Original GioPhysics extended practice
IB Physics · Extended Paper 2
A balanced original practice paper with long-form reasoning, calculations, and supplied-data graphs.
- Time
- 175 minutes
- Questions
- 10
- Marks
- 128
- Answers
- Complete teacher key
Instructions
- Answer every question and show all reasoning.
- State assumptions and use appropriate significant figures and units.
- Support evaluative conclusions with the supplied evidence and the relevant physics.
Every question is independently written by GioPhysics. This is not an awarding-body paper.
Analyse a triangular gas cycle
A monatomic ideal gas follows A(2.0 L, 100 kPa) → B(2.0 L, 250 kPa) → C(5.0 L, 100 kPa) → A, with B→C a straight line. Determine net work by the gas. Calculate temperatures at A, B and C for n = 0.100 mol, then find ΔU and Q on each leg. Use R = 8.31 J mol⁻¹ K⁻¹ and 1 kPa·L = 1 J. State whether the cycle operates as an engine.
Triangular ideal-gas cycle
The three state coordinates and return to A are exactly those stated in the prompt; straight segments define the paths.
Read the exact data as a table
| Volume / L | Pressure / kPa |
|---|---|
| 2 | 100 |
| 2 | 250 |
| 5 | 100 |
| 2 | 100 |
These are the supplied readings, not a worked answer. Lines join the supplied readings; they are not a fitted model.
Brake a rotating flywheel
A flywheel has moment of inertia 0.800 kg m² and rotates at 30.0 rad s⁻¹. A brake pad applies a constant tangential force of 12.0 N at radius 0.250 m until it stops. Determine torque, angular deceleration, stopping time, angle turned and thermal energy produced. If the measured stopping time is 9.0 s instead, infer the average additional resisting torque and discuss whether constant total torque is justified by one timing measurement.
Open the interactive answer workspaceModel a floating research buoy
A sealed cylindrical research buoy has mass 42.0 kg, external cross-sectional area 0.180 m² and total height 0.500 m. It floats upright in seawater of density 1025 kg m⁻³. Calculate its submerged depth and freeboard. A 12.0 kg instrument is added centrally; calculate the new freeboard and decide whether the buoy remains safe if 0.080 m freeboard is required. Discuss stability information missing from this one-dimensional model.
Open the interactive answer workspaceAudit a claimed heat-engine performance
An engine operates between reservoirs at 600 K and 300 K. Its manufacturer claims that each cycle absorbs 1200 J, produces 720 J of work and rejects 480 J. Check first-law consistency, calculate claimed efficiency and compare it with the Carnot limit. Calculate the total entropy change of the two reservoirs per claimed cycle and use it to diagnose the claim. State one way a real engine could legitimately produce 720 J of work per cycle.
Open the interactive answer workspaceResonance in an open air column
An open pipe of physical length 0.840 m resonates at consecutive frequencies 200 Hz, 400 Hz and 600 Hz. Explain the mode pattern, infer wave speed if end correction is negligible, and calculate the end correction per end if an independent temperature measurement gives sound speed 343 m s⁻¹. Predict the first three resonances if one end is then closed but the same effective length applies. Discuss a limitation of that assumption.
Open the interactive answer workspaceInfer source speed from two Doppler readings
A stationary observer hears 680 Hz as a siren approaches and 612 Hz after it passes. Take sound speed as 340 m s⁻¹ and model the source speed as constant, with no wind. Derive expressions for the two heard frequencies, determine source speed and emitted frequency, and explain how a steady wind from source toward observer would affect a naive calculation using ground-frame speed 340 m s⁻¹.
Open the interactive answer workspaceDiagnose a non-ohmic filament lamp
A filament lamp gives measured (current in A, potential difference in V) pairs (0, 0), (0.10, 0.60), (0.20, 1.40), (0.30, 2.70), and (0.40, 4.80). Determine the static resistance at 0.40 A and estimate dynamic resistance between 0.30 and 0.40 A. Explain the graph shape microscopically, calculate power at 0.40 A, and design a safe circuit and procedure to collect the data without overheating the lamp between readings.
Filament-lamp current–voltage observations
The five supplied measurements are shown without an extrapolation; static and local dynamic resistance must be distinguished.
