I have a question about a Newton's cradle type collision, but it has a twist to it. First, I will describe two well-known results, and then I'll add my twist.

First, consider a cradle with five balls (all of equal mass and size), and I move the leftmost ball up and release it. It strikes the row of balls and by conservation of KE (assuming this is an elastic collision) and momentum, the rightmost ball comes out swinging outwards. See picture below.

enter image description here

Second, consider the same setup, and I move the two leftmost balls up and release them together. They strike the row of balls and by conservation of KE (assuming this is an elastic collision) and momentum, the two rightmost balls come out swinging outwards together. See picture below.

enter image description here

Ok, now here is my twist. Suppose I lift the leftmost ball up, but now I add highly quick-acting adhesive glue to the ball so that it sticks to anything it touches. I release the ball and it hits the row of balls. My question is, how many balls should come out swinging on the other end and at what initial velocity? See images below for a rough depiction. (Note that no other balls on the right end have any glue introduced.)

enter image description here enter image description here enter image description here

Here are a couple of possible answers and their reasonings:

  1. Answer 1: One ball comes out with the same velocity. Since the leftmost ball hits the row of balls and it doesn't ever bounce off, the scenario is identical to that of the one without any glue. If you couldn't see the glue, you wouldn't be able to tell the difference, because in either case the initial ball doesn't bounce or anything, and both KE and momentum are transferred to the rightmost end as usual. This scenario is identical to that of releasing one ball from the left, so one ball comes out from the right.
  2. Answer 2: Two balls come out with half the velocity. We can imagine that when ball #1 (the initial leftmost one) strikes ball #2 (the second leftmost one), they combine and an inelastic collision occurs. This means we go from $(m_{1}, v_{1}) = (m, v)$ and $(m_{2}, v_{2}) = (m, 0)$ to $(m_{\text{combined}}, v_{\text{combined}}) = (2m, v/2)$. Momentum is preserved but KE is lost (because an inelastic collision occurs), and thus we now have two balls together impacting the row. Afterwards, everything as in a normal Newton's cradle would happen. Thus, this scenario is identical to that of releasing two balls from the left, so two balls come out from the right (albeit with half the velocity together).
  3. Answer 3: None of the others because it depends on the specifics like the glue type. Maybe the answer varies based on further details of the setup. Maybe you get a different answer if you are using glue vs using velcro vs using cold welding (in vacuum in space) vs who knows what. Maybe it depends on the specific sequence of events that takes place in the narrow time interval of nanoseconds, and maybe the result changes each trial or something.
  4. Answer 4: None of the others but it doesn't depend on further specifics. Maybe something weird like the rightmost two balls are launched, but they are launched with different velocities, and this result is independent of further details like glue type, ball material, collision time, etc.
  5. Answer 5: All the balls will swing in unison with one-fifth of the original velocity. The "fast-acting" glue creates an inelastic collision, but there is no reason it would reduce the kinetic energy by half only. In fact, it would lead to a maximally inelastic collision, and the only way to obtain this while conserving momentum is to have all the balls swing with velocity $v/5$ together with mass $5m$.

Initially, it seemed like Answer 1 and Answer 2 were equally plausible to me, but now Answer 5 seems the most likely. How do we decide which of the these is correct? I am intending to do this experiment myself in the future, but I am curious if anyone knows how to analyze it and make an accurate prediction ahead of time. Does anyone have any thoughts?

Whatever answer is provided, I'd like to see a strong argument in favor of it.

For example, @BobD has pointed out that the glue would be deformed upon impact, thereby kinetic energy would be partially lost in the process. This is a valid argument against Answer 1, but then the next question is, if one claims Answer 2 is correct, how do we know exactly that half the kinetic energy is lost and not some other fraction? In other words, why would Answer 2 be valid instead of Answer 3, 4, or 5?

An answer that shows this experiment being done in the real world would be acceptable as well.

