Tag: Physics

📚 Mastering Application Questions in A-Level Physics: 9630-PH03 Specimen Paper v4.2 | A-Level物理9630-PH03样卷应用题解题技巧

The 9630-PH03 International A-Level Physics Specimen Paper v4.2 represents a critical assessment of practical skills and synoptic application. It demands not just recall but the ability to interpret data, evaluate experimental procedures, and communicate physics reasoning clearly. This article guides you through proven techniques to tackle its application questions with confidence.

9630-PH03国际A-Level物理样卷v4.2是对实验技能和综合应用能力的一次关键测评。它不仅考查记忆，更要求考生能解读数据、评估实验过程并清晰地阐释物理推理。本文将带你掌握攻克其中应用题的可靠技巧，助你从容应对。

1. Understanding the Assessment Objectives | 理解评估目标

Questions in PH03 are built around AO2 (application of knowledge) and AO3 (experimental skills), with a smaller AO1 presence. This means you must select and apply physics concepts to novel situations, analyse data, and justify conclusions. Knowing that marks are heavily weighted towards process rather than final answers helps you focus on showing logical steps.

PH03的题目围绕AO2（知识应用）和AO3（实验技能）构建，AO1占比较少。这意味着你必须在陌生情境中选择并应用物理概念、分析数据并论证结论。了解评分侧重于过程而非最终答案，能帮助你专注于展示逻辑步骤。

2. Interpreting Tables of Experimental Data | 解读实验数据表格

Always start by examining column headings, units, and precision. Look for anomalous points that deviate from the overall trend. In the specimen paper v4.2, you may find repeated measurements – use these to calculate mean values and estimate random uncertainty via half the range. Pay attention to significant figures: the number of decimal places in a mean should match the original data.

始终先从列标题、单位和精密度入手。找出偏离总体趋势的异常点。在样卷v4.2中，你可能会遇到重复测量——利用它们计算平均值，并通过半极差法估算随机不确定度。注意有效数字：平均值的小数位数应与原始数据保持一致。

3. Mastering Graph Plotting | 掌握图形绘制

Choose scales that occupy at least half the grid in both directions and use simple multiples (1, 2, 5, 10). Label axes with quantity and unit, e.g., ‘t / s’. Plot data points as small crosses (x) or circled dots. Draw a best-fit line – not simply connecting dots – and consider whether the relationship forces the line through the origin. When a straight line is expected, use a transparent ruler to balance points above and below.

选择刻度使数据点占据至少一半的网格面，并采用简单倍数（1、2、5、10）。坐标轴标注量和单位，如“t / s”。数据点绘制成小十字（x）或带圈圆点。描画最佳拟合线——不要简单连点——并考虑物理关系是否强制通过原点。当预期为直线时，用透明直尺平衡上下方的点。

4. Calculating Gradients and Intercepts | 计算斜率和截距

When finding a gradient, select two widely spaced points on the best-fit line, not data points. Compute

gradient = Δy/Δx = (y₂ – y₁)/(x₂ – x₁)

and attach the correct units, typically y-unit/x-unit. The y-intercept is read directly where the line crosses the axis; never use a data point for this. Relate the gradient to a physical quantity using the linearised equation, e.g., for T² = (4π²/g) L, gradient = 4π²/g.

计算斜率时，在最佳拟合线上选取两个间距较大的点，而非原始数据点。按上式计算，并带上正确的单位，通常为y单位/x单位。y截距直接从直线与轴的交点读取，切勿使用数据点求截距。利用线性化方程将斜率与物理量联系起来，例如 T² = (4π²/g)L，斜率 = 4π²/g。

5. Handling Uncertainties | 处理不确定度

Specimen paper v4.2 frequently tests error analysis. Absolute uncertainty is the ± value in a measurement; percentage uncertainty = (absolute uncertainty/measurement) × 100%. For sums or differences, add absolute uncertainties; for products or quotients, add percentage uncertainties. When a quantity is raised to a power n, multiply the percentage uncertainty by n. Show your working when combining uncertainties to secure method marks.

样卷v4.2经常考查误差分析。绝对不确定度是测量值中的±值；百分不确定度 =（绝对不确定度/测量值）× 100%。对于加减运算，将绝对不确定度相加；对于乘除运算，将百分不确定度相加。若物理量被升至n次方，则将百分不确定度乘以n。合并不确定度时要展示运算过程以获取方法分。

6. Tackling “Explain” and “Suggest” Questions | 攻克“解释”与“建议”题

Use clear causal chains: ‘If X increases, Y increases because…’ or ‘The gradient is steeper due to…’. Back every statement with a physics principle or equation. When asked to suggest an improvement, comment on reducing a specific source of uncertainty, controlling a variable, or using more precise instruments. Avoid vague answers like ‘do it more carefully’.

使用清晰的因果链：“若X增加，Y增加，因为……”或“斜率更陡是由于……”。每句话都要有物理原理或方程支撑。当被问及改进建议时，要针对减少特定误差源、控制变量或使用更精密的仪器进行说明。避免“做得更仔细”这类模糊回答。

7. Applying Equations to Novel Scenarios | 在新情境中应用方程

PH03 application questions often embed unfamiliar equipment or configurations. Identify which physical principles (conservation of energy, Ohm’s law, Newton’s second law) apply. Rearrange the equation before inserting numbers, and always convert to SI base units. For instance, if a question gives a diameter in cm, convert to m before calculating cross-sectional area A = π(d/2)².

PH03应用题常引入陌生设备或构型。要先识别适用的物理原理（能量守恒、欧姆定律、牛顿第二定律）。代入数字前先重排方程，并始终转换为SI基本单位。例如，若题目给出的直径以cm为单位，应在计算截面积A = π(d/2)²前将之转换为m。

8. Validating Results with Physical Intuition | 用物理直觉验证结果

A quick sanity check prevents simple mistakes. Does your calculated value for the resistivity of a metal lie around 10⁻⁷ to 10⁻⁸ Ω m? Is the acceleration of a falling object roughly 9.8 m s⁻² in the absence of drag? Check orders of magnitude, signs, and directions. A negative acceleration when the object is slowing down makes sense; a negative mass does not.

快速合理性检查能避免简单错误。你算出的金属电阻率是否大致在10⁻⁷至10⁻⁸ Ω m范围内？无阻力时落体加速度是否约9.8 m s⁻²？检查数量级、正负号和方向。物体减速时加速度为负是合理的；质量为负则不然。

9. Avoiding Common Pitfalls | 避开常见陷阱

Many candidates lose marks by omitting units in final answers, using inconsistent significant figures, or misreading graph scales. Another frequent error is forcing a straight line through the origin when the physical model does not demand it, or extrapolating beyond the data range without justification. When calculating period from a timing over multiple oscillations, always divide by the number of oscillations.

很多考生因最后答案漏写单位、有效数字不一致或看错图形刻度而失分。另一个常见错误是当物理模型不要求时硬使直线通过原点，或不加说明地将数据范围外推。用多次振荡的时间求周期时，务必要除以振荡次数。

10. Using Mark Schemes Effectively | 有效利用评分方案

Study the specimen mark scheme alongside questions to internalise how answers earn marks. Notice that command words like ‘state’ require a short answer, while ‘explain’ demands a link between two ideas. Mark schemes often award marks for specific phrases such as ‘random error reduced by averaging’ or ‘systematic error from zero offset’. Practice writing in the style the mark scheme rewards.

将样卷评分方案与题目对照学习，深刻体会如何得分。注意指令词如“陈述”要求简短回答，而“解释”则需联系两个观点。评分方案常因特定短语而给分，例如“通过取平均减小随机误差”或“零点偏移引起的系统误差”。练习用评分方案所青睐的风格作答。

11. Synoptic Connections across Topics | 跨主题综合联系

PH03 is a synoptic paper, meaning questions can blend mechanics, materials, waves, and electricity. For example, an experiment may involve a spring-in-series arrangement with an electrical measurement of extension. Be ready to recall formulas from AS topics such as Hooke’s law, the Young modulus, or the oscilloscope time-base setting. Synoptic linkages are often tested through practical design – can you plan how to measure a physical quantity using a circuit or optical setup?

PH03是一份综合考查的试卷，意味着题目可能融合力学、材料、波动和电学。例如，某实验可能涉及串接弹簧并以电学方法测量伸长量。要准备好调取AS阶段的知识，如胡克定律、杨氏模量或示波器时基设置。综合联系常通过实验设计来考查——你能否规划用电路或光学装置测量一个物理量？

12. Time Management and Revision Strategy | 时间管理与复习策略

The specimen paper v4.2 allocates roughly 1.5 minutes per mark. Read through the entire question before attempting any parts, and leave time for graph plotting, which can consume 10–15 minutes. In revision, practise past application-style questions under timed conditions, and mark them yourself using the specimen mark scheme. Focus on areas where you repeatedly lose marks, such as uncertainty propagation or graph-work.

样卷v4.2大致按照每分1.5分钟分配时间。先通读整个题目再开始作答，并留出10–15分钟用于绘图。复习时，在有时间限制的条件下练习往届应用题，并使用样卷评分方案自行批改。重点关注反复丢分的环节，如不确定度传递或图形处理。

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 A-Level Physics Jun-18 Insert 5 Concept Analysis | A-Level物理Jun-18 Insert5概念解析

The June 2018 A-Level Physics examination included an insert (page 5) that introduced the principles of optical fibres and their application in medical endoscopy. This article unpacks the key physical concepts from that insert, offering a bilingual, exam-focused explanation of total internal reflection, fibre structure, signal degradation, and clinical imaging techniques. Mastery of these topics is essential for both the Medical Physics option and general wave optics questions across all major exam boards.

