📚 A-Level Physics June 2018 Paper 3 Formula Derivation | A-Level物理2018年6月卷3公式推导
The June 2018 Paper 3 for A-Level Physics often challenges students with data analysis and formula manipulation in a practical context. Whether you are handling a pendulum, free-fall, or circuit experiment, being able to derive and linearise the relevant equations is crucial for achieving high marks. This article walks you through the most common formula derivations that appeared in or are relevant to the June 2018 Paper 3, with clear steps and uncertainty treatment.
2018年6月的A-Level物理卷3常以实验情境考查数据处理和公式推导，令许多学生感到棘手。无论是处理单摆、自由落体还是电路实验，能够推导并将方程线性化是取得高分的关键。本文将带你一步步梳理与2018年6月卷3相关的最常见公式推导，包括清晰的步骤和不确定度处理方法。
1. The Role of Paper 3 in A-Level Physics | 卷3在A-Level物理中的角色
Paper 3 typically assesses practical skills, experimental design, data presentation, and evaluation. In many examination boards, June 2018 Paper 3 required candidates to derive straight-line relationships from fundamental physical laws, plot suitable graphs, and calculate quantities like gravitational acceleration g, internal resistance r, or Young modulus E from the gradient or intercept.
卷3通常评估实验技能、实验设计、数据呈现与评估。在多个考试局中，2018年6月的卷3要求考生从基本物理定律推导出直线关系，绘制合适的图像，并由斜率或截距计算重力加速度g、内阻r或杨氏模量E等量。
2. Core Equations to Remember | 需要牢记的核心方程
Before diving into derivations, ensure you can recall the fundamental forms: v = u + at, s = ut + ½at², T = 2π√(L/g) for a simple pendulum, V = IR, and V = ε – Ir for a cell. These serve as starting points for creating linear graphs.
在深入推导前，确保你能记起基本形式：v = u + at、s = ut + ½at²、单摆的T = 2π√(L/g)、V = IR以及电池的V = ε – Ir。这些是建立线性图像关系的起点。
3. Deriving the Simple Pendulum Period | 推导单摆周期公式
For a simple pendulum with small amplitude, the period T is given by T = 2π √(L / g). Squaring both sides we obtain T² = (4π²/g) L. This is the key linear relationship: plotting T² against L yields a straight line through the origin with gradient m = 4π²/g.
对于小振幅单摆，周期T由T = 2π √(L / g)给出。两边平方得到T² = (4π²/g) L。这是关键的线性关系：绘制T²-L图将得到一条通过原点的直线，斜率为m = 4π²/g。
Thus, the experimental value of g can be determined from the gradient: g = 4π² / m. In a typical June 2018 Paper 3 question, you might be asked to show this derivation, identify the axes, and explain how to find g from the best-fit line.
因此，g的实验值可由斜率求出：g = 4π² / m。在典型的2018年6月卷3试题中，你可能被要求展示这一推导过程，确定坐标轴，并解释如何由最佳拟合线求g。
4. Linearising the Free-Fall Equation | 自由落体方程的线性化
An object dropped from rest obeys s = ½ g t². However, when initial velocity is not zero or timing is uncertain, the full equation s = u t + ½ g t² is used. Dividing through by t gives: s/t = u + ½ g t. This is of the form y = c + m x, where y = s/t, x = t, gradient = ½ g, and intercept = u.
静止释放的物体遵循s = ½ g t²。然而当初速度不为零或计时不确定时，需使用完整方程s = u t + ½ g t²。两边除以t得：s/t = u + ½ g t，形式为y = c + m x，其中y = s/t，x = t，斜率为½ g，截距为u。
s/t = u + (½g) t
This transformation is extremely useful when practical measurements involve distance and time pairs. The June 2018 paper frequently expected candidates to rearrange equations into a linear form, explaining what to plot and how to extract g.
