Question

A ball of 2 kg moving at the velocity of 6 m per second strikes a wall and after striking it returns with the same speed . If the time of collision was 0.01 seconds find the force exerted by the ball on the wall .

Answer 1

acc.=v-u/t

a=0-6/0.01

a=-6/0.01

a=-600

f=ma

f=-600x2

hence f = -1200N

Answer 2

Impulse, F dt= change in momentum, ( p2-p1)………….(1)

If before the collision ball of mass m=4 kg moves with velocity v= 2 m/s, then initial momentum of the ball is p1=mv=(2)(6) =12

The momentum of the ball after collision is -mv= p2=-12.

Change in momentum(p2-p1)= (-12–12)

Using eq.(1), F=-24/(0.01) N ,=-2400N=2400N, because force on the ball is opposite to force on the wall.