Year 13 CIE Science: Case Study Practical Exercise | CIE 13年级科学：案例分析实战演练

📚 Year 13 CIE Science: Case Study Practical Exercise | CIE 13年级科学：案例分析实战演练

In CIE A Level Science papers, especially Paper 5 (Planning, Analysis and Evaluation), you are often asked to analyse a set of experimental data, process uncertainties, plot a linearised graph and draw a valid conclusion. This case study walks you through a real experiment to determine the acceleration due to gravity g using a free‑fall method. You will learn how to handle raw data, convert variables for a straight‑line graph, calculate absolute and percentage uncertainties, draw error bars, find best‑fit and worst‑fit lines, and finally quote your result with a proper uncertainty.

在CIE A Level科学试卷中，尤其是Paper 5（实验设计、分析与评估），你经常需要分析一组实验数据、处理不确定度、绘制线性化图线并得出有效结论。本案例分析带你亲历一个使用自由落体法测量重力加速度g的真实实验。你将学习如何处理原始数据、转换变量以获得直线图、计算绝对和相对不确定度、绘制误差棒、找出最佳拟合线和最差拟合线，最后以恰当的不确定度表述结果。

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1. The Scenario – Measuring g by Free Fall | 情景设定——用自由落体测量g

A Year 13 student releases a small steel ball from an electromagnet at different heights s above a timing plate. The time of fall t is recorded three times for each height, using a digital timer with a precision of 0.01 s. The student knows the equation of motion s = ½ g t² and plans to plot s against t² to obtain a straight line whose slope is ½ g.

一名13年级学生用一个电磁铁在不同高度s处释放小钢球，球落在计时板上。每个高度用精度为0.01 s的数字计时器记录三次下落时间t。该学生已知运动方程 s = ½ g t² ，并计划绘制 s 对 t² 的图线，得到一条直线，其斜率为½ g。

The raw data are shown in the table below. Your task as the analyst is to process these data, estimate uncertainties, construct the necessary graph, extract the value of g and evaluate the experiment critically.

原始数据如下表所示。你作为分析师的任务是处理这些数据、估算不确定度、构建必要的图线、提取g值并批判性地评估该实验。

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2. Raw Data Collection | 原始数据收集

s / m	t₁ / s	t₂ / s	t₃ / s
0.200	0.20	0.21	0.20
0.400	0.28	0.29	0.28
0.600	0.35	0.35	0.36
0.800	0.40	0.41	0.40
1.000	0.45	0.45	0.46

The instrument precision for the metre rule is ±0.5 mm, and for the timer it is ±0.01 s. However, the dominant uncertainty in t comes from the random spread of the readings, so we will use the half‑range as the absolute uncertainty in the mean time.

米尺的仪器精度为 ±0.5 mm，计时器的为 ±0.01 s。然而，t的主要不确定度来源于读数的随机分散，因此我们将使用半范围作为平均时间的绝对不确定度。

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3. Processing the Data – Mean and t² | 数据处理——平均值与t²

For each height, calculate the mean time t̄ and its absolute uncertainty Δt = (t_max − t_min) / 2. Then compute t̄². The uncertainty in t̄² is given by Δ(t̄²) = 2 t̄ × Δt.

对于每一个高度，计算平均时间 t̄ 及其绝对不确定度 Δt = ( 最大 t − 最小 t ) / 2 。然后计算 t̄²。t̄²的不确定度由下式给出：Δ(t̄²) = 2 t̄ × Δt。

s / m	t̄ / s	Δt / s	t̄² / s²	Δ(t̄²) / s²
0.200	0.203	0.005	0.0412	0.0020
0.400	0.283	0.005	0.0801	0.0028
0.600	0.353	0.005	0.1246	0.0035
0.800	0.403	0.005	0.1624	0.0040
1.000	0.453	0.005	0.2052	0.0045

Notice that the absolute uncertainty in t̄ is constant (±0.005 s) because the spread of readings is similar at all heights. Consequently Δ(t̄²) increases with t̄².

注意到 t̄ 的绝对不确定度是一个常数（±0.005 s），因为在所有高度上读数的分散程度类似。因此，Δ(t̄²) 随 t̄² 的增大而增大。

The uncertainty in s is negligible at ±0.5 mm, so we will place error bars only on the t̄² axis.

s的不确定度 ±0.5 mm 可以忽略不计，因此我们只把误差棒标在 t̄² 轴上。

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4. Linearising the Equation | 方程线性化

The expected relationship is s = ½ g t². Plotting s on the vertical axis and t² on the horizontal axis should give a straight line through the origin. The gradient m is equal to ½ g.

