# Does Newton's third law apply to momentum or to forces?

I read all the previous answers concerning the 3rd law and I have seen that it is definitely not universal, (Edit: but conservation of momentum is. If it is not universal it should be not a problem to exclude also gravity.)

Newton derived his law from the collisions in a primitive 'cradle'. Why do physicists strive to make that statement universally applicable? It seems generally agreed that it is a consequence of the conservation of momentum in collisions. What is the conceptual or practical necessity to extend this principle to situations where it is not applicable? Conservation of momentum is a universal law, but that does not imply that action must be always equal to reaction, which as a matter of fact does not happen.

• 1) any body can offer a max resistance due to its mass (oterwise called 'inertia') = k (reaction), any action on it which is < k will get an adequate reaction, but any action which is > k will obviously not get a sufficient reaction. If a ball hits a wall with p > k the wall will crumble.
• 2) can an object provide a reaction if the object never gets in contact with it?

Update: The answers, do not answer my question:

They did do just that from the 17th century to the 19th century, but that is no longer the case.

I was not referring to 19th century but to today's phisicists:

Alba,you don't quite understand Newton's laws of motion.... Newton's third law is about forces. - David Hammen

If 3rd law is about forces and not momenta*, please explain how it works when a skater is pushing at the rail or at a basketball: the reaction id different: The action in both cases does 210 J of work, but the reaction is different, what is equal is only the momenta. Please show how you get equal actions and reaction considering forces.

... ..Saying the third law should imply equal forces in any arbitrary pair of scenarios seems obviously nonsensical – Hypnosifl

Can someone please tell me if this is true? According to 3rd law, If I exert a force of k N on 10 different bodies, shouldn't I expect the same reaction of k N from all the 10 different bodies? If the reaction (force) of the rail and of the ball is the same, how can the same force produce a velocity times greater in the second case?

I have not changed the above discussion as it is necessary to unterstand the old answers. But it is now obsolete.

Bounty question

Consider a man in vacuum or frictionless environment pushing (F = kN) at one object (stone, ball etc.) with one hand or at two objects with two hands in opposite direction, show:

1) that the object(s) exerts an equal reaction (F = kN) and it is a force and not a (change of) momentum

2) that the same force exterted on an object of different mass provokes the same reaction .

• You seem to be asking multiple thing here. You should refine your question to ask only a few related things. Additional questions on top of those should be posted as new questions. – Jim Dec 9 '14 at 14:32
• @Jim, No, Jim, it is only one question, if it is true (not considering of course QM etc) that to the same action (of a force) always corresponds an equal reaction (of a force). They are all aspects of the same issue, but I'll surely post another question as this has been unjustly and unfairly misunderstood and targeted from the beginning by hasty readers (a dozen downvotes) – user59485 Dec 9 '14 at 15:02
• @Jim, that has to be proved. I understand what force is very well, and also impulse. And when I objected to John Rennie, he shifted ground from force to impulse. We have already experienced here the excruciating pain of verv,very... long (and wrong) answers. I your long answer can give a sensible explanation, well.., it is more than welcome! :) – user59485 Dec 9 '14 at 15:17
• @Jim, "..That question would be likely closed as a duplicate of this one". Have you seen the question in Wolfram's comment is identical to my original question. It has not been closed as a duplicate and it has not been downvoted, and there are many others around. Is it a matter of fortune, Jim? – user59485 Dec 13 '14 at 14:44
• @UserAnonymous, your comment is very acute!, probably it was punished because I dared downvote one answer, and naively declared it. Then I was the target of many hostile comments, now deleted. But you can still find some traces of condescension somewhere! :) – user59485 Dec 14 '14 at 6:33

Why do physicists strive to make that statement universally applicable?

They did do just that from the 17th century to the 19th century, but that is no longer the case. Electromagnetism, relativity theory, and quantum theory put an end to thinking that Newton's mechanics was universal.

It's a trivial matter to deduce Newton's third law by assuming that linear and angular momentum are conserved quantities, that interactions only occur between pairs of particles, and that these interactions are instantaneous. The first assumption becomes a bit difficult on the very large scale (millions of light years). There are multibody interactions in quantum mechanics that render the second assumption false. With regard to the third assumption, even Maxwell's electrodynamics is more than a bit problematic.

