Answer
(a) $\omega = \frac{6v}{19L}~rad/s$
(b) The ratio of the final kinetic energy to the initial kinetic energy is 0.158.
Work Step by Step
(a) Let $M$ be the mass of the bullet. We can use conservation of angular momentum to solve this part of the question.
$L_2 = L_1$
$I\omega = Mv(\frac{L}{2})$
$[\frac{1}{3}(4M)L^2+M(\frac{L}{2})^2]~\omega = Mv(\frac{L}{2})$
$[\frac{4L}{3}+\frac{L}{4}]~\omega = \frac{v}{2}$
$(\frac{19L}{12})~\omega = \frac{v}{2}$
$\omega = \frac{6v}{19L}~rad/s$
(b) We can find the initial kinetic energy.
$K_1 = \frac{1}{2}Mv^2$
We can find the final kinetic energy.
$K_2 = \frac{1}{2}I\omega^2$
$K_2 = \frac{1}{2}[\frac{1}{3}(4M)L^2+M(\frac{L}{2})^2]~\omega^2$
$K_2 = \frac{1}{2}[\frac{4ML^2}{3}+\frac{ML^2}{4}]~(\frac{6v}{19L})^2$
$K_2 = \frac{1}{2}(\frac{19M}{12})~(\frac{6v}{19})^2$
$K_2 = (\frac{M}{24})~(\frac{36v^2}{19})$
$K_2 = \frac{3Mv^2}{38}$
We can find the ratio of the final kinetic energy to the initial kinetic energy.
$\frac{K_2}{K_1} = \frac{\frac{3Mv^2}{38}}{\frac{1}{2}Mv^2}$
$\frac{K_2}{K_1} = \frac{3}{19} = 0.158$