2018年6月的A-Level物理考试中，插入材料（第5页）介绍了光纤原理及其在医用内窥镜中的应用。本文对这份材料中的核心物理概念进行解析，以中英双语、紧扣考纲的方式讲解全内反射、光纤结构、信号衰减以及临床成像技术。掌握这些内容，对于医学物理选修部分以及各大考试局通用的波动光学题目都至关重要。

1. Context and Role of the Insert | 插入材料的背景与作用

In A-Level Physics, exam inserts often provide scientific articles or data that simulate real-world applications. The Jun-18 Insert 5 focused on the physics behind fibre-optic endoscopes, asking students to apply knowledge of waves, optics, and materials to a clinical context. This mirrors how physicists and engineers design minimally invasive diagnostic tools.

在A-Level物理中，考卷的插入材料通常会提供模拟真实应用的科学文章或数据。2018年6月的Insert 5聚焦于光纤内窥镜背后的物理原理，要求学生将波动、光学和材料知识应用于临床情境。这反映了物理学家和工程师如何设计微创诊断工具。

2. Core Principle: Total Internal Reflection | 核心原理：全内反射

The fundamental mechanism that confines light within an optical fibre is total internal reflection (TIR). When light travels from a medium of higher refractive index n₁ into a medium of lower refractive index n₂, and the angle of incidence exceeds the critical angle θc, all the light is reflected back into the denser medium with no transmission. The critical angle is given by θc = sin⁻¹(n₂ / n₁), a direct consequence of Snell’s law.

将光约束在光纤内部的基本机制是全内反射（TIR）。当光从折射率较高的介质 n₁ 进入折射率较低的介质 n₂，且入射角超过临界角 θc 时，所有光线都会被反射回光密介质，没有透射。临界角由 θc = sin⁻¹(n₂ / n₁) 给出，这是斯涅尔定律的直接结果。

Snell’s law describes the relationship: n₁ sin θ₁ = n₂ sin θ₂, where θ₁ is the angle of incidence and θ₂ is the angle of refraction. For refraction at 90°, sin θ₂ = 1, leading to the critical angle expression. TIR only occurs when n₁ > n₂ and θ₁ > θc.

斯涅尔定律描述了这一关系：n₁ sin θ₁ = n₂ sin θ₂，其中 θ₁ 为入射角，θ₂ 为折射角。当折射角为 90° 时，sin θ₂ = 1，即可得到临界角公式。全内反射仅当 n₁ > n₂ 且 θ₁ > θc 时发生。

3. Optical Fibre Structure: Core, Cladding, and Buffer | 光纤结构：纤芯、包层与缓冲层

A standard step-index fibre consists of a cylindrical core made of high-purity silica glass with a refractive index n_core, surrounded by a cladding of slightly lower refractive index n_cladding. A plastic outer buffer coating provides mechanical protection and prevents the cladding from scratches that could cause light leakage. The core diameter in medical endoscopes is typically 50–200 µm, while the cladding adds an additional 10–20 µm thickness.

标准阶跃折射率光纤由高纯度石英玻璃制成的圆柱形纤芯（折射率 n_core）和折射率略低的包层（n_cladding）构成。外部塑料缓冲涂层提供机械保护，并防止包层被划伤而引起漏光。医用内窥镜中的纤芯直径通常为 50–200 µm，包层额外增加 10–20 µm 的厚度。

Using cladding with a lower refractive index is crucial because it increases the critical angle at the core–cladding interface, allowing more light rays to undergo TIR even at modest incident angles. Without cladding, contact with external materials (e.g., moisture or a finger) would drastically alter the effective n₂, causing unpredictable losses.

使用折射率较低的包层至关重要，因为它增大了纤芯-包层界面的临界角，即使入射角较小，更多光线也能实现全内反射。若无包层，与外界物质（如水分或手指）接触会显著改变有效 n₂，导致不可预测的损耗。

4. Numerical Aperture and Acceptance Angle | 数值孔径与接收角

The light-gathering ability of a fibre is quantified by its numerical aperture (NA). A larger NA means the fibre can accept light from a wider cone of rays. NA is defined as NA = √(n_core² − n_cladding²). The acceptance angle θa in air (nₐ = 1) satisfies sin θa = NA. In medical endoscopy, a high NA is desirable to capture maximum illumination from tissue surfaces, but this must be balanced against modal dispersion.

光纤的集光能力由其数值孔径（NA）来量化。NA 越大，光纤能够接收的光线锥角就越宽。NA 定义为 NA = √(n_core² − n_cladding²)。空气中的接收角 θa（nₐ = 1）满足 sin θa = NA。在医用内窥镜中，高 NA 有利于从组织表面获取最大限度的照明，但这必须与模式色散进行权衡。

Typical core and cladding indices in medical fibres are n_core = 1.62 and n_cladding = 1.52, yielding an NA of about 0.56 and an acceptance angle of approximately 34°. This provides efficient coupling of light from LED or laser sources.

医用光纤典型的纤芯和包层折射率分别为 n_core = 1.62 和 n_cladding = 1.52，得到的 NA 约为 0.56，接收角约 34°。这能有效地将 LED 或激光光源发出的光耦合进光纤。

5. Attenuation and the Decibel Scale | 衰减与分贝标度

As light propagates through a fibre, its intensity decreases exponentially due to absorption and scattering. Attenuation is commonly expressed in decibels per kilometre (dB km⁻¹) using the formula: Attenuation = (10/L) log₁₀(P₀/P), where P₀ is the input power, P is the output power after length L. A 3 dB loss corresponds to halving the power. Modern endoscope fibres exhibit attenuation as low as 0.2 dB km⁻¹ for infrared light.

光在光纤中传播时，由于吸收和散射，强度呈指数衰减。衰减通常以每千米分贝（dB km⁻¹）表示，公式为：衰减 = (10/L) log₁₀(P₀/P)，其中 P₀ 为输入功率，P 为经过长度 L 后的输出功率。3 dB 的损耗对应功率减半。现代内窥镜光纤对红外光的衰减可低至 0.2 dB km⁻¹。

The main attenuation mechanisms include Rayleigh scattering from microscopic inhomogeneities in the glass and absorption by OH⁻ ions and metal impurities. Operating wavelengths near 1300 nm and 1550 nm minimise these losses, which is why near-infrared light is often used in coherent fibre bundles for imaging.

主要的衰减机制包括玻璃中微观不均匀性引起的瑞利散射，以及 OH⁻ 离子和金属杂质的吸收。工作波长在 1300 nm 和 1550 nm 附近可使这些损耗最小化，因此近红外光常用于成像的相干光纤束。

6. Pulse Broadening and Dispersion | 脉冲展宽与色散

Digital information in fibres is carried by light pulses. Dispersion causes pulses to broaden as they travel, limiting the data rate and resolution in endoscopic imaging. Two main types of dispersion are material dispersion (chromatic) and modal dispersion. Material dispersion arises because different wavelengths travel at different speeds in the glass, while modal dispersion occurs because different ray paths (modes) take different times to traverse the fibre.

光纤中的数字信息由光脉冲承载。色散导致脉冲在传播过程中展宽，限制了内窥镜成像的数据速率和分辨率。两种主要的色散是材料色散（色度色散）和模式色散。材料色散源于不同波长在玻璃中的传播速度不同；模式色散则是因为不同的光线路径（模式）通过光纤所需的时间不同。

In a step-index multimode fibre, the time delay Δt between the axial ray and the ray at the maximum acceptance angle is approximately Δt ≈ (n_core L / c) × (n_core/n_cladding − 1). This can be reduced by using graded-index fibres, where the refractive index gradually decreases from the centre to the cladding, causing rays to follow curved paths and equalising transit times.

在阶跃折射率多模光纤中，轴向光线与最大接收角光线之间的时间延迟 Δt 约为 Δt ≈ (n_core L / c) × (n_core/n_cladding − 1)。使用渐变折射率光纤可以减少这一效应，其折射率从中心向包层逐渐降低，使光线沿曲线路径行进，从而平衡渡越时间。

7. Coherent and Incoherent Fibre Bundles | 相干与非相干光纤束

Medical endoscopes use two distinct types of fibre bundles. An incoherent bundle carries illumination light from the external source to the target tissue; the relative positions of individual fibres do not need to be preserved. A coherent bundle, however, transmits the image back from the tissue to the eyepiece or camera sensor. In a coherent bundle, the spatial arrangement of fibres is identical at both ends, so each fibre acts like a pixel, preserving the image.

医用内窥镜使用两种不同的光纤束。非相干光纤束将照明光从外部光源传送到目标组织；各个光纤的相对位置无需保持不变。而相干光纤束则把图像从组织传回目镜或相机传感器。在相干束中，光纤在两端按相同的空间排列，因此每根光纤就像一个像素，从而保留图像。

If fibres are randomly packed, a honeycomb-like pattern can appear in the transmitted image. Modern endoscopes use leached fibre bundles or lens-grin systems to reduce pixelation. The number of fibres in a coherent bundle typically ranges from 30 000 to 100 000, defining the image resolution.

如果光纤随机排列，传送的图像中可能会出现蜂窝状图案。现代内窥镜采用沥滤光纤束或透镜-梯度折射率系统来减少像素化。相干束中的光纤数量通常在 30 000 至 100 000 之间，决定了图像分辨率。

8. Clinical Application: Endoscopic Imaging and Biopsy | 临床应用：内窥镜成像与活检

During a gastroscopy or colonoscopy, the endoscope carries both illumination and imaging channels, along with a working channel for instruments. White light from a xenon or LED source is transmitted through the incoherent bundle to illuminate the mucosa. The reflected light is collected by the objective lens and focused onto the distal face of the coherent bundle, which carries the image to the proximal end for viewing or video capture.

在进行胃镜或结肠镜检查时，内窥镜同时带有照明通道、成像通道，以及用于器械的工作通道。来自氙灯或 LED 的白光通过非相干光纤束传送到粘膜上进行照明。反射光由物镜收集并聚焦到相干束的远端端面上，相干束将图像传输到近端供观察或视频捕捉。

Narrow-band imaging (NBI) takes advantage of the wavelength dependence of light absorption by haemoglobin. By filtering the illumination to blue (415 nm) and green (540 nm), clinicians can enhance the contrast of superficial capillaries and mucosal patterns, improving early cancer detection. This is a direct application of the wave optics and spectral analysis principles from the A-Level syllabus.

