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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 assume the collision to be elastic, the velocity of ball Q after collision will be:
(g=10 m/s² )

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4m/s

0.25m/s

2m/s

answer is D.

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Detailed Solution

$u = \sqrt{2 gh} = \sqrt{2 \times 10 \times 0.2} = 2 m / s$

since e=1 & masses are same
Velocity of Q =2 m/s.

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