Original GioPhysics extended practice
Cambridge AS & A Level Physics · Extended Paper 1
A balanced original practice paper with long-form reasoning, calculations, and supplied-data graphs.
- Time
- 165 minutes
- Questions
- 10
- Marks
- 122
- Answers
- Complete teacher key
Instructions
- Answer every question and show the complete physical argument.
- Use the data and relationships appropriate to your course.
- Give units, directions and justified assumptions where required.
Every question is independently written by GioPhysics. This is not an awarding-body paper.
Force-sensor calibration and uncertainty
Known masses give (mass in kg, force in N) readings (0.2, 1.97), (0.3, 2.95), (0.4, 3.92), (0.5, 4.91), (0.6, 5.88). Mass uncertainty is ±0.001 kg and force uncertainty ±0.02 N. (a) Find endpoint gradient and unit. (b) Estimate g. (c) Find percentage uncertainty in the first mass and force. (d) Test whether the intercept is resolved from zero. (e) Explain steepest/shallowest lines and one systematic error.
Force-sensor calibration
Force reading for independently known masses.
Read the exact data as a table
| Mass / kg | Force / N |
|---|---|
| 0.2 | 1.97 |
| 0.3 | 2.95 |
| 0.4 | 3.92 |
| 0.5 | 4.91 |
| 0.6 | 5.88 |
These are the supplied readings, not a worked answer. No fitted trend is drawn.
Rocket-cart ejection and resistance
A 5.0 kg cart initially at rest ejects a 0.80 kg package backward at 6.0 m/s relative to ground. Remaining cart then experiences 1.4 N resistance for 3.0 s. (a) Find immediate cart speed. (b) Find resistive impulse and later speed. (c) Find total kinetic energy after ejection and explain its source. (d) Explain Newton's third law and when momentum conservation is approximate.
Open the interactive answer workspaceWinch energy and instantaneous power
A winch lifts 420 kg through 18 m in 24 s, from rest to 3.0 m/s, losing 1.6 × 10⁴ J. Take g = 9.81 m/s². (a) Find GPE and KE gains. (b) Find electrical input and average power. (c) At final speed input power is 18.0 kW and resistance 900 N. Find acceleration. (d) Explain average versus instantaneous power.
Open the interactive answer workspaceWave transmission at a string boundary
A 12 Hz transverse wave of amplitude 3.0 cm travels at 4.8 m/s in a light string, then at 3.2 m/s in a heavier string. (a) Find both wavelengths. (b) Find period, angular frequency and maximum particle speed. (c) Explain unchanged frequency, changed wavelength and partial reflection. (d) State reflected-pulse phase change.
Open the interactive answer workspaceThermistor divider alarm
An NTC thermistor has (temperature in °C, resistance in kΩ) values (10, 8.1), (20, 5.4), (30, 3.7), (40, 2.6), (50, 1.9), (60, 1.4). It is series with 3.3 kΩ across 6.0 V; output is across the fixed resistor. (a) Find output at 20 °C and 50 °C. (b) Explain sensitivity direction. (c) Estimate trigger temperature for 3.5 V. (d) Explain voltmeter loading and a comparator improvement.
NTC calibration
Resistance at six temperatures.
Read the exact data as a table
| Temperature / °C | Resistance / kΩ |
|---|---|
| 10 | 8.1 |
| 20 | 5.4 |
| 30 | 3.7 |
| 40 | 2.6 |
| 50 | 1.9 |
| 60 | 1.4 |
These are the supplied readings, not a worked answer. Lines join the supplied readings; they are not a fitted model.
Quark bookkeeping in two decays
A free neutron undergoes beta-minus decay and a neutral pion decays into two photons. (a) Write both reactions, including the antineutrino. (b) Give proton and neutron quark compositions. (c) State the quark-level change and weak boson. (d) Check charge, baryon and lepton number. (e) Explain why photons carry no baryon number.
