Figure shows a smooth spherical ball of mass m striking two identical equilateral triangular wedges, each of mass m. The velocity of ball at the instant of impact is v₀ . If e is the coefficient of restitution, then velocity of either wedge after impact is

If v₁ and  v₂ are the velocity of ball and the wedge after collision, then

For ball: $2\mathrm{Jcos}{60}^{\circ }={\mathrm{mv}}_1-\left(-{\mathrm{mv}}_0\right)\text{⇒}\mathrm{J}={\mathrm{mv}}_1+{\mathrm{mv}}_0$....(1)

and $\mathrm{Jsin}{60}^{\circ }={\mathrm{mv}}_2$....(2)

from (1) and (2), we get 

$\frac{2{\mathrm{v}}_2}{\sqrt{3}}={\mathrm{v}}_1+{\mathrm{v}}_0\dots \dots \dots ..\left(\mathrm{i}\right)$

$\mathrm{e}=\frac{\left[{\mathrm{v}}_1\cos {60}^{\circ }-\left(-{\mathrm{v}}_2\cos {30}^{\circ }\right)\right]}{\left({\mathrm{v}}_0\cos {60}^{\circ }-0\right)}=\frac{{\mathrm{v}}_1+\sqrt{3}{\mathrm{v}}_2}{{\mathrm{v}}_0}$

$\Rightarrow {\mathrm{ev}}_0={\mathrm{v}}_1+\sqrt{3}{\mathrm{v}}_2.......\left(\mathrm{ii}\right)$
From (i) and (ii) we get

${\mathrm{v}}_2=\frac{\sqrt{3}}{5}(1+\mathrm{e}){\mathrm{v}}_0$