Newton's Cradle.by Donald Simanek
The physics toy and physics demo sold as "Newton's cradle" is also called "colliding balls", "Newton's spheres", "counting balls", "impact balls", "ball-chain", the "executive pacifier", and even, believe it or not, "Newton's balls." This document makes no claim to be a complete treatment of this device. The extensive literature references should be consulted by those who wish to delve deeper. We will attempt to shed a little light on things not often mentioned in elementary textbook treatments, and suggest some experiments one might do to test certain assumptions about the physics of elastic and inelastic collisions.
1. The apparatus.The apparatus usually consists of an odd number (7 is common) of identical steel balls each suspended by a bifilar suspension from a sturdy frame. The balls are carefully aligned along a horizontal line, just touching. If you don't have one of these and haven't seen one in action, please play a bit with this interactive flash animation demonstrating the idealized behavoir of this apparatus. When the ball on one end is pulled aside and allowed to swing as a pendulum, it hits the next ball. But the outcome is fascinating, the one ball on the far end is knocked away from the others with the same speed as the first ball had initially and all of the other balls remain nearly at rest. If you pull back two balls and let them strike the others, two balls are knocked from the other end, and all the other balls remain nearly at rest. Why does this happen? Why are these the only outcomes that occur? Why not others? We shall refer to this as the 'standard behavior' and the standard observed outcome, for the purposes of discussion. We are quite aware that this is an idealized outcome, and that the real apparatus doesn't quite achieve it, though it comes quite close. We are also aware that deviations from the ideal conditions (differences in mass, spherical vs. cylindrical masses, some balls touching, some not) can cause very interesting deviations from expected behavior and are severe tests for any model of system behavior. Some of these are discussed in the references at the end of this document. Why has this become a standard demonstration in physics courses? What important principle is it supposed to show? Usually it is "advertised" as a demonstration that momentum is conserved in collisions. Well, the outcome certainly does illustrate that, but so does every other mechanical interaction you might care to consider, whether it be elastic or inelastic. This particular apparatus, cleverly designed to be nearly elastic, is a special case, and the full generality of conservation of momentum is not demonstrated by it. Some books say that this demo shows that both energy and momentum are conserved in a collision. That's closer to the mark. But still, this apparatus is a special case: collisions between spherical nearly perfectly elastic balls of equal mass, size and composition. And those special conditions are responsible for the intriguing and special behavior we observe. Unfortunately this raises in the inquiring student mind all sorts of questions, such as "What if the balls were of different size, mass, shape or composition?" And that opens a whole can of worms that could lead one far astray and consume a lot of class time. It can easily lead into discussion of energy-dispersive systems and the importance of impedance matching. Seeing the demonstration raises a very sticky question: "How do the balls "know" that if you have N balls initially moving, that exactly N balls should swing out from the other end?" This is the feature of this apparatus that justifies the name "counting balls", for the system seems to "remember the number". This is the big question that the elementary accounts do not answer satisfactorily. Answering all of these questions would be fine for an upper-level university physics course, but are hardly suitable for a freshman or high school course. Sometimes textbooks suggest a simpler version of the apparatus: marbles rolling on a grooved track. This presents even stickier problems, for the result depends on conservation of energy (including linear and rotational kinetic energy terms), conservation of momentum and conservation of angular momentum. Still worse, friction, rolling resistance, slipping on the track and momentum exchange with the track during the collision all affect the outcome. Do not expect to find the answers in this web document. Consult the journal references at the bottom of this document if you want to delve deeper. As a point of departure we begin by limiting our discussion to the classic apparatus: identical perfectly elastic spherical balls. Only later will we address systems with various shapes, masses and materials.
2. Inadequate textbook treatments.
Textbooks and internet sites often tell lies about this demonstration. Here's a few of these lies, with my comments in square brackets:
Some fake explanations try to convince the reader by citing a particular example:
V -> balls at rest before: O OO V -> after: OO O at rest Consider three balls. Balls 2 and 3 are stationary. Ball 1 hits ball 2 with speed V. Ball 3 moves away with speed V leaving balls 1 and 2 stationary. [Yes, that's what happens in the real world.] Momentum and energy are both conserved. The math is easily checked, and every statement is true. But this not the end of the story by any means, nor is it a proof, nor does it give any insight into the problem. You do not prove the impossibility of all other outcomes by showing that one other outcome is impossible. That's an elementary fallacy of logic. Are there other outcomes that satisfy conservation of energy and momentum, yet are not observed to happen? Yes, and they are easily found. For simplicity, take all masses to be 1. Ball 1 has initial speed V, balls 2 and 3 are initially at rest, touching each other.
V=6 -----> Before: o oo Ball 1 moving at speed 6, 1 23 balls 2 and 3 are initially at rest. Contact: ooo -> All three moving with speed 2 After: oo o -----> Balls 1 and 2 at rest. Ball 2 has speed 6. This is the observed outcome, which we will call "case 1", summarized below. But why does case 2 not occur?
