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Newtons cradle, conservation of momentum

  1. May 21, 2014 #1
    1. The problem statement, all variables and given/known data
    I don't understand how momentum is conserved in Newtons cradle. If I look at the component vectors for the initial ball on the left that I raise, one component points down and one points to the right but then looking at the vector components on the ball that moves on the right end, it has one component up and one component to the right. The horizontal components are conserved but the vertical components aren't.


    2. Relevant equations
    M1v1i + m2v2i = m1v1f + m2v2f for x and y components


    3. The attempt at a solution

    I think the vertical component of momentum is not conserved due to gravity?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 21, 2014 #2

    Simon Bridge

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    You have used: ##\vec p=m\vec v##
    ... you have correctly noticed that momentum is a vector, but you seem to have forgotten that the initial and final momenta in your description are both zero (the balls start and end with v=0) - which conserves momentum over the entire journey.

    Note: conservation laws only apply to closed systems ... you have concentrated on the balls.
    Is the ball a closed system? Is the cradle a closed system?

    When the ball is released, you can draw a free-body diagram to understand it's motion.
    What is the total force on the ball? What does this do to the ball's momentum?
    Are there any external forces acting on the frame of the cradle?

    At the instant the ball impacts the others, what direction is it's momentum?
    At the instant the right-hand ball starts to move, what direction is it's momentum?

    ... you can then draw a free-body diagram to understand it's subsequent motion.
     
  4. May 22, 2014 #3

    BiGyElLoWhAt

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    Try looking up impulse on google
     
  5. May 22, 2014 #4

    rcgldr

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    Momentum and energy would ideally be conserved, but even the idealized reaction for Newton's cradle is that the pack of balls at rest will shift on each collision, and since they are supported by finite length lines to the frame, gravity somewhat opposes the pack shift, so some momentum is lost unless you include the frame and the earth that the frame rests on as parts of a closed system. Here's a link to a good article about Newton's cradle:

    https://www.lhup.edu/~dsimanek/scenario/cradle.htm
     
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