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Let's consider that we have a Newton's cradle in vacuum:

enter image description here

Considering that each ball has a mass of 100g or 0.1 kg we release the ball and at the time of contact, the ball has a final velocity of 1 m/s. So the momentum will be:

$$p = mv = 1*0.1 = 0.1 kg m/s$$

If the momentum is conserved, the ball at the other side should also come out with a speed of 1 m/s. Then it will come back with an equal magnitude of momentum and the first ball should again move back at 1 m/s. Note that there is another effect at play here, which is the 'pendulum effect' which reverses the direction of the momentum but perfectly conserves its magnitude. And this process should keep going. But since the collision is inelastic, the kinetic energy will not be conserved and even in a vacuum some energy will be lost in the form of heat, but not as sound because it is in a vacuum. But according to the law of conservation of momentum, the momentum should still be conserved even if the kinetic energy is not. But the problem here is that since some kinetic energy is lost, the speed of the balls should gradually decrease. At one point the balls should stop moving, and so they will have 0 velocity. And if that happens, the momentum will become 0, even though we started out with 0.1 kg m/s. Doesn't this seem to violate the law of conservation of momentum? I can't seem to make sense of it

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Dec 22, 2020 at 18:29

4 Answers 4

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I don't think this is a bad question at all - I don't understand the down votes. Issues like this used to confuse me too, and I don't feel that any of the comments or answers so far have really got to the bottom of the issue.

Let's think about your setup a bit. You have a Newton's cradle toy in a vacuum, so it's isolated from air resistance. But you still have gravity acting on it, which means it must be resting on a surface. You didn't specify whether that surface is frictionless or not, so I'll cover both cases.

Case 1: the cradle is resting on a frictionless surface

In this case, you have a ball of mass $0.1\,\mathrm{kg}$ moving at $1\,\mathrm{ms}^{-1}$, which hits an object of mass $0.4\,\mathrm{kg}$ moving at $0\,\mathrm{ms}^{-1}$, namely the remaining four balls. A small amount of energy is converted into heat during the collision, which means that the 5th ball comes out of the collision at a speed of $(1-\varepsilon)\,\mathrm{ms}^{-1}$, where $\varepsilon$ is some small number. In order for momentum to be conserved, that means the remaining four balls will not be completely stationary, but instead will be moving at a velocity of about $\varepsilon/4 \,\mathrm{ms}^{-1}$ in the same direction as the original ball. (Really we should worry about the mass of the frame and the dynamics of the strings and so on too, but I'll ignore all that.)

This will happen repeatedly until the balls come to rest relative to each other, at which point the combined system of 5 balls will have the momentum of the original ball at the moment of the first collision, meaning that when it comes to rest the whole system will be sliding at a rate of $1/5 \,\mathrm{ms}^{-1}$ along the frictionless surface.

Case 2: the surface has friction.

Now let's assume the friction is high enough that the cradle doesn't move relative to the surface it's sitting on. In this case, the ball is colliding with an object that consists of the four balls, the frame, the surface it's attached to, and the planet that's attached to that. Once the balls have come to rest, the entire Earth will have an little bit of extra momentum, but since its mass is so high we don't usually bother to account for that. So generally speaking, when we have an inelastic collision with a stationary object, we just treat momentum as not being conserved by that collision.

The actual dynamics are more complicated of course. Momentum first gets transferred from the moving to the other four balls in the same way as described above, then it gets transferred to the frame, then to the local area of the Earth's crust, where it will reverberate as seismic waves for a while before eventually becoming spread out evenly over the whole planet.

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  • $\begingroup$ finally a good answer. This makes a lot of sense. I am reflecting on it $\endgroup$
    – Neelim
    Dec 22, 2020 at 3:06
  • $\begingroup$ I had one question, in the case of a frictional surface, wouldn’t the initial force to bring the ball back make cradle move in the direction of the force and make the earth move in the opposite direction? Then it would cancel out the acceleration at the end when the balls stop and ultimately the cradle would have 0 velocity? Then the final momentum inside the system of the cradle would be 0, and the change in the earth’s momentum would also be 0. But in this case the total momentum at the start, considering the earth, would also be 0. So the momentum would be conserved, and it would check out $\endgroup$
    – Neelim
    Dec 22, 2020 at 10:36
  • $\begingroup$ @Neelim I'm not sure if I understand you correctly, but I think that's right. If we consider "time 0" to be the moment when you first let go of the ball, just as it starts to swing downwards, then the final change in momentum will indeed be 0. As it swings down the ball steals some momentum from the cradle (and hence the Earth), but by the time everything has stopped moving that momentum has all been exactly returned, and the total change is zero. $\endgroup$
    – N. Virgo
    Dec 22, 2020 at 10:39
  • $\begingroup$ Actually the ‘time 0’ in this case would have to be the moment I pull the ball at first, before letting it go. $\endgroup$
    – Neelim
    Dec 22, 2020 at 11:01
  • $\begingroup$ @Neelim that works too. In that case, when you're pulling on the ball, the ball and your arm are moving in one direction while the Earth moves very very slightly in the other direction, keeping the total momentum at zero. Then when you let go things proceed as in my previous comment $\endgroup$
    – N. Virgo
    Dec 22, 2020 at 11:03
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I think that to understand your misunderstanding we have to consider the conditions for conservation of momentum: momentum is conserved so long as there is no net external force acting on the system in question. Considering the simple case in a vacuum, even friction between balls or between the strings and the post is acting. There may even be other forces that aren't listed here that you could think of. But in essence, this is why momentum is not conserved in this case: because it is not ideal. Hence, energy is being dissipated away as heat or sound (if we insert a medium), causing the balls to lose energy and therefore velocity/linear momentum.

