Conservation of Linear Momentum

Spec mapping: OCR H556 Module 3.3 — Conservation of linear momentum for a closed system; derivation from Newton's third law; application to one-dimensional collisions and explosions. (Refer to the official OCR H556 specification document for exact wording.)

Conservation of linear momentum is one of the most powerful principles in physics. It says that, for any closed system (one with no external resultant force), the total vector momentum stays constant — forever, to whatever accuracy you can measure. This single idea is enough to predict the outcome of collisions, explosions, rocket launches, particle-physics reactions and even the dance of galaxies. Together with conservation of energy (Module 3.4) it forms one half of the two-pillar conservation framework that the rest of the A-Level course leans on.

This lesson states the principle precisely, derives it from Newton's third law, works through five 1-D applications, and gives an A* synoptic framework that lets you choose the right "system" for any exam question. It is the foundation for the elastic/inelastic/2-D/explosion lessons that follow — and the PAG 6 collision-on-air-track practical is built directly on this principle.

Key Definition: For a closed system on which no external resultant force acts, the total linear momentum vector $\sum \vec{p}$ is constant. In one dimension: $\sum_{i} m_{i} u_{i} = \sum_{i} m_{i} v_{i}$ , where $u_{i}$ are pre-interaction velocities and $v_{i}$ are post-interaction velocities of the bodies labelled $i$ . Crucially, the principle applies separately to each independent direction in space.

Statement of the Principle

A formal A-Level statement that will earn full credit on a "state the principle" question:

The total linear momentum of a closed system remains constant, provided the resultant external force on the system is zero.

Three pieces of small print, all of which OCR examiners look for:

"Closed system" — the boundary of the system must be drawn so that no matter enters or leaves during the interaction. For collisions and explosions the bodies are usually obvious (the two trolleys, the bullet and the block) but you must declare the system before applying the principle.
"Resultant external force is zero" — internal forces between members of the system are fine (in fact they are exactly what causes momentum to be exchanged). It is the external forces (gravity from outside the system, friction from the ground, contact from a wall) that must vector-sum to zero. Crucially, the condition need only hold for the component(s) you care about: on a smooth horizontal track, horizontal external forces are zero so horizontal momentum is conserved, even though vertical gravity and the normal contact force are both non-zero.
Momentum is a vector. Conservation holds independently for each perpendicular component. In 2-D you get two equations ( $p_{x}$ and $p_{y}$ conserved separately); in 3-D you get three.

A useful refinement for collision problems: even if a finite external force does act (e.g. gravity on a colliding ball), the impulse delivered during the brief collision interval $\Delta t$ is $F_{\text{ext}}\,\Delta t$ , which is negligible when $\Delta t$ is short and the internal impulsive forces are large. So momentum conservation is, to high accuracy, a statement about impulsive interactions: things happen too fast for gravity to matter.

Derivation From Newton's Third Law

Conservation of momentum is not an independent postulate of mechanics — it is a consequence of Newton's laws, and OCR expects candidates to be able to derive it.

Consider two bodies A and B that interact for a short time $\Delta t$ . During the interaction A exerts a force $\vec{F}_{AB}$ on B, and by Newton's third law B exerts a force $\vec{F}_{BA} = -\vec{F}_{AB}$ on A.

Apply Newton's second law (in its $\vec{F} = d\vec{p}/dt$ form) to each body:

$\vec{F}_{BA} = \frac{\Delta \vec{p}_A}{\Delta t} \quad \Longrightarrow \quad \Delta \vec{p}_A = \vec{F}_{BA}\,\Delta t$

$\vec{F}_{AB} = \frac{\Delta \vec{p}_B}{\Delta t} \quad \Longrightarrow \quad \Delta \vec{p}_B = \vec{F}_{AB}\,\Delta t$

Adding:

$\Delta \vec{p}_A + \Delta \vec{p}_B = (\vec{F}_{BA} + \vec{F}_{AB})\,\Delta t = \vec{0} \quad \text{(by N3)}$

Therefore:

$\Delta(\vec{p}_A + \vec{p}_B) = \vec{0} \quad \Longleftrightarrow \quad \vec{p}_A + \vec{p}_B = \text{constant}$

The total momentum of the two-body system is unchanged by the interaction. This argument extends straightforwardly to any number of internally interacting bodies — each Newton-III pair cancels in the sum — so as long as the external resultant force is zero, the total momentum of the whole system is conserved.

The derivation makes the chain of reasoning crystal clear: Newton's second law (definition of force as $dp/dt$ ) plus Newton's third law (action-reaction) gives conservation of momentum. There is no extra physics involved.

