📚 OCR A-Level Physics June 2023 Paper 1: Experimental Investigation | OCR A-Level物理2023年6月卷1：实验探究
Practical investigation questions form a central part of OCR A-Level Physics Paper 1, and in the June 2023 session, candidates were required to analyse an experiment designed to determine the acceleration due to gravity, g, using a freely falling object and light gates. This article dissects the experimental setup, data handling, graphical analysis, uncertainty propagation, and evaluation of the procedure, mirroring the expectations of the 2023 Paper 1 assessment. Whether you are reviewing your performance or preparing for future papers, understanding the precise demands of this investigation will sharpen your experimental skills.
实验探究题是OCR A-Level物理卷1的核心内容。在2023年6月的考试中，考生需要分析一个利用自由落体和光门测定重力加速度g的实验。本文详细拆解实验装置、数据处理、图像分析、不确定度传递以及实验方案评估，全面回应2023年卷1的评分要求。无论你是在复盘自己的答案，还是为未来的考试做准备，透彻理解这道实验探究题将显著提升你的实验技能。
1. Experimental Aim and Theory | 实验目的与原理
The investigation set out to determine a value for the acceleration of free fall, g, by measuring the time taken for a steel ball to fall through a measured vertical distance. The fundamental relationship for an object dropped from rest is s = ½gt², where s is the distance fallen, t is the time of fall, and g is the acceleration due to gravity. By ensuring the initial velocity is zero, a graph of s against t² yields a straight line through the origin, with gradient equal to ½g.
该探究旨在通过测量钢球下落固定竖直距离所需的时间来测定自由落体加速度g。物体从静止下落的基本关系式为s = ½gt²，其中s为下落距离，t为下落时间，g为重力加速度。在确保初速度为零的条件下，绘制s-t²图像应得到一条过原点的直线，其斜率等于½g。
The theory relies on the assumptions that air resistance is negligible and that the release mechanism imparts no initial downward velocity. In an A-Level context, candidates are expected to manipulate the equation to a linear form, recognise that the constant ½g can be extracted from the gradient, and discuss the implications of systematic and random errors on the final value.
该理论成立的前提是空气阻力可忽略，且释放机构不会给小球施加任何初始向下的速度。在A-Level阶段，考生需要将方程转化为线性形式，识别出斜率中包含了½g，并讨论系统误差和随机误差对最终结果的影响。
2. Apparatus and Setup Description | 仪器与装置描述
The apparatus listed in the June 2023 Paper 1 scenario included a metre rule with millimetre divisions, an electromagnet attached to a clamp stand, a steel ball bearing, two light gates connected to a data logger, and a plumb line. The electromagnet was used to hold and release the steel ball from rest. The first light gate was placed flush with the bottom of the ball when suspended from the electromagnet, to start the timer at the instant of release. A second light gate was positioned a known distance s below the first, to stop the timer as the ball passed through it.
2023年6月卷1题设中所列的仪器包括一把毫米刻度米尺、固定在铁架台上的电磁铁、一个钢球、两个连接至数据记录器的光门，以及一个铅垂线。电磁铁用于固定和释放钢球，确保它从静止开始下落。第一个光门紧贴在悬挂于电磁铁下方的钢球底部，以便在释放瞬间启动计时器。第二个光门放置在第一个光门下方已知距离s处，当钢球穿过该光门时停止计时器。
A crucial detail in the setup was the alignment of the apparatus using the plumb line to ensure the fall was exactly vertical and that the light gates were centred on the path of the ball. The data logger recorded the time interval t between the interruption of the first and second light beams. This arrangement minimises reaction time errors associated with manual stopwatch measurements.
装置细节中的一个关键点是用铅垂线校准仪器，确保下落路径完全竖直，且光门对准小球的运动路线。数据记录器记录下小球遮挡第一道与第二道光束之间的时间间隔t。这种布置最大程度减少了手动秒表计时的反应时误差。
3. Data Collection Procedure | 数据收集步骤
The experiment proceeded by setting a distance s between the two light gates, measuring this distance with the metre rule, and recording the time t displayed by the data logger. The procedure was repeated three times for each distance to obtain average t values. The separation s was varied in increments of 0.100 m from approximately 0.200 m to 1.000 m, providing a range of data points for a straight-line graph.
