Why do two balls launch from the right in a Newton's cradle instead of just one?

[member659127]

TL;DR

Newton's cradle

That question came from one of the students in my classical mechanics class, it took me unprepared at first but was able to give the correct answer in like half a minute. They liked it a lot and it initiated a useful discussion about laws of nature extending even to some particle physics concepts like Compton scattering. So I wanted to share it here also... Tutors don't give the answer so quickly please

The famous Newton cradle: You have several balls hanging. You release one from left, it hits the balls and the rightmost one launches. Everybody has seen this. Now, another case. You release 2 balls together from the left and you observe that 2 balls launch from right. The question is WHY? Why not a single ball from right but two to be more clear...

 

[osilmag]

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.

 

[fresh_42]

If the final was only one, it would go higher than the initial two.

And why can't this happen? There is no conservation law for velocity.

 

[russ_watters]

And why can't this happen? There is no conservation law for velocity.

Not per se, but there is a conservation of energy, for which velocity is the key component here. If the mass changes you couldn't satisfy both conservation of momentum and energy at the same time because of the V² term in KE.

 

[Dullard]

I predict a different result if the 2 'motive' balls are super-glued together.

 

[Nugatory]

I predict a different result if the 2 'motive' balls are super-glued together.

Why? Asking, not arguing.

 

[member659127]

Ok, for momentum and energy arguments here is a configuration which conserves both momentum and energy...

 

[Dullard]

Why? Asking, not arguing.

I may be confused, but:
There are are several results which satisfy conservation laws (1 ball really high, 2 balls same height, 3 balls less high...). The number of balls 'excited' is the result of the number of motive impacts.

 

[fresh_42]

I may be confused, but:
There are are several results which satisfy conservation laws (1 ball really high, 2 balls same height, 3 balls less high...). The number of balls 'excited' is the result of the number of motive impacts.

You have to solve both: $2m\vec{v}=m\sum_k \vec{v}_k$ and $m\vec{v}^2=\frac{m}{2}\sum_k \vec{v}_k^2$

 

[gmax137]

The "Newton's cradle" page on Wiki provides some interesting points, in particular the importance of the (small) separation between the balls.

 

[member659127]

I like to think about those sort of questions by asking myself what would be the case in an "ideal" universe. Here, what I mean by an ideal universe is the one in which all objects are perfectly rigid, zero elasticity so the interactions are instantaneous (and impulses are represented by dirac delta functions you might say). In such a universe (avoiding all sorts of complications regarding elasticity, pressure waves, etc...) the only way I can work my way through this one is to study the pairwise inteactions.

Quoting wikipedia:
"... when two balls are dropped to strike three stationary balls in a cradle, there is an unnoticed but crucial small distance between the two dropped balls, and the action is as follows: The first moving ball that strikes the first stationary ball (the second ball striking the third ball) transfers all its velocity to the third ball and stops. The third ball then transfers the velocity to the fourth ball and stops, and then the fourth to the fifth ball. Right behind this sequence is the first ball transferring its velocity to the second ball that just stopped, and the sequence repeats immediately and imperceptibly behind the first sequence, ejecting the fourth ball right behind the fifth ball with the same small separation that was between the two initial striking balls..."

Still not sure about the "small separation" argument, and curious what would happen in my ideal universe if there was zero separation. I can't convince myself that there will be one and only one path the system could take...

(just a crazy idea: the indeterminacy in quantum mechanics... could it be because the systems are similar to my "ideal" universe where things and rules are just "simple"...)

 

[Dale]

You have to solve both: $2m\vec{v}=m\sum_k \vec{v}_k$ and $m\vec{v}^2=\frac{m}{2}\sum_k \vec{v}_k^2$

Yes, but you have 2 equations and 5 unknowns, so there will be multiple solutions. One such solution is shown by @erbahar in post 7. This is a valid solution, but I have never seen it happen, so there is something more than just conservation involved.

 

[fresh_42]

Yes, but you have 2 equations and 5 unknowns, so there will be multiple solutions. One such solution is shown by @erbahar in post 7. This is a valid solution, but I have never seen it happen, so there is something more than just conservation involved.

Newton #1?

 

[vanhees71]

It's anything but trivial. The analysis with simple models shows that Newton's cradle is a system with (nearly) dispersionless "signal propagation" and that explains the observed behavior. The apparently simple explanation with energy-momentum conservation alone is of course wrong for the above discussed reasons. It only provides the "explanation" if you already assume the observation that you try to explain! For a very nice review, see

https://doi.org/10.1119/1.2344742
and references therein.

 

[A.T.]

I predict a different result if the 2 'motive' balls are super-glued together.

Has anyone tried it? Or replacing the 2 dropped balls with a single one of twice the mass?

 

[Peter Morgan]

https://doi.org/10.1119/1.2344742and references therein.

I haven't followed the references, because I presume it would have been mentioned in the article linked to if another significantly different explanation was contained therein.
My immediate reaction to the above discussion is that energy and momentum are not the only conserved quantities. There are five second-order differential equations in the $n$ variables $\theta_i$, say, with a force law $f(\theta_i-\theta_{i+1})$ for the interactions between the balls, giving us the $n$ equations $$\frac{d^2\theta_i}{dt^2}+\theta_i+(1-\delta_{i,1})f(\theta_{i-1}-\theta_i)+(1-\delta_{i,n})f(\theta_i-\theta_{i+1})=0,\quad 1\le i\le n,$$ assuming the $\theta_i$ are small and scaled so the mass is 1. That system of equations has more than just two conserved quantities. The $2n$ initial conditions would fix the subsequent evolution, which is presumably not very sensitive to the precise force law insofar as real-world examples are typically not finely engineered.

 

[DaveC426913]

Yeah, with two motive balls, why not one ball twice as high?

 

[fresh_42]

Yeah, with two motive balls, why not one ball twice as high?

Not per se, but there is a conservation of energy, for which velocity is the key component here. If the mass changes you couldn't satisfy both conservation of momentum and energy at the same time because of the V² term in KE.

 

[DaveC426913]

Does that ... answer my question?

 

[PeroK]

Yeah, with two motive balls, why not one ball twice as high?

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.

 

[fresh_42]

Yes, if momentum $mv$ is conserved and kinetic energy $\frac{1}{2}mv^2$ then mass cancels out, and an increased velocity cannot be both: linear and quadratic.

 

[biffus22]

Good Lordie, Isn't it amazing just how complicated even such a very "simple" application of the laws of Mechanics can be??

 

[Dale]

Yes, if momentum $mv$ is conserved and kinetic energy $\frac{1}{2}mv^2$ then mass cancels out, and an increased velocity cannot be both: linear and quadratic.

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.

 

[Vanadium 50]

But with 5 balls you have 5 unknowns and still just two equations.

But you do have constraints, and those constraints will provide additioonal equations (only one thing happens). For example, a ball cannot overtake another ball.

 

[Dale]

But you do have constraints, and those constraints will provide additioonal equations (only one thing happens). For example, a ball cannot overtake another ball.

The configuration in post 7 also satisfies those constraints.

 

[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.

 

[haruspex]

In another current thread on the subject I modeled the three ball case as three springs.
See post #11 at https://www.physicsforums.com/threads/conservation-of-momentum-elastic-collisions.972238
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.

 

[member659127]

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]

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.

 

[Vanadium 50]

The configuration in post 7 also satisfies those constraints.

Which is why I said "for example".