A uniform rod of length `l` and mass 2 m rests on a smooth horizontal table. A point mass `m` moving horizontally at right angles to the rod with velocity `v` collides with one end of the rod and sticks it. Then

Options

A. angular velocity of the system after collision is `(2)/(5)(v)/(l)`
B. angular velocity of the system after collision is `(v)/(2l)`
C. The loss in kinetic energy of the system as a whole as a result of the collision `(3)/(10)mv^(2)`
D. The loss in kinetic energy of the system as a whole as a result of the collision `(7mv^(2))/(24)`

Explanation

`C_(1)` is centre of mass of rod
`C_(2)` is centre of mass of both
`thereforemv=3mv_(0)impliesv_(0)=(v)/(3)` ..(i)
`L_(i)=L_(f)` about `C_(2)` we have
`mv(l)/(6)=I_(C_(2))omega`
`[((2ml^(2))/(12))+2m((l)/(3))^(2)+m((l)/(6))^(2)]`
`thereforeomega=(2)/(5)(v)/(l)impliesK_(i)=(1)/(2)mv^(2)`
`K_(f)=(1)/(2)(3m)((v)/(3))^(2)+(1)/(2)I_(C_(2))omega^(2)`
`=(1)/(6)mv^(2)+(1)/(2)((5)/(12)ml^(2))((2)/(5)(v)/(l))^(2)`
`=(1)/(5)mv^(2)`
`therefore` loss of kinetic energy `K_(i)-K_(f)=(3)/(10)mv^(2)`