Original GioPhysics extended practice
AP Physics · Extended Paper 1
A balanced original practice paper with long-form reasoning, calculations, and supplied-data graphs.
- Time
- 170 minutes
- Questions
- 10
- Marks
- 125
- 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.
Reconstruct a journey from a velocity graph
A cart moves along a straight track. Its measured velocity–time coordinates are (0 s, 0 m s⁻¹), (2 s, 4 m s⁻¹), (5 s, 4 m s⁻¹), and (7 s, −2 m s⁻¹); adjacent points are joined by straight lines. Determine the acceleration on each interval, the displacement and total distance from 0 to 7 s, and the time at which the cart reverses. Sketch how the position graph must curve and identify an experimental limitation of estimating velocity by frame-to-frame video differences.
Measured velocity of the cart
Four measured coordinates from the prompt joined by straight segments; the vertical-axis zero crossing is part of the supplied evidence.
Read the exact data as a table
| Time / s | Velocity / m s⁻¹ |
|---|---|
| 0 | 0 |
| 2 | 4 |
| 5 | 4 |
| 7 | -2 |
These are the supplied readings, not a worked answer. Lines join the supplied readings; they are not a fitted model.
Collision followed by spring compression
A 0.200 kg clay block moving at 6.00 m s⁻¹ strikes and sticks to a stationary 0.800 kg cart on a level low-friction track. The combined object then compresses a spring of constant 180 N m⁻¹ until momentarily at rest. Find the speed just after collision, the mechanical energy lost during impact, and the maximum compression. Explain why momentum is used during impact but mechanical energy is used during compression, and propose a measurement that tests the spring model.
Open the interactive answer workspaceEnergy and timing in a spring oscillator
A 0.400 kg cart attached to a horizontal spring of constant 64.0 N m⁻¹ oscillates with amplitude 0.0800 m on a low-friction track. Find the angular frequency, period, total energy, and speed when x = 0.0500 m. Starting at x = +A at t = 0, determine the first time the cart reaches x = 0.0500 m. Describe how modest damping would change the position–time graph without changing the equilibrium position.
Open the interactive answer workspaceField, potential and work between two charges
A +3.00 μC point charge is fixed at x = 0 and a −1.00 μC charge at x = 0.400 m. At the midpoint x = 0.200 m, calculate the net electric field and electric potential. Find the external work required to bring a +2.00 nC test charge slowly from infinity to the midpoint. State the initial direction of its acceleration if released, and explain why zero potential would not necessarily imply zero field.
Open the interactive answer workspaceSelect and bend a proton beam
Protons travel right through crossed fields: E = 8.00 × 10⁵ V m⁻¹ upward and B = 0.200 T out of the page. The undeflected beam then enters a region containing only the same magnetic field. Determine the selected speed, force directions in the selector, circular radius, orbital period, and whether doubling speed changes the period. Use proton mass 1.67 × 10⁻²⁷ kg and charge 1.602 × 10⁻¹⁹ C.
Open the interactive answer workspaceDetermine Planck's constant from stopping voltage
A metal is illuminated at several frequencies. Measured (frequency in 10¹⁴ Hz, stopping potential in V) pairs are (6.0, 0.482), (7.0, 0.895), (8.0, 1.309), and (9.0, 1.722). Treat the best-fit line through the rounded data as having gradient 0.4135 V per 10¹⁴ Hz and intercept −2.00 V. Determine Planck's constant, work function and threshold frequency. Predict the effects of doubling intensity above threshold, and explain why classical wave theory fails here. Use e = 1.602 × 10⁻¹⁹ C.
Stopping-potential observations
The four measurements from the prompt are shown without a fitted line so the gradient and intercept remain part of the analysis.
Read the exact data as a table
| Frequency / 10¹⁴ Hz | Stopping potential / V |
|---|---|
| 6 | 0.482 |
| 7 | 0.895 |
| 8 | 1.309 |
| 9 | 1.722 |
These are the supplied readings, not a worked answer. No fitted trend is drawn.
