A uniform bar of length 6a and mass 8 m lies on a smooth horizontal table. Two point masses m and 2m moving in the same horizontal plane with speed 2v and v respectively, strike the bar and stick to the bar after collision. Denoting angular velocity (about the centre of mass), total energy and centre of mass velocity by ω, E and vc respectively, we have after collision
vc=0
ω=v5a
E=35mv2
Using momentum conservation, 2m×v−m×2v=(8m+2m+m)vc⇒vc=0
Now, using conservation of angular momentum, 2m×v×a+m×2v×2a=Iω where I=8m(6a)212+2ma2+m(2a)2=30ma2
Solving we get, ω=v5a
Now, energy 12Iω2=35mv2