Conservation of Momentum

Momentum is one of the most powerful concepts in physics. While Newton’s second law relates force to acceleration, momentum gives us a way to analyse interactions between objects — collisions, explosions, and other events — without needing to know the details of the forces involved. The principle of conservation of momentum is one of the most fundamental laws in all of physics.

Momentum

Momentum is defined as the product of an object’s mass and velocity:

p = mv

where p is momentum in kg m/s, m is mass in kg, and v is velocity in m/s.

Momentum is a vector quantity — it has direction as well as magnitude. The direction of momentum is the same as the direction of velocity.

Worked Example 1

A 1200 kg car travels at 15 m/s due east. What is its momentum?

p = mv = 1200 x 15 = 18 000 kg m/s due east

Reference: Typical Momentum Values

Object	Mass	Speed	Momentum
Tennis ball served	0.058 kg	60 m/s	3.5 kg m/s
Cricket ball bowled	0.16 kg	40 m/s	6.4 kg m/s
Sprinter	80 kg	10 m/s	800 kg m/s
Family car at 30 mph	1200 kg	13 m/s	15 600 kg m/s
Lorry at 50 mph	10 000 kg	22 m/s	220 000 kg m/s
Supertanker at sea	300 000 000 kg	5 m/s	1.5 x 10⁹ kg m/s

The Principle of Conservation of Momentum

In a closed system (no external forces), the total momentum before an event equals the total momentum after the event.

Total momentum before = Total momentum after

This applies to all interactions: collisions, explosions, objects joining together, objects breaking apart. The key condition is that no external resultant force acts on the system (or that external forces are negligible compared to the internal forces during the event).

Why It Works

Conservation of momentum follows directly from Newton’s third law. When two objects interact, they exert equal and opposite forces on each other (Newton’s third law). These forces act for the same time, so the impulses are equal and opposite. One object gains momentum in one direction, the other gains the same amount in the opposite direction. The total change in momentum of the system is zero.

flowchart TD
    A["Two objects interact"]
    A --> B["Newton's 3rd Law:\nF on A = -F on B"]
    B --> C["Same contact time Δt"]
    C --> D["Impulse on A = FΔt\nImpulse on B = -FΔt"]
    D --> E["Δp_A = -Δp_B"]
    E --> F["Δp_A + Δp_B = 0"]
    F --> G["Total momentum\nis CONSERVED"]

Types of Collision

Summary Table

Type	Momentum Conserved?	KE Conserved?	Objects After
Elastic	Yes	Yes	Separate
Inelastic	Yes	No (KE lost)	Separate
Perfectly inelastic	Yes	No (maximum KE loss)	Stick together
Explosion	Yes (total = 0 if from rest)	No (KE gained from stored energy)	Separate

Elastic Collisions

An elastic collision is one in which both momentum and kinetic energy are conserved. No kinetic energy is lost to deformation, heat, or sound.

True elastic collisions are rare in everyday life but are a good approximation for collisions between hard objects like snooker balls and for atomic/molecular collisions.

Special case: For an elastic head-on collision between equal masses where one is stationary, the moving object stops and the stationary one moves off at the same speed. This is seen in Newton’s cradle.

Inelastic Collisions

An inelastic collision is one in which momentum is conserved but kinetic energy is not. Some kinetic energy is converted to other forms (heat, sound, deformation).

Most real collisions are inelastic. A car crash is a highly inelastic collision — the cars deform significantly, and kinetic energy is converted to heat, sound, and the work done in crushing metal.

Perfectly Inelastic Collisions

A perfectly inelastic collision is one in which the objects stick together after the collision and move as a single body. This is the collision that loses the maximum possible kinetic energy (while still conserving momentum).

Worked Example 2: Perfectly Inelastic Collision

A 2.0 kg trolley moving at 3.0 m/s collides with a stationary 4.0 kg trolley. They stick together. Find the velocity after the collision and the kinetic energy lost.

Momentum before = momentum after: m₁v₁ + m₂v₂ = (m₁ + m₂)v (2.0)(3.0) + (4.0)(0) = (2.0 + 4.0)v 6.0 = 6.0 v v = 1.0 m/s

KE before = (1/2)(2.0)(3.0²) = 9.0 J KE after = (1/2)(6.0)(1.0²) = 3.0 J KE lost = 9.0 - 3.0 = 6.0 J (converted to heat, sound, deformation)

Fraction of KE lost = 6.0/9.0 = 2/3 = 67%

Worked Example 3: Elastic Collision (Equal Masses)

A 1.0 kg ball A moving at 4.0 m/s collides head-on with a stationary 1.0 kg ball B. The collision is elastic. Find the velocities after the collision.

