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Newton's cradle, 5 ball system

  1. Oct 26, 2011 #1
    if two balls [1,2] are pulled away: oo ooo, two balls [4,5] on the opposite side: ooo→ oo swing out. What happens if ball # 3 or 4 or 5 has mass = 2 ?
    1) oo Ooo, 2) oo oOo, 3) oo ooO
     
    Last edited: Oct 26, 2011
  2. jcsd
  3. Oct 26, 2011 #2
    Both kinetic energy and momentum must be conserved, so if the 3rd ball were made heavier what effect do you think this would have?
     
  4. Oct 26, 2011 #3

    rcgldr

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    In previous threads on this, in addition to conservation of momentum and energy, it also helped to consider the balls to have spring like qualities, where the balls experience forces related to compressions, decelerations, and accelerations between balls during collision cycles.
     
  5. Oct 26, 2011 #4
    I suppose gaps between balls do not affect the outcome, only simplify solution.
    case 1 and 3 should have similar outcome, case 2 is more difficult to break into separate collisions, but I suppose balls should have a certain rebound. Is it so?
     
    Last edited: Oct 26, 2011
  6. Oct 29, 2011 #5
    I couldn't find previous thread, I can't see in what way that can help. If balls are touching we cannot consider it an ideal cradle, where initial and final velocity are the same.
    Gaps between balls not only avoid complications but they let us also comprehend outcomes that are counter-intuitive: consider the simple 3-ball system case: ooo, if we pull away 2 balls
    oo→ o , we are surprised the outcome is o oo→ (vI = vF) , but if you consider [just 1 mm] gaps between balls the otcome becomes obvious:

    oo2.........o , o o2o , o.... o2 o→, o→ o2 o→ , o1 o2o→

    ball 2 hits ball 3 and stops, is hit by ball 1 and swings out with ball 3
     
    Last edited: Oct 29, 2011
  7. Oct 29, 2011 #6

    rcgldr

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  8. Oct 30, 2011 #7
    link [Simanek] confirms what I said: {2.,2.}: "the balls aren't really touching, so a series of independent two-ball collisions occur", so in
    Case 1: vI = vF = 6
    Case 2 [vI 6 = 4+4-2] considers the outcome when balls 2 and 3 are tied, soldered, is it so?
     
  9. Nov 2, 2011 #8
    you are referring to this,
    oo→ Ooo, now, if balls 1,2 where tied, result would be [[itex]\equiv[/itex] O→ Ooo ] , ooO oo→, (because the 2 loose balls, when hit, behave as if they were tied). But they are not, the question is : what is their [2 falling-hitting balls] behaviour when they are tied/not-tied touching/not-tied-not-touching/?
    what do you think?
     
  10. Nov 3, 2011 #9
    I'll take a swing at it :P

    Tied: Wouldn't the hitting balls stop at the colliding point while the targeted balls swing to the right?

    Not Tied, Touching: The hitting balls both swing backwards to the left upon collision, while the targeted balls swing to the right

    Not Touching: The first ball (from the left) hits the second ball only after the second ball swings back from hitting the target ball (3rd ball).


    Just a guess :P
     
  11. Nov 4, 2011 #10
    No matter how you arrange this system what I said remains true, momentum and kinetic energy will remain constant. That means that if the middle ball has double the mass of all the others there is no effect on your problem, additionally you will get the same result whether or not the two balls you release are tied or touching.
     
  12. Nov 4, 2011 #11
    if you think so, please, show your solutions. Let's say vI = 9
     
    Last edited: Nov 4, 2011
  13. Nov 4, 2011 #12

    rcgldr

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    Note that at the end of {2.,2.} is this statement: the surprising outcome is essentially the same whether or not they are touching.

    Also mentioned throughout, is the fact that even if the middle balls end up at rest, they do get displaced slightly during each half cycle, but ideally end up back at their orignal position after a full two collision cycle. In reality, once the center balls are displaced, gravity will tend to recenter them. Using very long support lines would reduced the gravity / pendulum related effects on the middle balls.
     
    Last edited: Nov 4, 2011
  14. Nov 4, 2011 #13

    DaveC426913

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    Simply replace the m=2 ball with two m=1 balls and you've got the same thing.

    Just as oo oooo becomes oooo oo
    so does oo Ooo become ooO oo.
     
  15. Nov 4, 2011 #14
    Sure, rcglrd [btw: thanks for the interesting link], that is what I said: the outcome doesn't not change if balls are/aren't touching. Only math solution is more complicated; the value of vF, in [link-] case 1, wouldn't be exactly 6 but, say: 5.89735!
    Of course, in the end, gravity will recenter all the balls, but we are considering first oscillation.

    The purpose of this thread is to see if we can work out some practical, general rules, and see if there can be any exceptions to the principle of "equivalence of separate collisions" [,for example when two untied balls are falling at high speed].
    If we strictly apply that principle, all 3 cases in OP generate more or less confusion because:
    first general rule says that momentum passes on undisturbed if ball in the middle is smaller and not bigger, is it so?
     
    Last edited: Nov 4, 2011
  16. Nov 4, 2011 #15

    DaveC426913

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    A small ball combined with a regular ball is essentially the same as one larger ball.

    i.e: oo.oo is the essentially same as oOoo (or ooOo).

    That being said, it is possible, using small, regular and large balls to make fractional combinations:

    eg: oO oo begets oO . o . o
     
  17. Nov 6, 2011 #16
    ...aren't we forgetting [approach/separation] velocity ? [itex]\sum[/itex]v=[itex]\sum[/itex]'v
     
    Last edited: Nov 6, 2011
  18. Nov 6, 2011 #17
    Not if the balls have different masses, no.
     
  19. Nov 7, 2011 #18
    that is not true, can you find out the oucome of this simple case?:
    m1=2,v1=9
    m2=1 v2=0
    m=2O→v=9 o m=1

    v1±v2 = v'2-v'1
     
    Last edited: Nov 7, 2011
  20. Nov 7, 2011 #19
    Kinetic energy is conserved, velocity is not; if conservation of velocity were true in every case then conservation of momentum would not. Take your case above
    [tex]
    v_1 = v_1' + v_2'
    [/tex]
    [tex]
    m_1 v_1 = m_1 v_1' + m_2 v_2' \Rightarrow m_1 v_1' + m_1 v_2' = m_1 v_1' + m_2 v_2' \Rightarrow m_1 = m_2
    [/tex]
    Therefore the two cannot both be, in the general case, true.
     
  21. Nov 7, 2011 #20
    in every case in this thread, because all collisions are head-on.
    you haven't offered a single solution, so far.
     
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