📚 Mastering Formula Derivation for AS Physics PH02: Insights from the June 2022 Examiner Report | 掌握AS物理PH02公式推导：2022年6月考官报告解析
The AQA AS Physics Unit 2 (PH02) examination challenges students not only to recall formulae but to demonstrate a clear, logical derivation of key relationships. The June 2022 examiner report highlighted that many candidates lost marks by skipping logical steps, misusing vectors, or failing to state underlying principles. This article uses those insights to build a detailed revision guide on the essential formula derivations for PH02, presented in a side‑by‑side bilingual format to strengthen both understanding and exam technique.
AQA AS物理单元2（PH02）考试不仅要求学生记忆公式，更要求清晰、有逻辑地推导关键关系。2022年6月的考官报告指出，许多考生因跳过逻辑步骤、误用矢量或未阐明基本原理而失分。本文利用这些洞察，构建一份PH02核心公式推导的详细复习指南，并以中英对照形式呈现，以加深理解并提高应试技巧。
1. What Examiners Look For in Derivations | 考官在推导题中寻找的要点
Examiners expect a sequence that begins with a fundamental law or definition, moves through algebraic manipulation, and ends with the target formula. Every step must be justified, and diagrams or sign conventions must be explicitly linked to the mathematics. The June 2022 report noted that candidates often wrote final equations without showing substitution or rearrangement, losing method marks.
考官期待的推导过程始于基本定律或定义，经过代数操作，最终得到目标公式。每一步都需要有依据，图示或符号约定必须与数学运算明确关联。2022年6月的报告指出，考生常直接写出最终方程而未展示代入或变形过程，从而丢失方法分。
- Always state the starting physics principle (e.g., Newton's second law, conservation of energy). / 始终先陈述起始物理原理（如牛顿第二定律、能量守恒）。
- Define all symbols before using them. / 在使用每个符号前先给出定义。
- Include small remarks like “assuming constant acceleration” to show awareness of limitations. / 添加如“假设加速度恒定”的简短备注以体现对限制条件的认识。
2. Deriving the SUVAT Equations from Definitions | 从定义出发推导运动学方程
ntly jumbled in the exam, with many candidates failing to justify the use of average velocity.
从加速度定义出发：a = (v − u)/t，假设加速度恒定。整理得 v = u + at。求位移时，利用平均速度：s = ½(u + v) t。代入 v 得到 s = ut + ½at²。将 v = u + at 平方并与 s = ½(u + v) t 联立消去 t，即得 v² = u² + 2as。考试中这些步骤常被混淆，许多考生未能解释为何使用平均速度。
s = ½ (u + v) t → s = ut + ½at² → v² = u² + 2as
3. Deriving Conservation of Momentum from Newton's Laws | 从牛顿定律推导动量守恒
Consider two bodies A and B interacting. According to Newton's third law, the force FAB on A by B is equal and opposite to FBA on B by A. Apply Newton's second law in impulse form: F = Δp/Δt. Thus, ΔpA = FAB Δt and ΔpB = FBA Δt = −FAB Δt. Adding, ΔpA + ΔpB = 0, so total momentum pA + pB is constant. The examiner reported that candidates often omitted the step linking equal forces to equal and opposite changes in momentum.
考虑两个相互作用的物体 A 和 B。根据牛顿第三定律，A 受 B 的力 FAB 与 B 受 A 的力 FBA 大小相等、方向相反。应用冲量形式的牛顿第二定律：F = Δp/Δt。因此，ΔpA = FAB Δt，ΔpB = FBA Δt = −FAB Δt。两式相加得 ΔpA + ΔpB = 0，故总动量 pA + pB 恒定。考官报告称考生常遗漏将等大力与等大反向动量变化相联系的步骤。
4. Deriving Centripetal Acceleration a = v²/r | 向心加速度公式推导
/r directed radially inward. Many candidates in June 2022 lost credit by not mentioning the vector nature of velocity change or failing to explain why Δv is perpendicular to v in the limit.
对于以恒定速率 v 在半径 r 的圆周上运动的物体，取一小段时间 Δt。速度的变化量 Δv 指向圆心，大小约为 v Δθ，其中 Δθ = vΔt / r。因此 Δv ≈ (v²Δt)/r。加速度 a = Δv/Δt 即得 a = v²/r，方向径向向里。2022年6月考试中许多考生因未提及速度变化的矢量性或未解释为何 Δv 在极限下垂直于 v 而失分。
Δv = v Δθ = v (vΔt/r) → a = v²/r
5. Deriving the Young Modulus Equation | 杨氏模量公式推导
ation question may require you to start from the definitions of stress and strain and rearrange to show that E can be determined from the gradient of a force‑ imes confused original length with extension, so they advised clear labelling of L and ΔL.
杨氏模量 E 的定义是拉伸应力除以拉伸应变：E = (F/A) / (ΔL/L) = FL/(AΔL)。推导题可能要求从应力和应变的定义出发，通过变形证明 E 可由力‑伸长量图线的斜率乘以 L/A 求得。考官指出学生有时混淆原长与伸长量，因此建议清楚标注 L 和 ΔL。
E = (F/A) ÷ (ΔL/L) → F = (EA/L) ΔL
6. Deriving Snell's Law from Wave Speed | 从波速推导折射定律
/f and λ₂ = v₂/f, we get sinθ₁/sinθ₂ = v₁/v₂. Introducing refractive index n₁ = c/v₁ and n₂ = c/v₂ yields n₁ sinθ₁ = n₂ sinθ₂. The report highlighted that many candidates skipped the link to wavefront geometry and lost marks.
