Tag: Physics

📚 Complete Physics Experimental Investigations | 完整物理实验探究

Mastering experimental physics is essential for any student tackling practical examinations in IGCSE or A-Level science courses. This guide covers the complete process of planning, conducting, analysing, and evaluating experiments, with an emphasis on key skills such as controlling variables, handling uncertainties, and interpreting data. From mechanics to electricity and waves, you will gain a solid foundation in experimental methodology that can be applied across all topics.

掌握物理实验探究对于任何参加IGCSE或A-Level科学课程实践考试的学生都至关重要。本指南涵盖了从实验规划、实施、分析到评估的完整过程，重点在于控制变量、处理不确定度和解释数据等关键技能。从力学到电学再到波动，你将获得可应用于所有主题的实验方法的坚实基础。

1. Designing an Experiment | 设计实验

A well-designed experiment starts with a clear aim or research question. The aim should be specific, measurable, and based on a scientific hypothesis. For example: ‘Investigate how the length of a pendulum affects its period.’ Before writing a detailed method, identify the independent variable (the factor you change) and the dependent variable (the factor you measure).

一个设计良好的实验始于明确的实验目的或研究问题。该目的应具体、可测量并基于科学假说。例如：“探究摆长如何影响单摆的周期。”在撰写详细方法之前，应先确定自变量（你改变的因素）和因变量（你测量的因素）。

Next, list all equipment needed, including measuring instruments, and decide on a suitable range and number of readings. A minimum of five readings for the independent variable is recommended to establish a trend. Always include a fair test plan: keep control variables constant. For the pendulum, control variables might include the mass of the bob and the amplitude of swing.

接下来，列出所有所需设备，包括测量仪器，并确定合适的自变量范围和读数次数。建议至少采集五组自变量数据以确定趋势。始终安排公平测试方案：保持控制变量不变。对于单摆，控制变量可能包括摆锤的质量和摆动幅度。

2. Variables and Controls | 变量与控制

Independent variable: The factor you deliberately change. In a pendulum experiment, it is the length of the string. Plot this on the x-axis of a graph unless the relationship is not straightforward.

自变量：你有意改变的因素。在单摆实验中，它是绳长。除非关系不是直接的，否则通常将它绘制在图的x轴上。

Dependent variable: The factor you measure as a result. Here, it is the period (T). This is plotted on the y-axis.

因变量：你作为结果测量的因素。此处，它是周期（T）。这绘制在y轴上。

Control variables: Factors that must be kept the same to ensure a fair test. For the pendulum, keep the bob mass constant, the initial displacement (small angles only, less than 10°) constant, and avoid air drafts. Record the values of control variables so the experiment can be replicated.

控制变量：为确保公平测试而必须保持不变的因素。对于单摆，保持摆锤质量不变，初始摆角不变（仅小角度，小于10°），并避免气流。记录控制变量的值，以便实验可以重复。

3. Measurement Techniques | 测量技术

Use instruments appropriate to the required precision. A metre ruler is fine for pendulum length (±0.1 cm or ±0.5 mm depending on scale). For time, a digital stopwatch provides a resolution of 0.01 s, but human reaction time introduces a larger uncertainty, typically ±0.2 s.

使用适合所需精度的仪器。米尺适用于测量摆长（根据刻度，不确定度可为±0.1 cm或±0.5 mm）。对于时间，数字秒表提供0.01 s的分辨力，但人的反应时间会引入更大的不确定度，通常为±0.2 s。

To reduce random errors, repeat measurements and take an average. For the period, measure the time for 10 or 20 complete swings and then divide by the number of swings to find the period. This technique reduces the impact of reaction time uncertainty per swing.

为减少随机误差，应重复测量并取平均值。对于周期，可以测量10次或20次完整摆动的时间，然后除以摆动次数来求得周期。此技巧降低了每次摆动反应时间不确定度的影响。

For length, measure from the point of suspension to the centre of the bob. Mark the centre clearly. For small amplitudes, the effective length is the distance to the centre of mass.

对于长度，应从悬挂点测量到摆锤的中心。清楚标记中心。对于小振幅，有效长度是到质心的距离。

4. Data Collection and Recording | 数据收集与记录

Record raw data in a clearly labelled table, using correct units in headings (e.g., Length l / cm). Do not write units inside the data cells. Include columns for independent, dependent, and any derived quantities (e.g., time for 10 swings, period). Show repeated readings and averages.

在标注清晰的表格中记录原始数据，表头使用正确的单位（例如，长度 l / cm）。不要在数据单元格内写入单位。包括自变量、因变量和任何派生量（例如，10次摆动的时间、周期）的列。展示重复读数和平均值。

All data should be recorded to the same decimal place consistent with the measuring instrument’s precision. For example, if using a metre ruler with mm marks, record length as 25.0 cm, not 25 cm.

所有数据应记录到与测量仪器精度一致的小数位。例如，如果使用带毫米刻度的米尺，应将长度记录为25.0 cm，而非25 cm。

If anomalous results occur, identify them and repeat those measurements if time permits. Do not erase anomalous data; instead, circle them and note possible reasons.

如果出现异常结果，识别它们并在时间允许的情况下重复这些测量。不要擦除异常数据；而是将其圈出，并注明可能的原因。

5. Tables and Graphs | 表格与图表

When plotting a graph, use a sharp pencil for points and line of best fit. Label axes with quantity and unit (e.g., Period squared T² / s²). Choose a sensible scale that uses more than half of the graph paper in both directions, avoid awkward scales like intervals of 3 or 7.

在绘制图表时，用削尖的铅笔描点和画最佳拟合线。坐标轴标注物理量和单位（例如，周期平方 T² / s²）。选择合理的刻度，使图形在双向都占满半张坐标纸以上，避免使用像3或7这样棘手的间隔。

Plot data points as small crosses (×) or encircled dots. Draw a best-fit straight line or smooth curve. For a straight line, use a transparent ruler. If the line does not pass through the origin, it may indicate a systematic error or an intercept.

将数据点画为小叉号（×）或带圈的点。画一条最佳拟合直线或平滑曲线。对于直线，使用透明直尺。如果直线不通过原点，可能表明存在系统误差或截距。

Calculate gradient using a large triangle; do not use plotted data points for the gradient triangle. Show coordinates on the graph. State the gradient with units. If the relationship is linearised (e.g., T² vs l for a pendulum), the gradient has physical meaning: g = 4π² / gradient.

使用大三角形计算斜率；不要将绘制的数据点用于斜率三角形。在图上显示坐标。说明斜率并带单位。如果关系已线性化（例如，单摆的 T² 对 l），则斜率有物理意义：g = 4π² / 斜率。

6. Error Analysis and Uncertainty | 误差分析和不确定度

Uncertainty in a single measurement is often half the smallest scale division (e.g., ±0.5 mm for a metre ruler). For a digital instrument, it is the resolution (e.g., ±0.01 s for stopwatch), but remember human reaction time dominates for timing.

单次测量的不确定度通常是仪器最小刻度的一半（例如，米尺为±0.5 mm）。对于数字仪器，它是分辨力（例如，秒表为±0.01 s），但要记住对于计时，人的反应时间起主要作用。

For repeated measurements, uncertainty can be estimated as half the range: (max − min)/2. Express the result as (mean ± uncertainty) with units.

对于重复测量，不确定度可用极差的一半来估算：(最大值 − 最小值) / 2。将结果表示为（平均值 ± 不确定度）并带单位。

When calculating a derived quantity, propagate uncertainties. For addition/subtraction, add absolute uncertainties. For multiplication/division, add percentage uncertainties. For example, speed v = s/t, if s has an uncertainty of 2% and t has 3%, then v has 5% uncertainty.

在计算导出量时，需传递不确定度。对于加减，将绝对不确定度相加。对于乘除，将百分比不确定度相加。例如，速度 v = s / t，若 s 的不确定度为2%，t 为3%，则 v 的不确定度为5%。

Use percentage difference to compare an experimental value with an accepted value: % difference = |experimental − accepted| / accepted × 100%. Comment on accuracy and possible systematic errors.

使用百分比差异将实验值与公认值比较：%差异 = |实验值 − 公认值| / 公认值 × 100%。评价准确度和可能的系统误差。

7. Drawing Conclusions | 得出结论

State whether the results support the hypothesis. Refer directly to the graph, gradient, or data trend. For a pendulum, if T² ∝ l, then the period squared is directly proportional to length, in accordance with the theoretical equation T² = (4π²/g) l.

