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Question

A uniform rod of length L rests on a frictionless horizontal surface. The rod is pivoted about a fixed frictionless axis at one end. The rod is initially at rest. A bullet travelling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its centre and becomes embedded in it. The mass of the bullet is one-sixth the mass of the rod. The ratio of the kinetic energy of the system before the collision to the kinetic energy of the bullet after the collision is 1x. Find the value of x.

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Solution

Angular momentum about O remains constant just before and just after collision
Li=Lf or (m6vL2)=1ω
=[mL23+m6L24]ω
Solving, we get
ω=2v9L
KfKi=12Iω212(m6)v2
=6Iω2v2
=6[mL23+mL224](2v9L)2v2
=6×924×481=19

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