A 0.140-kg baseball is dropped from rest.It has a speed of 1.20 m/s just before it hits the ground,and it rebounds with an upward speed of 1.00 m/s.The ball is in contact with the ground for 0.0140 s.What is the average force exerted by the ground on the ball during the time of contact?
A)2.00 N upwards
B)2.00 N downwards
C)22.0 N upwards
D)22.0 N downwards
E)0.00 N

Question

Quizplus · Accepted Answer

Using the impulse-momentum theorem:
Impulse = Change in momentum
The initial momentum (before the ball hits the ground) is:
p = mv = (0.140 kg)(1.20 m/s) = 0.168 kg m/s downwards
The final momentum (after the ball rebounds) is:
p' = mv' = (0.140 kg)(1.00 m/s) = 0.140 kg m/s upwards
The change in momentum is:
Δp = p' - p = (0.140 kg m/s) - (-0.168 kg m/s) = 0.308 kg m/s upwards
The impulse is equal to the average force multiplied by the time:
Impulse = FaveΔt
Solving for the average force:
Fave = Δp/Δt = (0.308 kg m/s)/(0.0140 s) = 22.0 N upwards
Therefore, the answer is (C).

A 0140-Kg Baseball Is Dropped from Rest