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Definition of elastic collision

  1. Aug 29, 2008 #1
    I am a bit confused about the definition of elastic collision.
    What i have studied is, elastic collision is one in which the kinetic energy is conserved.

    Now, some books also say that elastic collision is one in which both momentum and kinetic energy is conserved. But, is it true that momentum is always conserved in an elastic collision.
    I have an example where in the collision is elastic but the momentum is still not conserved.

    A rod of mass M is hanging from the hinge of the ceiling. Now a point mass 'm' collides elastically with the rod. Surely the energy is conserved. But is the momentum conserved?
     
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  3. Aug 29, 2008 #2

    Defennder

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    Why isn't momentum conserved in that case?
     
  4. Aug 29, 2008 #3
    I have a dog that tells me that dogs don't talk. If a collision is to be elastic, then momentum and kinetic energy must be conserved by definition. Like defender stated, there is no reason why momentum would not be conserved. Remember that momentum does not only work linearly but rotationally as well.
     
  5. Aug 29, 2008 #4

    tiny-tim

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    Hi i_island0! :smile:

    Momentum in any direction is instantaneously conserved in all collisions in which the external forces have no component in that direction :smile:

    (either because there are no such forces, or because there are forces, but they are infinitesimal compared with the impulsive force of the collision)

    but there are plenty of exam problems in which you use momentum to find the velocity immediately after the collision, and then external forces take over, and you have to forget about momentum, and use energy. :rolleyes:
     
  6. Aug 29, 2008 #5
    Ok, so please follow this link and please comment on it. Please take the collision to be elastic in nature.
    http://www.flickr.com/photos/63184961@N00/2808272733/

    Can we write momentum conservation equation for the system shown?
    If yes, can you tell me the equation.
    Of course angular momentum can be conserved about the hinge, but how about the momentum.
     
  7. Aug 29, 2008 #6

    tiny-tim

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    … you're right …

    Hi i_island0! :smile:

    Yes … on second thoughts, you're right :approve: … linear momentum is not conserved.
    The collision is elastic, so the ball will bounce back with unknown speed V, and the rod will start to move with unknown angular momentum w.

    This is different from the usual exam problem, which I was expecting, of a ball or bullet hitting something on the end of a string.

    In that case, the external force is along the string, which is vertical, and so there will be conservation of horizontal components of momentum.

    In this case, the external force is at the hinge, and its direction is unknown, which prompted you to point out:
    … and you're right :smile: … momentum is not conserved (or to be precise, it is, but we don't know along which direction), but angular momentum about any axis through the hinge is.

    hmm :redface: … so I should have added to my last post:
    Angular momentum about any axis is instantaneously conserved in all collisions in which the external forces have no torque about that axis …​

    So you have two equations … conservation of energy, and of angular momentum … which should be enough to find the two unknowns. :smile:
     
  8. Aug 29, 2008 #7
  9. Aug 29, 2008 #8
    I would further want to comment on the definition of elastic collision.
    Elastic collision is one in which translation Kinetic energy is conserved. IF during collision, some portion of energy is converted into vibrational or even rotational KE, then such a collision cannot be called elastic.
     
  10. Aug 29, 2008 #9

    tiny-tim

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    Hi i_island0! :smile:

    I don't think I agree with that … "elastic" simply means that kinetic energy is conserved, and rotational KE is included.

    Of course, "elastic" isn't a word like "newton" that has an internationally recognised definition … it's just a customary word … but I'm pretty sure the people who set exams use it to include all KE.

    hmm … on the other hand, you were right the last time :rolleyes:

    so if you have any quotation to back up your claim, let's see it! :smile:
     
  11. Aug 29, 2008 #10
    sure i have, else why will i make such statements.
    I was just reading a book, i forgot its name, i will look back again. But i was also shocked to see that elastic collision was defined as relating to only translation KE.
    Currently i am reading feyman's lectures on this and if i have very supportiv proof, i shall certainly write back on this.
     