That the conservation laws do appear to hold locally lends some credence to Newton's laws, where "locally" means on the scale of a cluster of galaxy groups. It's not universal, but except for cosmologists, most physicists and engineers don't work at this huge scale. That true multibody interactions only appear at the quantum scale (e.g., a proton or the helium nucleus) also gives some credence to Newtonian mechanics.

With regard to the third assumption, electrodynamics represented the start of the downfall of Newtonian mechanics as a universal theory. There is no room for a finite velocity field that mediates a force in Newtonian mechanics. Newton's third law implicitly assumes an infinite propagation speed. Linear and angular momentum can still hold as conserved quantities, but Newton's third law fails in the case of a finite velocity field that mediates some force (e.g., electromagnetism).

What about gravitation? This was explicitly mentioned in the question. It turns out that Newton's law of gravitation is an extremely good approximation in the case of relative velocities much, much smaller than the speed of light and distances much, much larger than the Schwarzschild of the most massive body. Our solar system is a good example of these conditions. Mercury comes closest to violating these conditions, and even then, the general relativistic deviation of Mercury's orbit from that predicted by Newtonian mechanics is extremely small. The deviation is dominated by a 43 arcseconds per century precession.

While observable, this is an extremely small rate. This deviation was observed decades before Einstein developed his theory of general relativity. That general relativity post-dicted this deviation almost perfectly was one of the reasons general relativity was accepted so quickly by the scientific community.

## Update based on changes to the question

While Newtonian mechanics is not universally true, physics educators still teach Newtonian mechanics because it is extremely accurate across a wide number of practical applications and because understanding Newtonian mechanics is an essential stepping stone toward understanding more advanced physics.

Alb, your updates indicate that you don't quite understand Newton's laws of motion. Asking about momentum, kinetic energy, and potential energy just adds to your confusion. Newton's third law is about forces. In particular, it addresses forces between pairs of objects. Per this law, if object A exerts a force $\vec F$ on object B then object B exerts an equal but opposite force on object A.

Newton's law of gravitation obeys Newton's third law. The Earth's gravitational force exerts a force of 1 newton on a 102 gram apple and make the apple fall towards the Earth. The third law reaction to this force is an equal but opposite gravitational force of 1 newton exerted by the apple on the Earth. What if we add a falling 102 kg boulder to the mix? I'll assume the apple and boulder are side-by-side and are a bit over 26 centimeters apart, center-to-center. Now there are three third law reaction pairs:

• The Earth pulls the apple downward with a force of one newton and the apple pulls the Earth upward with a force of one Newton.
• The Earth pulls the boulder downward with a force of 1000 newtons and the boulder pulls the Earth upward with a force of 1000 Newton.
• The boulder pulls the apple sideways towards the boulder with a force of 10-8 newtons and the apple pulls the boulder sideways towards the apple with a force of 10-8 newtons.

The net force on the apple is one newton downward plus 10-8 newtons to the side. Dividing by the mass of the apple yields a downward acceleration of 9.8 m/s2, plus a tiny, tiny sideways acceleration of 9.8×10-8 m/s2 towards the boulder.

The net force on the boulder is 1000 newtons downward plus 10-8 newtons to the side. Dividing by the mass of the boulder yields a downward acceleration of 9.8 m/s2, plus a tiny, tiny sideways acceleration of 9.8×10-11 m/s2 towards the apple.

The net force on the Earth is 1001 newtons upwards. Dividing by the mass of the boulder yields an upward acceleration of 1.7×10-22 m/s2 towards the boulder and apple. This is of course unmeasurably small.

## Second update, based on further edits

First, a meta-comment. After a point, it's better to ask a different question as opposed to editing your question over and over.

Your key confusion is that you are focusing on the wrong objects. Newton's third law addresses forces between pairs of objects. You are confusing yourself by insisting on looking at extra objects, looking at different situations, and looking at extraneous concepts such as energy.

