📚 OCR A-Level Physics June 2023 Mark Scheme 1 Practical Investigation | OCR A-Level 物理 2023 年 6 月卷一评分方案：实验探究
Understanding how examiners award marks in Paper 1 practical questions is key to achieving top grades. This article breaks down the mark scheme for a typical experimental investigation in OCR Physics A, using a pendulum-based determination of g as the model. By pairing each English explanation with its Chinese counterpart, you will learn precisely what examiners expect regarding method, data handling, uncertainty analysis, and evaluation.
理解考官在卷一实验题中如何给分是取得高等级的关键。本文以典型的单摆法测重力加速度 g 为范例，逐项拆解 OCR 物理 A 实验探究题的评分方案。通过中英对照讲解，你将准确掌握考官在方法、数据处理、不确定度分析和实验评价等方面的评分要求。
1. Overview of the Practical Question | 实验题总览
In OCR Paper 1 ‘Modelling Physics’, one question typically tests practical skills through a structured investigation. The June 2023 variant presented a simple pendulum experiment: measure the period T for different pendulum lengths L, plot a suitable straight-line graph, and determine a value for the acceleration of free fall, g, alongside its absolute uncertainty.
在 OCR 卷一 ‘物理建模’ 中，通常有一道结构化探究题考察实验技能。2023 年 6 月试卷设置了一个单摆实验：测量不同摆长 L 下的周期 T，绘制合适的直线图，并求出重力加速度 g 及其绝对不确定度。
The mark scheme is not just a set of correct answers — it rewards precise control of variables, appropriate measurement techniques, valid graphical work, and critical evaluation of errors. Recognising these mark points turns a good practical into a secure, high-level answer.
评分方案并不是简单的标准答案——它会奖励对变量的精确控制、恰当的测量技术、有效的图像作业以及严谨的误差评价。认清这些给分点有助于将一个不错的实验转化为一份稳妥的高分回答。
2. Identifying and Controlling Variables | 识别与控制变量
The independent variable is the pendulum length L, measured from the point of suspension to the centre of the bob. The dependent variable is the period T, which is determined by timing multiple oscillations. At least two control variables must be stated clearly: amplitude of oscillation (kept small, e.g. less than 10°) and mass of the bob (or shape/size) — because the theory T = 2π√(L/g) only holds for small angles and is independent of mass only if air drag is negligible.
自变量是摆长 L，从悬点测量到摆锤中心。因变量为周期 T，通过测量多次摆动的时间得出。必须清楚说明至少两个控制变量：摆角振幅（保持很小，如小于 10°）和摆锤质量（或形状/尺寸）——因为理论式 T = 2π√(L/g) 仅在小角度成立，且只有在空气阻力可忽略时周期才与质量无关。
The mark scheme awards one mark for identifying the variables and one for a valid control method: ‘use a protractor to ensure initial displacement is always 5°–10°’ and ‘use the same pendulum bob throughout’. Omitting how to measure or control the angle will lose the method mark.
评分方案会为识别变量给一分，为有效的控制方法给一分：’用量角器确保初始位移始终为 5°–10°’ 以及 ‘全程使用同一个摆锤’。如果遗漏如何测量或控制摆角，就会失去方法分。
3. Apparatus and Measurement Techniques | 仪器与测量技术
Key apparatus includes a clamp stand, string, a small dense bob, a metre rule, a stopwatch (or light gate), and a protractor. The mark scheme credits using a metre rule with millimetre precision for L, measuring from the clamp to the middle of the bob, and using a fiducial marker (e.g. a pin at the equilibrium position) to help judge the centre of oscillation when counting.
主要仪器包括铁架台、细线、小而密度大的摆锤、米尺、秒表（或光闸）和量角器。评分方案鼓励使用精度达毫米的米尺测量 L，从夹具量至摆锤中心，并使用参考标记（如在平衡位置放置大头针）以帮助数摆动次数时判断振荡中心。
For timing, the scheme typically accepts either a stopwatch or a light gate, but insists on timing at least 20 oscillations to reduce the percentage uncertainty in T. The stopwatch should be started and stopped when the bob passes the fiducial marker moving in the same direction; human reaction time is then a random error that can be reduced by repeating.
