# Is momentum conservation more fundamental than newton's third law?

Let spring be attached with a block moving with $$v_1$$; when another block with velocity $$v_2(v_2 \gt v_1)$$ hits the spring from the rear end, the spring gets compressed until both the blocks have the same velocity; after that velocity will again change but this time the front block moves with greater velocity than the rear block.

Since there is no external force acting on this bock-spring-block system, the law of conservation of momentum holds for every instant. But, there is no application of Newton's third law: when the spring gets compressed, it provides restoring force & decelerates the rear block while accelerating the front block. The momentum of the system is conserved at every moment. Restoring force is not reactive force; that is when a mass attached with a spring accelerates, it imparts certain force, says $$F$$. This prompts the internal stress of the spring to decelerate the mass; only after a certain instant, the restoring force does balance the force on the mass; so before that time, the restoring force was not equal to the force the mass was provided on the spring; so it is not a reaction-force. So, how could I apply Newton's Third law of motion, here, then?? Or is it applicable here? I don't think so but whatever it is, momentum is always conserved.

Let a spring is attached with a mass $$m$$ on one end of it in a vertical configuration; under gravity, the spring would be stretched due to $$mg$$ till the restoring force balances $$mg$$. Before that instant, restoring force was not equal to $$mg$$; but for Newton's third law to be applicable, action must be always equal & opposite to reaction; here only after a certain time, did the restoring force become equal to $$mg$$. Hence, restoring force is not a reactive force. I then conclude that the third law is not applicable here at all although momentum of the system is conserved.

So, Newton's third law of motion does not apply to spring-body-motion, isn't it? But is it so? Or am I missing something about the third law in the spring examples above?

• To shorten Timaeus correct answer: your assumption that a massless spring can delay a force is simply wrong. You are postulating a system that is instantaneous while at the same time expecting it to have a finite speed of sound. If you want to treat this correctly with finite speed of sound, then you need to use continuum mechanics to model the spring and then everything will come out just fine. Sep 17 '15 at 19:43
• @CuriousOne: Hi, sir; after a long time!! .... Newton's third law always speaks of instantaneous effect; if I push you, you would would push me in the opposite direction, instantaneously; isn't it? This is Newton's third law & for this very instantaneous effect, it fails in many places where the propagation time of the force cannot be ignored in comparison with the time scale of motion as nothing propagates greater than the speed of light. Here the restoring force is delayed & not instantaneous; so how could it follow the third law?
– user36790
Sep 18 '15 at 3:54
• Yep, that's a legalistic reading that completely fails to understand what the law is trying to teach. One could, of course, argue that instantaneous effects don't exist, which doesn't invalidate Newtonian mechanics, it forces us simply to use the much more correct (and harder to apply) continuums version of it. In that case the problem doesn't even occur because all interactions are over infinitesimal distances, only. Sep 18 '15 at 3:58
• @CuriousOne: How should I apply continuum mechanics in this case? Could you please explain?
– user36790
Sep 18 '15 at 4:29
• Preferably by reading a book about it first? :-) Sep 18 '15 at 4:31

Newton's third law applies. However the way it applies has been complicated by the fact that you described the spring as massless.

If you had a spring with mass then when it exerts a force on something that something can exert a reaction force back on it.

Since you gave the spring so little mass the force of the object on the spring just makes the spring accelerate a lot, basically enough to always keep it in contact with the object no matter how fast the object moves, as the spring compresses. And slowing the spring down to stay in contact with the object as it expands.

You can approximate the reality a step further by connecting the two ends of the spring and saying that it pulls on each connection when it is overextended and pushes on both connections when it is overly compressed. In reality each end is exerting forces on the part of the spring next to it and each part of the spring is exerting forces on each part next to it and so on. And each part has mass. So really if you hit one end the other end wouldn't know about it for a while, but unless you want to track all that the force on the spring can just be a thing that is balanced by the forces on the two ends to give a net acceleration to the spring that in many cases is just zero and so you can ignore it.

Let's look at the weight of mass $M$ with the vertical spring of mass $m$. So like all problems you can solve it with a free body diagram. First no weight. Gravity pulls the spring down (it has a non-zero mass) so the spring stretches until it is finally overextended enough that it pulls the ceiling with a force of $mg$ which leads to a reaction force of the the ceiling pulling the spring up with a force $mg$.

