Original GioPhysics extended practice
AP Physics · Extended Paper 2
A balanced original practice paper with long-form reasoning, calculations, and supplied-data graphs.
- Time
- 185 minutes
- Questions
- 10
- Marks
- 136
- Answers
- Complete teacher key
Instructions
- Answer every free-response question in the space provided.
- Show the physics principles and mathematical steps that support each claim.
- Label diagrams and explain how experimental evidence would test a model.
Every question is independently written by GioPhysics. This is not an awarding-body paper.
Energy accounting for a cyclic heat engine
An ideal gas follows the clockwise rectangular cycle A(1.0 L, 100 kPa) → B(1.0 L, 300 kPa) → C(3.0 L, 300 kPa) → D(3.0 L, 100 kPa) → A. During B→C it absorbs 900 J, during D→A it rejects 300 J, during A→B it absorbs 100 J and during C→D it rejects 300 J. Use 1 kPa·L = 1 J. Determine the work on every leg, net work, changes in internal energy on B→C and D→A, total heat over the cycle, and efficiency. Explain the clockwise sign.
Specified pressure–volume cycle
The four vertices and closing point supplied in the prompt are joined in order; area and direction must be interpreted by the student.
Read the exact data as a table
| Volume / L | 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.
Rolling dynamics down an incline
A uniform solid cylinder of mass 2.00 kg rolls without slipping down a 25.0° incline through a vertical drop of 1.20 m. Its moment of inertia is I = ½mR². Determine its translational acceleration, the static-friction magnitude and direction, and its speed after the drop. Explain why static friction does no mechanical work in the ideal rolling model and compare the speed with that of a frictionless sliding block.
Open the interactive answer workspaceDesign and diagnose a Venturi flow meter
Water of density 1000 kg m⁻³ flows steadily through a horizontal pipe whose area narrows from 8.00 × 10⁻⁴ m² to 4.00 × 10⁻⁴ m². The upstream speed is 1.50 m s⁻¹ and pressure is 180 kPa. Model the flow as incompressible and non-viscous. Determine the volume flow rate, narrow-section speed and pressure. Then explain how a real pressure reading affected by viscosity would change the inferred flow rate, and outline a calibration procedure.
Open the interactive answer workspaceExtract an RC time constant from charging data
A 220 μF capacitor charges through an unknown resistor from a 12.0 V ideal supply. Recorded (time in s, capacitor voltage in V) pairs are (0, 0), (1, 4.72), (2, 7.59), (3, 9.32), and (4, 10.38). Determine the time constant and resistance, the charge at 4.0 s, and the current at 2.0 s. Explain a linearized graph that could test the exponential model and how voltmeter input resistance could bias the result.
Capacitor charging observations
Measured capacitor voltage values listed in the prompt; a smooth curve is shown only to connect sequential measurements.
Read the exact data as a table
| Time / s | Capacitor voltage / V |
|---|---|
| 0 | 0 |
| 1 | 4.72 |
| 2 | 7.59 |
| 3 | 9.32 |
| 4 | 10.38 |
These are the supplied readings, not a worked answer. Lines join the supplied readings; they are not a fitted model.
Image formation and an experimental focal length
A 2.00 cm object is 18.0 cm in front of a thin converging lens of focal length 12.0 cm. Determine the image location, magnification, image height and whether the image can be projected. The lens is then used in a classroom experiment with several object distances. Describe a graph that determines focal length without relying on a single trial, and identify two alignment controls.
Open the interactive answer workspaceClassify equilibria in a quartic potential
A 0.500 kg particle moves in one dimension with potential U(x) = ax⁴ − bx², where a = 20.0 J m⁻⁴ and b = 4.00 J m⁻². Determine all equilibrium positions and classify their stability. Find the small-oscillation angular frequency about either stable equilibrium, the barrier energy relative to a minimum, and the turning points for total energy E = −0.100 J. Explain why a particle with E = −0.100 J cannot cross from one well to the other.
Open the interactive answer workspaceEscape from a circular orbit
A spacecraft is in a circular orbit at radius r = 2R_E from Earth's centre. Use GM_E = 3.986 × 10¹⁴ m³ s⁻² and R_E = 6.37 × 10⁶ m. Find its circular speed, orbital period and specific mechanical energy. Determine the local escape speed and the minimum instantaneous tangential speed increase that places it on a zero-energy escape trajectory. Explain why the engine energy per kilogram is not simply ½(Δv)².
