Question

Part A If the velocity of a pitched ball has a magnitude of 47.5 m/s and the batted ball's velocity is 51.5 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Answer 1

Explanation:

Let us assume that the mass of a pitched ball is 0.145 kg.

Initial velocity of the pitched ball, u = 47.5 m/s

Final speed of the ball, v = -51.5 m/s (in opposite direction)

We need to find the magnitude of the change in momentum of the ball and the impulse applied to it by the bat. The change in momentum of the ball is given by :

$\Delta p=m(v-u)\\\\\Delta p=0.145\times ((-51.5)-47.5)\\\\\Delta p=-14.355\ kg-m/s$

So, the magnitude of the change in momentum of the ball is 14.355 kg-m/s.

Let the the ball remains in contact with the bat for 2.00 ms. The impulse is given by :

$J=\dfrac{\Delta p}{t}\\\\J=\dfrac{14.355}{2\times 10^{-3}}\\\\J=7177.5\ kg-m/s$

Hence, this is the required solution.