Physics · System Of Particles

JEE Main 2026 — 22 January, Evening Shift — Question 29

A uniform bar of length 12 cm and mass $20 m$ lies on a smooth horizontal table. Two point masses m and $2 m$ are moving in opposite directions with same speed of v and in the same plane as the bar, as shown in figure. These masses strike the bar simultaneously and get stuck to it. After collision the entire system is rotating with angular frequency $\omega$ . The ratio of v and $\omega$ is :

Question figure

Option A:
33
Correct
Option B:
$2 \sqrt{88}$
Option C:
66
Option D:
32

Answer: A

Step-by-step solution

Using angular momentum conservation about COM of rod : $\mathrm{L}_{\mathrm{i}}=\mathrm{L}_{\mathrm{f}}$

\begin{aligned} \mathrm{m} \times \mathrm{V} \times 4 & +2 \mathrm{~m} \times \mathrm{V} \times 2=\left(\frac{20 \mathrm{~m}(12)^{2}}{12}+\mathrm{m} \times 4^{2}+2 \mathrm{~m} \times 2^{2}\right) \omega 8 \mathrm{mV} & =(240 \mathrm{~m}+24 \mathrm{~m}) \omega 8 \mathrm{~V} & =264 \omega \frac{\mathrm{~V}}{\omega} & =33 \end{aligned}

Solution figure

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 22 January, Evening Shift
Subject: Physics
Chapter: System Of Particles
Topic: Conservation of Linear Momentum

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