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I How can work be done during collisions?

  1. Apr 8, 2016 #1
    Consider a collision between two point particles, with no external forces acting on the system.

    Linear and angular momentum of the system are always conserved, while the kinetic energy of the system is conserved only if internal forces acting in the collision are conservative. This last point is not clear to me.

    During the time interval [itex]\tau[/itex] of the collision we assume that the position of the two particles does not change, i.e. [itex]\vec{r_1}=\vec{r_2}=\vec{r}[/itex]. But the velocity of each particle (precisely its momentum) does change during the collision, and, from WE theorem, that means that a work is done on each particle, equal to the change in kinetic energy of the particle itself.

    Nevertheless work is force times a displacement, and we said that the position of particle is constant during the collision. So where is the displacement?

    Supposing that I didn't made mistake in this reasoning, if there is no displacement there is no work, so how can the velocity of particles change actually?

    And then if there is no work how can kinetic energy of each particle (and of the system) not be conserved?
     
  2. jcsd
  3. Apr 8, 2016 #2
    If we're talking contact forces at the macroscopic scale then clearly there is an issue with the point particle model. Consider though the collision between two billiard balls. During the contact time for the collision there is a displacement and hence one ball does work on the other.
     
  4. Apr 8, 2016 #3

    Dale

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    That is only true in the center of momentum frame where there is no work done on either particle.
     
  5. Apr 8, 2016 #4

    PeroK

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    You perhaps need to think about what assumptions are. This is a good example:

    Assumption a): there is no displacement (or time interval) for the collision.

    That assumption is all right up to a point, but as you have shown you cannot take it too literally.

    Assumption b): the displacement (and time interval) of the collision are small enough to be neglected

    That assumption is better, as it simply means you are not going to complicate your calculations by worrying about what happens during the brief duration of the collision itself.

    Almost all assumptions you'll come across have two forms similar to a) and b). E.g. "frictionless", as opposed to "neglible friction"; "massless", as opposed to "of negligible mass" etc.
     
  6. Apr 8, 2016 #5
    I would like to expand on what Brainpushups is saying. The billiard balls are deformable (elastic, say), and, as they are colliding (and their leading edges are in contact, and not moving) their centers of mass are still moving. So the contact force is being applied during a displacement of each center of mass. First the centers of mass get closer together, but then the compression of the balls is enough to fully stop the centers of mass and to cause them to start moving apart. Finally, the balls entirely separate from one another. But during the contact period, work is done on each ball. A very crude model for this would be two rigid balls moving together from opposite directions and then compressing a massless spring. The spring would do work on each ball, first slowing it, then stopping it, then speeding it off in the opposite direction.

    Chet
     
  7. Apr 8, 2016 #6

    Dale

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    Again, the contact patch is only stationary in the center of momentum frame (for identical balls). In other frames the patch is moving and work is done.

    In the center of momentum frame there is no work done on either ball. There is a force at the contact patch, but that contact patch is not moving so the energy transfered (work) at the contact patch is 0. Instead, energy is converted internally within each ball from kinetic, to elastic, and back to kinetic, without any work transferring energy from one ball to the other.
     
  8. Apr 8, 2016 #7
    That's a bad assumption for point particles. Actually, it takes some time for a collision to occur and it happens over some distance. For example, two electrons repel each other with a force that goes as 1/r^2, so actually the "collision" goes on over infinite range.
     
  9. Apr 8, 2016 #8
    The OP is asking about an inelastic collision, so K.E. of each ball changes. Let's assume equal masses, equal initial velocity toward each other and for simplicity assume a totally inelastic collision so the balls meet and stop moving. In the cm frame, the total momentum is zero before and after the collision but there is a change in total K.E. as well as K.E. of each ball. How can that happen without work being done?
     
  10. Apr 8, 2016 #9
    This is another good example to reaffirm our understanding of the fact that work is a frame-dependent quantity. As Dale correctly points out:

    From the frame of reference of the center of mass of one of the balls, work is done by the other ball at the interface. The interface is moving toward the center of mass of the ball during the initial part of the contact interval, and the interface is moving away from the center of mass during the latter part of the contact interval (and there is force acting at the interface during the entire contact interval). The amount of work at any time during the contact translates into stored elastic energy plus kinetic energy of the parts of the ball moving relative to the center of mass. However, if the system is symmetric, the net amount of work done on the ball during the total contact is close to zero.
     
