91Ó°ÊÓ

Particle A of mass 2.50 kg with velocity$\left(-3.00\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{i}}$

Particle B of mass 4.00 kg with velocity$\left(4.50\mathrm{m}/\mathrm{s}\right)\hat{\mathrm{j}}$

The collision occurs at$\left(-0.500\mathrm{m},-0.100\mathrm{m}\right)$

Using the formula,$\stackrel{\mathbf{→}}{\mathbf{L}}\mathbf{=}\mathbf{m}\mathbf{×}\stackrel{\mathbf{→}}{\mathbf{r}}\mathbf{×}\stackrel{\mathbf{→}}{\mathbf{v}}$, find the angular momentum of the stuck-together particles with respect to the origin.

Formula is as follow:

$\stackrel{→}{\mathrm{L}}=\mathrm{m}×\stackrel{→}{\mathrm{r}}×\stackrel{→}{\mathrm{v}}$

Where, m is mass, r is radius, vis velocity and L is angular momentum.

The total angular momentum before the collision is,

$$\begin{gathered} \stackrel{→}{{\mathrm{L}}_{\mathrm{i}}}={\mathrm{m}}_{\mathrm{A}}×\stackrel{→}{{\mathrm{r}}_{\mathrm{A}}}×\stackrel{→}{{\mathrm{v}}_{\mathrm{A}}}+{\mathrm{m}}_{\mathrm{B}}×\stackrel{→}{{\mathrm{r}}_{\mathrm{B}}}×\stackrel{→}{{\mathrm{v}}_{\mathrm{B}}} \\ \stackrel{→}{{\mathrm{L}}_{\mathrm{i}}}=\left[\left(0.5\mathrm{m}\right)\left(2.5\mathrm{kg}\right)\left(3.0\mathrm{m}/\mathrm{s}\right)+\left(0.1\mathrm{m}\right)\left(4.0\mathrm{kg}\right)\left(4.5\mathrm{m}/\mathrm{s}\right)\right]\hat{\mathrm{k}} \end{gathered}$$

The final angular momentum of the stuck-together particles after the collision, which is a measure relative to the origin is,

$\stackrel{→}{{\mathrm{L}}_{\mathrm{f}}}=\stackrel{→}{{\mathrm{L}}_{\mathrm{i}}}=\left(5.55\mathrm{kg}.{\mathrm{m}}^2/\mathrm{s}\right)\hat{\mathrm{k}}$

Hence,the angular momentum of the stuck-together particles with respect to the origin is$\left(5.55\mathrm{kg}.{\mathrm{m}}^2/\mathrm{s}\right)\hat{\mathrm{k}}$.

Therefore, using the formula, $\stackrel{→}{\mathrm{L}}=\mathrm{m}\stackrel{→}{\mathrm{r}}×\stackrel{→}{\mathrm{v}}$, and conservation of angular momentum, the angular momentum of the stuck-together particles with respect to the origin can be found.