Question

A ball of mass $m$ is released from the top of a smooth movable wedge of mass $m$. When the ball collides with the floor, velocity of the wedge is $v$. Then the maximum height attained by the ball after an elastic collision with the floor is : (Neglect any edge at the lower end of the wedge)

• A. $\frac{2v^2}{g}$
• B. $\frac{v^2}{4g}$
• C. $\frac{4v^2}{g}$
• D. $\frac{v^2}{2g}$

Answer 1

Answer: (A)

$\frac{2v^2}{g}$

Let $u$ be the velocity of the ball w.r.t. wedge when it reaches the floor.

Then, the x-component of velocity of the ball w.r.t. ground will be $(v-\frac{u}{\sqrt{2}})$ towards right.

By momentum conservation :
0 = m(v) + m $(v-\frac{u}{\sqrt{2}})$ $\Rightarrow u=2\sqrt{2}v$
Therefore, the y-component of velocity of the ball after the elastic collision with the floor will be u cos ${45}^{\circ }=\frac{u}{\sqrt{2}}=2v$ (upward)
$\therefore$ Maximum height = $\frac{{\left(2v\right)}^2}{2g}=\frac{2v^2}{g}$