A uniform bar of length 6α and mass 8 m lies on a smooth horizontal table. Two point masses m and 2 m moving in the same horizontal plane with speeds 2 v and v respectively strike the bar as shown in fig.and stick to the bar after collision. Denoting angular velocity about the centre of mass, total energy and centre of mass velocity of ω E and vc respectively, we have after collision

Applying the law of conservation a of angular momentum, we have2mva+m(2v)(2a)=2ma2ω+m(2a)2ω+8m(6a)212ω …………..(1)Here 2ma2,m2a2 and 8m6a2/12 are the moment of inertia of mass 2m, m and the bar respectively and ω is the angular velocity  bout the centre of mass. From eq. (1), we have6mva=2ma2ω+4ma2ω+24ma2ω=30ma2ω∴ ω=V5a…………………………………(2)(K'E') rotation=122ma2ω2+m(2a)2ω2+8m(6a)212ω2=1230ma2ω2=1230ma2V5a2=35mv2∴ E=35mv2……………………………….(3)The centre of mass does not charge in rotatory motion. The centre of mass is initially at rest. In the absence of external forces, the centre of mass continues in the same state of rest. Hence                    VC=0 …………………………….(4)