📚 A-Level Physics Unit 4 Jan 22 Mark Scheme: Experimental Investigation | A-Level 物理 Unit 4 2022年1月评分方案：实验探究精解
This article unpacks the experimental investigation question from the Edexcel International A-Level Physics Unit 4 (WPH14/01) January 2022 examination, using the official mark scheme as a lens. We decode the precise criteria examiners used to award marks for practical planning, measurement, data processing, graphical analysis and evaluation. Whether you are preparing for a future Unit 4 paper or consolidating your practical skills, understanding how marks are allocated is the key to securing top grades in practical-based questions.
本文以 Edexcel 国际 A-Level 物理第四单元（WPH14/01）2022 年 1 月考试的官方评分方案为镜，深度剖析实验探究题。我们将拆解考官在实验计划、测量、数据处理、图形分析和评估环节中使用的精确评分标准。无论你是在准备未来的 Unit 4 考试，还是在巩固实验技能，理解分数是如何分配的都是在实验题中拿到高分的钥匙。
1. Structure of the Experimental Question | 实验题的结构
The Unit 4 experimental question typically carries 12 to 16 marks and follows a predictable structure: recording raw data, tabulating results, processing data (often requiring a linearisation graph), determining a physical quantity, and evaluating the procedure. In the January 2022 mark scheme, marks were split across these sections with a heavy emphasis on valid conclusions and error handling.
第四单元的实验题通常占 12 到 16 分，且遵循一个可预测的结构：记录原始数据、列表整理结果、处理数据（通常需要绘制线性化图像）、确定某个物理量以及评估实验过程。在 2022 年 1 月的评分方案中，分数分散在这些板块，尤其侧重得出有效结论和处理误差的能力。
2. Command Words and What They Demand | 指令词及其要求
Command words in the question paper map directly onto the mark scheme. ‘Measure’ and ‘record’ require accurate values with appropriate units and precision. ‘Plot’ demands correctly labelled axes and sensible scales. ‘Determine’ expects a calculation with the final answer given to the correct number of significant figures. ‘Evaluate’ and ‘suggest’ push you to identify limitations, quantify their effects where possible, and propose realistic improvements. The Jan 22 scheme shows that vague evaluation without specific physics reasoning scores zero.
试卷中的指令词直接对应评分方案中的得分点。’Measure’ 和 ‘record’ 要求给出带有合适单位和精度的准确数值。’Plot’ 要求正确标记坐标轴并使用合理分度。’Determine’ 期望计算出结果，且最终答案保留正确的小数位数。’Evaluate’ 和 ‘suggest’ 则要求你指出局限性，尽可能量化其影响，并提出切合实际的改进措施。2022 年 1 月的评分方案显示，缺乏具体物理推论的模糊评价不会得分。
3. Designing a Valid Investigation: The Simple Pendulum | 设计合理的探究：单摆实验
In the Jan 22 paper, the experimental context involved determining the acceleration of free fall, g, using a simple pendulum. The independent variable was the pendulum length l (measured from the point of suspension to the centre of the bob). The dependent variable was the oscillation period T. Control variables included the amplitude of swing (kept small, below 10°), the mass of the bob, and the room temperature. A well-planned investigation always begins by naming these three variable types explicitly.
在 2022 年 1 月的试卷中，实验背景是用单摆测定自由落体加速度 g。自变量是摆长 l（从悬挂点到摆球中心的距离）。因变量是摆动周期 T。控制变量包括摆幅（始终保持小角度，低于 10°）、摆球质量和室温。一项设计良好的探究总是从明确命名这三类变量开始。
4. Control of Variables and Fair Testing | 变量控制与公平测试
To isolate the relationship between l and T, all other factors must be kept constant. The Jan 22 mark scheme rewarded students who stated that a small amplitude ensures the motion approximates simple harmonic motion, making the period independent of amplitude. Using the same bob eliminated mass as a variable. Conducting all trials in a short time window minimised temperature drift that could alter the length of the string. Marks were deducted if control measures were only listed but not justified in physics terms.
