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 bulled 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.
(a). What is the final angular velocity of the rod?
(b). What is the ratio of the kinetic energy of the system after the collision to the kinetic energy of te bullet the o collision?

Explanation

(a). Angular momentum about `O` remains constant just before and just after collision.
or `((m)/(6)v(L)/(2))=Iomega`
`=[(mL^(2))/(3)+(m)/(6)(L^(2))/(4)]omega`

Solving we get
`omega=(2v)/(9L)`
(b). `(K_(f))/(K_(i))=((1)/(2)Iomega^(2))`/((1)/(2)((m)/(6))v^(2))`
`=(6Iomega^(2))/(v^(2))`
`=(6[(mL^(2))/(3)+(mL^(2))/(24)]((2v)/(9L))^(2))/(v^(2))`
`=6xx(9)/(24)xx(4)/(81)`
`=(1)/(9)`