  • 1
    $\begingroup$ @BobD Sure, but why? What makes the collision inelastic? My question is intended to be about the detailed reasoning. $\endgroup$ Sep 3 at 18:15
  • 2
    $\begingroup$ @BobD In any other scenario, I would agree. However, whether or not the glue is present, the ball ends up with no rebound. So in a superficial sense, the glue makes no difference. Why is this reasoning wrong? This is the core of the confusion. $\endgroup$ Sep 3 at 18:17
  • 1
    $\begingroup$ OK so I assume we both agree that kinetic energy must be lost due to the inelastic deformation of the glue and that this must somehow be accounted for with regard to the motion of the other balls, with the understanding that momentum is conserved. Part of the problem is we don’t know the degree of inelasticity of the collision. $\endgroup$
    – Bob D
    Sep 3 at 19:24
  • 1
    $\begingroup$ This is a problem for you to solve but if it were me, I would assume a completely inelastic collision and consider first the simple two ball example where the moving ball sticks to the stationary ball and the two move together with a velocity that conserves momentum but where ¼ of the kinetic energy is lost. Good luck $\endgroup$
    – Bob D
    Sep 3 at 19:25
  • 1
    $\begingroup$ Interesting question. You could avoid the deformation of the glue by having some other mechanism to keep the two balls together. Maybe a mechanical hook with a timer that is built to snap into place just at the right time. There is no need to have something that deforms and dissipate energy $\endgroup$
    – Prallax
    Sep 3 at 19:50

3 Answers 3


I'm not good with experiments, so I decided to simulate it.

tl;dr: probably Answer 3. The result depends on the details of the sound wave propagation in the cradle.

I modeled each sphere as an array of identical springs with very high elastic constant to emulate the elasticity of the metal. Each spring is attached to the ones adjacent to it, unless they belong to a different sphere, as shown in the first figure.

a description of the setup

$N_b$ is the number of metal spheres and $N_s$ is the number of springs contained in a sphere. The equation of motion for the position of the center of mass of the spring $i$ is

$\frac{d^2 x_i}{dt^2} = F_i^+ - F_i^- - K_p(x_i-x_i^{\rm rest})$

For $i$ that goes from 1 to $N_b \times N_s$. Here $F_i^+$ and $F_i^-$ are the forces that act on either side of the spring.

The term $K_p(x_i-x_i^{\rm rest})$ is instead emulating the fact that the spheres are suspended in a pendulum. Each spring will have a rest position $x_i^{\rm rest}$, determined by where it is hanged, and will try to swing back there if moved away from it. Here I am exploiting the fact that for small oscillations a pendulum is close to an harmonic oscillator. For the sake of simplicity in this way I am assuming all the springs are individually attached to a pendulum, while in reality it is only maybe the central one, but this doesn't really make any difference for what we are concerned.

Now, if the rest length of the springs is $l_{\rm rest}$, the forces acting on the springs will be

$F_i^+ = F_{i+1}^- = K_s(x_{i+1}-x_i-l_{\rm rest})$

Unless, if the two strings are not attached (e.g. because they belong to different spheres) and $x_{i+1}-x_i-l_{\rm rest} \ge 0$, then $F_i^+ = F_{i+1}^- = 0$, meaning that the two spheres are not touching at the moment.

$K_s$ is the elastic constant (divided by the mass) of the springs that constitute the metal spheres. Since the metal is very rigid, it is important that $K_s \gg K_p$.

Let's look at what happens with this setup, with $N_b=5$ and $N_s=4$.

newton's cradle with 5 spheres and 4 springs per sphere

In this plot I am showing the position of center of mass of each spring $x_i$ as a function of time. Each line corresponds to one spring and they are colored base on which sphere they belong to. Initially, all the spheres are at rest except the blue one, which as been moved a bit. Once released, it swings back to its rest position and hits the others at $t \approx 1$. Then the blue one stops and most of the motion is transferred to the purple one, as expected for a Newton's cradle. Notice however that also the red one is moving a bit, and the blue one has an almost imperceptible recoil. This is not a problem with the simulation; all these effects are really present in real Newton's cradles, for instance here. I was quite satisfied when I saw this :)

If I increase the number of springs things get even more interesting, here is what happens with 20 springs per sphere. Since each spring is now representing a smaller amount of mass, I increase $K_s$ proportionally to the number of springs, so that the overall elasticity of the spheres is kept constant.

newton's cradle with 5 spheres and 20 springs per sphere

small digression

If you'll allow me a small digression, we can zoom in the region of the first collision to see what is happening there.

zoom in the collision region

After the spheres touch, there is a brief time in which all the five spheres are in contact, and then the purple one detaches.