Common Exam Mistake: Citing only Newton's third law without quoting Newton's second law in the form $F = \Delta p / \Delta t$ . Both are required to make the derivation work.

Diagram — A Closed System

flowchart LR
  subgraph System["Closed system (boundary drawn here)"]
    A((Body A)) <-->|"Internal N3 pair (cancels)"| B((Body B))
  end
  X((External world)) -. "External resultant force = 0 ⇒ Σp constant" .- System

The boundary is the analytical choice you make — not a physical wall. Inside the boundary, internal forces redistribute momentum between members but cannot change the total. Outside, anything is happening; the only thing that matters is the net external force on the boundary contents.

Practical Examples of "Closed"

Real exam questions test whether you can identify the closed-system condition. The phrase "a smooth horizontal surface" is code for no friction → no external horizontal force → horizontal momentum is conserved. But vertical momentum is not conserved unless gravity is balanced by something else (which on a horizontal table it is — by the normal contact force).

A summary of common scenarios:

Problem	Closed system	External forces (sum to zero in the relevant component?)
Two trolleys colliding on a smooth track	Both trolleys	Gravity vs normal force (vertical cancels); no horizontal forces ✓
Gun firing a bullet (both free)	Gun + bullet	Gravity acts vertically; horizontal momentum conserved ✓
Alpha decay of a nucleus	Parent nucleus	Negligible external forces on the femtosecond decay timescale ✓
Car braking to a stop	Car alone	Friction from the road is external ✗ — momentum not conserved
Ball bouncing off a wall	Just the ball	Wall force is external ✗ — for the ball alone, momentum is not conserved
Ball bouncing off a wall (extended system)	Ball + wall + Earth	All forces internal ✓ — momentum conserved (Earth's recoil is tiny but real)

The "ball + wall + Earth" row is the philosophically important one. Conservation always holds for a sufficiently enlarged closed system. The art of momentum problems is choosing the smallest boundary that still makes the external forces (or impulse-time products) negligible.

Worked Example 1 — Two Trolleys Colliding (1-D)

A $2.0$ kg trolley travels at $+4.0$ m s $^{-1}$ and collides with a stationary $3.0$ kg trolley on a smooth horizontal track. After the collision, the $2.0$ kg trolley moves at $+1.0$ m s $^{-1}$ . Find the velocity of the $3.0$ kg trolley.

System = both trolleys. Smooth horizontal track ⇒ no external horizontal force ⇒ horizontal momentum is conserved.

Take "right" (the direction of the initial moving trolley) as positive. Conservation of momentum:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

$(2.0)(4.0) + (3.0)(0) = (2.0)(1.0) + (3.0) v_2$

$8.0 = 2.0 + 3.0\,v_2 \quad \Longrightarrow \quad v_2 = \frac{6.0}{3.0} = +2.0 \text{ m s}^{-1}$

The $3.0$ kg trolley moves forward at $2.0$ m s $^{-1}$ . Both velocities are positive — both trolleys continue in the original direction, as expected when a lighter trolley hits a heavier stationary one without bouncing backwards.

Common Exam Mistake: Skipping the line $(2.0)(4.0) + (3.0)(0) = (2.0)(1.0) + (3.0) v_2$ and substituting directly. OCR mark schemes routinely award the "set up the conservation equation" mark separately from the "obtain $v_2 = 2.0$ m s $^{-1}$ " mark; missing the equation line costs that mark.

Worked Example 2 — Perfectly Inelastic Collision

A $1500$ kg car travels east at $20$ m s $^{-1}$ and collides with a $2500$ kg truck travelling east at $10$ m s $^{-1}$ . The vehicles lock together. Find their common velocity just after impact (assume a smooth horizontal road).

$p_{\text{before}} = (1500)(20) + (2500)(10) = 30\,000 + 25\,000 = 55\,000 \text{ kg m s}^{-1}$

$p_{\text{after}} = (1500 + 2500)\,v = 4000\,v$

Conservation:

$4000\,v = 55\,000 \quad \Longrightarrow \quad v = 13.75 \text{ m s}^{-1} \text{ east}$

The combined wreckage moves slower than the car and faster than the truck — a mass-weighted average. Note that kinetic energy is not conserved here (Lesson 8 explores why and computes the deficit), but momentum is.

Worked Example 3 — Recoil of a Gun

A $0.020$ kg bullet is fired from a $3.0$ kg rifle. The bullet leaves the muzzle at $400$ m s $^{-1}$ . Find the recoil velocity of the rifle, assuming both are initially at rest and the rifle is free.

