Conservation of Linear Momentum

Conservation of linear momentum is one of the most powerful principles in physics. It tells us that, for any isolated system (one with no external forces), the total momentum stays constant — forever, to arbitrary accuracy. This single idea is enough to predict the outcome of collisions, explosions, rocket launches, nuclear reactions, and even the motion of galaxies.

This lesson states the principle carefully, derives it from Newton's Third Law, works through several one-dimensional examples, and lays the foundation for the collision and explosion problems in Lessons 7–10. It corresponds to OCR Module 3.5.2.

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Statement of the Principle

The total linear momentum of a closed system is constant, provided no external resultant force acts on it.

Three things to note:

1. "Closed system" — the boundary of your chosen system must be drawn so that no matter enters or leaves.
2. "No external resultant force" — internal forces (between members of the system) are fine. External forces (gravity from outside the system, friction from the ground, air resistance from outside) must sum to zero for the component(s) you care about.
3. Momentum is a vector. Conservation holds separately for each perpendicular component.

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Derivation From Newton's Third Law

Consider two bodies A and B that interact (for example, collide). During the interaction, A exerts a force F_AB on B, and by Newton's Third Law B exerts a force F_BA = −F_AB on A.

Apply Newton's Second Law to each:

F_BA = Δp_A / Δt ⇒ Δp_A = F_BA Δt F_AB = Δp_B / Δt ⇒ Δp_B = F_AB Δt

Adding:

Δp_A + Δp_B = (F_BA + F_AB) Δt = 0 (by Newton's Third Law)

Therefore:

Δ(p_A + p_B) = 0

The total momentum does not change during the interaction. This argument extends to any number of internally interacting bodies — each Third Law pair cancels in the sum — so as long as the external forces are zero (or their vector sum is), the total momentum of the system is conserved.

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Diagram — A Closed System

flowchart LR
  subgraph System
    A((Body A)) <-->|"Internal forces: Third Law pair"| B((Body B))
  end
  X((External world)) -. "No external force ⇒ momentum conserved" .- System

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Practical Examples of "Closed"

In exam questions, the phrase "a smooth horizontal surface" often appears. This is code for: no friction, so no external horizontal force, so horizontal momentum is conserved. But vertical momentum is not conserved unless gravity is balanced by something else (usually the normal contact force from the surface).

Common system boundaries:

Problem	Closed system	External forces (zero sum?)
Two trolleys colliding on a smooth track	Both trolleys	Gravity = normal force (vertical cancels); no horizontal forces ✓
Gun firing a bullet (both free)	Gun + bullet	Gravity; assume horizontal component is isolated
Alpha decay of a nucleus	Parent nucleus	Negligible external forces on timescale of decay
Car braking to a stop	Car	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 (Lesson 5)

Always identify the system boundary clearly before claiming conservation.

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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. No external horizontal force ⇒ horizontal momentum is conserved.

p_total(before) = p_total(after) (2.0 × 4.0) + (3.0 × 0) = (2.0 × 1.0) + (3.0 × v) 8.0 = 2.0 + 3.0 v v = 6.0 / 3.0 = +2.0 m s⁻¹

The 3.0 kg trolley moves forwards at 2.0 m s⁻¹. Notice how both velocities are positive, meaning both trolleys continue in the original direction — this is what we expect when a lighter trolley hits a heavier stationary one without bouncing backwards.

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Worked Example 2 — Perfectly Inelastic (Sticking) 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_before = 1500 × 20 + 2500 × 10 = 30 000 + 25 000 = 55 000 kg m s⁻¹ p_after = (1500 + 2500) × v = 4000 v

Conservation:

4000 v = 55 000 v = 13.75 m s⁻¹ east

The combined wreckage is slower than the car but faster than the truck, as we would expect from an average weighted by mass. Note that kinetic energy has not been conserved here (see Lesson 8) — some has been converted to heat, sound, and deformation — but momentum has.

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Worked Example 3 — Recoil of a Gun