Disclaimer: while most things in this simulation don't change much if I change the number of springs, the duration of the contact phase is not one of those. I experimented with different numbers of springs and observed that the more springs I put, the longer is the contact phase.

It's even cooler if I color-code this plot according to the spring deformation $(x_{i+1}-x_i-l_{\rm rest})/l_{\rm rest}$

zoom in the collision region color coded with spring deformation

Here you can see how much each spring is deformed. Blue means it's compressed, while red means it's extended. I think it's quite cool to see all the sound waves propagating in the metal.

end of small digression

Consistency check: what happens when we throw two spheres at the same time?

newton's cradle with two spheres moving

Two spheres come in, two spheres get out, as expected. Now, back to the question.

What happens if two spheres are glued together?

To figure this out we need a practical definition of what it means to glue two spheres. Someone in the comments argued that the glue may deform during the impact and dissipate energy that way, but for the moment I will take a very simple definition. As soon as the blue sphere and the orange sphere touch, the two springs become attached, as if the two spheres had become a single piece of metal. Just that, the simplest kind of inelastic collision.

Under this simple assumption, I ran the simulation and this is the result.

newton's cradle with glue

Only the purple sphere is kicked. This seems to favor Answer 1. Interestingly, it seems like in the following collisions also the red sphere starts to be excited, but if I let the simulation run for longer to see what happens:

same as previous, but for a longer time

The situation descends rapidly into chaos, after which all the five spheres start oscillating together, like it sometimes happens with real Newton's cradles. For comparison, here's the same plot, but without glue

same as previous, but without glue

From the tests I made, it seems to be possible to delay the insurgence of chaos by increasing $K_s$, at the cost of an increased computational cost. All these tiny vibrating springs act in a way similar to the vibrating atoms in the metal. Even though the equations I am using conserve energy perfectly, the energy tends to move from kinetic energy of the spheres to incoherent thermal vibrations of the springs (that's the second law of thermodynamics in action, yo!). My intuition is that increasing $K_s$ I am making it harder for the springs to vibrate, and thus slowing down the inevitable increase of entropy.

One last experiment before concluding this already long answer. What happens if there is a small gap between the orange sphere and the green sphere? If they are not touching from the start, but there is some space between them, after the blue sphere hits the orange and becomes glued to it, orange and blue would move together for a brief time, and hit together the green one. And we know that in Newton's cradle, if two spheres hit, then two spheres are kicked. But, interestingly, we just saw that if there is no gap, then only one sphere is kicked out. The question is: is a tiny gap sufficient to subvert this conclusion?

The result is very interesting. I tried different gaps between orange and green. The next figure is obtained with a very small gap of 1/10 of the diameter of a sphere.

gap: 1/10

The result is similar as if there were no gap: one sphere is kicked. But if we increase the gap to 1/4, everything changes.

gap: 1/4

Here two spheres are kicked! Maybe there is a minimum gap size after which the cradle behaves as if two spheres had been put in motion. But another surprise awaits: if we increase the gap even more, to 1/2 of the sphere diameter we are back again at kicking out only one sphere!

gap: 1/2

To understand what's going on, it's useful to zoom in at the impact region, and look at the spring deformation.

First, the case with 1/10 gap:

resonance 1/10

When the blue and orange spheres are glued, there's some strong oscillation going on inside. And the oscillation seems to be in phase at the time of the collision with the green sphere. The result is a strong pulse of pressure waves that traverse the other spheres and kicks the purple one.

resonance 1/4

In case of gap 1/4 instead, the oscillations are not in phase with the time of collision with the green sphere. The result is a kick to the purple sphere, that only uses about half of the elastic energy. The rest of the energy does another oscillation inside the blue and orange sphere and is then released to kick the red sphere.