Newton's third law is conceptually very simple. If objects A and B interact by some mechanism, the interaction results in equal but opposite forces on the two objects. Third law interactions are always paired, and the two element of the pair always act on different bodies.

It's easy to get confused. For example, consider a book at rest on the surface of the Earth. The book is pulled downward by gravitation and pushed upward by the normal force. Do these constitute a third law pair? The answer is no. For one thing, there are two interactions here, gravitation and the normal force. There is but one interaction in a third law pair. For another, there is but one object here, the book. Third law pairs always act on two different objects. The third law counterpart to the downward gravitational force exerted by the Earth on the book is the upward gravitational force exerted by the book on the Earth. The third law counterpart to the upward normal force exerted by the surface of the Earth on the book is the downward normal force exerted by the book on the surface of the Earth.

One potential source of confusion is that introductory physics problems oftentimes ignore the third law counterpart to some force. Going back to the book, while the book is indeed pulling the Earth upward gravitationally, the acceleration of the Earth is so extremely small that it can be and is ignored.

• " Per this law, if object A exerts a force F⃗ on object B then object B exerts an equal but opposite force on object A." You should explain how a stone or a basketball can ever exert a force on the thrower. – user59485 Dec 14 '14 at 6:17
• dear David, i ve learnt a lot from your answer, very happy i came across it. If i may: how can one tell that Newton s 3rd law assumes infinite propagation speed? And do you happen to know of any introductory sources for learning the modern understanding of forces, namely mediated by finite velocity fields? Thanks a lot. By the way, this post: physics.stackexchange.com/questions/315395/… is in the same context and may interest you. – user929304 Mar 4 '17 at 11:47

"Conservation of momentum is probably a universal law, but that does not imply that action must be always equal to reaction, which as a matter of fact does not happen."

Sure it does, at least in Newtonian physics where forces act instantaneously (the link in your first sentence, with the comment that the 3rd law is "definitely not universal", deals with electromagnetism, a theory consistent with relativity's requirement that no information can travel faster than light; in relativistic theories the 3rd law no longer holds, though conservation of momentum still does). Force is just the first derivative of momentum $dp/dt$, meaning the force on a body at any given moment is just the rate of change of the body's momentum at that moment. This implies that if the force $F$ on a body is constant over some time interval $\Delta t$, then that body's change in momentum must be given by $\Delta p = F \Delta t$, just by the definitions of force and momentum. If you have two systems 1 and 2 with no external forces acting on them, conservation of momentum implies that over any time interval $\Delta t$ it must be true that the changes in momentum $\Delta p_1$ and $\Delta p_2$ were equal and opposite, i.e. $\Delta p_1 = -\Delta p_2$. So if the forces were constant over the time, it must be true that $F_1 \Delta t = -F_2 \Delta t$ (where $F_1$ is the force on system 1 and $F_2$ is the force on system 2), so $F_1 = -F_2$. Even if the force is non-constant, you can divide the time interval in question into a series of infinitesimal intervals of length $dt$ such that the force can be taken as constant over each one, and the above argument implies that between any time $t$ and $t + dt$, it must be true that $F_1 (t) = -F_2 (t)$.

By the same token, the force at each moment can be integrated over some time-interval to give the total change in momentum over that interval, and if the forces on two bodies are equal and opposite at each moment then each body must experience an equal and opposite change in momentum over any given time-interval.

You give this example:

"1) Any body can offer a max resistance = k (reaction), any action on it which is < k will get an adequate reaction, but any action which is > k will obviously not get a sufficient reaction. If a ball hits a wall with p > k the wall will crumble."

What does crumbling vs. not crumbling have to do with Newton's third law? Even if the ball punches a hole through the wall, it is still slowed down, and the ball's decrease in forward momentum must have been caused by forces from bits of the wall acting on the ball for the short time interval that it was passing through the wall. At every moment that the ball was passing through the wall, the force that the ball was exerting on each little bit of the wall was matched by an equal and opposite force from that little bit back on the ball.