计时方面，方案一般接受秒表或光闸，但强调至少测量 20 次完整摆动以减小 T 的百分不确定度。应在摆锤同方向经过参考标记时启动和停止秒表；人的反应时间此时属于随机误差，可通过重复测量予以降低。
One mark is reserved for stating that the length L should be measured with a metre rule (reading to ±1 mm) and that the stopwatch should read to 0.01 s — even if the effective uncertainty is larger due to reaction time.
其中一分专门留给说明：长度 L 应用米尺测量（可读至 ±1 mm），秒表应读到 0.01 s——即使由于反应时间，实际不确定度会更大。
4. Detailed Experimental Method | 详细实验步骤
A high-scoring sequence includes: set pendulum length L ≈ 0.500 m, measure L from suspension point to centre of bob, displace bob through a small angle (≈5°), release and simultaneously start timing 20 complete oscillations, stop timing at the 20th passage through the fiducial marker, record the total time t, repeat twice more for the same L, then calculate mean t and T = t/20. Repeat for 5–6 different lengths up to about 1.200 m.
高分步骤序列如下：设置摆长 L ≈ 0.500 m，从悬点量至摆锤中心，将摆锤拉离平衡位置一个小角度（≈5°），松开并同时开始计时 20 次完整摆动，在第 20 次经过参考标记时停止计时，记录总时间 t，对相同 L 再重复两次，计算平均 t 及 T = t/20。随后对 5–6 个不同长度（最大至约 1.200 m）重复上述过程。
The mark scheme rewards the explicit mention of repeats (for reliability), the use of averages, and the conversion to period. Many candidates lose a mark by timing only one oscillation or by failing to state the number of oscillations clearly.
评分方案会奖励明确提及重复实验（以提高可靠性）、使用平均值以及将总时间转换为周期。很多考生仅测量一次摆动或未能清晰说明摆动次数而丢失一分。
5. Recording Data and Table Design | 数据记录与表格设计
A well-designed table must include columns with quantity/symbol, unit, and appropriately headed raw data: L / m, t₁ / s, t₂ / s, t₃ / s, mean t / s, T / s. The mark scheme often checks that length L is recorded in metres (e.g. 0.500, not 50 cm) and that all time readings are given to the same number of decimal places consistent with the instrument — typically to two decimal places for a stopwatch (0.01 s).
设计良好的表格必须包含量及其符号、单位以及正确标题的原始数据栏目：L / m，t₁ / s，t₂ / s，t₃ / s，mean t / s，T / s。评分方案常检查长度 L 是否以米为单位记录（如 0.500 而非 50 cm），以及所有时间读数是否保留与仪器一致的小数位数——秒表通常保留两位小数（0.01 s）。
Calculations such as mean t and T must be shown clearly, and significant figures should be managed sensibly; for example, T usually has three significant figures given the precision of a manual stopwatch. A mark is available for correctly calculating period from averaged time values.
平均时间 t 和周期 T 等计算应清晰展示，且有效数字应合理处理；例如，鉴于手动秒表的精确度，T 通常取三位有效数字。正确地从平均时间值计算周期可获得一分。
6. Graph Plotting and Linearisation | 绘图与直线化
The relation T = 2π√(L/g) is non-linear. To obtain g from a straight-line graph, the mark scheme expects candidates to square both sides: T² = (4π²/g) L. Therefore plot T² on the y‑axis and L on the x‑axis. The graph should have suitable scales (occupying more than half the grid), labelled axes with units, and accurately plotted points.
关系式 T = 2π√(L/g) 是非线性的。为通过直线图求 g，评分方案期望考生将两边平方：T² = (4π²/g) L。因此以 T² 为纵轴、L 为横轴绘图。图形应采用合适标度（占据网格一半以上），标明坐标轴及单位，并精确描点。
A line of best fit should be drawn through the points, and the gradient m calculated using a large triangle shown on the graph. The scheme insists that the triangle used for gradient must cover at least half the line length; otherwise, the gradient determination will not attract full marks.
应通过各点画出最佳拟合直线，并利用图上画的大三角形计算斜率 m。方案强调用于求斜率的大三角形必须覆盖直线至少一半长度；否则，斜率测定无法获得满分。
Often the scheme accepts gradient values within a certain tolerance: e.g. m = 4.0 ± 0.2 s² m⁻¹. The final g is then g = 4π² / m, which yields a value likely around 9.5–10.0 m s⁻², considering typical experimental errors.
