📚 AS Physics Unit 3 Mark Scheme Jun19 Formula Derivation | AS 物理 Unit 3 Jun19 评分方案公式推导
Edexcel IAL Physics Unit 3 (WPH13) tests your ability to handle experimental data, linearise equations, and derive physical quantities from graph gradients. The June 2019 paper featured classic experiments including the simple pendulum and resistivity of a wire. The mark scheme rewards clear derivation steps, correct identification of slope as a combination of constants, and systematic uncertainty propagation. This article breaks down the key formula derivations, showing you exactly how to go from raw equations to final calculated values and their uncertainties.
Edexcel IAL 物理第三单元（WPH13）考查处理实验数据、将方程线性化以及从图像斜率推导物理量的能力。2019年6月的试卷涵盖了单摆和导线电阻率等经典实验。评分方案奖励清晰的推导步骤、正确识别斜率与常数的关系，以及系统的不确定度传递方法。本文拆解关键的公式推导，展示如何从原始方程出发，最终得出计算值及其不确定度。
1. The Simple Pendulum Equation | 单摆方程
The period T of a simple pendulum of length l is given by T = 2π √(l/g), where g is the acceleration of free fall. Because this is not a linear relationship, we cannot find g directly from a T–l graph. The necessary step is to square both sides.
长度为 l 的单摆周期 T 由 T = 2π √(l/g) 给出，其中 g 为自由落体加速度。由于这不是线性关系，我们无法直接从 T–l 图求 g。必须先将两边平方。
T² = (4π²/g) l
This is now in the form y = m x, with y = T², x = l, and gradient m = 4π²/g. The equation predicts a straight line through the origin.
这一形式为 y = m x，其中 y = T²，x = l，斜率 m = 4π²/g。该方程预图像为过原点的直线。
2. Determining g from the Gradient | 由斜率确定 g
Plot T² on the vertical axis and l on the horizontal axis. Draw the best-fit straight line and calculate its gradient m. The relationship m = 4π²/g rearranges to
将 T² 作在纵轴，l 作在横轴。画出最佳拟合直线并计算斜率 m。由 m = 4π²/g 整理得
g = 4π² / m
For example, if the best-fit line gives m = 4.05 s²/m, then g = 4π² / 4.05 ≈ 9.75 m/s². The mark scheme does not penalise small rounding differences as long as the method is clearly shown.
举例来说，若最佳拟合线斜率 m = 4.05 s²/m，则 g = 4π² / 4.05 ≈ 9.75 m/s²。只要步骤清晰，评分方案不会因微小的四舍五入差异而扣分。
3. Uncertainty in g from the Slope Uncertainty | 由斜率不确定度求 g 的不确定度
Unit 3 requires you to estimate the uncertainty in the gradient using worst-fit lines (lines passing through all error bars with the greatest or least slope). If the best gradient is m_best and the worst gradient is m_worst, then Δm = |m_best – m_worst|. Since g ∝ 1/m, the fractional uncertainty in g equals the fractional uncertainty in m:
Unit 3 要求用最差拟合线（穿过所有误差棒、斜率最大或最小的直线）估算斜率的不确定度。若最佳斜率为 m_best，最差斜率为 m_worst，则 Δm = |m_best – m_worst|。因为 g ∝ 1/m，g 的分数不确定度等于 m 的分数不确定度：
Δg/g = Δm/m
Hence Δg = g × (Δm/m). If m_best = 4.05 and m_worst = 4.20 s²/m, then Δm = 0.15 s²/m, giving Δg = 9.75 × (0.15/4.05) ≈ 0.36 m/s². The final result is expressed as g = 9.8 ± 0.4 m/s² to appropriate significant figures.