Edited to add a little more on this, since it's a particular point of contention: Although Alba is unwilling to change his/her mind on this point, for the benefit of other readers I just want to point out that what I said above is definitely the mainstream physics view, for example in my comments to Bert I pointed to the snippets from the solutions manual to a college physics textbook here and here (and some additional snippets which can be seen googling specific phrases) which pose the following question and answer:

A brick hits a window and breaks the glass. Since the brick breaks the glass,

(a) the force on the brick is greater than the force on the glass,

(b) the force on the brick is equal to the force on the glass,

(c) the force on the brick is smaller than the force on the glass,

(d) the force on the brick is in the same direction as the force on the glass.

Solution: According to Newton's third law, the answer is (b). Why isn't answer (c) correct since the brick breaks the glass? It takes a considerably smaller force to break a glass than to break a brick. When the brick hits the glass, they exert equal but opposite forces on each other, say 150 N. It may take only 100 N to break the glass and 1000 N to break the brick. So the glass breaks and the brick remains basically undamaged

Likewise, from the same page, there's this snippet:

There is another misconception concerning the third law. The third law states that the two forces are equal no matter what. For example, an egg and a stone collide with each other, the egg breaks and the stone is intact. Since the egg breaks, we often conclude that the force by the stone on the egg is greater than the force by the egg on the stone. This is not so. The forces are always equal. The egg breaks because it is simply easier to break.

More generally, if you go to google books perform this search for "Newton's third law" along with the word "always", you can find a large number of textbooks saying that the forces between two interacting objects are always equal and opposite, with no exceptions noted for breakage or any other classical scenario. And note that the people here who claim the forces would be unbalanced, like Alba and Bert, never cite any physics expert making this claim, nor do they offer any mathematical derivation of this claim or any experimental evidence to believe it's true--it's a basically crackpot style of argument which rejects scientific evidence/analysis and expert knowledge in favor of personal intuitions.

"2) can an object provide a reaction if the object never gets in contact with it?"

Yes, in Newtonian physics the gravitational force acts instantaneously at a distance, and the gravitational forces between a pair of objects at any given instant are always vectors of equal magnitude but opposite direction. In relativistic theories like classical or quantum electromagnetism, forces no longer act instantaneously but are limited by the speed of light, so the forces on a pair of objects at a given instant need not be equal and opposite--Newton's laws don't really apply in this form, although apparently there is a modified local form of the third law dealing with forces within the field itself, see p. 384 of Relativity Made Relatively Easy on google books here.

On the scenario with the girl pushing the rail and the ball: (Note: I'm leaving this explanation in for now even though you removed the detailed statement of the scenario seen in this version of your question, since you still say "explain how it works when a skater is pushing at the rail or at a basketball" in your question, but you didn't answer my question in comments about whether you still want a detailed analysis or something more conceptual, if you aren't interested in the gory details I can edit.) As I mentioned in comments, "every action has an equal and opposite reaction" is simply a colloquial way of describing the fact that a force from system A onto system B must be matched, at the same instant, by an equal and opposite force from B onto A. It doesn't mean anything more than that, so if you think you see any other implications from the phrase which can't be directly derived from the assumption of equal and opposite forces at each moment, you need to just accept that physicists don't always use words in a common-sense way. In particular, a misunderstanding to ask which force is the "action" and which is the "reaction"--there is no technical distinction between the two, again it's a purely colloquial way of describing the fact that the forces are equal and opposite at any given instant. It's also incorrect to interpret the "action" as an amount of work done rather than a force (work is force*distance), as you did in the first scenario with the girl where you said the "action" was the 210 J of work done--the third law only says that as long as the girl is pushing the ball or rail, the force from her hand on the object she's pushing is matched by an equal and opposite force from the object on her hand (which would move her backwards if she's standing on a frictionless surface, as seems to be assumed in the diagrams where she acquires a backwards momentum equal and opposite to the forward momentum acquired by the object). In my analysis below, I will assume that the forces between the girl and the object being pushed are equal and opposite in this way.

