Prove the law of conservation of linear momentum use it to find the recoil velocity of a gun when a bullet is fired from it

Explanation

`(i)` The force on each particle (Newton's second law) can be written as,
`vecF_(12)=(dvecp_(1))/(dt)` and `vecF_(21)=(dvecp_(2))/(dt)`
`(ii)` Here `vecp_(1)` is the momentum of particle `1` which changes due to the force `vecF_(12)` exerted by particle `2`. Further `vecp_(2)` is the momentum of particle `2`. This changes due to `vecF_(21)` exerted by particle `1`.
`(dvecp_(1))/(dt)=-(dvecp_(2))/(dt)`
`(dvecp_(1))/(dt)+(dvecp_(2))/(dt)=0`
`(d)/(dt)(vecp_(1)+vecp_(2))=0`
`(iii)` It implies that `vecp_(1)+vecp_(2)=` constant vector (always).
`(iv) vecp_(1)+vecp_(2)` is the total linear momentum of the two particles `(vecp_("tot")=vecp_(1)+vecp_(2))`. It is also called as total linear momentum of the system. Here, the two particles constitute the system.
`(v)` If there are no external forces acting on the system, then the total linear momentum of the system `(vecp_("tot"))` is always a constant vector. i.e. `vecF_(21)=vecF_(12)` .
Recoil velocity of gun
when the gun is fired ,the bullet moves forward and the gun recoil backward. Let `v_(b)` and `v_(g)` are their respective velocities , the total momentum of the bullet- gun system, after firing is `m_(b)v_(b)+m_(g)v_(g)`
According to the law of conservation of momentum total momentum before firing is equal to total momentum after firing.
i.e. `0=m_(b)v_(b)+m_(g)v_(g)` (or) `v_(g)=-(m_(b))/(m_(g))v_(b)`