Why, when one releases 2 balls in Newton's cradle, two balls on the opposite side bounce out at approximately the same speed as the 1st pair, rather than one ball at higher speed, or 3 balls at lower speed?

-

Contrary to what is stated in many textbooks, energy-momentum conservation alone cannot explain the behavior of Newton’s cradle. For N balls we have two equations and N final velocities to calculate. Hence, conservations laws can do the job only for N=2. This means that if we want to give an explanation of the cradle behavior based on conservation laws, we have to split the N-ball collision into a sequence of two-ball collisions, as done in some of the answers given to the question.

The problem with this approach is that it assumes that initially the balls are not touching one another, which is not the case in most cradles. One might argue that it doesn’t really matter that the balls are initially in contact, only that the collision time (the time during which momentum is transferred from one ball to the other) is much smaller than the time it takes for the mechanical perturbation to cross one ball. However, the typical collision time is of the order of 0.1 ms, much bigger than the propagation time of about 0.01 ms. Under such conditions one should expect wave propagation along the ball chain to play a significant role in the cradle’s dynamics.

There is a nice paper,

F. Herrmann and P. Schmälzle. Simple explanation of a well‐known collision experiment. Am. J. Phys. 49, issue 8, pp. 761 (1981). doi: 10.1119/1.12407. Open access version at F. Herrmann's Karlsruher Physikkurs website.

showing that wave propagation along the ball line must be dispersion-free if we are to find equal numbers of incoming and outgoing balls. The authors also provide an interesting picture of how the system decides how many balls it is going to send away. Basically, the first impact produces two wave pulses moving in opposite directions along the ball line. The pulses are reflected at the ends of the line and meet again at the point where the ball chain breaks up. It is easy to see that this break-up point is symmetrical (with respect to the middle of the line) to the impact point, which explains why the number of outgoing balls is equal to that of the incoming ones.

-
Thanks for the nice literature reference! It would be amusing to create a (Non-)Newtonian cradle from some dispersive material. – nibot Feb 1 '11 at 16:37
This link (physikdidaktik.uni-karlsruhe.de/publication/ajp/…) is open access -- for those who don't have a subscription to AJP. – Fabian Jul 7 '11 at 19:50
Open access link now dead :( – Qmechanic Sep 18 '12 at 10:25
I have found a link to the paper – Ali Caglayan Jun 18 at 16:43

It conserves both energy and momentum in the collision at the same time.

By design, when the balls collide the strings that hold them up are vertical (assuming balls are only swung from one side). This means there are no horizontal forces from the string on the balls so linear momentum in the direction of swing must be conserved in the collision. Energy is also nearly conserved provided not too much noise and heat are produced.

Consider the two-ball case where each ball has mass $m$ and velocity $v$ at the time of collision. The kinetic energy will be $E = mv^2$ and the momentum $p = 2mv$. So suppose $n$ balls were to fly off the other side with velocity $u$. The energy would be $\frac{n}{2} mu^2$ and the momentum $nmu$. So by conservation we must have $mv^2 = \frac{n}{2} mu^2$ and $2mv = nmu$. It is not difficult to see that the only solution to these equation together is $n=2$ and $u=v$.

A more complete analysis would rule out solutions with different balls leaving at different speeds, but I think this is enough to demonstrate the principle.

-
Thank you too Philip. This also is very insightful. – jack mosevich Jan 21 '11 at 22:50
It's not the only way to conserve momentum and kinetic energy. If you pull back two balls and drop them against three, for example, you could conserve momentum and energy by having the sets of balls act as solid blocks, so that the two balls bounced back the way they came from with some diminished speed and the three balls pushed ahead at another speed. – Mark Eichenlaub Jan 30 '11 at 20:02
@Mark is right, of course, but this analysis serves to show that the observed behavior is plausible, when many people find it counter intuitive. – dmckee Mar 10 '11 at 23:28
I agree with Mark and dmckee – Philip Gibbs - inactive Mar 11 '11 at 12:08

Let's start with an observation from billiards. Say the red ball is stationary and you hit it dead-on with the cue ball at speed $v$. The cue ball will stop and the red ball will continue on at speed $v$. Lubos gave a nice, simple description of this in his answer. You can see it happening at the beginning of this video.

