AQA A-Level Physics: Engineering 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 3.11 Engineering physics (3.11.1 Rotational dynamics and 3.11.2 Thermodynamics and engines) of the AQA AS and A-level Physics specification (7407/7408, version 1.4, AS and A-level exams June 2016 onwards). It is A-level only and one of the five options (sections 3.9 to 3.13), assessed in Paper 3 Section B (35 marks).

Learn the content with the study guide and the revision notes. Take R = 8.31 J mol⁻¹ K⁻¹. Treat all gases as ideal.

Questions

1. Two flywheels have equal mass and radius. One is a uniform disc; the other has most of its mass in its rim. Explain which stores more energy at the same angular speed. [2]

2. A light cross-shaped frame carries four small 0.25 kg masses, each 0.40 m from the axis of rotation.

(a) Calculate the moment of inertia of the frame about the axis. [1] (b) Calculate its rotational kinetic energy at 5.0 rad s⁻¹. [2]

3. A grinding wheel turns at 2400 revolutions per minute. When the motor is switched off, it slows uniformly and stops in 20 s.

(a) Calculate its initial angular speed. [1] (b) Calculate its angular acceleration. [2] (c) Calculate the number of revolutions it makes while stopping. [2]

4. A motor applies a constant torque of 9.0 N m to a flywheel with I = 1.6 kg m². A constant frictional torque of 1.0 N m acts. The flywheel starts from rest.

(a) Calculate its angular acceleration. [2] (b) Calculate the time it takes to reach 150 rad s⁻¹. [1] (c) At 150 rad s⁻¹ the motor is switched off. Calculate the time the flywheel takes to stop. [2] (d) Suggest why, in practice, the deceleration in (c) is not uniform. [1]

5. A skater spins at 1.5 rev s⁻¹ with arms out (I = 2.4 kg m²), then pulls her arms in (I = 0.90 kg m²).

(a) Calculate her new rate of rotation in revolutions per second. [2] (b) Explain why her kinetic energy increases although her angular momentum is constant. [2]

6. A wheel with I = 0.45 kg m² starts from rest. A constant torque acts for 2.0 s, after which ω = 12 rad s⁻¹.

(a) Calculate the angular impulse given to the wheel. [2] (b) Calculate the torque. [1] (c) Show that the work done by the torque equals the wheel’s final kinetic energy. [2]

7. A gas at constant pressure 1.5 × 10⁵ Pa expands from 2.0 × 10⁻³ m³ to 3.2 × 10⁻³ m³, receiving 450 J by heating.

(a) Calculate the work done by the gas. [1] (b) Calculate the change in its internal energy. [2] (c) The initial temperature is 300 K. Calculate the final temperature. [2] (d) State the relationship between Q and W for an isothermal change, and why. [1]

8. In a diesel engine, 6.0 × 10⁻⁴ m³ of air (γ = 1.4) at 1.0 × 10⁵ Pa and 290 K is compressed adiabatically to 3.0 × 10⁻⁵ m³.

(a) Calculate the amount of air in moles. [1] (b) Calculate the final pressure. [2] (c) Calculate the final temperature. [2] (d) Use the first law to explain why the temperature rises. [2]

9. A gas follows cycle ABCD. A→B: expands at 4.0 × 10⁵ Pa from 1.0 × 10⁻³ m³ to 2.5 × 10⁻³ m³. B→C: constant volume, to 1.0 × 10⁵ Pa. C→D: compressed at 1.0 × 10⁵ Pa to 1.0 × 10⁻³ m³. D→A: constant volume.

(a) Calculate the work done by the gas from A to B. [1] (b) Calculate the work done on the gas from C to D. [1] (c) Calculate the net work done per cycle. [1] (d) The cycle repeats 20 times per second. Calculate the power output. [1] (e) State the net change in internal energy per cycle. [1]

10. A six-cylinder, four-stroke diesel engine runs at 2400 rpm. Each indicator loop has area 520 J. Brake torque is 200 N m. Fuel (calorific value 4.5 × 10⁷ J kg⁻¹) is used at 4.0 × 10⁻³ kg s⁻¹.

