Q.

A uniform rod of length L rests on a frictionless horizontal surface. The rod is pivoted about a fixed frictionless axis at one and and is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed V0 strikes the rod at its centre and becomes embedded in it. The mass of the bullet is one-sixth of the mass of the rod. The final angular velocity of the rod is (ω)and the ratio of the kinetic energy of the system after collision to the kinetic energy of the bullet before collision is 1x where x is _________

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answer is 9.

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Detailed Solution

About A, angular momentum of the system (rod + bullet) is conserved
Li=Lfm6v0L2=mL23+m6L22ωω=2v09L
 

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Ki=12m6v02=mv0212Kf=12IAω2=12mL23+mL224ω2=mv02108KfKi=mv02/108mv02/12=19

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