通常方案允许斜率在某一容差范围内：如 m = 4.0 ± 0.2 s² m⁻¹。然后最终 g = 4π² / m，在考虑典型实验误差的情况下，得出的值大约在 9.5–10.0 m s⁻²。
7. Uncertainty Analysis in the Mark Scheme | 评分方案中的不确定度分析
The absolute uncertainty in the gradient is estimated from the difference between the steepest and shallowest plausible best-fit lines (worst lines). The mark scheme requires drawing these lines explicitly or, at minimum, calculating Δm = (m_max – m_min)/2. The absolute uncertainty in g is then found using the relationship Δg = g × (Δm / m).
斜率的绝对不确定度通过最陡和最平可能的最佳拟合线（最劣线）之差来估算。评分方案要求明确画出这些直线，或至少计算 Δm = (m_max – m_min)/2。然后利用关系式 Δg = g × (Δm / m) 求 g 的绝对不确定度。
For individual length measurements, the scheme credits using a ruler’s resolution (±1 mm) as the absolute uncertainty, but notes that the dominant uncertainty originates from timing, where percentage uncertainty = (average reaction time / total time) × 100%. A typical reaction time uncertainty per stopwatch start/stop is quoted as 0.2 s, and this must be correctly propagated for T.
对于单个长度测量，方案认可使用米尺的分辨率（±1 mm）作为绝对不确定度，但指出主要的不确定度来源于计时，其百分不确定度 =（平均反应时间 / 总时间）× 100%。按下/停秒表的典型反应时间不确定度常引作 0.2 s，这必须正确传递给 T。
The mark scheme penalises treating the uncertainty of a repeated set of measured times simply as the range/2 without justifying it as a measure of random spread. Examiners look for explicit comments such as ‘uncertainty = (max – min)/2 for a small set of repeats’.
若仅将重复测量时间组的不确定度简单处理为极差/2，而未说明这是衡量随机散布程度的方法，评分方案会扣分。考官喜欢看到明确的说明，如 ‘对于一小组重复值，不确定度 =（最大值 – 最小值）/2’。
8. Determining the Result and Presenting Final Answer | 确定结果并给出最终答案
The final value of g must be expressed as g = value ± absolute uncertainty to an appropriate number of significant figures. The scheme typically accepts g ≈ 9.8 m s⁻² ± 0.4 m s⁻². The units must be stated (m s⁻²). The consistency of the experimental value with the accepted value of 9.81 m s⁻² is assessed through percentage difference or by checking if the accepted value falls within the experimental range.
最终 g 值必须表示为 g = 数值 ± 绝对不确定度，并取适当的有效数字。方案通常接受 g ≈ 9.8 m s⁻² ± 0.4 m s⁻²。必须写明单位（m s⁻²）。通过百分差异或检查公认值是否落在实验范围内，来评估实验值与公认值 9.81 m s⁻² 的一致性。
Marks are awarded for a complete final statement: ‘The determined value of g is (9.8 ± 0.4) m s⁻², which agrees with the accepted value of 9.81 m s⁻² within experimental uncertainty.’
完整的最终陈述可获得分数：’测得的 g 值为 (9.8 ± 0.4) m s⁻²，在实验不确定度范围内与公认值 9.81 m s⁻² 吻合。’
9. Evaluation and Sources of Error | 评价与误差来源
The mark scheme distinguishes between systematic and random errors. A systematic error might arise from measuring length L from the clamp to the top of the bob rather than to its centre (always overestimating L), or a stopwatch that runs slow. Random errors include human reaction time in starting/stopping the timer and the difficulty of keeping the amplitude exactly constant throughout 20 swings.
评分方案会区分系统误差和随机误差。系统误差可能源自测量长度 L 时从夹具量至摆锤顶部而非其中心（总是高估 L），或秒表走慢。随机误差包括启停计时器时人的反应时间，以及在 20 次摆动中难以保持振幅完全恒定。
A mark is specifically available for suggesting a realistic improvement that reduces the dominant error. For the pendulum, using a light gate to measure the period removes human reaction time and allows a more precise determination of the equilibrium crossing. An alternative is to measure the time for 50 oscillations with a marker and video analysis.
专门有一分留给提出一个切实的改进措施以减少主要误差。对于单摆，使用光闸测量周期可消除人的反应时间，并更精确地确定经过平衡位置的时刻。另一种方法是配合标记和视频分析，测量 50 次摆动的时间。
The ability to link a specific error to its effect on the final calculated g is rewarded: ‘Measuring L to the wrong point produces a systematic shift in gradient, causing a proportional error in g, without affecting the linearity.’
