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Conservation of Momentum v.s. Conservation of Energy

  1. Nov 2, 2012 #1
    Hi, I'm new to this forum.

    So I do not understand how, in an inelastic system, momentum is conserved.

    Consider this problem:
    Tarzan swings from a branch 4 meters up and picks up jane how far can he swing up the other side?
    He weighs 100kg and is standing still on the branch. He jumps off and picks up 60kg Jane who is standing still on the ground. How high of a branch can he reach on the other side?

    At the top, Tarzan has total energy = mgh = 3920J

    When he goes down to pick up Jane, why doesn't this work:
    Ek = (0.5)(160)(v^2)
    v = 7m/s

    Why must we get velocity using Ek formula before Tarzan picks her up, and then use conservation of momentum to find new velocity, and then use Ek to get the new total energy (where there is a loss) and then find the new height. So like this:
    3920 = (0.5)(100)(v^2)

    Momentum Before pickup = Momentum after pickup
    v = 5.53m/s (as opposed to 7m/s)

    And the rest.

    WHERE is the energy lost?! Both mathematically and conceptually. I mean, we are in a closed system and assuming no friction, right?

    I am just perplexed by this whole unit.
    I understand that energy is lost through sound and heat, but it should not be going on here.
  2. jcsd
  3. Nov 2, 2012 #2


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    welcome to pf!

    hi 012anonymousx! welcome to pf! :smile:
    yes it should :wink:

    "rigid bodies" aren't as rigid as we like to pretend

    they vibrate a lot, and vibration requires energy

    the mystery is how, in elastic collisions, that energy isn't lost …

    in most collisions, at least some of it is lost (in the sense that it isn't returned to the kinetic energy of the body as a whole) :wink:
  4. Nov 2, 2012 #3
    Energy is lost from deformation of the bodies. At this level, just accept that elastic collisions conserve energy and momentum, whereas inelastic collisions only conserve momentum.
  5. Nov 2, 2012 #4
    Aha, so at my level... we aren't taught the whole story.
    In the example,
    Eg = Ek(before pickup) = Ek(afterpickup) + Evibration + Edeformation (I suppose)

    Now, I have a question. How does my simple momentum conservation formula,
    pbefore = pafter,
    mv = mv
    Know exactly how much energy is lost.
    Doesn't one need to know the exact rigidness and softnes of surfaces etc.?
    Is this an actual formula that is used or is it just something to teach me principle?

    So another hypothetical question based on above:
    A car crashes into a wall. This is a clear example of how the kinetic energy is lost.

    Now, how would momentum be conserved?
    pbefore = p after
    m(high v) = m(no v)

    Where is the momentum conservation in this case?
    Last edited: Nov 2, 2012
  6. Nov 3, 2012 #5


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    hi 012anonymousx! :smile:

    (just got up :zzz:)
    this is an actual exact formula that applies in all collisions in which there is no external force (in the relevant direction)

    (btw conservation of angular momentum also applies exactly, if there is no relevant external torque)

    (energy has nothing to do with it)

    there is an external horizontal force (the force keeping the wall fixed in the ground),

    so horizontal momentum is not conserved! :wink:
  7. Nov 3, 2012 #6
    law of conservation of linear momentum has never been violated where you talk about space or on the earth...In the car crash example very high force acts on car due to reduction in velocity from some value to zero in very small time(i.e deaceleration) so f=-ma acts on it.and due the force acting obviously the momentum will change.For such problems learn the concept of impulse of force(ΔP=FΔt) where ΔP is the change in momentum and Δt is a very small time
  8. Nov 3, 2012 #7
    Pi(momentum of tarzan moving with velocity Vi)=Pf(momentum of tarzan moving with jane at velocity Vf)
    so till the point he touches the ground he is alone so only his mass should be considered for initial case....
    Last edited: Nov 3, 2012
  9. Nov 3, 2012 #8
    there is no loss of energy bro you calculated it wrongly.When tarzan reaches the ground till that point its mass is only 100kg so u cant consider it as 160.....
  10. Nov 3, 2012 #9
    Okay, I understand conceptually. But now mathematically:

    The conservation of momentum formula says there is an inverse relationship between mass and velocity. When mass is picked up, velocity goes down by that same amount.

    So in the kinetic energy formula, when v goes down and mass goes up, because v is squared, there is a bigger difference. Therefore there is a "loss" of energy.

    And I suppose there is a "constant" for this loss, related to how much mass the momentum is distributed through.

