Law of conservation of linear momentum is a extremely important consequence of newton's third law of motion in combination with the second law of motion
Consider two particles of mass m1 and m2 interacting with each other and forces acting on these particles are only the ones they exert on each other.
Let F12 be the force exerted by the particle 2 on particle 1 having mass m1 and velocity v1 and F21=-F12 be the force exerted by the particle 1 on particle 2 having mass m2 and velocity v2
Applying newton second law of each particle on each particle
F12=m1(dv1/dt)
and F21=m2(dv2/dt)
from newton's third law of motion F21=-F12
or m1(dv1/dt) + m2(dv2/dt)=0
Since mass of the particles are not varying with time so we can write
(d/dt)(m1v1 +m2v2)=0
or m1v1 +m2v2=constant
--(13)
we have already defined the quantity mv as the momentum of the particle
Thus we conclude that when two particles are subjected only to their mutual interactions ,the sum of the momentums of the bodies remains constant in time or we can say the total momentum of the two particles does not change because of the any mutual interactions between them
For any kind of force between two particles then sum of the momentum ,both before and after the action of force should be equal i.e total momentum remains constant
We thus arrive to the statement of principle of conservation of linear momentum
" when no resultant external force acts on system ,the total momentum of the system remains constant in magnitude and direction"
In absence of external forces for a number of interacting particles,law of conservation of linear momentum can be expressed as
m1v1 +m2v2+m3v3+m4v4+...=constant
Law of conservation of linear momentum is one of the most fundamental and important principle of mechanics
This principle also holds true even if the forces between the interacting particles is not conservative
Once again ,the total momentum of two or any number of particles of interacting particles is constant if they are isolated
from outside influences (or no resultant external forces is acting on the particles)
Example -1
A bullet of mass $m$ is fired horizontally with a velocity $u$ on a wooden block of Mass $M$ suspended from a support and get embedded into it.
a. find the velocity of the wooden + block system after the collision
b. Find The Kinetic energy of the wooden + block system after the collision Solution
Initial velocity of bullet=u
Initial velocity of block=0
So net momentum before collision=$mu$
Let v be the velocity after collision
Then Net momentum after collision=$(M + m)v$
Now linear momentum is conserved in this collision
so
$mu=(M + m)v$
or $v=\frac {mu}{M+m}$
So kinetic energy after collision
=$\frac {1}{2}(M+m) v^2 = \frac {m^2u^2}{2(M+m)}$
Watch this tutorial for more information on Law of conservation of linear momentum
