A uniform thin rod of length$\mathbf{0}\mathbf{.500}\mathbf{}\mathbf{m}$ and mass$\mathbf{4.00}\mathbf{}\mathbf{kg}\mathbf{}$ can rotate in a horizontal plane about a vertical axis through its centre. The rod is at rest when a$\mathbf{3}\mathbf{.00}\mathbf{}\mathbf{g}$bullet travelling in the rotation plane is fired into one end of the rod. As viewed from above, the bullet’s path makes angle$\mathbf{\theta }\mathbf{=}\mathbf{}\mathbf{60.0}\mathbf{°}$with the rod (Figure). If the bullet lodges in the rod and the angular velocity of the rod is$\mathbf{10}\mathbf{}\mathbf{rad}\mathbf{/}\mathbf{s}$immediately after the collision, what is the bullet’s speed just before impact?

$\mathrm{r}=0.25\text{m}$

${\mathrm{m}}_{\mathrm{l}}=4.0\text{0kg}$

${\mathrm{m}}_{\mathrm{b}}=0.003\text{kg}$

$\text{θ}={60}^0$

Find the angular momentum of the bullet before the collision about the given point. Using the law of conservation of angular momentum, find the angular momentum, and finally, using the relationship between linear and angular velocity, find the speed.According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formula is as follow:

Initial angular momentum = Final angular momentum

According to the law of conservation of angular momentum,

$\text{Initialangularmomentum}=\text{Finalangularmomentum}$

$\begin{array}{c}{\mathrm{L}}_1={\mathrm{L}}_2\\ \mathrm{mvrsin}\left(\mathrm{\theta }\right)=\left(\frac{1}{12}{\mathrm{m}}_{\mathrm{l}}{\mathrm{L}}^2+{\mathrm{m}}_{\mathrm{b}}{\mathrm{r}}^2\right)\times \mathrm{\omega }\\ 0.003\times 0.25\times \mathrm{v}\times \sin \left(60\right)=\left(\frac{1}{12}\times 4\times {0.5}^2+0.003\times {0.25}^2\right)\times 10\end{array}$

Solving for v,

$\mathrm{v}=1300\text{m/s}$

Hence, the speed of the bullet just before impact is$\mathrm{v}=1300\text{m}/\text{s.}$

Therefore, by applying the law of conservation of angular momentum, the speed of the bullet just before impact can be found.