SOLUTIONS TO PROBLEMS: CHAPTER 9
P9.1
(b)
At maximum height v  0 , so p  0 .
(a)
Its original kinetic energy is its constant total energy,
Ki 
1 2 1
2
m vi   0.100 kg 15.0 m s  11.2 J.
2
2
At the top all of this energy is gravitational. Halfway up, one-half of it is gravitational and the other
half is kinetic:
1
 0.100 kg v2
2
2  5.62 J
 10.6 m s
0.100 kg
K  5.62 J
v
j
Then p  m v   0.100 kg10.6 m s ˆ
P9.2
For the system of two blocks p  0 ,
(a)
or
pi  pf
Therefore,
0  M vm   3M
Solving gives
  2.00 m s
vm  6.00 m s
(motion toward the left).
(b)
1 2 1
1
kx  M vM2   3M  v32M  8.40 J
2
2
2
FIG. P9.4
p  1.06 kg  m sˆ
j .
P9.3
I   Fdt area under curve
(a)


I
1
1.50  103 s 18 000 N   13.5 N  s
2
(b)
F
13.5 N  s
 9.00 kN
1.50  103 s
(c)
From the graph, we see that Fm ax  18.0 kN
FIG. P9.7
1 2
1
m v1  m gy1 . The rebound speed is given by m gy2  m v22 .
2
2
The impulse of the floor is the change in momentum,
*P9.4
The impact speed is given by
m v2 up  m v1 dow n  m  v2  v1  up
m


2gh2  2gh1 up

 0.15 kg 2 9.8 m s2

0.960 m  1.25 m
 up
 1.39 kg  m s upw ard
P9.5
Take x-axis toward the pitcher
pix  Ix  pfx :
(a)
 0.200 kg15.0 m s   cos45.0  Ix  0.200 kg 40.0 m s cos30.0
Ix  9.05 N  s
piy  Iy  pfy :
 0.200 kg15.0 m s   sin 45.0  Iy   0.200 kg 40.0 m s sin 30.0
I
(b)
1
1
0  Fm   4.00 m s  Fm  20.0 m s  Fm  4.00 m s

2
2
3
ˆ
ˆ
Fm  24.0  10 s  9.05i 6.12j N  s
I

Fm 
P9.6
 9.05ˆi 6.12ˆj N  s

 377iˆ 255ˆj N
Momentum is conserved
10.0  10
3

kg v   5.01 kg 0.600 m s
v  301 m s
P9.7
(a), (b)
Let vg and vp be the velocity of the girl and the
plank relative to the ice surface. Then we may say that vg  vp is the velocity
of the girl relative to the plank, so that
vg  vp  1.50
(1)
But also we must have m gvg  m pvp  0 , since total momentum of
the girl-plank system is zero relative to the ice surface. Therefore
45.0vg  150vp  0 , or vg  3.33vp
Putting this into the equation (1) above gives
3.33vp  vp  1.50 or vp  0.346 m s
FIG. P9.21
Then vg  3.33 0.346  1.15 m s
P9.8
Energy is conserved for the bob-Earth system between
bottom and top of swing. At the top the stiff rod is in compression and the
bob nearly at rest.
1
M vb2  0  0  M g2
2
vb2  g4 so vb  2 g
Ki U i  K f  U f :
Momentum of the bob-bullet system is conserved in the
FIG. P9.24
collision:
mv m

v
M 2 g
2

v
4M
m
g
P9.9
(a)
First, we conserve momentum for the system of two football players in the x direction (the
direction of travel of the fullback).
 90.0 kg 5.00 m s  0  185 kg V cos
where  is the angle between the direction of the final
velocity V and the x axis. We find
V cos  2.43 m s
(1)
Now consider conservation of momentum of the system in
the y direction (the direction of travel of the opponent).
 95.0 kg 3.00 m s  0  185 kg V sin 
which gives,
V sin   1.54 m s
(2)
Divide equation (2) by (1)
1.54
tan  
 0.633
2.43
  32.3
From which
Then, either (1) or (2) gives
V  2.88 m s
(b)
Ki 
1
1
90.0 kg 5.00 m s2  2 95.0 kg 3.00 m s2  1.55 103 J
2
Kf
1
2
185 kg 2.88 m s  7.67  102 J

