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2014-08-30T02:07:41+05:30

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Second law:
The rate of change of momentum is directly proportional to the applied force and takes place in the direction that the force acts.

        F α Δp / Δt = k Δp/Δt = k Δ(m v) Δt = k m Δv/Δt = k m a
                   as Δv/Δt = dv/dt = acceleration
       We choose k = 1   So F = ma

Let us say body1 exerts a force F1 on body 2 and the body 2 exerts a force F2 on body1. These two bodies are in isolation. Let their masses be m1 and m2 respectively.

From the second law and first law, when there is no external force acting on them (or the system of body1 and body2), the acceleration of the system is zero. So the total momentum of the system remains constant.

p1 = m1 u1 at time t1 and m1 v1 at t = t2.    p2 = m2 u2 at t = t1 and p2 = m2 v2 at t = t2.

          p1 + p2 = constant
      m1 v1 + m2 v2 = m u1 + m2 u2
      m1 (v1 - u1) = - m2 (v2 - u2)
       m1 (v1 - u1) / (t2 - t1) = - m2 (v2 - u2) / (t2 - t1)

       m1 Δv1 / Δt = - m2 Δv2 / Δt
         m1 a1 = - m2 a2
         F2 = - F1
This is the third law.

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