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Expert-verified Found in: Page 326 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # Two particles, each of mass ${\mathbf{2}}{\mathbf{.}}{\mathbf{90}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{4}}{\mathbf{}}{\mathbf{kg}}$ and speed 5.46 m/s, travel in opposite directions along parallel lines separated by 4.20 cm. (a) What is the magnitude L of the angular momentum of the two-particle system around a point midway between the two lines? (b) Is the value different for a different location of the point? If the direction of either particle is reversed, what are the answers for (c) part (a) and (d) part (b)?

1. The magnitude L of the angular momentum of the two-particle system around a point midway between the two lines is $\mathrm{L}=6.65×{10}^{-5}\mathrm{kg}.{\mathrm{m}}^{2}/\mathrm{s}$.
2. The value is not different for a different location of the point.
3. If the direction of either particle is reversed, the answer for part (a) is zero.
4. If the direction of either particle is reversed, then the value is different for different locations of a point.
See the step by step solution

## Step 1: Given Data

$\mathrm{m}=2.90×{10}^{-4}\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{V}=5.46\mathrm{m}/\mathrm{s}\phantom{\rule{0ex}{0ex}}\mathrm{d}=4.20\mathrm{cm}$

## Step 2: Determining the concept

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.

## Step 3: (a) Determining the magnitude L of the angular momentum of the two-particle system around a point midway between the two lines

To calculate angular momentum, use the following formula:

$\mathrm{L}=\mathrm{m}×\mathrm{v}×\mathrm{r}+\mathrm{m}×\mathrm{v}×\mathrm{r}\phantom{\rule{0ex}{0ex}}\mathrm{L}=2×\mathrm{m}×\mathrm{v}×\mathrm{r}\phantom{\rule{0ex}{0ex}}\mathrm{L}=\mathrm{m}×\mathrm{v}×\mathrm{d}\phantom{\rule{0ex}{0ex}}\mathrm{L}=2.90×{10}^{-4}×5.46×0.042\phantom{\rule{0ex}{0ex}}\mathrm{L}=6.65×{10}^{-5}\mathrm{kg}.{\mathrm{m}}^{2}/\mathrm{s}$

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

## Step 4: (b) Determining if value different for a different location of the point

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.

## Step 5: (c) Determining if the direction of either particle is reversed, what is the answer for part (a)

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

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

## Step 6: (d) Determining if the direction of either particle is reversed, what is the answer part (b)

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

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

Therefore, using the formula for angular momentum, angular momentum can be found with given conditions. ### Want to see more solutions like these? 