Conservation of Linear Momentum: Complete Notes for BSc 1st Year Physics

Introduction

Linear momentum and its conservation are among the most powerful ideas in all of physics. They don't just help you solve textbook problems — they explain why guns kick backward when fired, how rockets travel through the vacuum of space with nothing to push against, why crumple zones in cars save lives, and how atomic nuclei split and fly apart in predictable ways.

The law of conservation of linear momentum is a direct consequence of Newton's Third Law and is one of the most universally applicable principles in classical mechanics. Unlike energy, which can be converted into heat and "lost" from a mechanical system, momentum is conserved in every interaction between objects, provided no external force acts on the system. This makes it especially useful in collision and explosion problems where the forces between objects are complex but the momentum bookkeeping is always reliable.

These notes are written for BSc 1st Year Physics students preparing for university examinations. Every concept is explained from definition to derivation to application, with fully worked numerical examples, a formula quick-reference, exam-style important questions, and FAQs at the end.

1. What Is Linear Momentum?

Linear momentum is a vector quantity defined as the product of an object's mass and its velocity:

p = mv

where:

p = linear momentum (vector, same direction as velocity)
m = mass of the object (kg)
v = velocity of the object (m/s)

SI Unit: kg·m/s (kilogram metre per second)

Because momentum is a vector, both its magnitude and direction matter. An object moving at 10 m/s to the right and another moving at 10 m/s to the left have momenta of equal magnitude but opposite sign — they do not cancel unless you're adding them together in a system.

Key points:

Doubling mass at constant velocity → momentum doubles
Doubling velocity at constant mass → momentum doubles
A stationary object (v = 0) has zero momentum regardless of mass

Momentum vs. kinetic energy: Both depend on mass and velocity, but momentum (mv) is a vector while kinetic energy (½mv²) is a scalar. In collisions, you track momentum as a vector and energy as a scalar — they behave differently.

2. Impulse and the Impulse-Momentum Theorem

Before stating the conservation law, it's worth understanding how momentum changes — and that's through impulse.

Newton's Second Law in its most general form says:

F = dp/dt (Force equals rate of change of momentum)

For a constant force over a time interval Δt:

F × Δt = Δp = m(v − u)

This product F × Δt is called impulse (J):

J = F × Δt = change in momentum

Impulse-Momentum Theorem: The impulse applied to an object equals the change in its momentum.

Why impulse matters: It explains why:

A cricket ball hit with a bat gets such high velocity — large force over a short time
Car airbags and crumple zones reduce injury — they increase collision time (Δt), which reduces the force (F) needed to produce the same change in momentum
A skilled martial artist "goes with" a punch — increasing contact time reduces impact force

3. Law of Conservation of Linear Momentum

Statement

"If no net external force acts on a system of particles, the total linear momentum of the system remains constant (conserved), regardless of the internal forces between the particles."

Mathematically, for a system:

p_initial = p_final

m₁u₁ + m₂u₂ + ... = m₁v₁ + m₂v₂ + ...

where u = initial velocity, v = final velocity.

Derivation from Newton's Laws

Consider two bodies A (mass m₁, initial velocity u₁) and B (mass m₂, initial velocity u₂) that interact for a short time t. Their final velocities are v₁ and v₂.

Step 1: By Newton's Third Law, the force A exerts on B is equal and opposite to the force B exerts on A:

F_AB = −F_BA

Step 2: By Newton's Second Law, each force equals rate of change of momentum. For body A:

F_BA = m₁(v₁ − u₁) / t

For body B:

F_AB = m₂(v₂ − u₂) / t

Step 3: Substituting into the Third Law relation:

m₁(v₁ − u₁) / t = −m₂(v₂ − u₂) / t

Step 4: Multiply both sides by t:

m₁v₁ − m₁u₁ = −m₂v₂ + m₂u₂

Step 5: Rearrange:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

This proves that total momentum before interaction = total momentum after interaction. The derivation works for any pair of interacting bodies, and it extends by the same argument to any number of particles in an isolated system.

