A uniform rod of length 6a and mass 8m lies on a smooth horizontal table. Two particle of masses m and 2m , moving in the same horizontal plane but in opposite directions with speeds 2v and v respectively strike and rod normally as shown in figure and stick to the rod. Denoting angular velocity ( about the centre of mass), total energy and transnational velocity of centre of mass by `omega`, E and `v_(c)` respectively after the 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`