📚 AP Physics 2: Free-Response Questions (FRQ) Analysis | AP 物理2：真题（FR）解析
AP Physics 2 free-response questions challenge students to apply conceptual understanding, mathematical reasoning, and experimental design skills. This analysis walks through typical FRQ topics, offering step-by-step breakdowns of common question types, worked examples, and strategies to maximize your score.
AP 物理2 的自由回答题（FRQ）考查学生综合运用概念理解、数学推理和实验设计的能力。本文将通过典型 FRQ 主题，逐步解析常见题型，提供范例解答与高分策略。
1. AP Physics 2 FRQ Overview | AP物理2 FRQ 总览
The exam includes four free-response questions: one experimental design, one quantitative/qualitative translation, and two short-answer questions. You must explain physics principles, derive algebraic expressions, interpret graphs, and justify reasoning clearly.
考试包含四道自由回答题：一道实验设计题，一道定量/定性转换题，以及两道简答题。你需要清晰地解释物理原理、推导代数表达式、解读图像并论证推理。
FRQs are graded on a rubric that awards points for correct physics, consistent logic, and proper use of scientific notation. Partial credit is possible, so showing your work is essential.
FRQ 按评分标准给分，正确的物理概念、连贯的逻辑和规范的符号使用都能得分。允许部分得分，因此展示解题过程至关重要。
2. Fluids: Buoyancy and Continuity | 流体：浮力与连续性方程
Sample FRQ context: A wooden block of mass 0.50 kg and volume 8.0×10⁻⁴ m³ is placed in water (ρ = 1000 kg/m³). A metal piece of mass 0.20 kg is gently added on top. Determine whether the block sinks and calculate the submerged volume in each case.
FRQ 案例：质量为 0.50 kg、体积为 8.0×10⁻⁴ m³ 的木块放入水中（ρ = 1000 kg/m³）。将质量为 0.20 kg 的金属块轻轻放在木块顶部。判断木块是否沉没，并计算每种情况下木块浸没的体积。
Analysis: First, compare the block’s density with the fluid. Density of wood = m/V = 0.50 / (8.0×10⁻⁴) = 625 kg/m³, which is less than water, so it floats. When metal is added, the total mass is 0.70 kg. The maximum buoyant force equals ρ_water × V_block × g. The weight of the system is (0.70)g.
解析：首先，比较木块密度与流体密度。木块密度 = m/V = 0.50 / (8.0×10⁻⁴) = 625 kg/m³，小于水的密度，因此木块漂浮。加上金属块后，总质量为 0.70 kg。最大浮力为 ρ_water × V_block × g。系统总重为 0.70g。
Max F_b = 1000 × 8.0×10⁻⁴ × 9.8 ≈ 7.84 N, weight = 0.70 × 9.8 = 6.86 N
Since buoyant force can exceed weight, the block still floats. The submerged volume is found from F_b = weight: ρ_water × V_sub × g = (0.70)g → V_sub = 0.70 / 1000 = 7.0×10⁻⁴ m³. Thus, 7/8 of the block is underwater.
由于最大浮力可大于重力，木块仍漂浮。由浮力等于重力得浸没体积：V_sub = 0.70 / 1000 = 7.0×10⁻⁴ m³，即木块的 7/8 没入水中。
3. Thermodynamics: PV Processes and Efficiency | 热力学：PV 过程与效率
Typical FRQ: An ideal gas undergoes a cyclic process ABCA: A→B is isothermal expansion, B→C is isochoric cooling, C→A is adiabatic compression. Given P_A, V_A, and V_B = 2V_A, find the net work done and the thermal efficiency.
典型 FRQ：理想气体经历循环过程 ABCA：A→B 等温膨胀，B→C 等容降温，C→A 绝热压缩。已知 P_A, V_A, V_B = 2V_A，求净功及热效率。
The work for isothermal process: W_AB = nRT ln(V_B/V_A) = P_A V_A ln(2). For isochoric, W_BC = 0. For adiabatic, W_CA = -ΔU_CA, where ΔU = (3/2)nR(T_A – T_C). The net work is the area enclosed by the cycle on a PV diagram.
等温过程做功：W_AB = nRT ln(V_B/V_A) = P_A V_A ln(2)。等容过程 W_BC = 0。绝热过程 W_CA = -ΔU_CA，其中 ΔU = (3/2)nR(T_A – T_C)。净功等于 PV 图上循环所围面积。
Efficiency e = W_net / Q_absorbed. Heat is only absorbed during isothermal expansion: Q_AB = W_AB. Thus, e = W_net / (P_A V_A ln 2). Express all quantities in terms of P_A, V_A, and the adiabatic relation P_A V_A^γ = P_C V_C^γ to solve for temperatures.
