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AQA A-level Physics revision notes

Further mechanics and thermal physics (A-level only)

Section 3.6
Year 2
Year 2: this is content the exam board adds beyond the AS subject content, for the full A-level.
7 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against AQA 7408 section 3.6

Checked against AQA 7408 section 3.6. Review basis: the qualification registry sourced from the AQA A-level Physics (7408) specification; registry verification recorded 11 July 2026.

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In the exam: Data and formulae booklet provided · calculator allowed in every paper

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3.6.1.1

Circular motion

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • An object moving at constant speed in a circle accelerates because its velocity changes direction. Its instantaneous velocity is tangential, while acceleration and resultant force point towards the centre.
  • Angular speed is ω=v/r=2πf\omega=v/r=2\pi f in rad s1\text{rad s}^{-1}, with one complete revolution equal to 2π2\pi radians.
  • Therefore a=v2/r=ω2ra=v^2/r=\omega^2r and F=mv2/r=mω2rF=mv^2/r=m\omega^2r.
  • “Centripetal force” is not a new force: it is the inward resultant supplied by forces such as tension, friction or gravity.
  • The specification does not require a direction for the angular-velocity vector.
Velocity is tangential while centripetal acceleration and resultant force point towards the centre.
Worked example

A 0.50kg0.50\,\text{kg} object travels at 6.0m s16.0\,\text{m s}^{-1} in a circle of radius 2.0m2.0\,\text{m}. Find its acceleration and resultant force.

  1. 1.a=v2/r=6.02/2.0=18m s2a=v^2/r=6.0^2/2.0=18\,\text{m s}^{-2}.
  2. 2.F=ma=0.50×18=9.0NF=ma=0.50\times18=9.0\,\text{N}.
  3. 3.Both vectors point radially towards the centre.

Answer: Acceleration is 18 m s⁻² and resultant force is 9.0 N, both inwards.

Common mistakes

  • Don't draw the resultant force tangentially in the direction of motion.
  • Don't add a separate centripetal force as well as the real inward forces.
  • Don't use revolutions per second directly as angular speed without converting each revolution to 2π radians.

Exam tip

Identify the real forces first, then equate their inward resultant to the required centripetal force.

Tier 1 · Easy

ORIGINAL

A turntable rotates at 120rev min1120\,\text{rev min}^{-1}. Calculate its angular speed.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A 0.45kg0.45\,\text{kg} object moves at constant speed 6.0m s16.0\,\text{m s}^{-1} in a horizontal circle of radius 0.80m0.80\,\text{m}. Determine the resultant force and state its direction.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A 0.240kg0.240\,\text{kg} coin rests 0.350m0.350\,\text{m} from the centre of a horizontal rotating disc. The disc can exert a maximum frictional force of 0.990N0.990\,\text{N} on the coin. Determine the greatest rotation frequency for which the coin does not slide. Explain which force supplies the centripetal force.

[5 marks]

Total for this question: 5

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3.6.1.2

Simple harmonic motion (SHM)

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Simple harmonic motion has acceleration proportional to displacement and directed towards equilibrium: a=ω2xa=-\omega^2x.
  • A suitable displacement equation is x=Acosωtx=A\cos\omega t, and speed at displacement xx satisfies v=±ωA2x2v=\pm\omega\sqrt{A^2-x^2}; the sign follows the direction of motion.
  • Speed is greatest at equilibrium, vmax=ωAv_{\max}=\omega A, whereas acceleration magnitude is greatest at the extremes, amax=ω2Aa_{\max}=\omega^2A.
  • On time graphs, velocity is the gradient of displacement and acceleration is the gradient of velocity, so the three sinusoidal curves have fixed quarter-cycle phase differences.
  • Define the positive direction before assigning signs.
Displacement, velocity and acceleration in SHM are sinusoidal with fixed phase relationships.
Worked example

An oscillator has A=0.040mA=0.040\,\text{m} and ω=5.0rad s1\omega=5.0\,\text{rad s}^{-1}. At x=+0.024mx=+0.024\,\text{m} it moves towards equilibrium. Find acceleration and velocity.

  1. 1.a=ω2x=(5.0)2(0.024)=0.60m s2a=-\omega^2x=-(5.0)^2(0.024)=-0.60\,\text{m s}^{-2}.
  2. 2.v=5.00.04020.0242=0.16m s1|v|=5.0\sqrt{0.040^2-0.024^2}=0.16\,\text{m s}^{-1}.
  3. 3.Motion towards equilibrium from positive xx gives negative velocity.

Answer: Acceleration is −0.60 m s⁻² and velocity is −0.16 m s⁻¹.

