Edexcel IAL Physics: Materials — Revision Notes

Condensed for the final weeks. For the full explanation, use the Materials study guide.

Density and fluids

rho = m / V         upthrust = rho g V (weight of fluid displaced)

An object floats if its average density is less than the fluid’s.

Viscous drag — Stokes’ law:

F = 6 pi eta r v        (small sphere, laminar flow only)

Terminal velocity in a fluid occurs when weight = upthrust + viscous drag. Note that this is a three-force balance, not two — omitting upthrust is the standard error in the falling-sphere experiment.

Viscosity decreases as temperature increases for liquids, but increases with temperature for gases. That opposite behaviour is examined, and the reason differs: in liquids the intermolecular forces weaken; in gases the increased molecular motion raises momentum transfer between layers.

CORE PRACTICAL 2 uses a falling-ball method to determine a liquid’s viscosity, applying Stokes’ law together with the object’s weight and the upthrust acting on it.

Hooke’s law and elasticity

F = k x

Valid up to the limit of proportionality.

Order of features: limit of proportionality → elastic limit → yield point → breaking point.

Elastic deformation returns to original shape; plastic deformation is permanent.

A yield point is where some materials extend significantly with little or no increase in force, beyond the elastic limit. A stress-strain graph identifies the breaking stress — the maximum stress a material can withstand before it fractures — and allows direct comparison between different samples regardless of dimensions.

Stress, strain and the Young modulus

stress = F/A        strain = x/L        E = stress/strain = FL/Ax

The Young modulus is a property of the material, not the specimen. Two wires of the same metal have the same E regardless of length or thickness; only their stiffness k differs.

CORE PRACTICAL 3 determines the Young modulus of a wire by measuring extension against applied force and calculating stress and strain from its known cross-sectional area and original length.

Worked example. A wire of natural length 2.00 m and cross-sectional area 1.5 × 10⁻⁷ m² extends by 3.0 mm under a force of 45 N, within its limit of proportionality.

stress = 45 / (1.5 x 10^-7) = 3.0 x 10^8 Pa
strain = 0.0030 / 2.00      = 1.5 x 10^-3  (no units)
E = stress/strain = (3.0 x 10^8) / (1.5 x 10^-3) = 2.0 x 10^11 Pa

Experimental determination: use a long thin wire — long for a measurable extension, thin for a large stress from a modest load. Measure the diameter at several points with a micrometer and take a mean. E is the gradient of the stress–strain graph in the linear region.

Strain energy

E = 1/2 F x = 1/2 k x^2

This is the area under the force–extension graph — which is why you must find the area, not use the formula, when the graph is not linear.

For loading beyond the elastic limit, the loading and unloading curves differ and the area between them is energy dissipated, usually as heat. In rubber this is called hysteresis, and it is why a bouncing rubber ball warms up and does not return to its original height.

Material properties

Term	Meaning
Strong	High breaking stress
Stiff	High Young modulus
Tough	Absorbs much energy before fracture
Brittle	Breaks at the elastic limit with no plastic deformation
Ductile	Large plastic deformation; can be drawn into wire

These are five separate properties, and questions rely on distinguishing them. Glass is stiff and strong but brittle and not tough; copper is ductile and tough but less stiff.

Exam traps

• Omitting upthrust from the terminal velocity force balance in a fluid.
• Assuming viscosity behaves the same way with temperature in liquids and gases.
• Treating the Young modulus as a property of the object.
• Using ½Fx on a non-linear graph.
• Confusing strong, stiff and tough.
• Applying Stokes’ law to turbulent flow or to non-spherical objects.

Self-test

1. Give the three forces acting on a sphere falling at terminal velocity in a liquid.
2. How does viscosity respond to temperature in liquids and in gases?
3. Why do two wires of the same material have the same Young modulus?
4. What does the area between loading and unloading curves represent?
5. Distinguish stiff, strong and tough.
6. A wire of length 2.00 m and cross-sectional area 1.5 × 10⁻⁷ m² extends 3.0 mm under 45 N. Calculate the Young modulus.
7. What is a yield point?

Answers: 1. Weight downwards, upthrust upwards, and viscous drag upwards. 2. It decreases with temperature in liquids as intermolecular forces weaken, but increases with temperature in gases because faster molecular motion increases momentum transfer between layers. 3. The Young modulus is defined by stress and strain, which account for cross-sectional area and original length, so it depends only on the material. 4. The energy dissipated, usually as thermal energy — hysteresis. 5. Stiff means a high Young modulus; strong means a high breaking stress; tough means absorbing a large amount of energy before fracturing. 6. stress = 45 ÷ (1.5×10⁻⁷) = 3.0×10⁸ Pa; strain = 0.0030 ÷ 2.00 = 1.5×10⁻³; E = 2.0×10¹¹ Pa. 7. The point beyond the elastic limit where a material extends significantly with little or no increase in force.