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Special relativity and really fast trains

  1. May 4, 2015 #1
    1. The problem statement, all variables and given/known data

    Two passenger trains A and B, each 240 m long, pass a 60 m long railroad platform in Winnepeg. The trains are moving in opposite directions at equal speeds of 0.665c with respect to the ground. Train A is traveling west and all tracks are perfectly straight.

    A) From the point of view of a passenger on train A, how fast is train B moving? (Give your answer as a fraction of the speed of light, e.g. if you get 0.952c, you enter 0.952.)

    B) How long does it take train B to pass the passenger on train A?

    2. Relevant equations

    u' = ( u - v ) / ( 1 - ( u v ) / c^2 )

    u = ( u + v ) / ( 1 + (u v ) / c^2 )

    gamma = 1 / sqrt [ 1 - ( v / c ) ^2 ]

    l = l_proper / gamma

    3. The attempt at a solution

    I am studying for a final exam. The above was part of a multipart homework question. I got everytjing correct except for what is shown above. The homework has been returned and I know that the answer to part A is .922 and the answer to part B is 3.36e-7 s. I cant figure out how to get there.

    First I looked at the Galilean method. This put me at something like 1.32 c. obviously this is wrong and I didn't even attempt it.

    Then I tried to use the velocity equations stated above using u' and got u' = 0 / something

    I then figured out part a by using the u equation and substituting the values u = v = .665c. This got me to the .922 I was looking for.

    For part b I am still stuck. I thought what I should do is use this new speed I found in part a to calculate a new gamma

    from the gamma equation above; using v = .992c I find gamma = 2.59

    then I calculate length dilation of the train using gamma = 2.59 and l_proper = 240 m to find l = 92.82 m.

    then I use t = d / v = 112.15 / .665c = 4.65e-7.

    I really need some conceptual help on this one.

    I appreciate any response.
     
  2. jcsd
  3. May 4, 2015 #2

    BvU

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    Kudos for your stamina! You did well on part a. The two expressions are in fact one: u' expression is fine but own speed is u and other train speed is -v (it's moving in the other direction).

    In part b you use the right v to find gamma. All you have to do is use the same v on the contracted(*) length of 92.82 m.

    ( (*) in general we speak of time dilation and length contraction)
     
  4. May 5, 2015 #3
    Ahhh. I plugged in the old v. thanks for your help.
     
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