I'd be curious as to what source that diagram actually came from, though, because I think they made a slight error in their numbers for the case of the girl pushing the rail. I would presume they tried to keep things simple by assuming the girl exerted a constant force F on the rail before letting go after some time t. And if they want the total work to be 210 J, then the sum of work done on the rail and work done on the girl should be 210 J. And work is force * distance, so to find the work done on each, we just need to figure out the distance each moved while the force was being exerted. For an object with mass m that starts at rest, if it experiences a constant force F its displacement as a function of time is given by $d(t) = (1/2)(F/m)t^2$, so the work as a function of time is given by W(t)=F*d(t) or $W(t)=(1/2)(F^2/m)t^2$. Since the mass of the girl is 20 kg and the mass of the rail is 1000 kg, and the time and the magnitude of the force should be equal for each, then if we want the sum of work done on each to be 210 J, this gives us the equation $(1/2)(F^2/20)t^2 + (1/2)(F^2/1000)t^2 = 210$. This simplifies to $(51/2000)F^2 t^2 = 210$, and dividing both sides by (51/2000) and then taking the square root of both sides gives $F* t = 90.7485$. Force * time is equal to the impulse, which is just the change in momentum; so the change in momentum should be 90.7485, but they gave the change in momentum as 91.6 instead, I think they made a mistake in their math (or maybe it was just a roundoff error). So, from here on in the analysis I'll use my corrected value for the impulse to the rail rather than the one they give.

Now, if the girl exerted the same constant force F on both objects for two different times $t_1$ and $t_2$, then it would be true that $F * t_1 = 20$ and $F * t_2 = 90.7485$. The first equation can be rearranged as $F = 20/t_1$, and substituting that value for F into the second equation gives $20 (t_2/t_1) = 90.7485$, divide both sides by 20 to get $(t_2/t_1) = 4.5374$. So with the assumption she used the same constant force, she must have pushed the rail for 4.5374 times as long as she pushed the ball--if we imagine F=20 Newtons, then she could have pushed the ball for 1 second and the rail for 4.5374 seconds. Again, Newton's third law just says that during the time period she was pushing each object, the object was pushing backwards on her with an equal and opposite force of 20 Newtons in the other direction.

Are these numbers consistent with the idea that the total work done on both the girl and the object being pushed was 210 J in both cases? Again remembering that work is given by $W(t)=(1/2)(F^2/m)t^2$, let's look at each case:

For the work done on the ball, we have F=20, m=1, and t=1. So the work would be (1/2)(400/1)(1) = 200.

For the work done on the girl by the ball, we have F=20, m=20, and t=1. So the work would be (1/2)(400/20)(1) = 10.

For the work done on the rail, we have F=20, m=1000, and t=4.5374. So the work would be (1/2)(400/1000)(20.588) = 4.12

For the work done on the girl by the rail, we have F=20, m=20, and t=4.5374. So the work would be (1/2)(400/20)(20.588) = 205.88

So, you can see that in each case, the sum of the work done on the girl and the object does indeed work out to 210 J. And you can also see that this analysis was based on assuming that while the girl exerts a 20 N force on the object, the object is exerting a 20 N force back on her, in accord with Newton's third law.

Response to more new edits:

"Can someone please tell me if this is true? According to 3rd law, If I exert a force of k N on 10 different bodies, shouldn't I expect the same reaction of k N from all the 10 different bodies?"

Yes, you should. However, the statement of mine you are asking about, "Saying the third law should imply equal forces in any arbitrary pair of scenarios seems obviously nonsensical", was referring to any arbitrary pair of scenarios, if you restrict the comparison to scenarios where you apply the same force to both objects then in that case the force of the objects on you would be the same. But in the problem with the girl pushing the ball and the rail it wasn't specifically stated that the force she exerted on each was the same--in my analysis, I did assume for simplicity that the instantaneous force was the same, but I also assumed she exerted the force for a longer time period on one object, which was why the change in momentum was different in the two scenarios.

"Consider a man in vacuum or frictionless environment pushing (F = kN) at one object (stone, ball etc.) with one hand or at two objects with two hands in opposite direction, show:

1) that the object(s) exerts an equal reaction (F = kN) and it is a force and not a (change of) momentum

2) that the same force exterted on an object of different mass provokes the same reaction ."