为了孤立 l 与 T 之间的关系，所有其他因素必须保持不变。2022 年 1 月的评分方案奖励那些指出小振幅确保运动近似为简谐运动、从而使周期与振幅无关的学生。使用同一个摆球排除了质量的影响。在短时间内完成所有测量可以最大程度减少因温度变化导致绳长改变的可能。只罗列控制变量而未用物理术语加以论证会被扣分。
5. Measurement Techniques and Instrument Selection | 测量技术与仪器选择
Length l was measured using a metre rule with millimetre markings, and the diameter of the bob was obtained with vernier callipers; the effective length was taken as the distance from the clamp to the bottom of the bob plus its radius. To reduce reaction time uncertainty, the period T was determined by timing 20 complete oscillations with a digital stopwatch and then dividing by 20. The mark scheme accepted repeating each timing three times and averaging, provided the raw data were clearly shown.
摆长 l 使用带有毫米刻度的米尺测量，摆球直径用游标卡尺获取；有效摆长取悬挂点到摆球底端的距离再加上其半径。为了减小反应时间带来的不确定度，周期 T 通过用数字秒表测量 20 次全振动的时间再除以 20 得到。评分方案接受每个时间重复测量三次并取平均值，前提是原始数据必须清晰呈现。
6. Data Presentation and Tabulation | 数据呈现与制表
All recorded data must be entered into a neat table with quantity/unit headers, e.g. ‘l / m’ and ‘T / s’. The Jan 22 mark scheme insisted on consistent decimal places: if l was measured to ±0.001 m, all values in that column had to be shown to three decimal places. Calculated columns such as T² / s² were expected, with values rounded appropriately. No marks were awarded if raw data were missing or if units appeared only next to the numbers rather than in the header.
所有记录的数据都必须整齐地填入表格，并在表头注明量/单位，例如“l / m”和“T / s”。2022 年 1 月的评分方案强调小数点位数一致：若 l 的测量精度为 ±0.001 m，则该列所有数值必须保留三位小数。还应包含计算列，如 T² / s²，数值需适当四舍五入。若缺少原始数据，或单位只写在数字旁边而非表头中，将完全不得分。
7. Graph Plotting and Achieving Linearity | 绘图与实现线性化
The relationship T = 2π√(l/g) is non-linear, so linearisation is essential. Squaring both sides gives the straight-line equation:
T² = (4π² / g) l
Plotting T² on the vertical axis and l on the horizontal axis yields a straight line through the origin. In the Jan 22 scheme, marks were awarded for axes labelled with quantities and units, sensible scales that occupy more than half the grid, accurate plotting of points to within half a small square, and a sharp best-fit line. Failure to linearise correctly was a major mark-losing pitfall.
关系式 T = 2π√(l/g) 是非线性的，因此必须进行线性化。两边平方得到直线方程：
T² = (4π² / g) l
将 T² 作为纵轴、l 作为横轴绘图，会得到一条过原点的直线。2022 年 1 月的方案中，正确标注包含单位的坐标轴、选取超过半幅坐标纸的合理分度、描点精确到半小格以内，并画出清晰的“最佳拟合线”才能拿到绘图分。未能正确线性化是主要的失分陷阱。
8. Determining Gradient and Intercept | 确定斜率与截距
From the line of best fit, the gradient m = Δ(T²) / Δl is calculated using a large triangle that covers at least half of the drawn line. The Jan 22 mark scheme required the triangle vertices to be clearly shown on the graph, with coordinates read to the precision of the scale. The intercept was expected to be zero or very nearly so; a significantly non-zero intercept hinted at a systematic error in length measurement, such as misreading the zero of the metre rule.
在最佳拟合线上，选取一个覆盖所绘直线至少一半长度的大三角形，计算斜率 m = Δ(T²) / Δl。2022 年 1 月的评分方案要求图上清晰标出三角形顶点，坐标读数精确到分度值允许的范围。截距应接近于零；若截距显著不为零，则暗示长度测量中存在系统误差，例如米尺的零点读数错误。
9. Calculating Derived Quantities and Uncertainties | 计算导出量与不确定度
Once the gradient m was found, g was determined from g = 4π² / m. The mark scheme awarded the final mark only if the unit (m s⁻²) was given and the value was sensible (about 9.81 m s⁻²). Percentage uncertainty in g was derived by combining the percentage uncertainty in the gradient, which itself came from the difference between the best-fit and worst-acceptable-fit lines: Δm = |m_best – m_worst|. Many candidates lost marks because they attempted to estimate uncertainty using the scatter of points alone, without drawing the worst-fit line.
