Conservation of Momentum in 2D

At GCSE, conservation of momentum problems were one-dimensional — objects colliding or exploding along a single line. At A-Level, you must handle collisions where objects move in two dimensions. The key insight is that momentum is conserved independently along each axis. This lesson builds your fluency with component resolution, 2D collision solving, and explosion problems.

Momentum as a Vector

Momentum p = mv is a vector. This means it has both magnitude and direction. When we say momentum is conserved, we mean the total vector momentum of the system is unchanged — not just its magnitude. In two dimensions, this means:

$\sum p_x \text{ (before)} = \sum p_x \text{ (after)}$ $\sum p_y \text{ (before)} = \sum p_y \text{ (after)}$

Conservation of momentum applies separately along the x-axis and the y-axis. This gives us two independent equations, which is what makes 2D problems solvable.

Resolving Momentum into Components

If an object of mass m moves at speed v at angle θ to the x-axis, its momentum components are:

$p_x = mv\cos\theta$ $p_y = mv\sin\theta$

This is exactly the same as resolving any vector into perpendicular components.

flowchart TD
    A["2D Momentum Problem"] --> B["Step 1: Draw a diagram\nShow all velocities and angles\nbefore and after collision"]
    B --> C["Step 2: Choose coordinate axes\nAlign x with one object’s initial direction"]
    C --> D["Step 3: Resolve ALL momenta\ninto x and y components"]
    D --> E["Step 4: Apply conservation\nSeparately in x and y"]
    E --> F["Step 5: Solve for unknowns\nTwo equations → two unknowns"]
    F --> G["Step 6: Find resultant\nv = √(vx² + vy²)\nθ = tan⁻¹(vy/vx)"]
    G --> H["Step 7: Check\nVerify p_total is the same\nbefore and after"]

Worked Example 1: Snooker Ball Collision

A white cue ball of mass 0.17 kg travelling at 5.0 m s⁻¹ strikes a stationary red ball of the same mass. After the collision, the white ball moves at 3.0 m s⁻¹ at 40° to its original direction. Find the speed and direction of the red ball.

Before the collision:

White: pₓ = 0.17 × 5.0 = 0.85 N s, p_y = 0
Red: pₓ = 0, p_y = 0
Total: pₓ = 0.85 N s, p_y = 0

After the collision (white ball):

pₓ = 0.17 × 3.0 × cos 40° = 0.17 × 3.0 × 0.766 = 0.391 N s
p_y = 0.17 × 3.0 × sin 40° = 0.17 × 3.0 × 0.643 = 0.328 N s

Red ball after collision (using conservation):

pₓ = 0.85 − 0.391 = 0.459 N s → vₓ = 0.459 / 0.17 = 2.70 m s⁻¹
p_y = 0 − 0.328 = −0.328 N s → v_y = −0.328 / 0.17 = −1.93 m s⁻¹

Speed of red ball: $v = \sqrt{2.70^2 + 1.93^2} = \sqrt{7.29 + 3.72} = \sqrt{11.01} = 3.3 \text{ m s}^{-1}$

Direction of red ball: $\theta = \tan^{-1}\left(\frac{1.93}{2.70}\right) = 35.6°$

The red ball moves at 3.3 m s⁻¹ at 35.6° below the original direction of the white ball.

Oblique Collisions

In an oblique collision, two objects meet at an angle rather than head-on. The method remains the same:

Choose a coordinate system (typically align x with one object's initial direction)
Resolve all momenta into x and y components before the collision
Apply conservation of momentum in x and y separately
Solve the two equations simultaneously for the unknowns

Worked Example 2: Two-Car Collision at a Junction

Car A (mass 1200 kg) travels east at 15 m s⁻¹. Car B (mass 800 kg) travels north at 20 m s⁻¹. They collide and lock together. Find the speed and direction of the wreckage.

