A bar of mass M =1.00 kg and length L =0.20 m is lying on a horizontal frictionless surface. One end of the bar is pivoted at a point about which it is free to rotate. A small mass m =0.10 kg is moving on the same horizontal surface with 5.00 ms -1 speed on a path perpendicular to the bar. It hits the bar at a distance L / 2 from the pivoted end and returns back on the same path with speed v. After this elastic collision, the bar rotates with an angular velocity ω . Which of the following statement is correct?

ω=6.98 rad s -1 and v =4.30 ms -1

ω=3.75 rad s -1 and v =4.30 ms -1

ω=3.75 rad s -1 and v =10.0 ms -1

ω=6.80 rad s -1 and v =4.10 ms -1

Physics · Rotational Dynamics

JEE Advanced 2023 — Paper 1 — Question 20

A bar of mass $\mathrm{M}=1.00 \mathrm{~kg}$ and length $\mathrm{L}=0.20 \mathrm{~m}$ is lying on a horizontal frictionless surface. One end of the bar is pivoted at a point about which it is free to rotate. A small mass $\mathrm{m}=0.10 \mathrm{~kg}$ is moving on the same horizontal surface with $5.00 \mathrm{~ms}^{-1}$ speed on a path perpendicular to the bar. It hits the bar at a distance $\mathrm{L} / 2$ from the pivoted end and returns back on the same path with speed v. After this elastic collision, the bar rotates with an angular velocity $\omega$ . Which of the following statement is correct?

Option A:
$\omega=6.98 \mathrm{rad} \mathrm{s}^{-1}$ and $\mathrm{v}=4.30 \mathrm{~ms}^{-1}$
Correct
Option B:
$\omega=3.75 \mathrm{rad} \mathrm{s}^{-1}$ and $\mathrm{v}=4.30 \mathrm{~ms}^{-1}$
Option C:
$\omega=3.75 \mathrm{rad} \mathrm{s}^{-1}$ and $\mathrm{v}=10.0 \mathrm{~ms}^{-1}$
Option D:
$\omega=6.80 \mathrm{rad} \mathrm{s}^{-1}$ and $\mathrm{v}=4.10 \mathrm{~ms}^{-1}$

Answer: A

Step-by-step solution

$\mathrm{Li}=\mathrm{Lf}$

$m \times 5 \times \frac{L}{2}=\frac{M L^{2}}{3} \times \omega-m v \times \frac{L}{2}$

$5=\frac{4 \omega}{3}-v$ $\mathrm{v}_{2}-\mathrm{v}_{1}=\mathrm{e}\left(\mathrm{u}_{1}-\mathrm{u}_{2}\right)$

$\frac{L}{2} \omega-(-\mathrm{v})=1(5-0)$

$\frac{\omega}{10}+v=5$

Solving (1) & (2)

$\omega=6.98 \mathrm{rad} / \mathrm{sec}$

$\mathrm{v}=4.3 \mathrm{~m} / \mathrm{s}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2023
Paper: Paper 1
Subject: Physics
Chapter: Rotational Dynamics
Topic: Angular Momentum and its Conservation