As I asked in a comment, how do you want people to "show" this? Normally it's just assumed as a basic axiom, and the axioms of Newtonian physics are judged based on how predictions derived from these axioms agree with experiment. So do you want experimental evidence or what? You can derive equal and opposite forces from conservation of momentum, if you have two isolated objects that are the only momentum-carriers (no momentum carried by a field as in electromagnetism, for example). But I already did this in the first paragraph of my answer--if you find anything about that derivation to be problematic, please point to the specific step you have doubts about.

• "For the work done on the girl by the ball, we have F=20, work would be = 10. (J).." Are you really saying that the ball is producing 10J of work? it is generating KE? How can a ball in the air produce energy? where does it take from? does it burn calories, or what? – user59485 Dec 11 '14 at 14:12
• @Alba -- Any object A that exerts a force on some other object B and causes B to move in space is doing work on B, by the definition of "work" in physics. Do you agree that according to Newton's third law the ball is exerting a force back on the girl while she is pushing on it, and do you agree that she moves backwards (assuming she's skating on a frictionless surface) due to this force? Work = displacement * force in direction of displacement, it's that simple. – Hypnosifl Dec 11 '14 at 15:30
• Huh? The answer by bobie in the link seems to just be about the history of what Newton meant, I don't see anything in it that's a "technical" analysis of the modern mainstream conception of the third law, independent of what Newton may or may not have meant when he first wrote down his laws. If you think I'm missing something, please highlight the portion of bobie's argument that you think would show a problem with the modern third law, regardless of what Newton thought (or better yet, write a new question dealing with this argument alone). – Hypnosifl Mar 30 '15 at 15:16
• If anyone has bothered to read this, I think my suspicion of sockpuppetry was correct, since my "achievements" timeline shows that a bunch of negative votes I had gotten, including most of the ones on this answer, just vanished because five accounts were deleted, apparently including GreenRay's whose comments are now just labeled "user59485". – Hypnosifl Mar 31 '15 at 21:10

Note that force is just the momentum per unit of time:

$F=\frac{dP}{dt}$

I.e. the force is the speed of momentum transmission.

When two bodies interact, they will exchange momentum. The quantity of momentum that will be transmitted from one body will be the same quantity, received by another.

This is depicted by following picture which is the third Newton's law and means that momentum, was lost by one body is gained by another.

This law is applicable in each situation, when the time of transmission is not zero. I.e. this law is hardly applicable for instant collisions. During instant collisions, the significant amounts of momentum are transmitted at infinitesimally small amounts of time -- i.e. the forces are nearly infinite.

So, the third law is definitely applicable, when skater pushes something.

If she pushes rail with force $F$ and then pushes ball with the same force $F$, the she will have the same reaction of $-F$. This force will cause the same acceleration of skater $-F/m$.

The only difference will be at the other side. Fixed rail will stay, because it has infinite mass $+F/\infty = 0$. And in the case of ball it will be finite: $+F/m_{ball}$

UPDATE

Note, that it is hard to imagine, how is it possible for skater to push ball and rail with the same force. Since ball is very lightweight, to perform the same force on it, girl should move very fast. Physically it is possible, but impossible physiologically.

Instead of a ball you can imagine something heavier, like sled.

Your scenario of ball-girl-rail is not completely clear to me - but I can tell you this: both Newton's laws, and conservation of energy, are quite generally applicable.

When the girl pushes against a light object (like the ball), then that object will accelerate away from her. If she maintains a certain force for a certain time, she will impart the same momentum to herself (and equal and opposite momentum to the ball). However, she will NOT do the same amount of work, since work done = $F\cdot d$ and the distance will be greater if the force is maintained for the same length of time because the ball moves. And the difference is seen in the kinetic energy of the ball.

Mathematically, for two objects 1, 2 with mass $m_1$ and $m_2$, initially at rest, a force $F$ for a time $\Delta t$ will have the following effect:

Momentum: $$m_1\cdot v_1 = F\Delta t = -m_2 \cdot v_2$$ Energy: $$E = \frac{p^2}{2m}$$ As the second object (ball) goes from "very light" to "super heavy", you can see how the amount of energy it has after the interaction changes from "a lot" to "none".

