Chapter 9 Solutions
5.
Picture the Problem: The baseball drops straight down, gaining momentum due to the acceleration of gravity.
Strategy: Determine the speed of the baseball before it hits the ground, then use equation 2-12 to find the height from which it
was dropped.
p 0.780 kg  m/s

 5.20 m/s
m
0.150 kg
Solution: 1. Use equation 9-1 to find
the speed of the ball when it lands:
v
2. (b) Use equation 2-12 to solve
for y0 . Let y  0 and v0  0 :
v 2  v02  2 g  y  y0 
17.
 5.20 m/s 
v2

 1.38 m
2 g 2  9.81 m/s 2 
2
y0 
Picture the Problem: The marble drops straight down from rest and rebounds from the floor.
Strategy: Use equation 2-12 to find the speed of the marble just before it hits the floor. Use the same equation and the known
rebound height to find the rebound speed. Then use equation 9-6 to find the impulse delivered to the marble by the floor. Let
upward be the positive direction, so that the marble hits the floor with speed vi and rebounds upward with speed vf .
Solution: 1. (a) Find vi
using equation 2-12:
vi2  v02  2 g y
2. Find vf using equation 2-12 again:
v 2  vf2  2 g y
vi   v02  2 g y   02  2  9.81 m/s 2   1.44 m   5.32 m/s
vf  v 2  2 g y  02  2  9.81 m/s 2   0.640 m   3.54 m/s
3. Use equation 9-6 to find I:
I  mv  m  vf  vi 
  0.0150 kg  3.54   5.32  m/s   0.133 kg  m/s
4. (b) If the marble had bounced to a greater height, its rebound speed would have been larger and the impulse would have been
greater than the impulse found in part (a).
18.
Picture the Problem: The ball rebounds from the floor in the manner indicated
by the figure at right.
Strategy: The impulse is equal to the change in the y-component of the
momentum of the ball. Use equation 9-6 in the vertical direction to find the
impulse.
Solution: Apply equation 9-6
in the y direction:
65°
65°
I  p y  mv y
 m v0 cos 65  (v0 cos 65)   0.60 kg  5.4 m/s  2cos 65   2.7 kg  m/s
22.
Picture the Problem: The two skaters push apart and move in opposite directions without friction.
Strategy: By applying the conservation of momentum we conclude that the total momentum of the two skaters after the push is
zero, just as it was before the push. Set the total momentum of the system to zero and solve for m2 . Let the velocity v1 point in
the negative direction, v 2 in the positive direction.
p1x  p2 x  0  m1v1x  m2 v2 x
m1v1x   45 kg   0.62 m/s 
m2 

 31 kg
v2 x
0.89 m/s
Solution: Set ptotal  0 and solve for m2 :
27.
y
Picture the Problem: The vector diagram at right indicates the momenta of the three
pieces.
Strategy: Because the plate falls straight down its momentum in the xy plane is zero.
That means the momenta of all three pieces must sum to zero. Choose the motion of the
two pieces at right angles to one another to be in the x̂ and ŷ directions. Set the total
225°
x
momentum equal to zero and solve for v 3 .
Solution: 1. Set
p  0
and solve for v 3 :
 p  mv xˆ  mv yˆ  mv
v3 
32.
0
3
 mv  xˆ   mv  yˆ
m
 v 
2. Find the speed v3 :
v3 
3. Find the direction of v3 :
  tan 1 
2
  v  
2
  v  xˆ   v  yˆ
2v
 v3, y 
1  v 
  tan    45  180  225
 v 
 v3, x 
Picture the Problem: The bullet collides with the block, passing right through it and continuing in a straight line but at a slower
speed than it had initially. Afterwards the block moves with constant speed in the same direction as the bullet.
Strategy: Use conservation of momentum to find the speed of the bullet after the collision. Because this is an inelastic collision
we expect a loss of kinetic energy. Use equation 7-6 to find the initial and final kinetic energies and confirm the energy loss. Let
the subscripts b and B denote the bullet and the block, respectively.
Solution: 1. (a) Let pi  pf and solve for vbf :
mb vbi  0  mb vbf  mB vBf
m v  mB vBf
vbf  b bi
mb
 0.0040 kg  650 m/s    0.095 kg  23 m/s 

