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How do solids deform and how do objects move through fluids?

Hooke's law and the spring constant, stress, strain and the Young modulus, elastic strain energy, density and upthrust, and viscous drag with Stokes' law and terminal velocity.

A focused answer to the Edexcel 9PH0 materials and fluids content, covering Hooke's law, stress, strain and the Young modulus, elastic strain energy, density and upthrust, and viscous drag through Stokes' law and terminal velocity.

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What this dot point is asking

Edexcel wants you to apply Hooke's law and the spring constant, define stress, strain and the Young modulus and measure it, calculate elastic strain energy, use density and upthrust, and analyse viscous drag through Stokes' law and the idea of terminal velocity.

The answer

Hooke's law and the spring constant

The relationship is linear up to the limit of proportionality. Beyond the elastic limit a material is permanently deformed and does not return to its original length when the load is removed.

Stress, strain and the Young modulus

The Young modulus measures stiffness independent of the sample's shape. It is the gradient of the straight-line part of a stress-strain graph. The standard Edexcel core practical loads a long thin wire over a pulley, measures extension with a marker and ruler (or vernier), plots stress against strain, and takes the gradient. A long wire gives a measurable extension; a thin wire gives a high stress for modest loads.

Elastic strain energy

Density, upthrust and viscous drag

Density is ρ=mV\rho = \frac{m}{V}. An object immersed in a fluid feels an upthrust equal to the weight of fluid it displaces (Archimedes' principle), U=ρfVgU = \rho_f V g. A small sphere moving slowly through a viscous fluid feels a retarding drag given by Stokes' law:

As a falling sphere speeds up, the drag grows until it balances the net downward force; the sphere then falls at a constant terminal velocity. Setting weight equal to upthrust plus drag and solving for vv gives vterm=2r2g(ρsρf)9ηv_{\text{term}} = \frac{2r^2 g(\rho_s - \rho_f)}{9\eta}.

Examples in context

Bridge and building engineers choose materials by their Young modulus and yield stress so structures flex little under load and never reach the elastic limit. A bungee cord stores elastic strain energy that decelerates the jumper. Ships float because their hull displaces enough water for the upthrust to equal their weight. The falling-ball method measures the viscosity of oils and syrups using terminal velocity, and Stokes' law also governs how slowly fine dust or fog droplets settle through air.

Try this

Q1. Define the Young modulus. [1 mark]

  • Cue. The ratio of tensile stress to tensile strain in the linear (Hookean) region of the material.

Q2. A spring extends 5.05.0 cm under a 1010 N load. Find the spring constant. [2 marks]

  • Cue. k=Fx=100.050=200k = \frac{F}{x} = \frac{10}{0.050} = 200 N per metre.

Q3. State the condition for an object falling through a fluid to be at terminal velocity. [2 marks]

  • Cue. The net force is zero: the weight equals the upthrust plus the viscous drag, so the object falls at constant speed.

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20185 marksA metal wire of length 2.02.0 m and diameter 0.800.80 mm extends by 1.21.2 mm when a load of 5050 N is applied. Calculate the Young modulus of the metal.
Show worked answer →

Cross-sectional area: A=πr2=π(0.40×103)2=5.03×107A = \pi r^2 = \pi (0.40 \times 10^{-3})^2 = 5.03 \times 10^{-7} m squared.

Stress: σ=FA=505.03×107=9.94×107\sigma = \frac{F}{A} = \frac{50}{5.03 \times 10^{-7}} = 9.94 \times 10^{7} Pa. Strain: ε=ΔLL=1.2×1032.0=6.0×104\varepsilon = \frac{\Delta L}{L} = \frac{1.2 \times 10^{-3}}{2.0} = 6.0 \times 10^{-4}.

Young modulus: E=σε=9.94×1076.0×104=1.7×1011E = \frac{\sigma}{\varepsilon} = \frac{9.94 \times 10^{7}}{6.0 \times 10^{-4}} = 1.7 \times 10^{11} Pa.

Markers reward correct area from the diameter, stress and strain, and the value about 1.7×10111.7 \times 10^{11} Pa.

Edexcel 20214 marksA small sphere of radius 1.01.0 mm and density 25002500 kg per cubic metre falls through a liquid of viscosity 0.800.80 Pa s and density 900900 kg per cubic metre. Determine its terminal velocity.
Show worked answer →

At terminal velocity weight equals upthrust plus drag: 43πr3ρsg=43πr3ρfg+6πηrv\frac{4}{3}\pi r^3 \rho_s g = \frac{4}{3}\pi r^3 \rho_f g + 6\pi\eta r v.

Rearranging: v=2r2g(ρsρf)9η=2×(1.0×103)2×9.81×(2500900)9×0.80v = \frac{2r^2 g(\rho_s - \rho_f)}{9\eta} = \frac{2 \times (1.0 \times 10^{-3})^2 \times 9.81 \times (2500 - 900)}{9 \times 0.80}.

v=2×106×9.81×16007.2=0.03147.2=4.4×103v = \frac{2 \times 10^{-6} \times 9.81 \times 1600}{7.2} = \frac{0.0314}{7.2} = 4.4 \times 10^{-3} m per second.

Markers reward balancing weight, upthrust and Stokes drag, and the value about 4.4×1034.4 \times 10^{-3} m per second.

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