4. Conditions for Momentum Conservation

Momentum is conserved when:

No net external force acts on the system. Internal forces (between particles within the system) do not change the total momentum. External forces do. A "system" must be defined carefully — if you include everything that exerts force, momentum is conserved.
The system is isolated. Practically, this means the time of interaction is short enough that external forces (like gravity or friction) produce negligible impulse during the interaction. This is why collision problems use conservation of momentum even though gravity and friction technically exist.
The chosen reference frame is inertial (non-accelerating).

5. Types of Collisions

Collisions are the most common application of conservation of momentum. They're classified by what happens to kinetic energy:

Elastic Collision

Both momentum and kinetic energy are conserved.

Characteristics:

No deformation of the objects
No energy lost to heat or sound
Practically: billiard balls, elastic collision of gas molecules

Equations for a 1D elastic collision (m₂ at rest initially, u₂ = 0):

After collision:

v₁ = u₁(m₁ − m₂) / (m₁ + m₂)
v₂ = 2m₁u₁ / (m₁ + m₂)

Special cases:

Equal masses (m₁ = m₂): v₁ = 0, v₂ = u₁ — the first object stops and the second moves at the first's initial speed
m₁ >> m₂: v₁ ≈ u₁ (heavy object barely slows), v₂ ≈ 2u₁
m₁ << m₂: v₁ ≈ −u₁ (light object bounces back), v₂ ≈ 0

Perfectly Inelastic Collision

The objects stick together after collision. Maximum kinetic energy is lost.

Momentum is conserved; kinetic energy is not.

Equation:

m₁u₁ + m₂u₂ = (m₁ + m₂)v

v = (m₁u₁ + m₂u₂) / (m₁ + m₂)

Inelastic Collision

Momentum is conserved but kinetic energy is partially lost. Objects don't stick together but deform. Most real-world collisions are inelastic.

Comparison Table

Aspect	Elastic	Perfectly Inelastic	Inelastic
Momentum conserved?	✅ Yes	✅ Yes	✅ Yes
Kinetic energy conserved?	✅ Yes	❌ No (maximum loss)	❌ No (partial loss)
Objects stick together?	No	Yes	No
Real-life example	Billiard balls	Clay catching a ball	Car crash

6. Applications of Conservation of Momentum

Recoil of a Gun

When a gun fires a bullet, the system (gun + bullet) initially has zero momentum (both at rest). After firing:

0 = mv + MV

V = −mv / M

where m = bullet mass, v = bullet velocity, M = gun mass, V = recoil velocity.

The negative sign means the gun moves opposite to the bullet. A heavier gun recoils less than a lighter one. The muzzle velocity of competition rifles is controlled partly by gun weight — heavier rifles reduce felt recoil.

Rocket Propulsion

A rocket in space has no air to push against. It works entirely on conservation of momentum: expelling exhaust gases backward at high speed creates a forward reaction force on the rocket. No ground, no air, no problem — Newton's Third Law and momentum conservation are enough.

v_rocket × M_rocket = v_exhaust × m_exhaust (in magnitude, opposite directions)

This is why rocket fuel mass is such a critical engineering constraint — you need enormous fuel mass expelled at high velocity to accelerate a relatively small payload.

Explosion from Rest

An object at rest (p = 0) explodes into two or more pieces. Total momentum must still be zero:

0 = m₁v₁ + m₂v₂ + ...

The pieces fly in opposite directions with momenta that cancel. Heavier fragments move slower; lighter fragments move faster.

Jumping from a Boat

When you jump forward from a stationary boat, you push backward on the boat (and the boat pushes you forward). Total momentum = 0 before and after. The boat moves backward as you jump forward.

7. Conservation of Momentum in Two Dimensions

When a collision is not head-on (a "glancing collision"), the momentum must be conserved separately in each direction.

