Original GioPhysics extended practice
Cambridge IGCSE Physics · Extended Paper 2
A balanced original practice paper with long-form reasoning, calculations, and supplied-data graphs.
- Time
- 170 minutes
- Questions
- 10
- Marks
- 124
- Answers
- Complete teacher key
Instructions
- Answer every question and show all working.
- Use SI units unless a question states otherwise.
- Read graph scales carefully and give answers to a sensible number of significant figures.
Every question is independently written by GioPhysics. This is not an awarding-body paper.
Specific heat capacity from a heating curve
A 1.50 kg block is heated by a 300 W heater. Its (time in s, temperature in °C) readings are (0, 20), (60, 31), (120, 42), (180, 53), (240, 64). (a) Find mean temperature-rise rate. (b) Assuming no loss, calculate specific heat capacity. (c) The accepted value is 820 J/(kg °C). Calculate mean energy-loss rate. (d) Explain two improvements and why the straight line cannot continue indefinitely.
Block heating curve
Temperature under constant heater power.
Read the exact data as a table
| Time / s | Temperature / °C |
|---|---|
| 0 | 20 |
| 60 | 31 |
| 120 | 42 |
| 180 | 53 |
| 240 | 64 |
These are the supplied readings, not a worked answer. Lines join the supplied readings; they are not a fitted model.
Vehicle rebound and collision safety
A 1200 kg vehicle moving at 18 m/s strikes a barrier and rebounds at 2.0 m/s. Contact lasts 0.25 s. (a) Calculate momentum before and after, with initial direction positive. (b) Calculate impulse and average force. (c) Relate the force on the barrier to that on the vehicle. (d) A new barrier gives the same momentum change in 0.60 s. Calculate its average force, explain the safety advantage and state one model limitation.
Open the interactive answer workspaceHydraulic lift and rotational balance
A hydraulic lift has piston areas 4.0 cm² and 240 cm². The input force is 180 N and the output platform weighs 1200 N. (a) Calculate transmitted pressure. (b) Calculate output force and maximum added load in equilibrium. (c) Explain the output displacement. (d) An off-centre load rests on two supports. State the complete equilibrium conditions and explain how support forces are found.
Open the interactive answer workspaceHeating gas in a rigid canister
A rigid sealed canister contains gas at 100 kPa and 290 K. It is heated to 348 K. (a) Calculate final pressure. (b) Explain the change using particles. (c) Explain the difference with a freely moving piston. (d) State two reasons the ideal particle model may fail at very high pressure.
Open the interactive answer workspaceRefraction through a glass block
A ray enters glass from air at 50.0° to the normal and refracts at 30.0°. (a) Calculate refractive index. (b) Calculate critical angle. (c) Explain the parallel but displaced emerging ray. (d) Describe a ray-pin experiment, precautions and a graph method for refractive index.
Open the interactive answer workspaceFilament-lamp characteristic
A lamp gives (potential difference in V, current in A) readings (0, 0), (0.5, 0.18), (1, 0.35), (1.5, 0.49), (2, 0.6), (2.5, 0.68). (a) Calculate resistance at 0.50 V and 2.50 V. (b) Explain the change. (c) Find power at 2.0 V and energy in 5.0 min. (d) Describe a circuit and safe method for both signs of the characteristic.
Lamp current–voltage data
Current measured at increasing lamp potential differences.
Read the exact data as a table
| Potential difference / V | Current / A |
|---|---|
| 0 | 0 |
| 0.5 | 0.18 |
| 1 | 0.35 |
| 1.5 | 0.49 |
| 2 | 0.6 |
| 2.5 | 0.68 |
These are the supplied readings, not a worked answer. Lines join the supplied readings; they are not a fitted model.
Generator and high-voltage transmission
A 200-turn coil changes flux per turn from +3.0 × 10⁻⁴ Wb to −3.0 × 10⁻⁴ Wb in 0.020 s. (a) Find mean induced e.m.f. (b) Explain polarity and two output improvements. (c) An ideal transformer raises 120 V to 6000 V while transmitting 24 kW. Find both currents. (d) For cable resistance 8.0 Ω, compare high-voltage loss with direct 120 V transmission. (e) Explain the grid advantage.
