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Momentum & Collisions

How impacts, explosions and rockets conserve momentum. 45 topics, from beginner to advanced. Open the interactive version with labs and quizzes

Momentum is mass times velocity, and in an isolated system it is conserved. This area explains impulse, the change in momentum produced by a force acting over time, and applies conservation to collisions in one and two dimensions. It also covers elastic and inelastic collisions and the physics of rockets and recoil.

How the topics connect

Momentum and collisions connect force and motion over time. Impulse explains why a longer contact reduces the force, and conservation of momentum lets you predict the results of collisions and explosions without knowing the forces inside them. Angular momentum extends the same idea to spinning and orbiting bodies.

Real-world uses

Vehicle safety

Explain why airbags and crumple zones extend the time of a collision, reducing the force on occupants for the same change in momentum.

Sports equipment

Understand why a cricket bat, golf club or tennis racket transfers momentum efficiently to a ball, and how the coefficient of restitution affects the rebound.

Space and rocket propulsion

Calculate how much thrust a rocket gets from expelling exhaust gas at a given speed, without needing anything to push against.

Particle physics

Use momentum conservation to reconstruct the masses and speeds of particles produced in high-energy collisions.

Road safety

Crumple zones and airbags extend the time of a crash, which reduces the average force on the occupants.

Space travel

Rockets use momentum conservation to change direction and speed, and mission planners use the rocket equation to budget fuel.

Beginner

Linear momentum

The product of mass and velocity.

Key relationp = mv

Momentum p = mv combines how much matter moves with how fast. It is a vector pointing along the velocity, in kg·m/s, and is what a collision exchanges.

Worked example

A 0.16 kg ball at 30 m/s has p = 4.8 kg·m/s.

Common mistake

Momentum is not energy; two bodies with equal momentum can have very different kinetic energy.

Beginner

Momentum is a vector

It has direction, so opposite motions can cancel.

Because momentum has direction, opposite motions subtract. Add momenta component by component in two or three dimensions.

Worked example

A 2 kg cart at +3 m/s and a 3 kg cart at −2 m/s have total p = 0.

Common mistake

Don't add magnitudes.

Beginner

Impulse

Change in momentum caused by a force acting over time.

Key relationJ = FΔt

Impulse J = ∫F dt measures a force's effect over time and equals the change in momentum. The same impulse can come from a big force briefly or a small force for longer.

Worked example

A 1500 N kick lasting 0.02 s gives J = 30 N·s.

Common mistake

Impulse is a vector; use signs.

Beginner

Impulse–momentum theorem

Impulse equals the change in momentum.

Key relationFΔt = mΔv

FΔt = mΔv ties force and time to velocity change. It's the main tool for impacts when the force is unknown but time is known.

Worked example

A 0.2 kg ball changes from 10 to −10 m/s in 0.01 s: F_avg = 400 N.

Common mistake

Average force, not peak force.

Beginner

Why airbags work

Longer stopping time reduces the average force for the same impulse.

An airbag lengthens the stopping time, so the same momentum change needs a smaller average force and so less injury.

Worked example

Stopping 70 kg from 15 m/s in 0.05 s needs 21 kN; in 0.2 s only 5 kN.

Common mistake

Impulse stays the same; the force changes.

Beginner

Conservation of momentum

In an isolated system total momentum is constant.

Key relationm₁u₁+m₂u₂ = m₁v₁+m₂v₂

When no net external force acts, total momentum before equals after: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. It holds even when kinetic energy is lost.

Worked example

A 1 kg ball at 4 m/s hits a stationary 3 kg ball and sticks: v = 1 m/s.

Common mistake

Check the system is isolated at the moment of interest.

Beginner

Isolated system

No net external force acts on it.

A system is isolated when external forces sum to zero (or are negligible over the short interaction). Internal forces of any size don't change total momentum.

Worked example

Two ice skaters pushing off each other on frictionless ice.

