Option: Engineering physics – A-level Physics exam questions with mark scheme

1Core4 marks

A rotor in a laboratory demonstration consists of a hub and four small masses, each of mass 0.50 kg, fixed to light spokes. Each mass is 0.30 m from the axis of rotation. The moment of inertia of the hub about the axis is 0.012 kg m².

(a) Calculate the moment of inertia of the rotor about its axis. Treat the four masses as point masses. [2]

(b) The four masses are moved along the spokes until each is 0.15 m from the axis. Calculate the new moment of inertia of the rotor. [1]

(a) use of $I = \Sigma mr^2$ for the masses: $4 \times 0.50 \times 0.30^2 = 0.18$ (kg m²)(1)
(a) $I = 0.18 + 0.012 = 0.19$ kg m² (accept 0.192; do not allow 0.18 alone)(1)
(b) $4 \times 0.50 \times 0.15^2 + 0.012 = 0.057$ kg m² (allow ECF for omission of hub only if consistent with (a))(1)
(c) the distribution of the mass about the axis / distance of the mass from the axis / position of the axis of rotation (allow shape of the body; ignore 'size')(1)
Answer: (a) 0.19 kg m² (0.192) (b) 0.057 kg m² (c) distribution of mass relative to the axis
Examiner tip: The hub's moment of inertia must be added; students who only sum $mr^2$ for the four masses lose the second mark.

2Standard6 marks

A hybrid bus uses a flywheel to store energy recovered during braking. The flywheel is a uniform solid disc of mass 60 kg and radius 0.25 m. The moment of inertia of a uniform disc about its central axis is $\tfrac{1}{2}MR^2$.

When the bus pulls away, the flywheel slows from 18 000 rev min⁻¹ to 9000 rev min⁻¹ and transfers its lost kinetic energy to the wheels.

(a) Show that the kinetic energy released by the flywheel is about 2.5 MJ. [3]

(b) The energy is released in 10 s. Calculate the mean power delivered by the flywheel. [1]

(c) Engineers designing a flywheel for a vehicle often choose a flywheel with most of its mass concentrated at the rim, rotating at very high angular speed. Explain why this design is chosen. [2]

(a) $I = 0.5 \times 60 \times 0.25^2 = 1.875$ kg m² and conversion to rad s⁻¹: $18\,000 \times 2\pi/60 = 1885$ rad s⁻¹ (or 942 rad s⁻¹)(1)
(a) use of $E_k = \tfrac{1}{2}I\omega^2$ for both speeds: $3.33\times10^6$ J and $8.33\times10^5$ J (allow $\tfrac12 I(\omega_1^2-\omega_2^2)$ in one step)(1)
(a) $\Delta E_k = 2.50\times10^6$ J (seen to at least 3 sf, e.g. 2.50 MJ or 2 498 000 J)(1)
(b) $P = 2.5\times10^6/10 = 2.5\times10^5$ W (250 kW) (allow ECF from (a))(1)
(c) mass at rim / large radius gives large moment of inertia for a given (small) mass, so vehicle mass is kept low(1)
(c) $E_k \propto \omega^2$ so high angular speed stores a large amount of energy (for a given $I$) (allow energy stored per unit mass is high)(1)
Answer: (a) 2.50 MJ (b) 250 kW (c) rim mass gives high $I$ for low mass; $E_k\propto\omega^2$
Examiner tip: In a show-that, quote the answer to more significant figures than given (2.50 MJ, not 2.5 MJ) and convert rev min⁻¹ to rad s⁻¹.

3Standard6 marks

A bench grinding wheel has a moment of inertia of 0.042 kg m² about its axle. It rotates at 2900 rev min⁻¹. When the motor is switched off, the wheel decelerates uniformly and comes to rest in 35 s because of a constant frictional torque at the bearings.

(a) Calculate the magnitude of the angular deceleration of the wheel. [2]

(b) Calculate the number of revolutions the wheel makes while coming to rest. [2]

(d) Calculate the power the motor must supply to keep the wheel rotating at a steady 2900 rev min⁻¹ with no load applied. [1]

