Question

A(n) 0.239 kg baseball is thrown with a speed of 31.4 m/s. It is hit straight back at the pitcher with a final speed of 19.6 m/s.

(1)What is the magnitude of the impulse delivered to the ball? Answer in units of kg · m/s

(2)Find the average force exerted by the bat on the ball if the two are in contact for 0.00216 s. Answer in units of N..

Answer 1

The magnitude of the impulse is the change in momentum. Momentum is mass times velocity. Change in momentum is final momentum minus initial momentum.

Let's take the direction in which the ball is hit to be the positive direction - and then the other direction is the negative direction.

Final momentum = mass times velocity
= (0.239 kg)(31.4 m/s)
= 7.505kg m/s

Initial momentum = mass times velocity
= (0.239 kg)(-19.6 m/s)
= -4.6844 kg m/s

Impulse = change in momentum
= final momentum - initial momentum
= 7.505 kg m/s - -4.6844 kg m/s
= 12.1894 kg m/s.

2.

The average force exerted by the bat on the ball if the two are in contact for 0.00216 s. is F=impulse/time

F=12.1894/0.00216=5643 N