Everything below prints as one AP AP Physics 2 practice paper set: papers A & B with their answer keys, plus the full-length study package exam. Use the Download PDF / Print button (or Cmd/Ctrl+P) to save it.
Paper A
AP Physics 2 — Practice Paper A
Original unofficial practice questions · paper A · answer key on the last page
Total time: see section headers · No guessing penalty
| Section | Questions | Format |
|---|---|---|
| Section I: Multiple Choice | ||
| Section II: Free Response |
Section I — Multiple Choice
Coulombs: F = k q₁q₂/r². Doubling distance gives
A. 1/4 the forceB. twice the forceC. 4x forceD. same forceThe electric field points
A. away from positive chargeB. toward positiveC. in all directionsD. with field lines closedA fluid's buoyant force equals
A. weight of displaced fluidB. its massC. the object's weightD. pressure at the bottomThe ideal gas law is
A. PV = nRTB. P = nVC. PVT = nRD. P + V = nRTWave interference producing a stationary pattern requires
A. two coherent sourcesB. random sourcesC. absorptionD. reflection onlyThe speed of sound in air is roughly
A. 343 m/sB. 3 m/sC. 1500 m/sD. 300,000 m/sA converging lens forms a real image when the object is
A. beyond the focal pointB. inside the focal pointC. at the lensD. at infinity onlyEnergy stored in a capacitor with capacitance C and voltage V is
A. ½CV²B. CVC. C²VD. V/CElectromagnetic waves consist of
A. oscillating E and B fieldsB. sound wavesC. matter particlesD. static chargesFluids flow due to a
A. pressure differenceB. temperature onlyC. volume onlyD. densitySection II — Free Response
Explain how a convex lens forms a real image and compute the image position for f = 12 cm, object at 18 cm using 1/f = 1/do + 1/di.
7 points · rubric: Formula 2 pts, substitution 2 pts, position + type 3 pts.
Describe the physical meaning of total internal reflection and give the condition for it to occur.
5 points · rubric: Concept 3 pts, condition 2 pts.
Answer Key
1. 1/4 the force — Inverse-square law.
2. away from positive charge — Direction convention.
3. weight of displaced fluid — Archimedes.
4. PV = nRT — Equation of state.
5. two coherent sources — Coherent superposition.
6. 343 m/s — Room-temp value.
7. beyond the focal point — Lens equation behavior.
8. ½CV² — Stored energy formula.
9. oscillating E and B fields — Transverse fields.
10. pressure difference — Driving gradient.
Free response — rubric notes
1. Formula 2 pts, substitution 2 pts, position + type 3 pts. · model: 1/di = 1/12 - 1/18 = 1/36 → di = 36 cm, real and inverted.
2. Concept 3 pts, condition 2 pts. · model: Occurs going high to low index with angle beyond the critical angle; light reflects entirely.
Paper B
AP Physics 2 — Practice Paper B
Original unofficial practice questions · paper B · answer key on the last page
Total time: see section headers · No guessing penalty
| Section | Questions | Format |
|---|---|---|
| Section I: Multiple Choice | ||
| Section II: Free Response |
Section I — Multiple Choice
Coulombs: F = k q₁q₂/r². Doubling distance gives
A. same forceB. twice the forceC. 1/4 the forceD. 4x forceThe electric field points
A. toward positiveB. in all directionsC. with field lines closedD. away from positive chargeA fluid's buoyant force equals
A. its massB. weight of displaced fluidC. pressure at the bottomD. the object's weightThe ideal gas law is
A. P + V = nRTB. PVT = nRC. PV = nRTD. P = nVWave interference producing a stationary pattern requires
A. absorptionB. two coherent sourcesC. random sourcesD. reflection onlyThe speed of sound in air is roughly
A. 1500 m/sB. 300,000 m/sC. 343 m/sD. 3 m/sA converging lens forms a real image when the object is
A. inside the focal pointB. at infinity onlyC. at the lensD. beyond the focal pointEnergy stored in a capacitor with capacitance C and voltage V is
A. V/CB. ½CV²C. C²VD. CVElectromagnetic waves consist of
A. sound wavesB. matter particlesC. oscillating E and B fieldsD. static chargesFluids flow due to a
A. temperature onlyB. densityC. volume onlyD. pressure differenceSection II — Free Response
Explain how a convex lens forms a real image and compute the image position for f = 12 cm, object at 18 cm using 1/f = 1/do + 1/di.