So in the end it looks like the result is very sensitive to the geometry of the system, the sound speed and whether the pressure waves are in phase or not at the time of the collisions.

This result was most unexpected, and I feel I have only scratched the surface of what could be said about this topic. Thank you for asking such a nice question. I'm pasting my code here, in case anyone wants to play with it.

import numpy as np
import matplotlib as mpl
import matplotlib.pyplot as plt

from scipy.integrate import solve_ivp

# Define the parameters

N_balls = 5
N_springs_per_ball = 20
N_springs = N_springs_per_ball * N_balls
ball_diameter = 1

spring_rest_length = ball_diameter / N_springs_per_ball

# k_spring is the elastic coefficient divided by the mass of the spring
k_spring = 100000*N_springs_per_ball

# I am treating the swing of the pendulum as if it were another harmonic oscillator
# since the oscillations are small. k_pendulum is the elastic coefficient of the
# harmonic oscillator divided by the mass
k_pendulum = np.pi

# Rest positions of the springs
X_rest = np.linspace(0, (N_springs-1)*spring_rest_length, N_springs)

# Insert a gap between orange and green ball
gap_size = 1/10*ball_diameter
for i in range(N_springs_per_ball*2, N_springs):
    X_rest[i] += gap_size

# Boolean array that defines whether two adjacent springs can detatch or are
# glued together.
attached = np.ones(N_springs-1).astype(bool)
for i in range(1,N_balls):
    attached[i*N_springs_per_ball-1] = False
# When True, glue the first and second sphere together after the first collision
do_glue = True

# Set initial conditions

X0 = X_rest.copy()
X0[:N_springs_per_ball] -= 2*ball_diameter
V0 = np.zeros(X0.shape)
U0 = np.concatenate([X0, V0])

# Set time of the simulation
t_eval = np.linspace(0,1.3,1300)
#t_eval = np.linspace(0,10,1000)
t_span = (t_eval[0], t_eval[-1])

# Define the time of the first contact between the first and the second ball
# After this time, the two balls will be glued together.
# I will determine this time precisely during the integration.
t_contact = t_span[1]

# f is a function that gives the derivatives of position and velocities
def f(t,U):
    global t_contact
    attached_a = attached.copy()
    X = U[:N_springs]
    V = U[N_springs:]
    F = k_spring * (np.diff(X) - spring_rest_length)
    if do_glue == True:
        # Determine the time of first contact between the first and second sphere
        if F[N_springs_per_ball-1] < 0 and t<t_contact:
            t_contact = t
        # If we are after first contact, the first two balls are glued together.
        if t >= t_contact:
            attached_a[N_springs_per_ball-1] = True

    # Set to zero the force if the springs are detatched.
    F[(~attached_a) & (F>0)] = 0
    dVdt = np.diff(F, prepend=0, append=0) - k_pendulum * (X-X_rest)
    dXdt = V
    dUdt = np.concatenate([dXdt, dVdt])
    return dUdt

# Solve the system of equations
res = solve_ivp(f, t_span, U0, t_eval=t_eval)
t = res.t
U = res.y

# Plot the result
for i in range(N_springs):
    plt.plot(t, U[i,:], color = f"C{i//N_springs_per_ball}")

# Spring deformation plot
normalize = mpl.colors.Normalize(vmin=-5, vmax=5)
for i in range(800,1300):
    d = (np.diff(U[:N_springs,i])-spring_rest_length)/spring_rest_length*100
    tt = np.ones(d.shape)*t[i]
    plt.scatter(tt,U[:N_springs-1, i], c=d, norm=normalize, s=1, cmap = "coolwarm")
cb = plt.colorbar()
cb.set_label("spring deformation (%)")


In conclusion, I would say Answer 3. The number of balls launched and their velocity depends on the details of the pressure waves oscillations in the spheres. And therefore it depends on the geometry, position, timing, soundspeed.. etc. I would expect it also to depend on the glueing method: in this simulaion I assumed that the two glued spheres become a single piece of metal, i.e. the elastic constat of the spring joining them is equal to the ones of the other springs. But in reality the glue/velcrum/latch could have a different elasticity and this would affect the propagation of sound waves in the spheres.

