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When talking about collisions, we usually talk about momentum because it allows us to analyze the situation in an easy way, since it is conserved. However, forget for a moment about the momentum. I had a confusion about Newton's Third Law in the situation of a collision.

Consider the following scenario: in a 1D situation, a mass 'm' at the left moves with a velocity '+v', such that it collides with an equal mass 'm' at the right which moves with a velocity '-v'. At the moment of impact, how are the forces in this situation? I imagine two possible answers, but I don't know which one is the correct answer.

  1. The mass at the left pushes the mass at the right with a force F, and the mass at the right pushes the mass at the left with a force -F. Therefore, both masses feel a force |F|.
  2. The mass at the left pushes the mass at the right with a force F, and this causes a reaction, pushing the mass at the left with a force -F. Further, the mass at the right will also exert another push of force -F to the left, and this force causes a reaction, pushing the mass at the right with a force F. Therefore, both masses feel a force |2F|.

Which answer is correct and why? It seems like the first answer is correct because the force of one causes a force to that mass due to the other mass. But, the second possible answer would also be reasonable, because since the second mass was also moving, it will also cause a force on the first mass, which causes a force on the second mass due to the first mass.

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7 Answers 7

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The first scenario is correct. Remember that a force must have a cause, a source. In the first scenario, when object A hits object B, then A causes the force on B and B causes the force on A.

In the second scenario you can't identify the source, it seems. It seems that the force of A on B also "causes" the force on A itself. A force doesn't cause a force. This scenario seems off. Also, why stop there? Why wouldn't the second force, produced by the first force, produce a third force? Which in turn will produce a fourth force and so on? This thought experiment seems to have no limiting principle and would thus result in infinite force. Which is obviously not what we see in reality - so this scenario must be based on an incorrect assumption.

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Lets modify your question a bit. What happens when the right mass is stationary and the left mass moves towards it as you have described. Well the left mass stops and the right mass moves with the exact same velocity (before collision) of the left mass.

What we infer:
The reason LM stops - because of the opposite reaction force
The reason RM moves - because of the force applied by the LM

Coming back to your original question.

What we infer:
The reason LM stops moving to the left - because of the opposite(equal) reaction force from its collision with RM
The reason LM moves the left - because of the force applied by RM
Similarly for the right mass.
I hope this solves your doubt.

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Your question is like asking whether zero equals 4-4 or 2-2, the answer being that you can consider either to be correct.

@hawexrutile made the point that you could consider the case in the frame which the mass at the left is moving and the mass at the right is initially stationary. In that case you might imagine that the moving left hand mass imparts a force F to the stationary right hand mass, causing the right hand mass to accelerate, while the right hand mass imparts a reaction -F to the left hand mass causing it to decelerate.

That said, all movement is relative, so you could also consider the collision from the frame in which the left hand mass is stationary and the right hand mass is moving, in which case you might consider that the right hand mass imparts a force on the left hand.

You can also consider the collision in an infinite number of frames moving at other speeds relative to the centre of mass of the two objects, in which cases you might take different views about which mass is imparting which force.

So the maths work out regardless of which point of view you take. The two objects exert the same force on each other, and whether you take that to mean that one object provides all the force and the other all the reaction, or vice versa, or some hybrid of those two extremes, makes no difference.

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The first scenario is the correct one, but you should consider that also with your second scenario you get the same conclusion, there is an equal and opposite force on both masses ($2F$ or $F$ doesn't mind). The two forces appear simultaneously, you can't identify one as the consequence of the other. If you'll study electromagnetism you could have the same doubt with Faraday-Lenz induction law...

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Co-moving frame of left mass claims that the mass on the right side is the source of force, and left mass just push the right mass by $-F$.

Similarly, co-moving frame of right mass claims the opposite; and both are right and there is nothing wrong with that. After all, the choice of frame is arbitrary.

As for stationary observer, two mass push away the other one at the same time; thus an ambiguity (which was pointed out by the question) rises about the source of force. However, in this frame we can use the Newton law, namely $F = dp/dt$ to do physics. That is, this observer can measure the change in the momentum of each ball separately, and assign a $dp$ to it. Moreover, the time of collision is a measurable quantity. So the force exerted on left and right mass can be measured; which is more than enough to do physics.

I'd like to point out that $|F|$ in collision is not always a meaningful quantity. When two rigid bodies collide, $dt\to0$ and $F = A\delta(t-t_0)$ where $\delta(t-t_0)$ is the dirac distribution. And $|A\delta(t-t_0)|$ is meaningless, unless you integrate it with respect to time, which of course give you the conservation of momentum law, and we do physics with that, not the force directly.

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Newton's third law says every action has an equal an opposite reaction.

Your second scenario posits that the reaction to the force $F$ the left mass $m$ exerts on the right mass is $2F$ and the reaction to the force $F$ the right mass exerts on the left mass is $2F$. Neither reaction is equal to the action of the other and therefore does not follow Newton's third law.

Hope this helps.

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I would not say that trying to understand what object is the source of the force is justified. The third Newton's law tells us: "If two bodies exert forces on each other, these forces are equal in magnitude and opposite in direction." So there is nothing about the source. Instead, we simply know that due to bodies' deformation during the collision, there is some type of repulsion, and since change in momentum of the second body is directed to the right, the force acting on the second body must also be directed to the right; similarly, since change in momentum of the first body is directed to the left, the force acting on the first body must also be directed to the left.

Your first statement is actually correct. In the second statement, confusion arises when you try to add excessive reaction forces. Note that forces go "in pairs", so if you call a force to be the opposite of another, it cannot further be the opposite of some other force. For that reason, it is reasonable to consider just one pair of forces $F$ and $-F$ on the first and the second body respectively.

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