7 points · rubric: Formula 2 pts, substitution 2 pts, position + type 3 pts.
Describe the physical meaning of total internal reflection and give the condition for it to occur.
5 points · rubric: Concept 3 pts, condition 2 pts.
Answer Key
1. 1/4 the force — Inverse-square law.
2. away from positive charge — Direction convention.
3. weight of displaced fluid — Archimedes.
4. PV = nRT — Equation of state.
5. two coherent sources — Coherent superposition.
6. 343 m/s — Room-temp value.
7. beyond the focal point — Lens equation behavior.
8. ½CV² — Stored energy formula.
9. oscillating E and B fields — Transverse fields.
10. pressure difference — Driving gradient.
Free response — rubric notes
1. Formula 2 pts, substitution 2 pts, position + type 3 pts. · model: 1/di = 1/12 - 1/18 = 1/36 → di = 36 cm, real and inverted.
2. Concept 3 pts, condition 2 pts. · model: Occurs going high to low index with angle beyond the critical angle; light reflects entirely.
Full-length study package exam
AP Physics 2 Full Practice Exam
Time: 3 hours (90 min MCQ + 90 min FRQ) 50 Multiple Choice + 4 Free Response
Section 1: Multiple Choice (50 questions, 90 minutes)
Fluids (Questions 1–6)
1. A piece of cork (density 240 kg/m³) floats in water. What percentage of the cork's volume is below the water's surface? (A) 12% (B) 24% (C) 48% (D) 76%
2. Water flows through a pipe that narrows from diameter 6 cm to 3 cm. If the flow speed in the wide section is 1.5 m/s, what is the speed in the narrow section? (A) 0.75 m/s (B) 3.0 m/s (C) 6.0 m/s (D) 9.0 m/s
3. A hydraulic lift has a small piston of area 0.02 m² and a large piston of area 0.40 m². If a force of 100 N is applied to the small piston, the output force is most nearly: (A) 100 N (B) 500 N (C) 1000 N (D) 2000 N
4. A metal object weighs 50 N in air and 40 N when fully submerged in water. The density of the metal is most nearly: (A) 1000 kg/m³ (B) 2500 kg/m³ (C) 5000 kg/m³ (D) 10000 kg/m³
5. The gauge pressure at a depth of 10 m below the surface of a lake (ρ = 1000 kg/m³) is most nearly: (A) 1.0 × 10⁴ Pa (B) 9.8 × 10⁴ Pa (C) 1.0 × 10⁵ Pa (D) 2.0 × 10⁵ Pa
6. According to Bernoulli's equation, as fluid flows faster through a constriction, the pressure in the constriction: (A) Increases (B) Decreases (C) Remains the same (D) Depends on the fluid density
Thermodynamics (Questions 7–14)
7. A gas in a rigid container is heated. Which of the following increases? (A) Volume (B) Pressure (C) Both pressure and volume (D) Neither
8. An ideal gas expands isothermally from 2 L to 6 L. The final pressure compared to the initial pressure is: (A) Three times larger (B) The same (C) One-third as large (D) One-ninth as large
9. A Carnot engine operates between 400 K and 600 K. Its maximum efficiency is: (A) 25% (B) 33% (C) 50% (D) 67%
10. During an isochoric process, the work done by the gas is: (A) Positive (B) Negative (C) Zero (D) Cannot be determined
11. The root-mean-square speed of gas molecules in a sample at temperature T is v_rms. If the temperature is doubled, the new rms speed is: (A) 2v_rms (B) √2 v_rms (C) v_rms/√2 (D) 4v_rms
12. A heat engine absorbs 500 J from a hot reservoir and exhausts 350 J to a cold reservoir. Its efficiency is: (A) 15% (B) 30% (C) 50% (D) 70%
13. Which process has the steepest slope on a PV diagram? (A) Isothermal (B) Isobaric (C) Isochoric (D) Adiabatic
14. The second law of thermodynamics states that: (A) Energy is conserved in all processes
(B) The entropy of an isolated system never decreases
(C) All heat engines are 100% efficient
(D) Heat flows from cold to hot spontaneously
Electric Force, Field, and Potential (Questions 15–24)
15. Two charges, +2 μC and -2 μC, are 0.10 m apart. The force between them is: (A) Attractive, 3.6 N (B) Repulsive, 3.6 N (C) Attractive, 0.36 N (D) Repulsive, 0.36 N
16. The electric field inside a hollow conducting sphere is: (A) Zero (B) Maximum at the center (C) Uniform throughout (D) Proportional to r
17. A proton moves through a potential difference of 1000 V. Its gain in kinetic energy is: (A) 1000 J (B) 1.6 × 10⁻¹⁶ J (C) 1.6 × 10⁻¹⁹ J (D) 1000 eV
