Original GioPhysics extended practice
Cambridge IGCSE Physics · Extended Paper 1
A balanced original practice paper with long-form reasoning, calculations, and supplied-data graphs.
- Time
- 160 minutes
- Questions
- 10
- Marks
- 116
- 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.
Electric-bus speed–time evidence
An electric bus has these (time in s, speed in m/s) readings: (0, 0), (4, 12), (10, 12), (14, 4), (16, 4), (18, 0). Join successive points with straight lines. (a) Describe the motion. (b) Calculate acceleration from 0–4 s and 10–14 s. (c) Calculate total distance. (d) Calculate average speed. (e) Explain why speed data alone may not give displacement.
Electric-bus speed
Speed recorded over an eighteen-second journey.
Read the exact data as a table
| Time / s | Speed / m s⁻¹ |
|---|---|
| 0 | 0 |
| 4 | 12 |
| 10 | 12 |
| 14 | 4 |
| 16 | 4 |
| 18 | 0 |
These are the supplied readings, not a worked answer. Lines join the supplied readings; they are not a fitted model.
Density of an irregular fitting
A fitting has mass 156.4 g. Water in a measuring cylinder rises from 48.0 cm³ to 67.5 cm³ when it is submerged. (a) Calculate density in g/cm³ and kg/m³. (b) Explain how to make the measurements reliably. (c) A trapped bubble occupies 1.2 cm³. Calculate corrected density and explain the original error direction. (d) Explain why dissolving material needs another method.
Open the interactive answer workspacePumped-storage energy chain
A station raises 2.5 × 10⁶ kg of water through 120 m in 45 min. Pump efficiency is 78%; generation efficiency is 70%. Take g = 9.8 N/kg. (a) Calculate stored GPE. (b) Calculate pump input energy and power. (c) Calculate generated electrical output. (d) Calculate round-trip efficiency. (e) Explain why the system remains useful below 100% efficiency.
Open the interactive answer workspaceDesigning a fair insulation comparison
A covered beaker cools in a 20 °C room with (time in s, temperature in °C) readings (0, 80), (120, 68), (240, 59), (360, 52), (480, 47). (a) Find mean cooling rate over the first and final 240 s. (b) Explain the difference. (c) Design a fair comparison of two insulation materials, including controls, repeats and dependent variable. (d) Explain why equal thickness is not always the fairest engineering comparison.
Cooling curve
Covered-beaker temperature in a room at 20 °C.
Read the exact data as a table
| Time / s | Temperature / °C |
|---|---|
| 0 | 80 |
| 120 | 68 |
| 240 | 59 |
| 360 | 52 |
| 480 | 47 |
These are the supplied readings, not a worked answer. Lines join the supplied readings; they are not a fitted model.
Sound speed from wavelength evidence
A loudspeaker gives (frequency in Hz, wavelength in m) measurements (250, 1.36), (300, 1.13), (350, 0.97), (400, 0.85), (450, 0.76). (a) Estimate sound speed from two readings. (b) Test another reading for consistency. (c) Explain frequency and wavelength changes on entering warmer air. (d) Describe a stationary-wave wavelength method and uncertainty reduction.
Sound wavelength data
Wavelength at five frequency settings.
Read the exact data as a table
| Frequency / Hz | Wavelength / m |
|---|---|
| 250 | 1.36 |
| 300 | 1.13 |
| 350 | 0.97 |
| 400 | 0.85 |
| 450 | 0.76 |
These are the supplied readings, not a worked answer. No fitted trend is drawn.
Electromagnetic choices in rescue and medicine
Choose suitable electromagnetic regions for satellite communication, thermal imaging through smoke and imaging a broken bone, explaining each. (a) Calculate frequency of 3.0 cm satellite radiation using c = 3.0 × 10⁸ m/s. (b) Order the chosen regions by photon energy. (c) Explain one hospital safety control. (d) State what all three regions have in common when travelling through a vacuum.