Take east as x-positive, north as y-positive.

x-direction: $1200 \times 15 + 800 \times 0 = (1200 + 800) \times v_x$ $18{,}000 = 2000 \times v_x$ $v_x = 9.0 \text{ m s}^{-1}$

y-direction: $1200 \times 0 + 800 \times 20 = 2000 \times v_y$ $16{,}000 = 2000 \times v_y$ $v_y = 8.0 \text{ m s}^{-1}$

Speed: $v = \sqrt{9.0^2 + 8.0^2} = \sqrt{81 + 64} = \sqrt{145} = 12.0 \text{ m s}^{-1}$

Direction: $\theta = \tan^{-1}\left(\frac{8.0}{9.0}\right) = 41.6° \text{ north of east}$

Worked Example 3: Oblique Glancing Collision

A 3.0 kg puck A travels east at 8.0 m s⁻¹ and a 2.0 kg puck B travels north at 6.0 m s⁻¹. After collision, puck A moves at 5.0 m s⁻¹ at 30° north of east. Find the velocity of puck B.

x-direction: $3.0 \times 8.0 + 2.0 \times 0 = 3.0 \times 5.0\cos 30° + 2.0 \times v_{\text{Bx}}$ $24.0 = 12.99 + 2.0 v_{\text{Bx}}$ $v_{\text{Bx}} = \frac{24.0 - 12.99}{2.0} = 5.51 \text{ m s}^{-1}$

y-direction: $3.0 \times 0 + 2.0 \times 6.0 = 3.0 \times 5.0\sin 30° + 2.0 \times v_{\text{By}}$ $12.0 = 7.50 + 2.0 v_{\text{By}}$ $v_{\text{By}} = \frac{12.0 - 7.50}{2.0} = 2.25 \text{ m s}^{-1}$

Speed of B: v = √(5.51² + 2.25²) = √(30.36 + 5.06) = √35.42 = 5.95 m s⁻¹

Direction: θ = tan⁻¹(2.25/5.51) = 22.2° north of east

Explosions in 2D

Conservation of momentum also applies to explosions and separations in 2D. If a stationary firework explodes into three fragments, the total momentum of all three fragments must be zero (since the initial momentum was zero). This means the momentum vectors of the three fragments form a closed triangle when drawn tip-to-tail.

Worked Example 4

A 3.0 kg shell explodes into three fragments. Fragment A (1.0 kg) moves east at 30 m s⁻¹. Fragment B (1.2 kg) moves north at 25 m s⁻¹. Find the velocity of fragment C (0.8 kg).

Total initial momentum = 0 in both directions.

x-direction: 1.0 × 30 + 1.2 × 0 + 0.8 × vₓ = 0 → vₓ = −30/0.8 = −37.5 m s⁻¹

y-direction: 1.0 × 0 + 1.2 × 25 + 0.8 × v_y = 0 → v_y = −30/0.8 = −37.5 m s⁻¹

Speed: v = √(37.5² + 37.5²) = 37.5√2 = 53.0 m s⁻¹

Direction: θ = tan⁻¹(37.5/37.5) = 45° south of west (or 225° from east)

Worked Example 5: Two-Fragment Explosion from a Moving Object

A 5.0 kg projectile travelling east at 40 m s⁻¹ at the top of its trajectory explodes into two fragments. Fragment A (2.0 kg) moves at 60 m s⁻¹ at 30° above the eastward direction. Find the velocity of fragment B (3.0 kg).

Initial momentum: pₓ = 5.0 × 40 = 200 N s east, p_y = 0.

x-direction: $200 = 2.0 \times 60\cos 30° + 3.0 \times v_{\text{Bx}}$ $200 = 103.9 + 3.0 v_{\text{Bx}}$ $v_{\text{Bx}} = 32.0 \text{ m s}^{-1}$

y-direction: $0 = 2.0 \times 60\sin 30° + 3.0 \times v_{\text{By}}$ $v_{\text{By}} = -20.0 \text{ m s}^{-1}$

Speed of B: v = √(32.0² + 20.0²) = √(1024 + 400) = √1424 = 37.7 m s⁻¹

Direction: θ = tan⁻¹(20.0/32.0) = 32.0° below east

Key Exam Tips for 2D Momentum Problems

Always draw a diagram showing velocities and angles before and after
Define a clear coordinate system and stick to it throughout
Resolve every velocity into components — do not try to use magnitudes alone
Write two separate equations — one for x, one for y
Check your answer by verifying that total momentum in each direction is conserved
Units: momentum is in kg m s⁻¹ or equivalently N s

Common mistake: Using the magnitude of momentum directly in 2D problems. You cannot write "total momentum before = total momentum after" as a single scalar equation. You MUST resolve into components. The only exception is if all motion is along one line.

