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=1/3`

Answer

A, C, D

Explanation

The two point masses stick to the rod thus
`2m xx V-m xx 2V=0`
Therefore, the velocity of centre of mass of `V_(C)=0` or (A) correct.
Therefore, the velocity of centre of mass `V_(C)=0` or (A) correct. After sticking of masses, centre of mass remaining at C.
Now by conservation of angular momentum, about C, we get
`2mva + m(2v)(2a)=1/12 8m (6a^(2)) omega + 2m a^(2) omega (2a)^(@)omega`
`6mVa = 30 m a^(2) omega` or `omega = v/(5a)` , Thus, C is also correct.
Total moment of inertia about centre of mass C, `I=2ma^(2) + 4ma^(2) + 24 ma^(2) = 30 ma^(2)`
And total energy after collison is
`=1/2Iomega^(2) = 1/2 xx 30 ma^(2) xx (V/(5a))^(2) = 3/5 mv^(2)`