A uniform rod of mass 3 m and length `l` is lying on a smooth horizontal surface. Two particle of mass `m` and 2 m moving with speed `2v_(0)` and `v_(0)` in the opposite directly hit the rod perpendicularly at a distance `(l)/(3)` and `(l)/(6)` from the midpoint O of the rod ad stick to it after collision. if the loss in kinetic energy of the system due to collision is `(n)/(5)mv_(0)^(2)` find the value of n.

Answer

9

Explanation

From censervation of momentum
`2mv_(0)-2mv_(0)=6mv_(cm)`
`impliesv_(cm)=0`
`I_(cm)=(3ml^(2))/(12)+(2ml^(2))/(36)+(ml^(2))/(9)=(5)/(12)ml^(2)`
From conservation of angular momentum
`2mv_(0)(l)/(6)+2mv_(0)(l)/(3)=(5)/(12)ml^(2)omega`
`omega=(12v_(0))/(5l)`
loss in kinetic energy of system `k_(i)-k_(f)`
`=(1)/(2)mv_(0)^(2)+(1)/(2)mxx4v_(0)^(2)-(1)/(2)(5)/(12)ml^(2)((12v_(0))/(5l))^(2)=(9)/(5)mv_(0)^(2)`