Question

A pitcher applies an average force of 6.2 N in the forward direction on a baseball for 0.65 seconds. The baseball’s mass is 0.145 kg. and initial velocity is zero.

What momentum does the baseball have as it is released by the pitcher?
How fast is the baseball going after the impulse?
The catcher catches the baseball in the problem above. It takes 0.15 seconds to stop the ball. How much average force must the catcher apply during that time interval? (Ignore air resistance.)

Answer 1

1. The momentum of the baseball as it is released by the pitcher is 0.913 Ns.

2. The velocity of the baseball after the impulse is 9.3 m/s.

3. The average force that the catcher must apply during the 0.15 second time interval is 12.2 N.

Step-by-step explanation:

1. Momentum = Force x Time = 6.2 N x 0.65 s = 0.913 Ns

2. Velocity = Momentum / Mass = 0.913 Ns / 0.145 kg = 9.3 m/s

3. Force = Change in Momentum / Time = (0.913 Ns - 0) / 0.15 s = 12.2 N