说明结果是否支持假说。直接引用图表、斜率或数据趋势。对于单摆，如果 T² ∝ l，则周期平方与摆长成正比，符合理论方程 T² = (4π²/g) l。

Use the equation of the line (if linear) to extract constants, e.g., calculate g from gradient. Report the experimental value with its uncertainty and compare with 9.81 m/s².

使用直线方程（如果是线性的）提取常数，例如，从斜率计算 g。报告实验值及其不确定度，并与9.81 m/s² 比较。

Avoid definitive statements like ‘proves’; use ‘supports the theory’ or ‘within experimental uncertainty, the relationship is valid’.

避免使用“证明”等确定性的陈述；使用“支持该理论”或“在实验不确定度范围内，该关系成立”。

8. Evaluation and Improvements | 评估与改进

Identify sources of error: random errors (reaction time, parallax) and systematic errors (zero error on a measuring device, incorrectly marked scale). Suggest practical improvements: using a fiducial marker for timing oscillations, measuring length with a vernier caliper for greater accuracy, or using a light gate to remove reaction time.

识别误差来源：随机误差（反应时间、视差）和系统误差（测量装置零误差、刻度标记不正确）。提出切实可行的改进：使用基准标记来计时摆动，使用游标卡尺测量长度以获得更高准确度，或使用光闸消除反应时间。

Discuss reliability: more data points, repeated measurements, and checking consistency. If the line of best fit has outliers, suggest repeating those measurements. State whether the method could be modified to reduce uncertainty or extend the investigation.

讨论可靠性：更多数据点、重复测量和检查一致性。如果最佳拟合线有离群值，建议重复这些测量。说明是否可以修改方法以减少不确定度或扩展研究。

Always relate improvements to specific errors identified earlier. For instance, to minimise reaction time, use a light gate connected to a data logger; to improve length measurement, use a travelling microscope for small distances.

始终将改进与之前识别的具体错误联系起来。例如，为了使反应时间最小化，使用连接数据记录仪的光闸；为了改进长度测量，使用移测显微镜测量小距离。

9. Common Experiments – Mechanics | 常见实验——力学

Free fall to determine g: Drop a steel ball from various heights and use a trapdoor or light gates to measure time of fall. Plot height h against (time)², gradient = g/2. The equation h = ½ g t² is used if initial velocity is zero.

自由落体测定 g：从不同高度释放钢球，使用活动门或光闸测量下落时间。绘制高度 h 对 (时间)² 图，斜率 = g/2。如果初速度为零，使用方程 h = ½ g t²。

Newton’s second law: Use a dynamics trolley pulled by a falling mass over a pulley. Vary the accelerating force (mass hanger) while keeping total mass constant, or vary mass and keep force constant. Plot acceleration a against force F to obtain 1/m as gradient.

牛顿第二定律：使用一个由绕过滑轮的落体牵引的动力小车。改变加速力（砝码挂架）但保持总质量不变，或改变质量并保持力不变。绘制加速度 a 对力 F 的图，得到斜率为 1/m。

Investigate momentum or conservation of energy using light gates

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 Key Concepts from the January 2023 International Physics (PH05) Insert | 2023年1月国际物理(PH05)插入材料概念解析

The January 2023 International Physics (PH05) Insert provides a set of reference data, diagrams, and theoretical relationships that underpin the Unit 5 paper, covering thermodynamics, nuclear and particle physics, and astrophysics. This article unpacks the core physical principles embedded in that insert – from ideal gas behaviour to stellar evolution – guiding you through the essential equations, their meanings, and the conceptual links required to tackle examination questions with confidence.

2023年1月国际物理(PH05)考试的插页材料提供了一组参考数据、示意图和理论关系式，支撑着第五单元的试卷内容，涵盖热力学、核与粒子物理以及天体物理。本文拆解该插页中蕴含的核心物理原理——从理想气体行为到恒星演化——带领你梳理关键方程、其物理意义以及解题所需的概念联系，帮助你自信应对考试。

1. Ideal Gas Law and Kinetic Theory | 理想气体定律与分子运动论

The insert typically reminds candidates of the ideal gas equation pV = nRT and its microscopic counterpart pV = NkT, where N is the number of molecules, k the Boltzmann constant, and T the absolute temperature. These laws arise from a kinetic model assuming point-like molecules moving randomly, undergoing perfectly elastic collisions with container walls.

插页通常会提醒考生理想气体状态方程 pV = nRT 及其微观形式 pV = NkT，其中 N 为分子数，k 为玻尔兹曼常数，T 为绝对温度。这些定律源自一个动力学模型，该模型假设分子为质点并做无规则运动，与器壁发生完全弹性碰撞。

The kinetic theory also links macroscopic pressure to molecular motion through p = ⅓ ρ or p = ⅓ Nm / V. From this, the average translational kinetic energy per molecule is derived as ½ m = ³⁄₂ kT, illustrating that temperature is a direct measure of microscopic kinetic energy.

分子运动论也将宏观压强与分子运动联系起来，通过 p = ⅓ ρ 或 p = ⅓ Nm / V 描述。由此导出单个分子的平均平动动能为 ½ m = ³⁄₂ kT，表明温度是微观动能的直接量度。

When solving problems involving a fixed mass of gas, it is vital to recall that the number of moles n remains constant, allowing the combined gas law p₁V₁/T₁ = p₂V₂/T₂ to be applied for changes of state. The insert data often include values for the molar gas constant R = 8.31 J mol⁻¹ K⁻¹ and the Boltzmann constant k = 1.38 × 10⁻²³ J K⁻¹.

在解决涉及固定质量气体的问题时，务必牢记物质的量 n 恒定不变，从而可以利用组合气体定律 p₁V₁/T₁ = p₂V₂/T₂ 处理状态变化。插页数据通常给出摩尔气体常数 R = 8.31 J mol⁻¹ K⁻¹ 和玻尔兹曼常数 k = 1.38 × 10⁻²³ J K⁻¹。

2. First Law of Thermodynamics and Processes | 热力学第一定律与过程

The first law of thermodynamics, ΔU = Q − W, appears in the insert as a core relation for energy conservation in thermal systems. Here, ΔU is the change in internal energy, Q the thermal energy added to the system, and W the work done by the system on its surroundings. Sign conventions are critical: work done by the gas is positive W, and heat entering the system is positive Q.

热力学第一定律 ΔU = Q − W 作为热力系统能量守恒的核心关系式出现在插页中。其中 ΔU 是内能的变化，Q 是加入系统的热能，W 是系统对外界做的功。正负号约定至关重要：气体对外做功 W 取正值，系统吸热 Q 取正值。

For an isothermal expansion of an ideal gas, ΔU = 0, so Q = W; the gas absorbs heat and does an equal amount of work. In an adiabatic process, Q = 0, thus ΔU = −W; the gas cools as it does work. The insert may provide data on specific heat capacities at constant volume (C_V) and constant pressure (C_P), allowing calculation of energy changes in mole-based terms: ΔU = n C_V ΔT.

对于理想气体的等温膨胀，ΔU = 0，因此 Q = W；气体吸收热量并做等量的功。在绝热过程中，Q = 0，于是 ΔU = −W；气体做功时温度下降。插页可能提供定容热容 (C_V) 和定压热容 (C_P) 的数据，便于以摩尔为单位计算能量变化：ΔU = n C_V ΔT。

Real thermodynamic cycles, such as those in heat engines, rely on the first law together with the ideal gas law. The insert’s diagrams frequently show p–V loops, where the enclosed area represents the net work done per cycle. Interpreting such areas as energy transfers is a key skill.

真实的热力循环，例如热机中的循环，依赖于第一定律和理想气体定律的结合。插页中的示意图常常展示 p–V 回路，其包围的面积代表每个循环的净功。将这种面积解读为能量转移是一项关键技能。

3. Blackbody Radiation and Wien’s Displacement Law | 黑体辐射与维恩位移定律

Blackbody radiation curves provided in the insert illustrate how the spectral intensity distribution shifts with temperature. A perfect blackbody absorbs and emits all wavelengths, and the peak wavelength λ_max is inversely proportional to the absolute temperature T, as given by Wien’s law: λ_max T = 2.898 × 10⁻³ m K.

