A ball of mass $m$ hits a floor with a speed $v$ making an angle of incidence $\theta ={45}^{\circ }$ with the normal to the floor. If the coefficient of restitution is $e=\frac{1}{\sqrt{2}}$, find the speed of the reflected ball and the angle of reflection.

Since the floor exerts a force on the ball along the normal during the collision so horizontal component of velocity remains same and only the vertical component changes. 

Therefore,$v^{\mathrm{\prime }}\sin {\theta }^{\mathrm{\prime }}=v\sin \theta =\frac{v}{\sqrt{2}}$ and 

    $\begin{array}{r}{v}^{\mathrm{\prime }}\cos {\theta }^{\mathrm{\prime }}=ev\cos \theta =\frac{1}{\sqrt{2}}v\times \frac{1}{\sqrt{2}}=\frac{v}{2}\\ \Rightarrow v^{\mathrm{\prime }2}=\frac{v^2}{2}+\frac{v^2}{4}=\frac{3}{4}{v}^2\Rightarrow v^{\mathrm{\prime }}=\frac{\sqrt{3}}{2}v\\ and \tan {\theta }^{\mathrm{\prime }}=\sqrt{2}\Rightarrow {\theta }^{\mathrm{\prime }}={\tan }^{-1}\sqrt{2}.\end{array}$