A wedge (free to move) of mass M has one face making an angle $\alpha$ with horizontal and is resting on a smooth rigid floor. A particle of mass m hits the incline face of the wedge with a horizontal velocity $v_0$. If is observed that the particle rebounds in vertical direction after impact. Neglect friction between particle and the wedge and take $\mathrm{M}=2\mathrm{m},$${\mathrm{v}}_0=10{\mathrm{ms}}^{-1},$$\tan \alpha =2,$$\mathrm{g}=10{\mathrm{ms}}^{-2}$. Calculate the coefficient of restitution for the impact.

Along common tangent direction, the speed of the particle remains unchanged. So we have

${\mathrm{v}}_0\cos \alpha =\mathrm{vsin}\alpha \Rightarrow \mathrm{v}={\mathrm{v}}_0\cot \alpha ....\left(1\right)$

Also, we can conserve linear momentum of the ‘particle + wedge’ system along horizontal direction. So,

${\mathrm{mv}}_0+0=2\mathrm{mV}\Rightarrow \mathrm{V}=\frac{{\mathrm{v}}_0}{2}$

Coefficient of restitution

$\mathrm{e}=\frac{(\text{ Relative velocity of separation }{)}_{\mathrm{n}-\text{ line }}}{(\text{ Relative velocity of approach }{)}_{\mathrm{n}\text{-line }}}$

$\Rightarrow \mathrm{e}=\frac{\mathrm{Vsin}\alpha +\mathrm{vcos}\alpha }{{\mathrm{v}}_0\sin \alpha }=\frac{\mathrm{V}}{{\mathrm{v}}_0}+\frac{\mathrm{v}}{{\mathrm{v}}_0}\cot \alpha$

Using (1) and (2), we get

$\mathrm{e}=\frac{1}{2}+{\cot }^2\alpha$

$\text{ Given that }\tan \alpha =2$

$\Rightarrow \cot \alpha =\frac{1}{2}\Rightarrow \mathrm{e}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}=0.75$