1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Two colliding spaceships in Special Relativity

  1. Mar 15, 2012 #1
    1. The problem statement, all variables and given/known data

    ---------> A. B <----------
    v. v
    ------------------------------------ this is static frame of ref (D) the space station

    Where v=0.97c

    In the static frame of reference, ship A has the same length as ship B and are travelling towards eachother. Calculate the speed of ship A relative to ship B according to an observer on the space station. Express your answer as a multiple of vA.

    Use your answer as part of a short argument to show that the time required for the whole of ship A to pass through the whole of ship B is ΔtA, as observed from the space station.

    2. Relevant equations
    v'= u - v / 1-vu/c^2

    3. The attempt at a solution
    I think that v' = 0.97c - (-0.97c) / 1 - 0.97c*-0.97c/c^2 because they are travelling towards eachother and comes out approximately as 2.9995e8 ms^-1... Or am I barking up the wrong tree and how would you express as a multiple of vA?

    How would I work out the 2nd part though... I'm stumped?

    Help much appreciated... Thanks for reading
     
  2. jcsd
  3. Mar 15, 2012 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You should really get in the habit of using parentheses. What you've written means
    $$v' = 0.97 c - \frac{-0.97c}{1} - \frac{0.97c\times (-0.97c)}{c^2}$$ In any case, the velocity-addition formula doesn't apply here. What you found was the speed of A an observer at rest with respect to B would see. The question is asking you for the speed of A relative to B an observer in the rest frame of the space station would see. Hint: the answer will be greater than c.

    Say the tips of the ships meet at t=0 and x=0. Some time later, their tails will both be at x=0. How long does that take?

    Code (Text):
    A ------>.<------ B  at t=0
    B <------.------> A  at t=?
     
  4. Mar 15, 2012 #3
    I thought nothing could travel faster than c?
     
  5. Mar 15, 2012 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Nothing is traveling faster than c.
     
  6. Mar 16, 2012 #5
    ...it's just seen from the space station that the speed is faster than c?
     
  7. Mar 16, 2012 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    In the space station's rest frame, the relative speed of the ships is faster than c. This just means the distance between the ships is decreasing at a rate higher than c. This is the similar to sending two photons off in opposite directions. Both photons travel with speed c relative to you, but the distance between them increases at a rate of 2c.
     
  8. Mar 16, 2012 #7
    Ahhhh! Gotcha... so what I've worked out would be from the v' = 0.97c - (-0.97c) / 1 - 0.97c*-0.97c/c^2 is the speed in the frame of reference of one of the spaceships?

    btw... thank you very much for your help!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Two colliding spaceships in Special Relativity
Loading...