Question

A baseball approaches home plate at a speed of 47.0 m/s, moving horizontally just before being hit by a bat. The batter hits a pop-up such that after hitting the bat, the baseball is moving at 52.0 m/s straight up. The ball has a mass of 145 g and is in contact with the bat for 1.80 ms. What is the average vector force the ball exerts on the bat during their interaction?

Answer 1

Let the initial direction of motion of the ball as positive x axis and vertical up direction as the positive y axis, we can write the

Initial velocity of the ball, vi = (47) m/s

Final velocity of the velocity, vf = (52) m/s

So, change in velocity of the ball = vf - vi

= (52 - 47) m/s

Thus change in momentum = m(vf - vi)

= 0.145(52 - 47)

= (7.54 - 6.815) kg.m/s

From impulse - momentum theorem, we have,

Impulse = change in momentum

=> Fav*t = m(vf - vi)

=> Fav = m(vf - vi)/t

= [(7.54 - 6.815)/1.8*10^-3] N

= (418.89 - 378.62) N

So, average vector force = (418.89 - 378.62) N