Conservation of Momentum

The principle of conservation of linear momentum is the most powerful single idea in collision mechanics. Whenever no external resultant force acts on a system — a collision, an explosion, a recoil, a coupling of trucks — the total momentum is the same vector before and after, no matter how violent or complicated the interaction in between. It turns an unknowable instant of contact into a one-line equation.

1 Where this sits in AQA 7367

Conservation of momentum is core to the Mechanics option, Paper 3 (7367/3M), and underpins every collision question that follows. With Paper 3 weighted AO1 40% / AO2 25% / AO3 35%, examiners often pair this principle with restitution (next two lessons) or with energy, so the AO3 problem-solving load is high. It builds directly on the impulse–momentum principle of the previous lesson and on Newton's third law from A-Level Maths Mechanics. Because it most directly tests the application of a technique to model a physical interaction, it is a natural carrier of AO1 (using the method) and AO3 (solving the problem in context) marks; the AO2 reasoning marks come from justifying assumptions and from "show that" derivations. Expect it to appear both as a standalone short question and, more often, as the first stage of a longer multi-part collision problem.

2 Core theory — statement and proof

The principle

If no external resultant force acts on a system, the total linear momentum of the system is constant.

For two interacting particles this reads

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

with every velocity measured along one declared positive direction (and component-wise in two dimensions). The principle generalises immediately to any number of particles: the total momentum $\sum_i m_i \vec v_i$ before equals the total after, provided the only forces are internal to the system.

Why it is true (Newton's third law)

During contact, particle A pushes B with force $\vec F$ ; by Newton's third law B pushes A with $-\vec F$ . Over the contact time $\Delta t$ these deliver equal and opposite impulses:

\begin{aligned} \text{impulse on A} &= -\vec F\,\Delta t = m_1\vec v_1 - m_1\vec u_1, \\ \text{impulse on B} &= +\vec F\,\Delta t = m_2\vec v_2 - m_2\vec u_2. \end{aligned}

Adding, the right-hand impulses cancel, so $(m_1\vec v_1 + m_2\vec v_2) - (m_1\vec u_1 + m_2\vec u_2) = \vec 0$ — total momentum is unchanged. Crucially this needs no assumption about the collision being elastic; momentum is conserved even when kinetic energy is not.

The three collision types

Type	Momentum	Kinetic energy
Perfectly elastic	conserved	conserved
Inelastic	conserved	some lost
Perfectly inelastic (coalescence)	conserved	maximum loss

The vital point — and a frequent exam discriminator — is that the first column is always "conserved" as long as no external resultant force acts. Kinetic energy is the variable: it is conserved only in the perfectly elastic case $(e=1)$ and is maximally lost when the bodies coalesce. Never assume KE is conserved unless you are told the collision is elastic.

Kinetic energy lost

When KE is not conserved, the energy lost in a collision is

$\Delta KE = \Big(\tfrac12 m_1 u_1^2 + \tfrac12 m_2 u_2^2\Big) - \Big(\tfrac12 m_1 v_1^2 + \tfrac12 m_2 v_2^2\Big),$

i.e. total KE before minus total KE after. For the special case of coalescence with the second body at rest, substituting the common velocity $v = \dfrac{m_1 u_1}{m_1 + m_2}$ and simplifying gives the compact formula

$\Delta KE = \frac{m_1 m_2}{2(m_1 + m_2)}\,(u_1 - u_2)^2,$

which depends only on the masses and the relative approach speed $u_1 - u_2$ . This is well worth memorising as a check on coalescence problems.

3 Worked examples with mark scheme

Worked Example 1 — coalescence

A $4\ \mathrm{kg}$ object moving at $6\ \mathrm{m\,s^{-1}}$ strikes a $2\ \mathrm{kg}$ object at rest; they coalesce. Find the common velocity and the kinetic energy lost.

\begin{aligned} m_1 u_1 + m_2 u_2 &= (m_1 + m_2)v \\ 4(6) + 2(0) &= 6v \;\Rightarrow\; v = 4\ \mathrm{m\,s^{-1}}. \end{aligned}

Kinetic energy: before $=\tfrac12(4)(6^2) = 72\ \mathrm{J}$ ; after $=\tfrac12(6)(4^2) = 48\ \mathrm{J}$ ; so $\Delta KE = 24\ \mathrm{J}$ lost.

Mark scheme: M1 conservation equation with coalescence on the right; A1 $v=4$ ; M1 KE before $-$ KE after; A1 $24\ \mathrm{J}$ . (4 marks.)

