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Non-parallel Relativistic Velocity Addition

  1. Sep 26, 2013 #1
    1. The problem statement, all variables and given/known data

    (d) A spaceship moves with velocity (4/5)c relative to Earth. The astronaut throws his empty beer bottle out the window with velocity (3/5)c relative to the ship in the sideways direction. What is the velocity of the bottle (magnitude and direction) relative to the Earth (i) classically, (ii) relativistically?

    2. Relevant equations

    Velocity Addition (relativistic)

    [tex]u_x = \frac {u_x' + v}{1 + \frac{v}{c} \frac{u_x'}{c}} [/tex]

    [tex]u_y = \frac {u_y' + v}{\gamma (1 + \frac{v}{c} \frac{u_x'}{c})} [/tex]

    Pythagorean Theorem

    3. The attempt at a solution

    I just want to know if I've done this correctly. This is a problem on a practice exam.

    For the classical portion I just used the Pythagorean theorem [tex] \sqrt {(\frac{4c}{5})^2 + (\frac{3c}{5})^2} = c [/tex]

    For the relativistic portion I use the Velocity Addition equations. I use [tex]v = \frac{4}{5}c = u_x' [/tex]and [tex]u_y' = \frac{3}{5}c [/tex]

    This gives:

    [tex]u_y = \frac{15}{41}c [/tex]

    and using the Pythagorean theorem with u_x = (4/5)c I get [tex] \frac{13}{15}c = .866 c [/tex]


    I am embarrassed to say that I only think it's the right answer because it looks like it would be. Conceptually I have no idea why I would set both u_x' and v equal to 4/5 c. Can anybody explain if I did get it correct, and if I didn't how I might get the correct answer? Thank you!
     
  2. jcsd
  3. Sep 26, 2013 #2

    TSny

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    Homework Helper
    Gold Member

    This equation cannot be correct. For the case ##u_y' = 0## what would you expect ##u_y## to be?

    Other than your result for ##u_y##, I think the rest of your work looks good (as far as method).

    EDIT: I now see another error. You wrote
    Shouldn't ##u_x' = 0##? Relative to the ship, the bottle moves only in the y' direction.
    Maybe you meant to write [tex]v = \frac{4}{5}c = u_x [/tex]. That would give you the value ##u_x## = ##\frac{4}{5}c## that you used later.
     
    Last edited: Sep 26, 2013
  4. Sep 27, 2013 #3
    Ah nevermind, I see what I did wrong the first time when I tried it with u_x' = 0. I must've been tired and reduced the fraction incorrectly, making it seem as if my final result was incorrect. Also the U_y equation I posted was not the one I used, it's a typo from copying and pasted the latex code for U_x and forgetting to delete (+v). Thank you very much, I have the correct answer now to be, (481)^(1/2)c/25 = .877 c
     
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