As per the given figure, a small ball P slides down the quadrant of a circle and hits the other ball Q of…

Let $v_p$ be the velocity of ball $P$ just before collision. Therefore, applying conservation of energy for $P$ (between the moment it is released and the moment just before collision),

$mgl+0=\frac{1}{2}m{v_p}^2$

$\Rightarrow v_p=\sqrt{2gl}$

          $=\sqrt{2\times 10\times 0.2}$

         $=2 \mathrm{m} {\mathrm{s}}^{-1}$ 

Since both balls have equal masses and the collision is perfectly elastic, the velocities of both the balls will get interchanged.

Therefore, the velocity of ball $Q$ just after the collision is $2 \mathrm{m} {\mathrm{s}}^{-1}$.