插页中给出的黑体辐射曲线展示了光谱强度分布如何随温度变化。理想黑体吸收并发射所有波长，其峰值波长 λ_max 与绝对温度 T 成反比，由维恩定律给出：λ_max T = 2.898 × 10⁻³ m K。

This relation is extremely useful in astrophysics to estimate the surface temperatures of stars. A blue star with λ_max ≈ 100 nm implies a temperature around 29,000 K, while a red star peaking at 700 nm has a temperature near 4,100 K. The insert often provides the law and graphical data for such analysis.

这一关系在天体物理中极为有用，可用来估算恒星的表面温度。一颗 λ_max ≈ 100 nm 的蓝色恒星意味着温度大约为 29,000 K，而一颗峰值在 700 nm 的红色恒星温度接近 4,100 K。插页通常提供该定律和用于这类分析的曲线图数据。

The Stefan–Boltzmann law P = σ A T⁴, where σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴, is also commonly referenced. It expresses the total power radiated per unit area, enabling luminosity calculations for stars when combined with their radius. Both Wien’s and Stefan–Boltzmann’s laws are pillars for understanding stellar properties.

斯特藩-玻尔兹曼定律 P = σ A T⁴，其中 σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴，也常被引用。它给出了单位面积辐射的总功率，结合恒星半径即可进行光度计算。维恩定律和斯特藩-玻尔兹曼定律是理解恒星性质的两大支柱。

4. Photoelectric Effect and Photon Model | 光电效应与光子模型

The photoelectric effect demonstrates the particle nature of light, where an electron is emitted from a metal surface when the incident photon energy exceeds the work function φ. The key equation, h f = φ + E_k,max, is a central component of the insert’s formula sheet. Here h is Planck’s constant (6.63 × 10⁻³⁴ J s), f the frequency, and E_k,max the maximum kinetic energy of ejected electrons.

光电效应展示了光的粒子性，当入射光子能量超过逸出功 φ 时，电子会从金属表面发射出来。核心方程 h f = φ + E_k,max 是插页公式表中的重要组成部分。其中 h 为普朗克常数 (6.63 × 10⁻³⁴ J s)，f 为频率，E_k,max 为逸出电子的最大动能。

The threshold frequency f₀ is defined when E_k,max = 0, yielding φ = h f₀. Graphs of stopping potential V_s against frequency f produce a straight line with gradient h/e, and the x-intercept gives f₀. Such graphical skills are commonly required, and the insert provides useful constants and conversion factors for electronvolts.

阈值频率 f₀ 定义为 E_k,max = 0 时的频率，从而得到 φ = h f₀。遏止电压 V_s 对频率 f 的图线为一条斜率为 h/e 的直线，x 截距即 f₀。这种图形分析技能常常被考察，插页提供了有用的常数和电子伏特换算因子。

The photon model also connects to momentum: a photon has momentum p = h/λ. This appears in Compton scattering and particle interactions, but more importantly in the context of radiation pressure, which is sometimes explored in astrophysical contexts within Unit 5.

光子模型还关联到动量：光子的动量为 p = h/λ。这在康普顿散射和粒子相互作用中出现，但更重要的是在辐射压情境中，这有时会在第五单元的天体物理背景下进行探索。

5. Radioactive Decay and the Exponential Law | 放射性衰变与指数规律

Radioactive decay follows an exponential probability process, described by the equation N = N₀ e⁻ᵗ/τ or equivalently N = N₀ e^{−λ t}. The decay constant λ is the probability of decay per unit time, and τ = 1/λ is the mean lifetime. The insert provides both forms, along with the relationship between half-life t₁/₂ and decay constant: t₁/₂ = ln 2 / λ.

放射性衰变遵循指数概率过程，由方程 N = N₀ e⁻ᵗ/τ 或等价地 N = N₀ e^{−λ t} 描述。衰变常数 λ 是单位时间内的衰变概率，τ = 1/λ 为平均寿命。插页提供这两种形式，以及半衰期 t₁/₂ 与衰变常数的关系：t₁/₂ = ln 2 / λ。

Activity A = λ N represents the number of decays per second, measured in becquerels (Bq). The insert often includes a decay curve or a table of count rate data, allowing students to determine half-life graphically or through logarithmic manipulation. The exponential nature means that in one half-life, exactly half the nuclei remain, irrespective of the initial number.

活度 A = λ N 表示每秒衰变次数，单位为贝可勒尔 (Bq)。插页通常包含衰变曲线或计数率数据表格，使学生能够通过图像或对数运算确定半衰期。指数性质意味着在一个半衰期内，无论初始数量多少，正好剩下一半的原子核。

Applications include carbon-14 dating, medical tracers, and nuclear waste management, all of which rely on the predictable decrease in activity. Understanding that the decay curve never reaches zero is fundamental to the stochastic nature of quantum decay processes.

应用包括碳-14 测年、医用示踪剂以及核废料管理，这些都依赖于活度可预测的下降。理解衰变曲线永远不会触及零，是理解量子衰变过程随机性的基础。

6. Nuclear Binding Energy and Mass Defect | 核结合能与质量亏损

A central topic in the nuclear physics section is the mass defect and binding energy. The insert provides data tables of isotopic masses (in atomic mass units, u) and reminds candidates that 1 u = 931.5 MeV/c². The binding energy of a nucleus is the energy required to separate it into its constituent nucleons, calculated from Δm = (Z m_p + N m_n) − M_nucleus.

核物理部分的一个核心主题是质量亏损与结合能。插页提供同位素质量数据表（以原子质量单位 u 为单位），并提醒考生 1 u = 931.5 MeV/c²。原子核的结合能是将核子分开为单个核子所需的能量，通过 Δm = (Z m_p + N m_n) − M_nucleus 计算。

The binding energy per nucleon is a measure of nuclear stability. A graph of binding energy per nucleon against mass number A typically peaks around iron-56, explaining why energy can be released in both nuclear fusion (light nuclei combining) and nuclear fission (heavy nuclei splitting). The insert often provides a sketch of this curve, which is critical for interpreting energetics of fusion and fission.

每个核子的平均结合能是核稳定性的一种度量。平均结合能对质量数 A 的图线通常在铁-56 附近达到最高峰，这解释了为什么核聚变（轻核结合）和核裂变（重核分裂）都能释放能量。插页往往提供该曲线的示意图，这对于解释聚变与裂变的能量学至关重要。

Einstein’s mass–energy equivalence, E = mc², is the foundation here. When nucleons bind together, the total mass decreases, and the lost mass appears as energy released. This energy can be calculated directly from the mass defect using the conversion factor given in the insert.

爱因斯坦的质能等价方程 E = mc² 是此处的根基。核子结合在一起时总质量减少，亏损的质量以释放的能量形式出现。利用插页给出的换算因子，即可直接从质量亏损算出该能量。

7. Nuclear Fusion and Stellar Energy Generation | 核聚变与恒星能量生成

In stellar interiors, the proton–proton chain and the CNO cycle convert hydrogen into helium, releasing vast amounts of energy. The net reaction 4 ¹H → ⁴He + 2 e⁺ + 2 ν_e + energy is the dominant energy source in main-sequence stars. The insert may provide mass values for these particles, allowing students to compute the energy liberated per reaction.

在恒星内部，质子-质子链反应和碳氮氧循环将氢转化为氦，释放出巨大的能量。净反应 4 ¹H → ⁴He + 2 e⁺ + 2 ν_e + 能量 是主序星主要的能源。插页可能提供这些粒子的质量值，使学生能够计算每次反应释放的能量。

Positrons produced in the fusion reactions annihilate with electrons, contributing additional gamma-ray energy via e⁺ + e⁻ → 2γ. The total energy release per helium nucleus formed is about 26.7 MeV. Such calculations reinforce the connection between mass defect and the energy released in the core of the Sun.

聚变反应中产生的正电子与电子湮灭，通过 e⁺ + e⁻ → 2γ 贡献额外的伽马射线能量。每形成一个氦核总共释放约 26.7 MeV 的能量。这类计算巩固了质量亏损与太阳核心释放能量之间的联系。

The temperature and pressure required for fusion are explained by the need to overcome Coulomb repulsion. The insert may include information on the Gamow peak and tunnelling probability, linking nuclear physics to the probability of quantum tunnelling in hot, dense plasmas.

实现聚变所需的高温高压由克服库仑斥力的需要来解释。插页可能包含关于伽莫夫峰和隧穿概率的信息，将核物理与高温、高密度等离子体中的量子隧穿概率联系起来。

8. Stellar Spectra and the Hertzsprung–Russell Diagram | 恒星光谱与赫罗图

The insert often presents a simplified Hertzsprung–Russell (HR) diagram, plotting luminosity against temperature or spectral class. Main sequence stars form a diagonal band where they spend most of their lives fusing hydrogen. Once hydrogen in the core is exhausted, stars evolve into red giants, supergiants, or white dwarfs depending on their initial mass.

