B Question about Newton's 3rd law: Two boxes sliding toward each other and colliding

AI Thread Summary
When a 10 kg box is pushed with a 20 N force toward a stationary 5 kg box, it accelerates at 2 m/s² until they collide. Upon collision, the 5 kg box exerts a 10 N force on the 10 kg box, which is an internal force that cancels out with the equal force acting in the opposite direction. This results in a new system where the 10 kg box's acceleration reduces to 1.33 m/s², while the previously stationary 5 kg box accelerates at the same rate. The net external force acting on the 10 kg box is now 13.33 N after accounting for the internal forces. The interaction illustrates Newton's Third Law, where internal forces do not affect the overall acceleration of the system.
inuka00123
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imagine a person pushing a 10Kg cardboard box with 20N of force toward a 5Kg cardboard box, with the second law we can calculate it is moving at 2ms/s acceleration toward the 5Kg box, relative to the 10Kg box we can say the 5Kg box is moving toward the 10Kg box with 2ms/s acceleration with a 10N force, when the both boxes collide 5kg box is putting a 10N force to the surface of the 10Kg box. my question is how can we take it as a one system with a 15Kg of mass and 20N force and acceleration with 1.33ms/s, what happens to the opposite forces that act between the boxes when they collide?
 
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inuka00123 said:
what happens to the opposite forces that act between the boxes when they collide?
They're still there but they have equal magnitudes. So because they act in opposite directions, they cancel out and do not contribute to the overall acceleration of the two-block system. That's a consequence of Newton's Third law. Only external forces to the two-block system can accelerate it. Here, the external force is the 20 N.

Consider this. Someone can grab you by your collar and lift you up by applying an external force on you, but you cannot grab your own collar and lift yourself up. The upward force that your hand exerts on your collar is cancelled by the downward force that the collar exerts on your hand. Try it to see how it works, but be careful not to pull too hard lest you choke yourself or rip your shirt.
 
inuka00123 said:
imagine a person pushing a 10Kg cardboard box with 20N of force toward a 5Kg cardboard box, with the second law we can calculate it is moving at 2ms/s acceleration toward the 5Kg box,
No, it could be moving away, slowing down.
inuka00123 said:
my question is how can we take it as a one system with a 15Kg of mass and 20N force and acceleration with 1.33ms/s, what happens to the opposite forces that act between the boxes when they collide?
They are internal forces.
 
inuka00123 said:
... when both boxes collide 5kg box is putting a 10N force to the surface of the 10Kg box. my question is how can we take it as a one system with a 15Kg of mass and 20N force and acceleration with 1.33ms/s...
An instant after the boxes collide:
1) The value of the acceleration of the moving 10 kg box gets reduced to 1.33 m/s^2.
2) The value of the acceleration of the 5 kg box that was in repose gets increased to 1.33 m/s^2.
3) The value of the action-reaction contact forces becomes 6.66 N (of opposite directions).
4) The box of 10 kg alone is now being pushed by a net force of 20 N - 6.66 N = 13.33 N, which enables an acceleration of that box equal to 1.33 m/s^2.
5) The box of 5 kg alone is now being pushed by a net force of 6.66 N, which enables an acceleration of that box equal to 1.33 m/s^2.
 
kuruman said:
They're still there but they have equal magnitudes. So because they act in opposite directions, they cancel out and do not contribute to the overall acceleration of the two-block system. That's a consequence of Newton's Third law. Only external forces to the two-block system can accelerate it. Here, the external force is the 20 N.

Consider this. Someone can grab you by your collar and lift you up by applying an external force on you, but you cannot grab your own collar and lift yourself up. The upward force that your hand exerts on your collar is cancelled by the downward force that the collar exerts on your hand. Try it to see how it works, but be careful not to pull too hard lest you choke yourself or rip your shirt.
Lnewqban said:
An instant after the boxes collide:
1) The value of the acceleration of the moving 10 kg box gets reduced to 1.33 m/s^2.
2) The value of the acceleration of the 5 kg box that was in repose gets increased to 1.33 m/s^2.
3) The value of the action-reaction contact forces becomes 6.66 N (of opposite directions).
4) The box of 10 kg alone is now being pushed by a net force of 20 N - 6.66 N = 13.33 N, which enables an acceleration of that box equal to 1.33 m/s^2.
5) The box of 5 kg alone is now being pushed by a net force of 6.66 N, which enables an acceleration of that box equal to 1.33 m/s^2.
I undetstand.
 
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