📚 AS Physics Unit 1 Insert Experiment (Jan 20) – Free Fall Analysis | AS物理单元1 Jan20插入页实验 – 自由落体分析
The January 2020 AS Physics Unit 1 insert presented an experimental scenario designed to assess students’ practical skills in data analysis. In this investigation, the motion of a steel ball falling freely under gravity was studied using a pair of light gates. The insert provided measurements of time intervals and distances, which allowed the determination of the acceleration due to gravity, g. This article unpacks the experiment, explains the underlying physics, demonstrates step-by-step calculations, and guides through the evaluation of uncertainties and errors. It serves as a comprehensive revision resource for students preparing for their practical-based exam questions.
2020年1月的AS物理单元1插入页提供了一项实验场景，旨在考查学生的数据分析实践能力。该实验使用一对光闸研究钢球在重力作用下的自由落体运动。插入页提供了时间间隔和距离的测量数据，可用于测定重力加速度 g。本文拆解该实验，解释背后的物理原理，逐步展示计算过程，并引导学生评估不确定度与误差。对于备考实践类考题的学生而言，这是一份全面的复习资料。
1. Experimental Setup | 实验装置
A steel ball of known diameter was dropped from rest above a pair of light gates connected to a data logger. The light gates were placed a measured vertical distance apart. As the ball passed through the first gate, it interrupted the beam, allowing the timer to record the time for the ball to cross the beam. The same process occurred at the second gate. The data logger also recorded the time interval between the first and second interruptions.
一个已知直径的钢球从静止状态下落，穿过一对连接数据记录器的光闸。光闸间隔一定垂直距离放置。当球通过第一个光闸时，它遮断了光束，计时器记录下球遮光的时间。同样在第二个光闸处重复该过程。数据记录器还记录了两次遮光之间的时间间隔。
The ball’s diameter d was measured with a micrometer screw gauge, giving a value of 0.0200 m with an absolute uncertainty of ±0.0001 m. The distance s between the two light gates was measured using a metre ruler as 1.000 m ±0.001 m. The timer readings provided three time measurements: the interruption time at the first gate t₁, the interruption time at the second gate t₂, and the transit time T between the two gates.
钢球直径 d 用千分尺测量，数值为 0.0200 m，绝对不确定度 ±0.0001 m。两个光闸之间的距离 s 用米尺测量为 1.000 m ±0.001 m。计时器读数提供了三个时间量：第一个光闸的遮光时间 t₁、第二个光闸的遮光时间 t₂ 以及两光闸间的通行时间 T。
2. Understanding the Insert Data | 理解插入页数据
The insert provided the following measurements in a clear table. Students were required to extract these values and use them for subsequent calculations. Identifying the correct quantities and their uncertainties is the first critical skill tested.
插入页以清晰的表格形式提供了以下测量值。学生需要提取这些数值并用于后续计算。正确识别各物理量及其不确定度是考查的首项关键技能。
| Quantity | Symbol | Value | Absolute Uncertainty |
|---|---|---|---|
| Diameter of steel ball | d | 0.0200 m | ±0.0001 m |
| Time through first gate | t₁ | 0.0133 s | ±0.0001 s |
| Time through second gate | t₂ | 0.0043 s | ±0.0001 s |
| Vertical distance between gates | s | 1.000 m | ±0.001 m |
| Transit time between gates | T | 0.324 s | ±0.001 s |
Notice that the times recorded for the beam interruption are very short, especially at the second gate, because the ball is moving faster. The precision of the timer (±0.0001 s) is sufficient, but the relative uncertainty in t₂ will be relatively large. This is a common feature of such experiments and must be considered in the error analysis.
