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One dimensional Elastic collision of two identical particle

  1. Jan 30, 2016 #1
    Hi everyone. I've a question that i wondered since the high school. Let's take two identical particles (same mass) that collide frontally. Assume it's an elastic collision. We have to conservate both the momentum and kinetic energy:

    v_1 + v_2 = v'_1 + v'_1
    v^2_1 + v^2_2 = v'^2_1 + v'^2_1

    (where primes denotes the velocities after the collision). Now I do know the solution: the velocities are swapped among the two particles and here comes my question: since the one I wrote is a symmetric system, why should I not accept the solution where the two velocity are not swapped?
     
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  3. Jan 30, 2016 #2

    PeroK

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    You should accept that as a valid solution. What physical situation does it represent?
     
  4. Jan 30, 2016 #3
    Every book I checked does not take into account such a solution. In fact, in order to obtain the desired solution they dived by (v_1 - v'_1) (and same for the v_2). As a physical solution I imagined a moving particle colliding with a static one. In such a situation, the solution where each particle keeps its velocity is represented by the first particle hitting the second one that remains still while the first one goes on. Am I right? Is this possibile?
     
  5. Jan 30, 2016 #4

    PeroK

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    It's simpler than that. Imagine particle B is at rest and particle A is moving towards it at speed ##v##. Particle B closes its eyes and afterwards is still at rest, while particle is moving away at speed ##v##. What has happened? It's very simple!
     
  6. Jan 30, 2016 #5

    PeroK

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    If you haven't got it yet, here's another clue:

    4ea4e79503e29-image.jpg
     
  7. Jan 30, 2016 #6
    Ok thanks Perok that's very illuminating! :D
    In other words IF there is a collision THEN we can assume momentum is transferred, ELSE we have a miss, right?
     
  8. Jan 30, 2016 #7

    PeroK

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    Yes. I'm suprised none of the books mention this, as it's a good example of having to interpret what is, after all, a perfectly valid mathematical solution. The equations are equally satisfied with no change to either particle.
     
  9. Jan 30, 2016 #8
    I suppose that, being a 1D problem, you cannot admit a miss...
     
  10. Jan 30, 2016 #9
    Both of you got a point. Now you convinced me. Many many thanks.
     
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