System = rifle + bullet. Take "forwards" (direction of bullet) as positive. Initially both at rest, so $p_{\text{before}} = 0$ .

$0 = (0.020)(400) + (3.0)(v_{\text{rifle}})$

$8.0 + 3.0\,v_{\text{rifle}} = 0 \quad \Longrightarrow \quad v_{\text{rifle}} = -2.67 \text{ m s}^{-1}$

The negative sign confirms backward recoil: rifle moves at $2.67$ m s $^{-1}$ opposite to the bullet. This is exactly the physics of rocket propulsion — eject a small mass forward at high speed, and the large mass recoils slowly backward.

In practice, a rifle held against the shoulder transfers momentum into the shooter's whole body, so the rifle's recoil speed is much smaller — but this is an external impulse from the shooter's grip during the brief firing interval. If the shoulder transfer is included as internal to the (rifle + shooter) system, then that extended system's centre of mass remains at rest, and the shooter rocks backward measurably.

Worked Example 4 — Explosion From Rest

A $5.0$ kg object initially at rest explodes into two fragments of masses $2.0$ kg and $3.0$ kg. The $2.0$ kg fragment flies east at $15$ m s $^{-1}$ . Find the velocity of the $3.0$ kg fragment.

$0 = (2.0)(+15) + (3.0)\,v$

$30 + 3.0\,v = 0 \quad \Longrightarrow \quad v = -10 \text{ m s}^{-1}$

The fragments fly apart in opposite directions, with speeds inversely proportional to their masses. The momentum vectors sum to zero, exactly as they must when the initial total was zero. This is the archetype of every "explosion from rest" problem (Lesson 10 returns to this in detail).

Worked Example 5 — Choosing the System Carefully

A $0.50$ kg ball is dropped from $2.0$ m above the ground and bounces elastically off the floor, returning to its original height. Identify (a) whether momentum is conserved during the bounce if the "system" is the ball alone, and (b) what conserves momentum if you enlarge the system to ball + Earth.

(a) System = ball alone. During the bounce, the ball reverses velocity from $-v$ (downward, just before impact) to $+v$ (upward, just after, with $v = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 2.0} = 6.26$ m s $^{-1}$ ). The change in momentum is:

$\Delta p = m(v - (-v)) = 2mv = 2 \times 0.50 \times 6.26 = 6.26 \text{ kg m s}^{-1}$

This is not zero, so the ball's momentum is not conserved during the bounce. The external force is the impulsive contact force from the floor, which delivers an upward impulse $J = 2mv = 6.26$ N s.

(b) System = ball + Earth. The Earth (mass $5.97 \times 10^{24}$ kg) recoils downward when the ball pushes down on it during the bounce. By momentum conservation:

$m_{\text{ball}}(2v) = m_{\text{Earth}}\,v_{\text{Earth}}$

$v_{\text{Earth}} = \frac{2 \times 0.50 \times 6.26}{5.97 \times 10^{24}} \approx 1.05 \times 10^{-24} \text{ m s}^{-1}$

The Earth recoils at $\sim 10^{-24}$ m s $^{-1}$ — undetectable, but real. The total momentum of (ball + Earth) is unchanged. This is the canonical demonstration that enlarging the system restores conservation: every "external" force has a Newton-III partner that, if included, makes the system closed.

A* Discriminator — When Conservation "Almost" Holds

In real exam problems involving collisions on a rough surface, the friction force is external — but if the collision lasts only $\Delta t \sim 10^{-3}$ s, the friction impulse $\mu m g \Delta t$ is tiny compared with the internal collision impulse. So we still use momentum conservation as an excellent approximation, and worry about friction only over the macroscopic time after the collision when the bodies slide to rest.

The rule of thumb: if the interaction is impulsive (very brief, very large internal forces), external steady forces can be ignored during the interaction itself. This is why we apply conservation to the moment of impact, not to the entire experiment.

Conservation of Linear Momentum

Conservation of Linear Momentum

Statement of the Principle

Derivation From Newton's Third Law

Diagram — A Closed System

Practical Examples of "Closed"

Worked Example 1 — Two Trolleys Colliding (1-D)

Worked Example 2 — Perfectly Inelastic Collision

Worked Example 3 — Recoil of a Gun

Worked Example 4 — Explosion From Rest

Worked Example 5 — Choosing the System Carefully

A* Discriminator — When Conservation "Almost" Holds

Synoptic Links

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