插页常常提供一幅简化的赫罗图，以光度对温度或光谱型作图。主序星形成一个对角线带，它们在主序阶段大部分时间进行氢聚变。一旦核心的氢耗尽，恒星会根据其初始质量演化为红巨星、超巨星或白矮星。

The spectral class sequence O, B, A, F, G, K, M relates to surface temperature and absorption line features. The insert may supply typical lines for each class, such as Balmer lines of hydrogen and absorption lines of metals, enabling the classification of stars and the determination of chemical composition.

光谱型序列 O, B, A, F, G, K, M 与表面温度和吸收线特征相关。插页可能提供每个光谱型的典型谱线，如氢的巴耳末线和金属的吸收线，从而能够进行恒星分类和化学成分确定。

Luminosity L can be expressed as L = 4πR²σT⁴, which when combined with the HR diagram allows determination of stellar radii. Giants and supergiants lie above the main sequence because of their enormous radii despite relatively low surface temperatures. White dwarfs are faint but hot, occupying the lower-left region.

光度 L 可表达为 L = 4πR²σT⁴，结合赫罗图就能确定恒星的半径。巨星和超巨星位于主序上方，因为它们尽管表面温度较低，但半径巨大。白矮星暗弱但温度高，占据左下区域。

9. Hubble’s Law and Cosmic Expansion | 哈勃定律与宇宙膨胀

Hubble’s observational discovery that the recessional velocity v of galaxies is proportional to their distance d, v = H₀ d, is a central piece of evidence for the expanding Universe. The insert provides the current value of the Hubble constant H₀ ≈ 70 km s⁻¹ Mpc⁻¹, along with the concept of cosmological redshift z = Δλ / λ₀.

哈勃的观测发现——星系退行速度 v 与其距离 d 成正比，v = H₀ d——是宇宙膨胀学说的核心证据。插页给出哈勃常数 H₀ 的当前值约为 70 km s⁻¹ Mpc⁻¹，以及宇宙学红移 z = Δλ / λ₀ 的概念。

For relatively low velocities, v ≈ c z can be used to estimate recessional speed from observed redshift. The age of the Universe can be roughly estimated by 1/H₀, known as the Hubble time, yielding an order of 13.8 billion years. The insert often includes a conversion between parsecs and light-years for such estimations.

对于相对较低的速度，可以使用 v ≈ c z 从观测红移估算退行速度。宇宙的年龄可由 1/H₀ 粗略估算，即哈勃时间，得出约 138 亿年的数量级。插页通常包含秒差距与光年之间的换算，以便进行这类估算。

The shift of spectral lines towards longer wavelengths is analogous to the Doppler effect for light, but at cosmological scales it is interpreted as the expansion of space itself. Understanding this distinction is important, as the insert may show spectra of distant galaxies with their characteristic absorption lines shifted redward.

光谱线向长波方向的移动类似于光的多普勒效应，但在宇宙尺度上这被解释为空间本身的膨胀。理解这一区别很重要，因为插页可能展现遥远星系的光谱，其特征吸收线向红端移动。

10. Linking the Insert to Exam Questions | 将插页内容与考题联系起来

Effective use of the insert means not merely locating equations but recognising which principle applies to the scenario described. Begin by scanning the provided material for relevant constants, then identify the theoretical framework: whether it is a thermodynamic process, a nuclear reaction, or an astrophysical measurement. The insert serves as a prompt for your memory of the underlying physics.

有效使用插页不仅意味着找到方程，还要识别哪一条原理适用于所描述的情境。先快速浏览提供的材料找到相关常数，然后确定理论框架：是热力过程、核反应还是天体物理测量。插页是唤起你基础物理记忆的提示。

In calculations, always check units carefully – for example, converting electronvolts to joules when necessary (1 eV = 1.60 × 10⁻¹⁹ J). Make use of the data on the insert to express answers in the expected form, and when drawing graphs, label axes with quantities and units exactly as given in the insert’s notation.

在计算时，务必仔细检查单位——例如，必要时将电子伏特转换为焦耳 (1 eV = 1.60 × 10⁻¹⁹ J)。利用插页上的数据将答案表达为预期形式，并且在绘制图线时，用插页中的符号给坐标轴加上物理量和单位标签。

Finally, the conceptual links across topics are often tested: for instance, connecting the kinetic theory of gases to the thermal pressure supporting a star, or relating nuclear binding energy to the fusion energy that powers the Sun. The January 2023 insert, like others in this series, weaves these threads together, and your revision should reflect that interconnected understanding.

最后，不同主题之间的概念联系经常被考查：例如，将气体分子运动论与支撑恒星的热压力联系起来，或将核结合能与驱动太阳的聚变能联系起来。2023年1月的插页与本系列其他插页一样，将这些线索交织在一起，你的复习也应体现出这种相互联系的理解。

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 AS Physics Paper 2 January 2018 Mark Scheme Concept Analysis | AS物理2018年1月卷二评分标准概念解析

The January 2018 Edexcel AS Physics Paper 2 examined a wide range of core concepts, from mechanics and materials to waves and electricity. A close study of the mark scheme reveals exactly what examiners were looking for: precise use of scientific vocabulary, clear understanding of definitions, and the ability to apply fundamental principles to unfamiliar situations. This article dissects the key concepts tested, offering paired English–Chinese explanations that mirror the mark scheme’s required depth.

2018年1月爱德思AS物理卷二考查了从力学、材料到波动、电学的诸多核心概念。仔细研读评分标准会发现，考官要求的是精准的科学用语、清晰的定义理解，以及将基本原理应用于陌生情境的能力。本文逐项剖析被考到的关键概念，提供与评分标准深度相匹配的中英双语解释。

1. Momentum and Impulse | 动量与冲量

The mark scheme consistently awards marks for stating that impulse equals the change in momentum, often expressed as FΔt = Δp. In a collision or an explosion, the principle of conservation of momentum applies provided no external resultant force acts. Candidates needed to recall that momentum is a vector, so direction matters when adding or subtracting momenta.

评分标准反复给分的要点是：冲量等于动量的变化量，常写作 FΔt = Δp。在碰撞或爆炸问题中，只要没有外合力作用，动量守恒定律就适用。考生必须记住动量是矢量，因此动量相加或相减时必须考虑方向。

For a car crash scenario, the mark scheme expected the calculation of the force exerted on a passenger using the rate of change of momentum. The area under a force–time graph represents impulse, and examiners awarded marks for stating this relationship even if the numerical integration was not perfectly accurate.

在汽车碰撞场景中，评分标准要求用动量变化率计算乘客受到的力。力–时间图下的面积表示冲量；即使数值积分不够精确，只要表述出这层关系，考官就会给分。

2. Elastic and Inelastic Collisions | 弹性碰撞与非弹性碰撞

Marks were allocated for recognising that kinetic energy is not conserved in inelastic collisions. In a perfectly elastic collision, both momentum and kinetic energy are conserved. The mark scheme frequently checked whether candidates could calculate the speed after impact using conservation of momentum alone and then test whether the kinetic energy condition holds.

评分标准在非弹性碰撞中给分的依据是能否识别动能不守恒。在完全弹性碰撞中，动量和动能同时守恒。考生要能先用动量守恒求出碰撞后的速度，再检验动能是否满足守恒条件，这是得分的关键。

Examiners often penalised vague language; simply saying “energy is lost” was insufficient unless accompanied by “kinetic energy is transformed into thermal and sound energy”. Precise wording is crucial.

考官常因表述模糊而扣分；“能量损失”的说法不够，必须明确指出“动能转化为热能和声能”。精确的措辞至关重要。

3. Stress, Strain and the Young Modulus | 应力、应变与杨氏模量

The mark scheme demanded the definition of tensile stress as force per unit cross-sectional area (σ = F/A) and tensile strain as extension per unit original length (ε = ΔL/L₀). The Young modulus E was defined as the ratio of stress to strain within the limit of proportionality, and its unit was expected to be Pa or N m⁻².