If the force is applied for the same distance (rather than constant time), then the total energy for the two objects will be the same - with the lighter object having a greater share of the energy as seen from the energy-momentum equation above.

UPDATE complete worked example.

Girl 50 kg, ball 1 kg, force of 20 N for 0.2 seconds:

$$p = F\Delta t = 4 kg m/s\\ E_{girl} = \frac{p^2}{2m} = \frac{16}{100} = 0.16 J\\ E_{ball} = \frac{16}{2} = 8 J$$

Let's see whether this is equal to the work done:

$$a = \frac{F}{m}\\ x = \frac12 a t^2\\ x_{girl} = \frac12 \frac{20}{50} 0.2^2 = 0.5\cdot 0.4 \cdot 0.04 = 0.008 m\\ E_{girl} = F\cdot x_{girl} = 20 \cdot 0.008 = 0.16 J$$

Doing the same for the ball:

$$x_{ball} = \frac12 \frac{20}{1} 0.2^2 = 0.5 \cdot 20 \cdot 0.04 = 0.4 m\\ E_{ball} = 20 \cdot 0.4 = 8 J$$

So the ball gets most of the energy, being lighter. The same force acts on both - but they move by different amounts. Newton's law holds - and energy, momentum and work done all balance as expected.

• The girl is exerting a force of 20N on the ball, according to the law, the ball should exert a force of 20N on the girl, isn't this what it is usually stated? Therefore, in total, there should be 40N in the system. Please correct any mistake. – user59485 Dec 11 '14 at 17:04
• @Floris: Sir, I've posted an answer. May I earnestly urge you to please make a scrutiny of my ans and please tell me whether I am correct in my conception?? – user36790 Dec 13 '14 at 9:31
• "..a force F for a time Δt will have the following effect.." , F*t = [change of] momentum. Floris, you are a great guy, is it so hard to see that you are referring to momentum, and 3rd law refers to momentum and not to force ?, Newton did not mention a 'force', it is only the sons of wikipedia that say so! :) – bobie Dec 16 '14 at 6:48
• ".. I went to the source. Newton didn't day either "force" or "momentum". He used "actio".." I couldn't reply in a comment, I posted a new answer here, I don't want to criticize you but the urban legend :), if you want to discuss it, you can email me. – bobie Dec 18 '14 at 13:12
• "..Multiplying by time doesn't change the vector nature, and the sum of the vectors is zero." Have you read my post? Are you saying that there is a case when two vectors, because of their direction, do not require two forces to come into being? m=1, F*t = 20N*0.2s = 8J, one single force can never do 8.16 J of work – bobie Dec 19 '14 at 9:12

In a more general and loosely specified manner Newton's Third Law states that one cannot touch without being touched. It seems a very reasonable and obvious statement. However, Quantum Mechanics has shown it is false with such examples as Quantum Non-demolition Measurement.

"Quantum non-demolition (QND) measurements improve sensitivity by evading measurement back-action"

Unfortunately you got answers that are wrong in many aspects.

1) I read all the previous answers concerning the 3rd law and I have seen that it is definitely not universal

The third law is universal in classical mechanics. You stop to talk about forces in quantum field theory, but that is another issue.

2)Conservation of momentum is a universal law, but that does not imply that action must be always equal to reaction, which as a matter of fact does not happen.

This is wrong: the first law defines conservation of momentum for an isolated particle and the second law says this can fail if a force is applied (some consider this the definition of force). But, if you want to show that momentum is conserved in an isolated system of interacting particles, you need the third law. Momentum is conserved in a system if and only of the third law is correct. This is a mathematical theorem, so saying that the third law in not universal is equivalent to saying that the law of consevation of momentum fails in isolated systems consisting of more than one object.

If the reaction (force) of the rail and of the ball is the same, how can the same force produce a velocity 4.58 times greater in the second case?

Because of different reasons. 1) if the rail is heavier it would acquire a different speed from the same force. 2) the rail is also feeling other forces that stop it from moving, like the one from the floor to which it is attached.

