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 s⁻²]

Question

G1 Physics · Accepted Answer

Ball P descends through height h = 0.20 m, so just before collision its speed is v = √(2gh) = √(2×10×0.20) = 2 m s⁻¹. In a head-on elastic collision of equal masses, velocities are exchanged. Hence Q moves with 2 m s⁻¹.