91Ó°ÊÓ

$$\begin{gathered} \mathrm{m}=2.90×{10}^{-4}\mathrm{kg} \\ \mathrm{V}=5.46\mathrm{m}/\mathrm{s} \\ \mathrm{d}=4.20\mathrm{cm} \end{gathered}$$

Using the formula for angular momentum, determine the magnitude and direction of the angular momentum.

Formula is as follow:

$\mathrm{L}=\mathrm{m}×\mathrm{v}×\mathrm{d}$

Where, m is mass, d is distance, v is velocity and L is angular momentum.

To calculate angular momentum, use the following formula:

$$\begin{gathered} \mathrm{L}=\mathrm{m}×\mathrm{v}×\mathrm{r}+\mathrm{m}×\mathrm{v}×\mathrm{r} \\ \mathrm{L}=2×\mathrm{m}×\mathrm{v}×\mathrm{r} \\ \mathrm{L}=\mathrm{m}×\mathrm{v}×\mathrm{d} \\ \mathrm{L}=2.90×{10}^{-4}×5.46×0.042 \\ \mathrm{L}=6.65×{10}^{-5}\mathrm{kg}.{\mathrm{m}}^2/\mathrm{s} \end{gathered}$$

Hence,the magnitude Lof the angular momentum of the two-particle system around a point midway between the two lines is$\mathrm{L}=6.65×{10}^{-5}\mathrm{kg}.{\mathrm{m}}^2/\mathrm{s}$.

If location of point is changed then the angular momentum doesn’t change.

Hence, the value is not different for a different location of the point.

If the direction of any one of theparticleschanges, the angular momentum becomes zero; because angular momentum is vector quantity. As the direction changes,the angular momentumof both particles are in oppositedirections, so they cancel out each other.

Hence, if the direction of either particle is reversed, the answer for part (a) is zero.

Since, the result depends on the choice of axis, so yes, the value would be different for different locations ofpoints.

Hence, if the direction of either particle is reversed, then the value is different for differentlocationsof a point.

Therefore, using the formula for angular momentum, angular momentum can be found with given conditions.