  11. Apr 8, 2016 #10

    Dale

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    The KE is not transfered to another object it just changes form internally. If the collision is elastic then the KE goes into elastic potential energy and back to KE. If the collision is inelastic then it goes into thermal energy or acoustic energy or whatever. Either way no work is done because no energy is transfered in or out of the ball, it just changes form within the ball. (In the center of momentum frame)

    In frames where the contact patch is moving then work is done. Energy is lost by the trailing ball and gained by the leading ball. In such frames there is still an internal conversion of energy, but it is not complete and there is always some KE remaining.
     
  12. Apr 9, 2016 #11
    Thanks for the replies! Take as an example a collision between two balls. At the contact point there is a force [itex]f[/itex] acting on the one ball and a force [itex]-f[/itex] acting on the other. Supposing that these forces do work, then, since the two balls move at the same speed during the collision time interval and so their displacement is the same, can I say that [itex]dW_1+dW_2=0[/itex] (i.e. the two works done by the two forces are equal and opposite)? This would imply the conservation of kinetic energy for the system composed of the two balls. If this holds true, conservative or not would not make any difference. Newton third law (which I used here) is valid for non conservative forces too, so the two works should add up to zero in any case (conservative or not). I am missing something here but I don't see where.
     
  13. Apr 9, 2016 #12
    What do you think you are missing?
     
  14. Apr 9, 2016 #13

    A.T.

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    In inelastic collision the work done at the contact is not fully converted into work done internally at the center of mass. So the change in KE is different than the external work done at the contact.
     
  15. Apr 9, 2016 #14
    I don't know, maybe it is a wrong assumption to say that [itex]f[/itex] and [itex]-f[/itex] are the only forces acting in the collision, but, also in the case of friction forces between the two balls, the exact same force must be exerted on the one ball and on the other (in the opposite direction). So, provided the same displacement during collison, it should hold true that the works done add up to zero in any case..
     
  16. Apr 9, 2016 #15
    I think your real question is "in an inealstic collision, what is the physical mechanism for the partial irreversible conversion of usable mechanical energy (e.g. stored elastic energy and kinetic energy) to less usable internal energy (i.e., thermal energy) of the ball and surroundings?" Does this capture what your are asking?

    Chet
     
  17. Apr 9, 2016 #16

    Dale

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    Can you clarify please? Are you asking about a perfectly elastic collision, or a perfectly inelastic collision, or a partially inelastic collision? Also, are you asking about the center of momentum frame or some other frame? Also, are you considering identical balls or is one ball different than the other in some way?

    This is all correct. Except that KE is not necessarily conserved, only total energy.
     
  18. Apr 9, 2016 #17
    Thanks for the reply! I'm sorry I did not specify: actually I'm talking about a generic collision (inelastic or elastic) between the two balls, just to see how the works done by forces behave in general at the contact moment. Since, as you pointed out, work in com frame is zero I'm assuming an inertial laboratory frame where work is non zero and again for masses or other differences between the two balls, I'm thinking about a general case, so let's suppose that they have different masses.

    So the thing is that the works are always equal and "opposite" in every collision but they can have other effect besides the change in KE of particle (deformation, heat, sound), and this is what makes elastic and inelastic collision different. Did I get that right?
     
  19. Apr 9, 2016 #18

    ehild

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    During the time of collision when the colliding bodies touch each other, they do not behave as point masses. The forces acting between them deform them. the point of application of the forces moves, so there is work done. The bodies can regain KE after the collision, or some of the mechanical energy transforms to other kinds of energy. See the video:
     
  20. Apr 9, 2016 #19

    Dale

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    Yes that is correct.

    OK, that is the most general case, and so the equations are the most complicated. So I won't solve the initial and final velocities, I will just show you how the energy conservation works. I will model the balls as springs so that we can calculate the details of the forces and the location of the contact patch. Hopefully I can do that tomorrow.
     
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