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Three balls in the plane

  1. Mar 3, 2014 #1
    Hello everyone,

    I'm trying to find the exact velocities of three balls in the plane after they collide elastically. I'm assuming arbitrary positive masses and arbitrary positive radii. Of course, three balls can collide in many ways:
    1. in an equilateral triangle: one coming from north, one coming from roughly east south east, one coming from roughly west south west
    2. in a "Newton's cradle": two balls stationary, touching; third ball comes in and hits one of them
    3. two balls "sandwiching" the third
    In such a collision, I know that the sum of the initial momenta will equal the sum of the final momenta, and I know that the initial kinetic energy will equal the final kinetic energy. This gives me two equations to solve. A third equation is needed. This one usually comes from positional considerations and our intuition about the collision: in (1), the north ball will go north, etc; in (2), if L is the line passing through the stationary balls' centers, and the third ball is traveling along L, one ball will stand still (or maybe not), but if the third ball is perpendicular to L, the blow might be glancing or direct, and who knows what happens; in (3), the sandwiched ball will act like a buffer and the momentum equations will pretend it's not there (unless it's moving and happens to be sandwiched at exactly the right time).

    The details of each particular case are extensive. My intuition fails in more complicated cases. What are some general principles that I can use to come up with a recipe for solving this in general? For example, is it true that if a few balls are touching, they act like one solid object except they break off "at the extremities"?

    I am not satisfied with approximations (e.g. to avoid a 3-at-once collision, nudge one ball very slightly to turn the problem into two 2-at-once collisions). However, I am curious to see whether nudging a ball's position by a small vector ε, solving for the final velocities after the two collisions, and letting ||ε|| → 0+, (whether) the final velocities "make sense", as if they were the final velocities correctly calculated after one 3-at-once collision. For this, though, I need to know what the correct final velocities are, in order to actually check that these limiting velocities equal them.

    Thank you,
  2. jcsd
  3. Mar 3, 2014 #2


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    I think you will always have to consider the boundary conditions for all of the balls:

    1. Original position
    2. Original momentum

    That should give you the additional information required.
  4. Mar 4, 2014 #3
    UltrafastPED: I'm not sure I understand. Could you explain your answer in the context of the following setup?

    Let B1 be a ball of radius r1 = 1 and position x1(t) = (t, 0).
    Let B2 be a ball of radius r2 = √2 − 1 and position x2(t) = (t − 1, 2t − 3).
    Let B3 be a ball of radius r3 = 2 and position x3(t) = (5 − t, t − 1).

    If we define 3C2 = 3 functions dij(t) = ||xi(t) − xj(t)|| then the first instance of dij(t) = ri + rj occurs at t = 1. This is a 3-at-once collision. Click here for a picture.

    What happens now?
    Last edited: Mar 4, 2014
  5. Mar 4, 2014 #4


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    Objects already in contact can transmit momentum without moving.
  6. Mar 4, 2014 #5
    I still don't get it. Does B2 (the left ball) transmit all of its momentum to B1 (the middle ball)? Does this momentum "pass through" into B3 (the right ball)? How much momentum does B2 (left) receive from B1 (middle)? Does B2 (left) receive, indirectly, any momentum from B3 (right)?

    For the sake of computation, if any, assume the balls have uniform density (like 1/π) so that their masses equal the squares of their radii.
  7. Mar 4, 2014 #6


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    You said nothing about the spin of the balls, so I assume there is no spin initially, and the balls are frictionless so that no spin can be induced by the collision.

    Momentum transfer can happen:
    a) only between ball pairs that are in contact, as equal but opposite forces (Newtons 3rd)
    b) only along the contact normal (assuming frictionless balls)
    c) only with not negative initial closing speed along the contact normal
    d) only such that final closing speed along the contact normal is not positive

    So instead of formulating the conservation of total momentum, write down the momentum equations for each contacting pair (Newtons 3rd automatically ensures the conservation of total momentum). The contact normal vs velocity and kinetic energy give you additional constraints.

    Your variables are now the magnitudes of momentum transfer for each contacting pair (that satisfies c). For 3 balls this gives you maximally 3 variables, but eventually less than that.
    Last edited: Mar 4, 2014
  8. Mar 5, 2014 #7
    Dear A.T.,

    Thank you for the detailed response. How do your four principles explain Newton's cradle? The Wikipedia article consistently reiterates that a 3-at-once collision has many possible solutions (something which I did not know before) and that all the animations of Newton's cradle actually involve many 2-at-once collisions. Does my question have no answer, then? If Ci (for i = 2 and 3) are balls of radius 1 located at (2, 0) and (4, 0) with x'2(t) = x'3(t) = (0, 0), and if C1 is another ball of radius 1 with x1(t) = (t - 1, 0), then what happens at the collision at t = 1? Assume elastic collisions, no rotation, no friction, and equal unit mass throughout.
  9. Mar 6, 2014 #8


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    I don't know if they do. Newton's cradle involves the propagation of an impulse, or sequential collisions.
  10. Mar 7, 2014 #9
    Just to give this thread some closure, here is a paper which explains (in the special case of Newton's cradle) why a many-at-once collision theoretically has infinitely many solutions, but why in real life we only see one outcome: the ideal cradle is a dispersion-free system. And here is Newton's cradle in very slow motion. It appears to be a sequence of 2-at-once collisions, which gives me more faith in my ε idea.
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