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 speeds `2v` and `v`, respectively, strike the bar (as shown in the figure) 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)`, respectively, we have after collision

Options

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

Answer

A, C, D

Explanation

Applying conservation of linear momentum
`2m(-v)+m(2v)+8mxx0=(2m+m+8m)V_(c)`
`implies V_(c)=0`

Applying coservation of angular momentum about centre of mass we get
`2mvxxa+m(2v)xx2a=Iomega`
where `I=1/12(8m)(6a^(2))+2mxxa^(2)+mxx4a^(2)=30ma^(2)`
`6mva=30ma^(2)xxomegaimpliesv/(2a)=omega`
Energy after collision
`E=1/2 Iomega^(2)=1/2xx30 ma^(2)xx(v^(2))/(25a^(2))=(3mv^(2))/5`