Question

A 2.50 kgparticle that is moving horizontally over a floor with velocity$\left(3.00m/s\right)\hat{\mathbf{j}}$ undergoes a completely inelastic collision with a 4.00 kg particle that is moving horizontally over the floor with velocity$\left(4.50m/s\right)\hat{\mathbf{i}}$. The collision occurs at xycoordinates$\left(-0.500m,-0.100m\right)$. After the collision and in unit-vector notation, what is the angular momentum of the stuck-together particles with respect to the origin?

Short answer

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

Step-by-step solution

Step 1: Given Data

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$\left(-0.500\mathrm{m},-0.100\mathrm{m}\right)$

Determining the concept

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

Formula is as follow:

$\overrightarrow{\mathrm{L}}=\mathrm{m}\times \overrightarrow{\mathrm{r}}\times \overrightarrow{\mathrm{v}}$

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

Determining the angular momentum of the stuck-together particles with respect to the origin

The total angular momentum before the collision is,

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

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

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

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, $\overrightarrow{\mathrm{L}}=\mathrm{m}\overrightarrow{\mathrm{r}}\times \overrightarrow{\mathrm{v}}$, and conservation of angular momentum, the angular momentum of the stuck-together particles with respect to the origin can be found.