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Newtons Cradle 2 balls

  1. Dec 21, 2012 #1
    We have a Newtons cradle , when you let go of one ball at one end you know what happens. Rlease one ball and the other side has a ball osciallte.
    But if you release two balls (with a 5 ball cradle) , why is that two balls on the other side of the middle ball also rise in response. How does the system "Know" we have released two balls? Why doesn the other side just have one ball oscillating higher?
  2. jcsd
  3. Dec 21, 2012 #2

    Vanadium 50

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    Can you work out the one ball case mathematically?
  4. Dec 21, 2012 #3


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    Conservation of energy and momentum. This is, of course, assuming that all balls have the same mass.
  5. Dec 21, 2012 #4
    Well i get that the system is conserving the momentum and tranferring the energy along the lines of balls, but I dont understand why doesnt the enrgy just get transferred into a higher arc?
  6. Dec 21, 2012 #5


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    Have you tried wikipedia? It's got actually quite a good analysis of the subject.

    Long story short, you can treat the two-ball case as if it was a single ball hitting the stationary ones, and then another one hitting the now-stationary first one.
  7. Dec 21, 2012 #6


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    Conservation of momentum and energy don't limit the result to just one possible outcome. For example, the released balls could end up bouncing backwards a bit and momentum and energy will be conserved. In the real world, a single released ball will bounce backwards at a small velocity (slight rebound). Link to a web site with the details:

  8. Dec 23, 2012 #7


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    Keep in mind two things: (1) the balls are not perfectly rigid but deform slightly during a collision, and (2) the "information" about the collision takes some time to travel from one ball to the next.

    You can rethink what is going on if you imagine there is a very small gap between each pair of adjacent balls, including the two balls that are initially raised and simultaneously released. It's not a perfect model, but it builds a time delay into the system that would account for the time it takes balls to deform during a collision, or the time it takes the deformed wave to travel across the diameter of a ball.

    Thought about this way, every collision involves only two balls (of equal mass), one moving and one at rest. See if you can work out the entire process of these single-pair collisions, and what finally happens with the two balls at the other end. (Hint: this model predicts exactly what is observed. :smile:)
    Last edited: Dec 23, 2012
  9. Dec 23, 2012 #8
    Newton's cradle is usually used to demonstrate conservation of momentum.
    This means that the momentum IN will equal the momentum OUT.
    2 balls In will give 2 balls OUT
    3 balls in will give 3 balls OUT
    4 balls IN will give 4 balls OUT
    5 balls IN will give 5 balls OUT (think about it !!!!)
    Try all combinations !!!!
    If you could not see the 5 balls (a piece of card can be used to hide them) and you pretend that you can't hear any collision you would not realise that a collision had occurred.
    The fact that collisions may not be perfectly elastic complicates matters slightly (KE is not conserved) but can be safely ignored if all you want is to use the cradle for what it was intended to show.
  10. Dec 23, 2012 #9


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    Not from conservation of momentum alone. You could get 1 ball out at twice the speed of the 2 balls in, and you'd still conserve momentum.

    However, this would require an increase in kinetic energy.
  11. Dec 23, 2012 #10
    The conventional 'newtons cradle' consists of 5 freely hanging steel balls in a frame of some sort.
    My list of possible collisions is based on the given number of balls pulled together to one side and released together.
    I have never seen 2 balls (or 3 or 4 or5) released produce anything other than 2 (or 3 or 4 or 5) balls come out of the other side.
    This is very easy to confirm by experiment. You should try this before making any unequivocal comment.
    I am sure that with some trickery, like sticking balls together you can get other combinations.
    With no such trickery I think that I can guarantee that number IN = number OUT
    There are many possible combinations that should be tried....
    2 in from 1 side, 1 in from the other side........I won't list them all
  12. Dec 24, 2012 #11


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    Nobody is disputing that. Certainly I don't.

    My comment was directed at the statement that we only need conservation of momentum, and nothing else, to completely explain what we all agree happens in reality.

    Consider this scenario, which we agree does NOT happen:

    2 balls in → 1 ball, at TWICE the speed, out.​

    Fact: the above does NOT happen.
    Fact: momentum is conserved in the above scenario
    Last edited: Dec 24, 2012
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