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Length contraction of cars

  1. Aug 19, 2007 #1
    car 2 is twice as long as car 1 when they are at rest

    a stationary policeman observes that car 2 is the same length as car 1 as car 2 passes car 1 going through a speed trap.

    car 1 is going 1/2 the speed of light

    I tried to solve this with

    L 1/sqrt(1-(1/2)^2) = 2L/sqrt(1-v^2/c^2) and I got an imaginary anwer
  2. jcsd
  3. Aug 19, 2007 #2


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    your equation is wrong. the sqrt(1-v^2/c^2) should be in the numerator on the RHS

    [edit: and the LHS too]
    Last edited: Aug 20, 2007
  4. Aug 19, 2007 #3
    Why? I took the stationary policeman as my intertial reference frame. To him L 1/sqrt(1-(1/2)^2) is the length of car 1 and 2L/sqrt(1-v^2/c^2) is the length of car 2. These quantities must be equal, right?
  5. Aug 20, 2007 #4


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    As olgran said that sqrt should be in the numerator on both sides. The length contracts... Can you explain the steps you took to get it in the denominator?
  6. Aug 20, 2007 #5


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    Nope. For example, to him the length of car one is


    Which is *not* the same as what you have written.

    Don't get your gammas upsidedown.
  7. Aug 20, 2007 #6


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    remember length contracts when moving and proper length is always biggest. and since [tex]\gamma \geq 1[/tex] always, you can quickly check whether your answer make sense or not... for [tex]\frac{L}{\gamma} \leq L[/tex]
  8. Aug 20, 2007 #7
    I see. So the velocity of car 2 must be sqrt(13/16)c, right?
  9. Aug 20, 2007 #8


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