📚 AS Physics Unit 2 Mark Scheme January 2020: Key Concepts Explained | AS物理Unit 2 2020年1月评分标准核心概念解析
The January 2020 AS Physics Unit 2 (WPH12/01) mark scheme reveals precisely how examiners award marks for understanding of materials, waves and electricity. This article breaks down the most important concepts, definitions, graphical analyses and common pitfalls that appeared in that session, helping you to see why certain answers gain full credit and how to avoid losing marks on technical details.
2020年1月AS物理第二单元（WPH12/01）的评分方案精准地展示了考官如何对材料、波和电学部分的理解进行评分。本文拆解了该次考试中最核心的概念、定义、图像分析以及常见失分点，帮助你理解为什么某些答案能拿到满分，以及如何避免在技术细节上丢分。
1. Stress, Strain and the Young Modulus | 应力、应变与杨氏模量
The mark scheme insists on precise language when defining tensile stress. The force must be described as applied normally (or perpendicular) to the cross‑sectional area. Simply writing ‘force over area’ is often insufficient for the definition mark because it misses the directionality condition. The accepted equation is σ = F / A, where σ is stress, F is force and A is cross‑sectional area. Its unit is the pascal (Pa) or N m⁻².
评分方案在定义拉伸应力时要求措辞精确。必须说明力是垂直作用于横截面积。只写“力除以面积”往往拿不到定义分，因为它遗漏了方向条件。可接受的方程式是 σ = F / A，其中 σ 为应力，F 为力，A 为横截面积，单位是帕斯卡（Pa）或 N m⁻²。
Tensile strain is often forgotten to be a ratio. The mark scheme rewards stating that strain is extension per unit original length: ε = ΔL / L₀. Because it is a ratio of two lengths, strain has no unit. Many candidates incorrectly use the extended length in the denominator; the mark scheme explicitly penalises this unless the change correction is negligible in context.
拉伸应变常被忘记是一个比值。评分方案对描述应变是每单位原始长度的伸长量予以给分：ε = ΔL / L₀。因为它是两个长度之比，应变没有单位。很多考生错误地在分母中采用了伸长后的长度；除非在问题情境中变化可忽略，否则评分方案会明确扣分。
The Young modulus E links stress and strain in the linear region: E = σ / ε. For a mark, candidates must appreciate that E is a measure of stiffness, with units Pa. In the January 2020 paper, describing the Young modulus as the gradient of the linear part of a stress–strain graph was awarded marks, but only if the response specified ‘initial linear region’ and mentioned that the cross‑sectional area and original length are constant.
杨氏模量 E 在弹性阶段将应力与应变联系起来：E = σ / ε。为拿到分数，考生必须理解 E 是刚度的量度，单位为 Pa。在2020年1月的试卷中，将杨氏模量描述为应力‑应变曲线直线部分的斜率可以得分，但前提是答案指明“初始线性区域”并且提到横截面积和原始长度保持不变。
2. Waves and Polarisation | 波与偏振
Transverse and longitudinal waves are distinguished by the direction of oscillation relative to energy transfer. The mark scheme awards marks for stating that transverse waves oscillate perpendicular to the direction of propagation, while longitudinal waves oscillate parallel. Polarisation was tested as evidence for transverse waves: only transverse waves can be polarised because the oscillations occur in a plane perpendicular to the direction of travel.
横波与纵波通过振动方向相对于能量传递的方向来区分。评分方案对准确说明“横波的振动方向垂直于传播方向，纵波的振动方向平行于传播方向”给予分数。偏振被考察作为横波的证据：只有横波才能被偏振，因为其振动发生在一个垂直于传播方向的平面内。
When explaining polarisation using a microwave transmitter and receiver, the mark scheme required the metal grille to be rotated. A maximum signal when the rods are vertical (if the transmitted wave is vertically polarised) and zero signal when the rods are horizontal demonstrate polarisation. Candidates who simply said ‘rotate the transmitter’ without describing the orientation of the grille often lost a mark.
在解释用微波发射器和接收器演示偏振时，评分方案要求旋转金属栅格。如果发射的微波是垂直偏振的，当金属棒竖直时信号最大，水平时信号为零，这便证明了偏振。只写“旋转发射器”而没有说明栅格的取向，考生常常丢分。
3. Standing Waves: Theory and Experiment | 驻波：理论与实验
Standing waves are formed when two progressive waves of the same frequency travel in opposite directions and superpose. The mark scheme looks for the key phrase ‘superposition of oppositely travelling waves’. Nodes are points of zero amplitude due to destructive interference, while antinodes have maximum amplitude from constructive interference.
