Derivation of Newton third law of motion ??
 

Dear Student
"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: F_AB = – F_BA. The pair of forces in this law are often called an "action-reaction pair." Each force is said to be a "reaction" force of the other.

Suppose there is mutual interaction present between bodies A and B.
Let 
$$\begin{gathered} \overrightarrow{F_{BA}} is force exerted by A on B \\ and \\ \overrightarrow{F_{AB}} is force exerted by B on A \\ Now \\ \overrightarrow{F_{BA}} =\frac{d\overrightarrow{p_B}}{dt}, \overrightarrow{F_{AB}} =\frac{d\overrightarrow{p_A}}{dt} \\ Where \\ \overrightarrow{p_A} is momentum of A and \overrightarrow{p_B} is momentum of B. \\ Adding two forces \\ \overrightarrow{F_{BA}} +\overrightarrow{F_{AB}} =\frac{d\overrightarrow{p_B}}{dt}+\frac{d\overrightarrow{p_A}}{dt} =\frac{d\left(\overrightarrow{p_B}+\overrightarrow{p_A}\right)}{dt} \\ But total momentum of the system \left(\overrightarrow{p_B}+\overrightarrow{p_A}\right) has to be cons\tan t as no external \\ force is applied on it. \\ Hence any change or rate of change will be zero \\ \overrightarrow{F_{BA}} +\overrightarrow{F_{AB}} =\frac{d\left(\overrightarrow{p_B}+\overrightarrow{p_A}\right)}{dt}=0 \\ Thus \\ \overrightarrow{F_{BA}}=-\overrightarrow{F_{AB}} \end{gathered}$$
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