窄带成像（NBI）利用了血红蛋白对光吸收的波长依赖性。通过将照明过滤为蓝光（415 nm）和绿光（540 nm），临床医生可以增强浅表毛细血管和粘膜形态的对比度，从而提高早期癌症检测能力。这是对A-Level教学大纲中波动光学和光谱分析原理的直接应用。

9. Endoscope Specifications and Design Trade-offs | 内窥镜规格与设计权衡

Designing an endoscope for maximum diagnostic yield involves balancing NA, fibre diameter, flexibility, and cost. A larger core increases illumination but reduces flexibility and increases the risk of fibre breakage. A high NA collects more light but exacerbates modal dispersion, blurring the image. These trade-offs can be summarised in the following table:

设计一款具有最高诊断效能的内窥镜，需要平衡 NA、光纤直径、柔韧性和成本。纤芯越大，照明越强，但会降低柔韧性并增加光纤断裂风险。高 NA 可收集更多光线，但会加剧模式色散，使图像模糊。这些权衡可用下表总结：

These parameters are frequently explored in A-Level Physics data-response questions where students must justify design choices using physical principles.

这些参数在A-Level物理的数据分析题中经常出现，要求学生运用物理原理论证设计选择。

10. Exam Relevance and Key Formulae to Master | 考试关联与必须掌握的核心公式

Questions based on Insert 5 typically assess understanding of TIR, fibre structure, attenuation, and image transmission. Students should be confident applying the following equations:

基于Insert 5的题目通常会考查对全内反射、光纤结构、衰减和图像传输的理解。学生应能熟练运用以下方程：

• Critical angle: θc = sin⁻¹(n₂ / n₁)
• Numerical aperture: NA = √(n_core² − n_cladding²)
• Attenuation in dB: α = (10 / L) log₁₀(P₀ / P)
• Time dispersion (step-index): Δt ≈ (n_core L / c)(n_core / n_cladding − 1)

These should be memorised and used in conjunction with ray diagrams showing light entry cones and pulse propagation. Practice linking the physical parameters to the quality of the final medical image.

这些公式应当熟记，并结合光线进入锥角和脉冲传播的光路图来使用。练习将物理参数与最终医学图像的质量联系起来。

11. Misconceptions and Common Pitfalls | 常见误解与易错点

A frequent misconception is that TIR occurs at the core–buffer interface. Students must correctly identify that the reflection occurs at the core–cladding boundary, and the buffer merely protects the fibre. Another error is assuming that all light entering the fibre is automatically guided: only rays entering within the acceptance cone will undergo TIR. Furthermore, pupils often overlook that attenuation is an exponential process, not a linear one, and that doubling the distance more than doubles the loss in decibels due to the logarithmic relationship.

一个常见误解是全内反射发生在纤芯-缓冲层界面。学生必须正确识别反射发生在纤芯-包层边界，而缓冲层仅起到保护作用。另一个错误是认为所有进入光纤的光线都会被自动传导：只有进入接收锥内的光线才会发生全内反射。此外，学生经常忽略衰减是指数过程，而非线性过程，由于对数关系，距离加倍时分贝损耗的增加并非简单的两倍。

When calculating attenuation, ensure the power ratio P₀ / P is greater than 1, and the dB value is positive for a loss. In questions comparing multimodal and monomodal fibres, highlight that monomodal fibres have a very thin core (≈ 8 µm) and eliminate modal dispersion, but they are not used in conventional endoscopes because of the difficulty in constructing coherent bundles with such fine fibres.

计算衰减时，要确保功率比 P₀ / P 大于 1，且损耗的分贝值为正。在比较多模和单模光纤时，要强调单模光纤的纤芯极细（约 8 µm），可消除模式色散，但由于难以用如此细的光纤构建相干束，它们并不用于传统内窥镜。

12. Summary and Revision Strategy | 总结与复习策略

Insert 5 from the June 2018 A-Level Physics exam serves as an excellent resource for mastering applied optics. Focus your revision on drawing clear, labelled ray diagrams for step-index fibres, practising attenuation and NA calculations, and explaining how the physical specifications of a fibre bundle translate into clinical image quality. Use past-paper questions from the Medical Physics option to test your understanding, and always connect theory to the practical constraints of real endoscopes.

2018年6月A-Level物理考试的Insert 5是掌握应用光学的绝佳资源。复习时应着重绘制清晰的阶跃折射率光纤光线图并进行标注，练习衰减和数值孔径的计算，并解释光纤束的物理规格如何转化为临床图像质量。利用医学物理选修部分的历年真题测试自己的理解，并始终将理论与实际内窥镜的设计限制联系起来。

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 9630 International A-Level Physics Command Words: Applied Question Techniques | 9630国际A-Level物理指令词应用题技巧

Command words in 9630 International A-Level Physics are not just exam vocabulary – they are precise instructions that tell you exactly how to structure your answer and what the examiner is looking for. Misreading a ‘describe’ as an ‘explain’ can lose you most of the marks even if you know the physics. This guide breaks down the most common command words, explains the thinking behind each one, and gives you practical techniques to maximise your score in applied questions.

9630国际A-Level物理考试中的指令词不仅仅是考试词汇——它们是精确的指令，告诉你应该如何组织答案、考官究竟想看什么。如果把“描述”误读为“解释”，即使你懂物理原理也会丢掉大部分分数。本指南拆解最常见的指令词，解释每个指令背后的思维逻辑，并为你提供实用技巧，帮助你在应用题中拿到最高分。

1. Overview of Command Words | 指令词总览

Before tackling individual command words, it helps to see them grouped by the cognitive skill they demand. The table below summarises the key instructions that appear on paper 1, 2 and 3 of the 9630 specification. Knowing this map means you can read a sub‑question and immediately recall the expected style, length and depth of response.

在逐一攻克指令词之前，先按照它们所要求的认知技能进行分组会很有帮助。下表总结了9630考试大纲中试卷1、2和3出现的关键指令。熟悉这张地图意味着你一看到子问题，就能立即想到期望的答题风格、长度和深度。

2. State | 陈述

The ‘state’ command expects a short, factual answer without any development. You are being asked to retrieve a piece of knowledge from your memory – a constant, a law, a unit or a simple observation. With ‘state’, never waste time on extra sentences; one correct word or phrase is enough.

“陈述”指令要求你给出简短、事实性的答案，无需任何展开。它考察的是你从记忆中提取知识的能力——一个常数、一条定律、一个单位或一个简单的观测结论。遇到“陈述”时，不要在额外句子上浪费时间；一个正确的词语或短语就足够了。

Example: ‘State the SI base unit of force.’ The answer is simply ‘newton (N)’ or ‘kg m s⁻²’, not a full sentence. Technique tip: in data‑response questions, the answer is often printed in the given text – scan the stem before writing.

示例：“陈述力的国际单位制基本单位。”答案就是“牛顿 (N)”或“kg m s⁻²”，而不是一个完整的句子。技巧提示：在图表数据题中，答案常常就印在题目所给的信息里——动笔之前先快速扫读题干。

When ‘state’ appears with a value, e.g. ‘state the half‑life from the graph’, read axes carefully and include units. A common pitfall is writing ‘5’ when the time axis is in days; always attach the unit even if the question does not explicitly repeat ‘state the value and unit’.

当“陈述”和一个数值一起出现时，例如“从图中陈述半衰期”，要仔细阅读坐标轴并带单位。常见失分点是写出“5”而时间轴是以天为单位的；即使题目没有明确重复“陈述数值和单位”，也一定要带上单位。

3. Define | 定义

‘Define’ requires the exact meaning of a physical quantity or concept, usually in the form of a word equation or a precise sentence. Impress the examiner by starting with the formula if there is one, then stating the relationship in words. This covers two common mark schemes: one mark for the equation, one for the verbal description.

“定义”要求给出一个物理量或概念的精确含义，通常以文字方程或精确的句子的形式。如果有公式，先用公式开头，再用语言陈述关系，这能给考官留下好印象。这样涵盖两种常见的评分方案：方程一分，文字描述一分。

For instance, define power: ‘Power = work done / time taken, or the rate of doing work.’ A safe technique is to write: ‘Power is defined as P = W/t, where P is power, W is work done and t is time.’ Always define the symbols you introduce. Do not use the defined term in your definition; avoid ‘density is how dense something is’.

例如，定义功率：“功率=做功/所用时间，即做功的速率。”一个稳妥的技巧是写出：“功率定义为 P = W/t，其中 P 是功率，W 是做功，t 是时间。”一定要对你引入的符号进行定义。避免在定义中使用被定义的词本身；不要写“密度就是物体有多密”这种话。

For quantities like resistivity or electric field, use the defining equation from the formula sheet and explain it: ‘resistivity ρ = RA/L, where R is resistance, A is cross‑sectional area and L is length.’ If the question says ‘define in words’, you can still write the equation as a support, but the marks are for the statement.

对于像电阻率或电场强度这样的量，使用公式表里的定义方程并加以解释：“电阻率 ρ = RA/L，其中 R 是电阻，A 是横截面积，L 是长度。”如果题目说“用文字定义”，你仍然可以把方程作为辅助写出来，但分值是针对文字陈述的。

4. Calculate | 计算

‘Calculate’ questions test your ability to select the correct equation, substitute data and handle units. Full marks are given for a correct numerical answer with the appropriate unit, but marks are also awarded for working. Always show the formula, the substitution line and the final answer to three significant figures unless stated otherwise.

“计算”题考察你选择正确方程、代入数据和处理单位的能力。如果得出带适当单位的正确数值答案，能得到全部分数，但解题步骤也会得分。务必展示公式、代入行以及最终答案（一般取三位有效数字，除非另有说明）。

Write your working downwards in a logical order: ① relevant formula, ② rearranged form if needed, ③ numbers with units, ④ calculator display, ⑤ rounded answer. For example: v = u + at → v = 0 + (9.81)(2.5) → v = 24.525 → v = 24.5 m s⁻¹ (3 s.f.). Brackets help avoid substitution errors. Cross‑check your unit cancellation; if you expected m s⁻² and got N kg⁻¹, that can still be correct because N kg⁻¹ equals m s⁻².