Common mistake

Gravity and normal forces cancel on a level surface, so horizontal momentum is conserved.

Beginner

Recoil of a gun

Bullet and gun gain equal and opposite momenta.

Key relationM V = −m v

A gun and bullet start at rest, so total momentum stays zero: MV = −mv. The gun's recoil speed is small because its mass is large.

Worked example

0.02 kg bullet at 400 m/s, 4 kg gun: recoil 2 m/s.

Common mistake

The gun and bullet get equal and opposite momentum but very different kinetic energy.

Beginner

Elastic collision

Both momentum and kinetic energy are conserved.

Key relationcollision

Both momentum and kinetic energy are conserved, with e = 1. Objects bounce apart without permanent deformation.

Worked example

Billiard balls and gas molecules are nearly elastic.

Common mistake

Most real macroscopic collisions lose some kinetic energy.

Beginner

Inelastic collision

Momentum is conserved but kinetic energy is lost.

Key relationcollision

Momentum is conserved but kinetic energy decreases, converted into heat, sound or deformation. Most real collisions are inelastic.

Worked example

A car crash crumples the bodies.

Common mistake

"Inelastic" doesn't mean momentum is lost.

Beginner

Perfectly inelastic collision

The bodies stick together and move with a common velocity.

Key relationv = (m₁u₁+m₂u₂)/(m₁+m₂)

The bodies stick and move with a common velocity v = (m₁u₁ + m₂u₂)/(m₁ + m₂). This loses the maximum kinetic energy compatible with momentum conservation.

Worked example

A 2 kg and a 1 kg cart at 3 m/s and 0: v = 2 m/s.

Common mistake

Maximum loss, but not all kinetic energy unless the centre of mass is at rest.

Intermediate

Coefficient of restitution

Ratio of separation speed to approach speed.

Key relatione = (v₂−v₁)/(u₁−u₂)

e = relative speed of separation ÷ relative speed of approach. e = 1 is elastic, e = 0 perfectly inelastic.

Worked example

A ball dropped from 1 m bouncing to 0.64 m has e = 0.8.

Common mistake

e depends on materials and speed.

Intermediate

Elastic head-on collision velocities

Final velocities follow from momentum and energy conservation.

Key relationv₁ = ((m₁−m₂)u₁+2m₂u₂)/(m₁+m₂)

From the two conservation laws: v₁ = ((m₁ − m₂)u₁ + 2m₂u₂)/(m₁ + m₂) and similarly for v₂. These show how the mass ratio determines the outcome.

Worked example

m₁ = 2, m₂ = 1, u₁ = 3, u₂ = 0: v₁ = 1, v₂ = 4 m/s.

Common mistake

Check by confirming relative speed is reversed.

Intermediate

Equal masses collide elastically

They exchange velocities in one dimension.

With equal masses in a 1-D elastic collision the bodies exchange velocities. A moving ball hitting a stationary equal one stops dead, and the target moves off.

Worked example

Newton's cradle.

Common mistake

This holds only head-on and elastic.

Intermediate

Light hits heavy object

The light object rebounds with nearly the same speed.

A light body rebounds with almost its original speed, and the heavy target barely moves, like a ball bouncing off a wall.

Worked example

Tennis ball hitting a wall at 10 m/s returns at nearly 10 m/s.

Common mistake

The momentum change belongs to the heavy object, but its speed change is tiny.

Intermediate

Heavy hits light object

The heavy object continues almost unchanged while the light flies off at about twice the speed.

The heavy body continues nearly unchanged and the light one is knocked off at about twice the heavy body's speed.

Worked example

A truck hitting a ball sends it off at about 2v.

Common mistake

Energy is shared, but most stays with the heavy body.

Intermediate

Energy lost in inelastic collision

Kinetic energy loss in a perfect inelastic collision.

Key relationΔK = ½μ(u₁−u₂)²

The kinetic energy lost in a perfectly inelastic collision is ΔK = ½μ(u₁ − u₂)² with reduced mass μ. It depends only on relative speed.

