A thin uniform bar lies on a frictionless horizontal surface and is free to move in any way on the surface. Its mass is 0.16kg and length is 1.7m. Two particles of each of mass 0.08 kg are moving on the same surface and towards the bar in the direction perpendicular to the bar, one with a velocity of 10 m/s and the other with velocity 6 m/s. If collision between particles and bar is completely inelastic, both particles strike with bar simultaneously. The velocity of center of mass after collision is
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a
2 m/s
b
4 m/s
c
12 m/s
d
16 m/s
answer is B.
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Detailed Solution
see fig Applying the law of conservation of momentum, we getmv1+mv2=2m+m0vCM ∴ vCM=mv1+mv22m+m0 or vCM=0⋅08×10+0⋅08×6(0⋅16+0⋅16) =1⋅280⋅32=4m/s