AQA A-Level Physics: Further mechanics and thermal physics (7408) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 3.6 Further mechanics and thermal physics of the AQA AS and A-level Physics specification (7407/7408, version 1.4, AS and A-level exams June 2016 onwards): 3.6.1 Periodic motion and 3.6.2 Thermal physics, including Required practicals 7 and 8. It is A-level only: Paper 1 assesses 6.1 and Paper 2 assesses 6.2. It is core content, not one of the options (sections 3.9 to 3.13), of which you study one for Paper 3 Section B.

Learn the content first with the study guide. Take g = 9.81 m s⁻², c(water) = 4200 J kg⁻¹ K⁻¹, R = 8.31 J mol⁻¹ K⁻¹, k = 1.38 × 10⁻²³ J K⁻¹ and N_A = 6.02 × 10²³ mol⁻¹.

Questions

1. A wheel turns at 450 revolutions per minute.

(a) Calculate its angular speed. [2] (b) Calculate the speed of a point 0.30 m from the axle. [1]

2. A car of mass 1200 kg travels round a flat bend of radius 45 m at a constant 15 m s⁻¹.

(a) Calculate the centripetal force on the car. [2] (b) The maximum sideways friction the road can supply is 9000 N. Calculate the maximum speed for this bend. [2] (c) Name the force that provides it. [1]

3.

(a) State the condition for an object to move with simple harmonic motion. [2] (b) Explain how a v–t graph is obtained from an x–t graph, and state the phase relationship between a and x. [2]

4. A body moves in SHM with amplitude 0.060 m and period 0.80 s. Timing starts at x = +0.060 m.

(a) Calculate its maximum speed. [2] (b) Calculate its maximum acceleration. [1] (c) Calculate its speed when x = 0.036 m. [2] (d) Calculate the time at which it first reaches x = 0.030 m. [2]

5. In Required practical 7 you investigate how the period of a mass-spring system depends on the mass.

(a) Describe how to measure the period as accurately as possible. [3] (b) A graph of T² against m is a straight line through the origin with gradient 1.58 s² kg⁻¹. Calculate the spring constant. [2] (c) State why the pendulum part of the practical must use small amplitudes. [1]

6. A 0.10 kg mass on a spring of stiffness 40 N m⁻¹ oscillates with amplitude 0.050 m. Damping is negligible.

(a) Calculate the total energy of the oscillator. [1] (b) Calculate the kinetic energy when the displacement is 0.030 m. [2] (c) Calculate the maximum speed. [1] (d) Calculate the period of oscillation and state the period of the variation of potential energy with time. [2]

7.

(a) Distinguish between a free vibration and a forced vibration. [2] (b) Explain what is meant by resonance. [2] (c) Describe how increasing the damping changes a graph of amplitude against driving frequency. [2] (d) Give one example of resonance involving stationary waves. [1]

8. A 50 W heater warms 0.20 kg of solid wax; a data logger records the temperature. It rises by 20 K in 160 s, then stays constant for 640 s while the wax melts.

(a) Calculate the specific heat capacity of the solid wax. [2] (b) Calculate the specific latent heat of fusion of the wax. [2] (c) Explain, in terms of particles, why the temperature is constant while the wax melts. [2] (d) Suggest why both values are likely to be too high. [1]

9.

(a) Cooling water flows through an engine at 0.15 kg s⁻¹ and its temperature rises by 12 K. Calculate the rate at which energy is removed. [2] (b) The air in a bicycle pump warms when it is compressed quickly. Explain this using the first law of thermodynamics. [2]

10. Required practical 8 investigates the gas laws.

(a) For trapped air at 293 K, a graph of p against 1/V has gradient 12.2 Pa m³. Calculate the number of moles of air. [3] (b) At constant pressure, a gas has volume 50.0 cm³ at 0 °C and 68.3 cm³ at 100 °C. Use these values to estimate absolute zero in °C. [3] (c) State one variable that must be controlled in (b). [1]

11.

