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Homework Help: Simple SR question

  1. Apr 18, 2013 #1
    1. The problem statement, all variables and given/known data

    An observer at a station on the moon measures the time of a spacecraft passing
    with constant speed v. The front of the spacecraft passes him at 0 s and the rear
    at 0.2 μs. The observer measures the length of the spacecraft to be 1.5 m.
    Explain briefy the term proper length. What is the proper length and the speed
    v (in units of c) of the spacecraft?

    2. Relevant equations

    3. The attempt at a solution

    The definition for proper length we've been given is "The length Lo of an object measured in the rest frame of the object is the PROPER LENGTH".

    I've tried to think of it in terms of "events" and i'm getting the timings occur at the same place in space by the observer, but the measurements aren't as its at the back and front of the rocket (probably where i'm going wrong).
    So i'm saying L=1.5m and To=0.2μs, which from v=L/To gives 7.5E6 m/s or 2.5E-2 c.

    so to find Lo use L=Lo/γ→ Lγ=Lo then because its fairly slow speed use low speed approximation (because γ=1.0005 without it) of γ=1+(1/2)β^2

    which give the proper length to be 1.5007m and yeah that seems wrong.

    any help appreciated, and more so how to think when tackling these problems?
  2. jcsd
  3. Apr 18, 2013 #2


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    Hi Murgs2012! :smile:
    Looks ok, apart from the 7.
  4. Apr 18, 2013 #3
    so would i be correct in saying that because it is moving at non-relativistic speeds (0.025c) the length contraction is effectively zero so the proper length and the contracted length are both 1.5m?
  5. Apr 18, 2013 #4


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    no, i meant, i don't get 7
  6. Apr 18, 2013 #5
    Oh sorry, just re-did it and got 1.5005m?
  7. Apr 18, 2013 #6


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    that's what i get :smile:

    (is it right?)
  8. Apr 18, 2013 #7
    No idea, downside of university "problems "(past paper in this case) are the lack of answers because they love repeating questions across years :\.
    But if we both got to the same answer I'll take it as right :)

    Thanks for the help.
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