评分标准要求定义拉伸应力为力除以横截面积 (σ = F/A)，拉伸应变为伸长量除以原长 (ε = ΔL/L₀)。杨氏模量 E 是在比例极限内应力与应变之比，单位是 Pa 或 N m⁻²。

Questions on the force–extension graph of a copper wire required candidates to identify the elastic limit, yield point, and plastic region. The mark scheme awarded marks for linking the gradient of the linear portion to the Young modulus, using E = (slope) × (L₀/A).

铜丝的力–伸长图像相关题目要求考生识别弹性极限、屈服点和塑性区。评分标准给分点之一是将线性段的斜率与杨氏模量挂钩，即 E = (斜率) × (L₀/A)。

4. Energy Stored in a Stretched Material | 拉伸材料中储存的能量

The area under a force–extension graph represents the work done in stretching a material, which is stored as elastic strain energy. For a material obeying Hooke’s law, this energy equals ½FΔL or ½kΔL². The mark scheme explicitly rewarded candidates who explained that not all the work done is stored as strain energy if plastic deformation occurs.

力–伸长图下的面积代表拉伸材料所做的功，这部分功以弹性势能的形式储存。对于遵循胡克定律的材料，储存的能量等于 ½FΔL 或 ½kΔL²。评分标准特意奖励那些说明发生塑性形变时并非所有功都转化为应变能的考生。

When unloading curves were drawn, the area between the loading and unloading curves represented energy dissipated as heat. This concept frequently appeared in materials questions and required careful graph interpretation.

画出卸载曲线时，加载与卸载曲线之间的面积代表以热的形式散失的能量。这一概念频繁出现在材料相关题目中，要求仔细解读图像。

5. Potential Dividers and Sensor Circuits | 分压器与传感器电路

The potential divider equation V_out = V_in × R₂/(R₁+R₂) was tested both algebraically and in practical contexts like LDR and thermistor circuits. The mark scheme expected candidates to explain that as light intensity increases, the LDR resistance falls, so the share of voltage across the fixed resistor rises.

分压器公式 V_out = V_in × R₂/(R₁+R₂) 既以代数形式考查，也以 LDR 和热敏电阻电路的应用形式考查。评分标准期望考生解释：随着光照强度增大，LDR 电阻下降，因此固定电阻两端的电压份额增大。

Markers looked for statements linking the change in resistance to the output voltage and then to the switching action of a comparator circuit. The concept of a “threshold” voltage triggering an output high or low was frequently awarded marks.

考官寻找的是将电阻变化与输出电压联系起来，再与比较器电路的切换动作联系起来的表述。提到“阈值”电压触发输出高或低电平的表述通常得分。

6. Resistivity and Drift Velocity | 电阻率与漂移速度

The definition of resistivity ρ = RA/L was tested, with the mark scheme insisting on the meaning of each symbol and the unit Ω m. Examiners awarded marks for stating that resistivity depends on temperature and the material, but not on dimensions. The drift velocity equation I = nAve was often linked to resistivity through R = ρL/A.

电阻率 ρ = RA/L 的定义是考查重点，评分标准要求解释每个符号的含义和单位 Ω m。考官给分点在于说明电阻率取决于温度和材料，而非尺寸。漂移速度公式 I = nAve 常通过 R = ρL/A 与电阻率联系起来。

In the January 2018 paper, candidates were asked to explain why a thinner wire has a higher resistance. The mark scheme expected the reasoning: for the same applied p.d., a smaller cross-sectional area A gives a larger resistance R because R ∝ 1/A, and also the drift velocity increases for the same current, leading to more collisions with lattice ions.

在2018年1月的试卷中，考生要解释为何较细的导线电阻较高。评分标准期望的推理是：相同的电势差下，较小的横截面积 A 会导致较大的电阻 R（R ∝ 1/A），同时相同电流下漂移速度增大，电子与晶格离子的碰撞增多。

7. Wave Superposition and Standing Waves | 波的叠加与驻波

The mark scheme required a precise statement of the principle of superposition: when two waves meet, the resultant displacement is the vector sum of the individual displacements. Standing wave formation was explained as the superposition of two progressive waves of the same frequency, travelling in opposite directions, with nodes at points of zero displacement and antinodes at maximum amplitude.

评分标准要求精确表述叠加原理：两列波相遇时，合位移是各分位移的矢量和。驻波的形成解释为：两列频率相同、传播方向相反的行波叠加，形成位移为零的波节和振幅最大的波腹。

Examiners awarded marks for linking the distance between adjacent nodes (or antinodes) to half a wavelength (λ/2). Tight strings and open/closed pipes questions often required candidates to identify the harmonic number from a given standing wave pattern.

考生若能将相邻波节（或波腹）间距与半波长 (λ/2) 联系起来即可得分。弦和开/闭管类题目常要求根据驻波图案判断谐波次数。

8. Diffraction Grating and Interference | 衍射光栅与干涉

The grating equation d sinθ = nλ was central. The mark scheme explicitly required the definition of d as the grating spacing (1/lines per metre), θ as the angle between the zero-order line and the nth-order maximum, and n as the order number. Candidates were expected to convert between lines per mm and grating spacing in metres.

光栅方程 d sinθ = nλ 是核心。评分标准明确要求定义 d 为光栅常数（每米线条数的倒数），θ 为零级明纹与第 n 级明纹之间的夹角，n 为级数。考生要能进行“线/毫米”与“光栅常数（米）”之间的换算。

Questions on the maximum observable order tested understanding of the condition sinθ ≤ 1. The mark scheme also credited the explanation that a grating with more lines per metre gives larger angular separations, leading to more accurate wavelength measurements.

关于最高可观测级数的题目考查对 sinθ ≤ 1 条件的理解。评分标准还认可如下解释：单位长度线条数更多的光栅会产生更大的角间距，从而能更准确地测量波长。

9. Photoelectric Effect and Photon Model | 光电效应与光子模型

The mark scheme rewarded reference to the one-to-one interaction between a photon and an electron. Key points included: emission occurs only if photon energy hf ≥ work function φ; any excess energy becomes the electron’s maximum kinetic energy E_{k max} = hf − φ. The stopping potential V_s was linked to E_{k max} via eV_s.

评分标准奖励提及光子与电子之间一对一相互作用的答案。关键点是：只有当光子能量 hf ≥ 逸出功 φ 时才会发射电子；多余的能量转化为电子的最大动能 E_{k max} = hf − φ。截止电压 V_s 通过 eV_s 与 E_{k max} 关联。

Explanations of the independence of frequency on intensity, and the existence of a threshold frequency, were required to contrast with the wave theory. The mark scheme frequently asked for the Einstein equation to be expressed clearly, and marks were lost if candidates omitted the work function term.

解释光电子发射与光强无关、存在截止频率等现象时，要求考生对比波动理论。评分标准经常要求清晰写出爱因斯坦光电方程；若遗漏逸出功项，则会丢分。

10. Internal Resistance and EMF | 内阻与电动势

The definition of electromotive force (e.m.f.) as the energy transferred per unit charge when no current is drawn was often tested. The terminal p.d. V = ε − Ir was used to determine internal resistance r from a graph of V against I, where the gradient is −r and the y-intercept is ε.

电动势 (e.m.f.) 的定义——没有电流流过时单位电荷所转移的能量——经常被考查。端电压 V = ε − Ir 用于从 V–I 图像中确定内阻 r，图像的斜率为 −r，纵截距为 ε。

The mark scheme demanded that candidates describe why the terminal p.d. falls when current increases: due to the “lost volts” across the internal resistance. Practical investigations using a variable resistor to vary the current and record V were common, with marks for repeating readings and avoiding parallax errors.

评分标准要求考生说明为何端电压随电流增加而下降：因为内阻上存在“内电压降”。利用可变电阻改变电流并记录 V 的实验探究题很常见；重复读数、避免视差错误等细节均有得分点。

11. Charge, Current and Potential Difference | 电荷、电流与电势差

Current was defined as the rate of flow of charge, I = ΔQ/Δt. The mark scheme often rewarded the link to the number of charge carriers using Q = Ne and I = N e/t. Potential difference was defined as the energy transferred per unit charge, V = W/Q.

电流定义为电荷的流动速率，I = ΔQ/Δt。评分标准常奖励将其与载流子数目联系起来的表述：Q = Ne 及 I = N e/t。电势差的定义为每单位电荷转移的能量，V = W/Q。

In circuit analysis, the conservation of charge (Kirchhoff’s first law) and the conservation of energy (Kirchhoff’s second law) were required to be stated in precise terms. The mark scheme penalised casual language such as “current is used up” instead of “current is conserved at a junction”.

