A 08kg ball moving horizontally at 6ms strikes a vertical wa

A 0.8kg ball moving horizontally at 6m/s strikes a vertical wall and rebounds with speed of 2m/s. If the ball stays in contact with the wall for 2 ms,what is the magnitude of the average force exerted by the wall on the ball ?

Solution

Average force = change in momentum/time

so initial momentum= mv1= .8*6 = 4.8

final momentum = -.8*2 = -1.6

Change in momentum magnitude= 6.4 Kg ms-1

average force = 6.4/2*10^-3 = 3200 N