Chapter 9: Solutions
P9.55
Place the origin at Xcm.
me xe  mt xt
me  mt
mx
xe   t t
me
(72.5 kg)(555 ft)

5.97  1024 kg
X cm  0 
  6.74  1021 ft
The earth moves only 6.74 1021 ft .
P9.56
(a)Use conservation of momentum.
1
v 1
mv  m(0)  m    mv2
2
3 2
2
1
mv  mv2
3
2
4
v2 
v
3
1 2
mv
2
2
2
1 v
1  1  4 
Kf  m     m   v 
2 3
2  2  3 
1
1
2 4
 mv  mv 2  mv 2
18
9
2
Ki  Kf , therefore, the collision is elastic.
(b)
Ki 
P9.57
Use conservation of momentum to determine the horizontal speed of the bullet and block.
m  the mass of the bullet
M  the mass of the block
mv  M (0)  ( m  M )vf
 m 
vf  
v
mM 
Recall that d y  h  (1/ 2) gt 2 for a mass that is initially stationary. Find the horizontal
distance.
 m
x  vf t  
 mM
P9.59

0.0100 kg
m  2(0.750 m)
 2h 

 2.16 m
  725 
v
g  0.0100 kg  1.30 kg  
s
9.81 m2

s
Use the thrust equation to estimate the force.

kg




m
1000 3 (1 m 2 )(31 in.) 0.0254 in.
m
m

m
thrust 
v
10
  0.24 N
s
t
s 

(9 h) 3600 h

P9.60
(a)The change in momentum per second is the weight of the apple.
p
 F  mg  3.0 N
t
(b)
p  F t  W t  Wtf  (3.0 N)(1.5 s)  4.5 kg  m s
P9.61
Place the origin at the center of the wheel and the lead weight on the positive x-axis.
X cm  0 
mlw xlw  mwh X cm,prev
mlw  mwh
mlw
 0.0502 kg 
2
X cm,prev  
xlw   
 (25.0 cm)  3.54  10 cm
mwh
 35.5 kg 
Before the lead weight was added, the center of mass was 0.354 mm from the center of
the wheel.
P9.65
m

(a) Scale reading  W  (m1  m2 ) g  (0.150 kg  1.20 kg)  9.81 2   13.2 N
s 

(b)
In the absence of the liquid, the ball would fall with an acceleration equal to g.
The liquid is retarding the motion of the ball with a force of m1  g      m1g. So, the
4
4
g

scale reading is
3
m
3

3

m1 g  m2 g   m1  m2  g   (0.150 kg)  1.20 kg   9.81
  12.9 N
4
4

4

s2 
P9.68
kg 

m   0.125
 (0.500 m)  0.0625 kg
m


 m 
F  mg  
v
 t 
 m  
 mg    v  v
 L  
m  
kg 
m  
m

 (0.0625 kg)  9.81
 0.125
1.33   1.33 
2  
m 
s  
s 
s  

 0.834 N
P9.69
3
  
 
v
mv0  mr 20
2
Vcm  v0 
3
m  mr
2
1
(m  mr )  m  mr
3
2
2 1
 2
   mr   1   m
3 2
 3
mr  2m
P9.85