Common mistakes

  • Don't drop the minus sign in a=ω2xa=-\omega^2x.
  • Don't assume velocity is positive because displacement is positive.
  • Don't place maximum speed at an extreme rather than at equilibrium.

Exam tip

Use the stated direction of motion to choose the sign after calculating speed.

Tier 1 · Easy

ORIGINAL

For an oscillator, displacement to the right is positive. At one instant x=+0.030mx=+0.030\,\text{m} and ω=4.0rad s1\omega=4.0\,\text{rad s}^{-1}. Calculate its acceleration, including its sign.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

An oscillator has amplitude 0.080m0.080\,\text{m} and angular frequency 5.0rad s15.0\,\text{rad s}^{-1}. At an instant when x=+0.048mx=+0.048\,\text{m}, it is moving in the negative direction. Determine its velocity and acceleration. Take displacement in the positive direction as positive.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

An oscillator starts at its positive extreme at t=0t=0 and has x=0.120cos(4.00t)x=0.120\cos(4.00t), where xx is in metres and rightwards is positive. Determine its displacement, velocity and acceleration at t=0.350st=0.350\,\text{s}.

[5 marks]

Total for this question: 5

3.6.1.3

Simple harmonic systems

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A mass-spring oscillator has period T=2πm/kT=2\pi\sqrt{m/k}, while a simple pendulum at small angle has T=2πl/gT=2\pi\sqrt{l/g}. Other oscillators may be analysed from supplied information.
  • In ideal spring SHM, total energy is E=12kA2E=\tfrac12kA^2; elastic potential energy is Ep=12kx2E_{\text{p}}=\tfrac12kx^2 and the remainder is kinetic.
  • Thus kinetic energy is greatest at equilibrium, potential energy is greatest at the extremes and total energy is constant.
  • Damping transfers energy to the surroundings, reducing amplitude.
  • Required practical 7 measures how period depends on mass for a spring and on length for a pendulum.
Kinetic and potential energy exchange during ideal SHM while total energy remains constant.
Worked example

A 0.20kg0.20\,\text{kg} mass oscillates on a spring of stiffness 80N m180\,\text{N m}^{-1}. Calculate its period.

  1. 1.Use T=2πm/kT=2\pi\sqrt{m/k}.
  2. 2.T=2π0.20/80T=2\pi\sqrt{0.20/80}.
  3. 3.0.0025=0.050\sqrt{0.0025}=0.050, so multiply by 2π2\pi.

Answer: The period is 0.314 s.

Common mistakes

  • Don't use amplitude in the mass-spring period equation.
  • Don't apply the pendulum formula when the angle is not small.
  • Don't claim damping changes the equilibrium position rather than the amplitude.

Exam tip

For practical graphs, square the period to obtain a linear relation with mass or pendulum length.

Tier 1 · Easy

ORIGINAL

A 0.200kg0.200\,\text{kg} mass is attached to a spring of stiffness 80.0N m180.0\,\text{N m}^{-1}. Calculate the period of small vertical oscillations.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A simple pendulum has period 1.60s1.60\,\text{s}. Determine its length. Use g=9.81N kg1g=9.81\,\text{N kg}^{-1} and assume the oscillation angle is small.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A 0.300kg0.300\,\text{kg} mass oscillates on a spring of stiffness 120N m1120\,\text{N m}^{-1} with amplitude 0.0500m0.0500\,\text{m}. Determine its total energy and its speed at displacement 0.0300m0.0300\,\text{m}. Damping later reduces the amplitude to 0.0400m0.0400\,\text{m}. Calculate the percentage of the original energy that has been dissipated.

[6 marks]

Total for this question: 6

3.6.1.4

Forced vibrations and resonance

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A free vibration follows an initial disturbance and occurs at the system’s natural frequency. A forced vibration is maintained by a periodic driving force and settles at the driving frequency.
  • Resonance occurs when driving frequency equals natural frequency, causing maximum energy transfer and the greatest steady amplitude.
  • Increasing damping dissipates more energy per cycle, so the resonance peak becomes lower and broader: the resonance is less sharp.
  • Resonance is useful in selected mechanical systems and in producing stationary waves, but it can damage structures or machinery driven close to a natural frequency.
  • A complete explanation links frequency matching, energy transfer and amplitude.
Damping lowers and broadens the amplitude peak and shifts the resonant frequency slightly below the natural frequency.
Worked example

Two identical oscillators are driven through the same range of frequencies; one is more heavily damped. Compare their response curves.

  1. 1.Both show maximum response near the same natural frequency.
  2. 2.The more heavily damped oscillator loses more energy per cycle.
  3. 3.Its maximum amplitude is smaller and its peak is broader.

Answer: Greater damping produces a lower, broader and less sharp resonance peak.

