Original GioPhysics extended practice
Cambridge AS & A Level Physics · Extended Paper 2
A balanced original practice paper with long-form reasoning, calculations, and supplied-data graphs.
- Time
- 195 minutes
- Questions
- 10
- Marks
- 141
- 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.
Projectile trajectory reconstructed from data
A ball has (time in s, vertical position in m) data (0, 0), (0.4, 5.216), (0.8, 8.864), (1.2, 10.944), (1.6, 11.456), (2, 10.4), (2.4, 7.776) and constant horizontal velocity 9.0 m/s. (a) Use y = uᵧt − ½gt² to estimate uᵧ and g. (b) Find model maximum time and height. (c) Find time and range on return to launch level. (d) State accelerations. (e) Explain two departures from an ideal parabola.
Projectile vertical position
Vertical position sampled during flight.
Read the exact data as a table
| Time / s | Vertical position / m |
|---|---|
| 0 | 0 |
| 0.4 | 5.216 |
| 0.8 | 8.864 |
| 1.2 | 10.944 |
| 1.6 | 11.456 |
| 2 | 10.4 |
| 2.4 | 7.776 |
These are the supplied readings, not a worked answer. No fitted trend is drawn.
Submerged gate in equilibrium
A uniform horizontal gate of length 2.4 m and mass 90 kg is hinged at one end. Its centre is 1.8 m below liquid of density 850 kg/m³. Liquid gives 3.6 kN downward at 1.5 m from hinge; a vertical cable acts at far end. Take g = 9.81 m/s². (a) Find gauge pressure at centre. (b) Find cable tension. (c) Find hinge vertical force. (d) Explain centre of pressure and complete equilibrium.
Open the interactive answer workspaceStress–strain energy and elastic behaviour
A wire gives (strain, stress in MPa) readings (0, 0), (0.001, 80), (0.002, 160), (0.003, 240), (0.004, 300), (0.006, 330). Original length is 2.0 m, area 1.5 × 10⁻⁶ m² and volume 3.0 × 10⁻⁶ m³. (a) Find Young modulus. (b) Find force and extension at strain 0.003. (c) Find energy density and energy. (d) Identify departure from proportionality. (e) Describe a safe experiment and distinguish proportional from elastic limit.
Metal stress–strain data
Stress measured during tensile loading.
Read the exact data as a table
| Strain | Stress / MPa |
|---|---|
| 0 | 0 |
| 0.001 | 80 |
| 0.002 | 160 |
| 0.003 | 240 |
| 0.004 | 300 |
| 0.006 | 330 |
These are the supplied readings, not a worked answer. Lines join the supplied readings; they are not a fitted model.
Double-slit gradient and measurement design
For 640 nm light and slit separation 0.200 mm, measured (screen distance in m, fringe spacing in mm) data are (0.4, 1.28), (0.5, 1.6), (0.6, 1.92), (0.7, 2.24), (0.8, 2.56). (a) Find graph gradient and slit separation. (b) Compare with stated value. (c) Predict spacing at 1.20 m. (d) Explain covering one slit and reducing wavelength. (e) Explain multi-fringe measurement and a control.
Fringe spacing against screen distance
Bright-fringe spacing at several screen distances.
Read the exact data as a table
| Screen distance / m | Fringe spacing / mm |
|---|---|
| 0.4 | 1.28 |
| 0.5 | 1.6 |
| 0.6 | 1.92 |
| 0.7 | 2.24 |
| 0.8 | 2.56 |
These are the supplied readings, not a worked answer. No fitted trend is drawn.
Cell e.m.f., internal resistance and power
A cell gives (current in A, terminal p.d. in V) readings (0, 1.52), (0.5, 1.43), (1, 1.34), (1.5, 1.25), (2, 1.16). (a) Find e.m.f. and internal resistance. (b) At 1.50 A find external resistance, load power, internal heating and chemical power. (c) Verify conservation. (d) Find load for maximum power and its efficiency. (e) Explain why that operation is inefficient.
Cell terminal characteristic
Terminal potential difference as current changes.
Read the exact data as a table
| Current / A | Terminal p.d. / V |
|---|---|
| 0 | 1.52 |
| 0.5 | 1.43 |
| 1 | 1.34 |
| 1.5 | 1.25 |
| 2 | 1.16 |
These are the supplied readings, not a worked answer. Lines join the supplied readings; they are not a fitted model.
Circular orbit, energy and transfer
A 750 kg spacecraft circles a 6.0 × 10²⁴ kg planet at radius 8.0 × 10⁶ m. Use G = 6.67 × 10⁻¹¹ N m²/kg². (a) Find field strength, orbital speed and period. (b) Find angular speed and verify centripetal acceleration. (c) Find energy needed to move to a circular orbit of radius 1.20 × 10⁷ m using E = −GMm/(2r). (d) Explain why one forward burn first makes an ellipse.
