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B Momentum vs kinetic energy help

  1. Dec 17, 2017 #1
    I have a question about momentum vs kinetic energy.

    For example, a block C with velocity Vc and mass=2m hit a block B with mass=m at stand still on a LEVELED frictionless track( no change of potential energy). The two block stick together and move at velocity Vcb. Find the relation of Vc vs Vbc.

    If I use m1V1 = m2v2, then 2mVc=3mVbc, so final velocity Vbc = 2/3Vc.

    But if I use 1/2 (2m)Vc^2= 1/2 (3m) Vbc^2, then Vbc= sqrt( 2/3) Vc.

    Why is the two answer different using momentum and kinetic energy equation?

    Thanks
     
    Last edited: Dec 17, 2017
  2. jcsd
  3. Dec 17, 2017 #2

    Ibix

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    Is this an elastic collision or inelastic? What do you know about energy conservation and momentum conservation in the two cases?
     
  4. Dec 17, 2017 #3
    They stick together, so it's an inelastic collision.

    Thanks
     
  5. Dec 17, 2017 #4

    FactChecker

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    There is no rule of conservation of kinetic energy in inelastic collisions.
     
  6. Dec 17, 2017 #5
    Why is that? On a frictionless surface, the total kinetic energy before and after the collision should remains the same as long as the surface is absolutely flat ( no potential energy created).
     
  7. Dec 17, 2017 #6

    jbriggs444

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    In an inelastic collision, the extra energy often goes into heat. Kinetic energy is not conserved.
     
  8. Dec 17, 2017 #7
    But is that also true for momentum?
     
  9. Dec 17, 2017 #8

    Ibix

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    In reality, momentum and energy are both always conserved. The difference is that you frequently can't track where the energy went.

    As @jbriggs444 notes, you'll probably find that the blocks and/or air have heated up, or they may well have deformed somehow. The "missing" energy (the sum of the initial kinetic energies minus the sum of the final kinetic energies) went into doing that. If you enclose the whole experiment in a very sensitive calorimeter and work in a vacuum so that no energy can escape as sound, you'll be able to measure the "missing" energy. But we don't normally have a calorimeter in our experiment (it's probably not practical anyway) and we just say "energy is not conserved" and treat the heat as "lost".

    With momentum, there's no problem tracking it because there's nowhere for it to go. Vibrations and rotations, which can absorb energy, have zero net momentum. The only way you can "lose" momentum is if pieces break off your masses and fly away and you don't track them. But you haven't said anything breaks off, so momentum is conserved.
     
  10. Dec 17, 2017 #9

    FactChecker

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    And that is why it is so important to first distinguish between kinetic energy versus total energy and between elastic versus inelastic collisions.
     
  11. Dec 17, 2017 #10
    Thanks for the explanation.
     
  12. Jan 3, 2018 #11
    This conversation is about a thought experiment, not an actual trial. We want to understand and predict what happens to the energy in a collision between two blocks (or automobiles or football players or whatever), not set up an actual frictionless surface with calorimeters, etc.

    So let us just accept that no heat is generated since everything is frictionless, no sound escapes, energy is not radiated off into space somewhere, no potential energy change, no chemical reaction, etc. Keep it as simple as possible - it's a thought experiment.

    Two blocks of specified masses collide. They stick together and move off together, and we want to predict the resulting velocity. The collision investigator, yungman, intended to eliminate all cop-outs and complications by specifying that one block was at stand-still before the collision, and that all motion is at the same level (2 dimensions), so let me eliminate one more. There is no vibration resulting from the impact, since vibration is possible only if one block or the other has an elastic component.

    Now we have two blocks, one with a mass of 1000 Kg just sitting there (on frictionless, massless ball bearings), getting blindsided by a block with a mass of 2000 Kg, silently zipping along (on a frictionless cushion of air) at 15 meters per second. Suppose both blocks are solid spheres with radius of 1 meter (one is twice as dense as the other).

    We are physicists, so let's nail down the direction and impact angle. Looking down from above, imagine that each spherical block is marked like a clock, with a 9 on the left and a 3 on the right. The moving block is moving in the 12 o'clock direction, and its 1 o'clock mark strikes and becomes stuck to the 7 o'clock mark on the stationary block.

    Now the question becomes (since we know already that the two blocks, after an inelastic collision at the same elevation, move off stuck together moving horizontal in the same direction as the one block was already moving, but at 2/3 the velocity): Since kinetic energy is proportional to the mass times VELOCITY SQUARED, and we have ruled out all forms of energy that kinetic could convert into (like heat), WHERE IS THE MISSING KINETIC ENERGY?
    [2Mg x 1 = 3Mg x 4/9 + ?] Simplified: [2 = 4/3 + ?]
     
  13. Jan 3, 2018 #12

    A.T.

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    Let us just accept that I didn't already eat my cookie... WHERE IS MY COOKIE?
     
  14. Jan 3, 2018 #13

    Ibix

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    Your thought experiment is internally inconsistent. You posit a scenario (an inelastic collision) in which energy is lost by some dissipative mechanism while insisting that there are no such dissipative mechanisms. You can't have it both ways.
     
  15. Jan 3, 2018 #14

    CWatters

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    Both blocks incur acceleration which can't happen over zero distance as that would imply infinite acceleration, so some compression of at least one block(s) must occur. If it's an inelastic collision that distortion absorbs energy rather than storing it as would occur in an elastic collision.
     
  16. Jan 3, 2018 #15

    FactChecker

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    If the total kinetic energy is reduced, then some form of energy must have increased. You can't call all other forms of energy a "cop-out" if you then want to know where the energy went. That would be saying that you really do not want to know the answer.

    The fact is that conservation of momentum and conservation of energy can not co-exist if all forms of energy other than kinetic are ruled out. One depends on V and the other depends on V2.
     
  17. Jan 3, 2018 #16

    russ_watters

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    No; Clap your hands. Where did the kinetic energy of your hands go?

    [Edit: sigh, old thread]
     
  18. Jan 3, 2018 #17

    CWatters

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    Yeah so last year :-)
     
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