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Conservation of momentum in an *open* system

  1. Jan 15, 2016 #1
    i got in an argument with my physics teacher about a test question recently, and am still reluctant to abandon my logic. The question is a standard explosion problem, akin to this:

    A firework is placed in the midst of some motionless billiard balls. The firework goes off and the billiard balls scatter outward from the force of the explosion. The momentum of the billiard balls immediately after the explosion is: A more than before the explosion B less than before the explosion C no change in momentum

    The firework is not part of the system, only the force of its explosion. My point is that the momentum must have increased, as the balls were motionless before (so no momentum), and they were moving after (there is momentum). This does not break the rule of Pi=Pf because it's Ina n open system, and energy was added. The only was the momentum could have stayed 0 was of the balls' momentum balanced (the balls all had another ball go in the opposite direction at the same speed); but we can't know this for sure unless we already knew that the momentum did not change. My teacher says that we must assume the balls's momentum after all balanced/cancelled because Pi=Pf ((me again) but again, it was an open system!)

    Thanks for any help
     
    Last edited by a moderator: Jan 15, 2016
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  3. Jan 15, 2016 #2

    Filip Larsen

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    Welcome to PF!

    As I read your question I think you have missed that momentum has direction in addition to magnitude, that is, momentum is a vector.

    If the explosive charge and the billiard balls are all at rest (if it helps, imagine this to be in the center of a very large, but otherwise empty box representing the system) what is then the total momentum of the system?

    Assuming that all the particles from the explosive charge stick to the billiard ball during the explosion and that all the balls are moving at constant speed, what is then the total momentum of the system (note, that no force has acted through the "walls" of the system)?
     
  4. Jan 15, 2016 #3
    Hi Thomas:

    I also have doubts about the answer, but I think that you are correct. If you consider the entire environment as a closed system, including the material of the firework and the air occupying space in which the balls and the firework were before the explosion, the pre-explosion total momentum vector was zero, so the post-explosion total momentum vector must also be zero. If as your teacher suggests, the post-explosion total momentum vector for the balls is zero, then this would require that the post-explosion total momentum vector for the particles of the former firework together with all the air molecules is also zero. It seems to me that that would be quite an unexpected coincidence.

    Regards,
    Buzz
     
  5. Jan 15, 2016 #4

    PeroK

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    If, instead of a firework, you have someone with a snooker cue, I think you have your answer!
     
  6. Jan 15, 2016 #5

    Merlin3189

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    And if there had been only one ball?
     
  7. Jan 15, 2016 #6

    haruspex

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    True, but the correct conclusion is that, with incomplete information regarding the fate of the exhaust from the blast, the net momentum of the balls is indeterminate. That is not one of the three offered answers. What we can say is that it has not reduced in magnitude.
    While this makes it a dodgy question, Thomas' argument that the momentum of the balls (collectively) has necessarily increased just because they are moving is equally wrong.
     
  8. Jan 15, 2016 #7
    Hi haruspex:

    I cannot disagree with the main thrust of your point. However, there is perhaps a nit regarding "equally". Statistically, the probability that the total momentum vector for the moving balls is zero is infinitesimally tiny.

    Regards,
    Buzz
     
  9. Jan 15, 2016 #8

    haruspex

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    I was measuring equality in terms of soundness of argument, which is something of a value judgment. Thomas argued that because the balls go from stationary to moving they must have gained momentum. That indicates a misunderstanding of the principles. The teacher's error was in failing to consider the practical details and add sufficient wording to cover them.
     
  10. Jan 15, 2016 #9

    HallsofIvy

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    The kinetic energy, which is a scalar, increases. The momentum, which is a vector quantity does not.
     
  11. Jan 15, 2016 #10
    For whatever it's worth, I agree with Buzz. The final momentum can't be exactly zero.

    Chet
     
  12. Jan 18, 2016 #11

    CWatters

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    Just a badly worded problem in my opinion. Mind you in exams missiles manage to explode and conserve momentum without needing to mention details such as the gaseous products of the explosion.
     
  13. Feb 11, 2016 #12
    My point is not that because the balls have gained speed, that they have necessarily gained momentum; it is that they may have gained momentum, but we cannot be sure if they have or have not, unless we know the exact directions and speed at/in which they were pushed. This is because they momentums may have balanced, and the total momentum of all the halls is equal to zero, or they may not balance, and momentum was gained. We cannot assume that they must have equal momentum before and after because it (the balls) is not a closed system (introduction of kinetic energy (and therefore momentum) through the fire-cracker)
     
  14. Feb 11, 2016 #13

    haruspex

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    That argument is still not correct.

    You are right that momentum of the balls might not be zero, but not for the reasons you give.
    The reason, as others have posted, is that explosive devices work by releasing gases, and we cannot know the fate of those. They would have momentum too. The sum of the momenta of the balls and the gases would be zero, but each in itself is almost surely not. So it is not to do with introduced energy.

    If you replace the explosive with springs attached to the balls (each attached to one ball) then the question works as intended. The sum of the momenta of the balls is zero.
     
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