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Length contraction in an inelastic collision

  1. Jul 3, 2012 #1
    Preparing for a classical prelim by going over previous exams.

    1. The problem statement, all variables and given/known data
    A relativistic meter stick moves with speed v in the lab. It collides head on with an impenetrable wall completely inelastically, thereby coming to rest in the lab frame. What is the maximum length of the stick in the lab after the collision?


    2. Relevant equations
    [itex]p=\gamma mv[/itex]

    [itex]L_{0}=\gamma L[/itex]

    [itex]E^{2}=(pc)^{2}+(mc^{2})^{2}[/itex]

    3. The attempt at a solution
    Since the particle comes to rest after hitting the wall, I've assumed that the wall is infinitely massive. :\ not sure if that is a good idea.

    [itex]E_{0}=E_{1}[/itex]

    [itex](p_{0}c)^{2}+(m_{0}c^{2})^{2}=(m_{0}c^{2})^{2}+(m_{wall}c^{2})^{2}[/itex]

    [itex]p=\gamma m_{0}v_{0}=\infty[/itex]

    [itex]\gamma=\frac{L_{0}}{L}[/itex]

    [itex]\frac{L_{0}}{L}m_{0}v_{0}=\infty[/itex]

    [itex]L_{0}=\infty[/itex]

    Not sure if this is correct at all. Any hints or ideas?

    Thanks.
     
  2. jcsd
  3. Jul 4, 2012 #2

    cepheid

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    Hey, sorry for the dumb question, but I'm having trouble understanding the problem. If L0 is the proper length of the stick (the length that you measure in the rest frame of the stick), and if the stick comes to rest in the lab frame after the collision, then wouldn't the observer eventually just measure the length of the (now stationary) rod to be L0? And wouldn't this be larger than any length that the lab observer measured while the stick was in motion?
     
  4. Jul 4, 2012 #3
    That's not a dumb question at all, I'm having the same problem lol. That's what I thought, that it would be larger just not sure how much.
     
  5. Jul 4, 2012 #4

    cepheid

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    Figuring out the difference between the rest length and the length measured in the lab frame when the thing is moving with speed v is simple. It's just given by the length contraction formula. You've already written down the answer: they differ by a factor of gamma.

    What I'm unclear on is what the question is actually asking.
     
  6. Jul 5, 2012 #5

    gabbagabbahey

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    I agree with cepheid that this is a poorly worded question. Specifically, the "thereby coming to rest in the lab frame" part. Is such a scenario strictly possible? What about conservation of momentum?

    As far as your attempted solution goes, is energy conserved in an inelastic collision?
     
  7. Jul 5, 2012 #6
    if you read the section 3.3 Length contraction part of Rindler's book Relativity: Special, General, and Cosmological you will understand it well.

    question based on that information of hitting wall propagates at speed of light and so reaching information of hitting wall to the other end of rod takes time in the rod's frame. read that section it also has similiar example.
     
  8. Jul 5, 2012 #7

    HallsofIvy

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    This has nothing at all to do with relativity since everything happens in the lab frame.
    And, unfortunately, the crucial part, the makeup of the stick, which determines whether it will contract or even break when it hits the wall, is not given.
     
  9. Jul 5, 2012 #8

    cepheid

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    Well, the stick is described as relativistic in the problem, and it will be length contracted when it is in motion in the lab frame, and not length contracted when it is at rest in the lab frame. Can you elaborate on what you mean?
     
  10. Jul 6, 2012 #9
    Thanks for the help everyone. After looking at some of the replies and my solution I'm going to go back and read over the material as I think I may be missing a few crucial concepts.
     
  11. Jul 6, 2012 #10

    vela

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    This is fine, but it turns out to not be very useful. An infinitely massive wall can absorb momentum and energy without moving. It'll absorb whatever is needed so that conservation of energy and momentum hold.

    Based on what crimsonidol said, what the problem is asking makes sense to me now. When the front end of the meter stick hits the wall, it immediately comes to rest, but the rest of the meter stick can't know this has happened immediately. It takes time for this signal to travel back through the stick. Once a piece of the stick gets this info, it also comes to rest. When the signal finally reaches the back end, the entire stick will have come to rest. The end result is that in the lab frame, you end up with a smushed stick that is no longer 1 meter long.

    I found the final length in meters is given by ##\sqrt{\frac{1-\beta}{1+\beta}}## where ##\beta=v/c##.
     
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