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Homework Help: Special relativity spaceships time question

  1. Sep 30, 2006 #1
    Here is the problem and my solution,

    Two spaceships each measuring 100m in its own rest frame, pass by each other travelling in opposite directions. Instruments on board spaceship A determine that the front of spaceship B takes 5x10^-6 s to traverse the full length of A.
    a) what is the relative velocity of the two spaceships
    b) how much time elapses on a clock on spaceship B as it traverses the full length of A?

    So for this i thought that since A is doing the measurement, it takes 5 x 10^-6 s for B to pass 100 m. Since according to A, it is still 100m. then the velocity is just 100/5 x 10^-6. is this correct?

    then i took this speed and put it into the time dilation equation with the 5x10^-6s to get the answer for part b.

    is this correct and if not how should i go about this problem, thanks.
  2. jcsd
  3. Oct 1, 2006 #2
    The first one seems correct.
    In b if you by time dilation equation mean [itex] T = \gamma T_0[/itex] where T is the time obsered in b, then this will be incorrect. I had a similar problem (thread: Relativity). You must remember that the events don't occure at the same location, that is [itex] \Delta x \neq 0 [/itex] and [itex] \Delta x' \neq 0 [/itex]. You have [itex] \Delta x =100 [/itex] and [itex] \Delta t =5\times 10^{-6} [/itex] in A's reference frame, and since B is moving with a relative velocity v, you must use the Lorentztransformation to transform this time [itex] \Delta t [/itex] to B's referenceframe, which will be

    [tex] \Delta t' = \gamma(v)\left(\Delta t-\Delta x\,\frac{v}{c^2}\right) [/tex]

    You can also solve this problem by considering the event from B. In B referenceframe the length of A's spaceship will be [itex] L = L_0/\gamma(v) [/itex], and since A is moving with a relative velocity v with respect to B, the time it will take to for B to traverse THAT length of A, will be

    [tex] \Delta t' = \frac{L}{v} =\frac{L_0}{v\gamma(v)} [/tex]
    Last edited: Oct 1, 2006
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