因此 Δg = g × (Δm/m)。若 m_best = 4.05、m_worst = 4.20 s²/m，则 Δm = 0.15 s²/m，Δg = 9.75 × (0.15/4.05) ≈ 0.36 m/s²。最终结果用合适的有效数字表示为 g = 9.8 ± 0.4 m/s²。
4. Combining Uncertainties in Length and Period | 长度与周期不确定度的合成
The raw measurements have their own uncertainties. A typical metre rule gives Δl = ±1 mm, while a stopwatch has a reaction‑time uncertainty of about ±0.2 s. The percentage uncertainty in T² is twice that in T because squaring doubles the fractional uncertainty:
原始测量量各有其不确定度。米尺通常给出 Δl = ±1 mm，而秒表的反应时间不确定度约为 ±0.2 s。T² 的百分不确定度是 T 的两倍，因为平方会使分数不确定度翻倍：
%U(T²) = 2 × %U(T)
If %U(l) is very small compared to %U(T²), the overall uncertainty in g is dominated by timing errors. The mark scheme expects you to identify the largest source of uncertainty.
若 %U(l) 远小于 %U(T²)，则 g 的总不确定度主要由计时误差支配。评分方案期望你能指出最大的不确定度来源。
5. The Resistivity Equation for a Wire | 导线电阻率方程
The second experiment in the June 2019 paper involved measuring the resistivity ρ of a metal wire. The resistance R of a wire of length L, cross‑sectional area A, and resistivity ρ is
2019年6月试卷的第二个实验涉及测量金属丝的电阻率 ρ。长度为 L、横截面积为 A、电阻率为 ρ 的导线的电阻为
R = ρL / A
The area for a circular wire of diameter d is A = πd²/4. Substituting this into the resistance equation yields
对于直径为 d 的圆形导线，A = πd²/4。代入电阻方程得
R = (4ρ / πd²) L
This linear relation is the key to finding ρ from a graph.
这一线性关系是从图像求 ρ 的关键。
6. Linearising R = ρL/A | 将 R = ρL/A 线性化
Since the wire has a constant diameter, the factor (4ρ/πd²) is constant. Therefore plotting R on the y‑axis against L on the x‑axis gives a straight line through the origin. The gradient k of this line is
由于导线直径恒定，因子 (4ρ/πd²) 为常数。因此以 R 为纵轴、L 为横轴作图，得到过原点的直线。该直线的斜率 k 为
k = 4ρ / πd²
Rearranging, the resistivity ρ is given by
整理得电阻率 ρ 为
ρ = k π d² / 4
This derivation must be shown clearly in your answer to meet the mark scheme requirements.
作答时必须清晰展示这一推导，才能满足评分方案的要求。
7. Calculating Resistivity from Experimental Data | 由实验数据计算电阻率
Suppose the gradient of the R–L graph is k = 1.20 Ω/m and the diameter of the wire is d = 0.50 mm = 5.0 × 10⁻⁴ m. Then
假设 R–L 图的斜率 k = 1.20 Ω/m，导线直径 d = 0.50 mm = 5.0 × 10⁻⁴ m，则
ρ = 1.20 × π × (5.0 × 10⁻⁴)² / 4 ≈ 2.36 × 10⁻⁷ Ω·m
The mark scheme accepts answers around this value, provided the unit is given in ohm‑metres (Ω·m). Use the same number of significant figures as the least precise measurement.
评分方案接受该值附近的答案，只要单位是欧姆·米（Ω·m）。使用与最不精确测量量相同的有效数字位数。
8. Propagating Uncertainties for Resistivity | 电阻率不确定度的传递
The formula ρ = k π d² / 4 shows that ρ is proportional to k and to d². Therefore the fractional uncertainty in ρ is the sum of the fractional uncertainties in k and in d, with the contribution from d doubled:
公式 ρ = k π d² / 4 表明 ρ 正比于 k 和 d²。因此 ρ 的分数不确定度是 k 和 d 的分数不确定度之和，其中 d 的贡献翻倍：
Δρ/ρ = Δk/k + 2(Δd/d)
Δk is found from worst‑fit lines, while Δd is either the micrometer reading uncertainty or the standard deviation of several diameter measurements. For instance, if Δk/k = 3% and Δd/d = 1%, then Δρ/ρ = 3% + 2×1% = 5%. Hence Δρ = 0.05 × 2.36×10⁻⁷ = 1.2×10⁻⁸ Ω·m, giving ρ = (2.36 ± 0.12)×10⁻⁷ Ω·m.