当平面波穿过两介质界面时，频率 f 保持不变。波速 v = fλ。利用波阵面几何关系，入射角 θ₁ 和折射角 θ₂ 与波长有关：对于同一波阵面间距 d，有 sinθ₁ = λ₁/d、sinθ₂ = λ₂/d。又因 λ₁ = v₁/f、λ₂ = v₂/f，故 sinθ₁/sinθ₂ = v₁/v₂。引入折射率 n₁ = c/v₁、n₂ = c/v₂，即得 n₁ sinθ₁ = n₂ sinθ₂。报告强调，许多考生跳过波阵面几何的连接，导致失分。
7. Deriving Path Difference for Single‑Slit Diffraction Minima | 单缝衍射极小值的光程差推导
points yields a sinθ = nλ for n = 1, 2, 3... The 2022 report noted that candidates often omitted the reasoning behind dividing the slit, instead jumping to the final formula.
对于宽度为 a 的单缝，将缝分成上下两半。若上半部某点发出的波与下半部相距 a/2 的对应点发出的波路程差为 λ/2，则发生相消干涉。第一极小角度 θ 满足 (a/2) sinθ = λ/2，即 a sinθ = λ。推广至多点配对，得到 a sinθ = nλ（n = 1, 2, 3...）。2022年报告指出考生常略去分割单缝的推理过程，直接写出最终公式。
a sinθ = nλ (minima)
8. Deriving Double‑Slit Fringe Spacing Δx = λD/d | 双缝干涉条纹间距推导
les, sinθ ≈ tanθ ≈ y/D, where y is the distance from the central maximum and D is the slit‑to‑screen distance. Constructive interference occurs when d sinθ = nλ, so yn = nλD/d. The fringe spacing Δx = yn+1 − yn = λD/d. The June 2022 examiner report reminded teachers to stress the small‑angle approximation, as many candidates incorrectly applied it for large θ.
对于相距 d 的双缝，到亮纹方向的光程差为 d sinθ。小角度下 sinθ ≈ tanθ ≈ y/D，其中 y 为距中央明纹的距离，D 为缝与屏间距。相长干涉条件为 d sinθ = nλ，故 yn = nλD/d。条纹间距 Δx = yn+1 − yn = λD/d。2022年6月的考官报告提醒教师强调小角近似，因为许多考生在较大角度下仍错误使用这一近似。
Δx = λD / d
9. Deriving Kinetic Energy from Work Done | 从做功推导动能
Starting with the work‑energy principle, net work done W = F s. Using Newton' c equation v² = u² + 2as, eliminating s gives W = m (v² − u²)/2. Defining kinetic energy Ek e in kinetic energy. The report pointed out that candidates often used the final formula without linking it to the work done, losing the derivation marks.
从功能原理出发，合外力做的功 W = F s。运用牛顿第二定律 F = ma 及运动学方程 v² = u² + 2as，消去 s 得 W = m (v² − u²)/2。定义动能 Ek = ½mv²，则合外力做功等于动能的变化量。报告指出，考生常直接使用最终公式而未与做功相联系，丢掉推导分数。
W = Fs = ma × (v² − u²)/2a = ½mv² − ½mu²
10. Common Pitfalls and Examiner Recommendations | 常见错误与考官建议
Based on the June 2022 PH02 report, the most frequent errors were: failing to state the assumptions (e.g., constant acceleration, small angles), confusing scalars and vectors, not showing intermediate algebraic steps, and mismatching units. Examiners recommended practicing derivations with a focus on step‑by‑step logic, using annotated diagrams, and always concluding with the target formula neatly boxed.
根据2022年6月PH02报告，最常见的错误有：未能陈述假设条件（如匀加速度、小角度），混淆标量与矢量，未展示中间代数步骤，以及单位不匹配。考官建议练习推导时要注重步骤逻辑，使用注释清晰的图示，并始终以整洁的方框写出目标公式。
| Pitfall / 错误 | Why it costs marks / 为何失分 | How to avoid / 如何避免 |
| Skipping the initial principle / 省略初始原理 | No anchor for the derivation / 缺乏推导依据 | Always write the defining equation first / 始终先写定义方程 |
| Ignoring vector direction / 忽略矢量方向 | Sign errors in final expression / 最终表达式符号错误 | Use a clear sign convention and stick to it / 采用明确的符号约定并一以贯之 |
Derivation questions are an opportunity to display full mastery of the physics behind the equations. By internalising the logical flow from principle to formula and heeding the feedback from the June 2022 PH02 examiner report, you can transform a challenging task into a reliable source of marks. Practice writing out each derivation repeatedly, and you will soon find that the steps become second nature.
公式推导题是展示对公式背后物理原理全面掌握的绝佳机会。通过内化从原理到公式的逻辑流程，并吸取2022年6月PH02考官报告的反馈，你定能将挑战转化为稳定的得分点。反复练习写出每一个推导过程，这些步骤很快便会成为你的第二天性。
Published by TutorHao | Physics Revision Series | aleveler.com
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