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Conservation of Momentum and Energy

  1. Mar 3, 2014 #1
    1. The problem statement, all variables and given/known data

    This is the question show that when one of the steel balls, suspended by strings next to
    each other (as in a newton's cradle), is pulled to the left and released, only a single ball recoils to
    the right under ideal elastic-collision conditions. Assume that each ball has a mass m
    , and that the ball coming in from the left strikes the other balls with a speed v0. Now, consider the hypothetical case where one ball comes from the left but two balls recoil to the right.
    Determine the speed the two recoiling balls must have in order to satisfy
    (a)(3 marks) momentum conservation, and
    (b)(3 marks) energy conservation.

    2. Relevant equations
    momentum p =mv
    energy k = .5 mv^2


    3. The attempt at a solution
    I assumed in both cases I would put p1=p2, and k1=k2, where the masses of the two recoiling balls, p2 and k2, will be 2m, and so find v in that way but that doesn't work. don't know what else to try.
     
  2. jcsd
  3. Mar 3, 2014 #2

    BvU

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    What do you mean when you say "that doesn't work" ?
     
  4. Mar 3, 2014 #3
    Sorry didn't make myself clear. what I did was tried putting values on both sides, like this for p for instance.
    mv0= 2mv (here m cancel out). so the value of v in conservation of momentum according to this would be v0/2, essentially half of v0, which seemed pretty intuitive to me, but I've been told that's not right. I get the same speed for energy conservation.
     
  5. Mar 3, 2014 #4

    BvU

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    So if I understand you well, you get vo/2 for a). Intuitively that looks ok to me too...
    Can you show me how you get the same result for b) ?
     
  6. Mar 3, 2014 #5
    oh sorry, for B) I get v= √(v0^2)/2. start with .5 mv0^2= .5 2mv^2 (masses and halves cancel out).


    actually the formula of K isn't given. I am assuming "energy conservation" refers to Kinetic energy
     
  7. Mar 3, 2014 #6

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    OK, so a) and b) are answered. What, then, is the problem ?
     
  8. Mar 3, 2014 #7
    oh it is right? I was told by a fellow student that my answers weren't right and I couldn't work out any other solution so asked here.
     
  9. Mar 3, 2014 #8

    BvU

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    You have shown that the hypothetical case can't be solved: if you want both momentum balance and kinetic energy balance, the only way that can be realized if just one ball moves off to the right. Watch a few of the Newton's cradle youtube videos, or experiment with colliding one, two etc. identical coins with a row of five, four etc. of the same coins on a smooth table !
     
  10. Mar 3, 2014 #9
    OK. thanks very much for clearing this up
     
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