请注意，光束遮断时间非常短，尤其是在第二个光闸处，因为球运动得更快。计时器的精度 (±0.0001 s) 虽然足够，但 t₂ 的相对不确定度会比较大。这是此类实验的常见特点，在误差分析中必须加以考虑。
3. Key Physics Principles | 关键物理原理
The experiment relies on the equations of uniformly accelerated motion. Since the ball is dropped from rest and falls freely, its acceleration is the acceleration due to gravity, g, assuming negligible air resistance. The instantaneous speed of the ball as it passes through a light gate is found by dividing the diameter of the ball by the time the beam is interrupted: v = d / t.
本实验基于匀加速运动方程。由于球从静止下落并做自由落体运动，其加速度即为重力加速度 g，假设空气阻力可忽略不计。球通过光闸时的瞬时速度等于球的直径除以光束被遮断的时间：v = d / t。
Once the speeds at the two gates are known, the acceleration can be calculated directly using the definition of constant acceleration: a = (v₂ − v₁) / T, where T is the time taken to travel between the gates. Alternatively, if multiple sets of data are available, the equation v₂² = v₁² + 2 a s can be used to find g from a graph of v₂² against s, where the slope is 2g.
一旦获知球在两光闸处的速度，便可直接用匀加速度的定义计算加速度：a = (v₂ − v₁) / T，其中 T 是球在两光闸之间的运动时间。另外，如果有多组数据可用，还可利用方程 v₂² = v₁² + 2 a s，通过绘制 v₂²–s 图像求 g，此时图线斜率为 2g。
In this analysis, we use the direct calculation from the single set of data provided in the insert. We denote the speed at the first gate as u and at the second gate as v. The experimental acceleration is then compared with the standard value of g = 9.81 m s⁻².
在本分析中，我们采用插入页所提供的那一组数据进行直接计算。将球在第一个光闸处的速度记作 u，在第二个光闸处的速度记作 v。然后将实验所得的加速度与标准值 g = 9.81 m s⁻² 进行比较。
4. Calculating Initial and Final Velocities | 计算初速度和末速度
Using the interruption times and the ball’s diameter, we compute the instantaneous speeds. The underlying assumption is that the ball’s speed does not change significantly during the short time it takes to pass through the beam, which is valid for a small ball and narrow beam.
利用遮光时间和球的直径，我们可以算出瞬时速度。其基本假设是，球在穿过光束的短暂时间内速度变化可以忽略，这对于小尺寸钢球和窄光束是成立的。
Initial speed at first gate: u = d / t₁ = 0.0200 m / 0.0133 s = 1.5038 m s⁻¹. Rounding to an appropriate number of significant figures (three, based on the values), we record u = 1.50 m s⁻¹.
第一个光闸处的初速度：u = d / t₁ = 0.0200 m / 0.0133 s = 1.5038 m s⁻¹。根据有效数字（保留三位），我们记作 u = 1.50 m s⁻¹。
Final speed at second gate: v = d / t₂ = 0.0200 m / 0.0043 s = 4.6512 m s⁻¹. Again quoting to three significant figures, v = 4.65 m s⁻¹.
第二个光闸处的末速度：v = d / t₂ = 0.0200 m / 0.0043 s = 4.6512 m s⁻¹。同样保留三位有效数字，v = 4.65 m s⁻¹。
It is worth noting the dramatic increase in speed over the 1.000 m fall, from about 1.5 m s⁻¹ to 4.7 m s⁻¹, illustrating the effect of gravitational acceleration.
值得注意的是，在 1.000 m 的下落过程中，速度从约 1.5 m s⁻¹ 增加到 4.7 m s⁻¹，增幅显著，这体现了重力加速度的作用。
5. Determining Acceleration due to Gravity | 测定重力加速度
The acceleration a is determined from the change in velocity and the transit time T. The average acceleration over the interval between the gates is assumed constant and equal to g.
加速度 a 由速度变化量和通行时间 T 确定。光闸间距内的平均加速度视为恒定且等于 g。
a = (v − u) / T = (4.65 m s⁻¹ − 1.50 m s⁻¹) / 0.324 s = 3.15 m s⁻¹ / 0.324 s = 9.72 m s⁻²
This gives an experimental value of g = 9.72 m s⁻². Compared with the accepted value of 9.81 m s⁻², the percentage difference is |9.72 − 9.81| / 9.81 × 100% ≈ 0.92%. This small discrepancy suggests the experiment was carried out with reasonable precision, but we must still evaluate the uncertainty range to determine whether the result is consistent with the standard value.
