as the positive directions. Thus,
both a and a are positive in the indicated directions
and a = ra . We apply the appropriate form of Newton’s
second law to each diagram to obtain:
a
M
r
T
T
Rotation of Cylinder: t = Iα
⇒
rT sin 90° = Ia
or
T = Ia r
a
m
so
1⎛1
2⎞ ⎛ a⎞
⎜ Mr ⎟⎠ ⎜⎝ ⎟⎠
r⎝2
r
T=
T=
giving
1
Ma
2
mg
[1]
Translation of man:
264
Chapter 8
ΣFy = ma
⇒
mg − T = ma
T = m ( g − a)
or
[2]
Equating Equations [1] and [2] gives 12 Ma = m ( g − a ) , or
a=
(
a 3.92 m s 2
=
= 9.80 rad s 2
r
0.400 m
(b)
From a = ra , we have
(c)
As the rope leaves the cylinder, the mass of the cylinder decreases, thereby
decreasing the moment of inertia. At the same time, the weight of the rope leaving
the cylinder would increase the downward force acting tangential to the cylinder,
and hence increase the torque exerted on the cylinder. Both of these effects
will cause the acceleration of the system
+ (120 to
N )increase
( 0.320 mwith
) time. (The increase would
= 1.40 kN
−2
be slight in this case, given
the
large
mass
of
the
cylinder.)
2.80 × 10 m
a =
(
8.36
8.37
8.11
67352_ch08.indd 253
I=
t net rF sin 90° ( 0.330 m ) ( 250 N )
=
=
= 87.8 kg ⋅ m 2
a
a
0.940 rad s 2
t net = Ia
(b)
For a solid cylinder, I = Mr 2 2, so
(c)
⇒
)
(a)
M=
8.38
)
( 75.0 kg) 9.80 m s2
mg
=
= 3.92 m s2
m+M 2
75.0 kg + ( 225 kg 2 )
(
)
2
2I 2 87.8 kg ⋅ m
=
= 1.61 × 10 3 kg
2
r2
0.330
m
(
)
(
)
w = w 0 + a t = 0 + 0.940 rad s 2 ( 5.00 s ) = 4.70 rad s
(
The angular acceleration is a = w f − w i
) Δt
2/9/11 1:30:14 PM
254
70
8.14
8.12
8.41
266
Chapter 8
(
)
− 2.88 m s 2 = 138 N
mg is
(a) initial
As the
woman
walks to
theand
beam,
The
angular
velocity
of the right
wheelalong
is zero,
the final angular xvelocity
she will eventually reach a point where the beam
will start
rotate
v to
50.0
m sclockwise about the rightmost
wpivot.
=
= 40.0 rad s
f =
point,
r At this
1.25
m the beam is starting to lift up
L/2
n1
Mg
the leftmost
pivot and
Hence,off
theofangular
acceleration
is the normal force, n1,
exerted by that pivot will have diminished to zero.
w f − w i 40.0 rad s − 0
= y = 0 ⇒ 0 −=mg
83.3
rad+sn2 2 = 0
=
Then,
ΣF
− Mg
0.480 s
Δt
Chapter a
8
n2
n2 = ( m + M ) g
or
The torque acting on the wheel is t = fk ⋅r, so t = I a gives
2
2
I a (110 kg ⋅ m ) (83.3 rad s )
=
= 7.33 × 10 3 N
r
1.25 m
fk =
Thus, the coefficient of friction is
mk =
8.42
8.43
8.13
fk 7.33 × 10 3 N
2
2MR 2 5 w 2
=
=Iw 0.524
2 MR 2w 2
2
2
= 4 N2
=
=
=
n 1.40 × 10
2
2
2
2
2 2
2 2
7
Mv 2 + Iw 2 M ( Rw ) + 2MR 5 w
5 MR w + 2 MR w
(
Assuming the solid sphere starts from rest, and taking y = 0 at the level of the bottom
1 2 1
2
of
mvt =the (total
10.0 mechanical
kg )(10.0 menergy
s ) = 500 J
KEthe
t =incline,
2
2
(b)
KEr =
1 2 1⎛ 1
Iw = ⎜ m R 2
2
2⎝ 2
2
⎞ ⎛ vt ⎞
⎟⎠ ⎜ 2 ⎟
⎝ R ⎠
1
1
2
mvt2 = (10.0 kg )(10.0 m s ) = 250 J
4
4
(c)
KE total = KEt + KEr = 750 J
(a)
Treating the particles on the ends of the rod as point masses, the total moment of inertia of
2
2
the rotating system is I = I rod + I1 + I 2 = mrod C2 12 + m1 ( C 2 ) + m2 ( C 2 ) . If the mass of
2
the rod can be ignored, this reduces to I = 0 + ( m1 + m2 ) ( C 2 ) , and the rotational kinetic
energy is
KEr =
(b)
1
1
2
2
Iw 2 = ⎡⎣( 3.00 kg + 4.00 kg ) ( 0.500 m ) ⎤⎦ ( 2.50 rad s ) = 5.47 J
2
2
If the rod has mass mrod = 2.00 kg, the rotational kinetic energy is
KEr =
1 2 1⎡ 1
2
2
2
Iw = ⎢ ( 2.00 kg )(1.00 m ) + ( 3.00 kg + 4.00 kg )( 0.500 m ) ⎤⎥ ( 2.50 rad s )
2
2 ⎣ 12
⎦
or KEr = 5.99 J .
8.46
8.15
8.49
67352_ch08.indd 254
Rotational Equilibrium and Rotational Dynamics
269
2 is
The work
the grindstone
1
Using
Wnetdone
= KEon
f − KEi = 2 Iw f − 0, we have
wf =
8.50
)
(a)
(a)
=
8.45
8.14
(
)
2Wnet
=
I
2F ⋅s
=
I
2 ( 5.57 N ) ( 0.800 m )
= 149 rad s
4.00 × 10 −4 kg ⋅ m 2
Using conservation of mechanical energy,
2/9/11 1:30:17 PM
(
=
8.51
8.16
)
10 9.8 m s 2 ( 6.0 m ) sin 37°
36 rad sand Rotational Dynamics
Rotational=Equilibrium
7 ( 0.20 m )
2
71
255
The moment
inertia of the
cylinder
With
the bodyofpositioned
as shown
in is
Figure P8.15c, we use the following sketch to determine
the distance x for each body part:
1
1 ⎛ w⎞
1 ⎛ 800 N ⎞
I = MR 2 = ⎜ ⎟ R 2 = ⎜
(1.50 m )2 = 91.8 kg ⋅ m 2
2
2⎝ g⎠
2 ⎝ 9.80 m s 2 ⎟⎠
The angular acceleration is given by
a=
t F ⋅ R ( 50.0 N )(1.50 m )
=
=
= 0.817 rad s 2
I
I
91.8 kg ⋅ m 2
At t = 3.00 s, the angular velocity is
w = w i + a t = 0 + ( 0.817 rad s 2 )( 3.00 s ) = 2.45 rad s
and the kinetic energy is
1
1
2
KEr = Iw⎛22= (291.8
⎞ kg2⋅ m 2 ) ( 2.45 rad s ) = 2 276 J
L = Iw2 = ⎜ MR
w = ( 2.40 kg ) ( 0.180 m ) ( 35.0 rad s ) = 1.81 kg ⋅ m 2 s
⎟
2
⎝3
⎠
3
8.52
8.55
8.17
As themass
bucket
drops,
loses gravitational
potential
energy.
r = 0.500
m s about the center of the connecting
Each
moves
in ait circular
path of radius
The spool
rotational
rod.
Their gains
angular
speed iskinetic energy and the bucket gains
translational kinetic energy. Since the string does not slip on
the spool, v 5.00 m s
w= =
= 10.0 m s
r
0.500 m
Neglecting the moment of inertia of the light connecting rod, the angular momentum of this rotating system is
⎛ ⎞
⎛ 0.59 2AU ⎞
2
kmkg
s )+=3.00
0.91kgkm
s
⎤⎦ w⎟⎠=( 54
⎜⎝ =⎟⎠ ⎡⎣vmp 1=
L = Iw
r 2⎜⎝+ 35
m2 rAU
m ) (10.0 rad s ) = 17.5 J ⋅ s
( 4.00
) ( 0.500
8.18
8.61
8.56
The moment
Yes of inertia of the cylinder before the putty arrives is
(a)
Ii =
1
1
2
M R 2 = (10.0 kg )(1.00 m ) = 5.00 kg ⋅ m 2
2
2
After the putty sticks to the cylinder, the moment of inertia is
I f = I i + mr 2 = 5.00 kg ⋅ m 2 + ( 0.250 kg )( 0.900 m ) = 5.20 kg ⋅ m 2
2
Conservation of angular momentum gives I f w f = I i w i ,
or
8.63
8.19
⎛I
wf = ⎜ i
⎝ If
2
⎞
⎛ 5.00
⎡ kg ⋅ m ⎞ ( 7.00 rad s⎤) = 6.73 2rad s
2
=
w
⎟ i ⎜ ⎢
⎝ 5.20 kg ⋅ m 2 ⎟⎠ + 0.400 kg ⎥ (1.00 m ) ( 4.00 rad s ) = 1.73 kg ⋅ m s
⎠
⎣
⎦
The initial angular velocity of the puck is
wi =
( vt )i
ri
=
0.800 m s
rad
= 2.00
0.400 m
s
Since the tension in the string does not exert a torque about the axis of revolution, the angular
momentum of the puck is conserved, or I f w f = I i w i . Thus,
⎛I
wf = ⎜ i
⎝ If
2
2
⎛ mri2 ⎞
⎛ ri ⎞
⎞
⎛ 0.400 m ⎞
=
=
w
=
w
w
⎟ ( 2.00 rad s ) = 5.12 rad s
⎜ ⎟ i ⎜⎝
⎟ i ⎜
2⎟ i
0.250 m ⎠
⎝ mrf ⎠
⎝ rf ⎠
⎠
The net work done on the puck is
Wnet = KE f − KEi =
or
Wnet =
67352_ch08.indd 255
(
( 0.120 kg) ⎡( 0.250 m )2
2
)
(
)
1
1
1
m
I f w 2f − I i w i2 = ⎡⎣ mrf2 w 2f − mri2 w i2 ⎤⎦ = ⎡⎣ rf2w 2f − ri2w i2 ⎤⎦
2
2
2
2
⎣
(5.12
2
2
2
rad s ) − ( 0.400 m ) ( 2.00 rad s ) ⎤⎦
2/9/11 1:30:20 PM
This yields
72
256
2
⎛
⎜
⎝
Chapter 8
2
⎛ mri2 ⎞
⎛ ri ⎞
⎞
⎛ 0.400 m ⎞
=
=
w
=
w
w
⎟ ( 2.00 rad s ) = 5.12 rad s
⎜ ⎟ i ⎜⎝
⎟ i ⎜
2⎟ i
0.250 m ⎠
⎝ mrf ⎠
⎝ rf ⎠
⎠
The net work done on the puck is
Wnet = KE f − KEi =
or
Wnet =
(
( 0.120 kg) ⎡( 0.250 m )2
⎣
2
)
(
)
1
1
1
m
I f w 2f − I i w i2 = ⎡⎣ mrf2 w 2f − mri2 w i2 ⎤⎦ = ⎡⎣ rf2w 2f − ri2w i2 ⎤⎦
2
2
2
2
(5.12
2
2
2
rad s ) − ( 0.400 m ) ( 2.00 rad s ) ⎤⎦
(
)
2
2
This yields Wnet = 5.99 × 10=−2(1.12
J . ) 9.80 m s = 12.3 m s
2
8.64
8.20
8.65
With all
crew
members onofthe
rim ofmomentum,
the station, the
acceleration experienced is the cen(a)
From
conservation
angular
I f wapparent
f = Ii w i ,
tripetal acceleration,
so
⎛I
wf = ⎜ i
⎝ If
KE f =
⎛ I ⎞
⎛ I ⎞ 1
⎛ I ⎞
1
1
I f w 2f = ( I1 + I 2 ) ⎜ 1 ⎟ w o2 = ⎜ 1 ⎟ ⎡⎢ I1 w o2 ⎤⎥ = ⎜ 1 ⎟ KEi
2
2
⎦ ⎝ I1 + I 2 ⎠
⎝ I1 + I 2 ⎠
⎝ I1 + I 2 ⎠ ⎣ 2
⎞
⎛ I1 ⎞
wo
⎟wi = ⎜
⎝ I1 + I 2 ⎟⎠
⎠
2
(b)
or
KE f
KEi
=
I1
<1
I1 + I 2
Since this is less than 1.0, kinetic energy was lost.
8.66
8.73
8.21
The total moment of inertia of the system is
⎡
⎛ d⎞⎤
(a)
Li = 2 ⎢ M v ⎜ ⎟ ⎥ = M vd
⎝ 2⎠⎦
⎣
(b)
⎛1
⎞
KEi = 2 ⎜ M vi2 ⎟ = M v 2
⎝2
⎠
(c)
L f = Li = M vd
(d)
vf =
(
Lf
2 M rf
)
=
M vd
= 2v
2 M (d 4)
Rotational Equilibrium and Rotational Dynamics
279
2
⎛1
⎞
KEpage
M v 2f ⎟ = M ( 2 v ) = 4M v 2
continued(e)
on next
f = 2⎜
⎝2
⎠
8.75
8.22
(f)
Wnet = KE f − KEi = 3M v 2
(a)
Since the bar is in equilibrium, ΣFy = 0 giving
ᐉ
Fs = mg − F1 − F2
and
or
(b)
67352_ch08.indd 256
(
x
)
Fs = ( 2.35 kg ) 9.80 m s 2 − 6.80 N − 9.50 N
Fs = 6.73 N upward
F1
Fs
ᐉ
2
Fg = mg
F2
We require the sum of the torques about an axis perpendicular to the page and passing
through the left end of the bar be zero. This gives
2/9/11 1:30:22 PM
(b)
or
8.76
8.77
8.23
(a)
(a)
Equilibrium and
Rotational
Dynamics
We require the sum of the torques about anRotational
axis perpendicular
to the
page and
passing
through the left end of the bar be zero. This gives
( m − m ) gL) (cosq
( m − m ) g cosq
ΣFx = Σt
0 left
⇒ =R0sin15.0°
C2 2=)0+=F2 ⋅ C1= 0 2
=+ Fs 1⋅ x −−2(2Tmgsinq
end
m1 L + m2 L
( m1 + m2 ) L
⎡
2.35
kg
) 9.80 m s2 2 − 9.50 N ⎤⎦ (1.30 m )
⎡⎣( mg T2 )sinq
− F2 ⎤⎦ ⋅ C ⎣(
[1]
R
=
or angular
x = acceleration is a =
The
maximum in the horizontal position (q = 0°), where the gravisin15.0°
Fs
6.73 N
tational forces have maximum lever arms and exert the maximum torque on the system.
Also, note that a = 0 at q = 90°. This is understandable since the vertical orientation is a
and
x = 0.389 m = 38.9 cm
position of unstable equilibrium (t net = 0).
(
)
Taking PEg = 0 at the level of the horizontal
Since the pulley
is very light (so I ≈ 0) and
axis passing through the center of the rod, the
rotates without friction, the net torque about
total energy of the rod in the vertical position is
the axis of the pulley is
F
r
Σt = T1r − T2 r = Ia ≈ ( 0 )a = 0
From this, we see that T1 = T2, or the tension
in the rope has the same value T on both
sides of the pulley.
(b)
73
257
T1
T2
From the rotational form of Newton’s second law, Σt = ΔL Δt , we see that if Σt = 0 then
ΔL Δt = 0. This means that the angular momentum of the system will be constant at its initial value of zero at all times. Thus, the upward speeds of the monkey and the bananas must
always be equal.
Another way of arriving at this conclusion is to realize that the monkey and the bananas have
the same net upward force, ΣFy = T − Mg, acting on them. Thus, they have the same upward
acceleration and, both having started from rest, will
have the same upward speeds at all times.
clockwise or a = −1.67 rad s 2 , Tu = 135 N, M = 80.0 kg, R = 0.625 m,
and r = 0.230 m, we have
(c)
No, the monkey will not reach the bananas. The monkey and the bananas move upward at
the same speed, staying a fixed distance
apart (at least until the bananas become tangled in
(80.0 kg)( 0.625 m )2 −1.67 rad s2 = 22 N
pulley).
Tthe
=
135
N
+
l
2 ( 0.230 m )
(
)
(
8.79
8.24
)
We neglect the weight of the board and assume
that the woman’s feet are directly above the
point of support by the rightmost scale. Then,
the force diagram for the situation is as
shown at the right.
P
→
w
→
Fg1
x
2.00 m
→
Fg2
From ΣFy = 0, we have Fg1 + Fg2 − w = 0, or w = 380 N + 320 N = 700 N.
Choose an axis perpendicular to the page and passing through point P.
Then Σt = 0 gives w ⋅ x − Fg1 ( 2.00 m ) = 0 , or
x=
8.80
8.25
8.81
Fg1 ( 2.00 m )
w
=
( 380 N ) ( 2.00 m )
700 N
= 2vcg
= 1.09 m
(a) Since only conservative forces do work on the long
Choose
an we
axisuse
perpendicular
page for
andthis
passing
rod,
conservationtoofthe
energy
pure through
rotation
the center
of
the
cylinder.
Then,
applying
Σt
=
Ia
to the
about the fixed point1 O. The
rod
starts
from
rest
w = 0)
(
2
2
1
cylinder
gives
( 2T ) ⋅ Rof=gravity
( at RO.) , ori
2 M R at a
2MR
with
the center
the= level
of point
(
)
(
)
Choosing this level as the reference level for gravitational
1
Tpotential
= M at energy, we have
[1]
4
Now apply ΣFy = ma y to the falling objects to obtain
( 2m ) g − 2 T = ( 2m ) at , or
at = g −
(a)
67352_ch08.indd 257
T
m
[2]
Substituting Equation [2] into [1] yields
2/9/11 1:30:23 PM
74
258
8.81
Chapter 8
M at
[1]
Now
apply
ΣFy perpendicular
= ma y to the falling
to passing
obtain through
Choose
an axis
to the objects
page and
2mcenter
T =the
(the
) g − 2of
( 2m
) at , or Then, applying Σt = Ia to the
cylinder.
cylinder gives ( 2T ) ⋅ R = 12 M R 2 a = 12 M R 2 ( at R ) , or
T
at = g −
[2]
1 m
T=
4
(a) Substituting
Equation [2] into [1] yields
(
T=
67352_ch08.indd 258
)
M mg
.
M + 4m
From Equation [2] above,
at = g −
8.82
(
a
M
R
→
2T
→
Mg ⎛ M ⎞
T
−⎜
4
⎝ 4m ⎟⎠
which reduces to T =
(b)
)
= 2vcg
2T
→
at
2m
→
2mg
Mg
4mg
1 ⎛ M mg ⎞
= g−
=
M + 4m
M + 4m
m ⎜⎝ M + 4m ⎟⎠
(a)
A smooth (that is, frictionless) wall cannot exert a force parallel to its surface. Thus,
the only force the vertical wall can exert on the upper
end of the ladder is a horizontal normal force.
(b)
Consider the force diagram of the ladder given
at the right. If the rotation axis is perpendicular
to the page and passing through the lower end
of the Iladder, the lever arm of the normal
force n 2 that the wall exerts on the upper
end of the ladder is
2/9/11 1:30:26 PM
ࢠ ˈ঑ࠪࡪ঑
PROBLEM SOLUTIONS
9.1
(a)
If the particles in the nucleus are closely packed with negligible space between them, the
average nuclear density should be approximately that of a proton or neutron. That is
rnucleus ≈
mproton
Vproton
=
mproton
3
4p r 3
∼
(
3 1.67 × 10 −27 kg
(
4π 1 × 10
−15
m
)
)
∼ 4 × 1017 kg m 3
3
The density of iron is r Fe = 7.86 × 10 3 kg m 3 and the densities of other solids and liquids
are on the order of 10 3 kg m 3. Thus, the nuclear density is about 1014 times greater than
that of common solids and liquids, which suggests that atoms must be mostly empty space.
Chapter 9Solids and liquids, as well as gases, are mostly empty space.
(b)
294
9.2
9.3
9.2
Let the weight of the car be W. Then, each tire supports
(a) The total normal force exerted on the bottom acrobat’s shoes by the floor equals the total
weight of the acrobats in the tower. That is
(
)
n = mtotal g = ⎡⎣( 75.0 + 68.0 + 62.0 + 55.0 ) kg ⎤⎦ 9.80 m s 2 = 2.55 × 10 3 N
9.4
9.5
9.3
n
n
2.55 × 10 3 N
= 3.00 × 10 4 Pa
=
=
Atotal 2 Ashoe 2 ⎡ 425 cm 2 1 m 2 10 4 cm 2 ⎤
⎣
⎦
sole
(b)
P=
(c)
If the acrobats are rearranged so different ones are at the bottom of the tower, the total
weight supported, and hence the total normal force n, will be unchanged. However, the total
7.988 ×will
10 −3change
kg unless all the acrobats
area Atotal = 2 Ashoe sole,mand
wear the
total hence the pressure,
=
=
= 1.76 × 10 4 kg m 3
−7
3
same size shoes. VAu + VCu 3.79 × 10 m + 7.46 × 10 −8 m 3
(
)
(a) The mass of gold in the coin is
The average density of either of the two original worlds was
r 0=
M
M
3M
=
=
3
V
4p R 3 4p R3
The average density of the combined world is
r=
9.7
9.4
M total
2M
4 2 ( 2M )
32M
=
=
3 =
3
V′
4p ⎛ 3 ⎞
p 32 R3 9p R
⎜⎝ R ⎟⎠
3 4
( )
so
r ⎛ 32M ⎞ ⎛ 4p R3 ⎞ 128
=
=
= 4.74
r0 ⎜⎝ 9p R3 ⎟⎠ ⎜⎝ 3M ⎟⎠ 27
(a)
0.045
m )(×0.110
10 6 Nm )( 0.260 m )⎤⎦
Fatm = PA = Patm p r 2 = 8.04 × 10 4 Pa p ( 2.00⎡⎣(m
)2 = 01.01
or
(b)
Fg
67352_ch09.indd 288
r = 4.74r 0
( ) ((
) )
= mg = ( rV ) g = r ⎡⎣(p r ) h ⎤⎦ g
= ( 415 kg m ) ⎡⎣p ( 2.00 m ) (10.0 m )⎤⎦ ( 7.44 m s ) =
Solids and Fluids
295
2
3
(c)
or
2
2
3.88 × 10 5 N
Now, consider the thin disk-shaped region 2.00 m in radius at the bottom end of the column
of methane. The total downward force75on it is the weight of the 10.0-meter tall column of
methane plus the downward force exerted on the upper end of the column by the atmosphere.
Thus, the pressure (force per unit area) on the disk-shaped region located 10.0 meters below
the ocean surface is
2/9/11 1:32:15 PM
294
76
9.3
(
Chapter 9
(a)
(c)
) ⎡⎣p ( 2.00 m ) (10.0 m )⎤⎦ ( 7.44 m s ) =
2
9.10
9.11
9.6
Ftotal Fatm + Fg 1.01 × 10 6 N + 3.88 × 10 5 N
−=0
=200 MPa
= 1.11 × 10 5 Pa
5
= 1.00 ×p10
MPa
=2 1.00 × 10 5 10 6 Pa = 1.00 × 1011 Pa
A=
p r2
2.00
m
(
)
0.002 − 0
(
)
By definition, Young’s modulus is the ratio of the tensile stress to the tensile strain in an elastic
Two cross-sectional areas in the plank, with one directly above the rail and one at the outer end of the
material. Thus, Young’s modulus for this material is the slope of the linear portion of the graph in2
plank, separated by distance h = 2.00 m and each with area A = ( 2.00 cm ) (15.0 cm ) = 30.0 cm ,
the “elastic behavior” region. This
is
move a distance Δx = 5.00 × 10 −2 m parallel to each other. The force causing this shearing
effect in the plank is the weight of the man F = mg applied perpendicular to the length of the
plank at its outer end. Since the shear modulus is given by S = shear stress shear strain =
( F A) ( Δx h ) = Fh ( Δx ) A, the shear modulus for the wood in this plank must be
S=
296
3.88 × 10 5 N
The
normal
exerted on the
bottom
shoes
by the
floorend
equals
thecolumn
total
Now,total
consider
theforce
thin disk-shaped
region
2.00acrobat’s
m in radius
at the
bottom
of the
weight
of theThe
acrobats
in the tower.
That
of
methane.
total downward
force
onisit is the weight of the 10.0-meter tall column of
methane plus the downward force exerted on the upper end of the column by the atmosphere.
Thus, the pressure (force per unit area) on the disk-shaped region located 10.0 meters below
the ocean surface is
P=
9.8
9.9
9.5
2
Chapter 9
(80.0 kg) (9.80
(5.00 × 10
−2
)
)(
m s 2 ( 2.00 m )
)(
)
m ⎡⎣ 30.0 cm 2 1 m 2 10 4 cm 2 ⎤⎦
= 1.05 × 10 7 Pa
(a) In order to punch a hole in the steel plate, the
Young’s
modulusmust
is defi
ned as
stresswith
strain
= ( F A ) ( ΔL L0 ) = ( F ⋅ L0 ) ( A ⋅ ΔL ). Thus, the
superhero
punch
outYa=plug
crosselongation
of
the
wire
is
sectional area,
(
)
⎡( 200 kg ) 9.80 m s 2 ⎤ ⋅ ( 4.00 m )
F ⋅ L0
⎦
ΔL =
= ⎣
= 4.90 × 10 −3 m = 4.90 mm
−4
2
10
2
A ⋅Y
0.200 × 10 m 8.00 × 10 N m
.
(
9.12
9.7
9.13
(
)
4 ⎡⎣( 90 kg ) 9.80 m s 2 ⎤⎦ ( 50 m )
= 3.5 × 108 Pa
2
−2
p 1.0 × 10 m (1.6 m )
.
(
)
From Y = F L0 A ( ΔL )
From Y = stress strain = ( stress )( L0 ΔL ), the maximum compression the femur can withstand
before breaking is
ΔLmax =
9.9
9.19
)
The acceleration of the forearm has magnitude
F L0
with A = p d 2 4 and F = mg, we get
Using Y =
A ( ΔL )
Y=
9.14
9.15
9.8
)(
( stress )max ( L0 )
Y
(160 × 10
)
18 × 10 Pa
(
=
6
Pa ( 0.50 m )
= 4.4 × 10 −3 m = 4.4 mm
= 6.28 × 10 4 N
)
9
The tension and cross-sectional area are constant through the entire length of the rod, and the
total elongation is the sum of that of the aluminum section and that of the copper section.
ΔLrod = ΔLAl + ΔLCu =
F ( L0 )Al
AYAl
+
F ( L0 )Cu
AYCu
=
F ⎡ ( L0 )Al ( L0 )Cu ⎤
+
⎢
⎥
A ⎢⎣ YAl
YCu ⎥⎦
where A = p r 2 with r = 0.20 cm = 2.0 × 10 −3 m. Thus,
(5.8 × 10 N )
=
p ( 2.0 × 10 m )
3
ΔLrod
9.20
67352_ch09.indd 294
−3
2
1.3 m
2.6 m ⎤
⎡
−2
⎢⎣ 7.0 × 1010 Pa + 11× 1010 Pa ⎥⎦ = 1.9 × 10 m = 1.9 cm
Assuming the spring obeys Hooke’s law, the increase in force on the piston required to compress
the spring an additional amount
2/9/11 1:32:27 PM
(1.00 × 10
9.21
9.10
9.7
(
3
) ( 7.50 × 10 −3 m )
)(
) (
)
2
kg m 3 9.80 m s 2 ⎡⎢p 1.20 × 10 −2 m ⎤⎥
⎦
⎣
= 2.11 m
Solids and Fluids
77
295
We first find the absolute pressure at the interface between oil and water.
(a)
P1 = P0 + roil ghoil
(
)(
)
= 1.013 × 10 5 Pa + 700 kg m 3 9.80 m s2 ( 0.300 m ) = 1.03 × 10 5 Pa
This is the pressure at the top of the water. To find the absolute pressure at the bottom, we use
P2 = P1 + r water ghwater , or
⎤
⎛
⎞ 3 ⎡
1.00
× 10 3 kg2 m 3
3
5
P2 = 1.03 ×⎜ 10 5 Pa +⎟ h10
kg
m
9.80
m 3s ( 0.200
Pa cm
) = 1.05
⎢
⎥m
=
cm ) =× 10
0.490
( 20.0
w
3
⎝
⎠
⎢⎣ ( 3.00 ) 13.6 × 10 kg m ⎥⎦
(
9.11
9.23
)(
)
(
)
The density of the solution is r = 1.02 r water = 1.02 × 10 3 kg m 3. If the glucose solution is to flow
into the vein, the minimum required gauge pressure of the fluid at the level of the needle is equal
to the gauge pressure in the vein, giving
Pgauge = P − P0 = r ghmin = 1.33 × 10 3 Pa
and
300
9.24
9.25
9.12
Chapter 9
(a)
(a)
hmin =
1.33 × 10 3 Pa
1.33 × 10 3 Pa
=
= 0.133 m
rg
1.02 × 10 3 kg m 3 9.80 m s2
(
)(
)
From the definition of bulk modulus,
(
) , the change in volume of the
P = P0 + r gh = 101.3 × 10 3 Pa + 1.00 × 10 3 kg m 3 9.80 m s2 ( 27.5 m )
(
)(
)
= 3.71× 10 5 Pa
(b)
The inward force the water will exert on the window is
(
F = PA = P p r
9.26
9.13
9.29
(a)
2
2
⎛ 35.0 × 10 −2 m ⎞
4
= 3.71 × 10 Pa p ⎜
⎟⎠ = 3.57 × 10 N
2
⎝
) (
5
)
If
we assume
a vacuum
While
the system
floats, B = wtotal = wblock + wsteel, or
rwood gVsubmerged = rblock gVblock + msteel g
Solids and Fluids
301
When msteel = 0.310 kg, Vsubmerged = Vblock = 5.24 × 10 −4 m 3, giving
rblock =
rwoodVblock − msteel
m
= rwood − steel
Vblock
Vblock
= 1.00 × 10 3 kg m 3 −
(b)
0.310 kg
= 408 kg m 3
5.24 × 10 −4 m 3
If the total weight of the block+steel system is reduced, by having msteel < 0.310 kg, a
smaller buoyant force is needed to allow the system to float in equilibrium. Thus, the block
will displace a smaller volume of water and will be only partially submerged.
The block is fully submerged when msteel = 0.310 kg. The mass of the steel object can
increase slightly above this value without causing it and the block to sink to the bottom.
As the mass of the steel object is gradually increased above 0.310 kg, the steel object
begins to submerge, displacing additional water, and providing a slight increase in the
buoyant force. With a density of about eight times that of water, the steel object will
be able to displace approximately 0.310 kg 8 = 0.039 kg of additional water before
it becomes fully submerged. At this point, the steel object will have a mass of about
0.349 kg and will be unable to displace any additional water. Any further increase in
the mass of the object causes it and the block to sink to the bottom. In conclusion,
the block + steel system will sink if msteel ≥ 0.350 kg.
9.30
67352_ch09.indd 295
(a)
2/9/11 1:32:29 PM
msteel < 0.310 kg, a
smaller buoyant force is needed to allow the system to float in equilibrium. Thus, the block
will displace a smaller volume of water and will be only partially submerged.
78
296
9.11
302
9.14
9.31
9.30
The block is fully submerged when msteel = 0.310 kg. The mass of the steel object can
increase slightly above this value without causing it and the block to sink to the bottom.
As the mass of the steel object is gradually increased above 0.310 kg, the steel object
Chapter 9begins to submerge, displacing additional water, and providing a slight increase in the
buoyant force. With a density of about eight times that of water, the steel object will
be able to displace approximately 0.310 kg 8 = 0.039 kg of additional water before
Young’s
modulusfully
is defi
ned as
it becomes
submerged.
At this point, the steel object will have a mass of about
0.349 kg and will be unable to displace any additional water. Any further increase in
the mass of the object causes it and the block to sink to the bottom. In conclusion,
Chapter 9the block + steel system will sink if msteel ≥ 0.350 kg.
B additional water displaced
The boat sinks until the weight of the
the weight of the truck.
2
(a)
A⫽(2.00 m)2⫽4.00 mequals
Thus,
Survivor
wtruck = ⎡⎣ rwater ( ΔV ) ⎤⎦ g
t
d
m⎞
kg ⎞ ⎡
⎛
wr
w m ) 4.00 × 10 −2 m ⎤ ⎛⎜ 9.80
= ⎜ 10 3
⎟ ( 4.00 m )( 6.00
⎟,
⎦⎝
⎝
s2 ⎠
m3 ⎠ ⎣
(
(b)
or
9.33
9.15
)
Inwgeneral,
“sinkers”
be expected to be thinner with heavier bones, whereas
Since
the system
is3 inwould
equilibrium,
truck = 9.41 × 10 N = 9.41 kN
“floaters” would have lighter bones and more fat.
When held underwater, the ball will have three forces acting on it: a downward gravitational
force, mg = r ballVg = r ball 4p r 3 3 g; an upward buoyant force, B = rwater Vg = rwater ( 4p r 3 3) g;
and an applied force, F. If the ball is to be in equilibrium, we have (taking upward as
positive)
(
)
ΣFy = F + B − mg = 0
⎡
⎛ 4p r 3 ⎞
⎛ 4p r 3 ⎞
⎛ 4p r 3 ⎞ ⎤
F = mg − B = ⎢ rball ⎜
−
r
g
=
r
g
−
r
⎥
water
ball
water
⎜⎝ 3 ⎟⎠
⎜⎝ 3 ⎟⎠ g
⎝ 3 ⎟⎠ ⎦
⎣
(
or
)
giving
3
4p ⎛ 0.038 0 m ⎞
2
F = ⎡⎣( 0.084 0 − 1.00 ) × 10 3 kg m 3 ⎤⎦
⎜
⎟⎠ 9.80 m s
3 ⎝
2
= − 0.258 N
(
)
so the required applied force is F = 0.258 N directed downward .
height the balloon will come to equilibrium and go no higher.
9.35
9.16
(a)
The buoyant force is the difference between the weight in air and the apparent weight when
immersed in the alcohol, or B = 300 N − 200 N = 100 N. But, from Archimedes’ principle,
this is also the weight of the displaced alcohol, so B = ( ralcohol V ) g. Since the sample is fully
submerged, the volume of the displaced alcohol is the same as the volume of the sample.
This volume is
V=
(b)
304
At some
B
ralcohol g
=
(
100 N
= 1.46 × 10 −2 m 3
700 kg m 3 9.80 m s2
)(
)
The mass of the sample is
Chapter 9
m=
weight in air
300 N
=
= 30.6 kg
9.80 m s 2
g
and its density is
r=
9.36
The difference between the weight in air and the apparent weight when immersed is the buoyant
force exerted on the object by the fluid.
(a)
67352_ch09.indd 296
m
30.6 kg
=
= 2.10 × 10 3 kg m 3
V 1.46 × 10 −2 m 3
The mass of the object is
2/9/11 1:32:32 PM
Solids and Fluids
9.16
9.47
9.17
(a) Bernoulli’s
When at rest,
the tension
in the cable
the weight
the 800-kg
object, 7.84 × 10 3 N.
From
equation,
choosing
y = 0 equals
at the level
of the of
syringe
and needle,
2
2F L
1
1
P2 + 2Thus,
r v2 =from
P1 + Y2 r=v1 , so0 the
flow
speed
in the needle
is cable is
, the
initial
elongation
of the
A ( ΔL )
2 ( P − P2 )
v2 = v12 + F ⋅ L1
7.84 × 10 3 N ( 25.0 m )
0
r
ΔL =
=
= 2.5 × 10 −3 m = 2.5 mm
A ⋅Y
4.00 × 10 −4 m 2 20 × 1010 Pa
In this situation,
(b) When the load is accelerating upward, Newton’s second law gives the tension in the
F
2 .00 N
cable as
P1 − P2 = P1 − Patm = ( P1 )gauge =
=
= 8.00 × 10 4 Pa
A1 2 .50 × 10 −5 m 2
F − mg = ma y, or F = m g + a y
[1]
Thus, assuming v1 ≈ 0,
If m = 800 kg and a y = +3.0 m s 2 , the elongation of the cable will be
2 8.00 × 10 4 Pa
v2 = 0 +
= 12.6 m s
Chapter 9
1.00 × 10 3 kg m 3
(
(
)
)(
(
(
308
9.48
9.49
9.18
79
297
)
)
)
We apply Bernoulli’s equation, ignoring the very small change in vertical position, to obtain
(a) The1 volume flow rate is Av , and the mass flow rate is
P1 − P2 = 2
(
)(
)
r Av = 1.0 g cm 3 2.0 cm 2 ( 40 cm s ) = 80 g s
(b)
From the equation of continuity, the speed in the capillaries is
⎛ Aaorta ⎞
⎛ 2 .0 cm 2 ⎞
vcapiliaries = ⎜
40 cm s )
⎟ vaorta = ⎜
3
2 (
⎝ 3.0 × 10 cm ⎟⎠
⎝ Acapillaries ⎠
310
9.50
9.53
9.19
Chapter 9or
vcapiliaries = 2.7 × 10 −2 cm s = 0.27 mm s
(a) consider
The cross-sectional
areathe
ofviewpoint
the hose isof projectile motion to find the speed at which the
First,
the path from
water emerges from the tank. From Δy = v0 y t + 12 a y t 2 with v0 y = 0, Δy = −1.00 m, and a y = −g,
we find the time of flight as
t=
2 ( Δy )
=
ay
2.00 m
g
From the horizontal motion, the speed of the water coming out of the hole is
v2 = v0 x =
Δx
g
= ( 0.600 m )
=
t
2.00 m
( 0.600 m )2 g
2.00 m
=
(1.80 × 10
−1
)
m g
We now use Bernoulli’s equation, with point 1 at the top of the tank and point 2 at the level of the
hole. With P1 = P2 = Patm and v1 ≈ 0, this gives r g y1 = 12 r v22 + r g y2 , or
(
)
1.80 × 10 −1 m g
v22
= 9.00 × 10 −2 m = 9.00 cm
=
h = y1 − y2 =
2g
2g
9.54
9.20
9.55
Solids and Fluids
311
(a) For minimum input pressure so the water will just reach the level of the rim, the gauge
pressure atthe
theflupper
end inside
is zerothe
(i.e.,
the absolute
pressure inside the upper end of the pipe
First, determine
ow speed
larger
section from
is atmospheric pressure), and the flow rate is zero. Thus, Bernoulli’s equation,
flow
rate
1.80 × 10 −4 m 3 s
1
v1P=+ 2
=
= 0.367 m s
2
A1
p 2.50 × 10 −2 m 4
(
)
The absolute pressure inside the large section on the left is P1 = Patm + r gh1, where h1 is
the height of the water in the leftmost standpipe. The absolute pressure in the constriction is
P2 = Patm + r gh2 , so
67352_ch09.indd 297
2/9/11 1:32:34 PM
80
298
9.18
=
Chapter 9
1.80 × 10 −4 m 3 s
(
p 2.50 × 10 −2 m
)
2
4
= 0.367 m s
+ r gh
The
pressure
the the
large
section
on its
thecross-sectional
left is P1 = Patmarea
(a) absolute
The force
neededinside
to shear
bolt
through
is 1F, where
= ( A ) (hstress
1 is ), or
the height of the water in the leftmost standpipe. The absolute pressure in the constriction is
P2 = Patm + r gh2 , so
P1 − P2 = r g( h1 − h2 ) = r g ( 5.00 cm )
The flow speed inside the constriction is found from Bernoulli’s equation with y1 = y2 (since the
pipe is horizontal). This gives
v22 = v12 +
v2 =
or
2
( P1 − P2 ) = v12 + 2 g ( h1 − h2 )
r
( 0.367
(
)
m s ) + 2 ( 9.80 m s ) 5.00 × 10 −2 m = 1.06 m s
2
The cross-sectional area of the constriction is then
A2 =
flow rate 1.80 × 10 −4 m 3 s
=
= 1.70 × 10 −4 m 2
v2
1.06 m s
and the diameter is
d2 =
9.56
9.57
9.21
312
(a)
(
)
(
v0 y = 0 − 2 a y ( Δy )max = −2 −9.80 m s2
) ( 40.0 m ) =
28.0 m s
Because of the low density of air and the small change in altitude, atmospheric pressure at
the fountain top will be considered equal to that at the geyser vent. Bernoulli’s equation,
2
with vtop = 0, then gives 12 r vvent
= 0 + r g(ytop − yvent ), or
continued on next page
vvent = 2 g ytop − yvent = 2 9.80 m s 2 ( 40.0 m ) = 28.0 m s
(
(c)
)
(
4 1.704 ×2.35
10 −4 ×m102 −6 m 2
= 1.47=×1.73
10 −2 ×m10=−3 1.47
=
m = cm
1.73 mm
p
p
Apply
equation
with pointprojectile
1 at the open
of the
tank
point 2 at
For
theBernoulli’s
upward flight
of a water-drop
fromtop
geyser
vent
toand
fountain-top,
2
2
the
opening
of
the
hole.
Then,
v y = v0 y + 2 a y ( Δy ), with v y = 0 when Δy = Δymax , gives
Chapter 9
(b)
4 A2
=
p
)
(
)
Between the chamber and the geyser vent, Bernoulli’s equation with vchamber ≈ 0 yields
2
+ r g yvent , or
( P + 0 + r g y )chamber = Patm + 12 r vvent
⎡1 2
⎤
P − Patm = r ⎢ vvent
+ g ( yvent − ychamber ) ⎥
⎣2
⎦
2
⎤
kg ⎞ ⎡ ( 28.0 m s ) ⎛
m⎞
⎛
⎢
= ⎜ 10 3
+ ⎜ 9.80 2 ⎟ (175 m ) ⎥ = 2.11 × 10 6 Pa
3⎟
⎝
⎝
m ⎠⎢
2
s ⎠
⎥⎦
⎣
or
9.58
9.59
9.22
67352_ch09.indd 298
Pgauge = P − Patm = 2.11 MPa = 20.8 atmospheres
Solids and Fluids
313
(a) Since the tube is horizontal,
From ΣFy = T − mg − Fy = 0, the balance reading is found to be T = mg + Fy , where Fy is the
vertical component of the surface tension force. Since this is a two-sided surface, the surface tension force is F = g ( 2 L ) and its vertical component is Fy = g ( 2 L )cosf , where f is the contact
angle. Thus, T = mg + 2g Lcosf .
2/9/11 1:32:37 PM
Solids and Fluids
T = mg + 2g Lcosf .
9.22
9.60
9.23
9.61
(a)
81
299
tube fas=shown
(b)+ of
TConsider
= 0.40 Nthewhen
0° ⇒ in partmg
2gthe
L= 0.40 N
sketch at the right. The volume of water in the
armNis when f = 180° ⇒
Tright
= 0.39
mg − 2g L= 0.39 N
[1]
[2]
mw [2] from
100 [1]
g gives
SubtractingVEquation
=
= 100 cm 3
w =
ρw 1.00 g cm 3
0.40 N − 0.39 N 0.40 N − 0.39 N
g =
=
= 8.3 × 10 −2 N m
But, we also
= A×210
hw ,−2giving
= 1.5 × 10 5 Pa
4 Lknow that 4Vw3.0
m4
−3
p 0.15 × 10 m
Vlaw,
100pressure
cm 3
w
From Poiseuille’s
the
hw =number
= is
= difference
20.0 cm required to produce a given volume flow rate of
A2 5.00
cm 2
flThe
uid Reynolds
with viscosity
(
))
((
(b)
(
)
)
(
)
1 050
kg observe
m 3 ( 0.55
mthe
s ) 2.0
× 10 −2that
m has been forced
From r
the
above,
that
mercury
out of the right arm
vdsketch
RN
=
=
= 4.3 × 10 3
−3
2
into thehleft arm is
2.7 × 10 N ⋅s m
In this region (RN > 3 000), the flow is turbulent .
9.62
9.71
9.24
From Poiseuille’s law, the flow rate in the artery is
The total vertical component of the surface tension force must equal the weight of the column of
fluid, or F cosf = g (2p r) ⋅ cosf = r(p r 2 )h ⋅ g. Thus,
g =
9.72
9.73
9.25
316
9.74
9.75
9.26
((
)(
)
)(
)
From thelaw
defigives
nitionthe
of the
coefficient
Stokes’s
viscosity
of theofairviscosity,
as
Chapter h
9=
3.0 × 10 −13 N
Fr
=
= 1.4 × 10 −5 N ⋅s m 2
6p r v 6p 2.5 × 10 −6 m 4.5 × 10 −4 m s
(
)(
)
Fick’s law gives the diffusion coefficient as
The observed diffusion rate is (8.0 × 10 −14 kg) (15 s) = 5.3 × 10 −15 kg s. Then, from Fick’s law,
the difference in concentration levels is found to be
C2 − C1 =
=
9.76
9.79
9.27
)(
× 10 −2−3m 1 0802 kg m 3 9.80 m s 2 5.0 × 10 −4 m
hr gr 12.1
= 500 × 10 N ⋅ s m ⎡⎣( 0.010 m ) ( 0.040 m ) ⎤⎦ ( 0.30 m s )= 5.6 × 10 −2 N m
= 0.12 N
2 cosf =
2 cos 0°
1.5 × 10 −3 m
( Diffusion rate) L
DA
(5.3 × 10
(5.0 × 10
(
(
−10
)
)
kg s ( 0.10 m )
−15
= 1.8 × 10 −3 kg m 3
m s 6.0 × 10 9.80
m m s2 (1.20 m ) = 1.23 × 10 4 Pa
2
)(
)(
−4
2
)
)
Using
vt = 2 r g rarea
− rf of 9h
The
cross-sectional
the aorta is A1 = p d12 4 and that of a single capillary is Ac = p d 22 4.
If the circulatory system has N such capillaries, the total cross-sectional area carrying blood from
the aorta is A2 = NAc = Np d 22 4.
2
From the equation of continuity, A2 = (v1 v2 )A1, or
Np d 22 ⎛ v1 ⎞ p d12
=⎜ ⎟
4
⎝ v2 ⎠ 4
yielding
2
2
⎛ v ⎞⎛ d ⎞
⎛
1.0 m s ⎞ ⎛ 0.50 × 10 −2 m ⎞
= 2.5 × 10 7
N =⎜ 1⎟⎜ 1⎟ =⎜
−2
−6
⎝ v2 ⎠ ⎝ d 2 ⎠
⎝ 1.0 × 10 m s ⎟⎠ ⎜⎝ 10 × 10 m ⎟⎠
9.80
67352_ch09.indd 299
The object has volume
2/9/11 1:32:40 PM
300
82
9.28
9.85
9.25
Chapter 9
.
Consider the diagram and apply
(a)
Bernoulli’s equation to points A and B,
taking y = 0 at the level of point B, and
recognizing that v A ≈ 0. This gives
PA + 0 + rw g( h − L sinq ) = PB +
A
h
1
rw v B2 + 0
2
L
Recognize that PA = PB = Patm since both
points are open to the atmosphere. Thus,
we obtain v B = 2 g( h − L sinq ) .
B
␪
Valve
Now the problem reduces to one of projectile motion with v0 y = v B sinq . At the top of the arc
v y = 0, y = ymax , and v 2y = v02 y + 2 a y (Δy) gives
ymax − 0 =
or
9.29
9.87
v 2y − v02 y
2a y
=
2g ( h − L sinq ) sin 2 q
0 − v B2 sin 2 q
=
2g
2(−g)
(
)
3
⎡
− 0.179 kg m 3 ⎡ 4p ( 0.40 m ) 3⎤ − 0.25 kg ⎤
⎣ 2
⎦
⎢ ⎡
⎥ ( 2.0 m ) = 1.9 m
ymax⎢= ⎣10.0 m − ( 2.00 m ) sin 30.0° ⎤⎦ sin 30.0° = 2.25 m above the
⎥ level of point B
0.050 kg
⎢⎣
⎥⎦
Four forces are acting on the balloon: an upward buoyant force exerted by the surrounding air,
B = (rair Vballoon )g; the downward weight of the balloon envelope, Fg, balloon = mg; the downward
weight of the helium filling the balloon, Fg, He = (r HeVballoon )g; and the downward spring force,
Fs = k Δx . At equilibrium, Δx = L, and we have
ΣFy = 0
or
⇒
B − Fs − Fg, balloon − Fg, He = 0
Fs = kL = B − Fg, balloon − Fg, He = ( rair Vballoon ) g − mg − ( r HeVballoon ) g
⎡( rair − r He ) Vballoon − m ⎤⎦ g
L= ⎣
k
PB − PA = ( rwater − roil ) gL
⎡⎣(1.29 − 0.179 ) kg m 3 ⎤⎦ 5.00 m 3 − 2.00 × 10 −3 kg 9.80 m s2
This yields
L=
Combining Equations [1] and [2] yields 90.0 N m
and
{
or
9.89
9.30
L = 0.605 m
(
=
and
(
)(
2 (1 000 − 750 ) 9.80 m s2 5.00 × 10 −2 m
The pressure on the surface of the two
hemispheres is constant at all points, and the
force on each element of surface area is directed
along the radius of the hemispheres. The applied
force along the horizontal axis must balance the
net force on the “effective” area, which is the
cross-sectional area of the sphere,
Aeffective = p R
}(
)
2
F = Pgauge Aeffective = ( P0 − P ) p R 2
1.29
F
)=
R
)
[2]
13.8 m s
F
P
P0
Solids and Fluids
9.31
9.91
A water
When
droplet
msteel
emerging
= 0.310 from
kg, one of the holes
becomes a projectile with v0 y = 0 and v0 x = v.
The time for this droplet to fall distance h to the
floor is found from Δy = v0 y t + 12 a y t 2 to be
t = 2 h g.
The horizontal range is R = vt = v 2 h g.
321
301
83
3
2
h3
If the two streams hit the floor at the same spot,
it is necessary that R1 = R2 , or
h2
1
h1
v1
2 h1
2 h2
= v2
g
g
With h1 = 5.00 cm and h2 = 12.0 cm, this reduces
to v1 = v2 h2 h1 = v2 12.0 cm 5.00 cm , or
R1 ⫽ R2
v1 = v2 2 .40
[1]
Apply Bernoulli’s equation to points 1 (the lower hole) and 3 (the surface of the water). The
pressure is atmospheric pressure at both points and, if the tank is large in comparison to the size
of the holes, v3 ≈ 0. Thus, Patm + 12 r v12 + r gh1 = Patm + 0 + r gh3, or
v12 = 2 g( h3 − h1 )
[2]
Similarly, applying Bernoulli’s equation to point 2 (the upper hole) and point 3 gives
Patm + 12 r v22 + r gh2 = r atm + 0 + r gh3, or
v22 = 2 g( h3 − h2 )
[3]
Square Equation [1] and substitute from Equations [2] and [3] to obtain
2 g( h3 − h1 ) = 2 .40 ⎡⎣ 2 g( h3 − h2 ) ⎤⎦
Solving for h3 yields
h3 =
2 .40 h2 − h1 2 .40 (12 .0 cm ) − 5.00 cm
=
= 17.0 cm
1.40
1.40
so the surface of the water in the tank is 17.0 cm above floor level .
67352_ch09.indd 301
2/9/11 1:32:45 PM
1
ࢠ ࠊ‫י‬չଝ
PROBLEM SOLUTIONS
10.1
10.3
10.2
10.4
10.3
10.7
Introduction
(a)
TF =
9
9
TC + 32 = ( −273.15) + 32 = − 460°F
5
5
(b)
TC =
5
(TF − 32) = 95 (98.6 − 32) = 37°C
9
(c)
TF =
9
9
9
TC + 32 = ( TK − 273.15) + 32 = ( −173.15) + 32 = − 280°F
5
5
5
(a)
TC = TK − 273.15 = 20.3 − 273.15 = −253°C
(b)
TF =
9
9
TC + 32 = ( −253) + 32 = − 423°F
5
5
(100°C ) = 1.74 atm
(
)
When the
low-density
gas Celsius
is held constant,
pressure
temperature
Apply
TF volume
= 95 TC +of
32a to
two different
temperatures,
and
(TC )and
(TC )2, are related by
1
a linear equation
to obtain
[1]
(TF )1 = 95 (TC )1 + 32
(TF )2 = 95 (TC )2 + 32
and
[2]
Subtracting Equation [1] from [2] yields
(TF )2 − (TF )1 = 95 ⎡⎣(TC )2 − (TC )1 ⎤⎦
or
10.8
10.9
10.4
ΔTF = ( 9 5) ΔTC
(a) with
UsingT the
result of Problem 10.7 above,
Start
F = − 40°F and convert to Celsius.
TC =
330
10.10
10.5
10.13
5
5
TF − 32 ) = ( − 40 − 32 ) = − 40°C
(
9
9
Since Celsius and Fahrenheit degrees of temperature change are different sizes
(1 Celsius degree = 1.8 Fahrenheit degrees), this is the only temperature with the same numeric
Chapter 10
value on both scales.
(a)
Since the
temperature
on the Rankine
Fahrenheit
are identical, the temWe choose
radius as differences
our linear dimension.
Then,and
from
ΔL = a Lscales
0 ( ΔT ) ,
perature readings on the two thermometers must differ by no more than an additive constant
(i.e.,
L − L0
2.21 cm − 2.20 cm
ΔT = TPC2 − 20.0°C =
=
= 35.0°C
T2 =
a L0
⎡1.30 × 10 − 4 ( °C )−1 ⎤( 2.20 cm )
⎣
⎦
P1
or
TC = 55.0°C
84
67352_ch10.indd 322
2/9/11 1:34:04 PM
=
10.6
10.15
10.2
m V0
r0
m
m
=
=
=
V0 + ΔV V0 + b V0 ( ΔT ) 1 + b ( ΔT ) 1 + b ( ΔT )
Thermal Physics
327
85
5 in Celsius 5temperature in the underground tank and the tanker truck is
The difference
(a) TC = ( TF − 32 ) = (134 − 32 ) = 56.7°C , and
95
5 9
ΔTC = ( ΔTF ) = ( 95.0 − 52.0 ) = 23.9°C
5
9 ) = − 62.1°C
TC = 9( −79.8 − 32
9
If V is the volume of gasoline that fills the tank at 52.0°F, the volume this quantity of gas
52°F
would occupy on the tanker truck at 95.0°F is
(b)
V95°F = V52°F + ΔV = V52°F + b V52°F ( ΔT ) = V52°F [1 + b ( ΔT )]
1.000 × 10 −2
= −13.0 23.9°C
× 10 2 °C
=
⎡
= 1.00 ×b 10 gal⎡ 1 + 9.6
× 10 −4−1 °C
)⎤⎦ = 1.02 × 103 gal
⎣ × 10 −6 (°C ) (⎤⎦ ) (
3 ⎣11
(
10.17
10.7
(a)
10.18
10.19
10.8
(
)
Using the result of Problem 10.14, with b = 3a , gives
r=
(b)
)
3
r0
11.3 × 10 3 kg m 3
=
= 11.2 × 10 3 kg m 3
1+ b ( ΔT ) 1+ 3 ⎡ 29 × 10 −6 (°C )−1 ⎤ ( 90°C − 0°C )
⎣
⎦
No . Although the density of gold would be less on a warm day, the mass of the bar would
be the same, regardless of its temperature, and that is what you are paying for. (Note that
in the calculations of parts (a) and
the volume of the bar increases with increasing temperature, whereas its density decreases.
(b). Thus, the results obtained would not be changed if the initial length of the wire were
Its mass, however, remains constant.)
doubled.
(a) When a wire undergoes a decrease in temperature of magnitude ΔT
From ΔL = L − L0 = a L0 ( ΔT ) , the final value of the linear dimension is L = L0 + a L0 ( ΔT ).
To remove the ring from the rod, the diameter of the ring must be at least as large as the diameter
of the rod. Thus, we require that
LBrass = LAl , or ( L0 )Brass + a Brass ( L0 )Brass ( ΔT ) = ( L0 )Al + a Al ( L0 )Al ( ΔT )
ΔT =
This gives
(a)
( L0 )Al − ( L0 )Brass
a Brass ( L0 )Brass − a Al ( L0 )Al
If ( L0 )Al = 10.01 cm,
ΔT =
⎡19 × 10
⎣
−6
( °C )
−1
10.01− 10.00
= −199°C
⎤(10.00 ) − ⎡ 24 × 10 −6 ( °C )−1 ⎤(10.01)
⎦
⎣
⎦
so T = T0 + ΔT = 20.0°C − 199°C = − 179°C which is attainable
(b)
If ( L0 )Al = 10.02 cm
ΔT =
⎡19 × 10
⎣
−6
( °C )
−1
10.02 − 10.00
= −396°C
⎤(10.00 ) − ⎡ 24 × 10 −6 ( °C )−1 ⎤(10.02 )
⎦
⎣
⎦
and
T = T0 + ΔT = − 376°C which is below absolute zero and unattainable
.
10.9
10.21
The increase in temperature is ΔT = 35°C − ( −20°C ) = 55°C.
−1
Thus, ΔL = a L0 ( ΔT ) = ⎡⎣11 × 10 −6 ( °C ) ⎤⎦ ( 518 m ) ( 55°C ) = 0.31 m = 31 cm
67352_ch10.indd 327
2/9/11 1:34:12 PM
328
86
Chapter 10
.
10.10
10.23
(
)
−3
Therefore,
1.27amount
atm + the
4.70band
× 10(with
atm
°C Tlength
If
allowed to do Pso,= the
initial
L0 ) would contract as it cools to
C
37°C is ΔL = a L0 ΔT . Since the band is not allowed to contract, it will develop a tensile stress
(a) At
given
by absolute zero the gas exerts zero pressure
⎛ a L0 ΔT ⎞
⎛ ΔL ⎞
Stress = Y ⎜
=Y⎜
⎟ = Ya ΔT
⎟
L0
⎝ L0 ⎠
⎝
⎠
If A = (height ⋅ thickness) = (4.0 mm)(0.50 mm) = 2.0 × 10 −6 m 2 is the cross-sectional area of the
band, the tension in the band will be
N⎞
⎛
−6
−1
2
F = A ⋅ ( Stress ) = 2.0 × 10 −6 m 2 ⎜ 18 × 1010
⎟ 17.3 × 10 °C (80°C − 37°C ) = 2.7 × 10 N
⎝
m2 ⎠
(
10.11
10.25
10.24
(
)
⎛ 3.80
The drum and the carbon tetrachloride, both having
an initial
× 10 −3 volume
m 3 ⎞ of V0 = 50.0 gal, expand at
(a)
10.0 gal ) ⎜
= 27.7 kg
(
different rates as the temperature rises by ΔT = 20.0°C.
1 From
gal ΔV⎟⎠ = b V0 ( ΔT ) , with b = 3a as
⎝
the coefficient of volume expansion for the steel drum, we obtain
(b)
⎛
⎞
Vspillage = ΔVcarbon
− ΔVsteel = ⎜ b carbon
− 3a steel ⎟ V0 ( ΔT )
⎝ tetrachloride
⎠
tetrachloride
drum
(
or
10.12
10.27
)
)
(
Vspillage = ⎡5.81 × 10 −4 (°C ) − 3 11 × 10 −6 (°C )
⎣
periodic intervals.
−1
−1
20.0°Cin) =the0.548
gal at
)⎤⎦ (50.0 gal ) (placed
pipeline
(a)
The gap width is a linear dimension, so it increases in “thermal enlargement” as the
temperature goes up. The gap expands in the same way the material removed to create the
gap would have expanded.
(b)
At 190°C, the length of the piece of steel that is missing, or has been removed to create the
gap, is L = L0 + ΔL = L0 [1 + a ( ΔT )] . This gives
(
)
L = (1.600 cm ) ⎡⎣1 + 11 × 10 −6 °C−1 (190°C − 30.0°C )⎤⎦ = 1.603 cm
10.29
10.13
334
(a)
n=
(
)
1.013 × 10 5 Pa atm 1.0 × 10 −6 m 3
PV
=
= 4.2 × 10 −5 mol
RT
(8.31 J mol ⋅ K )( 293 K )
Thus,
Chapter 10
(b)
)(
.
molecules ⎞
⎛
19
N = n ⋅ N A = 4.2 × 10 −5 mol ⎜ 6.02 × 10 23
⎟ = 2.5 × 10 molecules .
⎝
mol ⎠
(
)
(
Since both V and T are constant, n2 n1 = P2 V2 RT2
) (P V
1
1
)
RT1 = P2 P1, or
continued on next page
⎛P ⎞
⎛ 1.0 × 10 −11 Pa ⎞
n2 = ⎜ 2 ⎟ n1 = ⎜
4.2 × 10 −5 mol = 4.1× 10 −21 mol
⎝ 1.013 × 10 5 Pa ⎟⎠
⎝ P1 ⎠
= 1.81 × 10 24 molecules
(
10.30
10.14
10.31
(a)
(
)
)
The volume
of gas
the gas
From
the ideal
law,isPV = nRT , we find P T = nR V . Thus, if both n and V are constant
as the gas is heated, the ratio P T is constant, giving Pf T f = Pi Ti , or
⎛ Pf ⎞
⎛ 3P ⎞
T f = Ti ⎜ ⎟ = ( 300 K ) ⎜ i ⎟ = 900 K = 627°C
⎝ Pi ⎠
⎝ Pi ⎠
(b)
If both pressure and volume double as n is held constant, the ideal gas law gives
⎛ ( 2Pi ) ( 2Vi ) ⎞
⎛ Pf V f ⎞
T f = Ti ⎜
= Ti ⎜
⎟ = 4Ti = 4 ( 300 K ) = 1 200 K = 927°C
⎟
PiVi
⎝ PiVi ⎠
⎝
⎠
10.32
67352_ch10.indd 328
The mass of the gas in the balloon does not change as the temperature increases. Thus,
2/9/11 1:34:16 PM
⎛
⎜
⎝
10.15
10.33
⎞
3 ⎛ 273 K ⎞
3
⎟ = 0.131 kg m
⎟ = 0.179 kg m ⎜⎝
373 K ⎠
⎠
(
)
87
329
(b)
We start
Kelvintemperatures
temperatureare
and convert to the Rankine temperature in several
The initial
andwith
finalthe
absolute
stages, using the Fahrenheit to Rankine conversion from part (a) above.
Ti = TC ,i + 273 = ( 25.0 + 5273) K = 298 K and 5T f = TC, f + 273 = ( 75.0 + 273) K = 348 K
TK = TC + 273.15 = ( TF − 32.00 ) + 273.15 = ⎡⎣( TR − 459.67 ) − 32.00 ⎤⎦ + 273.15
9
The volume of the tank is assumed
to be unchanged, or9V f = Vi . Also, two-thirds of the gas is
withdrawn, so n5f = ni 3. Thus, from the ideal
5 gas5 law,
5
= ( TR − 491.67 ) + 273.15 = TR − ( 491.67 ) + 273.15 = TR − 273.15 + 273.15
9
9
9
9
Pf V f ⎛ n ⎞f ⎛RT
T f ⎞ K ⎞ ⎛ 1 ⎞ ⎛ 3485 K ⎞
⎛ n f ⎞⎞⎛⎛516
Tf⎞
⎛ 5.98 mol
= ⎟ ⎜ f ⎟ Pi⇒
atm
4.28 atm
) =6 Pa
= ⎜ Pf = ⎜ ⎟⎟⎜⎜ ⎟ Pi ⎟= 9.50
1.28
× 10
⎜ ⎟×
⎜ 10 Pa⎟ (=11.0
⎜
⎝ 7.83 mol
Pi Vi ⎝ n⎠i ⎝RT
Ti ⎠ K ⎠ ⎝ 3 ⎠ ⎝ 298 K ⎠
Ti ⎠
⎝ ni ⎠⎠⎝⎝ 292
or
(
10.16
10.35
10.34
Thermal Physics
)
With n held constant, the ideal gas law gives
(a)
V1 ⎛ P2 ⎞ ⎛ T1 ⎞ ⎛ 0.030 atm ⎞ ⎛ 300 K ⎞
−2
=
=⎜
⎟⎜
⎟ = 4.5 × 10
V2 ⎜⎝ P1 ⎟⎠ ⎜⎝ T2 ⎟⎠ ⎝ 1.0 atm ⎠ ⎝ 200 K ⎠
Since the volume of a sphere is V = ( 4p 3)r 3, V1 V2 = ( r1 r2 ) .
3
Thus,
⎛V ⎞
r1 = ⎜ 1 ⎟
⎝V ⎠
2
10.17
10.37
10.36
13
⎛
−2
r2 = 4.5
⎜⎝ × 10
(
)
13
⎞
⎠
m 5 Pa = 449 kPa
( 20 ⎟m=) =4.497.1× 10
The
(a) pressure
With 100 m below the surface is found, using P1 = Patm + r gh , to be
(
)(
)
P1 = 1.013 × 10 5 Pa + 10 3 kg m 3 9.80 m s 2 (100 m ) = 1.08 × 10 6 Pa
The ideal gas law, with both n and T constant, gives the volume at the surface as
⎛ P ⎞
⎛P⎞
⎛ 1.08 × 10 6 Pa ⎞
V2 = ⎜ 1 ⎟ V1 = ⎜ 1 ⎟ V = ⎜
1.50 cm 3 = 16.0 cm 3
⎝ 1.013 × 10 5 Pa ⎟⎠
⎝ P2 ⎠
⎝ Patm ⎠
(
10.18
10.39
The average kinetic energy of the molecules of any ideal gas at 300 K is
KE =
10.41
10.40
10.19
)
continued on next page
1
3
3⎛
J⎞
−21
mv 2 = k BT = ⎜ 1.38 × 10 −23
⎟ ( 300 K ) = 6.21 × 10 J
.
2
2
2⎝
K⎠
One
any
substance
containsinAvogadro’s
number
andtemperature
has a mass equal
(a) mole
The of
rms
speed
of molecules
a gas of molar
massofMmolecules
and absolute
T is to the
molar mass, M. Thus, the mass of a single molecule is m = M N A .
For helium, M = 4.00 g mol = 4.00 × 10 −3 kg mol, and the mass of a helium molecule is
m=
4.00 × 10 −3 kg mol
= 6.64 × 10 −27 kg molecule
6.02 × 10 23 molecule mol
Thermal Physics
337
a helium
continued Since
on next
page molecule contains a single helium atom, the mass of a helium atom is
matom = 6.64 × 10 −27 kg
10.43
10.20
Consider a time interval of 1.0 min = 60 s, during which 150 bullets bounce off Superman’s
chest. From the impulse-momentum theorem, the magnitude of the average force exerted on
Superman is
Fav =
150 Δp bullet 150 m ( v − v0 )
I
=
=
Δt
Δt
Δt
(
)
150 8.0 × 10 −3 kg ⎡⎣( 400 m s ) − ( −400 m s ) ⎤⎦
= 16 N
=
60 s
10.44
67352_ch10.indd 329
(a)
The volume occupied by this gas is
2/9/11 1:34:19 PM
330
88
10.13
10.21
10.45
Chapter 10
We
the
radius
our linear
from k B = 1.38 × 10 −23 J K is Boltzmann’s
If vchoose
, we
mustashave
vrms = dimension.
3k BT m =Then,
vesc , where
rms = vesc
constant and m is the mass of a molecule (for helium, m = 6.64 × 10 −27 kg). Thus, the required
2
absolute temperature is T = mvesc
3k B.
(a)
To have vrms = vesc on Earth where vesc = 1.12 × 10 4 m s , the required temperature for the
helium gas is
(6.64 × 10 kg) (1.12 × 10
T=
3 (1.38 × 10
J K)
−27
4
ms
−23
(b)
)
−27
3
ms
−23
(a)
= 2.01 × 10 4 K
If vrms = vesc on the Moon where vesc = 2.37 × 10 3 m s, the temperature must be
(6.64 × 10 kg) ( 2.37 × 10
T=
3 (1.38 × 10
J K)
10.49
10.22
2
)
2
= 901 K
Thermal Physics
339
As the acetone undergoes a change in temperature ΔT = ( 20.0 − 35.0 ) °C = −15.0 °C, the
final volume will be
V f = V0 + ΔV = V0 + b V0 ( ΔT ) = V0 [1 + b ( ΔT )]
(
= (100 mL ) ⎡1 + 1.50 × 10 −4 (°C )
⎣
(b)
10.51
10.23
−1
)( −15.0 °C)⎤⎦ = 99.8 mL
When acetone at 35°C is poured into the Pyrex flask that was calibrated at 20°C, the
volume of the flask temporarily expands to be larger than its calibration markings indicate.
−1
However, the coefficient of volume expansion for Pyrex ⎡⎣ b = 3a = 9.6 × 10 −6 (°C ) ⎤⎦
−1
is much smaller than that of acetone ⎡⎣ b = 1.5 × 10 −4 (°C ) ⎤⎦ . Hence, the temporary
increase in the volume of the flask will be much smaller than the change in volume of
the acetone as the materials cool back to 20°C, and this change in volume of the flask
has negligible effect on the answer.
For a temperature change ΔTF = TF − TF ,0 on the Fahrenheit scale, the corresponding temperature
change on the Celsius scale is
ΔTC = TC − TC ,0 =
(
)
(
)
5
(TF − 32) − 95 TF ,0 − 32 = 95 TF − TF ,0 = 95 ( ΔTF )
9
Therefore, if L0 = 35.000 m and ΔTF = 90.000°F − 15.000°F = 75.000°F , the final length of the
beam is
5
L = L0 [1 + a ( ΔT )] = ( 35.000 m ) ⎡⎢1 + 11 × 10 −6 °C−1 ( 75.000 ) °C ⎤⎥ = 35.016 m
9
⎣
⎦
(
10.52
10.55
10.24
If
Pi number
is the initial
gaugeofpressure
of the isgas
cylinder,
the= initial
The
of moles
CO 2 present
n =inmtheM,
where m
6.50 gabsolute
and M =pressure
44.0 g is
mol. Thus,
5
P
Pi + Ptemperature
,
where
ati, abs
the=given
(
20.0°C
=
293
K
)
and
pressure
(1.00
atm
=
1.013
×
10
Pa),
the volume
0
will be
V=
10.56
10.25
10.59
)
nRT mRT ( 6.50 g ) (8.31 J mol ⋅ K ) ( 293 K )
=
=
= 3.55 × 10 −3 m 3 = 3.55 L
5
.
MP
P
( 44.0 g mol ) 1.013 × 10 Pa
(
)
Whenexpansion,
air trappedthe
in the
tube is
constant
volume
After
increase
in compressed,
the length ofatone
span istemperature, into a cylindrical
L
y
0.40 m long, the ideal gas law gives its pressure as
L0 ⫽ 125 m
ΔL = a L.0 ( ΔT )
−1
= ⎣⎡12 × 10 −6 ( °C ) ⎤⎦ (125 m ) ( 20.0°C ) = 3.0 × 10 −2 m
giving a final length of L = L0 + ΔL = 125 m + 3.0 × 10 −2 m
67352_ch10.indd 330
2/9/11 1:34:21 PM
ΔL = a L.0 ΔT
342
Chapter 10
Thermal Physics
−1
= ⎣⎡12 × 10 −6 ( °C ) ⎤⎦ (125 m ) ( 20.0°C ) = 3.0 × 10 −2 m
( )
89
331
is Young’s
the m
wire
material
= 125
+ 3.0
× 10 −2 and
m A is the cross-sectional area of the
givingwhere
a finalYlength
of L =modulus
L0 + ΔLfor
For the given
steel wire, with ΔT = −10.0 − 35.0 °C, the tension that develops in the
From wire.
the Pythagorean
theorem,
wire is
2
2
y = L2 − L20 = [(125 + 0.030 ) m ] − (125 m ) = 2.7 m
−1
10
2 was released.
−6
2
F = 20.0 × 10 N m 4.00 × 10 m ⎡⎣11.0 × 10 −6 ( °C ) ⎤⎦ ( 45.0°C ) = 396 N
(
10.63
10.26
10.60
(b)
(a)
(a)
(b)
)(
)
As
thethe
temperature
decreases
while the
is prevented
fromno
contracting,
the stress
thatthe
Yes,
angular speed
will increase
as wire
the disk
cools. Since
external torque
acts on
From the in
ideal
develops
the gas
wirelaw,
is
disk, the angular momentum of the disk will be conserved. As the disk cools, its radius, and
hence its moment of inertia will decrease. Then, in order to keep angular momentum
( L = Iw ) constant, the angular speed must increase.
Since angular momentum is conserved, Iw = I 0w 0 or w = ( I 0 I ) w 0 . Thus,
2
2
⎛
⎞
R0
⎛ 1 MR02 ⎞
⎛R ⎞
w0 = ⎜ 0 ⎟ w0 = ⎜
w = ⎜ 12
⎟ w0
2⎟
⎝ R⎠
⎝ 2 MR ⎠
⎝ R0 [1 + a ( ΔT )] ⎠
or
w=
10.64
67352_ch10.indd 331
w0
[1 + a ( ΔT )]
2
=
25.0 rad s
(
⎡1 + 17 × 10
⎣
−6
)
°C−1 ( 20.0 − 850 ) °C ⎤⎦
2
= 25.7 rad s
Let container 1 be maintained at
2/9/11 1:34:24 PM
ࢠ ࠊ˕ࢽ߾۰ࢂ߾οए
PROBLEM SOLUTIONS
11.1
As mass m of water drops from the top to the bottom of the falls, the gravitational potential
energy given up (and hence, the kinetic energy gained) is Q = mgh. If all of this goes into raising
the temperature, the rise in temperature will be
(
)
9.80 m s 2 (807 m )
Q
m gh
ΔT =
=
=
= 1.89°C
mcwater
m cwater
4 186 J kg ⋅ °C
and the final temperature is ⎛ TQf =⎞Ti + ΔT = 15.0°C4+1.89°C = 16.9°C3
= 0.25 2.8 × 10 stairs = 7.0 × 10 stairs
= 0.25 ⎜
⎝ mgh ⎟⎠
(
11.3
11.2
)
The mechanical energy transformed into internal energy of the bullet is
1
1⎛1
⎞ 1
Q = ( KEi ) = ⎜ mvi2 ⎟ = mvi2. Thus, the change in temperature of the bullet is
⎠ 4
2
2⎝2
1 m v2
Q
(300 3m s )
⎞ °C
= 4Pmax i =1.82 × 10 J s ⎛ 1=Cal176
mc = m clead = 4 (128 J kg ⋅ °C
⎜⎝ 4) 186 J ⎟⎠ = 1.74 Cal s
0.250
0.250
2
ΔT =
11.4
11.5
11.3
(a)
(a)
(
)
The
instantaneous
Q
= 0.600
ΔPEg =power
0.600 (ismgPh=) Fv
= 0.600 ⋅ m 9.80 m s2 ( 50.0 m )
(
)
Q = 294 m 2 s2 ⋅ m
or
From Q = mc(ΔT ) = mc(T f − Ti ), we find the final temperature as
(
)
294 m 2 s2 ⋅ m
Q
T f = Ti +
= 25.0°C +
= 25.8°C
mc
m ( 387 J kg ⋅ °C )
(b)
Observe that the mass of the coin cancels out in the calculation of part (a). Hence, the
Q
1.23 × 10 3ofJ the mass of the coin .
result
is independent
c=
=
= 234 J kg ⋅ °C = 0.234 kJ kg ⋅ °C
m ( ΔT ) ( 0.525 kg ) (10.0°C )
11.7
11.4
11.6
From ΔL = a L0 ( ΔT ), the required increase in temperature is found, using Table 10.1, as
ΔT =
⎛ 1 yd
ΔL
3.0 × 10 −3 m
=
⎜
−1
−6
a steel L0
11 × 10 (°C ) 13 yd ⎝ 3.0 ft
(
)(
)
⎞ ⎛ 3.281 ft ⎞
⎟ ⎜⎝ 1 m ⎟⎠ = 23°C
⎠
The mass of the rail is
m=
(
)
70 lb yd (13 yd ) ⎛ 4.448 N ⎞
w
2
=
⎜⎝ 1 lb ⎟⎠ = 4.1 × 10 kg
g
9.80 m s2
so the required thermal energy (assuming that csteel = ciron ) is
(
)
Q = mcsteel ( ΔT ) = 4.1 × 10 2 kg ( 448 J kg ⋅ °C ) ( 23°C ) = 4.2 × 10 6 J
11.8
The change in temperature of the rod is
90
67352_ch11.indd 344
2/9/11 1:37:52 PM
Energy
Energy in
in Thermal
Thermal Processes
Processes
11.5
11.9
70.
3
The mass
water
(a)
2.03 ×of10
J sinvolved is
72.
(a)
(b) 7.84 ft 2 ⋅ °F ⋅ h Btu
0.457 kg ⎛ 3 kg ⎞
11
3
14
m = rV = ⎜ 10
⎟ 4.00 × 10 m = 4.00 × 10 kg
⎝
m3 ⎠
(b) If the samples and inner surface of the insulation are preheated, nothing undergoes a temperature change during the test. Therefore, only the mass of the wax, which undergoes a
(a) Q = mc ( ΔT ) = 4.00 × 1014 kg ( 4 186 J kg ⋅ °C ) (1.00°C ) = 1.67 × 1018 J
change of phase, needs to be known.
(
)
(
(b)
)
The power input is P = 1 000 MW = 1.00 × 10 9 J s,
Q
1.67 × 1018 J ⎛
1 yr
⎞
=2m gh 29(1.50 ⎜kg ) 9.80 m7 s 2⎟ (=3.00
52.9
m )yr
⎝
b
P= 1.00 ×=10 J s 3.156 × 10 s ⎠
= 0.105°C
mw cw
( 0.200 kg) ( 4 186 J kg ⋅ °C)
The internal energy added to the system equals the gravitational potential energy given up by the
quantity
of energy
from
the water-cup combination in a time interval of 1 minute is
ΔPEg = 2m
2The
falling
blocks,
or Q =transferred
b gh. Thus,
(
t=
so
11.10
11.6
11.11
349
91
351
)
Q = ⎡⎣( mc )water + ( mc )cup ⎤⎦ ( ΔT )
⎡
⎛
⎛
J ⎞
J ⎞⎤
3
+ ( 0.200 kg ) ⎜ 900
= ⎢( 0.800 kg ) ⎜ 4 186
⎟⎠ ⎥ (1.5°C ) = 5.3 × 10 J
⎟
kg
⋅
°C
kg
⋅
°C
⎠
⎝
⎝
⎣
⎦
The rate of energy transfer is
P=
11.12
(a)
11.13
11.7
(a)
Q 5.3 × 10 3 J 6.7 × 10J 6 J ⎛ 1 h ⎞
=
= 88 = 88 W
Δt
60 s = 5 × 10 2s J s ⎜⎝ 3 600 s ⎟⎠ ∼ 4 h
From the relation between compressive stress and strain, F A = Y ( ΔL L0 ), where Y is
1 of the material.
