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Concept of conservation

  1. Nov 12, 2004 #1
    I had a conceptual question. Say I have a ball and I throw it against the wall and it bounces back. I have initial values for its velocity but not for the final velocity using the [tex]\vec{P}=m\vec{v}[/tex] I noticed that momentum is not conserved. Well if momentum is not conserved then that means it is not inelastic or elastic and the only option left is completely inelastic but..the wall cant move with the ball...whats going on? Is it just that you cant determine these kinds of collisions? Or do I have some misunderstandings of the whole concept of the elastic or inelastic collisions?
  2. jcsd
  3. Nov 12, 2004 #2


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    I'm not sure about the wording, but you basically have 3 choices: completely elastec, completely inelastic, and somewhere in between.
    Can't it? Why not? There is no such thing as a perfectly rigid object.

    Also think about what else is happening: is there any other energy that you can account for during the impact?
  4. Nov 12, 2004 #3
    The wall exherts a force back on the ball, so would that mean that the wall moves a tad bit? All I know is that there is a very small [tex]\Delta{t}[/tex] where there is contact btween the ball and the wall there is a mass and an initial velocity [tex]\vec{v_0}[/tex].Also I know the impulse, using the [tex]\vec{J}\int\sum\vec{F}dt[/tex] I found all the values that I needed. I saw that the initial [tex]KE[/tex] does not equal the final...and the same with the initial [tex]\vec{P}[/tex] it cant be elastic or inelastic but there would only leave 2 other option 1) I cant know 2)completly inelastic

    completely inelastic would mean that [tex]m\vec{v_0}=(M+m)\vec{v_f}[/tex]
    looking at it in terms of the equations I doesnt make sense..but it would if the the wall moved a littele...you said that there is no perfectly rigid body so a completly inelastic collision would have to be the safest answer..? Is that the correct reason or answer?
    Last edited: Nov 12, 2004
  5. Nov 12, 2004 #4
    you'll get a small amount of deformation in the wall at the time of impact. think about what happens to the ball at impact though.
  6. Nov 12, 2004 #5
    It will experiance a force from the wall, an equal force in magnitude.
  7. Nov 12, 2004 #6
    isn't that where e, the coefficient of restitution comes into play?
  8. Nov 12, 2004 #7
    might that force cause the ball to deform as well?

    as russ said earlier, any time you're talking about collisions in a problem, you have to specify what type of collision you are going to be working with. if your problem deals with elastic collisions, or if you have defined your objects to be completely rigid and non-deformable, then momentum and energy are conserved for your collision problem.

    if your problem deals with inelastic collisions, or if one or more of your objects are deformable, then momentum is not conserved (energy is always conserved). some of that momentum/energy is taken up deforming your object(s).

    you cannot talk about collisions without specifying the type of collision you want to work with, or the properties of your colliding objects.
  9. Nov 12, 2004 #8


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    Yeah, the problem of approaching the collision problem from the momentum side of things is that you need to now the characteristics of the collision and can't, at least as I see it, link it to the behavior of the problem itself... directly at least if you don't have any information on the actual event. If you'd want to incorporate the real deal what is going on in the collision you would have to do a dynamic contact analysis, where the deformations subjected to both the wall and the ball would be accounted for. That, however gets really easy really complex and at this point in time the usual way to solve it is to numerically solve Newton's 2nd incorporating stiffness of both parties (in this case if the ball is as soft as they usually are, in nonlinear sense), in some cases even damping would come to play.
  10. Nov 12, 2004 #9
    For an efficient ball (like a superball or a tennis ball) and a sufficiently smooth and dense, sturdy wall, I think you could approximate conservation of momentum to be true. There will be less afterwards but it is a small change compared to the total energy, if you throw it hard.
  11. Nov 12, 2004 #10
    All I know is this! I was given a problem I solved it and then I was asked what kind of collision it is. Im not making this problem up for "fun" it was given to me. The final and initial momentum are different (by a significant amount) and so is the kinetic energy :cry: what to do?
  12. Nov 12, 2004 #11
    if momentum is not conserved, then it is an inelastic collision.
  13. Nov 12, 2004 #12


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    Er... no. If the kinetic energy is not conserved, then it is an inelastic collision. If the momentum is not conserved, then there is an external force acting on the system.

  14. Nov 12, 2004 #13
    duh, that's right. my mistake. thanks for correcting me.
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