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Question #42445

Difficulty: 8/10 | 🟢 Free

The Question

A uniform rigid rod of mass $2m$ and length $L$ rests on a smooth horizontal surface. A particle of mass $m$ moving with a horizontal velocity $v_0$ perpendicular to the rod collides perfectly inelastically with the rod at a distance of $\frac{L}{4}$ from its center. What is the angular velocity $\omega$ of the system immediately after the collision?

Options

A.
$\frac{4v_0}{5L}$
B.
$\frac{12v_0}{11L}$
C.
$\frac{6v_0}{7L}$
D.
$\frac{3v_0}{2L}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Intuition_Breaker

Train your brain to resist "obvious" answers that are designed to deceive your intuition.

Concepts Behind This Question

Newton’s Third Law Pairing|Impulsive Torque and Angular Momentum|System vs Particle Displacement

Why This Question Is Tricky

Common Mistakes

This question tests the application of the conservation of angular momentum in off-center inelastic collisions.

A common JEE trap is calculating the angular impulse and moment of inertia about the geometric center of the rod rather than the newly formed system's center of mass.

Students failing to shift the reference point to the instantaneous CM will incorrectly conserve angular momentum about the rod's center, leading to the trap answer 12v_0/11L.

The physics dictates that for a free system, rotation occurs about the system's center of mass, making it the only valid non-accelerating reference point for angular momentum conservation immediately post-collision.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Laws of Motion - Class 11 Part 1
Concept	Newton’s Third Law Pairing
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free