1 From the discussion
2
Young’s modulus
on linear expansion, the strain due to
Wnet = ΔKE = m v 2f − v02 = ( 75 kg ) ⎡⎣(11.0 m s ) − 0 ⎤⎦ = 4.54 × 10 3 J → 4.5 × 10 3 J
thermal expansion
2 can be written
2 as ΔL L0
(
)
Wnet 4.54 × 10 3 J
= 9.1 × 10 2 J s = 910 W
=
5.0 s
Δt
(b)
P=
(c)
If the mechanical energy is 25% of the energy gained from converting food energy, then
Wnet = 0.25 ( ΔQ ) and P = 0.25 ( ΔQ ) Δt , so the food energy conversion rate is
ΔQ
P
⎛ 910 J s ⎞ ⎛ 1 Cal ⎞
= 0.87 Cal s
=
=⎜
⎟
Δt
0.25 ⎝ 0.25 ⎠ ⎜⎝ 4 186 J ⎟⎠
(d)
11.14
11.15
11.8
The excess thermal energy is transported by conduction and convection to the surface of the
skin and disposed of through the evaporation of perspiration. on which the block
slides.
The mechanical energy converted into internal energy of the block is
(a)
1
WhenQthermal
the water and aluminum will have a common temperature
= 0.85equilibrium
( KEi ) = 0.85is reached,
of T f = 65.0°C. Assuming that2 the water-aluminum system is thermally isolated from the environment, Qcold = −Qhot, so mw cw T f − Ti,w = −mAl cAl T f − Ti,Al , or
(
mw =
11.16
11.9
11.17
67352_ch11.indd 349
(
−mAl cAl T f − Ti,Al
(
cw T f − Ti,w
)
)
(
)
) = − (1.85 kg) (900 J kg ⋅ °C) (65.0°C − 150°C) =
( 4 186
0.845 kg
J kg ⋅ °C ) ( 65.0°C −Energy
25.0°Cin) Thermal Processes
353
The kinetic energy given up by the car is absorbed as internal energy by the four brake drums
Since
temperature
of iron).
the water
and the steel container is unchanged, and neither substance
(a
totalthe
mass
of 32 kg of
Thus,
undergoes a phase change, the internal energy of these materials is constant. Thus, all the energy
given up by the copper is absorbed by the aluminum, giving mAl cAl ( ΔT )Al = mC u cC u ΔT C u , or
2/9/11 1:38:03 PM
350
92
Chapter 11
(c)
(d)
(e)
11.18
11.19
11.10
Δv v − 0 11.0
ms
am = = ⎛ c=C u ⎞ ⎡ ΔT
= Cu ⎤ m
= 2.20 m s 2
⎢
⎥
Al Δt⎜
Cs
u
t
−
0
5.00
⎟
⎝ cAl ⎠ ⎣ ( ΔT )Al ⎦
2
P = ma 2387
t = ( 75.0
kg )−2.20
m s 2 t = 363 kg ⋅ m 2 s4 ⋅ t = ( 363 W s ) ⋅ t
85°C
25°C
⎛
⎞⎛
⎞
=⎜
( 200 g) = 2.6 × 10 2 g = 0.26 kg
⎝ 900 ⎟⎠ ⎜⎝ 25°C − 5.0°C ⎟⎠
Maximum instantaneous power occurs when t = tmax = 5.00 s, so
(
)
()
(
(
)
)
(
)
2
−Qhot
m
− Ti ,ws ) == mixture
−m
− sTi ,Fe isolated from the environment, we have
Qcold =Assuming
(a)
the
Pmax ⇒
=that
363
Jtin-lead-water
1.82Fe c×Fe10isT3 thermally
wc
w s T f( 5.00
f J
(
)
(
)
(
or ofmthe
T f of
− Tusing
=food
−mSnenergy,
cSn T f −that
Ti ,Snrate
−m
If thisQcorresponds
must
cold = −Qhot to 25.0%
w cwrate
i ,w
Pb cPbbeT f − Ti ,Pb
)
and since mSn = mPb = mm etal = 0.400 kg and Ti ,Sn = Ti ,Pb = Thot = 60.0°C, this yields
Tf =
=
mw cw Ti ,w + mm etal ( cSn + cPb ) Thot
mw cw + mm etal ( cSn + cPb )
(1.00 kg) ( 4 186 J kg ⋅ °C) ( 20.0°C) + ( 0.400 kg) ( 227
(1.00 kg) ( 4 186 J kg ⋅ °C) + ( 0.400 kg) ( 227
yielding
(b)
J kg ⋅ °C + 128 J kg ⋅ °C ) ( 60.0°C )
J kg ⋅ °C + 128 J kg ⋅ °C )
T f = 21.3°C
If an alloy containing a mass mSn of tin and a mass mPb of lead undergoes a rise in temperature ΔT, the thermal energy absorbed would be Q = QSn + QPb , or
(m
Sn
(
)
(
)
(
)
+ mPb ) calloy ΔT = mSn cSn ΔT + mPb cPb ΔT , giving calloy =
If the alloy is a half-and-half mixture, mSn = mPb , then this reduces to calloy =
and yields
(c)
354
calloy =
mSn cSn + mPb cPb
mSn + mPb
cSn + cPb
2
227 J kg ⋅ °C + 128 J kg ⋅ °C
= 178 J kg ⋅ °C
2
For a substance forming monatomic molecules, the number of atoms in a mass equal to the
molecular weight of that material is Avogadro’s number, N A. Thus, the number of tin atoms
in mSn = 0.400 kg = 400 g of tin with a molecular weight of MSn = 118.7 g mol is
Chapter 11
⎛m ⎞
⎞
⎛
400 g
N Sn = ⎜ Sn ⎟ N A = ⎜
(6.02 × 1023 mol −1 ) = 2.03 × 1024
⎝ 118.7 g mol ⎠⎟
⎝ MSn ⎠
⎛m ⎞
⎞
⎛
400 g
and, for the lead, N Pb = ⎜ Pb ⎟ N A = ⎜
(6.02 × 1023 mol −1 ) = 1.16 × 10 24
⎝ 207.2 g mol ⎟⎠
⎝ M Pb ⎠
continued on next page
(d)
We have
N Sn 2.03 × 10 24
=
= 1.75
N Pb 1.16 × 10 24
and observe that
cSn 227 J kg ⋅ °C
=
= 1.77
cPb 128 J kg ⋅ °C
from which we conclude that the specific heat of an element is
proportional to the number of atoms per unit mass of that element.
the showdown of the water cups.
11.11
11.21
67352_ch11.indd 350
The total energy absorbed by the cup, stirrer, and water equals the energy given up by the silver
sample. Thus,
2/9/11 1:38:05 PM
the showdown of the water cups.
11.21
The total energy absorbed by the cup, stirrer, and water equals the
energy
given up
by the silver351
93
Energy
in Thermal
Processes
sample. Thus,
11.9
The mass
of water involved is
⎣⎡ mc cAl + ms cCu + mw cw ⎦⎤ ( ΔT )w = [ mc ΔT
]
Ag
kgthe
⎛ 3 of
⎞ cup gives11 3
14
Solving
m =for
rVthe
= ⎜mass
10
⎟ 4.00 × 10 m = 4.00 × 10 kg
⎝
m3 ⎠
ΔT Ag
⎤
1 ⎡
− ms cCu − mw cw ⎥ ,
mc =
⎢ mAg cAg
(a)
cAl ⎣
( ΔT )w
⎦
(
(
mc =
or
11.22
11.12
11.25
)
)
⎤
1 ⎡
(87 − 32 )
− ( 40 g ) ( 387 ) − ( 225 g ) ( 4 186 ) ⎥ = 80 g
⎢( 400 g ) ( 234 )
900 ⎣
( 32 − 27 )
⎦
If
N conservation
pellets are used,
the mass
of the lead
is Nm
pellet . Since
The
of energy
equation
for this
process
is the energy lost by the lead must equal
the energy absorbed by the water,
( energy to melt ice ) + ( energy to warm melted ice to T ) = ( energy to cool water to T )
mice L f + mice cw ( T − 0°C ) = mw cw (80°C − T )
or
This yields T =
mw cw (80°C ) − mice L f
(m
ice
+ mw ) cw
so
J kg ⋅ °C ) (80°C ) − ( 0.100 kg ) ( 3.33 × 10 5 J kg )
= 65°C
3
⎞
0.403 kg
10⋅6°C
cm
3 J ⎛kg
3
1.1
kg×)10
4 −4186
(
(
)
=
403
cm
=
=
4.03
m
(
)
⎜⎝ 1 m 3 ⎟⎠
1 000 kg m 3
(1.0 kg) ( 4 186
T=
11.13
11.27
Remember that energy must be supplied to melt the ice before its temperature will begin to rise.
Then, assuming a thermally isolated system, Qcold = −Qhot, or
(
)
(
mice L f + mice cwater T f − 0°C = −mwater cwater T f − 25°C
)
and
Tf =
mwater cwater ( 25°C ) − mice L f
(m
5
J kg )
Assuming all work done against friction is used to melt snow, the energy balance equation is
f ⋅ s = msnow L f . Since f = m k ( mskier g ), the distance traveled is
s=
11.31
11.15
+ mwater ) cwater
(825 g ) ( 4 186 J kg ⋅ °C) ( 25°C) − ( 75 g ) (3.33 × 10
( 75 g + 825 g ) ( 4 186 J kg ⋅ °C)
22 × 10 −3 kg
T f = =16°C
= 0.067 or 6.7%
0.33 kg
yielding
11.29
11.14
ice
=
msnow L f
m k ( mskier
(1.0 kg) (3.33 × 10
⎛
g ) 0.20 ( 75⎜kg ) ( 9.80
⎝
=
5
J kg )
× 10 3 ⎞m⎛ 1=min
2.3⎞ km
breaths
⎞ ⎛ = 2.3
m ⎟⎠s⎜⎝2 22.0
⎟⎜
⎟ = 11.1 W
)
min ⎠ ⎝ 60 s ⎠
Assume that all the ice melts. If this yields a result T > 0, the assumption is valid, otherwise the
problem must be solved again based on a different premise. If all ice melts, energy conservation
(Qcold = −Qhot ) yields
mice ⎡⎣cice [ 0°C − ( −78°C )] + L f + cw ( T − 0°C )⎤⎦ = − ( mw cw + mcal cC u )( T − 25°C )
or
67352_ch11.indd 351
T=
(m
w
cw + mcal cC u )( 25°C ) − mice ⎡⎣cice ( 78°C ) + L f ⎤⎦
( mw + mice ) cw + mcal cC u
2/9/11 1:38:08 PM
352
94
Chapter 11
(e)
change
in temperature of the slab as it absorbs the thermal energy computed above is
With The m
w = 0.560 kg, mcal = 0.080 g, mice = 0.040 g, cw = 4 186 J kg ⋅°C,
J J kg ⋅°C, and L f = 3.33 × 10 5 J kg,
cC u = 387 J= kg ⋅°C, 6.7
cice ×=10
2 090
= 79°C
(96.0 kg) (880 J kg ⋅ °C)
6
this gives
(f) The rate the slab absorbs solar energy is
⎡( 0.560 )( 4 186 ) + ( 0.080 )( 387 )⎤⎦ ( 25°C ) − ( 0.040 ) ⎡⎣( 2 090 )( 78°C ) + 3.33 × 10 5 ⎤⎦
T=⎣
( 0.560 + 0.040 )( 4 186 ) + 0.080 ( 387 )
or
11.33
11.32
11.16
T = 16°C and the assumption that all ice melts is seen to be valid.
(a)
The mass
of 2.0is liters
of water is
mwof=terms:
rV = (10 3 kg m 3 ) ( 2.0 × 10 −3 m 3 ) = 2.0 kg.
The energy
required
the following
sum
The energy required to raise the temperature of the water (and pot) up to the boiling
point of water is
Qboil = ( mw cw + mAl cAl )( ΔT )
⎡
⎛
⎛
J ⎞
J ⎞⎤
5
+ ( 0.25 kg ) ⎜ 900
Qboil = ⎢( 2.0 kg ) ⎜ 4 186
⎟⎠ ⎥ (100°C − 20°C ) = 6.9 × 10 J
⎟
kg
kg
⎠
⎝
⎝
⎣
⎦
The time required for the 14 000 Btu h burner to produce this much energy is
or
t boil =
(b)
Qboil
6.9 × 10 5 J ⎛
1 Btu
=
⎜⎝
14 000 Btu h 14 000 Btu h 1.054 × 10 3
⎞
−2
⎟ = 4.7 × 10 h = 2.8 min
J⎠
Once the boiling temperature is reached, the additional energy required to evaporate all of
the water is
Qevaporate = mw Lv = ( 2.0 kg ) ( 2.26 × 10 6 J kg ) = 4.5 × 10 6 J
and the time required for the burner to produce this energy is
tevaporate =
11.34
11.35
11.17
Qevaporate
14 000 Btu h
=
4.5 × 10 6 J ⎛
1 Btu
⎞
h = 18 min
⎜⎝
⎟ = 0.30
in Thermal Processes
14 000 Btu h 1.054 × 10 3 Energy
J⎠
359
The total energy input required is
(a) The bullet loses all of its kinetic energy as it is stopped by the ice. Also, thermal energy is
transferred from the bullet to the ice as the bullet cools from 30.0°C to the final temperature. The sum of these two quantities of energy equals the energy required to melt part of
the ice. The final temperature is 0°C because not all of the ice melts.
(b)
The total energy transferred from the bullet to the ice is
Q = KEi + mbullet clead 0°C − 30.0°C =
(
= 3.00 × 10
−3
(
1
mbullet vi2 + mbullet clead ( 30.0°C )
2
⎡ 2.40 × 10 2 m s
kg ⎢
2
⎢
⎣
)
) + (128 J kg ⋅ °C) (30.0°C)⎤⎥ = 97.9 J
⎥
2
⎦
The mass of ice that melts when this quantity of thermal energy is absorbed is
m=
Q
(L )
f water
11.36
67352_ch11.indd 352
(a)
=
⎛ 10 3 g ⎞
97.9 J
= 2.94 × 10 −4 kg ⎜
= 0.294 g
5
3.33 × 10 J kg
⎝ 1 kg ⎟⎠
Observe that the equilibrium temperature will lie between the two extreme temperatures
2/9/11 1:38:11 PM
=
11.18
11.17
11.39
(5.669 6 × 10
−8
75 W
−5
Energy in Thermal
m 2 Processes
4 = 1.1 × 10
W m 2 ⋅ K 4 (1.0 ) ( 3 300 K )
)
95
353
Since the
andstabilizes,
the steel container
unchanged,
and neither
When
the temperature
temperature of
of the
the water
junction
the energyistransfer
rate must
be the substance
same for
undergoes
phaseorchange,
internal
energy
of
these
materials
is
constant.
Thus,
all
the
energy
each
of thearods,
PCu = Pthe
.
The
cross-sectional
areas
of
the
rods
are
equal,
and
if
the
temperaAl
givenofup
the copper
is absorbed
by the aluminum,
cAl ( ΔT
= mrod.
c ΔT C u , or
ture
thebyjunction
is 50°C,
the temperature
differencegiving
is ΔT m=Al50°C
for)Aleach
Cu Cu
⎛ ΔT
⎛ ΔT ⎞
⎤⎞
⎛ cC u ⎞ ⎡ ΔT
Cu
Thus, mAl =PCu
⎜⎝ L) ⎥⎟⎠ m=Cku Al A ⎜⎝ L ⎟⎠ = PAl , which gives
⎜⎝ c= k⎟⎠Cu⎢ (AΔT
Cu
Al
Al
Al ⎦
⎣
⎛ 387 ⎞ ⎛ 85°C − 25°C
⎞ W m ⋅ °C ⎞
⎛m238
2 ⎟(
= ⎜ ⎛ ⎟k ⎜Al ⎞
200 g ) = 2.6 × 10 2 g = 0.26 kg
2.0
cm ) = 9.0 cm
LAl
= ⎜ ⎠ ⎝ 25°C
=
L
⎝ 900
⎠( 37 K )
Cu− 5.0°C
⎟ (15
⎜
⎟
= 5.3
× 10 2 W
⎝ 397
⎝ k Cu ⎠ =
2 W m ⋅ °C ⎠
0.14 m ⋅ K W
(
11.18
11.19
11.41
)
The window will consist of the glass pane and a stagnant air layer on each side (see
Example 11.10 in text). From Tables 11.3 and 11.4, the R-values for these layers are
Rpane =
L
k glass
Rair = 0.17
and
layer
Thus,
=
0.40 × 10 −2 m
m 2 ⋅ °C
= 5.0 × 10 −3
0.80 W m ⋅ °C
W
2
ft 2 ⋅ °F ⎛ 1 Btu h ⎞ ⎛ 1 m ⎞ ⎛ 1 °C ⎞
m 2 ⋅ °C
= 0.030
⋅⎜
⎟
⎜
⎟
⎜
⎟
Btu h ⎝ 0.293 W ⎠ ⎝ 3.281 ft ⎠ ⎝ 9 5 °F ⎠
W
Rtotal = ΣRi = ( 0.030 + 5.0 × 10 −3 + 0.030 )
m 2 ⋅ °C
m 2 ⋅ °C
= 0.065
W
W
The energy loss through the window in a 12 hour interval is then
A ( Th − Tc )
( 2.0 m 2 )9( 20°C) ⋅ (12 h ) ⎛ 3 6003s ⎞ = 2.7 × 107 J
⋅ t = 1.5 × 102 J
=
=⎜ 39 m ⎟
°C WJ kcal ) ⎝ 1 h ⎠
ΣRi 9 300 kcal
0.065
m 3m( 4⋅186
Q = P⋅t =
11.43
11.20
(
)
The absolute temperatures of the two stars are TX = 5 727 + 273 = 6 000 K and
TY = 11 727 + 273 = 12 000 K. Thus, the ratio of their radiated powers is
(
)
)
−2
m
⎞ ) 2.0 ×
PY s AeTY4 ⎛( 58
TY W
4 10
=
=
=
2
=
16
(
)
=
= 7.3 × 10 −2 W m ⋅ °C
4
⎜
⎟
2
PX s AeTX 0.80
⎝ TX ⎠m ( 25°C − 5.0°C )
4
(
11.44
11.21
11.47
The rate of energy transfer through a compound slab is
⎛ 10 2 cm ⎞ ⎛ 4.186 J ⎞
J
cal
⎛ ΔT ⎞
P = k A⎜
, with k = 0.200
⎜⎝
⎟⎠ = 83.7
⎟
⎜
⎟
⎝ L ⎠
s ⋅ m ⋅ °C
cm ⋅ °C ⋅ s ⎝ 1 m ⎠ 1 cal
Thus, the energy transfer rate is
J
⎛ 200°C − 20.0°C ⎞
⎛
⎞
P = ⎜ 83.7
⎟
⎟ [(8.00 m ) ( 50.0 m )] ⎜⎝
⎝
s ⋅ m ⋅ °C ⎠
1.50 × 10 −2 m ⎠
(
)(
)
= 4.02 × 108 J s = 4.02 × 108 W
= ×402
8.64
10 4 MW
s = 5 × 108 J
11.48
11.51
11.22
Since 97% of the incident energy is reflected, the rate of energy absorption from the sunlight is
4
−8
The
total
power×radiated
the (Sun
P = s AeT
, where
s = of
5.669
× 10radiation.
W m 2 ⋅ K 4 , the emisPabsorbed
= 3.0%
0.030
I ⋅ Ais), where
I is the
intensity
the 6solar
( I ⋅ A) = by
2
sivity is e = 0.986, the surface area (a sphere) is A = 4p r , and the absolute temperature is
T = 5 800 K. Thus,
P = ( 5.669 6 × 10 −8 W m 2 ⋅ K 4 ) 4p ( 6.96 × 108 m ) ( 0.986 )( 5 800 K )
2
4
or P = 3.85 × 10 26 W. Thus, the energy radiated each second is
E = P ⋅ Δt = ( 3.85 × 10 26 J s )(1.00 s ) = 3.85 × 10 26 J
11.52
67352_ch11.indd 353
The energy added to the air in one hour is
2/9/11 1:38:13 PM
96
354
11.23
11.55
=
Chapter 11
2.0 × 10 4 J
= 4.5°C
(10 kg) ( 448 J kg ⋅ °C)
The conservation of energy equation is Qcold = −Qhot , or
and, for the lead,
(m c
)
+ mcup cglass ( T − 27°C ) = −mCu cCu ( T − 90°C )
N Sn
m
Cu cCu ( 90°C ) + mw cw + mcup cglass ( 27°C )
This gives
T = N Pb
, or
mw cw + mcup cglass + mCu cCu
w w
T=
11.56
11.57
11.24
(
)
( 0.200 )( 387 )( 90°C ) + ⎡⎣( 0.400 ) ( 4 186 ) + ( 0.300 )(837 )⎤⎦ ( 27°C )
= 29 °C
( 0.400 ) ( 4 186 ) + ( 0.300 )(837 ) + ( 0.200 )( 387 )
(a) With a body temperature of T = 37°C + 273 = 310 K and surroundings at temperature
The total energy needed is
(
)
Q = mLv = ( 2.00 kg ) 2.00 × 10 4 J kg = 4.00 × 10 4 J
and the time required to supply this energy is
Q 3.16
4.00 ×× 10
1074 JJ ⎛ 1 h ⎞ 3 ⎛ 1 min ⎞
=
= 4.00 × 10= s1.83
⎜⎝ h ⎟⎠ = 66.7 min
P = 410.0
60.0 s
800 JJ ss ⎜⎝ 3 600 s ⎟⎠
(a) The energy delivered to the heating element (a resistor) is transferred to the liquid nitrogen,
causing part of it to vaporize in a⎛ liquid-to-gas
The⎞ total energy delivered
80.0°C − T ⎞ phase transition.
⎛ T − 30.0°C
In the steady state, PAu = PAg, or k Au A ⎜
=k A
⎟⎠of 4.0 h.
to the element equals the product⎝ of theLpower⎟⎠ andAgthe ⎜⎝time interval
L
t=
11.58
11.61
11.25
⎡⎛
⎞ next page⎤
⎞
J ⎛ continued
J on
−8.00 °C ⎥
−20.0 °C + 3.33 × 10 5
+ ⎜ 2 090
⎢⎜
⎟
kg ⎝
kg ⋅ °C ⎟⎠
⎦
k Au (80.0°C ) ⎣+⎝k Ag ( 30.0°C ) ⎠ 314 (80.0°C ) + 427 ( 30.0°C
)
=
T=
= 51.2 °C
k +k
314 + 427
= 3.25 × 10 4 J = Au32.5 Ag
kJ
This gives
11.62
11.69
11.26
(a) The net rate of energy transfer by radiation between a body at absolute temperature
At a pressure of 1 atm, water boils at 100°C. Thus, the temperature on the interior of the copper
kettle is 100°C and the energy transfer rate through the bottom is
W ⎞
2 ⎛ 102°C − 100°C ⎞
⎛ ΔT ⎞ ⎛
P = k A⎜
= 397
⎟ ⎡p ( 0.10 m ) ⎤⎦ ⎜
⎝ L ⎟⎠ ⎜⎝
m ⋅ °C ⎠ ⎣
⎝ 2.0 × 10 −3 m ⎟⎠
= 1.2 × 10 4 W = 12 kW
11.70
67352_ch11.indd 354
(a)
The surface area of the stove is
2/9/11 1:38:17 PM
ࢠࠊࠇଝ‫ئ‬৕
PROBLEM SOLUTIONS
12.1
The constant pressure is P = (1.5 atm )(1.013 × 10 5 Pa atm ) and the work done on the gas is
W = −P ( ΔV ) .
(a)
378
12.3
12.2
12.2
ΔV = + 4.0 m 3 and
Chapter 12
W = −P ( ΔV ) = − (1.5 atm )(1.013 × 10 5 Pa atm ) ( + 4.0 m 3 ) = − 6.1× 10 5 J
(b)
ΔV = − 3.0 m 3, so
(c)
As seen from the areas under the paths in the5 PV diagrams above,
the higher pressure
during
3
W = − P (phase
ΔV ) =of− (the
1.5process
atm )(1.013
× 10
Pa atm
3.0 mby
= +gas
4.6in×(a)
10 5than
J in (b).
) ( −done
)the
the expansion
results
in more
work
(a)
The work
on (b)
the are
gas shown
in this below:
constant pressure process is
The sketches
fordone
(a) and
⎛ nRT f nRTi ⎞
= −nR T f − Ti
W = −P ( ΔV ) = −P ⎜
−
⎝ P
P ⎟⎠
(
or
)
W = − ( 0.200 mol )(8.31 J mol ⋅K )( 573 K − 293 K ) = − 465 J
(b)
12.4
12.5
12.3
The negative sign for work done on the gas indicates that
3.2 J
the gas does positive work on its=surroundings
= 0.17.
19 J
(a) The work done by the gas on the projectile is given by the area under the curve in the PV
In each
case, the
work
diagram.
This
is done on the gas is given by the negative of the area under the path on the
PV diagram. Along those parts of the path where volume is constant, no work is done. Note that
1 atm = 1.013 × 10 5 Pa and 1 Liter = 10 −3 m 3.
(a)
WIAF = WIA + WAF = −PI (VA − VI ) + 0
The Laws of Thermodynamics
= − ⎡⎣ 4.00 (1.013 × 10 5 Pa ) ⎤⎦ ⎡⎣( 4.00 − 2.00 ) × 10 −3 m 3 ⎤⎦ = − 810 J
(b)
379
WIF = − ( triangular area ) − ( rectangular area )
=−
Wenv
1
( PI − PB ) (VF − VB ) − PB (VF − VB ) = − 12 ( PI + PB ) (VF − VB )
2
1
= − ⎡⎣( 4.000 + 1.00 )(1.013
× 10 5 Pa ) ⎤⎦ (34.00 − 2.00 ) 3× 10 −3 m 3
= +P2( ΔV ) = ( 70.0 × 10 3 Pa ) ( −0.20 m
) = −14 × 10 J = −14 kJ
and (c) is the
= correct
− 507 J choice.
2.
When volume is constant, the work done on the gas is zero, so the first law of thermodynamWIB + Win
= 0 − PB energy
(c) gives
WIBFthe= change
(VF − Vas
BF internal
B )
ics
= − ⎡⎣1.00 (1.013 × 10 5 Pa ) ⎤⎦ ⎡⎣( 4.00 − 2.00 ) × 10 −3 m 3 ⎤⎦ = − 203 J
97
67352_ch12.indd 371
2/9/11 1:39:21 PM
98
378
12.4
12.7
12.3
Note
Chapter 12
that, to 2 significant figures, this equals the area enclosed within the process diagram
given above.
(c) P
As
seen from the areas under the paths in the PV diagrams above, the higher pressure during
With
f = Pi = P, the ideal gas law gives Pf V f − Pi Vi = P ( ΔV ) = nR ( ΔT ) , so the work done by
the expansion phase of the process results in more work done by the gas in (a) than in (b).
⎛ m ⎞
R T f − Ti
the gas is Wenv = +P ( ΔV ) = nR ( ΔT ) = ⎜
⎝ M He ⎟⎠ pressure process is
(a) The work done on the gas in this constant
(
)
If Wenv = 20.0 J, the mass of helium
in the
gas
sample is
nRT
⎛
i ⎞
= −nR T f − Ti
−
⎜⎝
⎟
P ⎠
W ( M He )
( 20.0 J )( 4.00 g mol )
=
= 0.0963 g = 96.3 mg
mor= env
(8.31 J mol ⋅K )(373 K − 273 K )
R T f − Ti
(
(
12.9
12.5
(a)
)
)
From the ideal gas law, nR = PV f T f = PVi Ti . With pressure constant this gives
⎛ Vf ⎞
T f = Ti ⎜ ⎟ = ( 273 K )( 4 ) = 1.09 × 10 3 K
⎝ Vi ⎠
(b)
The work done on the gas is
(
)
(
)
W = −P ( ΔV ) = − PV f − PVi = −nR T f − Ti = −nR ( 4Ti − Ti )
3
The10
Laws
= − (1.00 mol )(8.31 J mol ⋅K )[ 3 ( 273 K )] = − 6.81×
J of
= Thermodynamics
− 6.81 kJ
12.10
12.6
12.11
12.13
12.7
(a)
(a)
The work done on the fluid is the negative
⎛ 1.013 × 10 5 Pa ⎞ ⎛ 10 −3 m 3 ⎞
W = −P ( ΔV ) = − ( 0.800 atm )( −7.00 L ) ⎜
⎟⎠ ⎜⎝ 1 L ⎟⎠ = 567 J
1 atm
⎝
(b)
ΔU = Q + W = −400 J + 567 J = 167 J
(a)
Along the direct path IF, the work done on the gas is
381
.
W = − ( area under curve )
1
= − ⎡⎢(1.00 atm )( 4.00 L − 2.00 L ) + ( 4.00 atm − 1.00 atm )( 4.00 L − 2.00 L ) ⎤⎥
2
⎣
⎦
W = − 5.00 atm ⋅ L
Thus,
ΔU = Q + W = 418 J − 5.00 atm ⋅ L
⎛ 1.013 × 10 5 Pa ⎞ ⎛ 10 −3 m 3 ⎞
= 418 J − ( 5.00 atm ⋅ L ) ⎜
⎟⎠ ⎜⎝ 1 L ⎟⎠ = − 88.5 J
1 atm
⎝
(b)
Along path IAF, the work done on the gas is
⎛ 1.013 × 10 5 Pa ⎞ ⎛ 10 −3 m 3 ⎞
W = − ( 4.00 atm )( 4.00 L − 2.00 L ) ⎜
⎟⎠ ⎜⎝ 1 L ⎟⎠ = −810 J
1 atm
⎝
From the first law, Q = ΔU − W = − 88.5 J − ,( −
810 Jis) =a statement
722 J
which
of the first law of
thermodynamics.
12.8
12.15
From kinetic theory, the average kinetic energy per molecule is
KE m olecule =
3
3⎛ R ⎞
kB T = ⎜
T
2
2 ⎝ N A ⎟⎠
For a monatomic ideal gas containing N molecules, the total energy associated with random
molecular motions is
67352_ch12.indd 378
2/9/11 1:39:41 PM
3
3⎛ R
kB T = ⎜
2
2 ⎝ NA
=
m olecule
⎞
⎟⎠ T
The Laws of Thermodynamics
379
99
area ) the total energy associated with random
(b)
WIF = − ( triangular
) − ( rectangular
For a monatomic
ideal gasarea
containing
N molecules,
molecular motions is
1
1
= − ( PI − PB )3(V⎛ F N− V⎞B ) − PB 3(VF − VB ) = − ( PI + PB ) (VF − VB )
2 m olecule = ⎜
2
U = N ⋅ KE
RT
=
nRT
2 ⎝ N A ⎟⎠
2
12.16
12.17
12.9
1 for an ideal gas, the internal
−3
3ideal gas is found to be
=−
nRT
of a monatomic
Since PV =
⎡( 4.000 + 1.00 )(1.013 × 10 5 Pa ) ⎤energy
⎣
⎦ ( 4.00 − 2.00 ) × 10 m
given by U =232 PV .
⎛ 3
⎞
P0 V0 − ⎜ − P0 V0 ⎟ = 6P0 V0
⎝ 2
⎠
The work done on the gas is the negative since V f > Vi of the area under the curve on the PV
(a)
Theorchange in the volume occupied by the gas is
diagram,
(
(
)
)
ΔV = V f − Vi = A L f − Li = ( 0.150 m 2 )( −0.200 m ) = −3.00 × 10 −2 m 3
and the work done by the gas is
Wenv = +P ( ΔV ) = ( 6 000 Pa ) ( −3.00 × 10 −2 m 3 ) = −180 J
(b)
The first law of thermodynamics is ΔU = Q + W = −Qoutput − Wenv . Thus, if ΔU = −8.00 J,
the energy transferred out of the system by heat is
Qoutput = −ΔU − Wenv = − ( −8.00 J ) − ( −180 J ) = + 188 J
12.19
12.10
(a)
W = F ⋅ d = ( 25.0 × 10 3 N ) ⋅ ( 0.130 m ) = 3.25 × 10 3 J = 3.25 kJ
(b)
Since the internal energy of an ideal gas is a function of temperature alone, the change in
the internal energy in this isothermal process is ΔU = 0 .
(c)
From the first law of thermodynamics,
Q = ΔU − W = 0 − ( 3.25 kJ ) = −3.25 kJ
12.20
12.21
12.11
(d)
If the energy exchanged as heat is Q = 0 while the work done on the gas is positive (W > 0 ),
the first law of thermodynamics, ΔU = Q + W = 0 + W > 0, tells us that the internal energy of
the system must increase. Since the internal energy of an ideal gas is directly proportional to the
absolute temperature, the temperature must increase .
(a)
(a)
The
done on
the gasinvolving
as it goesan ideal gas, the work done on the gas is
In anwork
isothermal
process
from
point
A
to
point
C
is
the
W = −Wenv = −nRT ln V f Vi . But, when temperature is constant, the ideal gas law gives
negative
of the
areaand
under
PV = P V
= nRT
wethe
may write the work done on the gas as
(
i
384
Chapter 12
(b)
i
f
)
f
⎛ Vf ⎞
⎛ 1.25 m 3 ⎞
= − 4.58 × 10 4 J
W = −Pi Vi ln ⎜ ⎟ = − (1.00 × 10 5 Pa ) ( 0.500 m 3 ) ln ⎜
⎝ 0.500 m 3 ⎟⎠
⎝ Vi ⎠
The change in the internal energy of an ideal gas is ΔU = nCv ( ΔT ) , and for an isothermal
process, we have ΔU = 0. Thus, from the first law of thermodynamics, the energy transfer
by heat in this isothermal expansion is
Q = ΔU − W = 0 − ( − 4.58 × 10 4 J ) = + 4.58 × 10 4 J
(c)
12.22
67352_ch12.indd 379
(a)
ΔU = 0
[See part (b) above.]
The number of atoms per mole in any monatomic gas is Avogadro’s number
N A = 6.02 × 10 23 atoms mol
2/9/11 1:39:42 PM
380
100
12.25
12.8
12.12
ΔU AB and − WAB are
Chapter 12
positive.
(a)
original
volume
theasaluminum
is V0 = mP (kPa)
rAl , and the change in volume is
The work
done
by theofgas
it
ΔV = b Vfrom
) = ( 3a
expands
point
A to
( m rAlB)( ΔT ) . 400
0 ( ΔT
Al )point
is given by the area under the PV
B
The workbetween
done bythese
the aluminum
diagram
points. is then 300
Consider the sketch given at the
= +P ( ΔV
P ( 3a
Wenv observe
) =this
Al ) ( m rAl ) ( ΔT )
right and
that
area
B
D
200
can be broken into two rectangular
⎞
5.0 kg
−1 ⎛
5
−6
= (1.013
× 10 areas.
Pa ) 3 ⎡⎣The
24 × 10 ( °C ) ⎤⎦ ⎜ A
( 70°C ) = 0.95 J
areas and two
triangular
⎝ 2.70 × 10 3 kg m 3 ⎟⎠
100
total area is given by
A
(b)
The energy transferred by heat to the aluminum is
V (m3)
1
2
3
Q = mcAl ( ΔT ) = ( 5.0 kg ) ( 900 J kg ⋅°C )( 70°C ) = 3.2 × 10 5 J
(c)
5
6
The work done on the aluminum is W = −Wenv = − 0.95 J, so the first law gives
ΔU = Q + W = 3.2 × 10 5 J − 0.95 J = 3.2 × 10 5 J
12.13
12.29
4
The net work done by a heat engine operating on
the cyclic process shown in Figure P12. equals the
triangular area enclosed by this process curve. Thus,
The Laws of Thermodynamics
387
P (atm)
6.00
1
4.00
( 6.00 atm − 2.00 atm )( 3.00 m 3 − 1.00 m 3 )
2
( base )( altitude )
5
2.00
3 ⎛ 1.013 × 10 Pa ⎞
5
= 4.00
atm
⋅
m
⎟⎠ = 4.05 ×5 10 J
⎜⎝ 3 1 atm
1
= ⎡⎣( 4.00 − 1.00 ) m ⎤⎦ ⎡⎣( 6.00 − 2.00 ) × 10 Pa ⎤⎦
V (m3)
2
1.00 2.00 3.00
3
= 405 × 10 J = 405 kJ
FIGURE P12.13
= 6.00 × 10 5 Pa = 600 kJ
The net work done by a heat engine operating
Thethe
maximum
possible
efficiency
for aP12.30
heat engine operating between reservoirs with absolute
on
cyclic process
shown
in Figure
T
=
25°
+
273
=
298
K
temperatures
of
c
equals the triangular area enclosed by thisand Th = 375° + 273 = 648 K is the Carnot efficiency
process curve.TThis is 298 K
eC = 1− c = 1−
= 0.540
or1 54.0%
⎛
⎞
108 J ) ⎜ 1−
= 5.67 × 108 J
648× K
Th = ( 7.69
⎝ 3.80 ⎟⎠
(a) The coefficient of performance of a heat pump is COP = Qh
(a) The maximum efficiency possible is that of a Carnot engine operating between reservoirs
having absolute temperatures of Th = 1 870 + 273 = 2 143 K and Tc = 430 + 273 = 703 K.
Weng =
12.30
12.31
12.14
12.32
12.33
.1
Th − Tc
T
703 K
= 1− c = 1−
= 0.672 ( or 67.2%)
2 143 K
Th
Th
Weng
From e =
, we find Weng = e Qh = 0.420 (1.40 × 10 5 J ) = 5.88 × 10 4 J
Qh
eC =
(b)
Weng 5.88 × 10 4 J
1=800 J
= 5.88 × 10 4 W = 58.8 kW
= 0.43
or
43%
t=
1.00
s
4 200 J
The work done by a heat engine equals the net energy absorbed by the engine, or
The efficiency of a heat engine is e = Weng Qh , where Weng is the work done by the engine
and Qh is the energy absorbed from the higher temperature reservoir. Thus, if
so
12.34
12.35
.1
(a)
(a)
P=
Weng = Qh 4, the efficiency is e = 1 4 = 0.25 or 25% .
(b)
From conservation of energy, the energy exhausted to the lower temperature reservoir is
Qc = Qh − Weng. Therefore, if Weng = Qh 4, we have Qc = 3 Qh 4 or Qc Qh = 3 4 .
12.36
67352_ch12.indd 380
The work done by the engine equals the change in the kinetic energy of the bullet, or
2/9/11 1:39:44 PM
mgun ciron
12.11
12.37
.1
(a)
(a)
(b)
(b)
12.12
12.38
.1
12.41
=
(123 J )(1 0.011 0 − 1)
(1.80 kg) ( 448
J kg ⋅°C )
= 13.7°C
The Laws of Thermodynamics
381
101
5
−3
3
Weng
1.20 × 10 3 J L ) ⎛ 1.013 × 10 Pa ⎞ ⎛ 10 m ⎞ = 567 J
W
) =Q−c ( 0.800
e ≡= −P ( ΔV
= 1−
= 1− atm )( −7.00
=
0.294
or
29.4%
(
)
⎟
⎜
⎜
1 atm
⎠ ⎝ 1 L ⎟⎠
⎝
Qh
Qh
1.70 × 10 3 J
2
ΔU ==QQ+ W
= −400 J + 5673 J = 167 J 3
W
eng
h − Qc = 1.70 × 10 J − 1.20 × 10 J = 5.00 × 10 J
Weng the5.00
We treat
sprinter’s
as a thermodynamic system and apply the first law of ther× 10 2 body
J
P=
=
= 1.67 × 10 3 W = 1.67 kW
modynamics,
ΔU
=
Q
+
W.
Then,
with
ΔU) == −7.5
10 5 J and W = −4.8 × 10 5 J (negative
0.433
11.9×kJ
(1− e )s= ( 21.0 kJ )(1−
t = Qh 0.300
because the sprinter does work on the environment), the energy absorbed as heat is
(a) The absolute temperature of the cold reservoir is
The actual efficiency of the engine is
(a)
(c)
e = 1−
Qc
Qh
= 1−
300 J
= 0.400
500 J
If this is 60.0% of the Carnot efficiency, then
e
0.400 2
=
=
0.600 0.600 3
eC =
Thus, from eC = 1− Tc Th , we find
12.42
12.45
Tc
2
1
= 1− eC = 1− =
T
343 K
Th
= 17.2
( COP )hp,C =3 h3 =
Th − Tc 343 K − 323 K
(a) The Carnot efficiency represents the maximum possible efficiency. With
The thermal energy transferred to the room by the water as the water cools from 1.00 × 10 2 °C to
20.0°C is
Q = mcw ΔT = ( 0.125 kg ) ( 4 186 J kg ⋅°C )(80.0°C ) = 4.19 × 10 4 J
If the room has a constant absolute temperature of T = 20.0° + 273 = 293 K, the increase in the
entropy of the room is
Q 4.19 × 10 4 J6
=
× 10 =J 143 J K 3 J
T = 2.26
293 K
= 6.06 kJ K
= 6.06 × 10
K
373 K
The energy added to the water by heat is
A quantity of energy, of magnitude Q, is transferred from the Sun and added to Earth. Thus,
−Q
+Q
ΔSSun =
and ΔSEarth =
, so the total change in entropy is
TSun
TEarth
ΔS =
12.46
12.47
ΔStotal = ΔSEarth + ΔSSun =
12.49
12.50
67352_ch12.indd 381
(a)
Q
Q
−
TEarth TSun
⎞
⎛ 1
1
= (1 000 J ) ⎜
−
⎟ = + 3.27 J K
−mKice L 5f 700 K ⎠
Qenv⎝ 290
ΔSenv =
=
= − 79 J K
T
T
The table is shown below. On the basis of the table, the most probable result of a toss is
2 H and 2 T .
End Result
Possible Tosses
Total Number
of Same Result
All H
1T, 3H
2T, 2H
3T, 1H
All T
HHHH
HHHT, HHTH, HTHH, THHH
HHTT, HTHT, THHT, HTTH, THTH, TTHH
TTTH, TTHT, THTT, HTTT
TTTT
1
4
6
4
1
(b)
The most ordered state is the least likely. This is seen to be all H or all T .
(c)
The least ordered state is the most likely. This is seen to be 2H and 2T .
(a)
In a game of dice, there is only
2/9/11 1:39:46 PM
382
102
Possible Tosses
Total Number
of Same Result
All H
1T, 3H
2T, 2H
3T, 1H
All T
HHHH
HHHT, HHTH, HTHH, THHH
HHTT, HTHT, THHT, HTTH, THTH, TTHH
TTTH, TTHT, THTT, HTTT
TTTT
1
4
6
4
1
Chapter 12
(b)
(g)
(c)
(h)
12.50
12.51
End Result
to obtain a 7 with a pair of dice. The combinations that yield a 7 are:
The most ordered state is the least likely. This is seen to be all H or all T .
1 + 6, 2 + 5, 3 + 4, 4 + 3, 5 + 2, and 6 + 1. Note that 1 + 6 and 6 + 1 are different combinaThe least ordered state is the most likely. This is seen to be 2H and 2T .
tions
Since in that the 6 occurs on different members of the pair of dice in the two combinations.
(a) energy
In a game
of dice,from
therethe
is water
only by heat, and absorbed by the freezer, is
The
transferred
⎛
J ⎞
= 3.3 × 10 5 J
Q = mL f = ( rV ) L f = ⎡⎣(10 3 kg m 3 ) (1.0 × 10 −3 m 3 ) ⎤⎦ ⎜ 3.33 × 10 5
kg ⎟⎠
⎝
Thus, the change in entropy of the water is
392
Chapter
(a) 12ΔSwater
(d)
(b)
12.52
12.53
=
( ΔQ )
r
=
water
T
− 3.3 × 10 5 J
J
= −1.2 kJ K
= −1.2 × 10 3
K
273 K
and that of the freezer is
The magnitudes of the thermal energy transfers, appearing in the numerators, are the same
+ 3.3
10 5 Jreservoir necessarily has a smaller denominator. Hence,
r )freezer
for the two( ΔQ
reservoirs,
but
the× cold
ΔSfreezer =
=
= +1.2 kJ K
its positive change
T dominates.
273 K
The
entropylost
of by
a reservoir
The change
potentialinenergy
the log isistransferred away by heat, so the energy transferred from
the log to the reservoir is ΔQr = mgh. The change in entropy of the reservoir (universe) is then
Qh
2
+
ΔQr mgh ( 70.0 kg ) ( 9.80 m s )( 25.0 m )
ΔS =
=
=
= 57.2 J K Th
Tc
T
T
300 K
12.54
12.55
change in entropy
of a reservoir
The maximum
rate at which
the bodyiscan dissipate waste heat by sweating is
⎛
kg ⎞ ⎛
ΔQ ⎛ Δm ⎞
J ⎞⎛ 1 h ⎞
3
=⎜
= 1.0 × 10 3 W
⎟⎠ Lv = ⎜ 1.5
⎜ 2 430 × 10
⎟
⎟
⎝
Δt
Δt
h ⎠⎝
kg ⎠ ⎜⎝ 3 600 s ⎟⎠
⎝
If this represents 80% of the maximum sustainable metabolic rate
⎡⎣ i.e., ΔQ Δt = 0.80 ( ΔU Δt )m ax ⎤⎦ , then that maximum rate is
ΔQ is
Δtonly1.0
× 10 3 W
⎛ ΔU ⎞The body
cient
converting chemical energy to mechanical
= about 25% =effi
1.3
× 10in3 W
⎜⎝
⎟⎠ =
0.80
0.80
Δt m ax
energy.
12.56
12.57
(a) At the sleeping rate of 80.0 W, the time required for the body to use the 450 Cal of energy
Using the metabolic rates from Table 12.4, we find the change in the body’s internal energy (i.e.,
supplied by the bagel is
the energy consumed) for each activity, and the total change for the day, as:
Metabolic
Rate (W)
Activity
Internal Energy Change
Sleeping – 8.0 h
80
ΔU1 = − (80 J s )(8.0 h )( 3 600 s 1 h ) = −2.3 × 10 6 J
Light Chores – 3.0 h
230
ΔU 2 = − ( 230 J s )( 3.0 h )( 3 600 s 1 h ) = −2.5 × 10 6 J
Slow Walk – 1.0 h
230
Running – 0.5 h
465
ΔU 3 = − ( 230 J s )(1.0 h )( 3 600 s 1 h ) = −8.3 × 10 5 J
ΔU 4 = − ( 465 J s )( 0.50 h )( 3 600 s 1 h ) = −8.4 × 10 5 J
TotaleC′ = 0.644 = 3.01
eC
12.59
ΔU1 + ΔU 2 + ΔU 3 + ΔU 4 = − 6.5 × 10 6 J = − 6.5 MJ
0.214
The work output from the engine in an interval of one second is Weng = 1 500 kJ. Since the efficiency of an engine may be expressed as
e=
Weng
Qh
=
Weng
Weng + Qc
⎛1 ⎞
⎛ 1
⎞
the exhaust energy each second is Qc = Weng ⎜ − 1⎟ = (1 500 kJ ) ⎜
− 1 = 4.5 × 10 3 kJ
⎝e ⎠
⎝ 0.25 ⎟⎠
The mass of water flowing through the cooling coils each second is
(
)(60 L )(10
−3
m 3 1 L ) = 60 kg
so the rise in the temperature of the water is
67352_ch12.indd 382
2/9/11 1:39:49 PM
The Laws of Thermodynamics
⎛1 ⎞
⎛ 1
⎞
= Weng ⎜ − 1⎟ = (1 500 kJ ) ⎜
− 1 = 4.5 × 10 3 kJ
⎝e ⎠
⎝ 0.25 ⎟⎠
change
the internal
thiscooling
monatomic
ideal second
gas is is
The mass
of in
water
flowing energy
throughofthe
coils each
12.19
12.60
12.61
103
383
3 10 3 3kg m 3 (360 L ) 10 −3 m33 1 L = 60 kg
m
ΔU= =rV =
Pf(V f − Pi Vi =) ( 2P0 () (V0 ) − ( P0 ))( 2V0 ) = 0
2
2
2
2
so the rise in the temperature of the water is 3
3
= 0 − P0 V0 = − P0 V0 , or Q < 0
Then, from the first law, Q = ΔU − W
Qc
2
2
4.5 × 10 6 J
ΔT =
=
= 18°C
6 (N ⋅ m) 3
6
mc
J kg
8⋅×water
J ⋅°C
K
(60×kg10) (=4N8186
(a) W ==F
d 10
= ( 25.0
m⋅s
))= 3.25 × 103 J = 3.25 kJ
)×⋅ (100.130
K ⋅s
Assuming the gravitational potential energy given up by the falling water is transformed
(b)
Sincework
the
internal
of anhits
is
function
ofarea
temperature
alone,
(a) thermal
The
donewhen
byenergy
the
inideal
process
ABa equals
this
curvethe
onchange
the PV in
into
energy
thesystem
water
thegas
bottom
of the the
falls,
theunder
rate of
thermal
energy
the
internal
energy
in
this
isothermal
process
is
diagram.
production
is Thus,
Wenv = ( triangular area ) + ( rectangular area ) , or
Wenv = [ 12 ( 4.00 atm )( 40.0 L )
Pa ⎞ ⎛ 10 −3 m 3 ⎞
⎛
+ (1.00 atm )( 40.0 L )] ⎜ 1.013 × 10 5
⎟
⎝
atm ⎠ ⎜⎝
L ⎟⎠
= 1.22 × 10 4 J = 12.2 kJ
Note that the work done on the system is WAB = −Wenv = −12.2 kJ for this process.
(b)
The work done on the system (that is, the work input) for process BC is the negative of the
area under the curve on the PV diagram, or
Pa ⎞ ⎛ 10 −3 m 3 ⎞
⎛
WBC = − [(1.00 atm )(10.0 L − 50.0 L )] ⎜ 1.013 × 10 5
⎟
⎝
atm ⎠ ⎜⎝ 1 L ⎟⎠
= 4.05 kJ
(c)
The change in internal energy is zero for any full cycle, so the first law gives
Qcycle = ΔU cycle − Wcycle = 0 − (WAB + WBC + WCA )
processes.
12.63
(a)
bothkJ
parts (c) and (d) since both are cyclic
= 0 − ( −12.2 kJ + 4.05 kJ + 0 ) for
= 8.15
The energy transferred to the gas by heat is
J ⎞
⎛
3
Q = mc ( ΔT ) = (1.00 mol ) ⎜ 20.79
⎟ (120 K ) = 2.49 × 10 J = 2.49 kJ
⎝
mol ⋅K ⎠
(b)
Treating the neon as a monatomic ideal gas, Equation 12.3b gives the change in internal
energy as ΔU = 32 nR(ΔT ) , or
ΔU =
(c)
3
J ⎞
3
(1.00 mol ) ⎛⎜⎝ 8.31
⎟ (120 K ) = 1.50 × 10 J = 1.50 kJ
2
mol ⋅K ⎠
From the first law, the work done on the gas is
W = ΔU − Q = 1.50 × 10 3 J − 2.49 × 10 3 J = − 990 J
67352_ch12.indd 383
2/9/11 1:39:51 PM
⎛
⎜⎝
384
104
12.65
Chapter 12
(b)
(a)
⎞
⎟⎠ ( 273 K )
= 9.07 × 10 3 J = 9.07 kJ
the internal
energy
an ideal gas
The change in length,
due to
linearofexpansion,
of is
theΔU
rod=isnCv ( ΔT ) , and for an isothermal
process, we have ΔU = 0. Thus, from the first law of thermodynamics, the energy transfer
−1
by heat
expansion
10 −6 ( °Cis) ⎤⎦ ( 2.0 m )( 40°C − 20°C ) = 4.4 × 10 −4 m
ΔLin= this
a L 0isothermal
( ΔT ) = ⎡⎣11×
Q = exerts
ΔU − W
= 0 − (F−=4.58
J ) =kg+)4.58
The load
a force
mg ×=10
m 4 s J2 ) = 5.88 × 10 4 N on the end of the
(64 000
(9.80× 10
rod in the direction of movement of that end. Thus, the work done on the rod is
W = F ⋅ ΔL = ( 5.88 × 10 4 N ) ( 4.4 × 10 −4 m ) = 26 J
(b)
The energy added by heat is
⎛
J ⎞
Q = mc ( ΔT ) = (100 kg ) ⎜ 448
( 20 °C) = 9.0 × 105 J
kg ⋅°C ⎟⎠
⎝
(c)
12.67
(a)
5
5
5 + W = 9.0 × 10 J + 26 J = 9.0 × 10 J .
From the first law,
ΔU
=Q
1.83
× 10
J
=
=
0.55
kg
3.33 × 10 5 J kg
(10.0 × 103 Pa ) (1.00 m3 )
PV
TA = A A =
nR
(10.0 mol )(8.31 J mol ⋅K )
(10.0 × 10 Pa ) (6.00 m )
PV
TB = B B =
nR
(10.0 mol )(8.31 J mol ⋅K )
3
(b)
40.0
30.0
= 1.20 × 10 2 K
= 722 K
P (kPa)
20.0
A
3
B
10.0
1.00 2.00 3.00 4.00 5.00 6.00
V (m3)
As it goes from A to B, the gas is expanding and hence, doing work on the environment.
The magnitude of the work done equals the area under the process curve from A to B. We
subdivide this area into 2 rectangular and 2 triangular parts:
Wenv = ⎡⎣(10.0 × 10 3 Pa )( 6.00 − 1.00 ) m 3 ⎤⎦ + ⎡⎣( 40.0 − 10.0 ) × 10 3 Pa ⎤⎦ (1.00 m 3 )
1
+ 2 ⎡⎢ (1.00 m 3 )( 40.0 − 10.0 ) × 10 3 Pa ⎤⎥ = 1.10 × 10 5 J
⎣2
⎦
(c)
The change in the internal energy of a monatomic, ideal gas is ΔU = 32 nR ( ΔT ) , so
ΔU A→ B =
(d)
3
3
nR ( TB − TA ) = (10.0 mol )(8.31 J mol ⋅K )( 722 K − 120 K ) = 7.50 × 10 4 J
2
2
From the first law of thermodynamics, Q = ΔU − W , where W is the work done on the gas.
In this case, W = −Wenv = −1.10 × 10 5 J , and
Q = ΔU − W = 7.50 × 10 4 J − ( −1.10 × 10 5 J ) = 1.85 × 10 5 J
67352_ch12.indd 384
2/9/11 1:39:54 PM
ࢠऑѰ˕ળѰ
PROBLEM SOLUTIONS
13.1
(a)
Taking to the right as positive, the spring force acting on the block at the instant of
release is
Fs = −kA = − (130 N m ) ( +0.13 m ) = −17 N or 17 N to the left
(b)
At this instant, the acceleration is
Fs
−17 N
2
or
a = 28 m s 2 to the left
9.80mms s 2
( 2.30kgkg=)−28
Fa = m
mg= 0.60
−2
=
=
=
m = 1.54 cm
=
1.54
×
10
k
k
1.46 × 10 3 N m
The force compressing the spring is the weight of the object. Thus, the spring will be compressed
Assuming
the spring obeys Hooke’s law, the magnitude of the force required to displace the end
a distance of
a distance x from the equilibrium position (by either compressing or stretching the spring) is
F = k x , where k is the force constant of the spring.
(
13.2
13.3
1.
13.4
13.5
1.
)
(
F = k x = (137 N m ) ( 7.36 × 10
)
m ) = 10.1 N
(a)
If x = − 4.80 cm, the required force is F = k x = (137 N m ) 4.80 × 10 −2 m = 6.58 N
(b)
If x = + 7.36 cm, the
. required force is
−2
Fs mg
50 N
(a)
spring constant
is k = the
= tension
= in the−2spring
= 1.0 × 10 3 N m
WhenThe
the system
is in equilibrium,
x
x
5.0 × 10 m F = k x must equal the weight of
the object. Thus,
(
)
−2
k x ( 47.5 N m ) 5.00 × 10 m
k x = mg giving375
mN
=
= 0.242 kg
=
3
=
N m m s2
g= 1.25 × 10 9.80
0.300 m
1.
13.7
(a)
When the block comes to equilibrium, ΣFy = −ky0 − mg = 0, giving
(
)
(10.0 kg) 9.80 m s2
mg
=−
y0 = −
= − 0.206 m
k
475 N m
or the equilibrium position is 0.206 m below the unstretched position of the lower end of
the spring.
(b)
When the elevator (and everything in it) has an upward acceleration of a = 2.00 m s 2,
applying Newton’s second law to the block gives
ΣFy = −k ( y0 + y ) − mg = ma y
or
ΣFy = ( −ky0 − mg ) − ky = ma y
where y = 0 at the equilibrium position of the block. Since −ky0 − mg = 0 [see part (a)],
this becomes −ky = ma and the new position of the block is
y=
or
13.8
67352_ch13.indd 400
ma y
−k
=
(10.0 kg) ( +2.00
− 475 N m
m s2
) = − 4.21 × 10
−2
m = − 4.21 cm
4.21 cm below the equilibrium position .
(c)
When the cable breaks, the elevator and its contents will be in free-fall with a y = −g. The
new “equilibrium” position of the block is found from ΣFy = −ky0′ − mg = m ( −g ), which
yields y0′ = 0. When the cable snapped,
105 the block was at rest relative to the elevator at distance y0 + y = 0.206 m + 0.042 1 m = 0.248 m below the new “equilibrium” position. Thus,
while the elevator is in free-fall, the block will oscillate with amplitude = 0.248 m about
the new “equilibrium” position, which is the unstretched position of the spring’s lower end.
(a)
The work required to stretch the spring equals the elastic potential energy of the spring in
the stretched condition, or
2/9/11 1:41:24 PM
106
406
13.6
Chapter 13
.
The free-body
point in
theitscenter
of the
When
the cablediagram
breaks, of
thethe
elevator
and
contents
will be in free-fall with a y = −g. The
string“equilibrium”
is given at theposition
right. From
weissee
that from ΣFy = −ky0′ − mg = m ( −g ), which
new
of thethis,
block
found
yields y0′ = 0. When the cable snapped, the block was at rest relative to the elevator at distance ΣF
y0x +=y0= ⇒
0.206
+ sin
0.042
1 m==00.248 m below the new “equilibrium” position. Thus,
F −m2T
35.0°
while the elevator is in free-fall, the block will oscillate with amplitude = 0.248 m about
6
NN
m which is the
F5.00 × 10
375
the new
position,
unstretched position of the spring’s lower end.
k=“equilibrium”
x
=
3.16
× 10N−2 m = 2.23 m s
T
=
= 327
vor
=
i
i
1000
kg
2
sin
35.0°
2
sin
35.0°
m
(a) The work required to stretch the spring equals the elastic potential energy of the spring in
From conservation of mechanical energy,
the stretched condition, or
(b) Since
the bow
requires an applied horizontal force of 375 N to hold1 the
KE + PE
kxi2 ,string at 35.0° from
g + PEs f = KE + PE g + PEs i or 0 + mgh f + 0 = 0 + 0 +
2
the vertical, the tension in the spring must be 375 N when the spring is stretched 30.0 cm.
givingThus, the spring constant is
(a)
(c)
(
13.8
1.
13.11
(
k=
13.13
1.
) (
2 mgh f
(
)
m)
2 ( 0.100 kg ) 9.80 m s2 ( 0.600 m )
(b)
(
−2
M+m
⎛ m ⎞
= vi ⎜
⎝ M + m ⎟⎠
k
(10.0 × 10
)
M+m
=
k
mvi
( M + m) k
kg ( 300 m s )
= 0.478 m
1 N m)
kg2) (=19.6
( 2.01 kx
(575 N m )( 0.400 )2 = 46.0 J
2
Fm ax
230
N
k=
Assume
the rubber bands obey Hooke’s law. Then, the force constant of each band is
x m ax = 0.400 m = 575 N m
F
15 N
k= s =
= 1.5 × 10 3 N m
x 1.0 × 10 −2 m
yielding
(a)
=
)
= 2.94 × 10 3 N m
2
x
2
2.00
×
10
A
1 2
kA 2
⎛1
⎞
or
x=±
kA = 3 ⎜ kx 2 ⎟
⇒
x=±
⎠
⎝
3
2
2
3k 2
An unknown quantity of mechanical energy is converted into internal energy during the collision.
Thus, we apply conservation of momentum from just before to just after the collision and obtain
mvi + M ( 0 ) = ( M + m ) V, or the speed of the block and embedded bullet just after collision is
V = (m M + m)vi . We now use conservation of mechanical energy, (KE + PEs ) f = (KE + PEs )i ,
from just after collision until the block comes to rest. This gives 0 + 12 kx 2f = 12 ( M + m ) V 2 + 0, or
2
i
xf = V
13.14
13.15
1.
)
xf =
−3
Thus, when both bands are stretched 0.20 m, the total elastic potential energy is
2
⎛1
⎞
PEs = 2 ⎜ kx 2 ⎟ = 1.5 × 10 3 N m ( 0.20 m ) = 60 J
⎝2
⎠
(
(b)
Conservation of mechanical energy gives ( KE + PEs ) f = ( KE + PEs )i , or
1 2
mv + 0 = 0 + 60 J
2
13.16
1.
13.17
)
2 ( 60 J )
so
v=
= 49 m s
−3
50 × 10
kg
between the block
and surface.
(
)
(a)
= (83.8 N m ) 5.46 × 10 −2 m = 4.58 N
The maximum speed
occurs at the equilibrium position and is vmax = A k m. Thus,
(b)
kA2 (16.0 N m ) ( 0.200 m )
=
= 4.00 kg
2
vmax
( 0.400 m s )2
2
m=
and
(
13.18
1.
13.19
(
(
(a)
67352_ch13.indd 406
)
2
Fg = mg = ( 4.00
2p kg ) 9.80 m s = 39.2 N
=
= 3.14 rad s
2.00 s
2p r 2p ( 0.200 m )
(a)
v
=
=
= 0.628 m s
t
From conservation
T 2 of mechanical
2.00 s 2 energy, KE + PEg + PEs
2
1
1
1
2 mv + 0 + 2 kx = 0 + 0 + 2 kA , or
(b)
k 2
v=
A − x2
m
) = ( KE + PE
f
g
)
+ PEs , we have
i
)
The speed is a maximum at the equilibrium position, x = 0.
2/9/11 1:41:37 PM
Vibrations and Waves
407
107
mv + 0 + kx = 0 + 0 + kA , or
2
13.9
(a)
2
1
2
2
1
2
(b)
At maximum
from equilibrium, all of the energy is in the form of elastic
k displacement
v = energy,
A2giving
− x2
potential
m
2 ( 47.0 Jat) the equilibrium
The speed is a maximum
position, x = 0.
3
=
Nm
2 = 1.63 × 10
0.240 m )
(
k 2
(19.6 N m ) ( 0.040 m )2 = 0.28 m s
vmax =
A =
m position
At the equilibrium
( 0.40 kg)
(b)
When x = − 0.015 m,
(a)
(
(19.6 N m ) ⎡( 0.040 m )2 − − 0.015 m 2 ⎤ =
(
)⎦
( 0.40 kg) ⎣
v=
(c)
0.26 m s
When x = + 0.015 m,
(19.6 N m ) ⎡( 0.040 m )2 − + 0.015 m 2 ⎤ =
(
)⎦
( 0.40 kg) ⎣
v=
(d)
)
0.26 m s
If v = 12 vmax, then
(
)
k 2
1
A − x2 =
m
2
This gives A2 − x 2 =
1
4
k 2
A
m
A2, or
3 ⎛ 10.0 N m
3 ⎞
2
2
= 3.5 cm
= ( 4.0 cm )
⎜⎝ 50.0 × 10 −32 kg ⎟⎠ ⎡⎣( 0.250 m ) − ( 0.125 m ) ⎤⎦ = 3.06 m s
( Ax −= xA ) 2=
2
13.21
1.
2
(
1 2 1
kxi = (850 N m ) 6.00 × 10 −2 m
2
2
)
2
(a)
PEs,i =
(b)
Since the surface is frictionless, the total energy of the block-spring system is constant.
Thus, KE + PEs = KEi + PEs,i = 0 + 1.53 J. At the equilibrium position, PEs = 0, so the
2
kinetic energy must be KEmax = 12 mvmax
= 1.53 J, which yields
vmax =
(c)
2KEmax
=
m
= 1.53 J
2 (1.53 J )
= 1.75 m s
1.00 kg
At x = xi 2 = 3.00 cm, the elastic potential energy is PEs = 12 kx 2, and the conservation of
energy gives KE + PEs = E, or 12 mv 2 + 12 kx 2 = E and
(
13.22
1.
13.23
(a)
The angle of the crank pin is q = w t. Its
x-coordinate is x = A cosq = A cosw t, where
A is the distance from the center of the
wheel to the crank pin. The displacement of
the piston from its zero position (i.e., its
location when q = wt = p 2) is the same as
that of the crankpin, x(t) = A coswt. This is
of the correct form to describe simple harmonic
motion. Hence, one must conclude that the
motion is indeed simple harmonic.
67352_ch13.indd 407
(a)
2
␻t
A
x ⫽0
x ⫽0
13.24
)
2 (1.53 J ) − (850 N m ) 3.00 × 10 −2 m
2E − kx 2
v=
=
= 1.51 m s
⎛ 250 N1.00
−k ( − A ) ⎛ k ⎞
m ⎞ kg
0.035 m ) = 18 m s 2
amax = m
=⎜ ⎟ A=⎜
(
⎟
⎝ m⎠
m
⎝ 0.50 kg ⎠
x(t)
x(t)
The springs compress 0.80 cm when supporting an additional load of
2/9/11 1:41:41 PM
108
408
13.12
1.
13.25
Chapter 13
(a)
Vibrations and Waves
either
of the turning
points,
= ± energy
A
At the
equilibrium
position,
the xtotal
of the system is in the form of kinetic energy
2
and 12 mvmax
= E , so the maximum speed is
vmax =
(b)
13.26
1.
13.29
2E
=
m
2 ( 5.83 J )
= 5.98 m s
0.326 kg
The period of an object-spring system is T = 2p m k , so the force constant of the spring is
k=
(c)
2
4p 2 m 4p ( 0.326 kg )
=
= 206 N m
T2
( 0.250 s )2
At the turning points, x = ± A, the total energy of the system is in the form of elastic potential energy, or E = 12 kA2 , giving the amplitude as
2E
2 ( 5.83 J )
A =0.52 rad
=
= 0.238 m
0.52
k
206
N rad
m = 2.1 × 10 −2 s = 21 × 10 −3 s = 21 ms
t=
=
w1
8.0p rad s
1
(a)
f = =
= 1.89 Hz
0.528issfound from
The spring Tconstant
(b)
The period of oscillation of an object-spring2 system is T = 2p m k
F
mg ( 0.010 kg ) 9.80 m s
k= s =
=
= 2.5 N m
x
x
3.9 × 10 −2 m
(
)
(
)
When the object attached to the spring has mass m = 25 g, the period of oscillation is
( ) cos t k m
0.025 kg
m
= 2p
T = 2p
⎛ 250 N =m ⎞0.63 ⎡s
250 N m ⎤
2
= − 5.00k× 10 −2 m 2.5
N m ⎟ cos ⎢( 0.500 s )
⎥ = − 4.59 m s
⎜⎝ 0.500
kg ⎠
0.500 kg ⎦
⎣
(a) Your calculator must be in radians mode for part (a) of this problem.
Note:
(
13.30
13.31
1.
411
(a)
)
The angular frequency of this oscillation is w = k m and the displacement at time t is
x = A coswt . At t = 3.50 s, the spring force will be F = −kx = −kA cos(wt), or
⎡⎛ 5.00 N m ⎞
⎤
N⎞
⎛
3.50 s )⎥ = −11.0 N,
F = − ⎜ 5.00
3.00 m ) cos ⎢⎜
(
(
⎟
⎟
⎝
m⎠
⎢⎣⎝ 2.00 kg ⎠
⎦⎥
or
(b)
F = 11.0 N directed to the left
The time required for one complete oscillation is T = 2p w = 2p m k . Hence, the number
of oscillations made in 3.50 s is
N=
13.32
1.
13.35
L
Δt
Δt
3.50 s0.994
5.002 N m
=
= 50.881
= C =
= 1.001
T
2.00
7 kg
2p m k LT2p 0.992
2 written as
For a system executing simple harmonic motion, the total energy may
g Tbe
From T = 2π L g , the length of a pendulum with period T is L =
.
4p 2
2
9.8 m s 2 (1.0 s )
(a) On Earth, with T = 1.0 s , L =
= 0.25 m = 25 cm
4p 2
(
)
(3.7 m s ) (1.0 s)
2
(b)
If T = 1.0 s on Mars, L =
4p 2
2
= 0.094 m = 9.4 cm
(c) and (d) The period of an object on a spring is T = 2p m k , which is independent of the local
free-fall acceleration. Thus, the same mass will work on Earth and on Mars. This mass is
m=
67352_ch13.indd 408
kT2
4p 2
2/9/11 1:41:45 PM
, whichVibrations
is independent
of the local
and Waves
409
109
free-fall acceleration. Thus, the same mass will work on Earth and on Mars. This mass is
(d)
13.36
13.37
1.
(a)
(a)
(e)
(b)
2
At the equilibrium
position,
N m ) (PE
1.0s s=) 0, so the block has kinetic energy KE = E = 0.125 J and
k T 2 (10
m
=
=
speed
m = 0.25 kg
2
4p 2 = 2p 4p59.6
= 37.5 s
1.67 m s2
The height of the tower
is almost
2 ( 0.125
J ) the same as the length of the pendulum. From
=
= 1.00 m s
T = period
2p L ofg ,the
wependulum
obtain
The
0.250 kgis T = 2p L g . Thus, on the Moon where the free-fall
acceleration is smaller, the period will be longer and the clock will run slow .
If the surface was rough, the block would spend energy overcoming a retarding friction
force as it moved toward the equilibrium position, causing it to arrive at that position with
The ratio of the pendulum’s period on the Moon to that on Earth is
TMoon 2p L gMoon
=
=
TEarth
2p L gEarth
gEarth
gMoon
Hence, the pendulum of the clock on Earth makes gEarth gMoon “ticks” while the clock on
the Moon is making 1.00 “tick.” After the Earth clock has ticked off 24.0 h and again reads
12:00 midnight, the Moon clock will have ticked off
( 24.0 h )
gMoon
1.63 m s2
⎛ 60 min ⎞
= ( 24.0 h )
= 9.79 h = 9 h + ( 0.79 h ) ⎜
= 9 h + 47 min
⎝ 1 h ⎟⎠
gEarth
9.80 m s 2
and will read 2 9 : 47 AM .
2
⎛ 1.25 s ⎞
⎛ T ⎞
I = mgd ⎜
= ( 0.238 kg ) 9.80 m s 2 ( 0.180 m ) ⎜
= 1.66 × 10 −2 kg ⋅ m 2
⎟
⎟
⎝ 2p ⎠
⎝ 2p ⎠
The period is the time for one complete oscillation. Hence,
(
13.39
1.
(a)
T=
(b)
416
13.40
1.
13.41
2.00 min ⎛ 60 s ⎞ 120 s
⎜⎝
⎟=
82
1 min ⎠ 82.0
)
or
T = 1.46 s
The period of oscillation of a simple pendulum is T = 2p C g , so the local acceleration of
gravity must be
Chapter 13
g=
4p 2 C 4p 2 ( 0.520 m )
=
= 9.59 m s2
T2
(120 s 82.0 )2
The apparent free-fall acceleration is the vector sum of the actual free-fall acceleration and the
(a)
Theofdistance
from the
bottom of aTo
trough
to the
top an object that is hanging from a vertinegative
the elevator’s
acceleration.
see this,
consider
of
a
crest
is
twice
the
amplitude
of
the
wave.
Thus,
cal string in the elevator and appears to be at rest to the elevator passengers. These passengers
8.26 cminand
A = 4.13
cmnegative
.
believe2A
the= tension
the string
is the
of the object’s weight, or
8.26 cm
(b)
The horizontal distance from a crest to a
trough is a half wavelength. Hence,
l 2 = 5.20 cm and l = 10.4 cm
(c)
13.42
1.
13.43
1
1
=
= 5.56 × 10 −2 s
f 18.0 s −1
The wave speed is
−1
v = l f = (10.4 cm⎛)(18.0
⎞ s ) = 187 cm s = 1.87 m s
⎜⎝ t ⎟⎠
(a) The amplitude is the magnitude of the maximum
The speed
of the wave
is equilibrium
displacement
from
v=
67352_ch13.indd 409
FIGURE P13.18
The period is
T=
(d)
5.20 cm
Δx 425 cm
=
= 42.5 cm s
Δt
10.0 s
2/9/11 1:41:49 PM
110
410
13.20
13.44
13.45
1.
Chapter 13
Vibrations and Waves
and thekfrequency is number of vibrations occurring each second, or f = 40.0 vib 30.0 s.
v=
m
v
42.5 cm s
( 42.5 cm s )(30.0 s ) = 31.9 cm
Thus, l = =
=
5.0
m
s
f 40.0 vib
= 30.0 s = 0.2540.0
Hz vib
20 m
(a)
(a)
When
the boat
is at rest infor
theawater,
speed
of the
relative to
theitsboat
is the same
The speed
of propagation
wave the
is the
product
ofwave
its frequency
and
wavelength,
as
the
speed
of
the
wave
relative
to
the
water,
v = l f . Thus, the frequency must be
f=
418
417
(b) 13
The
Chapter
v 3.00 × 108 m s
=
= 5.45 × 1014 Hz
l
5.50 × 10 −7 m
period is
T=
1
1
=
= 1.83 × 10 −15 s
f 5.45 × 1014 Hz
13.46
13.49
1.
From F
v=
=
m
13.50
13.53
1.
The mass per unit length of the wire is
(a) The mass per unit length is
1 350 N
= 5.20 × 10 2 m s
5.00 × 10 −3 kg m
m=
m 0.0600 kg
=
= 1.20 × 10 −2 kg m
5.00 m
L
From v = F m , the required tension in the string is
F = v 2 m = ( 50.0 m s ) (1.20 × 10 −2 kg m ) = 30.0 N
2
13.57
1.
F
8.00 N
2
4n 2 ML
4n 2sL2 ⎞
=
25.8
⎞⎛m
⎛ ⎞ ⎛ 2nL ⎞ =⎛ M
m
1.20⎜⎝ × 10⎟⎠ −2⎜⎝ kg ⎟⎠m = ⎜⎝ ⎟⎠ ⎜ 2 ⎟ =
L ⎝ t ⎠
t2
t
(b)
v=
(a)
The tension in the string is F = mg = ( 3.00 kg ) ( 9.80 m s 2 ) = 29.4 N. Then, from
v = F m , the mass per unit length is
m=
(b)
F
29.4 N
=
= 5.10 × 10 −2 kg m
2
v
( 24.0 m s )2
When m = 2.00 kg, the tension is
F = mg = ( 2.00 kg ) ( 9.80 m s 2 ) = 19.6 N
and the speed of transverse waves in the string is
F
19.6 N 9
=
2.70 × 10 Pa= 19.6 m s
= × 10 −2 kg3 m
= 586 m s
m
5.10
7.86 × 10 kg m 3
If the tension in the wire is F, the tensile stress is Stress = F A, so the speed of transverse waves
in
theperiod
wire may
written asis T = 2p L g , so the length of the string is
The
of thebependulum
v=
13.58
1.
13.59
2
gT 2 ( 9.80 m s ) ( 2.00 s )
=
= 0.993 m
4p 2
4p 2
2
L=
The mass per unit length of the string is then
67352_ch13.indd 410
2/9/11 1:41:52 PM
=
(9.80
m s 2 ) ( 2.00 s )
2
4p 2
= 0.993 m
Vibrations and Waves
The mass per unit length of the string is then
13.25
(a)
411
111
At the
m equilibrium
0.060 0 kgposition, the total
kgenergy of the system is in the form of kinetic energy
mand
= mv
= 2 = E , so the
= 6.04
× 10 −2 speed is
maximum
L max
0.993 m
m
When the pendulum is vertical
and stationary,
the tension in the string is
2 ( 5.83
J)
=
= 5.98 m s
0.326 kg
F = M ball g = ( 5.00 kg ) 9.80 m s 2 = 49.0 N
(b) The period of an object-spring system is
and the speed of transverse waves in it is
F2 = F1 = F, the speed of waves in the second string is
(
v=
13.60
13.61
1.
If
(a)
)
F
49.0 N
⎛ F ⎞= 28.5 m s
=
2F
m = 6.04 ×=10 −22 ⎜kg m ⎟ = 2 v1 = 2 ( 5.00 m s ) = 7.07 m s
m1
⎝ m1 ⎠
Constructive interference produces the maximum amplitude
Am′ ax = A1 + A2 = 0.30 m + 0.20 m = 0.50 m
(b)
Destructive interference produces the minimum amplitude
x = 0.23 m, the speed is
Amin
m = 0.10 m
′ = A1 − A2 = 0.30 m − 0.20
(
) ( 0.25 m )2 − ( 0.23 m )2 = 0.12 m s
13.63
1.
(a)
The period of a vibrating object-spring system is T = 2p w = 2p m k , so the spring
constant is
k=
(b)
2
4p 2 m 4p ( 2.00 kg )
=
= 219 N m
2
T
( 0.600 s )2
If T = 1.05 s for mass m2, this mass is
kT 2 ( 219 N m ) (1.05 s )
=
= 6.12 kg
4p 2
4p 2
2
422
13.64
1.
13.67
Chapter 13
m2 =
(a)
WhenPEthe
gun is fired, the energy initially stored as elastic potential energy in the spring
Choosing
g = 0 at the initial height of the 3.00-kg object, conservation of mechanical energy
is transformed into kinetic energy of the bullet.
Assuming no loss of energy, we have
gives KE + PEg + PEs = KE + PEg + PEs , or 12 mv 2 + mg ( −x ) + 12 kx 2 = 0, where v is the
(
) (
f
)
i
speed of the object after falling distance x.
(a)
When v = 0, the non-zero solution to the energy equation from above gives
2
1
2 kx m ax = mgx m ax , or
k=
(b)
2
2 mg 2 ( 3.00 kg ) ( 9.80 m s )
=
= 588 N m
x m ax
0.100 m
When x = 5.00 cm = 0.050 0 m, the energy equation gives v = 2gx − kx 2 m , or
v = 2 ( 9.80 m s 2 ) ( 0.050 0 m ) −
13.68
13.69
1.
(588
N m ) ( 0.050 0 m )
= 0.700 m s
3.00 kg
2
(a) We apply conservation of mechanical energy from just after the collision until the
The maximum acceleration of the oscillating system is
block comes to rest. Conservation of energy gives ( KE + PEs ) f = ( KE + PEs )i or
am ax = w 2 A = ( 2p f ) A
2
The friction force,
67352_ch13.indd 411
2/9/11 1:41:56 PM
112
412
Chapter 13
am ax = w 2 A = ( 2p f ) A
2
(b)
The frequency
motion
is
The friction
force, fs ,ofacting
between
the two blocks must be
capable of accelerating block B at this rate. When block B is
1
on the verge of slipping,
fs ==4.0
m s nHz
= m s mg = mam ax
=
( fs )sm−1ax == 4.0
0.25
s
and we must have
(c) As discussed above, the amplitude of the motion is A = 5.2 cm
2
am ax = ( 2p f ) A = m s g
Thus, A =
424
1.