So the spring feels a force down of $mg$ from gravity and it exerts a force down on the ceiling of $mg$ which means it feels a reaction force up due to the ceiling. So there are two forces on the spring, by gravity downwards and by ceiling upwards. The action-reaction forces are that the spring pulls the earth up with a force of $mg$ (which is the action-reaction pair for gravity). And the spring pulls the ceiling down (which is the action-reaction pair of the contact connection between the spring and the ceiling).

In many classes this is considered the natural length, but note it is longer than if you laid the spring horizontally on a table. Now you attach the mass $M$ to the bottom. The spring is like a thousand less massive and less long springs all attached together end to end. Now originally for the stretched spring each layer of spring had a force of gravity due to that layer's mass and a force up from the stretched spring above it and a force down from the stretched spring below it. A stretched spring pulls both ends towards it as if wants to be not so stretched. So they are all balanced. Now you attach the mass $M$ to the bottom and for a moment there is more force acting on it so it accelerates downwards. But then it only had two choices, get longer or pull the spring above it downwards, or in reality do both. But that whole layer of spring really moved downwards because for a little bit there really was more force on the bottom layer. But as each layer of spring stretches they each pull their two ends harder. Eventually each layer is so stretched the bottom layer can pull the layer above it with force equal to $Mg$ and $g$ times its own mass $dm$ so pulls with a force $g(M+dm)$ and the one above can pull the one even above it with a force of $g(M+2dm)$ and the $N$th layer can pull the layer above it with a force of $g(M+Ndm).$ Thus the $Nth$ is pulled up with a force of $g(M+Ndm)$ and down by the spring below it with a force of $g(M+(N-1)dm)$ and by gravity with a force of $gdm.$

Do a free body diagram for each object. For each object I'll tell the force (and I'll mention the reaction force in parenthesis but since that is always on a different object it doesn't affect the force on this object).

So now the mass $M$ feels gravity pull down with force $Mg$ (reaction force is the earth pulled with force $Mg$) and it feels a force up from the spring of $Mg$ ( the reaction force is $Mg$ down that the spring feels from the mass $M$).

So now the spring of mass $m$ feels gravity pull down with force $mg$ (reaction force is the earth pulled with force $mg$) and it feels a force down from the spring of $Mg$ (the reaction force is $Mg$ up that the mass $M$ feels from the spring). And it also feels a force up from the ceiling of force $g(M+m)$ (the reaction force is the spring pulling the ceiling down with a force $g(M+m)$).

The ceiling feels a force down equal to its own weight $W$ (earth feels an equal force up) and feels a force down of $g(M+m)$ from the spring (spring feels an equal force up) and feels a pressure from the earth to give a force up of $W+g(M+m)$. (Earth feels a pressure pushing it down an equal amount).

The earth feels a force up of $mg$ from the spring, up of $Mg$ from the mass, up of $W$ from the gravitational weight of the ceiling, and a force down of $W+g(m+M)$ from the pressure of the (domed) ceiling.

It balances out. Always. And you can set $m=0$ at the end if you want.

The question is not whether there is another force equal & opposite to a given force which you've listed out nicely. Newton's third law says that if $A$ imparts force on $B$, $B$ instantaneously imparts reactive force on $A$.

When you read the first law scientifically it is a supplement to the second law because the second law actually fails to always tell us that an object stays at rest when there is no force and no velocity rather than merely have instanteously no acceleration. Staying at rest and have zero instantaneous acceleration are different, and you need the former to get unique solutions from $F=ma.$ And the second law says that when you know the Force Law you can get a total force and it says to restrict your models to ones where $m\vec a_i=\sum_j \vec F_{ij},$ where $\vec F_{ij}$ is the force on $i$ die to $j.$ So it generates a restricted set of motions when provided with force laws. Thus if your observations differ from the predicted once you can reject the force law. The third law says that tour force law must have $\vec F_{ij}=-\vec F_{ji}$ or else the force law should be rejected.

When you read them as science it is all about rejecting wrong laws. So you hypothesize a Force Law, then you check the third law and if it does not pass you reject the Force Law without even needing to compare with observation (that law is not Newtonian, so it isn't part of the theory so don't use it to make predictions within Newtonian Mechanics). Then if the Law is good so far use the second law to make some models and see how they compare to observations if you see observations that can't be fit by some model, reject the Force Law. If it does pass, then go on to use the first law to restrict the models even more, for instance the first law restricts the allowed motions enough that you can then use statistical methods and thus allow for experimental errors.

I know this isn't how most classes cover it. But these versions still have relativistic equivalents even when Relativity changed absolutely everything.

