Question

A sphere of mass $m$ hold at a height $2R$ between a wedge of same mass $m$ and a rigid wall, is released from rest. Assuming that all the surfaces are frictionless. Find the speed of sphere when it hits the ground.

• A. $\sqrt{2gR}\sin \alpha$
• B. $\sqrt{2gR}\cos \alpha$
• C. $\sqrt{gR}\sin \alpha$
• D. $\sqrt{gR}\cos \alpha$

Answer 1

The correct option is A $\sqrt{2gR}\sin \alpha$

Suppose speed of sphere is $v$ in downward direction when it hits ground and speed of wedge at this instant is $v_0$ in horizontal direction.


From geometry,

$\tan \alpha =\frac{v}{v_0}$

$\Rightarrow v_0=v\cot \alpha ........\left(1\right)$

Now, from work- energy theorem

Decrease in potential energy of sphere
= Increase in kinetic energy of both sphere and wedge

$\Rightarrow mgR=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}_0^2-(0+0)$

$\Rightarrow mgR=\frac{1}{2}m{v}^2+\frac{1}{2}m(v\cot \alpha {)}^2\text{[from Eq.(1)]}$

$\Rightarrow mgR=\frac{1}{2}m{v}^2(1+{\cot }^2\alpha )$

$\Rightarrow mgR=\frac{1}{2}m{v}^2{\text{cosec}}^2\alpha$

$\Rightarrow v=\sqrt{2gR}\sin \alpha$

So, option $\left(a\right)$ is correct.