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A rocket moves toward a mirror at .8c. My answer conflicts with chegg's

  1. Sep 29, 2012 #1
    This is an even numbered exercise so I am not able to get confirmation from the book on the correct answer.
    1. The problem statement, all variables and given/known data

    An observer in a rocket moves toward a mirror at speed v = .8c relative to the reference frame S. The mirror is stationary with respect to S. A light pulse emitted by the rocket travels toward the mirror and is reflected back to the rocket. The front of the rocket is a distance d from the mirror (as measured by observer in S) at the moment the light pulse leaves the rocket. What is the total travel time of the pulse as measured by observers in (a) the S frame and (b) the front of the rocket)


    2. Relevant equations
    L' = Lp*gamma


    3. The attempt at a solution
    For frame S:
    I started by finding the time it takes the light to travel the distance d. I found this to be d/c. I then found the distance the rocket travels in that time to be .8d. To find the total distance the light will travel I subtracted .8d from 2d and was left with 1.2d. The time it takes, then, for the light to travel this distance is 1.2d/c. Chegg said it was d/.9c

    For frame S: I did the same thing only used length contraction to find the distances.
     
  2. jcsd
  3. Sep 29, 2012 #2
    You computation is incorrect. 0.8d is how much the rocket moves while the light travels to the mirror. But it keeps moving after that.
     
  4. Sep 29, 2012 #3
    Ahhh true, thanks. How would I find the total time then? Would this work:
    .2d = .8c*t + c*t ... solve for t
    The distance that will be traveled is .2d, so the sum of the distances must be .2d. (Edited from .2*t)
    After I figure this out, can I use the same method, only changing the distances according to length contraction?
     
    Last edited: Sep 29, 2012
  5. Sep 29, 2012 #4
    The equation for the second leg of the journey seems correct. However, I don't understand what .2d*t means.

    After you have the time and the position of the return in one frame, you can transform those immediately into the time and position in the other frame. You don't need to redo the entire computation.
     
  6. Sep 29, 2012 #5
    Just fixed the .2d*t to .2d. Thanks a lot.
     
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