Read the exact data as a table
| Current / A | Potential difference / V |
|---|---|
| 0 | 0 |
| 0.1 | 0.6 |
| 0.2 | 1.4 |
| 0.3 | 2.7 |
| 0.4 | 4.8 |
These are the supplied readings, not a worked answer. No fitted trend is drawn.
Balance a current-carrying wire
A 0.300 m horizontal wire segment of mass 18.0 g lies perpendicular to a uniform horizontal magnetic field of 0.250 T. The wire is supported so that magnetic force may act vertically. Determine the current magnitude and direction needed for the magnetic force to balance its weight. If the current is 20% larger, find the initial vertical acceleration. Discuss why the complete circuit geometry matters and how a balance could measure B.
Open the interactive answer workspaceMass defect and stellar fusion energy
In the net proton–proton chain, four hydrogen atoms ultimately form one helium-4 atom plus other products. Use atomic masses m_H = 1.007825 u and m_He = 4.002603 u; electron masses cancel in this atomic-mass comparison. Calculate mass defect and energy released per reaction in MeV and joules. Estimate reactions per second needed for luminosity 3.83 × 10²⁶ W and mass converted to energy per second. Use 1 u = 931.5 MeV c⁻² and 1 eV = 1.602 × 10⁻¹⁹ J.
Open the interactive answer workspaceEnergy levels in an infinite quantum well
An electron is confined to a one-dimensional infinite well of width L = 0.500 nm. Derive the allowed energies from standing-wave boundary conditions, then calculate E₁, E₂ and the photon wavelength for a 2→1 transition. Write the normalized ground-state wavefunction, state the number of nodes for n = 3, and explain why the electron does not have a single classical trajectory between walls. Use h = 6.626 × 10⁻³⁴ J s and m_e = 9.11 × 10⁻³¹ kg.
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.
Analyse a triangular gas cycle
Answer: Net work = +225 J. Temperatures: 241 K, 602 K and 602 K. ΔU_AB = +450 J, ΔU_BC = 0, ΔU_CA = −450 J. Heats: +450 J, +525 J and −750 J. It is a heat engine.
- The clockwise triangular area is W_net = ½(250−100)(5−2) = 225 kPa L = 225 J, positive for work done by the gas.
- Using PV = nRT gives T_A = 200/[0.100(8.31)] = 241 K, T_B = 500/0.831 = 602 K and T_C = 500/0.831 = 602 K.
- For a monatomic ideal gas ΔU = (3/2)nRΔT. Thus ΔU_AB = +450 J, ΔU_BC = 0 and ΔU_CA = −450 J, summing to zero over the cycle.
- Works are W_AB = 0, W_BC = average pressure times ΔV = [(250+100)/2](3) = 525 J, and W_CA = 100(2−5) = −300 J.
- From Q = ΔU + W_by, heats are +450 J, +525 J and −750 J. Net Q = +225 J equals net work, and positive cycle work means the device acts as a heat engine.
Brake a rotating flywheel
Answer: Brake torque = 3.00 N m; α = −3.75 rad s⁻²; stop time = 8.00 s; angle = 120 rad; thermal energy = 360 J. A 9.0 s stop implies total torque 2.67 N m, so another effect averages +0.333 N m in the direction of rotation.
- The brake torque magnitude is τ = rF = (0.250)(12.0) = 3.00 N m, opposite the angular velocity.
- Using τ = Iα gives α = −3.00/0.800 = −3.75 rad s⁻². Constant α gives stopping time t = (0−30.0)/(−3.75) = 8.00 s.
- Angular displacement is the average angular speed times time: θ = (30.0+0)t/2 = 120 rad.
- Initial rotational energy ½Iω² = ½(0.800)(30.0)² = 360 J becomes thermal energy. It also equals τθ = (3.00)(120) J.
- A 9.0 s stop corresponds to α_avg = −3.33 rad s⁻² and τ_total,avg = −2.67 N m, 0.333 N m less opposing than the nominal pad torque. One elapsed time determines only an average; angular-speed data throughout the stop are needed to test torque constancy.
Model a floating research buoy
Answer: Initial submerged depth = 0.228 m and freeboard = 0.272 m. With the instrument, submerged depth = 0.293 m and freeboard = 0.207 m, so it meets the 0.080 m requirement.