2
Thus, the kinetic energy lost is 783 Jinto internalenergy.
P9.10
By conservation of momentum for the system of the two
billiard balls (with all masses equal),
5.00 m s 0   4.33 m s cos30.0  v2 fx
v2 fx  1.25 m s
0   4.33 m s sin 30.0  v2 fy
v2 fy  2.16 m s
v2 f  2.50 m s at  60.0
P9.11
FIG. P9.33
The x-coordinate of the center of mass is
xCM 
0 0 0 0
 m ixi 
m
2.
00
kg

3.
00
kg  2.50 kg  4.00 kg
 i 
xCM  0
and the y-coordinate of the center of mass is
yCM 
 m iyi   2.00 kg  3.00 m    3.00 kg  2.50 m    2.50 kg  0   4.00 kg  0.500 m 
2.00 kg  3.00 kg  2.50 kg  4.00 kg
mi
yCM  1.00 m
P9.12
Let A 1 represent the area of the bottom row of
squares, A 2 the middle square, and A 3 the top pair.
A  A1  A 2  A 3
M  M 1M 2M
3
M1 M

A1
A
A1  300 cm 2 , A 2  100 cm 2 , A 3  200 cm 2 ,
A  600 cm
2
 A  300 cm
M 1  M  1 
 A  600 cm
2
2
2
 A  100 cm
 M  2 
 A  600 cm
2
M
 A 3  200 cm

3  M 
 A  600 cm
2
M
2
2
M 
M
2
M 
M
6
M 
M
3
FIG. P9.41
SOLUTIONS TO PROBLEMS: CHAPTER 10
P10.1
(b)
 t0  5.00 rad
(a)
 t 0 
d
 10.0  4.00tt 0  10.0 rad s
dt t 0
 t 0 
d
 4.00 rad s2
dt t 0
 t3.00 s  5.00  30.0  18.0  53.0 rad
 t 3.00 s 
d
 10.0  4.00tt 3.00 s  22.0 rad s
dt t 3.00 s
 t 3.00 s 
d
 4.00 rad s2
dt t 3.00 s
i 
P10.2
 f  i
(a)
t
(b)
 f  t 
P10.3


100 rev  1 m in   2 rad  10
rad s,  f  0



1.00 m in  60.0 s  1.00 rev 
3
0  103
2.00
s 5.24 s
  f   i
 10
t 
rad

 6
2


  10 
s 
s  27.4 rad
  6 
1
2
 f  i  it  t2 and  f  i  t are two equations in two unknowns i and 
i   f  t:


1
2
1
2
 f  i   f  t t t2   ft t2
1
 2 rad 
2
37.0 rev 
 98.0 rad s 3.00 s    3.00 s

 1 rev 
2


232 rad  294 rad  4.50 s2  :

61.5 rad
 13.7 rad s2
4.50 s2
Given r 1.00 m ,   4.00 rad s2 ,  i  0 and i  57.3  1.00 rad
P10.4
(a)
 f  i  t 0  t
At t 2.00 s ,  f  4.00 rad s2  2.00 s  8.00 rad s
(b)
v  r  1.00 m  8.00 rad s  8.00 m s
ar  ac  r 2  1.00 m  8.00 rad s  64.0 m s2
2


at  r  1.00 m 4.00 rad s2  4.00 m s2
The magnitude of the total acceleration is:
a  ar2  at2 
64.0 m s    4.00 m s 
2 2
2 2
 64.1 m s2
The direction of the total acceleration vector makes an angle  with respect to the radius to point P:
 at 
 4.00
 tan1 
 3.58

