A 140 g baseball with a velocity of 25.0 m/s is hit by a baseball bat and leaves at 30.0 m/s in the opposite direction. If the ball was in contact with the bat for 12.0 ms, what was the average force on the ball?
A) 750 N
B) 642 N
C) 550 N
D) 482 N
E) 366 N

Question

Quizplus · Accepted Answer

We can use the impulse-momentum theorem to solve this problem:

impulse = force x time = change in momentum

The initial momentum of the ball is: 
p_initial = m * v_initial = 140 g * 25.0 m/s = 3.50 kg*m/s

The final momentum of the ball is:
p_final = m * v_final = 140 g * (-30.0 m/s) = -4.20 kg*m/s (negative because the ball is moving in the opposite direction)

The change in momentum is:
Δp = p_final - p_initial = (-4.20) - 3.50 = -7.70 kg*m/s

The time of contact is given as 12.0 ms = 0.012 s.

Therefore, the average force on the ball is:
F = Δp / t = (-7.70 kg*m/s) / (0.012 s) = -642 N

The negative sign in the force indicates that it is acting in the opposite direction to the motion of the ball (i.e. towards the bat), which makes sense because the bat is applying a force to the ball.

Note that the magnitude of the force is given by |-642 N| = 642 N, so the best choice is (B) with a magnitude of 642 N.

A 140 G Baseball with a Velocity of 25