把你的解题步骤自上而下地写出来，遵循逻辑顺序：① 相关公式，② 必要时重排公式，③ 带单位的数字，④ 计算器显示值，⑤ 四舍五入后的答案。例如：v = u + at → v = 0 + (9.81)(2.5) → v = 24.525 → v = 24.5 m s⁻¹（三位有效数字）。使用括号有助于避免代入错误。还要复核一下单位是否约掉；如果你预期得到 m s⁻² 却得到了 N kg⁻¹，这仍然可能正确，因为 N kg⁻¹ 就等于 m s⁻²。

In multi‑step calculations, carry extra significant figures through intermediate steps and only round at the end. Use the ‘Ans’ key on your calculator wisely. If the question provides values to 3 s.f., give your final answer to 3 s.f. – matching precision shows you understand uncertainty.

在多步计算中，中间步骤要保留额外的有效数字，只在最后一步进行舍入。合理使用计算器上的“Ans”键。如果题目给出的数值是三位有效数字，你的最终答案也要取三位有效数字——精度的匹配体现了你对不确定度的理解。

5. Describe | 描述

‘Describe’ questions ask ‘what’ happens, not ‘why’. You must report observations, trends from a graph, or stages of a process. Resist the urge to write about forces or energy unless the question explicitly says ‘describe and explain’. A pure describe answer should look like a witness statement: factual, sequential and free of reasoning.

“描述”题问的是“发生了什么”，而不是“为什么”。你必须报告观测结果、图表趋势或过程的各个阶段。除非题目明确说“描述并解释”，否则要克制去写力或能量的冲动。一个纯粹的描述答案应该像一份目击证词：事实性、按顺序、不含推理。

For a graph‑based describe, use the pattern: ‘As x increases from … to …, y increases/decreases linearly/non‑linearly, reaching a maximum/minimum at …’ Quote coordinates! A typical 3‑mark describe‑graph question expects: initial value, shape of curve and end behaviour. Never just say ‘it goes up then down’.

对于基于图线的描述，使用这样的句式：“当 x 从……增加到……时，y 线性/非线性地增加/减少，在……处达到最大/最小值。” 引用坐标！一道典型的3分图表描述题期望你写出：初始值、曲线形状和末端行为。千万不要只说“它先上升后下降”。

When describing an experiment’s procedure, use numbered bullet points if the exam board allows (check with your teacher). Include the apparatus, quantities measured, variables controlled and the sequence of actions. For example: ‘1. Measure mass m using a top‑pan balance. 2. Suspend the spring vertically and attach the mass…’

在描述实验步骤时，如果考试局允许，可以使用带编号的项目符号（可以向老师确认）。内容包括仪器、测量的量、控制的变量以及操作顺序。例如：“1. 用电子天平测量质量 m。2. 垂直悬挂弹簧并挂上质量……”

6. Explain | 解释

‘Explain’ is the most frequent command word for higher‑mark questions. You must link cause and effect using physical principles, laws and mechanisms. Your answer should contain the word ‘because’ or ‘therefore’ at least once, and it should make a clear causal chain: change → physical reason → outcome.

“解释”是高分题目里最常见的指令词。你必须运用物理原理、定律和机制来联系因果。你的答案至少应包含一次“因为”或“因此”，并且要呈现出一条清晰的因果链：变化 → 物理原因 → 结果。

Structure your explain response in three layers: (1) name the relevant law or principle, e.g. Faraday’s law; (2) state how a variable changes, e.g. flux increases; (3) state the consequence with reference to the given situation, e.g. hence a larger induced emf is recorded. Always anchor your explanation in the specific scenario – ‘cutting field lines’ is better than just ‘electromagnetic induction’.

将你的解释回答构建为三层结构：(1) 说出相关定律或原理，如法拉第定律；(2) 陈述某个变量如何变化，如磁通量增加；(3) 结合给定情境说明结果，如因此记录到更大的感应电动势。始终把解释锚定在具体场景中——“切割磁感线”要好过仅说“电磁感应”。

Use precise vocabulary: ‘internal energy increases because work is done on the gas’ beats ‘it gets hotter’. For particle physics, explain by referring to conservation laws: ‘the reaction cannot occur because strangeness is not conserved’. This precision impresses examiners and targets the mark points.

使用精确的词汇：“因为对气体做功，内能增加”优于“它变热了”。在粒子物理中，借助守恒定律来解释：“该反应不可能发生，因为奇异数不守恒。”这种精确性能打动考官，并直击得分点。

7. Suggest | 建议

‘Suggest’ invites you to apply your physics knowledge to a new, sometimes unfamiliar, situation. There is often more than one acceptable answer. The examiner wants to see a sensible physical idea, even if it is speculative. You do not need to prove it – just propose a plausible mechanism and link it to core physics.

“建议”邀请你把物理知识应用到一个新的、有时并不熟悉的情境中去。通常有不止一种可接受的答案。考官想看到的是合理的物理思想，哪怕是推测性的。你无需证明它——只需提出一个看似合理的机制，并与核心物理联系起来。

When faced with a suggest question, break the stem into physical quantities. Example: ‘Suggest why a longer pendulum might give less accurate period measurements.’ Think about amplitude decay, timing uncertainty or air resistance. Write: ‘A longer pendulum has a larger moment of inertia, so air resistance could damp oscillations more noticeably over several cycles, making the exact moment of passing the fiducial mark harder to judge.’

遇到建议题时，把题干分解成物理量。示例：“建议为什么更长的摆锤可能会导致周期测量的准确性降低。”思考振幅衰减、计时不确定度或空气阻力。写出：“更长的摆锤具有更大的转动惯量，因此空气阻力可能在若干个周期内更加明显地阻尼振荡，使得通过参考标记的准确时刻更难判断。”

Always include a physics‑based reason, never just a guess. Connect to a principle: energy dissipation, wave interference, quantum uncertainty, etc. If you are stuck, ask yourself ‘what could change the result?’ and work backwards. A vague ‘experimental error’ scores nothing; specify the error type and its effect.

始终要包含一个基于物理的理由，绝不能只是猜一下。联系到某个原理：能量耗散、波的干涉、量子不确定度等等。如果卡壳了，问自己“什么会改变结果？”然后逆向推导。模糊的“实验误差”不得分；要明确误差类型及其影响。

8. Evaluate | 评估

‘Evaluate’ questions demand a balanced judgement, supported by evidence. You must present both strengths and weaknesses of a model, method or conclusion, then state your overall opinion. This is usually the last part of a long question and carries the highest marks, so a structured answer is crucial.

“评估”题要求你做出平衡的判断，并有证据支持。你必须同时陈述一个模型、方法或结论的优点与不足，然后给出你的总体意见。这通常是一道大题的最后一问，分值最高，所以结构化的回答非常关键。

Begin by identifying at least two positive aspects and two limitations. Use physics arguments: ‘The method assumes the internal resistance of the voltmeter is infinite, which is not true in reality, so the measured terminal pd is slightly lower.’ Then give your conclusion: ‘Overall, this method is suitable for a preliminary investigation but insufficient for precise determination of emf.’

开头先指出至少两个正面方面和两个局限性。运用物理论证：“该方法假设电压表内阻无穷大，这在现实中并不成立，因此测得的端电压略微偏低。”然后给出结论：“总体而言，该方法适合初步调查，但不足以精确测定电动势。”

You can use comparison keywords: more/less reliable, greater/smaller systematic error, reasonable approximation. Link your evaluation to the context – lab‑based measurements, industrial safety, or astronomical observations. A clear final verdict that matches the evidence will capture the conclusion mark.

可以使用比较性关键词：更可靠/较不可靠、更大/更小的系统误差、合理的近似。把你的评估与背景联系起来——实验室测量、工业安全或天文观测。一个清晰的、与证据相符的最终结论会帮你拿到结论分。

9. Compare | 比较

‘Compare’ requires you to identify similarities and differences between two situations, graphs, or quantities. Do not just list features separately; use connective phrases like ‘both … whereas …’ or ‘X has a larger … than Y’. The examiner wants to see direct comparative language, not two separate descriptions.

“比较”要求你指出两种情境、图线或物理量之间的相似点和不同点。不要仅仅分别罗列特征；要使用像“两者都……而……”或“X 的……比 Y 更大”这样的连接短语。考官想看到直接比较的语言，而不是两个孤立的描述。

A reliable structure for a 4‑mark compare: (1) one similarity, (2) one difference, (3) another difference (often numerical), (4) a contrasting or linking statement. Example for two springs: ‘Both obey Hooke’s law for the range shown. Spring A has a higher spring constant (gradient) than spring B. Spring B stores more elastic energy at a given extension because the area under its F–x graph is larger.’

对于一道4分的比较题，可靠的结构是：(1) 一个相似点，(2) 一个不同点，(3) 另一个不同点（通常是数值上），(4) 一个对比性或联系性的陈述。以两根弹簧为例：“在所显示范围内，两者都遵守胡克定律。弹簧 A 的弹簧常数（斜率）比弹簧 B 大。在相同伸长量下，弹簧 B 储存的弹性势能更多，因为其 F–x 图下的面积更大。”

If graphs are involved, quote data and use terms like ‘steeper gradient’, ‘higher intercept’ or ‘non‑linear above …’. In circuit comparisons, compare resistance, power output or time constants. Always keep the focus on physics quantities, not superficial appearance.

如果涉及图表，引用数据并使用“更陡的斜率”“更高的截距”或“在……以上呈非线性”等术语。在电路比较中，比较电阻、输出功率或时间常数。始终聚焦于物理量，而非表面外观。

10. Determine | 确定

‘Determine’ signals that you must work out a quantity using given information, often from a graph or a data set. It is essentially a calculation, but the data may need to be extracted first (e.g. from a tangent or intercept). The key is to show how you obtained the value, not just to state it.