For an elastic head-on collision between equal masses where one is initially stationary, ball A stops and ball B moves off with ball A’s original velocity:

v_A = 0, v_B = 4.0 m/s

Check momentum: (1.0)(4.0) + (1.0)(0) = (1.0)(0) + (1.0)(4.0). 4.0 = 4.0. Conserved. Check KE: (1/2)(1.0)(4.0²) = (1/2)(1.0)(4.0²). 8.0 = 8.0. Conserved.

Worked Example 4: Head-On Collision (Both Moving)

A 3.0 kg ball A moves at 5.0 m/s to the right. A 2.0 kg ball B moves at 4.0 m/s to the left. They collide and A bounces back at 1.0 m/s to the left. Find the velocity of B and determine whether the collision is elastic.

Taking right as positive: Before: p = (3.0)(5.0) + (2.0)(-4.0) = 15 - 8.0 = 7.0 kg m/s After: (3.0)(-1.0) + (2.0)(v_B) = 7.0 -3.0 + 2.0 v_B = 7.0 v_B = 10.0/2.0 = 5.0 m/s to the right

KE before = (1/2)(3.0)(25) + (1/2)(2.0)(16) = 37.5 + 16.0 = 53.5 J KE after = (1/2)(3.0)(1.0) + (1/2)(2.0)(25) = 1.5 + 25.0 = 26.5 J

KE lost = 53.5 - 26.5 = 27.0 J. The collision is inelastic.

Explosions

In an explosion, objects that are initially at rest (or moving together) separate and move apart. No external force acts, so momentum is conserved.

If the system starts at rest, the total momentum is zero. After the explosion, the total momentum must still be zero — the momenta of the fragments must cancel.

Worked Example 5

A 3.0 kg trolley at rest on a smooth surface has a spring mechanism that pushes a 1.0 kg ball off it at 6.0 m/s to the right. What is the velocity of the trolley?

Total momentum before = 0 (at rest) Total momentum after = 0: (3.0)v_trolley + (1.0)(6.0) = 0 3.0 v_trolley = -6.0 v_trolley = -2.0 m/s (i.e. 2.0 m/s to the left)

The trolley recoils in the opposite direction.

KE gained = (1/2)(3.0)(2.0²) + (1/2)(1.0)(6.0²) = 6.0 + 18.0 = 24.0 J. This energy came from the elastic potential energy stored in the spring.

Worked Example 6: Three-Fragment Explosion

A stationary firework explodes into three pieces. Fragment A (0.10 kg) moves north at 20 m/s. Fragment B (0.15 kg) moves east at 10 m/s. Fragment C has mass 0.25 kg. Find its velocity.

North-south: 0.10 x 20 + 0 + 0.25 v_CN = 0 v_CN = -2.0/0.25 = -8.0 m/s (south)

East-west: 0 + 0.15 x 10 + 0.25 v_CE = 0 v_CE = -1.5/0.25 = -6.0 m/s (west)

Speed of C = sqrt(8.0² + 6.0²) = sqrt(64 + 36) = sqrt(100) = 10.0 m/s Direction: arctan(6.0/8.0) = 36.9° west of south (or bearing 216.9°)

Collisions in Two Dimensions

Although A-Level problems typically involve one-dimensional collisions, the principle extends to two dimensions. Momentum is conserved independently in each perpendicular direction:

Total momentum in the x-direction before = total momentum in the x-direction after
Total momentum in the y-direction before = total momentum in the y-direction after

Newton’s Cradle

Newton’s cradle demonstrates conservation of both momentum and kinetic energy. When one ball is lifted and released:

It collides elastically with the row. Momentum and KE are conserved.
One ball at the far end swings out with the same speed. Two balls in, two balls out, etc.
The intermediate balls barely move because the collision transmits momentum through them almost perfectly.

If two balls are lifted together, two balls swing out at the other end — not one ball at double speed, because that would conserve momentum but not kinetic energy.

