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31P

Expert-verifiedFound in: Page 1147

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A spaceship whose rest length is 350 m has a speed of 0.82 c with respect to a certain reference frame. A micrometeorite, also with a speed of 0.82c in this frame, passes the spaceship on an antiparallel track. How long does it take this object to pass the ship as measured on the ship?**

The time taken by the object to pass the ship is $1.2\mathrm{\mu s}$.

**Suppose a particle is moving with speed u ^{'} in x^{'} in an inertial frame S^{'}. If the frame S^{'} is moving with a velocity v with respect to another frame S, then the velocity of the particle with respect to frame S is given by,**

${\mathbf{u}}{\mathbf{=}}\frac{{\mathbf{u}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{v}}{\mathbf{1}\mathbf{+}\frac{{\mathbf{u}}^{\mathbf{\text{'}}}\mathbf{v}}{{\mathbf{c}}^{\mathbf{2}}}}$.....................(1)

**Here, the speed of the light is c.**

Rearrange the equation (1) for u^{'}.

$\begin{array}{rcl}\mathrm{u}\left(1+\frac{\mathrm{u}\text{'}\mathrm{v}}{{\mathrm{c}}^{2}}\right)& =& {\mathrm{u}}^{\text{'}}+\mathrm{v}\\ \frac{{\mathrm{uu}}^{\text{'}}\mathrm{v}}{{\mathrm{c}}^{2}}-{\mathrm{u}}^{\text{'}}& =& \mathrm{v}-\mathrm{u}\\ \mathrm{u}\text{'}\left(\frac{\mathrm{uv}}{{\mathrm{c}}^{2}}-1\right)& =& \mathrm{v}-\mathrm{u}\\ {\mathrm{u}}^{\text{'}}& =& \frac{\mathrm{v}-\mathrm{u}}{\left(\frac{\mathrm{uv}}{{\mathrm{c}}^{2}}-1\right)}\end{array}$

Substitute 0.82c for u, and -0.82c for v in equation (2).

${\mathrm{u}}^{\text{'}}=\frac{-0.82\mathrm{c}-0.82\mathrm{c}}{\frac{\left(-0.82\mathrm{c}\right)\left(0.82\mathrm{c}\right)}{{\mathrm{c}}^{2}}-1}\phantom{\rule{0ex}{0ex}}{\mathrm{u}}^{\text{'}}=0.98\mathrm{c}$

The expression to calculate the time is given by,

data-custom-editor="chemistry" $\u2206{\mathrm{t}}^{\text{'}}=\frac{\u2206\mathrm{x}}{{\mathrm{u}}^{\text{'}}}$

Substitute 350m for data-custom-editor="chemistry" $\u2206\mathrm{x}$ and 0.98c for u^{'} in equation (3).

$\begin{array}{rcl}\u2206{\mathrm{t}}^{\text{'}}& =& \frac{350\mathrm{m}}{0.98\mathrm{c}}\\ & =& \frac{350\mathrm{m}}{0.98\left(3\times {10}^{8}\mathrm{m}/\mathrm{s}\right)}\\ & =& 1.2\times {10}^{-6}\mathrm{s}\\ & =& 1.2\mathrm{\mu s}\end{array}$

Therefore, the time taken by the object to pass the ship is data-custom-editor="chemistry" $1.2\mathrm{\mu s}$.

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