📚 Pre-U OCR Physics: Case Study Practical Exercises | Pre-U OCR 物理：案例分析实战演练
Case studies in Pre-U Physics bridge the gap between theoretical principles and real-world experimental practice. You are expected not only to select and use appropriate equations, but also to evaluate uncertainties, identify systematic errors, and suggest meaningful improvements. This article works through ten targeted case studies, each designed to sharpen your ability to handle experimental data, analyse graphs, and write sound conclusions — exactly the skills tested in Component 03 and in the long-answer sections of your written papers.
Pre-U 物理中的案例分析是连接理论原理与真实实验实践的桥梁。你不仅需要选择并运用合适的公式，还要评估不确定度、识别系统误差并提出有意义的改进方案。本文通过十个有针对性的案例，每一个都旨在锻炼你处理实验数据、分析图像和写出可靠结论的能力——这正是 Component 03 和笔试长答题部分所考查的核心技能。
1. Finding g Using a Free‑Fall Timer | 自由落体计时法测 g
A student releases a steel sphere from rest at a measured height h above a switch pad, and the time of fall t is recorded. The expected relationship is h = ½ g t², so a graph of h against t² should yield a straight line with gradient ½ g. Data obtained: (t² = 0.045 s², h = 0.200 m), (0.091 s², 0.400 m), (0.138 s², 0.600 m), (0.182 s², 0.800 m). The gradient is calculated as (0.800 − 0.200) / (0.182 − 0.045) ≈ 4.38 m s⁻², giving g ≈ 8.76 m s⁻², which is about 11 % lower than the accepted 9.81 m s⁻². Systematic errors include the finite response time of the switch pad and air resistance. To improve, the student could use a photogate system to measure time directly at two positions, eliminating the initial‑release delay.
学生从静止释放一颗钢球，球下落到一个计时开关垫上方已测高度 h 处，记录下落时间 t。理论关系为 h = ½ g t²，因此绘制 h–t² 图像应得到一条斜率为 ½ g 的直线。实验数据：(t² = 0.045 s², h = 0.200 m), (0.091 s², 0.400 m), (0.138 s², 0.600 m), (0.182 s², 0.800 m)。计算斜率为 (0.800 − 0.200) / (0.182 − 0.045) ≈ 4.38 m s⁻²，得 g ≈ 8.76 m s⁻²，比公认值 9.81 m s⁻² 低约 11 %。系统误差包括开关垫的有限响应时间和空气阻力。改进方法：使用光电门在两个位置直接计时，消除起始释放延迟。
2. Determining the Young Modulus of a Wire | 测定金属丝的杨氏模量
A copper wire of original length L₀ = 2.500 m and diameter d = 0.36 mm is stretched by hanging masses. The extension ΔL is measured with a vernier scale. Stress σ = F/A and strain ε = ΔL/L₀ are computed. Plotting σ against ε gives a gradient E. One set of readings: mass = 2.00 kg, ΔL = 5.2 mm. Cross‑sectional area A = π (d/2)² = π (0.18 × 10⁻³ m)² ≈ 1.02 × 10⁻⁷ m². Stress = (2.00 × 9.81) / 1.02 × 10⁻⁷ ≈ 1.93 × 10⁸ Pa. Strain = 5.2 × 10⁻³ / 2.500 = 2.08 × 10⁻³. E = 1.93 × 10⁸ / 2.08 × 10⁻³ ≈ 9.28 × 10¹⁰ Pa. The accepted value for copper is ~1.1 × 10¹¹ Pa, so the measurement is low. Likely reasons: the wire may not be uniform, or the extension may include some slack in the clamping. Repeating with loading and unloading cycles can reduce that effect.
