Newton's cradle, 5 ball system

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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
 
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  • #2
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Both kinetic energy and momentum must be conserved, so if the 3rd ball were made heavier what effect do you think this would have?
 
  • #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.
 
  • #4
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... it also helped to consider the balls to have spring like qualities.
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?
 
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  • #5
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In previous threads..., 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.
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
 
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  • #6
rcgldr
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  • #7
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. If balls are touching we cannot consider it an ideal cradle, where initial and final velocity are the same.
The older threads eventually link onto this web page:http://www.lhup.edu/~dsimanek/scenario/cradle.htm
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?
 
  • #8
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1)oo Ooo, 2) oo oOo, 3) oo ooO
Both kinetic energy and momentum must be conserved, so if the 3rd ball were made heavier what effect do you think this would have?
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?
 
  • #9
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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?
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
 
  • #10
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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?
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.
 
  • #11
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That means that if the middle ball has double the mass of all the others there is no effect on your problem,.
if you think so, please, show your solutions. Let's say vI = 9
 
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  • #12
rcgldr
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Simanek confirms what I said: {2.,2.}: "the balls aren't really touching, so a series of independent two-ball collisions occur"
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.
 
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  • #13
DaveC426913
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if you think so, please, show your solutions.
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.
 
  • #14
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.. gaps between balls do not affect the outcome, only simplify solution...
Note that at the end of {2.,2.} is this statement: the surprising outcome is essentially the same whether or not they are touching.
... gravity will tend to recenter them. .
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?
 
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  • #15
DaveC426913
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first general rule says that momentum passes on undisturbed if ball in the middle is smaller and not bigger, is it so?
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
 
  • #16
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... kinetic energy and momentum must be conserved...
...momentum and kinetic energy will remain constant.' ..
...aren't we forgetting [approach/separation] velocity ? [itex]\sum[/itex]v=[itex]\sum[/itex]'v
 
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  • #17
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Not if the balls have different masses, no.
 
  • #18
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Not if the balls have different masses, no.
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
 
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  • #19
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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.
 
  • #20
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if conservation of velocity were true in every case then
in every case in this thread, because all collisions are head-on.
you haven't offered a single solution, so far.
 
  • #21
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Being "head-on" has nothing to do with this, as I have repeatedly stated the answers to all of the problems in the OP can be gained by two conservation laws; velocity being conserved is NOT one of them.
Not only that but above I proved that your case of the two balls does NOT conserve velocity.
 
  • #22
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if you work out properly my example you'll see that net velocity of separation equals net velocity of approach , (the same as conservation of momentum means net momentum): v' s = v a,do you have any solution to the "cradle"?
 
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  • #23
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I'd like to point out that your original formula (which was used in my last post) was changed after you made that post, but we might just chalk that up to a typo. In any case the person I was responding to stated that the sums of the velocities before/after the collision remained constant, that is not true and I was right to point that out.
 
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  • #24
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at hyperphysics you can find all explanations, I cannot do any better, sorry, you cannot work it out with a general formula, because sum depends on the outcome of each individual case, sometimes it is plus plus,sometimes it is plus minus, exactly like momentum, only the signs are reversed. If one knows the rules of elastic collision [net] sum of approach or separations, means something definite, like net [sum of] momentum:
(+) + (+), (+) + (-), (-)+(+), (-)+(-).
It was not a typo, I'm sorry if you did not appreciate it, I was only trying to help you when I saw from your reply that you did not know anything about it. It was just a hint, giving you the actual procedure in that particular example. bye.
 

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