Error	Why It Is Wrong	Correct Approach
Adding momentum magnitudes in 2D	Momentum is a vector; magnitudes do not add unless co-linear	Resolve into x and y components separately
Forgetting one object's y-component	All objects contribute to both axes	If an object moves at an angle, resolve both components
Using speed instead of velocity component	Speed ignores direction	Use v cos θ and v sin θ consistently
Wrong sign for deflected object	y-component direction depends on angle choice	If cue ball goes up, object ball goes down (and vice versa)

Summary

Momentum is conserved independently in each perpendicular direction
To solve 2D problems: resolve all momenta into x and y components, apply conservation in each direction separately
Oblique collisions require resolving both objects' momenta before and after
For explosions from rest, the total momentum in every direction must sum to zero
For explosions from a moving object, the total momentum must equal the initial momentum
Always define your coordinate system clearly and use consistent signs

A-Level Deep Dive: Conservation of Momentum in 2D

Spec mapping

Edexcel 9PH0 specification Topic 6 — Further Mechanics covers momentum as a vector quantity, conservation of linear momentum in two dimensions for systems of interacting bodies, and the analysis of collisions and explosions where motion is not confined to a single line (refer to the official specification document for exact wording). The 2D extension builds directly on the 1D conservation work earlier in Topic 6 and is examined predominantly on Paper 2 (Advanced Physics II), although collision and explosion problems also appear in synoptic Paper 3 contexts. The Edexcel formula booklet provides $p = mv$ but does not state that momentum is conserved as a vector — this is assumed knowledge. Candidates must therefore know that conservation applies independently in any two perpendicular directions, and must choose appropriate axes to exploit that independence.

Worked example with full mark scheme

Question (8 marks):

A snooker cue ball of mass $0.17\,\text{kg}$ travels at $3.0\,\text{m s}^{-1}$ along the positive $x$ -direction and strikes a stationary red ball of mass $0.16\,\text{kg}$ in a glancing collision. After the collision, the cue ball moves at $1.8\,\text{m s}^{-1}$ at $30°$ above the $x$ -axis.

(a) Calculate the speed of the red ball and its direction relative to the original line of motion of the cue ball. (6)

(b) State, with a calculation, whether the collision is elastic. (2)

Solution with mark scheme:

(a) Step 1 — define axes and write conservation equations.

Let positive $x$ be the cue ball's initial direction and positive $y$ be perpendicular (in the plane of the table). Let the red ball move with speed $v_r$ at angle $\phi$ below the $x$ -axis (since the cue ball deflects above, the red ball must deflect below to conserve $y$ -momentum from zero).

M1 — correct statement that momentum conserves separately in $x$ and $y$ , with axes defined.

Step 2 — conserve $x$ -momentum.

$m_c u_c = m_c v_c \cos 30° + m_r v_r \cos\phi$ $0.17 \times 3.0 = 0.17 \times 1.8 \cos 30° + 0.16 \, v_r \cos\phi$ $0.510 = 0.2650 + 0.16 \, v_r \cos\phi$ $v_r \cos\phi = 1.531\,\text{m s}^{-1}$

M1 — correct $x$ -equation including $\cos 30°$ on the cue-ball term. A1 — correct numerical value for $v_r \cos\phi$ .

Step 3 — conserve $y$ -momentum.

$0 = m_c v_c \sin 30° - m_r v_r \sin\phi$ $0 = 0.17 \times 1.8 \times 0.5 - 0.16 \, v_r \sin\phi$ $v_r \sin\phi = 0.9563\,\text{m s}^{-1}$

M1 — correct $y$ -equation with the negative sign reflecting the chosen direction convention.

Step 4 — combine to find $v_r$ and $\phi$ .

Square and add: $v_r^2 = 1.531^2 + 0.9563^2 = 2.344 + 0.9145 = 3.259$ $v_r = 1.81\,\text{m s}^{-1}$

Divide: $\tan\phi = \dfrac{0.9563}{1.531} = 0.6246 \implies \phi = 32.0°$

A1 — $v_r = 1.8\,\text{m s}^{-1}$ (2 s.f.) at $32°$ below the original line.

