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) 22.0 N upwards
B) 2.00 N upwards
C) 0.00 N
D) 2.00 N downwards
E) 22.0 N downwards

Question

Quizplus · Accepted Answer

The average force exerted by the ground on the ball can be calculated using the impulse-momentum theorem, which states that the impulse (change in momentum) equals the average force times the time interval. The change in momentum is the final momentum minus the initial momentum. The final momentum is $m \cdot v_{\text{up}} = 0.140 \cdot 1.00$ (upwards), and the initial momentum is $m \cdot v_{\text{down}} = 0.140 \cdot (-1.20)$ (downwards, hence the negative sign). The total change in momentum is $0.140 \cdot 1.00 - 0.140 \cdot (-1.20) = 0.308 \, \text{kg} \cdot \text{m/s}$. The time of contact is 0.0140 s, so the average force is $F = \Delta p / \Delta t = 0.308 / 0.0140 = 22.0 \, \text{N}$ upwards.

A 0140-Kg Baseball Is Dropped from Rest