Open the interactive answer workspaceNuclear equations and radiation evidence
²²⁶₈₈Ra emits alpha and its daughter then emits beta-minus. (a) Write equations using X and Y for daughters. (b) Explain the beta-minus nuclear change. (c) Compare alpha, beta and gamma ionisation and penetration. (d) Explain why gamma changes neither proton nor nucleon number and why heating does not trigger decay.
Open the interactive answer workspaceGalaxy recession and expansion age
Original teaching data give (distance in 10²² m, speed in km/s) (1, 720), (2, 1390), (3, 2110), (4, 2780), (5, 3510). (a) Find endpoint gradient. (b) Convert it to s⁻¹ and estimate 1/H in years using 3.16 × 10⁷ s/year. (c) Explain redshift evidence. (d) Explain sample scatter. (e) State limitations of treating 1/H as exact cosmic age.
Illustrative galaxy data
Original simplified recession-speed measurements.
Read the exact data as a table
| Distance / 10²² m | Speed / km s⁻¹ |
|---|---|
| 1 | 720 |
| 2 | 1390 |
| 3 | 2110 |
| 4 | 2780 |
| 5 | 3510 |
These are the supplied readings, not a worked answer. No fitted trend is drawn.
Household solar-storage audit
A home uses a 1.8 kW heater for 2.5 h, six 9 W lamps for 5.0 h, and a 120 W refrigerator for an effective 8.0 h daily. Solar supplies 6.2 kWh through a battery of 82% round-trip efficiency. (a) Find each and total demand. (b) Find useful battery energy and grid shortfall. (c) Convert demand to joules. (d) At 230 V find heater current and choose 3, 5, 10 or 13 A fuse. (e) Explain two reasons this balance does not prove off-grid operation.
Open the interactive answer workspaceTeacher copyDetailed answer and marking guideShow 10 answers
Teacher copy
Detailed answer and marking guide
Accept equivalent physics expressed clearly. Award method credit where the reasoning is valid.
Specific heat capacity from a heating curve
Answer: Rate 0.183 °C/s; uncorrected c = 1090 J/(kg °C); mean non-stored power about 74.5 W.
- Temperature rises 64 − 20 = 44 °C in 240 s, so mean rate is 44/240 = 0.183 °C/s. The equal eleven-degree rises each minute support approximate linearity over this interval.
- Heater supplies E = Pt = 300 × 240 = 72000 J. With no losses, c = E/(mΔT) = 72000/(1.50 × 44) = 1090 J/(kg °C).
- Using the accepted c, energy stored in block = 1.50 × 820 × 44 = 54120 J. Energy not stored = 72000 − 54120 = 17880 J, giving 17880/240 = 74.5 W.
- Insulate the block and improve heater/thermometer contact; record more points, fit a gradient and repeat. Include energy used to heat apparatus when refining the model.
- As temperature difference from the room grows, loss rate generally rises. Eventually heater power can balance losses, so temperature approaches a steady value instead of rising linearly forever.
Vehicle rebound and collision safety
Answer: Momenta +21600 and −2400 kg m/s; impulse −24000 N s; average forces −96000 N and −40000 N.
- Initial momentum is 1200 × 18 = +21600 kg m/s. Rebound is opposite the chosen positive direction, so final momentum is 1200 × (−2.0) = −2400 kg m/s.
- Impulse equals final minus initial momentum: −2400 − 21600 = −24000 N s. The negative sign is physically important because the impulse points opposite the original motion.
- Average resultant force is Δp/Δt = −24000/0.25 = −96000 N. This average does not reveal the usually larger instantaneous peak force.
- Newton's third law gives equal-magnitude opposite-direction interaction forces on vehicle and barrier. They act on different objects, so they must not be cancelled on one free-body diagram.
