Law of Conservation of Momentum: Definition and Formula with Worked Example

Let’s get started with the theoretical bit on the conservation of momentum law, first. 

Your NCERT book discusses this principle of momentum conservation in section 4.7. 

“Law of conservation of momentum: The total momentum of an isolated system of interacting particles is conserved.”

Simple Explanation -

When there is any collision or separation and there is no external push or pull (force) in a uniform space (isolated system), the total quantity of motion before the event must be exactly equal to the total quantity of motion after. The quantity of motion is simply mass multiplied by velocity.  

Core Idea in Conservation of Momentum

Why does this law hold? 

• It comes directly from Newton’s Second Law, and more directly, from the Third Law of Motion. 
• For every internal force one body exerts on another (like Block A pushing Block B), there is an equal and opposite reaction force (Block B pushing Block A). 
• Within the isolated system, these force pairs perfectly cancel each other out. So the total momentum does not change.

For your tests and annual exams, knowing the conservation of momentum formula with the law should be enough. You can score full marks, especially when you practice the numerical problems.

But, for competitive exams such as JEE Main, you must know when and how to apply it. Numerical problems on the conservation of momentum often come in combination with other principles that you learn in later chapters, including the work-energy theorem. 

Here is a table you can take note of while approaching these exams.

Aspect	CBSE Board (Class XI)	JEE Main & Advanced
Question Style	Define the principle. Solve simple 1-D collisions, mostly.	Multi-step, multi-body collisions with 2-D problems. Explosion or recoil scenarios.
Depth Required	Emphasise the formal statement and a single example, such as colliding carts.	Link momentum with energy conservation, or solve simultaneous equations quickly and accurately.
Misconceptions to avoid	Forgetting to check for external forces like friction.	Mixing up sign conventions for vector components in 2-D problems.

 

Find some key concepts related to the Law of Conservation of Momentum that are essential before approaching the practice problems below. 

Momentum is a Vector

Momentum, often represented as  $\vec{p}$ in equations, is a vector quantity. 
Mathematically, we can define it as the product of mass (a scalar) and velocity (a vector), which, as you may recall that this operation is a scalar multiplication. So, you can show that as

$\vec{p}=m\vec{v}$

Here $m$ is the mass (kg) and $\vec{v}$ is the velocity (m/s).

System Momentum: The total momentum of a system is the vector sum of individual momenta.

${\vec{p}}_{\text{total}}=\sum m_i{\vec{v}}_i$

For conservation, ${\vec{p}}_{\text{total}}$ before an event equals ${\vec{p}}_{\text{total}}$ after.

The Golden Rule: Isolated Systems

An isolated system is one where the net external force is zero. We can represent this as a condition

$\sum {\vec{F}}_{\text{ext}}=0$ .

The law of conservation of momentum applies only to isolated systems. 

Link to Newton's Laws

Conservation of momentum comes directly from Newton's Third Law.

For two objects A and B, the force ${\vec{F}}_{AB}$ (A on B) equals $-{\vec{F}}_{BA}$ (B on A).

Since force is the rate of change of momentum: ${\vec{F}}_{AB}=-{\vec{F}}_{BA}=\Rightarrow \frac{d{\vec{p}}_A}{dt}=-\frac{d{\vec{p}}_B}{dt}$

Thus, $\frac{d}{dt}\left({\vec{p}}_A+{\vec{p}}_B\right)=0$

This tells us that the total momentum is constant. 

The conservation of momentum formula tells us that the total momentum before a collision is the same as the total momentum after the collision. 

We can also say that the overall momentum of an isolated system remains constant before and after a collision. Just remember that no external forces are acting on the system. 

The Conservation of Momentum Formula is

m₁u₁ + m₂u₂ = m₁v₁' + m₂v₂'

Find the variables below. 

Masses

• m₁ represents the mass of the first object
• m₂ represents the mass of the second object
• These values remain constant throughout the collision

Initial Velocities (before interaction/collision)

• u₁ is the initial velocity of object 1 before the collision
• u₂ is the initial velocity of object 2 before the collision
• These velocities can be positive, negative, or zero. That would depend on the direction of motion

Final Velocities (after interaction/collision)

• v₁' is the final velocity of object 1 after the collision
• v₂' is the final velocity of object 2 after the collision

 

 

 

 

 

It’s better to have a grounded knowledge of the law of conservation of momentum when approaching the competitive exams for engineering. 