在电路分析中，要求用精确的语言表述电荷守恒（基尔霍夫第一定律）和能量守恒（基尔霍夫第二定律）。评分标准对诸如“电流被消耗”这样随意的语言予以扣分，正确的说法是“电流在节点处守恒”。

12. Practical Skills: Uncertainties and Graph Plotting | 实验技能：不确定度与绘图

Throughout Paper 2, marks were available for correct handling of uncertainties. Absolute and percentage uncertainties in repeated measurements, and the propagation of uncertainties when quantities were multiplied or divided, featured regularly. The mark scheme expected the scale of a graph to use at least half the grid paper, with accurately plotted points and a line of best fit that does not force through the origin unless physically justified.

整份卷二均有不确定性处理的得分点。重复测量的绝对和百分比不确定度，以及物理量相乘或相除时不确定度的传递，经常出现。评分标准要求作图时量标尺占用至少一半方格纸面积，数据点准确标绘，最佳拟合线不得强行通过原点，除非有物理依据。

Anomalous points were expected to be identified and ignored when drawing the best-fit line. Examination technique points from the mark scheme included labelling axes with quantity and unit, and using a sharp pencil for precise plotting. These practical marks collectively made up a significant portion of the total grade.

异常数据点需要被识别出来，在绘制最佳拟合线时予以忽略。评分标准中提示的考试技巧包括：用物理量和单位标注坐标轴，用尖细铅笔准确描点。这些实验技能分值合计占据了总分的相当一部分。

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 Mastering Formula Derivations in AS Physics Unit 2: Insights from the Jan 20 Exam Report | 掌握AS物理Unit 2公式推导：2020年1月考卷深度剖析

Mastering formula derivations is the key to truly understanding AS Physics Unit 2 topics such as mechanics, materials, waves, and electricity. In the January 2020 examination, many students lost marks because they could not derive key equations from first principles. This article walks you through the essential derivations that appear most often in exams, helping you move beyond rote memorisation and develop a deep conceptual understanding.

掌握公式推导是真正理解AS物理Unit 2中力学、材料、波和电学等主题的关键。在2020年1月的考试中，许多学生因为无法从基本原理推导出关键方程而失分。本文将带你逐一掌握考试中最常出现的核心推导，帮助你超越死记硬背，建立深刻的概念理解。

1. Understanding Acceleration to Get the First SUVAT Equation | 理解加速度并导出第一个运动学方程

Acceleration is defined as the rate of change of velocity. For uniform (constant) acceleration, the change in velocity is simply final velocity v minus initial velocity u over time t. The definition a = (v – u) / t is your starting point.

加速度定义为速度的变化率。对于匀加速度，速度变化量即为末速度v减初速度u除以时间t。定义式 a = (v – u) / t 是整个推导的起点。

By multiplying both sides by t and adding u, you obtain v = u + at, which is the first of the SUVAT equations. This simple rearrangement is often overlooked by students who try to memorise the result without appreciating its origin.

等式两边同乘以t并加上u，即可得到 v = u + at，这是匀加速直线运动方程组的第一个。这个简单的代数变换常被忽略，学生往往直接记忆结果却不知其来龙去脉。

v = u + at

2. Displacement via Average Velocity | 通过平均速度求位移

When acceleration is constant, velocity changes linearly, so the average velocity is the arithmetic mean of the initial and final speeds: v_avg = (u + v) / 2. Displacement s is then just average velocity multiplied by time t.

当加速度恒定时，速度呈线性变化，因此平均速度等于初末速度的算术平均值：v_avg = (u + v) / 2。位移 s 就等于平均速度与时间 t 的乘积。

Substituting v = u + at from the first equation into the average velocity formula, we get s = ((u + [u + at])/2) × t = (2u + at) × t / 2 = ut + ½at². This derivation links the two fundamental relationships neatly.

将第一式 v = u + at 代入平均速度公式，可得 s = ((u + [u + at])/2) × t = ut + ½at²。这个推导自然地将两个基本关系连接起来，而非凭空出现。

s = ut + ½at²

3. Eliminating Time: The Timeless Equation | 消去时间：与时间无关的运动方程

Sometimes we know initial and final speeds along with displacement, but not the time taken. To derive v² = u² + 2as, start with v = u + at and solve for t = (v – u)/a. Insert this t into s = ut + ½at² and simplify: s = u((v-u)/a) + ½a((v-u)/a)².

有时我们已知初末速度和位移，却不知道所用时间。要推导 v² = u² + 2as，可从 v = u + at 解出 t = (v – u)/a，然后代入 s = ut + ½at² 并化简：s = u((v-u)/a) + ½a((v-u)/a)²。

After careful algebraic manipulation, the terms combine to give v² – u² = 2as, or v² = u² + 2as. This equation is particularly useful in energy-related problems.

经过仔细的代数运算，项合并后得到 v² – u² = 2as，即 v² = u² + 2as。这个方程在处理与能量相关的问题时尤为有用。

v² = u² + 2as

4. Kinetic Energy and the Work-Energy Theorem | 动能及功能定理

Consider a constant force F acting on an object to displace it by s. The work done is W = F s. Using Newton’s second law F = ma and the equation v² = u² + 2as, we can express acceleration a as (v² – u²)/(2s). Substituting into the work equation yields W = m × (v² – u²)/(2s) × s = ½mv² – ½mu².

设一个恒定力F作用于物体，使之产生位移s，做功为 W = F s。利用牛顿第二定律 F = ma 以及 v² = u² + 2as，可将加速度 a 表示成 (v² – u²)/(2s)。代入做功式子得到 W = m × (v² – u²)/(2s) × s = ½mv² – ½mu²。

This shows that the net work transfers energy that appears as a change in kinetic energy. Hence, kinetic energy is defined as KE = ½mv². Understanding this derivation prevents the common mistake of forgetting the factor ½.

这表明净功转移的能量表现为动能的变化。由此定义动能 KE = ½mv²。理解这个推导能避免常见错误的出现，比如漏掉系数 ½。

KE = ½mv²

5. Conservation of Linear Momentum from Newton’s Third Law | 由牛顿第三定律推导动量守恒

Imagine two objects colliding. Object 1 exerts an impulse FΔt on object 2, and by Newton’s third law, object 2 exerts an equal and opposite impulse -FΔt on object 1. The impulse equals the change in momentum: for object 1, m₁v₁ – m₁u₁ = FΔt, and for object 2, m₂v₂ – m₂u₂ = -FΔt.

设想两个物体发生碰撞。物体1对物体2施加冲量 FΔt，根据牛顿第三定律，物体2同时对物体1施加等大反向的冲量 -FΔt。冲量等于动量变化：对物体1有 m₁v₁ – m₁u₁ = FΔt，对物体2有 m₂v₂ – m₂u₂ = -FΔt。

Adding the two impulse equations eliminates the force term, leaving m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. This is the principle of conservation of linear momentum, applicable when no external forces act.

将两式相加，力项被抵消，得到 m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂。这就是动量守恒定律，条件是系统不受外力作用。记住这个推导过程有助于理解动量是矢量，方向必须考虑。

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

6. Defining Young’s Modulus from Stress and Strain | 从应力与应变定义杨氏模量

Tensile stress σ is the force applied per unit cross-sectional area: σ = F/A. Tensile strain ε is the extension relative to the original length: ε = ΔL/L. Young’s modulus E quantifies a material’s stiffness and is defined within the elastic limit as the ratio of stress to strain.

拉应力 σ 是单位截面积上所受的力：σ = F/A。拉应变 ε 是伸长量相对于原始长度的比值：ε = ΔL/L。杨氏模量 E 量化材料的刚度，在弹性限度内定义为应力与应变之比。

Substituting the expressions for stress and strain, E = (F/A) / (ΔL/L) = FL / (A ΔL). This derived formula is frequently used in experiments to find E from a force-extension graph.

将应力和应变的表达式代入，E = (F/A) / (ΔL/L) = FL / (A ΔL)。这个推导出的公式常用于实验中，通过力-伸长图线来求杨氏模量。

E = FL / (A ΔL)

7. Relating Wave Speed, Frequency, and Wavelength | 波速、频率与波长的关系推导

A wave advances by one complete wavelength λ during one period T. Since frequency f is the number of cycles per second, f = 1/T. Therefore, the speed of the wave is distance divided by time: v = λ / T = λ × (1/T) = f λ.