当实验测量给出距离与时间数据对时，这种变换极为有用。2018年6月试卷常期望考生将方程变形为线性形式，解释需要绘制什么以及如何提取g。
5. Deriving g from a T²-L Graph with Uncertainty | 从含不确定度的T²-L图推导g
Once you plot T² against L, the best-fit gradient m has an absolute uncertainty Δm read from the worst-acceptable line. Since g = 4π²/m, the fractional uncertainty in g is Δg/g = Δm/m. This comes from the rule for powers: when a quantity is raised to the power -1, its fractional uncertainty is identical. More explicitly, using Δg = (dg/dm) Δm leads to the same result.
绘制T²-L图后，最佳拟合线斜率m的绝对不确定度Δm可从最差可接受线读出。因为g = 4π²/m，g的相对不确定度Δg/g = Δm/m。这来自幂次规则：当一个量以-1次幂出现时，其相对不确定度不变。更明确地，利用Δg = (dg/dm) Δm可得相同结果。
Δg = (4π² / m²) Δm, giving Δg/g = Δm/m
In Paper 3, you might need to express the final result as g ± Δg with appropriate significant figures and units, a skill tested extensively in the 2018 series.
在卷3中，你可能需要将最终结果表示为g ± Δg，并配以合适有效数字和单位，这一技能在2018年系列试卷中被广泛考查。
6. Deriving Internal Resistance from Terminal p.d. | 由端电压推导内阻
The terminal potential difference V across a cell delivering current I is given by V = ε – I r, where ε is the e.m.f. and r is the internal resistance. This is already a linear equation: plotting V (y-axis) against I (x-axis) yields a straight line with gradient -r and intercept ε.
电池输送电流I时的端电压V由V = ε – I r给出，其中ε为电动势，r为内阻。这已经是线性方程：将V（纵轴）对I（横轴）作图，得到斜率为-r、截距为ε的直线。
Sometimes the experiment measures V for different external loads R, and we use V = ε R / (R + r). Rearranging: 1/V = (1/ε) + (r/ε)(1/R). Thus plotting 1/V against 1/R gives a straight line with gradient r/ε and intercept 1/ε. This derivation frequently appeared in practical skills papers around 2018.
有时实验测量不同外接负载R下的V，我们利用V = ε R / (R + r)。整理得：1/V = (1/ε) + (r/ε)(1/R)。因此绘制1/V-1/R图得到斜率为r/ε、截距为1/ε的直线。这一推导经常出现在2018年前后的实验技能试卷中。
7. Ohm’s Law and Resistivity Linearisation | 欧姆定律与电阻率线性化
For a wire, resistance R = ρ L / A. To find resistivity ρ, one can measure R for different lengths L, using a constant cross-sectional area A. Plotting R against L gives a straight line of gradient ρ/A, hence ρ = gradient × A. If A is known, ρ is determined.
对于一根导线，电阻R = ρ L / A。要得到电阻率ρ，可测量不同长度L下的R，保持截面积A不变。绘制R-L图得到斜率为ρ/A的直线，因而ρ = 斜率 × A。已知A时即可求出ρ。
In the June 2018 Paper 3, a similar approach may have been needed, possibly requiring you to log the equation or manipulate it to find the dependent variable. Always identify what is constant and which variable is changed.
在2018年6月的卷3中，可能需要类似方法，或许还要求对方程取对数或加以变形以找出因变量。务必明确哪些量保持不变，哪个量被改变。
8. Combining Uncertainties in Derived Quantities | 推导量中不确定度的合成
Suppose a quantity Q is calculated from measured p, q with uncertainties Δp, Δq. The rules are: for Q = p ± q, ΔQ = Δp + Δq (absolute); for Q = p × q or Q = p/q, fractional uncertainty ΔQ/Q = Δp/p + Δq/q; for Q = pⁿ, ΔQ/Q = |n| Δp/p. These must be applied stepwise in derivations.
设量Q由具有不确定度Δp、Δq的测量值p、q计算得出。规则为：若Q = p ± q，ΔQ = Δp + Δq（绝对）；若Q = p × q或Q = p/q，相对不确定度ΔQ/Q = Δp/p + Δq/q；若Q = pⁿ，ΔQ/Q = |n| Δp/p。这些需逐步应用于推导中。
In Paper 3 context, after linearising, you often combine gradient and intercept uncertainties. For instance, ε = 1/intercept from 1/V vs 1/R plot, so Δε is derived from intercept uncertainty. Mastery of these rules was essential for the 2018 exam.