实验流程如下：设定两个光门之间的距离s，用米尺测量该距离，并记录数据记录器显示的时间t。每个距离下重复测量三次以获得平均t值。s的取值从大约0.200 m开始，以0.100 m为间隔递增，直至约1.000 m，为直线图像提供充足的数据点。
During data collection, candidates were instructed to ensure that the ball fell cleanly through both light gates without touching the sides, and that the release was smooth. The initial distance was measured from the lower edge of the clamped ball at the electromagnet to the upper edge of the second light gate, as this corresponded to the distance travelled while the timer was active.
在数据收集过程中，考生被要求确保钢球干净利落地穿过两个光门而不触碰边缘，同时释放动作要平稳。初始距离是从电磁铁夹持的钢球下缘量到第二个光门的上缘，因为这段距离对应于计时器工作期间小球运动的实际路程。
A table was constructed to record s, the three time readings t₁, t₂, t₃, the mean time t₍ₘₑₐₙ₎, and the calculated t². The use of a table with clear headings and consistent units, as shown below, is expected in a high-scoring response.
实验要求构建一个表格用于记录s、三次时间读数t₁、t₂、t₃、平均时间t₍ₘₑₐₙ₎以及计算出的t²。使用表头清晰、单位统一的表格是高分作答的必备条件，如下表所示。
| s / m | t₁ / s | t₂ / s | t₃ / s | t₍ₘₑₐₙ₎ / s | t² / s² |
|---|---|---|---|---|---|
| 0.200 | 0.202 | 0.199 | 0.201 | 0.201 | 0.0404 |
| 0.300 | 0.247 | 0.248 | 0.246 | 0.247 | 0.0610 |
| 0.400 | 0.286 | 0.285 | 0.287 | 0.286 | 0.0818 |
| 0.500 | 0.319 | 0.320 | 0.318 | 0.319 | 0.1018 |
| 0.600 | 0.350 | 0.351 | 0.349 | 0.350 | 0.1225 |
4. Graphical Plotting and Line of Best Fit | 作图与最佳拟合线
According to s = ½gt², plotting s on the y-axis and t² on the x-axis should produce a straight line passing through the origin. Candidates were expected to choose appropriate scales, label axes with quantities and units, and accurately plot the data points. A transparent ruler was used to draw a line of best fit that balanced the scatter of points about the line.
根据s = ½gt²，将s作为纵轴、t²作为横轴作图，应得到一条过原点的直线。考生需选择合适的坐标轴比例，标注物理量和单位，并准确描点。使用透明直尺画出最佳拟合线，使数据点在直线两侧均匀分布。
Since the expected relationship includes the origin, the line should ideally pass through (0,0). However, if small systematic errors are present, the line might intercept the t² axis slightly. In the 2023 Paper 1, a well-drawn graph would allow the examiner to assess the precision of the data. The gradient was extracted using a large triangle, typically spanning at least half the graph’s width, to minimise relative uncertainty.
由于预期关系包含原点，最佳拟合线理论上应通过(0,0). 若存在轻微的系统误差，直线可能在t²轴上有微小截距。在2023年卷1中，规范绘制的图像能让评分员评估数据的精密程度。选取一个足够大的三角形来计算斜率，通常至少跨越图像宽度的一半，以减小相对不确定度。
gradient = Δs / Δ(t²) = ½g
5. Determining g from the Graph | 从图像确定g值
Once the gradient of the s versus t² graph was obtained, the value of g was calculated by doubling the gradient: g = 2 × gradient. For instance, if the gradient was found to be 4.91 m s⁻² (a plausible value from typical data), then g = 9.82 m s⁻². The answer was quoted to an appropriate number of significant figures, usually three, to reflect the precision of the measurements.
