A-Level Physics Unit 3 Jan 20 Formula Derivation | A-Level物理单元三 2020年1月试卷公式推导

📚 A-Level Physics Unit 3 Jan 20 Formula Derivation | A-Level物理单元三 2020年1月试卷公式推导

In the January 2020 Unit 3 paper for International A-Level Physics, the ability to manipulate experimental equations and derive meaningful physical quantities from a linear graph was heavily tested. This article revisits a classic example – determining the Young modulus of a metal wire – and shows step by step how to derive the necessary expression, construct an appropriate straight-line graph, and propagate uncertainties correctly. Mastering this derivation will not only help you tackle similar questions but also strengthen your grasp of practical skills needed for high marks.

在2020年1月的国际A-Level物理单元三试卷中，从实验方程推导物理量并利用直线图求解的能力是考查重点。本文将通过一个经典例子——测定金属丝的杨氏模量——逐步展示如何推导所需表达式、建立合适的直线图以及正确传递不确定度。掌握这一推导过程不仅能帮助你应对类似题目，还能夯实获得高分所需的实验技能。

1. Understanding the Exam Focus | 理解考试重点

Unit 3 papers assess your competence in experimental planning, data analysis, and evaluation. A typical question provides raw measurements and asks you to derive an equation that allows a straight-line plot. From the gradient or intercept you then calculate a physical constant. The January 2020 paper featured a problem where candidates had to determine the Young modulus E of a copper wire by measuring its extension under different loads. The derivation of the linear relationship was essential.

单元三试卷考查实验设计、数据分析和评估能力。典型的题目会给出原始测量数据，要求推导出可用于绘制直线图的方程，然后由斜率或截距计算物理常数。2020年1月的试卷中就有一道题，要求通过测量铜丝在不同负载下的伸长量，测定其杨氏模量E。推导线性关系是解题核心。

2. Experimental Setup for Young Modulus | 杨氏模量实验装置

A long thin wire is clamped at one end and passes over a pulley at the other. Weights are added to the free end to apply a force F = mg. The original length L₀ is measured with a metre rule, and the diameter d is measured at several points using a micrometer. The extension ΔL is recorded for at least six different loads, ensuring the elastic limit is not exceeded.

长细金属丝一端固定，另一端绕过滑轮，通过在自由端添加砝码施加拉力F = mg。原长L₀用米尺测量，直径d用千分尺在多点测量。至少记录六组不同负载下的伸长量ΔL，并确保不超出弹性限度。

3. Fundamental Definition of Young Modulus | 杨氏模量的基本定义

Young modulus E is defined as the ratio of tensile stress to tensile strain within the elastic region. Mathematically:

E = stress / strain = (F/A) / (ΔL / L₀)

杨氏模量E定义为弹性范围内拉伸应力与拉伸应变之比。数学表达式为：

E = 应力 / 应变 = (F/A) / (ΔL / L₀)

Here F is the applied force, A is the cross-sectional area of the wire, ΔL is the extension, and L₀ is the original length.

其中F是施加的力，A是金属丝的横截面积，ΔL是伸长量，L₀是原始长度。

4. Expressing Cross-Sectional Area | 表达横截面积

For a wire with a circular cross-section, the area A is given by:

A = π d² / 4

对于圆形截面的金属丝，横截面积A为：

A = π d² / 4

Substituting this into the stress formula yields: stress = F / (π d² / 4) = 4F / (π d²).

将其代入应力公式得到：应力 = F / (π d² / 4) = 4F / (π d²)。

5. Deriving the Linear Graph Equation | 推导线性图方程

To obtain a straight-line graph, we rearrange the definition so that one variable is proportional to another. Starting from E = (F/A) / (ΔL / L₀), we can write:

为了得到直线图，我们需要重新整理定义式，使一个变量与另一个变量成正比。从 E = (F/A) / (ΔL / L₀) 出发，可得：

E = (F L₀) / (A ΔL)

Then multiply both sides by ΔL and divide by E to isolate ΔL:

然后两边乘以ΔL并除以E，解出ΔL：

ΔL = (L₀ / (A E)) × F

Since L₀, A, and E are constants for a given wire within the elastic limit, ΔL is directly proportional to the applied force F. Replacing A with πd²/4 gives:

由于在弹性限度内，对于给定的金属丝，L₀、A和E均为常数，因此ΔL与施加的力F成正比。将A替换为πd²/4得到：

ΔL = [4L₀ / (π d² E)] × F

This is the equation of a straight line passing through the origin, with gradient m = 4L₀ / (π d² E).