Now imagine a row of balls (red-orange-yellow-green-blue-violet, say) lined up perfectly.

You hit the red one dead-on with the cue, and the cue stops and the red ball goes ahead. After just a moment, the red ball hits the orange ball. The red one stops and the orange on goes on ahead. Then the orange one hits the yellow one and stops, etc. Ultimately the violet ball comes out of the line, and all the other balls in the line have just moved down the table a little way. One ball in, one ball out. You can see something pretty much like this happen 50 seconds into the same video (it's not quite perfect, though).

Now we do it again. Line up all the balls, but this time roll the cue ball towards the red ball, and roll the 8-ball right after it.

The cue ball strikes the red ball and stops; the red ball goes forward, as before. The red ball hits the orange ball. This time, that's not the only thing happening. Simultaneously with that collision, the 8-ball runs into the stopped cue ball. After these collisions there are two balls in motion - the orange ball in front, and the cue ball, which has now had two collisions, one in front and one from behind.

This process repeats, going down the line, always with two balls in motion with one stationary ball in between. Near the end, the two moving balls are the violet (last) ball and green (third-from-last) ball. The violet ball is free, but the green ball impacts the blue one, and that's the last collision. The result is that the blue and violet balls come away from the line up. Everything else is stopped. Two balls went in, and two come out.

From here it is not hard to see that we could send in three, four, or even 100 balls (in theory) and get the same number out again.

The difficulty with this explanation is that in Newton's cradle, the balls physically touch one another, and so the stop-start argument does not apply in the same way. This simple analysis should make Newton's cradle plausible, but a full argument relies on studying the continuum mechanics associated with the collisions between solid bodies. This paper, mentioned by Georg in the comments, provides such an analysis.

Note: I didn't watch the YouTube video all the way through and don't necessarily vouch for everything it says about physics. I just wanted examples of those shots. Also, the balls in billiards are rolling, which can sometimes make a difference compared to Newton's cradle, but here we will assume it does not.

-
Hello Mark, there is one weak point in Your otherwise brilliant explanation. I think it is not obvious, that having smaller and smaller distance between balls in the end leads to the real cradle. Being in contact is something different than having the tiniest distance. I think one has to go to the real mechanics within the spheres (compression, sound wave propagation) to get the "innermost truth" :=) I assume that when one analyses the elastic reaction within the sphere, the main result will be, that it takes enough time, to make the spheres act as if there was some distance. – Georg Jan 30 '11 at 11:53
@Georg I don't think there's a big problem with thinking about balls in contact as being essentially the same as balls separated by a small distance in this particular scenario. Even if we imagine the balls had infinite bulk modulus, the result would be the same. We could discuss things like the speed of sound in the balls and wave motion, and that would be interesting, but it would also be beyond the scope of the original question. Those considerations would be more important if the question were about how long after the impact on the left hand side the balls took to come out on the right. – Mark Eichenlaub Jan 30 '11 at 12:07
Hello Mark to me it is not so obvious. I'd like to see a "movie" of the strain inside the sphere, for me the efficency of momentum and energy transfer between steel or glas spheres still is a wonder. – Georg Jan 30 '11 at 13:39
@Georg It sounds like you're asking about why the collisions are elastic. I agree that may not be obvious, but it's a different issue. I wanted as simple an answer as possible here. It's helpful to note that the speed of sound in the balls is on the order of thousands of meters per second, while in these experiments the speed of the balls themselves is only meters per second. If we moved to larger, softer, less-elastic balls moving at higher speeds, we'd see a higher fraction of energy dissipated. – Mark Eichenlaub Jan 30 '11 at 18:51
@Georg After thinking about it some more, I realize there's more to the point you made than I originally acknowledged. If the balls are touching, one might wonder why, if you pull out one ball from the left and let it drop, you don't get all four of the other balls to bounce up together to the right and the single ball to bounce elastically back to the left, as would happen if the balls were welded together. – Mark Eichenlaub Jan 30 '11 at 19:00

First, I have a free iPhone app for Newton's cradle - called Kinetic Balls - which has about 500 skins. ;-)

The events in the cradle are composed of many steps although they occur quickly after each other. In particular,

if you have five balls, the first ball starts by colliding with the second ball, the second ball hits the third ball, the third ball hits the fourth ball, the fourth ball hits the fifth ball. OK?