在卷3中，线性化后你常需综合斜率和截距的不确定度。例如，从1/V-1/R图截距得到ε = 1/截距，因此Δε由截距不确定度导出。掌握这些规则对2018年试卷至关重要。
9. Deriving Young Modulus from a Load-Extension Graph | 由载荷-伸长图推导杨氏模量
For a wire of original length L and cross-sectional area A under force F, stress = F/A, strain = e/L, so Young modulus E = (F/A) / (e/L) = F L / (A e). Rearranging gives F = (E A / L) e. Hence a graph of F against extension e yields gradient = EA/L, from which E = gradient × L / A.
对于原长L、截面积A的金属丝，在力F下，应力 = F/A，应变 = e/L，因此杨氏模量E = (F/A)/(e/L) = F L / (A e)。整理得F = (E A / L) e。故F-e图斜率为EA/L，由此E = 斜率 × L / A。
This derivation is a classic practical paper task. Be prepared to identify which variable is manipulated, which is measured, and how to use the linear relationship to calculate E with uncertainty.
这一推导是经典的实验卷任务。要准备好识别哪个变量被改变、哪个被测量，以及如何利用线性关系计算E及其不确定度。
10. Handling Logarithmic Relationships in Capacitor Discharge | 处理电容器放电中的对数关系
For a capacitor discharging through a resistor, V = V₀ e^(-t/RC). To linearise, take natural log: ln V = ln V₀ – (1/RC) t. Plotting ln V against time t gives gradient -1/RC and intercept ln V₀. The time constant RC can be found from the gradient, a skill tested indirectly in June 2018.
电容器通过电阻放电时，V = V₀ e^(-t/RC)。为线性化，取自然对数：ln V = ln V₀ – (1/RC) t。绘制ln V-时间t图，斜率为-1/RC，截距为ln V₀。时间常数RC可由斜率求出，这一技能在2018年6月考试中也有所涉及。
ln V = ln V₀ – (t / RC)
Remember that RC has units of seconds; if t is in seconds, gradient magnitude directly gives 1/RC.
记住RC单位为秒；若t以秒为单位，斜率的大小直接给出1/RC。
11. Practical Tips for June 2018 Paper 3 Candidates | 给2018年6月卷3考生的实用提示
Always show your algebraic rearrangement step-by-step, clearly stating what you will plot on y and x axes. Label axes with quantity and unit, and include error bars when required. When explaining derivation, use words like ‘rearranging the equation to a linear form y = mx + c’ to gain full marks.
务必逐步展示代数变形，清楚说明纵、横轴分别绘制什么。坐标轴标注物理量和单位，必要时加上误差棒。在解释推导时，要用诸如“将方程变形为线性形式y = mx + c”等表述以获得满分。
Check your final derived quantity’s uncertainty makes sense; if experimental percentage uncertainty in g is above 20%, you likely made a measurement or calculation error. The 2018 examiners expected evidence of critical evaluation.
检验最终推导量的不确定度是否合理；如果g的实验百分不确定度超过20%，很可能存在测量或计算错误。2018年的考官期望看到批判性评估的证据。
12. Conclusion: Mastering Derivation for High Scores | 结语：掌握推导以获高分
Formula derivation in A-Level Physics Paper 3 is not just mathematical manipulation; it demonstrates deep understanding of the underlying physics. By practicing the linearisation techniques and uncertainty propagation outlined here, you will be well prepared to tackle any similar question, just as appeared in the June 2018 exam. Keep these derivations at your fingertips and you will approach the practical paper with confidence.
A-Level物理卷3中的公式推导不仅是数学操作，更体现出对基础物理的深刻理解。通过练习本文总结的线性化技巧和不确定度合成方法，你将能从容应对类似问题，正如2018年6月考试中出现的那样。把这些推导烂熟于心，你就能信心满满地面对实验卷。
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