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Q.

A thin uniform bar lies on a frictionless surface and is free to move in any way on the surface. Its mass is 0.16 kg and length 3 meters. Two particles each of mass 0.08 kg are moving on the same surface and towards the bar in a direction perpendicular to the bar, one with a velocity 10 m/s and the other with 6 m/s as shown in figure. The first particle strikes the bar at A and the other at the point B. The points A and B are at distances 0.5 m from centre of the bar. The particles strike the bar at the same instant of time and stick to the bar after collision. Calculate the loss of kinetic energy of the system in the above collision process. (In Joule)

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answer is 2.72.

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Detailed Solution

Loss in  KE=KfKi
Kf=12mvcm2+12Icmω2           Ki=12(0.08)(10)2+12(0.08)(6)2
Apply conservation of momentum and angular momentum to get  Vcm and  ω.

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