13.73
Chapter 13
→
n
B
→
fs
( 0.600 ) ( 9.80 m s2 )
ms g
=
= 6.62 × 10 −2 m = 6.62 cm
2
( 2p f )2
2p
1.50
Hz
)]
[ (
→
→
Fg ⫽ m g
Newton’s law of gravitation is
GMm
⎛4
⎞
F = − 2 , where M = r ⎜ p r 3 ⎟
⎠
⎝3
r
⎛4
⎞
F= − ⎜ p r Gm ⎟ r
⎝3
⎠
Thus,
m
M
F
r
RE
which is of Hooke’s law form, F = −k r , with
4
p r Gm ⎞
⎛ 15.0 mm ⎞
3 ⎛
⎜⎝
⎟⎠ vinner,m ax = ⎜⎝ 3.00 mm ⎟⎠ ( 0.25 cm s ) = 1.3 cm s
The inner tip of the wing is attached to the end of the spring and always moves with the same
k of the
500vibrating
N m spring. Thus, its maximum speed is
speed
as =
the end
(a) w
=
= 15.8 rad s
m
2.00 kg
k=
13.74
13.75
1.
(b)
Apply Newton’s second law to the block while the elevator is accelerating:
ΣFy = Fs − mg = may
With Fs = kx and a y = g 3 , this gives kx = m ( g + g 3), or
x=
13.76
67352_ch13.indd 412
(a)
2
4 mg 4 ( 2.00 kg ) ( 9.80 m s )
=
= 5.23 × 10 −2 m = 5.23 cm
3k
3 ( 500 N m )
Note that as the spring passes through the vertical position, the object is moving in a circular arc of radius
2/9/11 1:41:59 PM
ࢠࡸળ ‫ܕ‬չ
PROBLEM SOLUTIONS
14.1
1.
The Celsius temperature is TC =
air is
5
9
(T
F
− 32 ) =
5 114 − 32 ) =
9(
45.6°C and the speed of sound in the
. Considering the Earth’s crust to consist of a
very viscous fluid, our estimate of the average bulk modulus of the material in Earth’s crust is
T
45.6
v = ( 331 m s ) 1+ C = ( 331 m s ) 1+
= 358 m s
2
273 7 × 10 3 m s =273
(
)(
) 1× 1011 Pa
14.2
14.3
1.
The speed of longitudinal waves in a fluid is v = B r
(a) We ignore the time required for the lightning flash to arrive. Then, the distance to the lightning stroke is
d ≈ vsound ⋅ Δt = ( 343 m s )(16.2 s ) = 5.56 × 10 3 m = 5.56 km
No. Since vlight >> vsound, the time required for the flash of light to reach the observer is
negligible in comparison to the time required for the sound to arrive, and knowledge of the
2 (150
m ) speed of light is not needed.
actual value
of the
=
= 0.196 s
1 533 m s
The speed of sound in seawater at 25°C is 1 533 m s
(a) Because the speed of sound in air is vair = 343 m s while its speed in the steel rail is
(b)
14.4
14.5
1.
vsteel = 5 950 m s, the pulse traveling in the steel rail arrives first .
(b)
The difference in times when the two pulses reach the microphone at the opposite end of
the rail is
⎞
⎛
L
L
1
1
−2
2
Δt =( 9.80−m s 2 )=(1.95
−
(8.50s )m
)⎜
⎟ = 2.34 × 10 s = 23.4 ms
= vair vsteel
=⎝ 343
18.6 m
m s 5 950 m s ⎠
2
14.7
1.
At T = 27.0°C, the speed of sound in air is
v27 = ( 331 m s ) 1+
TC
27.0
= 347 m s
= ( 331 m s ) 1+
273
273
and at T = 10.0°C, the speed is
v10 = ( 331 m s ) 1+
TC
10.0
= 337 m s
= ( 331 m s ) 1+
273
273
Since v = λ f , the change in wavelength will be
Δl =
14.8
1.
14.9
v10 v27 v10 − v27 (337 − 347) m s
−
=
=
= −2.5 × 10 −3 m = −2.5 mm
f
f
f
4.00 × 10 3 Hz
Sound
433
At T = 27°C, the speed of sound in air is
Since the sound had to travel the distance between the hikers and the mountain twice, the time
required for a one-way trip was 1.50 s. The distance the sound traveled to the mountain was
d = ( 343 m s )(1.50 s ) = 515 m
14.10
The sound power incident on the eardrum is P = IA, where I is the intensity of the sound and
113
A = 5.0 × 10 −5 m 2 is the area of the eardrum.
(a)
67352_ch14.indd 426
At the threshold of hearing, I = 1.0 × 10 −12 W m 2
2/9/11 1:43:20 PM
114
432
14.6
14.11
1.
Chapter 14
(
) (5.0 × 10
−5
m 2 ) = 5.0 × 10 −5 W
T=
10.0°C,
At a temperature
of of
thekm
speed
sound
air is b = 50 dB) is
(a)
The intensity
sound
at 10
fromofthe
hornin(where
−12
W m 2 )10 5.0
= 1.0 × 10 −7 W m 2
I = I 0 10 β 10 = (1.0
TC × 10
10.0
1+
1+
= 337 m s
=
331
m
s
(
)
(
)
273
2732
Thus, from I = P 4p r , the power emitted by the source is
The elapsed time between
when the3 stone
was released
and2 when the sound
is heard is
the sum
2
1
4p r 2 I =for
4pthe
× 10 tomfall
× 10 −7h W
4p t× 10
W = for
1.3sound
× 10 2 W
(10stone
) (1.0
of the time Pt1 =required
distance
andmthe) =time
required
to
travel
2
distance h in air on the return up the well. That is, t1 + t2 = 2.00 s. The distance the stone falls,
(b) At r = 50 m, the intensity of the sound
gt 2 will be
starting from rest, in time t1 is
h= 1
2
P
1.3 × 10 2 W
I=
=
=
4.1×
10 −3 W m 2
2
4p r 2 4p ( 50 m )
and the sound level is
⎛ I ⎞
⎛ 4.1× 10 −3 W m 2 ⎞
9
= 10
log ( 4.1
b = 10 log ⎜ ⎟ = 10 log ⎜
96 dBgreater than
1 000
and the answer
to part
(a) ×is10
) = times
I0 ⎠
1.0 × 10 −12 W m 2 ⎟⎠
⎝
⎝
this, explaining why some airport employees must wear hearing protection equipment.
1.
14.12
14.15
100 W
P
2
I = level
= b = 10 log ( I2 I=0 ),7.96
× 10I 0−2 =W
(a)
1.00m× 10 −12 W m 2.
The decibel
where
2
4p r
4p (10.0 m )
(a) If b = 100 dB, then log ( I I 0 ) = 10, giving I = 1010 I 0 = 1.00 × 10 −2 W m 2 .
I
(b) b = 10 log
I
0
(b) If all three toadfi
sh sound at the same time, the total intensity of the sound produced is
I ′ = 3I = 3.00 × 10 −2 W m 2, and the decibel level is
⎛ 3.00 × 10 −2 W m 2 ⎞
b ′ = 10 log ⎜
⎝ 1.00 × 10 −12 W m 2 ⎟⎠
= 10 log ⎡⎣( 3.00 )(1010 ) ⎤⎦ = 10 [ log ( 3.00 ) + 10 ] = 105 dB
14.16
14.21
1.
436
(a) sound
The intensity
the musical
(b (=I 80
is I m usic = I 0 10 β 10
The
level for of
intensity
I is bsound
= 10 log
I 0 )dB)
. Therefore,
Chapter b
14
2
⎛ I
I0 ⎞
⎛I ⎞
⎛I ⎞
⎛ I2 ⎞
− b1 = 10 log ⎜ 2 ⎟ − 10 log ⎜ 1 ⎟ = 10 log ⎜ 2 ⋅
⎟ = 10 log ⎜ ⎟
⎝ I1 ⎠
⎝ I0 ⎠
⎝ I0 ⎠
⎝ I 0 I1 ⎠
Since I = P 4p r 2 = ( P 4p ) r 2 , the ratio of intensities is
continued on next page
I 2 ⎛ P 4p ⎞ ⎛ r12 ⎞ r12
=
=
I1 ⎜⎝ r22 ⎟⎠ ⎜⎝ P 4p ⎟⎠ r22
2
2
r 2 (100 m ) + ( 200 m )
2
= C2 =
2
⎛ r1 m
⎛= r1 5⎞
⎛r ⎞
⎞ )2
r
100
(
A
Thus, b 2 − b1 = 10 log ⎜ 2 ⎟ = 10 log ⎜ ⎟ = 20 log ⎜ 1 ⎟
⎝ r2 ⎠
⎝ r2 ⎠
⎝ r2 ⎠
1.
14.23
14.22
( P 4p
)observed frequency is
P source,
= 0 )source
hears is
a moving
the
When
a stationary
observer
( vO the
=
The
intensity
at distance
r from
I=
2
2
r
4p r
⎛ v + vO ⎞
⎛ v ⎞
= fS ⎜
(a) fO = fS ⎜
⎝ v − vS ⎟⎠
⎝ v − vS ⎟⎠
(a)
When the train is approaching, vS = + 40.0 m s, and
(f )
O
approach
⎞
⎛
343 m s
= 362 Hz
= ( 320 Hz ) ⎜
⎝ 343 m s − 40.0 m s ⎟⎠
After the train passes and is receding, vS = − 40.0 m s
67352_ch14.indd 432
2/9/11 1:43:32 PM
⎞
⎛
⎜⎝ 343 m s − 40.0 m s ⎟⎠ = 362 Hz
14.9
Sound
433
115
Since After
the sound
had passes
to travel
theisdistance
between
the hikers
vS = − 40.0
m s, and the mountain twice, the time
the train
and
receding,
required for a one-way trip was 1.50 s. The distance the sound traveled to the mountain was
⎤
⎡
ms
( fO )recede = (320 Hz ) ⎢ 343 m 343
⎥ = 287 Hz
s − ( − 40.0 m s ) ⎦
⎣
Thus, the frequency shift that occurs as the train passes is
ΔfO = ( fO )recede − ( fO )approach = −75 Hz, or it is a 75 Hz drop
(b)
to decrease
the denominator in Equation
As the train approaches, the observed wavelength
is
14.12, and thereby increase the calculated observed frequency.
v
343 m s
l =⎛
= 0.948
⎡ 1 533m m s + ( −3.00 m s ) ⎤
⎞ =
3
f
=
5.27
10 3 Hz ) ⎢
(
)
⎥ = 5.30 × 10 Hz
⎜⎝ O approach
⎟⎠ ( 362× Hz
1
533
m
s
−
+11.0
m
s
(
)
⎦
⎣
(a) From Equation 14.12 in the textbook,
With the train approaching the stationary observer (vO = 0) at speed vt , the source velocity is
vS = + vt and the observed frequency is
(h)
14.24
14.25
1.
⎛ 343 m s ⎞
465 Hz = fS ⎜
⎟
⎝ 343 m s − vt ⎠
[1]
As the train recedes, the source velocity is vS = − vt and the observed frequency is
⎛ 343 m s ⎞
441 Hz = fS ⎜
⎟
⎝ 343 m s + vt ⎠
[2]
Dividing Equation [1] by [2] gives
465 343 m s + vt
=
441 343 m s − vt
and solving for the speed of the train yields vt = 9.09 m s .
14.26
1.
14.27
= 32.0 m s
(a) Since the observer hears a reduced frequency,⎛ the
v + source
vO ⎞ and observer are getting farther
= fS ⎜
. Since each train moves toward the
Both source
and observer
are inismotion,
apart. Hence,
the cyclist
behindso
thefOcar
⎝ v − vS ⎟⎠
other, vO > 0 and vS > 0. The speed of the source (train 2) is
vS = 90.0
km ⎛ 1 000 m ⎞ ⎛ 1 h ⎞
⎜
⎟⎜
⎟ = 25.0 m s
h ⎝ 1 km ⎠ ⎝ 3 600 s ⎠
and that of the observer (train 1) is vO = 130 km h = 36.1 m s. Thus, the observed frequency is
⎛ 343 m s + 36.1
⎛
fO = ( 500 Hz ) ⎜
⎝ 343 m⎜ s − 25.0
⎝
14.28
14.29
1.
m s⎞
⎞ ⎟ = 596 Hz ⎛
⎞
1 500
m
⎟ =s(⎠2 000 029 Hz ) ⎜ 1 500 − 0.021 6 ⎟ = 2 000 058 Hz
⎠
⎝
⎠
(a)
For a source receding from a stationary observer,
⎛
v
fO = fS ⎜
⎜⎝ v − − vS
(
)
⎞
⎛ v ⎞
⎟ = fS ⎜
⎟
⎟⎠
⎝ v + vS ⎠
Thus, the speed the falling tuning fork must reach is
⎛ f
⎞
⎛ 512 Hz ⎞
vS = v ⎜ S − 1⎟ = ( 343 m s ) ⎜
− 1 = 19.1 m s
⎝ 485 Hz ⎟⎠
f
⎝ O
⎠
The distance it has fallen from rest before reaching this speed is
67352_ch14.indd 433
2/9/11 1:43:35 PM
434
116
⎛
⎜
⎜⎝
Chapter 14
(
)
⎞
⎛ v ⎞
⎟ = fS ⎜
⎟
⎟⎠
⎝ v + vS ⎠
Thus, the speed the falling tuning fork must reach is
−11
2
If b 2 − b1 = 30.0
⎛ f dB and
⎞ I1 = 3.0 × 10
Hzm , ⎞then
⎛ 512W
vS = v ⎜ S − 1⎟ = ( 343 m s ) ⎜
− 1⎟ = 19.1 m s
⎝
Hz ⎠
= 3.0 × 10 −8 W m 2
( ⎝ fO ⎠
) × 103.00 485
14.14
The distance it has fallen from rest before reaching this speed is
(a) The decibel level, b , of a sound is given b = 10 log I I 0
vS2 − 0 (19.1 m s ) − 0
=
= 18.6 m
2 ay
2 ( 9.80 m s 2 )
2
Δy1 =
Sound
439
The time required for the 485 Hz sound to reach the observer is
Δy1
18.6 m
continued on next
t = page
=
= 0.054 2 s
343 m s
v
During this time the fork falls an additional distance
Δy2 = vS t +
1 2
1
2
a y t = (19.1 m s ) ( 0.054 2 s ) + ( 9.80 m s 2 ) ( 0.054 2 s ) = 1.05 m
2
2
The total distance fallen before the 485 Hz sound reaches the observer is
440
1.
14.31
Δy = Δy1 + Δy2 = 18.6 m + 1.05 m = 19.7 m
Chapter 14
(a)
For a plane traveling at Mach 3.00,
the half-angle of the conical wave
front is
x
t ⫽0
⎛v
⎞
⎛ 1 ⎞
q = sin −1 ⎜ sound ⎟ = sin −1 ⎜
⎝ 3.00 ⎟⎠
v
⎝ plane ⎠
The distance the plane has moved
when the wave front reaches the
observer is x = h tanq , or
x=
u
t ⫽?
h
h
Observer hears
Observer
the “boom”
Observer
ver
Obser
a
20.0 km, or
= 56.6 km
tan ⎡⎣sin −1 (1 3.00 ) ⎤⎦
u
b
FIGURE P14.13
The time required for the plane to travel this distance, and hence the time when the shock
wave reaches the observer, is
t=
(b)
1.
14.33
14.32
x
vplane
=(
x
56.6 × 10 3 m
=, the observed frequency
= 56.3 sis
)
3.00vsound 3.00 ( 335 m s )
⎛ v+0 ⎞
⎛ 2⎞
= ( 5.00
kHz ) ⎜ ⎟above.
fO = (is5.00
kHz
= 3.33 kHz
) ⎜farther along
The plane
56.6
km
as computed
⎟
⎝ 3⎠
⎝ v + v 2⎠
x
1.25 m x
The wavelength of the waves generated
by
the
speakers
is
The general expression for the observed frequency of a sound when the source and/or the observer
are in motion is
v 343 m s
l= =
Speaker 1
Speaker 2
v + v= 0.429 m
ff = 800
fS Hz O
O
v − vS
For the waves from the two speakers to interfere destructively at some point, the difference in the path
lengths from the speakers to that point must be an odd multiple of a half-wavelength. Thus, along the
line connecting the two speakers, destructive interference (and minima in amplitude) occur where
(1.25 m − x ) − x = ( 2n + 1)
67352_ch14.indd 434
l
2
where n is any integer
2/9/11 1:43:39 PM
=
343 m s
= 0.429 m
800 Hz
For the waves from the two speakers to interfere destructively at some point, the difference in the path
117
Sound
435
lengths from the speakers to that point must be an odd multiple of a half-wavelength. Thus,
along the
line connecting the two speakers, destructive interference (and minima in amplitude) occur where
(b)
Sound
If 5 trumpets are sounded together,
the total intensity of the sound is
l
where n is any integer
(1.25 m − x ) − x = ( 2n + 1)
2
441
continued on next page
l
⎛ 2n + 1 ⎞
or where x = 0.625 m − ( 2n + 1) = 0.625 m − ⎜
( 0.429 m )
⎝ 4 ⎟⎠
4
This gives:
n=0
⇒ x = 0.518 m
n = −1 ⇒ x = 0.732 m
n = 1 ⇒ x = 0.303 m
n = −2
⇒ x = 0.947 m
n=2
n = −3
⇒ x = 1.16 m
⇒ x = 0.089 m
Thus, minima occur at distances of
0.089 m, 0.303 m, 0.518 m, 0.732 m, 0.947 m, and 1.16 m
from either speaker.
14.34
14.35
1.
.
v 343 m s
The wavelength of the sound emitted by the speaker is l = =
= 0.454 m, and a half
f
756 Hz
At
point
D,
the
distance
of
the
ship
from
point
A
is
A
B
wavelength is l 2
800 m
= d 2 + (800 m ) =
2
( 600 m )2 + (800 m )2 = 1 000 m
d
Since destructive interference occurs for the first time when
the ship reaches D, it is necessary that − d = l 2, or
as
l = 2 ( − d ) = 2 (1 000 m2 − 600 m ) = −3 800 m
(
) (6.143 × 10 kg m ) = 823.8 N
1.
14.41
D
When the string vibrates in the fifth harmonic (i.e., in five equal segments) at a frequency of
f5 = 630 Hz, we have L = 5(l5 2), or the wavelength is l5 = 2L 5. The speed of transverse waves
in the string is then
v = l5 f5 = (2L 5) f5
For the string to vibrate in three segments (i.e., third harmonic), the wavelength must be such that
L = 3(l3 2) or l3 = 2L 3. The new frequency would then be
f3 =
14.42
14.45
1.
(a)
(a)
v (2L 5) f5 3
3
=
= f5 = ( 630 Hz ) = 378 Hz
l3
5
2L 3
5
Sound
A string fixed at each end forms standing wave patterns with a node at each end and an
From the sketch at the right, notice that when d = 2.00 m,
d
integer number of loops, each loop of length l 2
5.00 m − d
L=
= 1.50 m,
2
L
and
⎛ d / 2⎞
q = sin −1 ⎜
= 41.8°
⎝ L ⎟⎠
Then evaluating the net vertical force on the lowest bit of string,
gives the tension in the string as
F=
(b)
67352_ch14.indd 435
445
→
F
q
→
F
→
mg
ΣFy = 2F cosq − mg = 0
(12.0 kg) (9.80 m s2 )
mg
=
= 78.9 N
2 cosq
2 cos ( 41.8° )
The speed of transverse waves in the string is
2/9/11 1:43:43 PM
118
436
=
Chapter 14
(12.0 kg) (9.80
m s2 )
2 cos ( 41.8° )
= 78.9 N
Since
I = Pspeed
4p r 2of transverse waves in the string is
(b) The
v=
F
=
m
78.9 N
= 2.81× 10 2 m s
0.001 00 kg m
For the pattern shown,
3 ( l 2 ) = d, so l =
2d 4.00 m
=
3
3
Thus, the frequency is
f=
14.46
1.
14.47
2
v 3 ( 2.81× 10 m s )
=
= 2.11× 10 2 Hz
l
4.00 m
.
(a)
standing wave
of 6 loops,
In the For
thirda harmonic,
the string
of length L
forms a standing wave of three loops, each
of length l 2 = L 3. The wavelength of the
wave is then
l=
(a)
2L 16.0 m
=
≈ 5.33 m
3
3
l/2
0
l/4
l
3l /4
3l/2
5l/4
The nodes in this string, fixed at each end, will occur at distances of
0, l 2 = 2.67 m, l = 5.33 m, and 3l 2 = 8.00 m from the end.
Antinodes occur halfway between each pair of adjacent nodes, or at distances of
l 4 = 1.33 m, 3l 4 = 4.00 m, and 5l 4 = 6.67 m from the end.
(b)
The linear density is
m=
m 40.0 × 10 −3 kg
=
= 5.00 × 10 −3 kg m
L
8.00 m
and
and the wave speed is
closed at the other.
F
49.0 N
v = of a= pipe with an−3 antinode= at
99.0
s end and a node at the closed end is oneThe length
the m
open
m
5.00 × 10 kg m
quarter of the wavelength of the fundamental frequency, so the length of this pipe must be
v 99.0 m s
= 18.6 Hz
Thus, the frequency
f = m =s
vsound is 343
=
= 1.72
=
5.33
mm
l
4 ( 50.0 Hz )
4 f1
The distance between adjacent nodes (and between adjacent antinodes) is one-quarter of the
circumference.
(a) The fundamental wavelength of the pipe open at both ends is l1 = 2 L = v f1. Since the
speed of sound is 331 m s at 0°C, the length of the pipe is
(c)
14.48
14.51
1.
L=
(b)
v
331 m s
=
= 0.552 m
2 f1 2 ( 300 Hz )
At T = 30°C = 303 K,
v = ( 331 m s )
TK
303
= ( 331 m s )
= 349 m s
273
273
and
f1 =
14.52
67352_ch14.indd 436
(a)
v
v
349 m s
=
=
= 316 Hz
l1 2 L 2 ( 0.552 m )
To form a standing wave in the tunnel, open at both ends, one must have an antinode at
each end, a node at the middle of the tunnel, and the length of the tunnel must be equal to
an integral number of half-wavelengths [
2/9/11 1:43:46 PM
(
448
)
TK
303
= ( 331 m s )
= 349 m s
273
273
Sound
437
119
and
The yellow
v submarine
v
349 m s
Chapter 14
=
=
= 316 Hz
f1 =
l1 2 L 2 ( 0.552 m )
(b)
14.52
1.
14.53
(a) To form a standing wave in the tunnel, open at both ends, one must have an antinode at
Hearing would be best at the fundamental resonance, so l = 4L = 4 ( 2.8 cm )
each end, a node at the middle of the tunnel, and the length of the tunnel must be equal to
an integral number of half-wavelengths [
v
343 m
⎛ s ⎛ 100 cm⎞ ⎞ = 3.1× 10 3 Hz = 3.1 kHz
and f = =
⎜
− 1⎟ ⎟⎠= 3.98 Hz
l 4 ( 2.8 ⎜⎝cm ) ⎝ 1 m
⎠
14.54
1.
14.57
(a) commuter,
In the fundamental
mode
of a pipe
both
ends,
thethe
distance
The
stationaryresonant
relative to
the station
andopen
the at
first
train,
hears
actualbetween
source
antinodes
is
d
=
l
2
= 180 Hz from the first train. The frequency the commuter hears from the
frequency fO,1 = fS AA
second train, moving relative to the station and commuter, is given by
(
)
Sound
449
fO,2 = fS ± fbeat = 180 Hz ± 2 Hz = 178 Hz or 182 Hz
(
)
continuedThis
on next
page observer ( v = 0 ) hears the lower frequency f = 178 Hz if the second train
stationary
O
O,2
is moving away from the station vS = − vS , so fO = fS [(v + v O ) (v − vS )] gives the speed of the
receding second train as
(
)
⎛ 343 m s + 0
178 Hz = (180 Hz ) ⎜
⎜⎝ 343 m s − − vS
(
)
⎞
⎛ 343 m s + 0 ⎞
⎟ = (180 Hz ) ⎜
⎟
⎟⎠
⎝ 343 m s + vS ⎠
or
⎛ 180 Hz ⎞
343 m s + vS = ( 343 m s ) ⎜
⎝ 178 Hz ⎟⎠
and
⎡⎛ 180 Hz ⎞ ⎤
vS = ( 343 m s ) ⎢⎜
⎟ − 1⎥ = 3.85 m s
⎣⎝ 178 Hz ⎠ ⎦
so one possibility for the second train is vS = 3.85 m s away from the station .
(
)
)
The other possibility is that the second train is moving toward the station vS = + vS and the
commuter is detecting the higher of the possible frequencies fO,2 = 182 Hz . In this case,
fO = fS [(v + v O ) (v − vS )] yields
(
⎛ 343 m s + 0 ⎞
182 Hz = (180 Hz ) ⎜
⎟
⎝ 343 m s − vS ⎠
and
⎛ 180 Hz ⎞
343 m s − vS = ( 343 m s ) ⎜
⎝ 182 Hz ⎟⎠
⎡ ⎛ 180 Hz ⎞ ⎤
vS = ( 343 m s ) ⎢1− ⎜
⎟ ⎥ = 3.77 m s
⎣ ⎝ 182 Hz ⎠ ⎦
⎛ ⎞
196 ⎞
= ( 30.0ofcmthe
) ⎛⎜⎝ second
⎟ = 29.7 cm
In this case, ⎜⎝the ⎟⎠velocity
198 ⎠ train is vS = 3.77 m s toward the station .
or
14.58
14.59
1.
By
her string,
theasecond
violinist
increases
its sound
fundamental
Thus,
(a) shortening
First consider
the wall
stationary
observer
receiving
from anfrequency.
approaching
source havf1′ = f1ing
+ fvelocity
2.00frequency
Hz. Since
density
are(vboth
) Hz = 198received
andthe
refltension
ected byand
the the
walllinear
is freflect
= fS [v
− va )].
va+. The
beat = (196
identical for the two strings, the speed of transverse waves, v = F m
Now consider the wall as a stationary source emitting sound of frequency freflect to an
observer approaching at velocity va . The frequency of the echo heard by the observer is
⎛ v ⎞ ⎛ v + va ⎞
⎛ v + va ⎞
⎛ v + va ⎞
= fS ⎜
fecho = freflect ⎜
⎟ = fS ⎜
⎜
⎝ v ⎟⎠
⎝ v − va ⎟⎠ ⎝ v ⎠
⎝ v − va ⎟⎠
Thus, the beat frequency between the tuning fork and its echo is
⎛ 2va ⎞
⎛ v + va
⎞
⎛ 2 (1.33) ⎞
= ( 256 Hz ) ⎜
fbeat = fecho − fS = fS ⎜
− 1⎟ = fS ⎜
= 1.99 Hz
⎟
⎝ 343 − 1.33 ⎟⎠
⎝ v − va ⎠
⎝ v − va
⎠
continued on next page
67352_ch14.indd 437
2/9/11 1:43:49 PM
120
438
⎛ v ⎞ ⎛ v + va ⎞
⎛ v + va ⎞
⎞
⎟⎠ = fS ⎜
⎟⎠ = fS ⎜
⎜⎝
⎟
v
⎝ v − va ⎠
⎝ v − va ⎟⎠
⎛
⎜⎝
Chapter 14
Thus, the beat frequency between the tuning fork and its echo is
450
giving
⎛ 2va ⎞
⎛ v + va
⎞
⎛ 2 (1.33) ⎞
= ( 256 Hz ) ⎜
fbeat = fecho − fS = fS ⎜
− 1⎟ = fS ⎜
= 1.99 Hz
⎝ 343 − 1.33 ⎟⎠
⎝ v − va ⎟⎠
⎝ v − va
⎠
⎛ 343 m s + vcar 3 ⎞
415 Hz = ( 440 Hz ) ⎜
⎟ and
343 from
m s +the
vcarwall,
When the student moves⎝ away
⎠ va changes sign so the frequency of the echo
heard is fecho = fS [(v − va ) (v + va )]. The beat frequency is then
Chapter 14
(b)
fbeat = fS − fecho
giving
continued on next page
⎛
⎛ 2 va ⎞
v − va ⎞
= fS ⎜
= fS ⎜ 1−
⎟
⎟
v + va ⎠
⎝
⎝ v + va ⎠
v fbeat
2 fS − fbeat
va =
The receding speed needed to observe a beat frequency of 5.00 Hz is
14.60
1.
14.61
(343 m s )(5.00 Hz ) = 3.38 m s
=
va 0.118
m
2 ( 256=Hz
) − 5.00
=
0.0295
m Hz
= 2.95 cm
4
The extra sensitivity of the ear at 3 000 Hz appears as downward dimples on the curves in
At
normal
body
temperature
Figure
14.29
of the
textbook.of T = 37.0°C, the speed of sound in air is
TC
37.0
v=
= ( 331
m s )in1+
(331= m
At T =
37°C
310s )K,1+
the speed
of sound
air is
273
273
and the wavelength of sound having a frequency of f = 20 000 Hz is
v
37.0
(331 m s ) ) 1+
the temperature
in the
= (
1+ TC 273,
= 1.76
× 10 −2 m = 1.76
cmcolder room is given by
f ( 20 000 Hz )
273
2
⎡⎛
⎤
⎡⎛ 339 m s ⎞ 2 ⎤
⎞
−
1
=
273°C
(
)
⎢
⎥
⎢⎜
Thus, the diameter of the⎜eardrum is⎟ 1.76 cm .
⎟ − 1⎥ = 13.4°C
⎠
⎢⎣⎝
⎥⎦
⎢⎣⎝ 331 m s ⎠
⎥⎦
l=
14.62
14.63
1.
(a) At
We assume that the average intensity of the sound is directly proportional to the number of cars
passing each minute. If the sound level in decibels is b = 10 ⋅ log ( I I 0 ), the intensity of the sound
is I = I 0 ⋅10 b 10, so the average intensity in the afternoon, when 100 cars per minute are passing, is
I100 = I 0 ⋅1080.0 10 = (1.00 × 10 −12 W m 2 ) ⋅108.00 = 1.00 × 10 −4 W m 2
The expected average intensity at night, when only 5 cars pass per minute, is given by the ratio
I 5 I100 = 5 100 =1 20, or
I5 =
I100 1.00 × 10 −4 W m 2
=
= 5.00 × 10 −6 W m 2
20
20
and the expected sound level in decibels is
⎛I ⎞
b 5 = 10 ⋅ log ⎜ 5 ⎟
⎝ I0 ⎠
1.
14.65
14.64
(a)
−6
m−62 ⎞ W m 2 ⎞
⎛
⎞ ⎛ 5.00 × 10
⎛ 9.15W
× 10
= 67.0 dB
= ⎜10 ⋅ log
= 69.6 dB
=
10
⋅
log
−12
2 ⎟
⎜⎝ 1.00
⎟⎠ ⎜⎝ 1.00 × 10
W
× 10m−12⎠ W m 2 ⎟⎠
⎝
With
a decibel
of 103
dB,
the intensity
At
point
C, the level
distance
from
speaker
A is of the sound at 1.60 m from the speaker is
found from b = 10 ⋅ log ( I I 0 ) as
I = I 0 ⋅10 b
10
= (1.00 × 10 −12 W m 2 ) ⋅1010.3 = 1.00 × 10 −1.7 W m 2
If the speaker broadcasts equally well in all directions, the intensity (power per unit area)
at 1.60 m from the speaker is uniformly distributed over a spherical wave front of radius
r = 1.60 m centered on the speaker. Thus, the power radiated is
P = IA = I ( 4p r 2 ) = (1.00 × 10 −1.7 W m 2 ) 4p (1.60 m ) = 0.642 W
2
(b)
67352_ch14.indd 438
efficiency
2/9/11 1:43:52 PM
Sound
439
121
r = 1.60 m centered on the speaker. Thus, the power radiated is
2
The time required
sound
the2 observer
P = IA for
= I (the
4p r485
× 10to−1.7reach
W m
) =Hz(1.00
) 4p (1.60 ism )2 = 0.642 W
14.66
1.
14.69
18.6 m
= P ⎛ = 0.642
0.054 W
2 s− 15.3 m s ⎞
output
343
1 204 Hz
= 0.004 3 ⎟or= 0.43%
(b) efficiency = m s⎜ =
150mWs − 20.6 m s ⎠
Pinput ⎝ 1 533
During this time the fork falls an additional distance
The
act as a pipe closed at one end (the bottom) and open at the other (the top).
If r1 well
and rwill
2 are the distances of the two observers from the speaker, the intensities of the sound at
The
resonant
frequencies
are the odd integer multiples of the fundamental frequency, or
their locations are
I1 =
P
4p r12
I2 =
and
P
4p r22
where P is the power output of the speaker. The difference in the decibel levels for the two
observers is
2
⎛I ⎞
⎛I ⎞
⎛I ⎞
⎛ r2 ⎞
⎛r ⎞
⎛r ⎞
b1 − b 2 = 10 log ⎜ 1 ⎟ − 10 log ⎜ 2 ⎟ = 10 log ⎜ 2 ⎟ = 10 log ⎜ 22 ⎟ = 10 log ⎜ 2 ⎟ = 20 log ⎜ 2 ⎟
⎝I ⎠
⎝r ⎠
⎝r ⎠
⎝r ⎠
⎝I ⎠
⎝I ⎠
0
0
1
1
1
1
Since b1 = 80 dB and b 2 = 60 dB, we find that 80 − 60 = 20 log(r2 r1 ). This yields
log(r2 r1 ) = 1.0
We also know that
and
r2 r1 = 10
Then, Equation [2] yields
[1]
[2]
11r1 = 36.0 m
r1 = 36.0 m 11 ≈ 3.3 m
or
r2 = above
36.0 mthe
− rsidewalk.
= 36.0 m − 3.3 m = 32.7 m
1
The speeds of the two types of waves in the rod are
vlong =
Y
=
r
Y
Y ( A⋅ L)
=
and vtrans =
mV
m
F
=
m
F
=
m L
(
F⋅L
m
)
Y ( A⋅ L)
⎛ F ⋅ L⎛⎞
− v − v ⎞ 2 fSisvO
= 64 ⎜
⎟⎠ , or the requiredO tension
⎜
⎟=
⎝
m ⎜
m
⎟⎠
v
v
⎝
2
10
2 ⎡
−2
⎤
6.80
×
10
N
m
p
0.200
×
10
m
)⎣ (
)⎦
Y⋅A (
= 1.34 × 10 4 N
F=
⎛ =fbeat ⎞
⎤
⎡
8.30
Hz
64
64
v=⎢
=⎜
so
⎥ ( 343 m s ) = 1.95 m s
730 Hz ) ⎦
⎝ 2 f ⎟⎠
⎣ 2 (for
From the definingS equation
the decibel level,
On the weekend, there are one-fourth as many cars passing per minute as on a weekday. Thus, the
intensity, I 2 , of the sound on the weekend is one-fourth of that, I1, on a weekday. The difference in
the decibel levels is therefore
Thus, if vlong = 8 vtrans , we have
14.72
1.
14.75
r2 = 10r1
r1 + r2 = 36.0 m
Substituting Equation [1] into [2] gives:
1.
14.71
or
⎛I ⎞
⎛I ⎞
⎛I ⎞
b1 − b 2 = 10 log ⎜ 1 ⎟ − 10 log ⎜ 2 ⎟ = 10 log ⎜ 1 ⎟ = 10 log(4) = 6 dB
⎝ I2 ⎠
⎝ Io ⎠
⎝ Io ⎠
so, b 2 = b1 − 6 dB = 70 dB − 6 dB = 64 dB
67352_ch14.indd 439
2/9/11 1:43:55 PM
5ࢠ ࢷ̛ԯ˕ࢷ̛ࢠ
PROBLEM SOLUTIONS
15.1
1.1
(a)
When the balls are an equilibrium distance apart, the tension in the string equals the magnitude of the repulsive electric force between the balls. Thus,
F=
15.2
15.3
1.
15.4
15.5
1.
9.
15.7
1.
2 (8.99 × 10 9 N ⋅ m 2 C2 )
2ke
= 2.36 × 10 −5 C = 23.6 mC
Q on particle B. By Newton’s third law, particle B will then
Particle A exerts a force toward the Q
right
(a) From Coulomb’s law, F = ke 1 2 2 , we have
exert a of equal magnitude force toward
r the left on particle A. The ratio of the final magnitude of
the force to the original magnitude of the force is
( 7.50 × 10 −9 C) ( 4.20 × 10 −9 C) = 8.74 × 10 −8 N
F = (8.99 × 10 9 N ⋅ m 2 C2 )
(1.80 m )2 −9
−9
(
) (32.0 × 10 C) (58.0 × 10 C) = 1.25 × 10−2 m = 1.25 cm
(b) Since these are like charges (both positive),
m ⎞ is repulsive .
⎛ the force
0.180 N − ( 7.50 × 10 −3 kg ) ⎜ 9.80 2 ⎟
⎝
s ⎠
(a) The gravitational force exerted on the upper sphere by the lower one is
(a) The spherically symmetric charge distributions behave as if all charge was located at the
negligible in comparison to the gravitational force exerted by the Earth and
centers of the spheres. Therefore, the magnitude of the attractive force is
the downward electrical force exerted by the lower sphere. Therefore,
12 × 10 −9 C ) (18 × 10 −9 C )
ke q1 q2
9
2
2 (
=
8.99
×
10
N
⋅
m
C
= 2.2 × 10 −5 N
(
)
r2
( 0.30 m )2
(b)
the spheres
by N
a conducting
Φ E = When
EA = ( 5.00
N C ) ( are
4.00connected
m 2 ) = 20.0
⋅ m 2 C . wire, the net charge
−9
qnet = q1 + q2 = −6.0 × 10 C will divide equally between the two identical spheres.
theq force
ChoiceThus,
(b). If
= 60°isinnow
Quick Quiz 15.7 above, then Φ E = EA cosq which yields
Φ E = ( 5.00 N C ) ( 4.00 m2 2 ) cos ( 60° ) = 10.0 N ⋅ m 2 C.
2
2
−6.0 × 10 −9 C )
ke ( qnet 2 )
⎛
9 N⋅m ⎞ (
F=
= 8.99 × 10
2
2
⎟⎠ through
Choice (d). Gauss’s rlaw
states⎜⎝ that the electric
C2 flux
4 ( 0.30any
m )closed surface is equal to the net
enclosed charge divided by the permittivity of free space. For the surface shown in Figure 15.28,
465
−7 Q = −6 C,
the netorenclosed
charge
which gives Φ E = Q ∈0Electric Forces and Electric Fields
F = 9.0
× 10is
N (repulsion)
In the new equilibrium position,
Fs
k=
or
(8.99 × 10
k=
d
=
ΣFy = Fs − Fe = kd − Fe = 0.
Fs = kd
q1
ke q1 q2 r 2 ke q1 q2
=
d
d ⋅ r2
Thus,
9
k = 49.3 N m
67352_ch15.indd 457
( 2.50 N ) r 2
The charges induce opposite charges in the bulkheads, but the induced charge in the bulkhead near ball B is greater, due to B’s greater charge. Therefore, the system
moves slowly towards the bulkhead closer to ball B .
.
F=
8.
q2 =
( 2.50 N )( 2.00 m )2
q=
or
(b)
ke q ( 2q )
= 2.50 N ⇒
r2
N ⋅ m 2 C2 ) ( 0.800 × 10 −6 C ) ( −0.600 × 10 −6 C )
(3.50 × 10
−2
m ) ( 5.00 × 10 −2 m )
2
r
Fe
q2
122
2/9/11 1:45:59 PM
(
× 10 C )
) ( 0.800 × 10 C) ( −0.600
Electric Forces and Electric Fields
(3.50 × 10 m ) (5.00 × 10 m )
−6
−6
−2
62.
15.9
1.
⎛
⎜⎝
k = 49.3 N m
(a)
(b)
−2
2
⎞
−1
⎟⎠ = tan (1.00 ) = 45° above the horizontal .
2
2
(8.99 × 109 N ⋅ m 2 C2 ) ⎡⎣ 4 (1.60 × 10 −19 C) ⎤⎦
ke ( 2e )
F=
=
= 36.8 N
2
r2
(5.00 × 10 −15 m )
The mass of an alpha particle is m = 4.002 6 u, where 1 u = 1.66 × 10 −27 kg is the unified
mass unit. The acceleration of either alpha particle is then
F
36.8 N
a= =
= 5.54 × 10 27 m s 2
−27
m
4.002
6
1.66
×
10
kg
(
)
The net force on the −2 mC charge is F−2 m C = F2 + F3 = 111 N to the left .
15.10
15.11
1.
123
463
.
Thetheforces
areatastheshown
sketch
below.of the
In
sketch
right,inFRthe
is the
resultant
forces F6 and F3 that are exerted on the charge at
the origin by the 6.00 nC and the –3.00 nC charges,
respectively.
−9
−9
⎛
N ⋅ m 2 ⎞ ( 6.00 × 10 C ) ( 5.00 × 10 C )
F6 = ⎜ 8.99 × 10 9
= 3.00 × 10 −6 N
C2 ⎟⎠
⎝
( 0.300 m )2
−9
−9
⎛
N ⋅ m 2 ⎞ ( 3.00 × 10 C ) ( 5.00 × 10 C )
F3 = ⎜ 8.99 × 10 9
= 1.35 × 10 −5 N
C2 ⎟⎠
⎝
( 0.100 m )2
The resultant is FR =
or
1.
15.13
(F ) + (F )
2
6
⎛F ⎞
= 1.38 × 10 −5 N at q = tan −1 ⎜ 3 ⎟ = 77.5°
⎝ F6 ⎠
2
3
FR = 1.38 × 10 −5 N at 77.5° below the − x-axis
Please see the sketch at the right.
F1 =
(8.99 × 10
or
F1 = 0.288 N
F2 =
(8.99 × 10
or
F2 = 0.503 N
9
N ⋅ m 2 C2 ) ( 2.00 × 10 −6 C ) ( 4.00 × 10 −6 C )
( 0.500 m )2
9
N ⋅ m 2 C2 ) ( 2.00 × 10 −6 C ) ( 7.00 × 10 −6 C )
( 0.500 m )2
Electric Forces and Electric Fields
467
The components of the resultant force acting on the 2.00 mC charge are:
Fx = F1 − F2 cos 60.0° = 0.288 N − ( 0.503 N ) cos 60.0° = 3.65 × 10 −2 N
and Fy = −F2 sin 60.0° = − ( 0.503 N ) sin 60.0° = − 0.436 N
The magnitude and direction of this resultant force are
F = Fx2 + Fy2 =
at
67352_ch15.indd 463
( 0.036 5 N ) + ( 0.436 N )
2
2
= 0.438 N
⎛ Fy ⎞
⎛ −0.436 N ⎞
= −85.2° or 85.2° below the + x-axis
q = tan −1 ⎜ ⎟ = tan −1 ⎜
⎝ 0.036 5 N ⎟⎠
⎝ Fx ⎠
2/9/11 1:46:13 PM
464
124
Chapter 15
( 0.036 5 N ) + ( 0.436 N )
=
2
2
= 0.438 N
k ⎛q−0.436 N ⎞
⎛ Fy ⎞
−1
(b) q ΣF
Fetan
= −1e⎜ 1
= −85.2° or 85.2° below the + x-axis
at
= tan
y = 0⎜ →⎟ =
⎝ 0.036 5 N ⎟⎠
⎝ Fx ⎠
1.
15.15
Consider the free-body diagram of one of the spheres given
at the right. Here, T is the tension in the string and Fe is the
repulsive electrical force exerted by the other sphere.
ΣFy = 0 ⇒ T cos 5.0° = mg,
or T =
mg
cos 5.0°
ΣFx = 0 ⇒ Fe = T sin 5.0° = mg tan 5.0°
At equilibrium, the distance separating the two spheres is r = 2 L sin 5.0°.
Thus, Fe = mg tan 5.0° becomes
q = ( 2 L sin 5.0° )
ke q 2
= mg tan 5.0° and yields
( 2 L sin 5.0°)2
mg tan 5.0°
ke
( 0.20
4.89 × 10
= [ 2 ( 0.300 m ) sin 5.0° ] =
15.17
1.
2
kg
N ) ( 9.80 m s −9) tan 5.0°
N m . = 7.2 nC
=
2.25
×
10
2.17 × 8.99
10 −8 ×m10 9 N ⋅ m 2 C2
−3
−17
−19
F qE (1.60 × 10 C ) ( 640 N C )
=
=
= 6.12 × 1010 m s 2
1.673 × 10 −27 kg
m mp
(a)
a=
(b)
t=
(c)
Δx =
(d)
KE f =
Δv
1.20 × 10 6 m s
=
= 1.96 × 10 −5 s = 19.6 ms
a
6.12 × 1010 m s 2
v 2f − v02
2a
(1.20 × 10 m s) − 0 =
2 ( 6.12 × 10 m s )
2
6
=
10
2
11.8 m
2
1
1
m p v 2f = (1.673 × 10 −27 kg )(1.20 × 10 6 m s ) Electric
= 1.20Forces
× 10 −15and
J Electric Fields
2
2
15.19
1.
15.18
The force on a negative charge is opposite to the direction of the electric field and has magnitude
(a)
to the right as
F = qTaking
E. Thus,
positive, the resultant
electric
field× 10
at point
−6
F = −6.00
C (P
5.25 × 10 5 N C ) = 3.15 N
is given by
, but the value of the ratio
and (i.e.,
F = the
3.15electric
N due fi
north
eld) would be unchanged.
15.21
1.
(a)
y
The electrical force must be directed up the incline and have
a magnitude equal to the tangential component of the
gravitational force.
or
Fe = Q E = mg sinq
and
E = mg sinq Q
n
Q,
ΣFx = 0 ⇒ Fe − mg sinq = 0
67352_ch15.indd 464
469
q
Fe
x
m
q
mg
2/9/11 1:46:16 PM
465
125
Electric Forces and Electric Fields
15.7
Fs −m
Fe s=2 )kd
Fe = 0.
In the newmg
equilibrium
position,
× 10 −3 ΣF
kg )y (=9.80
sin−25.0°
sinq ( 5.40
(b) E =
=
= 3.19 × 10 3 N C
Q
7.00 × 10 −6 C
Thus,
Since Fe must be directed up the incline and the electrical force on a negative charge is
directed opposite to the field, it is necessary to have the electric field directed down the
incline.
Thus,
15.22
15.23
1.
E = 3.19 × 10 3 N( C down the
) =incline
7.5 × 10 −2 N
tan 37°
(a)
In order to suspend the object in the electric field, the electric force exerted on the object by the
field must be directed upward and have a magnitude equal to the weight of the object. Thus,
Fe = qE = mg, and the magnitude of the electric field must be
(
)
3.80 g ( 9.80 m s 2 ) ⎛ 1 kg ⎞
mg
3
E=
=
⎜ 3 ⎟ = 2.07 × 10 N C
q
18 × 10 −6 C
⎝ 10 g ⎠
The electric force on a negatively charged object is in the direction opposite to that of the electric field.
−8
5
Since the electric force must
eld must
downward.
× 10electric
s ) = fi5.27
× 10be
mdirected
s
( be directed upward,
) (1.00 the
1.
15.25
From the figure at the right, observe that
2r cos 45.0° + d = 2d. Thus,
r=
+Q
d
d
d
=
=
2 cos 45.0° 2 2 2
2
(
)
E x = E2 − E1 cos 45.0° =
E y = E1 sin 45.0° + 0 =
15.26
1.
15.27
E1
45.0°
d
P
45.0°
45.0°
2d
E2
r
+Q
kQ
ke Q ke Q ⎛ 2 ⎞ ke Q
kQ ⎛ 2⎞
= 2 − 2e
= 1− 2 e 2
− 2 ⎜
2
⎟
⎜
⎟
d
r ⎝ 2 ⎠
d
d
( d 2) ⎝ 2 ⎠
(
kQ ⎛ 2⎞
ke Q ⎛ 2 ⎞
= e
=
r 2 ⎜⎝ 2 ⎟⎠ ( d 2 2 ) ⎜⎝ 2 ⎟⎠
2
)
ke Q
d2
(a) Observe the figure at the right:
If the resultant field is zero, the
contributions from the two charges must
be in opposite directions and also have
equal magnitudes. Choose the line
connecting the charges as the x-axis, with
the origin at the –2.5 µC charge. Then, the
two contributions will have opposite
directions only in the regions x < 0 and
x > 1.0 m. For the magnitudes to be equal, the point must be nearer the smaller charge. Thus, the
point of zero resultant field is on the x-axis at x < 0.
Requiring equal magnitudes gives
ke q1
k q
= e 22
2
r1
r2
Thus,
(1.0 m + d )
or
6.0 mC
2.5 mC
=
2
d
(1.0 m + d )2
2.5
=d
6.0
Solving for d yields
d = 1.8 m,
67352_ch15.indd 465
or
2/9/11 1:46:21 PM
126
466
=d
Chapter 15
Solving forkedq1yields
q3
F2 =
d = 1.8 m,
or
15.29
1.
1.8 m to the left of the − 2.5 mC charge
at 4.40° below the +x-axis
From the symmetry of the charge distribution, students
should recognize that the resultant electric field at the
center is
ER = 0
If one does not recognize this intuitively, consider:
E R = E1 + E 2 + E 3
15.30
15.33
1.
Ex = E1 x − E2 x =
and
E y = E1 y + E2 y − E3 =
Thus,
ER = Ex2 + E y2 = 0
15.34
1.
15.37
15.36
ke q
k q
k q
sin 30° + e 2 sin 30° − e 2 = 0
2
r
r
r
The magnitude of q2 is three times the magnitude of q1 because 3 times as many lines emerge
(a)
sketch for (a) is shown at
from qThe
2 as enter q1.
the right. Note that four times
as many lines should leave q1
as emerge from q2 although, for
clarity, this is not shown in this
sketch.
(b)
1.
15.35
ke q
k q
cos 30° − e 2 cos 30° = 0
2
r
r
so
The field pattern looks the same
here as that shown for (a) with
the exception that the arrows
are reversed on the field lines.
Note in the sketches at the right that
electric field lines originate on
(a) The electric field in the plane of the charges has the
positive charges and terminate on
general appearance shown at the right:
negative charges. The density of
lines is twice as great for the −2 q
charge in (b) as it is for the 1q
charge in (a).
13
⎡
⎤
or
r ∼ 1 mm
⎢
⎥ = 5.2 × 10 −7 m
(
)
⎢⎣
⎥⎦
(b) a uniformly
It is zero charged sphere, the field is strongest at the surface.
For
If the weight of the drop is balanced by the electric force, then mg = q
Thus,
Em ax =
ke qm ax
,
R2
6
R 2 Em ax ( 2.0 m ) ( 3.0 × 10 N C )
=
= 1.3 × 10 −3 C
ke
8.99 × 10 9 N ⋅ m 2 C2
2
or
15.38
67352_ch15.indd 466
(a)
qm ax =
The dome is a closed conducting surface. Therefore, the electric field is zero
2/9/11 1:46:26 PM
Electric Forces and Electric Fields
1.
15.39
127
467
475
The
of the× resultant
mC×charge
in the direction of the
(a) components
10 −27 kg )force
× 1012 on
m the
s 2 ) 2.00
= 2.54
10 −15 Nare:
Fe = ma = (1.67
(1.52 acting
acceleration, or radially outward.
Fx = F1 − F2 cos 60.0° = 0.288 N − ( 0.503 N ) cos 60.0° = 3.65 × 10 −2 N
(b) The direction of the field is the direction of the force on a positive charge (such as the proand ton).
Fy = −F
60.0°
= is
− (directed
0.503 N )radially
sin 60.0°outward
= − 0.436
N magnitude of the field is
Thus,
field
. The
2 sinthe
The magnitude and
are
F direction
2.54 × 10of−152this
N resultant force
E( = e = ) ( 3.50 m
= 1.59 × 10 4 N C 3 N ⋅ m 2 C
−19 ) cos 25.0° = 1.38 × 10
q 1.60 × 10 C
15.40
15.41
1.
When
an electric
field of
magnitude
E is incident
We choose
a spherical
Gaussian
surface,
concentric with the charged spherical shell and of
2
on
a
surface
of
area
A,
the
fl
ux
through
radius r. Then, ΣEA cosq = E ( 4p r ) costhe
0° = 4p r 2 E .
surface is
(a) For r > a (that is, outside the shell), the total charge enclosed by the Gaussian surface is
Q = +q − q = 0. Thus, Gauss’s law gives 4p r 2 E = 0, or E = 0 .
(b)
Inside the shell, r < a, and the enclosed charge is Q = +q.
Therefore, from Gauss’s law, 4p r 2 E =
476
15.42
15.43
1.
The
Chapter 15
kq
q
q
, or E =
= e2
r
∈0
4p ∈0 r 2
field for r < a is E = ke q r 2 directed radially outward .
(a) The surface of the cube is a closed surface which surrounds a total charge of Q = 1.70 × 10 2 mC.
From Thus,
Gauss’s
the law,
electric
flux through
any closed
by law,
Gauss’s
the electric
flux through
the full surface of the cube is
surface is equal to the net charge enclosed divided by ∈0 .
Thus, the flux through each surface (with a positive flux
coming outward from the enclosed interior and a negative
flux going inward toward that interior) is
For S1:
Φ E = Qnet ∈0 = ( +Q − 2Q ) ∈0 = −Q ∈0
For S2:
Φ E = Qnet ∈0 = ( +Q − Q ) ∈0 = 0
For S3:
Φ E = Qnet ∈0 = ( −2Q + Q − Q ) ∈0 = −2Q ∈0
∈ = (6.55
0 ) ∈0× =1005 N ⋅ m 2 C
Φ E = Qnet Φ
Φ1 face = 0E =
= 1.64 × 10 5 N ⋅ m 2 C
4
4
(a) From Gauss’s law, the total flux through this closed surface is
The area of the rectangular plane is A = ( 0.350 m ) ( 0.700 m ) = 0.245 m 2 .
For S4:
15.44
15.45
1.
(a)
When the plane is parallel to the yz-plane, its normal line is parallel to the x-axis and makes
an angle q = 0° with the direction of the field. The flux is then
Φ E = EA cosq = ( 3.50 × 10 3 N C ) ( 0.245 m 2 ) cos 0° = 858 N ⋅ m 2 C
15.46
15.47
1.
(b)
When the plane is parallel to the x-axis, q = 90° and Φ E = 0 .
(c)
Φ E = EA cosq = ( 3.50 × 10 3 N C ) ( 0.245 m 2 ) cos 40.0° = 657 N ⋅ m 2 C
(a) Gauss’s law states that the electric flux through any closed surface equals the net charge
−2.00 nC positioned at the center of the spherical shell, we have
Note that with the point charge
enclosed divided by ∈ . We choose to consider a closed surface in the form of a sphere,
complete spherical symmetry0 in this situation. Thus, we can expect the distribution of charge on
centered on the center of the charged sphere and having a radius infinitesimally larger than
the shell, as well as the electric fields both inside and outside of the shell, to also be spherically
that of the charged sphere. The electric field then has a uniform magnitude and is persymmetric.
pendicular to our surface at all points on that surface. The flux through the chosen closed
surface is therefore Φ = EA = E ( 4p r 2 ) , and Gauss’s law gives
(a) We choose a sphericalE Gaussian surface, centered on
the center of the conducting shell, with radius
r = 1.50 m < a as shown at the right. Gauss’s law gives
Φ E = EA = E ( 4p r 2 ) =
67352_ch15.indd 467
Qinside
∈0
2/9/11 1:46:31 PM
15.47
128
468
15.16
Note that with the point charge −2.00 nC positioned at the center of the spherical shell, we have
complete spherical symmetry in this situation. Thus, we can expect the distribution of charge on
the shell,
Chapter
15 as well as the electric fields both inside and outside of the shell, to also be spherically
symmetric.
The attractive
force
between Gaussian
the charged
ends tends
to compress
the molecule. Its magnitude is
(a)
We choose
a spherical
surface,
centered
on
the center of the conducting shell, with radius−19
2
2
× 10 lawCgives
) = 4.89 × 10 −17 N
⎞ (1.60
⋅ mright.
r = 1.50 m <= a⎛ 8.99
as shown
the
Gauss’s
9 atN
×
10
2
⎜⎝
C2 ⎟⎠ ( 2.17 × 10 −6 m )
Q
Qinside
kQ
Φ E = EA = E ( 4p r 2 ) = inside or E =
= e center
2
4p ∈0 r
r2
The compression of the “spring” is ∈0
−6
10 N
r = (×0.010
0 )⋅(m
2.17C× )10
m ) ×=10
2.17C×)10 −8 m ,
x = ( 0.010 0()8.99
( −2.00
so
E=
2
(1.50 m )
F
so the spring constant is k =
x negative sign means that the field is radial inward .
and E = −7.99 N C . The
9
2
2
−9
(b)
All points at r = 2.20 m are in the range a < r < b, and hence are located within the conducting material making up the shell. Under conditions of electrostatic equilibrium, the
field is E = 0 at all points inside a conducting material.
(c)
If the radius of our Gaussian surface is r = 2.50 m > b, Gauss’s law (with total spheriQinside
kQ
cal symmetry) leads to E =
= e inside
just as in part (a). However, now
4p ∈0 r 2
r2
Qinside = Qshell + Qcenter = +3.00 nC − 2.00 nC = +1.00 nC. Thus, we have
E=
(8.99 × 10
9
N ⋅ m 2 C2 ) ( +1.00 × 10 −9 C )
( 2.50 m )2
= +1.44 N C
with the positive sign telling us that the field is radial outward at this location.
(d)
Under conditions of electrostatic equilibrium, all excess charge on a conductor resides
entirely on its surface. Thus, the sum of the charge on the inner surface of the shell and that
on the outer surface of the shell is Qshell = +3.00 nC. To see how much of this is on the inner
surface, consider our Gaussian surface to have a radius r that is infinitesimally larger than
a. Then, all points on the Gaussian surface lie within the conducting material, meaning that
E = 0 at all points and the total flux through the surface is Φ E = 0. Gauss’s law then states
that Qinside = Qinner + Qcenter = 0, or
surface
Qinner = − Qcenter = − ( −2.00 nC ) = +2.00 nC
surface
The charge on the outer surface must be
Qouter = Qshell − Qinner = 3.00 nC − 2.00 nC = +1.00 nC − s = − s
surface
surface
∈0
2 ∈0
15.49
1.
Because of the spherical symmetry of the
charge distribution, any electric field
present will be radial in direction. If a field
does exist at distance R from the center, it is
the same as if the net charge located within
r ≤ R were concentrated as a point charge
at the center of the inner sphere. Charge
located at r > R does not contribute to the
field at r = R.
(a)
At r = 1.00 cm, E = 0 since static
electric fields cannot exist within
conducting materials.
(b)
The net charge located at r ≤ 3.00 cm is Q = +8.00 mC.
Thus, at r = 3.00 cm,
67352_ch15.indd 468
2/9/11 1:46:35 PM
15.49
15.19
Because of the spherical symmetry of the
charge distribution, any electric field
present will be radial in direction. If a field
does exist at distance R from the center, it is
the same as if the net charge located within
r ≤ R were concentrated as a point charge
at the center of the inner sphere. Charge
located at r > R does not contribute to the
field at r = R.
Electric Forces and Electric Fields
469
129
The
a negative
charge
is opposite
(a) force
At r on
= 1.00
cm, E
= 0 since
static to the direction of the electric field and has magnitude
F = qelectric fields cannot exist within
conducting materials.
F = −6.00 × 10 −6 C ( 5.25 × 10 5 N C ) = 3.15 N
(b) The net charge located at r ≤ 3.00 cm is Q = +8.00 mC.
and
Thus, at r = 3.00 cm,
E=
=
ke Q
r2
(8.99 × 10
9
N ⋅ m 2 C2 ) (8.00 × 10 −6 C )
(3.00 × 10
−2
m)
2
7
= 7.99
× 10Forces
N Cand
( outward
Electric
Electric )Fields
(c)
At r = 4.50 cm, E = 0 , since this is located within conducting materials.
(d)
The net charge located at r ≤ 7.00 cm is Q = + 4.00 mC.
479
Thus, at r = 7.00 cm,
E=
=
15.50
1.
15.51
ke Q
r2
(8.99 × 10
9
N ⋅ m 2 C2 ) ( 4.00 × 10 −6 C )
( 7.00 × 10
−2
m)
2
= 7.34 × 10 6 N C ( outward )
It is three
desired
that the electric
exert a retarding force on the electrons, slowing them
The
contributions
to thefield
resultant
down
and
bringing
them
to
rest.
For
the retarding force to have maximum effect, it should
electric field at the point of interest are
be
anti-parallel
to
the
direction
of
the
electron’s motion. Since the force an electric field exerts
shown in the sketch at the right.
on negatively charged particles (such as electrons) is in the direction opposite to the field,
The magnitude of the resultant field is
ER = −E1 + E2 + E3
ER = −
ke q1 ke q2 ke q3
+ 2 + 2 = ke
r12
r2
r3
⎡ q1
q3 ⎤
q2
⎢− 2 + 2 + 2 ⎥
r2
r3 ⎥⎦
⎢⎣ r1
⎛
N ⋅ m 2 ⎞ ⎡ 4.0 × 10 −9 C 5.0 × 10 −9 C 3.0 × 10 −9 C ⎤
+
+
ER = ⎜ 8.99 × 10 9
⎢−
⎥
2
C2 ⎟⎠ ⎢⎣ ( 2.5 m )
⎝
(1.2 m )2 ⎥⎦
( 2.0 m )2
15.52
15.53
1.
E R = + 24 N C or E(R = 24 N )Ctan15.0°
in the +x-direction
= +5.25 × 10 −6 C = + 5.25 mC
1.00 × 10 3 N C
Consider the force diagram shown at the right.
(a) At a point on the x-axis, the contributions by the two
charges to the resultant field have equal magnitudes
given by E 1 = E2 = ke q r 2.
The components of the resultant field are
⎛ k q⎞
⎛ k q⎞
E y = E1 y − E2 y = ⎜ e2 ⎟ sinq − ⎜ e2 ⎟ sinq = 0
⎝ r ⎠
⎝ r ⎠
and
67352_ch15.indd 469
2/9/11 1:46:39 PM
130
470
⎛
⎜⎝
Chapter 15
⎞
⎛ ke q ⎞
⎟⎠ sinq − ⎜⎝ 2 ⎟⎠ sinq = 0
r
Since F
⎡ k ( 2q ) ⎤
⎛ k q⎞
⎛ k q⎞
and Ex = E1 x + E2 x = ⎜ e2 ⎟ cosq + ⎜ e2 ⎟ cosq = ⎢ e 2 ⎥ cosq
⎝ r ⎠
⎝ r ⎠
⎣ r
⎦
mg
E=
q sinq b r b
b
cosq
Since
= 2 = 3 = 2
, the resultant field is
3 2
r
r
r2
( a + b2 )
ER =
ke ( 2q ) b
(a
2
+ b2 )
3 2
in the +x-direction
Note that the result of part (a) may be written as ER =
(b)
ke (Q ) b
(a
2
total charge in the charge distribution generating the field.
+ b2 )
3 2
, where Q = 2q is the
In the case of a uniformly charged circular ring, consider the ring to consist of a very large
number of pairs of charges uniformly spaced around the ring. Each pair consists of two
identical charges located diametrically opposite each other on the ring. The total charge of
pair number i is Qi . At a point on the axis of the ring, this pair of charges generates an elec3 2
tric field contribution that is parallel to the axis and has magnitude Ei = ke b Qi ( a 2 + b 2 ) .
The resultant electric field of the ring is the summation of the contributions by all pairs of
charges, or
⎡
kb
ER = ΣEi = ⎢ 2 e 2 3 2
⎢ (a + b )
⎣
15.54
1.
15.55
⎤
ke b Q
⎥ ΣQi =
3 2
2
⎥
(a + b2 )
⎦
where Q = ΣQi is the total charge
on the ring.
+5.20 × 10 −6 C m 2
=
= 2.94 × 10 5 N C
−12
2
2
2
8.85
×
10
C
N
⋅
m
(
)
k Qb
E R = 2 e 2 3 2 in the +x-direction
+ b vary
( a not
) with distance as long as the distance is small compared with the
(b) The field does
dimensions of the sheet.
(a) As shown in Example 15.8 in the textbook, the electric field due to a nonconducting plane
sheet
chargeathas
constant
magnitude of E = s 2 ∈0
F2
Consider
theofsketch
thearight
and observe:
⎛d⎞
q = tan −1 ⎜ ⎟ = 45.0°
⎝d⎠
+Q
q
F1
r
d
r = d2 + d2 = d 2
and
Thus,
+2Q
Fx = F1 x + F2 x = +
and
or
15.56
67352_ch15.indd 470
(a)
d
−Q
ke −Q Q
ke Q 2
ke Q 2 ⎛ 2 ⎞
2 ⎛ ke Q 2 ⎞
=
+
0.354
=
+
sin
45.0°
+
0
=
+
4 ⎜⎝ d 2 ⎟⎠
r2
2d 2 ⎜⎝ 2 ⎟⎠
d2
Fy = F1 y + F2 y = −
ke −Q Q
k (Q ) ( 2Q )
k Q 2 ⎛ 2 ⎞ 2ke Q 2
+
cos 45.0° + e
=− e 2 ⎜
2
2
r
d
2d ⎝ 2 ⎟⎠
d2
⎛
ke Q 2
2 ⎞ ke Q 2
Fy = ⎜ 2 −
=
+1.65
4 ⎟⎠ d 2
d2
⎝
The downward electrical force acting on the ball is
2/9/11 1:46:43 PM
482
1.
15.57
Electric Forces and Electric Fields
Chapter 15
471
131
k diagram
q k of the positively
Q gives a force
The sketch at the2k
right
Ex = e2 cos 30.0° 2− 2e 2 = 2e ( 2Q cos 30.0° − q )
charged sphere. Here,
r F1 = ke q r r is therattractive force
exerted by the negatively charged sphere, and F2 = qE is exerted by
9
2
2
the electric field. (8.99 × 10 N ⋅ m C )
⎡ 2 ( 3.00 × 10 −9 C ) cos 30.0° − 2.00 × 10 −9 C ⎤
=
2
⎦
−2
4.00
×
10
m
(
) ⎣ mg
ΣFy = 0 ⇒ T cos10° = mg or T =
cos10°
Ex = +1.80 × 10 4 N C
2
ΣF
= x2F1++ETy2 sin10° or qE =
Then,
E =F2 E
x = 0 ⇒
ke q
+ mg tan10°
r2
At equilibrium, the distance between the two spheres is r = 2 ( L sin10° ). Thus,
E=
=
ke q
mg tan10°
+
2
q
4 ( L sin10° )
(8.99 × 10
9
N ⋅ m 2 C2 ) ( 5.0 × 10 −8 C )
4 [( 0.100 m ) sin10° ]
or the needed electric field strength is
67352_ch15.indd 471
2
+
( 2.0 × 10
−3
kg ) ( 9.80 m s 2 ) tan10°
(5.0 × 10
−8
C)
E = 4.4 × 10 5 N C
2/9/11 1:46:46 PM
ࢠ ࢷ̛߾οएࠪࢷ̛ࡈԛ
PROBLEM SOLUTIONS
1.
16.1
(a)
Because the electron has a negative charge, it experiences a force in the direction opposite
to the field and, when released from rest, will move in the negative x-direction. The work
done on the electron by the field is
(
)
(
W = Fx ( Δx ) = ( qE x ) Δx = −1.60 × 10 −19 C ( 375 N C ) −3.20 × 10 −2 m
)
= 1.92 × 10 −18 J
(b)
The change in the electric potential energy is the negative of the work done on the particle
by the field. Thus,
ΔPE = −W = −1.92 × 10 −18 J
(c)
494
16.2
16.3
1.
Since the Coulomb force is a conservative force, conservation of energy gives
ΔKE + ΔPE = 0, or KE f = 12 me v 2f = KEi − ΔPE = 0 − ΔPE, and
Chapter 16
vf =
−2 ( ΔPE )
=
me
(
−2 −1.92 × 10 −18 J
9.11× 10
−31
kg
)=
2.05 × 10 6 m s in the − x-direction
(a) Using conservation of energy, ΔKE + ΔPE = 0, with KE f = 0 since the particle is “stopped,”
The work
done by the agent moving the charge out of the cell is
we have
Winput = −Wfield = − ( −ΔPEe ) = +q ( ΔV )
2
1.50 × 10 5 N C 8.00 × 10 −2 m
Er 2
11
−19
−3
==1.60 ×=10 C + 990 × 102 J2 C = 1.4 ×−19
10 −20 J= 6.67 × 10 electrons
e
ke e
8.99 × 10 N ⋅ m C 1.60 × 10 C
(
16.4
16.5
1.
(
(
)(
)(
) ()
)
)
Assuming the sphere is isolated, the excess charge on it is uniformly distributed over its surface.
ΔV
600
J C the electric fi
UnderEthis
symmetry,
the sphere is the same as if all the excess
(a)
= spherical
N outside
C
=
= 1.13 × 10 5 eld
d sphere
5.33were
× 10 −3
m
charge on the
concentrated
as a point charge located at the center of the sphere.
(
)
× 10 −19
J C )field is E = ke−14
(600
= 5.00
cm,Cthe
electric
Thus, at r = 8.00 qcmΔV
> Rsphere1.60
Q r2
=
N
=
1.80
×
10
(b) F = q E =
d
5.33 × 10 −3 m
(c)
W = F ⋅ s cosq
−1.98 × 10 −3 J
⎤⎦ cosV
m49.5
0° = 4.37 × 10 −17 J
= 1.80 × 10 −14 N ⎡⎣=( 5.33 − 2.90 )−6× 10=−3 −
+40.0 × 10 C
(
67352_ch16.indd 486
16.7
1.
E=
16.8
16.9
1.
(a)
(a)
)
25 × 10 3 J C
ΔV
=
= 1.7 × 10 6 N C
d
1.5 × 10 −2 m
Electrical Energy and Capacitance
495
We use conservation of energy,
Δ ( KE ) + Δ ( PEs ) + Δ ( PEe ) = 0, recognizing
that Δ(KE) = 0 since the block is at rest at
both the beginning and end of the motion.
The change in the elastic potential energy is
2
− 0, where x max
given by Δ ( PEs ) = 12 kx max
is the maximum stretch of the spring. The
change in the electrical potential energy
132
is the negative of the work the electric field does, Δ ( PEe ) = −W = −Fe (Δx) = − (QE ) x max.
2
Thus, 0 + 12 kx max
− (QE ) x max = 0, which yields
2/9/11 1:54:33 PM
Electrical Energy and Capacitance
16.9
64.
495
(a)
We use conservation of energy,
Δ ( KE ) + Δ ( PEs ) + Δ ( PEe ) = 0, recognizing
that Δ(KE) = 0 since the block is at rest at
both the beginning and end of the motion.
Electrical Energy and Capacitance
493
133
The change in the elastic potential energy is
2
given by Δ ( PEs ) = 12 kx max
− 0, where x max
ab
is
the
maximum
stretch
of
the
spring. The
(a) C =
ke ( bin−the
a ) electrical potential energy
change
is the negative of the work the electric field does, Δ ( PEe ) = −W = −Fe (Δx) = − (QE ) x max.
2
Thus, 0 + 12 kx max
− (QE ) x max = 0, which yields
x max =
(b)
(
)(
)
−6
4
2QE 2 35.0 × 10 C 4.86 × 10 V m
=
= 4.36 × 10 −2 m = 4.36 cm
k
78.0 N m
At equilibrium, ΣF = Fs + Fe = 0, or − kx eq + QE = 0. Therefore,
x eq =
QE 1
= x max = 2.18 cm
k
2
The amplitude is the distance from the equilibrium position to each of the turning points
( at x = 0 and x = 4.36 cm ), so A = 2.18 cm = xmax 2 .
(c)
2
From conservation of energy, Δ ( KE ) + Δ ( PEs ) + Δ ( PEe ) = 0, we have 0 + 12 kx max
+ QΔV = 0.
Since x max = 2A, this gives
2
kx max
k ( 2A )
=−
2Q
2Q
2
ΔV = −
1.
16.11
(a)
V =∑
i
496
Chapter 16
(b)
PE =
or
(
2kA2
ΔV = −
Q × 10 4 V = 40.2 kV
= 4.02
)
ke qi ⎛
N ⋅ m 2 ⎞ ⎛ 5.00 × 10 −9 C 3.00 × 10 −9 C ⎞
= ⎜ 8.99 × 10 9
−
= 103 V
0.175 m ⎟⎠
ri
C2 ⎟⎠ ⎜⎝ 0.175 m
⎝
(
)(
)
−9
−9
ke qi q2 ⎛
N ⋅ m 2 ⎞ 5.00 × 10 C − 3.00 × 10 C
= ⎜ 8.99 × 10 9
= − 3.85 × 10 − 7 J
r12
C2 ⎟⎠
0.350 m
⎝
−9
The negative sign means that
positive
work
must
done
to separate
⎞ charges by an
−15.0
× 10be
C 27.0
× 10 −9 Cthe
9
2
2 ⎛
=
8.99
×
10
N
⋅
m
C
+
= +10.8 kV
∑
−2zero potential energy).
−2
⎜⎝ 1.00
infinite distance (that is, bring them up to
a state
of
× 10 m 1.00 × 10 m ⎟⎠
(
i
16.13
1.
16.12
(a)
(a)
(b)
)
⎛ −15.0 × 10 −9 C 27.0 × 10 −9 C ⎞
Calling the 2.00
mC charge
× 10 9 qN3,⋅ m 2 C2 ⎜
+
= +5.39 kV
∑ = 8.99
⎝ 2.00 × 10 −2 m 2.00 × 10 −2 m ⎟⎠
i
⎛
⎞
kq
q
q q
V = ∑ e i = ke ⎜ 1 + 2 + 2 3 2 ⎟
⎜⎝ r1 r2
ri
r + r ⎟⎠
i
(
)
1
2
⎛
⎛
N ⋅ m 2 ⎞ 8.00 × 10 −6 C 4.00 × 10 −6 C
⎜
= ⎜ 8.99 × 10 9
+
+
C2 ⎟⎠ ⎜ 0.060 0 m
0.030 0 m
⎝
⎝
⎡
⎛
q2 ⎞ ⎤
⎢
⎜⎝ 6 + r ⎟⎠ ⎥
⎢⎣
V = 2.67
× 10 V 2 ⎥⎦
(b)
1.
16.15
( 0.060 0 )2 + ( 0.030 0 )2
⎞
⎟
m ⎟⎠
⎞
2 ⎛
−6
−6⎛
⎞ C 2.00
m−6
C
4.00 × 10 −6 C
× 10
Replacing
by ×−10
2.00
×⋅10
in part
(a) yields
9 N
⎟
⎜
8.99
+
= − 8.002.00
× 10×−610C ⎜C
2
2
C2 ⎟⎠ ⎜ 0.030 0 m
⎝
⎟⎠
+
0.060
0
m
0.030
0
(
)
(
)
⎝
V = 2.13 × 10 6 V
W = − 9.08 J
(
)
(
)(
)
)(
)=
8.99 × 10 9 N ⋅ m 2 C2 −1.60 × 10 −19 C
ke q
=
= −5.75 × 10 −7 V
rA
0.250 × 10 −2 m
(a)
VA =
(b)
−1.92 × 10 −7 V
0.750 × 10 m
ΔV = VB − VA = −1.92 × 10 −7 V − −5.75 × 10 −7 V = +3.83 × 10 −7 V
=
(8.99 × 10
9
N ⋅ m 2 C2 −1.60 × 10 −19 C
−2
(
67352_ch16.indd 493
2.00 × 10 −6 C
)
2/9/11 1:54:48 PM
134
494
16.3
=
Chapter 16
(8.99 × 10
9
)(
N ⋅ m 2 C2 −1.60 × 10 −19 C
0.250 × 10
(
−2
m
)(
−5.75 × 10 −7 V
)
The work done by the
of−19
theCcell is
8.99agent
× 10 9moving
N ⋅ m 2 the
C2charge
−1.60out
× 10
kq
(b) VB = e =
= −1.92 × 10 −7 V
−2
rB
0.750
m
q
q
q
ke q ×k10
k
3k
=
+ e + e = −7 e
−7
ΔV = VB − VA = −1.92 × 10a−7 V −a −5.75
a × 10 aV = +3.83 × 10 V
(
(c)
16.17
1.
)=
)
electron
will
be repelled
thetriangle.
negatively
charged
particle
No . Theaoriginal
Imagine
test charge
placed
at the
center ofbythe
Since
the field
is zerowhich
at thesudcenter,
denly
point
A. Unlessnothe
electronforce
is fixed
in place,
willfact
move
thepotential
oppositeis
the testappears
chargeatwill
experience
electrical
at that
point.itThe
thatinthe
direction,
from points
andwork
B, thereby
potential
between
not zero ataway
the center
means Athat
would lowering
have to bethe
done
by an difference
external agent
to move
these
a test points.
charge from infinity to the center.
The Pythagorean theorem gives the distance from the midpoint of the base to the charge at the
apex of the triangle as
r3 =
( 4.00 cm )2 − (1.00 cm )2 = 15 cm = 15 × 10 −2 m
Then, the potential at the midpoint of the base is V = ∑ ke qi ri, or
⎛
V = ⎜ 8.99 × 10 9
⎝
(
) (
i
) (
)
−9
−7.00 × 10 −9 C
+7.00 × 10 −9 C ⎞
N ⋅ m ⎞ ⎛ −7.00 × 10 C
+
+
⎜
⎟
C2 ⎟⎠ ⎝
0.010 0 m
0.010 0 m
15 × 10 −2 m ⎠
2
= −1.10 × 10 4 V = − 11.0 kV
= 0.800 m .
16.18
16.19
1.
(a)
(a)
See the sketch below:
Conservation of energy gives
⎛1 1⎞
KE f = KEi + PEi − PE f = 0 + ke q1q2 ⎜ − ⎟
⎝ ri rf ⎠
(
)
With q1 = +8.50 nC, q2 = −2.80 nC, ri = 1.60 mm, and rf = 0.500 mm, this becomes
⎛
N ⋅ m2 ⎞
1
1
⎛
⎞
KE f = ⎜ 8.99 × 10 9
−
8.50 × 10 −9 C −2.80 × 10 −9 C ⎜
⎝ 1.60 × 10 −6 m 0.500 × 10 −6 m ⎟⎠
C2 ⎟⎠
⎝
(
yielding
(b)
16.25
1.
KE f = 0.294 J
(
2 KE f
m
)=
2 ( 0.294 J )
qQ
⎛= 271 mQ s⎞
ke ⎟ = 4 2ke
8.00 × 10 −6 kg ⎜
⎝
a⎠
a
(a) If a proton and an alpha particle, initially at rest 4.00 fm apart, are released and allowed to
From conservation of energy, ( KE + PEe ) f = ( KE + PEe )i , which gives
recede to infinity, the final speeds of the two particles will differ because of the difference
0 + keQq rf = 12 ma vi2 + 0, or
in the masses of the particles. Thus, attempting to solve for the final speeds by use of conservation of energy alone leads to a situation of having one equation with two unknowns
2 k Qq 2 k ( 79e )( 2e )
rf = e 2 = e
ma vi
ma vi2
rf =
16.24
)
When r = rf = 0.500 mm and KE = KE f = 0.294 J, the speed of the sphere having mass
m = 8.00 mg = 8.00 × 10 −6 kg is
vf =
16.20
16.23
1.
