91Ó°ÊÓ

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

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

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

$\text{θ}={60}^0$

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,

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

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

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.