(a) State two assumptions of the kinetic theory model of an ideal gas. [2] (b) Derive pV = ⅓Nm(c_rms)² for N molecules, each of mass m, in a cube of side L. [6]

12. A 0.010 m³ cylinder holds argon (molar mass 0.040 kg mol⁻¹, an ideal monatomic gas) at 1.5 × 10⁷ Pa and 290 K.

(a) Calculate the number of moles of argon. [2] (b) Calculate the mass of argon. [1] (c) Calculate the mean kinetic energy of an argon atom. [1] (d) Calculate the root mean square speed of the atoms. [2] (e) Calculate the internal energy of the gas. [2] (f) Gas is released at constant temperature until the pressure is 6.0 × 10⁶ Pa. Calculate the mass released. [2]

Answers

1. (a) f = 450/60 = 7.5 Hz [1]; ω = 2π × 7.5 = 47 rad s⁻¹ [1] (b) v = ωr = 47.1 × 0.30 = 14 m s⁻¹ [1] [3] Examiner insight: Using 450 as f loses the first mark; later marks may be allowed as error carried forward.

2. (a) F = mv²/r = 1200 × 15²/45 [1] = 6000 N [1] (b) v = √(Fr/m) = √(9000 × 45/1200) [1] = 18 m s⁻¹ [1] (c) Friction between the tyres and the road [1] [5] Examiner insight: “Centripetal force” does not answer (c); name a real force.

3. (a) Acceleration is proportional to displacement from the equilibrium position [1] and is always directed towards the equilibrium position [1]. (b) The velocity at each instant is the gradient of the x–t graph [1]. Acceleration and displacement are in antiphase (π rad apart) [1]. [4] Examiner insight: Writing only “a = −ω²x” without explaining the minus sign may not earn the second mark in (a).

4. (a) ω = 2π/0.80 = 7.85 rad s⁻¹ [1]; v_max = ωA = 0.47 m s⁻¹ [1] (b) a_max = ω²A = 7.85² × 0.060 = 3.7 m s⁻² [1] (c) v = ω√(A² − x²) = 7.85 × √(0.060² − 0.036²) = 7.85 × 0.048 [1] = 0.38 m s⁻¹ [1] (d) 0.030 = 0.060 cos ωt, so ωt = π/3 rad [1]; t = (π/3)/7.85 = 0.13 s (T/6) [1] [7] Examiner insight: A calculator in degrees gives ωt = 60 and a wrong time.

5. (a) Time at least ten oscillations and divide [1]. Time from the equilibrium position, using a marker [1]. Repeat and take a mean [1]. (b) T² = (4π²/k)m, so gradient = 4π²/k [1]; k = 4π²/1.58 = 25 N m⁻¹ [1] (c) T = 2π√(l/g) depends on the small-angle approximation sin θ ≈ θ, so the motion is only simple harmonic for small angles [1]. [6] Examiner insight: Each mark in (a) needs a distinct technique; “be careful” earns nothing.

6. (a) E = ½kA² = ½ × 40 × 0.050² = 0.050 J [1] (b) E_k = ½k(A² − x²) = ½ × 40 × (0.050² − 0.030²) [1] = 0.032 J [1] (c) ½mv² = 0.050, so v = 1.0 m s⁻¹ [1] (d) T = 2π√(0.10/40) = 0.31 s [1]; potential energy varies with period 0.16 s (half of T) [1] [6] Examiner insight: Giving ½kx² (the potential energy) in (b) scores no accuracy mark.

7. (a) Free: after release, with no driving force, it oscillates at its natural frequency [1]. Forced: a periodic driving force makes it oscillate at the driving frequency [1]. (b) Driving frequency equals natural frequency [1]; amplitude is maximum as energy transfer is greatest [1]. (c) The peak is lower [1]; it is broader and at a slightly lower frequency [1]. (d) A string driven by a vibration generator gives large stationary waves at its harmonic frequencies [1]. [7] Examiner insight: “Lower and broader” are separate points in (c); one alone is unlikely to gain both marks.

