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Special Relativity Problem

  1. May 21, 2016 #1
    1. The problem statement, all variables and given/known data
    The figure shows a ship (attached to reference frame S') passing us (standing in reference frame S) with velocity http://edugen.wileyplus.com/edugen/shared/assignment/test/session.quest2564447entrance1_N1002E.mml?size=14&ver=1463885870814 [Broken] = 0.952chttp://edugen.wileyplus.com/edugen/shared/assignment/test/session.quest2564447entrance1_N1004A.mml?size=14&ver=1463885870814 [Broken]. A proton is fired at speed 0.976c relative to the ship from the front of the ship to the rear. The proper length of the ship is 760 m. What is the temporal separation between the time the proton is fired and the time it hits the rear wall of the ship according to (a) a passenger in the ship and (b) us? Suppose that, instead, the proton is fired from the rear to the front. What then is the temporal separation between the time it is fired and the time it hits the front wall according to (c) the passenger and (d) us?

    2. Relevant equations
    L=L0*sqrt(1-ß^2)
    t=t0/sqrt(1-ß^2)

    3. The attempt at a solution
    Part a and c are easy, I simply used time=distance/speed since for the passenger it is the rest frame. However I was not able to get part b or d correct. I calculated the contacted length L=760*sqrt(1-0.952^2)=232.6344979m. And then calculated the time interval to be t=L/0.976c=0.798 microsecond. This was not the correct solution so I combined the velocities using the formula:
    u + v
    w = ---------
    1 + uv/c2
    where u=0.952c and v=-0.976c, and got t=L/w=2.29microsec

    Please let me know which part I did wrong thanks heaps!
     

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    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. May 21, 2016 #2

    TSny

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    The proton doesn't travel the distance L relative to frame S. The ship moves to the right while the proton is traveling from the front to the rear of the ship.
     
  4. May 22, 2016 #3
    Would it be shorter than L? L=ϒ(x-vt)?
     
  5. May 22, 2016 #4

    TSny

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    I don't think so. But I don't know what values of x and t you intend to use here.

    Since the ship has length L in frame S, the proton is a distance L from the rear of the ship at the instant it is fired according to frame S. As the proton is traveling towards the rear at speed w (relative to S) the rear is traveling to the right (relative to S). When will the proton meet the rear of the ship?
     
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