Open the interactive answer workspaceIdeal-gas cycle and first law
One mole of monatomic ideal gas follows clockwise vertices (volume in 10⁻³ m³, pressure in kPa) (1, 100), (1, 300), (3, 300), (3, 100), (1, 100). (a) Find work on every leg and net work. (b) Find temperatures A, B and C using R = 8.31 J/(mol K). (c) Find ΔU and Q on A→B and B→C. (d) Verify net heat equals net work. (e) Explain orientation and heat rejection.
Rectangular gas cycle
Constant-volume and constant-pressure processes.
Read the exact data as a table
| Volume / 10⁻³ m³ | Pressure / kPa |
|---|---|
| 1 | 100 |
| 1 | 300 |
| 3 | 300 |
| 3 | 100 |
| 1 | 100 |
These are the supplied readings, not a worked answer. Lines join the supplied readings; they are not a fitted model.
Capacitor discharge and stored field energy
A capacitor discharges through 220 kΩ with (time in s, voltage in V) readings (0, 8), (10, 5.84), (20, 4.26), (30, 3.11), (40, 2.27). (a) Estimate time constant and capacitance. (b) Find initial charge and energy. (c) Find initial current and resistor power. (d) Find energy fraction at 40 s. (e) Explain voltmeter loading and a semilog test.
Capacitor discharge
Voltage during resistor discharge.
Read the exact data as a table
| Time / s | Voltage / V |
|---|---|
| 0 | 8 |
| 10 | 5.84 |
| 20 | 4.26 |
| 30 | 3.11 |
| 40 | 2.27 |
These are the supplied readings, not a worked answer. Lines join the supplied readings; they are not a fitted model.
Photoelectric stopping-potential evidence
A metal gives (frequency in 10¹⁴ Hz, stopping potential in V) readings (5.5, 0.18), (6, 0.39), (6.5, 0.59), (7, 0.8), (7.5, 1.01). Use e = 1.60 × 10⁻¹⁹ C and mₑ = 9.11 × 10⁻³¹ kg. (a) Find gradient and h. (b) Estimate threshold frequency and work function in eV. (c) At 7.0 × 10¹⁴ Hz find maximum speed. (d) Explain intensity below/above threshold. (e) Explain photon evidence.
Photoelectric stopping potential
Stopping potential at five frequencies.
Read the exact data as a table
| Frequency / 10¹⁴ Hz | Stopping potential / V |
|---|---|
| 5.5 | 0.18 |
| 6 | 0.39 |
| 6.5 | 0.59 |
| 7 | 0.8 |
| 7.5 | 1.01 |
These are the supplied readings, not a worked answer. No fitted trend is drawn.
Hubble plot and cosmological evidence
Galaxies give (distance in Mpc, recession speed in km/s) (20, 1380), (40, 2860), (60, 4210), (80, 5650), (100, 7040). Use 1 Mpc = 3.086 × 10¹⁹ km and 1 year = 3.156 × 10⁷ s. (a) Find H₀. (b) Convert to s⁻¹ and calculate 1/H₀ in years. (c) Predict speed at 150 Mpc. (d) Explain population fitting. (e) Explain standard candles and redshift. (f) Explain two age limitations and two other hot-Big-Bang observations.
Galaxy recession relation
Recession speeds with independently estimated distances.
Read the exact data as a table
| Distance / Mpc | Recession speed / km s⁻¹ |
|---|---|
| 20 | 1380 |
| 40 | 2860 |
| 60 | 4210 |
| 80 | 5650 |
| 100 | 7040 |
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.
Projectile trajectory reconstructed from data
Answer: uᵧ = 15.0 m/s; g = 9.80 m/s²; peak 1.53 s and 11.5 m; flight 3.06 s; range 27.6 m; accelerations 0 horizontally and −9.8 m/s² vertically.
- For example at 0.8 s: 8.864 = 0.8uᵧ − 0.32g, and at 2.0 s: 10.4 = 2uᵧ − 2g. Solving gives uᵧ = 15.0 m/s and g = 9.80 m/s².
- At the peak vᵧ = uᵧ − gt = 0, so t = 15.0/9.80 = 1.53 s. Height uᵧ²/(2g) = 225/19.6 = 11.5 m, consistent with nearby samples.
- The non-zero solution of y = 0 is t = 2uᵧ/g = 3.06 s. Horizontal range = 9.0 × 3.06 = 27.6 m.
- Neglecting resistance, horizontal acceleration is zero and vertical acceleration remains −9.8 m/s², including at maximum height where only vertical velocity is momentarily zero.
- Air resistance reduces horizontal speed and changes symmetry. Camera scale, parallax, timing offset or tracking the ball edge instead of its centre can also move points away from the model.
