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(C) $1$. The ball $P$ slides down a height $h = 20\,cm = 0.2\,m$.$2$. By the law of conservation of energy,the velocity $v_P$ of ball $P$ just before

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(C) $1$. The ball $P$ slides down a height $h = 20\,cm = 0.2\,m$.$2$. By the law of conservation of energy,the velocity $v_P$ of ball $P$ just before the collision is given by $v_P = \sqrt{2gh}$.$3$. Substituting the values: $v_P = \sqrt{2 	imes 10 	imes 0.2} = \sqrt{4} = 2\,m/s$.$4$. Since the collision is elastic and the masses of the two balls are equal,the velocities are interchanged after the collision.$5$. Before the collision,ball $P$ has velocity $v_P = 2\,m/s$ and ball $Q$ is at rest $(v_Q = 0)$.$6$. After the collision,ball $P$ comes to rest $(v_P' = 0)$ and ball $Q$ moves with the velocity that ball $P$ had before the collision $(v_Q' = v_P = 2\,m/s)$.

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