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Conservation of Linear Momentum


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.

What is Linear Momentum?

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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).

Principle of Conservation of Linear Momentum

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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 p1, p2, p3,.... are the momentum in a closed system then principle of conservation of momentum states that the total momentum is constant.
p1 + p2 + p3 + p4 + ....... = constant.

Conservation of Linear Momentum Equation

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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 $\frac{d \vec{v}}{dt}$ = $\frac{d\vec{p}}{dt}$
A special case is when $\vec{F}$ = $\vec{0}$ => $\vec{p}$ is constant.

Let us consider the system of particles then the momentum for the system of particles is given as
$\vec{P}$ = $\sum_i$ $\vec{p_i}$ = $\sum_i$ mi $\vec{v_i}$

To get the linear momentum of individual particles we go for the larger momentum $\vec{P}$ in a many particle system.
$\frac{d\vec{P}}{dt}$ = $\sum_i$ $\frac{d \vec{p}}{dt}$ = $\sum_i$ Fi = $\vec{F_{net}}$
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.

Law of Conservation of Linear Momentum

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Let us consider two bodies of masses m1 and m2 moving towards each other having initial velocities u1 and u2. Both collide with each other for time t. After the collision, they move with the velocities v1 and v2.

Conservation of Momentum

From newtons second law, Force applied by A on B is given by

FAB = $\frac{m_2 v_2 - m_2 u_2}{t}$

Similarly, Force applied by B on A
FBA = $\frac{m_1 v_1 - m_1 u_1}{t}$

From Newton’s third law we have

$\frac{m_2v_2 - m_2u_2}{t}$ = $\frac{-(m_1 v_1 - m_1u_1)}{t}$


m2v2-m2u2 = -m1v1+m1u1


m1u1 + m2u2 = m1v1 + m2v2

This means the total momentum before collision is equal to total momentum after collision.

Conservation of Linear Momentum Example

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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.

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.
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.
Mass of bullet m1 = 50 g
                         = $\frac{50}{1000}$
                         = 0.050 kg

Velocity of bullet v1 = 200 ms-1

Therefore, Linear momentum of bullet = m1v1
                               = (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 $\times$ v2 kg ms-1

By law of conservation of momentum, linear momentum before collision = Momentum after collision
10 kgms-1 = 1.050 v2 kgms-1

Final Velocity v2 = $\frac{10\ kgms^{-1}}{1.050 kg}$
                        = 9.52 ms-1