)(
(
)
(
2 8.99 × 10 9 N ⋅ m 2 C2 (158 ) 1.60 × 10 −19 C
(6.64 × 10
−27
)(
kg 2.00 × 10 m s
7
)
2
)
2
= 2.74 × 10 −14 m
Electrical Energy and Capacitance
501
The excess charge on the metal sphere will be uniformly distributed over its surface. In this
ΔV
20.0
V
spherically
situation,
electric
the electric
outside
the plate
sphere
(a)
toward the
negative
E = symmetric
=
= 11.1
kV m potential
=the
1.11
× 10 4 fiVeldmand
d if all
1.80
10 −3 mcharge were concentrated as a point charge at the center of the
is the same as
the×excess
sphere. Thus, for points outside
the sphere,
8.85 × 10 −12 C2 N ⋅ m 2 7.60 × 10 −4 m 2
(b)
=
= 3.74 × 10 −12 F = 3.74 pF
1.80 × 10 −3 m
(
)(
)
(c)
67352_ch16.indd 494
2/9/11 1:54:51 PM
=
16.9
20.0 V
Electrical Energy and Capacitance
= 1.11 × 10 4 V m = 11.1 kV m toward the negative plate
1.80 × 10 −3 m
(
)(
C=
16.28
16.29
1.
502
16.30
1.
16.31
)
(
)(
)
−12
C2 N ⋅ m 2 2.30 × 10 −4 m 2
k ∈0 A (1.00 ) 8.85 × 10
=
= 1.36 × 10 −12 F = 1.36 pF
1.50 × 10 −3 m
d
(
)
(b)
Q = C ( ΔV ) = 1.36 × 10 −12 F (12.0 V ) = 1.63 × 10 −11 C = 16.3 × 10 −12 C = 16.3 pC
(c)
E=
(a)
C = ∈0
(b)
Qmax = C ( ΔV )max = C ( Emax d ) = ∈0
Chapter 16
(a)
(
)
ΔV
12.0 V
m )=tan15.0°
3
=
8.00 × 10
NC
= 8.00 ×( 0.040
10 3 V0 m
= 1.23
× 10 3 V = 1.23 kV
−9
d
1.50 × 10 −3 m
30.0 × 10 C
(
)
6
2
C2 ⎞ 1.0 × 10 m
A ⎛
= ⎜ 8.85 × 10 −12
= 1.1 × 10 −8 F
N ⋅ m 2 ⎟⎠
d ⎝
(800 m )
(
(
)
A
Emax d = ∈0 AEmax
d
)(
)(
)
= 8.85 × 10 −12 C2 N ⋅ m 2 1.0 × 10 6 m 2 3.0 × 10 6 N C = 27 C
27.0 mC
= 3.00
mFlled (k = 1), the capacitance is
Assuming= the capacitor
is air-fi
9.00 V
C=
(b)
1.
16.33
16.32
)
We use
of−12
energy,
8.85 × 10
C2 N ⋅ m 2 7.60 × 10 −4 m 2
A
∈0 conservation
= 3.74 × 10 −12 F = 3.74 pF
ΔC( =
KE ) + Δ=( PEs ) + Δ ( PEe ) = 0, recognizing
1.80 × 10 −3 m
d
that Δ(KE) = 0 since the block is at rest at
theΔV
−12 of the motion.
(c) both
Q=C
× 10end
F ( 20.0 V ) = 7.48 × 10 −11 C = 74.8 pC on one plate and
( beginning
) = 3.74and
The change in the elastic potential energy is
−74.8by
pCΔ (on
other
1
⎞plate.
given
PE⎛the
s)= 2
⎜⎝
⎟⎠ = 31.0 Å
C = ∈0 A d
(a) The capacitance of this air-filled ( dielectric constant, k = 1.00 ) parallel-plate capacitor is
(a)
(b)
(
16.26
16.27
1.
495
135
(
)(
)
8.85 × 10 −12 C2 N ⋅ m 2 0.200 m 2
∈0 A
=
= 5.90 × 10 −10 F
d
3.00 × 10 −3 m
(
)
(b)
Q = C ( ΔV ) = 5.90 × 10 −10 F ( 6.00 V ) = 3.54 × 10 −9 C
(c)
E=
ΔV
6.00 V
=
= 2.00 × 10 3 V m = 2.00 × 10 3 N C
d
3.00 × 10 −3 m
(d)
s =
Q 3.54 × 10 −9 C
=
= 1.77 × 10 −8 C m 2
A
0.200 m 2
(e)
Increasing the distance separating the plates decreases the capacitance, the charge
stored, and the electric field strength between the plates. This means that all of the
previous 100
answers
mC will be decreased .
=
= 100 V
1.00 mF
(a)
(a)
From Q = C ( ΔV ), Q25 = ( 25.0 mF )( 50.0 V ) = 1.25 × 10 3 mC = 1.25 mC
and
(b)
Q40 = ( 40.0 mF )( 50.0 V ) = 2.00 × 10 3 mC = 2.00 mC
Since the negative plate of one capacitor was connected to the positive plate of the other,
the net charge stored in the new parallel combination is
Q = Q40 − Q25 = 2.00 × 10 3 mC − 1.25 × 10 3 mC = 750 mC
The two capacitors, now in parallel, have a common potential difference ΔV across them.
The new charges on each of the capacitors are Q25′ = C1 ( ΔV ) and Q40′ = C2 ( ΔV ). Thus,
67352_ch16.indd 495
2/9/11 1:54:53 PM
136
496
Chapter 16
Q = Q40 − Q25 = 2.00 × 10 3 mC − 1.25 × 10 3 mC = 750 mC
The negative
sign means
that
positive
work
must be potential
done to separate
theΔV
charges
an
two capacitors,
now in
parallel,
have
a common
difference
acrossby
them.
infi
is, bring
up to a are
state
potential
Thenite
newdistance
charges(that
on each
of thethem
capacitors
Q25′of=zero
C1 ( ΔV
′40 = C2 ( ΔV ). Thus,
) and Qenergy).
16.12
Q
∑25′ =
(a)
i
⎛ 25 mF ⎞
C1
5 ⎛ −15.0 × 10 −9 C 27.0 × 10 −9 C ⎞
′ = ⎜× 10 9 N⎟ ⋅Qm40′ 2 =C2 Q⎜40′
=Q40
8.99
= +5.39 kV
+
8 ⎝ 2.00 × 10 −2 m 2.00 × 10 −2 m ⎟⎠
C2
⎝ 40 mF ⎠
(
)
and the total change now stored in the combination may be written as
(b)
5
13
Q = Q40′ + Q25′ = Q40′ + Q40′ = Q40′ = 750 mC
8
8
giving Q40′ =
(c)
The potential difference across each capacitor in the new parallel combination is
ΔV =
16.34
1.
16.35
(a)
8
( 750 mC) = 462 mC and Q25′ = Q − Q40′ = ( 750 − 462) mC = 288 mC
13
Q
Q
750 mC
=
=
= 11.5 V
Ceq C1 + C2 65.0 mF
.
First, we replace the parallel combination
between points b and c by its equivalent
capacitance, Cbc = 2.00 mF + 6.00 mF = 8.00 mF.
Then, we have three capacitors in series
between points a and d. The equivalent
capacitance for this circuit is therefore
1
1
1
3
1
=
+
+
=
Ceq Cab Cbc Ccd 8.00 mF
giving
(b)
Ceq =
8.00 mF
= 2.67 mF
3
The charge stored on each capacitor in the series combination is
Qab = Qbc = Qcd = Ceq ( ΔVad ) = ( 2.67 mF )( 9.00 V ) = 24.0 mC
Then, note that ΔVbc = Qbc Cbc = 24.0 mC 8.00 mF = 3.00 V. The charge on each capacitor
in the original circuit is:
(c)
504
1.
16.37
67352_ch16.indd 496
On the 8.00 mF between a and b:
Q8 = Qab = 24.0 mC
On the 8.00 mF between c and d:
Q8 = Qcd = 24.0 mC
On the 2.00 mF between b and c:
Q2 = C2 ( ΔVbc ) = ( 2.00 mF )( 3.00 V ) = 6.00 mC
On the 6.00 mF between b and c:
Q6 = C6 ( ΔVbc ) = ( 6.00 mF )( 3.00 V ) = 18.0 mC
Note that ΔVab = Qab Cab = 24.0 mC 8.00 mF = 3.00 V, and that
ΔVcd = Qcd Ccd = 24.0 mC 8.00 mF = 3.00 V. We earlier found that ΔVbc = 3.00 V, so we
conclude that the potential difference across each capacitor in the circuit is
Chapter 16
(a)
ΔV8 = ΔV2 = ΔV6 = ΔV8 = 3.00 V
The equivalent capacitance of the series combination
in the upper branch is
2/9/11 1:54:57 PM
Electrical Energy and Capacitance
16.16
(a)
1 triangle, 1each of the
2 +identical
1
1
At the center
of the
=
+
=
charges
produce
a fimF
eld contribution
of magnitude
3.00
6.00 mF 6.00
mF
Cupper
E1 = ke q a 2 . The three contributions are oriented
as shown
or
Cupperat= the
2.00right
mF and the components of the
resultant field are:
Likewise, the equivalent capacitance of the series
combination in the C
lower =branch
is
2.00 mF
upper
1
Clower
1
1
2 +1
+
=
2.00 mF 4.00 mF 4.00 mF
137
497
3.00 mF 6.00 mF
2.00 mF 4.00 mF
1
1
2 capacitance
+1
or of the series
= Likewise,
+ the equivalent
=
2.00
mF 4.00
4.00branch
mF is
combination
in mF
the lower
Clower = 1.33 mF
90.0 V
Clower = 1.33 mF 1
Clower
These two equivalent capacitances are connected in parallel with each other, so the equivaThese two equivalent capacitances are connected
incapacitance
parallel withfor
each
the equivalent
the other,
entire so
circuit
is
lent capacitance for the entire circuit is
Ceq = Cupper + Clower = 2.00 mF + 1.33 mF = 3.33 mF
Ceq = Cupper + Clower = 2.00 mF + 1.33 mF = 3.33 mF
=
(b)
or
Note that the same potential difference, equal to the potential difference of the battery,
exists across both the upper and lower branches. The charge stored on each capacitor in the
series combination in the upper branch is
Q3 = Q6 = Qupper = Cupper ( ΔV ) = ( 2.00 mF )( 90.0 V ) = 180 mC
and the charge stored on each capacitor in the series combination in the lower branch is
Q2 = Q4 = Qlower = Clower ( ΔV ) = (1.33 mF )( 90.0 V ) = 120 mC
(c)
The potential difference across each of the capacitors in the circuit is:
ΔV2 =
Q2 120 mC
=
= 60.0 V
C2 2.00 mF
ΔV4 =
Q4 120 mC
=
= 30.0 V
C4 4.00 mF
Q216 mC
180 =mC27.0 V Note that ΔV +QΔV
180 mC
V as
ΔV3 == 3 =
= 60.0 V
ΔV6 8= 6 =24 = ΔV = 36.0
= 30.0
V it should.
mF
3.00
mF
6.00
mF
C8.00
C
3
6
16.39
16.38
1.
(a)
The equivalent capacitance of the series
combination in the rightmost branch of
the circuit is
The circuit may be reduced in steps as shown above.
Using Figure 3,
Qac = ( 4.00 mF )( 24.0 V ) = 96.0 mC
Qac 96.0 mC
Then, in Figure 2, ( ΔV )ab = C = 6.00 mF = 16.0 V
ab
and
( ΔV )bc = ( ΔV )ac − ( ΔV )ab = 24.0 V − 16.0 V = 8.00 V
Finally, using Figure 1,
Q1 = C1 ( ΔV )ab = (1.00 mF )(16.0 V ) = 16.0 mC ,
67352_ch16.indd 497
Q8 = (8.00 mF )( ΔV )bc = 64.0 mC ,
and
Q5 = ( 5.00 mF )( ΔV )ab = 80.0 mC
Q4 = ( 4.00 mF )( ΔV )bc = 32.0 mC
2/9/11 1:54:59 PM
498
138
Chapter 16
Qac = ( 4.00 mF )( 24.0 V ) = 96.0 mC
(b) At the point (x, 0), where
Q 0 < 96.0
mC m, the potential is
x < 1.20
Then, in Figure 2, ( ΔV )ab = ac =
= 16.0 V
Cab 6.00 mF
ke ( −2q )
kq
2⎞
⎛ V 1
ΔV )ac −=( ΔV
)ab = 24.0
and ( ΔV )bc = (∑
− ⎟
+ V −e 16.0 V
= k=e q8.00
⎜⎝
1.20
m
−
x
x⎠
x
1.20
m
−
x
i
Finally,
or using Figure 1,
Q1 = C1 ( ΔV )ab = (1.00 mF )(16.0 V ) = 16.0 mC ,
Q8 = (8.00 mF )( ΔV )bc = 64.0 mC ,
16.41
.
(a)
and
Q5 = ( 5.00 mF )( ΔV )ab = 80.0 mC
Q4 = (. 4.00 mF )( ΔV )bc = 32.0 mC
Capacitors in a series combination store the same charge, Q = Ceq (ΔV ), where Ceq is the
equivalent capacitance and ΔV is the potential difference maintained across the series combination. The equivalent capacitance for the given series combination is 1 Ceq = 1 C1 + 1 C2,
or Ceq = C1C2 (C1 + C2 ), giving
Ceq =
( 2.50 mF ) (6.25 mF ) = 1.79 mF
2.50 mF + 6.25 mF
and the charge stored on each capacitor in the series combination is
Q = Ceq ( ΔV ) = (1.79 mF )( 6.00 V ) = 10.7 mC
(b)
When connected in parallel, each capacitor has the same potential difference, ΔV = 6.00 V,
maintained across it. The charge stored on each capacitor is then
Q1 = C1 ( ΔV ) = ( 2.50 mF )( 6.00 V ) = 15.0 mC
For C1 = 2.50 mF:
The charges are 89.4 mC on the 20 mF capacitor, 63.0 mC on the 6 mF capacitor, and
For CmC
Q2 = C2 ( ΔV ) = ( 6.25 mF )( 6.00 V ) = 37.5 mC
6.25
mF:the 15 mF and 3 mF capacitors.
2 = on
26.3
both
1.
16.45
16.46
16.47
1.
Q2 1
1
2
2
Energy stored =
= C ( ΔV ) = ( 4.50 × 10 −6 F )(12.0 V ) = 3.24 × 10 −4 J
= 9.79 kg
2
2C 2
6
⎡⎣
+ 2.26 × 10 J kg ⎤⎦
The energy transferred to the water is
(a) The energy initially stored in the capacitor is
( Energy stored )
1
(b)
=
Qi2
1
1
2
2
= Ci ( ΔV )i = ( 3.00 mF )( 6.00 V ) = 54.0 mJ
2
2Ci 2
When the capacitor is disconnected from the battery, the stored charge becomes isolated
with no way off the plates. Thus, the charge remains constant at the value Qi as long as
the capacitor remains disconnected. Since the capacitance of a parallel-plate capacitor
is C = k ∈0 A d, when the distance d separating the plates is doubled, the capacitance is
decreased by a factor of 2 C f = Ci 2 = 1.50 mF . The stored energy (with Q unchanged)
becomes
(
( Energy stored )
(c)
2
⎛ Q2 ⎞
Qi2
Qi2
=
= 2 ⎜ i ⎟ = 2 ( Energy stored )1 = 108 mJ
2C f 2 (Ci 2 )
⎝ 2Ci ⎠
When the capacitor is reconnected to the battery, the potential difference between the plates
is reestablished at the original value of ΔV = ( ΔV )i = 6.00 V, while the capacitance remains
at C f = Ci 2 = 1.50 mF. The energy stored under these conditions is
( Energy stored )
3
67352_ch16.indd 498
=
)
=
1
1
2
2
C f ( ΔV )i = (1.50 mF )( 6.00 V ) = 27.0 mJ
2
2
2/9/11 1:55:00 PM
Electrical Energy and Capacitance
= 1.50 mF. The energy stored under these conditions is
16.49
1.
(a)
(c)
139
499
1 momentum,
1
From conservation of linear
( Energy stored )3 = C f ( ΔV )i2 = (1.50 mF )(6.00 V )2 = 27.0 mJ
2
2
2
Qi2
⎛ m⎛ pQ⎞i ⎞
va ==⎜2 ⎜ ⎟ v⎟p = 2 ( Energy stored )1 = 108 mJ
ma va + m p v p = 0
or=
[2]
2 (remains
Ci 2 ) ⎝constant
Note that the charge on the plates
m⎝a2C
⎠ i ⎠ at the original value, Q0 , as the dielectric is inserted. Thus, the change in the potential difference, ΔV = Q C, is due to a change
When
the capacitor
reconnected
to the
battery,
the potential
difference
between the plates
in
capacitance
alone.isThe
ratioEquation
of the
final
and
initial
capacitances
is
Substituting
Equation
[2] into
[1]
gives
is reestablished at the original value of ΔV = ( ΔV )i = 6.00 V, while the capacitance remains
f
at C f C
=C
2k ∈0 A d = k and C f = Q0 ( ΔV ) f = ( ΔV )i = 85.0 V = 3.40
i=
Ci
Q0 ( ΔV )i ( ΔV ) f 25.0 V
∈0 A d
Ci
Thus, the dielectric constant of the inserted material is k = 3.40 , and the material is
probably nylon (see Table 16.1).
1.
16.51
(b)
If the dielectric only partially filled the space between the plates, leaving the remaining
space air-filled, the equivalent dielectric constant would be somewhere between k = 1.00
9.50
× 10 −8 Fpotential
0.025 0difference
× 10 −3 m )would then lie somewhere between
(
)
(
(air) and k = 3.40. The
resulting
= 1.04 m
−12
2
and)((8.85
ΔV )×f 10
= 25.0
( ΔV )i = 85.0= (V3.70
CV.
N ⋅ m 2 ) ( 7.00 × 10 −2 m )
(a)
The dielectric constant for Teflon® is k = 2.1, so the capacitance is
−12
2
2
−4
2
k ∈0 A ( 2.1) (8.85 × 10 C N ⋅ m ) (175 × 10 m )
C=
=
d
0.040 0 × 10 −3 m
C = 8.1× 10 −9 F = 8.1 nF
(b)
For Teflon®, the dielectric strength is Em ax = 60 × 10 6 V m, so the maximum voltage is
ΔVm ax = Em ax d = ( 60 × 10 6 V m ) ( 0.040 0 × 10 −3 m
ΔVm ax = 2.4 × 10 3 V = 2.4 C
kV
p ∓
16.55
1.
)
1 2
C p − C p Cs
4
Since the capacitors are in series, the equivalent capacitance is given by
d
d + d 2 + d3
1
1
1
1
d
d
=
+
+
= 1 + 2 + 3 = 1
Ceq C1 C2 C3 ∈0 A ∈0 A ∈0 A
∈0 A
16.56
1.
16.57
= ( Energy stored )total
∈0 A
where d = d1 + d 2 + d3
or Ceq =
Thus, thed sums of the energies
stored in the individual capacitors equals the total energy
stored by the system.
(a) Please refer to the solution of Problem 16.37 where
In the the
absence
of a dielecfollowing
results were obtained:
tric, the capacitance of the
parallel-plate capacitor is
1
1
d
d
c1
k
k
3
3
C0 = ∈0 A d.
2
With the dielectric inserted,
d
d
3
it fills one-third of the gap
between the plates as shown
2
d
c2
in sketch (a) at the right. We
3
model this situation as consisting of a pair of capaci(a)
tors, C1 and C2, connected
in series as shown in sketch
(b)
(b) at the right. In reality,
the lower plate of C1 and the
upper plate of C2 are one
and the same, consisting of the lower surface of the dielectric shown in sketch (a). The capacitances in the model of sketch (b) are given by
C1 =
67352_ch16.indd 499
k ∈0 A 3k ∈0 A
=
d 3
d
and
C2 =
∈0 A 3 ∈0 A
=
2d 3
2d
The equivalent capacitance of the series combination is
2/9/11 1:55:03 PM
1
d
3
With the dielectric inserted,
k
With the dielectric inserted, it fills one-third of the gap
With the dielectric inserted,
it fills one-third of the gap between the plates as shown
2
it fills one-third of the gapbetweendthe platesdas shown in sketch (a) at the right. We
3
between the plates as shown
in sketch (a) at the right. We model this situation as conin sketch (a) at the right. We
model this situation as con- sisting of a pair of capacimodel this situation as consisting of a pair of capaci- tors, C1 and C2, connected
sisting
of
a
pair
of
capacitors, C1 and C2, connected in series as shown in sketch
Chapter 16
tors, C1 and C2, connected in series as shown in sketch (b) at the right. In reality,
(a)
the lower plate of C1 and the
in series as shown in sketch
(b) at the right. In reality,
(d)
of energy,
ΔKE +plate
ΔPEof=C0,1 and
gives
speed
of the
alpha
particle at infinity in
upper
plate
of C
(b) at Conservation
the right. In reality,
the lower
thethe
2 are one
2
1
the
situation
of
part
(c)
as
m
v
−
0
=
−ΔPE,
or
the
lower
plate
of
C
and
the
upper
plate
of
C
are
one
2
a surface
a
and the same, consisting
of the lower
of the dielectric shown in sketch (a). The capaci1
2
upper
C2 areofone
tancesplate
in theofmodel
sketch (b) are given by −13
−2 −2.17 × 10 J
−2 ( ΔPE )
=
= 8.08
× 10 6 m s
∈
A
∈−27
3∈
kva∈0=A 3k
0
0 A
0 A
m
6.64
×
10
kg
= a
and C2 =
=
C1 =
2d 3
2d
d 3
d
(e) When, starting with the three-particle system, the two protons are both allowed to escape
The equivalent
ofno
theremaining
series combination
is
to infinity,capacitance
there will be
pairs of particles
and hence no remaining potential energy. Thus, ΔPE = 0 − PEb = −PEb, and conservation of energy gives the change
⎞ ⎛. Since
2k + 1the
1 in kinetic
d
2d
1⎞ 1
⎞ d
⎛1
⎞ ⎛ d= +PE
⎛ 2k + 1 ⎞ d particles,
⎛ 2k +this
=b⎜
=
+energy as=ΔKE
= ⎜are identical
=⎜
⎟⎠ protons
⎜ +=2−ΔPE
⎟⎜
⎟⎠
⎟
⎟
1
⎝
⎝
⎠
⎝
⎝
k
3 ∈0 giving
Ceq increase
3k ∈0 Ain kinetic
3 ∈0 A energy
k is split
A ⎠ between
A
3k proton∈=0 2A
3k ⎠ C0
⎝ 3 ∈equally
them
KE
0
.
140
500
(
)
and Ceq = [ 3k (2k + 1) ]C0 .
1.
16.61
The stages for the reduction of this circuit are shown below.
16.58
For a parallel-plate capacitor,
Thus, Ceq = 6.25 mF
1.
16.63
Electrical Energy and Capacitance
515
For a parallel-plate capacitor with plate separation d,
ΔVm ax = Em ax ⋅ d
d=
or
ΔVm ax
Em ax
The capacitance is then
C=
⎛ E
⎞
k ∈0 A
= k ∈0 A ⎜ m ax ⎟
d
⎝ ΔVm ax ⎠
and the needed area of the plates is A = C ⋅ ΔVm ax k ∈0 Em ax, or
A=
( 0.250 a× 10
−6
F ) ( 4.00 × 10 3 V )
2
N∈⋅0ma2 ) ( 2.00 × 108 V m )
( 3.00 )(8.85 ×=10 −12=C4p
= 0.188 m 2
ke
16.64
16.65
1.
(a)
Due to initially
sphericalstored
symmetry,
charge on
The charges
on thethe
capacitors
areeach of the concentric spherical shells will be
uniformly distributed over that shell. Inside a spherical surface having a uniform charge
distribution,
the ×charge
on that surface is zero. Thus, in this region,
Q1 = C1 ( ΔV )the
= electric
250due
V )to
= 1.5
10 3 mC
(6.0 mFfi)(eld
i
the potential due to the charge on that surface is constant and equal to the potential at the
spherical
surface
charge distribution, the potential due
and surface.
Q2 = C2 Outside
mF )( 250
V ) =having
5.0 × 10a2 uniform
mC
( ΔV )i = a( 2.0
to the charge on that surface is given by V = ke q r
continued on next page
67352_ch16.indd 500
2/9/11 1:55:06 PM
516
16.25
Electrical Energy and Capacitance
Chapter 16
501
141
ΔV
20.0 V
When E
the= capacitors
are connected
in parallel,
negative
of one
connected
to the
(a)
toward
the negative
plate
=
m = the
11.1
kV m plate
= 1.11
× 10 4 Vwith
−3
d
1.80
×
10
m
positive pl ate of the other, the net stored charge is
Q = Q∈− A
Q = 1.5 × 10 3 mC − 5.0 × 10 2 mC = 1.0 × 10 3 mC
(b) C = 1 0 2
d
The equivalent capacitance of the parallel combination is Ceq = C1 + C2 = 8.0 mF. Thus, the final
potential difference across each of the capacitors is
( ΔV )′ =
Q 1.0 × 10 3 mC
=
= 125 V
Ceq
8.0 mF
and the final charge on each capacitor is
Q1′ = C1 ( ΔV )′ = ( 6.0 mF )(125 V ) = 750 mC = 0.75 mC
and
67352_ch16.indd 501
Q2′ = C2 ( ΔV )′ = ( 2.0 mF )(125 V ) = 250 mC = 0.25 mC
2/9/11 1:55:09 PM
1장 전զࠪࢷ̛ࢵତ
PROBLEM SOLUTIONS
17.1
1.
The resistance of a wire of length L and diameter d is R = rL A = rL (p d 2 4 ), giving
d 2 = 4rL p R. Using Table 17.1, the needed diameter is found to be
4 ( 2.82 × 10 −8 Ω ⋅ m )( 32.0 m )
4rL
=
= 6.78 × 10 −4 m = 0.678 mm
s 1 h )Ω )
( p R )(1 h )(3 600
p ( 2.50
21
= 3.4 × 10 electrons
1.60 × 10 −19 C electron
Since
I = ΔQ
Δt electron in its orbit is T = 2p r v, and the current represented by the orbiting
The period
of the
d=
17.2
1.
17.3
electron is I = ΔQ Δt = e T = v e 2p r, or
( 2.19 × 106 m s) (1.60 × 10 −19 C) = 1.05 × 10−3 C s = 1.05 mA With a greater
Inumber
=
of charge carriers
in motion, they do not have to move as fast to have a specified
2p ( 5.29 × 10 −11 m )
number of them
passing a given point each second.
1.
17.5
(a)
The charge that moves past the cross section is ΔQ = I ( Δt ), and the number of electrons is
n=
=
(b)
1.
17.9
17.6
ΔQ I ( Δt )
=
e
e
(80.0 × 10
−3
C s ) ⎡⎣(10.0 min )( 60.0 s min ) ⎤⎦
= 3.00 × 10 20 electrons
1.60 × 10 −19 C
The negatively charged electrons move in the direction
opposite to the conventional
= 0.130current
mm s .flow.
The
of electrons
in the
line is=vM
I nqA = I n e (p d 2 4 ). The time to travel the
d =N
The drift
massspeed
of a single
gold atom
is m
atom
A
length of the 200-km line is then Δt = L vd = Ln e (p d 2 ) 4I, or
( 200 × 103 m ) ( 8.5 × 1028 m3 ) (1.6 × 10−19 C) p ( 0.02 m ) = 27 yr
Δt =
7
Rubber-soled shoes4 (and
rubber
gloves
can
increase
1 000
A ) ( 3.156
× 10
s yr ) the resistance to current and help reduce
the likelihood of a serious shock.
2
(b)
17.10
17.11
1.
ΔV 1.20 × 10 2 V
=
= 13.0 Ω
I
9.25 A
(a)
(a)
From Ohm’s law,
From Ohm’s law,
(b)
Using R = rL A and data from Table 17.1, the required length is found to be
R=
2
−3
RA R (p r ) (13.0 Ω )p ( 0.791× 10 m )
Current and Resistance
L=
=
=
= 17.0 m
r
r
150 × 10 −8 Ω ⋅ m
2
17.12
17.13
1.
525
−6
Using
A
A) R
( ΔV )m axR == IrL
m ax R = ( 80 × 10
Thus, if R = 4.0 × 10 5 Ω, ( ΔV )m ax = 32 V
and if R = 2 000 Ω, ( ΔV )m ax = 0.16 V
142
67352_ch17.indd 518
2/9/11 1:53:10 PM
=
1.
17.15
17.16
1.
17.17
ΔV
( 6.0 V )( 0.050 m )
=
= 1.8 × 10 7 A
r ( L A )m ax
1.7 × 10 −8 Ω ⋅ m
Current and Resistance
523
143
ΔV
12 V
=
= 30 Ω
I
0.40 A
(a)
R=
(b)
From R = rL A,
4 (1.12 × 10 −7 m 3 )
=
p (1.82 ⎤m )
−2
⎡
R ⋅ A ( 30 Ω ) ⎣p ( 0.40 × 10 m ) ⎦
r=
=
= 4.7 × 10 − 4 Ω ⋅ m
= 2.8
L × 10 −4 m = 0.28
3.2mm
m
The volume of the copper is
If a conductor of length L has a uniform electric field E maintained within it, the potential
difference between
the ends of the conductor is ΔV = EL. But, from Ohm’s law, the
m
V =between the potential difference across a conductor and the current through it is
relation
density
ΔV = IR, where R = r L A. Combining these relations, we obtain
p (1.60 Ω )( 0.800 × 10 −3 m )
−8
ΔV = EL ==IR = I ( r L A )
or
E ==r3.22
= rJ Ω ⋅ m
( I A×) 10
4 ( 25.0 m )
2
17.19
1.
17.18
2
The
Fromvolume
R = rLofAmaterial, V = AL0 = (p r0 ) L0 , in the wire is constant. Thus, as the wire is
stretched to decrease its radius, the length increases such that p rf2 L f = (p r02 ) L0 giving
2
2
2
L f = r0 rf L0 = ( r0 0.25r0 ) L0 = ( 4.0 ) L0 = 16L0 . The new resistance is then
(
Rf = r
1.
17.21
17.20
( )
)
Lf
Af
=r
⎛ L ⎞
16L 2
= r = 1.20 ×0 102 =V256 ⎜ r 02 ⎟ = 256R0 = 256 (1.00 Ω ) = 256 Ω
=
8.89
pr
r0 ⎠
⎝ pΩ
p ( 0.25r
13.50 )A
Lf
2
f
From Ohm’s law, R = ΔV I , and R = rL A = rL (p d 2 4 ), the resistivity is
−3
p Rd 2 p ( ΔV ) d 2 p ( 9.11 V )( 2.00 × 10 m )
r=
=
= 1.59 × 10 −8 Ω ⋅ m
=
4L
4IL
4 ( 36.0 A )( 50.0 m )
41.4 Ω − 41.0 Ω
−1
=
= 1.1× 10 −3 (°C )
Then, from Table 17.1, we
seeΩthat
the wire
is made
− 20.0°C
( 41.0
)( 29.0°C
) of silver .
2
17.22
17.23
1.
If
41.0resistance
Ω at T = 20.0°C
and
(a)R = The
at 20.0°C
is R = 41.4 Ω at T = 29.0°C, then R = R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦ gives
the temperature coefficient of resistivity of the material making up this wire as
−8
L (1.7 × 10 Ω ⋅ m )( 34.5 m )
R0 = r =
= 3.0 Ω
2
A
p ( 0.25 × 10 −3 m )
and the current will be I =
(b)
ΔV 9.0 V
=
= 3.0 A
R0
3.0 Ω
Current and Resistance
527
At 30.0°C,
R = R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦
( ) − 1 = 20.0°C + ( 46.2 Ω 30.0 Ω) − 1 = 1.6 × 10 2 °C
−1
a
3.9 × 10 −3 (°C )
−1
−3
⎡
1+
3.9
×
10
°C
30.0°C
−
20.0°C
=
3.0
Ω
(
)
(
)
(
)⎤⎦ = 3.1 Ω
continued on next page
⎣
(b) The expansion of the cross-sectional area contributes slightly more than the expansion of
ΔV 9.0 V
Thus, the
the length
currentofisthe
I =wire, so
= the answer
= 2.9would
A . be slightly reduced.
R
3.1 Ω
(
17.24
17.27
1.
For aluminum, the resistivity at room temperature is
At 80°C,
I=
or
67352_ch17.indd 523
)
5.0 V
ΔV
ΔV
=
=
2
R
⎡
⎤
⎡
R0 ⎣1+ a ( T − T0 ) ⎦ ( 2.0 × 10 Ω ) ⎣1+ ( − 0.5 × 10 −3 °C −1 )(80°C − 20°C )⎤⎦
I = 2.6 × 10 −2 A = 26 mA
2/9/11 1:53:23 PM
524
144
=
Chapter 17
5.0 V
ΔV
=
2
R0 ⎡⎣1+ a ( T − T0 ) ⎤⎦ ( 2.0 × 10 Ω ) ⎡⎣1+ ( − 0.5 × 10 −3 °C −1 )(80°C − 20°C )⎤⎦
(d) or With
I =two
2.6 conduction
× 10 −2 A = electrons
26 mA per iron atom, the density of charge carriers is
1.
17.29
From Ohm’s law, ΔV = I i Ri = I f R f , so the current in Antarctica is
⎛ R ⎡1+ a ( T − T ) ⎤ ⎞
⎛ 1+ ⎡3.9 × 10 −3 ( °C )−1 ⎤ ( 58.0°C − 20.0°C ) ⎞
⎛ Ri ⎞
0 ⎣
i
0 ⎦
⎣
⎦
⎟ = (1.00 A ) ⎜
I f = Ii ⎜ ⎟ = Ii ⎜
⎟
⎜⎝ 1+ ⎡3.9 × 10 −3 ( °C )−1 ⎤ ( −88.0°C − 20.0°C ) ⎟⎠
⎜ R0 ⎡1+ a T f − T0 ⎤ ⎟
⎝ Rf ⎠
⎣
⎦
⎝ ⎣
⎦⎠
2
5.6 Ω
= p (1.50 × 10 −3 m )
= 26 m (Nichrome)
or I f = 2.0 A
150 × 10 −8 Ω ⋅ m
(
1.
17.31
)
The volume of the gold wire may be written as V = A ⋅ L = m rd , where rd is the density of gold.
Thus, the cross-sectional area is A = m rd L. The resistance of the wire is R = re L A, where re is
the electrical resistivity. Therefore,
−8
3
3
3
r r L2 ( 2.44 × 10 Ω ⋅ m ) (19.3 × 10 kg m ) ( 2.40 × 10 m )
rL
= e d =
R= e
m rd L
m
1.00 × 10 −3 kg
253.8 Ω − 217 Ω
20.0°C
+ MΩ
= 63.3°C
= 2.71
R = 2.71× 10=6 Ω
⎡⎣3.92 × 10 −3 ( °C )−1 ⎤⎦ ( 217 Ω )
giving
17.33
1.
(a)
The power consumed by the device is P = I ( ΔV ), so the current must be
I=
(b)
17.34
(a)
17.35
1.
(a)
17.36
1.
17.37
(c)
R=
ΔV 1.20 × 10 2 V
=
= 14.4 Ω
I
8.33 A
The power loss in the line is Ploss = I 2 R = (1 000 A ) ⎡⎣( 0.31 Ω km )(160 km )⎤⎦, or
1.7 × 10 −8 Ω ⋅ m )(1.00 m )
r ×L 10 74r
Ploss = 5.0
W =L 504 (MW
RC u = C u = C u2 =
= 5.2 × 10 −3 Ω
2
−2
A
pd
p ( 0.205 × 10 m )
2
PC u = I 2 RC u = ( 20.0 A ) ( 5.2 × 10 −3 Ω ) = 2.1 W
2
−8
rAl L 4rAl L 4 ( 2.82 × 10 Ω ⋅ m )(1.00 m )
RAl =
=
=
= 8.54 × 10 −3 Ω
2
−2
A
pd2
p ( 0.205 × 10 m )
PAl = I 2 RAl = ( 20.0 A ) (8.54 × 10 −3 Ω ) = 3.42 W
2
and
530
P
1.00 × 10 3 W
=
= 8.33 A
ΔV 1.20 × 10 2 V
From Ohm’s law, the resistance is
and
(b)
2
No , the aluminum wire would not be as safe. If surrounded by thermal insulation, it would
get much hotter than the copper wire.
Chapter 17
(a) The energy used by a 100-W bulb in 24 h is
2
The power dissipated in a conductor is P = ( ΔV ) R, so the resistance may be written as
2
R = ( ΔV ) P. Hence,
( ΔV )
PA
P
RB
=
⋅
= A =3
2
RA
PB
( ΔV ) PB
2
or
RB = 3RA
Since R = rL A = rL (p d 2 4 ), this result becomes
⎛ 4rL ⎞
= 3⎜
2 ⎟
pd
⎝ p dA ⎠
4rL
2
B
or
d A2
=3
d B2
and yields d A d B = 3 .
17.38
67352_ch17.indd 524
(a)
2/9/11 1:53:25 PM
Current and Resistance
=3
17.13
−6
ΔV yields
(and
)m ax = I mdax Rd= (=80 ×310
( )R
. A
A
B
17.43
1.
−1
461°C
Thus, if R = 4.0 × 10
m ax
(a)
P = ( ΔV ) I = ( 75.0 × 10 −3 V ) ( 0.200 × 10 −3 A ) = 1.50 × 10 −5 W = 15.0 × 10 −6 W = 15.0 mW,
and if making
R = 2 000
Ω, ineffi
= sources
0.16 Vof light.
( ΔV )cient
them
m ax
5
17.38
1.
17.41
) − (120 1.80 )
=
( °C ) ) (120 1.80 )
( 0.400 × 10
Ω, ( ΔV ) = 32 V
−3
145
525
The power loss per unit length of the cable is P L = (I 2 R) L = I 2 (R L). Thus, the resistance per
unit length of the cable is
R P L 2.00 W m
= 2 =
= 2.22 × 10 −5 Ω m
L
I
( 300 A )2
1.
From R = rL A, the resistance per unit length is also given by R L = r A. Hence, the
cross-sectional area is p r 2 = A = r ( R L ), and the required radius is
r
1.7 × 10 −8 Ω ⋅−8m
=
⎡ 4 (150 ×−510 Ω ⋅ m=)0.016
( 3.00 m )=⎤ 1.6 cm −4
⎡
⎤
p ( R⎢ L ) ⎥ p=( 2.22
× 10 Ω m ) 1
⎢
⎥ = 3.8 × 10 m = 0.38 mm
p ( 4.0 × 10 Ω )
⎢⎣
⎥⎦
⎣
⎦
r=
1.
17.47
1
2
1
2
The energy required to bring the water to the boiling point is
E = m c ( ΔT ) = ( 0.500 kg ) ( 4 186 J kg ⋅°C )(100°C − 23.0°C ) = 1.61× 10 5 J
The power input by the heating element is
Pinput = ( ΔV ) I = (120 V )( 2.00 A ) = 240 W = 240 J s
Current and Resistance
533
Therefore, the time required is
t=
17.48
1.
17.49
E
1.61× 10 5 J
⎛ 1 min ⎞
=
= 671 s ⎜
= 11.2 min
⎝ 60 s ⎟⎠
Pinput
240 J s
(
) = $0.079 = 7.9 cents
(a) The rating2 of the 12-V battery is I ⋅ Δt = 55 A ⋅ h. Thus, the stored energy is
From P = ( ΔV ) R, the total resistance needed is R = (ΔV )2 P = (20 V)2 48 W = 8.3 Ω.
Thus, from R = rL A, the length of wire required is
17.50
17.53
1.
8.3 Ω )( 4.0 × 10 −6 m 2 )
R ⋅ A (that
3
(b) LWe
the temperature
coeffi
cient
resistivity
for Nichrome remains constant
= assume
=
m
= 1.1 km
= 1.1×
10of
−8
r
3.0
×
10
Ω
⋅
m
over this temperature range.
The energy available in the battery is
The battery is rated to deliver the equivalent of 60.0 amperes of current (i.e., 60.0 C/s) for 1 hour.
This is
× 10 5 C
Q = I ⋅ Δt = ( 60.0 A )(1 h ) = ( 60.0 C s ) ( 3 600
) = 2.16
by a s4.00
A fuse.
(a)
17.56
17.59
1.
ΔV 9.00 V
=
= 6.00 A
18 C
R
1.50= Ω
= 3.6 A
5.0 s
Note: All potential differences in this solution have a value of ΔV = 120 V. First, we shall do a
2
2
The cross-sectional
of the
conducting
material
A = enter
p ( router
rinner
) . ed numeric values for
symbolic
solution forarea
many
parts
of the problem
andisthen
the− specifi
the cases of interest.
Thus,
From the marked specifications on the cleaner, its internal resistance (assumed constant) is
(b)
67352_ch17.indd 525
−8
rL
rL
4rL 4 ( 2.82 × 10 Ω ⋅ m )(15.0 m )
=
=
= 1.50 Ω
=
2
A pd2 4 pd2
p ( 0.600 × 10 −3 m )
17.55
1.
R=
I=
2/9/11 1:53:28 PM
=
146
526
17.59
Chapter 17
18 C
= 3.6 A
5.0 s
2
2
The cross-sectional area of the conducting material is A = p ( router
− rinner
).
4 (1.12 × 10 −7 m 3 ) ( 4 (1.12 ×) −101−7 m 3 )
d=
=
−1
−1
−2(1.8
p1010
L−35 (Ω°C⋅ m
4.5
×
−4.0
3.9) × 10 −3 ( °C )
)
(5.00
3.5
×
× 10p4.75
m) )m
(
)
(
rL
7
= 3.7 × 10 Ω = 37 MΩ
R=
=
2
2
−2
A = 2.8
−4
p ⎡×(1.2
× 10
− mm
0.502 × 10 −2 m ) ⎤
2 = m
)
(
10
m
0.28
= 20.0 °C +⎣1.3 × 10 °C = 1.5 × 10 °C
⎦
Thus,
(
17.60
17.17
1.
17.61
Using chemical symbols to denote the two different metals, the resistances are equal when
If a conductor of length L has a uniform electric field E maintained within it, the potential
The currentbetween
in the wire
is I =ofΔV
= 15.0 V is
0.100
= 150
A.from
Then,Ohm’s
from the
for the
difference
the ends
theRconductor
ΔV =ΩEL.
But,
law,expression
the
drift
velocity,
v
=
I
nqA,
the
density
of
free
electrons
is
relation betweend the potential difference across a conductor and the current through it is
ΔV = IR, where R = r L A
I
150 A
n=
=
2
2
−4
vd e (p r ) ( 3.17 × 10 m s ) (1.60 × 10 −19 C ) p ( 5.00 × 10 −3 m )
or
67352_ch17.indd 526
)
n = 3.77 × 10 28 m 3
2/9/11 1:53:32 PM
1장 ऐզୣԻ
PROBLEM SOLUTIONS
1.
18.1
(a)
Connect two 50-Ω resistors in parallel to get 25 Ω. Then connect that parallel
combination in series with a 20-Ω resistor for a total resistance of 45 Ω.
(b)
Connect two 50-Ω resistors in parallel to get 25 Ω.
Also, connect two 20-Ω resistors in parallel to get 10 Ω.
Then, connect these two parallel combinations in series to obtain 35 Ω.
18.2
18.3
1.
(a)
(a)
When the 8.00-Ω resistor is connected across the 9.00-V terminal potential difference of the
The
bulbthe
acts
as a 192-Ω
resistor
(seeresistor
below),and
so the battery
circuit diagram
is:
battery,
current
through
both the
is
(b)
For the bulb in use as intended, Rbulb = ( ΔV ) P = (120 V ) 75.0 W = 192 Ω .
2
2
Now, assuming the bulb resistance is unchanged, the current in the circuit shown is
I=
ΔV
120 V
=
= 0.620 A
Req 0.800 Ω + 192 Ω + 0.800 Ω
and the actual power dissipated in the bulb is
P = ΔV
I 2 Rbulb 12
= (V
0.620 A ) (192 Ω ) = 73.8 W
I=
=
= 1.3 A
R
9.0 Ω
2
1.
18.5
(a)
The equivalent resistance of the two parallel
resistors is
1 ⎞
⎛ 1
Rp = ⎜
+
⎝ 7.00 Ω 10.0 Ω ⎟⎠
−1
= 4.12 Ω
Thus,
546
Chapter 18
(b)
Rab = R4 + Rp + R9 = ( 4.00 + 4.12 + 9.00 ) Ω = 17.1 Ω
I ab =
Also,
Then,
67352_ch18.indd 539
( ΔV )ab
Rab
=
34.0 V
= 1.99 A, so I 4 = I 9 = 1.99 A
17.1 Ω
( ΔV ) p = I ab Rp = (1.99 A )( 4.12 Ω ) = 8.20 V
147
2/9/11 1:57:04 PM
544
148
Chapter 18
(c)
18.6
52.
1.
18.7
( ΔV ) p = I ab Rp = (1.99 A )( 4.12 Ω ) = 8.20 V
Only the circuit
of )Figure
P18.50c.
In the other circuits, the batteries can be combined into
( ΔV
8.20
V
p
I7 =
= while the
= 1.17
A and 8.00-Ω resistors remain in parallel with each
aThen,
single effective
battery
5.00-Ω
7.00 Ω
R7
other.
( ΔV ) p 8.20 V
(d) The
The
and powerI10is=lowest in=Figure P18.50c.
= 0.820
A circuits in Figures P18.50b and P18.50d have in
10.0 the
Ω current.
R10 driving
effect 30-V batteries
= 2.00 A
(a) The parallel combination of the 6.00-Ω and 12.0-Ω resistors has an equivalent resistance of
When connected in series, we have R1 + R2 = 690 Ω
[1]
which we may rewrite as R2 = 690 Ω − R1
1
1
1
+
=
R1 R2 150 Ω
When in parallel,
[1a]
R1 R2
= 150 Ω
R1 + R2
or
Direct-Current Circuits
547
[2]
Substitute Equations [1] and [1a] into Equation [2] to obtain:
R1 ( 690 Ω − R1 )
= 150 Ω
continued on next page
690 Ω
R12 − ( 690 Ω ) R1 + ( 690 Ω )(150 Ω ) = 0
or
[3]
Using the quadratic formula to solve Equation [3] gives
R1 =
548
18.8
1.
18.9
690 Ω ±
( 690 Ω )2 − 4 ( 690 Ω )(150 Ω )
2
with two solutions of
R1 = 470 Ω
and R1 = 220 Ω
Then Equation [1a] yields
R2 = 220 Ω
or
Thus,18the
Chapter
(a)
(a)
R2 = 470 Ω
two resistors have resistances of 220 Ω and 470 Ω .
The equivalent resistance of this first parallel
Using
the rules
combination
is for combining resistors in series and parallel, the circuit reduces as shown
below:
10.0 Ω
10.0 Ω
25.0 V
− +
I
b
a
25.0 V
− +
10.0 Ω
I
b
a
25.0 V
− +
10.0 Ω
10.0 Ω
I
5.00 Ω
I20
c
20.0 Ω
5.00 Ω
I20
a
b
2.94 Ω
5.00 Ω
Step 1
25.0 Ω
Step 2
Step 3
From the figure of Step 3, observe that
I=
18.10
67352_ch18.indd 544
25.0 V
= 1.93 A and ΔVab = I ( 2.94 Ω ) = (1.93 A )( 2.94 Ω ) = 5.67 V
10.0 Ω + 2.94 Ω
ΔVab
5.67 V
=
= 0.227 A
25.0 Ω 25.0 Ω
(b)
From the figure of Step 1, observe that I 20 =
(a)
The figures below show the simplification of the circuit in stages:
2/9/11 1:57:17 PM
Direct-Current Circuits
= ( 2e 3) 3R = 2e 9R = 2I 3
1.
18.11
149
545
battery has
r = 0.15 is
Ω Rand
maintains a terminal potential difference of ΔV = 9.00 V while
The equivalent
resistance
eq = R + R p, where R p is the total resistance of the three parallel
supplying the current found above, the emf of this battery must be
branches:
−1
−1
= 9.17
1 + 0.17
Ω )( RΩ+ 5.0 Ω )
⎛e =1ΔV + Ir 1= 9.00 V +1(1.13⎞ A )( 0.15
⎛ 1Ω ) = ( 9.00
⎞ ) Ω( 30
Rp = ⎜
+
+
=⎜
+
=
⎟
⎟
⎝ 120 Ω 40 Ω R + 5.0 Ω ⎠
⎝ 30 Ω R + 5.0 Ω ⎠
R + 35 Ω
Thus,
75 Ω = R +
( 30 Ω )( R + 5.0 Ω ) R 2 + ( 65 Ω ) R + 150 Ω2
=
R + 35 Ω
R + 35 Ω
which reduces to R 2 − (10 Ω ) R − 2 475 Ω2 = 0 or ( R − 55 Ω )( R + 45 Ω ) = 0.
Only the positive solution is physically acceptable, so R = 55 Ω .
18.12
1.
18.13
The sketch at the right shows the equivalent circuit
The
in the
circuit
can be
combined
stages shown below to yield an equivalent
whenresistors
the switch
is in
the open
position.
For in
thisthe
simple
resistance
of
R
=
63
11
Ω.
)
series circuit, ad (
Figure 1
Figure 2
Figure 4
Figure 3
From Figure 5,
( ΔV )ad
I=
Then, from Figure 4,
Rad
=
18 V
(63 11) Ω
= 3.1 A
Figure 5
Direct-Current Circuits
551
( ΔV )bd = I Rbd = ( 3.1 A )( 30 11 Ω ) = 8.5 V
Now, look at Figure 2 and observe that
I2 =
( ΔV )bd
3.0 Ω + 2.0 Ω
=
8.5 V
= 1.7 A
5.0 Ω
so
67352_ch18.indd 545
2/9/11 1:57:20 PM
546
150
Chapter 18
so
(b)
=
8.5 V
= 1.7 A
5.0 Ω
( ΔV )
34.0 V
Ω )A,
= 5.1
( ΔV
so V
I ab )=be = I 2 Rabbe == (1.7 A )(=3.0
1.99
Rab
17.1 Ω
Finally,
from Figure 1, I12 =
Also,
18.14
18.15
1.
18.16
1.
18.17
( ΔV )be
R12
=
5.1 V
= 0.43 A
12 Ω
(a) The resistor network connected to the battery in
From ΔV = I ( R + r ), the internal resistance is
Figure P18.14 can be reduced to a single equivalent
resistance in the following steps. The equivalent
ΔV
9.00 V
combination
of Ω
the 3.00 Ω
rresistance
=
− Rof= the parallel
− 72.0
Ω = 4.92
I
0.117
and 6.00 Ω resistorsAis
(a) The equivalent resistance of the parallel
Goingcombination
counterclockwise
around
theb upper
1
between
points
and e loop,
is
applying Kirchhoff’s
loop
rule,
gives
Rp
+15.0 V − ( 7.00 ) I1 − ( 5.00 )( 2.00 A ) = 0
or
I1 =
15.0 V − 10.0 V
= 0.714 A
7.00 Ω
e
From Kirchhoff’s junction rule, I1 + I 2 − 2.00 A = 0
so
I 2 = 2.00 A − I1 = 2.00 A − 0.714 A = 1.29 A
Going around the lower loop in a clockwise direction gives
+ e − ( 2.00 ) I 2 − ( 5.00 )( 2.00 A ) = 0
or
18.19
1.
e = ( 2.00 Ω )(1.29 A ) + ( 5.00 Ω )( 2.00 A ) = 12.6 V
= 1.82 V with point b at the lower potential.
Consider the circuit diagram at the right, in which
Kirchhoff’s junction rule has already been
applied at points a and e.
R
+
−
e
Applying the loop rule around loop abca gives
I1 − I
e − R ( I1 − I ) − 4RI1 = 0
or
a
1⎛ e
⎞
I1 = ⎜ + I ⎟
5⎝ R ⎠
4R
I1
2R
c
b
d
+
− 2e
3R
I
I2 + I
I2
e
[1]
Next, applying the loop rule around loop cedc gives
−3RI 2 + 2e − 2R ( I 2 + I ) = 0
or
I2 =
2⎛ e
⎞
⎜ − I⎟
5⎝ R ⎠
[2]
Finally, applying the loop rule around loop caec gives
−4RI1 + 3RI 2 = 0
or
4I1 = 3I 2
Substituting Equations [1] and [2] into Equation [3] yields
[3]
I=
e
5R
Thus, if e = 250 V and R = 1.00 kΩ = 1.00 × 10 3 Ω, the current in the wire between a and e is
I=
18.20
67352_ch18.indd 546
250 V
= 50.0 × 10 −3 A = 50.0 mA flowing from a toward e.
5 (1.00 × 10 3 Ω )
Following the path of I1 from a to b and recording
changes in potential gives
2/9/11 1:57:22 PM
547
151
Direct-Current Circuits
Substitute
Equations
[1]Rand
[1a]kΩ
into= 1.00
Equation
Thus, if e =
250 V and
= 1.00
× 10 3[2]
Ω,to
theobtain:
current in the wire between a and e is
554
18.20
18.21
1.
R ( 690 Ω
− RV1 )
250
2
or−3 A =R150.0
− ( 690
150 Ω )e.= 0
I 1=
= 50.0
mAΩ
flowing
fromΩa)(toward
= 150
Ω × 10
) R1 + ( 690
3
690
Ω
5 (1.00 × 10 Ω )
[3]
Chapter 18
Using
the quadratic
formula
Equation
[3] gives
Following
the path
of
I1 from
asolve
to b and
recording
a
b
c
d
(a)
The
circuit
diagram
at to
the
changes
in
potential
gives
right shows the assumed
2
+
+
−
690 Ωof± the
Ω) −
( 690
40.0 V
current
in4 ( 690 Ω )(150 Ω )
Rdirections
− 360 V + 80.0 V −
1 =
each resistor. Note that the
200 Ω
80.0 Ω
20.0 Ω
70.0 Ω
total current flowing out of
the section of wire connecting
I2
I3
points g and f must equal the
I1
I4
current flowing into that
h
g
f
e
section. Thus,
I 3 = I1 + I 2 + I 4
[1]
Applying the loop rule around loop abgha gives
−200I1 − 40.0 + 80.0I 2 = 0
or
I2 =
1
(5I1 + 1.00 )
2
[2]
Next, applying the loop rule around loop bcfgb gives
+360 − 20.0I 3 − 80.0I 2 + 40.0 = 0
I 3 = 20.0 − 4I 2
or
[3]
Finally, applying the loop rule around the outer loop abcdefgha yields
−80.0 + 70I 4 − 200I1 = 0
or
I4 =
1
( 20I1 + 8.00 )
7
[4]
To solve this set of equations, we first substitute Equation [2] into Equation [3] to obtain
I 3 = 20.0 − 2 ( 5I1 + 1.00 )
or
I 3 = 18.0 − 10I1
[3ⴕ]
Now, substitute Equations [2], [4], and [3⬘] into Equation [1] to find
5
20 ⎞
1.00 8.00
⎛
−
⎜⎝ 10 + + 1+ ⎟⎠ I1 = 18.0 −
2
7
2
7
and
I1 = 1.00 A
Substituting this result back into Equations [2], [4], and [3⬘] gives
I 2 = 3.00 A
(b)
I 4 = 4.00 A
I 3 = 8.00 A
and
The potential difference across the 200-Ω resistor is
ΔV200 = I1 R200 = (1.00 A )( 200 Ω ) = 200 V
with point a at a lower potential than point h.
18.22
1.
18.23
(a)
(a)
18.23
(a) We name the currents I1 , I 2 , and I 3 as shown.
The 30.0-Ω and 50.0We name the currents I1 , I 2 , and I 3 as shown.
Applying Kirchhoff’s loop rule to loop abcfa
+ e 1 Kirchhoff’s
− e 2 − R2 I 2 −loop
R1 I1 rule
= 0 to loop abcfa gives + e 1 − e 2 − R2 I 2 − R1 I1 = 0
Applying
gives
or 3I 2 + 2I1 = 10.0 mA
or
and
I[1]
1 = 5.00 mA − 1.50I 2
I1 = 5.00 mA − 1.50I 2
[1]
Applying the loop rule to loop edcfe yields
+ e 3 − R3 I 3 − e 2 − R2 I 2 = 0
67352_ch18.indd 547
and
or
3I 2 + 4I 3 = 20.0 mA
2/9/11 1:57:26 PM
152
548
18.9
Chapter 18
(a)
I1 = 5.00 mA − 1.50I 2
[1]
Using the the
rules
for rule
combining
in series and parallel, the circuit reduces as shown
Applying
loop
to loop resistors
edcfe yields
below:
+ e 3 − R3 I 3 − e 2 − R2 I 2 = 0 or 3I 2 + 4I 3 = 20.0 mA
e
e
e
and
[2]
I 3 = 5.00 mA − 0.750I 2
10.0 Ω
Finally, applying Kirchhoff’s junction rule at junction c gives
I 2 = I1 + I 3
25.0 V
− +
[3]
Substituting Equations [1] and [2] into [3] yields
I 2 = 5.00 mA − 1.50I 2 + 5.00 mA − 0.750I 2
2.94 Ω
3.25I 2 = 10.0 mA
or
Step 3
This gives I 2 = (10.0 mA) 3.25 = 3.08 mA . Then, Equation [1] yields
⎛ 10.0 mA ⎞
I1 = 5.00 mA − 1.50 ⎜
= 0.385 mA
⎝ 3.25 ⎟⎠
10.0 mA ⎞
and, from Equation [2], I 3 = 5.00 mA − 0.750 ⎛⎜
= 2.69 mA
⎝ 3.25 ⎟⎠
(b)
Start at point c and go to point f, recording changes in potential to obtain
V f − Vc = − e 2 − R2 I 2 = −60.0 V − ( 3.00 × 10 3 Ω ) ( 3.08 × 10 −3 A ) = −69.2 V
or
1.
18.25
ΔV
cf
= 0.147
69.2 V
W and point c is at the higher potential .
=
= 0.081 6 or 8.16%
1.80 W
(a)
No. Some simplification could be made by recognizing that the 2.0-Ω and 4.0-Ω resistors
are in series, adding to give a total of 6.0 Ω; and the 5.0-Ω and 1.0-Ω resistors form a
series combination with a total resistance of 6.0 Ω. The circuit cannot be simplified any
further, and Kirchhoff’s rules must be used to analyze the circuit.
(b)
Applying Kirchhoff’s junction rule at junction a gives
I1 = I 2 + I 3
[1]
Using Kirchhoff’s loop rule on the upper loop yields
+ 24 V − ( 2.0 + 4.0 ) I1 − ( 3.0 ) I 3 = 0
or
I 3 = 8.0 A − 2 I1
For the lower loop:
[2]
+12 V + ( 3.0 ) I 3 − (1.0 + 5.0 ) I 2 = 0
Using Equation [2], this reduces to
I2 =
12 V + 3.0 (8.0 A − 2 I1 )
6.0
or
I 2 = 6.0 A − I1
[3]
Substituting Equations [2] and [3] into [1] gives I1 = 3.5 A .
Then, Equation [3] gives
18.26
67352_ch18.indd 548
I 2 = 2.5 A , and Equation [2] yields I 3 = 1.0 A .
Using Kirchhoff’s loop rule on the outer
perimeter of the circuit gives
2/9/11 1:57:28 PM
Direct-Current Circuits
549
153
.
1.
18.27
(a)
No.that
This
multi-loop
and
across
R4 is circuit does not contain any resistors in series (i.e., connected so all
the current in one must pass through the other) nor in parallel (connected so the voltage
drop across one is always the same as that across the other). Thus, this circuit cannot be
simplified any further, and Kirchhoff’s rules must be used to analyze it.
(b)
Assume currents I1 , I 2 , and I 3 in the directions shown.
Then, using Kirchhoff’s junction rule at junction a gives
I 3 = I1 + I 2
[1]
Applying Kirchhoff’s loop rule on the lower loop,
+10.0 V − ( 5.00 ) I 2 − ( 20.0 ) I 3 = 0
or
I 2 = 2.00 A − 4 I 3
[2]
and for the loop around the perimeter of the circuit,
or
20.0 V − 30.0I1 − 20.0I 3 = 0
I1 = 0.667 A − 0.667I 3
[3]
Substituting Equations [2] and [3] into [1]: I 3 = 0.667 A − 0.667I 3 + 2.00 A − 4 I 3
which reduces to 5.67I 3 = 2.67 A and gives I 3 = 0.471 A .
.
The
negative
sign
the answer
for I12Ameans
that this
current[3],
flows
the opposite
Then,
Equation
[2]ingives
I 2 = 0.116
, and from
Equation
I1 =in0.353
A.
direction to that shown in the circuit diagram and assumed during this solution. That is,
the
current
thedirections
middle branch
of the
circuit
flows
from right
left and has a
All actual
currents
are ininthe
indicated
in the
circuit
diagram
giventoabove.
magnitude of 0.416 A.
(a) Going counterclockwise around the upper loop,
Applying
Kirchhoff’s
rule at
Kirchhoff’s
loopjunction
rule gives
junction a gives
(h)
18.28
1.
18.29
I 3 = I1 + I 2
[1]
Using Kirchhoff’s loop rule on the leftmost
loop yields
−3.00 V − ( 4.00 ) I 3 − ( 5.00 ) I1 + 12.0 V = 0
so
I1 = ( 9.00 A − 4.00I 3 ) 5.00
or
I1 = 1.80 A − 0.800 I 3
[2]
For the rightmost loop,
−3.00 V − ( 4.00 ) I 3 − ( 3.00 + 2.00 ) I 2 + 18.0 V = 0
and
I 2 = (15.0 A − 4.00I 3 ) 5.00
or
I 2 = 3.00 A − 0.800 I 3
[3]
Substituting Equations [2] and [3] into [1] and simplifying gives 2.60I 3 = 4.80 and I 3 = 1.846 A.
Then Equations [2] and [3] yield I1 = 0.323 A and I 2 = 1.523 A.
Therefore, the potential differences across the resistors are
ΔV2 = I 2 ( 2.00 Ω ) = 3.05 V , ΔV3 = I 2 ( 3.00 Ω ) = 4.57 V
ΔV4 = I 3 ( 4.00 Ω ) = 7.38 V , and ΔV5 = I1 ( 5.00 Ω ) = 1.62 V
67352_ch18.indd 549
2/9/11 1:57:30 PM
I 2 = 3.00 A − 0.800 I 3
[3]
Substituting Equations [2] and [3] into [1] and simplifying gives 2.60I 3 = 4.80 and I 3 = 1.846 A.
Then Equations [2] and [3] yield I1 = 0.323 A and I 2 = 1.523 A.
Therefore, the potential differences across the resistors are
154
550
1.
18.31
18.32
1.
18.33
Chapter 18
ΔV2 = I 2 ( 2.00 Ω ) = 3.05⎛V , ΔV3 =⎞I⎛2 (Coulombs
3.00 Ω ) =⎞ 4.57
V
⎞
⎛ Coulombs
⎟⎠ ⎜⎝ Volts ⎟⎠ = ⎜⎝ Amperes ⎟⎠
⎜⎝
WhenΔV
the =switch
is closed
position
b, resistor
is Ω = 1.62 V
I 3 ( 4.00
Ω ) = in
7.38
V , and
ΔV5 = I1R(35.00
4
⎛ 1⎞ )
−t t
shorted out, leaving
R
and
R
in
series
with
the
=
Q
1−
e
=
Q
1−
( 1 ) m2ax (
) m ax ⎜⎝ e ⎟⎠ = 114 mC
⎞
⎛ gives
battery as shown in Figure 3. This
Coulombs
= Second−6
=
3
(a) The time constant is: t = ⎜⎝RC
=
75.0
×
10
Ω
× 10 F ) = 1.88 s .
Coulombs
( Second)⎟⎠( 25.0
6.00 V
=
−6
2.00
10 −3 A
(b)
= t has
, q =units
0.632Q
F )(12.0 V ) = 1.90 × 10 −4 C .
or tAt= tRC
of ×time.
m ax = 0.632 ( Ce ) = 0.632 ( 25.0 × 10
The
time constant is t = RC. Considering units, we find
and
(a) The time constant of an RC circuit is t = RC. Thus,
t = (1.00 × 10 6 Ω ) ( 5.00 × 10 −6 F ) = 5.00 s
(b)
(c)
Qm ax = Ce = ( 5.00 mF )( 30.0 V ) = 150 mC
To obtain the current through the resistor at time t after the switch
is closed, recall that the charge on the capacitor at that time is
q = Ce (1− e − t t ) and the potential difference across a capacitor is
VC = q C. Thus,
VC =
Ce (1− e
−t t
C
) =e
(1− e )
−t t
e
R
+
−
C
S
Then, considering switch S to have been closed at time t = 0, apply Kirchhoff’s loop rule
around the circuit shown above to obtain
+e − iR − VC = 0
or
i=
e − e (1− e − t t
)
R
The current in the circuit at time t after the switch is closed is then i = (e R ) e − t t , so the
current in the resistor at t = 10.0 s is
⎛
⎞ − 5.00 s
5
= ( 30.0 mA
) e−2.00 = 4.06 mA
⎟e
Ω ⎠ ) = −3.00 × 10 Ω = 5.87 × 10 5 Ω = 587 kΩ
ln (1 − 4.00 V 10.0 V )
ln ( 0.600 )
At
time
t
after
the
switch
is
closed,
the
potential
(a) The charge remaining on the capacitor
afterdifference
time t is qbetween
= Qe − t t the
. plates of the initially
uncharged capacitor is
Thus, if q = 0.750Q, then e − t t = 0.750 and −t t = ln ( 0.750 ),
30.0 V
(i = ⎜⎝ 1.00 × 10
18.34
18.35
1.
or
(b)
18.36
1.
18.37
10.0 s
6
t = −t ln ( 0.750 ) = − (1.50 s ) ln ( 0.750 ) = 0.432 s
t = RC , so
C=
t
1.50 s
=
= 6.00 × 10 −6 F = 6.00 mF
(R 250 × 10)3=Ω1.3 × 102 mC
Assuming the capacitor is initially uncharged and the switch is closed at t = 0, the charge on the
capacitor
time
0 isresistance
q = Qm ax (1of− the
e − t t element is
From
P = at
( ΔV
)2 tR,> the
R=
( ΔV )2
P
=
( 240 V )2
3 000 W
=19.2 Ω
When the element is connected to a 120-V source, we find that
67352_ch18.indd 550
ΔV 120 V
=
= 6.25 A , and
R 19.2 Ω
(a)
I=
(b)
P = ( ΔV ) I = (120 V )( 6.25 A ) = 750 W
2/9/11 1:57:33 PM
=
1.
18.39
18.40
18.41
1.
(120 V )2
25 Ω
= 5.8 × 10 2 W
551
155
Direct-Current Circuits
The current
drawn by
connected to a 120-V source is I1 = P ΔV = 75 W 120 V.
= I Rbulb
Then,
from Figure
4, a single
( ΔV )bd75-W
bd = ( 3.1 A ) ( 30 11 Ω ) = 8.5 V
Thus, the number of such bulbs that can be connected in parallel with this source before the total
current
drawn
will equal
A is that
Now,
look
at Figure
2 and30.0
observe
The total current required exceeds the limit of the circuit breaker, so they cannot be
30.0 Asimultaneously.
120 V ⎞ with a 15 A limit, no two of these appliances could be
noperated
=
= ( 30.0 =A8.5
) ⎛⎜⎝ VIn=fact,
1.7
⎟ =A48
I1 at the same 5.0
75
W ⎠ tripping the breaker.
Ωwithout
operated
time
(a) The current drawn by each appliance operating separately is
so
(a) The area of each surface of this axon membrane is
CoffeeA Maker:
= L ( 2p r ) = ( 0.10 m ) ⎡⎣ 2p (10 × 10 −6 m ) ⎤⎦ = 2p × 10 −6 m 2
and the capacitance is
C = k ∈0
⎛ 2p × 10 −6 m 2 ⎞
A
= 1.67 × 10 −8 F
= 3.0 (8.85 × 10 −12 C2 N ⋅ m 2 ) ⎜
d
⎝ 1.0 × 10 −8 m ⎟⎠
In the resting state, the charge on the outer surface of the membrane is
Qi = C ( ΔV )i = (1.67 × 10 −8 F ) ( 70 × 10 −3 V ) = 1.17 × 10 −9 C → 1.2 × 10 −9 C
The number of potassium ions required to produce this charge is
NK =
+
Qi 1.17 × 10 −9 C
=
= 7.3 × 10 9 K + ions
1.6 × 10 −19 C
e
and the charge per unit area on this surface is
−20
2
Qi 1.17 × 10 −9 C ⎛
1e
1e
⎞ ⎛ 10 m ⎞
=
=
=
⎜
⎟
−6
2
−19
2
⎟
⎜
A 2p × 10 m ⎝ 1.6 × 10 C ⎠ ⎝ 1 Å ⎠ 8.6 × 10 4 Å 2
s =
562
1e
( 290 Å )
2
This
Chapter 18
corresponds to a low charge density of one electronic charge per square of side 290 Å,
compared to a normal atomic spacing of one atom every few Å.
(b)
In the resting state, the net charge on the inner surface of the membrane is
− Qi = −1.17 × 10 −9 C, and the net positive charge on this surface in the excited state is
Q f = C ( ΔV ) f = (1.67 × 10 −8 F ) ( +30 × 10 −3 V ) = + 5.0 × 10 −10 C continued on next page
The total positive charge which must pass through the membrane to produce the excited
state is therefore
ΔQ = Q f − Qi
= + 5.0 × 10 −10 C − ( −1.17 × 10 −9 C ) = 1.67 × 10 −9 C → 1.7 × 10 −9 C
corresponding to
N Na =
+
(c)
If the sodium ions enter the axon in a time of Δt = 2.0 ms, the average current is
I=
(d)
ΔQ 1.67 × 10 −9 C
=
= 8.3 × 10 −7 A = 0.83 mA
Δt
2.0 × 10 −3 s
When the membrane becomes permeable to sodium ions, the initial influx of sodium ions
neutralizes the capacitor with no required energy input. The energy input required to charge
the now neutral capacitor to the potential difference of the excited state is
W=
67352_ch18.indd 551
18.42
ΔQ
1.67 × 10 −9 C
=
= 1.0 × 1010 Na + ions
e
1.6 × 10 −19 C Na + ion
2
1
1
2
C ( ΔV ) f = (1.67 × 10 −8 F ) ( 30 × 10 −3 V ) = 7.5 × 10 −12 J
2
2
The capacitance of the 10 cm length of axon was found to be C = 1.67 × 10 −8 F in the solution of
Problem 18.41.
2/9/11 1:57:35 PM
156
552
(d)
1.67 × 10 −9 C
= 8.3 × 10 −7 A = 0.83 mA
2.0 × 10 −3 s
When
the membrane
becomes
permeable
sodium
The
potential
difference
between
points b to
and
e is ions, the initial influx of sodium ions
neutralizes the capacitor with no required energy input. The energy input required to charge
the now
to the potential difference of the excited state is
ΔVbeneutral
= Rbe Icapacitor
total = ( 8.0 Ω )( 3.0 A ) = 24 V
so
18.42
1.
18.45
=
Chapter 18
W=
2
1
1
2
C ( ΔV ) f = (1.67 × 10 −8 F ) ( 30 × 10 −3 V ) = 7.5 × 10 −12 J
2
2
.
−8
The
ofopen
the 10
length
ofb,axon
was found
F in the solution
Sincecapacitance
the circuit is
at cm
points
a and
no current
flowsto be C = 1.67 × 10 2.00
4.00 V of
Ω
+ −
Problem
18.41.
through the
4.00-V battery or the 10.0-Ω resistor. A current I
a
will flow around the closed path through the 2.00-Ω resistor,
(a)
the and
membrane
becomes
4.00-ΩWhen
resistor,
the 12.0-V
batterypermeable
as shown to
in potassium
the sketch ions, these ions
+ flow out of the axon
12.0 VTo maintain this4.00
with no
energy
input
until the capacitor is neutralized.
outflΩow of
at the right.
This
current
hasrequired
magnitude
−
potassium ions and charge the now neutral capacitor to the resting action potential requires
I
an energy
of V
ΔV input12.0
b
I=
=
= 2.00 A
Rpath 2.00 Ω + 4.00 Ω
10.0 Ω
Along the path from point a to point b, the change in potential that occurs is given by
ΔVab = +e 4 − IR4 = + 4.00 V − ( 2.00 A )( 4.00 Ω ) = − 4.00 V
(a)
The potential difference between points a and b has magnitude
ΔVab = 4.00 V
(b)
564
18.46
18.47
1.
Since the change in potential in going from a to b was negative, we conclude that
a is at the higher potential .
Chapter 18point
(a)
(a)
The resistors combine to an equivalent resistance of Req = 15 Ω as shown.
Figure 1
Figure 3
(b)
From Figure 5, I1 =
Figure 2
Figure 4
Figure 5
ΔVab 15 V
=
= 1.0 A
15 Ω
Req
Then, from Figure 4,
67352_ch18.indd 552
2/9/11 1:57:38 PM
15 V
= 1.0 A
15 Ω
=
Direct-Current Circuits
553
157
Substituting
Equations
Then,
from Figure
4, [2] and [3] into Equation [1] yields
1⎞
4
⎛
andand ΔVIcd2 == I0.909
A
( 6.0 −Ω ) = 6.0 V
⎜⎝ΔVac = +ΔVdb
⎟⎠ I=2 =I1 3.00
1 ( 3.0 Ω ) = 3.0 V
3
3
I2 = I3 =
From Figure 2,
ΔVed = I 3 ( 3.6 Ω ) = 1.8 V
Then, from Figure 1,
and
(c)
ΔVcd
3.0 V
=
= 0.50 A
6.0 Ω 6.0 Ω
From Figure 3,
I5 =
ΔV fd
=
9.0 Ω
I4 =
ΔVed
1.8 V
=
= 0.30 A
6.0 Ω 6.0 Ω
ΔVed
1.8 V
= 0.20 A
=
9.0 Ω 9.0 Ω
From Figure 2, ΔVce = I 3 ( 2.4 Ω ) = 1.2 V . All the other needed potential differences were
calculated above in part (b). The results were
ΔVac = ΔVdb = 6.0 V ; ΔVcd = 3.0 V ; and ΔV fd = ΔVed = 1.8 V
(d)
The power dissipated in each resistor is found from P = ( ΔV ) R with the following results:
2
Pac =
Ped =
Pcd =
18.48
1.
18.49
( ΔV )2ac
=
Rac
( ΔV )ed2
=
Red
( ΔV )cd2
=
Rcd
( 6.0 V )2
6.0 Ω
(1.8 V )2
6.0 Ω
( 3.0 V )2
6.0 Ω
= 6.0 W
Pce =
= 0.54 W
Pfd =
Pdb =
= 1.5 W
( ΔV )ce2
Rce
( ΔV )2fd
R fd
( ΔV )2db
Rdb
2
V)
(1.2
Direct-Current Circuits
=
2.4 Ω
=
=
(1.8 V )2
9.0 Ω
( 6.0 V )2
6.0 Ω
= 0.60 W
565
= 0.36 W
= 6.0 W
(a) From
P = ( ΔV
) R the equivalent resistance is Req = R1 + R2 + ⋅⋅⋅ + Rn = n R. Thus, the
When
connected
in series,
current is I s = ( ΔV ) Req = ( ΔV ) n R, and the power consumed by the series configuration is
2
Ps = I s2 Req =
( ΔV )2
(n R)
2
(n R) =
( ΔV )2
nR
For the parallel connection, the power consumed by each individual resistor is P1 = ( ΔV ) R, and
the total power consumption is
2
n ( ΔV )
R
2
Pp = nP1 =
Ps ( ΔV )
1
1
R
=
= 2 or Ps = 2 Pp
⋅
2
n
n
Pp 1 n R n ( ΔV )
R + 18 Ω =
2
a are connected
b
c
d the
Recognize that the 5.00-Ω and the 8.00-Ω resistors
in parallel
and that
Note
the assumed
of the combination is
effective
resistancedirections
of this parallel
+ I1
5.00 Ω
8.00 V
I2
I3
three distinct currents in the circuit
−
−
diagram at the right. Applying the
3.00 V
−
3.00 Ω
+
junction rule at point c gives
4.00 V
I
2
Therefore,
18.50
18.53
1.
(a)
(a)
k
I1 = I 3 − I 2
[1] 6.00 mF
h
Applying Kirchhoff’s loop rule to
loop gbcfg gives
+ 8.00 V + 3.00I 2 − 5.00I1 = 0
or
1
+
−
+
I1
I=0
g
5.00 Ω
I3
f
5.00I1 − 3.00I 2 = 8.00 V
e
[2]
Finally, applying Kirchhoff’s loop rule to loop fcdef yields
67352_ch18.indd 553
2/9/11 1:57:41 PM
554
158
18.21
Chapter 18
(a)
I1 = I 3 − I 2
[1]
The circuitKirchhoff’s
diagram at loop
the rule to
Applying
right gbcfg
showsgives
the assumed
loop
directions of the current in
each resistor.
that2 −the
+ 8.00 VNote
+ 3.00I
5.00I1 = 0
or
5.00I1 − 3.00I 2 = 8.00 V
total current flowing out of
the
section
of wireKirchhoff’s
connectingloop rule to loop fcdef yields
Finally,
applying
points g and f must equal the
current
flowing
into that
−3.00I
or
5.00I 3 + 3.00I 2 = 4.00 V
2 − 5.00I 3 + 4.00 V = 0
section. Thus,
Substituting Equation [1] into Equation [2] gives
[2]
[3]
5.00I 3 − 8.00I 2 = 8.00 V
and subtracting this result from Equation [3] yields
I 2 = − 4.00 V 11.0
Equation [3] then gives the current in the 4.00-V battery as
I3 =
(b)
4.00 V − 3.00 ( −4.00 V 11.0 )
= +1.02 A
5.00
or
I 3 = 1.02 A down
From above, the current in the 3.00-Ω resistor is
I2 = −
(c)
4.00 V
= − 0.364 A
11.0
or
I 2 = 0.364 A down
Equation [1] now gives the current in the 8.00-V battery as
I1 = 1.02 A − ( −0.364 A ) = +1.38 A
or
I1 = 1.38 A up
(d)
Once the capacitor is charged, the current in the 3.00-V battery is I = 0 because of the
open circuit between the plates of the capacitor.
(e)
To obtain the potential difference between the plates of the capacitor, we start at the negative
plate and go to the positive plate (noting the changes in potential) along path hgbak. The result is
ΔVhk = +8.00 V + 3.00 V = 11.0 V
so the charge on the capacitor is
Qhk = Chk ( ΔVhk ) = ( 6.00 mF )(11.0 V ) = 66.0 mC
18.54
1.
18.57
Using a single resistor → 3 distinct values: R1 = 2.0 Ω, R2 = 4.0 Ω, R3 = 6.0 Ω
When the two resistors are connected in series, the equivalent resistance is Rs = R1 + R2 and the
2 resistors
in Series
→ a2 current
additional
values:
R4 = the
2.0 series
Ω + 6.0
Ω = 8.0 Ω, isand
power
delivered
when
I = distinct
5.00 A fl
ows through
combination
R5 = 4.0 Ω + 6.0 Ω = 10 Ω. Note: 2.0 Ω and 4.0 Ω in series duplicates R3 above.