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  • $\begingroup$ For simplicity let's consider that the only energy lost is in the form of heat during collisions. In this case some of the kinetic energy will be transferred to heat. But momentum is still supposed to be conserved according to the law $\endgroup$
    – Neelim
    Dec 22, 2020 at 2:08
  • $\begingroup$ We can talk about energy conservation too if you want. When you pull back one ball, there is a set amount of energy in the system. Since the system loses energy as heat, the balls slow down, which is expected. But where did the heat come from, there has to be some interaction between the balls and an external system (maybe friction as mentioned above), which implies that the system momentum cannot remain the same. $\endgroup$ Dec 22, 2020 at 2:13
  • $\begingroup$ Here I am assuming there is no friction. And there is no energy lost to an external system either, except maybe in the form of infrared as it can travel through vacuum. Otherwise it is the balls themselves that will heat up $\endgroup$
    – Neelim
    Dec 22, 2020 at 2:18
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    $\begingroup$ Yes, but if we do this experiment and see the balls slow down, we are witnessing a nonideal case. Since momentum is obviously not conserved, there has to be some external force. That is the explanation for why the balls eventually come to rest. If we could make a perfect system such that there is no friction, sound is not made, etc. then the balls would in fact keep swinging, but that is not the case with what you're describing. Some external force is acting (even if we cannot see it with our own eyes directly), and there are hundreds of years of experimental evidence backing this up. $\endgroup$ Dec 22, 2020 at 2:26
  • $\begingroup$ the only external force here is gravity, but it is applying an equal force in both directions over time so the net force over time is 0 $\endgroup$
    – Neelim
    Dec 22, 2020 at 2:33
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Let's simplify a bit more. Suppose there are only two balls, and they are made of wet clay. One balls swings toward the other and sticks - totally inelastic.

You have probably already done the math for collisions like this. Momentum is conserved, kinetic energy is not. Both balls have the same velocity afterward. The outcome is the velocity is $0.5$ m/s. The momentum is the same as before. Half the kinetic energy has been converted to heat and half is left as kinetic energy. The final state is both balls swing back and forth together.

For the $5$ steel balls that gradually lose kinetic energy, the final state will be something like this too. All $5$ balls swinging back and forth together.

You could show this experimentally by sticking two balls in the middle together with a very small piece of gum. Normally the balls are like extremely stiff springs. They deform slightly and push on the next ball. It is such a small and quick deformation that you can't see it. But gum would deform and convert some of that energy to heat.

The situation is different if you consider air friction. Now the forces are not all between balls. Air is outside the system. The balls push air around and slow down. Air speeds up and carry away energy and momentum. We don't count the momentum outside the system. We see the momentum and kinetic energy of the system decrease. It is not conserved in the system because the system is not isolated. Eventually all the balls would stop.

Of course, you can always choose a bigger system that does count the air. You might have to work at it, but you put the whole thing in a rocket in space where there is no air outside. In that case, you would have an isolated system again. If you added up the momentum of the balls and air and other rocket parts, you would find momentum is conserved. The final state of this system is the balls are stopped, and the air stopped blowing around. All the kinetic energy is converted to heat.

Momentum is conserved in this rocket. As the balls swing back and forth, the whole rocket would move a little bit in the opposite direction. The total momentum doesn't change.

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  • $\begingroup$ The system is isolated here though, it is in a vacuum. The only external force acting here is gravity, which because of the pendulum effect will reverse the direction of the momentum, but the magnitude should still be perfectly conserved $\endgroup$
    – Neelim
    Dec 22, 2020 at 2:38
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    $\begingroup$ The magnitude of momentum does not satisfy a conservation rule. The momentum vector does. Ignoring this does not help your understanding. $\endgroup$
    – nasu
    Dec 22, 2020 at 4:00
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You seem to dismiss the fact that conservation of momentum means conservation of momentum vector. If you consider the initial state right before the first collision, on the left hand side, the initial momentum is 0.1 kg m/s in the direction horizonatl to the right. Once the last ball on the right is moving up the momentum is already changed. It will change both in magnitude, decreasing to zero and in direction, gaining a vertical component. Then there is a point where its momentum is zero and then the motion is reversed. So you have a continous change in momentum way before you see the effects of friction or other kind of dissipative effects. Why the momentum is not conserved? The condition for conservation is to have no external forces and this is not the case here. You can only expect conservation if you take the states just before collision and just after collision between two balls. Or even before the first collision and right before the last ball starts to rise. But after that the tension in the string and gravity act to change the momentum.

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  • $\begingroup$ but you are also disregarding the pendulum effect, which would cause the magnitude of the momentum to be conserved perfectly even if the directions switches. Ultimately the force of gravity cancels out so there is a net horizontal force of 0 over time in the system $\endgroup$
    – Neelim
    Dec 22, 2020 at 2:29
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    $\begingroup$ The momentum is not conserved if it's direction switches. Get used to it. Momentum is a vector. And do you mean by "pendulum effect"? Whatever it is, it does not change the fact that conservation of momentum means conservation of a vector quantity. $\endgroup$
    – nasu
    Dec 22, 2020 at 4:04

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