Question

A spaceship whose rest length is 350 m has a speed of 0.82c 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?

Short answer

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

Step-by-step solution

Describe the expression for the velocity of the particle

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,

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

Here, the speed of the light is c.

Determine the time taken by the object to pass the ship

Rearrange the equation (1) for u^'.

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

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

$$\begin{gathered} {\mathrm{u}}^{\text{'}}=\frac{-0.82\mathrm{c}-0.82\mathrm{c}}{{\displaystyle \frac{\left(-0.82\mathrm{c}\right)\left(0.82\mathrm{c}\right)}{{\mathrm{c}}^2}}-1} \\ {\mathrm{u}}^{\text{'}}=0.98\mathrm{c} \end{gathered}$$

The expression to calculate the time is given by,

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

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

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

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