Q 11-05-091JEE MainJEE Main 2023 (30 Jan, Shift 1)Easy
As per the given figure, a small ball $P$ slides down the quadrant of a circle and hits the other ball $Q$ of equal mass which is initially at rest. Neglecting the effect of friction and assuming the collision to be elastic, the velocity of ball $Q$ after collision will be ($g=10\ \text{m s}^{-2}$)
Answer: (C) $2\ \text{m s}^{-1}$
Speed of $P$ at the bottom: $v=\sqrt{2gR}=\sqrt{2\times10\times0.2}=2\ \text{m s}^{-1}$.
In an elastic head-on collision between equal masses the velocities are exchanged, so $Q$ moves off with $2\ \text{m s}^{-1}$.
Solution by Sreeraj P, M.Sc Physics