 64.0
a 
  tan1 
(c)
P10.5
1
2
 f  i  it t2  1.00 rad 
(b)

s 99.0 m

 341 rad  54.3 rev
r 0.290 m
f 
vf
r


1
2
4.00 rad s2  2.00 s  9.00 rad
2
s vt 11.0 m s  9.00 s  99.0 m
(a)

c
22.0 m s
 75.9 rad s  12.1 rev s
0.290 m
m 1  4.00 kg , r1  y1  3.00 m ;
P10.6
m 2  2.00 kg , r2  y2  2.00 m ;
m 3  3.00 kg , r3  y3  4.00 m ;
  2.00 rad s about the x-axis
(a)
Ix  m 1r12  m 2r22  m 3r32
Ix  4.00 3.00  2.00 2.00  3.00 4.00  92.0 kg  m
2
KR 
(b)
2
2
2
1
1
2
Ix 2   92.0 2.00  184 J
2
2
FIG. P10.20
1
1
2
m 1v12   4.00 6.00  72.0 J
2
2
1
1
2
K 2  m 2v22   2.00 4.00  16.0 J
2
2
1
1
2
K 3  m 3v32   3.00 8.00  96.0 J
2
2
v1  r1  3.00 2.00  6.00 m s
K1 
v2  r2  2.00 2.00  4.00 m s
v3  r3  4.00 2.00  8.00 m s
K  K 1  K 2  K 3  72.0  16.0  96.0  184 J 
1
Ix 2
2
P10.7
Treat the tire as consisting of three parts. The two sidewalls are each treated as a hollow
cylinder of inner radius 16.5 cm, outer radius 30.5 cm, and height 0.635 cm. The tread region is treated as a
hollow cylinder of inner radius 30.5 cm, outer radius 33.0 cm, and height 20.0 cm.
Use I


1
m R12  R22 for the moment of inertia of a hollow cylinder.
2
Sidewall:



2
2
m    0.305 m    0.165 m   6.35  103 m 1.10  103 kg m 3  1.44 kg


1
2
2
Iside  1.44 kg  0.165 m    0.305 m    8.68  102 kg  m 2


2
Tread:

2
2
m    0.330 m    0.305 m    0.200 m  1.10  103 kg m


1
2
2
Itread  11.0 kg  0.330 m    0.305 m    1.11 kg  m 2


2
3
  11.0 kg
Entire Tire:


Itotal  2Iside  Itread  2 8.68  102 kg  m 2  1.11 kg  m 2  1.28 kg  m 2
P10.8
Every particle in the door could be slid straight down into a high-density rod across its
bottom, without changing the particle’s distance from the rotation axis of the door. Thus, a rod 0.870 m long
with mass 23.0 kg, pivoted about one end, has the same rotational inertia as the door:
I
1 2 1
2
M L   23.0 kg  0.870 m   5.80 kg  m
3
3
2
.
The heightofthe dooris unnecessary data.
P10.9
   0.100 m 12.0 N   0.250 m  9.00 N   0.250 m 10.0 N  
3.55 N  m
The thirty-degree angle is unnecessary information.
FIG. P10.31
P10.10
I
1
1
2
m R 2  100 kg  0.500 m   12.5 kg  m 2
2
2
 i  50.0 rev m in  5.24 rad s
 f   i 0  5.24 rad s


 0.873 rad s2
t
6.00 s


  I  12.5 kg  m 2 0.873 rad s2  10.9 N  m
The magnitude of the torque is given by fR  10.9 N  m ,
where f is the force of friction.
10.9 N  m
0.500 m
Therefore,
f
yields
k 
and
f 21.8 N

 0.312
n 70.0 N
f  kn
FIG. P10.38
P10.11
(a)
For the counterweight,
 Fy  m ay
becomes:
 50.0
50.0  T  
a
 9.80
For the reel
   I
TR  I  I
a
R
where
I
reads
1
M R 2  0.0938 kg  m 2
2
We substitute to eliminate the
acceleration:
FIG. P10.45
 TR 2 
50.0  T  5.10

 I 
T  11.4 N
a
and
50.0  11.4
 7.57 m s2
5.10


v2f  vi2  2a xf  xi :
vf  2 7.57 6.00  9.53 m s
(b)
Use conservation of energy for the system of the object, the reel, and the Earth:
 K  U i   K  U  f :
m gh 
1 2 1 2
m v  I
2
2
 v2 
I

