I A nice instructive riddle

Funley

This is not a simple situation. See, e.g., Matthias Reinsch, "Dispersion-free linear chains," Am. J. Phys. 62, 271-278 (1994) and references therein. A good article at a lower level was in The Physics Teacher, 35, Oct. 1997 by J.D. Gavenda and J.R. Edgington. Hope this helps.

• vanhees71

haruspex

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In another current thread on the subject I modelled the three ball case as three springs.
As has been noted, the "perfect" Newton's Cradle result requires a small separation so that no three balls are in contact simultaneously. On the other hand, I suspect my model exaggerates the discrepancy by ignoring the time it takes for the shock wave to traverse a ball.

M

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Two assumptions:

1. The balls ARE touching.
2. The spheres are PERFECTLY rigid.

Then it seems we have to conclude that those assumptions are so oversimplifying for this problem that it can not be solved?

haruspex

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2. The spheres are PERFECTLY rigid.
Not a good assumption. That leads to infinite forces.
Such idealisations - inextensible strings, perfect elasticity etc. - are only valid as asymptotic limits. E.g. if we let the spring constant of a string be k, solve, let k tend to infinity and find the solution converges, then we can argue we have the solution for a string of negligible extensibility.
In the present case, that is what I did at the link I posted. The end result does not depend on the value of k, so represents the limit as the spheres tend to perfect rigidity.

• Klystron and jbriggs444

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The configuration in post 7 also satisfies those constraints.
Which is why I said "for example".

bobob

That works for a two ball collision where you have two unknowns and two equations. But with 5 balls you have 5 unknowns and still just two equations. So there are an infinite number of solutions.
Well, it really isn't a linear momentum problem. The balls are on strings and are therefore, pendula. However, it's not necessary to go through anything tedious because there is symmetry about the middle point of where the strings attach. For example, if there are 5 balls, then the line along the string of the middle ball is the symmetry axis. The initial and final motion should be symmetric about that axis. If there were 4 balls, the symmetry would be the line between (and parallel to) the two middle strings.

haruspex

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The balls are on strings and are therefore, pendula
Any displacement of the angle of the string during the period of interest can be made arbitrarily small.
The initial and final motion should be symmetric about that axis.
It is tempting to assume that, but as I show at the link in post #27 it is not true. What you can say is that the solution should be time-reversible.

A.T.

But you do have constraints, and those constraints will provide additioonal equations (only one thing happens).
"Only one thing happens" doesn't necessarily mean that modelling it as a single collision will yield a unique solution, even with all appropriate constraints.

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PeroK

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Two assumptions:

1. The balls ARE touching.
2. The spheres are PERFECTLY rigid.

Then it seems we have to conclude that those assumptions are so oversimplifying for this problem that it can not be solved?
I wonder whether the issue isn't whether the balls are touching but that the bonds inside a sphere are different from the "bond" between adjacent touching spheres. Certainly if you applied a force in the opposite direction, then two touching balls would come apart easily; whereas, it would be hard to split a ball.

You could try the experiment with the wires not quite vertical, so that the balls are held together slightly by gravity.

I wonder whether it would work the same with cubes? That would give more symmetry for for shock wave along its line of propagation. Then one could argue that the cubes would not be perfectly smooth, and this would again create a delay in the shock wave at the point of contact.

Finally, if you drop two balls, then assuming this delay the experiment breaks down nicely into an impact from the first ball that results in one ball being ejected at the far end; followed almost immediately by a second impact that results in a second ball being ejected.

If, however, a single larger ball is dropped, then there would be theoretically a geometric sequence of reducing shockwaves, which would result in all the other balls being ejected, each with less momentum than the ball before.