Submerged gate in equilibrium
Answer: Gauge pressure 15.0 kPa; cable tension 2.69 kN; hinge force 1.79 kN upward; both resultant force and moment must vanish.
- Gauge pressure p = ρgh = 850 × 9.81 × 1.8 = 1.50 × 10⁴ Pa = 15.0 kPa. Atmospheric pressure cancels if it acts on both sides.
- Gate weight is 90 × 9.81 = 883 N at 1.2 m. Taking moments about hinge: T(2.4) = 883(1.2) + 3600(1.5), so T = 2692 N = 2.69 kN.
- Vertical balance gives hinge force + 2692 − 883 − 3600 = 0, hence hinge force = 1791 N = 1.79 kN upward.
- Pressure increases with depth, so deeper areas contribute more strongly and the liquid resultant need not act at the geometric centre; its line is the centre of pressure.
- Complete static equilibrium requires vector sum of forces zero and sum of moments about any point zero. Moment balance alone does not prevent translation.
Stress–strain energy and elastic behaviour
Answer: E = 8.0 × 10¹⁰ Pa; force 360 N; extension 6.0 mm; energy density 3.6 × 10⁵ J/m³; energy 1.08 J.
- Initial Young modulus E = stress/strain = 240 × 10⁶/0.003 = 8.0 × 10¹⁰ Pa; earlier points give the same ratio.
- Force F = σA = 240 × 10⁶ × 1.5 × 10⁻⁶ = 360 N. Extension = strain × length = 0.003 × 2.0 = 0.0060 m.
- Linear-region energy density is triangular graph area ½σε = ½ × 240 × 10⁶ × 0.003 = 3.6 × 10⁵ J/m³. Total energy is 1.08 J.
- At strain 0.006, linear extrapolation predicts 480 MPa but observed stress is 330 MPa, so direct proportionality has ended.
- Clamp behind a shield, add small loads and measure extension with a long gauge length. The proportional limit is loss of linearity; the elastic limit requires unloading to test return to original length.
Double-slit gradient and measurement design
Answer: Gradient 3.20 mm/m gives slit separation 0.200 mm; predicted spacing at 1.20 m is 3.84 mm.
- From w = λD/a, gradient of w against D is λ/a. Endpoint gradient = (2.56 − 1.28)/(0.80 − 0.40) = 3.20 mm/m = 3.20 × 10⁻³.
- Thus a = λ/gradient = 640 × 10⁻⁹/(3.20 × 10⁻³) = 2.00 × 10⁻⁴ m = 0.200 mm, agreeing with the stated spacing.
- At D = 1.20 m, w = gradient × D = 3.20 × 1.20 = 3.84 mm.
- Covering one slit removes two-source interference, leaving a single-slit diffraction pattern. Reducing wavelength reduces fringe spacing directly if geometry is fixed.
- Measure across n fringe gaps then divide by n, reducing fractional ruler uncertainty. Keep wavelength, slit separation and screen alignment fixed while changing D.
Cell e.m.f., internal resistance and power
Answer: E.m.f. 1.52 V; r = 0.18 Ω; at 1.50 A: R = 0.833 Ω, load 1.875 W, loss 0.405 W, chemical 2.28 W; maximum power at R = r with 50% efficiency.
- From V = ε − Ir, intercept gives ε = 1.52 V and gradient (1.16 − 1.52)/2.0 = −0.18 V/A gives r = 0.18 Ω.
- At 1.50 A and 1.25 V, external R = 1.25/1.50 = 0.833 Ω and load power VI = 1.875 W.
- Internal power I²r = 1.50² × 0.18 = 0.405 W. Chemical power εI = 1.52 × 1.50 = 2.28 W; 1.875 + 0.405 = 2.280 W.
- Maximum load power occurs at R = r = 0.18 Ω. Then terminal p.d. is ε/2 and equal powers go to load and internal heating, so efficiency is 50%.
- Maximum power is not maximum efficiency: half the chemical power heats the cell, wasting energy and potentially reducing life. Efficient supply normally uses R much larger than r.
Circular orbit, energy and transfer
Answer: g = 6.25 m/s²; v = 7.07 km/s; T = 7.11 × 10³ s; ω = 8.84 × 10⁻⁴ rad/s; energy increase 6.25 × 10⁹ J.
- Field strength GM/r² = 6.67 × 10⁻¹¹ × 6.0 × 10²⁴/(8.0 × 10⁶)² = 6.25 m/s².
- Circular balance gives v = √(GM/r) = 7.07 × 10³ m/s. Period 2πr/v = 7.11 × 10³ s.
- Angular speed is 2π/T = 8.84 × 10⁻⁴ rad/s. Check ω²r = 6.25 m/s², equal to gravitational acceleration in the circular orbit.
- Energy increase is GMm/2(1/r₁ − 1/r₂) = 6.25 × 10⁹ J. The less negative final energy represents added mechanical energy.