Worked example

Two 1 kg masses meeting head-on at 2 m/s each lose 4 J, all of it.

Common mistake

Zero loss if relative velocity is zero.

Advanced

Reduced mass

Two-body problems simplify using an effective mass.

Key relationμ = m₁m₂/(m₁+m₂)

For two interacting bodies it is convenient to use μ = m₁m₂/(m₁ + m₂): the relative motion behaves like a single body of mass μ.

Worked example

Earth–Moon: μ ≈ 0.99 of the Moon's mass.

Common mistake

Used in orbits and atomic spectra.

Intermediate

Collisions in two dimensions

Conserve momentum separately along each axis.

Conserve momentum separately along x and y. With an unknown number of unknowns, you usually need extra info (angles) or energy conservation.

Worked example

A ball hits a stationary one and goes off at 30°; use the components to find the other's direction.

Common mistake

Momentum is a vector; don't use the magnitudes.

Intermediate

Billiard ball collisions

Equal-mass elastic collisions split paths at 90°.

In an off-centre elastic collision of equal masses with one at rest, the two balls leave at right angles to each other. The impact line determines the target's direction.

Worked example

Break shots in pool.

Common mistake

This needs equal masses and elastic, frictionless contact.

Intermediate

Ballistic pendulum

Measures bullet speed using momentum and then energy conservation.

Key relationv = (m+M)/m √(2gh)

A bullet embeds in a hanging block. Momentum conservation gives the speed just after, and energy conservation then gives the swing height: v = ((m + M)/m)√(2gh).

Worked example

m = 10 g, M = 2 kg, h = 5 cm: v ≈ 200 m/s.

Common mistake

Don't use energy conservation across the collision, it's inelastic.

Intermediate

Bouncing ball

Each bounce retains a factor e of speed.

Key relationh_n = e^{2n} h₀

Each bounce multiplies the speed by e and the height by e². The height after n bounces is hₙ = e²ⁿh₀, and the total time is a geometric series.

Worked example

h₀ = 1 m and e = 0.8: h₁ = 0.64 m.

Common mistake

The ball stops in finite time even though it bounces infinitely many times.

Intermediate

Centre of mass

Mean position weighted by mass.

Key relationr_cm = Σm r/M

The centre of mass is the average position weighted by mass; for continuous bodies integrate. It moves as if all external forces act on it.

Worked example

Rod with masses 2 kg at 0 and 1 kg at 3 m: x_cm = 1 m.

Common mistake

It need not lie within the object.

Intermediate

Velocity of centre of mass

Equals total momentum divided by total mass.

Key relationv_cm = P/M

v_cm = Σmv/M = P/M. If net external force is zero, v_cm is constant, even when parts explode.

Worked example

A firework explodes at the top of its path; the centre of mass continues along the parabola.

Common mistake

Internal forces can't change v_cm.

Advanced

Centre-of-mass frame

Frame where total momentum is zero, simplifying collisions.

In this frame total momentum is zero, so collisions look symmetrical: particles approach with equal and opposite momenta. Switching to this frame simplifies analysis and particle-physics calculations.

Worked example

A fixed-target collision is equivalent to a head-on collision in the CM frame.

Common mistake

The CM frame is inertial only if there is no net external force.

Intermediate

Explosions and fragmentation

Total momentum before and after the explosion remains the same.

An explosion converts internal energy into kinetic energy while total momentum stays unchanged. The fragments' momenta sum to the original momentum.

Worked example

A 6 kg shell at rest breaks into 2 kg and 4 kg pieces; the 2 kg piece at 6 m/s means the 4 kg moves 3 m/s oppositely.

Common mistake

Kinetic energy increases, momentum doesn't change.

Advanced

Variable mass: rocket equation

Velocity change depends on exhaust speed and mass ratio.