18. Equipotential lines are always: (A) Parallel to electric field lines (B) Perpendicular to electric field lines (C) Radial (D) Circular
19. The electric potential due to a negative point charge is: (A) Positive everywhere (B) Negative everywhere (C) Zero at infinity only (D) Positive at close range
20. A charge q experiences a force F in an electric field E. If the charge is replaced by 3q, the force becomes: (A) F/3 (B) F (C) 3F (D) 9F
21. Two identical positive charges are separated by distance d. The electric field at the midpoint is: (A) kQ/d² to the left (B) kQ/d² to the right (C) Zero (D) 2kQ/d²
22. The work done to move a +1 C charge between two points differing in potential by 5 V is: (A) 0.2 J (B) 1 J (C) 5 J (D) 25 J
23. A metal sphere with charge +Q is brought near a neutral conducting sphere. The neutral sphere is then grounded. After removing the ground and the charged sphere, the neutral sphere has: (A) No charge (B) Positive charge (C) Negative charge (D) Both positive and negative charge
24. The relationship between electric field and electric potential in one dimension is: (A) E = ΔV/Δx (B) E = -ΔV/Δx (C) E = V/x (D) E = ΔV × Δx
Electric Circuits (Questions 25–30)
25. Three 12 Ω resistors in parallel have an equivalent resistance of: (A) 36 Ω (B) 12 Ω (C) 6 Ω (D) 4 Ω
26. A 9 V battery with internal resistance 1 Ω is connected to a 4 Ω resistor. The terminal voltage is: (A) 7.2 V (B) 7.5 V (C) 8.0 V (D) 9.0 V
27. In an RC circuit, the capacitor is fully charged after approximately: (A) 1 RC time constant (B) 2 RC time constants (C) 3 RC time constants (D) 5 RC time constants
28. A 100 W bulb and a 200 W bulb, both rated for 120 V, are connected in series to a 120 V source. Which bulb is brighter? (A) The 100 W bulb (B) The 200 W bulb (C) They are equally bright (D) Neither lights up
29. Kirchhoff's junction rule is a statement of conservation of: (A) Energy (B) Charge (C) Momentum (D) Mass
30. The power dissipated by a 10 Ω resistor carrying 2 A of current is: (A) 4 W (B) 20 W (C) 40 W (D) 400 W
Magnetism and Electromagnetic Induction (Questions 31–36)
31. An electron moving north enters a magnetic field directed upward. The magnetic force is directed: (A) North (B) South (C) East (D) West
32. A wire carrying current I in a magnetic field B experiences maximum force when the angle between I and B is: (A) 0° (B) 45° (C) 90° (D) 180°
33. The magnetic flux through a loop is decreasing. Lenz's law says the induced current: (A) Creates a field opposing the decrease (B) Creates a field supporting the decrease (C) Is zero (D) Is clockwise
34. A transformer with 100 primary turns and 400 secondary turns has a primary voltage of 60 V. The secondary voltage is: (A) 15 V (B) 60 V (C) 120 V (D) 240 V
35. A charged particle moves in a circle in a uniform magnetic field. If the magnetic field strength is doubled, the radius of the circle: (A) Doubles (B) Halves (C) Quadruples (D) Is unchanged
36. Faraday's law of induction states that the induced emf is proportional to: (A) The magnetic flux (B) The rate of change of magnetic flux (C) The area of the loop (D) The resistance of the loop
Geometric and Physical Optics (Questions 37–44)
37. Light travels from water (n = 1.33) into diamond (n = 2.42). The light bends: (A) Toward the normal (B) Away from the normal (C) Not at all (D) Totally internally reflected
38. An object is placed 5 cm from a concave mirror with focal length 10 cm. The image is: (A) Real and inverted (B) Virtual and upright (C) At the focal point (D) At infinity
39. In Young's double-slit experiment, the fringe spacing is increased by: (A) Increasing the slit separation (B) Decreasing the wavelength (C) Decreasing the slit separation (D) Moving the screen closer
40. A ray of light in glass (n = 1.5) strikes the glass-air boundary at 30° from the normal. The critical angle for this interface is approximately: (A) 30° (B) 42° (C) 48° (D) 90°
