Newton's Cradle.by Donald Simanek
The physics toy and 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."
1. The apparatus.The apparatus usually consists of an odd number of identical steel balls (7 is common) 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 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 get you 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". 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. 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.
v=6 --> Before: o oo Ball 1 moving at speed 6, 1 23 balls 2 and 3 are at rest.
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, as the only outcome that happens?
3. Discussion (posted here July, 2001. Revisions, May 2003).We first assume that the collision is elastic, that is, the sum of system kinetic and potential energies is conserved. We need not fuss about the fact that collisions are never perfectly elastic. They can be close enough to perfect for this analysis to help us understand what's going on. This is an example of simplifying the problem as a method for gaining insight, then relaxing the simplification later if necessary.
The word "elastic" has two distinct meanings in mechanics. A "perfectly elastic material" is one that can be deformed and then return to its original condition exactly, without energy loss. A "perfectly elastic collision" is one that involves contact and deformation of two bodies for a short time. After the collision the energy of the two body system is the same as before the collision. But a collision between two perfectly elastic bodies may not be an elastic collision. Consider a bell, made of highly elastic steel struck by a clapper of similar material. The materials are both very elastic. But after the collision of clapper and bell, considerable energy remains in the bell, vibrational energy, which is slowly dissipated as radiated sound energy and some is dissipated as thermal energy, heating the material of the bell. The materials are very elastic, but the collision is not, for after it's all over, a considerable fraction of the mechanical energy is gone through these two processes of dissipation. The clapper separated from the bell before the stored elastic energy in the bell could be converted to kinetic energy. We begin with the assumption that elastic compressions occur simultaneously in all of the balls. This may be the most controversial assumption, for we know that it can't be exactly true. Yet if we can show that using this assumption we can model the observed behavior of the system, that goes a long way toward showing that time delays of a compression pulse moving down the line of balls are not the reason for this observed behavior in this device. They may, however, be responsible for small or subtle departures from the "ideal" behavior. The case of two balls isn't an issue, for in that case the observed result is the only one that satisfies conservation of energy, momentum, and the "impenetrability" constraint.
One of the two solutions of the energy and momentum equations would require the moving ball to penetrate the stationary one, passing through it like a ghost. We must disallow this solution if the balls actually touch. Clearly this would violate the constraint of impenetrability. If the moving ball passes by the stationary one without striking it, this solution is perfectly valid. This presents no problem for us, since we restrict our attention to collisions of impenetrable balls. We assume the reader has done such two-body collision problems. These are worked out in most physics textbooks. So, having seen the objections to standard attempts at explanation, can a better one be constructed by simple means? Can we argue, from what we observe and what we know about physics, to arrive at a quasi-conceptual model of what's going on that will satisfy our questions and correctly predict what we can observe? Again, we emphasize that we do not seek a pure and clean mathematical derivation of a result from basic physical and mathematical principles. That has been done elsewhere, and may be found by consulting the references at the end of this document. We'd like to see how far we can get toward modeling and understanding this system with the simplest of observations, assumptions and elementary physics principles. Preferably, as physicists like to say, "from first principles." Rather than present a distilled and sanitized, lean and mean, exposition, I shall construct this explanation as an unfolding story, trying to reflect the thought processes one goes through when analyzing a new physical situation, starting from what we know and observe, leading to something new that we didn't know, and perhaps to something we didn't (or couldn't) observe initially.
