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Momentum conservation in perfectly inelastic collision

  1. Oct 15, 2013 #1
    Is momentum really conserved in a perfectly inelastic collision?

    By definition there is no conservation of kinetic energy, but isn't conservation of momentum dependent on conservation of energy? Shouldn't a system whose model takes into account conversion of kinetic energy into heat and deformation, not take into account conservation of momentum? Or is it that the loss of momentum is so negligible compared to the loss in kinetic energy that we can assume it's not existent?
     
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  3. Oct 15, 2013 #2

    rude man

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    Momentum is always conserved, even in aperfectly inelastic collision.

    The thing is, you have to be carefuil to include all the mass involved.

    Example 1: a squishy pool ball hits the side of the table and sticks to it. In this case the momentum of th entire Earth must be included!

    Example 2: perfectly inelastic collision between two squishy balls on ice with no friction. Then the only mass involved is the two balls.
     
  4. Oct 15, 2013 #3

    Nugatory

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    Conservation of momentum does not depend in any way on conservation of kinetic energy. Even in an inelastic collision in which kinetic energy is converted into heat (or used to crumple sheet metal, or shatter solid materials, or whatever) momentum is completely absolutely 100% conserved.
     
  5. Oct 15, 2013 #4
    And why is that?

    EDIT: So in any collision ranging from perfectly elastic to perfectly inelastic is momentum conserved?
     
  6. Oct 15, 2013 #5

    Nugatory

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    Yes.

    For some collisions, it's easiest to visualize this if you first transform into a frame in which the total momentum is zero.
     
  7. Oct 15, 2013 #6

    jtbell

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    The change in an object's momentum when a force acts on it is given by the impulse of the force: basically force times the length of time during which the force acts.

    According to Newton's Third Law, the forces that the two colliding objects exert on each other are equal in magnitude and opposite in direction. These two forces obviously act for the same length of time.

    Therefore the impulses delivered to the two objects are equal in magnitude and opposite in direction. The objects' momenta change by equal magnitudes in opposite directions, therefore the total momentum must be the same before and after. The two changes "cancel out" as far as the total momentum is concerned.
     
  8. Oct 15, 2013 #7
    The condition for conservation of momentum is "no net force acting on the system".

    Take, for example, two different balls A and B which move towards each other and collide. During the duration of collision, A exerts a force on B and vice versa. The forces are equal in magnitude and different in direction. (Newton's 3rd law) And of course, the duration of the force acting on A and the force acting on B is the same.

    Therefore we have
    Screen_Hunter_01_Oct_16_10_13.jpg

    See, the conservation of momentum is derived from newton's third law. Hence, it doesn't matter if there is energy loss.
     
  9. Oct 15, 2013 #8

    rude man

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    Yes. The center of mass does not move unless an external force is applied. This immediately leads to the conservation of momentum.
     
  10. Oct 15, 2013 #9
    So does conservation of momentum not allow for a collision with complete loss of kinetic energy? Because if a particle A collides with a stationary particle B, and all the kinetic energy is lost during the collision, there will be no energy left for the displacement of either particles, meaning no conservation of momentum. Is it the case that this never happens?

    Also, are you telling me that the vibrations of particles in the atmosphere around the collision that cause all sorts of things, like sound waves and elevations in temperature, do not carry momentum? I guess I could see conservation of momentum being applicable if you took all of those into account, or if you used an approximation in which the loss of momentum is so negligible in this collisions that you could just completely ignore it.

    Are you guys sure that the latter is not the case?
     
  11. Oct 16, 2013 #10

    D H

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    Sure. Pick a frame of reference in which the total linear momentum prior to the collision is zero. It will be zero after a perfectly inelastic collision. Since the system now comprises but one object, that one object isn't moving. That means there is no (macroscopic) kinetic energy after the collision -- in that frame.

    Remember that both momentum and energy are frame dependent quantities.

    Suppose your stationary particle B is *huge*. Imagine a one gram ball of glue colliding with a one megaton asteroid. In a frame in which the asteroid is initially stationary, it's going to remain essentially stationary post-collision. (Note: This initially stationary asteroid defines our frame of reference.) Even if the glueball is moving a million meters/second pre-collision, the glueball+asteroid is moving at a micrometer per second post-collision. How much energy does this have? Not much. In this frame, the pre-collision glueball had 500 megaJoules of kinetic energy while the post-collision glueball+asteroid has 500 microJoules of kinetic energy.

    Momentum is always conserved. Kinetic energy is not.
     
  12. Oct 16, 2013 #11
    Which sentences you read were telling you that particles do not carry momentum? I read everything in this thread and found no places you could pick up this idea.

    The particles certainly carry momentum.

    The loss of momentum in the system is zero instead of "negligible".
     