2m gh  m v2  I 2   v2  m  2 


R
R
 
v
2m gh

m  I2
R
P10.12
1 2 1 2 1
I
m v  I  m  2  v2
2
2
2
R 
Also, U i  m gh , U f  0 ,
and
K
Therefore,
Thus,
For a disk,
2 50.0 N  6.00 m
5.10 kg 
0.0938
 0.250

where

 9.53 m s
2
v
since no slipping.
R
vi  0
1
I
m  2  v2  m gh

2
R 
2
gh
v2 
1 I 

m R2 

1
I m R 2
2
 
v2 
So
2gh
1 12
vdisk 
or
4gh
3
2gh
or
2
 vring , the disk reaches the bottom first
I m R 2 so v2 
For a ring,
Since vdisk
vring  gh
SOLUTIONS TO PROBLEMS: CHAPTER 11
A  B  3.00 6.00  7.00 10.0   4.00 9.00   124
P11.1
AB 
(a)
 A B 
cos1 
 cos1  0.979  168
 AB 
ˆ
i
(b)
 3.00 2   7.00 2   4.00 2   6.00 2    10.0 2   9.00 2  127
kˆ
ˆ
j
A  B  3.00
7.00 4.00  23.0ˆ
i 3.00ˆ
j 12.0kˆ
6.00 10.0
9.00
A B 
 23.0 2   3.00 2   12.0 2  26.1
 A B 
sin1 
 sin1  0.206  11.9 or 168°
 AB 
(c)
P11.2
Only the firstm ethod gives the angle between the vectors unambiguously.
L   m iviri
  4.00 kg 5.00 m s  0.500 m    3.00 kg 5.00 m s  0.500 m 
L  17.5 kg  m 2 s , and

L  17.5 kg  m
2

s kˆ
FIG. P11.11
P11.3

r  6.00ˆ
i 5.00tˆ
jm
so

v


dr
 5.00ˆ
jm s
dt
p  m v  2.00 kg 5.00ˆ
j m s  10.0ˆ
j kg  m s
ˆ
i
and
P11.4
(a)
kˆ
ˆ
j
L  r p  6.00 5.00t
0 10.0
60.0 kg  m s kˆ
2
0
0
The net torque on the counterweight-cord-spool system is:


  r F  8.00  102 m  4.00 kg 9.80 m s2  3.14 N  m
.
(b)
L  r m v  I
(c)

P11.5
(b)
L  Rm v 
1
M 
 v

M R2    R  m   v 
 R

2
2
0.400 kg  m  v
3.14 N  m
dL
 7.85 m s2
  0.400 kg  m  a a 
0.400 kg  m
dt
(a)
vi  vi xi
zero
At the highest point of the trajectory,
 visin  
v2 sin 2
1
x R  i
and y  hm ax 
2g
2
2g
vi
2

O
v2
R
L1  r1  m v1
FIG. P11.17
2
 v2 sin 2
visin   ˆ

i
ˆ
ˆ


i
j  m vxii
2g
 2g



 m  visin   vicos ˆ

k
2g
2
(c)
v2 sin 2
L2  R ˆ
i m v2 ,w here R  i
g


 m Rˆ
i vicos ˆ
i visin  ˆ
j
 m vi3 sin 2 sin  ˆ
  m Rvisin  kˆ 
k
g
(d)
P11.6
The downward force of gravity exerts a torque in the –z direction.
(a)
Let M  mass of rod and m  mass of each bead. From Iii  If f , we have
1
 12 M
2

1
 2m r12  i   M

 12
2

 2m r22   f

 0.500 m , r1  0.100 m , r2  0.250 m , and with other values as stated in the problem, we
When
find
 f  9.20 rad s .
(b)
Since there is no external torque on the rod,
L  constant and  is unchanged .
P11.7
I   m iri2
(a)
2
2
 4d
 d
 2d
m   m   m  
 3
 3
 3
2d
3
2
m
m
1
2
d2
 7m
3
d
Think of the whole weight, 3mg, acting at the center of gravity.
 
 
 d ˆ
  r F    i
 3m g ˆ
j
 3

3m gd
3g
counterclockw ise
7d
(c)