Open the interactive answer workspaceField and potential of a charged solid cylinder
A very long nonconducting solid cylinder of radius R = 0.0200 m has uniform volume charge density ρ = 4.00 × 10⁻⁶ C m⁻³. Neglect end effects. Derive E(r) for r < R and r > R, then calculate E at r = 0.0100 m and r = 0.0400 m. Find V(0) − V(R), and sketch the qualitative E–r graph through r = 2R. Use ε₀ = 8.854 × 10⁻¹² F m⁻¹.
Open the interactive answer workspaceCurrent growth in an LR circuit
An ideal 9.00 V battery, 15.0 Ω resistor and 0.300 H inductor are connected in series at t = 0. The inductor initially carries no current. Determine the time constant, initial current gradient, current and inductor voltage at t = 0.0400 s, and magnetic energy then. Explain the inductor voltage immediately after closing and long afterward, and describe one safe way to observe the transient.
Open the interactive answer workspaceDerive the field on the axis of a charged ring
A thin ring of radius R = 0.100 m carries total uniform charge Q = +4.00 nC. A field point lies on its axis x = 0.0500 m from the centre. Starting from Coulomb's law, derive the axial electric field and calculate its magnitude and the potential there, taking V = 0 at infinity. Show the x ≪ R approximation, compare it numerically here, and explain why the transverse field components cancel.
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.
Energy accounting for a cyclic heat engine
Answer: Leg works: 0, +600 J, 0, −200 J; net work = +400 J. ΔU_BC = +300 J; ΔU_DA = −100 J. Net heat = +400 J and efficiency = 40.0% for total input 1000 J.
- Isochoric legs A→B and C→D have ΔV = 0, so W_by = 0. At constant pressure, B→C gives W = 300(3−1) = +600 J.
- For D→A, W = 100(1−3) = −200 J. Net work is therefore +400 J, also the rectangular area (300−100)(3−1) in kPa·L.
- Using ΔU = Q − W_by, B→C has ΔU = 900−600 = +300 J. D→A has Q = −300 J and W = −200 J, giving ΔU = −100 J.
- A complete cycle returns to the initial thermodynamic state, so total ΔU = 0 and Q_net = W_net = +400 J. This also constrains the combined heat on the two vertical legs.
- The stated remaining absorption makes total heat input 900 + 100 = 1000 J. Efficiency is W_net/Q_in = 400/1000 = 40.0%. Clockwise traversal means high-pressure expansion and low-pressure compression, hence positive work by the gas.
Rolling dynamics down an incline
Answer: a = 2.76 m s⁻² down the slope; static friction = 2.76 N up the slope; final speed = 3.96 m s⁻¹. A sliding block would be faster at 4.85 m s⁻¹.
- Translation along the slope gives mg sinθ − f = ma. Rotation about the centre gives fR = Iα, and no slip gives α = a/R.
- With I = ½mR², the torque equation gives f = ½ma. Substitution into translation yields mg sinθ = 3ma/2, so a = (2/3)g sin25° = 2.76 m s⁻².
- The friction is f = ½ma = 2.76 N and points uphill. It supplies the uphill torque needed to increase the cylinder's angular speed while the centre accelerates downhill.
- Energy gives mgh = ½mv² + ½I(v/R)² = ¾mv². Therefore v = √(4gh/3) = 3.96 m s⁻¹.
- The instantaneous contact point is at rest relative to the fixed slope, so ideal static friction transfers no energy at that point. A frictionless slider has v = √(2gh) = 4.85 m s⁻¹ because none of its energy is rotational.
Design and diagnose a Venturi flow meter
Answer: Flow rate = 1.20 × 10⁻³ m³ s⁻¹, narrow speed = 3.00 m s⁻¹, and ideal narrow pressure = 176.6 kPa. Viscous loss enlarges the pressure drop and makes an ideal calibration overestimate flow.
- Continuity for steady incompressible flow requires A₁v₁ = A₂v₂. The volume rate is Q = (8.00 × 10⁻⁴)(1.50) = 1.20 × 10⁻³ m³ s⁻¹.
- The narrow speed is v₂ = Q/A₂ = (1.20 × 10⁻³)/(4.00 × 10⁻⁴) = 3.00 m s⁻¹.