Ps = I 2 Rs = ( 5.00 A ) ( R1 + R2 ) = 225 W
2 resistors
in Parallel → 3 additional distinct values:
2
Thus,
R1 + R2 =
225 W
25.0 A 2
giving
R1 + R2 = 9.00 Ω
When the resistors are connected in parallel, the equivalent resistance is Rp = R1 R2
the power delivered by the same current ( I = 5.00 A ) is
[1]
(R
1
+ R2 ) and
RR ⎞
2 ⎛
Pp = I 2 Rp = ( 5.00 A ) ⎜ 1 2 ⎟ = 50.0 W
⎝ R1 + R2 ⎠
67352_ch18.indd 554
2/9/11 1:57:44 PM
570
Direct-Current Circuits
555
159
ΔV
12 V
50.0supplied
W
R1 Rby
2
(b)
toorthis circuit
the
battery
givingTheRcurrent
= 2.00
Ω is I total = R = 60.0 Ω = 0.20 A .
p =
2
eq
25.0 A
R1 + R2
[2]
Chapter 18
2
(c)
The power
delivered
by Equation
the battery[2]
is yields
Ptotal = Req I total
= ( 60.0 Ω )( 0.20 A ) = 2.4 W .
Substituting
Equation
[1] into
2
(d)
The
difference
a and b is
Ω − R1between
R ( 9.00
) = 2.00 points
R1 Rpotential
2
= 1
Ω
R1 + R2
9.00 Ω
or
R12 − ( 9.00 Ω ) R1 + 18.0 Ω2 = 0 . This quadratic equation factors as
(R
1
− 3.00 Ω ) ( R1 − 6.00 Ω ) = 0
Thus, either R1 = 3.00 Ω or R1 = 6.00 Ω, and from Equation [1], we find that either R2 = 6.00 Ω or
R2 = 3.00 Ω. Therefore, the pair contains one 3.00 Ω resistor and one 6.00 Ω resistor.
18.58
67352_ch18.indd 555
Consider a battery of emf e connected between
points a and b as shown. Applying Kirchhoff’s
loop rule to loop acbea gives
2/9/11 1:57:47 PM
1장 ࢕̛
PROBLEM SOLUTIONS
PROBLEM SOLUTIONS
19.1
1.
19.1
Remember that the direction of the magnetic force exerted on the negatively charged electron
Remember that the direction of the magnetic force exerted on the negatively charged electron
is opposite to the direction predicted by right-hand rule number 1. The magnetic field near the
is opposite to the direction predicted by right-hand rule number 1. The magnetic field near the
Earth’s equator is horizontal and directed toward the north. The magnetic force experienced by a
Earth’s equator is horizontal and directed toward the north. The magnetic force experienced by a
moving charged particle is always perpendicular to the plane formed by the vectors representing
moving charged particle is always perpendicular to the plane formed by the vectors representing
the magnetic field and the particle’s velocity.
the magnetic field and the particle’s velocity.
(a)
(a)
(b)
(b)
(c)
(c)
(d)
(d)
19.2
19.3
1.
19.3
584
1.
19.7
19.4
When the velocity of a positively charged particle is downward, right-hand rule number 1
When the velocity of a positively charged particle is downward, right-hand rule number 1
predicts a magnetic force toward the east. Hence, the force experienced by the negatively
.
predicts a magnetic force toward the east. Hence, the force experienced by the negatively
charged electron (and also the deflection of its velocity) is directed toward the west .
charged electron (and also the deflection of its velocity) is directed toward the west
When the particle moves northward, its velocity is parallel to the magnetic field, and it will
.
When
the particle moves northward, its velocity
is parallel to the magnetic field, and it will
experience
zero force and zero deflection .
experience zero force and zero deflection
The direction of the force on the negatively charged electron (and the deflection of its
The direction of the force on the negatively
charged electron (and the deflection of its
velocity) will be vertically upward ..
velocity) will be vertically upward
The direction
direction of
of the
the force
force on
on the
the negatively
negatively charged
charged electron
electron (and
(and the
the defl
deflection
ection of
of its
its
The
velocity)
will
be
vertically
downward
.
velocity) will be vertically downward..
(a) For
a positively
charged particle, the
direction
of the force
is that predicted
by the rightSince
Since the
the particle
particle is
is positively
positively charged,
charged, use
use the
the right-hand
right-hand rule
rule number
number 1.
1. In
In this
this case,
case, start
start with
with
hand
rule
number
one.
These
are:
the
fi
ngers
of
the
right
hand
in
the
direction
of
v
and
the
thumb
pointing
in
the
direction
the fingers of the right hand in the direction of v and the thumb pointing in the direction of
of F.
F. As
As
you
start
closing
the
hand,
the
fi
ngers
point
in
the
direction
of
B
after
they
have
moved
90°.
you start closing the hand, the fingers point in the direction of B after they have moved 90°. The
The
(a⬘)
results
results are
are
Chapter
(a) 19
into
(a)
the page
into theforce:
page
Gravitational
(b)
toward the right
(c)
toward bottom of page
(b)
toward the right
(c)
toward bottom of page
−31 fingers in the direction
−30 that as you close your hand, the
Hold F
the
with
v so
=right
mg =hand
10the
kg ) ( 9.80 m s 2 ) = 8.93of
× 10
N downward
(9.11×
g
fingers move toward the direction of B. The thumb will point in the direction of the force (and
hence
theforce:
deflection) if the particle has a positive charge. The results are
Electric
(a)
Fe = qE = ( −1.60 × 10 −19 C ) ( −100 N C ) = 1.60 × 10 −17 N upward
Magnetic force:
Fm = qvB sinq = ( −1.60 × 10 −19 C ) ( 6.00 × 10 6 m s ) ( 50.0 × 10 −6 T ) sin ( 90.0° )
−17
qBr
v = = 4.80 × 10 N in direction opposite right hand rule prediction
m
Fm = 4.80 × 10 −17 N downward
.
1.
19.9
19.8
The magnetic
force experienced
moving charged
has magnitude Fm = qvB sinq ,
Since
the acceleration
(and henceby
thea magnetic
force) isparticle
in the positive
where
q
is
the
angle
between
the
directions
of
the
particle’s
velocity
x-direction, the magnetic field must be in the negative y-direction (seeand the magnetic field. Thus,
sketch at the right) according to right-hand −13
rule number 1.
Fm
8.20 × 10 N
sinq =
=
= 0.754
6
qvB
× 10from
m sF)m(1.70
T )sinq as
× 10 −19fiC
The magnitude
of the
magnetic
eld
is found
= qvB
(1.60
) ( 4.00
and
67352_ch19.indd 576
q = sin −1 ( 0.754 ) = 48.9°
or
160
q = 180° − 48.9° = 131°
2/9/11 1:59:27 PM
.
19.9
The magnetic force experienced by a moving charged particle has magnitude Fm = qvB sinq ,
where q is the angle between the directions of the particle’s velocity and the magnetic field. Thus,
=
19.11
1.
(1.60 × 10
−19
8.20 × 10 −13 N
= 0.754
C ) ( 4.00 × 10 6 m s )(1.70 T )
Magnetism
583
161
−1
Magnetism
(b)
negatively
the force
is exactly opposite
what the 585
and For
q =asin
= 48.9°particle,
or the direction
q = 180°of
− 48.9°
= 131°
( 0.754 )charged
right-hand rule number 1 predicts for positive charges. Thus, the answers for part (b) are
reversed from those given in part (a)
The gravitational force is small enough to be ignored, so the magnetic force must supply the
needed centripetal acceleration. Thus,
m
q Br
v2
= qvB sin 90°, or v =
, where r = RE + 1000 km = 7.38 × 10 6 m
m
r
v=
(1.60 × 10
−19
C ) ( 4.00 × 10 −8 T ) ( 7.38 × 10 6 m )
1.67 × 10 −27 kg
= 2.83 × 10 7 m s
If v is toward the west and B is northward, F will be directed downward as required.
19.12
19.13
1.
Hold the right hand with the fingers in the direction of the current so, as you close the hand, the
For minimum field, B should be perpendicular to the wire. If the force is to be northward, the
fingers move toward the direction of the magnetic field. The thumb then points in the direction of
fi
eldforce.
must The
be directed
downward .
the
results are
To
(a) keep the wire moving, the magnitude of the magnetic force must equal that of the kinetic
friction force. Thus, BIL sin 90° = m k ( mg ), or
2
m k ( m L ) g ( 0.200 )(1.00 g cm ) ( 9.80 m s ) ⎛ 1 kg ⎞ ⎛ 10 2 cm ⎞
=
⎜⎝ 10 3 g ⎟⎠ ⎜⎝ 1 m ⎟⎠ = 0.131 T
I sin 90°
(1.50 A )(1.00 )
. . Both must be
(a) known
The magnitude
is direction of the force can be determined. In this problem, you can only
before the
F = BIL sinq = ( 0.300 T )(10.0 A )( 5.00 m ) sin ( 30.0° ) = 7.50 N
say that the force is perpendicular to both the wire and the field.
B=
19.14
1.
19.17
1.
19.19
From F = BIL sinq , the magnetic field is
B=
F L
0.12 N m
=
= 8.0 × 10 −3 T
I sinq (15 A ) sin 90°
The direction of B must be the + z-direction to have F in the –y-direction when I is in the
if the force is to be
+x-direction.
upward when the magnetic field is into the page.
1.
19.21
19.20
(a)
The
magnetic
force must
directed
and
its magnitude
To
have
zero tension
in thebewires,
the upward
magnetic
force
per unit must equal mg , the weight
of the wire.
the net
force acting
on the
wireweight
will beper
zero, and it can move upward at
length
must Then,
be directed
upward
and equal
to the
constant
speed.
unit
length
of the conductor. Thus,
(b)
The magnitude of the magnetic force must be BIL sinq = mg and for minimum field
q = 90°. Thus,
Bm in =
2
mg ( 0.015 kg ) ( 9.80 m s )
=
= 0.20 T
IL
( 5.0 A )( 0.15 m )
For the magnetic force to be directed upward when the current is toward the left, B must be
directed out of the page .
(c)
1.
19.23
If the field exceeds 0.20 T, the upward magnetic force exceeds the downward force of
gravity, so the wire accelerates upward .
For each segment, the magnitude of the force is given by F = BIL sinq , and the direction is given
by the right-hand rule number 1. The results of applying these to each of the four segments, with
B = 0.020 0 T and I = 5.00 A, are summarized below.
Segment
67352_ch19.indd 583
L (m)
q
F (N)
Direction
2/9/11 1:59:40 PM
584
162
19.7
Chapter 19
, are summarized below.
Gravitational force:
Segment
L (m)
q
F (N)
Direction
y
−31
2
180°
0 m sno
ab
Fg = mg 0.400
= ( 9.11× 10
kg ) ( 9.80
8.93 × 10 −30 N downward
) =direction
d
0.400
90.0° 0.040 0 negative x
bc
Electric force:
cd
0.400 2 45.0° 0.040 0 negative z
da
588
19.24
1.
19.27
0.400 2
Chapter 19
parallel to xzplane at 45° to
90.0° 0.056 6
both +x- and
+z-directions
z
B
c
a
I
b
x
B
The magnitude of the force is
The area of the loop is A = p ab, where a = 0.200 m and b = 0.150 m. Since the field is parallel to
the plane of the loop, q = 90.0° and the magnitude of the torque is
t = NBIAsinq
= 8 ( 2.00 × 10 −4 T )( 6.00 A )[p ( 0.200 m )( 0.150 m )] sin 90.0° = 9.05 × 10 −4 N ⋅ m
The torque is directed to make the left-hand side of the loop move toward you and the right-hand
side move away.
.
19.28
1.
19.29
Note that
angle
between
field
the linemagnetic
perpendicular
plane of the
loop
is
(a)
The the
torque
exerted
on athe
coil
by and
a uniform
field istot the
= NBIAsinq
, with
maximum
q = 90.0°
−
30.0°
=
60.0°.
Then,
the
magnitude
of
the
torque
is
torque occurring when q = 90°. Thus, the current in the coil must be
t m ax
0.15 N ⋅ m
= 0.56 A
=
NBA ( 200 )( 0.90 T ) ⎡( 3.0 × 10 −2 m ) ( 5.0 × 10 −2 m ) ⎤
⎣
⎦
I=
19.31
1.
(b)
If I has the value found above and q is now 25°, the torque on the coil is
⎛
⎞
m 2 ⎟Asin
90.0° = 4.33 × 10 −3 N ⋅ m
t = NBIAsinq = ( 200 )( 0.90 T⎜⎝)( 0.56
⎠ )[( 0.030 m )( 0.050 m )] sin 25° = 0.064 N ⋅ m
(a)
Let q be the angle the plane of the loop makes
with the horizontal as shown in the sketch at the
right. Then, the angle it makes with the vertical is
f = 90.0° −q . The number of turns on the loop is
N=
L
4.00 m
=
= 10.0
circumference 4 ( 0.100 m )
The torque about the z-axis due to gravity is
⎛s
⎞
t g = mg ⎜ cosq ⎟ , where s = 0.100 m is the length
⎝2
⎠
of one side of the loop. This torque tends to rotate
the loop clockwise. The torque due to the magnetic
force tends to rotate the loop counterclockwise about
the z-axis and has magnitude t m = NBIAsinq . At equilibrium, t m = t g, or
NBI ( s 2 ) sinq = mg ( s cosq ) 2. This reduces to
tanq =
Magnetism
589
( 0.100 kg) (9.80 m s2 )
mg
=
= 14.4
2NBIs 2 (10.0 )( 0.010 0 T )( 3.40 A )( 0.100 m )
Since tanq = tan ( 90.0° − f ) = cotf , the angle the loop makes with the vertical at
equilibrium is f = cot −1 (14.4 ) = 3.97° .
(b)
At equilibrium,
t m = NBI ( s 2 ) sinq
= (10.0 )( 0.010 0 T )( 3.40 A )( 0.100 m ) sin ( 90.0° − 3.97° ) = 3.39 × 10 −3 N ⋅ m
2
67352_ch19.indd 584
2/9/11 1:59:42 PM
Magnetism
.
19.11
The
force is small enough to be ignored, so the magnetic force must supply the
(b) gravitational
At equilibrium,
needed centripetal acceleration. Thus,
t m = NBI ( s 2 ) sinq
2
v
m
2
= (10.0 )( 0.010 0 T )( 3.40 A )( 0.100 m ) sin ( 90.0° − 3.97° ) = 3.39 × 10 −3 N ⋅ m
r
19.33
1.
(a)
590
From KE = 12 me v 2, the speed of the electron is
Chapter 19
(b)
585
163
v=
2 ( KE )
=
me
2 ( 3.30 × 10 −19 J )
9.11× 10 −31 kg
= 8.51× 10 5 m
The magnetic force acting on the electron must provide the necessary centripetal acceleration.
Thus, me v 2 r = qvB sinq , which gives
r=
(9.11× 10−31 kg) (8.51× 105 m s)
me v
=
qB sinq (1.60 × 10 −19 C )( 0.235 T ) sin 90.0°
= 2.06 × 10 −5 m = 20.6 × 10 −6 m = 20.6 mm
19.34
1.
19.35
Since
centripetal
acceleration
is furnished
by thethe
magnetic
acting
the be
ions,
For thethe
particle
to pass
through with
no deflection,
net forceforce
acting
on iton
must
zero. Thus,
mv 2
the magnetic
force and the electric force must be in opposite directions and have equal magniqvB
=
r gives
tudes. This
(
)
Fm = Fe , or qvB = qE, which reduces
to ×v10=−4E mB = 0.150 mm
= 1.50
(1.60 × 10 −19 C)( 0.930 T )2
19.37
1.
1
From conservation of energy, ( KE + PE ) f = ( KE + PE )i, we find that mv 2 + qV f = 0 + qVi , or the
2
speed of the particle is
(
2 q Vi − V f
v=
m
)=
2 q ( ΔV )
=
m
2 (1.60 × 10 −19 C )( 250 V )
2.50 × 10 −26 kg
= 5.66 × 10 4 m s
mv 2
The magnetic force supplies the centripetal acceleration giving qvB =
r
, the mass of the particle is
or
1.
19.38
19.43
r=
−26
4
mv ( 2.50 × 10 kg ) ( 5.66 × 10 m 2s ) 2 2
−2
=
2
KE
q B =R1.77 × 10 m = 1.77 cm
(
)
−19
=
qB
T
)
(1.60= ×410
2C ) (20.500
2 ( KE )
( KE ) q B 2 R 2
(a) B = m 0 I 2p r, the required distance is
From
r=
× 10×−710T3 ⋅ m
m A1.50
( 4p(9.00
m 0 I 2p
)−3( )( 20×A10)−8= T2.4) =× 10675−3 mA = 2.4 mm
=
=
2p B
−7
2p4p(1.7
× 10
× 10
T ⋅Tm) A
19.44
1.
19.47
Imagine grasping the conductor with the right hand so the fingers curl around the conductor in the
Treat
the lightning
bolt asfiaeld.
long,
straight
field
is direction of the
direction
of the magnetic
The
thumbconductor.
then pointsThen,
alongthe
themagnetic
conductor
in the
Chapter
current. The
× 10 4 A
( 4pare× 10−7 T ⋅ m A ) (1.00594
) = 2.0019× 10−5 T = 20.0 mT
m 0 Iresults
=
594
Chapter B
19=
2p r
2p (100 m )
s 2 + s 2 = s 2, where s = 0.200 m.
(a)
toward
the left
2
2
The distance from each wire to point P is given by
s + s = s 2, where s = 0.200 m.19.49
19.48
Assume that the wire on the right is wire 1 and that on the
left
is2wire 2. Also, choose the positive
2
1.
19.49
The distance from each wire to point P is given At
by point
r = 12 P,sthe
+ smagnitude
= s 2, where
s = 0.200 fim.
of the magnetic
eld
direction
2 for 2the magnetic field to be out of the page and negative into the page.
1
rAt=point
s
+
s
=
s
2,
where
s
=
0.200
m.
produced
by
each
of
the
wires
is
P, the magnitude of the magnetic field
2
produced
by point
each of
wires
is
(a)
At the
halftheway
between
the two wires, At point P, the magnitude of the magnetic field
−7
At point
P,
the
magnitude
of
the
magnetic
produced by each of the wires is
m I ( 4p × 10 T ⋅ m A )( 5.00 A ) 2field
B
= 0 =by each of the wires is
= 7.07 mT
produced
2p r
2p ( 0.200 m )
67352_ch19.indd 585
Carrying currents into the page, the field A produces at P
is directed to the left and down at –135°, while B creates
a field to the right and down at – 45°. Carrying currents
toward you, C produces a field downward and to the
right at – 45°, while D’s contribution is down and to the
left at –135°. The horizontal components of these equal magnitude contributions cancel in pairs,
while the vertical components all add. The total field is then
2/9/11 1:59:45 PM
=
( 4p × 10
−7
T ⋅ m A )( 5.00 A ) 2
164
586
Chapter 19
= 7.07 mT
( 4p × 10
−7
T ⋅ m A )( 5.00 A ) 2
= 7.07 mT
2p ( 0.200 m )
To
find the
minimum
eld,fithe
field
should be perpendicular to the current
Carrying
currents
intopossible
the page,fithe
eldmagnetic
A produces
at P
currents into the page, the field A produces at PCarrying
= 90.0°currents
. Then,
the left
–135°,
while
B creates
(isq directed
)to
intoand
thedown
page,atthe
field A
produces
at P
d to the left and down at –135°, while B createsis
a fidirected
eld to the
andand
down
at –at45°.
Carrying B
currents
to right
the left
down
creates
⎡ –135°,g while
⎛ 1 kg ⎞ ⎛ 10 2 cm ⎞ ⎤ 9.80 m s 2
⎞ and gdown
the right and down at – 45°. Carrying currents atoward
produces
a
fi
eld
downward
and
to
the
field toyou,
the⎛⎜C
right
at
–
45°.
Carrying
currents
=
0.500
⎢
⎟⎠
⎜⎝ 10 3 g ⎟⎠ ⎜⎝ 1 m ⎟⎠ ⎥ ( 2.00 A )(1) = 0.245 T
⎝ while
I sin
90.0°
ou, C produces a field downward and to the toward
⎦
right at you,
– 45°,
D’s
iscm
down
and
to the
C produces
acontribution
field⎣downward
and
to the
45°, while D’s contribution is down and to the right
left atat–135°.
horizontal
components
of these
– 45°,The
while
D’s contribution
is down
and equal
to the magnitude contributions cancel in pairs,
To find the direction of the field, hold the right hand with the thumb pointing upward (direction of
35°. The horizontal components of these equal left
magnitude
contributions
cancel
pairs,
while
vertical
allinadd.
Theoftotal
is then
at the
–135°.
The components
horizontal
components
thesefield
equal
magnitude contributions cancel in pairs,
the force) and the fingers pointing southward (direction of current). Then, as you close the hand,
vertical components all add. The total field is then
while the vertical components all add. The total field is then
the fingers point eastward. The magnetic field should be directed eastward
2p ( 0.200 m )
=
Bnetof=the
4 ( 7.07
= 4 ( 7.07 mT ) sin 45.0° = 20.0 mT toward the bottom
page mT ) sin 45.0° = 20.0 mT toward the bottom of the page
Bnet = 4 ( 7.07 mT ) sin 45.0° = 20.0 mT toward the bottom of the page
1.
19.51
Call the wire along the x-axis wire 1 and the other wire 2. Also, choose the positive direction for
the magnetic fields at point P to be out of the page.
At point P, Bnet = +B1 − B2 =
or
19.55
1.
T ⋅ m A ) ⎛ 7.00 A 6.00 A ⎞
−7
−
⎜⎝
⎟ = +1.67 × 10 T
2p
3.00 m 4.00 m ⎠
= 0 . This is as expected, because right-hand rule 2
x=0
shows
that
at
the
midpoint
the
fi
eld
due
to
the
upper wire is toward the right, while that due to
Bnet = 0.167 mT out of the page
the lower wire is toward the left. Thus, the two fields cancel, yielding a zero resultant field.
Bnet =
(a)
( 4p × 10
m 0 I1 m 0 I 2 m 0 ⎛ I1 I 2 ⎞
−
=
−
2p r1 2p r2 2p ⎜⎝ r1 r2 ⎟⎠
−7
The magnetic force per unit length on each of two parallel wires separated by the distance d
and carrying currents I1 and I 2 has the magnitude
F
=
m 0 I1 I 2
2p d
In this case, we have
F
(b)
19.56
19.57
1.
=
( 4p × 10
−7
T ⋅ m A )(1.25 A )( 3.50 A )
2p ( 2.50 × 10 −2 m )
= 3.50 × 10 −5 N m
The magnetic forces two parallel wires exert on each other are attractive if their currents are
in the same direction and repulsive if the currents flow in opposite directions. In this case,
the currents in the two wires are in opposite directions, so the forces. are repulsive .
(a)order
The
per unittolength
that parallel conductors
exert
on each
other
is unit
F length on the
In
forforce
the system
be in equilibrium,
the repulsive
magnetic
force
per
top wire must equal the weight per unit length of this wire.
Thus,
F
= m 0 I1 I 2 2p d = 0.080 N m, and the distance between the wires will be
d=
( 4p × 10−7 T ⋅ m A )(60.0 A )(30.0 A )
m 0 I1 I 2
=
2p ( 0.080 N m )
2p ( 0.080 N m )
= 4.5 × 10 −3 m = 4.5 mm
19.58
1.
19.61
67352_ch19.indd 586
The magnetic forces exerted on the top and bottom segments of the rectangular loop are equal
(a) From R = rL A, the required length of wire to be used is
in magnitude and opposite in direction. Thus, these forces cancel, and we only need consider the
sum of the forces exerted on the right and left sides of the loop. Choosing to the left (toward the
long, straight wire) as the positive direction, the sum of these two forces is
2/9/11 1:59:48 PM
Magnetism
587
165
, the required length of wire to be used is
19.22
2
As shown in end view in5.00
the right,
hence, the
⎤
Ω ) ⎡p (at0.500
× 10 −3the
m )rod4(and
( the sketch
R
⋅
A
⎣
⎦
current) is horizontal,
while
the
magnetic
fi
eld
is
vertical.
For
=
L=
= 58 m
−8
r constant velocity
1.7 × 10(zero
Ω acceleration),
⋅m
the rod to move with
it is
necessary that
The total number of turns on the solenoid (that is, the number of times this length of wire
radius
is
Σwill
Fx =goFmaround
− fk = aBI1.00-cm
sin 90.0°
− m k cylinder)
n=0
N=
or
(b)
L
58 m
=
= 9.2 × 10 2 = 920
2p r 2p (1.00 × 10 −2 m )
From B = m 0 nI, the number of turns per unit length on the solenoid is
n=
B
4.00 × 10 −2 T
=
= 7.96 × 10 3 turns m
m 0 I ( 4p × 10 −7 T ⋅ m A )( 4.00 A )
Thus, the required length of the solenoid is
−3 2
N 20.09.2
×10
× 10
kgturns
s2 )
) (9.80 =m0.12
m
12 Tcm
= = (=
= =0.26
3
7.96
×10A )turns
n ( 50
)( 0.30
(5.0 ×m10 −2 m )
19.62
19.65
1.
The magnetic
field inside
a solenoid
= m 0 nIcurrent
= m 0 Nloop
L of radius R and carrying current I
(a)
The magnetic
field atofthe
center ofisa Bcircular
is B = m 0 I 2R. The direction of the field at this center is given by right-hand rule number 2.
Taking out of the page (toward the reader) as positive, the net magnetic field at the common
center of these coplanar loops has magnitude
−7
m 0 I 2 m 0 I1 ( 4p × 10 T ⋅ m A )
3.00 A
5.00 A
−
=
−
−2
2r2
2r1
2
9.00 × 10 m 12.0 × 10 −2 m
Bnet = B2 − B1 =
= −5.24 × 10 −6 T = 5.24 mT
(b)
Since we chose out of the page as the positive direction, and now find that Bnet < 0, we
conclude the net magnetic field at the center is into the page .
(c)
To have Bnet = 0, it is necessary that I 2 r2 = I1 r1, or
2
⎛ I ⎞( 0.100
)( 0.200
A ⎞ kg ) ( 9.80 m s )
⎛ 3.00
r2 = ⎜ 2=⎟ r1 = ⎜
12.0 cm ) = 7.20=cm3.92 × 10 −2 T
(
⎟
⎝ (5.00
⎝ I1 ⎠
10.0AA⎠)( 0.500 m )
19.66
19.67
1.
For the rail to move at constant velocity, the net force acting on, it must be zero. Thus, the magniAssume
wire
1 is along
themust
x-axis
and that
wireof2 the
is along
theforce,
y-axis.giving BIL = m ( mg ), or
tude of the
magnetic
force
equal
friction
k
(a)
Choosing out of the page as the positive field direction, the field at point P is
B = B1 − B2 =
600
Chapter 19
(b)
= 5.00 × 10 −7 T = 0.500 mT out of the page
At 30.0 cm above the intersection of the wires, the field
components are as shown at the right, where
By = −B1 = −
=−
and
67352_ch19.indd 587
−7
m 0 ⎛ I1 I 2 ⎞ ( 4p × 10 T ⋅ m A ) ⎛ 5.00 A
3.00 A ⎞
−
−
=
⎜⎝
⎟
⎜
⎟
2p ⎝ r1 r2 ⎠
2p
0.400 m 0.300 m ⎠
m 0 I1
2p r
( 4p × 10
−7
T ⋅ m A )( 5.00 A )
2p ( 0.300 m )
=
( 4p × 10
−7
= −3.33 × 10 −6 T
T ⋅ m A )( 3.00 A )
2p ( 0.300 m )
= 2.00 × 10 −6 T
With Bz = 0, the resultant field is parallel to the xy-plane and
2/9/11 1:59:51 PM
600
Chapter 19
(b)
588
166
19.27
At 30.0 cm above the intersection of the wires, the field
components are as shown at the right, where
Chapter 19
−7 ab, where a = 0.200 m and b = 0.150 m. Since the field is parallel to
=p
The area of the loop
is ×A10
4p
T ⋅ m A )( 5.00 A )
(
− q = 90.0° and the magnitude
= −3.33
10 −6 Tis
the plane of the= loop,
of the×torque
2p ( 0.300 m )
and
Bx = B2 =
−7
m 0 I 2 ( 4p × 10 T ⋅ m A )( 3.00 A )
=
= 2.00 × 10 −6 T
2p r
2p ( 0.300 m )
With Bz = 0, the resultant field is parallel to the xy-plane and
B = Bx2 + By2 = 3.88 × 10 −6 T
at
⎛ By ⎞
q = tan −1 ⎜ ⎟ = −59.0°
⎝ Bx ⎠
−5
or =BB=2 I3.88
mT parallel
and ×59.0°
thewire
+x-direction
Tto)(the
5.00xy-plane
A ) = 8.00
10 −5 clockwise
N directedfrom
toward
2
1 = (1.60 × 10
1.
19.71
Label the wires 1, 2, and 3, as shown in Figure 1.
Also, let B1 , B2 , and B3 , respectively, represent the
magnitudes of the fields produced by the
currents in those wires, and observe that
q = 45°.
(
)
At point A, B1 = B2 = m 0 I 2p a 2 , or
B1 = B2 =
( 4p × 10
−7
T ⋅ m A ) ( 2.0 A )
2p ( 0.010 m ) 2
= 28 mT
FIGURE 1
( 4p × 10 T ⋅ m A ) ( 2.0 A ) = 13 mT
m0 I
=
2p ( 3a )
2p ( 0.030 m )
−7
and
B3 =
These field contributions are oriented as shown in Figure 2.
Observe that the horizontal components of B1 and B2 cancel
while their vertical components add to B3. The resultant
field at point A is then
FIGURE 2
BA = ( B1 + B2 ) cos 45° + B3 = 53 mT, or
B A = 53 mT directed toward the bottom of the page
At point B, B1 = B2 =
−7
m 0 I ( 4p × 10 T ⋅ m A ) ( 2.0 A )
=
= 40 mT
2p a
2p ( 0.010 m )
FIGURE 3
m0 I
= 20 mT. These contributions are oriented as
2p ( 2a )
shown in Figure 3. Thus, the resultant field at B is
and B3 =
602
Chapter 19
B B = B3 = 20 mT directed toward the bottom of the page
(
)
At point C, B1 = B2 = m 0 I 2p a 2 , while
B3 = m 0 I 2p a. These contributions are oriented
as shown in Figure 4. Observe that the horizontal
components of B1 and B2 cancel, while their vertical
components add to oppose B3 . The magnitude of the
resultant field at C is
BC = ( B1 + B2 ) sin 45° − B3 =
19.72
67352_ch19.indd 588
(a)
FIGURE 4
m 0 I ⎛ 2sin 45° ⎞
−1⎟ = 0
⎜
2p a ⎝
⎠
2
Since one wire repels the other, the currents must be in opposite directions
2/9/11 1:59:53 PM
. These contributions are oriented
as shown in Figure 4. Observe that the horizontal
components of B1 and B2 cancel, while their vertical
components add to oppose B3 . The magnitude of the
resultant field at C is
19.72
19.73
1.
Magnetism
167
589
the z-axis and has magnitudemtIm ⎛=2sin
t m = t g, or
NBIAsinq
45° .⎞At equilibrium,
2
) (9.80
−1to⎟m
= s0 ) tan8.0°
BNBI
B21)+sinq
B2 ) sin
45°
B3 (=) 2.0 This
⎜ reduces
C = ((s
= mg
( s−cosq
= 68 A
2p a ⎝
⎠
2
2.0 × 10 −7 T ⋅ m A
kg ) ( 9.80
m be
s 2 )in opposite directions
( 0.100
(a) Since one wire repels the other,
the currents
must
=
= 14.4
Note: We solve part (b) before
part (a) for this problem.
2 (10.0 )( 0.010 0 T )( 3.40 A )( 0.100 m )
(b)
Since the magnetic force supplies the centripetal acceleration for this particle, qvB = mv 2 r,
Since tanq = tan ( 90.0° − f ) = cotf , the angle the loop makes with the vertical at
or the radius of the
path is r = mv qB = p qB , where
equilibrium is
p = mv = 2m(KE) = 2 (1.67 × 10 −27 kg ) ( 5.00 × 10 6 eV ) (1.60 × 10 −19 J eV )
= 5.17 × 10 −20 kg ⋅ m s
Consider the circular path shown at the right and
observe that the desired angle is
⎡ (1.00 m ) qB ⎤
⎛ 1.00 m ⎞
a = sin −1 ⎜
= sin −1 ⎢
⎥
⎝ r ⎟⎠
p
⎣
⎦
or
⎡ (1.00 m )(1.60 × 10 −19 C ) ( 0.050 0 T ) ⎤
a = sin −1 ⎢
⎥ = 8.90°
5.17 × 10 −20 kg ⋅ m s
⎢⎣
⎥⎦
(a)
Magnetism
603
The linear momentum of the particle has constant magnitude p = mv, and
its vertical component as the particle leaves the field is py = − psina , or
(12 Ω )( 5.9 × 10 −2 N )
Fm
py== − ( 5.17
× 10
− 8.00
) == 0.59
T × 10 −21 kg ⋅ m s
= −20 kg ⋅ m s ) sin (8.90°
−2
( ΔV R ) L ( 24 V )(5.0 × 10 m )
19.74
1.
19.75
The force constant of the spring system is found from the elongation produced by the weight
The magnetic
acting
alone. force is very small in comparison to the weight of the ball, so we treat the motion as
that of a freely falling body. Then, as the ball approaches the ground, it has velocity components
with magnitudes of
vx = v0 x = 20.0 m s, and
v y = v02 y + 2a y ( Δy ) = 0 + 2 ( −9.80 m s 2 )( −20.0 m ) = 19.8 m s
The velocity of the ball is perpendicular to the magnetic field and, just before it reaches the
ground, has magnitude v = vx2 + v y2 = 28.1 m s. Thus, the magnitude of the magnetic force is
Fm = qvB sinq
= ( 5.00 × 10 −6 C ) ( 28.1 m s ) ( 0.010 0 T ) sin 90.0° = 1.41× 10 −6 N
19.76
(a)
The magnetic force acting on the wire is directed upward and of magnitude
Fm = BIL sin 90° = BI.
Thus,
67352_ch19.indd 589
2/9/11 1:59:56 PM
장 ࡪѦࢷߏ˕ࡪѦ˃ܹ
PROBLEM SOLUTIONS
2
20.1
.
B = BA cosq = B(p r ) cosq , where q is the angle between the direction of the field and the
QUICK Φ
QUIZZES
normal to the plane of the loop.
(a)
If the field is perpendicular to the plane of the loop, q = 0°, and
B=
(b)
8.00 × 10 −3 T ⋅ m 2
ΦB
=
= 0.177 T
2
p r cosq p ( 0.12 m ) cos 0°
( )
2
If the field is directed parallel to the plane of the loop, q = 90°, and
Φ B = BA cosq = B p r 2 cosq
Φ B = BA cosq −3= BA cos 90° = 0−2
2
= 6.98 × 10 T p 3.00 × 10 m cos 0° = 1.97 × 10 −5 T ⋅ m 2
( )
) (
(
20.3
.
)
The angle between the direction of the constant field and the normal to the plane of the loop
is q = 0°, so
(
)(
)
Φ B = BA cosq = ( 0.50 T ) ⎡⎣ 8.0 × 10 −2 m 12 × 10 −2 m ⎤⎦ cos 0° = 4.8 × 10 −3 T ⋅ m 2
.
.
20.5
610
field line that comes up through the area A on one side of the wire goes back down
through area A on the other side of the wire. Thus, the net flux through the coil is zero .
(a) 20
Every
Chapter
(b)
20.6
.
20.7
The magnetic field is parallel to the plane of the coil, so q = 90.0°. Therefore,
Φ B = BA cosq = BA cos 90.0° = 0 .
.
continued on next page
The
flux through
the loop
given
by Φ
= BA cosq
(a) magnetic
The magnetic
flux through
an is
area
A may
beB written
as , where B is the magnitude of the
magnetic field, A is the area enclosed by the loop, and q is the angle the magnetic field makes
with the normal to the plane of the loop. Thus,
Φ B = ( B cosq ) A
= ( component of B perpendicular to10A ) ⋅m
A
Φ B = BA cosq = 5.00 × 10 −5 T 20.0 cm 2
1 cm
Thus, the flux through the shaded side of the cube is
(
−2
)
(
Φ B = Bx ⋅ A = ( 5.0 T ) ⋅ 2.5 × 10 −2 m
(b)
67352_ch20.indd 604
2
= 3.1 × 10 −3 T ⋅ m 2
Unlike electric field lines, magnetic field lines always form closed loops, without beginning
or end. Therefore, no magnetic field
lines originate
or
2 terminate within the cube and any
cos 0°
p 1.6
× 10 −3 from
m ⎤the
− 0 ) ⎡must
ΔΦ
A cosq
( ΔB ) the
lineB entering
cube at(1.5
oneTpoint
emerge
some other point. The net
⎣
⎦ cube= at
=
=
=
1.0 × 10 −4 V = 0.10 mV
−3 closed surface, is zero .
flΔt
ux through the
cube,
and
indeed
through
any
120 × 10 s
Δt
(
20.9
.
20.8
)
)
(a)
As loop A moves parallel to the long straight wire, the magnetic flux through loop A does
not change. Hence, there is no induced current in this loop.
(b)
As loop B moves to the left away from the straight wire, the magnetic flux through this
loop is directed out of the page, and is decreasing in magnitude. To oppose this change in
flux, the induced current flows counterclockwise around loop B producing a magnetic flux
directed out of the page through the area enclosed by loop B.
(c)
As loop C moves to the right away from the straight wire, the magnetic flux through this
loop is directed into the page and is decreasing in magnitude. In order to oppose this change
in flux, the induced current flows clockwise around loop C
168
2/9/11 2:00:56 PM
producing
a magnetic fl169
ux
Induced Voltages
and Inductance
611
directed out of the page through the area enclosed by loop B.
20.11
The magnitude
the induced
emf is
(c)
As loop Cofmoves
to the right
away from the straight wire, the magnetic flux through this
loop is directed into the page and is decreasing in magnitude. In order to oppose this change
ΔΦthe
Δ ( B cosq
) A flows clockwise around loop C producing a magnetic flux
in flux,
current
B induced
Induced Voltages and Inductance
611
=
=
directed
Δt into the page
Δt through the area enclosed by loop C.
.
20.11
20.10
The
induced
is is considered to point in the original direction of the
If
themagnitude
normal to of
thethe
plane
of theemf
loop
magnetic field, then q i = 0° and q f = 180°. Thus, we find
ΔΦ B
Δ ( B cosq ) A
e =
=
Δt
Δt
If the normal to the plane of the loop is considered to point in the original direction of the
magnetic field, then q i = 0° and q f = 180°. Thus, we find
cos180° − ( 0.30 T ) cos 0° p ( 0.30 m2)
( 0.20
ΔΦ
−2
B T )ΔB
mV
9.4 ××10
2.26
mV
A = ( 0.050 0 T s ) ⎡⎣p ( 0.120 m ) ⎤⎦ == 2.26
=
10 −3 VV== 94
1.5 s
Δt
Δt
2
e ==
20.12
.
20.13
With the
eld directed
perpendicular
to the
(a)
Thefiinitial
field inside
the solenoid
is plane of the coil, the flux through the coil is
Φ B = BA cos 0° = BA. As the magnitude of the field decreases, the magnitude of the induced emf
in the coil is
⎛ 300 ⎞
Bi = m 0 nI i = 4p × 10 −7 T ⋅ m A ⎜
( 2.00 A ) = 3.77 × 10 −3 T
⎝ 0.200 m ⎟⎠
(
(b)
)
The final field inside the solenoid is
⎛ 300 ⎞
B f = m 0 nI f = 4p × 10 −7 T ⋅ m A ⎜
( 5.00 A ) = 9.42 × 10 −3 T
⎝ 0.200 m ⎟⎠
(
)
(
)
2
(c)
The 4-turn coil encloses an area A = p r 2 = p 1.50 × 10 −2 m
(d)
The change in flux through each turn of the 4-turn coil during the 0.900-s period is
(
)(
= 7.07 × 10 −4 m 2
)
ΔΦ B = ( ΔB ) A = 9.42 × 10 −3 T − 3.77 × 10 −3 T 7.07 × 10 −4 m 2 = 3.99 × 10 −6 Wb
(e)
The average induced emf in the 4-turn coil is
⎛ 3.99 × 10 −6 Wb ⎞
⎛ ΔΦ B ⎞
−5
e = N2 ⎜
= 4⎜
⎟
⎟⎠ = 1.77 × 10 V
⎝ Δt ⎠
0.900 s
⎝
Since the current increases at a constant rate during this time interval, the induced emf at
any instant during the interval is the same as the average value given above.
(f)
612
20.15
.
20.16
The induced emf is small, so the current in the 4-turn coil will also be very small.
This means that the magnetic field generated by this current will be negligibly small
Chapter 20
in comparison to the field generated by the solenoid.
(a)
When the magnet moves to the left, the flux through the interior of the coil is directed
toward the right and is decreasing in magnitude. To oppose this change in flux, the
magnetic field generated by the induced current should be directed to the right along
the axis of the coil. The current must then be left to right through the resistor.
(b)
When the magnet moves to the right, the flux through the interior of the coil is directed
toward the right and is increasing in magnitude. To oppose this increasing flux, the
magnetic field generated by the induced current should be directed toward the left along
the axis of the coil. The current must then be right to left through the resistor.
When the switch is closed, the current from the battery produces a magnetic field directed toward
the right along the axis of both coils.
(a)
67352_ch20.indd 611
As the battery current is growing in magnitude, the induced current in the rightmost coil
opposes the increasing rightward directed field by generating a field toward to the left along
the axis. Thus, the induced current must be left to right
2/9/11 2:01:07 PM
612
170
20.15
20.17
.
20.18
.
20.21
Chapter 20
through the resistor.
(a)
When
the magnet
the left,flthe
uxcurrent
throughproduces
the interior
of thethe
coil
is directed
The current
is zeromoves
. The to
magnetic
ux flthe
through
right
side of the
toward
the right into
and the
is decreasing
magnitude.
To oppose
thisoutward-directed
change in flux, the
loop is directed
page, and in
is equal
in magnitude
to the
flux the
magnetic
field generated
should
the right
current produces
throughby
thethe
leftinduced
side of current
the loop.
Thus,be
thedirected
net fluxtothrough
thealong
loop has a
the
axis of
the coil.
Theand
current
then be
left to right
constant
value
of zero
does must
not induce
a current.
(b)
The flux through the loop due to the long wire is directed out of the page and is increasing
in magnitude. To oppose this increasing outward flux, the induced current must generate a
magnetic field that is directed into the page through
44 V the area enclosed by the loop. Thus, the
= . = 8.8 A .
induced current in the loop must be clockwise
R 5.0 Ω
The
magnetic
field inside
the solenoid
If
theinitial
magnetic
field makes
an angle
of 28.0° is
with the plane of the coil, the angle it makes with
the normal to the plane of the coil is q = 62.0°. Thus,
e =
=
20.22
.
20.23
N ( ΔΦ B ) NB ( ΔA ) cosq
=
Δt
Δt
(
)(
)
p r 2 mV
t
B pr2
= 1.02 × 10 −5 V NB
= 010.2
=
= 0
Req
tR
NR
With the magnetic field perpendicular to the plane of the coil, the flux through each turn of
Thecoil
vertical
of the Earth’s magnetic field is perpendicular to the horizontal velocity of
the
is Φ Bcomponent
= BA = B p r 2 . Since the area remains constant, the change in flux due to the
the
metallic
truck
body.
Thus,
the motional
induced across the width of the truck is
changing magnitude of the magnetic
field isemf
ΔΦ B = ( ΔB ) p r 2 .
( )
⎡
⎛ 1m ⎞⎤
e = B⊥ Cv = 35 × 10 −6 T ⎢( 79.8 in ) ⎜
(37 m s ) =. 2.6 × 10 −3 V = 2.6 mV
⎝ 39.37 in ⎟⎠ ⎥⎦
The induced emf is
⎣
The vertical component of the Earth’s magnetic field is perpendicular to the horizontal velocity of
Thewire.
motional
in aofmetallic
objectemf
of length
C moving
through
the
Thus,emf
theinduced
magnitude
the motional
induced
in the wire
is a magnetic field at
speed v is given by e = B⊥ vC, where B⊥ is the component of the magnetic field perpendicular to
the velocity of the object. Thus,
(
(a)
20.24
20.25
.
)(
200 50.0 × 10 –6 T ⎡⎣ 39.0 cm 2 1 m 2 10 4 cm 2 ⎤⎦ cos 62.0°
1.80 s
(
)
)
e = 35.0 × 10 −6 T ( 25.0 m
m ) = 1.31 × 10 −2 V = 13.1 mV
(15.0
, ins )the
loop.
.
20.27
20.26
(a)
(a)
Observe
thatpasses
only the
horizontal
component,
Bh the
, of area
Earth’s
magnetic
field
is effective
As the loop
position
A, the
flux through
enclosed
by the
loop
is
in
exerting
a
vertical
force
on
charged
particles
in
the
antenna.
For
the
magnetic
force,
directed right to left and is increasing in magnitude. The induced current must flow
Fcounterclockwise
positive
charges
in
the
antenna
to
be
directed
upward
and
have
m = qvBh sinq , onas
seen from the right end
maximum magnitude (when q = 90°), the car should move eastward through the
northward horizontal component of the magnetic field.
(b)
e = Bh Cv, where Bh is the horizontal component of the magnetic field.
⎡⎛
km ⎞ ⎛ 0.278 m s ⎞ ⎤
e = ⎡⎣ 50.0 × 10 −6 T cos 65.0° ⎤⎦ (1.20 m ) ⎢⎜ 65.0
⎥
⎟
h ⎠ ⎜⎝ 1 km h ⎟⎠ ⎦
⎣⎝
The magnitude and direction of the Earth’s field varies from one location to the other,
4
so=the4.58
induced
× 10 −voltage
V in the wire changes. Further, the voltage will change if the tether
cord changes its orientation relative to the Earth’s field.
(
20.28
20.29
.
)
(a) Since e = B⊥ Cv, the magnitude of the vertical component of the Earth’s magnetic field at
The metal
bar of length
this location
is C and moving at speed v through the magnetic field experiences an
induced emf of magnitude e = B⊥ Cv, where B⊥ = B cosq is the component of the magnetic field
perpendicular to the velocity of the bar as shown in figure (a) below.
continued on next page
67352_ch20.indd 612
2/9/11 2:01:08 PM
Induced Voltages and Inductance
20.20
y
The magnitude of the average emf is
e =
=
N ΔΦ B
Δt
end view
q
NBA Δ ( cosqof)bar
=
Δt
B
(
B
v
q
n
x
Fm
)
200 (1.1 T ) 100 B× 10 −4 m 2 cos180° − cos 0°
||
171
613
mg
0.10
q s
current
into page
= 44 V
q
(a)
Therefore, the average induced current is I =
(b)
e
As the bar slides down the rails, the magnetic flux through the conducting path formed by the
bar, the rails, and the resistor R is directed downward and is increasing in magnitude. Thus, the
induced current must flow counterclockwise around the conducting path to generate an upward
flux, opposing the increase in flux due to the field B. This current flows into the page as indicated
in figure (b) and has magnitude I = e R = BCv cosq R.
I
According to right-hand rule number 1, the bar will experience a magnetic force Fm directed
horizontally toward the left as shown in figure (b). The magnitude of this force is
B 2 C2 v cosq
⎛ BCv cosq ⎞
Fm = BIC = B ⎜
C=
⎟
⎝
⎠
R
R
I
Now, consider the free-body diagram of the bar in figure (b), where n is the normal force exerted
on the bar by the rails. If the bar is to move with constant velocity (i.e., be in equilibrium), it is
necessary that
and
mg
cosq
ΣFy = 0 ⇒ n cosq = mg
or
n=
ΣFx = 0 ⇒ Fm = n sinq
or
B 2 C2 v cosq ⎛ mg ⎞
=⎜
sinq = mg tanq
⎝ cosq ⎟⎠
R
Thus, the equilibrium speed of the bar is
v=
20.30
20.31
.
IR ) R ( 0.500
A )kg
Ω ) m s 2 ) tan 25.0° (1.00 Ω )
( 6.00
( 0.200
) (9.80
( mg=tanq
= =
= 1.00 m s
=
BC
B 2 C2 cosq
2.80 m s
BCv,
From eIn=the
the required
speed
is coil, the magnitude of the flux passing through the loop is
(a)
initial
orientation
of the
Φ B = NBA, where A is the area enclosed by the loop and N is the number of turns on the
loop. After the loop has rotated 90°, the magnetic field is now parallel to the plane of the loop
and the flux through the loop is zero. The average emf induced in the loop as it rotates is
(b)
(
)
−2
NBA − 0 28 (1.25 T ) 2.80 × 10 m
=
= 8.19 × 10 −2 V = 81.9 mV
Δt
Δt
0.335 s ⎡⎛
⎞ ⎛ 2p rad ⎞ ⎛ 1 min ⎞ ⎤
⎟⎠ ⎜⎝
⎟⎜
⎟
⎢⎜⎝
1 rev ⎠ ⎝ 60 s ⎠ ⎥⎦
⎣
e 81.9 mV
The average induced
current is I = =
= 105 mA .
= 1.3 × 10 −2 V = 13 mV
R 0.780 Ω
e=
.
20.33
2
2
( 2.50 T()0.500
(1.20 m
T ) (1.20 m ) cos 25.0°
ΔΦ B
=
2
Note the similarity between the situation in this problem and a generator. In a generator, one
normally has a loop rotating in a constant magnetic field so the flux through the loop varies
sinusoidally in time. In this problem, we have a stationary loop in an oscillating magnetic field,
and the flux through the loop varies sinusoidally in time. In both cases, a sinusoidal
emf e = e max sinwt, where e max = NBAw , is induced in the loop.
The loop in this case consists of a single band ( N = 1) around the perimeter of a red blood cell
with diameter d = 8.0 × 10 −6 m. The angular frequency of the oscillating flux through the area of
this loop is w = 2p f = 2p ( 60 Hz ) = 120p rad s
67352_ch20.indd 613
2/9/11 2:01:10 PM
⎢⎜⎝
⎣
⎟⎠ ⎜⎝
⎟⎜
⎟
1 rev ⎠ ⎝ 60 s ⎠ ⎥⎦
= 1.3 × 10 −2 V = 13 mV
20.33
614
172
Note the similarity between the situation in this problem and a generator. In a generator, one
normally has a loop rotating in a constant magnetic field so the flux through the loop varies
sinusoidally in time. In this problem, we have a stationary loop in an oscillating magnetic field,
and the
Chapter
20 flux through the loop varies sinusoidally in time. In both cases, a sinusoidal
emf e = e max sinwt, where e max = NBAw , is induced in the loop.
Imagine
yourconsists
right hand
with
north (the
The
loopholding
in this case
of ahorizontal
single band
= 1fi) ngers
aroundpointing
the perimeter
of adirection
red bloodofcell
( Nthe
−6
the wire’s
velocity),
when
closefrequency
your handofthe
ngers curl downward
(inthe
thearea of
m.
d = 8.0such
× 10that
The you
angular
thefioscillating
flux through
with
diameter
direction
ofwB⊥=).2p
Your
be pointing
westward.
By right-hand
rule
number 1, the
this
loop is
f =thumb
2p ( 60will
Hz )then
= 120p
rad s. The
maximum
induced emf
is then
magnetic force on charges in the wire would tend to move positive charges westward. Thus,
2
−6
the west end of the wire⎛ p
will
end
8.0east
× 10
m 120p s –1
1.0 ×relative
10 −3 T top the
⎞ positive
d 2 be
e max = NBAw = B ⎜
= 1.9 × 10 −11 V
w=
4
⎝ 4 ⎟⎠
.
(
20.34
.
20.37
) (
(a)
e max = NBAw = 500 ( 0.60 T ) [( 0.080 m ) ( 0.20 m )] ( 4p rad s ) = 60 V
(b)
Note that the calculator must be in radians mode for the next calculation.
(c)
618
.
20.41
(
)
From e = L ( ΔI Δt ),
Chapter 20
.
(
)
ΔI e
75 × 10 −3 V
=
=
= 38 A s
Δt
L 2.0 × 10 −3 H
In terms of its cross-sectional area A, length C, and number of turns N, the self inductance
of a solenoid is given as L = m 0 N 2 A C
From e = L ΔI Δt , we have
(a)
L=
20.42
.
20.43
p 2 rad p 2 rad
=
= 0.13 s
w
4p rad s
The units of NΦ B I
2
2
4p × 10 −7 T ⋅ m A ( 400 ) ⎡p 2.5 × 10 −2 m ⎤
m0 N 2 A
⎣
⎦ = 2.0 × 10 −3 H = 2.0 mH
(a) L =
=
C
0.20 m
(b)
20.40
⎞⎤
s⎟ ⎥ = 57 V
⎠⎦
The emf is first maximum when w t = p 2 radians, or when
t=
20.39
.
)
minmotor
rad ⎞are still stationary and the back emf
⎛
⎞ ⎛ 1 the
⎞ ⎛ 2p coils
(a) angular
Immediately
afteristhe
switch
isrev
closed,
The
frequency
w=
⎜⎝ 120
⎟⎜
⎟⎜
⎟ = 4p rad s.
min ⎠ ⎝ 60 s ⎠ ⎝ 1 rev ⎠
is zero. Thus,
⎡
⎛p
e = e max sin (w t ) = ( 60 V ) sin ⎢( 4p rad s ) ⎜
⎝ 32
⎣
20.38
)(
( )
)
−3
V ( 0.50 s )
e
e ( Δt ) −312H× (10
4.00 A )
= 2.40 × 10
=
= 4.0−5× 10 −3 2H = 4.0 mH
=
ΔI Δt
ΔI
2.0 A − 3.5=A 1.92 × 10 T ⋅ m
500
(
From e = L ( ΔI Δt ), the self-inductance is
(a) In the series circuit of Figure P20.20, maximum current occurs after the switch has been
closed for a very long time, when current has stabilized and the back emf due to the
inductance has decreased to zero. This maximum current is given by
I max =
(b)
The time constant of the RL circuit is
t =
(c)
e 24.0 V
=
= 5.33 A
R 4.50 Ω
L 12.0 H 12.0 Ω ⋅ s
=
=
= 2.67 s
R 4.50 Ω
4.50 Ω
If the switch in the RL circuit is closed at time t = 0, the current as a function of time is
given by I = I max 1 − e −t t .
(
)
Thus, e −t t = 1 − I I max, or t = −t ln (1 − I I max ) . With t = 2.67 s, the current in this circuit
will be I = 0.950I max at time
t = − ( 2.67 s ) ln (1 − 0.950 ) = 8.00 s
67352_ch20.indd 614
2/9/11 2:01:12 PM
(
will be I = 0.950I max at time
173
Voltages
and Inductance
= 2.67 s,
the current
in this circuit615
) . With t Induced
t = − ( 2.67 s ) ln (1 − 0.950 ) = 8.00 s
.
20.47
(
.
)
From I = I max 1 − e −t t , we obtain e −t t = 1 − I I max . If I I max = 0.900 at t = 3.00 s, then
e − (3.00 s) t = 0.100
t =
or
−3.00 s
= 1.30 s
ln ( 0.100 )
Since the time constant of an RL circuit is t = L R, the resistance is
620
20.49
.
20.48
Chapter 20
R=
L 2.50 H
=
= 1.92 Ω
1.30 s
t
(
(a)
As t → ∞, I → I max =
PE L →
(b)
(a)
e
24 V
=
= 3.0 A, and
R 8.0 Ω
1 2
1
2
LI max = ( 4.0 H ) (3.0 A ) = 18 J
2
2
(
)
At t = t , I = I max 1 − e −1 = ( 3.0 A ) (1 − 0.368 ) = 1.9 A, and
PE L =
20.51
.
20.50
)
L
The
t is I = I max 1 − e −t t , where I max = e R , and the energy stored
(a) current
t = in the circuit at time
2
R is PE L = 12 LI .
in the inductor
1
2
( 4.0
2
2 H1) (1.9 A ) = 7.2 J
2 LI = ( 4.44 mH ) ( 0.500 A ) = 0.555 mJ
2
2
energy stored
an inductor
is PEby
The inductance
of abysolenoid
is given
L 12=LIm20 ,Nso
Athe
C self inductance is
L =
(
)
−3
2 ( PE L ) 2 0.300 × 10 J
=
= 2.08 × 10 −4 H = 0.208 mH
L=
2
I2
(1.70 A )
(b)
If I = 3.0 A, the stored energy will be
PE L =
.
20.53
20.52
((
)
2
1 LI 22 =1 1 7.6 ×10−4
−4
−3
H (3.0
3.6 A)2 = 9.36
4.9 ×10
mJ mJ
2.08 ×10 H
×10 −4JJ== 4.9
0.936
max=
2
2
2
The flUse
ux due
to the
current
in loop
1 passes from
leftcopper
to right
through
(a)
Table
17.1
to obtain
the resistivity
of the
wire
and fithe
nd area enclosed by
loop 2. As loop 1 moves closer to loop 2, the magnitude of this flux through loop 2 is increasing.
The induced current in loop 2 generates a magnetic field directed toward the left through the area
it encloses in
the1.25
increasing
× 10 −3 Vflux from loop 1.−2This means that the induced current
ΔBordere to oppose
e
=
6.22the
× 10left end
T s of
= the
62.2rod.
mT s
=ow =counterclockwise
2 =from
2
in loop 2 must
fl
as viewed
Δt
A pr
p 8.00 × 10 −2 m
(
20.54
.
20.55
(a)
(a)
)
The clockwise induced current in the loop produces a flux directed into the page through
After the right end of the coil has
the area enclosed by the loop. Since this flux opposes the change in flux due to the external
entered the field, but the left end
field, the outward-directed flux due to the external field must be increasing in magnitude.
has not, the flux through the area
This means that the magnitude of the external field itself must be increasing
enclosed by the coil is directed into
the page and is increasing in magnitude. This increasing flux induces
an emf of magnitude
e =
ΔΦ B NB ( ΔA )
=
= NBwv
Δt
Δt
in the loop. Note that in the above equation, ΔA = wv is the area enclosed by the coil that
enters the field in time Δt. This emf produces a counterclockwise current in the loop to oppose
the increasing inward flux. The magnitude of this current is I = e
67352_ch20.indd 615
2/9/11 2:01:15 PM
616
174
20.32
Chapter 20
=
ΔΦ B NB ( ΔA )
=
= NBwv
Δt
Δt
Notein
that
vertical
of theequation,
magneticΔA
field
is always
parallel
to theby
plane
coil,
thethe
loop.
Note component
that in the above
is the area
enclosed
the of
coilthe
that
= wv
and can
never
to Δt.
the This
flux through
the coil.
The maximum induced
in loop
the coil
is then
enters
thecontribute
field in time
emf produces
a counterclockwise
current emf
in the
to oppose
the increasing inward flux. The magnitude of this current is I = e R = NBwv R. The right
rev toward the top of the
end of the loop is now a conductor, of−5length Nw, carrying
a current
2
e max = NBhorizontal Aw = 100 2.0 × 10 T ( 0.20 m ) 1500
page
through a field directed into the page. The field exerts a magnetic
force of magnitude
min
(
)
N 2 B2 w 2 v
⎛ NBwv ⎞
F = BI ( Nw ) = B ⎜
directed toward the left
Nw ) =
(
⎟
⎝ R ⎠
R
622
Chapter 20
on this conductor, and hence, on the loop.
(b)
When the loop is entirely within the magnetic field, the flux through the area enclosed by
the loop is constant. Hence, there is no induced emf or current in the loop, and the field
exerts zero force on the loop.
(c)
After the right end of the loop emerges from the field, and before the left end emerges,
the flux through the loop is directed into the page and is decreasing. This decreasing flux
induces an emf of magnitude e = NBwv in the loop, which produces an induced current
directed clockwise around the loop so as to oppose the decreasing flux. The current has
magnitude I = e R = NBwv R. This current flowing upward, through conductors of total
length Nw, in the left end of the loop, experiences a magnetic force given by
N 2 B2 w 2 v
⎛ NBwv ⎞
F = BI ( Nw ) = B ⎜
Nw ) =
(
⎟
⎝ R ⎠
R
energy within the resistor.
20.56
.
20.57
(a)
(a)
the left
,directed
which is toward
transformed
into internal
The motional emf induced in the bar must be e = IR, where I is the current in this series
The current in the solenoid reaches I = 0.632I max in a time of t = t = L R, where
the moving bar must be
circuit. Since e = B⊥ Cv, the speed of sol
(
)
(
)
4p × 10 −7 T ⋅ m A (12 500 ) 1.00 × 10 −4 m 2
m N2A
L= 0
=
= 0.280 H
7.00 × 10 −2 m
C
2
Thus, t = ( 0.280 H ) (14.0 Ω ) = 2.00 × 10 −2 s = 20.0 ms .
(b)
The change in the solenoid current during this time is
⎛ ΔV ⎞
⎛ 60.0 V ⎞
ΔI sol = 0.632I max − 0 = 0.632 ⎜
= 0.632 ⎜
= 2.71 A
⎝ R ⎟⎠
⎝ 14.0 Ω ⎟⎠
so the average back emf is
⎛ 2.71 A ⎞
⎛ ΔI ⎞
= 37.9 V
e back = L ⎜ sol ⎟ = ( 0.280 H ) ⎜
⎝ Δt ⎠
⎝ 2.00 × 10 −2 s ⎟⎠
(c)
The change in the magnitude of the magnetic field at the location of the coil is one-half the change
1 ⎡
in the magnitude of the field at the center of the solenoid. Thus,
ΔBVoltages
m 0 nsol
( ΔIsol )⎤⎦ , and623
coil = 2 ⎣ and
Induced
Inductance
the average rate of change of flux through each turn of the coil is
( ΔB )coil Acoil
⎛ ΔΦ B ⎞
=
continued on next page
⎜⎝
⎟⎠ =
Δt coil
Δt
=
(d)
I coil =
e coil
Rcoil
=
( 4p × 10
−7
1
2
⎡⎣ m 0 nsol ( ΔI sol ) ⎤⎦ Acoil m 0 N sol ( ΔI sol ) Acoil
=
Δt
2 C sol ⋅ ( Δt )
)
(
2 7.00 × 10
N coil ( ΔΦ B Δt )coil
Rcoil
(
T ⋅ m A (12 500 ) ( 2.71 A ) 1.00 × 10 −4 m 2
=
−2
)(
m 2.00 × 10
−2
s
)
)=
1.52 × 10 −3 V
(820 ) (1.52 × 10 −3 V )
24.0 Ω
= 0.051 9 A = 51.9 mA
20.58
67352_ch20.indd 616
(a)
The gravitational force exerted on the ship by the pulsar supplies the centripetal
GM pulsar mship
acceleration needed to hold the ship in orbit. Thus, Fg =
2
rorbit
2/9/11 2:01:17 PM
( ΔB )coil Acoil
⎞
=
⎟⎠ =
Δt
coil
⎛
⎜⎝
=
20.36
(a)
(d)
(d)
20.58
.
20.59
(a)
(a)
(b)
( 4p × 10
⎡⎣ m 0 nsol ( ΔI sol ) ⎤⎦ Acoil m 0 N sol ( ΔI sol ) Acoil
=
Δt
2 C sol ⋅ ( Δt )
1
2
)
(
)
2
T ⋅ m A (12 500 ) ( 2.71 Induced
A ) 1.00Voltages
× 10 −4 mand
Inductance −3 617
= 1.52 × 10 175
V
2 7.00 × 10 −2 m 2.00 × 10 −2 s
−7
(
)(
( )
)
)
−3
Using eemax = NBAw
we fiΔt
nd)
(820 ) 1.52=× 10
N coil (,ΔΦ
1.7 ×V
1010 V
B
coil
coil
I coil =
=
=
Rcoil
Rcoil
24.0
rev ⎞Ω⎛ 2p rad ⎞⎤
2 ⎡⎛
The very
induced
to powerful
electric discharges.
e maxlarge
= 1 000
T) would
0.10 mlead
⎜60
⎟ ⎜ spontaneous
⎟⎥
( 0.20emf
⎢
0.051electric
9 A = and
51.9magnetic
mA
⎠ ⎝ 1 rev
s disrupt
⎦ ow of ions in their bodies.
The =strong
fields⎣⎝would
the⎠fl
= 7.5 kV
= 7.5×10 3 Vexerted
The
gravitational
on the
by the of
pulsar
suppliesforce
the centripetal
To move
the bar atforce
uniform speed,
the ship
magnitude
the applied
must equal that of the
GM pulsar mship
magnetic
force
retarding
the
motion
of
the
bar.
Therefore,
The
magnitude
of the
=
BIC.
F
acceleration
needed
to hold
shipwhen
in orbit.
Fg = the
appcoil
The maximum
induced
emf the
occurs
the Thus,
flux through
is changing
the most
2
rorbit
induced
current
is
rapidly. This is when the plane of the coil is parallel to the magnetic field
(
(
I=
)
( ΔΦ B Δt ) = B ( ΔA Δt ) = B Cv
e
=
R
R
R
R
so the field strength is B = IR Cv, giving Fapp = ( IR Cv ) IC = I 2 R v, and the current is
I=
(b)
20.60
20.63
.
Fapp ⋅ v
R
=
(1.00 N ) ( 2.00 m s )
8.00 Ω
= 0.500 A
Pdissipated = I 2 R = ( 0.500 A )2 (8.00 Ω ) = 2.00 W
(c) Pinput = Fapp ⋅ v = (1.00 N ) ( 2.00 m s ) = 2.00 W
around the loop to
create additional flux directed into the page through the enclosed area.
Since the magnetic field outside the solenoid is negligible in comparison to the field inside the
solenoid, we shall
the
flux
⎡p
4 − 0 ⎤⎦ the single-turn square loop is the same as that
d 2 through
ΔΦ B assume
B ΔAthat B
⎣
(a)
= turn of= the solenoid.
=
throughe aveach
Then, the induced emf in the square loop is
Δt
Δt
Δt
(
=
)
( 25.0 mT ) p ( 2.00 × 10 –2 m )
(
4 50.0 × 10 −3 s
)
2
= 0.157 mV
continued on next page
As the inward-directed flux through the loop decreases, the induced current goes clockwise
around the loop to create additional inward flux through
the enclosed area. With positive 625
Induced Voltages and Inductance
charges accumulating at B, point B is positive relative to A .
(b)
e av
(
–2
ΔΦ B ( ΔB ) A [(100 − 25.0 ) mT ] p 2.00 × 10 m
=
=
=
Δt
Δt
4 4.00 × 10 −3 s
(
)
)
2
= 5.89 mV
As the inward-directed flux through the enclosed area increases, the induced current goes
counterclockwise around the loop in to create flux directed outward through the enclosed
area. With positive charges now accumulating at A, point A is positive relative to B .
20.64
67352_ch20.indd 617
The induced emf in the ring is
2/9/11 2:01:20 PM
장 ˬզୣԻࠪࢷ࢕̛ળ
PROBLEM SOLUTIONS
21.1
21.1
For an AC circuit containing only resistance (the filament of the lightbulb), the power dissipated
2
2
is P = I rm2 s R = ( ΔVrm s R ) R = ΔVrm2 s R = ΔVm ax 2 R.
(
(
)
(
)
170 V 2
ΔVrm2 s
R=
=
P
75.0 W
(a)
170 V 2
ΔVrm2 s
(b) R =
=
continued on next pageP
100.0 W
21.2
21.2
21.3
2
Alternating Current Circuits and Electromagnetic Waves
= 193 Ω
633
2
= 145 Ω
(
)( s ) =
radians .
The general form of the generator voltage is Δv = ( ΔVm ax ) sin (wt ), so by inspection
For a simple resistance, i ( t ) = v ( t ) R = ( ΔVm ax sinwt ) R = I m ax sinwt. Thus, if i = 0.600I m ax at
t = 7.00 ms, we have
(a)
wt = 2p ft = sin −1 ( i I m ax )
giving f =
21.4
21.3
21.5
)
sin
−1
( 0.600 )
2p t
=
and
f=
sin −1 ( i I m ax )
2p t
0.644 rad
= 14.6 s −1 = 14.6 Hz .
( 2p rad )( 7.00 × 10 −3 s )
= 144 Ω
All lamps are connected in parallel with the voltage source, so ΔVrm s = 120 V for each lamp.
The
connection)
= rmR1s + R2 = 8.20 Ω +10.4 Ω = 18.6 Ω, so the current
Also,total
for resistance
each bulb,(series
the current
is I rm s =isPavReqΔV
in the circuit is
I rm s =
ΔVrm s 15.0 V
=
= 0.806 A
18.6 Ω
Req
The power to the speaker is then Pav = I rm2 s Rspeaker = ( 0.806 A ) (10.4 Ω ) = 6.76 W .
.
2
21.4
21.7
I rm s =
f=
21.8
21.9
21.5
(a)
(a)
ΔVrm s
=2p f C ( ΔVrm s ) , so
XC
I rm s
0.30 A
=
= 4.0 × 10 2 Hz
The
charge
a maximum
when
2p C ( ΔV
2p (is4.0
× 10 −6 F )( 30
V ) the voltage is a maximum, but the voltage across a
rm s )
capacitor is 90° out of phase with the current.
The expression for capacitive reactance is XC = 1 2p fC. Thus, if XC < 175 Ω, it is
necessary that
f=
(b)
1
1
>
2p C XC 2p ( 22.0 ×10 −6 F ) (175 Ω)
or
f > 41.3 Hz
For C1 , the reactance is XC,1 = 1 2p fC1, while for C2 , XC, 2 = 1 2p fC 2. Thus, for the same
frequencies, the ratio of the reactance for the two capacitors is
XC, 2
XC,1
176
C1 = 22.0 mF, C2 = 44.0 mF, and XC,1 < 175 Ω, we have
⎛ 22.0 mF ⎞
XC, 2 < ⎜
⎟ (175 Ω)
⎝ 44.0 mF ⎠
67352_ch21.indd 627
I rm s =
2 ( ΔVrm s )
XC
or
XC, 2 < 87.5 Ω
2/9/11 2:04:17 PM
= 2 ( ΔVrm s ) 2p f C
(b)
177
Waves
633
For C1 , the reactance is XC,1 = 1 2pAlternating
fC1, whileCurrent
for C2Circuits
, XC, 2 and
= 1 Electromagnetic
2p fC 2. Thus, for
the same
frequencies, the ratio of the reactance for the two capacitors is
(b)
R=
ΔV
XC,rm22s ⎛ 1 ⎞ ⎛ 2p f C1 ⎞ C1
⎟⎜
⎟=
=⎜
XPC,1 ⎜⎝ 2p f C2 ⎟⎠ ⎜⎝ 1 ⎟⎠ C2
⎛C ⎞
XC, 2 = ⎜ 1 ⎟ XC,1
⎝ C2 ⎠
or
If C1 = 22.0 mF, C2 = 44.0 mF, and XC,1 < 175 Ω, we have
ΔVR , m ax = 170
V mF ⎞
⎛ 22.0
XC, 2 < ⎜
⎟ (175 Ω)
⎝ 44.0 mF ⎠
21.6
21.11
21.7
21.13
or
X
Ω and Electromagnetic Waves
Alternating
Current
Circuits
C, 2 < 87.5
I m ax =
ΔVm ax
= 2p fC ( ΔVm ax ) = 2p ( 90.0 Hz )( 3.70 × 10 −6 F )( 48.0 V )
XC
or
I m ax = 1.00 × 10 −1 A = 100 mA
=
(80π
635
1
= 2.03 × 10 −5 F = 20.3 mF
rad s )(196 Ω )
From L = N Φ B I (see Section 20.5 in the textbook), the total flux through the coil is
Φ B, total = N Φ B = L ⋅ I, where Φ B is the flux through a single turn on the coil. Thus,
(Φ
B, total
)
m ax
⎡ ΔV ⎤
= L ⋅ I m ax = L ⋅ ⎢ m ax ⎥
⎣ XL ⎦
2 ( ΔVrm s )
2 (120 V )
=
= 0.450 T ⋅ m 2
=L
fL A ) 2p
(60.0AHz )
I rm s = 2p
2 ( 6.19
= 8.75
21.14
21.8
21.15
(a)
(a)
I m ax =
L=
(b)
ΔVm ax ΔVm ax
=
, so
XL
2p fL
ΔVm ax
100 V
=
= 4.24 × 10 −2 H = 42.4 mH
2p f I m ax 2p ( 50.0 Hz )( 7.50 A )
I m ax = ΔVm ax X L = ΔVm ax w L , or I m ax is inversely proportional to w . Thus,
I m ax,1 I m ax, 2 = w 2 w 1 , or
⎛I
⎞
⎛ 7.50 A ⎞
w 2 = ⎜ m ax,1 ⎟ w 1 = ⎜
2p ( 50.0 Hz )] = 942 rad s
⎝ 2.50 A ⎟⎠ [
⎝ I m ax,2 ⎠
21.16
21.9
Given:
1
1
XC =
=
= 66.3 Ω
2p f C 2p ( 60.0 Hz )( 40.0 × 10 −6 F )
Z = R 2 + ( X L − XC ) =
2
( 50.0 Ω )2 + ( 0 − 66.3 Ω )2 = 83.0 Ω
ΔVrm s 30.0 V
=
= 0.361 A
83.0 Ω
Z
(a)
I rm s =
(b)
ΔVR , rm s = I rm s R = ( 0.361 A )( 50.0 Ω ) = 18.1 V
(c)
ΔVC , rm s = I rm s XC = ( 0.361 A )( 66.3 Ω ) = 23.9 V
(d)
⎛ X − XC ⎞
−1 ⎛ 0 − 66.3 Ω ⎞
f = tan −1 ⎜ L
⎟ = −53.0°
⎟⎠ = tan ⎜⎝
⎝
R
50.0 Ω ⎠
so the voltage lags behind the current by 53° .
21.20
67352_ch21.indd 633
X L = 2p fL = 2p ( 60.0 Hz )( 0.100 H ) = 37.7 Ω
2/9/11 2:04:30 PM
634
638
178
21.21
.
21.6
Chapter 21
(a)
X L R=, m2p
f L2= 2p ( 240 Hz )( 2.50 H ) = 3.77 × 10 3 Ω
ΔV
ax =
1
1
XC = 2
=
= 2.65 × 10 3 Ω
ΔVrm2 s
f C 2p ( 240 Hz )( 0.250 × 10 −6 F )
Pav = I2p
rm s R =
R
Z = R 2 + ( X L − XC ) =
( 900 Ω )2 + ⎡⎣( 3.77 − 2.65) × 103 Ω ⎤⎦ = 1.44 kΩ
2
2
640
.
21.29
ΔVm ax
140 V
=
= 0.097 2 A
1.44 × 10 3 Ω
Z
(b)
I m ax =
(c)
3
⎡
⎤
⎛ X − XC ⎞
−1 ( 3.77 − 2.65 ) × 10 Ω
=
tan
f = tan −1 ⎜ L
⎟
⎢
⎥ = 51.2°
⎝
900 Ω
R ⎠
⎣
⎦
(d)
f > 0 , so the voltage leads the current
Chapter 21
X L = 2p fL = 2p ( 50.0 Hz )( 0.185 H ) = 58.1 Ω
XC =
1
1
=
= 49.0 Ω
2p f C 2p ( 50.0 Hz )( 65.0 × 10 −6 F )
Z ad = R 2 + ( X L − XC ) =
( 40.0 Ω )2 + ( 58.1 Ω − 49.0 Ω )2 = 41.0 Ω
2
21.31
.
(
)
ΔVm ax 2
ΔVrm s
150 V
=
=
= 2.59 A
Z ad
Z ad
( 41.0 Ω ) 2
and
I rm s =
(a)
Z ab = R = 40.0 Ω, so ( ΔVrm s )ab = I rm s Z ab = ( 2.59 A )( 40.0 Ω ) = 104 V
(b)
Z bc = X L = 58.1 Ω, and ( ΔVrm s )bc = I rm s Z bc = ( 2.59 A )( 58.1 Ω ) = 150 V
(c)
Z cd = XC = 49.0 Ω, and ( ΔVrm s )cd = I rm s Z cd = ( 2.59 A )( 49.0 Ω ) = 127 V
(d)
Z bd = X L − XC = 9.10 Ω, so ( ΔVrmAlternating
Z bd = ( 2.59
A )and
Ω ) = 23.6 VWaves
( 9.10
s )bd = I rm s Current
Circuits
Electromagnetic
(a)
Pav = I rm2 s R = I rm s ( I rm s R) = I rm s ( ΔVR, rm s ), so I rm s =
Thus,
(b)
R=
ΔVR, rm s
I rm s
=
641
Pav
14 W
=
= 0.28 A
ΔVR , rm s 50 V
50 V
= 1.8 ×10 2 Ω
0.28 A
Z = R 2 + X L2 , which yields
2
2
⎛ ΔV ⎞
2
⎛ 90 V ⎞
X L = Z 2 − R 2 = ⎜ rm s ⎟ − R 2 = ⎜
− 1.8 × 10 2 Ω ) = 2.7 × 10 2 Ω
⎝ 0.28 A ⎟⎠ (
⎝ I rm s ⎠
and
L=
XL
2.7 × 10 2 Ω
=
= 0.72 H
2p f 2p ( 60 Hz )
21.32
21.35
.
(a)
Z=
ΔVrm s
104 V
=
= 208 Ω
0.500 A
I rm s
(b)
67352_ch21.indd 634
2/9/11 2:04:34 PM
=
21.11
Alternating Current Circuits and Electromagnetic Waves
635
179
ΔVm ax
P
10.0 W
I m ax =
(b)
PavX=C I rm2 s R gives R = 2av =
= 40.0 Ω
2
I rm s ( 0.500 A )
(c)
( 208 Ω )2 − ( 40.0 Ω )2 = 204 Ω
Z = R 2 + X L2 , so X L = Z 2 − R 2 =
L=
and
21.36
21.37
.
104 V
= 208 Ω
0.500 A
XL
204 Ω⎛
=
⎜⎝ ) =
2p f 2p ( 60.0 Hz
⎛ 90.0 V ⎞
⎞
0.541
⎟ cos ( −41.5° ) = 45.4 W
⎟⎠ cosfH= ( 0.953 A ) ⎜⎝
2 ⎠
(
)
ΔVm axLC= 90.0
V and
sin (wt ) =of
V ) sin
v = ΔVm axfrequency
( 90.0
( 350t
) , observe
= 350mHrad
Given
1 2p
. Thus,
if L w
= 1.40
ands
The resonance
a series
RLC
circuit
is f0 =that
the desired resonance frequency is f0 = 99.7 MHz, the needed capacitance is
1
1
C
= power
=
= 1.82 × 10 −12 F = 1.82 pF
2 a maximum
2 2 delivered
The
to
the
circuit
2
6
4p f0 L 4p ( 99.7 × 10 Hzis
1.40 × 10 −6 when
H ) the rms current is a maximum. This
)
(
occurs when the frequency of the source is equal to the resonance frequency of the circuit.
21.38
21.39
.