Integrate a position-dependent force
A 2.00 kg particle moves along +x with speed 1.00 m s⁻¹ at x = 0. From x = 0 to x = 3.00 m the net force is F(x) = 12x − 3x² newtons, with x in metres. Determine the work, final speed, position and value of the maximum force, and the impulse during the motion if the travel time is 1.20 s and the final velocity remains positive. Explain why the impulse cannot be found from the area under F(x) without time information.
Open the interactive answer workspaceAn Atwood machine with pulley inertia
Masses m₁ = 3.00 kg and m₂ = 2.00 kg are joined by a light non-slip cord over a uniform-disk pulley of mass 1.00 kg and radius 0.200 m. The axle is frictionless and the system is released from rest. Determine the acceleration, both cord tensions and pulley angular acceleration. After the heavier mass falls 0.800 m, find its speed using energy and show consistency with constant-acceleration kinematics.
Open the interactive answer workspaceCompare connected and isolated dielectric insertion
A 6.00 μF parallel-plate capacitor is charged to 12.0 V. A dielectric of constant κ = 3.00 completely fills the gap. Compare two cases: A, the battery is disconnected before insertion; B, the ideal 12.0 V battery remains connected. For each case find final capacitance, charge, voltage and stored energy. Determine the external work in case A for quasistatic insertion and explain the direction of the dielectric force.
Open the interactive answer workspaceMechanical power and motional emf
A conducting rod of length 0.500 m slides right at constant 4.00 m s⁻¹ on horizontal rails in a uniform 0.800 T magnetic field into the page. The total circuit resistance is 2.00 Ω. Determine the motional emf, current magnitude and direction, magnetic force on the rod, external force required, and mechanical power. Verify power conservation and explain what changes if the rod speed doubles.
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.
Reconstruct a journey from a velocity graph
Answer: Accelerations: +2.0, 0 and −3.0 m s⁻². Displacement = 18.0 m; distance = 19.3 m; reversal at 6.33 s. Position rises with changing gradient, becomes linear, reaches a maximum at reversal, then falls.
- Acceleration is the velocity–time gradient. The three gradients are (4−0)/(2−0) = 2.0 m s⁻², zero, and (−2−4)/(7−5) = −3.0 m s⁻².
- Displacement is signed area. The first triangle contributes 4 m, the rectangle contributes 12 m, and the final trapezium contributes (4 + −2)(2)/2 = 2 m, giving 18 m.
- On the final line v = 4 − 3(t−5). Setting v = 0 gives t = 6.33 s. Splitting that interval at the crossing gives 2.67 m forward and 0.67 m backward.
- Total distance is therefore 4 + 12 + 2.67 + 0.67 = 19.3 m. Distance exceeds the magnitude of displacement because the cart travels backward after the turning instant.
- The position graph's gradient equals velocity: it steepens initially, is straight from 2–5 s, flattens to a maximum at 6.33 s, then slopes downward. Finite video frame spacing averages velocity and can blur rapid acceleration; scale calibration and parallax also matter.
Collision followed by spring compression
Answer: The combined speed is 1.20 m s⁻¹, collision energy loss is 2.88 J, and maximum compression is 0.0894 m. Momentum is approximately conserved over the short collision; post-collision kinetic energy becomes spring energy afterward.
- During the brief collision, the spring has not yet compressed appreciably and external horizontal impulse is negligible. Momentum conservation gives (0.200)(6.00) = (1.000)v, hence v = 1.20 m s⁻¹.
- Initial kinetic energy is ½(0.200)(6.00)² = 3.60 J. Immediately after impact it is ½(1.000)(1.20)² = 0.720 J.
- The collision is perfectly inelastic, so the lost mechanical energy is 3.60 − 0.720 = 2.88 J. It becomes deformation, thermal energy and sound rather than disappearing.
- After impact, negligible track losses let 0.720 J convert into elastic energy: ½kx² = 0.720. Thus x = √(1.44/180) = 0.0894 m.
- Measure force against compression slowly with a calibrated force sensor. A straight F–x graph through the origin supports constant k; the area beneath that graph should match the stored energy used here.
Energy and timing in a spring oscillator
Answer: ω = 12.6 rad s⁻¹, T = 0.497 s, E = 0.2048 J, v = 0.790 m s⁻¹, and the first arrival is at 0.0709 s. Damping produces a decaying envelope and slightly longer period.
