A ball with a velocity of `5 ms^(-1)` impinges at angle of `60^(@)` with the vertical on a smooth horizontal plane. If the coefficient of restitution is 0.5, find the velocity and direction after the impact.

Explanation

The impluse on the ball acts perpendicular to the smooth plane.
(i) The component of velocity of ball parallel to the surface.
(ii) For the component of velocity of ball perpendicular to the surface. Apply law of restitution.

The component of velocity parallel to the surface will be changed.
`v cos alpha = u cos 60^(@)" "` ...(1)
`v cos alpha = 5xx(1)/(2)(5)/(2)`
According to law of restitution
`v sin alpha = "e u sin" 60^(@) " "` ....(2)
`v sin alpha = (1)/(2)xx5xx(sqrt(3))/(2)=5(sqrt(3))/(4)`
Squaring and adding (1) and (2)
`v^(2)(sin^(2)alpha+cos^(2)alpha)=[(25)/(4)+(25xx3)/(16)]`
`v^(2)=(25)/(4)+(75)/(16) " " [because cos^(2)alpha+sin^(2)alpha=1]`
`v^(2)=(100+75)/(16)=(175)/(16)`
`v^(2)=10.9`
`therefore v = 3.3 ms^(-1)`