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77P

Expert-verifiedFound in: Page 326

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Two particles, each of mass ${\mathbf{2}}{\mathbf{.}}{\mathbf{90}}{\mathbf{\times}}{{\mathbf{10}}}^{\mathbf{-}\mathbf{4}}{\mathbf{}}{\mathbf{kg}}$ 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)?**

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

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

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

Formula is as follow:

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

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

To calculate angular momentum, use the following formula:

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

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

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 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.

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.

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