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Probably a very easy length contraction problem

  1. Jun 2, 2005 #1
    Hi all,

    Can someone tell me what to expect and how to calculate the length contraction when two bodies move towards each other, rather than away?

    The particular question I'm stuck on is this:

    A garage is 5m long, a car approaches it at 0.6c. What is the length of the garage according to the car?

    I know the length would be longer in the car's frame if it was moving away from the garage, and I could calculate by how much.. but what happens when the car moves towards it?
     
  2. jcsd
  3. Jun 2, 2005 #2
    What does the Lorentz transformation say when you reverse the sign of [itex]v[/itex]? Is there a difference?
     
  4. Jun 2, 2005 #3
    No, because v is squared in the Lorentz contraction equation..

    So is the answer 6.25m?
     
  5. Jun 2, 2005 #4

    Doc Al

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    Staff: Mentor

    Better check your use of the formula. (It's called length contraction for a reason!)
     
  6. Jun 2, 2005 #5
    but 4m was wrong according to the web test I did :(


    ... I guess the test has the wrong answer stored?

    In anycase, so it doesn't matter whether the bodies are moving towards or away from each other, the length will contact by the same amount in a given frame according to

    x'.(1 - (v^2/c^2))^1/2 = x


    right?
     
  7. Jun 2, 2005 #6

    Doc Al

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    Staff: Mentor

    Apparently so.
    Right. An object moving at a speed v with respect to a frame will be measured by that frame to have a length of [itex]L = L_0 \sqrt{1 - v^2/c^2}[/itex], where [itex]L_0[/itex] is the object's length in its own frame.
     
  8. Jun 2, 2005 #7
    thanks

    *grumbles about wasted time spent trying to find an answer she already had*
     
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