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. (`i`),(`ii`)
B. (`i`), (`ii`), (`iii`)
C. (`ii`),(`iv`)
D. (`i`),(`iii`),(`iv`)

Explanation

The location of `c.m.` is same in both the cases.
By the conservation of angular momentum about `C`,
`2mvxxa+mxx2vxx2a=I_(c)omega`
`=[(8m(6a)^(2))/(12)+2ma^(2)+m(2a)^(2)]omega`
`6mva=(30ma^(2))omegaimpliesomega=(v)/(5a)`
By the conservation of linear momentum
`2mxxv-mxx2v=(2m+m+6m)V_(c)impliesV_(c)=0`
`E=(1)/(2)I_(c)omega^(2)=(1)/(2)xx30ma^(2)((v)/(5a))^(2)=(3)/(5)mv^(2)`