Key relationΔv = v_e ln(m₀/m_f)

Δv = vₑ ln(m₀/m_f): the speed gain grows only logarithmically with mass ratio, so very large fuel fractions are needed.

Worked example

vₑ = 3 km/s, ratio 20: Δv = 9 km/s.

Common mistake

Higher exhaust speed matters more than adding fuel.

Intermediate

Thrust

Force on the rocket from exhaust.

Key relationF = v_e dm/dt

The force from ejecting mass is F = vₑ(dm/dt) (plus pressure terms). It is the rate of change of momentum of the exhaust.

Worked example

1000 kg/s at 3000 m/s gives 3 MN.

Common mistake

Thrust is independent of air.

Advanced

Multi-stage rockets

Dropping empty stages improves the mass ratio and final speed.

Dropping empty tanks and engines improves the mass ratio of each later stage, so the total Δv is the sum of the stages' Δv.

Worked example

Three stages reach orbital speeds that a single stage can't.

Common mistake

The delta-v adds, not the mass ratios.

Advanced

Jet and conveyor problems

Force required when mass is added or removed continuously.

When material is continuously added, e.g., sand falling onto a moving belt, the force needed to maintain speed v is F = v dm/dt, and half the power goes into heat.

Worked example

Sand dropping at 2 kg/s on a belt at 3 m/s needs 6 N.

Common mistake

Apply F = dp/dt to the belt plus new material.

Intermediate

Angular momentum

Rotational analogue of momentum.

Key relationL = r×p

L = r × p measures the amount of rotation about a point. For a particle in a circle L = mvr and for a rigid body L = Iω.

Worked example

1 kg at 2 m/s on a 3 m radius: L = 6 kg·m²/s.

Common mistake

Angular momentum depends on the chosen origin.

Intermediate

Conservation of angular momentum

Torque-free systems keep L constant.

Key relationL = Iω

If the net torque is zero, L stays constant. Changing the radius or the moment of inertia changes the angular speed.

Worked example

A diver tucks to spin faster.

Common mistake

Torque from gravity about the centre of mass is zero for free-falling bodies.

Beginner

Ice-skater spin

Pulling arms in reduces I and increases ω.

Pulling her arms in lowers I, so ω must increase to keep Iω constant. Kinetic energy increases because the skater does work pulling inward.

Worked example

I from 4 to 1.5 kg·m² turns 2 rev/s into 5.3 rev/s.

Common mistake

Angular momentum is conserved, kinetic energy is not.

Advanced

Kepler's second law from L

Equal areas in equal times follows from conserved angular momentum.

For a central force the angular momentum is constant, so the area swept per unit time, dA/dt = L/2m, is constant. Planets speed up near the Sun.

Worked example

Earth moves faster in January (perihelion).

Common mistake

This holds for any central force, not just gravity.

Advanced

Collision of particles in physics labs

Particle physicists use conserved energy and momentum to find unknown masses.

Detectors reconstruct particle masses from conserved energy and momentum. Invariant mass M² = (ΣE)² − |Σp|² (c = 1) identifies decay products.

Worked example

The Higgs boson shows as a peak in the invariant mass of photon pairs near 125 GeV.

Common mistake

Missing momentum signals an undetected particle such as a neutrino.

Advanced

Compton scattering

Photon–electron collision conserves energy and momentum.

Key relationΔλ = h/mc(1−cosθ)

A photon scattering off an electron loses energy and its wavelength increases by Δλ = (h/mₑc)(1 − cosθ). Conservation of energy and momentum explains it with the photon as a particle.

Worked example

Maximum shift of 2.43×10⁻¹² m at 180°.

Common mistake

The effect needs short wavelengths such as X-rays.

Beginner

Newton's cradle

A row of steel balls transfers momentum from one end to the other.

Steel balls transfer momentum through the chain, so one ball swings out at the far end with the same speed. Both energy and momentum conservation are needed to explain why exactly one ball leaves.