  13. Oct 16, 2013 #12

    rude man

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    That is incorrect. Remember, momentum is a vector, i.e. with direction. Say your particles collide with zero remaining kinetic energy, i.e. v3 = 0. Assuming no external forces on the particles (frictionless plane, no air friction, etc.) then we have

    momentum before the collision: m1 v1 + (-m2 v2)
    momentum after the collision: (m1 + m2)v3 = 0

    So this indicates that the only way your blob after the collision would come to rest is if their prior separate momenta equated to zero. If m1 v1 ≠ m2 v2 then the blob would have some remaining velocity v3 and k.e. = 1/2 (m1 + m2)(v3)^2.

    If you introduce air friction into the collision, that constitutes an external force to the system defined by the two particles alone, and the momentum is changed: Δp = ∫F dt where p is momentum, F is the force of air friction on both masses, and t is time.

    If on the other hand you include the air molecules as part of your system, then the total momentum of the system would remain unchanged.

    So you have to be careul to define your system. As I pointed out before, the whole Earth must sometimes be considered part of the "system".
     
    Last edited: Oct 16, 2013
  14. Oct 16, 2013 #13
    While you can assume that in a perfectly inelastic collision kinetic energy is lost and therefore that energy conservation fails, in a better model you would have to specify where that energy went, because in an isolated system there will AWAYLS be conservation of energy. So for simplistic purposes some models can completely ignore conservation of energy, like the ones that deal with friction, and still give accurate enough results even though we know that is NOT what is actually happening.

    Obviously in an isolated system momentum is conserved, but then again so is energy, by definition of an isolated system.

    My point is, if your model takes into account loss of energy to things like heat and sound, and those things carry momentum, two particles that collide and loose energy to that will also loose momentum. I am accepting the fact that in most cases you can assume that loss of momentum is zero, but it's just hard to ignore the fact that you use heat and sound to justify the loss of kinetic energy but completely forget about those when talking about momentum.

    It just bothers me that the source of squander of energy is NOT a source of squander of momentum, even though the things that use up that energy clearly HAVE momentum.

    Are you completely sure that conservation of momentum in an inelastic collision isn't just a useful approximation?
     
  15. Oct 16, 2013 #14

    A.T.

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    As rude man said, it's because energy is a scalar that has no direction information, while momentum is a vector with a direction.

    Since KE doesn't care about the direction, you can convert macroscopic KE (bulk movement in one direction) into microscopic KE (chaotic movement in all directions).

    But this doesn't work for momentum, because it has direction. So you cannot diffuse and hide momentum that way, because conservation of momentum implies the conservation of the direction of the original bulk movement (center of mass).

    Also note that the macroscopic KE is converted into internal potential energy as well, which is not associated with movement and momentum. But there is no such other type of momentum.
     
  16. Oct 16, 2013 #15
    Don't forget that momentum is conserved if the system has no interaction with external systems.
    If it heats up the medium or produces sound, there is interaction, the system is not perfectly isolated.
    So for an approximately isolated system, the momentum is approximately conserved, sure.
    Conservation of momentum is not an unqualified property. It only works for perfectly isolated systems. If you want it to be "perfect".

    But you are right. If a moving body hits a body at rest and they stick together, the final speed cannot be exactly zero so kinetic energy is not completely converted in other forms.
     
  17. Oct 16, 2013 #16

    D H

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    Something that hasn't been mentioned yet: Collisions occur quickly. What this means is that even if external forces are present, these external forces can oftentimes be ignored in a collision event because the rapidity of the event makes FΔt rather tiny.

    In that particular frame of reference, in which one of the bodies was at rest. You can always find a frame in which the pre-collision total momentum is zero, and in this frame, kinetic energy *is* completely converted into other forms.
     
  18. Oct 16, 2013 #17

    sophiecentaur

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    If you want to understand a basic concept like the conservation of momentum then it is simple models that will give you that understanding. Merely introducing greater and greater complexity doesn't help and can tempt one to assume that the Momentum, somehow 'sneaks' out of the system. That is looking at things the wrong way round. If the principle works between two bodies then it will work everywhere.
    All successful Science experiments attempt to keep things as simple as possible and that also applies to scientific thinking.
     
  19. Oct 16, 2013 #18
    Yes, but tiny is not "perfectly" zero. And he wants to be perfectly conserved.

    I expected you will say this. :smile:
     
  20. Oct 16, 2013 #19
    Well, conservation of momentum is derived from newton's third law of motion. So, it holds true whenever newton's third law of motion is applicable. As far as I know, newton's third law of motion is true when the frame of reference is inertial. (There is another condition for 3rd law to be true but it is not the point of this discussion.)

    Momentum is conserved before and after collision. To say that it is approximately conserved is to say that action and reaction is approximately equal.
     
  21. Oct 17, 2013 #20
    Not quite. It's derived by applying Newton's third law to an isolated system.
    External forces can change a system's momentum without violating any law.

    Collisions satisfy the "isolated system" condition pretty well, most of the time.
    However a car's momentum is not usually conserved (if it were the car would be useless :smile:) even though every internal interaction satisfies Newton's third law.
     
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