- Equal heights allow P₁ + ½ρv₁² = P₂ + ½ρv₂². Hence P₂ = 180000 + 500(1.50²−3.00²) = 176625 Pa, or 176.6 kPa.
- Viscosity dissipates mechanical energy and produces an extra pressure loss beyond the ideal Bernoulli drop. Interpreting that larger drop with an ideal equation would infer a flow rate that is too high.
- Calibrate by collecting a measured volume for a measured time at several valve settings while recording pressure difference. Plot actual Q against √ΔP, repeat readings, include uncertainties, and use the empirical curve rather than assuming an ideal coefficient.
Extract an RC time constant from charging data
Answer: τ ≈ 2.00 s, R ≈ 9.09 kΩ, q(4 s) = 2.28 mC, and i(2 s) = 0.486 mA. Plot ln(12−V_C) against t; its gradient should be −1/τ.
- For charging, V_C = ε(1−e^(−t/τ)). At t = τ, V_C = 0.632ε = 7.58 V. The 2.0 s reading is 7.59 V, so τ ≈ 2.00 s.
- Since τ = RC, R = 2.00/(220 × 10⁻⁶) = 9.09 × 10³ Ω. The units seconds per farad reduce to ohms.
- At 4.0 s the charge is q = CV_C = (220 × 10⁻⁶)(10.38) = 2.28 × 10⁻³ C.
- Current is i = (ε/R)e^(−t/τ). At 2.0 s this is (12.0/9090)e^(−1) = 4.86 × 10⁻⁴ A.
- Rearrange to ln(ε−V_C) = ln ε − t/τ; a straight line tests the exponential model. A finite voltmeter resistance in parallel with the capacitor provides leakage, lowering the final voltage and distorting the inferred time constant unless included in the effective circuit.
Image formation and an experimental focal length
Answer: The image is 36.0 cm beyond the lens, magnification −2.00, height −4.00 cm; it is real, inverted and projectable. Plot 1/d_i against 1/d_o; intercept = 1/f.
- Use the thin-lens equation 1/f = 1/d_o + 1/d_i with the real-is-positive convention. Then 1/d_i = 1/12.0 − 1/18.0 = 1/36.0 cm⁻¹.
- Thus d_i = +36.0 cm. The positive image distance places a real image on the far side of the lens, so a screen can intercept it.
- Magnification m = −d_i/d_o = −36.0/18.0 = −2.00. The image height is mh_o = (−2.00)(2.00 cm) = −4.00 cm; the sign denotes inversion.
- Rearrange to 1/d_i = −1/d_o + 1/f. Plot y = 1/d_i against x = 1/d_o; an ideal gradient is −1 and the y-intercept is 1/f.
- Keep object, lens centre and screen centre on one horizontal optical axis, and keep each plane perpendicular to it. Repeat focus judgments in both directions to reduce subjective focusing bias.
Classify equilibria in a quartic potential
Answer: Equilibria: x = 0 unstable and x = ±0.316 m stable. Small-oscillation ω = 5.66 rad s⁻¹; barrier above either minimum = 0.200 J. For E = −0.100 J, turning points in x² are 0.0293 and 0.1707 m², so the two wells remain disconnected.
- Force is F = −dU/dx = −(4ax³−2bx). Equilibrium requires 2x(2ax²−b) = 0, giving x = 0 and x = ±√(b/2a) = ±0.316 m.
- Stability follows from U'' = 12ax²−2b. At x = 0, U'' = −8.00 N m⁻¹, a local maximum; at either nonzero point U'' = 4b = 16.0 N m⁻¹, a minimum.
- Near a minimum, U is approximately quadratic with effective spring constant U''. Hence ω = √(U''/m) = √(16.0/0.500) = 5.66 rad s⁻¹.
- The minimum potential is −b²/(4a) = −0.200 J while U(0) = 0, so the central barrier is 0.200 J above either minimum.
- Set 20x⁴−4x² = −0.100 and let y = x²: 20y²−4y+0.100 = 0 gives y = 0.0293 or 0.1707 m². Since E < U(0), the central region is classically forbidden and trajectories stay in one well.
Escape from a circular orbit
Answer: Circular speed = 5.59 km s⁻¹, period = 1.43 × 10⁴ s, specific energy = −1.56 × 10⁷ J kg⁻¹, escape speed = 7.91 km s⁻¹, and minimum Δv = 2.32 km s⁻¹.