Δk 通过最差拟合线求得，Δd 则是千分尺读数不确定度或多个直径测量值的标准偏差。例如，若 Δk/k = 3%、Δd/d = 1%，则 Δρ/ρ = 3% + 2×1% = 5%。因此 Δρ = 0.05 × 2.36×10⁻⁷ = 1.2×10⁻⁸ Ω·m，最终 ρ = (2.36 ± 0.12)×10⁻⁷ Ω·m。
9. Common Graph-Plotting Errors | 作图常见错误
The June 2019 mark scheme penalises several typical mistakes: plotting T against l instead of T² against l; forcing the best‑fit line through the origin when the intercept is not zero; omitting axis labels and units; and neglecting to draw error bars. Always check if a non‑zero intercept has physical meaning – for the pendulum, it might indicate a systematic error in length measurement.
2019年6月的评分方案会对以下典型错误扣分：绘制 T–l 图而非 T²–l 图；在截距不为零时强迫最佳拟合线过原点；遗漏坐标轴标签和单位；以及未画误差棒。务必检查非零截距是否具有物理意义——对于单摆，它可能指示长度测量中的系统误差。
10. Distinguishing Systematic and Random Uncertainties | 区分系统与随机不确定度
In the pendulum experiment, a zero error on the metre rule or measuring to the bottom of the bob instead of its centre produces a systematic shift. This appears as a non‑zero intercept on the T²–l graph. Random uncertainties arise from human reaction time and cause the data points to scatter. The mark scheme expects you to discuss both types and suggest improvements (e.g., timing 20 oscillations to reduce %U in T).
在单摆实验中，米尺的零点误差或测量摆球底部而非中心，都会产生系统偏移，在 T²–l 图上表现为非零截距。随机不确定度来源于人的反应时间，导致数据点散布。评分方案期望你讨论这两种类型并提出改进措施（例如，计时20个周期以降低 T 的百分不确定度）。
11. Applying the Method to Young Modulus | 将方法应用于杨氏模量
Although not explicitly in the June 2019 paper, a similar linearisation is used for the Young modulus E. From E = (F×L)/(A×e), where e is extension, plotting F against e gives a gradient = (E×A)/L. Rearranging, E = (gradient × L) / A. The derivation steps and uncertainty propagation are identical in structure.
尽管未直接出现在2019年6月试卷中，类似的线性化方法也用于杨氏模量 E。由 E = (F×L)/(A×e)，其中 e 为伸长量，作 F–e 图得斜率 = (E×A)/L。整理得 E = (斜率 × L) / A。其推导步骤和不确定度传递在结构上完全相同。
12. Summary of Exam Technique | 应考技巧总结
To secure full marks on Unit 3 derivation questions: start from the theoretical equation, rearrange it into y = mx + c form, state what the gradient and intercept represent, plot the appropriate quantities with units, draw both best and worst lines, find the gradient with its uncertainty, and finally calculate the desired quantity with its absolute and percentage uncertainty. The June 2019 mark scheme rewards logical working, correct unit handling, and sensible significant figures.
要在 Unit 3 推导题中获得满分，请从理论方程出发，整理成 y = mx + c 的形式，说明斜率和截距的物理意义，绘制带有单位的正确物理量，画出最佳与最差拟合线，求出斜率及其不确定度，最后计算所求量及其绝对和百分不确定度。2019年6月评分方案奖励逻辑清晰的步骤、正确的单位处理和合理的有效数字。
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