(a) Calculate the indicated power. [3] (b) Calculate the brake power. [2] (c) Calculate the friction power. [1] (d) Calculate the input power. [1] (e) Calculate the overall, thermal and mechanical efficiencies. [3] (f) Give two reasons why the real indicator diagram encloses a smaller area than the theoretical one. [2]

11. A power station uses steam at 840 K and rejects energy to a river at 300 K.

(a) Explain why a heat engine needs a cold sink. [2] (b) Calculate the maximum theoretical efficiency. [1] (c) It takes in 1250 MW and delivers 500 MW of work. Calculate its efficiency and the rate of energy rejection. [2] (d) Suggest how a combined heat and power scheme improves the use of energy. [1]

12. A heat pump keeps a house at 294 K by taking energy from the ground at 279 K.

(a) Calculate the maximum theoretical COP of the heat pump. [2] (b) The real pump has COP 3.5 and delivers 7.0 kW. Calculate the input power and the rate of extraction from the ground. [2] (c) A refrigerator keeps food at 276 K in a 294 K kitchen. Calculate its maximum theoretical COP. [1] (d) Show that, between the same two temperatures, COP_hp = COP_ref + 1. [1]

Answers

1. The rim flywheel has more mass far from the axis, so a larger moment of inertia (I = Σmr²) [1]. E_k = ½Iω², so at the same ω it stores more energy [1]. [2] Examiner insight: The masses are equal, so the first mark needs the distribution of mass, not “heavier”.

2. (a) I = 4 × 0.25 × 0.40² = 0.16 kg m² [1] (b) E_k = ½ × 0.16 × 5.0² [1] = 2.0 J [1] [3] Examiner insight: Error carried forward is usually allowed in (b) if your I from (a) was wrong.

3. (a) ω = 2400 × 2π/60 = 251 rad s⁻¹ [1] (b) α = (0 − 251)/20 [1] = −12.6 rad s⁻² [1] (c) θ = ½(251 + 0) × 20 = 2513 rad [1]; 2513/2π = 400 revolutions [1] [5] Examiner insight: In (b), give the negative sign or say “deceleration”; a bare 12.6 risks the accuracy mark.

4. (a) Resultant torque = 9.0 − 1.0 = 8.0 N m [1]; α = 8.0/1.6 = 5.0 rad s⁻² [1] (b) t = 150/5.0 = 30 s [1] (c) α = −1.0/1.6 = −0.625 rad s⁻² [1]; t = 150/0.625 = 240 s [1] (d) Frictional torque (for example from air resistance) depends on angular speed, so it changes as the wheel slows [1]. [6] Examiner insight: Using 9.0 N m alone in (a) gives 5.6 rad s⁻² and loses both marks.

5. (a) I₁f₁ = I₂f₂ [1]; f₂ = 2.4 × 1.5/0.90 = 4.0 rev s⁻¹ [1] (b) E_k = L²/2I, so with L constant and I smaller, E_k is larger [1]. The energy comes from the work she does pulling her arms in [1]. [4] Examiner insight: Rev s⁻¹ is fine in (a) because 2π cancels, but not in E_k = ½Iω².

6. (a) Angular impulse = Δ(Iω) = 0.45 × 12 − 0 [1] = 5.4 N m s [1] (b) T = 5.4/2.0 = 2.7 N m [1] (c) θ = ½(0 + 12) × 2.0 = 12 rad, so W = Tθ = 2.7 × 12 = 32.4 J [1]; E_k = ½ × 0.45 × 12² = 32.4 J, the same [1]. [5] Examiner insight: On a “show that”, calculate and compare both values; asserting equality earns nothing.

