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Answer» Derivation of Newton’s third law of motion from Newton’s second law of motion
Consider an isolated system of two bodies A & B mutually interacting with each other, provided there is no external force acting on the system.
Let FAB, be the force exerted on body B by body A and FBA be the force exerted by body B on A.
Suppose that due to these FORCES FAB and FBA, dp1/dt and dp2/dt be the rate of the CHANGE of MOMENTUM of these bodies respectively.
Then, FBA = d p 1 dt ---------- (i)
=> FAB = d p 2 dt ---------- (ii)
Adding equations (i) and (ii), we get,
FBA + FAB = d p 1 dt + d p 2 dt
⇒ FBA + FAB = d( p 1 + p 2 ) dt
If no external force acts on the system, then
d( p 1 + p 2 ) dt = 0
⇒ FBA + FAB = 0
⇒ FBA = - FAB---------- (iii)
the above equation (iii) represents the Newton's third law of motion (i.e., for every action there is equal and opposite REACTION)...
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