What Actually Happens If You Shoot a Ball at a Newton’s Cradle?
dot physics forces video analysis collisions physics projectile motion estimation
Rhett Allain Matt Burgess Andy Greenberg Emily Mullin Will Knight Jordana Cepelewicz Amit Katwala Sandra Upson Kat Friedrich Ars Technica
In a recent video, the Slo Mo Guys fired at the clicky-clacky desk toy with a superhigh-speed ball. It looks pretty cool—and the physics are even cooler.
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Newton's cradle is a fabulous toy. In case you aren't familiar with this device, it usually consists of five hanging metal balls that all line up horizontally. If you pull back a ball on one end and let it go, it swings down and hits the other balls—and the result is that the ball farthest away from it swings out on the other side.

Then that ball swings back down, hits the rest of the balls, and the one you started with now bounces away from the group. The whole thing keeps repeating. It looks like this:

Newton's cradle shows up in many business offices as a clicky-clacky desk toy that just makes noise. But it's not just for fun—it's for physics. It lets us ponder important questions, like: What if instead of pulling a ball back and letting it swing down, you use an air cannon to shoot another ball at superhigh speed right at the first ball? And what if you record this on video at 82,000 frames per second?

Well, that is exactly what the Slow Mo Guys, Gav and Dan, try in this video:

https://youtu.be/YcBg6os2dPY

Let’s start off with some basic collision physics. There are two very important quantities to consider with any collision. The first is momentum (represented by the symbol p). This is the product of an object's mass (m) and its velocity (v) vector. Because it's a vector, we have to consider the direction in which the object is moving.

Why do we care about momentum? Well, it's the best way to describe the net force on an object. The momentum principle says that the force is proportional to the rate of change of momentum. As an equation, it looks like this:

We can use this momentum principle to look at a collision between two balls. I'll call them ball A and ball B.

While these two balls are in contact, there is a force that ball B exerts on A. But since forces are always an interaction between two objects, this means that A also pushes on B with a force of the same magnitude, but in the opposite direction. With these forces, both balls change in momentum, according to the momentum principle. They also have the same contact time (Δt).

This means that the change in momentum for ball B is exactly the opposite of the change in momentum for ball A. Or you could say that the total momentum of ball A plus ball B before the collision is the same as the total momentum after the collision. We call this “conservation of momentum.”

Conservation of momentum is actually a very powerful tool. If we know the momentum of two objects before a collision, then we know something about the momentum after the collision. Let's use a subscript notation of "1" for before the collision, and "2" for after. That gives the following:

That equation not only looks cool, but there’s something important about what's not there. We started with two equations that had forces in them, and then algebraically eliminated the forces to make one equation. That’s actually a really useful thing, since those collision forces aren't something that you can just write down as an equation. That’s because they depend on the types of materials that are interacting and how much they get deformed.

But is momentum conserved in all collisions? Technically, no—but practically, yes. If the only forces are due to the interaction between the two objects, then momentum is conserved. However, if one of the balls has a rocket engine providing it with an external force, then its change in momentum will be different than the change in momentum of the other object.

But even if there is an external force (like a gravitational force), we can sometimes ignore this extra force and pretend like momentum is still conserved. Honestly, it's not a terrible approximation, especially in the case of collisions that last over a very short time interval. During such a short time frame, the external forces don't really have much time to change the momentum, so it's almost like they aren't even there. For pretty much with any collision you see in a physics textbook, you are going to be able to say that momentum is conserved.

The second quantity to consider is the kinetic energy (KE). Like momentum, this also depends on both the mass and velocity of the object. However, there are two big differences: It's proportional to the velocity squared, and it's a scalar value (with no direction).

Since velocity is a vector and you can't technically square a vector, you have to find the magnitude of it first and then square it. We normally skip this in the equation and just use v2, but I wanted to show you the full thing.

So, here’s the obvious next question: Is kinetic energy also conserved, just like momentum is conserved? The answer is: sometimes. For some collisions that we call "elastic collisions," both kinetic energy and momentum are conserved. In general, elastic collisions happen between very bouncy objects—like two rubber balls, or pool balls colliding. If we have an elastic collision in one dimension (meaning everything takes place in a straight line), then we have two equations that we can use: conservation of momentum and conservation of kinetic energy.

