Two-Dimensional Collisions — Resolving Momentum

Spec mapping: OCR H556 Module 3.3 — Conservation of linear momentum applied to two-dimensional interactions; resolution of momentum into perpendicular components; oblique and glancing collisions. (Refer to the official OCR H556 specification document for exact wording.)

So far (Lessons 6–8) we have treated collisions along a single line. Real collisions almost always occur in two or three dimensions — think of a snooker break, a glancing car crash at an intersection, an oblique scattering event between two air-track pucks, or an alpha particle deflecting off a heavy gold nucleus. In this lesson we extend momentum conservation to two dimensions by exploiting the fact that momentum is a vector, and therefore each perpendicular component is conserved independently.

This corresponds to OCR Module 3.3, and it is the place where a secure grasp of vector resolution from earlier lessons (Newton's first law applied to forces in 2-D, projectile motion) pays direct dividends. The technique is identical to resolving forces on an inclined plane: split each velocity into perpendicular components, write a separate conservation equation for each axis, and solve simultaneously.

Key Definition: For a closed system in two (or three) dimensions, the total linear momentum is conserved as a vector: $\sum_i m_i \vec{u}_i = \sum_i m_i \vec{v}_i$ . Equivalently, each independent perpendicular component is conserved separately: $\sum_i m_i u_{i,x} = \sum_i m_i v_{i,x}$ and $\sum_i m_i u_{i,y} = \sum_i m_i v_{i,y}$ .

The Key Principle

For a closed system (no external resultant force), each perpendicular component of momentum is conserved separately:

$\sum_i m_i u_{i,x} = \sum_i m_i v_{i,x} \qquad \text{(x-component)}$

$\sum_i m_i u_{i,y} = \sum_i m_i v_{i,y} \qquad \text{(y-component)}$

This gives you two independent equations per collision — one for each axis — and you treat each as a 1-D problem. The crucial setup step is to resolve each velocity into components along your chosen $x$ and $y$ axes (usually horizontal and vertical, or "along/perpendicular to the initial motion of one body").

Common Exam Mistake: Trying to combine $p_x$ and $p_y$ into a single scalar equation. Don't. Treat them as independent — solve $x$ and $y$ separately, then recombine the components at the end to find magnitudes and directions.

Diagram — Resolving Velocities

flowchart TD
  A[Each body has velocity v at angle θ] --> B[Resolve into components]
  B --> C["v_x = v cos θ"]
  B --> D["v_y = v sin θ"]
  C --> E[Σp_x before = Σp_x after]
  D --> F[Σp_y before = Σp_y after]
  E --> G[Solve simultaneous equations]
  F --> G
  G --> H["Recombine: |v| = √(v_x² + v_y²), tan θ = v_y/v_x"]

Choosing Axes

Pick axes that make the algebra as easy as possible. Common choices:

Line of initial motion: set $x$ along the direction of the incoming particle. This gives the target particle zero initial velocity in both components — the cleanest setup.
Standard horizontal/vertical (compass cardinals): useful when gravity is relevant, or when the natural problem axes are N-S and E-W.
Along and perpendicular to the line of contact at impact: sometimes used in oblique collisions when the geometry of the contact surface is given.

For most OCR A-Level questions, choice (1) is cleanest and gives the tidiest equations. The choice doesn't affect the physics; it just changes how messy the arithmetic looks.

Worked Example 1 — Glancing Equal-Mass Elastic Collision and the Perpendicularity Theorem

A $0.50$ kg ball moves at $4.0$ m s $^{-1}$ east and collides elastically with an identical stationary ball. After the collision, ball 1 moves at $30°$ north of east. Find the speeds of both balls and the direction of ball 2.

Take $x$ = east, $y$ = north. Masses $m_1 = m_2 = 0.50$ kg.

Perpendicularity theorem (a famous result, derived below): for an elastic collision between equal masses with one initially at rest, the two final velocity vectors are at right angles to each other.

Proof: Let $m_1 = m_2 = m$ . Momentum conservation (vector): $\vec{u}_1 = \vec{v}_1 + \vec{v}_2$ . Squaring (i.e. dotting with itself): $u_1^2 = v_1^2 + 2\,\vec{v}_1 \cdot \vec{v}_2 + v_2^2$ . KE conservation: $u_1^2 = v_1^2 + v_2^2$ . Subtracting: $2\,\vec{v}_1 \cdot \vec{v}_2 = 0$ , so $\vec{v}_1 \perp \vec{v}_2$ . ∎

So if ball 1 moves at $30°$ N of E, then ball 2 must move at $60°$ S of E. Let $v_1$ and $v_2$ be the unknown magnitudes.

$x$ -momentum: $0.50 \times 4.0 = 0.50 v_1 \cos 30° + 0.50 v_2 \cos 60°$ , i.e. $4.0 = 0.866 v_1 + 0.500 v_2$ ... (a)

$y$ -momentum: $0 = 0.50 v_1 \sin 30° - 0.50 v_2 \sin 60°$ , i.e. $0 = 0.500 v_1 - 0.866 v_2$ , so $v_1 = 1.732 v_2$ ... (b)

Substitute (b) into (a): $4.0 = 0.866(1.732 v_2) + 0.500 v_2 = 1.500 v_2 + 0.500 v_2 = 2.000 v_2$ , giving $v_2 = 2.0$ m s $^{-1}$ . Then $v_1 = 1.732 \times 2.0 = 3.464$ m s $^{-1}$ .

