A uniform rod of length 8 a and mass 6 m lies on a smooth horizontal surface. Two point masses m and 2 m moving in the same plane with speed 2 v and v respectively strike the rod perpendicular at distances a and 2a from the mid point of the rod in the opposite directions and stick to the rod. The angular velocity of the system immediately after the collision is

Options

A. `(6v)/(32a)`
B. `(6v)/(33a)`
C. `(6v)/(40a)`
D. `(6v)/(41a)`

Answer

(d)

Explanation

According to law of conservation of angular momentum.
`Iomega=I_(1)omega_(1)+Iomega_(2)`
`Rightarrow Iomega=[m_(1)r_(1)^(2)xx(v_(1))/(r_(1))]+[m_(2)r_(2)^(2)xx(v^(2))/(r^(2))]`
`Iomega=(m_(1)r_(1)v_(1)+m_(2)r_(2)v_(2)`
`omega=(m_(1)r_(1)v_(1)+m_(2)r_(2)v_(2))/(I)`
`Rightarrow omega=((m)(a)(2v)+(2m)(2a)(v))/(I)=(6mav)/(I)`
`I=(6m(8a)^(2))/(12)+ma^(2)+2m(2a)^(2)`
`32ma^(2)+ma^(2)+8ma^(2)=41ma^(2)`
`therefore omega=(6mav)/(41ma^(2))=(6v)/(41a)`