41. Unpolarized light passes through a single polarizing filter. The transmitted intensity is: (A) I₀ (B) I₀/2 (C) I₀/4 (D) 0
42. A converging lens forms a real, inverted image that is the same size as the object. The object must be at: (A) The focal point (B) Twice the focal length (C) Three times the focal length (D) Infinity
43. Light of wavelength 600 nm in air enters glass (n = 1.5). The wavelength in the glass is: (A) 200 nm (B) 400 nm (C) 600 nm (D) 900 nm
44. In single-slit diffraction, the central maximum is: (A) Narrower than the secondary maxima (B) The same width as secondary maxima (C) Twice as wide as secondary maxima (D) Four times as wide as secondary maxima
Quantum, Atomic, and Nuclear Physics (Questions 45–50)
45. The threshold frequency for a metal surface is 6.0 × 10¹⁴ Hz. The work function of the metal is most nearly: (A) 1.0 eV (B) 2.5 eV (C) 4.1 eV (D) 6.0 eV
46. A hydrogen atom transitions from n = 3 to n = 1. The emitted photon has energy: (A) 1.51 eV (B) 3.40 eV (C) 12.09 eV (D) 13.6 eV
47. An alpha particle is: (A) A high-energy photon (B) An electron (C) Two protons and two neutrons (D) A positron
48. A radioactive sample has a half-life of 10 years. After 30 years, the fraction remaining is: (A) 1/8 (B) 1/4 (C) 1/3 (D) 1/2
49. Which of the following has the longest de Broglie wavelength when moving at the same speed? (A) Proton (B) Electron (C) Alpha particle (D) Neutron
50. In nuclear fission, a heavy nucleus: (A) Combines with another nucleus (B) Splits into lighter nuclei (C) Emits a gamma ray only (D) Gains a proton
Section 2: Free Response (4 questions, 90 minutes)
FRQ 1: Experimental Design — Fluids and Pressure
Students design an experiment to determine the density of an unknown liquid. They have a graduated cylinder, a spring scale, an irregularly shaped metal object of known mass (0.250 kg), and the unknown liquid.
(a) Describe an experimental procedure the students could use to determine the density of the liquid. Include what measurements they should take.
(b) The metal object weighs 2.45 N in air and 1.70 N when fully submerged in water (density 1000 kg/m³). Calculate the volume of the metal object.
(c) When the same metal object is fully submerged in the unknown liquid, the spring scale reads 1.95 N. Calculate the density of the unknown liquid.
(d) Identify one source of experimental error in this procedure and describe how it would affect the calculated density.
(e) The students then want to measure the density of a gas. Explain why the method described above would not work for a gas, and suggest what property they could measure instead.
FRQ 2: Mathematical Routines — Electric Circuits with RC Component
A circuit contains a 12 V battery (internal resistance negligible), two resistors (R₁ = 8 Ω in parallel with R₂ = 12 Ω), and a switch that connects a capacitor (C = 50 μF) in series with a third resistor R₃ = 4 Ω. The capacitor and R₃ branch is in parallel with R₁ and R₂. Initially the switch is open and the capacitor is uncharged.
(a) Calculate the equivalent resistance of R₁ and R₂ in parallel.
(b) The switch is closed at t = 0. Calculate the initial current through R₃.
(c) Calculate the time constant for the RC portion of the circuit.
(d) Calculate the energy stored in the capacitor after a very long time.
(e) On the axes provided (sketch conceptually), graph the current through R₃ as a function of time from t = 0 until the capacitor is fully charged. Label the initial current, the final current, and the time constant on your graph.
FRQ 3: Translation Between Representations — Optics
A converging lens with focal length f = 15 cm is used to form an image of an object. The object is a vertical arrow of height 4.0 cm.
(a) The object is placed 25 cm from the lens. Calculate the image distance and the magnification. Describe the image (real/virtual, upright/inverted, enlarged/reduced).
(b) Draw a ray diagram for the situation in part (a), showing at least two principal rays. Label the focal points and the object and image distances.
