Q. A uniform rod of length 8$a$ and mass 6$m$ lies on a smooth horizontal surface. Two point masses $m$ and 2$m$ moving in the same plane with speed 2$v$ and $v$ respectively strike the rod perpendicular at distances $a$ and 2$a$ form the mid point 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) $\frac{6v}{41a}$
Explanation
(d) Applying conservation of angular
momentum about point $O$,
[image:7_image96.emf]
$m(a)(2v) + 2m(2a)(v) = I\omega$
or $\omega = \frac{6mav}{I}$ …(i)
Now,
$I = \frac{6m(8a)^{2}}{12} + m(a)^{2} + 2m(2a)^{2}$
$= 32ma^{2} + ma^{2} + 8ma^{2} = 41ma^{2}$
Hence, from Eq.(i)
$\omega = \frac{6mav}{41ma^{2}} = \frac{6v}{41a}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.