A ball of mass 1 kg is dropped from a height of 3.2 m on smooth inclined plane. The coefficient of restitution for the collision is e= 1//2. The ball's velocity become horizontal after the collision.

A ball of mass 1 kg is dropped from a height of 3.2 m on smooth inclin

1.	A ball of mass 1 kg is dropped from a height of 3.2 m on smooth inclined plane. The coefficient of restitution for the collision is e= 1//2. The ball's velocity become horizontal after the collision.
Answer» the angle `theta=tan^(-1)(1/sqrt2)` The speed of the ball after the collision`=4sqrt2ms^(-1)` The TOTAL loss in kinetic ENERGY during the collision is `8 J`. The ball hits the inclined plane again while travelling VERTICALLY DOWNWARD. Solution :`8sintheta=vcostheta` `(8costheta)/2=vsintheta` `implies tantheta=1/sqrt(2)` `v=8/sqrt(2)=3sqrt(2)MS^(-1)` `/_\k=1/2xx1[(4sqrt2)^(2)-8^(2)]=-16J`