Open the interactive answer workspaceMolecular interpretation of a heated gas
A rigid vessel contains 0.80 mol monatomic ideal gas at 300 K and is heated to 450 K. Use R = 8.31 J/(mol K), Nₐ = 6.02 × 10²³ mol⁻¹. (a) Find ΔU using U = 3nRT/2. (b) Find molecule number. (c) Find rms-speed factor. (d) Find pressure factor and explain molecularly. (e) Explain zero boundary work and contrast constant-pressure heating.
Open the interactive answer workspaceDriven-oscillator resonance
A driven 0.50 kg oscillator gives (frequency in Hz, steady amplitude in cm) (0.8, 1.2), (0.9, 2.1), (1, 4.8), (1.1, 2.4), (1.2, 1.4). (a) Estimate resonance and angular frequency. (b) Estimate spring constant. (c) Explain phase well below, near and above resonance. (d) Predict increased damping. (e) Explain why finite peak does not prove zero damping and discuss benefit and hazard.
Driven response
Steady amplitude at five frequencies.
Read the exact data as a table
| Driving frequency / Hz | Amplitude / cm |
|---|---|
| 0.8 | 1.2 |
| 0.9 | 2.1 |
| 1 | 4.8 |
| 1.1 | 2.4 |
| 1.2 | 1.4 |
These are the supplied readings, not a worked answer. Lines join the supplied readings; they are not a fitted model.
Velocity selector and alternating supply
Positive singly charged ions pass undeflected through E = 2.4 × 10⁴ V/m and B = 0.080 T, then curve at radius 0.120 m in B = 0.150 T. Charge is 1.60 × 10⁻¹⁹ C. (a) Find speed and mass. (b) Give force direction for east velocity and downward B. (c) Explain magnetic work. (d) A 50 Hz supply is 170 V peak: find rms voltage and explain why unsynchronised a.c. cannot select steadily.
Open the interactive answer workspaceNuclear tracer dose and gamma imaging
A tracer has half-life 6.0 h and initial activity 48 MBq. A 20 min scan begins 12 h later. (a) Find activity at scan start and end. (b) Estimate decays using mean activity. (c) For one 140 keV photon per decay, estimate emitted gamma energy using 1 eV = 1.60 × 10⁻¹⁹ J. (d) Explain suitable gamma and half-life properties. (e) Explain attenuation, collimation and optimisation. (f) Distinguish activity from absorbed dose.
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.
Force-sensor calibration and uncertainty
Answer: Gradient and g ≈9.78 N/kg; first-reading percentage uncertainties 0.50% and 1.02%; endpoint intercept about 0.015 N, smaller than stated force uncertainty.
- Endpoint gradient = (5.88 − 1.97)/(0.60 − 0.20) = 3.91/0.40 = 9.775 N kg⁻¹. The unit follows force divided by mass.
- For F = mg, the best-fit gradient estimates gravitational field strength, so g ≈9.78 N/kg. Agreement with expectation is evidence for calibration, not a reason to replace the measured value.
- First mass percentage uncertainty is 0.001/0.200 × 100 = 0.50%. Force percentage uncertainty is 0.02/1.97 × 100 = 1.02%, so force resolution contributes the larger fraction.
- Endpoint-line intercept is 1.97 − 9.775(0.20) = 0.015 N. Since this is smaller than ±0.02 N force uncertainty, the data do not resolve it from zero.
- Draw steepest and shallowest lines compatible with error bars; half their gradient range estimates uncertainty. A zero offset shifts intercept, whereas mass calibration or force scale error changes gradient.
Rocket-cart ejection and resistance
Answer: Immediate cart speed 1.14 m/s forward; impulse −4.2 N s; later speed 0.143 m/s; post-ejection kinetic energy about 17.1 J.
- Remaining mass is 5.0 − 0.80 = 4.20 kg. With negligible external impulse during ejection, 0 = 4.20v + 0.80(−6.0), giving v = 1.14 m/s forward.
- Resistance impulse is FΔt = −1.4 × 3.0 = −4.2 N s. Initial post-ejection cart momentum is 4.80 kg m/s, so later momentum is 0.60 kg m/s and speed 0.60/4.20 = 0.143 m/s.
