Chapter 37: Q9P (page 1146)

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?

Expert verified

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

Step by step solution

The rest length of spaceship is $L_{0} = 130 m$

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.

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

$L = L_{0} \sqrt{1 - {(\frac{v}{c})}^{2}}$

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

Substitute all the values in equation.

$L = (130 m) \sqrt{1 - {(\frac{0.740}{c})}^{2}} L = 87.4$

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

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

$t = \frac{L_{0}}{v}$

Substitute all the values in equation.

$t = \frac{87.4 m}{(0.740 c) (\frac{3 \times 10^{8} m / s}{c})} t = 39.34 \times 10^{- 8} s t = 394 n s$

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

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