- For 0.60 s, average force = −24000/0.60 = −40000 N. Greater stopping time reduces average force, but occupant safety also depends on peak acceleration, restraint, deformation distance and force history.
Hydraulic lift and rotational balance
Answer: Pressure 4.5 × 10⁵ Pa; output force 1.08 × 10⁵ N; added load 1.068 × 10⁵ N; volume displacement is conserved and both force and moment sums vanish.
- Convert input area: 4.0 cm² = 4.0 × 10⁻⁴ m². Pressure p = F/A = 180/(4.0 × 10⁻⁴) = 4.5 × 10⁵ Pa.
- Output area is 0.0240 m², so force = pA = 4.5 × 10⁵ × 0.0240 = 1.08 × 10⁵ N. Added load = 108000 − 1200 = 106800 N.
- For an incompressible liquid, displaced volumes approximately match: Aᵢdᵢ = Aₒdₒ. The sixty-times-larger output area therefore moves one-sixtieth of the input distance, preserving the work trade-off.
- Static equilibrium requires zero vector resultant force and zero resultant moment about any point. Either condition alone is insufficient.
- The two upward support forces sum to total weight. Taking moments about one support removes its unknown force and determines the other; an off-centre load generally gives unequal support forces.
Heating gas in a rigid canister
Answer: Final pressure 120 kPa; faster particles collide more often and transfer more momentum. A moving piston permits expansion.
- At fixed amount and rigid volume, p/T is constant using kelvin temperatures. Therefore p₂ = 100 × 348/290 = 120 kPa.
- Heating increases mean molecular kinetic energy and speed. Particles reach the walls more frequently and each collision generally transfers more momentum, increasing force per area.
- Kelvin temperature must be used because it is proportional to mean kinetic energy. A Celsius ratio would imply 0 °C means no molecular motion.
- A freely moving piston allows volume to increase and can keep pressure near external pressure during slow motion; some heating then supplies boundary work.
- At high pressure, molecular volume is not negligible and intermolecular forces affect motion, violating ideal assumptions of point particles and no forces except elastic collisions.
Refraction through a glass block
Answer: Refractive index 1.53; critical angle 40.8°. Opposite refractions at parallel faces restore direction while finite thickness creates lateral shift.
- For air to glass, n = sin i/sin r = sin 50.0°/sin 30.0° = 1.532, so n = 1.53.
- At a glass–air critical angle, refracted angle is 90° and sin c = 1/n. Thus c = sin⁻¹(1/1.532) = 40.8°.
- Parallel faces have parallel normals. Refraction toward the normal at entry is reversed by refraction away at exit, restoring direction; travel through finite thickness leaves lateral displacement.
- Trace the block, place two widely separated incident pins, align two image pins without parallax, remove the block and draw rays and normals accurately. Use thin vertical pins.
- Repeat several incidence angles and plot sin i vertically against sin r horizontally. A best-fit gradient through the origin estimates n; repeats and large pin separation reduce uncertainty.
Filament-lamp characteristic
Answer: Resistances 2.78 Ω and 3.68 Ω; power 1.20 W; five-minute energy 360 J; heating raises filament resistance.
- At 0.50 V, R = V/I = 0.50/0.18 = 2.78 Ω. At 2.50 V, R = 2.50/0.68 = 3.68 Ω, so resistance is not constant.
- Greater current heats the filament. Stronger lattice vibrations increase charge-carrier collisions, causing the curved non-ohmic characteristic.
- At 2.0 V and 0.60 A, P = VI = 1.20 W. Five minutes is 300 s, so E = Pt = 1.20 × 300 = 360 J.
- Place ammeter and variable resistor in series with the lamp and voltmeter in parallel. Adjust a low-voltage supply gradually and stay within the lamp rating.
- Reverse supply connections for negative values, take readings quickly and repeat. A filament characteristic should be approximately symmetric, but should not be forced into a straight line.
Generator and high-voltage transmission
Answer: Mean e.m.f. 6.0 V; currents 200 A and 4.0 A; cable losses 128 W and 3.20 × 10⁵ W respectively.