- For an ideal mass–spring oscillator, ω = √(k/m) = √(64.0/0.400) = 12.65 rad s⁻¹ and T = 2π/ω = 0.497 s.
- At a turning point all energy is elastic, so E = ½kA² = ½(64.0)(0.0800)² = 0.2048 J.
- Energy conservation at x gives ½mv² = ½k(A²−x²). Thus v = ω√(A²−x²) = 0.790 m s⁻¹; the first passage is toward negative x.
- With x = A cosωt, cosωt = 0.0500/0.0800 = 0.625. Therefore t = cos⁻¹(0.625)/12.65 = 0.0709 s.
- Damping removes energy each cycle, so successive extrema lie inside a decreasing envelope. The oscillation becomes slightly slower for modest damping, while the fixed spring's force still vanishes at the same equilibrium location.
Field, potential and work between two charges
Answer: E = 8.99 × 10⁵ N C⁻¹ to +x; V = 8.99 × 10⁴ V; external work = 1.80 × 10⁻⁴ J; a positive test charge accelerates toward +x.
- At the midpoint, the positive source repels a positive test charge to the right. The negative source attracts it to the right, so the two field magnitudes add rather than subtract.
- E = k(3.00 × 10⁻⁶ + 1.00 × 10⁻⁶)/(0.200)² = 8.99 × 10⁵ N C⁻¹ in the +x direction.
- Potential is scalar and includes charge signs: V = k[(3.00 × 10⁻⁶)/0.200 − (1.00 × 10⁻⁶)/0.200] = 8.99 × 10⁴ V.
- Slow transfer changes electric potential energy by qV, so the external work is (2.00 × 10⁻⁹)(8.99 × 10⁴) = 1.80 × 10⁻⁴ J.
- The released positive charge accelerates with E toward +x. Potential can cancel algebraically at a point while field vectors do not; field depends on the spatial gradient of potential, not simply its value.
Select and bend a proton beam
Answer: v = 4.00 × 10⁶ m s⁻¹; electric force upward and magnetic force downward; r = 0.209 m; T = 3.28 × 10⁻⁷ s. Doubling speed doubles radius but not period.
- For a positive proton, electric force follows E and points upward. With velocity right and B out of the page, v × B points downward, so magnetic force can cancel it.
- Undeflected motion requires qE = qvB. Charge cancels, giving v = E/B = (8.00 × 10⁵)/0.200 = 4.00 × 10⁶ m s⁻¹.
- In the magnetic-only region, qvB supplies centripetal force: qvB = mv²/r. Thus r = mv/(qB) = 0.209 m.
- The period is circumference divided by speed: T = 2πr/v = 2πm/(qB) = 3.28 × 10⁻⁷ s.
- Doubling v doubles r because magnetic force scales with v while required centripetal force scales with v²/r. The ratio r/v and therefore the non-relativistic period remain unchanged.
Determine Planck's constant from stopping voltage
Answer: h = 6.62 × 10⁻³⁴ J s, work function = 2.00 eV, threshold frequency = 4.84 × 10¹⁴ Hz. Doubling intensity doubles photocurrent under unsaturated collection but does not change stopping voltage.
- Einstein's equation hf = Φ + K_max and eV_s = K_max combine to V_s = (h/e)f − Φ/e.
- Because the horizontal axis unit is 10¹⁴ Hz, h/e = 0.4135/(10¹⁴) V s. Therefore h = e(4.135 × 10⁻¹⁵) = 6.62 × 10⁻³⁴ J s.
- The vertical intercept is −Φ/e = −2.00 V. Hence Φ = 2.00 eV = 3.204 × 10⁻¹⁹ J.
- At threshold V_s = 0, so f₀ = Φ/h, equivalently 2.00/(4.135 × 10⁻¹⁵) = 4.84 × 10¹⁴ Hz.
- Greater intensity means more photons per second, so unsaturated photocurrent doubles, but each photon retains energy hf and the stopping potential is unchanged. Classical intensity-based energy transfer cannot explain an immediate threshold frequency or intensity-independent maximum electron energy.
Integrate a position-dependent force
Answer: Work = 27.0 J, final speed = 5.29 m s⁻¹, maximum force 12.0 N at x = 2.00 m, and impulse = 8.58 N s.