能够将特定误差与其对最终计算 g 的影响联系起来也可得分：’从错误的点测量 L 会造成斜率的系统偏移，导致 g 等比例误差，但不影响线性关系。’
10. Mark Allocation and Common Pitfalls | 分值分配与常见失分点
This type of practical question typically carries 12–15 marks. A typical breakdown: variables and method (3), table and raw data (2), graph and line of best fit (3), gradient calculation and determination of g (2), uncertainty analysis (3), evaluation and improvement (2). Examiners report that candidates frequently lose marks by not stating control variables concretely, using inappropriate scales, confusing accuracy with precision, and presenting uncertainties without any explanation of their origin.
此类实验题通常占 12–15 分。典型分值分配为：变量与方法 (3)，表格与原始数据 (2)，图形与最佳拟合线 (3)，斜率计算和 g 的测定 (2)，不确定度分析 (3)，评价与改进 (2)。考官报告考生常因未具体说明控制变量、使用不合理的标度、混淆准确度与精确度，以及陈述不确定度时没有解释其来源而丢分。
| Marking Point | 评分要点 | Common Error |
|---|---|---|
| Variables controlled with measurable detail | 控制变量并有可测量的细节 | Vague ‘keep angle small’ without stating how |
| Table L in metres, times to 0.01 s | 表格中 L 以米为单位，时间精确至 0.01 s | Recording L in cm or inconsistent dp |
| T² vs L graph with correct axes | T²-L 图且坐标轴正确 | Plotting T vs L (non-linear) directly |
| Large triangle for gradient, show coordinates | 大三角形求斜率，标明坐标 | Using data points instead of points on line |
| Absolute uncertainty in g from worst lines | 用最劣线求出 g 的绝对不确定度 | Simply guessing uncertainty without calculation |
11. Using Mathematical Relationships from the Mark Scheme | 使用评分方案中的数学关系式
The fundamental pendulum equation is T = 2π√(L/g). Squaring gives T² = (4π²/g)L, so gradient m = 4π²/g. The mark scheme explicitly checks that candidates can rearrange this to g = 4π² / m. It also rewards substituting the calculated gradient value correctly, showing the working, and quoting g to a justified number of significant figures — normally three, given the input data.
基本单摆方程为 T = 2π√(L/g)。平方得 T² = (4π²/g)L，因此斜率 m = 4π²/g。评分方案明确检查考生能否将该式变换为 g = 4π² / m。对于正确代入算得的斜率值、展示运算步骤、并以合理有效数字（给定输入数据，通常取三位）报告 g，均给予分数。
Percentage uncertainty is another tested quantity: %U = (Δm / m) × 100% = (Δg / g) × 100%. Some mark schemes reward stating that since the gradient m and g are inversely proportional, their percentage uncertainties are equal.
百分不确定度是另一考点：%U = (Δm / m) × 100% = (Δg / g) × 100%。有些评分方案会奖励指出由于斜率 m 与 g 成反比，故两者的百分不确定度相等。
12. Examiner Expectations and Final Strategy | 考官期望与最终策略
To secure all marks, candidates must treat every instruction as a mark opportunity: underline headings in the answer booklet, write the number of oscillations timed, state the fiducial marker purpose, and consistently quote units. The mark scheme rewards precise language, not vague statements. For example, ‘use a larger triangle to reduce percentage uncertainty in gradient’ is ineffective because the triangle size doesn’t inherently reduce uncertainty — but ‘using data points that span a wider range of L reduces the impact of random scatter on the gradient’ would be credited.
要确保拿到全部分数，考生应将每个指令视为得分机会：在答题册中给各项加上小标题，写明计时的摆动总次数，说明参考标记的用途，并始终注明单位。评分方案奖励精确的语言而非模糊的陈述。例如，’使用更大的三角形可减小斜率的百分不确定度’ 并不有效，因为三角形大小本身并不减小不确定度——但 ‘使用跨越更宽 L 范围的数据点可降低随机散点对斜率的影响’ 则会得分。
Finally, time management during the exam is crucial: spend roughly 1.5 minutes per mark. This exercise should not dominate the paper, but a clear, systematic response aligned with the mark scheme will convert preparation into tangible results.
最后，考试中的时间管理至关重要：大约每分 1.5 分钟。该题不应占据试卷过多篇幅，但与评分方案一致的清晰、系统的作答会将备考成果转化为实实在在的分数。
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