在一个周期 T 内，波向前传播一个完整的波长 λ。因为频率 f 是每秒的周期数，f = 1/T。因此波速等于距离除以时间：v = λ / T = λ × (1/T) = f λ。

This derivation emphasises that the wave equation v = fλ is not an empirical law but follows directly from the definitions of frequency and wavelength. Make sure you can reproduce this logic in an exam.

这个推导强调波动方程 v = fλ 并非经验公式，而是直接源于频率和波长的定义。务必确保在考试中能清晰地写出这个逻辑过程。

v = f λ

8. Electrical Power and Ohm’s Law Combinations | 电功率与欧姆定律组合推导

Potential difference V is defined as the work done per unit charge: V = W/Q. Current I is charge per unit time: I = Q/t. Electrical power P is the rate of doing work, so P = W/t = (VQ)/t = V I. This gives the most fundamental power equation P = VI.

电势差 V 定义为每单位电荷做的功：V = W/Q。电流 I 是单位时间通过的电量：I = Q/t。电功率 P 是做功的速率，所以 P = W/t = (VQ)/t = V I。这就得到最基本的功率等式 P = VI。

Using Ohm’s law V = IR, you can substitute to obtain two other forms: P = I × (IR) = I²R, and P = (V/R) × V = V²/R. These expressions are useful for analysing energy dissipation in resistors.

结合欧姆定律 V = IR，可以代入得到另外两种形式：P = I × (IR) = I²R，以及 P = (V/R) × V = V²/R。这些表达式在分析电阻上的能量耗散时十分有用。

P = VI = I²R = V²/R

9. Equivalent Resistance in Series and Parallel Circuits | 串联与并联电路等效电阻推导

In a series circuit, the current I is the same through all components. The total voltage V_total equals the sum of individual voltages: V_total = V₁ + V₂ = IR₁ + IR₂ = I(R₁+R₂). Hence, the equivalent series resistance R_total = R₁ + R₂.

串联电路中，通过每个元件的电流 I 相同。总电压 V_total 等于各电压之和：V_total = V₁ + V₂ = IR₁ + IR₂ = I(R₁+R₂)。因此，等效串联电阻 R_total = R₁ + R₂。

In a parallel circuit, the voltage V across each branch is the same, and the total current I_total splits: I_total = I₁ + I₂. Express each current using Ohm’s law: I_total = V/R₁ + V/R₂ = V(1/R₁+ 1/R₂). Therefore, 1/R_total = 1/R₁ + 1/R₂.

并联电路中，各支路电压 V 相同，总电流 I_total 分流：I_total = I₁ + I₂。用欧姆定律表示各支路电流：I_total = V/R₁ + V/R₂ = V(1/R₁+ 1/R₂)。因此，1/R_total = 1/R₁ + 1/R₂。

Series: R_total = R₁ + R₂; Parallel: 1/R_total = 1/R₁ + 1/R₂

10. Deriving the Potential Divider Formula | 分压器公式推导

A potential divider usually consists of two resistors R₁ and R₂ in series across an input voltage V_in. The current in the series circuit is I = V_in / (R₁ +

Published by TutorHao | Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 AS Physics Paper 4 Exam Report: Key Formula Derivations | AS物理试卷4考试报告：核心公式推导

AS Physics Paper 4 often includes questions that require you to derive key formulas from first principles. Examiner reports consistently highlight that students lose marks by memorising results without understanding the logical steps or the assumptions behind them. This article breaks down the most commonly tested derivations in mechanics, fields, and electricity, using the same approach that earns full marks in exams. Each section pairs a clear English explanation with its Chinese equivalent so you can cross-reference terminology and deepen your understanding.

AS物理试卷4经常包含要求从基本原理推导关键公式的题目。考官报告一再指出，学生因只会机械记忆而不理解推导逻辑或前提假设而失分。本文拆解了力学、场和电路中最常考的推导，采用与考试满分答案相同的思路。每一节都将清晰的英文讲解与中文对应内容配对，方便你对照术语并加深理解。

1. Centripetal Acceleration: a = v²/r | 向心加速度推导：a = v²/r

Consider an object moving at constant speed v in a circle of radius r. Although the speed is constant, the velocity changes direction continuously, so there is an acceleration towards the centre. In a small time interval Δt, the object moves from point A to point B through an angle Δθ. The velocity vectors at A and B have the same magnitude v but different directions. The change in velocity Δv is the vector from the tip of v_A to the tip of v_B.

考虑一个物体以恒定速率v在半径为r的圆周上运动。虽然速率不变，但速度方向不断改变，因此存在指向圆心的加速度。在很小的时间间隔Δt内，物体从A点运动到B点，转过角度Δθ。A点和B点的速度矢量大小均为v，但方向不同。速度的变化量Δv是从v_A箭头指向v_B箭头的矢量。

For small Δθ, the angle between v_A and v_B is Δθ, and the triangle formed by the two velocity vectors is similar to the triangle formed by the two radii OA and OB. From similarity, the ratio of the change in velocity to the speed equals the ratio of the arc length AB to the radius: |Δv| / v = arc AB / r. Since arc AB ≈ vΔt, we have |Δv| / v = vΔt / r, so |Δv| = v²Δt / r. The magnitude of the centripetal acceleration is a = |Δv| / Δt = v²/r.

对于很小的Δθ，v_A和v_B之间的夹角为Δθ，由这两个速度矢量构成的三角形与由两条半径OA和OB构成的三角形相似。根据相似性，速度变化量的大小与速率之比等于弧长AB与半径之比：|Δv| / v = 弧长AB / r。由于弧长AB ≈ vΔt，可得|Δv| / v = vΔt / r，因此|Δv| = v²Δt / r。向心加速度的大小为 a = |Δv| / Δt = v²/r。

a = v² / r

Examiners report that candidates often fail to state the vector nature of the quantities and lose marks for confusing arc length with chord length. Emphasise that the derivation assumes constant speed and a very small time interval.

考官报告指出，考生常常没有明确矢量的性质，并因混淆弧长与弦长而失分。要强调该推导假设速率恒定，且时间间隔非常小。

2. Conservation of Momentum from Newton’s Laws | 从牛顿定律推导动量守恒

Consider two objects, A and B, that interact with no external forces. According to Newton’s third law, the force F_{A on B} exerted by A on B is equal in magnitude and opposite in direction to the force F_{B on A} exerted by B on A. Therefore, F_{A on B} = – F_{B on A}.

考虑两个不受外力的物体A和B。根据牛顿第三定律，A对B的作用力F_{A on B}与B对A的作用力F_{B on A}大小相等、方向相反。因此，F_{A on B} = – F_{B on A}。

Newton’s second law states that the net force on an object equals the rate of change of its momentum: F = Δp / Δt. Applying this to each object during a time interval Δt, we get F_{B on A} = Δp_A / Δt and F_{A on B} = Δp_B / Δt. Substituting these into the third-law relation gives Δp_A / Δt = – Δp_B / Δt, which simplifies to Δp_A + Δp_B = 0.

牛顿第二定律指出，物体所受的合力等于其动量变化率：F = Δp / Δt。将该定律应用于时间间隔Δt内的每个物体，可得F_{B on A} = Δp_A / Δt 和 F_{A on B} = Δp_B / Δt。将这两个式子代入第三定律关系式中，得到Δp_A / Δt = – Δp_B / Δt，化简为Δp_A + Δp_B = 0。

Hence, the total change in momentum of the system is zero, meaning the total momentum before the interaction equals the total momentum after the interaction. This is the principle of conservation of linear momentum: in a closed system, total momentum remains constant.

因此，系统总动量变化为零，即相互作用前的总动量等于相互作用后的总动量。这就是动量守恒定律：在一个封闭系统中，总动量保持不变。

Total momentum before = Total momentum after

Many exam reports highlight that students forget to explicitly state the absence of external forces as a necessary condition. Always begin your derivation by specifying ‘in a closed system with no external resultant force’.

许多考试报告强调，学生忘记明确指出“无外力”这一必要条件。推导时务必先声明“在一个无合外力的封闭系统中”。

3. Electric Field of a Point Charge: E = kQ / r² | 点电荷的电场：E = kQ / r²

Start from Coulomb’s law: the magnitude of the electrostatic force between two point charges Q and q separated by distance r is F = kQq / r², where k = 1/(4πε₀). The electric field strength E at a point is defined as the force per unit positive charge at that point: E = F / q.