Case 2 clearly satisfies conservation of energy and momentum. Yet it is not observed to happen. Suspicion focuses on the processes happening during the impact, when the three balls were in contact for a brief time interval. What's going on there? The balls deform elastically near the point of impact. Can you find any other hypothetical situations that would conserve energy and momentum but do not happen? If we can answer this question for the three-ball case we might gain insight into the N-ball general case. In fact, if we look carefully at the two-ball case we might learn something about how the elastic properties of the balls store and release energy. Some textbooks and web sites tell you nothing other than a description of the behavior of the system and note that this behavior satisfies the conservation of energy and momentum. Perhaps they are "playing it safe" by not attempting to answer the obvious questions that "inquiring minds want to know." So, with these trivial distractions out of the way, what is the explanation? Why is just one (of many) momentum and energy conserving outcomes selected by the laws of physics, to be the only outcome that happens?
DiscussionThe final velocities after collision depend not only on the initial energy and momentum of the balls, but on the nature of the impact. The impact causes deformation of the balls at the points of contact, and the temporary storage of energy as elastic energy. In the case of three balls, there are two contact interfaces. During a brief time interval compression occurs at both interfaces, and some of the initial energy is stored at each one. During this time interval all the balls are moving with nearly the same velocity, and conservation of momentum determines that velocity exactly.Then separation occurs after the elastic energy is relased back to the balls. Since there are two interfaces, each interface stores only a part of the total compressional energy. In the case of two equal mass balls with ball 2 initially at rest the observed outcome tells us that just enough energy is released at the contact interface to cause ball 1 to slow it to a stop. Not all of that interface energy goes to ball 1, for by Newton's law we know that there's equal size and oppositely directed forces at the contact point acting on balls 1 and 2 at all times. But the elastic force acting leftward on ball 1 is acting opposite to its speed, succeeding only in slowing that ball to a stop. The elastic force acting rightward on ball 2 is in the direction of its speed, so it accelerates ball 2 to higher velocity. Ball 1 loses kinetic energy and ball 2 gains the same amount of kinetic energy. This is why the hypothetical outcome in case 2 (above) does not occur in the perfectly elastic collision of two identical balls. Other energy and momentum conserving outcomes do not occur for more obvious reasons. Let's return to the three ball case. If ball 1 impacted stationary balls 2 and 3 at speed V, an outcome with ball 1 emerging at speed V with balls 2 and 3 still at rest would conserve energy and momentum. But it would mean that ball 1 never hit balls 2 and 3. Or, it passed right through them. Other considerations include the question of the speed of the compression impulse through the metal of the balls. Some of these matters are discussed in the bibliography references.
Modeling Newton's CradleWe have sketched out an argument that makes plausible the behavior of the Newton's cradle, in particular we addressed the question of why some situations that would conserve energy and momentum do not in fact occur. Be we haven't looked at what's going on within the string of balls to bring about this result.1. The successive impacts model. The simplest model to understand is one that invokes a "cheat". It assumes the N balls are initially not touching. The first ball is pulled back and strikes the second with speed. The first ball comes to rest and the second moves forward with speed V, hits the third ball, and so on down the line, till the last ball is ejected with speed V. This is valid when the balls are actually separated. But then some folks assume that the explanation is also valid when the balls are touching. Well, the results are nearly the same in both cases, but the dynamics of the processes are certainly different. We will move on to look at the interesting case, where all balls initially are touching each other. 2. The compression pulse model. This assumes that a compression pulse begins in the metal balls at the point of first impact, traveling through the balls with the speed of sound. The speed of sound in the material of which the balls are made is much greater than the speeds of ball motion in this device. So the pulse does its work before any of the stationary balls have moved. The pulse travels forward and backward, reflecting from the ends of the string of balls and meeting at a point. Where is that point? Well, if the pulse originated between the first two balls, the pulse meets between the last two balls, where it gives up its energy to the last ball. This sounds plausible at first. But there's a troulesome issue. It requires that all of the energy of the first ball is given up to a compression pulse and all of that energy ends up and one localized point at the point where the last two balls touch. How does it do that without dispersal of energy, for the pulse initially goes in all directions, forward, backward, up, down and all directions in between? They reflect off the ball surfaces (the balls are round after all) in very complex paths (and most of these paths are not equal in length from start to finish). Though it sounds good, it fails to convince the skeptical student. And how does this pulse give up all its energy to the last ball, instead of giving some back to the string of balls behind it? And does this model hold up if the balls have identical mass but different sizes? What if a larger ball were placed in the middle of the array, say twice the diameter, replacing two of the other balls? This should not affect the time of arrival of the pulse at the other end. But it certainly does affect the outcome of the experiment. Perhaps simpler, add mass to a ball in the middle of the array, by attaching a weight to its bottom. Why should this affect the compression pulse? But it does affect the outcome. 