A uniform bar of length 6a and mass 8m lies on a smooth horizontal table. Two point masses m and 2m moving with the same horizontal speed 2v and v respectively, strike the bar as shown in the figure and stick to the bar after collision. Denoting angular velocity (about the centre of mass), total energy, and centre of mass velocity after collision by `omega`, E and `v_(C)` respectively, we have after collision

Options

A. `v_(C)=0`
B. `omega=(3v)/(5a)`
C. `omega=v/(5a)`
D. `E=(3mv^(2))/5`

Answer

A, C, D

Explanation

Linear momentum is conserved,
`therefore2m(-v) + m(2v) + (8m xx 0) = (2m + m + 8m)v_(C)`
or `-2mv + 2mv + 0 = 11mv_(C)`
or `v_(C)=0` ….(i)
Again since angular momentum, about centre of mass is conserved, we have
`(2mvxxa)+(mxx2vxx2a)=Iomega`
or 2mva + 4mva
`=[(2mxxa^(2))+(mxx4a^(2))+(8mxx(6a)^(2))/12]omega`
or `6mva=30ma^(2)omega` or `omega=v/(5a)` ...(iii)
After collision, let rotational KE = E
`thereforeE=1/2Iomega^(2)`
or `E=1/2xx(30ma^(2))xx(v/(5a))^(2)=(30ma^(2)xxv^(2))/(2xx25a^(2))=(3mv^(2))/5`
or `E=(3mv^(2))/5` ...(iii)
Hence options (a), (c), (d) are correct.