So let's study what happens when the first ball from the left hits the second ball from the left. The speed of the first ball is $v$ before it collides with the second one. Now, it simplifies things to look at the collision from the center-of-mass system of the first two balls.

Needless to say, the center of mass is moving by the speed which is the average of the speeds of the first two (equally heavy) balls. Because the first one is moving by $v$ to the right and the second one has the speed equal to $0$, the average is $v/2$, OK?

Before the collision, the first ball is moving by the speed $v-v/2=v/2$ relatively to the center-of-mass frame, and the second ball is moving by the speed $0-v/2=-v/2$ relatively to the center-of-mass frame.

Now, what happens after the collision? The balls can't penetrate through each other so their relative speed has to change the sign: the speeds get simply reversed. They change the sign. The situation is completely symmetric with respect to the exchange of the two balls combined with the reflection of "left" and "right" in the space. Consequently, the two final velocities (in the center-of-mass frame) must still be equal up to the opposite sign. The overall magnitude is guaranteed to be the same as before the collision, to conserve the kinetic energy $2\times mv^2/2$. The signs have to be opposite to one another - but also opposite to the initial ones because the balls can't penetrate one another.

So in the center-of-mass system, after the collision, the left ball will be moving by the speed $-v/2$ and the right ball will be moving by $+v/2$. Transforming back from the center-of-mass frame to the lab frame, we have to add $v/2$ again. So the first ball final speed will be zero, while the second ball's final speed will be $v/2+v/2=v$. OK?

Now, the second ball is approaching the third ball at speed $v$ while the third ball is at rest. Take these two balls' rest frame.

Now, repeat the fairy-tale above four times - I could do it, it would be very exciting, but I want to save the server's hard disks - and you will end up with the situation in which the first four balls are at rest and the fifth one is moving by the speed $v$ to the right.

-
Thanks Lubos. Very insightful. – jack mosevich Jan 21 '11 at 22:50
Always pleasure, Jack. – Luboš Motl Jan 22 '11 at 8:51
I forgot to say that even if the balls are touching each other, the answer "1,2,3,4" at rest and "5" taking the original speed of "1" is the only right answer. Any distribution of the momentum among several balls would lower the total energy and violate its conservation law. I guess that it's what Phil is explaining in more detail. Well, the momentum $p$ is distributed to $k_ip$ where $k_i$ sum up to one. Then the energy is the sum of $k_i^2$ times the initial energy, but unless all of the $k_i$'s are zero but one of them which is one, the sum of squares is smaller than one - energy is lost. – Luboš Motl Jan 22 '11 at 8:53
You've made it pretty clear why, if you pull back one ball, one ball comes out. Since it only discussed that case, though, your answer doesn't address the question, which is why the same number of balls come out as go in. It should be easy to adapt the discussion to cover that point. – Mark Eichenlaub Jan 30 '11 at 9:52
You are modelling a system with one ball released, whereas the OP asked for one with two, so you are not answering the question. – Sklivvz Jan 30 '11 at 9:55

Given that the initial balls do not bounce back, the original question can be answered directly as the top response shows, but he needed to specify that he was assuming the other balls do not bounce back or otherwise move. I can explain it without math: more or fewer balls can't come off at a slower or faster speed because energy and momentum must be conserved, and they are each a different function of velocity. If you change the velocity, and if no other balls move, then you would be violating one of the principles of conservation. For example, if only 1 ball came off with twice the velocity, momentum would be the same (conserved) as the incoming 2 balls, but there would be twice as much energy in the outgoing ball as there was in the 2 incoming balls. Conversely, if 4 balls came off with 1/2 the velocity, momentum would be conserved, but the energy would be half of what was put it in.

However, it is much more difficult to answer the question if other balls are allowed to bounce back in the other direction.