(a) $\omega = 2900 \times 2\pi/60 = 304$ rad s⁻¹(1)
(a) $\alpha = 304/35 = 8.7$ rad s⁻² (accept 8.68)(1)
(b) $\theta = \tfrac12(\omega_1+\omega_2)t = \tfrac12 \times 304 \times 35 = 5.3\times10^3$ rad (or use of $\omega_2^2 = \omega_1^2 + 2\alpha\theta$)(1)
(b) number of revolutions $= 5315/2\pi = 850$ (accept 845–846; allow 1450 × 35/60 = 846 rev directly)(1)
(c) $T = I\alpha = 0.042 \times 8.68 = 0.36$ N m (allow ECF)(1)
(d) $P = T\omega = 0.364 \times 304 = 110$ W (accept 111 W; allow ECF)(1)
Answer: (a) 8.7 rad s⁻² (b) about 850 revolutions (c) 0.36 N m (d) 110 W
Examiner tip: Students forget to convert rev min⁻¹ to rad s⁻¹, or give the angular displacement in radians when asked for revolutions.

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4Standard6 marks

The graph shows how the angular velocity $\omega$ of a motor-driven rotor varies with time $t$. The motor is switched off at $t$ = 10 s and the rotor then slows down because of a constant frictional torque. The moment of inertia of the rotor is 0.080 kg m².

(a) Determine the angular acceleration of the rotor between $t$ = 0 and $t$ = 4.0 s. [1]

(b) Determine the total number of revolutions made by the rotor between $t$ = 0 and $t$ = 16 s. [2]

(d) Assuming the frictional torque is the same throughout, calculate the torque exerted by the motor between $t$ = 0 and $t$ = 4.0 s. [2]

(a) gradient $= 120/4.0 = 30$ rad s⁻²(1)
(b) area under graph $= \tfrac12(4)(120) + (6)(120) + \tfrac12(6)(120) = 1320$ rad(1)
(b) $1320/2\pi = 210$ revolutions (allow ECF from area)(1)
(c) deceleration $= 120/6.0 = 20$ rad s⁻², so $T = I\alpha = 0.080 \times 20 = 1.6$ N m(1)
(d) resultant torque $= I\alpha = 0.080 \times 30 = 2.4$ N m(1)
(d) motor torque $= 2.4 + 1.6 = 4.0$ N m (allow ECF from (c); 2.4 N m alone scores 1 mark)(1)
Answer: (a) 30 rad s⁻² (b) 210 rev (c) 1.6 N m (d) 4.0 N m
Examiner tip: In (d) the motor must supply the frictional torque as well as the resultant torque; giving $I\alpha$ alone is the common error.

5Hard7 marks

Two discs are mounted on the same vertical axle. Disc A, of moment of inertia 0.15 kg m², rotates freely at 40 rad s⁻¹. Disc B, of moment of inertia 0.25 kg m², is initially at rest. Disc B is released so that it drops onto disc A. The discs slip against each other for 0.60 s and then rotate together at the same angular velocity. Frictional torque at the axle is negligible.

(a) Calculate the common angular velocity of the discs. [2]

(b) Calculate the kinetic energy lost when the discs couple and state what happens to this energy. [3]

(a) angular momentum is conserved: $0.15 \times 40 = (0.15 + 0.25)\omega$(1)
(a) $\omega = 15$ rad s⁻¹(1)
(b) initial $E_k = \tfrac12 \times 0.15 \times 40^2 = 120$ J and final $E_k = \tfrac12 \times 0.40 \times 15^2 = 45$ J (allow ECF for $\omega$)(1)
(b) loss $= 75$ J(1)
(b) transferred to internal energy / heat (and sound) by friction between the discs (do not allow 'energy is lost' alone)(1)
(c) angular impulse = change in angular momentum of B: $T\Delta t = 0.25 \times 15 = 3.75$ N m s(1)
(c) $T = 3.75/0.60 = 6.3$ N m (accept 6.25; allow use of disc A: $0.15 \times 25/0.60$)(1)
Answer: (a) 15 rad s⁻¹ (b) 75 J lost, transferred to internal energy by friction (c) 6.3 N m
Examiner tip: Kinetic energy is not conserved in this coupling; only angular momentum is, so don't equate rotational kinetic energies.

6Hard6 marks

A diver leaves a springboard rotating slowly with her body straight. In mid-air she pulls her knees to her chest (a 'tuck') and rotates rapidly. Just before entering the water she straightens her body again.