预期的关系是 s = ½ g t²。将 s 作为纵轴，t² 作为横轴作图，应得到一条过原点的直线。斜率 m 等于 ½ g。

m = ½ g → g = 2m

This linearisation is essential for CIE analysis questions. It converts a curved relationship into a straight one, making it easy to identify the slope and test proportionality.

这种线性化对 CIE 分析题至关重要。它把曲线关系转化为直线关系，便于识别斜率并检验是否成正比。

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5. Plotting the Graph and Error Bars | 绘制图线与误差棒

Your graph should use at least half of the grid paper, with clearly labelled axes and units. Plot the points (t̄², s) and add horizontal error bars with length 2×Δ(t̄²) on each point. The vertical error bars are too small to show.

你的图应至少占用一半的坐标纸，并清楚标注坐标轴和单位。标出数据点（t̄², s），并为每一点加上长度为 2×Δ(t̄²) 的水平误差棒。垂直误差棒太小，无法显示。

You must then draw a best‑fit straight line that passes as closely as possible to all points, ideally through the origin. To find the uncertainty in the gradient, also draw the worst‑fit lines: the steepest and shallowest possible lines that still pass through the error bars.

接着你必须画一条最佳拟合直线，使其尽可能靠近所有点，最好通过原点。为求斜率的不确定度，还要画出最差拟合线：即仍然穿过误差棒的最陡和最浅的合理直线。

A common CIE marking point: if the line does not clearly go through the origin, you may need to calculate the y‑intercept and discuss systematic error.

CIE评分常见要点：如果直线未明显通过原点，你可能需要计算y轴截距并讨论系统误差。

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6. Determining the Gradient from the Best‑Fit Line | 由最佳拟合线确定斜率

Select two well‑separated points on the best‑fit line (not data points) and calculate the gradient m:

在最佳拟合线上选取两个间隔较远的点（非原始数据点），计算斜率 m：

m = Δs / Δ(t²) = (0.960 – 0.120) m / (0.200 – 0.025) s² ≈ 0.840 / 0.175 ≈ 4.80 m/s²

Here I chose convenient coordinates from a hypothetical best‑fit line. Your values may differ slightly. The key is to show the calculation clearly.

此处我根据假设的最佳拟合线选择了方便的坐标，你的数值可能略有差异。关键要清晰展示计算过程。

The value of g from the best fit is therefore g = 2 × 4.80 = 9.60 m s⁻².

因此由最佳拟合得出的 g 值为 9.60 m s⁻²。

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7. Finding the Uncertainty in g – Worst‑Fit Gradients | 求g的不确定度——最差拟合斜率

Draw the steepest plausible line through the error bars and pick two points on it to find the maximum gradient m_max. Repeat for the shallowest line to find m_min. Then the absolute uncertainty in the gradient is Δm = (m_max − m_min) / 2.

穿过误差棒画出可能的最陡直线，并取两点计算最大斜率 m_max。对最浅线重复此步骤，求得 m_min。则斜率的绝对不确定度为 Δm = (m_max − m_min) / 2。

Supposing m_max = 5.00 m/s² and m_min = 4.60 m/s², we get Δm = 0.20 m/s². Therefore, Δg = 2 × Δm = 0.40 m s⁻².

假设 m_max = 5.00 m/s²，m_min = 4.60 m/s²，得到 Δm = 0.20 m/s²。因此 Δg = 0.40 m s⁻²。

Your final result is quoted as g = 9.60 ± 0.40 m s⁻². Always match the significant figures of the value and its uncertainty.

你的最终结果应表达为 g = 9.60 ± 0.40 m s⁻²。数值与不确定度的有效数字位数必须匹配。

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8. Percentage Uncertainty and Comparison with Accepted Value | 相对不确定度及与公认值的比较

Calculate the percentage uncertainty: (0.40 / 9.60) × 100% ≈ 4.2%. The accepted value is 9.81 m s⁻². The discrepancy is |9.81 − 9.60| = 0.21 m s⁻², which lies within the experimental uncertainty. Therefore, the result is consistent with the accepted value within experimental error.

