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

Net external force on the system is zero. Therefore, linear momentum will remain conserved. Initially both the particles have equal momentum in opposite direction. Hence, initial momentum is zero i.e., final linear momentum is also zero or velocity of centre of mass `v_(c)=0`
Let be the angular velocity about centre of mass. Net external torque on system is also zero. Hence, angular momentum about centre of mass will also be conserved , i.e. `L_(i) = L_(g)`
or `(2mva 4mva) =Iomega ={8m (6a)^(2)/12 + (2m)(s)^(2) + m(2a)^(2)}omega` or `6mva = (24 ma^(2) + 2ma^(2) + 4ma^(2)) omega =(30 ma^(2)) omega` or `omega = v/(5a)`
Finally, `E=1/2 Iomega^(2) = 1/2(30 ma^(2)) (v/(5a))^(2) =3/5 mv^(2)`