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Homework Help: Special Relativity: Find speed of 3rd object.

  1. Nov 17, 2016 #1
    1. The problem statement, all variables and given/known data
    Rest length of Ship B is twice that of Ship A.
    Ship A travels at 0.5c relative to a "fixed" observer.
    Ship B travels at such a speed that the same "fixed" observer measures Length B = Length A.
    How fast is Ship B traveling?

    2. Relevant equations

    All given in statement

    3. The attempt at a solution
    I have no experience with Latex, so pardon my clumsy equations.
    I know that the answer is that Ship B travels at 0.9c. But when I go through the math, I get things wrong.
    1) BetaA=0.5
    2) LA = gammaA * L0A (Length A = gammaA * Rest length A)
    3) LB = gammaB * L0B.

    4) L0B=2 * L0A (rest length B = twice that of A)
    5) LB = LA (length B is measured to be same as that of A)

    6) So LB = LA = gammaB * (2 * L0A)

    Combining eqns 2) and 6), I get
    gammaA = 2 * gammaB.

    But is should be gammaA = 0.5 * gammaB.

    I'm clearly making a bad conceptual mistake. Can you point it out to me?
  2. jcsd
  3. Nov 18, 2016 #2


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    Watch out, it is the other way around (recall, length *contraction*) so LB=LOB /gammaB and LA=L0A/gammaA

  4. Nov 18, 2016 #3
    Oh, my goodness, of course. Blush! :)
    Thank you for getting me out of my rut!
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