    Now my question is:

    Given these formulas, given ANY situation, there will always be a loss of energy to other forms because of the way the formulas work (reabove).

    So can there be such a thing as a completely elastic system, in which Ek is preserved? The formulas won't allow it. Can someone give me an example of one?
  11. Nov 3, 2012 #10


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    Sure. Look at the Kinetic Theory of gasses. All collisions there are elastic.

  12. Nov 3, 2012 #11
    Ah, so different theories and formulas for gasses and things at the subatomic level, right?

    But there is no such thing as an elastic situation in real life.
  13. Nov 3, 2012 #12
    He is asking if a completely elastic collision really exists. Kinetic gas theory approximates the collisions as elastic, even though there is electrostatic forces at play. A completely elastic collision doesn't exist in nature as far as I know.

    To stop you from further troubles, just refer to my first post. At a higher level you will learn about other sources of loss, but usually the loss is so minor that the approximation of ignoring them is fine.
  14. Nov 3, 2012 #13


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    yes, just as there's no such thing in real life as a frictionless surface, an inextensible string, a massless pulley, and so on …

    but there very nearly are … the error in assuming elasticity is far less than all the other experimental errors!! :smile:

    (and there are in exam questions! :wink:)
  15. Nov 4, 2012 #14
    Okay, I understand now and thanks for your help, but:

    One last thing. This is on my homework sheet. I wanted to understand conceptually before doing it.

    Assume frictionless surface.

    There is a stationary cart and a cart going at 1m/s towards it. Both 1kg.
    They hit. There is no velcro or magnets.
    What happens?

    That is the scenerio I made. Really it is investigating the "Thunk" collision (types of collisions). The specific question is:
    Find a sample collision that helps us easily decide if a collision, where two carts collide neither magnetically nor with velcro, is inelastic or elastic.

    Another possible scenerio might be that they are going toward each other.

    It is obviously inelastic but I need to conserve momentum. and also find energy loss. I need specific values for that. Can someone help? Here is what I have:
    p = p
    mv + 0 = mv'+mv'
    1(1) = v(m+m)
    1 = 2v
    v = 0.5

    Of course, I don't think I should me making the final velocities the same. I think this is only allowed for collisions where they stick.
    Please and thank you!
    Last edited: Nov 4, 2012
  16. Nov 4, 2012 #15


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    Please note that the Drude Model of conduction electrons makes full use of such elastic collision. And if you carry through that model, you arrive at Ohm's Law!

    Now, you can argue that this is a "approximation" (and as a condensed matter physicist, I can talk all day about this till everyone turns blue), but the fact remains that in numerous situations that you and everyone else are using, the scenario of elastic collisions are perfectly valid!

  17. Nov 4, 2012 #16


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    i don't see how you can answer that, except by doing an experiment with toy cars, say, or by assuming a given coefficient of restitution :confused:

    i suggest you start by saying what would happen if energy was conserved, and then work out the precise result assuming a given coefficient of restitution :smile:
  18. Nov 4, 2012 #17
    When you hit the wall, the wall experiences a force equal and opposite to the one the car experiences. That force is communicated to the ground.

    F=ma for the car. Earth get's -F=MA. Or ma=-MA -> A=-ma/M. M is huge compared to to you car, so A is tiny. Earth does get a little change in momentum, it's just too small to perceive.
    Last edited: Nov 4, 2012
  19. Nov 4, 2012 #18
    That is not completely accurate. Most discussions of the subject include some reference to a container which exchanges energy with the molecules. Furthermore, if the molecules have any accessible degrees of freedom such as a diatomic molecule, the collisions are not necessarily elastic. At least from the view of center of mass.

    That is to say, a collision can impart angular kinetic energy to the molecule which is lost to the translational kinetic energy. More complex molecules can have even more degrees of freedom. Some of these degrees of freedom "freeze out" at lower temperatures because of their quantum nature.

    Two mono-atomic molecules colliding without exciting their electrons to higher energy states can be considered completely elastic.
  20. Nov 4, 2012 #19


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    The earth gains all of the momentum lost by the car, but since the earth is massive the change in earth's velocity (angular and linear) is tiny. The total momentum of what can be considered a closed system consisting of the earth and the car remains constant, ignoring any external forces (like gravity from the moon or sun).

    As for the loss in kinetic energy with in-elastic collisions, in most situations, it's converted into heat during the deformation process. The relative amount of loss depends on how in-elastic the collision is.
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