The answers based on separation are not correct and one poster hinted at how the device can really be explained. It is only a happen accident that the series of independent strikes calculation is correct for multiple balls of equal weight. Imagine 3 balls with the one in the middle that is much heavier. The first ball will bounce off, which is not what happens when they are touching. So the calculation is wrong for a fundamental reason. When they are touching, the middle heavy ball is in a compressed state, not yet pushing back on the first ball to make it bounce back before it is releasing energy to the last ball. This can't be modeled with one striking pair followed by another. To see everything more clearly, model the balls as big, weak springs.

From wiki: "At the point of collision, two shock waves propagate, one forward and one backward. They are carrying all the kinetic energy of the formerly-moving balls as potential energy in the elastic compression of the metal. The wave going backward is reflected forward when it reaches the beginning of the chain of balls, coming behind the wave that was moving forward first. The first wave reaches the end and is reflected back and it collides with the delayed wave at a point symmetrical to the initial point of collision, forcing the balls apart, releasing the potential energy as kinetic energy.[8] This explanation is more complicated if balls of equal weight are given different lengths or if the balls have different weights, but the solution for the final velocities can be found by examining the compression and expansion of the metals as a result of the shock waves. The conservation of energy and momentum are adequate to explain the system if the potential energy and transition time is included in the conservation of energy calculations."

Now to answer the original question: As the other poster said, the shock wave meets at a point symmetrical to the original collision. However, the answer is more complicated if the balls at the end are of equal weight but different in length. The shock waves will not meet to begin with, but the expanding system of metal will eventually cause the equal-weight balls to separate so that the initial balls do not bounce back.

Important Update/Correction:

In looking at this in much more detail, I've found that no one understands how the cradle works. The reflected shock wave theory comes from: http://www.physikdidaktik.uni-karlsruhe.de/publication/ajp/Ball-chain_part1.pdf

but these researchers followed this paper up with this: http://www.physikdidaktik.uni-karlsruhe.de/publication/ajp/Ball-chain_part2.pdf

in which they show the collision time is 10 times slower than sound (shock) wave propagation which indicates "hertzian compression" (not simply shock waves) describes the cradle by including the springiness of the interfaces between the balls. There are 4 interfaces in 5 balls, so energy and momentum give the solution for the final velocity of 2 balls and then the ratio of the 3 subsequent interfaces to the first interface gives the 3 variables needed to solve for the other 3 balls. They apply F=ma to each ball where it is in contact with its 1 or 2 neighbors, subject to a psuedo-spring force (hertzian) in the surface contact of F=kx^1.5 rather than F=kx. This gives a set interdependent differential equations which I'll state for the most general case of different masses and different surface spring constants (youngs modulus) at the end. This may agree with experiment for the case of 1 ball striking 4 that are in contact, but it gives a solution for the case of 2 striking 3 that is not correct. I solved this system of equations in excel and got the same result as this paper: http://www.damtp.cam.ac.uk/user/hinch/publications/PRSLA455_3201.pdf in which ball 4 and 5 velocities would theoretically be 0.80 and 1.14 times the velocity of the 2 incoming balls. The paper asserts that you can observe the difference in velocity, but they fail to point out that this is a 2 times difference in kinetic energy and therefore the height reached by the 5th ball is 2 times more than the 4th, which is not what happens at all. (If 5th ball comes out 30 degrees max, the 4th ball would theoretically reach only 21 degrees, which is not what happens: they both reach 30 degrees at the same time). Here are the F=ma equations that have to be solved with runge-kutta, if the Hertzian theory is correct. I did it in excel.
x is displacement from at-rest point

m1a1= - A*(x1-x2)^1.5

m2a2 = A*(x1-x2)^1.5 - B*(x2-x3)^1.5

m3a3 = B*(x2-x3)^1.5 - C*(x3-x4)^1.5

m4a4 = C*(x3-x4)^1.5 - D*(x4-x5)^1.5

m5a5 = D*(x4-x5)^1.5

where

A=(2/((1/k1)^(2/3)+(1/k2)^(2/3)))^1.5

B=(2/((1/k2)^(2/3)+(1/k3)^(2/3)))^1.5

C=(2/((1/k3)^(2/3)+(1/k4)^(2/3)))^1.5

D=(2/((1/k4)^(2/3)+(1/k5)^(2/3)))^1.5

For plugging into runge-kutta: a=dV(t)/dt and x=t*V(t)

A, B, C, D are the net effective spring constant between each pair of balls for the surface compression.