- Flux change magnitude per turn is 6.0 × 10⁻⁴ Wb. Faraday's law gives mean e.m.f. NΔΦ/Δt = 200 × 6.0 × 10⁻⁴/0.020 = 6.0 V.
- Lenz's law makes induced effects oppose the flux change, conserving energy. More turns, larger area, stronger field or faster rotation increases rate of flux-linkage change.
- Ideal power conservation gives primary current 24000/120 = 200 A and secondary current 24000/6000 = 4.0 A.
- High-voltage cable loss I²R = 4.0² × 8.0 = 128 W. At 120 V, current 200 A gives 200² × 8.0 = 3.20 × 10⁵ W.
- High voltage reduces current for specified power, so heating falls with current squared. The voltage is stepped down near consumers for insulation practicality and safety.
Nuclear equations and radiation evidence
Answer: ²²⁶₈₈Ra → ²²²₈₆X + ⁴₂He; ²²²₈₆X → ²²²₈₇Y + ⁰₋₁e + antineutrino.
- Conserve nucleon and proton numbers: alpha emission gives ²²⁶₈₈Ra → ²²²₈₆X + ⁴₂He.
- In beta-minus decay, a neutron becomes a proton while an electron and electron antineutrino are emitted: ²²²₈₆X → ²²²₈₇Y + ⁰₋₁e + anti-νₑ.
- Alpha is most ionising and least penetrating; beta is intermediate; gamma is least ionising per path length and most penetrating, requiring substantial shielding for reduction.
- Gamma is an energy photon with no proton or neutron, so nuclear numbers do not change. It can follow a decay that leaves an excited daughter.
- Radioactive decay is a spontaneous nuclear process. Ordinary heating changes atomic motion on energy scales far below nuclear transitions and does not schedule individual decays.
Galaxy recession and expansion age
Answer: Gradient 697.5 km s⁻¹ per 10²² m = 6.98 × 10⁻¹⁷ s⁻¹; illustrative expansion timescale 4.54 × 10⁸ years.
- Using endpoints, gradient = (3510 − 720)/(5 − 1) = 697.5 km s⁻¹ per 10²² m. Wide separation limits relative reading uncertainty.
- Convert: H = 697.5 × 1000/10²² = 6.975 × 10⁻¹⁷ s⁻¹. Its reciprocal is 1.434 × 10¹⁶ s = 4.54 × 10⁸ years.
- The dataset is intentionally simplified original teaching data, not a current cosmological measurement. Spectral lines shifted to longer wavelengths provide recession evidence by comparison with laboratory wavelengths.
- Galaxies have local peculiar velocities and distance uncertainty, so one v/d value is unreliable. A large sample and best-fit trend reveal the expansion relation.
- The expansion rate changes with cosmic contents and history, and distance calibration affects the gradient. Therefore 1/H is a model timescale, not automatically exact age.
Household solar-storage audit
Answer: Demand 5.73 kWh; battery output 5.084 kWh; shortfall 0.646 kWh; 2.06 × 10⁷ J; heater current 7.83 A, choose 10 A.
- Heater uses 1.8 × 2.5 = 4.50 kWh. Lamps total 54 W = 0.054 kW and use 0.270 kWh. Refrigerator uses 0.120 × 8.0 = 0.960 kWh.
- Total = 4.50 + 0.270 + 0.960 = 5.73 kWh. Battery delivers 0.82 × 6.2 = 5.084 kWh, leaving 0.646 kWh from grid.
- Using 1 kWh = 3.6 × 10⁶ J, demand = 5.73 × 3.6 × 10⁶ = 2.06 × 10⁷ J.
- Heater current = 1800/230 = 7.83 A. The smallest listed fuse above normal current is 10 A; a 5 A fuse would interrupt normal use.
- Daily totals omit timing, battery capacity and peak power. Weather, state of charge, inverter loss and appliance starting currents can prevent off-grid operation despite an average energy balance.