You can refer to our article on collisions before reading below. 

Elastic Collision and Momentum

Both momentum and kinetic energy are conserved. This is an ideal case, closely approximated by the collision of hard spheres like billiard balls.

Momentum Conservation here looks like this. 

m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f

Note: The i/f subscript system is commonly used. This is because it clearly highlights the difference between the initial and final states. 

Kinetic Energy Conservation, then, would be 

½m₁v₁ᵢ² + ½m₂v₂ᵢ² = ½m₁v₁f² + ½m₂v₂f²

Inelastic Collision and Momentum

Only momentum is conserved. Some kinetic energy is lost, usually converted into heat, sound, or deformation of the objects. 

Just remember that any real-world collision that doesn't stick together is inelastic.

Momentum Conservation here would be seen as

m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f

And, the Kinetic Energy relationship would be 

½m₁v₁ᵢ² + ½m₂v₂ᵢ² > ½m₁v₁f² + ½m₂v₂f²

Perfectly Inelastic Collision

This is a special case of an inelastic collision, where the objects stick together after impact and move with a common final velocity. 

This is where the maximum possible kinetic energy is lost.

Momentum Conservation

m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)vf

Variables used in the above equations

m₁, m₂ = masses of objects 1 and 2

v₁ᵢ, v₂ᵢ = initial velocities of objects 1 and 2

v₁f, v₂f = final velocities of objects 1 and 2

vf = common final velocity (for perfectly inelastic collisions)

Subscript "i" = initial

Subscript "f" = final

Formulas related to the law of conservation of momentum you can remember for exams.   

• Momentum: $\vec{p}=m\vec{v}$ .
• Conservation of Momentum Formula: $\sum m_i{\vec{v}}_i=\sum m_i{\vec{v}}_i^{\text{'}}$ (before = after).
• For a two-body collision: $m_1{\vec{v}}_1+m_2{\vec{v}}_2=m_1{\vec{v}}_1^{\text{'}}+m_2{\vec{v}}_2^{\text{'}}$ .
• Perfectly inelastic collision: $m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2}=\left(m_1+m_2\right)\overrightarrow{v^{\text{'}}}$ .

 

Problem: A 2 kg block moving at +4 m/s on a frictionless surface collides with a 3 kg block moving at –2 m/s. They stick together after the collision. Find their final velocity.

 

Solution:

You can draw a Collision Diagram before and after.

Before the collision, we had these parameters.

+4 m/s → -2 m/s ←

┌─────┐ ┌─────┐

│ 2kg │ → │ 3kg │

└─────┘ └─────┘

AFTER COLLISION:

+0.4 m/s →

┌─────────────┐

│ 5 kg │

└─────────────┘

→ = Positive direction (right)

Free Body Diagrams

Block 1 of 2 kg

↑ N₁

│

┌─────┐

│ 2kg │

└─────┘

│

↓ mg₁

Block 2 of 3 kg:

↑ N₂

│

┌─────┐

│ 3kg │

└─────┘

│

↓ mg₂

Step 1: Define the System and check for isolation

System: The two blocks (2 kg + 3 kg)

External Forces:

Gravity (mg₁, mg₂) and Normal Forces (N₁, N₂) are vertical. They also cancel out.

So, friction = zero (frictionless surface)

There are no horizontal external forces

Conclusion: The system is isolated in the horizontal direction. Momentum is thus conserved.

Step 2: Here, calculate the Initial Momentum

Define the positive direction, that is, to the right

pᵢ = (m₁v₁ᵢ) + (m₂v₂ᵢ)

pᵢ = (2 kg)(+4 m/s) + (3 kg)(–2 m/s) = 8 – 6 = +2 kg·m/s

Step 3: Set up the Final State

After sticking together, which is a perfectly inelastic collision, the blocks form a single object.

So,

Combined Mass (M) = 2 kg + 3 kg = 5 kg

By letting the final velocity be vf

pf = Mvf = 5vf

Step 4: Apply the law of conservation of mmomentum

pᵢ = pf

2 kg·m/s = 5vf

vf = 2/5 = +0.4 m/s

Interpretation: The combined mass moves to the right (positive direction) at 0.4 m/s.