Question

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

• A. ${\left(\frac{gR}{3\sqrt{2}}\right)}^{1/2}$
• B. $\sqrt{2gR}$
• C. ${\left(\frac{5gR}{2\sqrt{3}}\right)}^{1/2}$
• D. $\sqrt{\frac{3}{2}gR}$

Answer 1

The correct option is A ${\left(\frac{gR}{3\sqrt{2}}\right)}^{1/2}$

$h=R\cos 45=\frac{R}{\sqrt{2}}$

Change in potential energy =$mgh$

$\frac{1}{2}m{v}^2+\frac{1}{2}m{u}^2=mgh{v}^2+u^2=2gh⟶\left(1\right)$

By momentom conservation,

$mu=m\left(v\cos 45°-u\right)=m\frac{v}{\sqrt{2}}v=2\sqrt{2}u⟶\left(2\right)$

Equation $\left(1\right)$ becomes,

${2u}^2+u^2=2gh{u}^2=\frac{2}{3}gh$

$\frac{1}{2}m{u}^2+\frac{1}{2}m{\left(v\cos 45°-u\right)}^2+\frac{1}{2}m{\left(v\sin 45\right)}^2=mg\frac{R}{\sqrt{2}}{u}^2+{(\frac{v}{\sqrt{2}}-u)}^2+{\left(\frac{v}{\sqrt{2}}\right)}^2=\sqrt{2}gR⟶(usingv=2\sqrt{2}u)u^2+u^2+{4u}^2=\sqrt{2}gR6u^2=\sqrt{2}gRu=\sqrt{\frac{gR}{3\sqrt{2}}}$