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

`u=5m//s, theta=60^(@),e=0.5, v=??`

Formula:
`vcostheta=eucosphi`
`vcostheta=eucos60^(@)" "...(1)`
Horizontal comparison
`-vsintheta=usin60^(@)" "...(2)`
`(1)^(2)+(2)^(2)`implies
`v^(2)[sin^(2)theta+cos^(2)theta]=e^(2)u^(2)cos^(2)60^(@)+u^(2)sin^(2)60^(@)`
`v^(2)=u^(2)(e^(2)cos^(2)60^(@)+sin^(2)60^(@))`
`v=usqrt(e^(2)cos^(2)60^(@)+sin^(2)60^(@))`
`=5sqrt(((1)/(2)).((1)/(4))+3//4)`
`v=(5)/(2)sqrt(3.25)implies4.5m//s`
Substitute the value of v in eqn (1)
`4.5costheta=0.5xx5xx(1)/(2)`
`(9)/(2)costheta=(2.5)/(2)`
`costheta=(2.5)/(2)xx(2)/(9)`
`costheta=0.2778`
`=73^(@)54'`