计算相对不确定度：(0.40 / 9.60) × 100% ≈ 4.2%。公认值为 9.81 m s⁻²。偏差为 0.21 m s⁻²，落在实验不确定度范围内。因此，结果在实验误差范围内与公认值一致。

Even if the best‑fit value does not exactly match 9.81, a well‑drawn uncertainty range that captures the true value is considered a success in CIE analysis questions.

即使最佳拟合值不完全等于9.81，只要不确定度范围覆盖了真值，在CIE分析题中也被认为是成功的。

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9. Evaluation of the Experiment – Sources of Error | 实验评估——误差来源

The most significant random error is the reaction time when starting and stopping the timer manually, although in this case a digital timer and electromagnetic release reduce it. Nevertheless, the consistent scatter in t values suggests small random fluctuations remain, possibly from air currents or the ball not being released from exactly the same position.

最显著的随机误差是手动操作计时器时的反应时间，尽管本案例中使用了数字计时器和电磁释放装置以减少此项误差。然而，t值的持续散布表明仍有小的随机波动，可能源于气流或球并非每次从完全同一位置释放。

A possible systematic error is the residual magnetism in the electromagnet causing a slight delay in release, making the measured time longer and g smaller. This is consistent with our result being slightly lower than 9.81 m s⁻². Another systematic error is the ball’s acceleration not being perfectly vertical, adding a small horizontal component and reducing effective vertical acceleration.

一个可能的系统误差是电磁铁的剩磁导致释放时略有延迟，使测得的时间偏长、g值偏小。这与我们结果略微低于9.81 m s⁻²相吻合。另一个系统误差是球的加速度不完全竖直，产生微小水平分量，减小了有效竖直加速度。

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10. Suggested Improvements | 改进建议

To reduce the random scatter, increase the height s and use a longer fall so that reaction time uncertainty becomes a smaller fraction of the total time. A better approach is to use a set of light gates connected to a data‑logger, which eliminates the human reaction time entirely and gives a precision of 0.001 s or better.

为减小随机散布，可增大高度s并使用更长的下落距离，使得反应时间不确定度占总时间的比例更小。更好的方法是使用一组光门连接至数据采集器，这样完全消除了人的反应时间，精度可达0.001 s或更高。

To eliminate the electromagnetic delay, use a mechanical release mechanism or place a small non‑magnetic spacer between the ball and the magnet so that the release is instantaneous. Also, evacuating the tube or using a denser, more aerodynamic ball reduces air resistance.

为消除电磁延迟，可使用机械释放装置，或在球与磁铁间放置一个小小的非磁性垫片，使释放瞬时完成。此外，抽空管中的空气或使用密度更大、流线外形更好的球可减小空气阻力。

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11. Linking Back to the CIE Syllabus – Key Skills Assessed | 回归CIE考纲——考查的关键技能

This exercise integrates the main competencies of CIE A Level Science Paper 5: recording raw data with repeated readings, calculating means and uncertainties, constructing a table with derived quantities, using error bars on a graph, drawing best‑fit and worst‑fit lines, determining slope and its uncertainty, and evaluating the experiment critically. It also tests your ability to express results with correct significant figures and units.

本次练习综合了CIE A Level科学Paper 5的主要能力：记录带重复读数的原始数据、计算平均值与不确定度、构建包含导出量的表格、在图线上使用误差棒、画出最佳拟合与最差拟合线、确定斜率及其不确定度，以及批判性地评估实验。同时也考查了你以正确有效数字和单位表述结果的能力。

Mastering these steps will not only improve your practical investigation skills but also prepare you for the 15‑mark analysis questions that frequently appear in the written papers.

掌握这些步骤不仅会提升你的实验探究技能，也能为笔试中经常出现的15分分析题做好充分准备。

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12. Final Advice for Your Own Case‑Study Practice | 案例分析练习的最终建议

When you tackle a CIE case study, always begin by identifying the variables and the relationship you need to linearise. Convert the data into the appropriate columns, estimate uncertainties for the quantities you plot, and be meticulous about drawing error bars. Never forget to use a sharp pencil and a long transparent ruler for gradients. And finally, link your evaluation to the specific sources of error evident in the data – this shows true analytical depth.

当你处理CIE案例时，总是先确定变量以及需要线性化的关系。将数据转换成合适的列，估算你所绘制量的不确定度，并细致地画出误差棒。永远不要忘记使用削尖的铅笔和长的透明尺来求斜率。最后，将你的评估与数据中显而易见的特定误差来源联系起来——这展示了真正的分析深度。

Published by TutorHao | Science Revision Series | aleveler.com

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