The action by girl is doing in both cases does 210 J of work, but the reaction is different, what is equal is only the momenta.

Also, from watching the drawings, you seem to mix action and reaction from the third law with work or energy. The action and the reaction are forces and as such are measured in newtons. Work is an entirely different concept.

Can someone please tell me if this is true? According to 3rd law, If I exert a force of k N on 10 different bodies, shouldn't I expect the same reaction of k N from all the 10 different bodies?

Yes, it is true (I am assuming you are doing 10kN on each, for a total of 100kN).

the action by girl is doing in both cases does 210 J of work, but the reaction is different, what is equal is only the momenta. Please show how you get equal actions and reaction considering forces.

It is not clear what you mean by the girl is doing the same work: 210J, because Energy, or work transferred is W=F.d then only way she can do that work is if she applies a certain force F for a certain distance C. But in that case the reactive force will also be F. Do not confuse the internal chemical/mechanical energy that she spends to move herself to the energy she transmits to another object. The last one depends on the force. If you know the final speed of the ball you can calculate (assuming an elastic collision where heat is not dissipated) the change in limear momentum, and with this change, if you have how longe the force acted, you can calculate the force. $F=\frac{\Delta P}{\Delta t}$. You cannot put by hand the energy the force and the momentum independently, that is inconsistent because they qre related. You give some variables and calculate the others. I am not sure if you drawings (that I find a little confussing), suggest that the girls makes a force of 8kgf to the ball and the ball reacts with 2 kgf? if that is what is suggesting it is incorrect. The forces will always be the same.

UPDATE: example requested by Alba Suposse F=10N and remains constant while the boy is in contact with the ball during the distance $d$, that takes a time $\Delta t$. The momentum gained by the ball will be the same as that gained by the boy (backwards):

$\Delta p_{ball}=-F.\Delta t=mv_{ball}$

$\Delta p_{boy}=F.\Delta t=Mv_{boy}$

Momentum is always conserved. Now, the work made by the force gives energy to the objects. The work energy theorem says that, for rigid bodies, the net work is transformed into kinetic energy of the object. If the object is not rigid and any of the forces acting on it deforms the object, then the Work-Energy Theorem will no longer be valid. Some of the energy transferred to the object has gone into deforming the object and is no longer available to increase or decrease the object's Kinetic Energy. This is what happens in our case, at least one of the objects is not rigid (the boy). Thus as Floris mentioned in his answer, the two objects will end up with different kinetic energies. Because the momentums are the same, and because $E_k = \frac{p^2}{2m}$ the lighter object will end up not only with a larget speed, but also with a larger kinetic energy.

• @Alba Actually I already answered that here: physics.stackexchange.com/questions/150335/… let me know if you have sppecific points you want clarified. – Wolphram jonny Dec 11 '14 at 15:20
• @Alba I put the example you wanted, I didnt add that the ball doesnt burn calories, but in case you do not know, it doesnt. Only the boy does to generate internal forces that act on the ball with force F – Wolphram jonny Dec 12 '14 at 1:46
• "...kinetic energy gained by the ball will be the same as that gained by the boy..." This is not correct, see Floris' example, 8, 0.16J : KE of the ball will be much greater than the boy's. The momenta are equal and this rules out that forces and F*d = KE be equal. That is why I asked this question and started the bounty. – user59485 Dec 12 '14 at 7:38
• @Alba Floris is correct, I forgot that this cannot be seen as an elastic interaction, so there will be energy lost as heat (similar to what happens in any inelastic collision). I updated my answer. – Wolphram jonny Dec 12 '14 at 16:01
• In inelastic collisions the impacting body sticks to the other body, this cannot happen here, neither does the hand rebound as it is not elstic. The reaction recoil of the hand on the body is therefore negligible, and, on the other hand, the ball has no energy or momentum of its own to give to the man. How can the momentum given to the man ever be equal to the momentum acquired by the ball? – user59485 Dec 14 '14 at 6:27

## protected by Community♦Dec 13 '14 at 6:09

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