Respuesta :

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.
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Answer:

Consider an isolated system of 2 bodies A & B. An isolated system means where there is no external force acting.Now let F be the force acting on B by A & F be the force acting on A by B.Now rate of change of momentum of A =dpA/dt and rate of change of momentum of B =d pB/dt

thus FAB=d pB/dt (i)......

FBA=dpA/dt (ii)........

Adding 1 and 2 we get FAB+FBA=d pB/dt+dpA/dt =d(pB+pA)/dt

But if no force is applied momentum will also be 0 because no velocity will be there so rate of change of momentum will also be 0

thus d(pA+pB)/dt=0

therefore FAB+FBA=0

or,FAB=-FBA

(3RD LAW OF MOTION)

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