7. (a) W = pΔV = 1.5 × 10⁵ × 1.2 × 10⁻³ = 180 J [1] (b) ΔU = Q − W [1] = 450 − 180 = 270 J [1] (c) At constant p, V/T is constant [1]; T₂ = 300 × 3.2/2.0 = 480 K [1] (d) Q = W, because T is constant so ΔU = 0 [1]. [6] Examiner insight: Writing ΔU = Q + W gives 630 J and loses both marks in (b).

8. (a) n = pV/RT = (1.0 × 10⁵ × 6.0 × 10⁻⁴)/(8.31 × 290) = 0.025 mol [1] (b) p₂ = p₁(V₁/V₂)^γ = 1.0 × 10⁵ × 20^1.4 [1] = 6.6 × 10⁶ Pa [1] (c) T₂ = T₁p₂V₂/(p₁V₁) = 290 × (6.63 × 10⁶ × 3.0 × 10⁻⁵)/(1.0 × 10⁵ × 6.0 × 10⁻⁴) [1] = 961 K [1] (d) Q = 0 and work is done on the gas, so W is negative [1]; ΔU = −W is positive, so T rises [1]. [7] Examiner insight: Using pV = constant (isothermal) in (b) gives 2.0 × 10⁶ Pa, which scores nothing.

9. (a) W = pΔV = 4.0 × 10⁵ × 1.5 × 10⁻³ = 600 J [1] (b) W = 1.0 × 10⁵ × 1.5 × 10⁻³ = 150 J [1] (c) Net work = 600 − 150 = 450 J (the loop area) [1] (d) P = 450 × 20 = 9.0 kW [1] (e) Zero: the gas returns to its starting state [1]. [5] Examiner insight: Constant-volume stages involve no work; adding work for them loses (c).

10. (a) 2400 rpm = 40 rev s⁻¹; four-stroke, so 20 cycles per second per cylinder [1]. Indicated power = 520 × 20 × 6 [1] = 62.4 kW [1] (b) ω = 2π × 40 = 251 rad s⁻¹ [1]; P = Tω = 200 × 251 = 50.3 kW [1] (c) 62.4 − 50.3 = 12.1 kW [1] (d) 4.5 × 10⁷ × 4.0 × 10⁻³ = 180 kW [1] (e) Overall = 50.3/180 = 0.28 [1]; thermal = 62.4/180 = 0.35 [1]; mechanical = 50.3/62.4 = 0.81 [1] (f) Any two of: burning takes time; valves do not move instantly; energy is lost through cylinder walls; the gas is not ideal [1] [1]. [12] Examiner insight: Using 40 cycles per second doubles the indicated power; later parts usually allow error carried forward.

11. (a) An engine cannot turn all the energy it takes in by heating into work [1]; it must reject some, Q_C, to a colder sink [1]. (b) (840 − 300)/840 = 0.64 [1] (c) Efficiency = 500/1250 = 0.40 [1]; rejected = 1250 − 500 = 750 MW [1] (d) Rejected energy heats nearby buildings instead of being wasted [1]. [6] Examiner insight: These temperatures are already in kelvin; adding 273 again loses (b).

12. (a) COP_hp(max) = T_H/(T_H − T_C) = 294/15 [1] = 19.6 [1] (b) W = 7.0/3.5 = 2.0 kW [1]; Q_C rate = 7.0 − 2.0 = 5.0 kW [1] (c) COP_ref(max) = 276/18 = 15.3 [1] (d) Q_H/W = (Q_C + W)/W = Q_C/W + 1 [1]. [6] Examiner insight: A COP greater than 1 is not an error; heat pumps move energy rather than create it.

Where marks are usually lost

• Leaving ω in rpm, or using applied instead of resultant torque.
• Assuming kinetic energy is conserved when a skater changes shape.
• Sign errors in Q = ΔU + W (W is work done by the gas).
• Using pV = constant for an adiabatic change.
• Counting one cycle per revolution in a four-stroke engine.
• Swapping thermal and mechanical efficiency, or inverting a COP formula.

Next steps

• Recap with the revision notes.
• Re-read the study guide.
• 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 3.11 Engineering physics (A-level only option).