Explain, in terms of angular momentum, moment of inertia and rotational kinetic energy, how the diver controls her rate of rotation during the dive. Your answer should include an explanation of where any change in rotational kinetic energy comes from. [6]

once in the air there is no external torque about her centre of mass (air resistance negligible, weight acts through centre of mass)(1)
so her angular momentum $I\omega$ is conserved / constant(1)
tucking brings her mass closer to the axis of rotation, so her moment of inertia decreases(1)
so her angular velocity increases (in inverse proportion to $I$) and she completes rotations quickly(1)
straightening increases $I$ again, so $\omega$ decreases, allowing a controlled vertical entry (less splash)(1)
angular momentum (the initial rotation) comes from the torque / angular impulse of the board's reaction force acting off-centre during take-off(1)
rotational $E_k = \tfrac12 I\omega^2 = L^2/2I$ so with $L$ constant, $E_k$ increases when $I$ decreases(1)
the extra kinetic energy comes from work done by her muscles pulling her limbs inwards (against the tendency to move outwards)(1)
when she extends, $E_k$ decreases; energy is transferred to internal energy in her muscles (does negative work)(1)

Level	Marks	What it looks like
Level 3	5–6	Clear, logically ordered explanation linking zero external torque to conservation of angular momentum, the change of moment of inertia on tucking and straightening to the change in angular velocity, AND explaining that rotational kinetic energy increases on tucking because the diver's muscles do work.
Level 2	3–4	Links conservation of angular momentum to the change in moment of inertia and angular velocity for tucking and/or straightening, with some reference to kinetic energy, but the energy explanation is incomplete or the reasoning has gaps.
Level 1	1–2	Some relevant statements, e.g. tucking makes her spin faster or angular momentum is conserved, with little or no linking of the ideas.
Level 0	0	No relevant content.

Answer: Angular momentum is conserved in flight (no external torque); tucking lowers $I$ so $\omega$ rises; straightening raises $I$ so $\omega$ falls. $E_k = L^2/2I$ rises on tucking because her muscles do work pulling the mass inwards.
Examiner tip: Many answers say 'energy is conserved so she spins faster' — rotational kinetic energy is NOT constant here; angular momentum is.

7Hard7 marks

Air in the cylinder of a diesel engine is compressed rapidly so that no heat is transferred to or from the air. At the start of the compression the air has pressure $1.00\times10^5$ Pa, volume $4.5\times10^{-4}$ m³ and temperature 290 K. It is compressed to a volume of $5.0\times10^{-5}$ m³. Treat the air as an ideal gas with $\gamma = 1.40$.

(a) Calculate the pressure of the air at the end of the compression. [2]

(b) Calculate the temperature of the air at the end of the compression. [2]

(a) use of $p_1V_1^{\gamma} = p_2V_2^{\gamma}$: $p_2 = 1.00\times10^5 \times (4.5\times10^{-4}/5.0\times10^{-5})^{1.40}$(1)
(a) $p_2 = 2.2\times10^6$ Pa (accept 2.17 MPa)(1)
(b) use of $pV/T$ = constant (or $TV^{\gamma-1}$ = constant): $T_2 = 290 \times \dfrac{2.17\times10^6 \times 5.0\times10^{-5}}{1.00\times10^5 \times 4.5\times10^{-4}}$(1)
(b) $T_2 = 700$ K (accept 698 K; allow ECF from (a))(1)
(c) adiabatic so $Q = 0$(1)
(c) work is done on the air, so $W$ (work done by the gas) is negative; from $Q = \Delta U + W$, $\Delta U = -W$ is positive(1)
(c) internal energy of an ideal gas is the kinetic energy of its molecules, so mean kinetic energy increases and temperature rises(1)
Answer: (a) $2.2\times10^6$ Pa (b) 700 K (c) $Q=0$, $W<0$ so $\Delta U>0$; higher molecular KE means higher temperature
Examiner tip: Sign convention matters: AQA's W is work done BY the gas, so in a compression W is negative and ΔU = −W is positive.

8Hard8 marks

A four-cylinder, four-stroke petrol engine runs at a constant 3600 rev min⁻¹. The area of the indicator diagram for one cylinder corresponds to 410 J of work per cycle.

