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NEWTON'S III LAW OF MOTION: ORIGINAL PROOF
           A GIFT (ON TEACHERS' DAY) FOR ALL MY TRAINEES
by Abhishek Alankar on Monday, September 5, 2011 at 10:07pm ·



Newton's III Law of Motion states, " To every action there is an equal and opposite reaction and
both take effect on two different bodies and they don't balance each other."


Let us consider a system of two bodies having masses M1 and M2 and position vectors R1 and R2.


The position vector of the centre of mass of the system will be given by


R= M1R1 + M2R2
     M1 + M2


Differentiating twice with respect to time t, we get the acceleration of centre of mass of the system


A = M1A1 + M2A2
      M1 + M2


i.e. (M1 + M2)A = M1A1 + M2A2


or   F = F1 + F2


From Newton's II Law of Motion we deduce that


F = 0 (As there are no external forces acting on the system acceleration of its centre of mass A = 0).


Therefore,


F1 + F2 = 0.


i.e. F1 = - F2.


This proves the proposition stated by Newton's III Law of Motion.

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rare and original proof of newton's III law of motion

  • 1. NEWTON'S III LAW OF MOTION: ORIGINAL PROOF A GIFT (ON TEACHERS' DAY) FOR ALL MY TRAINEES by Abhishek Alankar on Monday, September 5, 2011 at 10:07pm · Newton's III Law of Motion states, " To every action there is an equal and opposite reaction and both take effect on two different bodies and they don't balance each other." Let us consider a system of two bodies having masses M1 and M2 and position vectors R1 and R2. The position vector of the centre of mass of the system will be given by R= M1R1 + M2R2 M1 + M2 Differentiating twice with respect to time t, we get the acceleration of centre of mass of the system A = M1A1 + M2A2 M1 + M2 i.e. (M1 + M2)A = M1A1 + M2A2 or F = F1 + F2 From Newton's II Law of Motion we deduce that F = 0 (As there are no external forces acting on the system acceleration of its centre of mass A = 0). Therefore, F1 + F2 = 0. i.e. F1 = - F2. This proves the proposition stated by Newton's III Law of Motion.