A uniform bar of length `6a` and mass `8m` lies on a smooth horizontal table. Two point masses `m` and `2m` moving in the same horizontal plane with speeds `2v` and `v` respectively, strike the bar (as shown in figure) and stick to the bar after collision. Calculate (a) velocity of the centre of mass (b) angular velocity about centre of mass and (c) total kinetic energy, just after collision.

Explanation

The location of the center of mass remains same after collision.
By the conservation of linear momentum
`2mv-mxx2v=(2m+m+6m)v_(c.m.)implies v_(c.m.)=0`
By the conservation of angular momentum about the center of mass `C`
`2mvxxa+mxx2vxx2axx2a=I_(0)omega=[(8m(6a)^(2))/(12)+2ma^(2)+m(2a)^(2)]omega`
`6mva=30ma^(2)omegaimpliesomega=(v)/(5a)`
After collision, total energy
`E_(f)=K_(f)=(1)/(2)I_(0)omega^(2)=(1)/(2)(30ma^(2))((v)/(5a))^(2)`
`=(3)/(5)mv^(2)`