由此得出重力加速度的实验值 g = 9.72 m s⁻²。与公认值 9.81 m s⁻² 相比，百分差约为 0.92%。这个微小的偏差表明实验具有相当的精确度，但我们仍需评估不确定度范围，以判断该结果是否与标准值一致。
6. Uncertainty Analysis | 不确定度分析
To appreciate the reliability of the measurement, we calculate the percentage uncertainty in each measured quantity and then combine them to find the overall uncertainty in g. For a quantity Q with absolute uncertainty ΔQ, the percentage uncertainty is (ΔQ / Q) × 100%.
为了正确评价测量的可靠性，我们计算每个直接测量量的百分不确定度，然后合成得到 g 的总不确定度。对于绝对不确定度为 ΔQ 的量 Q，其百分不确定度为 (ΔQ / Q) × 100%。
| Quantity | Value | Absolute Uncertainty | % Uncertainty |
|---|---|---|---|
| d | 0.0200 m | ±0.0001 m | 0.5% |
| t₁ | 0.0133 s | ±0.0001 s | 0.75% |
| t₂ | 0.0043 s | ±0.0001 s | 2.3% |
| T | 0.324 s | ±0.001 s | 0.31% |
The velocities u and v are derived from d and the times, so their percentage uncertainties are found by adding the % uncertainty in d to the % uncertainty in each time. For u: 0.5% + 0.75% = 1.25%. For v: 0.5% + 2.3% = 2.8%.
速度 u 和 v 是由 d 和各时间计算得出的，因此它们的百分不确定度等于 d 的百分不确定度加上相应时间的百分不确定度。u 的不确定度：0.5% + 0.75% = 1.25%。v 的不确定度：0.5% + 2.3% = 2.8%。
The change in velocity Δv = v − u has an absolute uncertainty obtained by adding the absolute uncertainties of v and u. However, when combining percentage uncertainties for a quotient, we add the % uncertainty of Δv to that of T. The absolute uncertainty in u is 1.25% of 1.50 = 0.0188 m s⁻¹, and in v is 2.8% of 4.65 = 0.130 m s⁻¹. So Δv = 3.15 ± 0.149 m s⁻¹, giving a % uncertainty of (0.149 / 3.15) × 100% ≈ 4.7%. Adding the % uncertainty of T (0.31%) gives a total % uncertainty in g of about 5.0%.
速度变化量 Δv = v − u 的绝对不确定度由 v 与 u 的绝对不确定度相加得到。但在乘除运算中组合百分不确定度时，我们将 Δv 的百分不确定度与 T 的百分不确定度相加。u 的绝对不确定度为 1.25% × 1.50 = 0.0188 m s⁻¹，v 的绝对不确定度为 2.8% × 4.65 = 0.130 m s⁻¹。因此 Δv = 3.15 ± 0.149 m s⁻¹，其百分不确定度约为 4.7%。再与 T 的百分不确定度 (0.31%) 相加，得到 g 的总百分不确定度约为 5.0%。
Hence, g = 9.72 ± 0.49 m s⁻² (rounding to two significant figures in the uncertainty). The interval 9.2–10.2 m s⁻² comfortably includes the accepted value 9.81 m s⁻², so the result is consistent within experimental uncertainty. The dominant contribution comes from the short interruption time t₂, highlighting why timing fast events is a challenge.