- Floating equilibrium requires buoyant force equal weight: ρgAh = mg. Gravity cancels, so h = m/(ρA).
- Initially h = 42.0/[1025(0.180)] = 0.228 m. Freeboard is 0.500−0.228 = 0.272 m.
- With the instrument, total mass is 54.0 kg. Then h = 54.0/[1025(0.180)] = 0.293 m and freeboard is 0.207 m.
- Because 0.207 m exceeds the required 0.080 m, the buoy passes the stated vertical-displacement criterion with 0.127 m margin.
- This model says nothing about tipping. Safety also depends on centre of mass, centre of buoyancy, metacentric height, wave loading and whether the instrument shifts. A tall central mass can reduce rotational stability even while freeboard remains adequate.
Audit a claimed heat-engine performance
Answer: Energy is conserved, but claimed efficiency 60% exceeds the 50% Carnot limit. Reservoir entropy change is −1200/600 + 480/300 = −0.400 J K⁻¹, violating the second law. More heat input or a hotter source/lower sink is required.
- First-law accounting gives Q_H = W + Q_C: 1200 = 720 + 480 J, so the numerical energy balance is internally consistent.
- Claimed efficiency is W/Q_H = 720/1200 = 0.600. The maximum reversible efficiency is 1−T_C/T_H = 1−300/600 = 0.500.
- The hot reservoir entropy change is −Q_H/T_H = −1200/600 = −2.00 J K⁻¹. The cold reservoir gains +480/300 = +1.60 J K⁻¹.
- Total reservoir entropy change is −0.400 J K⁻¹. A cyclic engine returns to its own state, so it contributes no net entropy; the negative total violates the second law and confirms the claim is impossible.
- At the same temperatures, even a reversible engine needs Q_H ≥ W/η_C = 720/0.5 = 1440 J. Alternatively, a hotter source or colder sink could raise the Carnot limit, subject to real irreversibility lowering actual efficiency.
Resonance in an open air column
Answer: The open-pipe modes are n = 1, 2, 3 and imply 336 m s⁻¹ without correction. Effective length for 343 m s⁻¹ is 0.8575 m, so correction ≈ 8.75 mm per end. Closed-pipe resonances ≈ 100, 300 and 500 Hz under the same-length assumption.
- An open pipe has displacement antinodes at both ends and frequencies f_n = nv/(2L_eff). Integer multiples 200, 400 and 600 Hz identify consecutive n = 1, 2 and 3 modes.
- Ignoring end correction, v = 2Lf₁ = 2(0.840)(200) = 336 m s⁻¹.
- Using independent v = 343 m s⁻¹ gives L_eff = v/(2f₁) = 343/400 = 0.8575 m. The excess 0.0175 m shared by two open ends is 8.75 mm per end.
- With one end closed, only odd harmonics occur: f = v/(4L_eff), 3v/(4L_eff), 5v/(4L_eff). The assumed same effective length gives 100, 300 and 500 Hz.
- Closing an end changes the boundary geometry and its end correction; using the open-pipe effective length is therefore an approximation. Directly measure the new resonance series and check its odd-integer spacing.
Infer source speed from two Doppler readings
Answer: For a moving source, f_app = f c/(c−u) and f_rec = f c/(c+u). The data give u = 17.9 m s⁻¹ and f = 644 Hz.
- A moving source compresses wavelength ahead to λ_app = (c−u)/f and stretches it behind to λ_rec = (c+u)/f. A stationary observer measures c divided by these wavelengths.
- Thus f_app/f_rec = (c+u)/(c−u). Solving gives u = c(f_app−f_rec)/(f_app+f_rec).
- Substitution gives u = 340(680−612)/(680+612) = 17.9 m s⁻¹.
- Using either branch, f = f_app(c−u)/c = 680(340−17.9)/340 = 644 Hz; the receding branch provides the same value as a check.
- Wind changes wave speed relative to the ground and can advect both source-generated wavefronts and observer. The simple still-air ground formula may bias inferred u; a consistent derivation must use velocities relative to the medium rather than replacing c in only one place.
Diagnose a non-ohmic filament lamp
Answer: Static resistance at 0.40 A is 12.0 Ω; interval dynamic resistance is 21.0 Ω; power is 1.92 W. Rising temperature increases lattice vibration and resistivity, so the slope grows.
