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For example, I have a force of $3N$ accelerating a $1$kg block $A$ on a frictionless surface to the right. $A$ accelerates at a rate of $3$ $m/s^2$. Say it collides with another block $B$, weighing $2$kg.

By adding the weights and treating it as one larger object weighing $3$kg, the acceleration would now equal $1$ $m/s^2$. However, when $A$ collides with $B$, isn't there an opposing force of equal magnitude applied on $A$, causing the net force on $A$ to be $0N$?

Can anyone please explain the mistake in my logic? I know the final acceleration is $1N$ but I'm not sure why my logic of the objects colliding doesn't give that answer.

Edit: Maybe a better question is, how is force applied from object $A$ to object $B$ calculated? I thought it was $F=ma$, but if object $A$ has $m=1$ and $a=3m/s^2$, doesn't that apply a force from $A$ to $B$ of $3N$?

Qmechanic
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Luther Grusovin
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    Does this answer your question? Given Newton's third law, why are things capable of moving? – John Rennie Aug 24 '22 at 09:00
  • You can't add forces in Newton force pair of third law, because they targets different bodies, $A\to B$ and $B \to A$. Force summation applies only to forces acting on same body. – Agnius Vasiliauskas Aug 24 '22 at 09:02
  • @AgniusVasiliauskas But the original force that I mentioned is being applied on $A$, so there is a force pushing on $A$ and the normal force from $B$ pushing against it. – Luther Grusovin Aug 24 '22 at 09:04
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    @LutherGrusovin But the force that B exerts on A does not have to have a magnitude of 3 N. Indeed, we can see that the force that B exerts on A and the equal and opposite force that A exerts on B both have a magnitude of 2N. – gandalf61 Aug 24 '22 at 09:08
  • @gandalf61 Could you explain the reason for the force $A$ exerts on $B$ being $2N$? I don't doubt it, and I can see that $3N - 1N = 2N$, but if force $F$ applied on an object = $mass * acceleration$, doesn't that give $3N$ applied from $A$ to $B$? – Luther Grusovin Aug 24 '22 at 09:18
  • If I've understood correctly, A is accelerated by a 3 N force and then collides with B. There will then be normal contact forces between A and B. These forces will always be equal and opposite, but will rise in magnitude from zero to a peak value and then decrease. The magnitude will depend on factors which include the elastic properties of A and B. If A and B are fairly rigid, the bodies will undergo large accelerations when they collide, and the peak magnitude of the contact forces will be large, probably much larger than the externally applied 3 N force that accelerated A. – Philip Wood Aug 24 '22 at 10:09
  • @PhilipWood A constantly has the force 3 N on it, but if the normal contact force results in an equal and opposite force of 3 N on A, then shouldn't A stop moving? – Luther Grusovin Aug 24 '22 at 10:21
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    "Maybe a better question is, how is force applied from object to object calculated? I thought it was =, but if object has =1 m and = 3 m s$^{–2}$, doesn't that apply a force from to of 3N?" No, because when the bodies are colliding, A will no longer have an acceleration of 3 m s$^{–2}$, because it will be acted upon both by the 3 N external force and a normal contact force from B. The $\mathbf F$ in $\mathbf F =m\mathbf a$ is the RESULTANT force acting on the body. – Philip Wood Aug 24 '22 at 10:37
  • "but if the normal contact force results in an equal and opposite force of 3 N on A ..." But it won't. Suggest you read my first comment again. – Philip Wood Aug 24 '22 at 10:40
  • @PhilipWood So A has the initial force of 3 N and the normal force pushing back on it after it comes into contact with B, how is this normal force calculated? Thanks for all the replies btw, I'm just trying to have a complete understanding of how objects interact with eachother. – Luther Grusovin Aug 24 '22 at 10:56
  • Think about a brick of mass 1 kg falling from a height on your foot. Why does it hurt, whereas it wouldn't hurt if it were simply placed on your foot? The answer is that when the falling brick hits your foot it undergoes a large (upward) acceleration (larger than the original 10 m s${^–2}$ of the falling brick). The large acceleration is due to the large normal contact force (at its peak value) between the brick and your foot. [Strictly we should use the resultant of the contact force and the 10 N pull of gravity to calculate the brick's acceleration on hitting the foot.] – Philip Wood Aug 24 '22 at 10:57
  • "how is this normal force calculated? " By considering the elastic properties of the colliding bodies. It's not very easy, but a simple model would treat the bodies as masses fitted with buffers on springs that get compressed when the bodies collide. You could then apply simple harmonic motion theory. – Philip Wood Aug 24 '22 at 11:02
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    @Moderators: Please reopen this question or link to a different, more relevant answer, as this question is not about the 3NL pair. True, the scenario is a collision of two objects A & B, but the question is asking about the resulting force on the collider A that is itself being accelerated at the moment of collision (so the resultant of an external accelerating influence on A + the collision reaction force on A). – Filip Milovanović Aug 24 '22 at 17:08

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