Worked Example 2 — both bodies moving

A $3\ \mathrm{kg}$ ball at $5\ \mathrm{m\,s^{-1}}$ overtakes and strikes a $2\ \mathrm{kg}$ ball moving at $1\ \mathrm{m\,s^{-1}}$ in the same direction. Afterwards the $3\ \mathrm{kg}$ ball moves at $3\ \mathrm{m\,s^{-1}}$ . Find the velocity of the $2\ \mathrm{kg}$ ball.

\begin{aligned} 3(5) + 2(1) &= 3(3) + 2v \\ 17 &= 9 + 2v \;\Rightarrow\; v = 4\ \mathrm{m\,s^{-1}}. \end{aligned}

Mark scheme: M1 full conservation equation; A1 $v=4\ \mathrm{m\,s^{-1}}$ . A quick physical check: the $2\ \mathrm{kg}$ ball ends faster than the $3\ \mathrm{kg}$ ball $(4>3)$ , so they have separated — consistent with a real collision.

Worked Example 3 — explosion / recoil

A $5\ \mathrm{kg}$ shell at rest bursts into a $2\ \mathrm{kg}$ fragment moving at $10\ \mathrm{m\,s^{-1}}$ to the right and a $3\ \mathrm{kg}$ fragment. Find the velocity of the $3\ \mathrm{kg}$ fragment.

Total momentum before $= 0$ (at rest):

0 = 2(10) + 3v \;\Rightarrow\; 3v = -20 \;\Rightarrow\; v = -\tfrac{20}{3} \approx -6.67\ \mathrm{m\,s^{-1}}.

The $3\ \mathrm{kg}$ fragment moves at $6.67\ \mathrm{m\,s^{-1}}$ to the left. Mark scheme: M1 total momentum $=0$ ; A1 $-\tfrac{20}{3}\ \mathrm{m\,s^{-1}}$ . In an explosion KE is created (from chemical energy), yet momentum is still conserved — a point worth stating.

Worked Example 4 — head-on coalescence

Two particles, each of mass $m$ , travel toward each other at the same speed $v$ and coalesce on impact. Find the speed of the combined particle.

Take one direction as positive, so the velocities are $+v$ and $-v$ :

$m(v) + m(-v) = 2m\,V \;\Rightarrow\; 0 = 2mV \;\Rightarrow\; V = 0.$

The combined mass is stationary — all the kinetic energy $\left(2 \times \tfrac12 mv^2 = mv^2\right)$ is lost (converted to heat and deformation). Mark scheme: M1 signed momentum equation; A1 $V = 0$ with the comment that all KE is lost. This is the extreme of inelasticity: equal and opposite momenta cancel exactly.

Worked Example 5 — 2-D coalescence with energy

A particle of mass $2\ \mathrm{kg}$ with velocity $(4\vec i + 3\vec j)\ \mathrm{m\,s^{-1}}$ coalesces with a $3\ \mathrm{kg}$ particle at rest. Find the common velocity, the speed, and the fraction of KE lost.

Momentum is conserved component-wise, so as a vector:

$2(4\vec i + 3\vec j) + 3(\vec 0) = 5\vec v \;\Rightarrow\; \vec v = \tfrac15(8\vec i + 6\vec j) = (1.6\vec i + 1.2\vec j)\ \mathrm{m\,s^{-1}}.$

Speed $= \sqrt{1.6^2 + 1.2^2} = \sqrt{4} = 2\ \mathrm{m\,s^{-1}}$ . KE before $= \tfrac12(2)(4^2+3^2) = 25\ \mathrm{J}$ ; after $= \tfrac12(5)(2^2) = 10\ \mathrm{J}$ ; fraction lost $= \tfrac{15}{25} = 60\%$ . Mark scheme: M1 conservation as a vector; A1 $(1.6\vec i + 1.2\vec j)$ ; B1 speed $2$ ; M1 KE before and after; A1 $60\%$ . The fraction matches $\dfrac{m_2}{m_1+m_2} = \dfrac{3}{5}$ — the standard coalescence result.

Worked Example 6 — finding an unknown mass

A particle of mass $m$ moving at $8\ \mathrm{m\,s^{-1}}$ strikes a stationary particle of mass $4\ \mathrm{kg}$ ; they coalesce and move off together at $3\ \mathrm{m\,s^{-1}}$ . Find $m$ .

\begin{aligned} m(8) + 4(0) &= (m + 4)(3)\\ 8m &= 3m + 12 \;\Rightarrow\; 5m = 12 \;\Rightarrow\; m = 2.4\ \mathrm{kg}. \end{aligned}

Mark scheme: M1 conservation equation with the unknown mass on both sides; M1 rearrange to isolate $m$ ; A1 $m = 2.4\ \mathrm{kg}$ . This type — where the mass, not a velocity, is unknown — tests whether you can treat the conservation equation as a linear equation in any one symbol, not just a velocity.

Worked Example 7 — both moving, with energy lost

A $5\ \mathrm{kg}$ block moving at $4\ \mathrm{m\,s^{-1}}$ catches up with and strikes a $3\ \mathrm{kg}$ block moving at $1\ \mathrm{m\,s^{-1}}$ in the same direction; they coalesce. Find the common velocity and the kinetic energy lost.

\begin{aligned} 5(4) + 3(1) &= (5 + 3)v\\ 23 &= 8v \;\Rightarrow\; v = 2.875\ \mathrm{m\,s^{-1}}. \end{aligned}

KE before $= \tfrac12(5)(4^2) + \tfrac12(3)(1^2) = 40 + 1.5 = 41.5\ \mathrm{J}$ ; KE after $= \tfrac12(8)(2.875^2) = \tfrac12(8)(8.266) = 33.06\ \mathrm{J}$ ; KE lost $\approx 8.44\ \mathrm{J}$ .

