56.
Picture the Problem: A hammer strikes a rock with an elastic collision. The collision drives the rock in the forward direction
and slows down the speed of the hammer.
Strategy: This is a one-dimensional, elastic collision where one of the objects (the rock) is initially at rest. Therefore, we can
use the results of the analysis of elastic collisions to find the speed of the rock after being struck by the hammer. In this case the
subscript 1 refers to the hammer and the subscript 2 refers to the rock.
Solution: Use the results of the analysis
of elastic collisions to find the final
speed of the rock, v2,f :
v2,f
2m1
v
m1 m2
2 0.55 kg
0.55
0.19 kg
4.5 m/s
6.7 m/s
Insight: The elastic collisions analysis can also be used to find the velocity of the hammer after the collision:
v1,f
m1
m1
m2
v
m2
0.55 0.19 kg
4.5 m/s
0.55 0.19 kg
2.2 m/s. Note that the hammer slows down but is still going in the forward
direction after the collision. If the collision were instead considered to be inelastic (the hammer remains in contact with the rock
and they move together) the final velocity would be 3.3 m/s.
106. Picture the Problem: A truck strikes a 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,
the equations for the velocities after an elastic collision apply 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 v be the initial speed of the truck.
Solution: 1. Use the elastic collision
equation to find v1,f :
v1,f
m1
m1
m2
v
m2
2. Use the elastic collision equation to
find v2,f :
v2,f
2m1
v
m1 m2
1720 732 kg
15.5 m/s
1720 732 kg
2 1720 kg
1720
732 kg
15.5 m/s
6.25 m/s
vtruck
21.7 m/s
vcar
Insight: The elastic collision produces a bigger jolt for the car. If the collision were instead inelastic and the two vehicles stuck
together, the final speed of the car (and the truck) would be 10.9 m/s.
109. Picture the Problem: Two curling stones collide dead center on an ice rink. Stone 1 moves toward the north and collides with
stone 2, which is initially at rest.
Strategy: This is a one-dimensional, elastic collision where one of the objects (stone 2) is initially at rest. Therefore, the
equations for the velocities after an elastic collision apply and can be used to find the final speeds of the two stones. Let m1 be
the mass of stone 1, m2 be the mass of stone 2, and v be the initial speed of stone 1.
2 21 kg
Solution: 1. (a) Use the elastic
collision equation to find v2,f :
v2,f
2m1
v
m1 m2
21 16 kg
2. (b) Use the elastic collision
equation to find v1,f :
v1,f
m1
m1
21 16 kg
1.7 m/s
21 16 kg
m2
v
m2
1.7 m/s
1.9 m/s
0.23 m/s
Insight: If the two stones had equal mass, stone 1 would completely come to rest and stone 2 would have a final velocity of 1.7
m/s toward the north. One interesting fact about elastic collisions is that the velocity difference
before and after the collision is the same, even if the masses are not. In this case, the initial velocity difference of
1.70 – 0 m/s = 1.70 m/s is the same as the final velocity difference of 1.93 – 0.23 m/s = 1.70 m/s.
110. Picture the Problem: Two curling stones collide dead center on an ice rink. Stone 1 moves toward the north and collides with
stone 2, which is initially at rest.
Strategy: This is a one-dimensional, elastic collision where one of the objects (stone 2) is initially at rest. Therefore,
the equations for the velocities after an elastic collision apply and can be used to find the mass of stone 2 and the final speed of
stone 1. Let m1 be the mass of stone 1, m2 be the mass of stone 2, and v be the initial speed of stone 1.
v2,f
2. Multiply both sides by m1  m2 and
m1  m2 
divide both sides by v2, f v :
3. Subtract m1 from both sides and substitute
numerical values to find m2:
4. (b) Use the elastic collision equation to find v1,f :
m2 
v1,f
v2,f
2m1
v
m1 m2
Solution: 1. (a) Use the elastic collision
equation to find v2,f :
v

2m1
m1  m2
2m1
v2,f v
v
1.5 m/s
 2m1   m1  0.69 m/s  2  16 kg   16 kg  54 kg
v2,f
m1
m1
m2
v
m2
16 54 kg
1.5 m/s
16 54 kg
0.81 m/s
Insight: The less massive stone 1 rebounds backward after colliding with stone 2. One interesting fact about elastic collisions is
that the velocity difference before and after the collision is the same, even if the masses are not. In this
case, the initial velocity difference of 1.50 – 0 m/s = 1.50 m/s is the same as the final velocity difference of
0.69 − (− 0.81) m/s = 1.50 m/s.
113. Picture the Problem: The cart 4m collides with the cart 2m, which is given
kinetic energy as a result and later collides with the cart m.
v
Strategy: In each case a moving cart collides with a cart that is at rest, so that
the elastic collision equations given in the textbook will yield the final
velocities of all the carts. First apply the equations to the collision between
carts 4m and 2m, then to the collision between 2m and m. Let the 4m cart be
called cart 4, the 2m cart be called cart 2, and the m cart be called cart 1:
Solution: 1. (a) Apply the elastic collision
equations to the first collision:
2. Apply the elastic collision equation to the
second collision. In this case cart 2 has an initial
speed of 34 v :
3. (b) Verify that Ki  Kf by applying the definition of kinetic
energy and dividing both sides by mv02 :
v4,f
m4
m4
v2,f
2m4
v4,i
m4 m2
4m
2m
v2,f
m2
m2
2m
2m
m
m
v1,f
2m2
v2,i
m2 m1
2
4m
4m
2m
v
2m
2 4m
m1
v2,i
m1
 4m v 2  12  4m  13 v 
?
1
2
m2
v4,i
m2
2 2m
2m
2
m
1
3
4
3
v
4
3
4
3
v
v
v
v
4
9
v
16
9
v
 12  2m  94 v   12  m  169 v 
2
2
2 16 256 36 32 256 324
 




2
9 81 162 162 162 162 162
Insight: Note that due to the transfer of kinetic energy via collisions, the cart with the smallest mass ends up with the largest
speed and the largest kinetic energy.