效率 e = W_net / Q_吸入。仅等温膨胀过程吸热：Q_AB = W_AB。因此 e = W_net / (P_A V_A ln 2)。利用绝热关系 P_A V_A^γ = P_C V_C^γ 以及 V_C = V_A 可解出温度。
4. Electrostatics: Point Charges and Gauss’s Law | 静电场：点电荷与高斯定律
FRQ scenario: Three point charges are fixed at the vertices of an equilateral triangle. Calculate the net electric field at the center and the electric potential energy of the configuration. Discuss how Gauss’s law applies to a spherical surface enclosing these charges.
FRQ 场景：三个点电荷固定在等边三角形的顶点。求三角形中心处的合电场强度及该体系的电势能。讨论包围这些电荷的球面高斯定律如何应用。
Electric field is a vector sum; due to symmetry, the field at the center may cancel or add depending on charge signs. For three identical charges, the field at the centroid is zero. For different values, calculate the components: E_i = k|q|/r², direction from charge toward point. r = side/√3 for center.
电场为矢量和；根据对称性，中心处电场可能为零或非零。若三个电荷相同，中心场强为零。若不同，需计算分量：E_i = k|q|/r²，方向由电荷指向中心点。中心到顶点的距离 r = 边长/√3。
r = s/√3
Potential energy U = k(q₁q₂/r₁₂ + q₁q₃/r₁₃ + q₂q₃/r₂₃). Use r = side for all pairs. Gauss’s law states Φ = Q_enclosed/ε₀. The flux depends only on the net enclosed charge; surface shape does not matter.
电势能 U = k(q₁q₂/r₁₂ + q₁q₃/r₁₃ + q₂q₃/r₂₃)。由于等边三角形，每对电荷间距均为边长。高斯定律 Φ = Q_enclosed/ε₀，电通量只取决于封闭面内的净电荷，与面形状无关。
5. Circuits: RC Time Constants and Kirchhoff’s Rules | 电路：RC 时间常数与基尔霍夫定律
FRQ example: A capacitor C is initially charged to V₀ and then connected in series with a resistor R. Derive the expression for charge as a function of time and the time constant τ. Then explain how to use Kirchhoff’s loop rule to find the current.
FRQ 示例：电容 C 初始充电至 V₀，然后与电阻 R 串联构成回路。推导电荷随时间变化的表达式及时间常数 τ。并说明如何用基尔霍夫回路定则求电流。
By Kirchhoff’s loop rule: -q/C – iR = 0, but during discharge, i = -dq/dt, giving q/C + R dq/dt = 0. Solving the differential equation yields q(t) = Q₀ e^{-t/RC}, where τ = RC. Current i(t) = (V₀/R) e^{-t/τ}.
由基尔霍夫回路定则：放电过程中，q/C + R dq/dt = 0（其中 i = -dq/dt）。解微分方程得 q(t) = Q₀ e^{-t/RC}，时间常数 τ = RC。电流 i(t) = (V₀/R) e^{-t/τ}。
| t | q(t)/Q₀ | V across C |
| 0 | 1 | V₀ |
| τ | 0.37 | 0.37V₀ |
| 5τ | 0.0067 | ~0 |
Students must be able to sketch discharge graphs and interpret the half-life t_½ = τ ln 2.
考生须能画出放电曲线并理解半衰期 t_½ = τ ln 2。
6. Magnetism: Charged Particles in Fields | 磁学：电荷在磁场中的运动
FRQ task: A proton enters a uniform magnetic field B perpendicular to its velocity v. Determine the radius of its circular path, the period, and explain how the trajectory changes if the field also has a parallel component.
FRQ 任务：质子以速度 v 垂直进入匀强磁场 B。求其圆周运动半径、周期，并解释若磁场有平行分量，轨迹如何变化。
Magnetic force F = qvB provides centripetal force: qvB = mv²/r → r = mv/(qB). Period T = 2πr/v = 2πm/(qB), which is independent of speed. If velocity has a parallel component, the path becomes a helix with constant pitch.
磁力 F = qvB 提供向心力：qvB = mv²/r → r = mv/(qB)。周期 T = 2πm/(qB)，与速率无关。若速度有平行磁场分量，轨迹变为螺旋线，螺距恒定。
r = (m v_perp) / (q B), T = 2πm/(q B)
FRQs often ask to derive the work done by the magnetic field: zero, because the force is always perpendicular to displacement.
FRQ 常要求证明磁场对运动电荷做功为零，因为磁力始终垂直于位移。
7. Geometric Optics: Ray Tracing and the Lens Equation | 几何光学：光线追迹与透镜公式
FRQ scenario: A converging lens with focal length f = 10 cm is placed 30 cm from an object. Use ray tracing to locate the image. Then calculate image distance and magnification using the thin-lens equation. Describe the image characteristics.
