JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Conservation of Momentum

As shown below, bob A of a pendulum having massless string of length ' R ' is released from

6 0^{\circ}

to the vertical. It hits another bob B of half the mass that is at rest on a friction less table in the centre. Assuming elastic collision, the magnitude of the velocity of bob A after the collision will be (take g as acceleration due to gravity)

(A) $\frac{1}{3} Rg$ ✓ Correct
(B) $Rg$
(C) $\frac{4}{3} R g$
(D) $\frac{2}{3} Rg$

Answer: (A)

Solution

Bob A swings down from

60°

to the vertical. Its height drop is

h = R - R cos 60° = \frac{R}{2}

By energy conservation, the speed of A just before impact is

u = 2 g h = g R

Let

v_{1}

and

v_{2}

be the velocities of A (mass

m

) and B (mass

m /2

) right after the elastic collision. Conservation of momentum gives

m u = m v_{1} + \frac{m}{2} v_{2} ⟹ 2 u = 2 v_{1} + v_{2} ...(i)

For an elastic collision, the coefficient of restitution is 1:

v_{2} - v_{1} = u ...(ii)

From (ii),

v_{2} = u + v_{1}

. Substituting into (i):

2 u = 2 v_{1} + u + v_{1} ⟹ u = 3 v_{1} ⟹ v_{1} = \frac{u}{3} = \frac{1}{3} g R

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Conservation of Momentum

Solution

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