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Energy conservation in superball collision

  1. Dec 16, 2016 #1
    1. The problem statement, all variables and given/known data
    There are two elastic "superballs" of mass M and m placed on top of eachother with a smal distance. The lighter ball of mass m is on top of the bigger ball of mass M. The balls are released from a height h and have velocity u when they hit the ground. Prove that the top ball will have velocity ## v = \frac{3M-m}{M+m} \cdot u ## when it leaves the bigger ball.

    2. Relevant equations
    momentum conservation ## p_{before} = p_{after} ##
    energy conservation ##E_{before} = E_{after} ##
    zero momentum frame velocity: ##v_{ZMF} = \frac{p_{tot}}{m_tot} ##

    3. The attempt at a solution
    I didn't find it problematic to prove the requried result (see attached solution). What I find problematic is the result itself. I must have made some very silly misstake in my reasoning, but the end result I get is that the speed of the top ball after all collisions is ## u + 2v_{ZMF}## and of the bottom ball ##u##. However, before the two balls collide, they both had velocity ##u##. Clearly, energy conservation cannot be satisfied!

    Can somebody please point out my probably very stupid misstake to me? Thank you! :)
     

    Attached Files:

  2. jcsd
  3. Dec 16, 2016 #2

    BvU

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    Perhaps the big ball loses some speed when colliding with the smaller ball ? How did you find this ##u## for the bottom ball ?
     
  4. Dec 16, 2016 #3
    Oh wait, I really messed it up! The speed of the big ball should be ## u - 2v_{ZMF}##, shouldn't it?
     
  5. Dec 16, 2016 #4
    ##E_{ki} = 1/2mu^2 + 1/2Mu^2##
    ##E_{kf} = 1/2m(u^2 + 2(2v_{ZMF})u + (2v_{ZMF})^2) \\+ 1/2M(u^2 + 2(2v_{ZMF})u - (2v_{ZMF})^2)##
    ##E_{kf} = (1/2M+1/2m)u^2 + m(2v_{ZMF})u + 2m(v_{ZMF})^2 + M(2v_{ZMF})u - 2M(v_{ZMF})^2##
    ##E_{kf} = (1/2M+1/2m)u^2 + 2(v_{ZMF})^2(m - M) + (M + m)(2v_{ZMF})u##
    ##E_{ki} \ne E_{kf}##
    I think I did this correctly.
     
  6. Dec 16, 2016 #5
    Sooo, ##u - 2v_{ZMF}## is not right either?
     
  7. Dec 16, 2016 #6
    I think using momentum conservation will give right answer.
     
  8. Dec 16, 2016 #7
    As far as I am concerned ZMF is meant to make it quicker and easier, so I would like to use that method. But yes, momentum conservation should always give the right answer.
     
  9. Dec 16, 2016 #8
    No that's not what I meant.
    I mean that we can get right velocity for big ball after collision using momentum conservation.

    ##mu + Mu = m(u + 2v_{ZMF}) + M\color{red}{V}##
     
    Last edited: Dec 16, 2016
  10. Dec 16, 2016 #9

    haruspex

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    Yes, and with that I think you will find the energy is conserved in the lab frame.
     
  11. Dec 19, 2016 #10
    Yes. Thank you for helping me get it right! :)
     
  12. Dec 19, 2016 #11
    That looks correct but I did not get it.

    Initial velocity in lab frame :-
    Big mass - u
    Small mass - u

    Final velocity in lab frame :-
    Big mass - ##u - 2v_{ZMF}##
    Small mass - ##u + 2v_{ZMF}##
    As per OP,

    That is not going to get conserved. Either I am getting final velocity of small mass wrong or the big mass.
    Can you help me please ? :oldconfused::oldconfused:
     
  13. Dec 19, 2016 #12

    haruspex

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    Are you sure, or are you just guessing?
     
  14. Dec 19, 2016 #13
    I checked if it is going to conserve in post number 4.
    Can you please check my calculations ?
     
  15. Dec 19, 2016 #14

    haruspex

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    You have some signs wrong in the second equation.
     
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