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Q75P

Expert-verifiedFound in: Page 388

Book edition
9th Edition

Author(s)
Raymond A. Serway, John W. Jewett

Pages
1624 pages

ISBN
9781133947271

**Two identical particles, each of mass ${\mathbf{1000}}{\mathbf{}}{\mathit{k}}{\mathit{g}}$, are coasting in free space along the same path, one in front of the other by ${\mathbf{20}}{\mathbf{}}{\mathit{m}}$. At the instant their separation distance has this value, each particle has precisely the same velocity of role="math" localid="1663837344267" ${\mathbf{800}}\hat{\mathbf{i}}{\mathbf{}}{\mathbf{\text{m/s}}}$. What are their precise velocities when they are 2.0 m apart?**

The trailing particle is moving at $\left(800+1.73\times {10}^{-4}\right)\hat{i}\text{m/s}$ and the leading particle moves at the velocity $\left(800-1.73\times {10}^{-4}\right)\hat{i}\text{m/s}$.

The mass of each particle is ${m}_{1}={m}_{2}=m=1000kg$.

The initial distance between the particle is ${r}_{i}=20.0m$

The final distance between the particles is ${r}_{f}=2.0m$

- The direction of the gravitational force between two particles lies along the x-direction only.
- The velocity of the centre of mass does not change.

The formula for the velocity of the centre of mass is given by

${\overrightarrow{V}}_{CM}=\frac{{m}_{1}{\overrightarrow{v}}_{1}+{m}_{2}{\overrightarrow{v}}_{2}}{{m}_{1}+{m}_{2}}$

Here, ${m}_{1}={m}_{2}=m$, therefore

${\overrightarrow{V}}_{CM}=\frac{m{\overrightarrow{v}}_{1}+m{\overrightarrow{v}}_{2}}{m+m}\phantom{\rule{0ex}{0ex}}=\frac{{\overrightarrow{v}}_{1}+{\overrightarrow{v}}_{2}}{2}$

As the two particles are approaching at the same speed (say ) with respect to the centre of mass towards each other. This implies that the velocity of the trailing particle increases by and the velocity of the leading particle decreases by. Therefore,

${\overrightarrow{V}}_{CM}=\frac{\left({\overrightarrow{v}}_{1}+\Delta v\right)+\left({\overrightarrow{v}}_{2}-\Delta v\right)}{2}\phantom{\rule{0ex}{0ex}}=\frac{{\overrightarrow{v}}_{1}+{\overrightarrow{v}}_{2}}{2}\phantom{\rule{0ex}{0ex}}$

Thus, it is proved that the velocity of the centre of mass remains unchanged.

- In the frame of CoM (centre of mass), we can consider the initial velocity to be zero for both particles. And velocity when they are apart.
- The momentum is conserved for the system which implies that the velocity of both particles with respect to the centre of mass is equal in magnitude and opposite in direction.

Applying conservation of energy of the system in the CoM frame,

${U}_{gi}+{K}_{1i}+{K}_{2i}={U}_{gf}+{K}_{1f}+{K}_{2f}\phantom{\rule{0ex}{0ex}}-\frac{G{m}_{1}{m}_{2}}{{r}_{i}}+\frac{1}{2}{m}_{1}{v}_{1i}^{2}+\frac{1}{2}{m}_{2}{v}_{2i}^{2}=-\frac{G{m}_{1}{m}_{2}}{{r}_{f}}+\frac{1}{2}{m}_{1}{v}_{1f}^{2}+\frac{1}{2}{m}_{2}{v}_{2f}^{2}$

Here, ${m}_{1}={m}_{2}=m;\u200a\u200a{v}_{1i}={v}_{2i}=0\text{and}{v}_{1f}={v}_{2f}=v$ therefore,

$-\frac{G{m}^{2}}{{r}_{i}}+0+0=-\frac{G{m}^{2}}{{r}_{f}}+\frac{1}{2}m{v}_{}^{2}+\frac{1}{2}m{v}_{}^{2}\phantom{\rule{0ex}{0ex}}\frac{G{m}^{2}}{{r}_{f}}-\frac{G{m}^{2}}{{r}_{i}}=m{v}^{2}\phantom{\rule{0ex}{0ex}}\frac{Gm}{{r}_{f}}-\frac{Gm}{{r}_{i}}={v}^{2}\phantom{\rule{0ex}{0ex}}v=\sqrt{Gm\left(\frac{1}{{r}_{f}}-\frac{1}{{r}_{i}}\right)}$

Inserting the values for $G=6.67\times {10}^{-11}$ and mass of the particle; initial and final distance from the given data, we get

$v=\sqrt{\left(6.67\times {10}^{-11}\right)\left(1000\right)\left(\frac{1}{2.0}-\frac{1}{20.0}\right)}\phantom{\rule{0ex}{0ex}}=\sqrt{\left(6.67\times {10}^{-11}\right)\left(1000\right)\left(\frac{9.0}{10.0}\right)}\phantom{\rule{0ex}{0ex}}=1.73\times {10}^{-4}$

Hence, in the CoM frame, both particles are moving at speed $1.73\times {10}^{-4}\text{m/s}$ toward each other. And in the lab frame, the trailing particle is moving at $\left(800+1.73\times {10}^{-4}\right)\hat{i}\text{m/s}$ and the leading particle moves at velocity $\left(800-1.73\times {10}^{-4}\right)\hat{i}\text{m/s}$.

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