驻波由两列频率相同、传播方向相反的波叠加而成。评分方案寻找的关键短语是“反向传播波的叠加”。波节是因相消干涉而振幅为零的点，而波腹则因相长干涉而振幅最大。
For the microwave standing wave experiment, a metal reflector is placed in front of the transmitter, and a probe moved along the line detects maxima and minima. In the January 2020 scheme, measuring the distance between adjacent nodes to determine the wavelength (λ = 2 × node‑node distance) was a common assessment point. Candidates also had to explain why a metal plate reflects microwaves strongly, ensuring a well‑defined standing wave pattern.
对于微波驻波实验，在发射器前方放置一块金属反射板，沿直线移动探针检测极大值和极小值。在2020年1月的评分方案中，测量相邻波节间的距离以确定波长（λ = 2 × 波节间距）是一个常见考点。考生还需解释为什么金属板能强烈反射微波，从而形成清晰的驻波图样。
4. Diffraction and Interference Patterns | 衍射与干涉图样
Diffraction is the spreading of a wave when it passes through a gap or around an obstacle. The mark scheme expects the degree of spreading to increase as the gap width approaches the wavelength. Answers that simply stated ‘waves spread out’ were not enough; a comparison between gap width and wavelength was necessary for full marks.
衍射是波通过狭缝或绕过障碍物时发生的扩散现象。评分方案期望考生指出：当缝宽接近波长时，衍射程度会增大。仅仅写“波扩散开”是不充分的；必须对缝宽与波长进行比较才能拿到满分。
Young’s double‑slit experiment produces bright and dark fringes due to interference. The fringe spacing Δy is given by Δy = λD / d, where λ is wavelength, D is the distance from slits to screen and d is the slit separation. The January 2020 mark scheme rewarded stating that reducing d increases fringe separation, and that using a longer wavelength light has the same effect. Explicit reference to the formula was often required to justify the predicted change.
杨氏双缝实验因干涉而产生明暗条纹。条纹间距 Δy 由 Δy = λD / d 给出，其中 λ 是波长，D 是双缝到屏幕的距离，d 是双缝间距。2020年1月的评分方案对说明减小 d 会增大条纹间距，以及使用更长波长的光有同样效果给予分数。通常要求明确提出公式来证明预测的变化。
5. The Diffraction Grating Equation | 衍射光栅方程
For a diffraction grating, the relationship between the grating spacing d, the diffraction angle θ and the wavelength λ is d sin θ = nλ, where n is the order number. In the mark scheme, a common mark is allocated for converting the number of lines per millimetre into d in metres: d = 1 × 10⁻³ m / N. Missing the conversion to metres often led to an arithmetic penalty.
对于衍射光栅，光栅常数 d、衍射角 θ 和波长 λ 之间的关系为 d sin θ = nλ，其中 n 是级数。在评分方案中，常见的一分是授予将每毫米线数转换为以米为单位的 d：d = 1 × 10⁻³ m / N。缺少这一单位换算经常导致计算错误而扣分。
When measuring θ, the January 2020 mark scheme required candidates to describe using a spectrometer: first align the telescope with the straight‑through beam (θ = 0°), then rotate the telescope to the n‑th order image and read the angle from the vernier scale. Answers that merely said ‘measure the angle with a protractor’ received no credit because they lacked the precision expected for an A‑level practical technique.
在测量 θ 时，2020年1月的评分方案要求考生描述如何使用分光计：首先将望远镜对准直射光束（θ = 0°），然后旋转望远镜对准第 n 级明纹，并从游标尺读出角度。仅写“用量角器测量角度”的答案不得分，因为这不符合A‑level对实验技术精度的要求。
6. Electrical Circuits and Internal Resistance | 电路与内电阻
The emf ε of a source is the total energy delivered per unit charge, while the terminal pd V is the energy delivered to the external circuit. The mark scheme frequently tests the circuit equation ε = V + Ir or ε = I(R + r). Candidates must be able to rearrange it to the linear form V = –rI + ε for graphical analysis. Plotting V on the y‑axis and I on the x‑axis yields a straight‑line with gradient –r and y‑intercept ε.
电源电动势 ε 是每单位电荷提供的总能量，而端电压 V 是外电路得到的能量。评分方案经常考察电路方程 ε = V + Ir 或 ε = I(R + r)。考生必须能够将其变形为直线形式 V = –rI + ε 以进行图像分析。将 V 作为纵轴、I 作为横轴绘图，可得一条斜率为 –r、纵截距为 ε 的直线。
In the January 2020 examination, a common error was misidentifying the intercept. The mark scheme clearly stated that when the current is zero, V = ε, so the intercept on the voltage axis gives the emf. However, to find internal resistance r, the gradient must be taken as a negative value; many candidates simply quoted the magnitude without the negative sign, which was penalised if the working did not show the correct substitution into ε = V + Ir.
