As per the given figure, a small ball P slides down the quadrant 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),

$mgl + 0 = \dfrac{1}{2}m{{v}_{p}}^{2}$

$⇒ {v}_{p} = \sqrt{2gl}$

$= \sqrt{2 \times 10 \times 0 . 2}$

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

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 $2m{s}^{- 1}$.