Question

A ball of mass 0.25 kg moving horizontally with a velocity 20 ms-1 is struck by a bat. The duration of contact is 10-2s. After leaving the bat, the ball moves at a speed of 40 ms-1 in a direction opposite to the original direction of motion. Calculate the average force exerted by the bat.

Answer 1

Given:

m = 0.25kg

v_0x = 20m/s

t = 10^-2s

v_x = -40m/s

Find:

F

Solution:

$F_x = \frac{\Delta p_x}{t} = \frac{mv_x-m{v_0}_x}{t} = \\ = \frac{0.25kg*(-40m/s-20m/s)}{10^{-2}s} = -1500N$

Answer: F = 1.5 kN