(c) The object is now moved to a position 10 cm from the lens (inside the focal point). Calculate the new image distance and magnification. Describe the image.
(d) A student claims that if the object is placed exactly at the focal point, the image forms at infinity. Explain why this claim is correct using the thin lens equation.
(e) Describe how the image characteristics change as the object is continuously moved from very far away (d_o → ∞) to very close to the lens (d_o → 0).
FRQ 4: Qualitative/Quantitative Translation — Electromagnetic Induction
A rectangular conducting loop (width w = 0.15 m, height h = 0.25 m, resistance R = 3.0 Ω) is positioned partly inside a region of uniform magnetic field B = 0.5 T directed into the page. The field region extends 0.40 m in the horizontal direction. The loop is pulled to the right with constant velocity v = 4.0 m/s. At t = 0, the left edge of the loop is at the left boundary of the field.
(a) Calculate the magnetic flux through the loop as a function of the distance x that the left edge of the loop has moved into the field (for x < 0.40 m).
(b) Calculate the induced emf while the loop is entering the field (0 < x < 0.15 m).
(c) Determine the direction of the induced current in the loop while entering the field. Justify using Lenz's law.
(d) Calculate the force required to maintain constant velocity while the loop is entering the field.
(e) The loop is now pulled entirely out of the field (starting fully inside). Compare the magnitude of the induced emf during exit to the magnitude during entry. Explain your reasoning.
END OF EXAM
Answer Key & Rubric
AP Physics 2 Full Practice Exam — Answer Key
Section 1: Multiple Choice Answers and Explanations
Fluids (1–6)
1. (B) 24% — Fraction submerged = ρ_object/ρ_fluid = 240/1000 = 0.24 = 24%.
2. (C) 6.0 m/s — A₁v₁ = A₂v₂. A₁/A₂ = (r₁/r₂)² = (6/3)² = 4. So v₂ = 4(1.5) = 6.0 m/s.
3. (D) 2000 N — F₂ = F₁(A₂/A₁) = 100(0.40/0.02) = 100(20) = 2000 N.
4. (C) 5000 kg/m³ — F_buoy = 50 - 40 = 10 N. V = 10/(1000 × 9.8) = 1.02 × 10⁻³ m³. m = 50/9.8 = 5.10 kg. ρ = 5.10/1.02 × 10⁻³ = 5000 kg/m³.
5. (B) 9.8 × 10⁴ Pa — P_gauge = ρgh = 1000 × 9.8 × 10 = 98,000 Pa ≈ 9.8 × 10⁴ Pa.
6. (B) Decreases — Bernoulli's: P + ½ρv² = constant (horizontal pipe). Higher v → lower P.
Thermodynamics (7–14)
7. (B) Pressure — Rigid container → constant V. P/T = const, so P increases with T.
8. (C) One-third as large — Isothermal: P₁V₁ = P₂V₂ → P₂ = P₁(2/6) = P₁/3.
9. (B) 33% — η = 1 - T_C/T_H = 1 - 400/600 = 1/3 ≈ 33%.
10. (C) Zero — Isochoric: ΔV = 0, so W = PΔV = 0.
11. (B) √2 v_rms — v_rms ∝ √T. Double T → multiply by √2.
12. (B) 30% — η = (Q_H - Q_C)/Q_H = (500 - 350)/500 = 0.30 = 30%.
13. (D) Adiabatic — Adiabatic curves are steeper than isothermal on a PV diagram because temperature changes (cooling on expansion, heating on compression) amplify pressure changes.
14. (B) The entropy of an isolated system never decreases — This is the second law. Option (A) is the first law.
Electric Force, Field, and Potential (15–24)
15. (A) Attractive, 3.6 N — F = k|q₁q₂|/r² = (8.99 × 10⁹)(4 × 10⁻¹²)/0.01 = 3.6 N. Opposite charges → attractive.
16. (A) Zero — Electric field inside any conductor in electrostatic equilibrium is zero.
17. (D) 1000 eV — KE = qΔV = (1e)(1000 V) = 1000 eV.
18. (B) Perpendicular to electric field lines — By definition; no work along an equipotential requires F ⊥ displacement.
19. (B) Negative everywhere — V = kQ/r with Q < 0 gives V < 0 at all finite r.
20. (C) 3F — F = qE. Triple the charge → triple the force (field is unchanged).