0.0040 kg
 1.0 102 m/s
2. (b) The final kinetic energy is less than the initial kinetic energy because energy is lost to the heating and deformation of the
bullet and block.
 0.0040 kg  650 m/s 
3. (c) Use equation 7-6 to find Ki :
K i  12 mb vbi 2 
4. Use equation 7-6 to find K f :
K f  12 mb vbf 2  12 mB vBF 2

1
2
1
2
 0.0040 kg 104 m/s 
2
2
 850 J
 12  0.095 kg  23 m/s   47 J
2
34.
Picture the Problem: The putty is thrown horizontally, strikes the side of the block, and sticks to it. The putty and the block
move together in the horizontal direction immediately after the collision, compressing the spring.
Strategy: Use conservation of momentum to find the speed of the putty-block conglomerate immediately after the collision,
then use equation 7-6 to find the kinetic energy. Use conservation of energy to find the maximum compression of the spring
after the collision.
Solution: 1. (a) No, the mechanical energy of the system is not conserved because some of the initial kinetic energy of the putty
will be converted to heat, sound, and permanent deformation of material during the inelastic collision.
2. Set pi  pf and solve for vf :
 mp
mp vp   mb  mp  vf  vf  
m m
p
 b
K after  0  0  U rest
3. Set Eafter  Erest after the collision:
1
2
4. Solve the resulting expression for xmax :
37.

 vp

 m
 mb  mp   m pm
p
 b
xmax
2
 2 1 2
 vp  2 kxmax

 mp2 vp2
   0.0500 kg 2  2.30 m/s 2

 
 

 k  mp  mb     20.0 N/m  0.430  0.0500 kg  
 0.0371 m  3.71 cm
2
Picture the Problem: The truck strikes the car from behind. The collision sends the car lurching forward and slows down the
speed of the truck.
Strategy: This is a one-dimensional, elastic collision where one of the objects (the car) is initially at rest. Therefore, equation 912 applies and can be used to find the final speeds of the vehicles. Let m1 be the mass of the truck, m2 be the mass of the car,
and v0 be the initial speed of the truck.
45.
Solution: 1. Use equation 9-12 to find v1,f :
 m  m2 
 1720  732 kg 
v1,f   1
 v0  
 15.5 m/s   6.25 m/s  vtruck
 1720  732 kg 
 m1  m2 
2. Use equation 9-12 to find v2,f :
 2 1720 kg  
 2m1 
v2,f  
 v0  
 15.5 m/s   21.7 m/s  vcar
 m1  m2 
1720  732 kg 
Picture the Problem: The bricks are configured as shown at right.
Strategy: Use equation 9-14 to calculate the x-coordinate of the center of mass.
Assume that the mass of each brick is m and that the mass of each brick is
distributed uniformly. Reference all positions from the left side of the bottom
brick, as indicated in the figure.
Solution: Apply equation 9-14 directly:
X cm 
 mx  m x  m x
 m3 x3 m  L2  L  54L  1  11L  11

 
L

m1  m2  m3
3m
3  4  12
1 1
M
2 2
80.
Picture the Problem: The hockey player tosses the helmet in one direction and recoils in the other.
Strategy: The total horizontal momentum of the hockey player m p and helmet mh is conserved because there is no friction.
The total horizontal momentum remains zero both before and after the helmet is tossed. Use this fact to find the mass of the
player. Let the helmet be tossed in the positive direction.
Solution: Set pi,x  pf,x and solve for m p :
p
 0  mp vp, x  mh vh, x
mh vh, x
1.3 kg  6.5 m/s  cos11
mp  

 33 kg
vp, x
 0.25 m/s
x