And, as above, you could argue that no matter how rigid the balls are, the delay in the sequence of shock waves is still finite, so the same behaviour would persist in the limit.

tech99

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Conservation of momentum and energy are why two balls are knocked. If the final was only one, it would go higher than the initial two. The PE is transferred to KE to maintain velocity.
I don't think energy is conserved in a collision. I hear a click.

vanhees71

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In another current thread on the subject I modelled the three ball case as three springs.
As has been noted, the "perfect" Newton's Cradle result requires a small separation so that no three balls are in contact simultaneously. On the other hand, I suspect my model exaggerates the discrepancy by ignoring the time it takes for the shock wave to traverse a ball.
There's more to it! To model it with point particles connected by strings you have to adjust the spring constants and masses such as to guarantee a dispersion-free wave (see the AJP papers cited in #26). Newton's craddle is by far not as simple as it looks!

haruspex

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There's more to it! To model it with point particles connected by strings you have to adjust the spring constants and masses such as to guarantee a dispersion-free wave (see the AJP papers cited in #26). Newton's craddle is by far not as simple as it looks!
Dimensional analysis suggests adjusting only the spring constants and masses, but keeping the balls identical, is not going to change the qualitative behaviour. Changing the initial conditions, e.g. to construct a phonon, sounds more helpful.

pinball1970

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I don't think energy is conserved in a collision. I hear a click.
The click is sound which is particles in the air transferring energy to your ear drum. That energy is lost with each impact as is the friction caused between the balls and the air when they swing back and forth.
Also some energy will be lost between the balls?
So the cycle will decay and E is conserved it just gets less each time.
I cannot follow the arguments why two balls produce two balls, not one ball higher.
If I pushed one ball much harder could I eventually produce two?
P=mv so increasing the velocity is essentially the same as increasing the mass?

haruspex

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I cannot follow the arguments why two balls produce two balls, not one ball higher.
If I pushed one ball much harder could I eventually produce two?
P=mv so increasing the velocity is essentially the same as increasing the mass?
Two balls mass m in at speed v have momentum 2mv and KE mv2. If only one ball comes out and momentum is conserved it has speed 2v and KE 2mv2.
If that works, patent it quickly.

vanhees71

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Dimensional analysis suggests adjusting only the spring constants and masses, but keeping the balls identical, is not going to change the qualitative behaviour. Changing the initial conditions, e.g. to construct a phonon, sounds more helpful.
Yes, read the article. It's really surprising that you need this condition of $k \propto \omega$. It's obviously what's realized to good approximation with the real cradle. I wonder, how one can understand this properly from the theory of bouncing metallic spheres!

pinball1970

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One key point is that if a ball (moving) impacts another ball (at rest) of equal mass, then the first ball stops and the second moves with the velocity/momentum/energy of the first.

The height of the ball, therefore, only determines this momentum.

If, however, a more massive ball impacts a less massive ball, then the larger ball will continue with some of the momentum.
Say you have the initial ball mass 2m hits 4 balls all mass m will 2 balls be produced?

• Dale

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Say you have the initial ball mass 2m hits 4 balls all mass m will 2 balls be produced?
The simplest model is what I posted in post #34. If you imagine small gaps between the balls then all four balls will move:

The large ball hits the first ball, which hits the second ball and stops, which hits the third and stops, which hits the fourth and stops; the fourth being propelled.

Meanwhile, the large ball still has some momentum from the first collision, so it hits the first ball again, but with 1/3 of the momentum this time. By the same process as above, the third ball is propelled, with 1/3 of the momentum of the fourth.

This process repeats twice more, with each ball having 1/3 of the momentum of the previous. And, the large ball still has some residual momentum as well.

That's the simplest theoretical analysis that I can see.

PS see post #44 below from @stevendaryl for the same idea for the two-ball case.

PPS this video at 1:15 shows the experiment with a larger sphere. It actually matches quite accurately the analysis above!

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• strangerep

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There is a prior thread about this, link below. It's complicated. Some key issues brought up in the prior thread and linked to articles. Momentum and energy conservation aren't sufficient enough to restrict the result to a single outcome. The center pack shifts. Over time, the central pack will be visibly swinging due to the supporting strings having finite length. Compression of the balls is a key factor, and separation isn't needed.

