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 = (3)/(5) (v)/(a)`
C. `omega = (v)/(5a)`
D. `E = (3)/(5)mv^(2)`

Answer

A, C, D

Explanation

By the law of conservation of angular momentum, `2mva + m xx 2v xx 2a = 2ma^(2) omega + m (2a)^(2) omega + 8 m(6a)^(2) omega//12.` K.E. of reaction ` = (1//2) [2ma^(2)omega^(2) + m(2a)^(2)omega^(2) + 8 m(6a)^(2) omega^(2)//12]`
` (1///2) (30 ma^(2)omega^(2)) = (3//5) mv^(2)`