获得s-t²图像的斜率后，将其乘以2即可计算出g值：g = 2 × 斜率。例如，若测得斜率为4.91 m s⁻²（典型数据中的合理值），则g = 9.82 m s⁻²。答案需给出合适的有效数字位数，通常为三位，以体现测量的精密程度。
Candidates were also required to compare their experimental value with the accepted standard value of 9.81 m s⁻² and calculate the percentage difference. A calculation such as |9.82 – 9.81| / 9.81 × 100% ≈ 0.1% would demonstrate excellent agreement. This comparison forms part of the evaluation and helps identify the significance of systematic errors.
考生还需将自己得到的实验值与公认标准值9.81 m s⁻²进行比较，并计算百分差。例如，|9.82 – 9.81| / 9.81 × 100% ≈ 0.1%，表明数据吻合度极好。这种对比是评估环节的一部分，有助于判断系统误差的显著程度。
6. Uncertainty Analysis in Measured Quantities | 测量量的不确定度分析
The distance s was measured using a metre rule with a resolution of ±1 mm, giving an absolute uncertainty of ±0.001 m for each reading. However, since s involves the difference between two positions (start and end of fall), the uncertainty in s was often taken as ±2 mm to account for parallax and alignment errors. The time t was recorded electronically with a precision of ±0.001 s, but the repeatability of the release mechanism often introduced a larger random uncertainty, estimated from the spread of repeated readings.
距离s使用毫米刻度的米尺测量，其分辨力为±1 mm，因此每次读数的绝对不确定度为±0.001 m。但s涉及两个位置（下落的起点与终点）之差，为兼顾视差和对准误差，s的不确定度通常取±2 mm。时间t由电子计时器记录，精度为±0.001 s，但释放机构的重复性往往会引入更大的随机不确定度，该值可通过重复读数的分散程度来估算。
For each distance, the uncertainty in the mean time t₍ₘₑₐₙ₎ was found using the half-range method: (max t – min t) / 2. A specimen calculation for s = 0.200 m gives (0.202 – 0.199)/2 = 0.0015 s. The percentage uncertainty in t² was then approximated as twice the percentage uncertainty in t, since the power rule applies. These percentage uncertainties were used to discuss the reliability of the gradient and thus the uncertainty in g.
对每个距离，平均时间t₍ₘₑₐₙ₎的不确定度采用半范围法计算：(最大t – 最小t) / 2。以s = 0.200 m为例，(0.202 – 0.199)/2 = 0.0015 s。t²的百分不确定度近似为t的百分不确定度的两倍，这是由幂次规则决定的。这些百分不确定度用以讨论斜率的可靠性，进而得出g值的不确定度。
A detailed answer in the 2023 Paper 1 required candidates to combine uncertainties to estimate the uncertainty in g, typically using worst-case upper and lower gradient lines or a simple propagation of the percentage uncertainty in the gradient. Expressing the final result as g = 9.82 ± 0.05 m s⁻² would secure marks for uncertainty quantification.
2023年卷1的高分回答要求考生综合不确定度以估算g的不确定度，通常采用最差情况上下限斜率法，或对斜率的百分不确定度进行简单传递。将最终结果表述为g = 9.82 ± 0.05 m s⁻²，可确保获得不确定度量化部分的分数。
7. Identification and Discussion of Systematic Errors | 系统误差的识别与探讨
A key systematic error in this experiment arises from the residual magnetism of the electromagnet. When the circuit is broken, the electromagnet may not release the ball instantaneously, causing a small delay that increases the measured time. This would lead to a smaller gradient, and thus a systematically low value of g. The 2023 mark scheme rewarded identification of this effect and suggestions to minimise it, such as reversing the current momentarily to degauss the core or using a mechanical release.
