prove that tird law of motion is containing in second law of motion....
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Mill Consider two bodies, I will refer to them as blue ball and red ball, in an empty space interacting with nothing else but themselves with constant velocities vblue→vblue→ and vred→vred→ .
Their initial linear momentum is given by mbluevblue→+mredvred→mbluevblue→+mredvred→ .
Now, suppose that the blue ball starts creating a force F⃗ blue→redF→blue→red on the red ball and so the red ball is accelerated. What happens to blue ball?
It is really important to visualize that the system’s(red ball +blue ball) momentum has NOT changed, since there are no external forces acting on them and the second law of newtonian motion states that the force is the change in linear momentum.
So we can conclude that, F⃗ =dp⃗ dt=0⃗ F→=dp→dt=0→ and, taking the second derivative dF⃗ dt=0⃗ dF→dt=0→ . Or, in other words, the net force on the system does not change.
Reckoning that the force doesn’t change, the initial force is the same as the final force. Since in the beginning we said initial force is equal to zero, then so must be the final force. Hence: F⃗ final=F⃗ blue→red+F⃗ red→blue=0F→final=F→blue→red+F→red→blue=0 . Or in other words: F⃗ blue→red=−F⃗ red→blueF→blue→red=−F→red→blue. There you go!
Which means that the simple fact that the blue made a force on the red ball, had the consequence (so that the momentum would be conserved) that the red ball made another force with the same intensity on the blue ball to compensate.
Their initial linear momentum is given by mbluevblue→+mredvred→mbluevblue→+mredvred→ .
Now, suppose that the blue ball starts creating a force F⃗ blue→redF→blue→red on the red ball and so the red ball is accelerated. What happens to blue ball?
It is really important to visualize that the system’s(red ball +blue ball) momentum has NOT changed, since there are no external forces acting on them and the second law of newtonian motion states that the force is the change in linear momentum.
So we can conclude that, F⃗ =dp⃗ dt=0⃗ F→=dp→dt=0→ and, taking the second derivative dF⃗ dt=0⃗ dF→dt=0→ . Or, in other words, the net force on the system does not change.
Reckoning that the force doesn’t change, the initial force is the same as the final force. Since in the beginning we said initial force is equal to zero, then so must be the final force. Hence: F⃗ final=F⃗ blue→red+F⃗ red→blue=0F→final=F→blue→red+F→red→blue=0 . Or in other words: F⃗ blue→red=−F⃗ red→blueF→blue→red=−F→red→blue. There you go!
Which means that the simple fact that the blue made a force on the red ball, had the consequence (so that the momentum would be conserved) that the red ball made another force with the same intensity on the blue ball to compensate.
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