21. (C) Zero — Equal and opposite fields from symmetric charges cancel at the midpoint.
22. (C) 5 J — W = qΔV = (1 C)(5 V) = 5 J.
23. (C) Negative charge — Induction: +Q attracts electrons to near side, grounding allows electrons to flow in, removing ground traps excess electrons → negative charge.
24. (B) E = -ΔV/Δx — Field points in direction of decreasing potential.
Electric Circuits (25–30)
25. (D) 4 Ω — 1/R = 3/12 = 1/4 → R_eq = 4 Ω.
26. (A) 7.2 V — I = 9/(4+1) = 1.8 A. V_term = 9 - 1.8(1) = 7.2 V.
27. (D) 5 RC time constants — After 5τ, Q = 99.3% of Q_max.
28. (A) The 100 W bulb — R₁₀₀ = 144 Ω, R₂₀₀ = 72 Ω. In series, same I. P = I²R → larger R (100W bulb) dissipates more power.
29. (B) Charge — Junction rule: current in = current out → conservation of charge.
30. (C) 40 W — P = I²R = (2)²(10) = 40 W.
Magnetism and Electromagnetic Induction (31–36)
31. (C) East — For a positive charge: v north, B upward. Fingers point north, curl up → thumb points east. But this is an electron (negative), so force is west. Wait — let me redo: v = north (+ŷ), B = upward (+ẑ). For positive charge: v × B = ŷ × ẑ = +x̂ (east). For electron: force is opposite = west. The answer should be (D) West.
Correction for Q31: (D) West — The electron is negative, so the force direction is opposite to the right-hand rule result.
32. (C) 90° — F = BIL sin θ. Maximum when sin θ = 1, i.e., θ = 90°.
33. (A) Creates a field opposing the decrease — Lenz's law: induced current opposes the change in flux. Decreasing flux → current tries to maintain it.
34. (D) 240 V — V_s = V_p(N_s/N_p) = 60(400/100) = 240 V.
35. (B) Halves — r = mv/(qB). Double B → halve r.
36. (B) The rate of change of magnetic flux — Faraday's law: ε = -NΔΦ/Δt. The emf is proportional to the rate of change, not the flux itself.
Geometric and Physical Optics (37–44)
37. (A) Toward the normal — Going from lower n (1.33) to higher n (2.42), light bends toward the normal.
38. (B) Virtual and upright — d_o = 5 cm < f = 10 cm (object inside focal point of converging lens/mirror) → virtual, upright, magnified image.
39. (C) Decreasing the slit separation — Fringe spacing Δy = λL/d. Decreasing d increases Δy.
40. (B) 42° — θ_c = sin⁻¹(n₂/n₁) = sin⁻¹(1.0/1.5) = sin⁻¹(0.667) = 41.8° ≈ 42°.
41. (B) I₀/2 — A single polarizer transmits half the intensity of unpolarized light.
42. (B) Twice the focal length — For m = -1: d_i = d_o and 1/f = 1/d_o + 1/d_i = 2/d_o → d_o = 2f.
43. (B) 400 nm — λ_glass = λ_air/n = 600/1.5 = 400 nm. Frequency is unchanged.
44. (C) Twice as wide as secondary maxima — The central maximum spans from m = -1 to m = +1, while each secondary maximum spans between consecutive minima separated by one minimum position.
Quantum, Atomic, and Nuclear Physics (45–50)
45. (B) 2.5 eV — φ = hf₀ = (6.626 × 10⁻³⁴)(6.0 × 10¹⁴)/(1.6 × 10⁻¹⁹) = 2.48 eV ≈ 2.5 eV.
46. (C) 12.09 eV — ΔE = E₁ - E₃ = -13.6 - (-13.6/9) = -13.6 + 1.51 = -12.09 eV. Photon energy = 12.09 eV.
47. (C) Two protons and two neutrons — An alpha particle is ⁴₂He.
48. (A) 1/8 — 30 years = 3 half-lives. Fraction = (½)³ = 1/8.
49. (B) Electron — λ = h/(mv). The electron has the smallest mass, so it has the longest wavelength at the same speed.
50. (B) Splits into lighter nuclei — This is the definition of nuclear fission.
Section 2: Free Response Answers
FRQ 1: Fluids — Experimental Design
(a) Experimental procedure:
- Measure the weight of the metal object in air using the spring scale: W_air.