  • $\begingroup$ My answer got deleted, but it is necessarily answer 1 and then answer 4, precisely as I had said it has to be. Your simulation is very nice. $\endgroup$ Sep 5 at 0:56
  • $\begingroup$ @naturallyInconsistent I added a conclusion to explain that I think it's answer 3 $\endgroup$
    – Prallax
    Sep 5 at 7:09
  • $\begingroup$ I have to disagree with that, based upon basic kinematics, momentum and energy conservation. The properties of gluing is easy to conceptually understand to be a rather tiny alteration to the energy conservation, and so, again, it is necessarily answer 1 first, before going for answer 4. $\endgroup$ Sep 5 at 8:29
  • $\begingroup$ Beautiful answer. However I too think your simulations point to "answer 1", not 3. Answer 1 was "One ball comes out with the same velocity.". It's about the first time something came out on the other side, and that indeed was one ball. I agree it can be different in later osculations because the left side no longer has a single ball to jump out. $\endgroup$ Sep 5 at 10:12

Newton's cradle is probably the simplest, yet most overlooked idea in classical mechanics.

If two balls are released with velocity v, it results in two balls coming out with the same velocity on the right. Right? But Why?

Why not just a single ball came out with velocity 2v?

Things become much easier to imagine if you consider that there are small gaps between the balls. I know this has been repeated over a thousand times already, but an image can speak a thousand words.

enter image description here

Image credits: Wikipedia

Without glue case: The first ball moves with velocity v and hits second ball, transfers momentum completely and comes to rest. The second ball now moves with velocity v and hits third ball, transfers momentum and comes to rest. The process goes on till last ball, which moves outward with velocity v with no one to hit next.

If you have two balls initially moving with velocity v, you can imagine a cascade of momentum transfers, where the momentum travels one ball at a time towards the right, until the two balls at the right end have velocity v.

With glue case:

The first ball moves with velocity v, and hits the second ball. Both balls now move together with velocity v/2. These two balls are now like a single object of mass 2m. This object hits the third ball, and imparts it a velocity of 2v/6. But here, the object doesn't come to rest as in before cases. It moves back with v/6 speed. This causes more complications, as after the first instant, this object will swing back and cause further collisions. So to keep things simple, we can predict that immediate result of the glued case is this:

The last ball moves out with velocity 2v/6.

This third ball hits the fourth ball and comes to rest, and so on.

The exact dynamics will depend on the ratio of propagation time of the momentum from left to right and the time it takes for the object to oscillate back and cause further collisions.


The key is to come up with a precise model of the gluing process. One way to do this is to imagine the glue is actually an instantaneous "latch", which can absorb an arbitrary amount of energy in the process of latching the first two balls together. (I see that @Prallax suggested this in the comments.)

If this were the case, the way to solve this would be to imagine that there were only 2 balls in the chain, and compute how much energy would be absorbed by the latch. Evidently, in this case, in order to conserve momentum, after the first ball came in with velocity $v$, the two latched balls would then keep moving together at velocity $v/2$. From this you could compute the amount of energy lost to the latch.

Then, add the last three balls back in. Now you know how much energy is lost to the latch, and of course conservation of momentum must also hold. You can put these two facts together to calculate exactly what must happen. The result is your Answer 2; losing the very same amount of energy leads to the result where the right-most two balls fly out at velocity $v/2$.

Change the latch properties, so that it absorbs a different amount of energy, and you'll change the answer. I suppose that a slightly-less-than-instantaneous latching process would be affected by the presence of the other three balls, at which point the above analysis would fail. But to answer this question, you'd really need a detailed model of the latch itself, I imagine -- basically your Answer 3.


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