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A uniform bar of length 6a and mass 8m l...

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 the figure) and stick to the bar after collision. Denoting angular velocity (about the centre of mass), total energy and centre of mass velocity by `omega`, `E` and `V_(C)`, respectively, we have after collision

A

`V_(C)=0`

B

`omega = (3v)/(5a)`

C

`omega =v/(5a)`

D

`E=1/3`

Text Solution

Verified by Experts

The correct Answer is:
A, C, D
The two point masses stick to the rod thus
`2m xx V-m xx 2V=0`
Therefore, the velocity of centre of mass of `V_(C)=0` or (A) correct.
Therefore, the velocity of centre of mass `V_(C)=0` or (A) correct. After sticking of masses, centre of mass remaining at C.
Now by conservation of angular momentum, about C, we get
`2mva + m(2v)(2a)=1/12 8m (6a^(2)) omega + 2m a^(2) omega (2a)^(@)omega`
`6mVa = 30 m a^(2) omega` or `omega = v/(5a)` , Thus, C is also correct.
Total moment of inertia about centre of mass C, `I=2ma^(2) + 4ma^(2) + 24 ma^(2) = 30 ma^(2)`
And total energy after collison is
`=1/2Iomega^(2) = 1/2 xx 30 ma^(2) xx (V/(5a))^(2) = 3/5 mv^(2)`
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