- Submerge the metal object completely in the unknown liquid and measure the apparent weight: W_liquid.
- Calculate the buoyant force: F_b = W_air - W_liquid.
- Calculate the volume of the metal object (known from the water measurement or from its mass and known density).
- Use F_b = ρ_liquid × V × g to find ρ_liquid = F_b/(Vg).
(b) Volume of metal object: F_buoy_water = 2.45 - 1.70 = 0.75 N V = F_buoy/(ρ_water × g) = 0.75/(1000 × 9.8) = 7.65 × 10⁻⁵ m³
(c) Density of unknown liquid: F_buoy_unknown = 2.45 - 1.95 = 0.50 N ρ_unknown = F_buoy_unknown/(V × g) = 0.50/(7.65 × 10⁻⁵ × 9.8) = 667 kg/m³ (The unknown liquid is less dense than water, consistent with a smaller buoyant force.)
(d) Source of error: Air bubbles clinging to the metal object when submerged would displace additional fluid, increasing the apparent buoyant force. This would cause the calculated density to be higher than the true value, since ρ = F_b/(Vg) and F_b would be artificially large.
(e) Why the method fails for gases: Gases have very low density (about 1000 times less than liquids), so the buoyant force on the metal object would be negligibly small — likely smaller than the precision of the spring scale. Instead, one could use a balloon or collect a known volume of gas and measure its mass using a precise balance, then calculate ρ = m/V.
FRQ 2: Electric Circuits — Mathematical Routines
(a) Equivalent resistance of R₁ and R₂: 1/R₁₂ = 1/8 + 1/12 = 3/24 + 2/24 = 5/24 R₁₂ = 24/5 = 4.8 Ω
(b) Initial current through R₃: At t = 0, the uncharged capacitor acts like a wire (V_C = 0). R₁₂ and R₃ are effectively in parallel across the 12 V battery. V across R₃ = 12 V (same as battery, since V_C = 0) I₃_initial = V/R₃ = 12/4 = 3.0 A
(c) Time constant: The capacitor charges through the Thevenin resistance seen by the capacitor. Short the battery: R₁₂ is in parallel with R₃. R_equiv = (4.8)(4)/(4.8 + 4) = 19.2/8.8 = 2.18 Ω. τ = RC = (2.18)(50 × 10⁻⁶) = 1.09 × 10⁻⁴ s ≈ 0.109 ms
(d) Energy stored: After a long time, the capacitor is fully charged and no current flows through R₃. The full 12 V appears across the capacitor. U = ½CV² = ½(50 × 10⁻⁶)(12)² = ½(50 × 10⁻⁶)(144) = 3.6 × 10⁻³ J = 3.6 mJ
(e) Graph description: The current through R₃ starts at 3.0 A (at t = 0) and decreases exponentially toward 0 A. The curve is I(t) = I₀e^(-t/τ) = 3.0e^(-t/0.109ms). At t = τ ≈ 0.11 ms, the current drops to 3.0 × 0.368 = 1.10 A. After 5τ ≈ 0.55 ms, the current is essentially zero. The graph is a decaying exponential starting at 3.0 A on the y-axis and asymptotically approaching zero.
FRQ 3: Optics — Translation Between Representations
(a) Object at 25 cm: 1/f = 1/d_o + 1/d_i → 1/15 = 1/25 + 1/d_i 1/d_i = 1/15 - 1/25 = 5/75 - 3/75 = 2/75 d_i = 37.5 cm
m = -d_i/d_o = -37.5/25 = -1.5
Image height = |m| × h_o = 1.5 × 4.0 = 6.0 cm
The image is real (d_i > 0), inverted (m < 0), and enlarged (|m| > 1).
(b) Ray diagram:
- Ray 1: Parallel to the axis from the object tip → refracts through the far focal point F'
- Ray 2: Through the center of the lens → continues straight through
- Ray 3: Through the near focal point F → refracts parallel to the axis
- All three rays converge at the image point at d_i = 37.5 cm on the far side of the lens. The image is an inverted, enlarged arrow of height 6.0 cm.
(c) Object at 10 cm (inside focal point): 1/15 = 1/10 + 1/d_i 1/d_i = 1/15 - 1/10 = 2/30 - 3/30 = -1/30 d_i = -30 cm
m = -d_i/d_o = -(-30)/10 = +3.0
Image height = 3.0 × 4.0 = 12 cm
The image is virtual (d_i < 0), upright (m > 0), and enlarged (|m| > 1). This is a magnifying glass configuration.