这是一个过原点的直线方程，斜率 m = 4L₀ / (π d² E)。

6. Calculating Young Modulus from the Graph | 从图表计算杨氏模量

If we plot ΔL on the y-axis and F on the x-axis, the best-fit line should pass through the origin. The gradient m can be determined from the graph. Then we rearrange to find E:

若将ΔL作为y轴，F作为x轴作图，最佳拟合线应过原点。可由图求出斜率m，然后重新整理求E：

E = 4L₀ / (π d² m)

Be careful to use consistent SI units: L₀ in metres, d in metres, m in m N⁻¹, and E will be in Pa.

需注意统一使用国际单位：L₀以米为单位，d以米为单位，m以米每牛顿(m N⁻¹)为单位，则E的单位为帕斯卡(Pa)。

7. Uncertainty Propagation in the Derived Formula | 推导公式中的不确定度传播

The Jan 20 paper often requires you to calculate the percentage uncertainty in E. Assuming independent measurements, the fractional uncertainty in E is obtained by adding the relative uncertainties of the factors, with the exponent of each factor multiplied. For E = 4L₀ / (π d² m):

2020年1月试卷通常要求计算E的百分比不确定度。假设各测量相互独立，E的相对不确定度由各因子的相对不确定度相加得到，每个因子的指数需乘入。对于 E = 4L₀ / (π d² m)：

ΔE / E = ΔL₀ / L₀ + 2(Δd / d) + Δm / m

The constant 4/π has no uncertainty. The diameter appears squared, so its relative uncertainty is doubled. The gradient uncertainty Δm is found from the difference between the worst-acceptable line and best-fit line, or the standard error if available. Multiply the fractional uncertainty by 100 to get percentage uncertainty.

常数4/π没有不确定度。直径以平方形式出现，因此其相对不确定度加倍。斜率不确定度Δm可由最差可接受线与最佳拟合线之差求得，若有标准误差也可使用。将相对不确定度乘以100即得百分比不确定度。

8. Worked Example Simulating Jan 20 Data | 模拟2020年1月试卷数据的工作实例

Imagine a student obtained the following data for a steel wire: L₀ = 2.000 ± 0.002 m, d = 0.500 ± 0.010 mm. A series of forces were applied and the extensions measured, producing a ΔL vs F graph. The gradient of the best-fit line was found to be m = 0.0250 mm N⁻¹ ± 0.0010 mm N⁻¹. Convert everything to metres: d = 0.500 × 10⁻³ m, m = 0.0250 × 10⁻³ m N⁻¹ = 2.50 × 10⁻⁵ m N⁻¹.

假设某学生对钢丝测得以下数据：L₀ = 2.000 ± 0.002 m，d = 0.500 ± 0.010 mm。施加一系列力并测量伸长量，绘制出ΔL-F图。最佳拟合线斜率 m = 0.0250 mm N⁻¹ ± 0.0010 mm N⁻¹。将所有数据转换为米：d = 0.500 × 10⁻³ m，m = 0.0250 × 10⁻³ m N⁻¹ = 2.50 × 10⁻⁵ m N⁻¹。

First calculate E:

首先计算E：

E = 4 × 2.000 / (π × (0.500×10⁻³)² × 2.50×10⁻⁵) = 8.000 / (π × 2.50×10⁻⁷ × 2.50×10⁻⁵) = 8.000 / (π × 6.25×10⁻¹²) ≈ 4.07×10¹¹ Pa

Now uncertainties: ΔL₀/L₀ = 0.002/2.000 = 0.001; Δd/d = 0.010/0.500 = 0.02; Δm/m = 0.0010/0.0250 = 0.04. Thus ΔE/E = 0.001 + 2×0.02 + 0.04 = 0.001 + 0.04 + 0.04 = 0.081, so percentage uncertainty = 8.1%. Absolute uncertainty ΔE ≈ 0.081 × 4.07×10¹¹ = 3.3×10¹⁰ Pa; the result can be quoted as (4.1 ± 0.3) × 10¹¹ Pa.