Worked example

Two balls in swing out two.

Common mistake

Real cradles gradually lose energy.

Advanced

Momentum of light

Photons carry momentum even without mass.

Key relationp = h/λ

Photons carry p = E/c = h/λ even with no mass. Absorbing or reflecting light transfers momentum, giving radiation pressure.

Worked example

Solar sails use it.

Common mistake

Momentum doesn't require mass.

Advanced

Radiation pressure

Light exerts pressure when absorbed or reflected.

Key relationP = I/c

Light exerts pressure P = I/c if absorbed and 2I/c if reflected. At Earth sunlight (1361 W/m²) gives only 4.5 μPa.

Worked example

A 1 km² sail gets about 9 N.

Common mistake

Tiny but constant, so useful in space.

Advanced

Momentum in fluids

Fluid flow transfers momentum to surfaces causing forces.

Fluid jets change the momentum of the fluid hitting a surface, giving a force F = ρQv (rate of momentum change). This explains water jets, wind forces and jet engines.

Worked example

A 2 kg/s water jet at 10 m/s stopping on a wall exerts 20 N.

Common mistake

Force depends on mass flow rate and speed.

Intermediate

Crash test physics

Crumple zones extend collision time to reduce force.

Crumple zones, seat belts and airbags extend the stopping time and spread the force. Using impulse: F_avg = mΔv/Δt.

Worked example

A 1500 kg car at 15 m/s stops in 0.1 s with 225 kN; in 0.3 s only 75 kN.

Common mistake

Rigid cars aren't safer.

Beginner

Sports impulse examples

Follow-through in a swing increases contact time and the ball's momentum.

Follow-through increases contact time with a bat or racquet, so for the same force the impulse and ball speed are bigger. Catching a ball by moving the hand backward reduces the force.

Worked example

A golfer's long follow-through.

Common mistake

Impulse and not force alone sets the ball's speed.

Intermediate

Momentum and kinetic energy relation

Kinetic energy in terms of momentum.

Key relationK = p²/2m

K = p²/2m links them. Same momentum means the lighter body has more kinetic energy; same kinetic energy means the heavier one has more momentum.

Worked example

A 0.01 kg bullet and a 1 kg ball both with p = 5: K = 1250 J vs 12.5 J.

Common mistake

Don't equate momentum and energy.

Advanced

Hamiltonian momentum

Generalised momentum is derived from the Lagrangian.

Key relationp = ∂L/∂q̇

In advanced mechanics the generalised momentum is pᵢ = ∂L/∂q̇ᵢ. It reduces to mv for simple cases but can include extra terms, such as qA in a magnetic field.

Worked example

For a charge in a magnetic field p = mv + qA.

Common mistake

Canonical momentum is not always mv.

Advanced

Momentum conservation and symmetry

Translational symmetry of space implies conservation of momentum.

Noether's theorem: if the laws don't change when the system is shifted in space (translation symmetry), total momentum is conserved.

Worked example

This is why momentum is conserved in all known interactions.

Common mistake

Conservation can fail in an external potential that breaks the symmetry.

Frequently asked questions

Why do airbags reduce injuries?

The airbag increases the time over which your momentum changes to zero. A longer time means a smaller average force for the same impulse, which lowers the risk of injury.

Is momentum conserved in every collision?

Yes, total momentum is conserved in any collision when no external forces act on the system. Kinetic energy is conserved only in elastic collisions.

How does a rocket accelerate in space?

A rocket pushes exhaust gas backward at high speed. By conservation of momentum, the rocket gains momentum forward. It does not need air or ground to push against.

Is momentum always conserved?

Momentum is conserved in an isolated system with no net external force. In a system where external forces act, the momentum changes by the impulse of those forces.

Why do elastic collisions and inelastic ones differ?

Both conserve momentum, but elastic collisions also conserve kinetic energy. In inelastic collisions, some kinetic energy turns into heat, sound or deformation.

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