Open the interactive answer workspaceSafe parallel lighting circuit
Three 12 V, 24 W lamps are parallel across 12 V. (a) Find current in each and total current. (b) Find equivalent resistance. (c) Explain why parallel is preferable to series. (d) Choose a fuse from 3 A, 5 A and 10 A. (e) One lamp fails open: describe others and total current. (f) Explain live-wire fuse and switch placement in a mains version.
Open the interactive answer workspaceHalf-life from count-rate data
With a source present, a detector gives (time in h, count/min) readings (0, 820), (2, 430), (4, 235), (6, 138), (8, 89). Background is 40 counts/min. (a) Correct the readings and determine half-life. (b) Predict measured rate at 12 h. (c) Explain scatter and a reliable method. (d) Distinguish irradiation from contamination and state precautions.
Count-rate decay
Detector readings including background.
Read the exact data as a table
| Time / h | Count rate / min⁻¹ |
|---|---|
| 0 | 820 |
| 2 | 430 |
| 4 | 235 |
| 6 | 138 |
| 8 | 89 |
These are the supplied readings, not a worked answer. Lines join the supplied readings; they are not a fitted model.
Circular satellite orbit
A satellite orbits at radius 4.2 × 10⁷ m and speed 3.08 × 10³ m/s. Earth's radius is 6.4 × 10⁶ m. (a) Find altitude. (b) Find period in hours. (c) Identify centripetal force. (d) Explain larger-radius circular speed and period changes. (e) Explain astronaut apparent weightlessness.
Open the interactive answer workspacePendulum model investigation
A student measures (length in m, period in s) (0.4, 1.27), (0.6, 1.55), (0.8, 1.8), (1, 2.01), (1.2, 2.2). Model: T = 2π√(L/g). (a) Explain a T²-against-L plot. (b) Use L = 1.00 m to estimate g. (c) Give a reliable complete method and controls. (d) Explain graph extraction of g and a non-zero intercept. (e) Explain why large amplitude invalidates the model.
Pendulum period
Mean period at five effective lengths.
Read the exact data as a table
| Length / m | Period / s |
|---|---|
| 0.4 | 1.27 |
| 0.6 | 1.55 |
| 0.8 | 1.8 |
| 1 | 2.01 |
| 1.2 | 2.2 |
These are the supplied readings, not a worked answer. No fitted trend is drawn.
Teacher 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.
Electric-bus speed–time evidence
Answer: Accelerations +3.0 m/s² and −2.0 m/s²; distance 140 m; average speed 7.78 m/s; displacement also needs direction.
- The rising line from 0–4 s is uniform acceleration from rest, 4–10 s is constant speed, 10–14 s is uniform slowing, 14–16 s is constant slower motion and 16–18 s is uniform slowing to rest.
- Acceleration is graph gradient: (12 − 0)/4 = +3.0 m/s² initially and (4 − 12)/(14 − 10) = −2.0 m/s² during the first slowing interval. The sign describes the chosen direction.
- Distance is area below the graph: ½(4)(12) + (6)(12) + ½(12 + 4)(4) + (2)(4) + ½(2)(4) = 24 + 72 + 32 + 8 + 4 = 140 m.
- Average speed is total distance divided by total time: 140/18 = 7.78 m/s. Averaging the highest and lowest speeds would ignore the unequal durations spent at each speed.
- Speed has no sign. If the bus reverses, distance still accumulates positively but displacement must subtract travel in the opposite direction; a velocity–time graph or explicit directions are required.
Density of an irregular fitting
Answer: 8.02 g/cm³ = 8.02 × 10³ kg/m³; corrected density 8.55 g/cm³. A bubble makes apparent volume too large and density too small.
- The fitting volume is the water displacement, 67.5 − 48.0 = 19.5 cm³. It must be fully submerged without overflow because this difference is being equated to object volume.
- Density is ρ = m/V = 156.4/19.5 = 8.02 g/cm³. Since 1 g/cm³ equals 1000 kg/m³, the SI density is 8.02 × 10³ kg/m³.
