CP10 – 2D Collisions | Edexcel A Level Physics

Theory — Conservation of Momentum in 2D

When two spheres collide on a flat surface, momentum is conserved in both x and y directions independently.

Objectives

Observe 2D collisions between ball bearings on a flat surface recorded from above
Measure velocities and angles using path traces (as in Tracker software)
Verify that momentum is conserved in both x and y directions
Determine whether kinetic energy is conserved (elastic) or partly lost (inelastic)
Confirm that for equal-mass elastic collisions, θ₁ + θ₂ = 90°

2D Momentum Conservation

Momentum is a vector — it has direction. With ball 2 initially stationary, taking the original direction of ball 1 as the x-axis:

x: m₁u₁ = m₁v₁cosθ₁ + m₂v₂cosθ₂ y: 0 = m₁v₁sinθ₁ − m₂v₂sinθ₂

u₁ = speed of ball 1 before · v₁,v₂ = speeds after · θ₁ = angle of ball 1 from original direction · θ₂ = angle of ball 2 from original direction

The y-equation says the y-components of outgoing momenta are equal and opposite — they cancel. This is a key check: if py_after ≠ 0, there is a systematic error in your angle measurements.

Elastic vs Inelastic

KE before = ½m₁u₁² KE after = ½m₁v₁² + ½m₂v₂² % KE lost = (KE_before − KE_after) / KE_before × 100

Hard steel ball bearings: nearly elastic (~5% KE lost). Softer or different-mass balls: more inelastic.

Special Result — Equal Masses, Elastic

When m₁ = m₂ and the collision is elastic, conservation of momentum and KE together require:

θ₁ + θ₂ = 90° (always, regardless of impact parameter)

This follows from the fact that v₁ · v₂ = 0, which means the velocity vectors after are perpendicular.

Procedure

Using a camera overhead and Tracker software to analyse 2D collisions.

Equipment

Small steel ball bearings (two sizes) · Flat surface · Digital camera (overhead mount) · Computer with Tracker software · Ruler (for calibration) · Micrometer · Mass balance · Graph paper

1

Measure mass and diameter of each sphere

Weigh each ball on a balance. Measure diameter with a micrometer at several orientations. Record m₁ (moving ball) and m₂ (stationary ball).

💡 Steel ball bearings of equal diameter will have equal mass. Use different diameters to get a mass ratio ≠ 1.

2

Mount camera directly overhead

Fix the camera pointing straight down at the surface. Place a ruler in frame for calibration. The camera must be truly level — any tilt introduces parallax errors in angle measurements.

💡 In Tracker, use the calibration stick tool set to the ruler length. This converts pixel distances to metres.

3

Calibrate with a 1D head-on collision

Roll ball 1 along a ruler edge so it strikes ball 2 head-on. All motion stays in one line. This lets you establish the speed scale before attempting 2D collisions.

💡 In a perfectly elastic head-on collision between equal masses, ball 1 stops completely and ball 2 moves off at ball 1's original speed — a useful check.

4

Record 2D collisions at different angles

Place ball 2 (stationary) in the centre of the frame. Roll ball 1 along different approach lines so it strikes ball 2 off-centre at various impact parameters. After each collision record three tracks: ball 1 incoming, ball 1 outgoing, ball 2 outgoing.

💡 Mark ball 2's position on the paper between runs. Roll ball 1 from the same start position each time for consistent speeds.

5

Analyse with Tracker

Load each video. Use "point mass" to track each ball frame by frame. Tracker calculates velocities automatically. The path length between successive frames is proportional to speed. Use velocity overlay to show direction vectors.

💡 Track at least 5 frames before and after the collision for each ball. Average these to get reliable velocity values.

6

Measure angles and verify conservation

Draw lines through the path points before and after the collision. Measure θ₁ and θ₂ from the original direction of ball 1 using a protractor. Calculate px and py before and after. Check conservation.

💡 In this simulation, the angles are shown automatically on the protractor below the canvas. In real Tracker analysis you measure these from the path trace lines.

      🎱 Aim: Press and drag from the blue ball to set direction — release to fire.
    

Mass Ratio m₁ : m₂

Last Collision

—fire a collision

—

% KE lost

Collisions recorded: 0

Results Data Table

Each row = one collision. Check momentum conservation in x and y. For equal-mass elastic, θ₁ + θ₂ ≈ 90°.