- Work is the spatial integral: W = ∫₀³(12x−3x²)dx = [6x²−x³]₀³ = 54−27 = 27.0 J.
- The initial kinetic energy is ½(2.00)(1.00)² = 1.00 J. Work–energy gives K_f = 28.0 J, so v_f = √(2K_f/m) = 5.29 m s⁻¹.
- Differentiate the force: dF/dx = 12−6x. The interior maximum occurs at x = 2.00 m and has F = 12(2)−3(2²) = 12.0 N.
- Impulse equals momentum change, not spatial force area: J = m(v_f−v_i) = 2.00(5.29−1.00) = 8.58 N s.
- The area under F against x is work because dx is displacement. Impulse requires ∫F dt; the supplied travel time alone would not determine it from an F–x curve, but the momentum result does.
An Atwood machine with pulley inertia
Answer: a = 1.78 m s⁻²; T₁ = 24.1 N, T₂ = 23.2 N; α = 8.92 rad s⁻²; speed after 0.800 m = 1.69 m s⁻¹.
- For the descending mass, m₁g−T₁ = m₁a. For the rising mass, T₂−m₂g = m₂a. The unequal tensions provide the pulley's angular acceleration.
- Pulley torque is (T₁−T₂)R = Iα with I = ½MR² and α = a/R, hence T₁−T₂ = ½Ma.
- Combining equations gives (m₁−m₂)g = (m₁+m₂+M/2)a. Thus a = 9.81/5.50 = 1.78 m s⁻².
- T₁ = m₁(g−a) = 24.1 N and T₂ = m₂(g+a) = 23.2 N. Their difference supplies the required torque; α = a/R = 8.92 rad s⁻².
- The net lost gravitational energy (m₁−m₂)gh becomes ½(m₁+m₂)v² + ½I(v/R)² = ½(m₁+m₂+M/2)v². This gives v = 1.69 m s⁻¹, equal to √(2ah).
Compare connected and isolated dielectric insertion
Answer: Both cases C_f = 18.0 μF. A: Q = 72.0 μC, V = 4.00 V, U = 144 μJ, W_ext = −288 μJ. B: Q = 216 μC, V = 12.0 V, U = 1.296 mJ. The dielectric is pulled inward.
- Full insertion multiplies capacitance by κ, so C_f = 3.00(6.00 μF) = 18.0 μF. Initial charge is C₀V₀ = 72.0 μC and initial energy is ½C₀V₀² = 432 μJ.
- In case A, isolated charge remains 72.0 μC. Thus V_f = Q/C_f = 4.00 V and U_f = Q²/(2C_f) = 144 μJ.
- For quasistatic insertion with no battery exchange, W_ext = ΔU = 144−432 = −288 μJ. The negative sign means the field does +288 μJ of mechanical work while pulling the dielectric inward.
- In case B, voltage remains 12.0 V. Then Q_f = C_fV = 216 μC and U_f = ½C_fV² = 1.296 mJ; the battery supplies additional charge and energy.
- Insertion increases capacitance. At fixed charge this lowers field energy; at fixed voltage the battery–capacitor system also favors insertion after battery work is included. In both physical cases the net electromagnetic force draws the dielectric farther between the plates.
Mechanical power and motional emf
Answer: emf = 1.60 V; current = 0.800 A counterclockwise; magnetic force = 0.320 N left; external force = 0.320 N right; power = 1.28 W. Doubling speed doubles current and force and quadruples power.
- The swept-loop flux increases into the page. Lenz's law requires an induced field out of the page, so the current is counterclockwise.
- Motional emf is ε = BLv = (0.800)(0.500)(4.00) = 1.60 V. Ohm's law gives I = ε/R = 0.800 A.
- The rod's current direction and field give a magnetic force opposing motion. Its magnitude is F = BIL = (0.800)(0.800)(0.500) = 0.320 N left.
- Constant speed requires an external 0.320 N force right. Mechanical input power is Fv = (0.320)(4.00) = 1.28 W, equal to I²R = (0.800)²(2.00) = 1.28 W.
- Since ε and I scale with v, magnetic drag scales with v and required mechanical power Fv scales with v². Doubling speed therefore gives twice the current and force but four times the heating power.