So let's get to the third law. How does that get used. For gravity you have state that the gravitational force on A due to B at an instant t is equal to $GMm/r^2$ pointing from A's instantaneous location to B's instantaneous location where $m$ and $M$ are the masses and $r$ is the instantaneous distance between them and $G$ is an adjustable parameter that is the same for all objects, and it says that the force does act on all objects. Since it acts on all objects and because the magnitude is the same for the force on A and on B and the direction is equal on opposite for the force on A due to B versus the force on B due to A, the Force Law passed the third law, and is not rejected until maybe the second or first law get a crack at it, which will then require comparison with observation.

Where is this instantaneous effect??

Every single Newtonian force on A due to B has another force (given by that same Force Law) that acts on B due to A, which is equal and opposite. The forces come (from the Force Law) in pairs.

You hang a mass $M$ on the spring, it exerts $Mg$ on the spring-take it as action.

Your Force Law either already passed the third Law or it didn't. If it didn't stop and make another. If it did, the only point of labeling a force as "an action" is to remind you that later when you are working on forces on other objects that there better be an equal and opposite force later. And they just come in pairs.

Let's not add the mass, since that might make it even more complicated if we have to discuss the forces that hang the mass slowly or the friction forces that slow it down. Let's instead have a vacuum and have the mass oscillating on the spring. Let's say the spring is 10 cm long and sometimes 12cm long and take 1 second to go between the two lengths. That top part isn't moving at all so the total force on the top layer of the spring is zero. But the bottom layer of the spring is moving two cm down for a second, then moving two cm up for a second, then moving two cm down for a second, and so on. So it is accelerating which means there is an imbalance of forces acting on it. The middle layer of the spring is moving one cm down for a second, then moving one cm up for a second, then moving one cm down for a second, and so on. So it too is accelerating which means there is an imbalance of forces acting on it too. But it has the same mass so it is feeling one half the total force imbalance the layer at the bottom feels.

It turns out there is a moment when the spring is totally at that magic length where the forces would balance. Each layer feeling a total force from gravity and from the layer below and from the layer above that balances out and if you had attached the mass when it was this length it would hang there motionless. But in this example none of the layers of the spring except the top are at rest. And they are all moving at different speeds, and the speed is proportional to how far it is from top to bottom. This holds at all times. But when the layer below you moves down faster than you do and the layer above you moves down less fast than you do then each layer gets stretched. The stretching determines the forces each layer exerts on the layers next to it. They come in action-reaction pairs. They do not balance on a part when the spring has its parts accelerating. If you want to ignore the mass of the spring you can do that, and the result is much simpler when you do that.

The spring would provide restoring force & after a considerable stretch, the restoring force becomes equal to $Mg$.

At each moment, the force is determined by the Force Law. When the spring is finally stretched long enough, the force the bottom layer pulls the mass $M$ up will be $Mg.$ And the third law just says that the same Force Law will say there is a force on the spring due to the mass equal to $Mg$ and going down. It says that if your Force Law doesn't you shouldn't have used it and that if it does you should remember that force when discussing the forces in the spring.

And here's the thing. If you are ignoring the mass of the spring then you aren't going to bother computing the total force on the spring so you aren't going to care about the force on the spring due to the mass $M.$ So this whole question was answered in my first paragraph.

But before that, was restoring force equal to the action?? No!

I have no idea what you are trying to say, forces come from force laws. Contact force laws, friction force laws, gravitational force laws, Hooke's Law, etc.

So, how could it be reaction to the action??

The forces are due to the force laws. They don't always balance on an object when the object accelerates. They do always come in pairs between two objects. It isn't more complicated than that.

for Newton's third law to be applicable, action must be always equal & opposite to reaction;

Newton's third law is about pairs of forces. Forces on A due to B and Forces on B due to A. Sometimes you can use it to infer what one of those forces needs to be, but they really did just already exist. And when you aren't in statics the total force on an object A due to everything doesn't have to balance. For instance there is an instant where that spring has each part have zero net force on it, but if the parts have different velocities then it will stretch and change that and the forces will change as it stretches. I think you have a misconception about what is going on. When you ignore the mass of the spring, the forces on the spring are pointless and of no use. So you aren't going to use the third law.

By instantaneous, I meant at the same time. When the spring starts providing restoring force, at the beginning it is not equal to $Mg$, isn't it?

If you attach a mass to a spring the spring will lengthen based on the mass you add to it. When you hold it vertical it will lengthen compared to when horizontal based on its own mass. The lengthening is caused by imbalanced forces on each part.