一根原长 L₀ = 2.500 m、直径 d = 0.36 mm 的铜丝，通过悬挂砝码被拉伸，用游标尺测量伸长量 ΔL。计算应力 σ = F/A 和应变 ε = ΔL/L₀。绘制 σ–ε 图像，斜率即为 E。一组读数：质量 = 2.00 kg，ΔL = 5.2 mm。截面积 A = π (d/2)² = π (0.18 × 10⁻³ m)² ≈ 1.02 × 10⁻⁷ m²。应力 = (2.00 × 9.81) / 1.02 × 10⁻⁷ ≈ 1.93 × 10⁸ Pa。应变 = 5.2 × 10⁻³ / 2.500 = 2.08 × 10⁻³。E = 1.93 × 10⁸ / 2.08 × 10⁻³ ≈ 9.28 × 10¹⁰ Pa。铜的公认值约为 1.1 × 10¹¹ Pa，因此测量值偏低。可能原因：金属丝不均匀，或伸长量包含了夹具的松弛。通过加载–卸载循环重复测量可减弱这种影响。
3. Internal Resistance of a Cell by the Load Method | 用负载法测电池内阻
A student connects a variable resistor R across a cell and records terminal p.d. V and current I. The equation is V = ε − I r, where ε is the e.m.f. and r the internal resistance. Data: (I = 0.10 A, V = 1.48 V), (0.30 A, 1.35 V), (0.50 A, 1.22 V), (0.70 A, 1.08 V). Plotting V against I gives a straight line with y‑intercept ≈ 1.56 V (ε) and gradient ≈ −0.69 Ω, so r ≈ 0.69 Ω. The main systematic error is the resistance of the ammeter, which slightly reduces the recorded V. Using a high‑impedance voltmeter directly across the cell terminals helps minimise this. Additionally, temperature rise in the cell during high‑current draws can alter r.
学生在电池两端连接可变电阻器 R，记录端电压 V 和电流 I。方程 V = ε − I r，其中 ε 为电动势，r 为内阻。数据：(I = 0.10 A, V = 1.48 V), (0.30 A, 1.35 V), (0.50 A, 1.22 V), (0.70 A, 1.08 V)。绘制 V–I 图像，得到 y 截距 ≈ 1.56 V (ε)，斜率 ≈ −0.69 Ω，故 r ≈ 0.69 Ω。主要系统误差是电流表的内阻，它会略微降低记录的 V 值。用高阻抗电压表直接跨接电池两端可最小化此误差。此外，大电流放电时电池内部温升也会改变 r。
4. Wavelength of Laser Light Using a Double Slit | 双缝干涉测激光波长
Laser light passes through two slits of separation a = 0.50 mm and forms interference fringes on a screen at distance D = 1.50 m. The fringe spacing Δx is measured as the distance across ten fringes. Ten fringes span 14.2 mm, so Δx = 1.42 mm. The wavelength λ = (a Δx) / D = (0.50 × 10⁻³ m × 1.42 × 10⁻³ m) / 1.50 m ≈ 4.73 × 10⁻⁷ m = 473 nm. The laser label states 470 nm ± 10 nm, so the result is within tolerance. The largest uncertainty usually comes from measuring Δx: using a travelling microscope instead of a ruler improves the precision. Also, ensure the screen is perpendicular to the beam to avoid an oblique projection that systematically widens Δx.
激光通过间距 a = 0.50 mm 的双缝，在距离 D = 1.50 m 的屏幕上形成干涉条纹。条纹间距 Δx 通过测量十根条纹的总宽度得到。十根条纹宽度为 14.2 mm，故 Δx = 1.42 mm。波长 λ = (a Δx) / D = (0.50 × 10⁻³ m × 1.42 × 10⁻³ m) / 1.50 m ≈ 4.73 × 10⁻⁷ m = 473 nm。激光标签标称 470 nm ± 10 nm，结果在允差范围内。最大不确定度通常源自 Δx 的测量：使用移测显微镜代替直尺可提高精度。同时确保屏幕与光束垂直，避免倾斜投影使 Δx 系统偏大。
5. Spring Constant and Energy Conservation | 弹簧常数与能量守恒
A mass m is attached to a vertical spring and displaced downwards. The period T is measured for oscillations, and the spring constant k is found from T = 2π √(m/k). Alternatively, by loading the spring statically, k = mg/extension. In one test, m = 0.150 kg extends the spring by 0.085 m, giving k = (0.150 × 9.81) / 0.085 ≈ 17.3 N m⁻¹. When the same mass is set into oscillation, T = 0.56 s is measured. Using the dynamic method, k = 4π² m / T² = 4π² × 0.150 / 0.56² ≈ 18.9 N m⁻¹. The discrepancy arises from the effective mass of the spring itself, which is not zero. For a spring of mass m_s, the corrected formula substitutes an effective mass m_eff = m + m_s/3. If m_s = 0.020 kg, m_eff = 0.157 kg, then k ≈ 4π² × 0.157 / 0.56² ≈ 19.8 N m⁻¹, much closer to the static value, showing the importance of accounting for spring inertia.
