Question

A uniform rod of length $8a$ and mass $6m$ lies on a smooth horizontal surface. Two point masses $m$ and $2m$ moving in the same plane with speed $2v$ and $v$ respectively strike the rod perpendicularly at distance $a$ and $2a$ from the midpoint of the rod in the opposite directions and stick to the rod. The angular velocity of the system immediately after the collision is:

• A. $\frac{6v}{32a}$
• B. $\frac{6v}{33a}$
• C. $\frac{6v}{40a}$
• D. $\frac{6v}{41a}$

Answer 1

The correct option is D $\frac{6v}{41a}$

Using The conservation of angular momentum about CM of rod

$2mva+4mva=\frac{6m(8a{)}^2}{12}\omega +m{a}^2\omega +2m(2a{)}^2\omega$ (directly adding the terms of impulse and its effect on rod)

$\omega =\frac{6v}{41a}$