The resonance frequency is
1
1
f0 =
, so C =
4p 2 f02 L4 500 V
2p LC
=
= 15.5 A
290 Ω
For f0 = ( f0 )m in = 500 kHz = 5.00 × 10 5 Hz
The maximum input power is then
1
C = Cm ax =
= 5.1× 10 −8 F = 51 nF
2
2
5
−6
5.00
×
10
Hz
2.0
×
10
H
4p
)
=(( ΔVrm s ) ( I rm s )m ax) (
Pinput
(
)
m ax
For f0 = ( f0 )m ax = 1600 kHz = 1.60 × 10 6 Hz
= ( 4.50 × 10 3 V )(15.5 A ) = 6.98 × 10 4 W = 69.8 kW
1
−9
C = Cism infar= short2 of meeting
= 4.9
× 10
= 4.9
nF is lost in the
This
and
all ofF this
power
6 the 2customer’s
H)
4p (1.60 × 10 Hz ) ( 2.0 × 10 −6request,
transmission line.
21.40
.
21.45
(a) At resonance,
The power input to the transformer is
(P )
av
input
(
)
= ΔV1, rm s I1, rm s = ( 3 600 V )( 50 A ) = 1.8 × 10 5 W
For an ideal transformer, ( Pav )ouput = ( ΔV2, rm s ) I 2, rm s = ( Pav )input, so the current in the long-distance
power line is
I 2 , rm s =
(P )
av
(ΔV
input
2, rm s
)
=
1.8 ×10 5 W
= 1.8 A
100 000 V
The power dissipated as heat in the line is then
Plost = I 22, rm s Rline = (1.8 A ) (100 Ω ) = 3.2 × 10 2 W
2
The percentage of the power delivered by the generator that is lost in the line is
⎛ 3.2 × 10 2 W ⎞
Plost
× 100% = 0.18%
× 100%
=⎜
⎛
⎞
Pinput
1.8) ⎛× 6.0
10 5 V
W⎞⎟⎠= 20 turns
⎝400
=
(
⎜⎝
⎟
⎜⎝
⎟⎠
120 V ⎠
(a) Since the transformer is to step the voltage down from 120 volts to 6.0 volts, the secondary
The maximum
must haveoutput voltage ( ΔVm ax )2 is related to the maximum input voltage ( ΔVm ax )1 by the
N
expression ( ΔVm ax )2 = 2 ( ΔVm ax )1, where N1 and N 2 are the number of turns on the primary coil
N1
and the secondary coil, respectively. Thus, for the given transformer,
% Lost =
21.46
.
21.47
67352_ch21.indd 635
2/9/11 2:04:38 PM
636
180
( ΔV ) , where N
Chapter 21
m ax 1
1
and N 2 are the number of turns on the primary coil
and the secondary coil, respectively. Thus, for the given transformer,
Also by inspection,
1 500
(170 V ) = 1.02 × 103 V
m ax 2 =
250
ΔV
3
ΔVL, rm s = L, m ax
ΔV
2 the secondary is ( ΔVrm s ) = ( m ax )2 = 1.02 × 10 V = 721 V .
and the rms voltage across
2
2
2
21.48
.
21.49
(b)
( ΔV )
(a)
(a)
The output voltage of the transformer is
The solar energy incident each second on 1.00 m 2 of the surface of Earth’s atmosphere is
U total = 1 370
W
Js
N⋅m
N
= 1370 2 = 1370 2 = 1370
2
m
m
m ⋅s
m ⋅s
Of this, 38.0% is reflected and 62.0% is absorbed. From Equation 21.29 and Equation 21.30
in the textbook, the radiation pressures P1 due to the reflected radiation and P2 due to the
absorbed radiation are given by
P1 =
2 U reflected 2 ( 0.380 U total ) 0.760 U total
=
=
c
c
c
P2 =
and
U absorbed 0.620 U total
=
c
c
The total radiation pressure is then
Prad = P1 + P2 =
Prad =
or
(b)
( 0.760 + 0.620 ) U total
c
1.38 (1 370 N m ⋅s ) Alternating Current
2
and× Electromagnetic
Waves
= 6.30
10 −6 Pa
= 6.30 × 10 −6 N mCircuits
3.00 × 108 m s
647
In comparison, atmospheric pressure at the surface of the Earth is
Patm
101× 10 3 Pa
= 2 ( 3.84−6× 108 =m1.60
× 1010 times greater than the radiation pressure
) = 2.56
Prad 6.30
= × 10 Pa
s
3.00 × 108 m s
continued on next page
21.50
If I 0 is the incident
of Ea light beam, and I is the intensity
c Bm2 ax of the beam after passing
Emintensity
ax Bm ax
m ax
.
and
=
c,
we
fi
nd
Intensity
=
.
21.55
From
Intensity
=
through length L of2amfluid having
Bm axconcentration C of absorbing
2 m 0 molecules, the Beer-Lambert law
0
states that log10 I I 0
Thus,
Bm ax =
2 m0
( Intensity ) =
c
2 ( 4p × 10 −7 T ⋅ m A )
3.00 × 108 m s
(1 370
W m 2 ) = 3.39 × 10 −6 T
and Em ax = Bm ax c = ( 3.39 × 10 −6 T ) ( 3.00 × 108 m s ) = 1.02 × 10 3 V m because radio waves
travel so much faster than sound waves.
.
21.61
650
(a)
For the AM band,
Chapter 21
lm in =
lm ax =
(b)
fm ax
c
fm in
=
3.00 × 108 m s
= 188 m
1 600 × 10 3 Hz
=
3.00 × 108 m s
= 556 m
540 × 10 3 Hz
For the FM band,
lm in =
67352_ch21.indd 636
c
continued on next page
c
fm ax
2/9/11 2:04:42 PM
=
(b)
21.62
21.63
.
3.00 × 108 m s Alternating Current Circuits and Electromagnetic Waves
= 556 m
540 × 10 3 Hz
637
181
an inductor
always 90° or a quarter cycle out of phase with the instantaneous current.
For
the FM is
band,
Thus, when c
3.00 × 108 m. s
=
= 2.78 m
lm in =
fm ax
108 × 10 6 Hz
Kirchhoff’s loop
rule always applies to the instantaneous voltages around a closed path.
Thus, for thiscseries3.00
circuit
× 108Δvmsource
s = ΔvR + ΔvL , and at this instant when
= 8 m =s
lm ax× 10
= 3.4 m
3.00
6
=
=8811.0
f
× 10m
Hz
27.33 × 10m6 inHz
c 3.00 × 108 m s
⎤
14
l = = c 3.006 × 108= m11.0
s m ⎡
−12
⎢ × 10
⎥ = 6.003 6 × 10 Hz
= × 10 Hz 19
m
=
6.00
pm
(a) f l =27.33
= 6.00
8
⎣ 3.000 0 × 10 m s ⎦
f 5.00 × 10 Hz
21.63
(b)
(a)
(b)
The change in frequency
is
c 3.00 × 108 m s
l= =
= 7.50 × 10 −2 m = 7.50 cm
9
14
fΔf =4.00
Hz
fO −×f10
Hz − 6.000 0 × 1014 = 3.6 × 1011 Hz
S = 6.003 6 × 10
21.64
.
21.65
(a) Since
thethe
space
and
theare
ship
are moving
toward
oneother
another,
the frequency
afterat
Since
you and
car station
ahead of
you
moving
away from
each
(getting
farther apart)
being
Doppler
shifted
is
f
=
f
1+
u
c
S
a rate of u = 120 km h − 80 kmO h = 40
km h, the Doppler-shifted frequency you will detect is
fO = fS (1− u c ), and the change in frequency is
⎛ 40 km h ⎞ ⎛ 0.278 m s ⎞
⎛ u⎞
= − 1.6 × 10 7 Hz
Δf = fO − fS = − fS ⎜ ⎟ = − ( 4.3 × 1014 Hz ) ⎜
8
⎟
⎜
⎟
⎝ c⎠
1
km
h
⎠
⎝ 3.0 × 10 m s ⎠ ⎝
The frequency you will detect will be
7
14
Current
Circuits
and
Electromagnetic
Waves
fO = fS + Δf = 4.3 × 1014 Hz − 1.6 × 10Alternating
Hz = 4.299
999
84 × 10
Hz
.
21.67
651
The energy incident on the mirror in time Δt is U = Plaser ⋅ Δt, where Plaser is the power transmitted
by the laser beam
Plaser = 25.0 × 10 −3 W = 25.0 × 10 −3 J s = 25.0 × 10 −3 N ⋅ m s
From Equation 21.30 in the textbook, the rate of change in the momentum of the mirror as the
beam reflects from it is
Δp 2U c 2Plaser
=
=
Δt
Δt
c
The impulse-momentum theorem then gives the force exerted on the mirror as F = Δp Δt = 2Plaser c ,
and the radiation pressure on the mirror is
Prad =
F 2Plaser c 2Plaser
=
=
A
A
cA
where A = p r 2 = p d 2 4 is the area of the mirror illuminated (i.e., the cross-sectional area of the
laser beam). Thus,
Prad =
or
21.68
.
21.69
8 ( 25.0 × 10 −3 N ⋅ m s )
2Plaser
8Plaser
=
=
2
c (p d 2 4 ) p cd 2 p ( 3.00 × 108 m s ) ( 2.00 × 10 −3 m )
Prad = 5.31× 10 −5 N m 2 = 5.31× 10 −5 Pa
= 3R .
For a parallel-plate capacitor, C = ∈ A d
(a) From Equation 21.30 in the textbook, the momentum imparted in time Δt to a perfectly reflecting sail of area A by normally incident radiation of intensity I is Δp = 2U c = 2 ( IAΔt ) c.
From the impulse-momentum theorem, the average force exerted on the sail is then
67352_ch21.indd 637
2/9/11 2:04:46 PM
182
638
Chapter 21
= 2 ( IAΔt ) c.
21.21
(a)
From the impulse-momentum theorem, the average force exerted on the sail is then
1
2
4
2
XC =
Δp 2 ( IAΔt ) c 2IA 2 (1 340 W m ) ( 6.00 × 10 m )
fC =
F2p
= 0.536 N
=
=
av =
Δt
3.00 × 108 m s
Δt
c
(b)
aav =
(c)
From Δx = v0 t +
Fav 0.536 N
=
= 8.93 × 10 −5 m s 2
m 6 000 kg
2 ( Δx )
=
aav(
t=
.
21.71
R=
( ΔV )DC
I DC
=
2
21.72
21.73
.
2 ( 3.84 × 108 m )
1d
⎛
⎞
= ( 2.93 × 10 6 s )5⎜
⎟ = 33.9 d
⎝
8.93 × 10
s⎠ 6 J
) ( 0.4 mm s) (3 600 s) = 6 × 10 J8.64or× 104~10
−5
2 2
12.0 V
= 19.0 Ω
0.630 A
Z = R 2 + ( 2p f L ) =
Thus, L =
1 2
at , with v0 = 0, the time is
2
Z 2 − R2
=
2p f
ΔVrm s
24.0 V
=
= 42.1 Ω
0.570 A
I rm s
2
2 2
42.1ΩΩ)2) −−( 4.0
(6.0
(19.0
Ω )Ω )
−2
= 9.97
= mH
99.7 mH
= 1.2
× 10×−210H =H12
2p ( 60
60.0
2p
HzHz
) )
(a)
(a)
Em ax
E
0.20 × 10 −6 V m
= c, so Bm ax = m ax =
= 6.7 × 10 −16 T
Bm ax
c
3.00 × 108 m s
(b)
Intensity =
(c)
⎡p d2 ⎤
Pav = (Intensity) ⋅A = (Intensity) ⎢
⎥
⎣ 4 ⎦
−6
−16
Em ax Bm ax ( 0.20 × 10 V m ) ( 6.7 × 10 T )
=
= 5.3 × 10 −17 W m 2
2m 0
2 ( 4p × 10 −7 T ⋅ m A )
⎡ p ( 20.0 m )2 ⎤
−14
= ( 5.3 × 10 −17 W m 2 ) ⎢
⎥ = 1.7 × 10 W
4
⎣
⎦
21.75
.
Z = R 2 + ( XC ) = R 2 + ( 2p f C )
2
=
−2
( 200 Ω )2 + ⎡⎣ 2p ( 60 Hz )( 5.0 × 10 −6 F )⎤⎦ = 5.7 × 10 2 Ω
−2
⎛ ΔV ⎞
⎛ 120 V ⎞
Pav = I rm2 s R = ⎜ rm s ⎟ R = ⎜
( 200 Ω ) = 8.9 W = 8.9 × 10 −3 kW
⎝ 5.7 × 10 2 Ω ⎟⎠
⎝ Z ⎠
2
Thus,
and
2
cost = ΔE ⋅ ( rate ) = Pav ⋅ Δt ⋅ ( rate )
= (8.9 × 10 −3 kW ) ( 24 h ) (8.0 cents kWh ) = 1.7 cents
21.76
67352_ch21.indd 638
(a)
The intensity of radiation at distance r from a point source, which radiates total power P, is
I=P A
2/9/11 2:04:50 PM
장 ‫ࢂڂ‬؆‫ࢸ˵ࠪی‬
PROBLEM SOLUTIONS
22.1
.
(a)
1 eV
⎛
E = hf = ( 6.63 × 10 −34 J ⋅s ) ( 5.00 × 1017 Hz ) ⎜
⎝ 1.60 × 10 −19
(b)
E = hf =
−34
8
hc ( 6.63 × 10 J ⋅ s ) ( 3.00 × 10 m s ) ⎛ 1 nm ⎞
−19
=
⎜⎝ −9 ⎟⎠ = 6.63 × 10 J
2
l
3.00 × 10 nm
10 m
1 eV
⎛
⎞
E = 6.63 × 10 −19 J ⎜
= 4.14 eV
⎝ 1.60 × 10 −19 J ⎟⎠
22.2
22.3
.
⎞
3
⎟ = 2.07 × 10 eV = 2.07 keV
J⎠
Reflection and Refraction of Light
659
(a)
Thedistance
energy of
photon
is is
The total
thea light
travels
⎛
⎞
Δd = 2 ⎜ Dcenter to − REarth − RM oon ⎟
⎝ center
⎠
= 2 ( 3.84 × 108 − 6.38 × 10 6 − 1.76 × 10 6 ) m = 7.52 × 108 m
Therefore,
.
22.5
22.4
660
22.6
22.7
.
v=
Δd 7.52 × 108 m
=
= 3.00 × 108 m s
2.51 s .
Δt
The
(a) speed of light in a medium with index of refraction n is v = c n , where c is its speed in
vacuum.
(a)
n(hc 800 nm
3.00 × 108 m s
For water, n = 1.333, and v =
= 2.25 × 108 m s .
1.333
(b)
For crown glass, n = 1.52, and v =
(c) For diamond, n = 2.419, and v =
Chapter 22
3.00 × 108 m s
= 1.97 × 108 m s .
1.52
3.00 × 108 m s
= 1.24 × 108 m s .
2.419
(a)
photon
is E = hfThus,
= hcwhen
l q = 45.0° and the first medium is air (n = 1.00),
liquid energy
From nThe
Snell’s
law, of
n2 asinq
1
2 = n1 sinq 1 .
1
nairsinq 2 = (1.00 ) sin 45.0° n 2 .
we have
(a)
⎛ (1.00 ) sin 45.0° ⎞
For quartz, n2 = 1.458, and q 2 = sin −1 ⎜
⎟⎠ = 29.0° .
⎝
1.458
(b)
⎛ (1.00 ) sin 45.0° ⎞
For carbon disulfide, n2 = 1.628, and q 2 = sin −1 ⎜
⎟⎠ = 25.7° .
⎝
1.628
(c)
⎛ (1.00 ) sin 45.0° ⎞
For water, n2 = 1.333, and q 2 = sin −1 ⎜
⎟⎠ = 32.0° .
⎝
1.333
22.8
183
67352_ch22.indd 654
2/9/11 2:06:04 PM
658
184
Chapter 22
, or parallel to the incident ray
40.
.
22.9
(a) index
34.2°of refraction of zircon
(b) is34.2°
The
n = 1.923.
42.
c 3.00 × 10 m s
(c) andv(d)
index
affects the result.
(a)
= =Neither thickness=nor1.56
ms
× 10of8 refraction
n
1.923
(a) 24.42°
(b) See Solution.
l
632.8 nm
(b) The wavelength in the zircon is ln = 0 =
= 329.1 nm .
n refl1.923
(c) 33.44°
(d) Total internal
ection still occurs.
8
3.00 × 108 m s
rotate
clockwiseis f = v(f)= c2.9°
The frequency
=
= 4.74 × 1014 Hz .
ln l0 632.8 × 10 −9 m
2
h = ( 2.10 × 10 m ) tan ( 26.9° ) = 107 m
4.54 m
In the sketch at the right, observe that the law of reflection is
obeyed
as thelaw,
ray reflects from each of the mirrors. Also, note
From
77.5° Snell’s
that the normal lines to the two mirrors intersect at a right
⎡ n1 sinq1 ⎤
⎡ (1.00)sin 40.0°
⎤ = 29.4°
−1
angle
the
are−1perpendicular
to each
other.
⎢two mirrors
⎥ = sin
2 = sin
(a) qsince
⎢⎣
⎥⎦
n
1.309
2
⎣
⎦
Considering the right triangle formed by the two normal
lines
and
the
ray,
and
recalling
that
the
sum
of
the
interior
angles
of
and from the law of reflection, f = q1 = 40.0°.
any triangle is 180°, we find that
(e)
(c)
44.
22.10
.
22.13
46.
48.
Hence, the angle between the reflected and refracted rays is
a = 180° −q 2 − f = 180.0° − 29.4° − 40.0° = 110.6°
22.15
.
.
n1 sin q1 = n2 sin q 2
sin q1 = 1.333 sin 45.0°
q1 = sin −1 (1.333 sin 45.0° ) = 70.5°
Thus, the sun appears to be 19.5° above the horizontal .
.
.
22.17
22.16
The
reaches
the the
left-hand
distance
The incident
sketch atlight
the right
shows
path ofmirror
the rayatinside
the glass slab.
Considering the reflections at points A and B, the law of reflection
d that
2 = (a1.00
0.087we
5 mobserve that
) tanb5.00°
and
= q B=. Also,
tells us
= qm
A
above its bottom edge. The reflected light first reaches the right-hand mirror at height
Reflection and Refraction of Light
d = 2 ( 0.087 5 m ) = 0.175 m
663
It bounces between the mirrors with distance d
between points of contact with a given mirror.
Since the full 1.00 length of the right-hand mirror is
available for reflections, the number of reflections
from this mirror will be
N right =
1.00 m
= 5.71 →
0.175 m
5 full reflections
Since the first reflection from the left-hand mirror
occurs at a height of d 2 = 0.087 5 m, the total
number of that can occur from this mirror is
N left = 1+
22.18
67352_ch22.indd 658
(a)
1.00 m − 0.087 5 m
= 6.21 →
0.175 m
6 full reflections
From Snell’s law, the angle of refraction at the first surface is
2/9/11 2:06:11 PM
664
22.3
22.19
.
Chapter 22
659
185
Reflection and Refraction of Light
The total
From
Snell’s
distance
law, the angle
light travels
of incidence
is
at the air-oil interface is
⎛
⎞
Δd = 2 −1
D
⎡ center
⎤ − RM−1oon⎡ (⎟1.48 ) sin 20.0° ⎤
noil sinq
to − R
⎜
oilEarth
q = sin ⎝ ⎢ center
⎥ = sin ⎢ ⎠
⎥ = 30.4°
1.00
⎦
⎣
⎣ nair
⎦
6
6
8
= 2 ( 3.84
× 108 − 6.38
× 10light
− 1.76
× 10
m = 7.52
and the angle
of refraction
as the
enters
the) water
is × 10 m
⎡ n sinq oil ⎤
−1 ⎡ (1.48 ) sin 20.0° ⎤
Therefore,
( 60.0°
q ′ = sin −1 ⎢ oil (1.00
= sin
⎥) sin
⎥ = 22.3°
⎢ ) = 1.333
=
1.22
n
⎦
⎣
water
⎣
⎦
sin ( 45.0° )
22.20
22.21
.
Since the light ray strikes the first surface at normal inciApplying Snell’s law where the ray first enters the
dence, it passes into the prism without deviation. Thus,
glass gives
the angle of incidence at the second surface (hypotenuse
of the triangular
prism)
as shown in the
⎛ nwater sinq
⎞ is q1 −1=⎡45.0°
(1.333
) sin 42.0°
−1
1
sketch
the
right.
The
angle
of
refraction
is ⎤⎥ = 35.9°
f = sinat
=
sin
⎜ n
⎟
⎢
1.52
⎦
⎣
⎝
⎠
glass
q1
water, n = 1.333
d
f
d
y
Thus, d = 90.0° − f = 90.0° − 35.9° = 54.1°.
d
(a)
The distance down to point P, where the ray
emerges from the glass, is now seen to be
Crown glass,
n = 1.52
P
q2
y = d tand = ( 3.50 cm ) tan 54.1° = 4.84 cm
(b)
The angle of refraction as the ray leaves the block
is given by Snell’s law as
⎛ nglass sind ⎞
⎡ (1.52 ) sin 54.1° ⎤
= sin −1 ⎢
q 2 = sin −1 ⎜
⎥ = 67.5°
1.333
⎝ nwater ⎟⎠
⎦
⎣
(
) = 16.3°
22.22
22.25
.
From
Snell’s
law,right, q1 + b +q 2 = 180°.
As
shown
at the
When b = 90°, this gives q 2 = 90° −q1.
Then, from Snell’s law
sinq1 =
ng sinq 2
nair
= ng sin ( 90° −q1 ) = ng cosq1
Thus, when b = 90° ,
67352_ch22.indd 659
( )
sinq1
= tanq1 = ng or q1 = tan −1 ng Refl
. ection and Refraction of
cosq1
22.26
.
22.29
The index
of refraction
Using
Snell’s
law gives of the atmosphere decreases
with increasing altitude because of the decrease in
density of the atmosphere
altitude.
⎛ nair sinqwith
⎞ increasing
−1
−1 ⎛ (1.000)sin83.00° ⎞
i
= sin
(a)
q blue = sin
⎜⎝ right, the Sun ⎟⎠ = 47.79°
As indicated
in the
diagram
at
the
⎜⎝ ray
⎟
n
1.340
⎠
located at S below the blue
horizon appears to be located
at S′.
⎛ n sinq i ⎞
⎛ (1.000)sin83.00° ⎞
= sin −1 ⎜
(b) q red = sin −1 ⎜ air
⎟⎠ = 48.22°
⎝
1.331
⎝ nred ⎟⎠
22.30
The angles of refraction for the two wavelengths are
22.27
(a)
From Snell’s law, n2 =
Light
667
n1 sinq1
sinq 2
2/9/11 2:06:13 PM
186
660
22.7
.
22.31
Chapter 22
Thus,
whenisq1 = 45.0° and the first medium is air (n1 = 1.00),
From
law,ofnincidence
(a) Snell’s
The angle
the 1fi.rst
surface
2 sinq 2 = nat
1 sinq
nair = 1.0
=,(and
1.00the
45.0°ofnrefraction is
we have
) sinangle
= 30°
q1isinq
2
⎛ n sinq1i ⎞
⎛ (1.00 ) sin 45.0° ⎞
.
q 2 = sin −1 ⎜
q1r = sin −1 ⎜ air
⎟⎠q= =29.0°
⎟
⎝
1.458
1i 30°
⎝ nglass ⎠
(b)
−1 ⎛ 1.0 sin 30° ⎞
= sin
⎜⎝ de, n2 = 1.628,
⎟⎠ = 19°
For carbon
disulfi
and
1.5
60°
b
a
q1r q2i
q2r
nglass = 1.5
Also, a = 90° −q1r = 71° and b = 180° − 60° − a = 49°.
Therefore, the angle of incidence at the second surface is q 2 i = 90° − b = 41° . The angle of
refraction at this surface is
⎛ n sinq 2i ⎞
−1 ⎛ 1.5sin 41° ⎞
q 2r = sin −1 ⎜ glass
⎟ = sin ⎜
⎟ = 80°
⎝ 1.0
⎠
n
⎠
⎝
air
(b)
The angle of reflection at each surface equals the angle of incidence at that surface. Thus,
(q )
1 reflection
22.33
.
63.1°
58.6°==q 4.5°
2 r red, =
=violet
q1i −q
= 30°
and
(q 2 )−reflection
2i = 41°
Using Snell’s law gives
⎛ n sinq i ⎞
⎛ (1.000)sin 50.00° ⎞
= sin −1 ⎜
q red = sin −1 ⎜ air
⎟⎠ = 31.77°
⎟
⎝
n
1.455
⎝
⎠
red
and
⎛ n sinq i ⎞
⎛ (1.000)sin 50.00° ⎞
= sin −1 ⎜
q violet = sin −1 ⎜ air
⎟⎠ = 31.45°
⎟
⎝
1.468
⎝ nviolet ⎠
Thus, the dispersion is q red −q violet = 31.77° − 31.45° = 0.32°
( .
.
22.35
22.34
)=
40.4° .
(a) light
Snell’s
be written
as sinq
thea critical
incidence
As
goes law
fromcan
a medium
having
a refractive
index
n1 to
mediumangle
withof
refractive
index
(q1 = q c ),
1 sinq 2 = v
1 v2. At
of refraction
is 90°,
sinq
=
v
v
.
At
the
concrete-air
n2 < n1the
, theangle
critical
angle is given
byand
the Snell’s
relationlaw
sinqbecomes
=
n
n
c
1
2
c
2
1
boundary,
⎛v ⎞
⎛ 343 m s ⎞
= 10.7°
q c = sin −1 ⎜ 1 ⎟ = sin −1 ⎜
v
⎝ 1 850 m s ⎟⎠
⎝ 2⎠
22.36
22.37
.
(b)
Sound can be totally reflected only if it is initially traveling in the slower medium. Hence,
at the concrete-air boundary, the sound must be traveling in air .
(c)
Sound in air falling on the wall from most directions is 100% reflected , so the wall is a
good mirror.
Using
Snell’s
law, the index
of refraction
of the
liquid
is found
The angle
of incidence
at each
of the shorter
faces
of the
prismto
is be
45°, as shown in the figure at the right. For total internal reflection to
occur at these faces, it is necessary that the critical angle be less than
45°. With the prism surrounded by air, the critical angle is given by
sinq c = nair nprism = 1.00 nprism, so it is necessary that
sinq c =
or
22.38
67352_ch22.indd 660
nprism >
1.00
< sin 45°
nprism
1.00
1.00
=
=
sin 45°
2 2
2
The critical angle for this material in air is
2/9/11 2:06:16 PM
⎛
⎜⎝
⎞
−1 ⎛ (1.36 ) sin 42.7° ⎞
⎟⎠ = 67.3°Reflection and Refraction of
⎟⎠ = sin ⎜⎝
1.00
Light
661
187
.
22.11
22.39
the Sun
is 28.0°
horizon,from
the angle
of
When light
attempts
to above
cross athe
boundary
one medium
of refractive index n1 into a new
⎛
⎞
⎛ nw (1.00 nw ) ⎞
⎛ 1.00 ⎞
−1
−1 occur
−1 ⎛ 1.00
incidenceoffor
sunlightindex
at thenair-water
boundary
is
medium
refractive
<
n
,
total
internal
refl
ection
will
if the angle
of ⎞incidence
2
⎜
⎟ =1 sin ⎜ −1 n
⎟ = sin ⎜ n ⎟ = sin ⎜⎝ 1.78 ⎟⎠ = 34.2°
⎝
⎝ ( n2 g n1 ). ⎠
⎝ g ⎠
exceeds the critical angle
given⎠by q c = sin
q1 = 90.0° − 28.0° = 62.0°
⎛ 1.00
⎞
(c) andIf
(d)
Observe
of partq c(b)
that−1 all
of the interven(a)
n1angle
= 1.53
nin2 the
= naircalculation
= sin
= 40.8°properties
.
Thus,
the
ofand
refraction
is= 1.00, then
⎜⎝ the⎟⎠physical
1.53
ing layer (water in this case) canceled, and the result of part (b) is identical to that of
part (a). This will always be true when the upper and lower surfaces of the interven⎛ 1.333
⎞
−1
ing layer
parallel to each then
other. qNeither
the
thickness
nor the index of refraction
and nare
(b) If n1 = 1.53
⎜⎝
⎟⎠ = 60.6° .
2 = nwater = 1.333,
c = sin
1.53
of the intervening layer affects the result.
.
22.40
22.41
(a)
angle
forrefractive
which index n1 into water (n2 = 1.333), the critical
WhenThe
lightminimum
is coming
fromofaincidence
medium of
−1
total
internal
refl
ection
occurs
is
the
critiangle is given by q c = sin (1.333 n1 ).
cal angle. At the critical angle, the angle of
90°, asn1shown
in the
figure
the−1 (1.333 1.458 ) = 66.1° .
(a) refraction
For fused is
quartz,
= 1.458,
giving
q c at
= sin
right. From Snell’s law,
(b) In going from polystyrene (n1 = 1.49) to water, q c = sin −1 (1.333 1.49 ) = 63.5° .
(c)
22.42
.
22.43
⎛
⎞
)⎤
−1 ⎡ 2.419sin (1.6°
From sodium chloride
(n1 = 1.544)
q c = sin −1 (⎥1.333
1.544 ) = 59.7° .
= 2.9°
⎢
⎜⎝
⎟⎠ = sinto water,
1.333
⎦
⎣
(a) The index of refraction for diamond is ndiam ond = 2.419, and the critical angle at a diamondIf q c = 42.0° at the boundary between the prism
glass
q1
air boundary is
Surrounding
and the surrounding medium, then sinq c = n2 n1 gives
medium, nm
nm
= sin 42.0°
nglass
nglass
From the geometry shown in the figure at the right,
a = 90.0° − 42.0° = 48.0°, b = 180° − 60.0° − a = 72.0°
b 60.0°
qr
qc = 42.0° a
Surface 2
and q r = 90.0° − b = 18.0°. Thus, applying Snell’s law at the first
surface gives
⎛ sinq r ⎞
⎛ nglass sinq r ⎞
−1 ⎛ sin18.0° ⎞
= sin −1 ⎜
q1 = sin −1 ⎜
⎟ = sin ⎜⎝ sin 42.0° ⎟⎠ = 27.5°
⎟
nm
⎝
⎠
⎝ nm nglass ⎠
22.44
22.45
.
672
The
circular
must cover
the law
areagives
of thethe
surface
At the
air-iceraft
boundary,
Snell’s
angle of refraction in the ice as
through which light from the diamond could emerge.
Thus, it must form
the1i base of a cone (with apex at the
n sinq
Chapter 22
= air half angle
sinq1rwhose
diamond)
is q, where q is greater than
nice
or equal to the critical angle.
Since the sides of the ice layer are parallel, the angle of incidence at the ice-water boundary is
angle
at the
water-air
boundary
qThe
=critical
q1r . Then,
from
Snell’s
law, the
angle ofisrefraction in the water is
2i
⎡ nice ⎛ n sinq ⎞ ⎤
⎛ n sinq 2 i ⎞
⎛ nice sinq1r ⎞
−1
−1
air
1i
⎢
=
sin
=
sin
q 2 r = sin −1 ⎜ ice
⎜
⎟⎥
⎜⎝ n
⎟⎠
⎝ nwater ⎟⎠
⎢⎣ nwater ⎝ nice ⎠ ⎥⎦
water
or
⎛ n sinq1i ⎞
⎡ (1.00 ) sin 30.0° ⎤
= sin −1 ⎢
q 2 r = sin −1 ⎜ air
⎥ = 22.0°
⎟
1.333
⎝ nwater ⎠
⎦
⎣
Note that all of the properties of the ice canceled out in the above calculation, and the result is the
same as if the ice had not been present. This will always be true when the intermediate medium
has parallel sides.
22.46
Applying Snell’s law to this refraction, recognizing that nair = 1.00, gives
nglass sinq 2 = nair sinq1 = sinq1
If q1 = 2q 2 , this becomes
67352_ch22.indd 661
2/9/11 2:06:18 PM
662
188
.
22.14
22.47
⎛
⎜⎝
Chapter 22
⎞
−1 ⎛ 1.56 ⎞
⎟⎠ = 2 cos ⎜⎝ 2 ⎟⎠ = 77.5°
Consider
thethat
sketch
the right
and in
note
the at
incident
hori(a)
Given
the at
angle
q shown
thethat
figure
the right
is
zontal30.0°,
ray is the
parallel
to the distance
surface of
2. Thus,
thefrom
angle
maximum
themirror
observer
can be
the
the incident
raycontinue
makes with
mirror
1 mustedge
be aon=the
q1 =opposite
50.0°.
pool and
to see
the lower
Since side
the ray
must
obey
of the
pool
is the law of reflection at mirror 1,
the angle b must be b = a = 50.0°. Recalling that the
sum of the interior
a triangle
is m
always 180.0°,
= 1.07
d = (1.85angles
m ) tanof30.0°
we find that
d
Mirror 1
a
q
1.85 m
b
q1
g
q
q2
g = 180.0° −q1 − b = 180.0° − 50.0° − 50.0° = 80.0°
Mirror 2
If the pool is now completely filled with water, the
light ray coming to the observer’s eye from the lower
d´
Hence, in order to obey the law of reflection at mirror 2, the angle the outgoing reflected ray
opposite edge of the pool will refract at the surface of
makes with the surface of mirror 2 must be q 2 = g = 80.0°
the water as shown in the second figure. The angle of
air
f
1.85 m
refraction is
n
(b)
air
⎛ n sin 30.0° ⎞
−1 ⎡ (1.333 ) sin 30.0° ⎤
f = sin −1 ⎜ water
⎥ = 41.8°
⎟⎠ = sin ⎢
nair
1.00
⎝
⎦
⎣
30.0°
nwater
The maximum distance the observer can now be from the pool and still see the same
boundary is
d ′ = (1.85 m ) tanf = (1.85 m ) tan 41.8° = 1.65 m
22.49
.
(a)
.
From the geometry of the figure at the
right, observe that q1 = 60.0°. Therefore,
a = 90.0° −q1 = 30.0°
(q
and
2
+ 90.0° ) + a + 30.0° = 180.0°
Thus, q 2 = 180.0° − 120.0° − a = 30.0°
nglass
Since the prism is immersed in water,
n2 = 1.333 and Snell’s law gives
⎛ nglass sinq 2 ⎞
−1 ⎛ (1.66 ) sin 30.0° ⎞
q 3 = sin −1 ⎜
⎟⎠ = 38.5°
⎟⎠ = sin ⎜⎝
n2
1.333
⎝
(b)
For refraction to occur at point P, it is necessary that q c > q1 .
Thus,
⎛ n ⎞
q c = sin −1 ⎜ 2 ⎟ > q1
⎝ nglass ⎠
which gives n2 > nglass sinq1 = (1.66 ) sin 60.0° = 1.44 . .
22.50
.
22.51
When light coming from the surrounding medium is
In the figure at the right, observe that
incident on the surface of the glass slab, Snell’s law
b = 90° −q1 and a = 90° −q1 . Thus,
gives ng sinq
= ns sinq i , or
r
b =a.
Similarly, on the right side of the
prism, d = 90° −q 2 and g = 90° −q 2 ,
giving d = g .
67352_ch22.indd 662
Next, observe that the angle between
the reflected rays is B = (a + b ) + (g + d ) ,
so B = 2 (a + g ) . Finally, observe that
the left side of the prism is sloped at
angle a from the vertical, and the right side is sloped at angle g .
Thus, the angle between the two sides is
A = a + g , and we obtain the result B = 2 (a + g ) = 2 A .
2/9/11 2:06:21 PM
Reflection and Refraction of Light
d =g .
Next,
observe
that the
between
It bounces
between
theangle
mirrors
with distance d
the
reflected
rays
B = (awith
+ b )a+given
between
points
ofiscontact
(g + dmirror.
),
so B = 2 (a + g ) . Finally, observe that
the
leftthe
side
the prism
at
Since
fullof1.00
lengthisofsloped
the right-hand
mirror is
angle
a from
theections,
vertical,the
andnumber
the right
is sloped at angle g .
available
for refl
of side
reflections
Thus,
the mirror
angle between
from this
will be the two sides is
A = a + g , and we obtain the result B = 2 (a + g ) = 2 A .
Reflection and Refraction of
22.52
.
22.53
Light
663
189
675
= 5.71 → 5 full reflections
(a) Observe in the sketch at the right that a ray originally
Consider
light which
leaves
theedge
lower
endhave
of the
and angle
traveling
along the
inner
will
thewire
smallest
travelsof
parallel
to ection
the
wire
theouter
benzene.
incidence
when
it while
strikes
the
edge If
ofthe
thewire
fiber
Since
the
first refl
from
thein
left-hand
mirror
appears
toof
andobserver
along
the dry portion
inatstraight
the
curve.
Thus,
if 0.087
this looking
ray
is totally
internally
reflected,
occurs
a height
2=
5 m,
the
total
of the all
wire,
thiscan
rayoccur
from
the
end
the wire must enter
the
others
are from
alsolower
totally
reflofected.
number
ofofthat
this mirror
is
the observer’s eye as he sights along the wire. Thus, the ray
must refract
to the wirerefl
inected
air. The
of
For thisand
raytravel
to be parallel
totally internally
it isangle
necessary
refraction
that is then q 2 = 90.0° − 30.0° = 60.0°. From Snell’s law,
the angle of incidence was
⎛ n sinq 2 ⎞
⎛ (1.00 ) sin 60.0° ⎞
= sin −1 ⎜
q1 = sin −1 ⎜ air
⎟⎠ = 35.3°
⎝
1.50
⎝ nbenzene ⎟⎠
and the wire is bent by angle q = q 2 −q1 = 60.0° −q1 = 60.0° − 35.3° = 24.7° .
22.54
In the sketch at the right, the angle of incidence
at A is the same as the prism angle at point O.
This is true because tipping the line OA up
angle q from the horizontal necessarily tips its
normal line over at angle q from the vertical.
Given that q = 60.0°, application of Snell’s law
at point A gives
1.50 sin b = (1.00 ) sin 60.0° or
b = 35.3°
From triangle AOB, we calculate the angle of incidence and reflection, g , at point B:
67352_ch22.indd 663
2/9/11 2:06:24 PM
장 ʠࡌ˕Ԩइ
PROBLEM SOLUTIONS
.
23.1
.2
23.5
23.2
(1)
The first image in the left-hand mirror is 5.00 ft behind the mirror
or 10.0 ft from the person .
(2)
The first image in the right-hand mirror serves as an object for the left-hand mirror. It is
located 10.0 ft behind the right-hand mirror, which is 25.0 ft from the left-hand mirror.
Thus, the second image in the left-hand mirror is 25.0 ft behind the mirror
or 30.0 ft from the person .
(3)
The first image in the left-hand mirror serves as an object for the right-hand mirror. It is
located 20.0 ft in front of the right-hand mirror and forms an image 20.0 ft behind that
mirror. This image then serves as an object for the left-hand mirror. The distance from this
object to the left-hand mirror is 35.0 ft. Thus, the third image in the left-hand mirror is
⎛
⎞
= 4.58
m from the person .
35.0 ft behind⎜⎝the mirror⎟⎠ or
40.0 ft
When
an object
is inlocated
front of1.0
a plane
mirror,
mirror forms
upright,
that is
(a) With
the palm
m in front
of that
the nearest
mirror,anthat
mirrorvirtual
forms image
an image,
the same size as the object and as far behind the mirror as the object is in front of the mirror.
This statement is true even if the mirror is rotated, as shown in the ray diagrams given below. In
figure (a), a real object O1 is distance p1 in front of the upper mirror in the periscope. This mirror
forms the virtual image I1 at distance p1 behind the mirror. As shown in figure (b), this image
serves as the object for the lower mirror in the periscope, and is distance p2 = p1 + h in front
of the lower mirror. The lower mirror then forms the final image I2, an upright, virtual image,
located distance p2 = p1 + h behind this mirror.
O2 = I1
I1
Upper
mirror
Lower
mirror
| q1| = p1
O1
p2 = p1 + h
I2
p < f.
| q2| p
True. An upright, virtual image is formed when
=<
p2 f=, pwhile
1 + h an inverted, real image is
p
formed
when
p
>
f
.
1
Chapter 23
(b)
686
(a)
67352_ch23.indd 679
(b)
(c)
(a)
False. A magnified, real image is formed if 2 f > p > f , and a magnified, virtual image is
formed
if in the above ray diagrams, the final image is distance q = p + h behind the
As shown
2
1
lower mirror, where p1 is the distance from the original object to the upper mirror and h is
the vertical distance between the two mirrors in the periscope.
(b)
The final image is behind the mirror and is virtual .
(c)
As seen from the ray diagrams, the final image I2 is oriented the same way as the original
object, and is therefore upright . 190
(d)
The overall magnification is
2/9/11 2:07:40 PM
.
23.3
685
191
(c)
nearest
This
then the
serves
an object
the nearest
mirror,
an
oriented
the same
waywhich
as theforms
original
As
seenmirror.
from the
rayimage
diagrams,
finalasimage
I2 isfor
image
I
,
located
P
3
object, and is therefore upright .
(d)
The overall magnification is
(a)
⎛ q ⎞ ⎛ q ⎞ ⎛ − p ⎞ ⎛ −all
⎞ virtual images .
p are
M = M1 M 2 = ⎜ − 1 ⎟ ⎜ − 2 ⎟ = ⎜ − 1 ⎟ ⎜ − 2 ⎟ = ( +1)( +1) = +1
⎝ p ⎠ ⎝ p2 ⎠ ⎝ p1 ⎠ ⎝ p2 ⎠
Due to the finite value of1 the speed
of light, the light arriving at your eye must have
reflected from your face
191 at a slightly earlier time. Thus, the image viewed in the mirror
The final image is therefore upright and the same size as the original object.
(e)
23.6
.
23.7
Mirrors and Lenses
No . The images formed by plane mirrors are upright in both directions. Just as plane
mirrors do not reverse up and down, neither do they reverse left and right.
.
The image
wasvirtual
initially
upright
but became
inverted
whenare
Dina
was more
enlarged,
images
formed
by a concave
mirror
upright,
so Mthan
> 0.30 cm from the
mirror. From this information, we know that the mirror must be concave
q h ′ 5.00 cm
Thus, M = − = =
= + 2.50, giving
p h 2.00 cm
q = − 2.50 p = − 2.50 ( + 3.00 cm ) = −7.50 cm
The mirror equation then gives
f=
23.8
23.9
.
( 3.00 cm )( − 7.50 cm )
pq
=
= + 5.00 cm
p+q
3.00 cm − 7.50 cm
Mirrors and Lenses
687
The lateral magnification is given by M = − q p
(a) Since the mirror is convex, R < 0. Thus, R = − 0.550 m and f = R 2 = − 0.275 m. The mirror
equation then yields
q=
pf
( +10.0 m )( −0.275 m )
=
= − 0.268 m = −26.8 cm
p − f +10.0 m − ( −0.275 m )
The image is located 26.8 cm behind the mirror .
(b)
(c)
23.10
.
23.11
(a)
(a)
The magnification of the image is M = − q p. Since p > 0 and q < 0 in this case, we see that
M > 0. Therefore, the image is upright .
q
( −26.8 cm )
( −26.8 cm )
(−10.0 cm)= +2.68 × 10 −2 = + 0.026 8
=−
=−
= − × 10 2 cm = +10.0 .
p
10.0 m
10.0
1.00 cm
Since the object is in front of the mirror, p > 0. With the image behind the mirror,
The center of curvature of a convex mirror is behind the mirror. Therefore, the radius of
curvature, and hence the focal length f = R 2, is negative. With the image behind the
mirror, the image is virtual and q = −10.0 cm. The mirror equation then gives
M=−
p=
qf
( −10.0 cm )( −15.0 cm )
=
= +30.0 cm
q − f −10.0 cm − ( −15.0 cm )
The object should be placed 30.0 cm in front of the mirror .
(b)
The magnification of the mirror is
q
−10.0 cm
M = − q = − ( −10.0 cm ) = + 0.333
M = − p = − +30.0 cm = + 0.333
p
+30.0 cm
Therefore, the image is upright and one-third the size of the object.
Therefore, the image is upright and one-third the size of the object.
23.12
67352_ch23.indd 685
(a)
Since the mirror is concave, R > 0, giving R = +24.0 cm and f = R 2
2/9/11 2:07:54 PM
192
688
.
23.13
Chapter 23
.
(b)
The needed
ray diagram,
object 8.00 cm in front of the mirror, is shown below:
The image
is upright,
so M > 0,with
and the
we have
M=−
q
= + 2.0, or q = − 2.0 p = − 2.0 ( 25 cm ) = − 50 cm
p
The radius of curvature is then found to be
23.14
.
23.15
2 1 1
1
1
2 −1
= + =
−
=
, or R = 2 ⎛ 0.50 m ⎞ = 1.0 m
⎜⎝ +1 ⎟⎠
R p q 25 cm 50 cm 50 cm
Object
Distance, p
Image Distance, q
.
(a) For spherical mirrors, both concave and convex, the radius
of
curvature
is related
3.00 m
0.600tomthe
The focal
theRmirror
may bethe
found
from
given object
imagemirror,
distances
as
focallength
lengthofby
= 2 f . Hence,
radius
of the
curvature
of thisand
concave
having
±∞
0.500 m
1 f = 1f p=++30.0
1 q , or
cm, is
0
0
pq
(152 cm )(18.0 cm )
Mirrors and Lenses
689
f=
=
= +16.1 cm
p + q 152 cm + 18.0 cm
an upright
continued For
on next
page image twice the size of the object, the magnification is
M = − q p = +2.00 , giving q = − 2.00 p
Then, using the mirror equation again, 1 p + 1 q = 1 f becomes
1
2 −1
1
1 1 1
+ = −
=
=
p q p 2.00 p 2.00 p f
.
23.17
23.16
f
16.1 cm
=
= 8.05 cm
2.00
2.00
or
p=
(a)
(a)
1 of curvature of a concave mirror is in front of the mirror. Therefore, both the
The center
From
radius pof curvature and the focal length, f = R 2, are positive. Since the image is virtual,
the image distance is negative and q = −20.0 cm. With R = + 40.0 cm and f = +20.0 cm, the
mirror equation gives
p=
= 0.782 s
( −20.0 cm )( + 20.0 cm )
qf
=
= +10.0 cm
q − f −20.0 cm − ( + 20.0 cm )
Thus, the object should be placed 10.0 cm in front of the mirror .
(b)
The magnification of the mirror is
M=−
q
( −20.0 cm )
=−
= + 2.00
p
+10.0 cm
3 − 1the size of the object.
Therefore, the image is upright and twice
=
, or R = 60.0 cm .
30.0 cm
23.19
.
(a)
An image formed on a screen is a real image. Thus, the mirror must be concave since, of
mirrors, only concave mirrors can form real images of real objects.
(b)
The magnified, real images formed by concave mirrors are inverted, so M < 0 and
M =−
q 5.0 m
q
= 1.0 m
= − 5 , or p = =
5
5
p
The object should be 1.0 m in front of the mirror .
(a — revisited) Now that we have both the object distance and image distance, we can use the
mirror equation to be more specific about the required mirror. The needed focal length of
the concave mirror is given by
67352_ch23.indd 687
2/9/11 2:07:58 PM
.
23.9
687
193
(a — rSince
evisited)
that
we have Rboth
object
image
distance,
we can use the
(a)
the Now
mirror
is convex,
< 0.the
Thus,
R =distance
− 0.550 and
m and
f =R
2
mirror equation to be more specific about the required mirror. The needed focal length of
the concave mirror is given by
f=
23.20
.
23.21
Mirrors and Lenses
pq
(1.0 m )( 5.0 m )
=
= 0.83 m
p + q 1.0 m + 5.0 m
Mirrors and Lenses
691
(a)
The mirror
is convex,
f < 0, distance,
and we have
− f the transparent sphere from air
p → f∞=) enter
As parallel
rays from
the Sunso( object
( n1 = 1.00 ) , the center of curvature of the surface is on the side the light is going toward (back
side). Thus, R > 0. It is observed that a real image is formed on the surface opposite the Sun,
giving the image distance as q = +2R. Then,
n1 n2 n2 − n1
+
=
p q
R
becomes
0+
n − 1.00
n
=
R
2R
which reduces to n = 2n − 2.00, and gives n = 2.00 .
.
23.22
.
23.23
The location
of the
image formed
refraction
thisside the light comes from, R < 0, giving
Since
the center
of curvature
of thebysurface
is onatthe
n1 n2 byn2Equation
− n1
spherical surface is described
23.7 from the
R = − 4.0 cm. Then, +
=
becomes
textbook, which statespn1 pq
R
1.50
1.00 1.00 − 1.50
=
−
, or q = − 4.0 cm
− 4.0 cm
4.0 cm
q
Thus, the magnification M =
⎛n ⎞q
h′
= − ⎜ 1 ⎟ gives
h
⎝ n2 ⎠ p
⎛ nq⎞
1.50 ( −4.0 cm )
h′ = − ⎜ 1 ⎟ h = −
( 2.5 mm ) = 3.8 mm
1.00 ( 4.0 cm )
⎝ n2 p ⎠
23.24
.
23.27
below ground level.
Light scattered from the bottom of the plate undergoes two refractions, once at the top of the plate
In the drawing at the right, object O (the jellyfish)
and once at the top of the water. All surfaces are planes ( R → ∞ ) , so the image distance for each
is located distance p in front of a plane water-glass
refraction is q = − n n1
interface. Refraction2 at that
interface produces a
virtual image I ′ at distance q ′ in front it. This image
serves as the object for refraction at the glass-air
interface. This object is located distance p′ = q ′ + t
in front of the second interface, where t is the thickness of the layer of glass. Refraction at the glass-air
interface produces a final virtual image, I, located
distance q in front of this interface.
From n1 p + n2 q = (n2 − n1 ) R with R → ∞
for a plane, the relation between the object and
image distances for refraction at a flat surface is
q = − ( n2 n1 ) p. Thus, the image distance for the refraction at the water-glass interface is
q ′ = − ng nw p. This gives an object distance for the refraction at the glass-air interface of
p′ = (ng nw ) p + t and a final image position (measured from the glass-air interface) of
(
q=−
(a)
67352_ch23.indd 688
)
na
n
p′ = − a
ng
ng
⎡⎛ n ⎞
⎛ na ⎞ ⎤
⎡ ⎛ ng ⎞
⎤
a
⎢⎜ ⎟ p + t ⎥ = − ⎢⎜ ⎟ p + ⎜ ⎟ t ⎥
⎝ ng ⎠ ⎥⎦
⎢⎣ ⎝ nw ⎠
⎥⎦
⎢⎣⎝ nw ⎠
If the jellyfish is located 1.00 m (or 100 cm) in front of a 6.00-cm thick pane of glass, then
p = +100 cm, t = 6.00 cm, and the position of the final image relative to the glass-air interface is
2/9/11 2:08:01 PM
(a)
688
194
If the jellyfish is located 1.00 m (or 100 cm) in front of a 6.00-cm thick pane of glass, then
p = +100 cm, t = 6.00 cm, and the position of the final image relative to the glass-air interChapter 23
face is
(b)
The needed⎡⎛ray
diagram,
with the
object
1.00
⎤ front of the mirror, is shown below:
⎛ 1.00
⎞ 8.00 cm in
⎞
q = − ⎢⎜
⎟⎠ ( 6.00 cm )⎥ = −79.0 cm
⎟⎠ (100 cm ) + ⎜⎝
⎝
1.50
⎣ 1.333
⎦
It appears to be in the water, 79.0 cm back of the outer surface of the glass pane .
(b)
If the thickness of the glass is negligible (t → 0), the distance of the final image from the
glass-air interface is
q=−
na
ng
⎡ ⎛ ng ⎞
⎤
⎛ na ⎞
⎛ 1.00 ⎞
⎢ ⎜ ⎟ p + 0 ⎥ = − ⎜ ⎟ p = − ⎜⎝
⎟ (100 cm ) = −75.0 cm
1.333 ⎠
n
n
⎝ w⎠
⎢⎣ ⎝ w ⎠
⎥⎦
Now, it appears to be 75.0 cm back of the outer surface of the glass pane .
(c)
23.28
.
23.29
We
the diagram
2.00ray
cmshown
and Rin
Withassume
R1 = + the
2 = + 2.50 cm, the lens maker’s equation gives the focal length as
at the right is a paraxial ray so q1 and q 2
are both
⎛ 1 small
⎞
1 sufficiently
1 ⎞to allow us to⎛
1
1
= + 0.050 0 cm −1
= ( n −law
1) ⎜as − ⎟ = (1.50 − 1) ⎜
−
⎟
write Snell’s
R
f
R
2.00
cm
2.50
cm
⎠
⎝
⎝ 1
2 ⎠
or
.
23.31
Comparing the results of parts (a) and (b), we see that the 6.00-cm thickness of the glass
in part (a) made a 4.00-cm difference in the apparent position of the jellyfish. We conclude
nq
that the thicker
= − 1 the glass, the greater the distance between the final image and the outer
n2 glass.
p
surface of the
(a)
f=
1
= + 20.0 cm
0.050 0 cm −1
=−
( − 40.0 cm ) =
+20.0 cm
+2.00
The real image case is shown in the ray
diagram. Notice that p + q = 12.9 cm, or
q = 12.9 cm − p. The thin-lens equation, with
f = 2.44 cm, then gives
1
1
1
+
=
p 12.9 cm − p 2.44 cm
or
p2 − (12.9 cm ) p + 31.5 cm 2 = 0
Using the quadratic formula to solve gives
p = 9.63 cm or p = 3.27 cm
Both are valid solutions for the real image case.
(b)
The virtual image case is shown in the second
diagram. Note that in this case, q = − (12.9 cm + p ) ,
so the thin-lens equation gives
1
1
1
−
=
p 12.9 cm + p 2.44 cm
or
p2 + (12.9 cm ) p − 31.5 cm 2 = 0
Mirrors and Lenses
695
The quadratic formula then gives p = 2.10 cm or p = −15.0 cm .
Since the object is real, the negative solution must be rejected, leaving p = 2.10 cm .
23.32
67352_ch23.indd 689
Consider the figure at the right showing an object O located
distance L in front of a screen. A convergent lens is positioned to focus an image I on the screen. Observe that the
sum of the object and image distances is p + q = L, giving
p = L − q. Also, from the thin-lens equation, we have
2/9/11 2:08:04 PM
=−
.
23.9
23.33
(ii)
q=
(b)
q<0
(c)
M = −q p > 0
(d)
M=−
(a)
q=
(b)
q<0
(c)
M = −q p > 0
Chapter 23
(d)
(c)
( 40.0 cm )( −20.0 cm )
= −13.3 cm
40.0 cm − ( −20.0 cm )
(a)
(iii) (a)
23.34
23.35
.
687
195
(a) a divergent
Since the lens,
mirrorthe
is focal
convex,
R <is0.negative.
Thus, R =Hence,
− 0.550f m
and f =cm
R in
2 this case. The
For
length
= −20.0
thin-lens equation gives the image distance as q = pf ( p − f ) , and the magnification is given by
M = − q p.
(i)
696
13.8 cm
and Lenses
= − 0.381Mirrors
.
36.2 cm
⇒
( −13.3 cm )
+ 40.0 cm
⇒
(c)
M = −q p > 0
⇒
6.67 cm in front of the lens
upright image
= + 0.500
(10.0 cm ) (−20.0 cm ) = −6.67 cm
10.0 cm − (−20.0 cm )
⇒
10.0 cm in front of the lens.
= + 0.333
⇒
+ 20.0 cm
q<0
⇒
upright image
virtual image
( −10.0 cm )
(b)
13.3 cm in front of the lens.
virtual image
( 20.0 cm )( −20.0 cm )
= −10.0 cm
20.0 cm − ( −20.0 cm )
M=−
q=
⇒
⇒
virtual image
⇒
cm )
( −12.3
upright
image
=−
= + 0.615 .
+20.0 cm
( − 6.67 cm ) = + 0.667
(d) ray
M diagram
=−
The
for cm
this arrangement is shown above.
+10.0
(a) This is a real object, so the object distance is p = +20.0 cm. The thin-lens equation
The focal length of a converging lens is positive, so f = +10.0 cm. The thin-lens equation then
gives the image distance aspf
p (10.0 cm )
.
=
yields an image distance of q =
p − f p −10.0 cm
(a)
When p = +20.0 cm, q =
( 20.0 cm )(10.0 cm )
20.0 cm − 10.0 cm
= +20.0 cm, and M = −
q
= −1.00, so the
p
image is located 20.0 cm beyond the lens , is real (q > 0) , is inverted (M < 0) , and is
the same size as the object ( M = 1.00 ) .
(b)
When p = f = +10.0 cm, the object is at the focal point and no image is formed . Instead,
parallel rays emerge from the lens .
(c)
When p = 5.00 cm, q =
( 5.00 cm )(10.0 cm )
q
= +2.00, so the
p
5.00 cm − 10.0 cm
image is located 10.0 cm in front of the lens , is virtual (q < 0) , is upright (M > 0) , and is
= −10.0 cm, and M = −
twice the size of the object ( M = 2.00 ) .
23.36
We must first realize that we are looking at an upright, enlarged, virtual image. Thus, we
have a real object located between a converging lens and its front-side focal point, so
q < 0, p > 0, and f > 0.
The magnification is M = −
67352_ch23.indd 690
q
p
2/9/11 2:08:08 PM
698
688
196
.
23.39
Chapter 23
(b) The needed ray diagram, withpthe
8.00
in front
cmcm
f object
cm ) of the mirror, is shown below:
( 4.00
)(8.00
From the thin-lens equation, q1 = 1 1 =
= − 8.00 cm .
p1 − f1
4.00 cm − 8.00 cm
The magnification by the first lens is M1 = −
q1
( − 8.00 cm ) = + 2.00.
=−
p1
4.00 cm
The virtual image formed by the first lens is the object for the second lens, so
p2 = 6.00 cm + q1 = +14.0 cm, and the thin-lens equation gives
q2 =
(14.0 cm )( −16.0 cm )
p2 f2
=
= − 7.47 cm
p2 − f2 14.0 cm − ( −16.0 cm )
q2
( − 7.47 cm ) = + 0.533, so the overall
=−
p2
14.0 cm
magnification is M = M1 M 2 = ( + 2.00 ) ( + 0.533) = +1.07.
The magnification by the second lens is M 2 = −
The position of the final image is 7.47 cm in front of the second lens , and its height is
h ′ = M h = ( +1.07 )(1.00 cm ) = 1.07 cm .
Mirrors .and Lenses
Since M > 0, the final image is upright ; and since q2 < 0, this image is virtual
23.40
.
23.41
699
(a) From the thin-lens equation, the image distance for the first lens is
The thin-lens equation gives the image position for the first lens as
q1 =
p1 f1
( 30.0 cm )(15.0 cm )
=
= + 30.0 cm
p1 − f1
30.0 cm − 15.0 cm
and the magnification by this lens is M1 = −
q1
30.0 cm
=−
= −1.00.
30.0 cm
p1
The real image formed by the first lens serves as the object for the second lens, so
p2 = 40.0 cm − q1 = +10.0 cm. Then, the thin-lens equation gives
q2 =
p2 f2
(10.0 cm )(15.0 cm )
=
= − 30.0 cm
p2 − f2
10.0 cm − 15.0 cm
and the magnification by the second lens is
M2 = −
q2
( − 30.0 cm ) = + 3.00
=−
p2
10.0 cm
Thus, the final, virtual image is located 30.0 cm in front of the second lens ,
and the overall magnification is M = M1 M 2 = ( −1.00 ) ( + 3.00 ) = − 3.00 .
23.42
.
23.43
.
(a) With
Since the light rays incident to the first lens are parallel, p1 = ∞, and the thin-lens equation gives
q1 = f1 = −10.0 cm.
The virtual image formed by the first lens serves as the object for the second lens, so
p2 = 30.0 cm + q1 = 40.0 cm. If the light rays leaving the second lens are parallel, then q2 = ∞,
and the thin-lens equation gives f2 = p2 = 40.0 cm .
67352_ch23.indd 691
2/9/11 2:08:12 PM
Mirrors and Lenses
.
23.9
23.45
.
687
197
(a) Since
the mirror
convex,
R <are
0. Thus,
= − 0.550
and f = R
2 To avoid this, we
Note:
Final answers
to is
this
problem
highlyRsensitive
to m
round-off
error.
retain extra digits in intermediate answers and round only the final answers to the correct number
of significant figures.
Since the final image is to be real and in the film plane, q2 = + d .
Then, the thin-lens equation gives p2 =
q2 f 2
d (13.0 cm )
=
.
q2 − f2 d − 13.0 cm
The object of the second lens ( L2 ) is the image formed by the first lens ( L1 ) , so
13.0 cm ⎞
d2
⎛
q1 = (12.0 cm − d ) − p2 = 12.0 cm − d ⎜ 1+
=
12.0
cm
−
⎟
⎝
d − 13.0 cm ⎠
d − 13.0 cm
If d = 5.00 cm, then q1 = +15.125 cm; and when d = 10.0 cm, q1 = + 45.3 cm .
From the thin-lens equation, p1 =
q1 f1
q (15.0 cm )
= 1
.
q1 − f1 q1 − 15.0 cm
When q1 = +15.125 cm ( d = 5.00 cm ) , then p1 = 1.82 × 10 3 cm = 18.2 m .
When q1 = + 45.3 cm ( d = 10.0 cm ) , then p1 = 22.4 cm = 0.224 m .
since both p and f are positive in this
Thus, situation).
the range of focal distances for this camera is 0.224 m to 18.2 m .
23.47
.
In the sketch at the right, the center of curvature of
incoming
the left side of the biconcave lens is on the side of the light
incoming light. Thus, by the convention of Table 23.2,
the radius of curvature of this side is negative while
the radius of curvature of the right side is positive.
If R1 = −32.5 cm and R2 = 42.5 cm, the lens maker’s
equation gives the focal length of the lens as
| R1|
R2
⎞
⎛
1
1
1
= (1− n )( 5.43 × 10 −2 cm −1 )
= ( n − 1) ⎜
−
f
⎝ − 32.5 cm 42.5 cm ⎟⎠
(a)
For a very distant object ( p → ∞), the thin-lens equation gives the image distance as q = f .
Thus, if the index of refraction of the lens material is n = 1.53 for violet light,
q= f =
1
= −34.7 cm
(1− 1.53)( 5.43 × 10 −2 cm −1 )
and the image of violet light is formed 34.7 cm to the left of the lens .
(b)
If n = 1.51 for red light, the image distance for very distance source emitting red light is
q= f =
1
= −36.1 cm
(1− 1.51)( 5.43 × 10 −2 cm −1 )
The image of the very distant red light source is formed 36.1 cm to the left of the lens .
67352_ch23.indd 692
2/9/11 2:08:15 PM
198
688
.
23.49
Chapter 23
.
(b) The
raywhen
diagram,
with the
8.00
cm in front of the mirror, is shown below:
Since
q = +needed
8.00 cm
p = +10.0
cm,object
we find
that
1 1 1
1
1
18.0
= + =
+
=
f p q 10.0 cm 8.00 cm 80.0 cm
Then, when p = 20.0 cm,
18.0
1
18.0 − 4.00
14.0
1 1 1
= − =
−
=
=
80.0 cm
80.0 cm
q f p 80.0 cm 20.0 cm
or
q=
80.0 cm
= + 5.71 cm
14.0
Thus, a real image is formed 5.71 cm in front
mirror
of of
thethe
lens
from .the object.
23.50
23.51
.
(a)
Wepasses
are given
p = through
5 f , withthe
both
p and
beingposition
positive.isThe
thin-lens
equation
As light
left that
to right
lens,
the fimage
given
by
then gives
p f
(100 cm )(80.0 cm )
q1 = 1 1 =
= + 400 cm
p1 − f1
100 cm − 80.0 cm
This image serves as an object for the mirror with an object distance of p2 = 100 cm − q1 = −300 cm
(virtual object). From the mirror equation, the position of the image formed by the mirror is
q2 =
p2 f2
( − 300 cm ) ( − 50.0 cm ) = − 60.0 cm
=
p2 − f2 −300 cm − ( − 50.0 cm )
Mirrors and Lenses
703
image
is the object for the lens as light now passes through it going right to left. The object
continued This
on next
page
distance for the lens is p3 = 100 cm − q2 = 100 cm − ( − 60.0 cm ) , or p3 = 160 cm. From the
thin- lens equation,
q3 =
p3 f3
(160 cm )(80.0 cm )
=
= +160 cm
p3 − f3
160 cm − 80.0 cm
Thus, the final image is located 160 cm to the left of the lens .
⎛ q ⎞⎛ q ⎞⎛ q ⎞
The overall magnification is M = M1 M 2 M3 = ⎜ − 1 ⎟ ⎜ − 2 ⎟ ⎜ − 3 ⎟ , or
⎝ p1 ⎠ ⎝ p2 ⎠ ⎝ p3 ⎠
⎛ 400 cm ⎞ ⎡ ( − 60.0 cm ) ⎤ ⎛ 160 cm ⎞
−
= − 0.800
−
M = ⎜−
⎝ 100 cm ⎟⎠ ⎢⎣ ( −300 cm ) ⎥⎦ ⎜⎝ 160 cm ⎟⎠
Since M < 0, the final image is inverted .
23.52
.
23.53
.
(a) Since the object is midway between the lens and mirror, the object distance for the mirror is
A hemisphere is too thick to be described
p = +12.5 cm. The mirror equation gives the image position as
as a thin1 lens. The light is undeviated on
entry into the flat face. We next consider
the light’s exit from the curved surface,
for which R = − 6.00 cm.
The incident rays are parallel, so p = ∞.
Then,
67352_ch23.indd 693
n1 n2 n2 − n1
1.00 1.00 − 1.56
+
=
, from which q = 10.7 cm .
=
becomes 0 +
− 6.00 cm
p q
R
q
2/9/11 2:08:19 PM
=
.
23.61
23.9
( 30.0 cm )( −12.0 cm )
30.0 cm − 12.0 cm
= −20.0 cm
The diagram
at the
rightis shows
a bubble
O, located
cmm and f = R 2
(a)
Since the
mirror
convex,
R < 0. Thus,
R = −5.00
0.550
n2 = 1.00
above the center of a glass sphere having radius R = 15.0 cm.
This bubble is an actual distance of p = 10.0 cm below the
| q|
p
refracting surface separating the glass and air. Refraction at
this surface forms a virtual image I at distance q below the
surface as shown. The object distance p and image distance q 5.00 cm
are related by
n1 n2 n2 − n1
+
=
p q
R
or
687
199
Mirrors and Lenses
n2 n2 − n1 n1
=
−
q
R
p
I
O
R
n1 = 1.50
Thus, with R < 0 for the concave surface the light crosses going from glass into the air, we have
R = −15.0 cm, and
1.50
+1.00 − 4.50
−3.50
1.00 1.00 − 1.50
=
−
=
=
−15.0 cm 10.0 cm
30.0 cm
30.0 cm
q
or
q=
30.0 cm
= − 8.57 cm
− 3.50
Therefore, the bubble has an apparent depth of 8.57 cm below the glass surface.
23.62
67352_ch23.indd 694
The inverted image is formed by light that leaves the object and goes directly through the lens,
never having reflected from the mirror. For the formation of this inverted image, we have
2/9/11 2:08:21 PM
장 ળѰ˝ଝ
PROBLEM SOLUTIONS
24.1
.
The screen position of the mth order bright fringe of wavelength l in a double-slit interference
pattern is ym = (lL d)m. Then, Δy = (lL d)(m2 − m1 )is the separation between the bright fringes of
orders m1 and m2 . If Δy = 0.552 mm for the first and second order bright fringes when L = 1.75 m,
and the slit separation is d = 0.552 mm, the wavelength incident upon the set of slits is
l=
.
24.3
(a)
( 0.552 × 10
( Δy )(d1) 620
= × 10 m
(
L ( m2=− m1 )
Δy =
(a)
(a)
m ) ( 2.10 × 10 −3 m )
= 6.62 × 10 −7 m = 662 nm
m )(×2 10
− 1−)6 m = 2.40 mm
= 2.40
(1.75
m =1
− ybright
m=0
=
lL
,
d
or
−9
lL ( 546.1 × 10 m ) (1.20 m )
=
= 2.62 × 10 −3 m = 2.62 mm
d
0.250 × 10 −3 m
The distance between the first and second dark bands is
⎛
Δy =⎜⎝ydark
24.4
24.5
.
sin15.0°
)
−3
The distance between the central maximum and the first order bright fringe is
Δy = ybright
(b)
−9
5
⎞
lL) = 1.47 × 10 3 nm = 1.47 mm
l y= ( 589= nm
⎟
−
= 2.62 mm as in (a) above.
dark
⎠
m =1
2 m=0
d
For a bright fringe of order m, the path difference is d = ml, where
From d sinq = ml, the angle for the m = 1 maximum for the sound waves is
⎡m ⎛ v
⎡
⎞⎤
⎛ 354 m s ⎞ ⎤
1
⎛m ⎞
q = sin −1 ⎜ l ⎟ = sin −1 ⎢ ⎜ sound ⎟ ⎥ = sin −1 ⎢
⎟ ⎥ = 36.2°
⎜
⎝d ⎠
⎣ d ⎝ f ⎠⎦
⎣ 0.300 m ⎝ 2 000 Hz ⎠ ⎦
(b)
For 3.00-cm microwaves, the required slit spacing is
d=
(c)
ml (1)( 3.00 cm )
=
= 5.08 cm
sinq
sin ( 36.2° )
The wavelength of the light is l = d sinq m , so the frequency is
(1)( 3.00 × 108 m s )
c
mc
f
=
=
=
= 5.08 isn’t
× 1014soHz
angles, thel approximation
breaks
down
and36.2°
the spacing
regular.
d sinq
1.00
× 10 −6
m sin
(
24.6
24.7
.
. At larger
)
(a) The angular position of the bright fringe of order m is given by d sinq = ml, or
As indicated in−1the sketch at the right, the path difference
q = sin (ml d
betweenm waves from the two antennas that travel toward
the car is given by d = d sinq . When d = ml, where
m = 0, 1, 2,… , constructive interference produces the
maximum of order m. Destructive interference produces the
q
minimum of order m when d = ( m + 12 ) l.
d
q
x
(a)
At the m = 2 maximum, d = d sinq = 2 l, or
⎞ ( 300 m ) ⎡
d⎛
y
⎢
= ⎜ 2
⎟ = 200
⎢
2 ⎝ x + y 2 ⎠ 2
⎣
(b)
67352_ch24.indd 712
y
O
d =dsinq
400 m
(1000m ) + ( 400 m )
2
2
⎤
⎥ = 55.7 m
⎥
⎦
The next minimum encountered is the
2/9/11 2:09:55 PM
(
) l.
Wave Optics
(a)
At the m = 2 maximum, d = d sinq = 2 l, or
(b)
l=
found at
(b)
(c)
24.7
24.8
.
24.9
d sinq d ⎛
y
[ m =)l2 d⎜] , m2 = 0,2
+y
x
⎝
719
201
⎤
)l, or
⎞ ( 300 m ) ⎡
400 m
⎢
⎥
=
=
55.7
m are
±
1,
±
2,….
If
d
=
25l
,
the
fi
rst
three
dark
fringes
⎟
⎢ (1000m )2 + ( 400 m )2 ⎥
2
⎠
⎣
⎦
5
⎛ 23 2 minimum,
⎞
2 ⎞
−1 =
−1 ⎛ d
where
2, or
The next minimum encounteredqis =the
sinm
q 2 = sin
andit occurs
⎜⎝ ⎟⎠ = 3.4° , and
⎜⎝ ⎟⎠==5l5.7°
1
25
25
⎛ 5l ⎞
−1 ⎡ 5 ( 55.7 m ) ⎤
= 27.7°
q = sin −1are
⎜⎝ evenly
⎟⎠ = sin
⎢ because ⎥ the
angles are small and q ≈ sinq . At larger
The answers
spaced
2d
⎣ 2 ( 300 m ) ⎦
angles, the approximation breaks down and the spacing isn’t so regular.
At this point, y = x tanq = (1 000 m ) tan 27.7° = 525 m, so the car must travel an additional
(1) ( 6.0 × 10 −7 m )
125 min. =
As indicated
the sketch at the right,
the×path
m = 2.9 mm
= 2.9
10 −6difference
between waves from thesin12°
two antennas that travel toward
The
angular
position
bright
of (measured
order m is from
giventhe
byposition
d sinq =ofml
the m = 1 is
the
is given
=ofdthe
sinq
. When
Thecar
location
of by
thedbright
fringe
offringe
order m
the. Thus,
centralifmaximum)
2
−7
bright
fringe
is
located
at
q
=
12°
when
l
=
6.0
×
10
nm
=
6.0
×
10
m,
the
slit
spacing
is is
(ybright )m = (lL d)m, m = 0, ± 1, ± 2, …. Thus, the spacing between successive bright fringes
(
Δybright = ybright
)
m +1
(
− ybright
)
= ( lL d ) ( m + 1) − ( lL d ) m = lL d
m
The wavelength of the laser light must be
( Δy ) d = (1.58 × 10
l=
24.10
.
24.11
L
m ) ( 0.200 × 10 −3 m )
5.00 m
lL ⎛
⎜m +
d ⎝
Thus, d =
1⎞
lL
=
⎟⎠
2 m=0 2 d
lL ( 3.00 m ) (150 m )
=
= 11.3 m .
2y
2 ( 20.0 m )
( 4 − 2 ) = 2.13 × 10 −2 m = 2.13 cm
In ashown
double-slit
thepath
screen position of the mth order maximum for
As
in theinterference
figure at thepattern,
right, the
wavelengthinlthe
is waves
ym = (lL
d
difference
reaching
the telescope is
d = d 2 − d1 = d 2 (1 − sina ) . If the first minimum
occurs when q = 25.0°, then
(d = l 2) ml
d=
sinq
a = 180° − (q + 90.0° + q ) = 40.0° , and
d2 =
d
l 2
( 250 m 2) = 350 m
=
=
1− sina 1− sina 1− sin 40.0°
=
Thus, h = d 2 sin 25.0° = 148 m .
24.14
.
24.17
= 6.32 × 10 −7 m = 632 nm
(a)
The distance between the central maximum (position of A) and the first minimum is
y=
24.12
24.13
.
−2
bright
600 nm
= 233 nm .
2 (1.29 )
The angular
from the
of thewaves
centralrefl
maximum
is given
There
will bedeviation
a phase change
ofline
the radar
ecting from
both by
surfaces of the polymer,
giving zero net phase change due to reflections. The requirement for destructive interference in
the reflected waves is then
1⎞
⎛
2 t = ⎜ m + ⎟ ln
⎝
2⎠
or
t = ( 2m + 1)
l
where m = 0, 1, 2,…
4 nfilm
If the film is as thin as possible, then m = 0, and the needed thickness is
t=
l
3.00 cm
=
= 0.500 cm
4 nfilm
4 (1.50 )
This anti-reflectance coating could be easily countered by changing the wavelength of the
radar—to 1.50 cm—now creating maximum reflection!
24.18
67352_ch24.indd 718
(a)
Phase changes are experienced by light reflecting at either surface of the oil film, a upper
air-oil interface and a lower oil-water interface. Under these conditions, the requirement
for constructive interference is
2/9/11 2:10:09 PM
722
202
24.21
24.18
.
24.19
=
Chapter 24
3.00 cm
= 0.500 cm
4 (1.50 )
Light
reflecting
from coating
the firstcould
(glass-iodine)
a phasethe
change,
but light
This anti-refl
ectance
be easilyinterface
counteredsuffers
by changing
wavelength
of refl
theecting
at
the
second
(iodine-glass)
interface
does
not
have
a
phase
change.
Thus,
the
condition
for conradar—to 1.50 cm—now creating maximum
, or 2.5 mrefl=ection!
2m2 + 1 .
structive interference in the reflected light is 1
)l, with m = 0,1,2,…. The smallest
fi(a)
lm thickness
of
strongly
ecting
is surface
Phase
by
light
reflincident
ecting
atlight
either
of therefl
oilecting
film, afrom
upper
Since
the
thinchanges
ficapable
lm hasare
airexperienced
on both refl
sides
of
it,the
there
is a phase
change
for light
the
air-oil
interface
and
a
lower
oil-water
interface.
Under
these
conditions,
the
requirement
first surface but no change for light reflecting from the second surface. Under these conditions,
for constructive
interference
the requirement
to be met
if wavesisreflecting from the two sides are to produce constructive
interference and a strong reflection is
1⎞
⎛
2nfilm t = ⎜ m + ⎟ l
⎝
2⎠
or
l=
4nfilm t
2m + 1
m = 0, 1, 2,…
With nfilm = 1.473 and t = 542 nm, the wavelengths that produce strong reflections are given by
l = 4 (1.473)( 524 nm ) (2m + 1), which yields
722
24.20
.
24.21
m = 0 : l = 3.09 × 10 3 nm
m = 1: l = 1.03 × 10 3 nm
m = 2 : l = 617 nm
m = 3 : l = 441 nm
m = 4 : l = 343 nm
m = 5 : l = 281 nm
and many other shorter wavelengths. Of these, the only ones in the range 300 nm to 700 nm are
441 nm, and 343 nm .
Chapter
617 24
nm,
(a)
a phase
change
in the
reflection atinterface
the outersuffers
surface
of thechange,
soap film
no refl
change
Light With
reflecting
from
the first
(glass-iodine)
a phase
butand
light
ecting
on
refl
ection
from
the
inner
surface,
the
condition
for
constructive
interference
in
at the second (iodine-glass) interface does not have a1phase change. Thus, the condition the
for light
con= (mt+=2(m + 1 )l, with m = 0,1,2,…. The smallest
reflinterference
ected from the
bubble
2nfilm
structive
in soap
the refl
ected is
light
is t 2n
2
film
film thickness capable of strongly reflecting the incident light is
tm in =
24.22
.
24.25
(m
+ 1 2 ) l ( 0 + 1 2 ) λ 6.00 × 10 2 nm
=
=
= 85.4 nm
will be seen.
2nfilm
4 (1.756 )
2nfilm
m in
(a) With
nair is< inserted
nwater < nbetween
oil , reflections
When
the hair
one at the air-oil interface experience a phase change, while
refl
ections
at
the
oil-water
interface
experience no phase change. With B
one phase change at
end of the glass plates, which have
the
surfaces,
the
condition
for
constructive
interference
in
the
light
refl
ected
A
Δt by the film is
length L = 14.0 cm, 1a wedge-shaped
d
2n
t
=
(m
+
2
film
tm+1
air film is created as shown at the
tm
Δx
right. The maximum thickness of this
air wedge is equal to the diameter d
L = 14.0 cm
of the hair. If the bright interference
m
fringe of order occurs at point A
(where the air wedge has thickness tm ) and the adjacent bright fringe of order m + 1 occurs at B
(where the thickness is tm +1 ), we may use the properties of similar triangles to write
d Δt tm +1 − tm
=
=
L Δx
Δx
or
⎛ Δt ⎞
L
d =⎜
⎝ Δ x ⎟⎠
[1]
Since nair < nglass , light waves reflecting at the upper surface of the air film do not undergo a phase
change, but those reflecting from the lower surface do experience such a change. Thus, the condition for constructive interference to produce a bright fringe is 2nair t = (m + 12 )l, or the thickness
of the film at the location of the bright fringe of order m is tm = (m + 12 )l 2nair . The change in
thickness between the locations of adjacent bright fringes is then
Δt = tm +1 − tm =
[( m + 1) + 1 2] l − [ m + 1 2] l =
2nair
2nair
l
l
=
2nair 2
If the observed spacing between adjacent bright fringes is Δx = 0.580 mm when using light of
wavelength l = 650 nm, Equation [1] gives the diameter of the hair as
(650 × 10−9 m ) (14.0 × 10 −2 m ) = 7.84 × 10 −5 m = 78.4 mm
lL
⎛ l 2⎞
d =⎜
=
L
=
⎝ Δx ⎟⎠
2 ( Δx )
2 ( 0.580 × 10 −3 m )
24.26
67352_ch24.indd 719
With a phase change due to reflection at each surface of the magnesium fluoride layer, there is
zero net phase difference caused by reflections. The condition for destructive interference is then
2/9/11 2:10:14 PM
[
] l − [ m + 1 2] l =
2nair
2nair
l
l
=
2nair 2
If the observed spacing between adjacent bright fringes is Δx = 0.580 mm when
using
light of 719
203
Wave
Optics
wavelength l = 650 nm, Equation [1] gives the diameter of the hair as
724
24.26
.
