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Special Relativity and Relative motion

  1. Dec 4, 2007 #1
    1. The problem statement, all variables and given/known data

    An observer on Earth sees two spaceships moving in opposite directions and finally they collide. At t=0 the observer on Earth says that the spaceship 1 which moves to the right with Ua=0.8c is at the point A and the spaceship 2 which moves to the left with Ub=0.6c is at the point B. The distance AB=L=4,2.10^8 m.
    When do the two spaceships collide to the earth frame of reference?
    What is the velocity of the spaceship 2 to the frame reference of the spaceship 1?
    What is the velocity of the spaceship 1 to the frame reference of the spaceship 2?
    When does the collision happen to the frame reference of the spaceship 1 and to the frame reference of the spaceship 2?

    2. Relevant equations



    3. The attempt at a solution

    Let D be the point of the collision and AD=x, so DB=L-x
    The velocity is constant.
    spaceship 1: AD=x=0,8c.t
    spaceship 1: DB=L-x=0,6c.t
    By addition
    L=AD+DB
    L=(0,8c+0,6c).t
    t=4,2.10^8/(1,4.3.10^8)
    t=1sec
    Is it ok so far?
     
  2. jcsd
  3. Dec 4, 2007 #2

    Shooting Star

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    So far, correct.
     
  4. Dec 5, 2007 #3
    The velocity of the spaceship 2 to the frame reference of the spaceship 1 is the same as the velocity of the spaceship 1 to the frame reference of the spaceship 2.
     
  5. Dec 5, 2007 #4

    Shooting Star

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    Numerically, yes. Opp signs because in opp directions. But that shouldn't bother you.
     
  6. Dec 5, 2007 #5
    Using Lorentz transformations

    0,8c=(U21-0,6c)/(1-U21.0,6c/c^2)

    U21=1,4c/1,48

    Correct?
     
  7. Dec 5, 2007 #6
    When does the collision happen to the frame reference of the spaceship 1 and to the frame reference of the spaceship 2?

    Now we must use Lorentz transformation for the length?
     
  8. Dec 5, 2007 #7

    Shooting Star

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    (Is your keyboard very old, so you can't tell the , from the .? Just kidding...)

    Yes. You could have got it also by (0.8+0.6)c/(1+0.8*0.6).
     
  9. Dec 5, 2007 #8

    Shooting Star

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    The question is "when", not where. So, use the formula for time.
     
  10. Dec 5, 2007 #9
    Thank you very much for your time.
     
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