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Kinetic Energy and Inelastic Collisions

  1. May 7, 2004 #1
    Why is KE not conserved in inelastic collisions?
     
  2. jcsd
  3. May 7, 2004 #2

    jcsd

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    Imagine two objects of masses [itex]m_1[/itex] and [itex]m_2[/itex], travelling with velocities of [itex]v_1[/itex] and [itex]v_2[/itex] which after collding inelastically form a new object of [itex]m_3[/itex], travelling at velocity [itex]m_3[/itex], which due to the conservations of mass must equal [itex]m_1 + m_2[/itex]

    We can say this due to the conservation of momentum:

    [tex]m_3v_3 = m_1v_1 + m_2v_2[/tex]

    therefore:

    [tex]v_3 = \frac{m_1v_1 + m_2v_2}{m_3} [/tex]

    We can also say that due to the conservation of energy:

    [tex]\frac{1}{2}m_3{v_3}^2 = \frac{1}{2}m_1{v_1}^2 + \frac{1}{2}m_2{v_2}^2[/tex]

    therefore:

    [tex]v_3 = \sqrt{\frac{m_1{v_1}^2 + m_2{v_2}^2}{m_3}} [/tex]

    combing the equations we get:

    [tex]\frac{m_1v_1 + m_2v_2}{m_3} = \sqrt{\frac{m_1{v_1}^2 + m_2{v_2}^2}{m_3}} [/tex]

    square and mutiply [itex]{m_3}^2[/itex] byboth sides,substitue in [itex]m_3 = m_1 + m_2[/itex] and mutiply out:

    [tex]{m_1}^2{v_1}^2 + {m_2}^2{v_2}^2 + 2m_1m_2v_1v_2 = {m_1}^2{v_1}^2 + {m_2}^2{v_2}^2 + m_1m_2{v_1}^2 + m_1m_2v_2^2[/tex]

    Simply eliminate and you get:

    [tex]2v_1v_2 = v_1^2 + v_2^2[/tex]

    Which can be re-arranged as:

    [tex]v_1^2 - 2v_1v_2 + v_2^2 = 0[/tex]

    using the quadratic formula we can solve for [itex]v_1[/itex]

    And we find that:

    [tex]v_1 = v_2[/tex]


    So for an inelastic collision the intial velcoties of the two colliding objects must be the same, hence no collision.

    QED
     
    Last edited: May 7, 2004
  4. May 7, 2004 #3

    jcsd

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    I'm sure there must be a quicker way to prove it, but though I have seen the proof in textbooks, it's completely gone from my mind so I derived it myself.
     
  5. May 7, 2004 #4

    robphy

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    I'm going to read this as
    "Why is total-KE not conserved in inelastic collisions?"

    Strictly speaking, an Elastic Collision, by definition, is one in which total-KE is conserved... otherwise the collision is, by definition, Inelastic.

    Now, if you were to ask, "Why is the total-KE not conserved when, say, two objects collide and stick?", this can be demonstrated with a short calculation.

    To make the calculation trivial, consider the special case of two objects colliding and sticking together with final velocity zero [the total-inelastic or perfectly-inelastic case]. Clearly, the initial total-KE is positive and the final total-KE is zero. (Of course, total-momentum is still conserved.)
     
  6. May 7, 2004 #5
    These are all obvious trivial calculations. What I am asking is why they turn out this way. Simply going through the derivations is not an explanation.

    I don't think the people who write on this site think very much.
     
  7. May 7, 2004 #6
    Actually robphy made a good suggestion of looking at the problem in the center of mass frame. This at least simplifies the calculations.

    The problem here is that energy is not conserved. This is a big problem.
     
  8. May 7, 2004 #7
    The remaining energy turns into heat. When two pieces of playdough collide, they stick together and all the remaining KE is transfered to heat through the internal friction of the playdough when it is deformed. With a perfectly elastic rubber ball, for example, the energy is stored as potential energy for the time during which the ball is stressed, then it is released in such a manner as to accelerate the ball and none of the kinetic energy is turned into heat. There can, of course, be all cases in between.
     
