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

Options

A. `v_c = 0`
B. `omega = (3v)/(5a)`
C. `omeag = (v)/(5a)`
D. `E = (3mv^2)/(5)`

Answer

A, C, D

Explanation

(a,c,d) Applying conservation of linear momentum
`2m(-v)+m(2v)+ 8m xx 0 = (2m + m 8m) v_c`
`rArr v_c = 0`

Applying conservation of angular momentum about centre
of mass
`2mv xxa +m (2v) xx2a = I omega ...(i)`
Where `I = (1)/(12) (8m) xx (6a)^2 + 2m xx a^2 + m xx 4a^2`
`I = 30ma^2 ...(iii)`
`2mv (a) + m (2v)xx2a = 30 ma^2 xx omega rArr omega = (v)/(5a)`
Energy after collision, `E = (1)/(2) I omega^2`
`= (1)/(2)xx30 ma^2 xx(v^2)/(25a^2) = (3mv^2)/(5)`