在2020年1月的考试中，一个常见错误是混淆了截距。评分方案明确指出：当电流为零时 V = ε，因此电压轴上的截距即为电动势。然而，求内阻 r 时必须将梯度取为负值；很多考生只给出了大小而缺少负号，如果计算过程没有正确代入 ε = V + Ir 的证据，就会受到扣分。
7. Ohm’s Law and I–V Characteristics | 欧姆定律与伏安特性曲线
Ohm’s law states that, for a metallic conductor at constant temperature, the current I is directly proportional to the potential difference V. The mark scheme insists on the condition of constant temperature; without it, the definition is incomplete. The I–V graph of an ohmic conductor is a straight line passing through the origin.
欧姆定律指出，对于处于恒定温度下的金属导体，电流 I 与电势差 V 成正比。评分方案坚持要求写明恒温条件；否则该定义就不完整。欧姆导体的 I–V 曲线是一条通过原点的直线。
The I–V characteristics of a filament lamp and a diode were also examined. For the filament lamp, the resistance increases as the current increases because the metal filament heats up, causing more frequent collisions between electrons and lattice ions. The diode conducts only when forward‑biased above its threshold voltage, showing a very low resistance once conducting. In the mark scheme, describing the curve as ‘non‑linear’ and correctly linking shape to physical process was needed for the explanation marks.
白炽灯和二极管的伏安特性也受到了考查。对于白炽灯，随着电流增大电阻增加，因为金属灯丝温度升高，导致电子与晶格离子间的碰撞更频繁。二极管只有在正向偏压超过阈值电压时才导通，导通后呈现极低的电阻。在评分方案中，为拿到解释分，需要将曲线描述为“非线性”并正确地将形状与物理过程联系起来。
8. Potential Dividers and Sensor Circuits | 分压器与传感器电路
A potential divider uses two resistors in series to provide a fraction of the input voltage. The output voltage across resistance R₂ is Vout = Vin × (R₂ / (R₁ + R₂)). The January 2020 paper asked candidates to explain how a thermistor can be used in a potential divider to create a temperature‑sensing circuit. The mark scheme required stating that when the thermistor’s resistance decreases, the pd across it drops, while the pd across the fixed resistor rises, which can be used to trigger a switch.
分压器使用两个串联电阻来提供输入电压的一部分。电阻 R₂ 两端的输出电压为 Vout = Vin × (R₂ / (R₁ + R₂))。2020年1月的试卷要求考生解释如何利用热敏电阻在分压器中构建温度传感电路。评分方案要求说明：当热敏电阻的阻值降低时，其两端电压下降，而固定电阻两端电压升高，这可用于触发开关。
A common pitfall in these questions is describing the direction of change of Vout without linking it to the change in resistance. The mark scheme specifically rewards the logic: thermistor temperature up → Rthermistor down → Vthermistor down → Vfixed up. Simply memorising the formula is not enough; you must be able to reason through the chain of cause and effect.
这类问题中的一个常见陷阱是描述了 Vout 的变化方向却没有与电阻变化联系起来。评分方案明确给分的逻辑链条是：热敏电阻温度升高 → R热敏电阻 降低 → V热敏电阻 降低 → V固定电阻 升高。仅仅记住公式还不够；你必须能够沿着因果链进行推理。
9. Measurement Uncertainties and Significant Figures | 测量不确定度与有效数字
Every measurement carries an uncertainty, and the January 2020 mark scheme demanded consistent use of absolute or percentage uncertainties in combined calculations. For a single reading, the absolute uncertainty is typically ± half the smallest scale division. When two readings are taken for a length, the uncertainty in the length may be doubled if the start and end are read independently, but the scheme often allowed ± the smallest division if justified.
每次测量都带有不确定度，2020年1月的评分方案要求在组合计算中一致地使用绝对不确定度或百分比不确定度。对于单次读数，绝对不确定度通常是 ± 最小刻度值的一半。当对一个长度进行两次读数时，若起点和终点分别读取，则长度的不确定度可能加倍，但若能提供合理解释，评分方案通常也接受 ± 一个最小刻度值。
When combining uncertainties, the mark scheme looked for correct handling: for addition or subtraction, add absolute uncertainties; for multiplication or division, add percentage uncertainties. Candidates frequently lost marks by mixing absolute and percentage uncertainties or failing to round the final uncertainty to appropriate significant figures. The number of significant figures in the answer must reflect the precision of the input data.