从库仑定律出发：两个点电荷Q和q相距r时，静电力大小为F = kQq / r²，其中k = 1/(4πε₀)。电场强度E定义为该点处单位正电荷所受的力：E = F / q。

If a test charge q is placed at a distance r from the source charge Q, the force on it is F = kQq / r². Dividing this force by q gives the field strength due to Q at that location: E = F / q = (kQq / r²) / q = kQ / r². The direction of the field is radially outward from a positive Q and radially inward toward a negative Q.

如果把检验电荷q放在距离源电荷Q为r的位置，它所受到的力为F = kQq / r²。将力除以q就得到Q在该位置产生的电场强度：E = F / q = (kQq / r²) / q = kQ / r²。电场方向对于正Q沿径向向外，对于负Q沿径向向内。

E = kQ / r²

Be careful to distinguish between the source charge Q and the test charge q. A common exam error is to leave q in the final expression or to forget that E is a vector that depends on the sign of Q.

注意区分源电荷Q和检验电荷q。考试中常见的错误是在最终表达式中留下q，或是忘记E是矢量，其方向取决于Q的符号。

4. Uniform Electric Field and Potential: V = Ed | 匀强电场与电势：V = Ed

A uniform electric field exists between two parallel plates connected to a battery. The field strength E is constant in magnitude and direction. The potential difference V between the plates is defined as the work done per unit positive charge in moving a charge from one plate to the other against the field.

连接电池的两块平行板之间存在匀强电场。场强E的大小和方向恒定。两极板间的电势差V定义为将单位正电荷从一块极板逆着电场方向移动到另一块极板时所做的功。

Work done W on a charge q moving a distance d parallel to the field is W = Fd, where F = qE. Therefore, W = qEd. Since potential difference V = W / q, we get V = (qEd) / q = Ed. This relation holds strictly when the displacement is parallel to the field lines and the field is uniform.

对电荷q施加的力为F = qE，当它平行于电场移动距离d时，做功W = Fd，因此W = qEd。因为电势差V = W / q，可得V = (qEd) / q = Ed。这一关系在位移平行于电场线且电场均匀时严格成立。

E = V / d

The derivation shows that electric field strength can also be expressed in volts per metre (V m^-1), which is equivalent to newtons per coulomb (N C^-1). Many students fail to convert between these units correctly.

该推导表明电场强度也可表示为伏特每米（V m^-1），等同于牛顿每库仑（N C^-1）。许多学生未能正确换算这两种单位。

5. Force on a Current-Carrying Conductor: F = BIL sinθ | 载流导线所受的力：F = BIL sinθ

Place a straight conductor of length L carrying a current I in a uniform magnetic field of flux density B. The current consists of moving charges. A single charge q moving with drift velocity v experiences a magnetic force F_q = Bqv sinθ, where θ is the angle between v and B.

将长L、通有电流I的直导线置于磁通密度为B的匀强磁场中。电流由大量运动电荷构成。单个电荷q以漂移速度v运动，受到的磁力为F_q = Bqv sinθ，其中θ为v与B的夹角。

The total force on all charges in the conductor is the sum of the forces on each charge. The number of charges in the conductor is nAL, where n is the number density of charge carriers and A is the cross-sectional area. Current is given by I = nqvA. The total force is therefore F = (nAL) × (Bqv sinθ) = B (nqvA) L sinθ = BIL sinθ.

导线中所有电荷所受的总磁力等于各个电荷受力之和。导线中的电荷总数为nAL，其中n为载流子数密度，A为横截面积。电流由I = nqvA给出。总力因此为F = (nAL) × (Bqv sinθ) = B (nqvA) L sinθ = BIL sinθ。

F = BIL sinθ

The angle θ is taken between the direction of the current (which follows the drift velocity of positive charges) and the magnetic field. If the conductor is perpendicular to the field, sinθ = 1 and F = BIL. This microscopic derivation is a favourite in Paper 4, as it links several fundamental quantities.

θ角取电流方向（沿正电荷漂移速度方向）与磁场方向之间的夹角。若导线与磁场垂直，sinθ = 1，则F = BIL。这一微观推导是试卷4中的常见考点，因为它将多个基本量串联了起来。

6. Force on a Moving Charge: F = Bqv sinθ | 运动电荷所受的力：F = Bqv sinθ

Starting from the macroscopic formula F = BIL sinθ, we can derive the force on a single free charge. In a conductor, I = nqvA, and L can be expressed in terms of drift velocity v and time t: L = vt. Substituting these into F = B (nqvA) L sinθ gives the total force on all charges in a segment of length L.

从宏观公式F = BIL sinθ出发，可以推导出单个自由电荷所受的力。在导线中，I = nqvA，且L可通过漂移速度v和时间t表示为L = vt。将这些代入F = B (nqvA) L sinθ，得到长度为L的导线段内所有电荷所受的总力。

The number of charges in the segment is N = nAL = nA(vt). The total force is F = (BIL sinθ) = B (nqvA) (vt) sinθ = (nAvt) × (Bqv sinθ). Dividing the total force F by the number of charges N gives the force per charge: F_q = F / N = Bqv sinθ.

该段内的电荷总数为N = nAL = nA(vt)。总力为F = (BIL sinθ) = B (nqvA) (vt) sinθ = (nAvt) × (Bqv sinθ)。将总力F除以电荷总数N即可得到单个电荷所受的力：F_q = F / N = Bqv sinθ。

F = Bqv sinθ

Alternatively, this can be explained directly: the magnetic force on a charge is always perpendicular to both its velocity and the magnetic field, and its magnitude is proportional to q, v, B, and sinθ. Many examiner reports note that candidates struggle to identify θ correctly when the velocity is not perpendicular to the field.

也可以直接解释：电荷在磁场中所受的力始终垂直于其速度和磁场，大小与q、v、B以及sinθ成正比。许多考官报告指出，当速度不与磁场垂直时，考生往往难以正确确定θ角。

7. Faraday’s Law of Electromagnetic Induction: ε = – N ΔΦ / Δt | 法拉第电磁感应定律：ε = – N ΔΦ / Δt

Faraday’s law can be introduced through the concept of changing magnetic flux. Magnetic flux Φ through a coil of area A is defined as Φ = BA cosθ, where θ is the angle between the magnetic field B and the normal to the area. When the flux linking a coil changes, an electromotive force (emf) is induced.

法拉第定律可通过变化的磁通量概念引出。穿过面积为A的线圈的磁通量定义为Φ = BA cosθ，其中θ为磁场B与面积法线之间的夹角。当与线圈交链的磁通量发生变化时，就会感应出电动势（emf）。

Experiments show that the induced emf in a coil is directly proportional to the rate of change of flux linkage. For a coil of N turns, flux linkage is NΦ. The magnitude of the induced emf is therefore |ε| = N |ΔΦ| / Δt. To include the direction of the induced emf, Lenz’s law states that the induced current opposes the change in flux, giving the negative sign: ε = – N ΔΦ / Δt.

实验表明，线圈中感应出的电动势与磁通链的变化率成正比。对于N匝线圈，磁通链为NΦ。因此，感应电动势的大小为|ε| = N |ΔΦ| / Δt。为了包含感应电动势的方向，楞次定律指出感应电流总是阻碍磁通量的变化，从而引入负号：ε = – N ΔΦ / Δt。

ε = – N ΔΦ / Δt

The wording ‘rate of change of flux linkage’ is crucial in the derivation. A common pitfall in exams is to write simply ε = NΦ/t without the change symbol, or to omit the negative sign without referring to Lenz’s law. Always state Lenz’s law explicitly.

“磁通链变化率”这一表述在推导中至关重要。考试中常见的错误是写成ε = NΦ/t而没有表示变化的符号，或忽略负号且不提楞次定律。务必明确写出楞次定律。

8. Acceleration of a Charged Particle in an Electric Field | 电场中带电粒子的加速

When a particle of charge q accelerates from rest through a potential difference V, the work done by the electric field is W = qV. If there are no energy losses, this work is converted entirely into kinetic energy. The initial kinetic energy is zero, so the final kinetic energy is E_k = ½ mv².

当一个电荷量为q的粒子在电势差V中从静止加速时，电场做的功为W = qV。如果没有能量损失，这些功完全转化为动能。初动能为零，因此末动能为E_k = ½ mv²。

By energy conservation: qV = ½ mv². Solving for the final speed gives v = √(2qV / m). This derivation is valid for non-relativistic speeds and assumes the electric field is conservative. It is used extensively in problems involving electron guns and cathode ray tubes.

由能量守恒：qV = ½ mv²。解得末

Published by TutorHao | AS Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）