In addition to elastic, there are two other kinds of collisions. When two objects collide and stick together, like a lump of clay hitting a block, then we call this a completely “inelastic” collision. In that case, momentum is still conserved and we also know that the final velocity of the two objects is the same, because they stick together.

Finally, there is the case where two objects collide but don't stick together and don't conserve kinetic energy. We just call these “collisions,” since they aren't one of the two special cases (elastic and inelastic). But remember that in all these cases, momentum is conserved as long as the collision takes place over a short time interval.

OK, now let's consider a problem that's very much a part of Newton's cradle. Suppose I have two metal balls of equal mass (m), ball A and ball B. Ball B starts off at rest, and ball A is moving toward it with some velocity. (Let's call it v1.)

Before the collision, the total momentum would be mv1 + m×0 = mv1 (since ball B starts off at rest). After the collision, the total momentum must still be mv1. This means that both balls could be moving with a velocity of 0.5v1 or some other combination—as long as the total momentum is mv1.

But there's another constraint. Since it's an elastic collision, the kinetic energy must also be conserved. You can do the math (it's not too hard), but it turns out that in order to conserve both KE and momentum, there are only two possible outcomes. The first is that ball A ends up with a velocity v1 and ball B is still stationary. This is exactly what would happen if ball A missed ball B. The other possible result is that ball A stops and then ball B has a velocity of v1. You might have seen this happen when a pool ball hits a stationary one head on. The moving ball stops, and the other ball moves.

This is basically what happens with Newton's cradle. If the collisions between balls are elastic (that's a fair approximation) and everything is lined up (so that it's one dimensional), then the only solution for a ball on one side hitting the stack is for it to stop and for another ball to move instead. That's the only way to conserve both kinetic energy and momentum. If you want all the details in that derivation, here is a video for you:

What about an inelastic collision? It's fairly easy. Since both balls have the same mass and same velocity (because they stick together), the only solution is for them both to be moving at 0.5v1 after the collision. In the case of a plain collision (that's neither elastic or inelastic), both balls will have some velocity between 0 and v1.

Just as a demonstration, here are three colliding balls. The top shows an elastic collision, the bottom is inelastic, and the middle is somewhere in between.

I think that just looks cool.

Video Analysis of the Superfast Cradle

There are a couple of things that make the collision from the Slow Mo Guys video different from the action of a normal Newton’s cradle. Instead of five balls in the setup, there is a sixth, the ball that is shot out of the air cannon. This ball is moving super fast—but it also looks slightly smaller than the other balls in the cradle, which means it has a different mass.

And as you can see in the video, instead of the ball at the end of the column simply bouncing outward, four of the five balls snap off their strings entirely and fly away as the base falls over. This won’t work as a nice clicky-clacky office toy (and it might put a hole in your wall).

Let’s find out what’s going on here. Remember, for collisions that happen over a very short time interval, momentum should be conserved. The total momentum of everything before the collision should be equal to the total momentum of everything after the collision. Let's check. I'm going to assume all the balls have the same density. That means that by measuring the diameter of both the launched and target balls, I can calculate the volume and the masses of all the balls. (For this first round of analysis, I’m going to assume each one is a standard 3/4-inch ball bearing.) Then, I can find the velocity of all the balls both before, during, and after the collision.

To do this, I am going to use Tracker video analysis. The idea is to look at the location of an object in each frame of the video. If I know the time between frames, I can use this to get both position and time data for all the balls.

But … there is a small issue. The Slow Mo Guys recorded the impact at 82,000 frames per second. Of course, if the video is played back that fast, it would just look like normal speed. So, the playback is actually at 50 frames per second, which means that the time between frames is actually 6.1 microseconds.

After a bunch of clicking through frames I can get horizontal position data for all six balls. Here's what that plot looks like:

All these lines are horizontal position (x) vs. time. Since the horizontal velocity is the change in position divided by the change in time (vx = Δx/Δt), then the slope of the line will be the velocity. With that, the launched ball has a speed of 114.69 meters per second. If you convert this speed to different units, you get 256.6 miles per hour. That's pretty close to the value in the video listed at 270 miles per hour. The difference could be from my initial calibration of the video using a 3/4-inch ball—but it's not a big deal.