Common mistakes

  • Don't define resonance only as a large amplitude without matching frequencies.
  • Don't say a forced oscillator always vibrates at its natural frequency.
  • Don't claim damping makes the resonance peak taller.

Exam tip

When explaining a resonance hazard, name the driver, the natural frequency and how damping changes energy loss.

Tier 1 · Easy

ORIGINAL

State what is meant by resonance in a forced oscillator.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

Describe how increased damping changes an amplitude-against-driving-frequency resonance curve.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

A platform of effective oscillating mass 800kg800\,\text{kg} behaves like a spring of stiffness 3.20×105N m13.20\times10^5\,\text{N m}^{-1}. A machine drives it at 3.20Hz3.20\,\text{Hz}. Determine the natural frequency and explain why adding a damper reduces the risk of a large vibration amplitude.

[5 marks]

Total for this question: 5

3.6.2.1

Thermal energy transfer

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Internal energy is the sum of randomly distributed particle kinetic and potential energies. Heating a system or doing work on it increases internal energy; energy transferred from the system or work done by it decreases internal energy.
  • For a temperature change, Q=mcΔθQ=mc\Delta\theta.
  • During a change of state, Q=mlQ=ml and temperature stays constant while particle potential energy changes rather than average kinetic energy.
  • Continuous-flow heating often uses P=m˙cΔθP=\dot{m}c\Delta\theta when losses are negligible.
  • Practical determinations of specific heat capacity or specific latent heat must account for energy transferred to the apparatus and surroundings, electrical measurements and rate or mass uncertainties.
During a change of state, energy increases while temperature remains constant.
Worked example

Find the energy needed to heat 0.50kg0.50\,\text{kg} of aluminium through 20K20\,\text{K} when c=900J kg1K1c=900\,\text{J kg}^{-1}\text{K}^{-1}.

  1. 1.This is a temperature change, so use Q=mcΔθQ=mc\Delta\theta.
  2. 2.Q=0.50×900×20Q=0.50\times900\times20.
  3. 3.Keep mass in kilograms and temperature change in kelvin.

Answer: The energy required is 9.0 × 10³ J.

Common mistakes

  • Don't use the latent-heat equation for a temperature change.
  • Don't say particle kinetic energy increases during a constant-temperature change of state.
  • Don't treat electrical input energy as entirely transferred to the sample in a practical.

Exam tip

Split a multi-stage heating process into temperature-change and state-change energy terms before adding them.

Tier 1 · Easy

ORIGINAL

Calculate the energy required to raise the temperature of a 0.250kg0.250\,\text{kg} block of specific heat capacity 900J kg1K1900\,\text{J kg}^{-1}\,\text{K}^{-1} by 20.0K20.0\,\text{K}.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A 120W120\,\text{W} heater melts 0.0800kg0.0800\,\text{kg} of ice already at its melting temperature. Only 80.0%80.0\% of the electrical energy reaches the ice. Calculate the melting time. The specific latent heat of fusion is 3.34×105J kg13.34\times10^5\,\text{J kg}^{-1}.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

Water flows through an electric heater at 0.0180kg s10.0180\,\text{kg s}^{-1}. Its temperature rises from 18.0C18.0\,^{\circ}\text{C} to 42.0C42.0\,^{\circ}\text{C}. The electrical input power is 2.20kW2.20\,\text{kW}. Determine the rate of increase of the water's internal energy and the heater efficiency. Explain the energy transfer at particle level. Use c=4200J kg1K1c=4200\,\text{J kg}^{-1}\,\text{K}^{-1}.

[6 marks]

Total for this question: 6

3.6.2.2

Ideal gases

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For a fixed gas mass, empirical gas laws relate pressure, volume and absolute temperature: Boyle’s law applies at constant temperature and Charles’s law at constant pressure.
  • Absolute zero motivates the kelvin scale, so convert using T/K=θ/C+273.15T/\text{K}=\theta/^{\circ}\text{C}+273.15.
  • The ideal-gas equation is pV=nRTpV=nRT for nn moles or pV=NkTpV=NkT for NN molecules; distinguish molar mass from one molecule’s mass.
  • At constant pressure, work done by an expanding gas is W=pΔVW=p\Delta V.
  • Required practical 8 investigates Boyle’s and Charles’s laws, with pressure, volume and temperature uncertainties and control variables stated explicitly.
At constant temperature, pressure varies inversely with volume for a fixed ideal-gas mass.
Worked example

Find the pressure of 0.20mol0.20\,\text{mol} of ideal gas at 300K300\,\text{K} in 5.0×103m35.0\times10^{-3}\,\text{m}^3 using R=8.31J mol1K1R=8.31\,\text{J mol}^{-1}\text{K}^{-1}.

