Elastic collisions and Newton's third law

In summary: Then show that the net force on box A is always greater than the net force on box B. This is because the force on A is always applied through the distance between you and A, while the force on B is applied through the distance between you and B.
  • #1
Renato776
1
0
Hi everyone, I am new to this forum, and I'm having a hard time understanding Newton's third law and collisions, first of all I want to say that it is not homework and that I do know the basics of physics, vectors, energy, work, and momentum I also know and understand Newton's first and second law and I'm currently in my course of multivariable calculus, so derivatives and integrals are not a problem. Having saying that, I want you guys to consider the following system:

I am in space and I'm an entity so I can apply force with my mind and be everywhere, having stated that, I want to move box A and box B at the same time, both boxes are in repose in space, and in contact with each other, since I am quite lazy I want to apply force only on box A and I do so with a constant force F. This is what my current knowledge and understanding of physics tells me that its going to happen:

To box A is being applied a constant force F forward, and box A is aplying the same force to box B, according to Newton's third law box B is applying the same force backwards to box A, so in the instant I start applying force to box A this is what's going to happen: box A will remain in repose due to F-Fb/a=0 since F is from me and Fb/a is the reaction of box B over box A, and box B will start moving since the only force applied to it is Fa/b, because of this box B will start moving and box A wont, but in the moment they loose contact there is no longer any force acting on box B so it will start moving at a constant velocity V1 at the same time since the only force acting on box A is my force, (since they lost contact there is no longer the reaction force Fb/a acting on box A), so now box A will start moving, however it will not move at a constant velocity since I'm applying force to it, so box A will start accelerating and will eventually catch up with box B, this time a collision between boxes A and B will happen, assuming that both boxes have the same mass "m" and that the collision is perfectly elastic, both will chage their velocities and box B will get a higher velocity after the collision, however once the collision is over, it will keep moving at a constant velocity V2, now box A will change its velocity during the collision but since I'm not stopping my force, (never) then box A will keep accelerating after the collision and will catch up to box B once again, but this time at a higher velocitiy, they will collide once again, and after this, box B will get even higher velocity, then box A is going to accelerate and catch up eventually, collide again, and so on and this is going to repeat over and over again and the gap between the boxes will just keep increasing.

So the question is: is this behavior correct? Is this really how the system should move according to Physics and the initial information I gave? if it is not and the boxes should move together like a giant box (A+B) why is this happening? wouldn't that behavior (of both boxes acting like a giant one) be against Newton's third law? is there a concept that I'm missing?
 
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  • #2
Renato776 said:
To box A is being applied a constant force F forward, and box A is aplying the same force to box B
What makes you think that the force of you on box A is the same as the force of box A on box B?
 
  • #3
If Boxes A and B are in contact with each other, and are otherwise free to move with no other forces acting on them, then the force you apply to A cannot be equal in magnitude to the force B applies to A. If the two boxes have the same mass then the force B applies to A has to be half the force you apply to A. As long as A and B are in contact they must have the same acceleration, and if they have the same mass that means they each have the same net force applied to them.

I suggest you draw two well-separated free-body diagrams. One of A and one of B. Show that the net force on each box equals its mass times its acceleration.
 

Related to Elastic collisions and Newton's third law

1. What is an elastic collision?

An elastic collision is a type of collision between two objects where there is no loss of kinetic energy. This means that the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

2. How is momentum conserved in an elastic collision?

In an elastic collision, momentum is conserved because the total momentum of the system (the two colliding objects) remains the same before and after the collision. This is due to Newton's third law, which states that for every action, there is an equal and opposite reaction.

3. What is Newton's third law and how does it relate to elastic collisions?

Newton's third law states that for every action, there is an equal and opposite reaction. In the case of an elastic collision, this means that the force exerted by one object on the other is equal in magnitude but opposite in direction. This allows for the conservation of momentum and the preservation of kinetic energy in the collision.

4. Can an elastic collision occur between more than two objects?

Yes, an elastic collision can occur between multiple objects as long as the total momentum and kinetic energy of the system remain conserved. Each object will experience an equal and opposite force from the other objects involved in the collision.

5. How is an elastic collision different from an inelastic collision?

An inelastic collision is one where there is a loss of kinetic energy due to deformation or other non-conservative forces. This means that the total kinetic energy after the collision is less than the total kinetic energy before the collision. In contrast, an elastic collision is one where there is no loss of kinetic energy and the total kinetic energy before and after the collision is the same.

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