质量 m 悬挂在竖直弹簧下方，并向下位移。测量振荡周期 T，由 T = 2π √(m/k) 求弹簧常数 k。也可用静态加载法：k = mg/伸长量。一次测试中，m = 0.150 kg 使弹簧伸长 0.085 m，得 k = (0.150 × 9.81) / 0.085 ≈ 17.3 N m⁻¹。当同一质量做振荡时，测得 T = 0.56 s。用动态法，k = 4π² m / T² = 4π² × 0.150 / 0.56² ≈ 18.9 N m⁻¹。差异来源于弹簧自身的有效质量不为零。对于质量为 m_s 的弹簧，修正公式采用有效质量 m_eff = m + m_s/3。若 m_s = 0.020 kg，则 m_eff = 0.157 kg，得 k ≈ 4π² × 0.157 / 0.56² ≈ 19.8 N m⁻¹，与静态值更接近，表明考虑弹簧惯性十分重要。
6. Radioactive Decay Simulation and Half‑Life | 放射性衰变模拟与半衰期
Using one hundred dice to represent radioactive nuclei, a student removes any die showing a ‘6’ after each throw to simulate decay. The number remaining N after each throw models N = N₀ e^(−λ t), where t is the number of throws. Data: throws 0‑6, counts: 100, 84, 70, 58, 49, 41, 35. Plotting ln(N) against t yields a straight line. The half‑life in throws is the time for N₀ to halve: from 100 to 50 occurs between t = 3 and 4 throws, so T₁/₂ ≈ 3.5 throws. The decay constant λ = ln 2 / T₁/₂ ≈ 0.198 per throw. This simulation demonstrates the random nature of decay and the statistical fluctuations inherent when the sample is small. Increasing the number of dice would reduce the percentage fluctuation.
用一百颗骰子代表放射性原子核，学生每次投掷后移除所有出现“6”的骰子来模拟衰变。每次投掷后剩余数量 N 模拟 N = N₀ e^(−λ t)，其中 t 为投掷次数。数据：投掷 0–6 次，计数分别为 100, 84, 70, 58, 49, 41, 35。绘制 ln(N) 对 t 图像得到一条直线。衰变半衰期（以投掷次数计）是 N₀ 减半所需的时间：从 100 到 50 发生在 t = 3 至 4 次之间，故 T₁/₂ ≈ 3.5 次。衰变常数 λ = ln 2 / T₁/₂ ≈ 0.198 每投掷次。这个模拟展示了衰变的随机性以及样品较小时固有的统计涨落。增加骰子数量会减小百分比涨落。
7. Measuring the Speed of Sound Using Standing Waves | 用驻波法测声速
A loudspeaker connected to a signal generator is placed at one end of a tube, and a microphone is moved along the tube to detect nodes and antinodes. For a closed‑end tube, resonance occurs when L = (2n−1)λ/4. Frequencies are recorded for the first three resonances at f₁ = 210 Hz, f₂ = 630 Hz, f₃ = 1050 Hz. The frequency difference Δf = 420 Hz corresponds to the fundamental frequency of an open tube of the same length? Actually, for a closed tube, successive harmonics satisfy f_n = (2n−1)f₁, so Δf = 2 f₁ = 420 Hz, giving f₁ = 210 Hz, which matches. The wavelength λ = 4 L for the fundamental. If the length of the tube is L = 0.400 m, then λ = 1.60 m, and speed v = f₁ λ = 210 × 1.60 = 336 m s⁻¹. The accepted value at room temperature is about 343 m s⁻¹. The 2 % error can arise from the end correction (the antinode is slightly outside the open end) which effectively increases L. Applying an end correction of 0.3 × tube diameter improves accuracy.
