Derivation of Law of Conservation of Momentum: Formula and Derivation

The law of conservation of momentum is a fundamental principle in classical mechanics that states the total momentum of a closed system remains constant if no external forces act on it. Momentum, the product of an object's mass and velocity, is a vector quantity, meaning it has both magnitude and direction. The law is derived from Newton’s laws of motion, particularly the second and third laws, and is widely applicable in systems ranging from microscopic particles to celestial bodies.

This Story also Contains

1. Law of Conservation of Momentum
2. Derive the Mathematical Formula of Conservation of Momentum
3. Derivation of Momentum or Law of Conservation of Momentum in One-Dimensional
4. Derivation of the Law of Conservation of Momentum in Two-Dimensional
5. Examples of Conservation of Momentum
6. Exam-wise Weightage For the Derivation of the Law of Conservation of Momentum

This principle plays a key role in understanding collisions and interactions between objects, making it essential in fields like physics, engineering, and even astrophysics. In this derivation, we will explore how the conservation of momentum arises naturally from Newton’s laws and illustrate its application through various scenarios.

Law of Conservation of Momentum

If there is no external force acting on an isolated system, its overall momentum remains constant. As a result, if a system's total linear momentum remains constant, the resultant force exerted on it is zero. In the absence of external torque, angular momentum is conserved as well. The law of conservation of momentum is derived from Newton's third law of motion.

Conservation of momentum states that the momentum of the system is always conserved, i.e., the initial momentum and final momentum of the system are always conserved. We can also say that the total momentum of the system is always constant. The product of an object's velocity and mass is the object's momentum. It's a quantity with a vector. The overall momentum of an isolated system is conserved, according to conservation of momentum, a fundamental law of physics. In other words, if no external force acts on a system of objects, their overall momentum remains constant during any interaction. The vector sum of individual momentum is the overall momentum. In any physical process, momentum is conserved.

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Derive the Mathematical Formula of Conservation of Momentum

Newton's third law states that when object A produces a force on object B, object B responds with a force of the same magnitude but opposite direction. Newton derived the mathematical formula for the conservation of momentum.

Let us consider two moving balls, $A$ and $B$ of masses $m_1$ and $m_2$ and having initial velocities $u_1$ and $u_2$ such that $u_2<u_1$.

Suppose the balls collide at some point, and there is no external force acting on this system.

Let their final velocities be $v_1$ and $v_2$ respectively.
According to Newton's third law of motion,
Force on ball $B$ due to $A=-$ Force on ball $A$ due to $B$.

Or, $F_{A B}=-F_{B A} \ldots \ldots \ldots \ldots \ldots \ldots(i)$

Total initial momentum before collision $\left(p_i\right)=m_1 u_1+m_2 u_2$.

Total final momentum after collision $\left(p_f\right)=m_1 v_1+m_2 v_2$.
According to Newton's second law,

$F_{B A}=\frac{p_A^{\prime}-p_A}{t}=\frac{m_1 v_1-m_1 u_1}{t} \ldots \ldots \ldots(i i)$

$F_{A B}=\frac{p_B^{\prime}-p_B}{t}=\frac{m_2 v_2-m_2 u_2}{t} \ldots \ldots \ldots(i i i)$

From $(i),(i i)$ and $(i i i)$,

$
\begin{aligned}
& \frac{m_1 v_1-m_1 u_1}{t}=-\frac{m_2 v_2-m_2 u_2}{t} \\
& \Rightarrow m_1 v_1-m_1 u_1=-\left(m_2 v_2-m_2 u_2\right) \\
& \Rightarrow m_1 v_1+m_2 v_2=m_1 u_1+m_2 u_2 \\
& \Rightarrow \text { Final momentum }\left(p_f\right)=\text { Initial momentum }\left(p_i\right)
\end{aligned}
$

This is called the law of conservation of momentum.
As a result, the equation of the law of conservation of momentum is as follows: $m_1 u_1+m_2 u_2$ represents the total momentum of particles A and B before the collision, and $m_1 v_1+m_2 v_2$ represents the total momentum of particles A and B after the collision.

Derivation of Momentum or Law of Conservation of Momentum in One-Dimensional

A one-dimensional collision of two objects can be used to explain momentum conservation. Two objects with masses of $m_1$ and $m_2$ collide while moving in a straight line at velocities of $u_1$ and $u_2$, respectively. They gain velocities $v_1$ and $v_2$ in the same direction after colliding.

