Question

A bullet of mass m 1 is fired into a rod of length L and mass m 2 which is pivoted on one end and rests on a frictionless horizontal surface. The bullet's velocity v → i is horizontal, perpendicular to the rod, and a distance 3 L 4from the pivot. The bullet passes through the rod and continues moving in a straight line. Its kinetic energy after emerging from the rod is 1 2 of its initial kinetic energy. Determine the angular speed of the rod, about the pivot point, immediately after the bullet passes through. Express your answer in terms of m 1 , m 2 , v i , L, and constant(s).

Answer 1

The angular speed of the rod is
`w=((9m_(1)v_(i) )/(4m_(2)L ) )(1-(1)/(√(2) ) )`

Step-by-step explanation:

The final kinetic energy is:

`(1)/(2) mv_(f)^(2) =(1)/(2)((1)/(2) mv_(i)^(2) )`

Clearing vf:

`v_(f) =(1)/(√(2) ) v_(i)`

The conservation of angular momentum before and after collision is:

`m_(1) v_(i) ((3L)/(4) )=Iw+m_(1)v_(f) ((3L)/(4) )`

Clearing w:

`w=((9m_(1)v_(i) )/(4m_(2)L ) )(1-(1)/(√(2) ) )`