AQA A-level Physics (7408) · 3.6 Further mechanics and thermal physics
Mini-Lesson
Further mechanics and thermal physics
This mini-lesson covers AQA 3.6 — Further mechanics and thermal physics: circular motion (ω, a = v²/r), simple harmonic motion, springs and pendulums, damping and resonance, then thermal energy (Q = mcΔθ, latent heat), the ideal gas law and molecular kinetic theory.
Section 3.6 is compulsory and appears on Paper 2 (and the synoptic Paper 3).
Where this sits: AQA sections 3.1–3.8 are compulsory for every A-level Physics student. (Sections 3.9–3.13 are the five options — you study just one of those.)
Work through each screen, answer the questions as you go (many are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
3.6.1 · circular motion
Circular motion
An object moving in a circle at constant speed is still accelerating, because its velocity (a vector) is constantly changing direction.
ω = 2πf = v ÷ r · a = v² ÷ r = ω²r · F = mv² ÷ r = mω²rω = angular speed in rad s−1 · the acceleration points towards the centre
The centripetal force is not a new force — it is the resultant of the real forces (tension, gravity, friction, the normal contact force).
It acts perpendicular to the velocity, so it does no work and the speed does not change.
Remove it (cut the string) and the object flies off along the tangent, not radially outwards.
Worked example
A 0.20 kg ball on a 0.80 m string moves at 4.0 m s−1.
F = mv²/r = (0.20 × 4.0²) ÷ 0.80 = (0.20 × 16) ÷ 0.80 = 4.0 N.
Common error: there is no outward centrifugal force on the object. The outward push you feel on a roundabout is the reaction you exert — on the roundabout.
Calculate
Your turn — calculation 1
1A 0.20 kg ball is whirled in a horizontal circle of radius 0.80 m at a constant speed of 4.0 m s−1. Calculate the centripetal force.
N
Hint: F = mv² ÷ r = (0.20 × 16) ÷ 0.80.
Calculate
Your turn — calculation 2
2A wheel rotates at a frequency of 2.0 Hz. Calculate its angular speed, to 3 significant figures.
rad s⁻¹
Hint: ω = 2πf = 2 × π × 2.0.
3.6.2 · SHM
Simple harmonic motion
SHM definition: acceleration is proportional to displacement from equilibrium and directed towards equilibrium.
a = −ω²x · x = A cos(ωt) · v = ±ω√(A² − x²)maximum speed = ωA (at x = 0) · maximum acceleration = ω²A (at x = ±A)
x = A cos(ωt) assumes the object starts at maximum displacement (t = 0 at the amplitude).
The period is independent of the amplitude — this is isochronous motion, and it is why pendulum clocks work.
Energy swaps continuously between kinetic (maximum at the centre) and potential (maximum at the extremes). With no damping, the total is constant.
mass–spring: T = 2π√(m ÷ k) · simple pendulum: T = 2π√(l ÷ g)the pendulum period does not depend on the mass of the bob
Worked example
Pendulum of length 0.50 m: T = 2π√(0.50 ÷ 9.81) = 2π × √0.05097 = 2π × 0.2258 = 1.42 s.
Calculate
Your turn — calculation 3
3A simple pendulum has length 0.50 m. Calculate its period. (g = 9.81 m s−2)
s
Hint: T = 2π√(l ÷ g) = 2π × √(0.50 ÷ 9.81).
Calculate
Your turn — calculation 4
4An object oscillates with SHM of amplitude 0.050 m and frequency 2.0 Hz. Calculate its maximum speed.
m s⁻¹
Hint: vmax = ωA, and ω = 2πf = 12.57 rad s−1. So vmax = 12.57 × 0.050.
Quick check
Think it through
?An object oscillates with SHM described by x = A cos(ωt). Where is the object at t = 0, and what is its speed there?
3.6.2 · damping & resonance
Damping, forced vibrations and resonance
Damping: resistive forces dissipate energy, so the amplitude decreases over time. The period is almost unchanged for light damping.
Light damping → amplitude decays slowly. Critical damping → returns to equilibrium in the shortest time without oscillating (car suspension, door closers). Heavy (over) damping → returns slowly, no oscillation.
Free vibration: oscillation at the system’s natural frequency f0. Forced vibration: a periodic driver imposes its own frequency.
Resonance: when the driving frequency = the natural frequency, energy is transferred most efficiently and the amplitude reaches a maximum (the driver is always π/2 ahead in phase).
More damping → a lower, broader resonance peak, shifted very slightly to a lower frequency.
Real examples: resonance is useful in an MRI scanner and a radio tuning circuit; it is dangerous in bridges and buildings, which is why the Millennium Bridge needed dampers fitted.
Sort it
SHM, damping or resonance?
Tap a statement, then tap the box it belongs in.
〰️ SHM (always true)
🛑 Damping
📣 Resonance
3.6.3 · thermal energy
Specific heat capacity and specific latent heat
Q = mcΔθ · Q = mLc = specific heat capacity (J kg−1 K−1) · L = specific latent heat (J kg−1)
Specific heat capacity: the energy needed to raise the temperature of 1 kg by 1 K. (Water’s is huge — 4200 J kg−1 K−1 — which is why the sea warms slowly.)
Specific latent heat: the energy needed to change the state of 1 kg with no temperature change (fusion = solid↔liquid; vaporisation = liquid↔gas).
During a change of state the temperature stays constant: the energy goes into breaking bonds — increasing potential energy, not kinetic energy. That is why the graph has a plateau.
Internal energy = the sum of the randomly distributed kinetic and potential energies of the molecules.
Worked example
Heating 0.30 kg of water (c = 4200 J kg−1 K−1) by 25 K:
Q = mcΔθ = 0.30 × 4200 × 25 = 31 500 J (31.5 kJ).
Calculate
Your turn — calculation 5
5Calculate the energy needed to raise the temperature of 0.30 kg of water by 25 K. (c = 4200 J kg−1 K−1) Give your answer in joules.
J
Hint: Q = mcΔθ = 0.30 × 4200 × 25.
Quick check
Think it through
?Ice at 0 °C is heated steadily until it has all melted. During the melting, the temperature stays at 0 °C. Why?
3.6.4 · ideal gases
The ideal gas law and kinetic theory
pV = nRT = NkTn = moles · R = 8.31 J mol−1 K−1 · N = number of molecules · k = 1.38×10−23 J K−1 · T in kelvin
The gas laws combine: Boyle (pV constant at constant T), Charles (V ∝ T), and the pressure law (p ∝ T).
pV = ⅓Nm(crms)² · ½m(crms)² = (3/2)kTthe mean kinetic energy of a molecule depends only on the absolute temperature
Kinetic theory assumptions: many identical molecules in random motion; negligible volume compared with the container; collisions are perfectly elastic and of negligible duration; no intermolecular forces except during collisions; Newton’s laws apply.
The root mean square speed crms is the square root of the mean of the squared speeds — not the mean speed.
Raise T → molecules move faster → they hit the walls harder and more often → pressure rises (at fixed volume).