(b) Step 1 — compute KE before and after.

$\text{KE}_\text{before} = \tfrac{1}{2}(0.17)(3.0)^2 = 0.7650\,\text{J}$ .

$\text{KE}_\text{after} = \tfrac{1}{2}(0.17)(1.8)^2 + \tfrac{1}{2}(0.16)(1.81)^2 = 0.2754 + 0.2620 = 0.5374\,\text{J}$ .

M1 — KE calculation for both states.

A1 — KE is not conserved ( $\Delta\text{KE} \approx -0.23\,\text{J}$ ); the collision is inelastic.

Total: 8 marks (M4 A4).

Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A firework of mass $0.40\,\text{kg}$ travelling vertically upwards at $12\,\text{m s}^{-1}$ explodes at the top of its trajectory into two unequal fragments. Fragment A, of mass $0.15\,\text{kg}$ , is observed moving at $25\,\text{m s}^{-1}$ on a bearing $60°$ east of vertical-up. Determine the velocity (magnitude and direction) of fragment B immediately after the explosion. (6)

Mark scheme decomposition by AO:

M1 (AO1.1) — recognising momentum conserves as a vector and applying conservation separately in the vertical and horizontal directions.
M1 (AO2.1) — resolving fragment A's velocity: $25\sin 60° = 21.65\,\text{m s}^{-1}$ horizontal; $25\cos 60° = 12.5\,\text{m s}^{-1}$ vertical.
M1 (AO1.2) — vertical conservation: $0.40 \times 12 = 0.15 \times 12.5 + 0.25 \, v_{B,y}$ , giving $v_{B,y} = 11.7\,\text{m s}^{-1}$ upward.
M1 (AO1.2) — horizontal conservation: $0 = 0.15 \times 21.65 + 0.25 \, v_{B,x}$ , giving $v_{B,x} = -13.0\,\text{m s}^{-1}$ (i.e. $13.0\,\text{m s}^{-1}$ west).
A1 (AO2.2) — magnitude $|v_B| = \sqrt{11.7^2 + 13.0^2} = 17.5\,\text{m s}^{-1}$ .
A1 (AO2.5) — direction: $\arctan(13.0/11.7) \approx 48°$ west of vertical-up, with sign and reference clearly stated.

Total: 6 marks split AO1 = 3, AO2 = 3. Explosion problems are favoured on Paper 2 because they reward clean axis choice (vertical/horizontal here) and punish candidates who try to work in a single direction or who forget that the parent body had non-zero momentum at the instant of explosion.

Synoptic links

Connects to:

Topic 6 — 1D conservation of momentum: the 1D case is the special case where all velocities lie along a single chosen axis, so the perpendicular component equation reduces to $0 = 0$ . Every 2D problem decomposes into two independent 1D problems — confidence with the 1D method is a prerequisite, not an alternative.
Topic 4 — Mechanics, vectors and resolution: resolving velocities into perpendicular components using $v_x = v\cos\theta$ , $v_y = v\sin\theta$ is the core technique imported from kinematics. The same trigonometric resolution that handles inclined-plane forces handles inclined-velocity collisions.
Topic 9 — Thermal physics, kinetic theory: the derivation of $pV = \tfrac{1}{3}Nm\overline{c^2}$ rests on momentum change at a wall in 2D/3D — only the component perpendicular to the wall reverses, while parallel components are unaffected. Conservation of momentum in components is exactly the principle invoked there.
Topic 12 — Particle physics, scattering: alpha-particle scattering off gold nuclei is a 2D conservation-of-momentum problem at heart. The deflection angle of the alpha particle and the recoil direction of the nucleus are linked by exactly the equations used above, with the added subtlety that elastic scattering from a much heavier target reflects the alpha with little energy loss.
Topic 13 — Nuclear physics, decay kinematics: when a stationary nucleus emits an alpha and a recoil daughter, the two products fly apart back-to-back in the centre-of-momentum frame. This is a 1D explosion. When the parent is moving (e.g. a beam of unstable particles), the 2D conservation equations of this lesson take over.