现在计算不确定度：ΔL₀/L₀ = 0.002/2.000 = 0.001；Δd/d = 0.010/0.500 = 0.02；Δm/m = 0.0010/0.0250 = 0.04。因此ΔE/E = 0.001 + 2×0.02 + 0.04 = 0.081，百分比不确定度为8.1%。绝对不确定度ΔE ≈ 0.081 × 4.07×10¹¹ = 3.3×10¹⁰ Pa；结果可表示为 (4.1 ± 0.3) × 10¹¹ Pa。

9. Common Errors and How to Avoid Them | 常见错误及避免方法

A frequent mistake is forgetting to convert the diameter from mm to m before computing area, leading to an error of factor 10⁶. Another is using the wrong pair of variables for the graph – some students plot force against extension, but then the gradient becomes A E / L₀, which is equally valid but requires careful rearrangement. Always check that the derived gradient expression matches your axis labels. Also, if the extension axis does not start at zero due to an initial tightening error, the intercept should be analysed rather than forced through zero; the question may specify whether the line should pass through the origin.

一个常见错误是在计算面积前忘记将直径从毫米转换为米，这将导致10⁶倍的误差。另一个错误是图形变量选取不当——有些学生绘制力-伸长量图，此时斜率变为A E / L₀，虽然同样有效但需要仔细重整。务必确保推导出的斜率表达式与坐标轴标签一致。另外，若由于初始拉紧误差导致伸长量轴不始于零，则应分析截距，而不应强制过原点；题目可能会明确说明直线是否应过原点。

10. Alternative Derivation: Using a Log-Log Plot | 替代推导：使用双对数图

Sometimes the question may test understanding of logarithmic relationships. For instance, if the relationship were ΔL = k Fⁿ, taking logs gives log(ΔL) = log k + n log F. The gradient of a log-log plot then gives the exponent n. However, for the Young modulus investigation, the expected linear relationship is a direct proportion, so a simple ΔL vs F graph suffices. Being able to derive the linear form in both cases demonstrates strong analytical skills.

有时题目会考查对对数关系的理解。例如，若关系式为ΔL = k Fⁿ，取对数后可得 log(ΔL) = log k + n log F。双对数图的斜率即为指数n。但在杨氏模量实验中，预期为直接正比关系，简单的ΔL-F图即可满足。能在两种情况下推导线性形式，可展现扎实的分析能力。

11. Summary of Key Steps | 关键步骤总结

Start from the definition equation and substitute geometric quantities.
从定义方程入手，代入几何量。
Rearrange to express the measured variable (here ΔL) as a function of the controlled variable (F) in the form y = mx + c.
重整方程，将测量量（此处为ΔL）表示为控制变量（F）的函数，写成 y = mx + c 的形式。
Identify the gradient and relate it to the desired constant E.
明确斜率，并将其与所求常数E关联。
Use the best-fit gradient and the measurements of L₀ and d to calculate E.
利用最佳拟合斜率及L₀、d的测量值计算E。
Propagate uncertainties using the formula derived from the expression for E.
利用根据E的表达式推导出的公式进行不确定度传递。

12. Final Tips for Unit 3 Success | 单元3成功的最后提示

Always annotate your derivation steps clearly in the exam. Show the substitution, the rearrangement, and the final linear equation. Label your graph axes with the correct quantities and units, and write the gradient expression next to the graph. When calculating uncertainties, quote the percentage and absolute uncertainty with the correct number of significant figures. With disciplined practice of derivations like the one described, you can approach the January 2020 and similar papers with confidence.

考试时务必清晰注释推导步骤。写出代入、整理和最终的线性方程。在坐标轴上标记正确的物理量及单位，并在图旁写明斜率表达式。计算不确定度时，以正确的有效数字报告百分比与绝对不确定度。通过有素的练习，如上述推导所示，你将能自信应对2020年1月及类似试卷。

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A-Level Physics Unit 3 Jan 20 Formula Derivation | A-Level物理单元三 2020年1月试卷公式推导