本实验的一个关键系统误差源于电磁铁的剩磁。电路断开时，电磁铁可能无法瞬间释放钢球，造成微小延迟，导致测得的t值偏大。这会使斜率变小，从而系统性地低估g值。2023年的评分方案对识别这一影响并建议减小措施的回答给予奖励，例如瞬间反向通电对铁芯消磁，或改用机械释放方式。
Another systematic error relates to the definition of s. If the distance is measured from the lower edge of the ball to the upper edge of the second light gate, but the timer starts slightly after release, the effective s is smaller than measured, again reducing the gradient. A carefully positioned start trigger, perhaps using a pressure-sensitive pad at release, would reduce this discrepancy. The paper also accepted discussion of air resistance, though its effect at these speeds and masses is very small.
另一个系统误差与s的定义相关。若距离是从钢球下缘量到第二个光门上缘，但计时器在释放后稍晚才启动，则有效s将小于测量值，同样导致斜率减小。精心布置启动触发器，例如使用释放点处的压敏垫，可减小这种偏差。试卷也接受对空气阻力的讨论，但在这些速度与质量下，其影响微乎其微。
8. Random Errors and Technique Refinements | 随机误差与操作改进
Random errors were primarily introduced by variations in the release mechanism and slight wobbles in the ball’s trajectory. Repeating each measurement and calculating a mean time is the standard approach to minimise random uncertainties. Additionally, the paper encouraged candidates to note that using a computer data logger with a high sampling rate could detect the precise moment of beam interruption more consistently than a human-triggered release.
随机误差主要来自释放动作的差异以及钢球轨迹的轻微晃动。对每个距离重复测量并计算平均时间是减小随机不确定度的标准方法。此外，试卷鼓励考生指出，使用高采样率计算机数据记录器，能比人为触发的释放更稳定地检测光束遮挡的精确时刻。
Suggesting a longer fall distance to increase t, thereby reducing the percentage uncertainty in the time measurement, was also credited. However, candidates needed to acknowledge that distances beyond about 1.2 m might introduce noticeable air resistance, so an optimum range must be selected. The plumb line alignment and clamping rigidly were emphasised as simple yet effective ways to reduce random trajectory deviations.
建议使用更长的下落距离以增大t，从而降低时间测量的百分不确定度，这一想法也获得认可。但考生需认识到，超过约1.2 m的距离可能引入显著的空气阻力，因此必须选取最佳量程。使用铅垂线校准和刚性夹持这类简单而有效的方式，被强调为减少随机轨迹偏离的手段。
9. Evaluation of the Experimental Procedure | 实验方案评估
The evaluation section in June 2023 Paper 1 required candidates to judge the overall quality of the data. Using properly labelled error bars on the graph of s against t², or computing the absolute uncertainty in the gradient, revealed whether the points were consistent enough to claim a reliable value for g. Any anomalous points, likely due to a mistimed release or the ball brushing against a light gate, were identified and excluded with justification.
2023年6月卷1的评估部分要求考生判断数据的整体质量。在s-t²图中使用规范标注的误差棒，或计算斜率绝对不确定度，可以揭示数据点是否足够一致，从而声称所得g值可靠。任何异常点——很可能是由于释放时间错误或钢球擦碰光门造成的——都应被识别出来，并在说明理由后剔除。
An ideal evaluation did more than simply list weaknesses; it explained the impact of each identified limitation on the derived value of g. For example, ‘The residual magnetism would cause a delayed release, making t slightly larger, which reduces the gradient and gives a g value below the true value.’ This degree of clarity and scientific reasoning earns top marks.
理想的评估不止于罗列不足，而是要解释每项已识别限制对推导出的g值有何影响。例如，“剩磁会导致延迟释放，使t略微偏大，从而减小斜率，造成g值低于真值。”如此清晰的程度与科学推理能赢得最高分数。
10. Examiner Insights and Common Mistakes | 评分员洞察与常见错误
From examiner reports on similar investigations, common pitfalls include forgetting to square the time before plotting, mislabelling the graph axes, and failing to convert distances into standard units. Another frequent error is stating that the gradient is equal to g, without halving it. In the table, inconsistencies in significant figures or failing to calculate t2 for each row before plotting can lose marks.
