Question

prove the law conservation of momuntum on FAB=FBA​

Answer 1

The law of conservation of linear momentum easily follows from the third law of motion.
Consider a system of two particles, A and B. Let's say they interact with each other and 'A' experts a Force Fba on 'B' and in reaction, 'B' exerts a force Fab on 'A'.
Now, according to the second law of motion,
Fba= Mb x d/dt(Vb) = d/dt(Mb x Vb)
Fab= Ma x d/dt(Va) = d/dt(Ma x Va)
where Mb and Ma are masses of 'B' and 'A' respectively, and similarly Vb and Va their velocities in that order. d/dt denotes the derivative.
Now, adding the above two equations,
Fab + Fba = d/dt (Ma x Va) + d/dt(Mb x Vb)
Note that according to the third law of motion, these forces Fab and Fba are equal and opposite, so Fab= -Fba
Hence, 0= d/dt(Ma x Va) + d/dt(Mb x Vb) = d/dt(Pa + Pb)
where Pa and Pb are the linear momenta of 'A' and 'B' respectively.
Hence, Pa + Pb= constant (since the derivative is zero)
Therefore, The sum of the linear momenta of the bodies is constant.
The above calculations assume no external forces act upon the system. Hence, if there is zero net external force, the linear momentum of a system remains conserved, i.e. unchanged.