因此，g = 9.72 ± 0.49 m s⁻²（不确定度修约至两位有效数字）。该数值范围 9.2–10.2 m s⁻² 很宽裕地包含了公认值 9.81 m s⁻²，因此在实验不确定度范围内结果是一致的。其中最大的贡献来自极短的遮光时间 t₂，这也凸显了测量快速事件为何是一个挑战。
7. Graphical Method for Better Accuracy | 图形法提高精度
In a more extended investigation, rather than relying on a single pair of light gates, students would vary the height of release or the position of the gates to obtain multiple values of s and the corresponding final speed v. The equation v² = u² + 2 g s, with u being the speed at a fixed first gate, suggests a graph of v² against s would yield a straight line with slope = 2g.
在更深入的探究中，学生并不依赖单一的这对光闸，而是通过改变释放高度或光闸位置，获得多组 s 及相应的末速度 v。由方程 v² = u² + 2 g s，其中 u 为球在固定第一个光闸处的速度，绘制 v² 对 s 的图像将得到一条直线，斜率为 2g。
This graphical approach reduces the impact of random errors and allows for identification of systematic errors through the intercept. The intercept on the v² axis should equal u², providing an internal check. While the insert only supplied one set of data, exam questions often ask how an experiment could be improved, and the use of a graphical method is the standard recommendation.
这种图形方法可减少随机误差的影响，并可通过截距来识别系统误差。v² 轴上的截距应等于 u²，由此进行内部验证。虽然该插入页仅提供了一组数据，但考试问题常会问到如何改进实验，而采用图形法就是最标准的建议。
8. Sources of Error and Improvements | 误差来源与改进
Several factors may have contributed to the uncertainty beyond the instrument limitations. One systematic error is the assumption that the ball passes through the light gate exactly at its widest point. If the beam is not aligned with the centre of the ball, the effective diameter might be slightly different. Using a ball with a smaller diameter or multiple light beams can mitigate this.
除仪器限制外，有几个因素可能增大了不确定度。一项系统误差是假设球恰在其直径最宽处穿过光闸。若光束未与球心对齐，有效直径可能会略有不同。使用更小直径的球或采用多光束光闸可以减轻这种影响。
Air resistance is generally negligible for a dense steel ball over short distances, but if very high precision is sought, performing the experiment in a vacuum would eliminate drag. Parallax when measuring the distance s with a metre rule can be reduced by using a mounted ruler and ensuring the line of sight is perpendicular.
对于密度大的钢球在短距离内下落，空气阻力通常可忽略不计，但如果要追求极高精度，在真空中进行实验将消除空气阻力。用米尺测量 s 时存在的视差可以通过使用固定直尺并确保视线垂直来减小。
Human reaction time is not a factor here since timers are triggered electronically. However, any dirt on the ball could increase its diameter slightly. Cleaning the ball and repeating the experiment multiple times to average the results are simple but effective improvements.
本实验中由于计时器是电子触发的，人的反应时间不是影响因素。不过，钢球表面的污垢可能会略微增加其直径。清洁钢球并多次重复实验取平均值，是简单而有效的改进方法。
Another improvement is to use light gates connected to a computer that can automatically record many data points and compute statistics, allowing for more reliable mean values and uncertainty estimation from the spread of data.
另一项改进是使用与计算机连接的光闸，它可自动记录多个数据点并计算统计量，通过数据分散程度获得更可靠的平均值及不确定度估计。
9. Practical Skills Assessed | 考查的实践技能
The AS Unit 1 insert questions are designed to assess the following areas of practical competency: extracting and tabulating data, converting raw measurements into derived quantities, using appropriate significant figures, calculating and combining uncertainties, and evaluating the validity of conclusions. Being able to recognise that the instrumental precision is not the only source of error is a mark of deeper understanding.
AS 单元 1 插入页的题目旨在评估下述实践能力：提取并列表数据、将原始测量值转化为导出量、正确使用有效数字、计算并合成不确定度，以及评价结论的有效性。能够认识到仪器精度并非误差的唯一来源，是深层理解的标志。
The experiment also tests the candidate’s ability to identify realistic improvements and to link theoretical equations with practical measurements. For instance, justifying the use of the equation v = d / t relies on
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