24.27
(650 × 10−9 m ) (14.0 × 10 −2 m ) = 7.84 × 10 −5 m = 78.4 mm
lL
⎛ l 2⎞
(b)
d
=
=
L
=
⎟
Chapter 24 ⎜⎝
Δx ⎠
2 ( Δx )
2 ( 0.580 × 10 −3 m )
With aisphase
change
duedue
ection
at ⎞each
surface
ofofthe
uoride
layer,
⎛to refl
There
a phase
change
to refl
ection
at= the
bottom
themagnesium
air film butflnot
at the
top there
of theis
3.00
l condition
⎜
⎟
zero
net
phase
difference
caused
by
refl
ections.
The
for
destructive
interference
is then
film. The requirement for ⎝a dark fringe is⎠ then
2n
mlpath difference
where m
0, 1,position
2,… is a whole number of wavelengths (to three sigSince
for= this
air t =the
nificant figures), the waves interfere constructively and produce a maximum
At the 19th dark ring (in addition to the dark center spot), the order number is m = 19, and the
thickness of the film is
(c)
−9
, will produce destructive interference and hence
ml 19 ( 500 × 10 m )
tminimum
=
= reflected light for wavelength
= 4.75 × 10 −6l =
m540
= 4.75
nm. mm
For m = 1, 2, 3,… , we obtain
2nair
2 (1.00 )
thicknesses of 293 nm, 489 nm, 685 nm, … .
24.28
.
24.31
In a single-slit diffraction pattern, with the slit having width a, the dark fringe of order m occurs
at angle q m , where sinq m = m(l a) and m = ±1, ± 2, ± 3,…. The location, on a screen located
distance L from the slit, of the dark fringe of order m (measured from y = 0 at the center of the
central maximum) is (ydark )m = L tanq m ≈ L sinq m = ml ( L a ) .
(a)
The central maximum extends from the m = −1 dark fringe to the m = +1 dark fringe, so the
width of this central maximum is
⎛ lL ⎞
⎛ lL ⎞ 2lL
Central max. width = (ydark )1 − (ydark )−1 = 1⎜
− ( −1) ⎜
=
⎝ a ⎟⎠
⎝ a ⎟⎠
a
=
(b)
2 ( 5.40 × 10 −7 m )(1.50 m )
0.200 × 10 −3 m
The first order bright fringe extends from the m = 1 dark fringe to the m = 2 dark fringe, or
( Δy ) = ( y ) − ( y )
bright
dark
1
=
726
24.32
.
24.33
= 8.10 × 10 −3 m = 8.10 mm
2
(5.40 × 10
dark
−7
1
⎛ lL ⎞
⎛ lL ⎞ lL
− 1⎜
=
= 2⎜
⎝ a ⎟⎠
⎝ a ⎟⎠
a
m ) (1.50 m )
0.200 × 10 −3 m
= 4.05 × 10 −3 m = 4.05 mm
Note that the width of the first order bright fringe is exactly one-half the width of the
maximum.
Chapter 24
central
In a single-slit diffraction pattern, minima are found where
The locations of the dark fringes (minima) mark the edges of the maxima, and the widths of the
maxima equals the spacing between successive minima.
At the locations of the minima, sinq m = m(l a), and
⎡
⎛ 500 × 10 −9 m ⎞ ⎤
ym = L tanq m ≈ L sinq m = m ⎡⎣ L ( l a ) ⎤⎦ = m ⎢(1.20 m ) ⎜
⎥ = m (1.20 mm )
⎝ 0.500 × 10 −3 m ⎟⎠ ⎦
⎣
Then, Δy = Δm(1.20 mm), and for successive minima, Δm = 1. Therefore, the width of each
maximum, other than the central maximum, in this interference pattern is
width = Δy = (1)(1.20 mm ) = 1.20 mm
67352_ch24.indd 720
2/9/11 2:10:17 PM
204
722
24.21
.
24.35
Chapter 24
=
2 ( 0.500 m )( 680 × 10 −9 m )
3.00 × 10 −3 m
= 2.27 × 10 −4 m = 0.227 mm
Light
reflscreen
ecting from
the fiof
rstthe
(glass-iodine)
interface
a phase change, but light reflecting
With the
locations
dark fringe of
order msuffers
at
at the second (iodine-glass) interface does not have a phase change. Thus, the condition for constructive
interference
light is
)l, ±with
= 0,1,2,…. The smallest
(ydark
)m = L tanqinm the
≈ Lrefl
sinqected
for m = ±1,
2, ±m3,…
m = m(lL a)
film thickness capable of strongly reflecting the incident light is
the width of the central maximum is Δycentral
= (ydark )m = +1 − (ydark )m = −1 = 2(lL a), so
m axim um
⎛
⎞
a ⎜ Δycentral ⎟
−3
−3
⎝ m axim um ⎠ ( 0.600 × 10 m ) ( 2.00 × 10 m )
l=
=
= 4.62 × 10 −7 m = 462 nm
2L
2 (1.30 m )
24.36
24.37
.
Note: The small angle approximation does not work
sinq = m(l
(a)
(minima)
where proceed
well inDark
this bands
situation.
Rather,occur
you should
as a). For the first minimum, m = 1, and the
distance
from
the
center
of
the
central
maximum
is y1 = L tanq ≈ L sinq = L ( l a ) . Thus,
follows.
the needed distance to the screen is
At the first order minimum,
⎛ 0.75 × 10 −3 m ⎞
⎛ a⎞
= 1.1 m
L = y1 ⎜ ⎟ = ( 0.85 × 10 −3 m ) ⎜
⎝l⎠
⎝ 587.5 × 10 −9 m ⎟⎠
−9
⎛ 1( 640 × 10 m ) ⎞
⎛
⎞
= sin −1 ⎜
= 7.35°
⎜
⎟
(b) The width
maximum−6is 2 y⎟ = 2 ( 0.85 mm ) = 1.7 mm .
⎠ central
⎝ of the
⎝ 5.00 × 10 m ⎠1
.
24.41
The grating spacing is d =
1.00 cm ⎛ 1.00 m ⎞
1.00 m
.
⎜
⎟=
4 500 ⎝ 10 2 cm ⎠ 4.50 × 10 5
From d sinq = ml, the angular separation between the given spectral lines will be
Δq = sin −1 ⎡⎣ m lred d ⎤⎦ − sin −1 ⎡⎣ m lviolet d ⎤⎦ , or
⎡ m ( 434 × 10 −9 m ) ( 4.50 × 10 5 ) ⎤
⎡ m ( 656 × 10 −9 m ) ( 4.50 × 10 5 ) ⎤
Δq = sin −1 ⎢
⎥
⎥ − sin −1 ⎢
1.00 m
1.00 m
⎢⎣
⎥⎦
⎢⎣
⎥⎦
The results obtained are: for m = 1, Δq = 5.91° ; for m = 2, Δq = 13.2° ; and for m = 3,
Δq = 26.5° . Complete orders for m ≥ 4 are not visible.
24.47
.
1 cm
10 −2 m
=
= 3.64 × 10 −6 m. From d sinq = ml , or
2 750 2.75 × 10 3
q = sin −1 ( ml d ) , the angular positions of the red and violet edges of the second-order spectrum
are found to be
The grating spacing is d =
⎛ 2 ( 700 × 10 −9 m ) ⎞
⎛ 2l ⎞
q r = sin −1 ⎜ red ⎟ = sin −1 ⎜
⎟ = 22.6°
−6
⎝ d ⎠
⎝ 3.64 × 10 m ⎠
and
⎛ 2 ( 400 × 10 −9 m ) ⎞
⎞
⎛ 2l
q v = sin −1 ⎜ violet ⎟ = sin −1 ⎜
⎟ = 12.7°
−6
⎝ d ⎠
⎝ 3.64 × 10 m ⎠
Note from the sketch at the right that yr = L tanq r
and yv = L tanq v , so the width of the spectrum on the
screen is Δy = L ( tanq r − tanq v ) .
Since it is given that d = ( m + 1 2 ) l, the distance
from the grating to the screen must be
1.75 cm
Δy
=
L=
tanq r − tanq v tan ( 22.6° ) − tan (12.7° )
Screen
Grating
Δy
yr
qv
qr
yv
L
= 9.17 cm
24.48
67352_ch24.indd 721
The grating spacing is d = 1 cm 1 200
2/9/11 2:10:21 PM
(
.
24.53
24.54
24.57
.
)=
Wave Optics
54.7°
719
205
The more general expression for Brewster’s angle is given in problem P24. as
(b)
tanq p = n2 n1
(a)
⎛n ⎞
⎛ 1.52 ⎞
When n1 = 1.00 and n2 = 1.52, q p = tan −1 ⎜ 2 ⎟ = tan −1 ⎜
= 56.7° .
⎝ 1.00 ⎟⎠
⎝ n1 ⎠
(b)
⎛n ⎞
⎛ ⎞
⎛ 1.52 ⎞
1.77 ⎞
and−1n⎛ = 1.52,
q p = tan −1 ⎜ 2 ⎟ = tan −1 ⎜
When n1 = 1.333
= 48.8° .
⎜⎝ ⎟⎠ = tan ⎜⎝2 1.00 ⎟⎠ = 60.5°
⎝ 1.333 ⎟⎠
⎝ n1 ⎠
2
q , where
From Snell’s
Malus’slaw,
law,the
theangles
intensity
of the light
transmitted
polarizer
I = I=0 cos
n2 sinq
From
of incidence
and
refraction by
arethe
related
by n1issinq
1
2.
I 0 is the intensity of the incident light, and q is the angle between the direction of the plane
ofthe
polarization
of the incident
and the
transmission
disk.
Thus,
If
angle of incidence
is the light
polarizing
angle
(that is, qaxis
= qof), the
the polarizing
refracted ray
is perpendicular
1
p
to the reflected ray (see Figure 24.2 in the textbook), and the angles of incidence and refraction
are also related by q p + 90° +q 2 = 180°, or q 2 = 90° − q p .
Substitution into Snell’s law then gives
(
)
n1 sinq p = n2 sin 90° − q p = n2 cosq p
.
24.59
24.58
or
sinq p cosq p = tanq p = n2 n1
2
Frompolarizing
Malus’s law,
light transmitted
by the
The
angletheforintensity
light in of
airthe
striking
a water surface
is first polarizer is I1 = I i cos q1 .
The plane of polarization of this light is parallel to the axis of the first plate and is incident
on the second plate. Malus’s law gives the intensity transmitted by the second plate as
I 2 = I1 cos 2 (q 2 −q1 ) = I i cos 2 q1 cos 2 (q 2 −q1 ) . This light is polarized parallel to the axis of the
second plate and is incident upon the third plate. A final application of Malus’s law gives the
transmitted intensity as
I f = I 2 cos 2 (q 3 − q 2 ) = I i cos 2 q1 cos 2 (q 2 − q1 ) cos 2 (q 3 − q 2 )
With q1 = 20.0°, q 2 = 40.0°, and q 3 = 60.0°, this result yields
I ′ − I 0.500I2 0 − 0.336I 0 2
I f = (10.0
units
) cos2 ( 20.0°) = 6.89 units
= ) cos ( 20.0° ) cos =( 20.0°
0.164
I0
I0
.
24.61
(a)
If light has wavelength l in vacuum, its wavelength in a medium of refractive index n is
ln = l n. Thus, the wavelengths of the two components in the specimen are
ln =
l 546.1 nm
=
= 413.7 nm
n1
1.320
ln =
l 546.1 nm
=
= 409.7 nm
n2
1.333
1
and
(b)
2
The numbers of cycles of vibration each component completes while passing through the
specimen of thickness t = 1.000 mm are
N1 =
t
1.000 × 10 −6 m
=
= 2.417
ln
413.7 × 10 −9 m
1
and
N2 =
t
1.000 × 10 −6 m
=
= 2.441
ln
409.7 × 10 −9 m
2
Thus, when they emerge, the two components are out of phase by N 2 − N1 = 0.024 cycles.
Since each cycle represents a phase angle of 360°, they emerge with a phase difference of
67352_ch24.indd 722
2/9/11 2:10:25 PM
722
206
24.21
.
24.62
24.65
=
Chapter 24
1.000 × 10 −6 m
= 2.441
409.7 × 10 −9 m
−9 of phase by N − N = 0.024 cycles.
Thus, when they emerge, the two components
out
6 ( 450are
× 10
m phase
m ) 2 but1 light reflecting
)(1.40 change,
Light refl
ecting
first (glass-iodine)
interface
suffers
Since
eachfrom
cyclethe
represents
a phase angle
theyaemerge
with a phase difference of
= of 360°,
at the second (iodine-glass) interface does not have a0.150
phase×change.
10 −3 m Thus, the condition for constructive interference
in×the
reflmected
light
is ) = 8.6°
)l, with m = 0,1,2,…. The smallest
2.52
10 −2
2.52
cm
Δf ==( 0.024
cycles
cycle
) (=360°
film thickness capable of strongly reflecting the incident light is
There
will be a phase
changepattern,
associated
with theinterference (or minima) occur where
In
the single-slit
diffraction
destructive
refl
surface
but The
no change
sinqection
= m (at
l one
a ) for
m = of
0, the
± 1, fi±lm
2,….
screen locations, measured from the center
at the other
of theoffilm.
Therefore,
theatconof
centralsurface
maximum,
these
minima are
dition for a dark fringe (destructive interference) is
ym = L tanq m ≈ L sinq m = m ( lL a )
If we assume the first order maximum is halfway between the first and second order minima, then
its location is
y=
y1 + y2 (1+ 2 ) ( lL a ) 3lL
=
=
2
2
2a
and the slit width is
a=
24.67
.
−9
3lL 3 ( 500 × 10 m )(1.40 m )
=
= 3.50 × 10 −4 m = 0.350 mm
2y
2 ( 3.00 × 10 −3 m )
dark fringes are now observed.
The source and its image, located 1.00 cm below the mirror, act as a pair of coherent sources.
This situation may be treated as double-slit interference, with the slits separated by 2.00 cm, if
it is remembered that the light undergoes a phase change upon reflection from the mirror. The
existence of this phase change causes the conditions for constructive and destructive interference
to be reversed. Therefore, dark bands (destructive interference) occur where y = m ( lL d ) for
m = 0, 1, 2,….
The m = 0 dark band occurs at y = 0 (that is, at mirror level). The first dark band above the mirror
corresponds to m = 1, and is located at
−9
⎛ lL ⎞ ( 500 × 10 m ) (100 m )
=
y = (1) ⎜
= 2.50 × 10 −3 m = 2.50 mm
⎝ d ⎟⎠
2.00 × 10 −2 m
24.69
.
24.68
sinq > 1.
In
thewavelength
figure at the
observe
The
is right,
l = vsound
f that the path difference between
the direct and the indirect paths is
d = 2x − d = 2 h 2 + ( d 2 ) − d
2
736
With a phase change (equivalent to a half-wavelength shift)
occurring upon reflection at the ground, the condition for
constructive
interference is d = ( m + 1 2 ) l, and the condition for
Chapter
24
destructive interference is d = m l . In both cases, the possible
values of the order number are m = 0, 1, 2,….
(a)
d
The wavelengths that will interfere constructively are l =
. The longest of these is
m +1 2
for the m = 0 case and has a value of
l = 2d = 4 h 2 + ( d 2 ) − 2d = 4
2
(b)
The wavelengths that will interfere destructively are l = d m , and the largest finite one of
these is for the m = 1 case. That wavelength is
l = d = 2 h2 + ( d 2) − d = 2
2
24.70
67352_ch24.indd 723
( 50.0 m )2 + ( 300 m )2 − 2 ( 600 m ) = 16.6 m
( 50.0 m )2 + ( 300 m )2 − 600 m = 8.28 m
From Malus’s law, the intensity of the light transmitted by the first polarizer is
2/9/11 2:10:28 PM
= cos 2 0° cos 2 ( 45° − 0° ) cos 2 ( 90° − 45° ) = 0.25
24.71
.
Wave Optics
207
719
If the signal from the antenna to the
(b)
receiver
station is to be completely
polarized by reflection from the water,
the angle of incidence where it strikes
the water must equal the polarizing angle
from Brewster’s law. This is given by
⎛n
⎞
q p = tan −1 ⎜ water ⎟ = tan −1 (1.333) = 53.1°
⎝ nair ⎠
From the triangle RST in the sketch, the
horizontal distance from the point of
refection, T, to shore is given by
x = ( 90.0 m ) tanq p = ( 90.0 m )(1.333) = 120 m
and from triangle ABT, the horizontal distance from the antenna to point T is
y = ( 5.00 m ) tanq p = ( 5.00 m )(1.333) = 6.67 m
The total horizontal distance from ship to shore is then x + y = 120 m + 6.67 m = 127 m .
.
24.73
Dark fringes (destructive interference) occur where d sinq = ( m + 1 2 ) l for m = 0, 1, 2,….
Thus, if the second dark fringe ( m = 1) occurs at q = (18.0 min )(1.00° 60.0 min ) = 0.300°, the slit
spacing is
−9
1⎞ l
⎛
⎛ 3 ⎞ ( 546 × 10 m )
= 1.56 × 10 −4 m = 0.156 mm
d = ⎜m + ⎟
=⎜ ⎟
⎝
2 ⎠ sinq ⎝ 2 ⎠ sin ( 0.300° )
24.74
67352_ch24.indd 724
As light emerging from the glass reflects from the top of the air layer, there is no phase change
produced. However, the light reflecting from the end of the metal rod at the bottom of the air
layer does experience a phase change. Thus, the condition for constructive interference in the
reflected light is
2/9/11 2:10:31 PM
장 ˝ଝ̛̛
=
25.3
.
(180 mm )(175 mm )
180 mm − 175 mm
= 6.30 × 10 3 mm = 6.30 m
The thin-lens equation, 1 + 1 = 1 , gives the image distance as
p q f
q=
pf
(100 m )( 52.0 mm )
=
= 52.0 mm
p − f 100 m − 52.0 × 10 −3 m
Optical Instruments
743
thepage
magnitude of the lateral magnification, M = h ′ h = −q p , where the height of the
continuedFrom
on next
image is h ′ = 0.092 0 m = 92.0 mm, the height of the object (the building) must be
h = h′ −
.
25.4
25.5
744
25.6
25.7
.
25.8
.
25.9
2
2
p ⎛ ⎞ t = ⎡⎢ ( f2 -number
100 m) ⎤⎥ t = ⎛ 4.0 ⎞ ⎛ 1 s⎞ ≈ 1 100 s
⎜
⎟
⎜
⎟
=
92.0
mm
−
=
177
m
(
)
⎜ ⎟ 1
q⎝ ⎠
52.0 mm
⎢⎣ ( f1 -number
)2 ⎥⎦ 1 ⎝ 1.8 ⎠ ⎝ 500 ⎠
(a)
The intensity
is a measure
the is defined to be the ratio of focal length of the lens to its
The f-number
(or focal
ratio) ofofa lens
rate Therefore,
at which energy
is received
diameter.
the f-number
of by
the given lens is
the film per unit area of the image,
or I ∝ 1 A f
28 cm
= 7.0
f -number =image =
Chapter 25
D 4.0 cm
Consider
coming
Since
the rays
exposure
timefrom
is unchanged, the intensity of the light reaching the film must be doubled
opposite
edges
of the object
if
the energy
delivered
is to be doubled. Using the result of Problem 25.4 (part a), we obtain
and passing undeviated through
⎛I ⎞
22
2
2
⎛ 1⎞
the center
of the lens
f22 -number
= ⎜as 11shown
f1 -number ) = ⎜ ⎟ (11) = 61, or f2 -number = 61 = 7.8
(
)
(
⎟
⎝2⎠
at the right. For a very
⎝ Idistant
22 ⎠
object, the image distance
55 mm
= 46
on mm
the camera.
Thus,
use the
f =8.01.2setting
equalsyou
the should
focal length
of the
lens. If the angular width of the
f -numbers
If
a camera
haslens
a lens
with
focal
of virtual
55 mmimage
and can
operate
that(i.e.,
range from
object
is q, the
full
image
width
The
corrective
must
form
anlength
upright,
at the
nearatpoint
of the eye
fq =1.2
on
the
fi
lm
is
− 60.0 cm in this case) for objects located 25.0 cm in front of the eye (p = +25.0 cm). From
the thin-lens equation, 1 p + 1 q = 1 f , the required focal length of the corrective lens is
f =
pq
( 25.0 cm )( −60.0 cm )
=
= + 42.9 cm
p+q
25.0 cm − 60.0 cm
and the power (in diopters) of this lens will be
P=
.
25.11
25.10
(a)
(a)
(b)
1
fin meters
1
1= + 2.33 diopters
=
= − 3.70 diopters
+ 0.429− m
0.270 m
The lens should form an upright, virtual image at the far point ( q = − 50.0 cm ) for very
distant objects ( p ≈ ∞ ) . Therefore, f = q = − 50.0 cm, and the required power is
1
1
P= =
= − 2.00 diopters
pq f − 0.500 m
f =
p +lens
q is to form an upright, virtual image at the near point of the unaided eye
If this
( q = −13.0 cm ) , the object distance should be
p=
67352_ch25.indd 738
=
qf
( −13.0 cm ) ( − 50.0208cm ) = 17.6 cm
=
q − f −13.0 cm − ( − 50.0 cm )
2/9/11 2:11:42 PM
=
(b)
(d)
.
25.13
Optical
Instruments
If this lens is to form an upright, virtual image at the near point of the
unaided
eye
( q = −13.0 cm ) , the object distance should be
1 q = of
1 fa,very
the required
focal length
thehis
corrective
His lens must form an upright, virtual +image
distant object
(p ≈ ∞of) at
far point,lens is
80.0 cm in front of the eye. Therefore, the focal length is f = q = −80.0 cm.
pq
f =
+ q a virtual image at his near point (q = −18.0 cm), the object distance must be
If this lens is to pform
qf
( −18.0 cm ) (1− 80.0 cm ) = 23.2 cm
=
=
= −1.18 diopters
q − f −18.0 cm
− ( − 80.0
− 0.850
m cm )
The
lens should
an upright,
virtual
image at the woman’s far point
(a) corrective
Yes, a single
lens canform
correct
the patient's
vision
(q = − 40.0 cm) for a very distant object (p → ∞). The thin-lens equation gives the required
focal length as f = q = − 40.0 cm = − 0.400 m. Since f < 0, it is a diverging lens , and the
required power is
P=
25.16
25.17
.
1
1
1
=
= − 2.50
= diopters= + 0.67 diopters .
fin meters − 0.400 m
+1.5 m
(a)
The upper
portionformed
of the lens
should
form
upright,
virtual
image
of very lens,
distant
Considering
the image
by the
cornea
as aanvirtual
object
for the
implanted
the object
objects
≈ ∞ )isatpthe
point
thethe
eyeimage
m ). isThe
equation
then
distance
for this
= −far
5.33
cm,ofand
distance
q =thin-lens
+ 2.80 cm.
The thin-lens
( p lens
( q = −1.5
gives
q = the
−1.5focal
m, so
the needed
power is lens as
equation
thenf =
gives
length
of the implanted
f =
pq
( − 5.33 cm ) ( 2.80 cm ) = + 5.90 cm
=
p+q
− 5.33 cm + 2.80 cm
so the power is P =
25.18
.
25.19
747
209
f be( −13.0
( − 50.0
The image qmust
locatedcm
in)front
of cm
the)corrective lens, so it is a virtual image
p=
=
= 17.6 cm
.
q − f −13.0 cm − ( − 50.0 cm )
p=
.
25.15
25.14
1
= − 2.00 diopters
− 0.500 m
(a)
(a)
1
11
==
diopters .
==++17.0
2.50 diopters
f ++0.059
0.4000mm
The simple
person magnifi
is farsighted
The
er (a converging lens) is to form an upright, virtual image located 25 cm
in front of the lens ( q = −25 cm ) . The thin-lens equation then gives
p=
qf
( −25 cm )( 7.5 cm )
=
= +5.8 cm
q− f
−25 cm − 7.5 cm
so the stamp should be placed 5.8 cm in front of the lens .
(b)
748
25.20
25.21
.
When the image is at the near point of the eye, the angular magnification produced by the
simple magnifier is
Chapter 25
m = mmax = 1 +
(a)
Whenthe
a converging
lens is used as a simple magnifier, maximum magnification is obtained
From
thin-lens equation,
when the upright, virtual image is formed at the near point of the eye ( q
( 3.50 cm ) ( − 25.0 cm )
pq
f =
=
= + 4.07 cm
p+q
3.50 cm − 25.0 cm
(b)
With the image at the normal near point, the angular magnification is
m = mmax = 1 +
25.22
67352_ch25.indd 742
25 cm
25 cm
= 1+
= 4.3
f
7.5 cm
(a)
25.0 cm
25.0 cm
= 1+
= + 7.14
f
4.07 cm
When the object is at the focal point of the magnifying lens, a virtual image is formed at
infinity and parallel rays emerge from the lens. Under these conditions, the eye is most
relaxed and the magnification produced is
2/9/11 2:11:49 PM
742
210
42.
.
25.23
44.
46.
=
Chapter 25
( −25 cm )( 5.0 cm )
−25 cm − 5.0 cm
= + 4.2 cm
38
(a) cmFrom the thin-lens equation, a real inverted image
image
= h ′ishformed
= −q pat, an
where
thedistance
height ofofthe
image is h ′ = 0.092 0 m = 92.0 mm, the height of the object (the building) must be
2.2 × 1011 m
pf
( 71.0 cm )( 39.0 cm )
=
q=
= + 86.5 cm
p− f
71.0 cm − 39.0 cm
1.7 m
so the lateral magnification produced by the lens is
48.
q
86.5 cm
h′
=− =−
= −1.22
p
71.0 cm
h
M=
and the magnitude is M = 1.22 .
(b)
If h is the actual length of the leaf, the small-angle approximation gives the angular width of the
leaf when viewed by the unaided eye from a distance of d = 126 cm + 71.0 cm = 197 cm as
q ≈
h
h
=
d 197 cm
The length of the image formed by the lens is h ′ = M h = 1.22 h , and its angular width
when viewed from a distance of d ′ = 126 cm − q = 39.5 cm is
q′ ≈
h′
1.22 h
=
d ′ 39.5 cm
Optical Instruments
749
The angular magnification achieved by viewing the image instead of viewing the leaf
directly is
continued on next page
q ′ 1.22 h 39.5 cm 1.22 (197
25.0cm
cm)
≈
= = 1+
== 6.08
6.00
h 197 cm
q
39.5
cmcm
5.00
25.24
25.25
.
25.26
.
25.33
(a)
A simple
magnifi
er produces
whenMit1 is
forms
an upright,
virtual
me = M1 (magnifi
25 cm cation
fe ) , where
the lateral
magnifi
cation
The overall
magnifi
cation
is m = Mmaximum
1.0 cm 1 1.0
cm
image
near point
of Therefore,
the=eye (25.0
for a normal
eye). If
lengthisof the
produced
byatthetheobjective
thecm
required
focal length
forthe
thefocal
eyepiece
= lens.
fo the 1required
500 cm object distance for maximum magnification with a
magnifier is f = +5.00 cm,
normal
M eye
( 25 iscm ) ( −12 ) ( 25 cm )
fe = 1
= ⎛
⎞
⎞ ⎛ =1 2.1
mi cm
2
m
or
⎜⎝ −140 ⎟⎠ ⎜⎝ 1 609 m ⎟⎠ = 1.6 × 10 mi
Note: We solve part (b) before answering part (a) in this problem.
The lens for the left eye forms an upright, virtual image at qL = − 50.0 cm when the object
distance
pL = 25.0forms
cm, so
the thin-lens
(b) Theisobjective
a real,
invertedequation gives its focal length as
image, diminished in size, of a very
( 25.0 cm ) ( − 50.0 cm )
pobject
L qL
fdistant
=at q1 = fo . This image = 50.0 cm
L =
pL + qLobject25.0
cmeyepiece
− 50.0 cm
is a virtual
for the
at
pe = − fe
Similarly for the other lens, qR = −100 cm when pR = 25.0 cm, and f R = 33.3 cm.
(a)
Using the lens for the left eye as the objective,
m=
(b)
fo
f
50.0 cm
= L =
= 1.50
fe
f R 33.3 cm
Using the lens for the right eye as the eyepiece and, for maximum magnification, requiring
that the final image be formed at the normal near point ( qe = − 25.0 cm ) gives the object
distance for the eyepiece as
pe =
qe fe
( − 25.0 cm ) (33.3 cm ) = +14.3 cm
=
qe − fe
− 25.0 cm − 33.3 cm
The maximum magnification by the eyepiece is then
me = 1 +
67352_ch25.indd 743
25.0 cm
25.0 cm
= 1+
= +1.75
fe
33.3 cm
2/9/11 2:11:51 PM
=
(b)
(d)
f L 50.0 cm
=
= 1.50
f R 33.3 cm
Using the lens for the right eye as the eyepiece and, for maximum magnification, requiring
cm ) gives
the object747
that the final image be formed at the normal near point ( qe = − 25.0Optical
Instruments
211
distance for the eyepiece as
q f be located
( − 25.0 incmfront
) (33.3
The image must
of cm
the)corrective lens, so it is a virtual image
pe = e e =
= +14.3 cm
qe − fe
− 25.0 cm − 33.3 cm
1 qeyepiece
= 1 f , the
required focal length of the corrective lens is
The maximum magnification by +the
is then
pq 25.0 cm
25.0 cm
f e== 1 +
= 1+
= +1.75
m
p + q fe
33.3 cm
Optical Instruments
753
and the image distance for the objective is
q1 = L − pe = 10.0 cm − 14.3 cm = − 4.3 cm
The thin-lens equation then gives the object distance for the objective as
p1 =
q1 f1
( − 4.3 cm ) (50.0 cm ) = + 4.0 cm
=
q1 − f1
− 4.3 cm − 50.0 cm
The magnification by the objective is then
M1 = −
q1
( − 4.3 cm ) = +1.1
=−
p1
4.0 cm
objective
lens. magnification is m = M1me = ( +1.1) ( +1.75) = 1.9 .
and
the overall
25.34
.
25.35
(a)
For
a refracting
telescope,
theq overall
and thecation
magnifi
produced
= flateral
From
the thin-lens
equation,
bycation
the objective
= p f ( length
p − f ),issoLthe
o + fe , magnifi
is m is
=M
fo =feh ′ h = − q p = − f ( p − f ). Therefore, the image size will be
lens
h′ = M h = −
754
25.36
.
25.37
fh
fh
=
p− f
f−p
fh
.
p
(b)
If p >> f , then f − p ≈ − p and h ′ ≈ −
(c)
Suppose the telescope observes the space station at the zenith. Then,
Chapter 25
h′ ≈ −
fh
( 4.00 m )(108.6 m )
=−
= −1.07 × 10 −3 m = −1.07 mm
p
407 × 10 3 m
The approximate overall magnification of a compound microscope is given by
The
m = angular
−(L fo separation of the lights is q = d h , where d = 1.00 m is their linear separation, and
h is the altitude
of the satellite. If the lights are just resolved according to the Rayleigh criterion,
then q = q min = 1.22 (l D) , where l is the wavelength of the light, and D is the diameter of the
lens. Thus, the altitude of the satellite must be
h=
25.38
25.39
.
toward the
d
d
d⋅D
0.300nm
m)
(1.00 m )(656.2
=
=
=
= 4.92
1035 lines
m = . 492 km
=
1.8 ××10
−9
q 1.22(l D) 1.22l 1.22 5002 (×0.18
10 nm
m)
(
)
l
The resolving power of a diffraction grating is
If just resolved, the angular separation of the objects is q = q min = 1.22
D
⎡
⎛ 500 × 10 −9 m ⎞ ⎤
and s = r q = 8.0 × 10 7 km ⎢1.22 ⎜
⎥ = 9.8 km
⎝ 5.00 m ⎟⎠ ⎦
⎣
(
25.40
67352_ch25.indd 744
(a)
)
The wavelength of the light within the eye is ln = l n
2/9/11 2:11:54 PM
742
212
Chapter 25
Optical Instruments
l
= 0.60 mrad.
D
42.
.
25.43
38 cm
The limit of resolution in air is q m in
44.
2.2 × 1011 m
In oil, the limiting angle of resolution will be
1.7 m
( l noil ) = ⎛ 1.22 l ⎞ 1
l
q min oil = 1.22 oil = 1.22
⎜⎝
⎟
D
D
D ⎠ noil
46.
48.
or
25.44
.
25.45
q min oil =
q min air
noil
=
air
= 1.22
755
(
)
(
)
× 1015 m ly ⎤⎦ 1.22 575 × 10 −9 m
0.60 ⎡⎣mrad
( 23 ly=) 9.461
0.40
mrad
=
= 2.2 × 1011 m
1.5
0.68 m
The
separation
two6 stars
q = ×d 10
r −4 cm = 1.67 × 10 −6 m, and the highest order of
The angular
grating spacing
is dof=the
1 cm
000 =is1.67
600 nm light that can be observed is
mmax =
(
)
1.67 × 10 −6 m (1)
d sin 90°
=
= 2.78 → 2 orders
l
600 × 10 −9 m
The total number of slits is N = (15.0 cm ) ( 6 000 slits cm ) = 9.00 × 10 4 , and the resolving power
of the grating in the second order is
(
)
Ravailable = Nm = 9.00 × 10 4 2 = 1.80 × 10 5
The resolving power required to separate the given spectral lines is
Rneeded =
l
600.000 nm
=
= 2.0 × 10 5
Δl
0.003 nm
(
)(
)
1.22 9.50 × 10 3 m 575 × 10 −9 m
= with this grating. −3
= 1.7 m
These lines cannot be separated
4.0 × 10 m
.
25.47
25.46
A
fringe
shiftseparation
occurs when
the objects
mirror moves
l 4 .isThus,
The
angular
of two
seen ondistance
the ground
q = dthe
h distance moved (length of
the bacterium) as 310 shifts occur is
⎡ −9 600⎞ × 10 −9 m ⎤
⎛ l⎛⎞ ⎞ N ⎛ 650
= 1×+10⎢ m = 5.04−2 × 10⎥−5(160
= 1.000 5
ΔL = N shifts ⎜ ⎜⎟ = 310
⎟
shifts
⎟⎠ × 10 m ⎥ m =) 50.4 mm
⎝ 4⎝⎠ ⎠ ⎜⎝
4 ⎢⎣ 4 5.00
⎦
(
25.49
.
A fringe shift occurs when the mirror moves distance l 4 . Thus, if the mirror moves distance
ΔL = 0.180 mm, the number of fringe shifts observed is
N shifts =
25.50
25.51
.
)
(
)
−3
4 0.180
×−910m
ΔL⎛ ⎞4 ( ΔL ) ⎛ 632.8
⎞m = 1.31 × −510 3
× 10
=
=
= 250 ⎜
m = 39.6 mm
−9 ⎟ = 3.96 × 10
⎜
⎟
l 4⎝ ⎠ l
⎝ 550 4× 10 m⎠
A
fringe
occurs
thearm
mirror
distance
l 4
When
theshift
optical
pathwhen
in one
of amoves
Michelson’s
interferometer
increases by one
wavelength, four fringe shifts will occur (one shift for every quarter-wavelength change
in path length).
The number of wavelengths (in a vacuum) that fit in a distance equal to a thickness t is
N vac = t l . The number of wavelengths that fit in this thickness while traveling through
the transparent material is N n = t ln = t ( l n ) = nt l . Thus, the change in the number
of wavelengths that fit in the path down this arm of the interferometer is
ΔN = N n − N vac = ( n − 1)
t
l
and the number of fringe shifts that will occur as the thin sheet is inserted will be
# fringe shifts = 4 ( ΔN ) = 4 ( n − 1)
67352_ch25.indd 745
25.52
(a)
⎛ 15.0 × 10 −6 m ⎞
t
= 4 (1.40 − 1) ⎜
= 40
l
⎝ 600 × 10 −9 m ⎟⎠
When the central spot in the interferometer pattern goes through a full cycle from bright
to dark and back to bright, two fringe shifts have occurred, and the movable mirror has
moved a distance of 2(l 4
2/9/11 2:11:57 PM
Optical Instruments
747
213
and
number
fringe
that
the thin sheet
be image
(d) theThe
imageofmust
be shifts
located
in will
frontoccur
of theascorrective
lens,issoinserted
it is a will
virtual
⎛ 15.0 × 10 −6 m ⎞
t
# fringe shifts = 4 ( ΔN ) = 4 ( n − 1) = 4 (1.40 − 1) ⎜
= 40 Instruments
10 −9 length
m ⎟⎠ Optical
⎝ 600 ×focal
+ l1 q = 1 f , the required
of the corrective lens757
is
25.52
25.53
.
(a)
(a)
When
theshould
central
spot an
in upright,
the interferometer
pattern
goes
through
a full
cycleqfrom
bright
pq form
The lens
virtual image
at the
near
point of
the eye,
= −75.0
cm,
f =and
to
dark
back
to
bright,
two
fringe
shifts
have
occurred,
and
the
movable
mirror
has
p + q distance is p = 25.0 cm. The thin-lens equation then gives
when the object
moved a distance of 2(l 4
f =
pq
( 25.0 cm )( −75.0 cm )
=
= 37.5 cm = 0.375 m
p+q
25.0 cm − 75.0 cm
so the needed power is P =
(b)
1
1
=
= + 2.67 diopters .
f 0.375 m
If the object distance must be p = 26.0 cm to position the image at q = −75.0 cm, the actual
focal length is
f =
and P =
pq
( 26.0 cm )( −75.0 cm )
=
= 0.398 m
p+q
26.0 cm − 75.0 cm
1
1
=
= + 2.51 diopters
f 0.398 m
The error in the power is
( −16 cm )( −25 cm )
ΔP = ( 2.67= − 2.51) diopters = 0.16
diopters
= 44
cm too low
−16 cm − ( −25 cm )
25.57
.
(a)
For a refracting telescope, the magnification is m = fo fe , where fo and fe are the focal
lengths of the objective lens and the eyepiece, respectively. Thus, when the Yerkes
telescope uses an eyepiece with fe = 2.50 cm, the magnification is
m=
(b)
.
25.59
25.58
Standard astronomical telescopes form inverted images. Thus, the observer Martian polar
caps are upside down .
Optical Instruments
759
With
485 lines equally spaced in a height C, the distance separating adjacent lines is d = C 485.
We use
When the screen is viewed from a distance L, the angular separation between adjacent lines
is q = d L . If the individual lines are not to be seen (i.e., the lines are to be unresolved), this
angular separation must be less than the minimum angle of resolution, q min = 1.22(l D) by the
Rayleigh criterion. That is, we must have
q =
or
25.60
67352_ch25.indd 746
fo
20.0 m
=
= 8.00 × 10 2 = 800
fe 2.50 × 10 −2 m
(a)
d C 485
1.22 ⋅ l
=
< q min =
L
L
D
L
D
5.00 × 10 −3 m
>
=
= 15.4
C 485 (1.22 ⋅ l ) 485 (1.22 ) 550 × 10 −9 m
(
)
The image must be formed on the back of the eye (retina), so we must have q = 2.00 cm
when p = 1.00 m = 100 cm. The thin-lens equation gives the required focal length as
2/9/11 2:12:00 PM
742
214
42.
25.61
.
44.
46.
48.
=
Chapter 25
1.96 cm
= 9.80
0.200 cm
38
cm a converging lens forms a real image of a very distant object, the image distance equals the
When
focal length
of the lens. Thus, if the scout started a fire by focusing sunlight on kindling 5.00 cm
11
m f = q = 5.00 cm.
2.2
× 10
from
the lens,
1.7
(a) m When the lens is used as a simple magnifier, maximum magnification is produced when the
upright, virtual image is formed at the near point of the eye (q = −15 cm in this case). The
object distance required to form an image at this location is
p=
qf
( −15 cm )( 5.0 cm ) 15 cm
=
=
q− f
−15 cm − 5.0 cm
4.0
−15 cm
q
=−
= + 4.0 . Note
15 cm 4.0
p
that adapting Equation 25.5 for use with this “abnormal” eye would give an angular
magnification of mmax = 1 + q f = 1 + 15 cm 5.0 cm = + 4.0 .
and the lateral magnification produced is M = −
(b)
When the object is viewed directly while positioned at the near point of the eye, its angular
size is q 0 = h 15 cm . When the object is viewed by the relaxed eye while using the lens
as a simple magnifier (with the object at the focal point so parallel rays enter the eye), the
angular size of the upright, virtual image is q = h f . Thus, the angular magnification gained
by using the lens in this manner is
m=
67352_ch25.indd 747
h f
15 cm 15 cm
q
=
=
=
= + 3.0
f
5.0 cm
q 0 h 15 cm
2/9/11 2:12:03 PM
장 ۘо‫ࢇ۽‬Խ
PROBLEM SOLUTIONS
26.1
6.
(a)
Observers on Earth measure the time for the astronauts to reach Alpha Centauri as
Δt E = 4.42 yr. But these observers are moving relative to the astronaut’s internal biological
clock and hence experience a dilated version of the proper time interval Δt p measured on
that clock. From Δt E = g Δt p , we find
Δt p = Δt E g = Δt E 1− ( v c ) = ( 4.42 yr ) 1− ( 0.950 ) = 1.38 yr
2
Relativity
765
2
(b)
The astronauts are moving relative to the span of space separating Earth and Alpha Centauri.
Hence, they measure a length-contracted version of the proper distance, L p = 4.20 ly. The
continued on next
page measured by the astronauts is
distance
L = L p g = L p 1− ( v c ) = ( 4.20 ly ) 1− ( 0.950 ) = 1.31
= 30lybeats min .
2
26.2
26.3
6.
2
(a)
(a)
The
time
for 70 beats,
as measured
the astronaut
any observer
at rest with respect
To the
observer
on Earth,
the muonby
appears
to have and
a lifetime
of
to the astronaut, is Δt p = 1.0 min. The observer in the ship then measures a rate of
70 beats min
4.60 × 10 3 m
d
= 1.55 × 10 −5 s
Δt = =
v 0.990 ( 3.00 × 108 m s )
(b)
g =
(c)
To an observer at rest with respect to the muon, its proper lifetime is
1
1 − ( v c)
Δt p =
(d)
2
=
1
1− ( 0.990 )
2
= 7.09
Δt 1.55 × 10 −5 s
=
= 2.19 × 10 −6 s
g
7.09
The muon is at rest relative to the observer traveling with the muon. Thus, the muon travels
zero distance as measured by this observer. However, during the observed lifetime of the
muon, this observer sees Earth move toward the muon a distance of
( )
d ′ = v Δt p = ⎡⎣ 0.990 ( 3.00 × 108 m s ) ⎤⎦ ( 2.19 × 10 − 6 s ) = 6.50 × 10 2 m = 650 m
(e)
766
26.4
6.
26.5
As the third observer travels toward the incoming muon, his speed relative to the muon
is greater than that of the observer at rest on Earth. Thus, his observed gamma factor
(Δt = g Δt p) is higher, and he measures the muon’s lifetime as longer than that measured
Chapter 26
by the observer at rest with respect to Earth.
(a) trackside
The length
of the sees
meterstick
measured
by the observer as
moving at speed v = 0.900 c relative
The
observer
the supertrain
length-contracted
to the meterstick is
L = L p 1− ( v c ) = (100 m ) 1− ( 0.95) = 31 m
2
2
The supertrain appears to fit in the tunnel with 50 m − 31 m = 19 m to spare .
26.6
(a)
As measured by observers in the ship (that is, at rest relative to the astronaut), the time
required for 75.0 beats is
215
67352_ch26.indd 761
2/9/11 2:15:20 PM
75.0 1− ( 0.990 )
= 10.6 min
1.00 min
2
=
768
216
6.
26.7
26.15
That is, the life span of the astronaut (reckoned by the total number of his heartbeats) is
much longer as measured by an Earth clock than by a clock aboard the space vehicle.
Chapter 26
by the Earth-based
observer
is
The
contracted
length
of the ship,
= Lmomenta
Momentum
must
be conserved,
soLthe
ofmeasured
the two fragments
must add to
zero. Thus,
p − ΔL, as
2
2
their
magnitudes
must
be
equal,
or
g
m
v
=
g
m
v
.
This
gives
2
2
2
1
1
1
L = L p − ΔL = L p 1− ( v c ) . This yields 1− ( v c ) = 1− ΔL L p , and solving for the speed v of
(
(
)
)
the ship, we find v = c 1−( 2.50
1− ΔL
L .kg
Thus,
if the) proper length of the ship is L p = 28.0 m, and
× 10
) ( 0.893c
=
the observed contraction
is ΔL = 0.150 m, the2 speed of the ship must be
1− ( 0.893)
( )2
2
−28
p
(
)
2
0.150 ⎛m ⎞ 35.0 m s ⎞ ⎛ 60.0 s ⎞
⎛
v = c 1− ⎜ 1−
= 26.0 min + 1.06 × 10 −11 s .
⎟⎠ = 0.103c
8
⎜
⎟
⎟
⎜
⎝
28.0
m
⎝
⎠
2
3.00
×
10
m
s
1
min
⎠
⎝
and reduces to
26.8
26.9
6.
2
The driver is the observer at rest with respect to the clock measuring the 26.0-min time interval.
The proper
length ofmeasures
the fasterthe
ship
is three
slower ship
L pf = 3L
, yet they
Thus,
this observer
proper
timetimes
Δt p , that
and of
thethe
Earth-based
observer
measures
ps
bothdilated
appeartime
to have
contracted
length,
L.
Thus,
the
Δt =the
g Δtsame
=
[1
−
(v
c
p
(
(
L = L ps 1− ( vs c ) = 3L ps
2
)
(
)
(
1− v f c , or 1− ( vs c ) = 9 − 9 v f c
2
2
)
)
2
Relativity
767
This gives
2
c 8 + ( vs c )
8 + ( 0.350 )
vf =
= 3.0 s
c = 0.950 c
3
3
=
=
5.0 s
continued on next page
2
1− ( 0.80 )
( )2
Δt p
26.10
2.6 × 10 −8 s
26.11
(a) Δt = g Δt p =
=
= 1.3 × 10 −7 s
6.
2
2
1− ( 0.98 )
1− ( v c )
2
(b)
(c)
26.12
6.
26.13
d = v ( Δt ) = ⎡⎣ 0.98 ( 3.0 × 108 m s ) ⎤⎦(1.3 × 10 −7 s ) = 38 m
( )
8
2.6(×
10 −8)2 s=) =1.27.6
d ′ = v Δt p = ⎡⎣ 0.98 ( 3.02×=102.0
0.80
mm
( ) ( mms))⎤⎦(1−
Length
contraction
onlyfor
in the
dimension
parallel
to the= mv, while the relativistic expres(a) The
classical occurs
expression
linear
momentum
is pclassical
motion.
2
sion is p = g mv, where g = 1 1 − (v c) . Thus, if p = 3p
, it is necessary that the
classical
(a)
768
6.
26.15
26.14
(b)
gamma factor have a value g = 3. Solving for the speed of the particle gives
The sides labeled L2 and L3 in the figure at the right are
unaffected, but the side labeled L1 will appear contracted,
1
1 c 8
2c 2
= c box
1 − a 2rectangular
= c 1− =
=
= 0.943c
givingv the
g
9
3
3
Observe that the calculation above did not depend on the mass of the particle involved.
Thus, the result is the same for a proton or any other particle.
Chapter 26
Momentum
must be conserved, so the momenta of the two fragments must add to zero. Thus,
(a) Classically,
their magnitudes must be equal, or g 2 m2 v2 = g 1 m1 v1 . This gives
p = mv = m ( 0.990c ) = (1.67 × 10 −27 kg )( 0.990 ) 3.00 × 108 m s
(1.67 × 10−27 kg) v = ( 2.50 × 10−28 kg) ( 0.893c )
2
2
1− ( 0.893)
1− ( v c )
2
⎡ (1.67 × 10 −27 kg ) 1− ( 0.893)2 ⎤ v
⎛ ⎞
⎛ v⎞
⎢
⎥
=
1−
and reduces to
⎜ ⎟
⎜⎝ ⎟⎠ , or
c
⎢ ( 2.50 × 10 −28 kg )( 0.893) ⎥ ⎝ c ⎠
0.950c − 0.750c ⎦
⎣
=
= + 0.696c
2
( 0.950c )( 0.750c )
12.3 ( v c ) = 1, and yields 1−
v = 0.285c 2.
c
26.16
26.17
6.
We choose the direction of the spaceship’s motion relative to Earth as the positive direction.
Taking to the right as the positive direction, the velocity of the electron relative to the laboratory
Then, the spaceship’s velocity relative to Earth is vSE = + 0.750c. It is desired to have the velocity
is vEL = + 0.90c, and the velocity of the proton relative to the electron is vPE = − 0.70c. Thus, the
of the
rocket relative to Earth be vR E = + 0.950c. The relativistic relative velocity relation
relativistic addition of velocities (Equation
26.8 in the textbook) gives the velocity of the proton
(Equation 26.7 in the textbook) then gives the required velocity of the rocket relative to
relative to the laboratory as
the ship as
=
( − 0.70c ) + 0.90c
c
26.18
67352_ch26.indd 765
=
+ 0.20c
( −0.70c )( 0.90c ) 1− 0.63
1+
= + 0.54c = 0.54c toward the right
2
We take to the right as the positive direction. Then, the velocities of the two ships relative to
Earth are vR E = + 0.70c and vLE = − 0.70c. The velocity of ship L relative to ship R is given by the
relativistic relative velocity relation (Equation 26.7 in the textbook) as
2/9/11 2:15:28 PM
0.950c − 0.750c
= + 0.696c
( 0.950c )( 0.750c )
1−
+ 0.20c
( − 0.70c ) +c0.90c
2
=
=
= + 0.54c = 0.54c toward the right
( −0.70c )( 0.90c ) 1− 0.63
1+
c2
Taking to the right as the positive direction,
the velocity of the electron relative to the laboratory
=
+
0.90c,
and
the
velocity
of
the
proton
relative
the electron
is two
vPE =ships
− 0.70c.
Thus,
is
v
We ELtake to the right as the positive direction. Then,
theto
velocities
of the
relative
to the
relativistic
addition
of
velocities
(Equation
26.8
in
the
textbook)
gives
the
velocity
of
the
proton
217
765
Earth are vR E = + 0.70c and vLE = − 0.70c. The velocity of ship L relative to ship R Relativity
is given by the
relative
to
the
laboratory
as
relativistic relative velocity relation (Equation 26.7 in the textbook) as
=
26.17
26.18
The astronauts
to the span
of space separating Earth and Alpha Centauri.
vPE + vEL are moving
0.20c
( − 0.70crelative
))+− 0.90c
0.70c = +−1.40c
vHence,
== (a−0.70c
+−0.54c
== distance,
0.54c
PL =
they
measure
length-contracted
version
of==the
proper
L p = the
4.20
ly. The
=
0.94c
0.94c toward
toward
theright
left
vPE vEL
((−0.70c
))((0.90c
)) 1−
0.63
−0.70c
0.70c
1+
0.49
1+
1+
2
distance measured
by
the
astronauts
is
2
1−
c
cc 2
We
takeaway
to thefrom
rightEarth
as theaspositive
direction.
Then,the
thevelocity
velocities
of the
shipstorelative
Taking
the positive
direction,
of ship
A two
relative
Earth isto
Earth
are
v
=
+
0.70c
and
v
=
−
0.70c.
The
velocity
of
ship
L
relative
to
ship
R
is
given by the
RE
LE
vAE = + 0.800c,
and the velocity
of ship B relative to Earth is vB E = + 0.900c. The relativistic
relativistic
relative
velocity
relation
(Equation
26.7
in
the
textbook)
as
relative velocity relation (Equation 26.7 in the textbook) gives the velocity of ship B relative
(b)
26.18
6.
26.19
to ship A (and hence, the speed with which B is overtaking A) as
vB A =
26.20
6.
26.21
We firsttodetermine
velocity of
thegiven
pulsarthat
relative
to the rocket.
Earth as the
Taking
the right the
as positive,
it is
the velocity
of the Taking
rocket toward
relativethe
to observer
A is
positive
direction,
the
velocity
of
the
pulsar
relative
to
Earth
is
v
=
+
0.950c,
and
the
PE
vR A = + 0.92c. If observer B observes the rocket to have a velocity
vR B = − 0.95c, the velocity
velocityof
ofthe
rocket
relative
to
Earth
is
v
=
−
0.995c.
The
relativistic
relative
velocity
relation
(Equation
26.7
RE
observer B relative to the rocket
is vB R = + 0.95c. The relativistic velocity addition relation thenin
the
textbook)
gives
the
velocity
of
thestationary
pulsar relative
to the
rocket as
gives the velocity of B relative to the
observer
A as
vB A =
6.
26.25
26.22
vB E − vAE
0.900c − 0.800c
=
= + 0.357c
v1B E vAE
0.900c )( 0.800c )
(Hz
1−
=
=
0.161
1−
6.20c 2s
c2
v B R + vR A
+ 0.95c + 0.92c
=
= + 0.998c
or 0.998c⎞toward the right
⎛
v B R vR A
( 0.95c )( 0.92c )
1+
1+
⎟⎠ + 0 = 177.893 MeV
⎜
2
1u
⎝
c
c2
The nonrelativistic expression for kinetic energy is KE = 12 mv 2, while the relativistic expression is
(a)
2
KE = E − E R = (g − 1)ER = (g − 1)mc 2, where g = 1 1 − ( v c ) . Thus, when the relativistic kinetic
energy is twice the predicted nonrelativistic value, we have
⎞
⎛
1
1
⎟ mc 2 = 2 ⎛⎜ mv 2 ⎞⎟
⎜
−
1
2
⎝
⎠
2
⎟⎠
⎜⎝ 1 − ( v c )
2
2
1 = ⎡⎣1+ ( v c ) ⎤⎦ 1 − ( v c )
or
Squaring both sides of the last result and simplifying gives
( v c ) ⎡⎣( v c ) + ( v c )
2
4
2
− 1⎤⎦ = 0
Ignoring the trivial solution v c = 0, we must have ( v c ) + ( v c ) − 1 = 0. This is a quadratic
2
equation of the form x 2 + x − 1 = 0,with x = ( v c ) . Applying the quadratic formula gives
2
x = −1 ± 5 2. Since x = ( v c ) , we ignore the negative solution and find
4
(
2
)
−1+ 5
⎛ v⎞
x=⎜ ⎟ =
= 0.618
⎝ c⎠
2
⎛
1
−
which yields v =⎜⎜ c 0.618 = 0.786 c1−. ( 0.900 )2
⎝
2
6.
26.29
26.26
⎞
⎟ ( 0.511 Mev ) = 2.45 MeV
⎟⎠
2
Themrelativistic
totalofenergy
is E = gmoving
ER = g mc
the momentum
p =mass
g mv,having
where a speed of
be the2 mass
the fragment
at v,21 and
= 0.987c,
and m2 beisthe
Let
1
2
2
2
2
2 2
g = 1 1 − v c . Thus, E c = g m c , and p = g 2 m 2 v 2, so subtracting yields
⎛
⎞
E2
1
2 2
2
2
2 2
− p2 = g 2 m 2 ( c 2 − v 2 ) = g 2 m 2 c 2 (1− v 2 c 2 ) = ⎜
⎟ m c 1− v c = m c
2
2
2
c
⎝ 1− v c ⎠
(
)
Rearranging, this becomes
E2
= p2 + m 2 c 2
c2
67352_ch26.indd 766
or
E 2 = p2 c 2 + m 2 c 4
2/9/11 2:15:30 PM
772
218
768
Chapter 26
26.15
6.
26.31
26.32
6.
26.33
10 h
=
2
E
20.0 GeV ⎛ 10 MeV ⎞
4
(a)
g =
= must be equal,
their magnitudes
g 2 m2⎟⎠v=2 =3.91×
g 1 m1 v10
. This gives
⎜⎝ 1or
1
ER on0.511
GeV
The observers
EarthMeV
see the
clock
moving away
at 0.75c and compute the distance traveled
c 8 + vs c
before vthe
alarm sounds
as
f =
2.50 × 10 −28 kg ) ( 0.893c )
L p 3.00 × 10=3 (m
(b) L =
= 2
= 7.67 × 10 −2 m2 = 7.67 cm
s⎞
1− ( 0.893
g ( ⎡3.91×
) ( 10 4
13
)⎤⎦ (15 h) ) ⎛⎜⎝ 3600
⎟ = 1.2 × 10 m
⎣
1h ⎠
The clock, at rest in the ship’s frame of reference, will measure a proper time of Δt p = 10 h before
KE =reduces
E − E Rto= (g − 1) ER
and
sounding.
Observers on Earth move at v = 0.75c relative to the clock and measure an elapsed
time of
KE
1
1
so g = 1+
=
giving
v = c 1−
2
2
ER
1+
KE
ER )
(
1− ( v c )
3
(a)
The speed of an electron having KE = 2.00 MeV will be
ve = c 1−
1
(1+ KE E )
2
= c 1−
R ,e
(b)
1
= 0.979c
2
1+
2.00
0.511)
(
For a proton with KE = 2.00 MeV, the speed is
v p = c 1−
1
(1+ KE E )
2
= c 1−
R,p
6.
26.35
= 15 h
1− ( 0.75)
( )
Momentum must be conserved, so the momenta of the two fragments must add to zero. Thus,
2
1
(1+ 2.00 938)
2
= 0.065 2c
(c)
ve − v p = 0.979c − 0.065 2c = 0.914 c
(a)
Observers on Earth measure the distance to Andromeda to be
d = 2.00 × 10 6 ly = (2.00 × 10 6 yr)c. The time for the trip, in Earth’s frame
of reference, is Δt = g Δt p = 30.0 yr 1− (v c)2 . The required speed is then
Relativity
773
( )
v=
( 2.00 × 106 yr ) c
d
=
Δt 30.0 yr 1− ( v c )2
which gives (1.50 × 10 −5 ) ( v c ) = 1− ( v c ) . Squaring both sides of this equation and solv2
ing for v c yields v c = 1
1 + 2.25 × 10 −10 . Then, the approximation 1 1 + x = 1− x 2 gives
v
2.25 × 10 −10
≈ 1−
= 1− 1.13 × 10 −10
c
2
(b)
KE = (g − 1) mc 2, and
g =
1
1− ( v c )
2
=
1
1− (1− 1.13 × 10
−10
)
2
=
1
2.26 × 10 −10
Thus,
⎛
⎞
2
1
− 1⎟ (1.00 × 10 6 kg ) ( 3.00 × 108 m s ) = 5.99 × 10 27 J
KE = ⎜
−10
⎝ 2.26 × 10
⎠
(c)
cost = KE × rate
⎡
⎛ 1 kWh
= ⎢( 5.99 × 10 27 J ) ⎜
⎝ 3.60 × 10 6
⎣
26.36
67352_ch26.indd 767
⎞⎤
20
⎟⎠ ⎥ ( $0.13 kWh ) = $2.16 × 10
J ⎦
From KE = (g − 1) ER , we find
2/9/11 2:15:34 PM
774
26.37
6.
Chapter 26
Relativity
765
219
(b) observer
An
The astronauts
who moves
are moving
at speedrelative
v relative
to the
to an
span
object
of space
(or span
separating
of space)
Earth
having
and Alpha
properCentauri.
length
2
Hence,
they
measure
a
length-contracted
version
of
the
proper
distance,
L
=
4.20
ly.
p
L p sees a contracted length given by L = L p g = L p 1− (v c) . Thus, if the proper distanceThe
to the
measured
by the
astronauts
is a contracted value of L = 2.00 ly in the reference
star is distance
L p = 5.00
ly, and this
length
is to have
frame of the spacecraft, the speed of the spacecraft relative to the star must be
2
2
⎛ L⎞
⎛ 2.00 ly ⎞
v = c 1− ⎜ ⎟ = c 1− ⎜
=⎝ 5.00
s ) ( 6.17 × 10 − 4 ) = 1.85 × 10 5 m s = 185 km s
(3.00ly× ⎟⎠108= m0.917c
⎝ Lp ⎠
26.38
6.
26.39
(a)
For anDina
electron
at at
ve rest
= 0.750c,
the gamma
is
Since
and moving
Owen are
in the same
frame factor
of reference
(S′), they both see the ball
traveling in the negative x ′ direction with speed vball
velocity of 775
′ = 0.800c . Note that theRelativity
the ball relative to Dina is vB D = − 0.800c.
(b)
The distance between Dina and Owen, measured in their own rest frame, is L p = 1.80 × 1012 m.
Therefore, the time required for the ball to reach Dina, measured on her own clock, is
Lp
1.80 × 1012 m
2.25 × 1012 m
continued on next page
Δt p =
=
=
= 7.50 × 10 3 s
0.800c
3.00 × 108 m s
vball
′
(c)
Ed sees a contracted length for the distance separating Dina and Owen. According to him,
they are separated by a distance L = L p g = L p 1− (v c)2 , where v = 0.600c is the speed
of the S′ frame relative to Ed’s reference frame, S. Thus, according to Ed, the ball must
travel a distance
L = (1.80 × 1012 m ) 1− ( 0.600 ) = 1.44 × 1012 m
2
(d)
The ball has a velocity of vB D = − 0.800c relative to Dina, and Dina moves at velocity
vDE = + 0.600c relative to Ed. The relativistic velocity addition relation (Equation 26.8 from
the textbook) then gives the velocity of the ball relative to Ed as
vB D + vDE
− 0.800c + 0.600c
−0.200c
=
=
= − 0.385c
vB D vDE 1+ ( − 0.800 )( 0.600 )
0.520
1+
c2
3 MeV
GeV
× 10
Thus, the speed of the ball caccording=to4.60
Ed is
vball = cvB E == 4.60
0.385c
. c.
vB E =
26.40
6.
26.43
(a)
As seen by
an observer
at rest relative
to the pmirror
in and
frame
the light
must travelmv, is
The difference
between
the relativistic
momentum,
= g mv,
theS,
classical
momentum,
distance
d
before
it
strikes
the
mirror
and
then
a
distance
d
−
d
back
to
the
ship after
1
Δp = g mv − mv = (g − 1)mv.
reflection. Here, distance d1 = vt is the distance the ship moves toward the mirror in the
timeerror
t between
when
the Δp
pulse
emitted
from
the ship
and 0g
when
ected pulse
0.010
0, or (g
− 1)mv
= 0.010
mv.the
Thisrefl
gives
(a) The
is 1.00%
when
p =was
2
2
was
received
by
the
ship.
Since
all
observers
agree
that
light
travels
at
speed
c, the total
g = 1 0.990 , or 1− ( v c ) = ( 0.990 ) , and yields v = 0.141c .
travel time for the light is
2
2
2
m ) g = 1 0.900
( error is 10.0%,
)( 0.15
(b) When the
we have
−17, and 1− ( v c ) = ( 0.900 ) . In this case, the
=
2.5
×
10
kg
2
2
speed of the
2 (particle
3.00 × 10is8 vm= sc) 1− (0.900) = 0.436c .
26.44
26.45
6.
(a)
(a)
Yes. at rest, muons have a mean lifetime of Δt = 2.2 ms. In a frame of reference where
When
p
they move at v = 0.95c, the dilated mean lifetime of the muons will be
( )
t = g Δt p =
(b)
Δt p
1− ( v c )
2
=
2.2 ms
1− ( 0.95)
2
= 7.0 ms
In a frame of reference where the muons travel at v = 0.95c, the time required to travel
3.0 km is
t=
d
3.0 × 10 3 m
=
= 1.05 × 10 −5 s = 10.5 ms
v 0.95 ( 3.00 × 108 m s )
If N 0 = 5.0 × 10 4 muons started the 3.0 km trip, the number remaining at the end is
N = N 0 e − t t = ( 5.0 × 10 4 ) e − 10.5 m s 7.0 m s = 1.1× 10 4
67352_ch26.indd 768
26.46
(a)
The proper lifetime is measured in the ship’s reference frame, and Earth-based observers
measure a dilated lifetime of
2/9/11 2:15:38 PM
(
220
768
26.15
)
2
=
2.2 ms
1− ( 0.95)
2
= 7.0 ms
a frame of reference where the muons travel at v = 0.95c, the time required to travel
3.0 km is
(b) 26
In
Chapter
3
Momentum must
two fragments must add to zero. Thus,
d be conserved,
3.0 × 10so
mthe momenta of the
t
=
=
= 1.05 × 10 −5 s = 10.5 ms
8 g m v = g m v . This gives
their magnitudes
must
be
equal,
or
2
2
2
1
1
1
v 0.95 ( 3.00 × 10 m s )
× 10 thekg3.0
) the number remaining at the end is
( 2.50started
) ( 0.893c
If N 0 = 5.0 × 10 4 muons
km trip,
=
2
2
1− ( 0.893)
( )
N = N 0 e − t t = ( 5.0 × 10 4 ) e − 10.5 m s 7.0.m s = 1.1× 10 4
−28
26.46
6.
26.47
(a) reduces
The proper
and
to lifetime is measured in the ship’s reference frame, and Earth-based 2observers
The length of the space ship, as measured by observers on Earth, is L = L p 1− ( v c ) . In Earth’s
measure a dilated lifetime of
frame of reference, the time required for the ship to pass overhead is
L L p 1− ( v c )
Δt = =
= Lp
v
v
2
1
1
− 2
2
c
v
2
6.
26.49
Thus,
2
⎛ 0.75 × 10 −6 s ⎞
1
1 ⎛ Δt ⎞
1
s2
−17
=
+
=
+
=
1.74
×
10
2
⎜
⎟
2
2
⎟
⎜
v
c ⎝ Lp ⎠
m2
(3.00 × 108 m s) ⎝ 300 m ⎠
or
v (=
⎞
c
⎛
8 m⎞ ⎛
= 0.80 c
= ⎜ 2.4
× 10
1.99
× 10 30 ⎟⎠kg
8
)
(
)
⎜
3
⎝
s ⎝ 3.00
× 10× 10
m sm⎟⎠ = 1.47 km
s
=
1.47
−17
2
1.74 × 103.00 × 2108 m s
( m
)
1
2
Note: Excess digits are retained in some steps given
below to more clearly illustrate the method of
solution.
We are given that L = 2.00 m and q = 30.0° (both
measured in the observer’s rest frame). The
components of the rod’s length as measured in the
observer’s rest frame are
Lx = L cosq = ( 2.00 m ) cos30.0° = 1.732 m
and
L y = L sinq = ( 2.00 m ) sin 30.0° = 1.00 m
The component of length parallel to the motion has been contracted, but the component perpendicular to the motion is unaltered. Thus, L py = L y = 1.00 m and
L px =
(a)
Lx
1− ( v c )
2
=
1.732 m
1− ( 0.995)
= 17.34 m
The proper length of the rod is then
L p = L2px + L2py =
(b)
2
(17.34 m )2 + (1.00 m )2 = 17.4 m
The orientation angle in the rod’s rest frame is
⎛ L py ⎞
−1 ⎛ 1.00 m ⎞
q p = tan −1 ⎜
⎟ = tan ⎜⎝ 17.34 m ⎟⎠ = 3.30°
⎝ L px ⎠
26.50
(a)
Taking toward Earth as the positive direction, the velocity of the ship relative to Earth is
vSE = + 0.600c, and the velocity of the lander relative to the ship is v = + 0.800c. The relativistic velocity addition relation (Equation 26.8 in the textbook) then gives the velocity of
the lander relative to Earth as
LS
67352_ch26.indd 769
2/9/11 2:15:41 PM
2장 ߦ࢕‫י‬չଝ
PROBLEM SOLUTIONS
.
27.1
784
The energy of a photon having frequency f is given by Ephoton = hf, where Planck’s constant has a
value of h = 6.63 × 10 −34 J ⋅s. This energy may be converted to units of electron volts by use of the
conversion factor 1 eV = 1.60 × 10 −19 J.
(a)
1 eV
⎛
Ephoton = hf = ( 6.63 × 10 −34 J ⋅s ) ( 620 × 1012 Hz ) ⎜
⎝ 1.60 × 10 −19
⎞
⎟ = 2.57 eV
J⎠
(b)
1 eV
⎛
Ephoton = hf = ( 6.63 × 10 −34 J ⋅s ) ( 3.10 × 10 9 Hz ) ⎜
⎝ 1.60 × 10 −19
⎞
−5
⎟ = 1.28 × 10 eV
J⎠
continued on next page
Chapter 27
1 eV
⎞ −7
−7
0.289 68 × 10⎛− 2 m ⋅K
Ephoton = hf = ( 6.63 × 10 −34 J ⋅s ) ( 46.0
⎟⎠ =101.91×
= × 10 Hz ) ⎜⎝31.60 × 10
m =10501eV
nm
= −19
5.01×
J
5.78 × 10 K
(a) The power radiated by an object with surface area A and absolute temperature T is given by
4
= 6.63
× 10 −34
is Stefan-Boltzmann
Planck’s constant.
The energy
of law
a photon
is given11byofEthe
= hf
= hc l ,aswhere
Stefan’s
(see Chapter
textbook)
P = shAeT
, where
s Jis⋅sthe
−8
2
4
constant equal to 5.669 6 × 10 W m ⋅K
− 34
8
hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m s ) ⎛ 1.00 eV ⎞
−5
(a) E =
=
⎜⎝
⎟ = 2.49 × 10 eV
l
5.00 × 10 −2 m
1.60 × 10 −19 J ⎠
(c)
27.2
.
27.3
(b)
E=
− 34
8
hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m s ) ⎛ 1.00 eV
=
⎜⎝
−9
l
500 × 10 m
1.60 × 10 −19
− 34
8
hc ( 6.63 × 10− 2 J ⋅s ) ( 3.00 × 10 m s ) ⎛ 1.00 eV
=
E = 0.289
8 × 10 m ⋅K −9
⎜
−19
−1
5.00 × 10
m × 10 −10 m⎝ 1.60
10
lm ax =l
~10×
nm
= 2.898
7
10 K
Using Wien’s displacement law,
From Wien’s displacement law,
(c)
27.4
.
27.5
(a)
27.6
27.9
.
−11
lm ax
−2
−3
2
2
−2
−34
−9
−6
3
8
−34
−19
8
−18
1.02
J
−9 × 10
⎟⎠ =× 10
685
m
−19
(a)
f = 6.35 eV ⎜
⎝
(b)
(b)
The energy of a photon having the cutoff frequency or cutoff wavelength equals the work
function of the surface, or Ephoton = hfc = hc lc = f . Thus, the cutoff frequency of a surface
having a work function of f = 6.35 eV is
(c)
1 eV
⎛ 1.60 × 10 −19 J ⎞
f
6.35 eV
15
=
−34
⎟⎠ = 1.53 × 10 Hz
h 6.63 × 10 J ⋅s ⎜⎝
1 eV
The cutoff wavelength is
lc =
67352_ch27.indd 780
⎞
⎟ = 249 eV
Jg⎠ -rays
500 ×10 m
m⋅K
8.5×10
m ) m4⎤⎥⋅K
×108 × 10J s ⋅m
(4.0
) ⎡⎣⎢p= (0.289
0.289
8 × 10
⎦ ( = 9.47 × 10) m = 9.47 mm (infrared)
==
= 5.7 ×10 photons s
T(6.63×10 J ⋅ s) (3.00
306×10
K m s)
× 10 J ⎞J ⋅s ) ( 3.00 × 10 m s )
(6.63× 10
⎛ 1.60
=
J photon
= 2.90 × 10
fc =
27.10
⎞
⎟ = 2.49 eV
J⎠
hc c 3.00 × 108 m s
= =
= 1.96 × 10 −7 m = 196 nm
f
fc 1.53 × 1015 Hz
(d)
KEm ax = Ephoton − f = 8.50 eV − 6.35 eV = 2.15 eV
(e)
eVs = KEm ax , so the stopping potential is Vs = KEm ax e = 2.15 eV e = 2.15 V .
(a)
At the cutoff wavelength, KEm ax = 0, so the photoelectric effect equation (KEm ax = hc l
221
2/9/11 2:32:43 PM
786
222
.
27.11
27.10
=
Chapter 27
(b)
(d)
Them axlowest
frequency
of light
electrons
KE
= Ephoton
− f = 8.50
eV −that
6.35will
eV free
= 2.15
eV from the material is fc = c lc
(e)
1.24is×V10=−11KEm e = 2.15 eV (c)
eVs = KEm ax , so the stopping potential
e = 2.15 V .
s
m ax
(a)
(a)
From
photoelectric
effect
functioneffect
is f =equation
hc l − KE
= 0, so the
the work
photoelectric
(KE
At thethe
cutoff
wavelength,
KEmequation,
m axm,axor= hc l
ax
f=
(b)
27.12
.
27.13
c 3.00 × 108 m s
=
= 1.96 × 10 −7 m = 196 nm
fc 1.53 × 1015 Hz
lc =
(6.63 × 10
J ⋅s ) ( 3.00 × 108 m s ) ⎛
1 eV
⎞
⎜⎝
⎟ − 1.31 eV = 2.24 eV
−9
350 × 10 m
1.60 × 10 − 19 J ⎠
− 34
− 34
8
hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m s ) ⎛
1 eV
=
⎜⎝
f
2.24 eV
1.60 × 10 − 19
⎞
−7
⎟ = 5.55 × 10 m = 555 nm
J⎠
c 3.00 × 108 m s
= KE
=eV5.41× 1014 Hz
−9
m
ax10 0.216
lΔV
555
×
m
c
= 0.216 V
=
s =
e
e
(a) The maximum kinetic energy of the ejected electrons is related to the stopping potential
The two
frequencies
of the
allowed to strike the surface are
by the
expression
KEmlight
ax = e ( ΔVs ) . Thus, if the stopping potential is Vs = 0.376 V when
the incident light has
wavelength l = 546.1 nm, the photoelectric effect equation gives the
8
cfunction
3.00 ×of10this
mmetal
s as
work
=
f1 =
= 1.18 × 1015 Hz = 11.8 × 1014 Hz
−9
l1
254 × 10 m
(c)
fc =
and
f2 =
c 3.00 × 108 m s
=
= 6.88 × 1014 Hz
l2
436 × 10 −9 m
Quantum Physics
787
The graph you draw should look
continued somewhat
on next page
like that given at the
right.
The desired quantities, read
from the axes intercepts of the
graph line, should agree within
their uncertainties with
fc = 4.8 × 1014 Hz and f = 2.0 eV
.
27.15
Assuming the electron produces a single photon as it comes to rest, the energy of that photon is
Ephoton = ( KE )i = e ( ΔV ). The accelerating voltage is then
ΔV =
Ephoton
e
=
−34
8
hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m s ) 1.24 × 10 −6 V ⋅ m
=
=
el
l
(1.60 × 10 −19 C) l
For l = 1.0 × 10 −8 m, V =
1.24 × 10 −6 V ⋅ m
= 1.2 × 10 2 V
1.0 × 10 −8 m
and for l = 1.0 × 10 −13 m, V =
27.16
27.17
.
1.24 × 10 −6 V ⋅ m
= 1.2 × 10 7 V
1.0 ×when
10 −13the
m potential difference, ΔV , increases.
A photon
photon of
of maximum
maximum energy
energy or
andminimum
minimumwavelength
wavelengthisisproduced
producedwhen
whenthe
theelectron
electrongives
givesupup
A
all
its
kinetic
energy
in
a
single
collision,
or
all of its kinetic energy in a single collision within the target. Thus, Em ax = hc lm in = KEe = e ΔV.
If lm in = 70.0 pm = 70.0 × 10 −12 m, the required accelerating voltage is
ΔV =
67352_ch27.indd 783
.
(6.63 × 10−34 J ⋅s) (3.00 × 108 m s) = 1.78 × 104 V = 17.8 kV
hc
=
elm in
(1.60 × 10 −19 C) ( 70.0 × 10 −12 m )
2/9/11 2:32:51 PM
=
(1)( 0.129 nm )
2sin8.15°
Quantum Physics
= 0.455 nm
14.
.
27.19
(a) only lithium
(b) 0.81 eV
The interplanar spacing in the crystal is given by Bragg’s law as
16.
(a)
27.20
.
27.21
27.22
27.23
.
783
223
8.29 × 10 −11 m
ml
(1)( 0.140 nm ) (b) because it is mathematically impossible for the sinq to
=
= 0.281 nm
d=
be greater2sinq
than one. 2sin14.4°
6.63 × 10 −34 J ⋅s
1−
cosq
=
cos55.0°maximum
(
)
(1−
)
From
Bragglaw,
equation,
2d 9.11×
sinq =10
at10
which
diffraction
of order m
−31 the to
8
Usingthe
Bragg’s
the wavelength
isml,
found
be ×
kg )angle
3.00
m sthe
(
(
)
will be found when the first order maximum is at q1 = 12.6°, is given by
2 d sinq 2 ( 0.296
nm ) sin 7.6°
l = = 1.03 × 10
= −12 m ⎛ 1 nm ⎞ = 1.03= ×0.078
nm
10 −3 nm
⎜⎝ −91 ⎟⎠
m
10 m
From
the Compton
shift
equation,
the wavelength
x-rays is
Incident
The figure
at the right
shows
the situation
before shift of the scattered
photon
and after the scattering process. Note that the scattering angle is q = 180°, so the Compton equation
Before
l, p, E
gives
l′ − l =
(a)
+x
pe, KEe
l′, p′, E′
2h
h
(1− cos180°) =
me c
me c
After
Scattered
photon
l′ = l +
Electron
at rest
Recoiling
electron
2 (6.63×10 −34 J ⋅ s)
2h
−9
= 0.110 ×10 −9 m +
= 0.115×10
m
Quantum Physics
me c
(9.11×10−31 kg) (3.00 ×108 m s)
+x
789
The momentum of the incident photon is p = h l, while that of the scattered photon is
p′ = − h l ′ (the negative sign is included since momentum is a vector quantity and the
scattered photon travels in the negative x-direction). Thus, conservation of momentum
gives pe − h l ′ = h l + 0, or the momentum of the recoiling electron is
pe =
⎛
⎞
1
1
h h
+
+ = (6.63×10 −34 J ⋅ s) ⎜
⎟
−9
−9
⎝
0.110 ×10 m 0.115×10 m ⎠
l l′
= 1.18 ×10 −23 kg ⋅ m s
(b)
Assuming the recoiling electron is nonrelativistic, its kinetic energy is given by
(1.18 ×10−23 kg ⋅ m s) = 7.64 ×10−17 J ⎛⎜ 1 eV ⎞⎟ = 478 eV
p2
KEe = e =
⎝ 1.60 ×10 −19 J ⎠
2me
2 (9.11×10 −31 kg)
2
−12 rest energy of an
Note that this energy is very small in comparison
Thus,
= 4.85 ×to10the
m = 4.85 × 10 −3 electron.
nm
our assumption that the recoiling
electron) would be non-relativistic is seen to be valid.