• vanhees71

stevendaryl

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The behavior becomes pretty understandable if, instead of assuming that the balls are touching, you assume that there is a tiny space between adjacent balls. (And then consider the touching case as the limiting case as the gap size goes to zero.) If there is a tiny space between the balls, then that means that we only need to consider two-ball collisions. In a linear two-ball collision, it's pretty obvious that conservation of momentum and energy implies that in a collision between a ball of speed $v$ with a ball at rest, then afterward, the first ball is at rest and the second ball has speed $v$. If we assume that the collision takes negligible time, then the behavior follows. Here's an illustration with two moving balls hitting four at rest. In scene 1, balls number 1 and 2 are moving, and the others are at rest. In scene 2, after the first collision, balls number 1 and 3 are moving. In scene 3, balls number 2 and 4 are moving. Etc. Finally, in scene 6, balls number 5 and 6 are moving.

So the number of balls moving is always two.

Let the total falling mass be m and the velocity of impact be v. Let the total rising mass be M and its initial velocity be V.
From conservation of momentum we can write:

mv = MV

Because of the rigidity of the balls and the overall structure of the cradle the collision will be fairly elastic so most of the kinetic energy is conserved. Ignoring losses due to sound etc we can write:

mv2 = MV2
From the above v =V and m =M

PeroK

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Let the total falling mass be m and the velocity of impact be v. Let the total rising mass be M and its initial velocity be V.
How do you know that all the rising mass has the same velocity?

And how do you know that the falling mass doesn't rebound?

• member659127

How do you know that all the rising mass has the same velocity?

And how do you know that the falling mass doesn't rebound?
You don't know and my answer is an approximation only based on the assumption that the collision is perfectly elastic. It's the same sort of analysis you could apply to mono atomic gases where the collisions can be perfectly elastic.

PeroK

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You don't know and my answer is an approximation only based on the assumption that the collision is perfectly elastic. It's the same sort of analysis you could apply to mono atomic gas molecules which can collide perfectly elastically.
It's not an approximation. If you drop a ball onto the Earth it rebounds. And, in fact, in any elastic collision where a smaller mass impacts a larger mass at rest, the smaller mass will rebound.

That's another possible solution in the case of Newton's cradle. The falling ball rebounds and the others all move off.

In fact, the impact ball only stops in the special where the object is of the same mass. Which is why we have analysed the problem as though there are small gaps; or, in any case, as a sequence of simple collisions.

vanhees71

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I think #44 is a convincing elementary explanation. The only assumption you have to make is that the collisions are sufficiently elastic.

It's not an approximation. If you drop a ball onto the Earth it rebounds. And, in fact, in any elastic collision where a smaller mass impacts a larger mass at rest, the smaller mass will rebound.

That's another possible solution in the case of Newton's cradle. The falling ball rebounds and the others all move off.

In fact, the impact ball only stops in the special where the object is of the same mass. Which is why we have analysed the problem as though there are small gaps; or, in any case, as a sequence of simple collisions.
The Newton's cradle is not the earth or a solid lump but is made out of separate parts which can be put into contact with each other. I think the experimental results speak for themselves. If you haven't already done so look at the you tube videos on the cradle. There is no noticeable rebound.

You may want to try the following:
Place a small mass coin in contact with a large mass coin on a smooth surface. Now get an identical small mass coin and slide it towards the the large mass coin so that there is a collision. I just used two ten pence pieces and a pile of four two pound coins for the large mass. There was no appreciable rebound.

Also, my analysis is an approximation because it assumes a perfectly elastic collision which it is not. Using your bouncing ball analogy a perfectly elastic collision with the floor would result in the ball bouncing back to the same height it was released from. Of course that doesn't happen ........... you can't get perfectly elastic collisions between macroscopic objects. But the smaller the distortion resulting from the collision the more elastic it will be.

"A nice instructive riddle"

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