- Kinetic energy is ½(4.20)(1.1429²) + ½(0.80)(6.0²) = 2.74 + 14.4 = 17.1 J. Internal chemical or elastic energy supplies it; kinetic energy was not conserved.
- Cart and package exert equal opposite forces for equal times, giving opposite impulses. Those third-law forces act on different bodies and are internal to the combined system.
- If resistance gives appreciable impulse during the ejection interval, system momentum changes externally. Conservation is then approximate and must be justified by comparing external impulse with ejection momenta.
Winch energy and instantaneous power
Answer: GPE 7.42 × 10⁴ J; KE 1.89 × 10³ J; input 9.21 × 10⁴ J; average 3.84 kW; final acceleration about 2.33 m/s² upward.
- GPE gain = mgh = 420 × 9.81 × 18 = 7.42 × 10⁴ J. KE gain = ½mv² = ½ × 420 × 3.0² = 1.89 × 10³ J.
- Input energy supplies both gains plus losses: 74164 + 1890 + 16000 = 9.21 × 10⁴ J. Average power = 92054/24 = 3.84 kW.
- At final speed, motor force follows P = Fv: F = 18000/3.0 = 6000 N upward.
- Net force = 6000 − mg − 900 = 6000 − 4120 − 900 = 980 N upward, so acceleration = 980/420 = 2.33 m/s².
- Instantaneous power depends on force and speed at one instant; average power divides total energy by whole time, including low-speed starting and varying losses.
Wave transmission at a string boundary
Answer: Wavelengths 0.400 m and 0.267 m; period 0.0833 s; angular frequency 75.4 rad/s; maximum particle speed 2.26 m/s; reflected pulse is inverted.
- From v = fλ, wavelengths are 4.8/12 = 0.400 m and 3.2/12 = 0.267 m.
- Period T = 1/12 = 0.0833 s and angular frequency ω = 2πf = 75.4 rad/s.
- A string point has maximum transverse speed ωA = 75.4 × 0.030 = 2.26 m/s, distinct from propagation speed.
- The source fixes frequency and boundary displacement remains continuous. Lower transmitted speed therefore gives shorter wavelength; impedance mismatch requires reflected and transmitted components.
- Reflection from the higher-impedance, more fixed-like boundary is inverted, a phase change of π. Energy is shared, not duplicated, between reflected and transmitted waves.
Thermistor divider alarm
Answer: Outputs 2.28 V and 3.81 V; output rises with temperature; 3.5 V corresponds to about 43.4 °C.
- Vout = VsRfixed/(Rfixed + Rthermistor). At 20 °C, Vout = 6.0 × 3.3/(3.3 + 5.4) = 2.28 V.
- At 50 °C, Vout = 6.0 × 3.3/(3.3 + 1.9) = 3.81 V. NTC resistance falls as temperature rises, increasing fixed-resistor voltage share.
- At 3.5 V, solve 3.5/6.0 = 3.3/(3.3 + R), giving R = 2.36 kΩ.
- Interpolate between 2.6 kΩ at 40 °C and 1.9 kΩ at 50 °C: T ≈40 + (2.6 − 2.36)/(2.6 − 1.9) × 10 = 43.4 °C.
- A finite voltmeter lies parallel with 3.3 kΩ and changes the divider. A high-input-resistance comparator with adjustable reference and hysteresis produces a sharper, stable switch.
Quark bookkeeping in two decays
Answer: n → p + e⁻ + electron antineutrino; π⁰ → γ + γ; neutron udd, proton uud; d → u + W⁻, then W⁻ → e⁻ + antineutrino.
- The reactions are n → p + e⁻ + anti-νₑ and π⁰ → γ + γ. Two photons can conserve momentum in the pion rest frame by leaving oppositely.
- Neutron valence composition is udd and proton is uud. Converting one d to u raises the hadron charge by one elementary unit.
- At the weak vertex d → u + W⁻; the virtual W⁻ produces e⁻ + anti-νₑ. The electron is created rather than stored inside the neutron.
