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

[inuka00123]

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?

 

[kuruman]

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.

 

[Herman Trivilino]

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.

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.

 

[Lnewqban]

... 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.

 

[inuka00123]

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.

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.