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Question on Momentum

  1. Feb 10, 2013 #1
    We know that momentum of a closed or isolated system is always conserved .
    Now for INELASTIC collision, lets make It simple,consider a system with only two moving particles. If they collide inelastically, we know that energy is surely lost! now the system is no longer a closed system right?? The total momentum of d system must be decreasing right?
    However we could still use d principle of conservation of momentum equation is because during collision the impulsive force is totally exerted onto both colliding objects and is not to external factor right??
  2. jcsd
  3. Feb 10, 2013 #2
    Inelastic means that kinetic energy is lost. Energy is still constant if the system is closed. Momentum is also conserved.
  4. Feb 10, 2013 #3


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    No, total energy is conserved in an isolated system. Macroscopic kinetic energy can be reduced by converting into other forms of energy.

    Converting macroscopic kinetic energy into thermal energy creates radiation which leaves the two bodes. If you include that radiation in the system, then it is still closed, and both momentum and total energy are conserved.
  5. Feb 12, 2013 #4
    Now, I gt another question, for a closed.system, if inelastic collision where kinetic energy is lost occurs between particles, surely the total kinetic energy before must be less than after one right?? the kinetic energy might be lost to the system as other form of energy, bt in this case the total kinetic energy is no longer conserved right? ( even the total energy of the closed system is conserved. ) so accordingly the total momentum should also decreases right??
  6. Feb 12, 2013 #5

    Doc Al

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    Why do you think that? Just because the total macroscopic kinetic energy decreases in an inelastic collision does not mean that the total momentum decreases.
  7. Feb 12, 2013 #6


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    If two equal lumps of mud are moving towards one another, the total momentum is Zero (Earth frame of reference) but the the energy of each is mv2/2. After they collide and stick, the KE becomes zero (Energy has warmed up the mud) but the Momentum is unchanged - still zero. This trivial example shows how KE and Momentum behave differently so you can't necessarily expect them to behave the same, in any collision.
  8. Feb 12, 2013 #7
    Hmm, now there still seems like somethings is absent. It there a theoretical prove that total momentum is conserved also???
  9. Feb 12, 2013 #8


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    'Proof' by virtue of loads of experimental evidence where results confirm that a model, based on conservation of momentum, works. It's one of the most basic of ideas that hasn't been shaken yet ifaik.
    If you were to suggest a World in which momentum is not conserved the you would have to start pretty much from scratch and 'explain' all our observations in a different way.
    I assume that we are talking Newtonian Physics here? Let's not try to run before we can walk.
  10. Feb 12, 2013 #9


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    Is there "proof" that total momentum isn't conserved?

  11. Feb 12, 2013 #10
    You might consider "Noether's Theorem" to be a theoretical basis which implies momentum conservation. "Prove" is a loaded word though.
  12. Feb 12, 2013 #11


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    Yes, if we assume that the laws of physics are invariant under spatial translation (loosely speaking, that the laws of physics don't depend on location), then Noether's Theorem can be used to prove that total linear momentum is conserved.

    This of course begs the question, "how do we prove that the laws of physics are invariant under spatial translation?" :uhh:

    In the end, all "proof" in physics is based on experimental observation and is never 100% absolute. That's why we continue to do more and more precise experiments to test the things that we think we know. Sometimes we get a surprise that leads us to new physics!
  13. Feb 18, 2013 #12
    Well, qualitatively one might say that all the molecules of the body collide during inelastic collision. If you could "see" them one by one you would see that momentum is conserved in each collision. Yet, the molecules' momenta are not constrained in the z-direction (direction of motion of the projectiles). Thus they will conserve momentum in all directions. Their collisions are chaotic, thus increasing the temperature of the system-> heat is radiated.
    One could say that the nr of molecules that go upwards after collision is equal to those going down. The momenta are roughly equal, thus the transverse momentum of the system after the collision is zero.
    In analogy, if a molecule with momentum p collides with another with p=0, in the final state, p will be distributed between the two. However their vector sum will always be p.
    Therefore the total momentum parallel to the projectile's motion will be conserved.
    The above two result in momentum conservation, but mechanical energy non-conservation.
  14. Feb 18, 2013 #13


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    That post is confusing momentum conservation with Equipartion of Energy arguments.
  15. Feb 18, 2013 #14
    Yes it does!
    But if you want "proof" about the conservation of momentum in an inelastic collision, I don't see how you would answer it elseway!
    In my point of view, the momenta gained after the collisions span in x,y,z , but the net effect is that they are cancelled out leaving the total momentum identical with the initial one.
  16. Feb 18, 2013 #15
    e.g if you assume a frictionless pools table, and the white ball stricking the others while beginning the game, the argument would be as follows:

    -Assume that the collisions between the balls are elastic.
    -Assume that the white ball immediately stops after collision.
    -The collision is perfectly central.
    -The white ball's momentum is P.
    The rest of balls are scattered in all directions. However, the total transverse momentum of the balls going up is same as the transverse momentum of the balls going down (why shouldn't it be?). Thus these two components cancel out. What remains is the momentum parallel to the white ball's momentum.
    Each ball, j, carries some momentum p(j). If you add up these momenta you cannot find more than P , neither less because this would mean that there was kinetic energy lost which cannot be since collisions are assumed to be elastic.
    thus the p(1)+p(2)+....=P
    Now, if you take the example in the molecular scale, same goes for the molecules that collide when we have inelastic collisions.
    Transverse momenta are balanced due to the statistical nature of the effect.
    However, there is net amount of energy transfered to the transverse directions equal to E(j)=p(j)^2/2m(j), for each molecule. So the system's internal energy increases and there is heat radiated. The p^2 does the trick here.
    Doesn't this toy model make sense?
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