- Zero the balance, dry the fitting before weighing, read the cylinder at eye level at the bottom of the meniscus and repeat. A narrower graduated cylinder can reduce the percentage volume uncertainty.
- The bubble is not metal, so corrected volume = 19.5 − 1.2 = 18.3 cm³. Corrected density = 156.4/18.3 = 8.55 g/cm³; the original oversized denominator biased density downward.
- If material dissolves, both specimen mass and liquid composition change, and final volume is not simply water plus original solid volume. Use an inert displacement liquid or geometric method instead.
Pumped-storage energy chain
Answer: Stored 2.94 × 10⁹ J; pump input 3.77 × 10⁹ J; power 1.40 MW; output 2.06 × 10⁹ J; round-trip efficiency 54.6%.
- Stored gravitational energy is mgh = 2.5 × 10⁶ × 9.8 × 120 = 2.94 × 10⁹ J.
- Efficiency is useful output divided by input, so pump input = 2.94 × 10⁹/0.78 = 3.77 × 10⁹ J. Dividing is required because input exceeds stored energy.
- Forty-five minutes is 2700 s. Average input power = 3.77 × 10⁹/2700 = 1.40 × 10⁶ W = 1.40 MW.
- Generated output = 0.70 × 2.94 × 10⁹ = 2.06 × 10⁹ J. Round-trip efficiency = output/original input = 0.78 × 0.70 = 0.546 or 54.6%.
- Losses become thermal energy and sound through turbulence, friction and resistance. Storage remains valuable because surplus electricity can be shifted to high-demand periods and released quickly to stabilise supply.
Designing a fair insulation comparison
Answer: Rates 0.0875 °C/s and 0.0500 °C/s; cooling slows as temperature difference falls. Use matched containers and repeated, defined cooling measures.
- First 240 s: temperature drop is 80 − 59 = 21 °C, so mean rate = 21/240 = 0.0875 °C/s. Final 240 s: drop is 59 − 47 = 12 °C, giving 0.0500 °C/s.
- Thermal-transfer rate depends on temperature difference. The water–room difference falls from 60 °C toward 27 °C, reducing conduction, convection and net radiation rates.
- Use identical lidded containers with equal water mass and initial temperature, equal exposed area and simultaneous placement away from draughts. Change only insulation material.
- Use matched probes, record fixed intervals, compare time to a chosen temperature or drop over a chosen time, repeat and compare means and spread. Swap probes to check sensor bias.
- Equal thickness can mean unequal mass, cost or environmental burden because materials have different density. The controlled basis should match the design constraint: thickness, mass per area, price or available space.
Sound speed from wavelength evidence
Answer: Sound speed about 340 m/s; 350 Hz gives 339.5 m/s and is consistent. Source frequency stays fixed while speed and wavelength depend on medium.
- Use v = fλ. The 250 Hz reading gives 250 × 1.36 = 340 m/s; the 400 Hz reading gives 400 × 0.85 = 340 m/s.
- At 350 Hz, v = 350 × 0.97 = 339.5 m/s, only 0.5 m/s different. This is consistent with measurement precision rather than evidence of a speed change.
- Source oscillation fixes frequency, which remains continuous across a boundary. If wave speed changes in warmer air, wavelength changes because λ = v/f.
- Create resonance in a tube or reflection from a wall and locate successive nodes with a microphone. Adjacent nodes are λ/2 apart, so double the spacing.
- Measure across several node gaps and divide by their number; repeat in both directions, keep frequency and temperature controlled and reduce unwanted room reflections.
Electromagnetic choices in rescue and medicine
Answer: Microwave, infrared and X-ray respectively; 3.0 cm gives 1.0 × 10¹⁰ Hz; energy order X-ray > infrared > microwave.
- Microwaves in suitable bands pass through the atmosphere and can be directed with dishes for satellite communication. Convert 3.0 cm to 0.030 m before calculation.
- Frequency f = c/λ = 3.0 × 10⁸/0.030 = 1.0 × 10¹⁰ Hz.