扬声器连接信号发生器置于管的一端，传声器沿管移动以探测波节和波腹。对于闭管，共振条件为 L = (2n−1)λ/4。记录下前三个谐振频率：f₁ = 210 Hz，f₂ = 630 Hz，f₃ = 1050 Hz。频率差值 Δf = 420 Hz 对应同长开管的基频？实际上，闭管相邻谐波满足 f_n = (2n−1)f₁，故 Δf = 2 f₁ = 420 Hz，得 f₁ = 210 Hz，相符。基频波长 λ = 4 L。若管长 L = 0.400 m，则 λ = 1.60 m，声速 v = f₁ λ = 210 × 1.60 = 336 m s⁻¹。室温下的公认值约为 343 m s⁻¹。2% 的误差可归因于末端效应（波腹实际位于管口外略远处），这相当于有效增大 L。施加 0.3 倍管径的末端修正可提高准确度。
8. Capacitor Discharge and Time Constant | 电容放电与时间常数
A 100 μF capacitor is charged to 6.0 V and discharged through a 47 kΩ resistor. The p.d. V across the capacitor is recorded every 2 s: (0 s, 6.0 V), (2 s, 4.05 V), (4 s, 2.72 V), (6 s, 1.83 V), (8 s, 1.23 V). The time constant τ = RC = 47 × 10³ Ω × 100 × 10⁻⁶ F = 4.7 s. The discharge equation V = V₀ e^(−t/RC) leads to ln(V) = ln(V₀) − t/RC. Plotting ln(V) against t gives a gradient of −1/RC. From data, ln(V) values: 1.79, 1.40, 1.00, 0.60, 0.21. The gradient ≈ (0.21 − 1.79) / (8 − 0) = −0.1975 s⁻¹, so τ_measured = 1/0.1975 ≈ 5.06 s, slightly higher than 4.7 s. The discrepancy indicates either the capacitor has a larger actual capacitance (tolerance of electrolytics is often ±20 %) or the voltmeter’s input impedance is not infinite, altering the effective resistance.
一个 100 μF 电容充电至 6.0 V，通过 47 kΩ 电阻放电。每 2 秒记录电容两端电压 V：(0 s, 6.0 V), (2 s, 4.05 V), (4 s, 2.72 V), (6 s, 1.83 V), (8 s, 1.23 V)。时间常数 τ = RC = 47 × 10³ Ω × 100 × 10⁻⁶ F = 4.7 s。放电方程 V = V₀ e^(−t/RC) 推导出 ln(V) = ln(V₀) − t/RC。绘制 ln(V) 对 t 图像，斜率为 −1/RC。由数据得 ln(V) 值：1.79, 1.40, 1.00, 0.60, 0.21。斜率 ≈ (0.21 − 1.79) / (8 − 0) = −0.1975 s⁻¹，因此 τ_measured = 1/0.1975 ≈ 5.06 s，略高于 4.7 s。差异表明电容实际容量可能偏大（电解电容容差常为 ±20 %），亦或电压表输入阻抗并非无穷大，改变了等效电阻。
9. Lens Formula Verification with an Optical Bench | 光具座上透镜公式验证
A convex lens of focal length ~10 cm is used. Object distance u and image distance v are measured for several positions, and 1/u and 1/v are calculated. Data: (u = 15.0 cm, v = 30.0 cm), (20.0 cm, 20.0 cm), (25.0 cm, 16.7 cm), (30.0 cm, 15.0 cm). The lens formula 1/f = 1/u + 1/v is tested. For each pair, compute 1/f: for (15, 30), 1/f = 1/15 + 1/30 = 0.100 cm⁻¹, so f = 10.0 cm. The other three pairs give 0.100 cm⁻¹ as well, verifying the formula. A graph of 1/v against 1/u is a straight line with gradient −1 and intercepts on both axes equal to 1/f. The main source of uncertainty is locating the sharpest image; using a frosted screen and a parallax‑free method reduces this. The lens should be placed so that its principal axis is aligned with the screen centre to minimise off‑axis aberration.
使用一块焦距约 10 cm 的凸透镜。测量不同位置下的物距 u 和像距 v，并计算 1/u 和 1/v。数据：(u = 15.0 cm, v = 30.0 cm), (20.0 cm, 20.0 cm), (25.0 cm, 16.7 cm), (30.0 cm, 15.0 cm)。检验透镜公式 1/f = 1/u + 1/v。对每组计算 1/f：对于 (15, 30)，1/f = 1/15 + 1/30 = 0.100 cm⁻¹，故 f = 10.0 cm。其余三组均给出 0.100 cm⁻¹，验证了公式。绘制 1/v 对 1/u 图像，得到一条斜率为 −1、两轴截距均为 1/f 的直线。主要不确定度来自最清晰成像的判定；使用毛玻璃屏和消视差法可降低该误差。透镜的主光轴应与屏幕中心对齐，以降低离轴像差。
10. Specific Heat Capacity of Aluminium by Electrical Heating | 电热法测铝的比热容
A 0.500 kg aluminium block with an embedded heater and thermometer is supplied with electrical power P = V I = 10.0 V × 2.00 A = 20.0 W for t = 600 s. The temperature rises from θ₁ = 21.5 °C to θ₂ = 33.5 °C. The thermal energy supplied is Q = P t = 20.0 × 600 = 12000 J. The specific heat capacity c = Q / (m Δθ) = 12000 / (0.500 × 12.0) = 2000 J kg⁻¹ K⁻¹. The accepted value is 897 J kg⁻¹ K⁻¹, a very large discrepancy. The main error is heat loss to the surroundings: without insulation, most of the electrical energy is dissipated. To improve, the block should be wrapped in lagging, and the temperature recorded only after the heater is switched off to avoid direct heating of the thermometer. A correction by cooling‑curve extrapolation can also be applied.