“确定”意味着你必须利用所给信息（通常来自图线或数据集）算出一个量。它本质上是一种计算，但数据可能需要先提取（例如从切线或截距中）。关键在于展示你是如何得到这个值的，而不仅仅是陈述它。

When determining the gradient of a curve, draw a large tangent triangle on the graph print‑out if paper‑based, or calculate Δy/Δx clearly. Label the points you use and write the coordinates. A typical instruction: ‘Determine the acceleration at t = 2.0 s from the velocity–time graph.’ Show: a = (v₂ − v₁) / (t₂ − t₁) = (15 − 5) / (3.0 − 1.0) = 5.0 m s⁻².

当需要确定曲线的斜率时，如果是纸质考试，在打印出的图线上画一个大的切线三角形，或者清晰地计算 Δy/Δx。标记你使用的点并写出坐标。典型的指令是：“通过速度–时间图确定 t = 2.0 s 时的加速度。”展示：a = (v₂ − v₁) / (t₂ − t₁) = (15 − 5) / (3.0 − 1.0) = 5.0 m s⁻²。

For intercept‑based determinations, e.g. finding the internal resistance r from a V–I graph, write the equation ε = V + Ir, rearrange to V = −r I + ε, so r = −gradient. Show clearly how you relate the mathematical result to the physical constant. Units must emerge from your manipulation, not from guesswork.

对于基于截距的确定，例如从 V–I 图求内阻 r，写出方程 ε = V + Ir，重排为 V = −r I + ε，因此 r = −斜率。清晰地展示你如何将数学结果与物理常数联系起来。单位应该从你的推导中自然得出，而不是靠猜测。

11. Show that | 证明

‘Show that’ questions give you the answer, but you must prove it. Marks are for the working, not the result. Start from a fundamental equation, rearrange it logically, substitute the given values, and arrive exactly at the stated expression or number. Rounding errors can make you ‘fail to show’ – use exact values or fractions.

“证明”题已给出答案，但你必须证明它是正确的。分数是给过程的，而不是给结果的。从基本方程出发，逻辑地重排，代入给定数值，精准地得到所陈述的表达式或数字。舍入误差会让你“无法证明”——请使用精确值或分数。

For instance: ‘Show that the kinetic energy of the electron is about 3.2 × 10⁻¹⁶ J.’ Write: v = √(2eV/m) = √(2 × 1.60×10⁻¹⁹ × 2000 / 9.11×10⁻³¹) = 2.65×10⁷ m s⁻¹, then Eₖ = ½mv² = 0.5 × 9.11×10⁻³¹ × (2.65×10⁷)² = 3.20×10⁻¹⁶ J. The phrase ‘about’ in the question allows you to round to a reasonable precision, but clearly show the unrounded intermediate.

例如：“证明电子的动能约为 3.2 × 10⁻¹⁶ J。”写出：v = √(2eV/m) = √(2 × 1.60×10⁻¹⁹ × 2000 / 9.11×10⁻³¹) = 2.65×10⁷ m s⁻¹，然后 Eₖ = ½mv² = 0.5 × 9.11×10⁻³¹ × (2.65×10⁷)² = 3.20×10⁻¹⁶ J。题目中的“约”允许你舍入到合理的精度，但要清晰地展示未舍入的中间值。

Also use ‘show that’ for formula derivations. Link each step to a definition or a law. If you need to prove P = I²R, state: ‘P = IV from the definition of power, and V = IR from Ohm’s law. Substituting V gives P = I(IR) = I²R.’ Label the equations you combine, and your derivation will gain full marks.

还可以用“证明”来做公式推导。每一步都要与定义或定律关联起来。如果你需要证明 P = I²R，陈述：“由功率的定义得 P = IV，由欧姆定律得 V = IR。代入 V 给出 P = I(IR) = I²R。”为你所用的方程加上标注，这样你的推导就会获得全部分数。

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 A-Level Physics: June 2018 Mark Scheme 1 Concepts Explained | 2018年6月A-Level物理试卷1评分标准概念解析

Mark schemes are far more than simple answer lists – they reveal exactly how examiners expect you to apply physics ideas to unfamiliar contexts. The June 2018 Paper 1 mark scheme highlights recurring themes in mechanics, waves, electricity, particles and practical skills. By dissecting the mark allocation and common errors, this article unpacks the key concepts you must master to turn a decent answer into a full-mark response.

评分标准不仅仅是答案清单：它确切地展示了考官希望你如何将物理思想应用于陌生的情境。2018年6月试卷1的评分方案揭示了力学、波、电学、粒子和实验技能中反复出现的主题。通过剖析分值分配和常见错误，本文将解读你必须掌握的关键概念，帮助你把一个不错的答案变成满分答案。

1. Measurements and Uncertainties | 测量与不确定度

The paper opened with a practical skills question requiring a micrometer reading and a calculation of percentage uncertainty in a derived quantity. The mark scheme awarded the first mark for a reading consistent with the diagram (e.g. 5.22 mm) and then expected candidates to recognise that the absolute uncertainty in a single digital reading is half the smallest division, while for a repeated measurement it is the half-range from the spread of results.

试卷以一道实验技能题开场，要求读取螺旋测微器的示数并计算导出量的百分不确定度。评分标准对与示意图一致的读数（例如5.22 mm）给第一分，然后要求考生认识到单次数显读数的绝对不确定度为最小分度的一半，而对重复测量而言则是结果分散范围的一半。

The most frequent mistake was failing to convert all lengths to the same unit before combining uncertainties. In the mark scheme, a clear statement of the percentage uncertainty formula – (absolute uncertainty / measured value) × 100% – was essential, followed by careful propagation when quantities were multiplied or divided.

最常见的错误是在合成不确定度之前未能将所有长度转换为相同单位。在评分标准中，清楚地写出百分不确定度公式——（绝对不确定度 / 测量值）×100%——至关重要，当物理量相乘或相除时还需仔细进行误差传递。

Percentage uncertainty = (Δx / x) × 100%; for z = xy, %U_z = %U_x + %U_y

百分不确定度 = (Δx / x) × 100%；对于 z = xy，%U_z = %U_x + %U_y

2. Motion Graphs and Kinematics | 运动图像与运动学

A velocity–time graph featured prominently, with marks allocated for interpreting the area under the graph as displacement and the gradient as acceleration. The mark scheme required candidates to distinguish clearly between the distance travelled (the total area) and the change in displacement (the net signed area).

速度–时间图占据了显著位置，分数分配给将图下面积解释为位移、斜率解释为加速度。评分标准要求考生明确地区分经过的路程（总面积）与位移的变化量（净带符号面积）。

In one part, a straight‑line section from t = 2 s to t = 5 s had to be analysed using the equation v = u + at. The mark scheme insisted on a clear substitution of values, with the final acceleration expressed in m s⁻². Skipping the equation statement cost a mark in some boards.

在某一问中，从 t = 2 s 到 t = 5 s 的直线段必须用 v = u + at 进行分析。评分标准坚持要求清晰地代入数值，并将最终的加速度以 m s⁻² 表示。在某些考试局中，省略写出方程会被扣分。

v = u + at    s = ut + ½at²    v² = u² + 2as

v = u + at； s = ut + ½at²； v² = u² + 2as

3. Newton’s Laws and Momentum | 牛顿定律与动量

A collision problem tested the principle of conservation of linear momentum. The mark scheme rewarded candidates who defined a positive direction and wrote a momentum conservation equation: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. Merely quoting the law without applying it to the specific scenario earned no marks.

一道碰撞题考查了线性动量守恒原理。评分标准给分给那些定义了正方向并写出动量守恒方程的考生：m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂。仅仅引用定律而没有将其应用于具体情景则不能得分。

An impulse question linked momentum change to the area under a force–time graph. The mark scheme highlighted the vector nature of impulse: if the force reversed direction, the area below the axis had to be subtracted. Many June 2018 candidates treated impulse as a scalar and lost the final mark.

一道冲量题将动量变化与力–时间图下的面积联系起来。评分标准强调了冲量的矢量特性：如果力反向，轴以下的面积必须减去。2018年6月的许多考生把冲量当作标量处理，因而丢掉了最后一分。

Impulse = Δp = F_avg × Δt = area under F–t graph

冲量 = 动量变化 = F_avg × Δt = F–t 图下面积

4. Work, Energy and Power | 功、能量与功率

A vehicle travelling up an incline required the application of the work–energy principle. The mark scheme expected a breakdown of energy changes: gain in gravitational potential energy (mgh) plus work done against friction equals the work done by the engine force (F × d).

一辆沿斜坡行驶的车辆需要用功能原理来解答。评分标准期望对能量变化进行分解：重力势能的增加量 (mgh) 加上克服摩擦所做的功等于发动机力所做的功 (F × d)。

The question then asked for the useful power output. The mark scheme awarded a mark for recognising that power is the rate of doing work, P = Fv, with careful unit conversion to watts. A common pitfall was confusing the total power with the useful power after subtracting losses.

接着这问要求计算有用输出功率。评分标准对认识到功率是做功的速率 P = Fv 给分，并要求仔细地将单位换算为瓦特。一个常见的陷阱是将总功率与减去损耗后的有用功率搞混。

ΔE_k = ½m(v² − u²)    ΔE_p = mgΔh    P = ΔW / Δt = Fv

动能变化 = ½m(v² − u²)； 重力势能变化 = mgΔh； 功率 = ΔW / Δt = Fv

5. Superposition and Stationary Waves | 波的叠加与驻波

June 2018 Paper 1 contained a question on a stretched string vibrating in its second harmonic. The mark scheme required the drawing of a stationary wave pattern with nodes at the fixed ends and an antinode at the centre, correctly labelled with wavelength λ = L (the length of the string).