#	m₁:m₂	u₁ / m/s	v₁ / m/s	θ₁ / °	v₂ / m/s	θ₂ / °	px before / kg m/s	px after / kg m/s	py after / kg m/s	KE lost %	θ₁+θ₂ / °
No collisions recorded yet.

Momentum Vector Diagram

Tip-to-tail vector addition of outgoing momenta must equal the incoming momentum vector. Select a collision to display.

Collision:

Vector magnitudes

p before (m₁u₁)—

p₁ after (m₁v₁)—

p₂ after (m₂v₂)—

Conservation check

px before—

px after—

py after (→0)—

% error in px—

Energy

KE before—

KE after—

% KE lost—

θ₁ + θ₂—

Discussion Questions

Write your answers and reveal model answers when ready.

Q1

For equal-mass elastic collisions, theory predicts θ₁ + θ₂ = 90°. Prove this result using conservation of momentum and kinetic energy.

Let m₁ = m₂ = m. Conservation of momentum (vector): mu₁ = mv₁ + mv₂, so u₁ = v₁ + v₂. Conservation of KE: ½mu₁² = ½mv₁² + ½mv₂², so u₁² = v₁² + v₂². Take the dot product of the momentum equation with itself: u₁·u₁ = (v₁+v₂)·(v₁+v₂) = v₁²+2v₁·v₂+v₂². Substituting u₁² = v₁²+v₂² gives 0 = 2v₁·v₂. Since the dot product is zero, v₁ and v₂ are perpendicular, so θ₁+θ₂ = 90°.

Q2

Explain why the y-component of total momentum must be zero after the collision, and what a non-zero py_after would indicate.

Before the collision, ball 1 moves along the x-direction only (by definition of our coordinate system) and ball 2 is stationary. So py_before = 0. By conservation of momentum, py_after = 0 also. This means m₁v₁sinθ₁ = m₂v₂sinθ₂ — the y-components of the two outgoing momenta are equal and opposite. If the measured py_after is significantly non-zero, likely causes include: the camera was not perfectly overhead (parallax), the angles were measured from the wrong reference, or the surface was tilted giving a gravitational y-component.

Q3

A ball bearing of mass 14 g moving at 0.35 m/s strikes a stationary ball of mass 14 g. After the collision ball 1 moves at 0.25 m/s at 45° and ball 2 moves at 0.25 m/s at 45° on the other side. Verify momentum conservation and determine whether the collision is elastic.

px_before = 0.014 × 0.35 = 4.90 × 10⁻³ kg m/s. px_after = 0.014 × 0.25 × cos45° + 0.014 × 0.25 × cos45° = 2 × 0.014 × 0.25 × 0.707 = 4.95 × 10⁻³ kg m/s ≈ 4.90 × 10⁻³ ✓. py_after = 0.014 × 0.25 × sin45° − 0.014 × 0.25 × sin45° = 0 ✓. KE_before = ½ × 0.014 × 0.35² = 8.575 × 10⁻⁴ J. KE_after = 2 × ½ × 0.014 × 0.25² = 8.75 × 10⁻⁴ J. Note KE_after > KE_before slightly — this is rounding error; in practice KE_after ≤ KE_before. With θ₁+θ₂ = 90° and approximately equal speeds, this is consistent with an elastic collision.

Q4

Explain how Tracker software uses the video frames to calculate the speed of each ball before and after the collision.

Tracker first calibrates the scale using a ruler of known length in the frame — converting pixels to metres. The user marks the centre of each ball in successive frames. Since the time between frames is 1/framerate (e.g. 1/30 s), and the displacement between marks is measured in metres, the velocity is displacement/time. The speed is the magnitude of this velocity vector. For accuracy, several frames before and after the collision are tracked and averaged. The direction of motion is the angle of the velocity vector measured from the original direction of ball 1.

Q5

Predict what happens in a head-on elastic collision (impact parameter = 0) between equal-mass balls. Confirm your prediction by trying this in the simulation.

For a perfectly head-on elastic collision between equal masses: ball 1 comes to a complete stop (v₁ = 0) and ball 2 moves off in exactly the original direction of ball 1 (θ₂ = 0°) at ball 1's original speed (v₂ = u₁). This follows from conservation of momentum: m·u₁ = m·v₁ + m·v₂, and conservation of KE: u₁² = v₁² + v₂². The only solution is v₁ = 0, v₂ = u₁. In the simulation, aim the cue directly at ball 2 (zero offset) — ball 1 should stop at the collision point and ball 2 should continue in a straight line.