X-direction: m₁u₁ₓ + m₂u₂ₓ = m₁v₁ₓ + m₂v₂ₓ

Y-direction: m₁u₁ᵧ + m₂u₂ᵧ = m₁v₁ᵧ + m₂v₂ᵧ

Velocities must be resolved into components before applying conservation. This often requires trigonometry (sinθ and cosθ components) and the Pythagorean theorem to find the resultant final velocity.

8. Solved Numerical Problems

Problem 1 — Gun Recoil

A rifle of mass 5 kg fires a bullet of mass 0.05 kg at a velocity of 300 m/s. Find the recoil velocity of the rifle.

Given: m = 0.05 kg, v = 300 m/s, M = 5 kg

Initial momentum = 0 (both at rest)

Using conservation: 0 = mv + MV

0 = (0.05 × 300) + (5 × V)

0 = 15 + 5V

V = −15/5 = −3 m/s

The rifle recoils at 3 m/s opposite to the bullet's direction.

Problem 2 — Perfectly Inelastic Collision

A 2 kg ball moving at 5 m/s collides with a stationary 3 kg ball and they stick together. Find their common velocity after collision.

Given: m₁ = 2 kg, u₁ = 5 m/s, m₂ = 3 kg, u₂ = 0

Using: m₁u₁ + m₂u₂ = (m₁ + m₂)v

(2 × 5) + (3 × 0) = (2 + 3) × v

10 = 5v

v = 2 m/s (in the original direction of motion)

Problem 3 — Explosion from Rest

A shell of mass 20 kg at rest explodes into two pieces of mass 8 kg and 12 kg. If the 8 kg piece moves at 30 m/s, find the velocity of the 12 kg piece.

Initial momentum = 0 (at rest)

Conservation: 0 = m₁v₁ + m₂v₂

0 = (8 × 30) + (12 × v₂)

0 = 240 + 12v₂

v₂ = −240/12 = −20 m/s

The 12 kg piece moves at 20 m/s in the opposite direction to the 8 kg piece.

Problem 4 — Kinetic Energy Lost in Collision

A 4 kg object moving at 6 m/s collides with a stationary 2 kg object and they stick together. Find: (a) final velocity, (b) kinetic energy lost.

Given: m₁ = 4 kg, u₁ = 6 m/s, m₂ = 2 kg, u₂ = 0

(a) v = (m₁u₁) / (m₁ + m₂) = (4 × 6) / (4 + 2) = 24/6 = 4 m/s

(b) KE_initial = ½m₁u₁² = ½ × 4 × 36 = 72 J

KE_final = ½(m₁ + m₂)v² = ½ × 6 × 16 = 48 J

KE lost = 72 − 48 = 24 J

This energy was converted to heat, sound, and deformation during the collision.

Problem 5 — Impulse and Momentum Change

A cricket ball of mass 0.15 kg moving at 40 m/s is struck by a bat and returns at 50 m/s in the opposite direction. The contact time is 0.01 s. Find: (a) the change in momentum, (b) the average force exerted by the bat.

Taking the original direction as positive:

Initial momentum: p₁ = 0.15 × 40 = 6 kg·m/s

Final momentum: p₂ = 0.15 × (−50) = −7.5 kg·m/s

(a) Change in momentum = p₂ − p₁ = −7.5 − 6 = −13.5 kg·m/s

The magnitude is 13.5 kg·m/s in the direction of the bat's hit.

(b) F = Δp / Δt = 13.5 / 0.01 = 1,350 N

The bat exerts an average force of 1,350 N on the ball during contact.

9. Formula Quick-Reference

Concept	Formula
Linear momentum	p = mv
Impulse	J = F × Δt
Impulse-Momentum Theorem	F × Δt = m(v − u)
Conservation of momentum (2 bodies)	m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Perfectly inelastic collision	(m₁ + m₂)v = m₁u₁ + m₂u₂
Gun recoil	V = −mv/M
Elastic collision (m₂ at rest)	v₁ = u₁(m₁−m₂)/(m₁+m₂); v₂ = 2m₁u₁/(m₁+m₂)
KE lost in collision	ΔKE = KE_initial − KE_final
2D conservation (x)	m₁u₁cosθ₁ + m₂u₂cosθ₂ = m₁v₁cosφ₁ + m₂v₂cosφ₂