4. Three ball case.Consider a system with only three equal mass elastic balls. This approach is an example of the important scientific method of reduction: analyzing the system by (of N balls) by breaking it into simpler parts and solving them (divide and conquer), then putting the parts together to understand the larger system. The balls are steel and have equal mass. We shall begin by assuming they are perfectly elastic. Ball 1 initially moves with speed V; balls 2 and 3 are at rest. The problem, with given initial velocities, has two constraint equations (energy and momentum) and three unknowns (the final velocities). There are other constraints. For example, we assume the balls cannot penetrate, so the velocities after collision must satisfy v_{1} < v_{2}<v_{3}. The velocities during the brief time that all balls are in contact must satisfy v_{1} > v_{2}>v_{3}. Now let's look at the physics. What happens during the collision? We assumed the collision to be elastic, so we expect that the steel balls compress slightly at the areas of contact. If the compression is elastic, these compressions later relax. Any energy stored during the compression is converted to the kinetic energy the balls have after collision. If the balls are initially touching, we hypothesize that during some brief time interval both interfaces have compression. Very quickly this compression relaxes and the balls separate. Let's simplify the problem by assuming this scenario: As ball 1 impacts ball 2, compression occurs at both interfaces simultaneously (or nearly so). This temporarily stores energy, and the microscopic elastic forces sum to forces of equal size and opposite direction at each interface. That is, ball 1 exerts a force to the right on ball 2 and ball 2 exerts a force of equal size and opposite direction on ball 1 (to the left). Ditto for balls 2 and 3. This is an application of Newton's third law. Observation of the experiment shows us the one energy and momentum-conserving outcome that actually happens: Ball 1 comes to rest and ball 3 moves away with speed V. What does this observed fact tell us about the interactions at the interfaces? It tells us that the moving ball 1 experienced forces that brought it to rest. It tells us that ball 2, initially at rest, ends up at rest. So ball 2 experienced no net change of momentum. From this we conclude that ball 2 experienced equal size and oppositely directed impulses due to compressional forces at its two faces. Sometimes the key to solving a problem is in looking at the thing that seems most innocent. In this case, we must keep our eye on the middle ball (No. 2). You may say "But it doesn't do anything!" True, it is initially at rest and ends up at rest. But what happened in between those events? It must move forward during the compression. So it must experience an unbalanced force (and impulse) during that time. This accelerates it forward. But it ends up at rest, so there must also be an acceleration in the opposite direction, slowing it. This deceleration occurs during decompression. So we conclude that the net work on ball 2 is zero (or nearly so) since it had negligible displacement. But during compression the force on it from interface 2,3 is slightly greater than the force on it at interface 1,2. During decompression, the force imbalance reverses. One model suggested from this is that a compression pulse passes through the two balls, peaking a bit later at interface 2,3 than at 1,2. The time delay between interfaces is very small compared to the duration of the compression. These facts taken together tell us that the compression forces at the two interfaces rise and fall in about the same way and by about the same amount, and that the amount of stored energy is equal at the two interfaces during the time all balls were in contact. This is consistent with the observed facts about what happens to balls 1 and 3. The release of stored energy at interface 1,2 is sufficient to bring ball 1 to rest from its initial speed V. The release of stored energy at interface 2,3 is sufficient to take ball 3 from rest to speed V. These require equal amounts of energy. So we tentatively assume the following hypothesis: In the case of identical, perfectly elastic balls, the energy stored at the interfaces during contact is equal, or very nearly equal. Also, we note an obvious fact that stored energy at an interface is not released equally to the balls on either side of the interface. In this case, no net stored energy is released to ball 2. If one ball has negligible displacement during the collision, negligible energy was released to it. The casual reader may skip the next two sections on first reading, and may wish to return to them later.
5. Finer details.(This section necessarily repeats some points made in the previous section.) Let's tighten this up a bit. During the time two balls are in contact, the size of the contact forces continually changes. What's important is the time-average of the force over the time they were in contact. This determines the impulse, the time integral of Fdt, is equal to F_{ave}dt. Impulses obey Newton's third law also: The impulse of A on B is of equal size and opposite direction to the impulse of B on A. The impulse acting on a body is equal to the body's change of momentum, and the conservation of momentum law arises from consideration of impulses. The contact forces do work in compressing the balls near the impact points. This results in energy stored as potential energy at the interfaces. This energy is released as the compression relaxes, doing work on the balls. During impact and compression, conservation of momentum tells us that the three balls move with nearly equal speeds V/3, where V is the initial speed of ball 1. Their total kinetic energy is, however, (1/2)(3m)(V/3)^{2}, or (m/6)V^{2}. Ball 1 