A rubber ball of mass 50 g falls from a height of 5m and rebounds to a height of 1.25 m. Find the impulse and the average force between the ball and the ground if the time for which they are in contact was `0.1s`

Explanation

Velocity of the ball just before collision: `v^(2)=0+2gh`
`v=v_(1)=sqrt(2gh)=sqrt(2 xx 10 xx5)=10 ms^(-1)`
Therefore, momentum of the ball before collision,
`vec(P)_(i)=mvec(v)_(i)`
`=0.050 xx(-10) hat(j) Ns=-0.50 hat(j)Ns`
Velocity of the ball just after collision
`v_(f)=sqrt(2gh)=sqrt(2 xx 10 xx1.25)=5.0 ms^(-1)`
Therefore, momentum of the ball just after collision,
`vec(P)_(f)=mvec(v)_(f) =0.050 xx (5.0 hat(j))=0.25 hat(j)Ns`
Now impulse imparted by the ground on the ball
`vec(I)=Delta vec(P)=vec(P)_(f)-vec(P)_(i)=0.25 hat(j)-(-0.50hat(j))=0.75 hat(j)Ns`
Requried force , `F=(Delta P)/(Delta t)=(0.75)/(0.1)=7.5 N`.