在合并不确定度时，评分方案关注处理的正确性：对于加减运算，应合并绝对不确定度；对于乘除运算，应合并百分比不确定度。考生常常因混用绝对和百分比不确定度，或未将最终不确定度修约到合适的有效数字而失分。答案中有效数字的位数必须反映输入数据的精确程度。
10. Graph Skills: Line of Best Fit and Anomalous Points | 图像技能：最佳拟合线与异常点
Drawing an appropriate line of best fit was allocated marks in several parts of the Unit 2 paper. The mark scheme specified that the line should have a roughly equal number of points on either side and must ignore clear anomalies. Anomalous points were to be circled and labelled, but they should not influence the position of the line.
在Unit 2试卷的多个部分中，绘制合适的最佳拟合线是有分值分配的。评分方案规定，直线两侧的散点数量应大致相等，并必须忽略明显的异常点。异常点应用圆圈标出并注明，但不应影响直线的位置。
When calculating the gradient, the scheme requires selecting two points on the line that are as far apart as possible to minimise the percentage error. The coordinates of these points must be shown on the graph or stated clearly. Using data points from the table instead of points on the line is a serious error that can invalidate the gradient reading. The mark scheme explicitly examines whether the gradient triangle covers at least half of the line’s length.
在计算斜率时，评分方案要求尽可能选取直线上相距最远的两个点，以减小百分比误差。这些点的坐标必须在图上标出或清晰陈述。使用表格中原始数据点而非直线上的点是一个严重错误，会导致斜率读数无效。评分方案会明确审查梯度三角形是否涵盖了直线长度的一半以上。
11. Key Definitions and Standard Phrases | 关键定义与标准表述
The January 2020 mark scheme repeatedly rewarded the exact phrasing found in the specification. For example, the definition of the ‘threshold frequency’ in photoelectric emission is ‘the minimum frequency of electromagnetic radiation required to eject an electron from a metal surface’. Similarly, ‘coherent sources’ must be described as having ‘a constant phase difference and the same frequency’. Omitting ‘constant’ or ‘same’ often cost a mark.
2020年1月的评分方案再三对考纲中的精确表述给予奖励。例如，光电发射中“阈频率”的定义是“使电子从金属表面逸出所需电磁辐射的最小频率”。同理，“相干光源”必须被描述为具有“恒定的相位差和相同的频率”。漏掉“恒定”或“相同”常常会丢掉一分。
Another phrase that appeared was ‘plane polarised wave’, defined as a transverse wave in which the oscillations occur in one plane only. The mark scheme also examined the understanding that unpolarised light can be polarised by a polarising filter, and the intensity of the transmitted light is halved for complete polarisation of un‑polarised light. Full marks required linking this to Malus’s law I = I₀ cos²θ when a second filter is introduced, though Malus’s law is more formally A2 content; its conceptual basis was tested.
另一个出现的短语是“平面偏振波”，定义为振动只发生在一个平面内的横波。评分方案还考查了对非偏振光可以被偏振片偏振的理解，并知道对非偏振光完全偏振后，透射光强度会减半。当引入第二片偏振片时，要拿到满分需要将其与马吕斯定律 I = I₀ cos²θ 建立联系，尽管马吕斯定律更多属于A2的内容，但其概念基础受到了测试。
12. Practical Techniques and Evaluating Errors | 实验技术与误差评估
Questions on the Young modulus experiment required describing measurements: a micrometer to measure the diameter of the wire taken in several places and orientations to average; a tape measure for the original length; and a travelling microscope or Vernier scale for the small extension. The mark scheme credits statement that ‘mass hangers should be added and removed to check for elastic limit’ and ‘the wire should be thin to produce measurable extension for small loads’.
关于杨氏模量实验的问题要求描述测量：用千分尺在多个位置和方向测量金属丝的直径并取平均值；用卷尺测量原始长度；用移测显微镜或游标尺测量微小的伸长量。评分方案给分的陈述包括“应增减钩码以检查弹性极限”以及“金属丝应较细，以便在小负载下产生可测的伸长”。
To reduce uncertainty, the scheme rewarded repeating measurements and plotting a stress‑strain graph to determine the Young modulus from the gradient rather than from a single pair of values. Candidates were also expected to identify systematic errors, such as a zero error in the micrometer, and random errors, such as parallax when reading the extension. The distinction between reducing the effect of random errors (e.g. by averaging) and eliminating systematic errors (e.g. by calibration) was crucial for high marks on the evaluation question.
为减小不确定度，评分方案对重复测量并绘制应力‑应变图、从斜率求出杨氏模量而不是依赖单组数据进行奖励。考生还应能识别系统误差（如千分尺的零误差）和随机误差（如读取伸长量时的视差）。区分降低随机误差影响（如通过取平均值）和消除系统误差（如通过校准）对在评价题中拿高分至关重要。
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