KE check: Before $= \frac{1}{2}(0.50)(16) = 4.0$ J. After $= \frac{1}{2}(0.50)(12) + \frac{1}{2}(0.50)(4) = 3 + 1 = 4$ J ✓.

Both conservation laws hold. Final answer: ball 1 at $3.46$ m s $^{-1}$ at $30°$ N of E; ball 2 at $2.0$ m s $^{-1}$ at $60°$ S of E.

The takeaway: in elastic equal-mass collisions with one body initially at rest, the outgoing angles are not independent — once one is fixed, the other follows from perpendicularity. Snooker players know this in their bones: a centre-line straight shot on a stationary cue ball gives the cue ball a stop and the target ball at full speed forward; a glancing shot makes the two balls fly off at a perfect right angle.

Worked Example 2 — Inelastic 2-D Collision (Pucks on Ice)

A $3.0$ kg puck moves north at $2.0$ m s $^{-1}$ and collides with a $4.0$ kg puck moving east at $1.5$ m s $^{-1}$ . They stick together. Find the common final velocity (magnitude and direction) and the KE lost.

Take $x$ = east, $y$ = north.

Before:

$p_x = (3.0)(0) + (4.0)(1.5) = 6.0$ kg m s $^{-1}$
$p_y = (3.0)(2.0) + (4.0)(0) = 6.0$ kg m s $^{-1}$

After (combined mass $7.0$ kg, common velocity $\vec{v}$ ):

$p_x$ : $7.0\,v_x = 6.0 \Rightarrow v_x = 0.857$ m s $^{-1}$
$p_y$ : $7.0\,v_y = 6.0 \Rightarrow v_y = 0.857$ m s $^{-1}$

Magnitude: $|\vec{v}| = \sqrt{0.857^2 + 0.857^2} = 0.857 \sqrt{2} = 1.212$ m s $^{-1}$ .

Direction: $\tan^{-1}(0.857/0.857) = 45°$ , so 45° north of east.

KE before: $\frac{1}{2}(3.0)(4.0) + \frac{1}{2}(4.0)(2.25) = 6.0 + 4.5 = 10.5$ J.

KE after: $\frac{1}{2}(7.0)(1.212)^2 = \frac{1}{2}(7.0)(1.470) = 5.14$ J.

KE lost: $10.5 - 5.14 = 5.36$ J, about $51\%$ of the original. As expected for a perfectly inelastic collision, a large fraction of the kinetic energy has been lost to sound and deformation in the pucks.

Worked Example 3 — Glancing Car Crash at an Intersection

A $1500$ kg car moving north at $20$ m s $^{-1}$ collides at an intersection with a $2000$ kg car moving east at $15$ m s $^{-1}$ . They lock together. Find the common velocity immediately after the collision and the KE lost.

Take $x$ = east, $y$ = north.

Before: $p_x = (1500)(0) + (2000)(15) = 30\,000$ kg m s $^{-1}$ ; $p_y = (1500)(20) + (2000)(0) = 30\,000$ kg m s $^{-1}$ .

After (combined $3500$ kg): $v_x = 30\,000/3500 = 8.57$ m s $^{-1}$ ; $v_y = 8.57$ m s $^{-1}$ .

Speed: $\sqrt{2 \times 8.57^2} = 8.57\sqrt{2} = 12.1$ m s $^{-1}$ .

Direction: $45°$ north of east (since $v_x = v_y$ ).

KE before: $\frac{1}{2}(1500)(400) + \frac{1}{2}(2000)(225) = 300\,000 + 225\,000 = 525\,000$ J.

KE after: $\frac{1}{2}(3500)(12.1)^2 = \frac{1}{2}(3500)(146.4) = 256\,000$ J.

KE lost: $525\,000 - 256\,000 \approx 268\,000$ J $\approx 51\%$ of the original.

Half a megajoule of kinetic energy has gone into deforming the vehicles, heating the metal panels, and generating the bang of the collision. This is why both cars are typically write-offs after a moderate-speed side-impact crash — most of the macroscopic KE has been converted to permanent deformation.

A forensic-physics application: if the wreckage's final velocity (direction and speed) can be measured by skid marks, momentum-conservation can be inverted to deduce the impact speeds of the two vehicles. This is exactly how accident reconstruction works — it's an A-Level physics technique used in real courtroom evidence.

When the Angles Are Given (Two Unknowns, Two Equations)

Some exam questions give you the final directions of both particles and ask for both final speeds. You have:

2 unknowns: $v_1$ and $v_2$ (the magnitudes of the outgoing velocities).
2 equations: $x$ -momentum and $y$ -momentum conservation.