FRQ 场景：焦距 f = 10 cm 的会聚透镜置于物体前方 30 cm 处。用光线追迹法定位像，再利用薄透镜公式计算像距和放大率，并描述像的性质。
Three principal rays: a ray parallel to the axis passes through the focal point on the other side; a ray through the center goes straight; a ray through the near focal point emerges parallel. Ray tracing gives a real, inverted image beyond 2F.
三条主光线：平行于主轴的光线经透镜后过另一侧焦点；过光心的光线直线传播；过物方焦点的光线出射后平行于主轴。光路图显示在 2F 之外形成倒立实像。
Lens equation: 1/f = 1/d_o + 1/d_i → 1/10 = 1/30 + 1/d_i → d_i = 15 cm. Magnification m = -d_i/d_o = -15/30 = -0.5. Image is real, inverted, reduced.
透镜公式：1/10 = 1/30 + 1/d_i → d_i = 15 cm。放大率 m = -15/30 = -0.5。像为倒立、缩小的实像。
8. Modern Physics: Photoelectric Effect and Atomic Spectra | 现代物理：光电效应与原子光谱
FRQ task: Light of frequency f is incident on a metal surface with work function φ. Derive Einstein’s photoelectric equation and sketch the stopping potential vs. frequency graph. Explain how to extract Planck’s constant and the work function from the graph.
FRQ 任务：频率为 f 的光照射在功函数为 φ 的金属表面。推导爱因斯坦光电方程，画出遏止电压随频率变化的关系图，并说明如何从图中得出普朗克常量和功函数。
Photon energy hf = K_max + φ, and e V_stop = K_max. So V_stop = (h/e)f – φ/e. The slope = h/e, and intercept = -φ/e. This is a linear relation. FRQs require plotting data, calculating slope, and interpreting results.
光子能量 hf = K_max + φ，且 e V_stop = K_max。故 V_stop = (h/e)f – φ/e。斜率为 h/e，截距为 -φ/e，呈线性关系。FRQ 要求画数据图、计算斜率并解释结果。
Atomic spectra: Energy level transitions produce discrete photon energies. The Balmer series involves transitions ending at n=2. FRQs provide energy levels and ask to calculate wavelength of emission lines: E_photon = E_i – E_f = hc/λ.
原子光谱：能级跃迁产生分立光子能量。巴尔末系终态 n=2。题目给出能级，要求计算发射谱线波长：E_photon = E_i – E_f = hc/λ。
9. Experimental Design: Graphing and Error Analysis | 实验设计：作图与误差分析
FRQ example: Design an experiment to determine the refractive index of a glass block using Snell’s law. Specify procedure, variables, data table, graph, and how to analyze uncertainties.
FRQ 示例：设计一个实验，利用斯涅尔定律测定玻璃砖的折射率。说明步骤、变量、数据表格、图表及不确定度分析。
Measure incident angle θ₁ and refracted angle θ₂ for several values. Plot sin θ₁ vs. sin θ₂. The slope equals the refractive index n = sin θ₁ / sin θ₂. Use a protractor and ray box; estimate uncertainty in angle (±0.5°). Propagate to slope using max-min lines.
测量多组入射角 θ₁ 和折射角 θ₂。作 sin θ₁ – sin θ₂ 图，斜率即为折射率 n = sin θ₁ / sin θ₂。用量角器和光源箱；估计角度不确定度（±0.5°），用最大/最小梯度线传递不确定度。
Points are awarded for clearly identifying independent/dependent variables, describing how to control others, and stating how to linearize the data. Always include a well-labeled diagram.
明确自变量、因变量，说明控制变量方法，指出如何将数据线性化，并附有清晰标注的示意图，方可得分。
10. Strategies and Common Mistakes | 备考策略与常见错误
Key advice: Read the entire question before starting. Note the verbs: “derive” means show mathematical steps; “justify” means provide physics reasoning; “calculate” means numerical answer with units. Never leave a question blank—write a starting equation or diagram.
关键建议：答题前通读整道题。注意指令词：“derive”（推导）要求展示数学步骤；“justify”（论证）要求给出物理推理；“calculate”（计算）要求带单位数值答案。切勿空题，可写开始方程或示意图。
Common mistakes: confusing electric potential with electric potential energy, forgetting to convert units, misapplying the right-hand rule, using the lens equation sign conventions incorrectly, and omitting directions for vector quantities. Practice with official FRQs and review scoring guidelines to internalize the rubric.
常见错误：混淆电势与电势能、单位换算遗漏、右手定则用错、透镜公式符号规则应用不当、矢量缺少方向。使用官方 FRQ 真题练习并研读评分标准，内化得分要点。
Time management: Allocate about 25 minutes per question. Sketch graphs before making final plots. Label axes, show slopes, and use linearization techniques (e.g., plot T² vs. L for a pendulum to find g).
时间管理：每道题分配约25分钟。先画草图再正式绘图。标注坐标轴，展示斜率，运用线性化技巧（如绘制 T²–L 图求重力加速度 g）。
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