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

Given : velocity of ball: `5ms^(-1)`
Angle of inclination with vertical : `60^(@)`
Cofficient of restitution =0.5
Notes: Let the angle reflection is and the speed after collision is `v^(.)` the floor exerts a force on the ball along the normal during the collision. There is no force parallel to the surface. Thus, the parallel component of the velocity of the hall remains unchanged . Tihis gives
`v^(.)sintheta^(.)=vsintheta` ...(1)
Vertical component with respect to floor `=v^(.)costheta^(.)` (velocity of separation)
`velocity of approach=vcostheta`
`"coefficient of restituting e" =("velocity of separation")/("velocity of approach")`
`e=(v^(.)costheta^(.))/(v cos theta) therefore v^(.)costheta^(.)=evcostheta` ...(2)
from (1) and (2)
(ii) `v^(.)sqrt((1-sin^(.2)theta))=evcostheta`
`v^(2)(1-sin^(.2)theta)=e^(2)v^(2)cos^(2)theta`
`v^(.2)-v^(.2)sin^(.2)theta=e^(2)v^(2)cos^(2)theta`
`v^(.2)-v^(.2)sin^(2)theta=e^(2)v^(2)cos^(2)theta [v^(.)sintheta^(.)=vsintheta]`
`v^(.)=sqrt(v^(2)sin^(2)theta+e^(2)v^(2)cos^(2)theta)`
`v^(.)=vsqrt(sin^(2)theta+e^(2)cos^(2)theta)`
The speed after collision `v(.)sqrt(sin^(2)theta+e^(2)cos^(2)theta)`
`v^(.)=5sqrt(sin^(2)(60^(@))+(0.5)^(2)cos^(2)60^(@))=5sqrt(3/4+0.25xx1/4)`
`=5/2-sqrt(3.25)=2.5xx1.8=4.5ms^(-1)`
`"Angle of reflection" theta^(.)=tan^(-1)((tantheta)/e)=tan^(-1)((tan60^(@))/0.5)=tan^(-1)(sqrt3/0.5)`
`=tan^(-1)(3.464)=73.9^(@)`