3. The balls-and-springs model. This model imagines a linear string of balls with small springs between them. It treats the system as a lattice array. It turns out that to make this work as a simulation of the spherical ball Newton's cradle, the springs do not obey Hooke's law, F=-kx, but rather obey the Hertzian spring law, F = -kx^{1.5}. This, it is argued, is a result of the balls being spherical. A linear array of objects of different shape, say cylinders, would behave differently. While interesting, this model is not an exact simulation, for its predictions do not quite match the real behavior. 4. The interface compression model. We know that at impact the balls, being elastic bodies, must deform at the point of impact. This stores compressional energy. In the N-ball case we assume that all balls are in contact and moving at nearly the same speed. So conservation of momentum allows us to calculate that speed. (In the case of N balls with one ball initially moving at speed V, the balls move at V/N during the contact phase.) Then we can calculate the kinetic energy of this composite body. It will be less than the initial kinetic energy of the system, because some energy is temporarily stored as elastic energy at the interfaces. This energy is distributed at all of the interfaces, but not equally (except in a few special cases). During decompression, work is done as this energy is released at the interfaces. (In the case of one ball initially moving at speed, it is slowed to a stop, as are all the others except the last ball in the string, which moves away with speed V.) It all seems plausible, but doesn't answer the question of how the stored energy is distriuted at the N-1 interfaces. That depends on the initial conditions. Consider three balls, with ball 1 moving at speed V, ball 2 at rest, and ball 3 moving at speed -V. From the symmetry, we expect that the compressional energy is equal at the two interfaces. But in the case with ball 1 moving at speed V, with ball 2 and 3 at rest, we might suppose that twice as much energy is stored at interface 12 as at interface 23. Why? During the compression phase the force at interface 12 acts on the two other balls (total mass 2), accelerating them and doing work on them. The force at interface 23 acts on ball 3 (mass 1), accelerating it by nearly the same amount. So the force at 12 must be twice as large as the force at 23. Since these forces act over the same distance, the work they do is also in ratio 2:1. The stored energies at the interfaces are therefore in 2:1 ratio. But if the initial situation were speed V for ball 1, zero for ball 2 and -V for ball 3, then the forces at the interfaces are equal, and the energies stored there are equal. Testing the models. Any model we might devise ought to be successful for the N-ball case with identical spherical elastic balls. It also ought to be successuful in the case where the balls are of mixed sizes, shapes, and masses. The very fact that new models are proposed every year, in professional journals, is evidence that there's no fully successful model yet, certainly no simple one suitable for elementary physics classes. The bibliography references at the end of this web page is further evidence of that. A particularly interesting variation of this toy is made with several equal balls, but including one ball of larger mass than the others. Consider the three-ball version:
Can you predict the outcomes? Can you predict which ball moves faster, and whether one or more balls remain stationary after impact. Here's an AVI movie of case 4. It's size is 600 kB. You need the Windows media player or some other AVI player to view this video clip. Answers: Doing the experiment we observe that cases 1 and 5 have the same outcome: the middle ball remains at rest after collision and the last ball in line comes out with the same speed the first one had. It's as if the middle ball didn't participate. In case 2 all three balls move forward after collision. In case 3, the first ball rebounds, the other two move forward, the smallest one moving fastest. How well did you do? A naive conclusion is that in any 3 ball collision with the middle ball initially at rest, the middle ball would remain stationary after collision, the other two behaving exactly as in the two ball case when they hit each other. In cases 1 and 5 that's exactly what happens. But in case 2 it does not. Case 2 seems to be the reverse of case 4, but its outcome isn't. There's a profound clue in these results. Remember that we noted above that the 2-ball case is such that only one solution results from application of the laws of energy and momentum conservation. The predicted result is the one observed. Also, Newton's cradles with initial separation between the balls, produce nearly the same observed results as the kind with balls in contact, and may be modeled as successive 2-ball collisions. This gives important insight into the case where the balls are initially touching. If we model this case by saying that separation occurs first at the left-most interface, and then successively at the other interfaces, we can treat this problem as a succession of two-body interactions, and such a model predicts one unique outcome, which is the one observed. Commercial versions sold as scientific apparatus for physics demonstrations often include one ball that is 3 or 4 times the mass of the others. What do you suppose this is intended to demonstrate? In case 4 above we can calculate, by the successive collisions model that the outcome should be: V=6 -----> Before: o oO Ball 1 moving at speed 6, 1 23 balls 2 and 3 are initially at rest. Contact: ooO -> All three moving with speed 1. After: o o O Ball 2 at rest. <---- -----> 3.6 2.4 This is the observed outcome, as seen in the video above and summarized below.
Final exam: Let's look at the second case above. Check these results.
V=6 -----> Before: O oo Ball 1 moving at speed 6, 1 23 balls 2 and 3 are initially at rest. Contact: Ooo -> All three moving with speed 24/6 = 4 After: O o o ----> ------> --------> 2.16 5.76 9.6 This is summarized below, calclated by the successive collisions model.
Now do the experiment to see whether the actual device, with balls initially touching, matches these results from the successive collisions model. Any model of these devices must successfully predict situations with unequal mass balls as well as with unequal size and shapes.
Bibliography.These are all from the journals:
July, 2001. Revised 2002, 2003, 2003. Major revisions, August 2014) Input and suggestions are welcome at the address to the right. When commenting on a specific document, please reference it by name or content. Return to Donald Simanek's page. |