In physics, there are a number of conservation laws like linear momentum, angular momentum etc. These are fundamental and symmetric in nature and are verified by experiments. The law of conservation of linear momentum was proposed by Newton three centuries ago that became a basic law in physics. Let us know more about it.

Momentum in general is the measure of the quantity of motion. The linear momentum is the momentum in linear path. It is also defined as a body's translation motion and is measure of the product of mass m and velocity v, given as
p = mv
The SI unit of linear momentum is kilogram meters per second (kgms^{1}).
The linear momentum is the product of the mass of an object and the velocity of an object. Let take any two objects moving with some velocities towards each other. Then the momentum will be conserved. Hence,
the principle of conservation of linear momentum states that the sum of linear momentum in a closed system is constant. If p_{1}, p_{2}, p_{3},.... are the momentum in a closed system then principle of conservation of momentum states that the total momentum is constant.
p_{1} + p_{2} + p_{3 }+ p_{}_{4} + ....... = constant.
We know that the linear momentum is given as$\vec{p}$ = m $\vec{v}$
Using it we can rewrite newtons second law as$\vec{F}$ = m$\vec{a}$ = m $\frac{d \vec{v}}{dt}$ = $\frac{d\vec{p}}{dt}$
Let us consider the system of particles then the momentum for the system of particles is given asA special case is when $\vec{F}$ = $\vec{0}$ => $\vec{p}$ is constant.
$\vec{P}$ = $\sum_i$ $\vec{p_i}$ = $\sum_i$ m_{i} $\vec{v_i}$
To get the linear momentum of individual particles we go for the larger momentum $\vec{P}$ in a many particle system.
$\frac{d\vec{P}}{dt}$ = $\sum_i$ $\frac{d \vec{p}}{dt}$ = $\sum_i$ F_{i} = $\vec{F_{net}}$
If no net force acts on the system of particles then we get
$\vec{F_{net}}$ = $\vec{0}$ => P is a constant
Thus, the linear momentum of the system of particles is conserved when no net force acts on the system.
Let us consider two bodies of masses m_{1} and m_{2} moving towards each other having initial velocities u_{1} and u_{2}. Both collide with each other for time t. After the collision, they move with the velocities v_{1} and v_{2}.
From newtons second law, Force applied by A on B is given by
F_{AB} = $\frac{m_2 v_2  m_2 u_2}{t}$
Similarly, Force applied by B on A
F_{BA} = $\frac{m_1 v_1  m_1 u_1}{t}$
From Newtonâ€™s third law we have
F_{AB} = F_{BA}
Hence,
$\frac{m_2v_2  m_2u_2}{t}$ = $\frac{(m_1 v_1  m_1u_1)}{t}$
Or
m_{2}v_{2}m_{2}u_{2} = m_{1}v_{1}+m_{1}u_{1}
Or
m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}
This means the total momentum before collision is equal to total momentum after collision.
Lets go through some examples on linear momentum:
Bullet and gun : when a bullet is fired from the total linear momentum of the system remains conserved. On firing the gun the bullet moves forward whereas gun recoils backward.
Motion of a rocket : The upward motion of rocket can be explained using law of conservation of linear momentum. In a rocket the burnt gases comes out of the nozzle in downward direction with tremendous velocity. As a result the rocket gains an equal linear momentum to move in opposite direction that makes the rocket move with greater speed.
Boy and boat : A boy sitting in a boat suddenly jumps into the water that makes the boat move in opposite direction. Hence momentum is conserved.
Motion of a rocket : The upward motion of rocket can be explained using law of conservation of linear momentum. In a rocket the burnt gases comes out of the nozzle in downward direction with tremendous velocity. As a result the rocket gains an equal linear momentum to move in opposite direction that makes the rocket move with greater speed.
Boy and boat : A boy sitting in a boat suddenly jumps into the water that makes the boat move in opposite direction. Hence momentum is conserved.
Lets go through some sample numerical examples on conservation of linear momentum:
Solved Examples
Question 1: A body has mass of 2 kg moves with velocity 3ms^{1}. Calculate its linear momentum.
Solution:
Given :
Mass m = 2 kg, Velocity v = 3 ms^{1}
The linear momentum is given by
p = mv
= 2 kg $\times$ 3 ms^{1}
= 6 kgms^{1}.^{}
Solution:
Given :
Mass m = 2 kg, Velocity v = 3 ms^{1}
The linear momentum is given by
p = mv
= 2 kg $\times$ 3 ms^{1}
= 6 kgms^{1}.^{}
Question 2: A bullet of mass 50 g moves with a velocity of 200 ms^{1} collides with a freely suspended bag of mass 1 kg and gets struck in it. Calculate the velocity the sand bag acquires.
Solution:
Given:
Mass of bullet m_{1} = 50 g
= $\frac{50}{1000}$
= 0.050 kg
Velocity of bullet v1 = 200 ms1
Therefore, Linear momentum of bullet = m_{1}v_{1}
= (0.050 kg) $\times$ (200 ms^{1})
= 10 kg ms^{1}
As the bullet gets struck into the sand bag (mass M = (0.050 + 1)kg = 1.050 kg) the linear momentum after collision = 1.05 $\times$ v_{2} kg ms^{1}
By law of conservation of momentum, linear momentum before collision = Momentum after collision
10 kgms^{1 }= 1.050 v_{2} kgms^{1}
Final Velocity v_{2} = $\frac{10\ kgms^{1}}{1.050 kg}$
= 9.52 ms^{1}
Solution:
Given:
Mass of bullet m_{1} = 50 g
= $\frac{50}{1000}$
= 0.050 kg
Velocity of bullet v1 = 200 ms1
Therefore, Linear momentum of bullet = m_{1}v_{1}
= (0.050 kg) $\times$ (200 ms^{1})
= 10 kg ms^{1}
As the bullet gets struck into the sand bag (mass M = (0.050 + 1)kg = 1.050 kg) the linear momentum after collision = 1.05 $\times$ v_{2} kg ms^{1}
By law of conservation of momentum, linear momentum before collision = Momentum after collision
10 kgms^{1 }= 1.050 v_{2} kgms^{1}
Final Velocity v_{2} = $\frac{10\ kgms^{1}}{1.050 kg}$
= 9.52 ms^{1}