2018年6月试卷1包含一道关于张紧的弦在其第二谐振频率下振动的题目。评分标准要求画出驻波图案，两端为波节，中心为波腹，并正确地标出波长 λ = L（弦的长度）。

The explanation of how a stationary wave is formed was worth two marks: one for the idea of two progressive waves of the same frequency and amplitude travelling in opposite directions, and one for the superposition producing points of zero displacement (nodes) and maximum displacement (antinodes).

解释驻波是如何形成的值两分：一分是说明两列频率和振幅相同但传播方向相反的行波这一概念，另一分是叠加产生位移为零的点（波节）和位移最大的点（波腹）。

For a string fixed at both ends: f_n = n × (v / 2L), n = 1, 2, 3…

两端固定的弦：f_n = n × (v / 2L)，n = 1, 2, 3…

6. Refraction and Critical Angle | 折射与临界角

A ray of light passing from glass to air provided the context for Snell’s law and total internal reflection. The mark scheme insisted on stating n₁sinθ₁ = n₂sinθ₂, with angles measured from the normal. A common error was taking sinθ_c = n₁/n₂ incorrectly; the correct expression is sinθ_c = n₂/n₁ when n₂ < n₁.

一束从玻璃射向空气的光线提供了斯涅尔定律和全内反射的背景。评分标准坚持要求写出 n₁sinθ₁ = n₂sinθ₂，且角度要从法线量起。一个常见错误是错误地使用 sinθ_c = n₁/n₂；正确的表达式在 n₂ < n₁ 时应为 sinθ_c = n₂/n₁。

The mark scheme then linked critical angle to the refractive index. An exam tip emerging from the June 2018 mark scheme is that when asked to “explain the condition for total internal reflection”, candidates must mention both that the incident angle must exceed the critical angle and that light must be travelling from an optically denser medium to a less dense one.

随后评分标准将临界角与折射率联系起来。从2018年6月评分标准中得出的考试技巧是，当被要求“解释全内反射的条件”时，考生必须同时提到入射角必须超过临界角，并且光必须从光密介质射向光疏介质。

n = sin i / sin r    sin θ_c = 1 / n    (for light going from medium to vacuum)

n = sin i / sin r； sin θ_c = 1 / n （适用于光从介质射向真空）

7. I–V Characteristics and Resistance | 电流–电压特性与电阻

An electricity item presented the I–V graph of a filament lamp and asked why it is non‑ohmic. The mark scheme awarded a mark for stating that the temperature of the filament increases as the current increases, and another mark for linking the increased temperature to greater lattice ion vibrations, which cause more frequent collisions with electrons and thus higher resistance.

一道电学题目给出了白炽灯的 I–V 图，并问为什么它是非欧姆性的。评分标准对说明灯丝温度随电流升高而升高给一分，另一分则将温度升高与晶格离子振动加剧联系起来，离子振动加剧导致与电子的碰撞更加频繁，从而电阻变大。

A second sub‑question required calculation of resistance at a specific voltage using R = V / I from the graph. The mark scheme accepted a tangent or chord method, provided the values were read correctly. Failure to double the uncertainty when converting from mA to A was penalised.

第二小问要求利用图线用 R = V / I 计算特定电压下的电阻。评分标准接受切线法或弦线法，前提是读数正确。从 mA 转换为 A 时未能换算正确、或者忽略了倍率都会扣分。

For an ohmic conductor: V ∝ I, R constant; for a filament lamp: R increases with V.

对欧姆导体：V ∝ I，R 恒定；对白炽灯：R 随 V 增大而增大。

8. Resistivity and Circuit Analysis | 电阻率与电路分析

A practical context – measuring the resistivity of a metal wire – required the combination of R = ρL / A with the measurements of length, diameter and resistance. The mark scheme specified that the diameter must be averaged from several readings taken along the wire and then halved to give the radius for the cross‑sectional area A = πr².

一个实验情景——测量金属丝的电阻率——需要结合 R = ρL / A 与长度、直径和电阻的测量值。评分标准明确指出，直径必须沿导线多次测量取平均值，然后除以二得到半径，再计算截面积 A = πr²。

In the circuit analysis question, a potential divider with a thermistor was used. The mark scheme expected candidates to explain that as temperature rises, the thermistor’s resistance falls, so the share of the supply voltage across the fixed resistor increases, while the voltage across the thermistor decreases. If the candidate merely stated “resistance changes” without explaining the effect on voltmeter readings, only partial credit was given.

在电路分析题中，用到了一个包含热敏电阻的分压器。评分标准期望考生解释：随着温度升高，热敏电阻的阻值下降，因此固定电阻两端的电压份额增加，而热敏电阻两端的电压减小。如果考生仅仅说“电阻改变”而没有解释对电压表读数的影响，只能得到部分分数。

ρ = RA / L    A = πd² / 4    V_out = (R₂ / (R₁ + R₂)) × V_in

ρ = RA / L； A = πd² / 4； V_out = (R₂ / (R₁ + R₂)) × V_in

9. Quantum Phenomena: Photoelectric Effect | 量子现象：光电效应

A graph of maximum kinetic energy against frequency appeared, requiring interpretation using Einstein’s photoelectric equation. The mark scheme awarded marks for identifying the gradient as Planck’s constant h and the intercept on the frequency axis as the threshold frequency f₀. Candidates who mislabelled the work function Φ as the y‑intercept lost the interpretation mark.

一道关于最大动能–频率图线的题目需要用爱因斯坦光电方程来解释。评分标准给分给将斜率认定为普朗克常量 h、频率轴上的截距认定为截止频率 f₀。那些误将与 y 轴截距标为功函数 Φ 的考生则丢掉了释义分。

The explanation of why intensity does not affect the maximum kinetic energy was worth two marks. According to the mark scheme, intensity determines the photon arrival rate, but each photon’s energy depends solely on frequency (E = hf). One photon can only liberate one electron; if hf < Φ, no electrons are emitted regardless of intensity.

解释为什么光强不影响最大动能值两分。根据评分标准，光强决定光子到达的速率，但每个光子的能量仅取决于频率 (E = hf)。一个光子只能释放一个电子；若 hf < Φ，无论光强多大都不发射电子。

E_k_max = hf − Φ    Φ = hf₀

E_k_max = hf − Φ； Φ = hf₀

10. Particle Physics: Conservation Laws | 粒子物理：守恒定律

A particle interaction question tested the recognition of conservation of baryon number, lepton number and charge. The June 2018 mark scheme gave clear marks for stating the baryon numbers of proton (1) and neutron (1), and the lepton numbers of electron (1) and electron neutrino (1). Anti‑particles were assigned opposite numbers.

一道粒子相互作用题考查了对重子数、轻子数和电荷守恒的识别。2018年6月的评分标准明确地给分给写出质子（1）和中子（1）的重子数，以及电子（1）和电子中微子（1）的轻子数。反粒子则赋予相反的数值。

Particularly tricky was a strand involving strangeness, which is not conserved in weak interactions. The mark scheme rewarded candidates who identified the weak interaction signature (change in quark flavour) and stated that strangeness can change by ±1 in such processes.

特别棘手的一小问涉及奇异数，它在弱相互作用中不守恒。评分标准给分给那些识别出弱相互作用特征（夸克味变）并说明在这种过程中奇异数可以变化 ±1 的考生。

11. Practical Skills: Graph Plotting and Gradient Analysis | 实验技能：绘图与斜率分析

Throughout the June 2018 mark scheme, practical skills were assessed via graph questions. To gain full marks, axis labels had to include quantity and unit in the format “voltage / V”. The scale had to use more than half of the grid, and plotting points had to be to the nearest half‑square. Any “anomalous” point had to be circled and ignored in the line of best fit.

在2018年6月的评分标准中，实验技能通过图表题目进行了考察。要拿到满分，坐标轴标签必须以“电压 / V”的格式包含物理量和单位。刻度必须占据超过一半的网格，描点要精确到最近的半格。任何一个“异常”点都必须圈出，并在最佳拟合线中将其忽略。

When calculating the gradient, the mark scheme demanded a large triangle drawn on the graph, with coordinates explicitly stated. The final value had to be given to an appropriate number of significant figures, typically mirroring the data’s precision. Substituting the gradient into a physical formula (e.g. g = 4π² / gradient for a pendulum) was a step many candidates omitted.

在计算斜率时，评分标准要求在图上画一个大三角形，并明确写出坐标。最终数值的有效数字位数必须合适，通常与数据的精度相匹配。将斜率代入物理公式（例如摆的 g = 4π² / 斜率）是许多考生遗漏的步骤。

Gradient = Δy / Δx (with units) | 斜率 = Δy / Δx（带单位）

12. Tackling the Mark Scheme Mindset | 把握评分标准的思维模式

The June 2018 Paper 1 mark scheme consistently rewarded precise scientific language. For example, using “work done per unit charge” for potential difference rather than a vague “voltage”. Similarly, “rate of change of momentum” earned the mark for force, while “force times distance” alone was insufficient without stating the condition “in the direction of the force”.

2018年6月试卷1的评分标准始终如一地奖励精确的科学语言。例如，用“每单位电荷所做的功”来描述电势差，而不是模糊的“电压”。类似地，“动量的变化率”能得到力的那分，而仅仅说“力乘以距离”而不说明“在力的方向上”是不够的。

To succeed, you must treat the mark scheme as a checklist of key phrases. In longer explanations, the June 2018 examiners expected a logical sequence: state the underlying principle, apply it to the given numbers, and then articulate the outcome with a correct unit. Breaking a six‑mark question into bullet points of the three key steps can prevent the omission of vital linking statements.

要想成功，你必须把评分标准当作关键短语的核查清单。在较长的解释题中，2018年6月的考官期望看到逻辑的次序：陈述基本原理，将其应用于给定的数值，然后用正确的单位阐明结果。将一个6分的大题拆解成三个关键步骤的要点可以防止遗漏关键的连接语句。

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 Practical and Analytical Skills in A-Level Physics: Key Concepts Explained | A-Level物理实践与分析技能核心概念解析

Practical and analytical skills are the backbone of experimental physics. In the Oxford AQA International A-Level Physics course, students must not only understand theoretical concepts but also demonstrate the ability to design experiments, collect and process data, evaluate uncertainties, and draw valid conclusions. This article unpacks the essential concepts you need to master for the practical and analytical skills component.

