Question

In fig. 37-9, the origins of the two frames coincide at $\mathit{t}\mathbf{=}{\mathit{t}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{0}$ and the relative speed is $\mathbf{0}\mathbf{.}\mathbf{950}\mathit{c}$. Two micrometeorites collide at coordinates $\mathit{x}\mathbf{=}\mathbf{100}\mathit{k}\mathit{m}$ and $\mathit{t}\mathbf{=}\mathbf{200}\mathit{\mu }\mathit{s}$ according to an observer in frame $\mathit{S}$. What are the (a) spatial and (b) temporal coordinate of the collision according to an observer in frame ${\mathit{S}}^{\mathbf{\text{'}}}$ ?

Short answer

1. The spatial coordinate in $S^{\text{'}}$ is $138km$.
2. The temporal coordinate in $S^{\text{'}}$ is $-37.4\mu s$.

Step-by-step solution

The length contraction equation 

The length contraction equation states that $x-vt=x^{\text{'}}\sqrt{1-{\left(\frac{v}{c}\right)}^2}$, where the expression$x-vt$ is equal to the length measured in the inertial reference frame.

The spatial coordinate in frame 

By using the above given formula, we can calculate the spatial coordinate, $x^{\text{'}}$, of the event in the moving observer’s time frame.

Substitute the given values in the above formula:

$x-vt=x^{\text{'}}\sqrt{1-{\left(\frac{v}{c}\right)}^2}$

$$\begin{gathered} 100\times {10}^3-0.950.3\times {10}^8.200\times {10}^{-6}=x^{\text{'}}\sqrt{1-{\left(\frac{0.950c}{c}\right)}^2} \\ {x}^{\text{'}}=1.38\times {10}^{5}m \\ {x}^{\text{'}}=138km \end{gathered}$$

Thus, the spatial coordinate in $S^{\text{'}}$ is $138km$.

The temporal coordinate in frame

Here, we use time dilation equation, $t^{\text{'}}=\frac{t-{\displaystyle \frac{x}{v}}}{\sqrt{1-{\left({\displaystyle \frac{v}{c}}\right)}^2}}$, where $t-\frac{x}{v}$ shows that time measured in the reference frame.

Substitute the given values in the above equation:

$$\begin{gathered} t^{\text{'}}=\frac{t-{\displaystyle \frac{x}{v}}}{\sqrt{1-{\left({\displaystyle \frac{v}{c}}\right)}^2}} \\ {t}^{\text{'}}=\frac{200\times {10}^{-6}-{\displaystyle \frac{100\times {10}^3}{0.950.3\times {10}^8}}}{\sqrt{1-{\left({\displaystyle \frac{0.950c}{c}}\right)}^2}} \\ {t}^{\text{'}}=-3.74\times {10}^{-4}s \\ {t}^{\text{'}}=-374\mu s \end{gathered}$$

Thus, the temporal coordinate in $S^{\text{'}}$ is $-374\mu s$.