Question

A small ball of mass $m$ is released from rest from the position shown. All contact surfaces are smooth. The speed of the ball when it reaches its lowest position is

• A. $\sqrt{\frac{2gR}{3}}$
• B. $\sqrt{\frac{4gR}{3}}$
• C. $\sqrt{gR}$
• D. $\sqrt{2gR}$

Answer 1

Answer: (B)

Let the speed of ball be $v$ and speed of the wedge be $v_{2m}$
using conservation of momentum
$0=mv+2m{v}_{2m}$
$v_{2m}=-\frac{v}{2}$
Now, using work-energy theorem on the system,
$mgR=\frac{1}{2}{\left(mv\right)}^2+\frac{1}{2}(2m)\times {\left(\frac{v}{2}\right)}^2$
$\Rightarrow v=\sqrt{\frac{4gR}{3}}$