  1. 1.Rearrange pV=nRTpV=nRT to p=nRT/Vp=nRT/V.
  2. 2.p=(0.20)(8.31)(300)/(5.0×103)p=(0.20)(8.31)(300)/(5.0\times10^{-3}).
  3. 3.All quantities are already in SI units.

Answer: The pressure is 1.0 × 10⁵ Pa to two significant figures.

Common mistakes

  • Don't substitute a Celsius temperature into the ideal-gas equation.
  • Don't pair molecule number with the molar gas constant, or amount in moles with the Boltzmann constant.
  • Don't apply the constant-pressure work equation when pressure is changing.

Exam tip

Write the fixed quantity beside each gas law and convert temperature to kelvin before calculating.

Tier 1 · Easy

ORIGINAL

Convert 25.0C25.0\,^{\circ}\text{C} to kelvin.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

A sealed container holds 0.120mol0.120\,\text{mol} of an ideal gas at 35.0C35.0\,^{\circ}\text{C} in a volume of 2.50×103m32.50\times10^{-3}\,\text{m}^3. Calculate the pressure. Use R=8.31J mol1K1R=8.31\,\text{J mol}^{-1}\,\text{K}^{-1}.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

An ideal gas initially has pressure 1.00×105Pa1.00\times10^5\,\text{Pa}, volume 2.40×103m32.40\times10^{-3}\,\text{m}^3 and temperature 17.0C17.0\,^{\circ}\text{C}. It is compressed to 1.50×103m31.50\times10^{-3}\,\text{m}^3 and heated to 77.0C77.0\,^{\circ}\text{C}. Determine the number of molecules and the final pressure. Use k=1.38×1023J K1k=1.38\times10^{-23}\,\text{J K}^{-1}.

[5 marks]

Total for this question: 5

3.6.2.3

Molecular kinetic theory model

Notes
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Brownian motion is random motion of visible particles caused by unequal molecular impacts and supports the molecular model. Kinetic theory assumes many identical point molecules in random motion, negligible intermolecular forces except during perfectly elastic collisions, and negligible collision duration.
  • Momentum transfer to container walls gives pV=13Nm(crms)2pV=\tfrac13Nm(c_{\text{rms}})^2; random isotropic motion produces the factor 1/31/3.
  • Combining this with pV=NkTpV=NkT gives 12m(crms)2=32kT=3RT/(2NA)\tfrac12m(c_{\text{rms}})^2=\tfrac32kT=3RT/(2N_{\text{A}}).
  • For an ideal gas, internal energy is molecular kinetic energy because intermolecular potential energy is neglected.
  • This theory explains empirical gas laws and illustrates how evidence changes scientific models.
Random molecular collisions transfer momentum to the container walls and produce gas pressure.
Worked example

Calculate the average translational kinetic energy of one ideal-gas molecule at 300K300\,\text{K} using k=1.38×1023J K1k=1.38\times10^{-23}\,\text{J K}^{-1}.

  1. 1.Use Ek=32kT\overline{E_{\text{k}}}=\tfrac32kT.
  2. 2.Ek=1.5(1.38×1023)(300)\overline{E_{\text{k}}}=1.5(1.38\times10^{-23})(300).
  3. 3.This is the average for one molecule, not the whole gas.

Answer: The average translational kinetic energy is 6.21 × 10⁻²¹ J.

Common mistakes

  • Don't describe Brownian particles as individual gas molecules.
  • Don't confuse rms speed with mean speed.
  • Don't use the average energy per molecule as the total energy of every molecule in the gas.

Exam tip

In the pressure derivation, show the momentum change, collision interval and isotropic 1/31/3 step explicitly.

Tier 1 · Easy

ORIGINAL

Calculate the average translational kinetic energy of one ideal-gas molecule at 300K300\,\text{K}. Use k=1.38×1023J K1k=1.38\times10^{-23}\,\text{J K}^{-1}.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

A gas occupies 1.50×102m31.50\times10^{-2}\,\text{m}^3 at pressure 1.20×105Pa1.20\times10^5\,\text{Pa}. It contains 3.00×10233.00\times10^{23} molecules, each of mass 4.65×1026kg4.65\times10^{-26}\,\text{kg}. Use kinetic theory to determine crmsc_{\text{rms}}.

[4 marks]

Total for this question: 4

Tier 3 · Hard

ORIGINAL

A cubical container of side LL holds NN identical ideal-gas molecules of mass mm. Derive pV=13Nm(crms)2pV=\dfrac13Nm(c_{\text{rms}})^2 from molecular collisions with a wall. Hence show that the average translational kinetic energy of one molecule is 32kT\dfrac32kT.

[6 marks]

Total for this question: 6

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