(d)
2g
 3g  2d
a r     
up
 7d  3 
7
I

2
7m d

 m gd kˆ
The angular acceleration is not constant, but energy is.
 K  U  i  E   K  U  f
1
 d
0   3m  g    0  I 2f  0
 3
2
(e)
maximum kinetic energy  m gd
(f)
f 
(g)
L f  I f 
6g
7d
7m d2
3
6g
 14g 
 
 3 
7d
12
m d3 2
3
d
FIG. P11.45
(b)
m
P
(h)
vf   fr 
6g d

7d 3
2gd
21
P11.8
  r F  r F sin180  0
(a)
Angular momentum is conserved.
L f  Li
m rv  m rv
i i
v
(b)
rv
i i
r
m  rv
m v2
i i
T

r
r3
(c)
2
The work is done by the centripetal force in the negative direction.
Method 1:
r
W   F  d    Tdr   
m  rv
i i
 r 3
ri
2
dr 
m  rv
i i
2 r 
2 r
2
ri
m  rv
1
1 2  ri2 
i i  1


m vi  2  1
 2
2
2
2
 r ri 
r

2

Method 2:
W  K 
(d)
1 2 1 2
1 2  ri2 
m v  m vi 
m vi  2  1
2
2
2
r

Using the data given, we find
v  4.50 m s
P11.9
(a)
W  0.450 J
 d
Li  m 1v1ir1i  m 2v2ir2i  2m v 
 2
Li  2 75.0 kg 5.00 m s  5.00 m
Li  3 750 kg  m
(b)
T  10.1 N
2

s
1
1
m 1v12i  m 2v22i
2
2
2
 1
K i  2  75.0 kg 5.00 m s  1.88 kJ
 2
Ki 
(c)
Angular momentum is conserved: Lf  Li  3750 kg  m
(d)
vf 
Lf
 
2 m rf

3 750 kg  m 2 s
 10.0 m s
2 75.0 kg  2.50 m 
FIG. P11.51
2
s
(e)
2
 1
K f  2   75.0 kg10.0 m s  7.50 kJ
 2
(f)
W  K f  K i  5.62 kJ
P11. 10
(a)

Li  2  M

 d 
v    M vd
 2 
(b)
1

K  2 M v2  M v2
2

(c)
Lf  Li  M vd
FIG. P11.52
Lf
vf 
(e)
1

2
K f  2 M v2f  M  2v  4M v2
2

(f)
W  K f  K i  3M v2
2M rf

M vd
(d)
2M
 d4 
 2v
SOLUTIONS TO PROBLEMS: CHAPTER 12
P12.1
Take torques about P.
m bg
m 1g




  p  n0  2  d  m 1g  2  d  m bgd  m 2gx  0
d

2
m1
We want to find x for which n0  0 .
m 2g
O
m2
P
CG
x
x
 m 1g  m bg d  m 1g 2   m 1  m b  d  m 1 2
m 2g
nO
nP
m2

FIG. P12.3
P12.2
1 are
The coordinates of the center of gravity of piece
x1  2.00 cm and y1  9.00 cm .
The coordinates for piece 2 are
x2  8.00 cm and y2  2.00 cm .
The area of each piece is
A1  72.0 cm 2 and A 2  32.0 cm 2 .
FIG. P12.5
And the mass of each piece is proportional to the
area. Thus,
xCG 
2
2
 m ixi  72.0 cm   2.00 cm    32.0 cm   8.00 cm  
72.0 cm 2  32.0 cm 2
mi
3.85 cm
and
yCG
P12.3
2
2
m iyi  72.0 cm   9.00 cm    32.0 cm   2.00 cm 




104 cm 2
mi
Let the fourth mass (8.00 kg) be placed at (x, y), then
xC G  0 
x 
 3.00 4.00  m 4  x
12.0  m 4
12.0
 1.50 m
8.00
6.85 cm
yCG  0 
Similarly,
 3.00 4.00  8.00 y
12.0  8.00
y  1.50 m
P12.4
Relative to the hinge end of the bridge, the cable is attached
horizontally out a distance
distance
x   5.00 m  cos20.0  4.70 m and vertically down a
y   5.00 m  sin 20.0  1.71 m . The cable then makes the following angle
with the horizontal:
  12.0  1.71 m 
  71.1 .
4.70 m


  tan1 
(a)
Take torques about the hinge end of the bridge:
R x  0  R y  0  19.6 kN  4.00 m  cos20.0
T cos71.1  1.71 m   T sin 71.1  4.70 m

