Two-Dimensional Collisions — Resolving Momentum

So far (Lessons 6–8) we have treated collisions along a single line. Real collisions usually occur in two or three dimensions — think of a snooker break, a glancing car crash, or an alpha particle scattering from a 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 3.5.2, and it is the place where a secure grasp of vector resolution from Lesson 1 (Newton's First Law) and projectile motion pays dividends.

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The Key Principle

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

Σp_x(before) = Σp_x(after) Σp_y(before) = Σp_y(after)

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

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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

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Choosing Axes

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

1. Line of initial motion: set x along the direction of the incoming particle. This gives the target particle zero initial velocity in both components.
2. Standard horizontal/vertical: useful when gravity is relevant.
3. Along and perpendicular to a line of contact: sometimes used in oblique collisions when you know the geometry of the impact.

For most OCR A-Level questions, choice (1) is cleanest.

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Worked Example 1 — Glancing Elastic Collision (Equal Masses)

A 0.50 kg ball moves at 4.0 m s⁻¹ east and collides elastically with an identical stationary ball. After the collision, the first ball moves at 3.0 m s⁻¹ at 30° north of east. Find the velocity (magnitude and direction) of the second ball.

Take x = east, y = north. Masses: m₁ = m₂ = 0.50 kg.

Before the collision:

• p_x = 0.50 × 4.0 + 0.50 × 0 = 2.0 kg m s⁻¹
• p_y = 0.50 × 0 + 0.50 × 0 = 0

After the collision:

• Ball 1 at 3.0 m s⁻¹, 30° N of E:
  • v₁_x = 3.0 cos 30° = 2.598 m s⁻¹
  • v₁_y = 3.0 sin 30° = 1.500 m s⁻¹
• Ball 2: unknown v₂_x, v₂_y.

Conservation of p_x:

2.0 = 0.50 × 2.598 + 0.50 × v₂_x 2.0 = 1.299 + 0.50 v₂_x v₂_x = (2.0 − 1.299) / 0.50 = 1.402 m s⁻¹

Conservation of p_y:

0 = 0.50 × 1.500 + 0.50 × v₂_y 0 = 0.750 + 0.50 v₂_y v₂_y = −1.500 m s⁻¹ (i.e. southwards)

Magnitude:

|v₂| = √(1.402² + 1.500²) = √(1.966 + 2.250) = √4.216 = 2.053 m s⁻¹

Direction:

θ = tan⁻¹(1.500 / 1.402) = 46.9° south of east

(Remember: the y component is negative, i.e. south.)

Check KE conservation (elastic):

• KE_before = ½ × 0.50 × 16 = 4.0 J
• KE_after = ½ × 0.50 × 9 + ½ × 0.50 × 4.216 = 2.25 + 1.054 = 3.304 J

That is not 4.0 J, so the collision as specified is not actually elastic — the question stated that it was, but the numbers don't quite work out. (This discrepancy illustrates an important point: in a genuinely elastic collision between equal masses with one initially at rest, the final velocities are perpendicular to each other. Let's see why, and then redo the problem correctly.)

Perpendicularity Result

For an elastic collision between equal masses with one initially at rest, the two final velocities are always at right angles to each other. This is a famous and exam-relevant result, easily verified from momentum and KE conservation.

Proof: Let m₁ = m₂ = m. Momentum conservation: u₁ = v₁ + v₂ (vectors). Squaring: u₁² = v₁² + 2 v₁·v₂ + v₂². KE conservation: u₁² = v₁² + v₂². Subtracting: 2 v₁·v₂ = 0, so v₁ and v₂ are perpendicular. ∎

So in a properly elastic equal-mass collision where ball 1 goes off at 30° N of E, ball 2 must go off at 60° S of E (the complement of 30°). Let's redo the example with this constraint.

Redone Example (Correctly Elastic)

With v₁ at 30° N of E and v₂ at 60° S of E, we have two unknown magnitudes (v₁ and v₂). The constraint is momentum conservation along x:

2.0 = 0.50 v₁ cos 30° + 0.50 v₂ cos 60° 4.0 = v₁ × 0.866 + v₂ × 0.500 ...(a)

Along y:

0 = 0.50 v₁ sin 30° − 0.50 v₂ sin 60° 0 = v₁ × 0.500 − v₂ × 0.866 v₁ = v₂ × (0.866 / 0.500) = 1.732 v₂ ...(b)

Substituting (b) into (a):

4.0 = 1.732 v₂ × 0.866 + 0.500 v₂ = 1.500 v₂ + 0.500 v₂ = 2.000 v₂ v₂ = 2.0 m s⁻¹

Then v₁ = 1.732 × 2.0 = 3.464 m s⁻¹.

Check KE: ½ × 0.50 × 12 + ½ × 0.50 × 4 = 3 + 1 = 4 J ✓

So both conservation laws hold, and the final speeds are 3.464 m s⁻¹ and 2.0 m s⁻¹ at 30° N of E and 60° S of E respectively. The take-away is that in elastic equal-mass collisions, the outgoing angles are not independent — once one is fixed, the other follows from the perpendicularity rule.

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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).

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 v):

• p_x: 7.0 × v_x = 6.0 ⇒ v_x = 0.857 m s⁻¹
• p_y: 7.0 × v_y = 6.0 ⇒ v_y = 0.857 m s⁻¹

Magnitude: |v| = √(0.857² + 0.857²) = 0.857 × √2 = 1.212 m s⁻¹ Direction: tan⁻¹(0.857/0.857) = 45°, so 45° north of east.

KE before:

• ½ × 3.0 × 4.0 + ½ × 4.0 × 2.25 = 6.0 + 4.5 = 10.5 J

KE after:

• ½ × 7.0 × 1.212² = ½ × 7.0 × 1.470 = 5.14 J