initially had kinetic energy (m/2)V^{2} = (3m/6)V^{2}. So we conclude that 1/3 the initial kinetic energy is temporarily stored at each interface, while 1/3 of it is the kinetic energy of the forward motion of the center of mass of the three balls. [Student: verify these results.] The force at interface 1,2 acting on ball 1 is just sufficient to decelerate ball 1 to rest from speed V. The force at interface 2,3 is the same, and just enough to propel ball 3 from rest to speed V. The two forces on ball 2 are equal in size and oppositely directed, producing zero acceleration, so it remains at rest. We have learned some important things about action-reaction force pairs here. Looking at it another way, the stored potential energy at the interfaces does work on the balls touching there. The work done on ball 1 is just enough to bring it to rest, the work done by the equal size and oppositely directed forces on ball 2 results in no change of its kinetic energy, and the force on ball 3 accelerates it from rest to speed V. Note that the stored energy at interface 1,2 gets released to ball 1. The stored energy at interface 2,3 gets released to ball 3. The stored energy does not get released equally to adjacent balls. The reason is that ball 2 is begins and ends at rest, and its displacement during the collision is zero (or nearly so). Work is the product of force and displacement, Fdx, so if there's no displacement, no net work is done. To summarize: The fact that ball 2 begins and ends up at rest tells us that the interface impulses 1,2 and 2,3 are equal in size. The fact that ball 2 has zero net displacement tells us that the work done on it by compression forces was of equal size and opposite sign at its two faces. This tells us that the amounts of stored elastic energy at the interfaces were equal. Finally, these compressions at the two interfaces did work of equal size and opposite sign on balls 1 and 2. These are important facts for understanding this case and others to follow. Can it really be this simple?
6. More pesky details.We've glossed over some possibly important details. Just after impact the three touching balls don't just sit there. Their combined center of mass moves forward with speed V/3. Also, they are being dynamically deformed. In fact ball 1 moves faster than 2 and 2 faster than 3 as they compress. As they decompress, ball 3 moves faster than 2 and 2 faster than 1. This necessary fact of compression/decompression tells us that during compression the force at the 1,2 interface is greater than the force at the 2,3 interface (since ball 2 is accelerating forward) and during decompression the force at the 2,3 interface is greater than at the 1,2 interface (since ball 2 is then decelerating). It is only the time-average net force on ball 2 during the entire process of compression/decompression that is zero (we know that because the net momentum change of ball 2 is zero). And the net work done on ball 2 is zero (we know that because the net displacement of ball 2 is zero). To simplify the language of the following discussion, we will sometimes use the word force to refer to the time-averaged force over the duration of the compression.
7. The case with two of three balls initially movingNow consider a new and crucial experiment. Three identical balls in line, as before, but this time pull back balls 1 and 2 and let them hit ball 3 with speed V. Observed outcome: balls 2 and 3 come away with the same speed, V, leaving ball 1 at rest. How does our simplified (perhaps simple-minded?) model fare on this one?
V -> ball at rest before: oo o V/3 -> during: ooo V -> after: o oo at rest Our model fares very well, and we needn't assume anything new. Ball 2 experiences a net impulse of zero, according to our assumption of identical compression forces at the two interfaces, so the momentum of ball 2 doesn't change. It was moving initially with speed V so it must therefore come away with speed V. The impulse acting to the left on ball 1 is enough to slow it to rest, and that same impulse acting to the right on ball 3 is enough to take it from rest to speed V. What about the work done by the compression forces? The displacement of ball 2 isn't zero, but the compression forces acting on its two faces are at all times of nearly equal size and opposite sign, and therefore ball 2 experiences no net force. During compression, ball 2 slows a bit, but speeds up a bit on decompression. That was too easy!
8. N identical balls, one initially moving.Does the same analysis work for N balls? In the case where only one ball is pulled back, all balls except the end ones experience equal sign and oppositely directed compressional forces from the adjacent touching balls, so, while they do move a bit, they start at rest and end at rest. No problem here either.
9. N identical balls, two initially moving.Consider a string of N equal mass elastic balls, all touching. Pull back two and propel them at speed V toward the others.
V -> balls at rest before: oo oooooooo during: oooooooooo after: oooooooo oo at rest V -> On contact, compression forces are initiated. Are they of equal size at all interfaces? No, they can't be. Both balls 1 and 2 experience the same deceleration. If ball 1 receives an impulse I from interface 1,2, ball 2 receives the same impulse from that interface, and an impulse 2I from interface 2,3. This ensures that balls 1 and 2 receive equal size NET impulses. So in the case of four balls, with two initially moving, we expect the compression forces at the three interfaces to be in ratio 1:2:1. With five balls the ratios are 1:2:2:1. With N balls the ratios are 1:2:2:....:2:2:1. We spare the reader a detailed analysis of this. But we note that the stored energies are in these same ratios, and that only the first two and the last two balls in the string receive any net work.