实践与分析技能是实验物理的基石。在Oxford AQA国际A-Level物理课程中，学生不仅需要理解理论概念，还必须展示设计实验、收集与处理数据、评估不确定度并得出有效结论的能力。本文将解析你必须掌握的关键概念，以应对实践与分析技能部分的要求。

1. Variables and Experimental Design | 变量与实验设计

In any experiment, you must identify the independent variable (the one you change), the dependent variable (the one you measure), and the control variables (those kept constant). A well-designed experiment changes only one independent variable at a time to establish a clear cause-and-effect relationship.

在任何实验中，你必须确定自变量（你改变的变量）、因变量（你测量的变量）以及控制变量（需保持不变的变量）。设计良好的实验每次只改变一个自变量，以建立清晰的因果关系。

Use preliminary investigations to determine suitable ranges and intervals for measurements. For example, when investigating the period of a pendulum, you might test different lengths and decide to take measurements every 10 cm to obtain enough data points for a reliable graph.

通过预实验确定测量的合适范围和间隔。例如，在研究单摆周期时，你可以测试不同摆长，并决定每隔10厘米测量一次，以获取足够的数据点绘制可靠的图形。

Always record your experimental plan clearly, including a labelled diagram of the apparatus and a step-by-step method. This allows another person to replicate your experiment.

始终清晰地记录实验计划，包括带标注的仪器示意图和分步方法。这使他人能够重复你的实验。

2. Measurement Techniques and Instrumentation | 测量技术与仪器

Select instruments with appropriate precision and resolution. For instance, a micrometer screw gauge can measure to 0.01 mm, while a metre ruler typically has a resolution of 1 mm. The choice depends on the required accuracy.

选择具有适当精密度和分辨率的仪器。例如，千分尺可测量到0.01 mm，而米尺的分辨率通常为1 mm。选择取决于所需的准确度。

To reduce random error, take multiple readings and calculate the mean. For a pendulum’s period, measure the time for 10 oscillations and then divide by 10; this diminishes the impact of reaction time.

为了减少随机误差，进行多次测量并计算平均值。对于单摆的周期，测量10次全振动的时间再除以10；这能减小反应时间的影响。

Understand the difference between precision (the spread of repeated measurements) and accuracy (how close a measurement is to the true value). A precise set of data may still be inaccurate if there is a systematic error.

理解精密度（重复测量的离散程度）与准确度（测量值接近真值的程度）的区别。如果存在系统误差，一组精密的数据仍可能不准确。

3. Types of Errors and Uncertainty | 误差类型与不确定度

Random errors cause readings to be scattered around the true value. They can be reduced by averaging many measurements. Examples include human reaction time or fluctuations in environmental conditions.

随机误差导致读数分散在真值附近。可通过多次测量取平均值来减小。例子包括人的反应时间或环境条件的波动。

Systematic errors shift all measurements in the same direction, often due to faulty equipment or flawed technique. They cannot be eliminated by averaging but can be identified by using a different method or calibrating instruments.

系统误差使所有测量值朝同一方向偏移，通常源于仪器故障或方法缺陷。无法通过取平均值消除，但可通过使用不同方法或校准仪器来识别。

The absolute uncertainty of a single measurement is usually taken as the smallest division of the instrument (or half of it, depending on the context). For digital instruments, it is the resolution of the display, e.g., ±0.01 V for a voltmeter reading to two decimal places.

单次测量的绝对不确定度通常取仪器最小刻度值（或根据情况取一半）。对于数字仪器，则为显示屏的分辨率，例如读数为两位小数的电压表为 ±0.01 V。

4. Combining Uncertainties | 不确定度的合成

When you add or subtract quantities, the absolute uncertainties add. If A = B + C, then ΔA = ΔB + ΔC.

当物理量相加或相减时，绝对不确定度相加。若 A = B + C，则 ΔA = ΔB + ΔC。

For multiplication or division, percentage (or fractional) uncertainties add. If R = X × Y or R = X / Y, then the percentage uncertainty in R is the sum of the percentage uncertainties in X and Y.

对于乘除运算，百分比不确定度（或相对不确定度）相加。若 R = X × Y 或 R = X / Y，则 R 的百分比不确定度等于 X 和 Y 的百分比不确定度之和。

常见不确定度合成规则如下表所示。注意对于幂函数，百分比不确定度乘以指数。

Common uncertainty combination rules are shown in the table below. Note that for a power, the percentage uncertainty is multiplied by the exponent.

Always convert to absolute uncertainty at the end if needed. For example, if a length L = 2.00 ± 0.01 m and width W = 0.50 ± 0.01 m, the area A = L × W = 1.00 m². The % uncertainties are 0.5% and 2% respectively, so total % uncertainty in area is 2.5%, giving an absolute uncertainty of 0.025 m². Thus A = 1.00 ± 0.03 m² (rounded appropriately).

若需要，最终转换为绝对不确定度。例如，长度 L = 2.00 ± 0.01 m，宽度 W = 0.50 ± 0.01 m，面积 A = L × W = 1.00 m²。百分比不确定度分别为0.5%和2%，则面积的百分比不确定度为2.5%，绝对不确定度为0.025 m²，因此 A = 1.00 ± 0.03 m²（适当修约后）。

5. Graphical Representation of Data | 数据的图形表示

Plot the independent variable on the x-axis and the dependent variable on the y-axis. Use sensible scales that spread the data over more than half of the graph paper in each direction. Label axes with quantity and unit, e.g., ‘Time t / s’.

将自变量标绘在 x 轴上，因变量标绘在 y 轴上。使用合理的标度，使数据点占据每个方向超过一半的图纸面积。坐标轴标注物理量和单位，如’时间 t / s’。

Plot data points with small crosses or encircled dots, and immediately draw error bars if uncertainties are known. The length of an error bar represents the absolute uncertainty in that measurement. If uncertainty in x is negligible, you may only need vertical error bars.

用小十字或带圈的圆点标绘数据点，如果已知不确定度，立即画出误差棒。误差棒的长度代表该测量的绝对不确定度。若 x 的不确定度可忽略，可能只需垂直误差棒。

A line of best fit should be a straight line or smooth curve that passes through as many error bars as possible and has a balanced number of points on either side. Do not force the line through the origin unless there is a theoretical reason.

最佳拟合线应为一条尽可能穿过更多误差棒且两侧点数均衡的直线或光滑曲线。除非有理论依据，否则不要强制直线通过原点。

6. Linearising Graphs and Logarithmic Plots | 线性化图形与对数作图

Many physical relationships are not linear, e.g., y = kx². To test such a relationship, you can plot y against x²; if the graph is a straight line, the relationship is verified. This process is called linearising.

许多物理关系并非线性，例如 y = kx²。为检验这种关系，可绘制 y 对 x² 的图；如果图形为一条直线，则关系得证。这一过程称为线性化。

For exponential or power-law relationships, logarithmic plots are useful. If y = axⁿ, then log y = log a + n log x. Plotting log y against log x yields a straight line with gradient n and intercept log a. Semi-log plots (log y vs x) test exponential decay.

对于指数或幂律关系，对数作图很有用。若 y = axⁿ，则 log y = log a + n log x。绘制 log y 对 log x 的图会得到一条斜率为 n、截距为 log a 的直线。半对数图（log y 对 x）可检验指数衰减。

Use natural logs (ln) or log base 10; the gradient will change accordingly. Ensure you label axes as ‘ln (y)’ and ‘ln (x)’, and be careful with units.

使用自然对数（ln）或以10为底的对数；斜率会相应变化。确保坐标轴标注为’ln (y)’和’ln (x)’，并注意单位。

7. Determining Gradients and Intercepts | 梯度和截距的确定

To find the gradient of a straight-line graph, choose two points on the line of best fit that are far apart, not data points. Calculate gradient = (y₂ − y₁) / (x₂ − x₁). Read coordinates directly from the line.

要计算直线图形的梯度，在最佳拟合线上选取两个相距较远的点（非数据点），计算梯度 = (y₂ − y₁) / (x₂ − x₁)。直接从线上读取坐标。

The y-intercept is found by extending the line to x = 0 and reading the y-value, or by substituting a coordinate into the line equation y = mx + c. This intercept often gives a physical

Published by TutorHao | A-Level Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 A-Level Physics Unit 3 Mark Scheme Jan 2020: Concept Analysis | A-Level 物理第3单元2020年1月评分方案概念解析

The Unit 3 paper for Edexcel International A-Level Physics (WPH13) focuses on assessing practical skills through a written examination. The January 2020 mark scheme reveals the specific concepts and conventions that examiners expect students to understand and apply when handling experimental data, analysing results, and evaluating procedures. This article delves into these concepts, providing a clear breakdown to support effective revision.

爱德思国际A-Level物理第3单元（WPH13）的考试以笔试形式评估实验技能。2020年1月的评分方案揭示了考官期望学生理解和应用的具体概念与规范，涉及实验数据处理、结果分析和程序评估等方面。本文深入解析这些概念，提供清晰的分解以助力高效复习。

1. Table Design and Recording Data | 表格设计与数据记录

When recording measurements, the independent variable is placed in the left column and the dependent variable in the right. Each column heading must include the quantity and its unit, separated by a forward slash, e.g., ‘Time t / s’.

记录测量值时，自变量置于左列，因变量置于右列。每个列标题必须包含物理量及其单位，用斜杠分隔，例如“时间 t / s”。

All raw readings should be recorded to the same number of decimal places, reflecting the precision of the measuring instrument. For example, if a ruler reads to 1 mm, lengths should be stated as 12.3 cm, not 12 cm.

所有原始读数应记录相同的小数位数，以反映测量仪器的精度。例如，若一把尺子可读至1 mm，长度应记录为12.3 cm，而非12 cm。

Repeated measurements are vital to improve reliability. The mean should be calculated and recorded to an appropriate number of significant figures, usually the same or one more than the raw data.

