JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Conservation of Momentum

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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 / s^{2})

(A)0
(B) $0.25 m / s$
(C) $2 m / s$ ✓ Correct
(D) $4 m / s$

Answer: (C)

Solution

Ball P descends the quadrant of radius

20 cm

, dropping a height

h = 0.2 m

, so it reaches the bottom with speed

v = 2 g h = 2 \times 10 \times 0.2 = 2 m/s

. In a head-on elastic collision between equal masses the moving ball stops and passes its whole velocity to the stationary one. Ball Q thus moves off at

2 m/s

JEE Main · 2023 · 30 Jan 2023 · Shift 1 · Single correct option

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