Question

A(n) 0.47 kg softball is pitched at a speed of 10 m/s. The batter hits it back directly at the pitcher at a speed of 29 m/s. The bat acts on the ball for 0.019 s. What is the magnitude of the impulse imparted by the bat to the ball? Answer in units of N · s.

Answer 1

Answer:

18.33 Ns

Explanation:

As the pitch back speed has the opposite direction as before, the change in velocity would be

$\Delta v = v_2 - v_1 = 29 - (-10) = 39 m/s$

So the change in momentum of the ball would be the product of its velocity change and its mass

$\Delta p = \Delta v m = 39 * 0.47 = 18.33 kgm/s$

This is equals to the impulse acted on the ball by the bat, which is 18.33 Ns