A ball of mass m is released from A inside a smooth wedge of mass m as shown in figure. What is the speed of the wedge when the ball reaches point B? 

Let the velocity of wedge be v.

Loss in PE = Gain in KE

           $\mathrm{mg} \mathrm{R} \cos {45}^{\circ }=\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}\mathrm{m}{\left({\mathrm{v}}_1 \cos {45}^{\circ }-\mathrm{v}\right)}^2+\frac{1}{2}\mathrm{m}{\left({\mathrm{v}}_1 \sin {45}^{\circ }\right)}^2$

From conservation of linear momentum

           $\mathrm{m}\left({\mathrm{v}}_1 \cos {45}^{\circ }-\mathrm{v}\right)=\mathrm{mv}$

Here v₁ is the velocity of ball w.r.t. wedge. Solve to get

          $\mathrm{v}={\left(\frac{\mathrm{gR}}{3\sqrt{2}}\right)}^{1/2}$