根据评分员报告，类似探究题的常见误区包括：作图前忘记将时间平方、坐标轴标注错误，以及未能将距离转换为标准单位。另一个常见错误是声称斜率等于g，而忘记除以2。表格中有效数字位数的前后不一致，或画图前没有计算每行的t2值，都会导致失分。
Examiners also noted that candidates struggled to relate the intercept to the start timing error. An intercept on the t2 axis would suggest that the timer started after the ball had already fallen slightly, implying an overestimate of t. Good answers connected graph features to physical meaning. Students who merely wrote ‘systematic error’ without specifying its practical origin missed out on analysis marks.
评分员还指出，考生难以将截距与起始计时误差联系起来。t2轴上的截距表明计时器在钢球已下落一小段距离后才启动，意味着t被高估。好的作答会将图像特征与物理意义相联系。仅仅写一句“系统误差”而不指出其实际来源的考生，会错失分析分。
11. Practical Skills Checklist for Revision | 复习必备的实验技能清单
To master such experimental investigation questions, ensure you can do the following: accurately tabulate raw and processed data with correct headings and units; select and draw appropriate scales for a graph; plot points precisely using small crosses; draw a best-fit straight line; calculate gradient and intercept; manipulate equations into y = mx + c form; handle uncertainties in repeated measurements using half-range or standard deviation; propagate uncertainties for derived quantities; and evaluate experimental procedures in terms of accuracy, precision, and systematic versus random errors.
要掌握此类实验探究题，请确保你能做到以下各点：准确地将原始数据和处理数据整理在表格中，并配有正确的表头与单位；为图像选取和绘制合适的坐标轴比例；用小叉号精确描点；画出最佳拟合直线；计算斜率和截距；将方程变形为y = mx + c的形式；利用半范围或标准差处理重复测量值的不确定度；对导出量进行不确定度传递；从准确度、精密度以及系统与随机误差的角度评价实验方案。
The June 2023 Paper 1 required about 20 minutes of focused work on this investigation. Practising with past papers and writing timed responses will build fluency. Pay particular attention to the ‘plan, implement, analyse, evaluate’ structure that underpins OCR’s practical assessment ethos. Every detail in the question matters—do not ignore the small text about the data logger model or the material of the ball; examiners place it there for a reason.
2023年6月卷1要求在这道探究题上集中投入约20分钟。利用往年试卷练习并限时作答，能够培养流畅度。尤其要注意“计划、实施、分析、评估”这一支撑OCR实验考核理念的结构。题中的每个细节都很重要——不要忽略关于数据记录器型号或钢球材质的小字说明；评分员把它们放在那里是有原因的。
12. Extending the Concept to Other Paper 1 Scenarios | 将概念延伸至其他卷1情境
While the specific experiment in June 2023 focused on free fall to determine g, the overarching skills are transferable. Any linearisation task, such as using a pendulum to find g via T² versus l plots, or investigating the force–extension behaviour of a spring, will require similar tabulation, graphing, and uncertainty analysis. The ability to diagnose ‘non-zero intercepts’ and link them to systematic timing or length measurement errors recurs across OCR papers.
尽管2023年6月的具体实验聚焦于用自由落体测定g，但所有核心技能都是可迁移的。任何线性化任务，例如用单摆通过T²-l图像求g，或探究弹簧的力-伸长行为，都需要类似的表格编制、作图与不确定度分析。诊断“非零截距”并将其与系统性的计时或长度测量误差相关联的能力，在OCR试卷中反复出现。
Embracing the full experimental cycle, from apparatus selection to final comparison with a reference value, will not only prepare you for Section B of Paper 1 but also for the Practical Endorsement and the paper 3 synoptic questions. This deep engagement transforms a potentially intimidating question into a highly structured opportunity to demonstrate your scientific literacy.
全面拥抱从仪器选择到与参考值最终比较的完整实验周期，不仅能让你为卷1 B部分做好准备，也有助于应对实践认证和卷3的综合题。这种深度参与能将一道可能令人生畏的题目，转变为一个高度结构化的机会，充分展示你的科学素养。
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