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Kinetic energy transfer from large mass to smaller mass

  1. Sep 19, 2012 #1
    First: Theoretically, is it possible to transfer all of an object's KE to another object? e.g. (ignoring gravity, friction, sound, etc.) an object with mass=M and 1eV collides with another object of mass=m and 0eV-- afterwords, object M has 0eV and object m has 1eV (conservation of energy)? It seems like that should be possible but then...
    Second: (ignoring gravity, friction, sound, etc.) Imagine a pool ball colliding with a ping pong ball that is at rest; the pool ball is stopped completely and the ping pong ball moves away with a velocity that that conserves the pool ball's momentum. So would all of the kinetic energy be transfered as well? I can't imagine that happening in my head, the pool ball still keeps rolling on my imaginary pool table... Am I missing something really easy that explains why that wouldn't happen or is my imagination just that of a lesser quality?
     
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  3. Sep 19, 2012 #2

    mathman

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    Assuming an elastic collision, you need to satisfy both conservation of energy and conservation of momentum. When you work out the algebra, you will find exactly 2 solutions. One of them corresponds to the big ball missing the small ball. The other (assuming I didn't make a mistake) has the small ball going with a speed = 2V/(r+1), where V is the initial speed of the big ball (assuming the small ball at rest) and r =m/M, where m is the mass of the small ball and M is the mass of the big ball.

    I'll lrt you work out the energies involved.
     
  4. Sep 20, 2012 #3

    jbriggs444

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    As I understand the problem, the small ball is initially at rest and the large ball is initially moving. After an elastic collision the requirement is that the large ball be at rest with the small ball moving.

    I get a number of solutions. But they do not appear to match the ones you have in mind.

    Let V = big ball velocity before collision, v = little ball velocity after collision, M = big ball mass, m = little ball mass.

    One family of solutions has V = 0, M = whatever, m = whatever, v = 0 and no collision at all.

    Another family of solutions has M=0, m=whatever, V = whatever, v = 0 and
    no collision at all.

    If M and m are both zero then you can have your collision. V = whatever, v = whatever.

    Discarding those trivial solutions and considering only non-zero m, M and V...

    Conservation of momentum requires that MV before the collision is identical to mv after the collision.

    That, in turn means that v = M/m V (it also means that m has to be non-zero).

    That means that kinetic energy after the collision is given by e = 1/2 m ( M/m V )^2
    Kinetic energy before the collision is given by E = 1/2 M V^2

    For an elastic collision, e = E and so...

    1/2 m (M/m V)^2 = 1/2 M V^2
    m (M/m)^2 = M
    M^2/m = M
    M/m = 1
    M = m

    This is the solution embodied in Newton's Cradle.

    None of these solutions fit the requirement that there be a big ball, a little ball and an elastic collision.
     
  5. Sep 20, 2012 #4
    sure.

    Imagine a bat hitting a ball and stopping at the moment of impact.

    There will be relativistic parameter limits....for example, you can't send the ball off faster than 'c'.
     
  6. Sep 20, 2012 #5

    jbriggs444

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    It's easy to imagine stuff. It's harder to make sure that your imagination is compatible with the laws of physics.

    For a bat that is heavier than the ball, this scenario either requires a ball that has a non-zero impact velocity or a "super-elastic" collision that generates additional kinetic energy (e.g. from a coating of explosives on the bat).
     
  7. Sep 20, 2012 #6

    haruspex

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    Typically the bat would be rotating, and the ball would make contact at a part of the bat that's moving faster than the bat's c of g. Seems to me that with the right combination of mass ratios, contact point and linear and rotational velocities, a ball could bring a heavier bat to a dead stop.
     
  8. Sep 20, 2012 #7

    mathman

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    Symbols: V = speed of large ball before. X and Y speeds of large and small balls after collision. M and m masses of large and small balls. r = m/M.
    momentum: MX + mY = MV (small ball starts at rest)
    energy: MX2 + mY2 = MV2
    ------------------------------------
    X = V - rY (momentum)
    (V - rY)2 + rY2 = V2 (energy)
    V2 -2rVY + (rY)2 + rY2 = V2
    {-2V + rY + Y}rY = 0

    Solutions: Y = 0 or Y = 2V/(r+1)
     
  9. Sep 21, 2012 #8

    jbriggs444

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    OK. You have solved for the possible velocities of the small ball given conservation of energy and of momentum.

    Now the challenge is to take this solution and apply it to the goal of the problem. Under what circumstances can the small ball's final kinetic energy be equal to the large ball's initial kinetic energy?

    That constraint requires that:

    1/2 V2 = 1/2 rY2

    substituting Y = 2V/(r+1) gives:

    1/2 V2 = 1/2 (2V/(r+1))2

    Dividing out 1/2 V2 yields:

    1 = 4/(r+1)2
    1 = (r+1)2/4
    4 = (r+1)2
    r + 1 = 2 | r + 1 = -2
    r = 1 | r = -3

    Let us discard the solution with a negative mass ratio as non-physical. So r = 1.

    Back-substituting, into the earlier equation, Y = 2V/(r+1), this means that Y = V. So it seems that our solutions are the same after all.
     
  10. Sep 21, 2012 #9

    mathman

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    It is easier if you use the momentum equation. Assume momentum (and energy) of large ball = 0:
    V = (rY =) 2rV/(1+r) or 2r = 1+r or r = 1 and so Y = V (only solution).
     
  11. Oct 20, 2012 #10
    wow, thanks for all the input! You guys put much more time and effort into it than I thought you would.
     
  12. Oct 20, 2012 #11
    It is well known or can be easily calculated if mass M with velocity V collides with mass m at rest. (M>>m), then after collision, velocity of M remains approximately V, but velocity of m becomes approximately 2V.

    But that doesn't mean all energy of M is transferred to m, but just a small part is transferred, which is ignored in approximation. If by some magic we can transfer all energy of M to m during collision, then velocity of m after collision must be

    v = V√(M/m).

    Interesting, which means we can transform a low speed into high speed.
     
  13. Oct 20, 2012 #12

    Nugatory

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    Although the "some magic" that we're using would be magically suspending the law of conservation of momentum.... So not likely.
     
  14. Oct 20, 2012 #13

    AlephZero

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    Remember that in classical mechanics, "kinetic energy" depends on the reference frame you measure it in. So in that sense you can always transfer "all the kinetic energy" from one object to another. Just work in the reference frame where its velocity after the collision is zero.

    Of course for a situation like Newton's cradle, that reference frame is the obvious one to choose for other reasons - but that is just a coincidence.
     
  15. Oct 20, 2012 #14

    Nugatory

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    Ah - good point. It's worth adding that if the masses of the two objects are very different, and we choose the reference frame in which the kinetic energy of the more massive object is zero after the collision... It will be near as no never mind zero in that frame before the collision as well. Think bouncing a ball off a brick wall, in the frame in which the wall and the earth it's attached to are at rest.
     
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