得到斜率 m 后，由 g = 4π² / m 计算 g。评分方案只有在给出正确单位（m s⁻²）且数值合理（约 9.81 m s⁻²）时才给最后一步答案分。g 的百分不确定度通过组合斜率的百分不确定度得出，而斜率的不确定度来自最佳拟合线与最差可接受拟合线之差：Δm = |m_best – m_worst|。许多考生因为仅凭点的散布估算不确定度，而没有画出最差拟合线而失分。
10. Evaluating Sources of Error and Limitations | 评估误差来源与局限性
The Jan 22 evaluation section required candidates to distinguish between random and systematic errors. Random errors arose from human reaction time in starting and stopping the stopwatch, which could be reduced by timing more oscillations. Systematic errors included the finite amplitude (even below 10°) causing the period to deviate from the small-angle approximation, and the string possessing some mass and elasticity. High-scoring answers quantified the possible effect, e.g., ‘a 5° amplitude increases T by roughly 0.05%, introducing a systematic underestimate of g by about 0.1%’.
2022 年 1 月的评估部分要求考生区分随机误差和系统误差。随机误差主要来自启动和停止秒表时的反应时间，可通过测量更多个周期的总时间来减小。系统误差包括即使摆角小于 10° 仍会偏离小角度近似，以及细绳本身具有质量和弹性。高分答案能够量化可能的影响，例如“5° 的摆幅会使 T 增加约 0.05%，从而使 g 被系统性低估约 0.1%”。
11. Suggesting Improvements and Further Work | 提出改进措施与后续工作
Suggestions must be practical and linked to the identified limitations. Replacing the stopwatch with a light gate and electronic timer would eliminate human reaction time and improve precision by an order of magnitude. Using a thin, light, inextensible string and a small dense spherical bob would better satisfy the assumptions of the simple pendulum model. The mark scheme rejected vague suggestions like ‘use better equipment’ and required specific physics reasoning for each proposal.
所提建议必须切实可行，并与已指出的局限性相关联。用光门和电子计时器替代秒表可以消除人为反应时间，使精度提高一个数量级。使用细、轻且不可伸长的绳子和一个小而密的球形摆球，能更好地满足单摆模型的假设。评分方案拒绝了“使用更好的设备”这类模糊建议，要求每项提议都要有具体的物理论证。
12. Applying the Mark Scheme: A Worked Example | 应用评分方案：一个实例解析
Imagine a student’s data for l = 0.500, 0.600, 0.700, 0.800, 0.900 m yielded T = 1.42, 1.55, 1.68, 1.79, 1.90 s. The table included T² = 2.02, 2.40, 2.82, 3.20, 3.61 s² with consistent significant figures. The graph of T² against l was a straight line passing close to the origin. The gradient was read as 4.02 s² m⁻¹, giving g = 4π² / 4.02 ≈ 9.82 m s⁻². The worst-fit line gave a gradient of 3.90 s² m⁻¹, leading to Δg/g = (4.02 – 3.90)/4.02 ≈ 3.0%. The evaluation noted that the string was a thick cord, adding significant mass and shifting the centre of mass, explaining why the intercept was not exactly zero. According to the Jan 22 mark scheme, such an answer would secure full marks in every section.
假设一名学生的数据：l = 0.500, 0.600, 0.700, 0.800, 0.900 m，对应 T = 1.42, 1.55, 1.68, 1.79, 1.90 s。表格中附有 T² = 2.02, 2.40, 2.82, 3.20, 3.61 s²，并保持了统一的有效数字。T² 对 l 的图为一条接近过原点的直线。读出的斜率为 4.02 s² m⁻¹，得到 g = 4π² / 4.02 ≈ 9.82 m s⁻²。最差拟合线的斜率为 3.90 s² m⁻¹，由此得出 Δg/g = (4.02 – 3.90)/4.02 ≈ 3.0%。评估部分指出，使用的绳子是一根粗绳，增加了显著质量并移动了质心，这解释了截距为何不恰为零。按照 2022 年 1 月的评分方案，这样一份答案在每个部分都能拿到满分。
Published by TutorHao | Physics Revision Series | aleveler.com
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