- Static resistance at an operating point is V/I = 4.80/0.400 = 12.0 Ω.
- Dynamic resistance is the local gradient dV/dI. Over the final interval it is approximated by (4.80−2.70)/(0.40−0.30) = 21.0 Ω.
- As current rises, electrical heating raises filament temperature. Stronger lattice vibrations increase electron scattering and resistivity, so the V–I graph becomes progressively steeper.
- Power at 0.40 A is VI = (4.80)(0.400) = 1.92 W, consistent with I²R using the static resistance at that operating point.
- Use a low-voltage variable supply, series current-limiting resistor and switch, ammeter in series and voltmeter across the lamp. Increase current briefly, record simultaneous values, open the switch between readings, repeat in both directions, and do not exceed the rated current.
Balance a current-carrying wire
Answer: Required current = 2.35 A, directed so I × B is upward. At 20% larger current the initial acceleration is 1.96 m s⁻² upward.
- For a straight segment perpendicular to B, magnetic force magnitude is F_B = BIL. Balance requires BIL = mg.
- I = mg/(BL) = (0.0180)(9.81)/[(0.250)(0.300)] = 2.35 A. Use the vector rule I × B to choose whichever wire direction makes force upward.
- At 1.20I_balance, magnetic force is 1.20mg. Net upward force is 0.20mg, so initial acceleration is 0.20g = 1.96 m s⁻².
- Other circuit segments can experience magnetic forces too. Their directions, lengths within the field and mechanical supports determine whether they contribute to the measured vertical force or cancel.
- Place the field segment on a balance, reverse current to remove constant offsets, and use half the difference in readings. A graph of force against IL has gradient B when the segment is perpendicular; control temperature because wire resistance and current can drift.
Mass defect and stellar fusion energy
Answer: Mass defect = 0.028697 u; energy ≈ 26.73 MeV = 4.282 × 10⁻¹² J. About 8.94 × 10³⁷ reactions s⁻¹ and 4.26 × 10⁹ kg s⁻¹ are required.
- Initial atomic mass is 4(1.007825) = 4.031300 u. Subtracting helium mass gives Δm = 4.031300−4.002603 = 0.028697 u.
- Energy is Δmc² = (0.028697)(931.5) = 26.73 MeV.
- Convert to joules: 26.73 × 10⁶ eV multiplied by 1.602 × 10⁻¹⁹ J eV⁻¹ gives 4.282 × 10⁻¹² J per net reaction.
- Reaction rate for luminosity L is L/E = (3.83 × 10²⁶)/(4.282 × 10⁻¹²) = 8.94 × 10³⁷ s⁻¹.
- Mass converted to radiant energy is L/c² = 3.83 × 10²⁶/(2.998 × 10⁸)² = 4.26 × 10⁹ kg s⁻¹. This is mass defect, not the much larger mass of hydrogen processed; neutrino losses would refine the luminosity link.
Energy levels in an infinite quantum well
Answer: E_n = n²h²/(8mL²). E₁ = 1.50 eV, E₂ = 6.02 eV; the 2→1 photon has energy 4.51 eV and wavelength about 275 nm. ψ₁ = √(2/L)sin(πx/L); n = 3 has two internal nodes.
- Zero wavefunction at x = 0 and L requires an integer number of half wavelengths: L = nλ/2, so momentum magnitude p = h/λ = nh/(2L).
- Non-relativistic kinetic energy is p²/(2m), yielding E_n = n²h²/(8mL²). For L = 0.500 nm, E₁ = 2.41 × 10⁻¹⁹ J = 1.50 eV and E₂ = 4E₁ = 6.02 eV.
- Transition energy is E₂−E₁ = 3E₁ = 4.51 eV. Photon wavelength is hc/ΔE ≈ 275 nm.
- Normalization gives ψ₁(x) = √(2/L)sin(πx/L) inside and zero outside. State n has n−1 internal nodes, so n = 3 has two.
- The wavefunction provides a probability amplitude; |ψ|² predicts position distributions while energy states are stationary. A classical path with definite position and momentum at every instant is incompatible with the spatially extended state and uncertainty relation.