9.80 kN  7.00 m  cos20.0  0
which yields
(b)
T  35.5 kN
 Fx  0  Rx  T cos71.1  0
R x   35.5 kN  cos71.1  11.5 kN  right
or
(c)
FIG. P12.20
 Fy  0  R y  19.6 kN  T sin71.1  9.80 kN
0
Thus,
R y  29.4 kN   35.5 kN  sin 71.1  4.19 kN
 4.19 kN dow n
P12.5
(a)
 Fx  f nw
0
 Fy  ng  800 N  500 N
0
Taking torques about an axis at the foot of the ladder,
 800 N  4.00 m  sin 30.0   500 N 7.50 m  sin 30.0
nw  15.0 cm  cos30.0  0
Solving the torque equation,
 4.00 m  800 N    7.50 m  500 N   tan 30.0
nw  
 268 N .
15.0 m
Next substitute this value into the Fx equation to find
f  nw  268 N
in the positive x direction.
FIG. P12.13
Solving the equation
 Fy  0 ,
ng  1300 N
(b)
In this case, the torque equation
in the positive y direction.
  A  0 gives:
 9.00 m  800 N  sin30.0  7.50 m  500 N  sin30.0  15.0 m   nw  sin60.0  0
or
nw  421 N .
Since f  nw  421 N and f  fm ax  ng , we find

fm ax
421 N

 0.324 .
ng
1300 N
P12.6 (a) Consider the torques about an axis perpendicular to the
page and through the left end of the horizontal beam.
   T sin30.0 d 196 N  d  0 ,
T
30.0°
H
196 N
d
V
giving T  392 N .
(b)
FIG. P12.12
From  Fx  0 , H  T cos30.0  0 , or H   392 N  cos30.0  339 N to the right .
 Fy  0 , V  T sin30.0  200 N
From
 0 , or V  196 N   392 N  sin30.0  0 .
When x  xm in , the rod is on the verge of slipping, so
P12.7
f   fsm ax  sn  0.50n .
From
 Fx  0 , n  T cos37  0, or n  0.799T .
Thus, f  0.50 0.799T   0.399T
From
 Fy  0 ,
2.0 m
37°
n
f
x
2.0 m
Fg
Fg
FIG. P12.23
f T sin37  2Fg  0, or 0.399T  0.602T  2Fg  0 , giving T  2.00Fg .
Using
   0 for an axis perpendicular to the page and through the left end of the beam
 
gives  Fg  xm in  Fg  2.0 m    2Fg sin 37   4.0 m   0 , which reduces to xm in  2.82 m


.
P12.8
(b)
(a)
See the diagram.
If x  1.00 m , then
  O   700 N 1.00 m    200 N  3.00 m 
  80.0 N  6.00 m

  T sin 60.0 6.00 m   0
FIG. P12.43
Solving for the tension gives: T  343 N .
From
 Fx  0 ,
From
(c)
Rx  T cos60.0  171 N .
 Fy  0 ,
R y  980 N  T sin60.0  683 N .
If T  900 N :
 O   700 N  x   200 N  3.00 m    80.0 N  6.00 m    900 N  sin60.0  6.00 m   0 .
Solving for x gives: x  5.13 m .
P12.9
x
3L
4
If the CM of the two bricks does not lie over the
edge, then the bricks balance.
L
over the edge, then
4
3L
the second brick may be placed so that its end protrudes
over
4
the edge.
If the lower brick is placed
FIG. P12.24
P12.10
Using
 Fx   Fy     0 , choosing the origin at
the left end of the beam, we have (neglecting the weight of the beam)
 Fx  Rx  T cos  0,
 Fy  R y  T sin  Fg  0,
and
   Fg  L  d  T sin  2L  d  0 .
Solving these equations, we find:
(a)
T
(b)
Rx 
Fg  L  d
sin   2L  d
Fg  L  d cot
Ry 
2L  d
FIG. P12.45
FgL
2L  d
P12.11
Count the wires. If they are wrapped together so that all support nearly equal stress, the
number should be
20.0 kN
 100 .
0.200 kN
Since cross-sectional area is proportional to diameter squared, the diameter of the cable will
be
1 m m 
P12.12
100 ~ 1 cm .
F
L
Y
A
Li
L 
FLi
 200 9.80 4.00