So for 2 balls or more balls striking the others, no one knows how it works. Even as late as 2004, these papers were incorrectly described as the complete solution: http://www.maths.tcd.ie/~garyd/Publications/Delaney_2004_AmJPhys_Rocking_Newtons_Cradle.pdf

-
OK, after all that I think I found the answer: in a real cradle there must be at least 40 microns separating the balls in order to treat them as simple independent collisions that can be solved simply with conservation of energy and momentum. This is dispersion free energy propagation. I can combine this initial separation with the Hertzian differential equations and get the right answer, but the Hertzian complexity is redundant if they are indeed separated by at least 40 microns and the initial velocity is less than 1 m/s. Less than 40 microns requires the full hertzian solution – zawy Jul 7 '11 at 20:53

People answering questions like this make the dreadful mistake of falling back to the simplistic maths found in physics books, forgetting that the simplistic maths does not represent either atomic mechanisms or the actual way in which momentum works in the real world.

Newton's cradle forces us to consider some very interesting consequences of physics in the real world. The fact that the cradle uses spheres means, for instance, that energy transfer occurs through a surface contact point that (in theory) tends to zero. The spheres are designed to approximate a perfectly elastic material.

Here is a question. What would happen if one sphere at the end was replaced with an identical sized sphere of twice the mass? Lifting this ball would surely be like lifting two ordinary ones, but would the collision lead to two spheres at the other end lifting off?

The solution lies in the theory of waves (shock waves travelling through metal, just as sound waves travel through air). Normally, an n-ball collision causes n-balls to lift at the other end, because the shock wave can be thought of as n-balls long, travelling down the row of balls once the initial collision occurs. It is important to note that the ball collision speed must always be much lower than the speed of propagation of the shock wave through the metal, in order for the cradle to behave as expected.

To go back to the question of the heavier ball, well in this case the greater mass in the same sized sphere implies a higher density, which will change the speed at which a shockwave propagates through the material. What happens when a sound-wave moves between materials of different density? The pitch and the wavelength change. The same process gives us refraction in a prism for the wave that is light.

At an atomic level, the shock-wave concept is going to be difficult to conceive, since the energy in this case enters each new sphere through a tiny point of contact. The 'width' of the energy travelling down the line of balls will determine how many balls depart at the other end, the energy being essentially kinetic in nature. Inertia prevents the kinetic impulse from significantly displacing the balls until only the end balls contain the kinetic energy, and thus can begin moving.

Every physicist should have one mantra they repeat over and over! "The model is not reality." The model is a mathematical tool we construct to predict behaviour under certain constraints (constraints we may not even as yet understand). We see people struggle with the Newton Cradle problem because they insist on believing that simplistic models must never be superseded by more sophisticated understanding. Just look at the nonsense that demands the balls have a minimum gap between one another so the simple model can even be applied in the first place. Look at the fools who think it is a n-ball problem (disallowing balls of differing sizes or density), as if the underlying physics has anything whatsoever to do with macro-collections of trillions of atoms in a particular Human pleasing shape.

Wave theory is fundamental to modern physics. People who care to contemplate the dynamics of moving bodies should always be prepared to switch to a deeper understanding of energy propagation through various materials when the problem clearly switches from very simple mathematical models, to models that must take into account underlying mechanisms caused by the nature of atoms (and their bonds). Newtonian physics at the macro level will only provide correct solutions if the key assumptions remain valid. Newton's cradle is an example of an experiment where Newtonian physics must be considered at an atomic level instead (the creation of the shock-wave). Macro level maths (a very simplified model of reality) will not give correct results. Hacking the wrong model (by requiring a x micron gap between the balls) to try to get the wrong model to kind of work, maybe, in some very limited set of conditions, is a scientific atrocity.

-

## protected by Qmechanic♦Nov 19 '14 at 12:17

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).