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