(a) Calculate the indicated power of the engine. [2]

(b) The engine delivers a torque of 110 N m at the output shaft. Calculate the brake power. [1]

(d) Petrol of calorific value $4.6\times10^7$ J kg⁻¹ is used at a rate of $3.6\times10^{-3}$ kg s⁻¹. Calculate the mechanical efficiency and the overall efficiency of the engine. [3]

(e) Give one reason why the area of the indicator diagram of a real engine is smaller than the area of its theoretical cycle. [1]

(a) four-stroke so one cycle per two revolutions: $3600/60/2 = 30$ cycles s⁻¹(1)
(a) indicated power $= 410 \times 30 \times 4 = 4.9\times10^4$ W (49 kW)(1)
(b) $P = T\omega = 110 \times (3600 \times 2\pi/60) = 4.1\times10^4$ W (accept 41.5 kW)(1)
(c) friction power $= 49.2 - 41.5 = 7.7$ kW (allow ECF)(1)
(d) input power $= 4.6\times10^7 \times 3.6\times10^{-3} = 1.66\times10^5$ W(1)
(d) mechanical efficiency $= 41.5/49.2 = 0.84$ (84 %) (allow ECF)(1)
(d) overall efficiency $= 41.5/166 = 0.25$ (25 %) (allow ECF)(1)
(e) any one: valves do not open/close instantaneously / combustion (heat input) is not instantaneous so corners are rounded / a pumping loop (negative work) during induction and exhaust / heat is lost through the cylinder walls / working substance is not an ideal gas and its quantity changes(1)
Answer: (a) 49 kW (b) 41 kW (c) 7.7 kW (d) mechanical 84 %, overall 25 % (e) e.g. valves do not open or close instantaneously
Examiner tip: Using 60 cycles per second (forgetting a four-stroke cycle takes two revolutions) doubles the indicated power; many lose that mark.

9Standard7 marks

A fixed mass of ideal gas is taken round the cycle A → B → C → D → A shown on the p–V diagram.

(a) During process A → B, 420 J of energy is supplied to the gas by heating. Calculate (i) the work done by the gas and (ii) the change in internal energy of the gas. [3]

(b) Determine the net work done by the gas in one complete cycle. [2]

(c) Identify the two processes in which energy is transferred from the gas to the surroundings by heating. Explain your choice for one of these processes. [2]

(a)(i) constant pressure: $W = p\Delta V = 2.0\times10^5 \times (1.8-1.2)\times10^{-3}$(1)
(a)(i) $W = 120$ J(1)
(a)(ii) $\Delta U = Q - W = 420 - 120 = 300$ J (increase) (allow ECF from (i))(1)
(b) net work = area enclosed by loop $= (2.0-0.8)\times10^5 \times 0.6\times10^{-3}$(1)
(b) $= 72$ J(1)
(c) B → C and C → D (both needed; list principle: an extra wrong process loses the mark)(1)
(c) explanation for either: in B → C no work is done and the temperature falls (since $p$ falls at constant $V$), so $\Delta U < 0$ and $Q < 0$ / in C → D work is done on the gas and temperature falls, so $Q = \Delta U + W$ is negative(1)
Answer: (a) 120 J; ΔU = +300 J (b) 72 J (c) B → C and C → D
Examiner tip: Students read the volume axis as m³ rather than 10⁻³ m³ — always check the axis multiplier before calculating pΔV.

10Standard6 marks

(a) The steam in a power station turbine enters at 820 K and the cooling water is at 300 K. Calculate the maximum theoretical efficiency of the turbine as a heat engine. [1]

(b) State two reasons why the efficiency of a practical heat engine is less than this maximum. [2]

(c) A ground-source heat pump heats a house. Heat is extracted from the ground at 5 °C and delivered to the radiator water at 45 °C.

(i) Calculate the maximum theoretical coefficient of performance of the heat pump. [1]

(ii) The actual coefficient of performance is 3.2 and the heat pump delivers 9.6 kW to the house. Calculate the rate at which energy is extracted from the ground. [2]

(a) $(T_H - T_C)/T_H = 520/820 = 0.63$ (63 %)(1)
(b) any two from: friction (between moving parts) / heat losses to the surroundings (other than to the sink) / the working substance does not undergo ideal reversible processes / energy needed to run pumps, auxiliary equipment / incomplete combustion of fuel(1)
(b) second reason from the list (do not allow 'energy is wasted' without a mechanism)(1)
(c)(i) $T_H/(T_H - T_C) = 318/40 = 8.0$ (accept 7.95; temperatures must be in K)(1)
(c)(ii) work input $W = Q_H/\text{COP} = 9.6/3.2 = 3.0$ kW(1)
(c)(ii) $Q_C = Q_H - W = 9.6 - 3.0 = 6.6$ kW(1)
Answer: (a) 0.63 (b) e.g. friction; heat losses to surroundings (c)(i) 8.0 (ii) 6.6 kW
Examiner tip: Using Celsius temperatures in the efficiency or COP formulas (e.g. 45/40) is the most common way to lose these marks.

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