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Find the velocity (in a simple and timely manner)

  1. Jul 22, 2014 #1
    I'm preparing for the GRE, and I've encountered a practice problem that I can solve, but not in a manner practical for the test.

    1. The problem statement, all variables and given/known data

    Two spaceships approach Earth with equal speeds,
    as measured by an observer on Earth, but from
    opposite directions. A meterstick on one spaceship
    is measured to be 60 cm long by an occupant
    of the other spaceship. What is the speed of each
    spaceship, as measured by the observer on Earth?
    (A) 0.4c
    (B) 0.5c
    (C) 0.6c
    (D) 0.7c
    (E) 0.8c


    Well, the problem seems easy. And it is: Start with length contraction followed by Einstein Velocity Addition:
    2. Relevant equations

    Length Contraction
    [itex]\acute{d}=\frac{d}{\gamma}[/itex]

    Einstein Velocity Addition
    [itex]\acute{V}=\frac{V_{1}+V_{2}}{1+\frac{V_{1}V_{2}}{C^{2}}}[/itex]

    3. The attempt at a solution

    Well, I did what you would expect. I first solved for the velocity of the two spaceships relative to each other. This gives [itex]0.8c[/itex]. But, you need the speed of the ship relative to the Earth! So I used the addition formula. This gives [itex]0.5c[/itex], which is the correct answer!

    So, the problem? This is a GRE question. I don't have a calculator or that much time to solve the question.

    When you do the math, After a fair amount of algebraic manipulation, the final equation looks like this:
    [itex]v^{4}+2(1-2 \frac{d^{2}}{\acute{d^{2}}})v^{2}+1=0[/itex]

    And that's before throwing in the quadratic equation!

    Of course, you can just calculate [itex]\acute{v}[/itex] and just solve
    [itex]\acute{V}=\frac{2v}{1+\frac{v^{2}}{C^{2}}}[/itex], but that still ends in a quadratic equation and a lot of arithmetic. It's not a practical solution.

    I've tried manipulating the equations to create a simple, easy, equation to solve, to no luck. What do you think?
     
  2. jcsd
  3. Jul 22, 2014 #2

    maajdl

    User Avatar
    Gold Member

    My guess:
    The contraction factor from ship1 to ship2 is the product of the contraction factor from ship1 to earth and the contraction factor from earth to ship2.
    Isn't it?
    To be checked.
     
  4. Jul 22, 2014 #3
    Why would that be?
     
  5. Jul 27, 2014 #4
    Hi Narroo! On the physics GRE you don't have to get every question correct, it's probably best to find all the ones you can do quickly and easily first and then leave ones like this until last.

    But if you can remember the formula for [itex]\gamma[/itex] and the velocity addition formula, this one shouldn't take you more than 2-3 minutes.

    I agree - you know that [itex]\gamma = \frac{5}{3} = \frac{1}{\sqrt{1-\beta^2}}[/itex] so [itex]\beta = \frac{4}{5}[/itex].

    (it's a 3-4-5 triangle)

    [1 minute so far]

    Again I agree, you just solve [itex]\frac{2\beta}{1 + \beta^2} = \frac{4}{5}[/itex] for [itex]\beta[/itex], [itex]\beta = \frac{1}{2}[/itex].

    [2 minutes total :smile:]
     
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