Proof: Two balls of mass m at speed v have momentum 2mv and KE = 2 x (1/2)mv² = mv². If one ball at speed 2v: momentum = m(2v) = 2mv (conserved), but KE = (1/2)m(2v)² = 2mv² (doubled, NOT conserved). So two balls at v is the only option that conserves both.

Common Mistakes and Misconceptions

“Momentum is always conserved”: Momentum is conserved only in a closed system with no external resultant force. If a ball hits a wall fixed to the Earth, the ball-only system does not conserve momentum (the wall exerts an external force). The ball + Earth system does, but Earth’s recoil is immeasurably small.
Confusing conservation of momentum with conservation of KE: Momentum is always conserved in collisions (no external forces). Kinetic energy is only conserved in elastic collisions.
Forgetting the direction of momentum: Momentum is a vector. When objects move in opposite directions, their momenta have opposite signs. A 2.0 kg ball at 5.0 m/s right and a 3.0 kg ball at 5.0 m/s left do not have the same momentum.
Writing the wrong equation for explosions: Start from total momentum = 0 (if initially at rest), not from KE conservation.
Not checking whether a collision is physically possible: If your answer gives one object moving faster after a collision in the original direction, check that the other object’s velocity is consistent.

Exam Tips for Edexcel Physics

Always state "by conservation of momentum" before writing the momentum equation. This earns a method mark.
Define a positive direction clearly before setting up equations.
To determine whether a collision is elastic, calculate KE before and KE after separately, then compare. Do not assume.
For explosion questions, if the system starts at rest, immediately write "total momentum = 0."
If asked to "show that momentum is conserved," calculate total momentum before and after separately and show they are equal.
Edexcel often asks for the fraction or percentage of KE lost — always calculate both KE before and KE after as separate steps.

Key Points

Momentum = mv (vector, units: kg m/s).
Conservation of momentum: total momentum before = total momentum after (no external forces).
Elastic collisions: both momentum and KE conserved.
Inelastic collisions: momentum conserved, KE not conserved.
Perfectly inelastic: objects stick together; maximum KE loss.
Explosions: system starts at rest, so total momentum after = 0 (fragments have equal and opposite momenta).
Conservation of momentum is a direct consequence of Newton’s third law.
For equal-mass elastic collisions with one object at rest, the objects exchange velocities.

A-Level Deep Dive: Conservation of Momentum

Spec mapping

Edexcel 9PH0 specification, Topic 2 — Mechanics, sub-topic 2.6 establishes the conservation of linear momentum in a closed system, requiring candidates to apply it to collisions and explosions in one and (later) two dimensions, and to distinguish elastic from inelastic interactions (refer to the official Pearson Edexcel specification document for exact wording). Although momentum is introduced here in the context of contact forces, the conservation principle recurs across Topic 6 (Further mechanics) where it is extended to oblique two-dimensional collisions and impulse–momentum integrals, Topic 7 (Electric and magnetic fields) for charged-particle motion in cyclotrons and mass spectrometers, Topic 8 (Nuclear and particle physics) for alpha decay recoil and inelastic scattering, and Topic 9 (Thermodynamics) where the kinetic theory derivation of pressure rests on momentum change at the wall. The Edexcel data, formulae and relationships booklet supplies $p = mv$ but does not state $\Delta p_\text{total} = 0$ explicitly — the conservation statement must be quoted in working.

Worked example with full mark scheme

Question (8 marks):

A bullet of mass 12.0 g is fired horizontally into a stationary wooden block of mass 1.488 kg suspended as a pendulum bob from a light inextensible string of length 1.20 m. The bullet embeds in the block, and the bullet–block combination subsequently swings to a maximum height $h$ above its initial level.

(a) State the principle of conservation of momentum and explain why it can be applied to the embedding stage but not to the subsequent swing. (2)

(b) The bob is observed to rise through a vertical height $h = 0.180$ m. Use energy methods to determine the speed of the bullet–block combination immediately after the bullet has embedded. Take $g = 9.81$ m s⁻². (2)

(d) Show that approximately 99% of the bullet's initial kinetic energy is dissipated during the embedding. (1)

Solution with mark scheme:

(a) Statement. In a closed system on which no external resultant force acts, the total linear momentum is constant.

B1 — correct statement including the closed-system / no-external-force condition.

