Physics · System Of Particles

JEE Main 2025 — 29 January, Morning Shift — Question 31

As shown below, bob A of a pendulum having massless string of length ' R ' is released from $60^{\circ}$ 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)

Option A:
$\frac{1}{3} \sqrt{\operatorname{Rg}}$
Correct
Option B:
$\sqrt{\mathrm{Rg}}$
Option C:
$\frac{4}{3} \sqrt{\operatorname{Rg}}$
Option D:
$\frac{2}{3} \sqrt{\mathrm{Rg}}$

Answer: A

Step-by-step solution

Velocity of a just before hitting : $u=\sqrt{2 g \frac{R}{2}}=\sqrt{g R}$

Just after collision, let velocity of A and B are $\mathrm{v}_{1}$ and $\mathrm{v}_{2}$

respectively $\therefore$ by COM: $\mathrm{mu}=\mathrm{mv}_{1}+\frac{\mathrm{m}}{2} \mathrm{v}_{2}$

$2 \mathrm{v}_{1}+\mathrm{v}_{2}=2 \mathrm{u}$

$\mathrm{e}=1=\frac{\mathrm{v}_{2}-\mathrm{v}_{1}}{\mathrm{u}}$

$\Rightarrow \mathrm{v}_{2}-\mathrm{v}_{1}=\mathrm{u}$

From (i) -(ii) $\Rightarrow 3 \mathrm{v}_{1}=\mathrm{u} \Rightarrow \mathrm{v}_{1}=\frac{\mathrm{u}}{3}=\frac{1}{3} \sqrt{\mathrm{gR}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 29 January, Morning Shift
Subject: Physics
Chapter: System Of Particles
Topic: Collisions in One Dimension