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

Answer:

The angular speed of the rod is $w=(\frac{9m_{1}v_{i} }{4m_{2}L } )(1-\frac{1}{\sqrt{2} } )$

Explanation:

The final kinetic energy is:

$\frac{1}{2} mv_{f}^{2} =\frac{1}{2}(\frac{1}{2} mv_{i}^{2} )$

Clearing vf:

$v_{f} =\frac{1}{\sqrt{2} } v_{i}$

The conservation of angular momentum before and after collision is:

$m_{1} v_{i} (\frac{3L}{4} )=Iw+m_{1}v_{f} (\frac{3L}{4} )$

Clearing w:

$w=(\frac{9m_{1}v_{i} }{4m_{2}L } )(1-\frac{1}{\sqrt{2} } )$