Abaseball has a mass of 0.140 kg, and it is pitched towards home plate at a speed of 52.0 m/s. assume that the ball is pitched horizontally and that the ball is hit horizontally back in the opposite direction. if the bat exerts an average force of 8900 n for 1.90 ms, what is the final speed of the ball in m/s?

final speed, v2 = 68.79m/s

Explanation:

Given:

Mass, m = 0.140kg

Initial speed, v1= 52.0m/s

Force, F = 8900N

time, t= 1.90ms = 0.0019

Step 1: Using the law of conservation of momentum.

Impulse = summation of momentum

Ft = mv1 + mv2

v2= (Ft-mv1)/m

Step 2: substituting the values

v2 = (8900×0.0019 - 0.14×52)/0.14

v2 = 9.63/0.14

v2 = 68.79m/s

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach