What Is Momentum?

Momentum is, at its simplest, a measure of how much "motion" an object has — and how difficult it is to stop. Formally, the linear momentum of an object is defined as:

p = mv
Momentum (kg·m/s) = mass (kg) × velocity (m/s). Momentum is a vector — it has direction.

A 1,500 kg car travelling at 20 m/s has the same momentum as a 30 kg person running at 1,000 m/s — both have p = 30,000 kg·m/s. What's remarkable isn't the definition, but what happens when objects interact: the total momentum of an isolated system never changes.

This is the conservation of momentum — one of the most powerful and universally applicable principles in all of physics. It applies to billiard balls, subatomic particles, rocket engines, supernovae, and galaxy collisions. Wherever there's an interaction, momentum tells you what the outcome must be.

The Conservation Law

In an isolated system (no external forces), the total momentum before any interaction equals the total momentum after: p_total(before) = p_total(after).

Why Is Momentum Conserved? The Derivation from Newton's Third Law

Conservation of momentum isn't a separate postulate — it follows directly from Newton's Third Law: "for every action, there is an equal and opposite reaction." This derivation is one of the most elegant in classical physics, and it's worth going through carefully.

Consider two objects A and B that interact (collide, push, attract each other). By Newton's Third Law, the force that A exerts on B is equal and opposite to the force that B exerts on A:

F_AB = −F_BA
Newton's Third Law: action-reaction pair, equal in magnitude, opposite in direction.

By Newton's Second Law, force equals rate of change of momentum: F = dp/dt. So:

dp_B/dt = −dp_A/dt  ⟹  d(p_A + p_B)/dt = 0
The total momentum of the system doesn't change with time — it's conserved.

If the total rate of change of momentum is zero, then total momentum is constant. That's it. The conservation of momentum is Newton's Third Law in integral form. They're the same statement about physics expressed differently.

This connection also reveals when conservation of momentum breaks down: when there are external forces acting on the system. A ball rolling on the ground isn't isolated — friction is an external force, so its momentum decreases. The system you need to consider must include everything that exerts a force on anything inside it. Expand the boundary of your system until no external forces remain, and momentum is always conserved.

Impulse: The Bridge Between Force and Momentum Change

The impulse-momentum theorem connects the force acting on an object to the change in its momentum:

J = FΔt = Δp = m(v_f − v_i)
Impulse J (N·s) = Force × time interval = Change in momentum. Also a vector.

Impulse is why airbags save lives. A car crash changes your momentum by a fixed amount regardless of whether an airbag deploys. But Δp = FΔt: by increasing the time Δt over which the momentum change occurs (airbag extends the collision from ~10 ms to ~100 ms), the peak force F on your body decreases by roughly 10×. Same impulse, far smaller force, far less injury.

The same principle explains why martial artists break boards with bare hands: they aim to maximise force by minimising contact time (very hard, very fast strike), maximising F for a given impulse.

Insight: Average vs Peak Force

The impulse J = FΔt uses the average force over the collision interval. The peak force can be much higher. Engineers designing crumple zones, helmets, and packaging must consider the force-time profile, not just the impulse.

Types of Collisions: Elastic, Inelastic, and Perfectly Inelastic

Momentum is conserved in all collisions (for isolated systems). What distinguishes collision types is whether kinetic energy is also conserved:

Collision TypeMomentumKinetic EnergyReal-World Examples
ElasticConserved ✓Conserved ✓Billiard balls, atomic/molecular collisions, Newton's cradle (approximately)
InelasticConserved ✓Not conserved ✗Most macroscopic collisions — car crashes, dropped balls, bat hitting ball
Perfectly InelasticConserved ✓Maximum loss ✗Car crash where vehicles stick together, clay catching a ball, bullet embedding in block

In inelastic collisions, the "lost" kinetic energy doesn't disappear — it converts into heat, sound, and deformation (as required by energy conservation from the First Law of Thermodynamics). When a car crumples in a crash, the deformation absorbs kinetic energy deliberately — that's crash engineering working as designed.

Elastic Collision Equations (Two Objects)

For two objects with masses m₁, m₂ and initial velocities u₁, u₂, conserving both momentum and kinetic energy gives:

v₁ = (m₁-m₂)u₁ + 2m₂u₂ / (m₁+m₂)Final velocity of object 1
v₂ = (m₂-m₁)u₂ + 2m₁u₁ / (m₁+m₂)Final velocity of object 2

These look complicated, but check the special cases: if m₁ = m₂, then v₁ = u₂ and v₂ = u₁ — the objects swap velocities. This is exactly what billiard balls do when one strikes a stationary ball head-on.

Five Worked Examples

Example 1 — Basic

Perfectly inelastic collision — two carts

GivenCart A: m₁ = 2 kg, u₁ = 6 m/s (right)  |  Cart B: m₂ = 3 kg, u₂ = 0 m/s  |  They stick together.
01

Conservation of momentum: m₁u₁ + m₂u₂ = (m₁+m₂)v

02

2(6) + 3(0) = (2+3)v  →  12 = 5v  →  v = 2.4 m/s

03

KE before: ½(2)(36) = 36 J. KE after: ½(5)(5.76) = 14.4 J. Energy lost: 36 − 14.4 = 21.6 J (to heat/sound)

✓ Final velocity: 2.4 m/s (right)  |  KE lost: 21.6 J
Example 2 — Standard

Elastic collision — verify kinetic energy is conserved

Givenm₁ = 4 kg, u₁ = 5 m/s  |  m₂ = 2 kg, u₂ = −2 m/s (moving left)  |  Elastic collision.
01

Total initial momentum: p_i = 4(5) + 2(−2) = 20 − 4 = 16 kg·m/s

02

Using elastic collision formula: v₁ = [(4−2)(5) + 2(2)(−2)]/(4+2) = [10 − 8]/6 = 2/6 = 0.33 m/s

03

v₂ = [(2−4)(−2) + 2(4)(5)]/(6) = [4 + 40]/6 = 44/6 = 7.33 m/s

04

Check momentum: 4(0.33) + 2(7.33) = 1.33 + 14.67 = 16 ✓. Check KE: ½(4)(25)+½(2)(4) = 54 J before; ½(4)(0.109)+½(2)(53.7) = 0.22+53.7 ≈ 54 J ✓

✓ v₁ ≈ 0.33 m/s  |  v₂ ≈ 7.33 m/s  |  Both conservation laws verified.
Example 3 — Standard

Explosion — gun recoil

GivenRifle mass: M = 3.5 kg  |  Bullet mass: m = 0.015 kg  |  Bullet velocity: v_b = 900 m/s  |  System initially at rest.
01

Initial momentum = 0. By conservation: 0 = mv_b + MV_rifle

02

V_rifle = −mv_b / M = −(0.015 × 900)/3.5 = −13.5/3.5 = −3.86 m/s

03

The rifle recoils backward at 3.86 m/s — roughly 233 times slower than the bullet (as expected from the mass ratio).

✓ Rifle recoil velocity: 3.86 m/s (backward)
Example 4 — Harder

Bullet embeds in block — find initial speed

GivenBullet: m = 0.01 kg  |  Block: M = 2 kg (at rest, on frictionless surface)  |  After: block + bullet move at v_f = 1.2 m/s. Find: bullet's initial velocity u.
01

Perfectly inelastic collision: mu = (m+M)v_f

02

u = (m+M)v_f / m = (2.01 × 1.2)/0.01 = 2.412/0.01 = 241.2 m/s

03

Energy check — KE before: ½(0.01)(241.2²) = 291 J. KE after: ½(2.01)(1.44) = 1.45 J. Energy converted to heat/deformation: 289.5 J — nearly all of it, as expected for a perfectly inelastic collision with a huge mass ratio.

✓ Bullet initial velocity: 241.2 m/s
Example 5 — 2D

Two-dimensional glancing collision

GivenPuck A: 0.3 kg at 4 m/s in x-direction  |  Puck B: 0.3 kg at rest  |  After: A moves at 30° above x-axis, B moves at θ below x-axis. Find: v_A, v_B, and θ.
01

Equal masses in elastic collision: by the elastic result, total KE is conserved, so v_A² + v_B² = u_A² = 16 m²/s²

02

x-momentum: 0.3(4) = 0.3 v_A cos30° + 0.3 v_B cosθ → 4 = v_A(0.866) + v_B cosθ

03

y-momentum: 0 = 0.3 v_A sin30° − 0.3 v_B sinθ → v_A(0.5) = v_B sinθ

04

For equal-mass elastic collisions, the two post-collision velocities are perpendicular: θ = 90° − 30° = 60°. Then: v_A = 4 cos30° = 3.46 m/s; v_B = 4 sin30° = 2.0 m/s. Check: 3.46² + 2.0² = 12 + 4 = 16 ✓

✓ v_A = 3.46 m/s at 30°  |  v_B = 2.0 m/s at 60° below x-axis  |  Perpendicularity confirmed.

Centre of Mass

The centre of mass (CoM) of a system is the single point where, for the purposes of overall motion, all the mass can be treated as concentrated. For a two-body system:

x_CoM = (m₁x₁ + m₂x₂) / (m₁ + m₂)
Centre of mass position. Generalises to any number of bodies or continuous distributions.