(d) Object at focal point (d_o = f = 15 cm): 1/15 = 1/15 + 1/d_i → 1/d_i = 0 → d_i = ∞ When the object is at the focal point, the refracted rays are parallel and never converge (or diverge). The image forms at infinity — it is neither real nor virtual in the usual sense, and magnification approaches infinity.
(e) Continuous motion from ∞ to 0:
- d_o → ∞: Image is at the focal point (d_i → f), real, inverted, reduced (m → 0)
- d_o > 2f: Real, inverted, reduced (between f and 2f)
- d_o = 2f: Real, inverted, same size (m = -1), at d_i = 2f
- f < d_o < 2f: Real, inverted, magnified (beyond 2f)
- d_o = f: Image at infinity
- d_o < f: Virtual, upright, magnified (on the same side as the object)
- d_o → 0: Image approaches the object position, virtual, upright, m → 1
FRQ 4: Electromagnetic Induction
(a) Flux as a function of x (entering, x < 0.40 m): While entering: the area of the loop inside the field is A = x × h (for 0 < x < 0.15 m, where x is the distance the left edge has moved and the full width w is entering). Actually, let me reconsider: the area inside the field grows as the loop enters.
For 0 ≤ x ≤ w (0 ≤ x ≤ 0.15 m): the left edge has moved x into the field, so the area inside = x × h. Φ = B × (x × h) = (0.5)(x)(0.25) = 0.125x Wb
For w ≤ x ≤ 0.40 m (0.15 ≤ x ≤ 0.40 m): the entire loop width is inside. Area = w × h. Φ = B × w × h = (0.5)(0.15)(0.25) = 0.01875 Wb (constant)
(b) Induced emf while entering (0 < x < 0.15 m): Φ = 0.125x → dΦ/dx = 0.125 Since dx/dt = v = 4.0 m/s: ε = dΦ/dt = (dΦ/dx)(dx/dt) = (0.125)(4.0) = 0.50 V
(c) Direction of induced current: As the loop enters the field, the flux into the page is increasing. By Lenz's law, the induced current opposes this increase by creating a magnetic field out of the page inside the loop. By the right-hand rule, this requires a counterclockwise current (when viewed from above, i.e., looking in the direction of B).
(d) Force to maintain constant velocity: The induced current I = ε/R = 0.50/3.0 = 0.167 A. The force on a current-carrying wire in a magnetic field: Only the left side of the loop (length h = 0.25 m) is in the field and carrying current. F = BIh = (0.5)(0.167)(0.25) = 0.0208 N
By Lenz's law, this magnetic force opposes the motion (pulls the loop back into the field), so to maintain constant velocity, an equal and opposite external force of 0.021 N must be applied to the right.
(e) Emf during exit compared to entry: The magnitude of the induced emf during exit is the same as during entry (0.50 V). During exit, the flux is decreasing at the same rate it was increasing during entry (the loop dimensions, field strength, and velocity are the same). Faraday's law gives ε = |ΔΦ/Δt|, and the rate of change of flux has the same magnitude for entry and exit — only the sign (direction) changes. The exit emf is 0.50 V with opposite polarity.
Quick Reference: MCQ Answer Key
| Q | Ans | Q | Ans | Q | Ans | Q | Ans | Q | Ans |
|---|---|---|---|---|---|---|---|---|---|
| 1 | B | 11 | B | 21 | C | 31 | D | 41 | B |
| 2 | C | 12 | B | 22 | C | 32 | C | 42 | B |
| 3 | D | 13 | D | 23 | C | 33 | A | 43 | B |
| 4 | C | 14 | B | 24 | B | 34 | D | 44 | C |
| 5 | B | 15 | A | 25 | D | 35 | B | 45 | B |
| 6 | B | 16 | A | 26 | A | 36 | B | 46 | C |
| 7 | B | 17 | D | 27 | D | 37 | A | 47 | C |
| 8 | C | 18 | B | 28 | A | 38 | B | 48 | A |
| 9 | B | 19 | B | 29 | B | 39 | C | 49 | B |
| 10 | C | 20 | C | 30 | C | 40 | B | 50 | B |