- A short forward burn changes velocity at one location, not position everywhere, so it creates an ellipse with the burn point at lower apsis. A second burn at the raised side circularises; the final larger circular speed is lower despite added energy.
Ideal-gas cycle and first law
Answer: Work 0, +600, 0, −200 J; net +400 J. Tₐ = 12.0 K, Tᵦ = 36.1 K, T𝒄 = 108 K. A→B: ΔU = Q = 300 J; B→C: ΔU = 900 J, Q = 1500 J.
- Constant-volume work is zero. B→C work = 300000(0.003 − 0.001) = +600 J; D→A work = 100000(0.001 − 0.003) = −200 J. Net is +400 J.
- Using pV = RT: Tₐ = 100000(0.001)/8.31 = 12.0 K; Tᵦ = 36.1 K; T𝒄 = 300000(0.003)/8.31 = 108 K.
- For monatomic gas ΔU = 3RΔT/2. A→B gives about 300 J and W = 0, so Q = 300 J.
- B→C gives ΔU about 900 J. With W = +600 J by gas, Q = ΔU + W = 1500 J. Cooling legs reject a total about 1400 J.
- Over a cycle ΔU = 0, so net Q = net W = +400 J. Clockwise orientation means positive enclosed work; heat rejection occurs on the falling-temperature C→D and D→A legs.
Capacitor discharge and stored field energy
Answer: τ ≈31.8 s; C ≈145 μF; Q₀ ≈1.16 mC; U₀ ≈4.64 mJ; I₀ = 36.4 μA; P₀ = 0.291 mW; energy fraction ≈0.0805.
- From V = V₀e^(−t/τ), use 5.84/8.0 at 10 s: τ = −10/ln(5.84/8.0) = 31.8 s.
- C = τ/R = 31.8/220000 = 1.45 × 10⁻⁴ F = 145 μF. Initial charge CV = 1.16 × 10⁻³ C and energy ½CV² = 4.64 × 10⁻³ J.
- Initial current V/R = 8.0/220000 = 36.4 μA. Initial power V²/R = 2.91 × 10⁻⁴ W.
- At 40 s, voltage fraction = 2.27/8.0 = 0.28375. Since U ∝V², energy fraction = 0.28375² = 0.0805.
- A finite-resistance voltmeter adds a parallel path and reduces τ. A straight ln V against t plot with gradient −1/τ tests single-exponential decay.
Photoelectric stopping-potential evidence
Answer: Gradient 4.15 × 10⁻¹⁵ V s; h = 6.64 × 10⁻³⁴ J s; threshold 5.07 × 10¹⁴ Hz; work function 2.10 eV; maximum speed 5.30 × 10⁵ m/s.
- Endpoint gradient = (1.01 − 0.18)/[(7.5 − 5.5) × 10¹⁴] = 4.15 × 10⁻¹⁵ V s.
- From eVs = hf − φ, gradient = h/e. Thus h = 1.60 × 10⁻¹⁹ × 4.15 × 10⁻¹⁵ = 6.64 × 10⁻³⁴ J s.
- Threshold f₀ = 5.5 × 10¹⁴ − 0.18/(4.15 × 10⁻¹⁵) = 5.07 × 10¹⁴ Hz. Work function is gradient × f₀ = 2.10 eV.
- At 7.0 × 10¹⁴ Hz, Vs = 0.80 V and Kmax = eVs. Then v = √(2eVs/m) = 5.30 × 10⁵ m/s.
- Below threshold, more intensity gives no ideal emission; above it, intensity raises electron rate but not maximum energy. Immediate emission, threshold and linear energy–frequency relation support photons hf.
Hubble plot and cosmological evidence
Answer: H₀ ≈70.8 km s⁻¹ Mpc⁻¹ = 2.29 × 10⁻¹⁸ s⁻¹; Hubble time 1.38 × 10¹⁰ years; predicted 150 Mpc speed 1.06 × 10⁴ km/s.
- Endpoint gradient H₀ = (7040 − 1380)/(100 − 20) = 70.75 km s⁻¹ Mpc⁻¹; full regression would use all points.
- Convert H₀ = 70.75/(3.086 × 10¹⁹) = 2.29 × 10⁻¹⁸ s⁻¹. Reciprocal is 4.36 × 10¹⁷ s = 1.38 × 10¹⁰ years.
- At 150 Mpc, v = H₀d = 70.75 × 150 = 1.06 × 10⁴ km/s under the fitted linear relation.
- Peculiar velocities and distance uncertainty make one galaxy unreliable. Standard candles infer distance from known luminosity and measured flux; shifted spectral lines provide recession evidence.
- Expansion rate changes through history and distance calibration affects H₀, so 1/H₀ is not exact age. Cosmic microwave background and primordial light-element abundances independently support a hot early Universe.