重复测量对提高可靠性至关重要。应计算平均值并以适当的有效数字记录，通常与原始数据位数相同或多一位。

Below is a comparison of poor and good table design:

下面对比不良与良好的表格设计：

2. Plotting Graphs and Choosing Scales | 绘制图表与选择标度

Graphs must have clearly labelled axes with quantities and units. The scale should be chosen such that the plotted points occupy more than half of the grid in both directions, and the scale should be linear (not broken) unless a false origin is justified.

图表必须具有清晰标注的坐标轴，包含物理量和单位。标度应使数据点在两个方向上占网格的一半以上，且标度应为线性（无中断），除非有合理理由使用假原点。

Plotting points accurately with small crosses or encircled dots is essential; the mark scheme penalises blobs or inaccurate placement. Use a sharp pencil and check each coordinate.

用小十字或圈点准确标记数据点至关重要；评分方案会扣掉涂鸦或位置不准的分数。应使用削尖的铅笔，并检查每个坐标。

The dependent variable is conventionally plotted on the vertical axis. The title of the graph should indicate the relationship, e.g., ‘Graph of T² against l’.

通常将因变量绘制在纵轴上。图标题应说明关系，例如“T² 对 l 的图”。

3. Drawing Lines of Best Fit | 绘制最佳拟合线

A line of best fit must be a single, thin, straight line (if the trend is linear) that balances the points above and below. It does not have to pass through the origin unless the physics predicts that and the data strongly supports it.

最佳拟合线必须是一条细而直的单一线条（若趋势为线性），并平衡上下各点的分布。除非物理理论预测如此且数据有力支持，否则不必通过原点。

Identify and mark any anomalous points that deviate significantly from the overall trend. These should be excluded from the line-fitting but labelled on the graph.

识别并标出明显偏离总趋势的异常点。这些点应排除在拟合之外，但在图上标注。

The line of best fit must not be forced to curve; if the data show a curve, a smooth curve should be drawn with equal numbers of points on either side where possible.

最佳拟合线不应强行画成曲线；若数据呈现曲线，则应用平滑曲线绘制，尽量让两侧点数相等。

4. Calculating the Gradient | 计算斜率

The gradient must be calculated from the line of best fit, not from data points. Use a large triangle, reading coordinates from the graph grid as accurately as possible. Show the triangle and the coordinates used.

斜率必须根据最佳拟合线计算，而非从数据点计算。应使用大三角形，尽可能准确地从图网格读取坐标。必须显示三角形及所用坐标。

gradient = (y₂ − y₁) / (x₂ − x₁)

Express the gradient with its appropriate unit, which is the unit of the y-axis divided by the unit of the x-axis. For example, if y is T² / s² and x is l / m, then the gradient unit is s² m⁻¹.

用适当的单位表示斜率，即 y 轴单位除以 x 轴单位。例如，若 y 为 T² / s²，x 为 l / m，则斜率单位为 s² m⁻¹。

The mark scheme often awards marks for stating the coordinates of the points taken from the best-fit line and for showing the subtraction steps clearly.

评分方案通常奖励从最佳拟合线上取点坐标并清晰展示减法步骤的做法。

5. Determining the y-Intercept | 确定 y 截距

The y-intercept is the value where the line of best fit crosses the y-axis. If the x-axis starts at a non-zero value, you may need to calculate the intercept using the line equation: y = mx + c → c = y − mx.

y 截距是最佳拟合线与 y 轴相交的值。若 x 轴从非零值开始，可能需要用直线方程计算截距：y = mx + c → c = y − mx。

Always quote the y-intercept with its correct unit. The intercept can provide information about systematic errors or fixed values in the experiment.

始终以正确单位报出 y 截距。截距可提供关于系统误差或实验中固定值的信息。

In the mark scheme, credit is given for reading a valid point on the line and using correct substitution into the equation.

在评分方案中，从线上读取有效点并正确代入方程即可得分。

6. Uncertainty Analysis from Graphs | 从图表分析不确定度

To estimate the uncertainty in the gradient, draw the maximum and minimum slope lines that still fit within the error bars or the scatter of points. The uncertainty is given by:

Δgradient = (gradient_max − gradient_min) / 2

为估算斜率的不确定度，应绘制两条仍能拟合误差棒或数据点散布的最大和最小斜率线。不确定度由下式给出：

Δ梯度 = (最大斜率 − 最小斜率) / 2

When error bars are not given, use the worst-fit line method: a line that passes through all the error bars or, if absent, one that deviates most from the best-fit while still being plausible. The percentage uncertainty is then calculated.

在未给出误差棒的情况下，使用最差拟合线法：一条穿过所有误差棒的线，或若没有误差棒，一条偏离最佳拟合而仍然合理的线。然后计算百分比不确定度。

The y-intercept uncertainty can be found similarly by noting the intercepts of the extreme lines. Always express the final result as: best value ± uncertainty.

类似地，可通过记录极端线的截距求得 y 截距的不确定度。最终结果始终表示为：最佳值 ± 不确定度。

7. Percentage Uncertainty and Derived Quantities | 百分比不确定度与导出量

Percentage uncertainty of a measurement is: %U = (absolute uncertainty / measured value) × 100%. This is useful when comparing precision of different measurements.

测量的百分比不确定度为：%U = (绝对不确定度 / 测量值) × 100%。这有助于比较不同测量的精密度。

When a quantity is derived from multiplication or division, the maximum percentage uncertainty is the sum of the individual percentage uncertainties. For a quantity Q = ab/c:

%U(Q) = %U(a) + %U(b) + %U(c)

当导出量来自乘法和除法时，最大百分比不确定度等于各百分比不确定度之和。对于 Q = ab/c：

%U(Q) = %U(a) + %U(b) + %U(c)

If the quantity has an exponent, multiply the percentage uncertainty by that exponent. For example, if volume V = (4/3)πr³, then %U(V) = 3 × %U(r).

若物理量含指数，则将百分比不确定度乘以该指数。例如，若体积 V = (4/3)πr³，则 %U(V) = 3 × %U(r)。

The mark scheme often requires comparing an experimental value with an accepted value by calculating percentage difference: % difference = |experimental − accepted| / accepted × 100%.

评分方案常要求通过计算百分比差异来比较实验值与公认值：% 差异 = |实验值 − 公认值| / 公认值 × 100%。

8. Systematic and Random Errors | 系统误差与随机误差

Systematic errors shift all readings in one direction (e.g., zero offset, poorly calibrated instrument). They affect accuracy but not precision. The mark scheme rewards identifying a specific systematic error and explaining how it skews the result.

系统误差使所有读数向同一方向偏移（例如零位偏差、仪器校准不良）。它们影响准确度，但不影响精密度。评分方案奖励指出具体系统误差并解释其如何使结果产生偏差。

Random errors cause scatter about the true value (e.g., human reaction time, fluctuations in readings). They affect precision and can be reduced by taking many repeated measurements and averaging.

随机误差导致读数在真值附近散布（例如人的反应时间、读数波动）。它们影响精密度，可通过多次重复测量并取平均值来减小。

Being able to distinguish between the two types of errors and suggest relevant examples from the experiment is a key assessment objective. For instance, parallax error from reading a scale at an angle is a random error if not consistent, but may become systematic if always read from the same wrong angle.

能够区分两种误差类型并结合实验情境举例是一项关键的评估目标。例如，从一定角度读数产生的视差误差，若方向不固定则为随机误差，但若总是从同一错误角度读数，则变为系统误差。

9. Evaluating the Experiment | 评估实验

A strong evaluation goes beyond simply stating that an error exists. It explains the source of the error, its effect on the final measured quantity (e.g., ‘this would cause an overestimate of g’), and suggests a realistic improvement.

出色的评估不止于指出存在误差，还需说明误差的来源、其对最终测量量的影响（例如“这将导致 g 值被高估”），并提出切实可行的改进措施。

Common evaluation points include: time measurements for a single oscillation being too short (use multiple oscillations), difficulty in judging the exact point of release (use a fiducial marker), or friction introducing systematic errors.

常见的评估要点包括：单次振荡的时间测量过短（可用多次振荡）、难以判断精确释放点（可用参考标记）、或摩擦力引入系统误差。

The mark scheme credits linking the evaluation to the specific apparatus used and identifying the most significant source of uncertainty, often by comparing percentage uncertainties of different measured quantities.

评分方案奖励将评估与使用的具体仪器联系起来，并通过比较不同测量量的百分比不确定度来识别最重要的不确定度来源。

10. Suggesting Improvements and Safety | 提出改进与安全措施

Improvements must be practical and detailed. Instead of saying ‘use more accurate equipment’, specify a named instrument (e.g., ‘use a light gate connected to a data logger to eliminate human reaction time’).

改进措施必须具体且切实可行。不应说“使用更精确的设备”，而应指定具体仪器名称（例如“使用带数据记录仪的光电门以消除人的反应时间”）。

Safety considerations are occasionally credited. For instance, when stretching a wire, wear safety goggles to protect eyes from snapping; when using free-fall apparatus, use a soft landing pad to avoid breaking the impact sensor.

安全考量偶尔也会赋分。例如，在拉伸金属丝时，佩戴护目镜以防金属丝断裂伤眼；在使用自由落体装置时，使用软着陆垫以避免损坏撞击传感器。

The mark scheme rewards an understanding that reducing the largest percentage uncertainty in a measurement will most improve the overall experiment. Hence, students should identify which measurement contributes the dominant uncertainty and target improvements there.

评分方案奖励理解减小测量中最大的百分比不确定度最能改进整体实验。因此，学生应确定哪个测量贡献了主要的不确定度，并针对其进行改进。

Finally, always ensure that any suggested improvement does not introduce new significant hazards. A brief justification of why the improvement works is often required for full marks.

最后，始终确保任何建议的改进不会引入新的重大危险。通常需要简要说明改进有效的原因，才能得到满分。

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）