 4.90 m m
AY
0.200  104 8.00  1010



SOLUTIONS TO PROBLEMS: CHAPTER 13
m 2  5.00 kg  m 1
m 1  m 2  5.00 kg
P13.1
FG

m 1m 2
 1.00  108 N  6.67  1011 N  m
2
r
 5.00 kg m 1  m 12 
1.00  10
8
N
2
kg2
m 5.00 kg  m
  0.200 m  
0.040 0 m   6.00 kg
6.67  1011 N  m
1
1
2
2
2
2
kg2
Thus, m 12   5.00 kg m 1  6.00 kg  0
or
 m 1  3.00 kg m 1  2.00 kg  0
giving m 1  3.00 kg,so m 2  2.00 kg . The answer m 1  2.00 kg and m 2  3.00 kg is
physically equivalent.
P13.2
The force exerted on the 4.00-kg mass by the 2.00-kg mass
is directed upward and given by
F24  G

m 4m 2 ˆ
j 6.67  1011 N  m
2
r24
2
kg2

4.00 kg 2.00 kg ˆ
j
 3.00 m  2
 5.93  1011 ˆ
jN
The force exerted on the 4.00-kg mass by the 6.00-kg mass
is directed to the left
FIG. P13.5
F64  G
m 4m 6
2
r64
 iˆ   6.67  10
11
N m
2
kg2

4.00 kg 6.00 kg ˆ
i
 4.00 m  2
 10.0  1011 ˆ
iN
Therefore, the resultant force on the 4.00-kg mass is
F4  F24  F64 
 10.0iˆ 5.93ˆj  10
11
N
   4  G R
G  43R
GM
g 2 
R
R2
P13.3
3
3
If
gM
1
 
gE 6
4G M R M
3
4 G E RE
3
then
M  gM

E  gE
  RE   1
2
  R    6  4  3 .
 M
P13.4
(a)
so
(b)
Gm M
At the zero-total field point,
E
rE2

Gm M
M
rM2
MM
r
7.36  1022
 rE
 E
ME
5.98  1024 9.01
r
rE  rM  3.84  108 m  rE  E
9.01
3.84  108 m
rE 
 3.46  108 m
1.11
rM  rE
At this distance the acceleration due to the Earth’s gravity is
gE 
GM
rE2
 6.67  10
11
E

N m
2

kg2 5.98  1024 kg
 3.46  10 m 
8

2
gE  3.34  103 m s2 directed tow ard the Earth
P13.5
g
so
g


Gm ˆ Gm ˆ Gm
ˆ sin 45.0ˆ
i 2 j 2 cos45.0i
j
l2
l
2l
g
 
GM 
1  ˆ ˆ
1
 i j or
2 
2 2
l 
Gm 
1
2   tow ard the opposite corner
2 

2
l
FIG. P13.23
P13.6
g1  g2 
MG
r  a2
2
g1y   g2y
gy  g1y  g2y
g1x  g2x  g2 cos
cos 
 
r
a  r 
2
2 12
ˆ
g  2g2x  i
or
g
2M Gr
r  a 
2
2 32
tow ard the centerofm ass
FIG. P13.25
The height attained is not small compared to the radius of the Earth, so U  m gy does not
GM 1M 2
apply; U  
does. From launch to apogee at height h,
r
P13.7
K i  U i  Em ch  K f  U f :
GM EM
1
M pvi2 
2
RE
p
 0  0
GM EM
p
RE  h

 
2
1
10.0  103 m s  6.67  1011 N  m
2

  6.67  1011 N  m
 5.00  10
7
m
2

 5.98  1024 kg 
kg2 

6
 6.37  10 m  h

2
 
s2  6.26  107 m
2
3.99  1014 m
3
s2
1.26  107 m
2
s2
6.37  106 m  h 
 5.98  1024 kg 
kg2 

