Question

Derive Newton's third law

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Answer 1

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Derive Newton's third law

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The Third Law

• It asserts that "If body A exerts a force on body B, then B exerts a force of equal size and opposite direction on A." It can be written: FAB = – FBA. ... If that point or interface is treated as a "body" of mass zero, then Newton's second law tells us that Fnet = 0a, so Fnet = 0
• 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. the above equation (iii) represents the Newton's third law of motion (i.e., for every action there is equal and opposite reaction).

Answer 2

$\tt { \red{Derivation \: of \: Newton’s \: third \: law \: of \: motion}} \\ \tt \: from \: Newton’s \: second \: law \: of \: motion$

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.
• FBA be the force exerted by body B on A.

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Suppose that due to these forces FAB and FBA.

$\tt \: \frac{dp_1}{dt} \: and \: \frac{ dp_2}{dt}$be the rate of the change of momentum of these bodies respectively. 

$\tt \: Then \: \: FBA \: \frac{dp_1}{dt} = (i) \\ \\ \tt \: FAB = \frac{ dp_2}{dt} = (ii)$

• Adding equations (i) and (ii), we get, 

$\tt \: FBA + FAB = \frac{dp_1}{dt} + \frac{dp_2}{dt} \\ \\ \tt \: ⇒ FBA + FAB = \frac{d(p1 + p2)}{dt}$

• If no external force acts on the system, then 

$\tt \: ⇒ FBA + FAB = \frac{d(p1 + p2)}{dt} = 0 \\ \tt \: ⇒ FBA + FAB = 0 \\ \tt \: ⇒ FBA = - FAB----- (iii)$

The above equation (iii) represents the Newton's third law of motion.

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