Question

A uniform rod is resting freely over a smooth horizontal plane. A particle moving horizontally strikes at one end of the rod normally and gets stuck. Then

• A. the momentum of the particle is shared between the particle and the rod and remains conserved
• B. the angular momentum about the mid-point of the rod before and after the collision is equal.
• C. the angular momentum about the centre of mass of the combination before and after the collision is equal
• D. the centre of mass of the rod particle system starts to move translationally with the original momentum of the particle

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A the momentum of the particle is shared between the particle and the rod and remains conserved
B the angular momentum about the mid-point of the rod before and after the collision is equal.
C the angular momentum about the centre of mass of the combination before and after the collision is equal
D the centre of mass of the rod particle system starts to move translationally with the original momentum of the particle

The centre of mass of the rod lies at O initially and that of the system lies at point C finally.

The system (rod + particle) moves translationally with linear velocity $V_{CM}$ and also rotates about its COM (point C) with an angular velocity $w$

Now as net external force is Zero, thus linear momentum is conserved. i.e $P=constant$

$P_i=P_f$

$(M+m)V_{CM}=mv$

Hence COM of system moves translationally with initial momentum of particle.

Also, $F_{external}=0⟹{\tau }_{external}=0$ about all points.

Thus angular moment of the system is conserved about all points before and after the collision.

$L_i=L_f$ (for all points)