- Warm surfaces emit infrared related to temperature, so infrared cameras map heat. X-rays are absorbed more strongly by bone than soft tissue, producing detector contrast.
- Photon energy E = hf increases with frequency, so X-ray has greatest energy, then infrared, then microwave. All travel at c in vacuum.
- Ionising X-ray exposure is minimised using the lowest diagnostic exposure, beam collimation, shielding, distance and avoiding unnecessary repeats; benefit must justify risk.
Safe parallel lighting circuit
Answer: Each 2.0 A; total 6.0 A; equivalent 2.0 Ω; choose 10 A. One failed branch leaves two normal lamps and 4.0 A total.
- Each branch has 12 V, so I = P/V = 24/12 = 2.0 A. Junction conservation gives total current 6.0 A.
- Equivalent resistance is V/Itotal = 12/6.0 = 2.0 Ω. It is below each operating lamp resistance, as expected for parallel paths.
- Parallel branches each receive rated voltage and operate independently. Series lamps would share voltage, be dimmer and all extinguish if one filament opened.
- The lowest listed fuse above normal 6.0 A is 10 A. A 5 A fuse would interrupt normal operation.
- An open branch removes only one path. Two lamps remain at 12 V and normal brightness; total current becomes 4.0 A.
- Put switch and fuse in live so operation disconnects dangerous live potential. A neutral-only fuse could open while internal components remain live relative to Earth.
Half-life from count-rate data
Answer: Corrected rates 780, 390, 195, 98, 49; half-life ≈2.0 h; predicted 12 h measured rate ≈52 counts/min.
- Subtract background to obtain 780, 390, 195, 98 and 49 counts/min. The small rounding later is compatible with random decay statistics.
- Corrected rate halves from 780 to 390 in 2 h and to about 195 after another 2 h, so half-life is approximately 2.0 h.
- From 8 h to 12 h is two half-lives: corrected rate ≈49/4 = 12.25. Add background to predict about 52 counts/min.
- Decay is random, so finite counts scatter. Count longer, repeat equal intervals and average while keeping geometry fixed to reduce fractional uncertainty.
- Irradiation is external exposure and stops when source is removed; contamination is radioactive material present and can spread. Minimise time, maximise distance, shield appropriately and use tongs/storage protocols.
Circular satellite orbit
Answer: Altitude 3.56 × 10⁷ m; period 8.57 × 10⁴ s = 23.8 h; gravity is inward force; larger orbit is slower with longer period.
- Altitude is orbital radius minus Earth radius: 4.2 × 10⁷ − 6.4 × 10⁶ = 3.56 × 10⁷ m.
- Period T = circumference/speed = 2π(4.2 × 10⁷)/(3.08 × 10³) = 8.57 × 10⁴ s = 23.8 h.
- Earth's gravitational attraction supplies the inward centripetal resultant. Centripetal force is a role of gravity here, not an extra force added to it.
- At larger radius gravity is weaker, so circular speed is lower. The path is longer and speed lower, producing a longer period, consistent with T² proportional to r³.
- Spacecraft and astronauts fall together under gravity, leaving little normal contact force from the floor. Apparent weight vanishes although gravitational force does not.
Pendulum model investigation
Answer: At 1.00 m, g ≈9.77 m/s²; T²–L gradient is 4π²/g, so g = 4π²/gradient.
- Squaring gives T² = (4π²/g)L, a straight line through the origin. This linearisation makes model testing and extracting g clearer than a curved T–L plot.
- At L = 1.00 m and T = 2.01 s, g = 4π²L/T² = 9.77 m/s².
- Measure pivot to bob centre, use a small fixed release angle without push, time 10–20 complete oscillations from the same directional crossing, divide by count and repeat each length.
- Use several well-spaced lengths, a rigid stand, fiducial marker and no draught. Plot T² against L; gradient m gives g = 4π²/m, with steep/shallow acceptable lines estimating uncertainty.
- A non-zero intercept suggests timing offset, effective-length zero error or miscounting. At large angles, period depends on amplitude because the small-angle approximation fails.