Check with the relative-speed formula: $\dfrac{m_1 m_2}{2(m_1+m_2)}(u_1-u_2)^2 = \dfrac{15}{16}(4-1)^2 = \dfrac{15}{16}(9) = 8.4375\ \mathrm{J}$ ✓. Mark scheme: M1 conservation; A1 $v = 2.875$ ; M1 KE before − KE after; A1 $\approx 8.44\ \mathrm{J}$ . The independent check via the relative-speed formula is the kind of self-verification examiners love to see.

Worked Example 8 — recoil of a person on a trolley

A person of mass $60\ \mathrm{kg}$ stands on a stationary trolley of mass $40\ \mathrm{kg}$ on smooth rails. The person jumps off horizontally at $2\ \mathrm{m\,s^{-1}}$ relative to the ground. Find the recoil speed of the trolley.

The person-plus-trolley system starts at rest, so total momentum is zero throughout (smooth rails, no external horizontal force). Taking the person's jump direction as positive:

$0 = 60(2) + 40 V \;\Rightarrow\; 40V = -120 \;\Rightarrow\; V = -3\ \mathrm{m\,s^{-1}}.$

The trolley recoils at $3\ \mathrm{m\,s^{-1}}$ in the opposite direction. Mark scheme: M1 total momentum $= 0$ ; M1 signed equation with the person and trolley in opposite directions; A1 $3\ \mathrm{m\,s^{-1}}$ backwards. This is conceptually identical to the gun-and-shell recoil — the internal "push" of the jump conserves total momentum, sending the lighter trolley off faster than the heavier person, exactly as the mass ratio dictates.

Worked Example 9 — explosion into three fragments

A $6\ \mathrm{kg}$ shell at rest bursts into three fragments. A $1\ \mathrm{kg}$ fragment flies off at $8\ \mathrm{m\,s^{-1}}$ along $+\vec i$ , and a $2\ \mathrm{kg}$ fragment at $5\ \mathrm{m\,s^{-1}}$ along $+\vec j$ . Find the velocity of the remaining $3\ \mathrm{kg}$ fragment.

Total momentum is zero, so the three momenta sum to $\vec 0$ :

\begin{aligned} \vec 0 &= 1(8\vec i) + 2(5\vec j) + 3\vec v\\ 3\vec v &= -(8\vec i + 10\vec j) \;\Rightarrow\; \vec v = -\tfrac83\vec i - \tfrac{10}{3}\vec j\ \mathrm{m\,s^{-1}}. \end{aligned}

Speed $= \dfrac13\sqrt{8^2 + 10^2} = \dfrac13\sqrt{164} = \dfrac{2\sqrt{41}}{3} \approx 4.27\ \mathrm{m\,s^{-1}}$ , directed into the third quadrant (opposing the resultant of the other two). Mark scheme: M1 total momentum $=\vec 0$ as a vector; M1 solve each component; A1 $\vec v = -\tfrac83\vec i - \tfrac{10}{3}\vec j$ ; A1 speed $\approx 4.27\ \mathrm{m\,s^{-1}}$ . Three-fragment explosions are pure vector bookkeeping: the third momentum is minus the resultant of the first two, so it must point "between and opposite" the other two fragments — a useful direction check.

4 Specimen-style exam question

(Specimen-style.) A particle $P$ of mass $2\ \mathrm{kg}$ moving with velocity $(4\vec i + 3\vec j)\ \mathrm{m\,s^{-1}}$ collides with a particle $Q$ of mass $3\ \mathrm{kg}$ at rest. They coalesce. (a) Find the common velocity. (b) Find the speed of the combined particle. (c) Find the kinetic energy lost.

(a) Momentum is conserved in each component, so as a vector:

2(4\vec i + 3\vec j) + 3(\vec 0) = 5\vec v \;\Rightarrow\; \vec v = \tfrac15(8\vec i + 6\vec j) = (1.6\vec i + 1.2\vec j)\ \mathrm{m\,s^{-1}}.

Conservation of Momentum

Conservation of Momentum

1 Where this sits in AQA 7367

2 Core theory — statement and proof

The principle

Why it is true (Newton's third law)

The three collision types

Kinetic energy lost

3 Worked examples with mark scheme

Worked Example 1 — coalescence

Worked Example 2 — both bodies moving

Worked Example 3 — explosion / recoil

Worked Example 4 — head-on coalescence

Worked Example 5 — 2-D coalescence with energy

Worked Example 6 — finding an unknown mass

Worked Example 7 — both moving, with energy lost

Worked Example 8 — recoil of a person on a trolley

Worked Example 9 — explosion into three fragments

4 Specimen-style exam question

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