Really and truly. Draw a free body diagram for each object. Sum the forces on the object and equate to ma for that object. Repeat for each object. Then check to make sure that any forces two objects in your system exert on each other come in pairs of equal and opposite forces that a force on A due to B is equal and opposite to the force on B due to A when both A and B are in your system. There isn't anything else to do. If the forces don't come in those pairs then either you forgot a force, your bad. Or else your force is bad. Or else one of those objects isn't part of your system and so you didn't care about forces on it, so no big deal (the force would be there if you included that object too).

When the block exerts force $Mg$ on the spring, the spring exerts an opposing force immediately but that is not equal to $Mg$ at the beginning;

This is false and since it is false I can't figure out where it comes from. Have you done simpler problems like pushing a 5kg block into a 10kg block on a frictionless level surface with 15N of force so the whole thing accelerates at 1 m/s/s? It happens because the blocks exert equal and opposite 10N forces on each other so the 5kg block feels a net 5N force and the 10kg block feels a net 10N force on it.

The mass feels a force of Mg due to gravity. It feels a force from the spring that could be anything but that is related to the acceleration it feels. But whatever force the mass feels from the spring, the spring feels an equal and opposite force from the mass. You have to figure out what that force is.

it is after the spring reaches 'the magical length', the opposing force becomes equal to $Mg$. Only at that time, $$\text{spring force}= -{Mg}$$. But before that, it was not equal to $Mg$. So, how could it form the pair where one force is equal & opposite to the other

They aren't pairs. The pairs are the force on the spring due to the mass and the force on the mass due to the spring. That is the pair, and that is what the third law is about. Which is why it is pointless if you ignore the mass of the spring, because then who cares what force the spring feels. Another pair is the force of the mass on the earth and the force of the earth on the mass. They are also equal and opposite but if you are going to ignore the acceleration of the earth (because it is so tiny) then who cares about the force of the mass on the earth, so who cares about the third law.

The third law doesn't say that forces are related if they are equal and opposite, so it isn't the case that something magically happens when two forces on the same object become equal and opposite, they are still forces on the same object so they have zero to do with the third law.

The third law only says that when a force is exerted on A due to B then there is also a force on B due to A that is equal and opposite.

...before the spring reaches the 'magical length' that is before $t$? Newton's third law speaks of simultaneity of interaction, isn't it?

Newton's third law says that a force on A due to B has a corresponding equal and opposite force on B due to A. Yes that happens at all times, just like any other law. It doesn't say that gravity acting on A cares what a spring does to A or that a spring acting on B cares what gravity does to B. That is nothing. Gravity exerts equal and opposite forces on objects, so A feels an equal and opposite force of Gravity due to B as B feels a force of Gravity due to A. The pair both had the word gravity and they had different objects in the different order of "feel" and "due to."

Contact forces exerts equal and opposite forces on objects, so A feels an equal and opposite force of contact due to B as B feels a force of contact due to A. The pair both had the word contact and they had different objects in the different order of "feel" and "due to."

Friction forces exerts equal and opposite forces on objects, so A feels an equal and opposite force of friction due to B as B feels a force of friction due to A. The pair both had the word friction and they had different objects in the different order of "feel" and "due to."

Are you understanding now my point?

No. You just say wrong things. An action-reaction pair always has the two forces in the pair act on different objects. In your case you didn't give the spring a mass so one of the forces was pointless. And then you talk about two forces acting on the same object which is unrelated to Newton's third law.

And you insist on calling them weird names instead of naming the force according to the force law from which it comes. Which is just hiding the physics. Hiding physics isn't a point, it's a bad habit.

The mass M feels a weight of Mg down and feels a Hooke's Law force up or down or zero depending on what is going on. The spring feels a weight of mg down, and spring forces up and down or possibly both or neither from the ends being attached, depending on what is going on.

So, there is only one force in the spring, the Hooke's Law force; it act both as Third-law pair with $Mg$ from the block as well as act to restore the natural length that is internal stress & so when I remove the block, the Hooke's Law force only act as the restoring force; am I right now?

No, you are wrong. The spring can exert a force on the block and the block will exert an equal and opposite force on the spring. The spring can exert a force on the ceiling and the ceiling will exert an equal and opposite force on the spring. In equilibrium, if the spring has mass the force on the ceiling from the spring will be larger in magnitude than the force on the block from the spring.

• Comments are not for extended discussion; this conversation has been moved to chat. Sep 22 '15 at 7:18
• This has to be one of the biggest answers that I've ever seen .. would be nice if was made more concise Sep 26 '20 at 16:11