10. Important Exam Questions

Long Answer (8–10 Marks)

State and prove the law of conservation of linear momentum from Newton's Laws.
What is impulse? State and explain the impulse-momentum theorem with applications.
Distinguish between elastic and inelastic collisions. Derive expressions for velocities after a one-dimensional elastic collision.
Explain the recoil of a gun using conservation of momentum. A gun of mass 10 kg fires a bullet of 50 g at 500 m/s — find the recoil velocity.
Explain rocket propulsion using the principle of conservation of momentum.
How is momentum conserved in a two-dimensional collision? Explain with equations.

Short Answer (3–5 Marks)

State the law of conservation of linear momentum. Under what conditions does it hold?
Define impulse and state its SI unit.
What is a perfectly inelastic collision? Give one example.
Differentiate between elastic and inelastic collisions.
A body at rest explodes into two pieces. Explain how momentum is conserved.
What is the vector nature of momentum? Why does it matter in 2D problems?

Numerical Questions

A 3 kg ball moving at 8 m/s collides with a 5 kg ball at rest. They stick together. Find final velocity.
A 0.1 kg bullet is fired at 600 m/s from a 4 kg gun. Find recoil velocity.
A 10 kg object explodes into 6 kg and 4 kg pieces. The 6 kg piece moves at 5 m/s. Find the speed of the 4 kg piece.

Key Points Summary

Momentum p = mv is a vector with SI unit kg·m/s
Total momentum is conserved in an isolated system (no net external force)
Derived directly from Newton's Second and Third Laws
Elastic collision: both momentum and KE conserved
Perfectly inelastic: momentum conserved, maximum KE lost, objects stick together
Impulse = F × Δt = change in momentum
In 2D problems, apply conservation separately in x and y directions
Applications: gun recoil, rocket propulsion, explosions, all collision types

Frequently Asked Questions

Is momentum always conserved? Momentum is conserved whenever the net external force on the system is zero. In most collision problems, even though gravity and friction exist, the collision time is so short that these forces produce negligible impulse — making momentum conservation accurate enough to use.

Why is kinetic energy not always conserved in collisions? KE can be converted to other energy forms — heat from deformation, sound from impact, internal energy from permanent deformation. Momentum has no equivalent "loss" mechanism because it's tied to Newton's Third Law, not to energy conversion.

Can a system's momentum be conserved even if individual objects change speed? Yes. Individual objects' momenta change due to forces between them (internal forces). But these internal forces come in action-reaction pairs that cancel in the total — so the system's total momentum is unchanged. Only external forces change total system momentum.

What is the difference between impulse and force? Force is instantaneous (N). Impulse is force integrated over time (N·s = kg·m/s). A small force over a long time can produce the same impulse (same momentum change) as a large force over a short time. Airbags work by increasing the time of collision, reducing the peak force for the same total momentum change.

Why does rocket propulsion work in space where there's nothing to push against? Rockets don't push against the medium — they push against their own exhaust gases. When exhaust is expelled backward, the reaction force pushes the rocket forward. Newton's Third Law and momentum conservation work in a vacuum just as well as in air.

In a gun-bullet system, who experiences more force — the gun or the bullet? Both experience the same force magnitude (Newton's Third Law — equal and opposite). But the bullet accelerates much more because it has much less mass (a = F/m). The gun accelerates less because of its much greater mass.

Conclusion

The conservation of linear momentum is one of physics' most reliable tools precisely because it doesn't depend on the complexity of the forces involved — only on whether those forces are internal to the system. A collision between two asteroids, an explosion in a laboratory, a rocket in deep space, and a subatomic particle decay all follow exactly the same principle: in the absence of external forces, total momentum is constant.

For exam preparation: master the derivation from Newton's Laws (it's frequently asked), practice resolving 2D problems into x and y components, and always check your sign conventions — momentum is a vector, and getting directions wrong is the most common error in collision numericals.