Third Law of Motion

When a bullet is fired from a gun, the gun gets some velocity in the opposite direction. The velocity of the gun, commonly called recoil velocity, can be calculated by applying the law of conservation of momentum.
    •  When one object exerts a force on another object, the second object instantaneously exerts a force back on the first.
    •  These two forces are always equal in magnitude but opposite in direction.
    •  These forces act on different objects but never on the same object.
    •  Every action has an equal and opposite reaction.

There are several applications of Newton’s third law of motion; the launching of satellites is one among them.



Derivation of law of conservation of momentum from Newton's third law:

Law of conservation of momentum statesthat if no external force acts on a system of particles, the algebraic sum of the linear momenta of the particles remains conserved."

Consider two particles A and B, Which collide head on as shown in figure The particles move in a straight line before and after collision.



Let particle A have initial velocity u₁ and particle B has initial velocity u₂ . The two particles will collide, if u₁ > u₂ . Let after collision the final velocities of A and B becomes v₁ and v₂ respectively . The two particles will seperate after collision, if v₂ > v₁ . Let the two particles A and B have m₁ and m₂ respectively.

Initial momentum of particle A is m₁u₁
Initial momentum of particle B is m₂u₂
Final momentum of particle A is m₁v₁  
Final momentum of particle B is m₂v₂

Rate of change of momentum of particle  A = $\frac{{\text{m}}_{\text{1}}{\text{v}}_{\text{1}}\text{-}{\text{m}}_{\text{1}}{\text{u}}_{\text{1}}}{\text{t}}$   = $\frac{{\text{m}}_{\text{1}}\text{(}{\text{v}}_{\text{1}}\text{-}{\text{u}}_{\text{1}}\text{)}}{\text{t}}$    ........(1)
Rate of change of momentum of particle  B = $\frac{{\text{m}}_{\text{2}}{\text{v}}_{\text{2}}\text{-}{\text{m}}_{\text{2}}{\text{u}}_{\text{2}}}{\text{t}}$   = $\frac{{\text{m}}_{\text{2}}\text{(}{\text{v}}_{\text{2}}\text{-}{\text{u}}_{\text{2}}\text{)}}{\text{t}}$   ........(2)

Let $\overrightarrow{{\text{F}}_{\text{AB}}}$ and $\overrightarrow{{\text{F}}_{\text{BA}}}$ , be the forces exerted by particle B on A and particle A on B respectively.
According to Newton's second law of motion

$\overrightarrow{{\text{F}}_{\text{AB}}}$   =   $\frac{{\text{m}}_{\text{1}}{\text{v}}_{\text{1}}\text{-}{\text{m}}_{\text{1}}{\text{u}}_{\text{1}}}{\text{t}}$   = $\frac{{\text{m}}_{\text{1}}\text{(}{\text{v}}_{\text{1}}\text{-}{\text{u}}_{\text{1}}\text{)}}{\text{t}}$   and    $\overrightarrow{{\text{F}}_{\text{BA}}}$ =  $\frac{{\text{m}}_{\text{2}}{\text{v}}_{\text{2}}\text{-}{\text{m}}_{\text{2}}{\text{u}}_{\text{2}}}{\text{t}}$   = $\frac{{\text{m}}_{\text{2}}\text{(}{\text{v}}_{\text{2}}\text{-}{\text{u}}_{\text{2}}\text{)}}{\text{t}}$

Then, by Newton's third law of motion, we have

F_AB + F_BA  = 0
F_AB = - F_BA
ie, m₁v₁-m₁u₁ = -(m₂v₂-m₂u₂)
⇒ m₁u₁+m₂u₂ = m₁v₁+m₂v₂

or initial momentum of the system = final momentum of the system, which is the law of conversation of momentum.