The key theorem: the centre of mass of an isolated system moves at constant velocity (including at rest, if total momentum is zero). No internal forces — no matter how violent the explosion or collision — can shift the CoM's trajectory.

This is why an astronaut floating in space can't propel themselves by waving their arms wildly. Their CoM stays fixed; arms moving one way means the rest of the body moves the other. To actually move, you need to throw something (a wrench, exhale gas) — expelling mass in one direction to recoil in the other.

Angular Momentum: Momentum's Rotational Cousin

Just as linear momentum p = mv describes translational motion, angular momentum L describes rotational motion:

L = Iω (rigid body)    or    L = r × p (point particle)
I = moment of inertia; ω = angular velocity; r = position vector; × = cross product

Angular momentum is conserved when no net external torque acts on a system. This explains the figure skater spin: when she pulls her arms in, her moment of inertia I decreases, so her angular velocity ω must increase to keep L = Iω constant. The spin-up isn't powered by her muscles — it's conservation of angular momentum. For the full picture, see our Classical Mechanics guide on rotational dynamics.

Momentum Conservation in the Real World

  • Rocket propulsion: A rocket accelerates by ejecting mass (exhaust) at high speed backward. The reaction (Newton's Third Law) pushes the rocket forward. Momentum is conserved: the backward momentum of exhaust = forward momentum gain of rocket.
  • Particle physics: At CERN, when two protons collide, physicists measure the momenta of all outgoing particles. If the total doesn't match the input, they know a particle has escaped undetected — this is how neutrinos and dark matter candidates are inferred.
  • Jet engines and helicopters: Both work by throwing air backward/downward, gaining an equal and opposite forward/upward momentum. There's no magic — just Newton's Third Law.
  • Car safety: Crumple zones extend the collision time to reduce peak force (same impulse, smaller F). See the energy article for the energy side of crash physics.

Frequently Asked Questions

Momentum is conserved in an isolated system — one with no net external forces. In practice, you choose your system carefully. A collision between two cars on a frictionless surface: momentum conserved. The same cars on a rough road: friction acts externally, so the cars-alone system isn't isolated. Extend the system to include the Earth (which absorbs the friction force) and momentum is again conserved — but now you're dealing with Earth's momentum too.
Momentum (p = mv) is a vector — it has direction. Kinetic energy (KE = ½mv²) is a scalar. A ball moving left has positive KE but negative momentum (if leftward is negative). Crucially, momentum is conserved in all collisions; kinetic energy is only conserved in elastic collisions. When two cars crash and stick together, their total kinetic energy drops sharply but their total momentum remains the same.
Not in an isolated system. In modern physics, conservation of momentum is connected via Noether's theorem to the translational symmetry of space — the fact that the laws of physics are the same everywhere. As long as space is uniform (which it is, to extraordinary precision), momentum is conserved. It's one of the deepest conservation laws in nature.
Set m₁ = m₂ = m in the elastic collision formulas: v₁ = (0)u₁ + 2m·u₂/(2m) = u₂, and v₂ = u₁. The objects literally exchange velocities. Newton's cradle demonstrates this perfectly — the incoming ball stops and the outgoing ball moves at the same speed.
A rocket and its exhaust form an isolated system (in space). Initially, both are at rest: total momentum = 0. The rocket engine expels mass (exhaust) at high speed in one direction. By conservation of momentum, the rocket gains equal and opposite momentum — it accelerates. No ground to push off, no air to push against: just conservation of momentum. The Tsiolkovsky rocket equation quantifies this: Δv = v_e · ln(m₀/m_f).

Next: Energy and the Bigger Picture

Momentum tells you velocities. Energy tells you what's possible and what's lost. Read the companion article on kinetic and potential energy.

What Is Energy? → Full Mechanics Guide

Sources & Further Reading

  1. Halliday, D., Resnick, R., & Krane, K. S. (2002). Physics (5th ed., Vol. 1). Wiley. Chapter 9: Systems of Particles and Chapter 10: Collisions.
  2. Goldstein, H., Poole, C., & Safko, J. (2002). Classical Mechanics (3rd ed.). Addison-Wesley. §3.1: Noether's theorem and conservation laws.
  3. Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers (10th ed.). Cengage. Chapter 9: Linear Momentum and Collisions.
  4. Tipler, P. A., & Llewellyn, R. A. (2012). Modern Physics (6th ed.). W. H. Freeman. Momentum conservation in particle physics.
  5. Knight, R. D. (2016). Physics for Scientists and Engineers (4th ed.). Pearson. Chapter 11: Impulse and Momentum.