The ball of mass $m$ approaches a wall of mass $M$ with speed $6\text{m/s}$, perpendicular to the wall. The speed of the wall is $3\text{m/s}$ towards the ball. What will be the speed of the ball after collision with the wall, considering the collision to be perfectly elastic? (Assume $M>>>m$)

The correct option is D $12\text{m/s}$ away from the wall
Given, mass of wall is very large compared to the mass of the ball, hence after collision, velocity of the wall will remain unchanged.
Initial velocity of wall $=u_1$
Final velocity of wall $=v_1$

Considering the direction of motion of the ball before collision as $+ve$
Initial velocity of ball is $u_2=+6\text{m/s}$
and $u_1=v_1=-3\text{m/s}$


Let the final velocity of the ball be $v_2\text{m/s}$ after collision.


Then, $e=\frac{\text{speed of seperation}}{\text{speed of approach}}$
$\Rightarrow e=\frac{|v_2|-|v_1|}{|u_2|+|u_1|}=\frac{v_2-3}{6+3}$
$[\because e=1$ for elastic collision]
$\Rightarrow 1=\frac{v_2-3}{6+3}$
$\therefore v_2=9+3=12\text{m/s}$
Hence ball will move away from the wall after collision at a speed of $12\text{m/s}$.