Reasoning. During the embedding (timescale $\sim$ milliseconds), the impulsive contact force between bullet and block is many orders of magnitude larger than gravity or string tension, so external forces contribute negligible impulse and momentum is effectively conserved. During the subsequent swing (timescale $\sim$ second), gravity and tension are not negligible — they do work on the system and rotate the velocity vector — so momentum is no longer conserved. Energy methods are appropriate here because gravity is conservative.

B1 — distinguishes timescales / impulse magnitudes for the two stages.

(b) Step 1 — energy conservation during the swing. All initial kinetic energy of the combination converts to gravitational potential energy at the highest point (string remains taut, no air resistance assumed):

$\tfrac{1}{2}(M + m)V^2 = (M + m)gh$

The combined mass cancels:

$V = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 0.180}$

M1 — correct use of $\tfrac{1}{2}v^2 = gh$ (or equivalent).

$V = \sqrt{3.5316} = 1.879 \text{ m s}^{-1} \approx 1.88 \text{ m s}^{-1}$

A1 — $V = 1.88$ m s⁻¹ to 3 sf.

(c) Step 1 — apply conservation of momentum to the embedding. Take the direction of the bullet's initial motion as positive. Before: only the bullet moves, with speed $u$ . After: the combination of mass $(M + m) = 1.500$ kg moves with speed $V$ .

$mu = (M + m)V$

M1 — correct momentum equation, with the bullet–block treated as a single mass after embedding (perfectly inelastic).

Step 2 — solve for $u$ .

$u = \dfrac{(M + m)V}{m} = \dfrac{1.500 \times 1.879}{0.0120}$

M1 — correct rearrangement and substitution with consistent units (mass in kg).

$u = \dfrac{2.819}{0.0120} = 235 \text{ m s}^{-1}$

A1 — $u = 235$ m s⁻¹ to 3 sf.

(d) Initial KE $= \tfrac{1}{2}mu^2 = \tfrac{1}{2}(0.0120)(235)^2 = 331$ J. Final KE (immediately after embedding) $= \tfrac{1}{2}(1.500)(1.879)^2 = 2.65$ J. Fractional loss $= (331 - 2.65)/331 = 0.992$ , i.e. $\approx 99.2\%$ of the initial KE is dissipated as heat, sound, and deformation work in the wood.

B1 — explicit before/after KE comparison reaching $\approx 99\%$ .

Total: 8 marks (B3 M3 A2).

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A railway truck of mass 4.0 x 10³ kg moves along a level straight track at 3.0 m s⁻¹ and couples with a stationary truck of mass 6.0 x 10³ kg. The trucks then move together.

(a) Explain why the trucks can be modelled as a closed system during the coupling, and state the conservation principle that follows. (2)

(b) Calculate the common velocity of the trucks immediately after coupling. (2)

Mark scheme decomposition by AO:

Part	AO weighting	Earned by
(a) B1	AO1.2	States that horizontal external forces (rolling resistance, air drag) are negligible during the brief coupling impulse.
(a) B1	AO1.1a	Quotes conservation of linear momentum: total momentum before = total momentum after.
(b) M1	AO1.1b	Sets up $m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$ , i.e. $(4000)(3.0) + 0 = (10000)v$ .
(b) A1	AO1.1b	$v = 1.2$ m s⁻¹ in the original direction of motion.
(c) M1	AO2.1	Computes KE before $= \tfrac{1}{2}(4000)(3.0)^2 = 18\,000$ J and KE after $= \tfrac{1}{2}(10000)(1.2)^2 = 7200$ J.
(c) A1	AO3.1a	KE lost $= 10\,800$ J, dissipated as heat / sound / deformation work in the coupling mechanism.

Total: 6 marks split AO1 = 4, AO2 = 1, AO3 = 1. This is a textbook perfectly-inelastic collision with a short qualitative tail. Edexcel uses coupled-trucks problems repeatedly because the geometry is one-dimensional, the masses are clean numbers, and the AO3 mark for energy accounting separates surface candidates from those who treat momentum conservation and energy conservation as independent diagnostics.

Synoptic links

Newton's third law (sub-topic 2.2): conservation of momentum is not an independent postulate but a direct consequence of $\mathbf{F}_{AB} = -\mathbf{F}_{BA}$ . Equal and opposite contact forces acting for the same time produce equal and opposite impulses, so the changes in momentum cancel and the system total is constant.