- Charge is 0 = +1 − 1; baryon number remains 1; electron and antineutrino lepton numbers +1 and −1 cancel. Energy and momentum also remain conserved.
- Photons are gauge bosons, not three-quark baryons, so baryon number is zero. The neutral pion is a quark–antiquark meson and also has baryon number zero.
Molecular interpretation of a heated gas
Answer: ΔU = 1.50 kJ; N = 4.82 × 10²³; rms-speed factor 1.225; pressure factor 1.50; rigid-volume boundary work is zero.
- ΔU = 3nRΔT/2 = 1.5 × 0.80 × 8.31 × 150 = 1496 J = 1.50 kJ.
- Molecule number is nNₐ = 0.80 × 6.02 × 10²³ = 4.82 × 10²³.
- Mean translational kinetic energy is proportional to T, so rms speed changes by √(450/300) = √1.5 = 1.225.
- At fixed amount and volume, p ∝ T, giving factor 1.50. Faster molecules collide more often and transfer more momentum to vessel walls.
- Rigid volume has ΔV = 0, so boundary work is zero and heating equals ΔU. A constant-pressure piston expands, does work and needs additional heat for the same temperature rise.
Driven-oscillator resonance
Answer: Resonance about 1.0 Hz; ω = 6.28 rad/s; k ≈19.7 N/m; greater damping lowers and broadens the peak.
- Largest sampled amplitude is 4.8 cm at 1.0 Hz, so resonance is near 1.0 Hz; finer frequency steps are needed for precision.
- Angular frequency ω = 2πf = 6.28 rad/s. For light damping, k ≈mω² = 0.50 × 6.28² = 19.7 N/m.
- Displacement is approximately in phase with drive well below resonance, lags about π/2 near resonance and approaches antiphase well above it.
- Greater damping lowers and broadens the peak and can shift it slightly lower, reducing frequency selectivity.
- A finite steady peak actually requires dissipation balanced by input. Resonance supports tuning and sensing but can produce destructive structural motion, managed through damping or frequency separation.
Velocity selector and alternating supply
Answer: Speed 3.0 × 10⁵ m/s; mass 9.60 × 10⁻²⁷ kg; force north; Vrms ≈120 V; alternating field changes selection with entry phase.
- Undeflected condition qE = qvB gives v = E/B = 2.4 × 10⁴/0.080 = 3.0 × 10⁵ m/s.
- In analyser, qvB = mv²/r, so m = qBr/v = 9.60 × 10⁻²⁷ kg.
- For positive charge, v × B with east velocity and downward field points north; negative charge reverses direction.
- Magnetic force stays perpendicular to velocity, so F·v = 0. It changes direction without speed or kinetic-energy change.
- Vrms = 170/√2 = 120 V. An unsynchronised a.c. electric field changes magnitude and direction during ion entry, so balance and deflection depend on phase rather than selecting one steady speed.
Nuclear tracer dose and gamma imaging
Answer: Start 12.0 MBq; end 11.55 MBq; about 1.41 × 10¹⁰ decays; emitted gamma energy about 3.16 × 10⁻⁴ J.
- Twelve hours is two half-lives, so start activity = 48/4 = 12.0 MBq. Over 1/3 h, decay factor 2^(−1/18) = 0.9622, giving 11.55 MBq.
- Mean activity ≈11.775 MBq = 1.1775 × 10⁷ s⁻¹. Over 1200 s, decays ≈1.41 × 10¹⁰.
- Photon energy = 140000 × 1.60 × 10⁻¹⁹ = 2.24 × 10⁻¹⁴ J; total emitted energy ≈3.16 × 10⁻⁴ J.
- Gamma penetrates tissue to an external detector; an hours-long half-life supports preparation and imaging but then falls, reducing prolonged exposure.
- Attenuation lowers counts; collimation maps direction but rejects photons. Optimise administered activity and scan time while staff minimise time, maximise distance and shield.
- Activity is decays per second. Absorbed dose is deposited energy per unit mass and also depends on energy, geometry and absorption; not every emitted photon is absorbed.