A uniform bar of length 6a and mass 8m lies on a smooth horizontal table. Two point masses m and 2m moving in the same horizontal plane with speed 2v and v, respectively, strike the bar [as shown in the fig.] and stick to the bar after collision. Denoting angular velocity (about the centre of mass), total energy and centre of mass velocity by `omega`, E and `v_c` respecitvely, we have after collison

Explanation

The entire situation is shown in Fig.

(a) Let `V` be the velocity of centre of mass. As no external force is applied, linear momentum is conserved.
`:. (8m + m + 2m)V = 2m(-v) +m(2v) + 0`
`:. V = 0`
(b) As no external torque is applied, angular momentum of the system is conserved.
i.e., `2m(v) a+m (2v) 2a = [I_(1)+ I_(2) + I_(b)]omega`
`6m v a = [2m(a)^(2) + m(2a)^(2) + (8m(6a)^(2))/(12)]omega`
`= 30 ma^(2) omega`
`omega = (6 m v a)/(30m a^(2)) = (v)/(5a)`
As the system has only rotational motion, therefore,
Total `KE = KE` of rotation
`= (1)/(2)I omega^(2) = (1)/(2)(30ma^(2)).((v)/(5a))^(2)`
`E = (3)/(5)m v^(2)`