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

[member659127]

TL;DR Summary

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 seperation" argument, and curious what would happen in my ideal universe if there was zero seperation. 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".

 

[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]

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.

 

[PeroK]

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]

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]

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.

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]

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.
I'll try to get access to the article.

 

[pinball1970]

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]

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]

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.
I'll try to get access to the article.

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]

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?

 

[PeroK]

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!

 

[rcgldr]

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.

https://www.physicsforums.com/threads/a-funny-remark-about-Newtons-cradle.913222

 

[stevendaryl]

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.

 

[Dadface]

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]

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?

 

[Dadface]

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]

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]

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

 

[Dadface]

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.