8. (a) c = Pt/(mΔθ) = (50 × 160)/(0.20 × 20) [1] = 2000 J kg⁻¹ K⁻¹ [1] (b) l = Pt/m = (50 × 640)/0.20 [1] = 1.6 × 10⁵ J kg⁻¹ [1] (c) The energy increases the potential energies of the particles [1]; their mean kinetic energy, and so the temperature, does not change [1]. (d) Energy is lost to the surroundings, so more energy is counted than the wax absorbed [1]. [7] Examiner insight: Swapping the units of c and l usually costs the accuracy mark.

9. (a) P = (m/t)cΔθ = 0.15 × 4200 × 12 [1] = 7.6 × 10³ W [1] (b) Work done on the air increases its internal energy [1]. Compression is quick, so little energy leaves by heating, and the temperature rises [1]. [4] Examiner insight: “The molecules move faster” alone misses the work-done link the first law needs.

10. (a) pV = nRT, so p = nRT × (1/V) and gradient = nRT [1]. n = 12.2/(8.31 × 293) [1] = 5.0 × 10⁻³ mol [1] (b) Gradient = (68.3 − 50.0)/100 = 0.183 cm³ K⁻¹ [1]. V = 0 when θ = −50.0/0.183 [1] = −273 °C [1] (c) Pressure (or the mass of gas) [1] [7] Examiner insight: In (a), “gradient = nRT” is the method step; using °C loses the accuracy mark.

11. (a) Any two: random motion; elastic collisions; negligible molecular volume; no intermolecular forces except in collisions; negligible collision time [1] [1]. (b) A molecule with velocity component u hits a wall and rebounds: Δp = 2mu [1]. Time between collisions with that wall = 2L/u [1]. Force = rate of change of momentum = 2mu/(2L/u) = mu²/L [1]. For N molecules, total force = Nm(mean u²)/L [1]. With random motion, mean u² = ⅓(c_rms)², since c² = u² + v² + w² [1]. p = F/L² = Nm(c_rms)²/3L³, and V = L³, giving pV = ⅓Nm(c_rms)² [1]. [8] Examiner insight: Each derivation step must be visible; an unexplained ⅓ loses that mark.

12. (a) n = pV/RT = (1.5 × 10⁷ × 0.010)/(8.31 × 290) [1] = 62 mol [1] (b) Mass = 62.2 × 0.040 = 2.5 kg [1] (c) (3/2)kT = 1.5 × 1.38 × 10⁻²³ × 290 = 6.0 × 10⁻²¹ J [1] (d) Atom mass = 0.040/6.02 × 10²³ = 6.64 × 10⁻²⁶ kg [1]; c_rms = √(3kT/m) = 425 m s⁻¹ [1] (e) U = (3/2)nRT = (3/2)pV [1] = 1.5 × 1.5 × 10⁷ × 0.010 = 2.3 × 10⁵ J [1] (f) n remaining = (6.0 × 10⁶ × 0.010)/(8.31 × 290) = 24.9 mol [1]; mass released = (62.2 − 24.9) × 0.040 = 1.5 kg [1] [10] Examiner insight: Carry unrounded values of n through (f); early rounding can shift the final answer.

Where marks are usually lost

• Naming “centripetal force” instead of the real force that provides it.
• Omitting “towards equilibrium” from the SHM condition.
• Confusing ½kx² (potential energy) with the kinetic energy at displacement x.
• Mixing up the units of specific heat capacity and specific latent heat.
• Using °C instead of kelvin in pV = nRT.
• Using molar mass instead of the mass of one atom when finding c_rms.

Next steps

• Recap with the revision notes.
• Re-read the study guide.
• Practise uncertainties with the measurements practice.
• See every topic on the AQA A-level Physics hub and the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

AQA AS and A-level Physics (7407/7408) specification, version 1.4, July 2026, AS and A-level exams June 2016 onwards, published by AQA. Section 3.6 Further mechanics and thermal physics (A-level only).