  12. Aug 29, 2008 #11

    tiny-tim

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    That's what I thought! :smile:
    Yeah … sometimes books get it wrong! :rolleyes:

    I'll be interested to see which book it is! :tongue2:
     
  13. Aug 29, 2008 #12
    I can mail you if i know your email address
     
  14. Aug 29, 2008 #13
    Quite interesting to note that there are two different definitons of elastic collision. In classical mechanic we always begin with Newton laws while the conservation laws come later.

    What are the quantities that we can calculate from this problem? I think we can always cross check the answers for consistency using Newton and consevation laws.
     
  15. Aug 30, 2008 #14

    atyy

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    I wonder if the book that defined an elastic collision to include only translational KE had something like this in mind: If you have an inelastic collision in which the KE is lost to the heating of a gas, then you can also consider it to be an elastic collision if you count the translational and vibrational motion of the gas.
     
  16. Aug 30, 2008 #15

    Doc Al

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    I would say that an elastic collision between macroscopic bodies usually means one in which macroscopic kinetic energy is conserved. That includes translational and rotational KE of the bodies as a whole (treating them as rigid bodies or point masses), but excludes anything that implies internal degrees of freedom (like vibrational modes or internal energy).
     
  17. Aug 30, 2008 #16

    atyy

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    Yes, I was thinking along those lines as to how different definitions may come about. The part that I got stuck at was that the vibrational modes, say modelled as internal springs and masses, would consist of potential energy, so KE wouldn't be conserved anyway, so it didn't seem like the additional stipulation was necessary. If we did away with internal springs, then all internal energy would be kinetic, and we would need the additional stipulation, but then this would be more like an ideal gas, so it wouldn't be a rigid body.
     
  18. Aug 30, 2008 #17
    i would agree with you both... because its not at all making sense to exculde rotational energy.
    So, finally shall we conclude this thread with that:
    elastic collision is one in which KE (rotational + translation) is conserved.
    If some portion of KE goes in vibrational energy, then such collision shouldn't be called elastic.
     
  19. Jan 7, 2011 #18
    Re: … you're right …

    I know all of you probably thought this thread will be dead for ever and I apologize for brining it back to life, but I am investigating the subject of rigid body collisions, focusing on restrained/constrained collision conditions and that is why I refer to the hinge & rod example.

    Why would you say that the impulsive force exerted on the rod at the hinge is unknown? My understanding of the problem takes several assumptions but I am pretty sure they are already made and won't be repeating them. Here is my perception of things:
    - the hinge joint is a motion constraint which makes the attached end of the rod immobile in translational terms
    - the hinge will do its job of keeping the attached end of the rod immobile by applying an appropriate force in order to e.g. compensate the influence of gravitation
    - no matter what happens to the rod, what impulsive and non-impulsive forces and their moments are applied to it, the hinge will act correspondingly doing anything that is needed to keep the attached end of the rod immobile
    - now, an object hits the rod and thus exerts an impulsive force in a known direction; all non-impulsive forces are negligible since we're dealing with an impulse; due to the impulse, the rod will naturally try to gain velocity and the hinge will have to prevent that; therefore - the impulse at the hinge will be an exact opposite of the one at the collision point

    Voila.

    I would be grateful if you could comment on my interpretation of this thing and here is why: I came across this thread searching for angular momentum conservation at impact and found it very useful and making me think about it in a different way but I'm not really able to agree with all the remarks, which makes me feel uneasy.

    Thanks, k.
     
  20. Jan 8, 2011 #19
    in an elastic collision, energy doesnt transport from the particles.And in order to better understand stuff like this ,we should see why they are defined, for example ellastic collission I think is defined to give us the abillity to use the kinetic theory of gases (the whole PV=nRT stuff)
     
  21. Jan 8, 2011 #20
    sentinel - I assume you've expressed your general opinion about elastic collisions, but haven't really replied to my post, right?

    This is perfectly okay with me, I just wanted to know whether you want me to reply in turn :).
     
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