一块 0.500 kg 的铝块内置加热器和温度计，输入电功率 P = V I = 10.0 V × 2.00 A = 20.0 W，通电 t = 600 s。温度从 θ₁ = 21.5 °C 升至 θ₂ = 33.5 °C。提供的热能 Q = P t = 20.0 × 600 = 12000 J。比热容 c = Q / (m Δθ) = 12000 / (0.500 × 12.0) = 2000 J kg⁻¹ K⁻¹。铝的公认值为 897 J kg⁻¹ K⁻¹，误差极大。主要错误是向周围环境散热：无保温时，大部分电能散失。改进方法：用隔热层包裹铝块，并在停止加热后记录温度，避免加热器直接烘烤温度计。还可通过冷却曲线的外推进行修正。
11. Oscilloscope Measurement of an Unknown Frequency | 示波器测未知频率
A dual‑trace oscilloscope displays a known 500 Hz sine wave on channel A and an unknown signal on channel B. By adjusting the time base until the 500 Hz wave shows exactly one complete cycle over 4 horizontal divisions, the time‑base setting is verified as 0.5 ms/div. The unknown wave shows 2.5 cycles over the same 4 divisions, so its period is (4 × 0.5 ms) / 2.5 = 0.8 ms, and its frequency f = 1/(0.8 × 10⁻³) = 1250 Hz. To improve accuracy, the student can use Lissajous figures: feed the known frequency to the X‑input and the unknown to the Y‑input, then adjust until a stable 1:1 ellipse appears, or use the pattern’s shape to determine the exact frequency ratio. Care must be taken to ensure the signals are not over‑driven, causing clipping.
双踪示波器在通道 A 显示已知的 500 Hz 正弦波，通道 B 显示未知信号。调节时基，使 500 Hz 波形在 4 个水平格内恰好显示一个完整周期，验证时基设置为 0.5 ms/格。未知信号在相同 4 格内显示 2.5 个周期，因此其周期为 (4 × 0.5 ms) / 2.5 = 0.8 ms，频率 f = 1/(0.8 × 10⁻³) = 1250 Hz。为提高准确度，学生可使用李萨如图形：将已知频率输入 X 输入端，未知频率输入 Y 输入端，调节至出现稳定的 1:1 椭圆，或根据图形确定精确的频率比。必须注意信号不要过激励，以免削波。
12. Systematic Error in a Micrometer and Zero Correction | 螺旋测微器的系统误差与零位校正
A student notices that when the micrometer is fully closed, the barrel scale reads +0.03 mm instead of 0.00 mm. This zero error must be subtracted from all readings. The diameter of a thin wire is measured five times in different places: 0.42, 0.44, 0.43, 0.42, 0.44 mm (after zero correction). The mean diameter d̄ = 0.430 mm, and the random uncertainty can be estimated as half the range = (0.44 − 0.42)/2 = 0.01 mm. Thus, the result is d = 0.43 ± 0.01 mm. The cross‑sectional area A = π d²/4, and the percentage uncertainty in A is twice that in d (≈ 2 × 2.3 % ≈ 4.6 %). Always check zero error before and after a measurement session, as thermal expansion from handling can introduce a drift.
学生发现螺旋测微器完全闭合时，套筒刻度读数为 +0.03 mm 而非 0.00 mm。该零位误差必须从所有读数中扣除。一根细丝的直径在不同位置测量五次（已做零位修正）：0.42, 0.44, 0.43, 0.42, 0.44 mm。平均直径 d̄ = 0.430 mm，随机不确定度可用半量程估计 = (0.44 − 0.42)/2 = 0.01 mm。因此结果为 d = 0.43 ± 0.01 mm。截面积 A = π d²/4，A 的百分不确定度是 d 的两倍（≈ 2 × 2.3 % ≈ 4.6 %）。测量前后务必检查零位误差，因为操作过程中的热膨胀可能引起零位漂移。
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