- Circular gravity supplies centripetal acceleration, so v_c = √(GM/r) = √[3.986 × 10¹⁴/(1.274 × 10⁷)] = 5.59 × 10³ m s⁻¹.
- The period is T = 2πr/v_c = 2π√(r³/GM) = 1.43 × 10⁴ s, about 3.97 h.
- Specific orbital energy is ε = v_c²/2 − GM/r = −GM/(2r) = −GM/(4R_E) = −1.56 × 10⁷ J kg⁻¹.
- Zero total energy at the burn point requires v_esc = √(2GM/r) = √2v_c = 7.91 km s⁻¹. A tangential prograde burn therefore needs Δv = 7.91−5.59 = 2.32 km s⁻¹.
- Kinetic-energy change is ½(v_esc²−v_c²) = v_cΔv + ½(Δv)². The cross term matters because the spacecraft already has substantial orbital velocity before the burn.
Field and potential of a charged solid cylinder
Answer: Inside E = ρr/(2ε₀); outside E = ρR²/(2ε₀r). E(0.0100 m) = E(0.0400 m) = 2.26 × 10³ N C⁻¹. V(0) − V(R) = 45.2 V.
- Choose a coaxial Gaussian cylinder of length L. Cylindrical symmetry makes E radial and constant on its curved surface; flux through the flat ends is zero.
- For r < R, E(2πrL) = ρπr²L/ε₀, so E = ρr/(2ε₀). At 0.0100 m this is 2.26 × 10³ N C⁻¹.
- For r > R, the enclosed charge is ρπR²L. Therefore E = ρR²/(2ε₀r), giving 2.26 × 10³ N C⁻¹ at 0.0400 m.
- Potential difference is V(0)−V(R) = ∫₀ᴿE(r)dr = ρR²/(4ε₀) = 45.2 V. Potential falls outward along the outward field.
- The E–r graph starts at zero, rises linearly to ρR/(2ε₀) at R, stays continuous there, then falls as 1/r. A jump would incorrectly imply a surface-charge sheet added at R.
Current growth in an LR circuit
Answer: τ = 0.0200 s; (di/dt)_0 = 30.0 A s⁻¹; i(0.0400 s) = 0.519 A; V_L = 1.22 V; U_L = 0.0404 J.
- The LR time constant is τ = L/R = 0.300/15.0 = 0.0200 s. The final current is ε/R = 0.600 A.
- At t = 0 the current is zero, so the resistor drop is zero and L(di/dt) = ε. Therefore the initial gradient is 9.00/0.300 = 30.0 A s⁻¹.
- Current growth follows i = (ε/R)(1−e^(−t/τ)). At t = 0.0400 s = 2τ, i = 0.600(1−e⁻²) = 0.519 A.
- The inductor voltage is εe^(−t/τ) = 9.00e⁻² = 1.22 V. Stored magnetic energy is ½Li² = 0.0404 J.
- Initially self-induction takes the full battery voltage and opposes rapid current change; at steady state di/dt = 0 and its voltage tends to zero. Observe voltage with a properly rated data logger and include a flyback path before opening the circuit to suppress a damaging voltage spike.
Derive the field on the axis of a charged ring
Answer: E_x = kQx/(x²+R²)^(3/2) = 1.29 × 10³ N C⁻¹ away from the ring; V = kQ/√(x²+R²) = 322 V. The small-x approximation gives 1.80 × 10³ N C⁻¹ and is not accurate at x/R = 0.5.
- Every ring element is the same distance s = √(x²+R²) from the field point. An element dq contributes magnitude k dq/s².
- Components perpendicular to the axis cancel pairwise for diametrically opposite elements. The axial component is dE_x = (k dq/s²)(x/s) = kx dq/s³.
- Integrating around the ring gives E_x = kQx/(x²+R²)^(3/2). Substitution yields 1.29 × 10³ N C⁻¹ in the +x direction for positive Q and positive x.
- Potential adds as a scalar: V = ∫k dq/s = kQ/√(x²+R²) = 322 V. Differentiating −dV/dx reproduces the field expression.
- For x ≪ R, the denominator is approximately R³, so E_x ≈ kQx/R³ = 1.80 × 10³ N C⁻¹. Here x/R = 0.5 is not very small, and the approximation overestimates the exact field by about 40%.