Now that I have all the velocities before the collision and after, from the slopes of the other lines, I can see if momentum is actually conserved. I need the mass of the balls. Let's go with a standard 3/40inch ball bearing having a mass of 28.2 grams and assume that all the balls have the same mass. With that, the launched ball has a momentum of 3.23 kgm/s, and all the stuff after the collision has a momentum of 39.9 kgm/s.

Those two values are different—and I said that momentum should be conserved. What could be going wrong? It must be that I calculated with the assumption that all the balls have the same mass. But remember, the ball that’s shot out of the air cannon seems to be a little smaller than the others, so they should actually have different masses. So let’s try it again.

Let’s use the difference in the balls’ diameters to estimate the mass of the hanging balls. If I assume the launched ball has a diameter of 1.905 cm (that's 3/4 inch), then the cradle balls look like they are 1.77 centimeters. If they have the same density as the launched ball, then their mass would be 22.6 grams. Using this new mass, the final momentum is 3.29 kgm/s, which is much closer to the initial value of 3.23 kgm/s. I'm much happier now that physics does indeed work.

(If you want a homework assignment, you can check for the conservation of momentum in the vertical direction. It will be fun, trust me.)

But what about the kinetic energy? If it's a real Newton's cradle with perfectly elastic collisions, then the kinetic energy of the launched ball should be equal to the total kinetic energy of all the stuff moving after the impact.

A quick note: In order to calculate the kinetic energy, I need to know both the horizontal and the vertical velocity of each ball. Fortunately, I already did my homework, so I have those values. Using my two different masses of balls, I get an initial kinetic energy of 185.5 joules and a final kinetic energy of 108.9 joules. Clearly, kinetic energy is not conserved.

But we already knew that, because after the collision, the Slow Mo Guys show that the launched ball ends up with a giant dent in it. That deformation takes energy, and it means that the kinetic energy of the initial ball can't all go into the kinetic energy of the balls after the collision. It's not an elastic collision.

Now there are some other interesting questions that I need to answer, like: Why did the strings that hold up the balls on the Newton's cradle break?

In a normal situation, where the balls just swing back and forth like they are supposed to, the string pulls upwards on the final ball as it moves to the right. This upward-pulling force is perpendicular to the motion of the ball, so we can call it a “sideways” force. These sideways forces just change the direction of the ball. If the ball is moving at a normal speed (like 1 meter per second), then the force needed to turn it is rather small.

But what if the ball is moving much faster, like at 40 meters per second? In that case, the tension in the string also needs to be much larger in magnitude to get that ball to turn. However, strings have limits. They can only pull with a certain force before exceeding their breaking points. Clearly in this case, the strings aren't up to the task of getting that ball to turn—so they break.

Why does the whole Newton's cradle, including the base and supports, also move after the collision? You might think that the base would just stay put; I mean, the launched ball only hits the other balls and not the base. But let's consider a moment in time when that ball on the farthest side is moving to the right before that string breaks. Here is a force diagram of that situation:

At this instant, the ball is moving to the right, but the tension is pulling slightly up and to the left. I can break this force into two perpendicular components (labeled Tx and Ty). The Ty force is perpendicular to the motion of the ball and makes it turn. But the other component (Tx) is pulling to the left in the opposite direction to the motion of the ball.

Remember: Forces are always an interaction between two objects. So, if the string pulls to the left on the ball, then the ball pulls back on the string to the right. This is Newton's third law of motion: For every force there is an equal and opposite force. We could do the same thing with the forces on the string to show that the string pulls on the rest of the base to the right. It is this right-pulling force that gets the base to move and eventually fall over.

What about gravity—is it really OK to ignore the downward-pulling gravitational force in this case? Let's consider the time interval from the instant the launched ball touches the first ball on the cradle until the time when the balls are no longer in contact—that's the whole collision. Looking at the times from the video, this is an interval of just 61.5 milliseconds.