(
27.24
27.25
.
Atthe
point
A, the incident
photon
scatters
If
scattered
photon has
energy
equalattoangle
the kinetic energy of the recoiling electron, the energy
qoffrom
the
fi
rst
electron.
The
shift
in wavethe incident photon is divided equally
between them. Thus,
length that occurs in this scattering process is
given by the Compton
as
( E0 )photon equation
hc
hc
Ephoton =
⇒
=
, so l = 2 l0 and Δl = 2 l0 − l0 = l0 = 0.001 60 nm
l 2 l0
2
The Compton scattering formula, Δl = lC (1− cosq ) , then gives the scattering angle as
⎛ Δl ⎞
⎛ 0.001 60 nm ⎞
−1
q = cos−1 ⎜1−
⎟ = cos ⎜1−
⎟ = 70.0°
⎝ 0.002 43 nm ⎠
⎝ lC ⎠
67352_ch27.indd 784
2/9/11 2:32:54 PM
786
224
27.27
.
Chapter 27
(a)
(b)
Quantum Physics
791
The lowest
requiredfrequency
electron momentum
The
of light that iswill free electrons from the material is fc = c lc
p=
h 6.63 × 10 −34 J ⋅s ⎛
1 keV
⎞
−7 keV ⋅s
=
⎜
⎟ = 4.1× 10
l
m
1.0 × 10 −11 m ⎝ 1.60 × 10 −16 J ⎠
and the total energy is
E=
p2 c 2 + ER2
2
keV ⋅s ⎞
2
⎛
8
= ⎜ 4.1× 10 −7
⎟⎠ ( 3.00 × 10 m s ) + ( 511 keV ) = 526 keV
⎝
m
2
The kinetic energy is then
KE = E − ER = 526 keV − 511 keV = 15 keV
(b)
27.28
.
27.29
Ephoton
× 10 −34
⋅s ) (J3.00
× 108× 10
m 8s )m⎛ s ) 1 keV
× 10J −34
⋅s ) ( 3.00
hc ( 6.63( 6.63
⎞−7
=
= =
keV
10 2 nm
= ×414
⎜⎝ = 4.14−16× 10⎟⎠ = m1.2
−11
−19
l
1.0
×
10
m
1.60
×
10
J
( 3.00 eV )(1.60 × 10 J 1 eV )
(a) A 3.00 eV electron is nonrelativistic
E 2 − ER2 and its hmomentum
hc is p = 2me KE
and l = = , 2
For relativistic particles, p =
.
p
c
E − ER2
h
2
= E = KE
1−+(EvR c=) 3.00 MeV + 0.511 MeV = 3.51 MeV, so
For 3.00 MeV electrons,
mp v
l=
27.30
27.31
.
(6.63 × 10
J ⋅s ) ( 3.00 ×−34
108 m s ) ⎛ 1 MeV ⎞
2
−13
7
3.58
6.63
×
10
J ⋅s )2 ⎜
⎛−132.00⎟⎠ ×= 10
(
m× 10
s⎞ m
2
⎝
10 J
= 1.98 × 10 −14 m
(=3.51 MeV ) −27− ( 0.511 MeV ) 7 1.60 ×1−
(1.67 × 10 kg) ( 2.00 × 10 m s) ⎜⎝ 3.00 × 108 m s ⎟⎠
The de Broglie wavelength of a particle of mass m is l = h p
(a) From l = h p = h mv, the speed is
v=
(b)
27.32
.
27.33
−34
l=
h
6.63 × 10 −34 J ⋅s
=
= 1.46 × 10 3 m s = 1.46 km s
me l ( 9.11× 10 −31 kg ) ( 5.00 × 10 −7 m )
h
6.63 × 10 −34 J ⋅s
=
= 7.28 × 10 −11 m
me v ( 9.11× 10 −31 kg ) (1.00 × 10 7 m s )
.
(a)
conservation
of energy,
the increase
in kineticinenergy
must equalofthe
From From
the uncertainty
principle,
the minimum
uncertainty
the momentum
thedecrease
electroninis
potential energy, or KE − 0 = q ΔV. Since the particle is nonrelativistic, its kinetic energy
−34
and momentum
are6.63
related
the
= p2 2m
h
× 10by
J ⋅s expression KE
−25
Δpx =
=
=
5.3
×
10
kg ⋅ m s
4p ( Δx ) 4p ( 0.10 × 10 −9 m )
so the uncertainty in the speed of the electron is
Δvx =
Δpx 5.3×10 −25 kg ⋅ m s
= 5.8 ×10 5 m s or ~ 10 6 m s
=
me
9.11×10 −31 kg
If the speed is on the order of the uncertainty in the speed, then v ~ 10 6 m Quantum
s.
Physics
27.34
.
27.35
793
(a) With uncertainty Δx in position, the minimum uncertainty in the speed is
The uncertainty in the magnitude of the velocity of each particle is
Δvx = vx ⋅ ( 0.010 0%) = ( 500 m s ) ( 0.010 0 × 10 −2 ) = 5.00 × 10 −2 m s
If the mass is known precisely, the uncertainty in momentum is Δpx = m(Δvx ), and the minimum
uncertainty in position is
67352_ch27.indd 785
2/9/11 2:32:56 PM
(
14.
16.
) ( 0.010 0 × 10 ) = 5.00 × 10
−2
−2
Quantum Physics
225
783
ms
(a)
0.81 eVin momentum is Δpx = m(Δvx ), and the minimum
, where
If theonly
masslithium
is known precisely, the (b)
uncertainty
l
is
the
cutoff
wavelength
(calculated
above).
Thus,
uncertainty
in position is Δx m in = h 4p (Δpx ) = h 4p m(Δvx ).
c
(a) 8.29 × 10 −11 m
(b)
8
c
3.00
×
10
m
s
For the electron:
fc =
=
= 1.04 × 1015 Hz
−7
lc
2.88 × 10 m
h
6.63×10 −34 J ⋅ s
(c) Δx
The
photoelectric
effect
equation
may
be written as−2KEm ax = Ephoton − f. Therefore, if the
=
=
m in
−31
) this
me (Δvxon
kg
m s) eV, the maximum kinetic energy
4p surface
(9.11×10have
) (5.00E×10 = 5.50
photons4p
incident
energy
photon
of the ejected electrons is
= 1.16 ×10 −3 m = 1.16 mm
KEm ax = Ephoton − f = 5.50 eV − 4.31 eV = 1.19 eV
For the bullet:
Δx m in =
27.36
.
27.37
−34
h
h 6.63 × 10 J ⋅s 6.63×10 −34 J ⋅ s
==
= 5.28 × 10−6−32 m= 23 m s
=
−2
−31
4p m(Δvx ) 4p
5.00
×
10
m
s
0
kg
)
Δx
4p (m0.020
4p
9.11×10
kg
(
)
(
)
(
) (2.5×10 m )
e
m ax
Assuming
the electron
nonrelativistic,
the uncertainty
momentum
Δp = to
methe
Δv.lifetime
From the
The maximum
time oneis can
use in measuring
the energyinofitsthe
particle isisequal
uncertainty
principle,
h 4pform of the uncertainty principle is ΔE Δt ≥ h 4p . Thus, the
x ≥ One
of the particle,
or Δt Δx≈Δp
2 ms.
m ax
minimum uncertainty one can have in the measurement of a muon’s energy is
ΔEm in =
27.38
27.39
.
794
2
h
6.63 × 10 −34−20J ⋅s
= = ( 5.3 × 10 −6 kg=⋅ m3 ×s )10⎛−29 J1 MeV
4p Δtm ax
s )−27 kg ) ⎜⎝ 1.60 × 10 −13
4p 2( 2(1.67
× 10 × 10
⎞
⎟ = 5.3 MeV
J⎠
(a)
For a nonrelativistic
particle,
KE = 12 acceleration for the electrons, so m(v 2 r) = qvB, or
The magnetic
force supplies
the centripetal
p = mv = qrB. The maximum kinetic energy is then KEm ax = p2 2m = q 2 r 2 B 2 2 m , or
Chapter 27
KE
m ax
(1.60 × 10
=
−19
J ) ( 0.200 m ) ( 2.00 × 10 −5 T )
2
2
2
2 ( 9.11× 10 −31 kg )
= 2.25 × 10 −19 J
The work function of the surface is given by f = Ephoton − KEm ax = hc l − KEm ax , or
f=
27.40
27.41
.
(6.63 × 10
−34
J ⋅s ) ( 3.00 × 108 m s )
450 × 10 −9 m
− 2.25 × 10 −19 J
(
)
2
continued on next page
= 1, and
1 eV
⎛
⎞
= 1.36 eV
= c2.17 ×c102−19 J ⎜
⎝ 1.60 × 10 −19 J ⎟⎠
v=
=
2
2
The deIf Broglie
wavelength
is l = h has
p wavelength l = 1.00 × 10 −8 m, the kinetic energy of the
(a)
the single
photon produced
electron was
−34
8
hc (6.63×10 J ⋅ s) (3.00 ×10 m s)
= 1.99 ×10 −17 J = 124 eV
=
KEe =
−8
1.00 ×10 m
l
This electron is nonrelativistic and its speed is given by
v=
(b)
2 (1.99 ×10 −17 J) ⎛
⎞
2 ( KEe )
c
=
⎜
⎟ = 0.022 0 c
−31
8
me
9.11×10 kg ⎝ 3.00 ×10 m s ⎠
When the single photon produced has wavelength l = 1.00 × 10 −13 m,
KEe =
−34
8
hc (6.63×10 J ⋅ s) (3.00 ×10 m s) ⎛ 1 MeV ⎞
=
⎜
⎟ = 12.4 MeV
⎝ 1.60 ×10 −13 J ⎠
1.00 ×10 −13 m
l
the electron is highly relativistic and KE = (g − 1) ER , giving
g = 1+
KE
12.4 MeV
= 1+
= 25.3
ER
0.511 MeV
Then, v = c 1− 1 g
67352_ch27.indd 786
27.42
2
= c 1− 1 ( 25.3) = 0.999 2c .
2
From the photoelectric effect equation, KEm ax = Ephoton − f = hc l
2/9/11 2:33:00 PM
226
786
=
Chapter 27
(b)
(6.63×10
J ⋅ s) (3.00 ×108 m s) ⎛ 1 MeV ⎞
⎜
⎟ = 12.4 MeV
⎝ 1.60 ×10 −13 J ⎠
1.00 ×10 −13 m
−34
givingthe material is fc = c lc
the
is highly relativistic
KEfree
= (gelectrons
− 1) ER , from
Theelectron
lowest frequency
of light thatand
will
g = 1+
KE
12.4 MeV
= 1+
= 25.3
ER
0.511 MeV
There, when a
surface is illuminated2 with light above 2the threshold frequency of the surface, photoelectrons
Then, v = c 1− 1 g = c 1− 1 ( 25.3) = 0.999 2c .
are seen to be ejected with a delay of less than 10 −9 s.
27.42
27.45
.
796
27.46
.
27.47
−34
From the photoelectric
ax = Ephoton − f = hc l
h 6.63 × 10effect
J ⋅sequation, KEm−32
(a)
p= =
kg ⋅ m s
= 2.21× 10
−2
l
3.00 × 10 m
c 3.00 × 108 m s
=
= 1.00 × 1010 Hz
l
3.00 × 10 −2 m
(b)
f=
(c)
E = hf = ( 6.63 × 10 −34 J ⋅s ) (1.00 × 1010 Hz )
Chapter 27
1 eV
⎛
⎞
= 6.63 × 10 −24 J ⎜
= 4.14 × 10 −5 eV
⎝ 1.60 × 10 −19 J ⎟⎠
hc
and
is l0 = hc E0
The
of the incident
photon
is E0 =KE6.20 =keV,
Fromenergy
the photoelectric
effect
equation,
Ephoton
− fits=wavelength
− f.
m ax
l
For l = l0 , KEm ax = 1.00 eV,
For l =
27.48
.
27.49
l0
, KEm ax = 4.00 eV,
2
so
1.00 eV =
hc
−f
l0
[1]
giving
4.00 eV =
2 hc
−f
l0
[2]
Multiplying Equation [1] by a factor of
2 and subtracting the result from Equation [2] gives the
(1)( 2.49 × 10 −11 m )
work function as f = 2.00
= eV .
= 2.9 × 10 −10 m = 0.29 nm
2sin ( 2.5° )
(a) Minimum wavelength photons are produced when an electron gives up all its kinetic energy
The total
energy
of this
object
is 000 eV and
in amechanical
single collision.
Then,
Ephoton
= 50
Em ech = KE f = PEi = mgyi = ( 2.0 kg ) ( 9.80 m s 2 )( 5.0 m ) = 98 J
A photon of light having wavelength l = 5.0 × 10 −7 m has an energy of
Ephoton =
−34
8
hc ( 6.63 × 10 J ⋅s ) ( 3.00 × 10 m s )
=
= 4.0 × 10 −19 J
l
5.0 × 10 −7 m
Thus, if all the initial gravitational potential energy of the object were converted to light with
l = 5.0 × 10 −7 m, the number of photons that would be produced is
n=
27.50
67352_ch27.indd 787
(a)
Em ech
98 J
=
= 2.5 × 10 20 photons
−19
Ephoton 4.0 × 10
J photon
From v 2 = v02 + 2a y ( Δy ) , Johnny’s speed just before impact is
2/9/11 2:33:03 PM
2장 ࡕ࢕‫י‬չଝ
PROBLEM SOLUTIONS
.
28.1
(a)
The electrical force supplies the centripetal acceleration of the electron, so
me v 2 r = ke e 2 r 2 or v = ke e 2 me r .
(8.99 × 10 N ⋅ m C ) (1.60 × 10 C)
(9.11 × 10 kg) (1.0 × 10 m )
9
v=
(b)
(c)
No.
2
−19
2
−31
−10
2
= 1.6 × 10 6 m s
v 1.6 × 10 6 m s
=
= 5.3 × 10 −3 << 1, so the electron is not relativistic.
c 3.00 × 108 m s
The de Broglie wavelength for the electron is l = h p = h me v , or
6.63 × 10 −34 J ⋅ s
= 4.5 × 10 −10 m2= 0.45 nm
9
−31
N10
⋅ m6 2 mC2s) (1.60 × 10 −19 C )
158
kg×)10
1.6 ×
× 10(8.99
(9.11
(
=
5.0 MeV (1.60 × 10 −13 J MeV )
Yes. The wavelength
and the atom are roughly the same size.
= 4.5 × 10 −14 m = 45 fm
l=
(d)
28.2
.
28.3
Assuming a head-on collision, the a-particle comes to rest momentarily at the point of closest
(a) From Coulomb’s law,
approach. From conservation of energy,
k qq
(8.99 × 109 N ⋅ m 2 C2 ) (1.60 × 10 −19 C)
F = e 21 2 =
2
r
(1.0 × 10 −10 m )
802
Chapter 28
(b)
28.4
28.5
.
3.
4.
= 2.3 × 10 −8 N
The electrical potential energy is
9
2
−19
⋅3,m there
C2 )are
−1.60
× 10 −19 values
C ) (1.60
×the
10orbital
C ) quantum
(
ke q1 q2 (8.99 × 10 n N
=
3
possible
of
PE =
=
−10
× 10 quantum
m
number, = 0, 1, 2.rThere are a total of 2 ( 2 + 11.0
states for each value of ;
) possible
2 + 1 possible values of the orbital magnetic quantum number m , and 2 possible spin orientations (ms = ± 12 ) for each
there
10 3d states (having n = 3, = 2), 6 3p
1. Thus,
eV
nm ) ( 410
( 520 value
)
⎛of mnm
⎞ are
3
= − 2.3 × 10 −18 J ⎜
nmeV
= 1.94 mm
=−19
1.94⎟n×=
10−14
states (with n = 3, = =520
1), and
3s
states
⎝ 1.60
× 10(with
J ⎠ = 3, = 0), giving a grand total of 10 + 6 + 2 = 18
nm 2− 410
nm
n = 3 states, and the correct choice is (e).
If
(a) The wavelengths in the Lyman series of hydrogen are given by 1 l = RH (1 − 1 n 2 ), where
Since the electron is in some bound quantum state of the atom, the atom7is not
ionized, and
n = 2, 3, 4,… , and the Rydberg constant is RH = 1.097 373 2 × 10 m −1 . This can also be
choice (a) is false. The fact that
the
electron
is
in
a
d
state
means
that
its
orbital
quantum number
⎡ 2 ( n 2 −since
so the
first three
wavelengths
series(e)areis false.
as l (b)
= (1isRfalse.
1) ⎤⎦ , the
H ) ⎣ n Also,
is = written
2, so choice
maximum
value
of is nin
− 1,this
choice
Finally, the ground state of hydrogen is a 1s 2state, so choice (d) is false, leaving (c) as the only
⎛ 2 ⎞
1
true statement
= 1.215 × 10 −7 m = 121.5 nm
l1 =in the list of choices.
7
−1 ⎜
1.097 373 2 × 10 m ⎝ 22 − 1 ⎟⎠
Wavelengths of the hydrogen spectrum are given by 1 l = RH (1 n 2f − 1 ni2 ), where the Rydberg
constant is RH = 1.097 373 12 × 10 7 m −1 .⎛ Thus,
n f = 3 and ni = 5,
⎞
32 with
= 1.025 × 10 −7 m = 102.5 nm
l2 =
7
−1 ⎜ 2
⎟
1.097 373 2 × 10 m ⎝ 3 − 1 ⎠
1⎞
1
⎛ 1
= 1.097 373 2 × 10 7 m −1 ⎜ 2 − 2 ⎟ = 7.80 × 10 5 m −1
⎝3
⎠
5
l
⎛ 42 ⎞
1
= 9.720 × 10 −8 m = 97.20 nm
l3 =
7
−1 ⎜
−1 2 × 10 m
373
⎝ 4 2 − 1 ⎟⎠
and l = 1 (7.80 ×1.097
10 5 m
)
(b)
67352_ch28.indd 797
2
These wavelengths are all in the far227
ultraviolet region
2/9/11 2:50:39 PM
228
804
⎛ 42 ⎞
−8
⎜⎝ 4 2 − 1 ⎟⎠ = 9.720 × 10 m = 97.20 nm
Chapter 28
28.11
The
the emittedare
photon
(b) energy
Theseof
wavelengths
all inisthe far ultraviolet region of the spectrum.
of the spectrum.
28.6
.
28.7
(a)
The wavelengths
Paschen
series of hydrogen are given by 1 l
The energy
absorbed byin
thethe
atom
is
Ephoton = E f − Ei =
⎛ 1
−13.6 eV (−13.6 eV)
1⎞
−
= +13.6 eV ⎜ 2 − 2 ⎟
2
2
nf
ni
⎝ ni n f ⎠
Atomic Physics
803
1⎞9
2
⎛ 1
Ephoton = 13.6 eV ⎜ (8.99
− ×10
=N
2.86
⋅ m 2eVC2 ) (1.60 ×10 −19 C)
2
2 ⎟
⎝
⎠
2
5
=−
continued on next page
(0.052 9 ×10−9 m )
1⎞
⎛ 1
(b) Ephoton = 13.6 eV ⎜ 2 − 2 ⎟ = 0.472 eV
⎠ eV
⎝ 4J = 6− 27.2
= − 4.35×10 −18
(a)
28.8
28.9
.
(a)
(a)
2
With
the electrical
force supplying
The energy
of the emitted
photon isthe centripetal acceleration, me vn rn
1⎞
⎛ 1
Ephoton = E4 − E2 = −13.6 eV ⎜ 2 − 2 ⎟ = 2.55 eV
⎝4
2 ⎠
This photon has a wavelength of
l=
(b)
804
28.10
.
28.11
hc
Ephoton
=
(6.63 × 10
−34
J ⋅ s ) ( 3.00 × 108 m s )
( 2.55 eV ) (1.60 × 10
−19
J eV )
= 4.88 × 10 −7 m = 488 nm
Since momentum must be conserved, the photon and the atom go in opposite directions
with equal magnitude momenta. Thus, p = matom v = h l , or
Chapter 28
v=
h
matom l
=
6.63 × 10 −34 J ⋅ s
= 0.814 m s
(1.67 × 10 −27 kg) ( 4.88 × 10 −7 m )
The
supplies
the centripetal
acceleration, so mv 2 r
The magnetic
energy offorce
the emitted
photon
is
Ephoton =
−34
8
hc ( 6.626 × 10 J ⋅s ) ( 2.998 × 10 m s ) ⎛
1 eV
⎞
=
⎜⎝
⎟ = 1.89 eV
l
656 × 10 −9 m
1.60 × 10 −19 J ⎠
This photon energy is also the difference in the electron’s energy in its initial and final orbits. The
energies of the electron in the various allowed orbits within the hydrogen atom are
En = −
13.6 eV
n2
where
n = 1, 2, 3,…
giving E1 = −13.6 eV, E2 = −3.40 eV, E3 = −1.51 eV, E4 = −0.850 eV,….
1⎞
⎛1
= 13.6 eVwas
− 2 ⎟ the
= n10.2
⎜⎝ 2 from
Observe that Ephoton = E3 − E2 , so the transition
= 3eV
orbit to the n = 2 orbit .
1
2 ⎠
28.12
28.13
.
(a)
(a)
The
absorbed
n 2 a0 yields
r2 = is
4 ( 0.052 9 nm ) = 0.212 nm
rn = energy
(b)
With the electrical force supplying the centripetal acceleration, me vn2 rn = ke e 2 rn2, giving
vn = ke e 2 me rn , and pn = me vn = me ke e 2 rn .
Thus,
me ke e 2
p2 =
=
r2
(9.11×10
−31
kg) (8.99 ×10 9 N ⋅ m 2 C2 ) (1.6 ×10 −19 C)
2
0.212 ×10 −9 m
= 9.95×10 −25 kg ⋅ m s
(c)
67352_ch28.indd 801
2/9/11 2:50:55 PM
Atomic Physics
.
44.
229
801
Thus,
−31
−19
2
kg2 ) (C8.99
×10 9 ×N10
⋅ m−192 C2) )(1.60
× 10 9 N ⋅ m
× 10 −19 CC))
(9.11×10
(1.6 ×10
) ( −1.60
me ke e 2 (8.99
p2 =
==
−10 −9
r2
0.212
1.0
× 10×10
mm
2
1 eV
⎛
== −
9.95×10
2.3 × 10−25−18 kg
J ⎜⋅ m s
⎝ 1.60 × 10 −19
46.
28.4
⎞
⎟ = −14 eV
J⎠
(a)
If
(c)
⎛ 6.63
135 eV ⎛ h ⎞
(b) × 10 −34 J ⋅ s ⎞
−34
Ln = n ⎜
→ L2 = 2 ⎜
⎟
⎟⎠ = 2.11 × 10 J ⋅ s
⎝ 2p ⎠
2p
⎝
(d)
(9.95 × 10−25 kg ⋅ m s) = 5.44 × 10−19
1
p2
KE2 = mv22 = 2 =
2
2me
2 ( 9.11 × 10 −31 kg )
(e)
PE2 =
2
(8.99 × 109 N ⋅ m 2 C2 ) (1.60 × 10−19 C)
ke ( −e ) e
=−
r2
( 0.212 × 10−9 m )
1 eV
⎛
⎞
Atomic Physics
805
J⎜
= 3.40 eV
⎝ 1.60 × 10 −19 J ⎟⎠
2
continued on next page
= −1.09 × 10 −18 J = − 6.80 eV
6.63 × 10 −34 J ⋅ s ) ( 3.00 × 108 m s ) ⎛
(
1 eV
⎞
= = 3.40 eV − 6.80 −9
⎜⎝
⎟ = 0.970 eV
(f ) E2 = KE2 + PE
2
1 281 × 10 eVm= − 3.40 eV
1.60 × 10 −19 J ⎠
28.17
.
Since the electrical force supplies the centripetal acceleration,
me vn2 ke e 2
= 2
rn
rn
or
vn2 =
k2 e 2
me rn
From Ln = me rn vn = n , we have rn = n
vn2 =
me vn , so
k2 e 2 ⎛ me vn ⎞
⎜
⎟
me ⎝ n ⎠
which reduces to vn = ke e 2 n .
.
28.19
28.18
28.20
28.21
.
(a)
For He + , Z = 2 and r1 = 0.052 9 nm 2 = 0.026 5 nm .
(b)
For Li 2+ , Z = 3 and r1 = 0.052 9 nm 3 = 0.017 6 nm .
(c)
. r1 = 0.052 9 nm 4 = 0.013 2 nm .
For Be 3+ , Z = 4 and
(a)
Starting
n = 94 nm)
state,⇒
there
possible
transitions
the electron returns to the
rn = n 2 a0from
= n 2 the
(0.052
r3 =are
32 6(0.052
9 nm)
= 0.476asnm
ground (n = 1) state. These transitions are: n = 4 → n = 1, n = 4 → n = 2, n = 4 → n = 3,
n =the
3→
n =model,
1, n = 3the
→circumference
n = 2, and n =of
2→
= 1. Since
there
is abedifferent
change
in of
In
Bohr
an nallowed
orbits
must
an integral
multiple
energy
associated
with eachforofthe
these
transitions,
be rn = nl. Thus, the wavelength
the
de Broglie
wavelength
electron
in thatthere
orbit,will
or 2p
of the electron when in the n = 3 orbit in hydrogen is
l=
67352_ch28.indd 802
807
The radii for atomic number Z are rn = n 2 ( 2 me ke e 2 ) Z = n 2 a0 Z , so r1 = a0 Z , where
(a)
a0 = 0.052 9 nm is the radius of the first Bohr orbit in hydrogen.
(b)
28.22
Atomic Physics
(a)
2p r3 2p ( 0.476 nm )
=
= 0.997 nm
3
3
The energy levels in a single electron atom with nuclear charge +Ze are
En = − Z 2 (13.6 eV) n 2
2/9/11 2:50:57 PM
230
804
Chapter 28
of the spectrum.
28.11
.
28.23
28.8
The energy of the emitted
1 ⎞ is to ionize the hydrogen atom, the electron in this atom must
⎛ 1photon
(a) For
the=absorption
Ephoton
13.6 eV ⎜ of2 the
− 2photon
⎟ = 2.86 eV
⎝ 2 having
5 ⎠ an ionization energy less than or equal to the photon energy.
be in an excited state
That is, we must have
Eionization
1 ⎞ = −En ≤ Ephoton = 2.28 eV, or En ≥ −2.28 eV. The state with
⎛ 1
Ephoton
=
13.6
eV
−
= 0.472
eV
(b) the
2 ⎟
smallest value⎜⎝ of
this requirement
is the n = 3 state, with E3 = −1.51 eV.
⎠
4 2 n 6meeting
(b)
(a)
After the electron spends 1.51 eV of energy to escape from the atom,
it will retain the
With the electrical force supplying the centripetal acceleration, m vn2 rn
remaining absorbed photon energy (2.28 eV − 1.51 eV = 0.77 eV)e as
kinetic energy. Its
speed will then be
28.24
28.27
.
2 ( 0.77 eV ) ⎛ 1.60 × 10 −19 J ⎞ h 2
= 5.2 × 10 5 m s = 520 km s
En kg
= n⎜⎝2 E0 1where
9.11or× 10 −31
eV E0 ⎟⎠= 8mL2
2KE
=
me
v=
(a)
Coulombn force
supplies
centripetal
hold the electron in orbit, so
In theThe
3d subshell,
= 3 and
l = 2. the
Thenecessary
10 possible
quantumforce
statestoare
me vn2 rn
n=3
l=2
ml = + 2 ms = + 12
n=3
l=2
ml = + 2 ms = − 12
n=3
l=2
ml = +1 ms = + 12
n=3
l=2
ml = +1 ms = − 12
n=3
l=2
ml = + 0 ms = + 12
n=3
l=2
ml = + 0 ms = − 12
n=3
l=2
ml = −1 ms = + 12
n=3
l=2
ml = −1 ms = − 12
n=3
l=2
ml = − 2 ms = + 12
ml = − 2 ms = − 12
ml = − 3, − 2, −1, 0, +1, + 2, and + 3.
(a) For a given value of the principal quantum number n, the orbital quantum number l varies
The 3dfrom
subshell
0 to has n = 3 and l = 2 . For r-mesons, we also have s = 1. Thus, there are
15 possible quantum states, as summarized in the table below.
n=3
28.28
.
28.29
n
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
l
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
ml
+2
+2
+2
+1
+1
+1
0
0
0
–1
–1
–1
–2
–2
–2
+1
0
–1 +1
electron independently.
0
–1
+1
0
–1
states,+1
six for0each –1
+1 possible
0
–1
ms
28.31
.
28.32
28.33
.
67352_ch28.indd 803
l=2
In the table of electronic configurations (Table 28.4), or the periodic table on the inside back
cover of the text, look for the element whose last electron is in a 3p state and which has three
electrons outside a closed shell. Its electron configuration will then end in 3s 2 3p1. You should
Atomic Physics
find that the element is aluminum .
811
(a)
The electronic
configuration
for nitrogen
Observe
the electron
configurations
given in the periodic table on the inside back cover
of the textbook. Zirconium, with 40 electrons, has 4 electrons outside a closed krypton
core. The krypton core, with 36 electrons, has all states up through the 4 p subshell filled.
Normally, one would expect the next 4 electrons to go into the 4d subshell. However,
an exception to the rule occurs at this point, and the 5s subshell fills (with 2 electrons)
before the 4d subshell starts filling. The two remaining electrons in zirconium are in an
incomplete 4d subshell. Thus, n = 4 and = 2 for each of these electrons.
(b)
For electrons in the
2/9/11 2:51:00 PM
Atomic Physics
for each of these electrons.
44.
(b)
(c)
801
231
For electrons in the 4d subshell, with = 2, the possible values of m are
m = 0, ± 1, ± 2 , and those for ms are ms = ± 1 2 .
(6.626 × 10−34 J ⋅s) ( 2.998 × 108 m s) ⎛ 1 eV ⎞ = 1.89 eV
=
⎜
⎟
We have 40 electrons, so 656
the electron
× 10 −9 mconfiguration⎝ is:
1.60 × 10 −19 J ⎠
2
6
6
2
2
2s 2 2ispalso
3s 2 the
3p
3d 10 4s 2 4in
p6 the
4d 2electron’s
5s 2 = [Kr]4d
5sin
This photon 1s
energy
energy
its initial and final orbits. The
. difference
energies of the electron in the various allowed orbits within the hydrogen atom are
28.35
.
46.
(a)
NoteeV
that Z = 83 for bismuth.
135
(b)With a vacancy in the L-shell, an electron in the M-shell is
shielded from the nuclear charge by a total of 9 electrons, 2 electrons in the filled K-shell
and 7 electrons in the partially filled L-shell. Thus, our estimate for the energy of this
electron while it is in the M-shell is
− ( Z − 9 ) (13.6 eV ) − ( 74 ) (13.6 eV )
=
= −8.27 × 10 3 eV = −8.27 keV
32
9
2
EM ≈
2
When this electron drops down to fill the vacancy in the L-shell, it will continue to be
shielded from the nuclear charge by the 2 electrons in the filled K-shell. Our estimate for
the energy of this electron in the L-shell is then
− ( Z − 2 ) (13.6 eV ) − (81) (13.6 eV )
=
= −2.23 × 10 4 eV = −22.3 keV
22
4
2
EL ≈
2
Therefore, the estimate for the transitional energy, and hence the energy of the photon
produced, in a M- to L-shell transition in bismuth is
Ephoton = EM − EL ≈ −8.27 keV − ( −22.3 keV ) = 14 keV
(b)
The wavelength of the photon produced in the M- to L-shell transition should be approximately
108 m s )−11⎛
(6.63(× 10−34 J ⋅ s) (3.00) =× 3.11
hc
1 keV
⎞
−11
× 10 ⎜ m = 0.0311
nm
≈ −19
⎟⎠ = 8.9 × 10 m
−16
3
⎝
E
14
keV
1.60
×
10
J
1.60
×
10
C
40.0
×
10
V
( photon
)(
)
l=
.
28.37
N-shell
The transitions that produce the three longest
wavelengths in the K series are shown at the right.
The energy of the K-shell is EK = − 69.5 keV.
M-shell
L-shell
Thus, the energy of the L-shell is
EL = EK +
or
hc
l3
EL = − 69.5 keV +
(6.63 × 10
J ⋅ s ) ( 3.00 × 10 m s ) ⎛
1 keV
⎞
⎜⎝
⎟
0.021 5 × 10 −9 m
1.60 × 10 −16 J ⎠
−34
K-shell
8
= − 69.5 keV + 57.8 keV = −11.7 keV
Similarly, the energies of the M- and N-shells are
and
EM = EK +
(6.63 × 10−34 J ⋅ s) (3.00 × 108 m s) = −10.0 keV
hc
= − 69.5 keV +
l2
( 0.020 9 × 10−9 m ) (1.60 × 10 −16 J keV )
EN = EK +
(6.63 × 10−34 J ⋅ s) (3.00 × 108 m s) = − 2.30 keV
hc
= − 69.5 keV +
l1
( 0.018 5 × 10 −9 m ) (1.60 × 10 −16 J keV )
The ionization energies of the L-, M-, and N-shells are
11.7 keV, 10.0 keV, and 2.30 keV, respectively
67352_ch28.indd 804
2/9/11 2:51:03 PM
804
232
28.11
.
28.39
(6.63 × 10
−34
( 0.018 5 × 10
−9
J ⋅ s ) ( 3.00 × 108 m s )
m ) (1.60 × 10 −16 J keV )
= − 2.30 keV
The
of the
emitted photon is
The energy
ionization
k energies
( −e ) e of the L-, M-, and N-shells are
PE2 = e 1.00 × 10 −6 m
= r2
≈ 137
−11 2.30 keV, respectively
11.7 keV, 10.0
keV,
5.29
× 10and
m
2
2 ( 2.54 × 10 74 ) (1.05 × 10 −34 J ⋅ s )
=
2
6.67 × 10−3−11 N ⋅ m 2 kg−92 ) (1.991 × 10 30 kg ) ( 5.98 × 10 24 kg )
(
P ( ΔE Δt ) 4 ( 3.00 × 10 J 1.00 × 10 s )
= 4.24 × 1015 W m 2
(a) I = =
=
2
A =p1.18
d 2 4× 10 −63 m p ( 30.0 × 10 −6 m )
(d)
(b)
.
28.41
= − 69.5 keV +
Chapter 28
(a)
2 ⎤
W ⎞above
⎡ p 0.600
15 (d)
−9
The
is much
smaller
atomic
nucleus
E = result
I−15
A ( Δtcomputed
m ) than
1.20
× 10 −12
J
10 −9 s )of= an
× 10
) = ⎛⎜⎝ 4.24 ×in10part
(
(1.00the× radius
2 ⎟ ⎢
⎠
m ⎣ 4 quantized orbits of⎥⎦ the Earth is too small to observe.
(∼ 10 m), so the distance between
With one vacancy in the K-shell, an electron in the L-shell has one electron shielding it
from the nuclear charge, so Z eff = Z − 1 = 24 − 1 = 23. The estimated energy the atom gives
up during a transition from the L-shell to the K-shell is then
ΔE ≈ Ei − E f = −
2
⎡1
Z eff
1⎤
(13.6 eV ) ⎡ Z eff2 (13.6 eV ) ⎤ 2
− ⎢−
⎥ = Z eff (13.6 eV ) ⎢ 2 − 2 ⎥
2
2
ni
nf
⎢⎣
⎥⎦
⎢⎣ n f ni ⎥⎦
or
1
1
2
ΔE ≈ ( 23) (13.6 eV ) ⎡⎢ 2 − 2
2
⎣1
(b)
⎤ = 5.40 × 10 3 eV = 5.40 keV
⎥⎦
With a vacancy in the K-shell, we assume that Z − 2 = 24 − 2 = 22 electrons shield the
outermost electron (in a 4s state) from the nuclear charge. Thus, for this outer electron,
Z eff = 24 − 22 = 2, and the estimated energy required to remove this electron from the atom is
⎡ Z 2 (13.6 eV ) ⎤ 22 (13.6 eV )
Eionization = E f − Ei = 0 − Ei ≈ − ⎢ − eff
= 3.40 eV
⎥=
ni2
42
⎣
⎦
28.43
.
(c)
16
3
8.84
photons
KE = ΔE − Eionization = 5.40 keV − =3.40
eV×≈105.40
keV mm
(a)
ΔE = E2 − E1 = −13.6 eV (2)2 − ( −13.6 eV (1)2 ) = 10.2 eV
(b)
The average kinetic energy of the atoms must equal or exceed the needed excitation energy,
or 32 kB T ≥ ΔE, which gives
T≥
28.45
.
28.44
−19
J eV )
2 ( ΔE ) 2 (10.2 eV ) (1.60 × 10
=
= 7.88 × 10 4 K
−23
3kB
3 (1.38 × 10
J K)
In the Bohr model,
2 4
1 ⎞ ⎤ 4p 2 me ke2 e 4 ⎡ 1
1⎤
En − En −1 1 ⎡ −me ke e ⎛ 1
=
−
− 2⎥
=
⎢
f =
⎢
2 ⎟⎥
2
2
2
3
⎜
h ⎣⎢ 2
n ⎦
2h
( n − 1) ⎠ ⎥⎦
⎝n
h
⎣ ( n − 1)
which reduces to
67352_ch28.indd 805
f =
2p 2 me ke2 e 4 ⎛ 2n − 1 ⎞
⎜⎝ ( n − 1)2 n 2 ⎟⎠ .
h3
2/9/11 2:51:06 PM
2장 ଦ‫י‬չଝ
PROBLEM SOLUTIONS
.
29.1
From M E = rnuclear V = rnuclear (4p r 3 3), we find
⎛ 3 ME ⎞
r=⎜
⎝ 4p rnuclear ⎟⎠
agree.
29.2
.
29.3
29.4
29.5
.
13
⎡ 3 ( 5.98 × 10 24 kg ) ⎤
2
=⎢
1.8 cm
× 108.21
m cm = 0.961 , so they do
⎥ ==7.89
17
3
⎢⎣ 4p ( 2.3 × 10 kg m ) ⎥⎦
13
(a) From conservation of energy, ΔKE
= − ΔPE, or 1
The average nuclear radii are r = r0 A1 3, where r0 = 1.2 ×2 10 −15 m = 1.2 fm and A is the mass
number.
r = (1.2 fm )( 2 ) = 1.5 fm
13
(a)
For 12 H,
(b)
For
60
27
(c)
For
197
79
Au,
r = (1.2 fm )(197 ) = 7.0 fm
(d)
For
239
94
Pu,
r = (1.2 fm )( 239 ) = 7.4 fm
(a)
(a)
At the point of closest approach, PE f = KEi , so ke ( 2e ) ( 79e ) rm in = ma vi2 2, or
Co,
vi =
=
(b)
r = (1.2 fm )( 60 ) = 4.7 fm
13
13
13
Nuclear Physics
821
2ke ( 2e )( 79e )
ma rm in
316 (8.99 × 10 9 N ⋅ m 2 C2 ) (1.60 × 10 −19 C )
(6.64 × 10−27 kg) (3.2 × 10−14 m )
2
= 3.42 × 10 7 m s = 1.9 × 10 7 m s
2
9
2
2 ⎡
−19
⎤
ke q1 q2 (8.99 × 10 N ⋅ m C ) ⎣( 2 )( 79 )(1.60 × 10 C ) ⎦
KEi = PE f =
=
3.2 × 10 −14 m
rm in
⎛ 1 MeV ⎞
= 7.1 MeV
= 1.14 × 10 −12 J ⎜
⎝ 1.60 =
× 10
(1.2−13×J10⎟⎠ −15 m )( 238)1 3 = 7.4 × 10−15 m = 7.4 fm
2
9
2
2 ⎡
−19
⎤
ke q1 q2 (8.99 × 10 N ⋅ m C ) ⎣( 2 )( 6 )(1.60 × 10 C ) ⎦
= 2 =
= 27.6 N
2
rm in
(1.00 × 10−14 m )
29.6
(a)
For
29.7
.
(a)
Fm ax
(b)
am ax =
(c)
PEm ax
Fm ax
27.6 N
=
= 4.16 × 10 27 m s 2
6.64 × 10 −27 kg
mα
2
9
2
2 ⎡
−19
⎤
ke q1 q2 (8.99 × 10 N ⋅ m C ) ⎣( 2 )( 6 )(1.60 × 10 C ) ⎦ ⎛ 1 MeV ⎞
=
=
⎜⎝
⎟,
1.00 × 10 −14 m
rm in
1.60 × 10 −13 J ⎠
yielding PEm ax = 1.73 MeV .
29.8
67352_ch29.indd 816
If a star with a mass of two solar masses collapsed into a gigantic nucleus by converting all of its mass
into neutrons, the total number of nucleons (all
233neutrons), and hence the atomic number, would be
2/9/11 2:36:16 PM
824
234
29.20
.
29.9
Chapter 29
= (1.2 × 10 −15 m ) ( 2.38 × 10 57 )
13
= 1.6 × 10 4 m = 16 km
From
in825
the utextbook,
the 665
fraction
at ut=)=m
will
vi2 yr
2,285
or u,beand
Na, Δm = (29.4a)
11(1.007
u ) − (remaining
22.989 770
=5.00
For 1123 Equation
) + 12 (1.008
a0.200
23
for 12 Mg, Δm = 12 (1.007 825 u ) + 11(1.008 665 u ) − ( 22.994 127 u ) = 0.195 088 u. The
difference in the binding energy per nucleon for these two isobars is then
(a)
(a)
2
ΔEb ⎡⎣( Δm ) Na − ( Δm ) M g ⎤⎦ c
[ 0.200−19285 u2 − 0.195 088 u ](931.5 MeV u )
1.60
)
(
= (
= × 10 C )
= 3.4223× 10 7 m s = 1.9 × 10 7 m s
A
A−27
−14
6.64
×
10
kg
3.2
×
10
m
(
)(
)
= 0.210 MeV nucleon
23
11
The binding energy per nucleon is greater by 0.210 MeV nucleon in 1123 Na. This is attributable
to less proton repulsion
in 1123 Na than
in 1223 Mg
.
MeV
(
) = 7.93
(b)
(b)
29.13
.
23
12
(
2
The total binding energy for 24
12 Mg is Eb = ( Δm ) c = 12m
age binding energy per nucleon is
(a)
1
H
+ 12mn − m
) c , and the aver2
24
Mg
Eb ⎡⎣12 (1.007 825 u ) + 12 (1.008 665 u ) − 23.985 042 u ⎤⎦ ( 931.5 MeV u )
=
A
24
= 8.26 MeV nucleon
(
85
Rb, Eb = ( Δm ) c 2 = 37m
For 37
(b)
1
H
+ 48mn − m
) c , yielding
2
85
Rb
⎡3732
825 u ) 200
+ 48 (1.008
789 u ⎤⎦ ( 931.5 MeV u )
(1.00758.933
0.555665
355u ) − 84.911
8.768
E27
b
=⎣
A
85
Therefore, 56
Fe
has
a
greater
binding
energy
per
nucleon
than its neighbors. This gives us finer
26
= 8.70
MeV nucleon
detail than is shown
in Figure
29.4.
59
27
.
29.15
Co
From R = lN = lN 0 e − l t = R0 e − l t , with R = ( 0.842 ) R0 , we find e − l t = R R0 , and lt = − ln ( R R0 ) .
Since the half-life may be expressed as T1 2 = ln 2 l, this yields
t ln 2
( 2.00 d ) ln 2
T1 2 = − ( ) ln 2 = −
= 8.06 d
4
ln ( R R0 =
) (1.1×ln10( 0.842
Bq ))e − ( 2.0 h 6.05 h ) ln 2 = 8.7 × 10 3 Bq
29.16
.
29.17
− lt
The activity
is R activity
= lN = is
lNR00e=
= R0mCi,
e − l t , where
the h,
activity
at time
= 0, and
the decay
(a)
The initial
10.0
and at R
t=
4.00
R = 8.00
mCi.t Then,
from
0 is
− lt
− lt
constant
l ==lnlN
2 0Te1 2 = R0 e , the decay constant is
R =islN
l=−
ln ( R R0 )
t
=−
and the half-life is T1 2 =
(b)
ln ( 0.800 )
= 5.58 × 10 −2 h −1
4.00 h
ln 2
ln 2
=
= 12.4 h
l
5.58 × 10 −2 h −1
−3
10
−1
R0 (10.0 × 10 Ci ) ( 3.70 × 10 s 1 Ci )
=
= 2.39 × 1013 nuclei
l
(5.58 × 10−2 h−1 ) (1 h 3 600 s)
( ) = − T ⎡ ln (−R5.58R×100 ) ⎤ h= −( 30(h5) 730 yr ) ⎡ ln ( 0.125) ⎤ = 1.72 × 10 4 yr
⎥ )
1 2 ⎢
⎢ ln 2 ⎥
R = R0 e −ll t = (10.0 mCi
= 1.9 mCi
)e (
⎦
⎣
⎣⎢ ln 2 ⎦⎥
N0 =
−2
(c)
.
29.19
The mass of radon present at time t is equal to m = matom N = matom N 0 e − l t = m0 e − l t , where matom is
the mass of a single radon atom, N is the number or radon nuclei (and hence, atoms) present, and
N 0 is the number present at time t = 0, making m0 the mass of radon present at t = 0. The decay
constant for radon is l = ln 2 T1/ 2 = ln 2 (3.83 d), yielding
− t
m = m0 e − l t = m0 e (
67352_ch29.indd 821
−1
)
T1 2 ln 2
= ( 3.00 g ) e − (1.50 d 3.83 d ) ln 2 = 2.29 g
2/9/11 2:36:32 PM
− t T
ln 2
= e − l t = e ( ) = e − (123.3 yr 12.33 yr ) ln 2 = e − (10.0 ) ln 2 = 9.77 × 10 − 4 .
12
(d)
29.5
.
29.21
(a)
(a)
(b)
(b)
(c)
(d)
No. The decay model depends on large numbers of nuclei. After some long but finite time,
235
Nuclear
821
only one undecayed nucleus will remain. It is likely that the decay of this
finalPhysics
nucleus will
occur before infinite time.
4
⎛ 8.64
× 10approach,
s⎞
At the point of
closest
PE f = 5KEi , so ke ( 2e ) ( 79e ) rm in
T1 2 = 8.04 d ⎜
⎟⎠ = 6.95 × 10 s
1d
⎝
( Δm ) c 2 ( 0.002 388 u ) ( 931.5 MeV u )
=
=
= 1.11 MeV nucleon
Aln 2
2 −1
ln 2
−7
l=
=
s
=
9.97
×
10
T1 2 6.95 × 10 5 s
4
For 2 He, Δm = 2 (1.007 825 u ) + 2 (1.008 665 u ) − ( 4.002 603 u ) = 0.030 377 u, and
⎛ 3.7 × 1010 Bq ⎞
4
R = 0.500 mCi = ( 0.500 × 10 −6 Ci ) ⎜
⎟⎠ = 1.9 × 10 Bq
1 Ci
⎝
From R = lN, the number of radioactive nuclei in a 0.500 mCi of 131 I is
R
1.9 × 10 4 s −1
=
= 1.9 × 1010 nuclei
l 9.97 × 10 −7 s −1
N=
(e)
The number of half-lives that have elapsed is n = t T1 2 = 40.2 d 8.04 d = 5.00 , so the
remaining activity of the sample is
R0
R0
6.40 mCi
= 5.00
=
= 0.200 mCi
n
32.0
2
2
R=
29.22
29.23
.
Nuclear Physics
825
90
3
0
The
Sr 3nuclei
initially
present is
In
thenumber
decay 1of
H38→
2 He + −1 e +n e , the antineutrino is massless. Adding 1 electron to each side of
3
−
the decay gives ( 1 H + e ) → ( 32 He +2 e − ) +n e , or 13 H atom → 32 Heatom +n e . Therefore, using neutral
atomic masses from Appendix B, the energy released is
(
E = ( Δm ) c 2 = m
3
H
−m
3
He
)c
2
= ( 3.016 049 u − 3.016 029
( u ) (931.5 MeV
) u)
==0.018
MeV = 18.6 keV
4.28 6MeV
29.25
.
From R = lN = lN 0 e − l t = R0 e − l t, and T1 2 = 5 730 yr for 14 C (Appendix B), the age of the
sample is
t=−
29.26
.
29.27
ln ( R R0 )
λ
40
20
ln 2
= − ( 5 730 yr )
Ca + 20e − ) → e + + ( 1940 K + 19e − ) + e −
(
Then, Q = m
or
(b)
40
20 C a atom
−m
40
19 K atom
ln ( 0.600 )
= 4.22 × 10 3 yr
→
ln 2
14
7
N+
0
−1
e +n
or
40
Ca atom → e + + 40 K atom + e −
)
− 2me c 2 = ⎡⎣39.962 591 u − 39.963 999 u − 2 ( 0.000 549 u ) ⎤⎦ c 2
Q = ( −0.002 506 u ) c 2 < 0
so the decay cannot occur spontaneously .
In the decay 144
Nd →42 He + 140
Ce, we may add 60 electrons to each side, forming all neutral
60
58
atoms, and use masses from Appendix B to find
(
Q= m
or
67352_ch29.indd 822
ln ( R R0 )
2
We
thea decay
caseamount
by requiring
thatreleased
both the in
total
number
and isthe
Thecomplete
Q-value of
decay,formula
Q = ( Δmin) ceach
, is the
of energy
themass
decay.
Here, Δm
total
charge number
be the
on the
the original
two sides
of the and
equation.
the difference
between
the same
mass of
nucleus
the total mass of the decay products.
If Q > 0, the decay may occur spontaneously.
(a)
+
40
(a) For the decay 40
20 Ca → e + 19 K, the masses of the electrons do not automatically cancel.
Thus, we add 20 electrons to each side of the decay to yield neutral atoms and obtain
(
29.28
= − T1 2
144
60
Nd
−m
4
2 He
−m
) c = (143.910 083 u − 4.002 603 u − 139.905 434 u) c
2
140
58 C e
2
Q = ( +0.002 046 u ) c 2 > 0 so the decay can occur spontaneously .
(a)
2/9/11 2:36:36 PM
(
236
824
29.29
.
29.20
Chapter 29
29.32
.
29.33
= 0.186 MeV = 186 keV
=case
Δm (by
931.5
MeV remaining
u ) Aboth the
We
the missing
nuclide
in each
that
mass
(a) identify
From Equation
(29.4a)
in the
textbook,
therequiring
fraction
at ttotal
= 5.00
yr number
will be and the
total charge number be the same on the two sides of the decay equation.
(⎤in MeV ) 144 4
95
4
140
Kr → 37
Rb + −10 e +n ⎦ = (c)
2.81× 10
yr → 2 He + 58 Ce
60 Nd
ln 2
The initial activity of
the 1.00-kg carbon sample would56 have been
The more massive 56
27 Co decays into the less massive 26 Fe. To conserve charge, the charge of the
emitted particle must be +1e. Since the parent and the daughter have the same mass number, the
emitted particle must have essentially zero mass. Thus, the decay must be positron emission or
56
0
e + decay . The decay equation is 56
27 Co → 26 Fe + +1 e + n e .
(a)
29.30
.
29.31
)
212
83
Bi →
⎡
Tl + 42 ⎣He
208
81
(b)
95
36
We complete the reaction equation in each case by requiring that both the total mass number and
We identify the missing particles by requiring that both the total mass number and the total
the total charge number be the same on the two sides of the equation.
charge number be the same on the two sides of the equation and by remembering that some form
of neutrino always accompanies the emission of a beta particle or positron.
(a)
21
10
(a)
(b)
235
92
Ne + 42 He → 1224 Mg + 10 n
U + 10 n → 90
Sr +
38
144
54
Xe + 2 10 n
(8.99 × 109 N ⋅ m 2 C2 ) (1.60 × 10 −19 C) ⎛ 1 MeV ⎞ = 1.88 × 10 −15 m
2 11 H → 12 H + =+10 e + n e
⎜⎝
⎟
1.60 × 10 −13 J ⎠
( 6.258 − 5.494 ) MeV
2
(c)
29.34
.
29.35
(a)
(a)
(b)
For the first reaction:
Determine the product of the reaction by requiring that both the total mass number and the
total charge number be the same on the two sides of the equation. The completed reaction
equation is 73 Li + 42 He → 105 B + 01 n.
(
Q = Δmc 2 = ⎡⎢ m
⎣
7
3 Li
+m
4
2 He
) − (m
10
5B
+m
1
0n
)⎤⎦⎥ c
2
= ⎡⎣( 7.016 004 u + 4.002 603 u ) − (10.012 937 u + 1.008 665 u ) ⎤⎦ ( 931.5 MeV u )
⎛
mp ⎞
⎛
1.007 825 u ⎞
⎛
⎞
−1.64 MeV = 1.88 MeV
Q = ⎜ 1+
⎟ Q = ⎜ 1+
⎜
⎟
= −2.79
⎝ MeV⎠
⎜⎝
⎟
7.016 004 u ⎟⎠
m Li ⎠
⎝
7
3
29.37
.
29.36
(a)
(a)
Requiring that both charge and the number of nucleons (atomic mass number) be conserved,
the reaction is found to be 197
Au +10 n → 198
Hg + −10 e +n e . Note that the antineutrino has been
79
80
included to conserve electron-lepton number, which will be discussed in the next chapter.
(b)
We add 79 electrons to both sides of the reaction equation given above to produce neutral atoms
1
198
so we may use mass values from Appendix B. This gives 197
79 Au atom + 0 n → 80 Hgatom +n e , and,
remembering that the antineutrino is massless, the Q-value is found to be
(
Q = ( Δm ) c 2 = m
197
79 Au
+ mn − m
198
80 Hg
)c
2
= (196.966 552 u + 1.008 665 u − 197.966 750 u ) ( 931.5 MeV u ) = 7.89 MeV
The kinetic energy carried away by the daughter nucleus is negligible. Thus, the energy
released may be split in any manner between the electron and antineutrino, with the
maximum kinetic energy of the electron being 7.89 MeV .
29.39
.
We determine the product nucleus by requiring that both the total mass number and the total
charge number be the same on the two sides of the reaction equation. The completed reaction
equations are given below:
(a)
67352_ch29.indd 823
10
5
B + 42 He → 11 H +
13
6
C
(b)
13
6
C + 11 H → 42 He +
10
5
B
2/9/11 2:36:39 PM
0.13 rem yr
.
29.41
29.5
≈ 38 times background levels
Nuclear Physics
237
821
2
J of energy
each
kilogram
For each
rad point
of radiation,
10 −approach,
(a)
At the
of closest
PE f is= delivered
KEi , so keto( 2e
) ( 79e
) rm in of absorbing material.
Thus, the total energy delivered in this whole body dose to a 75.0-kg person is
− t T
ln 2
== e − l t = e ( ) = e − ( 5.00 yr 12.33 yr ) ln 2 = 0.755
100 mrad h
J kg ⎞
⎛
m ) kg ) = 18.8 J = 3.2 m
E = ( 25.0 rad ) ⎜ 10 − 2 = (1.0
⎟⎠ ( 75.0
⎝
10
mrad h
rad
− t T
ln 2
(b) At t = 10.0 yr,
= e − l t = e ( ) = e − (10.0 yr 12.33 yr ) ln 2 = 0.570 .
(a) rate
Theofdose
(in rem)
received
in time
Δt is given
by
The
delivering
energy
to each
kilogram
of absorbing
material is
(c) At t = 123.3 yr,
J kg
⎛ − 2 J kg ⎞
⎛ E m⎞
⎟ = 0.10
⎜⎝
⎟ = (10 rad s ) ⎜⎝ 10
rad ⎠
s
Δt ⎠
12
12
29.42
.
29.43
The total energy needed per unit mass is
⎛
J ⎞
E m = c ( ΔT ) = ⎜ 4 186
( 50°C ) = 2.1 × 105 J kg
kg ⋅°C ⎟⎠
⎝
so the required time will be
Δt =
29.44
.
29.47
⎞
⎟ = 24 d
s⎠
− lt
For equal
amounts
biological
the two doses (in rem units) must be equal, or
From
R=R
, theofelapsed
timedamage,
is
0e
t=−
29.48
.
29.49
5
2.1 ×J10kg
1d
energy needed 2.00
J kg
⎛
s⎜
2.1××10
10−64 °C
= =
4.78
= =
⎝
8.64 × 10 4
deliverycrate 4 186
0.10
J ⋅°C
kg ⋅s
J kg
(a)
ln ( R R0 ()
l
ln (d20.0
60.6 200
2 000mCi
) ln (mCi
) lnT( R lnR ()R R ) = − (d8.04
) = ) 40.6
= − T1 =2 − 1 2 0 = −0 (14.0
= 46.5
)
d d
ln 2 ln 2
l
ln 2ln 2
40
11
K atom is
The mass of a single
6 C atom is
matom = (11.011 u)(1.661× 10 −27 kg u) = 1.829 × 10 −26 kg = 1.829 × 10 −23 g
and the mass of 1 mole of 116 C is
M = matom NA = (1.829 × 10 −23 g atom ) ( 6.022 × 10 23 atoms mol ) = 11.01 g mol
The number of moles in a 3.50 mg sample is then
n=
(b)
msam ple
M
=
3.50 × 10 −6 g
= 3.18 × 10 −7 mol
11.01 g mol
The number of nuclei in the original sample is
N0 = nNA = ( 3.18 × 10 −7 mol ) ( 6.022 × 10 23 atoms mol ) = 1.91× 1017 atoms
or alternatively, N0 =
msam ple
matom
=
3.50 × 10 −6 g
Nuclear
Physics
.
= 1.91× 1017 atoms
1.829 × 10 −23 g atom
831
(c) The initial activity is
continued on next page
⎛ ln 2 ⎞
ln 2
17
14
R0 = lN 0 = ⎜
⎟ N 0 = 20.4 min 60.0 s min (1.91× 10 ) = 1.08 × 10 Bq
T
(
)
(
)
⎝ 12 ⎠
(d)
The activity after an elapsed time of t = 8.00 h = 480 min will be
R = R0 e − l t = R0 e
29.50
67352_ch29.indd 824
− t ln 2 T1 2
= (1.08 × 1014 Bq ) e − ( 480 m in ) ln 2 ( 20.4 m in ) = 8.92 × 10 6 Bq
From R = lN = lN0 e − l t, if we have an activity of R after time t has passed, the number
of unstable nuclei that must have been present initially is given by N0 = Re l t l
2/9/11 2:36:45 PM
238
824
29.20
29.51
.
⎛
⎜⎝
Chapter 29
(a)
(a)
87 be
From
Equation
the
theoffraction
at tthe
= 5.00
n e , textbook,
If
we assume
all(29.4a)
the 87 Srinnuclei
in a gram
materialremaining
came from
decayyrofwill
Rb nuclei,
87
10
9
the original number of Rb nuclei was N0 = 1.82 × 10 + 1.07 × 10 = 1.93 × 1010. Then,
from N = N 0 e − l t, the elapsed time is
t =−
(b)
(c)
.
29.55
⎞ ⎛ 10 6 mg ⎞
⎟⎠ ⎜⎝ 1 kg ⎟⎠ = 12 mg
T ln ( N N 0 )
ln ( N N 0 )
=− 12
=−
l
ln 2
(4.8 ×10
10
⎛ 1.82 ×1010
yr ) ln ⎜
10
⎝ 1.93×10
ln 2
⎞
⎟
⎠
= 4.1×10 9 yr
It could be no older. It could be younger if some 87 Sr were initially present.
⎛
⎞
⎛ 1.008 665 u ⎞
= ⎜all
1+ 87 Sr present
(17.6 MeV ) = 22.0 MeV 87
⎜ assumed
⎟ Qthat
We have
⎜⎝
⎟⎠
4.002 603 u ⎟⎠ came from the decay of Rb.
⎝ the
The decay constant for 235 U is l235 = ln 2 T1 2 = ln 2 0.70 × 10 9 yr = 9.9 × 10 −10 yr −1 , while
that for 238 U is l238 = ln 2 4.47 × 10 9 yr = 1.55 × 10 −10 yr −1 . Assuming there were N 0 nuclei of
each isotope present initially, the number of each type still present should be N 235 = N 0 e − l t
and N 238 = N 0 e − l t . With the currently observed ratio of 235 U to 238 U being 0.007, we have
− l − l )t
= 0.007, or our estimate of the elapsed time since the release of the elements
N 235 N 238 = e (
forming our Earth is
235
238
235
t=−
29.56
67352_ch29.indd 825
(a)
238
ln ( 0.007 )
ln ( 0.007 )
=−
= 5.9 × 10 9 yr
l235 − l236
9.9 × 10 −10 yr −1 − 1.55 × 10 −10 yr −1
Obtaining the mass of a single 239
94 Pu atom from the table of Appendix B in the text, we find
2/9/11 2:36:48 PM
장 핵߾οएࠪ‫ܕ‬վ࢕
PROBLEM SOLUTIONS
30.1
30.1
(a)
With a specific gravity of 4.00, the density of soil is r = 4.00 × 10 3 kg m 3. Thus, the mass
of the top 1.00 m of soil is
2
kg ⎞ ⎡
⎛ 1m ⎞ ⎤
⎛
m = rV = ⎜ 4.00 × 10 3
1.00 m )( 43 560 ft 2 ) ⎜
= 1.62 × 10 7 kg
(
⎟
3 ⎟ ⎢
⎝ 3.281 ft ⎠ ⎥⎦
⎝
m ⎠⎣
At a rate of 1 part per million, the mass of uranium in this soil is then
mU =
(b)
m 1.62 × 10 7 kg
=
= 16.2 kg
10 6
10 6
235
Since 0.720% of naturally occurring uranium is 235
92 U, the mass of 92
( U in the soil of) part (a) is
= ( 7.20 × 10 −3 ) mU = ( 7.20 × 10 −3 ) (16.2 kg ) = 0.117 kg = 117 g
m MeV
= 192
U
235
92
30.3
30.2
The energy released in the reaction 10 n +
Q = ( Δm ) c 2 = ⎡ m
⎣
235
92 U
− 11mn − m
88
38 Sr
235
92
U →
−m
136
54 Xe
88
38
1
Sr + 136
54 Xe + 12 0 n is
⎤ c2
⎦
= ⎡⎣ 235.043 923 u − 11(1.008 665 u ) − 87.905 614 u − 135.907 220 u ⎤⎦ ( 931.5 MeV u )
838
30.4
30.3
30.5
Chapter 30
= 126 MeV
Withenergy
a power
output ofbyPouta 100-W
= 1.00 ×lightbulb
10 9 W and
ciency
e = 0.400,
The
consumed
in aeffi
1.0-h
timeofperiod
is the total energy input
required each day is
⎛ 3 600 s ⎞
E = P ⋅ Δt = (100 J s )(1.0 h ) ⎜
= 3.6 × 10 5 J
⎝ 1 h ⎟⎠
The number of fission events, yielding an average of 208 MeV each, required to produce this
quantity of energy is
5
3.6
E⎞
MeV ⎞ ⎛
⎛ × 10
2.4 J×⎛⎜10 24 1atoms
=
n= ⎛
1.1× g1016⎞ = 9.4 × 10 2 g
⎟⎠ ⎜=235
−13
=
M
⎟
m
ol
⎝
23
⎟
⎜⎝ 6.02
⎜⎝ MeV
⎟⎠
208
208
MeV
1.60
×
10
J
⎝
mol ⎠
× 10 atoms mol ⎠
30.4
30.7
30.6
The total energy required
for one year is
released was
E = ( 2 000 kWh month ) ( 3.60 × 10 6 J kWh )(12.0 months ) = 8.64 × 1010 J
The number of fission events needed will be
N=
E
Eevent
=
8.64 × 1010 J
= 2.60 × 10 21
−13
( 208 MeV ) (1.60 × 10 J MeV )
and the mass of this number of 235 U atoms is
⎛ N ⎞
⎛ 2.60 × 10 21 atoms ⎞
m=⎜
=
M
m ol
⎜⎝ 6.02 × 10 23 atoms mol ⎟⎠ ( 235 g mol ) = 1.01 g
⎝ N A ⎟⎠
30.8
67352_ch30.indd 833
(a)
The mass of 235 U in the reserve is
239
2/9/11 2:39:56 PM
840
240
30.11
30.5
Chapter 30
(a)
4
2
He + 42 He →
8
4
(b)
8
4
Be + 42 He →
12
6
(c)
(b)
30.12
30.13
30.6
Be + g
C +g
(
)
The total energy released in this pair of fusion reactions is
= 8.68 MeV
Q = ( Δm ) c 2 = ⎡⎣3m He − m C ⎤⎦ c 2
The proton and the boron
nucleus both have positive charges but must come very close to
one another
in
order
for
fusion
to occur.
theyMeV
mustuhave
suffiMeV
cient kinetic energy to
⎤⎦ ( 931.5
= ⎡3 ( 4.002 603 u ) − 12.000
000 uThus,
) = 7.27
overcome⎣the repulsive Coulomb force one
exerts on the other.
4
12
(a) The energy released in the reaction 11 H + 115 B → 3( 42 He) is
The energy released in the reaction 21 H + 21 H → 31 H + 11 H is Q = ⎡ 2m
⎣
2
1H
−m
3
1H
− m H ⎤ c 2, or
⎦
1
1
Q = ⎡⎣ 2 ( 2.014 102 u ) − 3.016 049 u − 1.007 825 u ⎤⎦ ( 931.5 MeV u+) =42 He
4.03 MeV
30.14
30.7
30.15
(a)
With the deuteron and triton at rest, the total momentum before the reaction is zero. To
conserve momentum, the neutron and the alpha particle must move in opposite directions with
equal magnitude momenta after the reaction, or pa = pn . Neglecting relativistic effects, we
use the classical relationship between momentum and kinetic energy, KE = p2 2m, and write
2ma KEa = 2mn KEn , or KEa = (mn ma )KEn .
To conserve energy, it is necessary that the kinetic energies of the reaction products satisfy
the relation KEn + KEa = Q = 17.6 MeV. Then, using the result from above, we have
KEn + (mn ma )KEn = 17.6 MeV, or the kinetic energy of the emerging neutron must be
KEn =
30.16
30.8
30.17
17.6 MeV
= 14.1 MeV
1+ (1.008 665 u ) ( 4.002 603 u )
(
) = 5.49 MeV
The energy released in the reaction 11 H + 12 H → 32 He + g is
Note that pair production cannot occur in a vacuum. It must occur in the presence of a massive
particle which can absorb at least some of the momentum
of Physics
the photon
allow total
linear 841
Nuclear
and and
Elementary
Particles
momentum to be conserved.
When a particle-antiparticle pair is produced by a photon having the minimum possible frequency,
continuedand
on next
hencepage
minimum possible energy, the nearby massive particle absorbs all the momentum of
the photon, allowing both components of the particle-antiparticle pair to be left at rest. In such an
event, the total kinetic energy afterwards is essentially zero, and the photon need only supply the
total rest energy of the pair produced.
The minimum photon energy required to produce a proton-antiproton pair is Ephoton = 2 ( ER )proton.
Thus, the minimum photon frequency is
f=
and l =
p
30.9
30.19
30.18
Ephoton
h
2 ( ER )proton
h
=
2 ( 938.3 MeV )(1.60 × 10 −13 J MeV )
6.63 × 10 −34 J ⋅s
= 4.53 × 10 23 Hz
c 3.00 × 108 m s
=
= 6.62 × 10 −16 m = 0.662 fm
=f KEtotal
213 MeV − 95.0 MeV = 118 MeV
4.53− ×KE
10p 23= Hz
The total rest
energy
of the
p 0the
is converted
into energy
of the photons. Since the total momentum
kinetic
energy
after
pair production
is
was zero before the decay, the two photons must go in opposite directions with equal magnitude
momenta (and hence equal energies). Thus, the rest energy of the p 0 is split equally between the
two photons, giving for each photon
Ephoton =
67352_ch30.indd 837
=
ER ,p
2
0
=
135 MeV
= 67.5 Mev
2
2/9/11 2:40:07 PM
135 MeV
= 67.5 Mev
2
=
44.
Nuclear Physics and Elementary Particles
241
837
(a)
conservation
Ephotonof charge
pphoton =
= 67.5 MeV c
c
(b) conservation of electron-lepton number and conservation of muon-lepton number
Ephoton
67.5 MeV ⎛ 1.60 × 10 −13 J ⎞
= 1.63 × 10 22 Hz
and conservation
f=
(c)
of=baryon number
6.63 × 10 −34 J ⋅s ⎜⎝ 1 MeV ⎟⎠
h
30.20
.
(a)
Reaction
p + p → m + + e−
(b)
p +p→ p+p
(c)
p+p→ p+p +
−
+
30.22
30.23
.
Le : ( 0 + 0 → 0 + 1); and L m : ( 0 + 0 → −1+ 0 )
Charge, Q:
( −1+ 1 → + 1+ 1)
(1+ 1 → 1+ 0 )
Baryon Number, B:
(d) p + p → p + p + n
(e)
Conservation Law Violated
g +p→ n+p 0
Baryon Number, B:
Charge, Q:
(1+ 1 → 1 + 1 + 1)
(0 + 1 → 0 + 0)
via any interaction.
Observe0 that the given
reactions involve only mesons and baryons. With no leptons before or after
−
(a) reactions,
Λ → p we
+ pdo
the
not have to consider the conservation laws concerning the various lepton
numbers. All interactions always conserve both charge and baryon numbers. The strong interaction
Strangeness,
S:
0+0
⇒ ΔS ≠by
0 the weak
Not interaction
Conserved
also conserves
strangeness.
Conservation−1of→strangeness
may be violated
but never− by more 0than one
unit. With these facts in mind consider the given interactions:
(b) p + p → Λ + K 0
Strangeness, S:
(c)
0 + 0 → +1− 1
⇒ ΔS = 0
Conserved
0 + 0 → 0 −1
Nuclear Physics and Elementary Particles
843
⇒ ΔS ≠ 0
Not Conserved
−2 → −1+ 0
⇒ ΔS ≠ 0
Not Conserved
−2 → 0 + 0
⇒ ΔS ≠. 0
Not Conserved
Ξ− → Λ 0 + p −
Strangeness, S:
(f)
Conserved
p − + p → p − + Σ+
Strangeness, S:
(e)
⇒ ΔS = 0
p + p → Λ0 + Λ0
Strangeness, S:
(d)
0 + 0 → −1+ 1
Ξ0 → p + p −
Strangeness, S:
30.25
.
Reaction
Conservation Law
Conclusion
Baryon Number - violated: ( 0 + 1 → 0 )
Cannot Occur
(a)
p + p → 2h
(b)
K − + n → Λ0 + p −
All conservation laws observed.
May occur via the Strong
Interaction
(c)
K →p +p
0
Strangeness violated by 1 unit
( −1 → 0 + 0 ). All other conservation
laws observed.
Can occur via Weak
Interaction, but not
by Electromagnetic or
Strong Interactions.
(d)
Ω → Ξ +p
0
Strangeness violated by 1 unit
( −3 → −2 + 0 ). All other conservation
laws observed.
Can occur via Weak
Interaction, but not
by Electromagnetic or
Strong Interactions.
−
−
−
−
−
0
67352_ch30.indd 838
2/9/11 2:40:09 PM
continued on next page
242
840
30.11
Chapter 30
(e)
Strangeness violated by 1 unit
J ⎛→ −2 1+ yr
0 ). All ⎞⎟other
conservation
Ω → Ξ =+ p2.6 × 10 ( −3
= 5.5
yr
⎜
× 10 7 observed.
s⎠
1.5 × 1013 J s ⎝ 3.156 laws
Can occur via Weak
Interaction, but not
by Electromagnetic or
Strong Interactions.
K →p +p
−
0
−
0
21
Fission alone cannot meet the world’s energy needs
h 0 → 2g
(e)
All conservation laws observed.
Allowed via all interactions, but photons are the
mediator of the electromagnetic interaction and
thecontinued
lifetime of
h 0 page
is
onthe
next
consistent with decay by
that interaction.
.
(a)
Compare the given quark
⎛ 2 states
⎞ ⎛ to 2the⎞ entries
⎛ 1 in⎞ Table 30.4:
uud: charge = ⎜ − e ⎟ + ⎜ − e ⎟ + ⎜ + e ⎟ = − e . This is the antiproton .
⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3 ⎠
(a) suu = Σ +
(b) ud = p −
(b)
(c)
.
30.29
Can occur via Weak
Interaction, but not
by Electromagnetic or
Strong Interactions.
−
(a)
(d)
30.26
.
30.27
Strangeness violated by 1 unit
( −1 → 0 + 0 ). All other conservation
laws observed.
−
sd = K 0
(d) ssd = Ξ −
0
The number of water molecules in one liter (mass = 1 000 g) of water is
⎛ 1 000 g ⎞
N =⎜
(6.02 × 1023 molecules mol ) = 3.34 × 10 25 molecules
⎝ 18.0 g mol ⎟⎠
Each molecule contains 10 protons, 10 electrons, and 8 neutrons. Thus, there are
N e = 10 N = 3.34 × 10 26 electrons , N p = 10 N = 3.344 ×10 26 protons,
and
N n = 8 N = 2.68 × 10 26 neutrons
Each proton contains 2 up quarks and 1 down quark, while each neutron has 1 up quark and 2
down quarks. Therefore, there are
N u = 2 N p + N n = 9.36 × 10 26 up quarks , and N d = N p + 2 N n = 8.70 × 10 26 down quarks
0 .
.
30.31
The reaction is Σ 0 + p → Σ + + g + X, or on the quark level, uds + uud → uus + 0 + x.
The left side has a net 3 u, 2 d, and 1s. The right side has 2 u, 0 d, and 1s, plus the quark composition
of the unknown particle. To conserve the total number of each type of quark, the composition of the
unknown particle must be x = udd . Therefore, the unknown particle must be a neutron
.
.
30.32
.
30.37
With the kaon initially at rest, the total momentum was zero before the decay. Thus, to conserve
If a neutron starts with kinetic energy KEi = 2.0 MeV and loses one-half of its kinetic energy in
momentum, the two pions must go in opposite directions and have equal magnitudes of momeneach collision with a moderator atom, its kinetic energy after n collisions will be KE f = KEi 2n.
tum after the decay. Since the pions have equal mass, this means they must have equal speeds and
hence, equal energies. The rest energy of the kaon is then split equally between the two pions, and
The average kinetic energy associated with particles in a gas at temperature T = 20.0°C = 293 K
the energy of each is
(see Chapter 10 of the textbook) is
KE f =
3
3
1 eV
⎛
kB T = (1.38 × 10 −23 J K )( 293 K ) ⎜
⎝
2
2
1.60 × 10 −19
⎞
⎟ = 0.0379 eV
J⎠
Thus, the number of collisions the neutron must make before it reaches the energy associated
with a room temperature gas is n ln 2 = ln(KEi KE f ), or
n=
67352_ch30.indd 839
1
ln 2
2/9/11 2:40:12 PM
Nuclear Physics and Elementary Particles
837
243
), or
44.
(a)
.
30.39
30.38
1
To the
reaction
nuclei,
+0 is32 He → 42 He + +01 e + n e , we add three electrons to both sides to
(c)
conservation
baryon
number
The
reaction
p −for
+ pof
→
K 0 1+HΛ
1
3
4
obtain 1 H atom + 2 Heatom → 2 Heatom + −01 e + +01 e + ne. Then we use the masses of the neutral atoms
from Appendix B of the textbook to compute
conservation of charge6
s quark afterwards . This reaction does not
⎛ 1 ⎞ ⎛ 2.0 × 10 eV ⎞
= 26
n=⎜
⎟⎠ lnnumber
⎟⎠type
⎜⎝ 0.0379
⎝the
conserve
net
of
each
quark.
ln
2
eV
(b) conservation of electron-lepton number and conservation of muon-lepton number
Q = ( Δm ) c 2 = ⎡ m
⎣
1
1H
+m
3
2
He
−m
4
2
He
− 2 me ⎤ c 2
⎦
= ⎡⎣1.007 825 u + 3.016 029 u − 4.002 603 u − 2 ( 0.000 549 u ) ⎤⎦ ( 931.5 MeV u )
⎞
3.61× 10 30 J ⎛
1 yr
= 1.63 × 108 yr
=
= 18.8 MeV7.00 × 1014 J s ⎜⎝ 3.156 × 10 7 s ⎟⎠
30.41
.
Conserving energy in the decay p − → m − +n m , with the p − initially at rest gives E m + En = E R ,p , or
−
E m + En = 139.6 MeV
[1]
Since the total momentum was zero before the decay, conservation of momentum requires
the muon and antineutrino to recoil in opposite directions with equal magnitude momenta,
or pm = pn . The relativistic relation between total energy and momentum of a particle gives
for the antineutrino:
En = pn c, or pn = En c. The same relation applied to the muon gives
2
2
E m2 = pm c + ER2 , m . Since we must have pm = pn , this may be rewritten as E m2 = ( pn c ) + E R2 , m , and
using the fact that
pn c = En , we have E m2 = En 2 + E R2 , m . Rearranging and factoring then gives
( )
(
E m2 − En2 = E m + En
)(E
m
)
− En = E R2 , m = (105.7 MeV )
2
and
E m − En =
(105.7 MeV )2
[2]
E m + En
Substituting Equation [1] into [2] gives E m − En = (105.7 MeV)2 139.6 MeV, or
Nuclear Physics and Elementary Particles
E m − En = 80.0 MeV
849
[3]
Subtracting Equation [3] from Equation [1] yields 2En = 59.6 MeV, and
continued on next
E page
= 29.8 MeV
$174 000.
n
.
30.43
To compute the energy released in each occurrence of the reaction
4p + 2e − → a + 2n e + 6g
we add two electrons to each side to produce neutral atoms and obtain
4( 11 H atom ) → 42 Heatom + 2n e + 6g . Then, recalling that the neutrino and the photon both have zero
rest mass, and using the neutral atomic masses from Appendix B in the textbook gives
Q = ( Δm ) c 2 = ⎡ 4m
⎣
1
1 H atom
−m
4
2 He atom
⎤ c2
⎦
= ⎡⎣ 4 (1.007 825 u ) − 4.002 603 u ⎤⎦ ( 931.5 MeV u )
= 26.7 Mev
Each occurrence of this reaction consumes four protons. Thus, the energy released per proton
consumed is E1 = 26.4 MeV 4 protons
67352_ch30.indd 840
2/9/11 2:40:15 PM
(
840
244
30.11
Chapter 30
)
= 26.7 Mev
Each
(a) occurrence of this reaction consumes four protons. Thus, the energy released per proton
consumed is E1 = 26.4 MeV 4 protons = 6.68 MeV proton.
Therefore, the rate at which the Sun must be fusing protons to provide its power output is
rate =
P
E1
=
3.85 × 10 26 J s ⎛ 1 MeV ⎞
= 3.60 × 10 38 proton s
6.68 MeV proton ⎜⎝ 1.60 × 10 −13 J ⎟⎠
=
2 ( ER )proton
h
=
2 ( 938.3 MeV )(1.60 × 10 −13 J MeV )
6.63 × 10 −34 J ⋅s
= 4.53 × 10 23 Hz
and
67352_ch30.indd 841
2/9/11 2:40:19 PM