Question

A spaceship of rest length 130 m races past a timing station at a speed of 0.740c. (a) What is the length of the spaceship as measured by the timing station? (b) What time interval will the station clock record between the passage of the front and back ends of the ship?

Short answer

The time interval recorded by the station clock is 3.94 ns.

Step-by-step solution

Identification of given data

The rest length of spaceship is $L_0=130m$

The speed of the spaceship is v=0.740c

The length contraction is used to find the length measured by the observer S’ for the tube.

Determination of length of spaceship measured by timing station(a)

The length of spaceship measured by timing station is given as:

$L=L_0\sqrt{1-{\left(\frac{v}{c}\right)}^2}$

Here, is the speed of light and its value is $3\times {10}^{8}m/s$.

Substitute all the values in equation.

$$\begin{gathered} L=\left(130m\right)\sqrt{1-{\left(\frac{0.740}{c}\right)}^2} \\ L=87.4 \end{gathered}$$

Therefore, the length of spaceship measured by timing station is 87.4 m .

Determination of time interval recorded by the station clock(b)

The time interval recorded by the station clock is given as:

$t=\frac{L_0}{v}$

Substitute all the values in equation.

$$\begin{gathered} t=\frac{87.4m}{\left(0.740c\right)\left({\displaystyle \frac{3\times {10}^{8}m/s}{c}}\right)} \\ t=39.34\times {10}^{-8}s \\ t=394ns \end{gathered}$$

Therefore, the time interval recorded by the station clock is$394ns$ .