Question:medium

A rod of mass M and length 2L lies horizontally. A particle of mass m, moving with velocity v in the vertical plane, strikes the end B of the rod and sticks to it. Then the velocity of point B just after collision is:

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In rod–particle sticking collisions, always prefer angular momentum conservation about the system center of mass for clean calculations.
Updated On: Jun 19, 2026
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The Correct Option is C

Solution and Explanation

Step 1: Collision description.
A particle of mass m strikes the end of a uniform rod (mass M, length 2L) and sticks, constituting a perfectly inelastic collision followed by combined translation and rotation.

Step 2: Conservation principle.

External impulsive torque about the system's center of mass is negligible during impact, so angular momentum about the CM is conserved.

Step 3: Initial angular momentum.

Only the particle contributes: L_i = m v L (about rod center).

Step 4: Post-collision dynamics.

The combined system rotates about its CM with angular velocity ω; total moment of inertia I accounts for rod and particle.

Step 5: Angular momentum conservation equation.

m v L = I ω → ω = m v L / I.

Step 6: Velocity of point B after impact.

v_B = ω × distance from CM to B; simplification yields the expression corresponding to Option (3).
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