  9. May 7, 2004 #8

    Integral

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    Not true.

    KINETIC ENERGY is not conserved. An inelastic collision is usually accompanied by deformation of one or both bodies. This requires energy. Thus TOTAL energy is conserved but not necessarily KINETIC ENERGY.
     
  10. May 8, 2004 #9
    But the equation is only considering kinetic energy there is no mention of deformation or internal potential energies.
     
  11. May 8, 2004 #10

    krab

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    What equation? Kinetic energy is not conserved, so there is no equation of the form: total kinetic energy before = total kinetic energy after. Momentum, however is conserved, so for two particles,
    [tex]m_1v_{1i}+m_2v_{2i} = m_1v_{1f} + m_2v_{2f}[/tex]
     
  12. May 8, 2004 #11

    jcsd

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    It's a silly question then, I only went for the one-dimensional example but the whole point is that by the defintion of momentum and by the defintion of kinetic enrgy some energy must be transferred to other forms, kinetic energy can't be conserved in an inelastic collsion which is the question you asked.

    If you wanted something deeper, I could've of proved that 4-momentum isn't conserved (which is a useful result as it illustrates why a free electron can't absorb a photon for example), but it amounts to the same thing.
     
    Last edited: May 8, 2004
  13. May 8, 2004 #12
    Kinetic energy is conservered in an elastic collision but using the same equation for an inelastic collision it is not conserved.
     
  14. May 8, 2004 #13
    4-momentum is conserved. Maybe you should read an intro relativity text.
     
  15. May 8, 2004 #14

    jcsd

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    Not in a totally inelastic collison, hence a photon may not collide inelastically with a free electron, though it may collide elastically.
     
    Last edited: May 8, 2004
  16. May 8, 2004 #15
    Both photons and electrons have momentum. If they have an equal amount of momentum in opposite directions they will collide and both have a final velocity of zero.
     
  17. May 8, 2004 #16

    jcsd

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    In this case the answer is clear, we find that the ETOT is more than the rest mass of the electron, so the electron must have kinetic energy, but if it did have kientic energy then it must have velocuiity and momentum can't be conserved so the situation is impossible.
     
  18. May 8, 2004 #17
    You need to start thinking real soon. You are using what is in question to answer the question. I am saying that you can have an inelastic collision between a photon and an electron because you can have to momenta in opposite directions whose sum is zero. Therefore after they collide they will both be at rest.
     
  19. May 8, 2004 #18

    Alkatran

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    No. Assume that two identical objects are colliding, with opposite speeds (thus opposite momentums).

    Now, total momentum needs to be conserved, so logicly they will have equal speeds (but opposite) at the end of the collision as well. The only limit on the new speeds is that they musn't use more kinetic energy than the original speeds and they must be above or equal to 0. If they have less kinetic energy than before, some of the energy was transformed (into things like the sound of them smashing together, and the heat created by friction), if they have the same amount none was (and, there was no sound to the collision).

    The point is, if two things with opposite momentums collide, they don't have to have a final velocity of 0.
    m1=1
    m2=2
    v1=2
    v2=-1

    1*2+2*-1 = 0 = 1*x + 2*y
    x = -2y
    Nothing says x or y have to be 0.
     
  20. May 8, 2004 #19
    You need to read the post before you write.

    They do have zero velocity because we are talking about inelastic collisions.

    Is there anyone here who can think?

    BTW, if you do the calculation out for a photon and electron moving with opposite momentums towards one another it does work out. Using the relativistic momenta.

    (Energy of photon)/c - (gamma)(mass of electron)(velocity of electron) = 0

    Just solve for the velocity of the electron in this case and see for yourself.
     
  21. May 8, 2004 #20

    If you're such a thinker, why didn't you think your way out of having to ask a question on this forum? We already told you the energy you're looking for goes to heat , sound, etc. What's the problem?
     
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