A relativistic train with a rest length of 500.0 m takes 780 ns to pass a stationary observer standing on the train platform, as measured by the stationary observer. (a) What is the speed of the train?

elisegayezm 2022-09-29 Answered
A relativistic train with a rest length of 500.0 m takes 780 ns to pass a stationary observer standing on the train platform, as measured by the stationary observer.
(a) What is the speed of the train? (Hint: Remember to account for the Lorentz contraction of the spaceship.)
If S is the inertial frame with respect to the observer and S is the inertial frame with respect to the train, then the front of the train and an observer line up at t = t = 0 and x = x = 0,
Then, in the S frame, the front of the train sees itself at x = 500 m when t = 780 × 10 9 s e c o n d s
The correct answer is:
2.72 × 10 8 m s = 0.907 c
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Answers (1)

Emmanuel Russo
Answered 2022-09-30 Author has 9 answers
You should know that Lorentz-FitzGerald contraction is (in units of c = 1):
L = L 0 1 v 2 ,
where you are given that L 0 = 500 m = 1.6678 μ s and that in the inertial frame the contracted length L passes by the observer in Δ t = 780 ns. Therefore v = L / Δ t and:
L 2 = L 0 2 ( 1 L 2 Δ t 2 ) .
If you solve this for L algebraically and plug in the values that you are given, you should then get v = 0.906 from the above relationship to v. Close enough up to roundoff error.
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