A-Level Physics mechanics and materials: motion, momentum, energy and the Young modulus

Where mechanics and materials sit

In Edexcel IAL, impulse, momentum in two dimensions and circular motion are in Unit 4 (WPH14). In Cambridge 9702, circular motion is A Level topic 12.

The key ideas

For constant acceleration, v = u + at, s = ut + (1/2)at^2, v^2 = u^2 + 2as and s = (1/2)(u + v)t. A projectile has constant horizontal velocity (ignoring air resistance) and constant downward acceleration g, so treat horizontal and vertical motion separately and link them through time.

Newton's second law in its general form is F = Δp/Δt, the rate of change of momentum; for constant mass this is F = ma. Momentum p = mv is conserved in any collision if no external force acts. Kinetic energy is conserved only in elastic collisions. For equilibrium, the resultant force is zero and the sum of clockwise moments about any point equals the sum of anticlockwise moments.

Stress is force per unit cross-sectional area (σ = F/A) and strain is extension per unit original length (ε = Δx/L). The Young modulus E = σ/ε is a property of the material, not the sample. The area under a force-extension graph is the work done, which equals the elastic strain energy stored if the deformation is elastic: E = (1/2)FΔx for a material obeying Hooke's law.

Worked example 1: a ball is kicked horizontally at 15 m s^-1 off a 20 m cliff (g = 9.81 m s^-2)

1. Vertical motion: u = 0, s = 20 m, a = 9.81 m s^-2. From s = (1/2)at^2, t = sqrt(2 × 20 ÷ 9.81) = 2.02 s.
2. Horizontal motion: constant velocity, so distance = 15 × 2.02 = 30.3 m.
3. Vertical speed on landing: v = 9.81 × 2.02 = 19.8 m s^-1. Combine with the horizontal 15 m s^-1 to get the landing speed sqrt(15^2 + 19.8^2) = 24.8 m s^-1.

Worked example 2: finding the Young modulus of a wire

1. A wire 2.00 m long and 0.50 mm in diameter extends by 2.0 mm under a load of 40 N.
2. Area: A = π(d/2)^2 = π × (0.25 × 10^-3)^2 = 1.96 × 10^-7 m^2. Convert millimetres to metres first.
3. Stress = 40 ÷ 1.96 × 10^-7 = 2.04 × 10^8 Pa.
4. Strain = 2.0 × 10^-3 ÷ 2.00 = 1.0 × 10^-3 (no units).
5. E = stress ÷ strain = 2.04 × 10^11 Pa, which is typical of steel.

Worked example 3: a collision

1. A 2.0 kg trolley moving at 3.0 m s^-1 hits a stationary 1.0 kg trolley and they stick together.
2. Momentum before = 2.0 × 3.0 = 6.0 kg m s^-1. After: 3.0 × v = 6.0, so v = 2.0 m s^-1.
3. Kinetic energy before = (1/2)(2.0)(3.0)^2 = 9.0 J; after = (1/2)(3.0)(2.0)^2 = 6.0 J.
4. Kinetic energy is not conserved (3.0 J transferred to other stores), so the collision is inelastic, while momentum is still conserved.

Common mistakes that cost marks

• Using the cliff height for horizontal distance, or mixing horizontal and vertical quantities.
• Not converting mm to m, or using diameter instead of radius in the area.
• Forgetting that momentum is a vector, so velocities in opposite directions need opposite signs.
• Stating that energy is 'lost' in an inelastic collision without saying it is transferred, for example to internal energy.
• Incomplete definitions. For example, the moment of a force is force times the perpendicular distance from the pivot to the line of action of the force.
• Giving the Young modulus with units of strain or omitting Pa.

Exam technique and how a tutor helps

Learn the definitions from your specification word for word, because many 1 or 2 mark questions are definitions. In calculations, write the equation, substitute with units, and give the answer to a sensible number of significant figures with its unit. Draw a sketch with directions marked for any momentum or projectile question.

On Cambridge Paper 1 there are 40 multiple-choice questions in 1 hour 15 minutes, so speed with standard results matters. On Edexcel WPH11, longer questions often combine mechanics with an experimental context, such as how to measure g with light gates.

In one-to-one lessons a tutor drills the core definitions and unit conversions, then works through mixed past-paper questions from the student's board on the shared whiteboard, showing how each step earns credit.

Self-check: can you do these?

• Define the Young modulus. (Answer: the ratio of tensile stress to tensile strain)
• A 0.50 kg ball hits a wall at 8.0 m s^-1 and rebounds at 6.0 m s^-1. Find the change in momentum. (Answer: 7.0 kg m s^-1, away from the wall)
• A 60 kg student climbs 4.0 m in 5.0 s (g = 9.81). Find the useful power. (Answer: 471 W)
• What is the horizontal acceleration of a projectile with no air resistance? (Answer: zero)
• What does the area under a force-extension graph represent? (Answer: work done, or elastic strain energy if elastic)