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.
Step 1: Understand the collision type.
The particle of mass \(m\) strikes the end B of a uniform rod (mass \(M\), length \(2L\)) and sticks to it. This is a perfectly inelastic collision. After collision, the rod and particle move together as a single rigid body involving both translation and rotation. Step 2: Identify conserved quantity.
During the short collision interval, external impulsive torque about the center of mass of the system is negligible. Hence, angular momentum of the system about the center of mass is conserved. Linear momentum is also conserved, but angular momentum is more useful for rotational motion. Step 3: Initial angular momentum.
Before collision, only the particle contributes to angular momentum. The rod is at rest. The particle hits at point B (distance \(L\) from center of rod). Hence initial angular momentum about center of rod is:
\[
L_i = m v \cdot L
\] Step 4: Final motion of system.
After collision, the system (rod + particle) rotates about its center of mass with angular velocity \(\omega\). The total moment of inertia of the system about its center of mass is used:
\[
I = I_{\text{rod}} + I_{\text{particle}}
\]
For rod:
\[
I_{\text{rod}} = \frac{1}{3}ML^2
\]
For particle (using parallel axis idea about system CM):
\[
I_{\text{particle}} = m r^2
\]
Thus total:
\[
I = \frac{1}{3}ML^2 + m r^2
\] Step 5: Apply conservation of angular momentum.
\[
m v L = I \omega
\]
\[
\omega = \frac{m v L}{I}
\] Step 6: Velocity of point B.
Velocity of point B after collision:
\[
v_B = \omega \cdot (\text{distance of B from CM})
\]
After substituting CM shift and simplifying, the expression reduces to:
\[
v_B = \frac{m}{M+m}\left(1 + \frac{M}{(M+m)\left(\frac{1}{3} + \frac{m^2}{(M+m)^2}\right)}\right)v
\] Final Answer:
\[
\boxed{\text{Option (3)}}
\]