Before the impact, the total momentum

$$
\mathrm{p}_{\mathrm{i}}=\mathrm{m}_1 \mathrm{u}_1+\mathrm{m}_2 \mathrm{u}_2
$$
After the impact, the total momentum

$$
\mathrm{p}_{\mathrm{f}}=\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2
$$
Total momentum is conserved if no other force acts on the system of two objects. Therefore,

$$
\begin{aligned}
& \mathrm{p}_{\mathrm{i}}=\mathrm{p}_{\mathrm{f}} \\
& \mathrm{m}_1 \mathrm{u}_1+\mathrm{m}_2 \mathrm{u}_2=\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2
\end{aligned}
$$

NCERT Physics Notes:

• NCERT Notes class 11 physics
• NCERT Notes class 12 physics
• NCERT Notes for all subject

Derivation of the Law of Conservation of Momentum in Two-Dimensional

The overall momentum is $p_{i x}=p_1=m_1 u_1$ along the X-axis and $p_{\mathrm{iy}}=\mathrm{m}_2 \mathrm{u}_2$ along the Y-axis before the collision. The overall momentum after the collision is $\mathrm{p}_{\mathrm{fx}}=(\mathrm{m}+\mathrm{M}) \mathrm{ucos} \theta$ along the X-axis and $\mathrm{p}_{\mathrm{fy}}=(\mathrm{m}+\mathrm{M}) \mathrm{usin} \theta$ along the Y-axis, where $m$ is mass and $(m+M)$ is the resultant mass when particles get trapped inside it

where $m$ is mass and $(m+M)$ is the resultant mass when particles get trapped inside it.

$
\begin{aligned}
& \mathrm{p}_{\mathrm{ix}}=\mathrm{P}_{\mathrm{fx}} \\
& \mathrm{m}_1 \mathrm{v}_1=(\mathrm{m}+\mathrm{M}) \mathrm{ucos} \theta(1)
\end{aligned}
$
$
m_2 v_2=(m+M) u \sin \theta(2)
$
As a result of squaring and adding equations (1) and (2),

$
\begin{aligned}
& \left(\mathrm{m}_1 \mathrm{v}_1\right)^2+\left(\mathrm{m}_2 \mathrm{v}_2\right)^2=(\mathrm{m}+\mathrm{M})^2 \mathrm{u}^2\left(\cos ^2 \theta+\sin ^2 \theta\right) \\
& u=\frac{\sqrt{m_1^2 v_1^2+m_2^2 v_2^2}}{m+M}
\end{aligned}
$

It is the combined object's speed.

$
\tan \theta=\frac{m_2 v_2}{m_1 v_1}
$
This determines the direction of the velocity.

Examples of Conservation of Momentum

There are several examples that make the explanation for the law of conservation of momentum. Some of the most common examples of the Conservation of Momentum are

1. Recoil of a Gun

• When a bullet is fired from a gun, the bullet moves forward, and the gun recoils backward.
• The forward momentum of the bullet is equal and opposite to the backward momentum of the gun, keeping the total momentum conserved.

2. Collision of Billiard Balls

• When one billiard ball strikes another, the momentum of the moving ball is transferred to the second ball.
• The total momentum of both balls before and after the collision remains constant.

3. Rocket Propulsion

• A rocket expels gas backward at high speed, and as a result, the rocket moves forward.
• The backward momentum of the expelled gas balances the forward momentum of the rocket, ensuring the total momentum of the system is conserved.

4. Bomb exploding

Before exploding, the bomb is at rest, so its initial momentum is zero because v = 0. During an explosion, the bomb breaks into several fragments, which fly off in different directions and gain some momentum based on their masses and velocity. After the explosion, even though the individual fragments have their own momentum, the vector sum of all their momenta is zero.

5. A satellite and an astronaut

The satellite and the astronaut are in the same system. Initially, the astronaut is at rest. So, initially the momentum = 0.

Let the astronaut push against the satellite to move away, gaining momentum in one direction, while the satellite gains momentum in another direction (this is because the satellite and astronaut are in the same system, and push is an internal force).

$\begin{aligned} & m_a=\text { mass of astronaut } \\ & v_a=\text { velocity of astronaut after push } \\ & m_s=\text { mass of satellite } \\ & v_s=\text { velocity of satellite after push }\end{aligned}$

Since initial momentum = 0

$m_a v_a+m_s v_s=0 \Rightarrow m_a v_a=-m_s v_s$

They have the same product of mass and velocity but move in opposite directions. Hence, the momentum is Conserved.

Exam-wise Weightage For the Derivation of the Law of Conservation of Momentum

Apart from the Derivation of the Law of Conservation of Momentum in class 9, we study this formula in our higher class, such as class 11, and this is very important for competitive exams like JEE and NEET. Given the below table is the exam-wise weightage of this concept.