6
 6.37  10 m 
2

s2 
3.99  1014 m
3
s2
6
6.37  10 m  h
 3.16  107 m
h  2.52  107 m
P13.8
(a)
 G m 1m 2 
U Tot  U 12  U 13  U 23  3U 12  3 
r12 

U Tot  
(b)
P13.9

3 6.67  1011 N  m
2


kg2 5.00  103 kg
2
0.300 m
A tthe center of the equilateral triangle
 1.67  1014 J
 F  m a yields
Applying Newton’s 2nd Law,
Fg  m ac for each
star:
GM M
 2r 2

M v2
r
M 
or
4v2r
.
G
We can write r in terms of the period, T, by considering the time
and distance of one complete cycle. The distance traveled in one orbit is the
circumference of the stars’ common orbit, so 2 r  vT . Therefore
FIG. P13.13
4v2r 4v2  vT 
M 



G
G  2 
P13.10
For both circular orbits,
 F  m a:
G M Em
r2

m v2
r
v
(a)
GM E
r
FIG. P13.60
The original speed is
 6.67  10 N  m kg  5.98  10
 6.37  10 m  2  10 m 
11
vi 
2
6
2
24
5
kg

7.79  103 m s .
(b)
The final speed is
 6.67  10
11
vi 
N m

2

kg2 5.98  1024 kg
6
6.47  10 m


7.85  103 m s .
The energy of the satellite-Earth system is
K U g 
(c)
GM E m
1 2 GM E m 1 GM E GM E
mv 
 m


2
r
2
r
r
2r
Originally
6.67  10
E 
11
i
(d)
Finally
2


6.67  10
N m
2


kg2 5.98  1024 kg 100 kg
6
2 6.57  10 m
11
Ef  
N m



kg2 5.98  1024 kg 100 kg
6
2 6.47  10 m

 3.04  109 J .
 3.08  109 J .
(e)
Thus the object speeds up as it spirals down to the planet. The loss of gravitational energy is so large
that the total energy decreases by


Ei  E f  3.04  109 J 3.08  109 J  4.69 107 J .
(f)
The only forces on the object are the backward force of air resistance R, comparatively very small in
magnitude, and the force of gravity. Because the spiral path of the satellite is not perpendicular to the
gravitational force, one com ponentofthe gravitationalforce pulls forward on the satellite to do positive
work and make its speed increase.
P13.11
(a)
The work must provide the increase in gravitational energy
W  U g  U gf  U gi


GM EM
p
rf
GM EM
p
RE  y


GM EM
p
ri
GM EM
p
RE
 1
1 
 GM EM p 


R
R
 E
E  y
 6.67  1011 N  m 2 
1
1


24



 5.98  10 kg 100 kg 
2
6
kg
6.37  10 m 7.37  106 m 




W  850 M J
(b)
In a circular orbit, gravity supplies the centripetal force:
GM EM
p
2
 RE  y
Then,

M pv2
 RE  y
1
1 GM E M p
M pv2 
2
2  RE  y
So, additional work  kinetic energy required




11
2
5.98  1024 kg 100 kg
1 6.67  10 N  m
2
kg2 7.37  106 m
 

W  2.71 109 J
P13.12
(a)
The gravitational force exerted on m 2 by the Earth (mass m 1 ) accelerates m 2 according to:
G m 1m 2
. The equal magnitude force exerted on the Earth by m 2 produces negligible acceleration of
m 2 g2 
r2
the Earth. The acceleration of relative approach is then
g2 
Gm1
2
r
6.67  10

11
N m
2

kg2 5.98  1024 kg
1.20  10 m 
7
2

2.77 m s2 .
Again, m 2 accelerates toward the center of mass with g2  2.77 m s2 . Now the Earth accelerates
toward m 2 with an acceleration given as
(b)
m 1g1 
g1 
G m 1m 2
r2



6.67  1011 N  m 2 kg2 2.00  1024 kg
Gm 2

 0.926 m s2
2
7
r2
1.20  10 m


The distance between the masses closes with relative acceleration of