A ball of a mass m hits the floor with as speed v making an angle of incidence `theta` with the normal. The coefficient of restitution is e. Find the speed of the reflected ball and the angle of reflection of the ball.
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Explanation

Suppose the angle of reflection is `theta'` and the speed after the collision is v' (shown figure) 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 ball remains unchanged. This gives `v'sintheta'=vsintheta` ....(i)
For the components normal to the floor, the velocity of separation is v' cos `theta'` and the velocity of approach is v cos `theta`.
Hence, v' `cos theta'` = ev cos `theta`

From (i) and (ii), v' = v `sqrt(sin(2)theta+e^(2)cos^(2)theta)`
Hence, `tantheta'=(tantheta)/e`
For elastic collision, e = 1, so that `theta' = theta` and v' = v.