This is always solvable as a simple simultaneous linear system. (If the collision is also given as elastic and the angles are not both specified, then you have three constraint equations and four unknowns — the angles must be self-consistent, or the system is over-determined.)

Worked Example 4 — Both Angles Given (Particle-Physics Scattering)

A particle of mass $4m$ travelling at $1.5 \times 10^7$ m s $^{-1}$ collides with a stationary particle of mass $m$ . After the collision, the heavier particle moves at angle $15°$ to its original direction with unknown speed $v_1$ , and the lighter particle moves at angle $60°$ on the opposite side with unknown speed $v_2$ . Find $v_1$ and $v_2$ . (Use only momentum conservation.)

Take $x$ along the original direction of the heavy particle; $y$ perpendicular.

Before: $p_x = 4m \times 1.5 \times 10^7 = 6.0m \times 10^7$ ; $p_y = 0$ .

After:

Heavy particle: $(4m\,v_1 \cos 15°, 4m\,v_1 \sin 15°) = (3.864 m v_1, 1.035 m v_1)$
Light particle: $(m\,v_2 \cos 60°, -m\,v_2 \sin 60°) = (0.500 m v_2, -0.866 m v_2)$

$x$ -conservation: $6.0 \times 10^7 = 3.864 v_1 + 0.500 v_2$ ... (a)

$y$ -conservation: $0 = 1.035 v_1 - 0.866 v_2 \Rightarrow v_2 = 1.195 v_1$ ... (b)

Substitute (b) into (a): $6.0 \times 10^7 = 3.864 v_1 + 0.500(1.195) v_1 = 4.462 v_1$ , giving $v_1 = 1.345 \times 10^7$ m s $^{-1}$ . Then $v_2 = 1.195 \times 1.345 \times 10^7 = 1.607 \times 10^7$ m s $^{-1}$ .

KE check: Before $= \frac{1}{2}(4m)(1.5 \times 10^7)^2 = 4.5 \times 10^{14}\,m$ . After $= \frac{1}{2}(4m)(1.345 \times 10^7)^2 + \frac{1}{2}(m)(1.607 \times 10^7)^2 = (3.619 + 1.291) \times 10^{14}\,m = 4.910 \times 10^{14}\,m$ .

The KE after is $\approx 9\%$ greater than before — meaning these specific angles are not fully consistent with a purely elastic collision. (In a real exam this would either be flagged as inelastic, or the question would specify different angles.) The momentum solution is still valid.

Worked Example 5 — Heavy Stationary Target (Rutherford-Like Scattering)

An alpha particle of mass $m_\alpha \approx 6.64 \times 10^{-27}$ kg moves at $2.0 \times 10^7$ m s $^{-1}$ and deflects through $90°$ after scattering off a stationary gold nucleus (mass $\approx 197 m_p \gg m_\alpha$ ). Find the alpha's final speed (assuming elastic scattering) and the magnitude of momentum transferred to the gold nucleus.

In the limit $m_{\text{target}} \to \infty$ , an elastic collision merely reflects the projectile off the target without changing its speed (like a tennis ball off a brick wall). So:

$v_{\alpha,\text{final}} = 2.0 \times 10^7 \text{ m s}^{-1}$

The direction has rotated by $90°$ — the alpha now moves perpendicular to its original velocity. The change in alpha momentum is the vector:

$\Delta \vec{p}_\alpha = \vec{p}_{\text{after}} - \vec{p}_{\text{before}} = m_\alpha v\,(\hat{j} - \hat{i})$

with magnitude $|\Delta \vec{p}_\alpha| = m_\alpha v \sqrt{2}$ :

$|\Delta \vec{p}_\alpha| = (6.64 \times 10^{-27})(2.0 \times 10^7)(\sqrt{2}) = 1.88 \times 10^{-19} \text{ kg m s}^{-1}$

By momentum conservation, this same magnitude has been transferred to the gold nucleus (in the opposite direction). The gold nucleus's speed is $1.88 \times 10^{-19} / (197 \times 1.66 \times 10^{-27}) = 5.75 \times 10^5$ m s $^{-1}$ — tiny on the scale of the alpha velocity, but in principle measurable.

This is precisely the experiment Rutherford's team ran in 1909–1911. From the angular distribution of alpha-particle deflections (some scattered through 90°, occasionally even straight back through 180°), they deduced that the positive charge of the atom must be concentrated in a tiny, massive nucleus — overturning the "plum-pudding" model and founding nuclear physics.

Two-Dimensional Collisions — Resolving Momentum

Two-Dimensional Collisions — Resolving Momentum

The Key Principle

Diagram — Resolving Velocities

Choosing Axes

Worked Example 1 — Glancing Equal-Mass Elastic Collision and the Perpendicularity Theorem

Worked Example 2 — Inelastic 2-D Collision (Pucks on Ice)

Worked Example 3 — Glancing Car Crash at an Intersection

When the Angles Are Given (Two Unknowns, Two Equations)

Worked Example 4 — Both Angles Given (Particle-Physics Scattering)

Worked Example 5 — Heavy Stationary Target (Rutherford-Like Scattering)

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