10. The successive collisions model.Many books treat Newton's cradle by assuming that the balls are initially not touching. Ball 1 hits 2, and rebounds before 2 hits 3, and so on down the line. This gives essentially the same outcome as the case where all balls initially touch. But we have seen that sequence of events for the touching balls is different: Ball 1 hits 2, then all balls move forward together for a short time interval as compression and relaxation occur. Then ball 1 and N separate from the others, then ball 2 and N-1 and so on. Both scenarios satisfy conservation of energy and momentum at all steps. But to assume that the sequence of separations is the same in both cases is seems to me unjustified. High speed motion picture studies could settle this issue once and for all.
11. What about those "spurious" outcomes?Returning full cycle to the 3-ball case with one initially moving, what about those energy and momentum conserving solutions that didn't happen? We wondered why they didn't happen. Now we understand this well enough to answer that question. The solutions that would require rebound of ball 1 don't happen because there simply isn't enough stored potential energy at the 1,2 interface to make that happen. There is just enough energy stored at that interface to bring ball 1 to rest. The hypothesis of equal or near-equal stored potential energy at the interfaces is sufficient to rule out all but the observed 'standard' result in the system with equal mass balls. This is true even if the separation of both interfaces is simultaneous, and even if separation occurs at interface 2,3 before it occurs at interface 1,2. Reality check 1: Flansburg and Hudnut [1979] found experimentally and through mathematical modeling that the actual experiment does show a slight rebound of ball 1. Typically final velocities of three balls (one initially moving) were -0.12, +0.15 and +0.98 for air-track gliders, and -0.06, +0.09 and +0.97 for steel spheres, compared to the "ideal" result of 0, 0, and 1. This suggests that the energies stored at the interfaces are not exactly equal, but differ by a very small amount, the interface 1,2 having slightly larger energy than interface 2,1. Thus ball 3 separates away from the others when there's still a little stored energy at interface 1,2, just enough more than there was at 2,3 to cause slight rebound of ball 1, and a slightly lower than ideal velocity for ball 3. Reality check 2: We assumed instantaneous propagation of compression, even though we knew better. Why does this seem to work? Hermann and Schmälzle [1986] note that the collision time for 5 cm diameter steel balls is 0.2ms. This is much longer than the propagation time of a compression pulse through steel, which is about 10^{-2}ms.
12. Testing the model.Is this the complete story? Is this the last word about this apparatus? Certainly not! No matter how initially satisfying, a physical model must be tested in new situations, and tested skeptically. We may find this model inadequate to deal with cases with less symmetry, for example, when the balls have unequal masses. Try to think up some experimental tests that will challenge our model.
13. Summary and conclusions.So we have seen that the usual outcome of the Newton's cradle (string of identical spherical balls) can be understood from a few basic principles:
A challenge. Now how might we modify the apparatus to allow one or more of the other energy and momentum-conserving solutions to happen? Hmmm...
14. Beyond the cradle: Newton's asymmetric ballsAny successful model of Newton's cradle ought to correctly deal with the case of unequal mass balls, and even collision of objects of different shape or density. A particularly interesting variation of this toy is made with several equal balls, but including one ball of mass four times each of 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. Finally, is there something special about the cases where the ball masses are in ratio 3:1? Fair warning! Much of the above discussion dealt with identical balls. By identical we meant that they have the same shape (spherical) the same size and are made of the same material. When two colliding objects have differences in any one of these, there's the possibility that one will retain some vibrational energy after separation, that dissipates to the surrounding medium. A collision between two bodies made of perfectly elastic materials may in fact be an inelastic collision (not conserving kinetic energy). [Any collision that produces a sound (a click, clunk or ringing sound) loses energy, and isn't a perfectly elastic collision.] Anyone who wishes to carry this investigation further can consult a useful simulation and theory discussion at Computer Animations of Physical Processes website.
Bibliography.These are all from the journals:
Revised June 6, 2002, Jan 18, 2003, May 13, 2003. 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. |