📚 AS Physics Unit 3 Insert (June 2019): Experimental Investigation | AS物理第三单元插入材料(2019年6月)：实验探究
The June 2019 AS Physics Unit 3 insert is a classic example of how Edexcel assesses practical skills without students being physically in the lab. It supplies raw experimental data, tables to complete and clear instructions to plot a graph, then asks you to determine a physical quantity – often the Young modulus, resistivity or g – and to evaluate uncertainties rigorously. This article unpacks a typical investigation based on that insert, focusing on measuring the Young modulus of a metal wire, and teaches you how to handle every calculation, graph and uncertainty step confidently.
2019年6月的AS物理第三单元插入材料是艾德思评估实验技能的典型方式——学生无需进入实验室。它提供了原始实验数据、待填写的表格以及绘图的清晰指令，然后要求你测定一个物理量（常见的有杨氏模量、电阻率或重力加速度g），并严格评估不确定度。本文基于该材料的典型实验——测量金属丝的杨氏模量——展开分析，教你如何处理每一个计算、绘图和不确定度步骤，做到心中有数。
1. Understanding the Insert Context | 理解插入材料背景
In Paper 3 (IAS Physics Unit 3: Practical Skills in Physics I), the Insert is a separate booklet containing the scenario, measurement data and a partly completed table. Candidates must use this to fill in blanks, plot a graph, calculate a gradient with its uncertainty and finally determine a target quantity. The Jun19 insert presented an experiment in which a steel wire was stretched by adding slotted masses, with extension measured using a travelling microscope or a marker on a scale. The raw data included wire length, diameter and readings of force against extension.
在试卷三（IAS物理第三单元：物理实验技能I）中，插入材料是一份独立的小册子，包含了实验情境、测量数据和一张部分完成的表格。考生必须据此填补空缺、绘制图像、计算斜率及其不确定度，最终测定目标物理量。2019年6月的插入材料展示了一个实验：通过增加槽码来拉伸一根钢丝，并使用移测显微镜或标尺上的标记测量伸长量。原始数据包括金属丝的长度、直径以及力与伸长量的读数。
The investigation is heavily scaffolded – you are told which quantity to plot on each axis and often given the equation. Your job is to execute the practical reasoning: correct any zero errors, choose appropriate scales, draw a line of best fit (LBF) and worst fit lines (LWF), calculate percentage uncertainty, and then combine uncertainties to report the final result with its absolute uncertainty. The insert thus transforms a hands-on experiment into a written application of practical theory.
该实验探究极具引导性——会明确告知每个坐标轴画什么物理量，并常常给出相关公式。你的任务是完成实践推理：修正零点误差、选择合适的刻度、绘制最佳拟合线与最差拟合线、计算百分比不确定度，并最终合成不确定度，报告带有绝对不确定度的最终结果。因此，插入材料将一个动手操作的实验转化为书面形式的实践理论应用。
2. Key Measurements and Their Precision | 关键测量及其精度
n x for each added force F. The original length L was measured using a metre rule with a precision of ±1 mm, but the uncertainty is usually quoted as ±2 mm due to the difficulty of aligning the rule exactly with the wire ends. This gives a large percentage uncertainty for L unless the wire is very long (typically about 2 m).
在2019年6月的插入材料中，有三项基本测量必须仔细审视：金属丝的原长L、直径d，以及每增加一个力F对应的伸长量x。原长L使用米尺测量，精度为±1 mm，但由于难以使尺子与金属丝两端精确对齐，不确定度通常取为±2 mm。除非金属丝特别长（通常约2 m），否则L的百分比不确定度较大。
ading to 0.01 mm (10 μm). A common feature in insert exams is the presence of a zero error – ute uncertainty in a single micrometer reading is ordinarily ±0.01 mm, but because the wire may have slight variations along its length, several readings are taken in different places and perpendicular directions, and the uncertainty becomes the half-range (max − min)/2. This diameter uncertainty feeds directly into the cross‑sectional area A = πd²/4.
直径d使用千分尺测量，标称读数为0.01 mm（10 μm）。插入材料考试中常见的一个特征是零点误差的存在——当千分尺完全闭合时，读数可能为+0.02 mm。你必须从每个直径读数中减去这个零点误差才能得到真实直径。单次千分尺读数的绝对不确定度通常为±0.01 mm，但由于金属丝沿长度方向可能存在细微差异，需要在不同位置和垂直方向上多次读数，其不确定度则取半范围（最大值−最小值）/2。这个直径不确定度将直接传递到横截面积A = πd²/4的计算中。
gs in mm, with an absolute uncertainty of ±1 mm from the scale, though a set square or fiducial marker can reduce this. You must consider parallax error when aligning the marker with the scale, and the fact that the wire may go slightly slack at the start, requiring a 'pre‑load' to remove kinks.
伸长量x是借助金属丝上的一个标记在毫米刻度尺上移动测得的，或者使用移测显微镜测量。插入材料通常以mm为单位给 出读数，绝对不确定度取自刻度尺的±1 mm，但使用三角尺或基准标记可以降低误差。你必须考虑标记与刻度尺对齐时的视差，以及金属丝在开始时可能略微松弛，需要施加一个“预载荷”来消除弯曲。
3. The Experiment Setup: Young Modulus of a Wire | 实验设置：金属丝杨氏模量
The experimental arrangement involved clamping a long metal wire firmly at one end, passing it over a pulley at the other end, and attaching a mass hanger. Weights were added in steps, and the extension of the wire was recorded each time. The original length L was the distance between the fixed clamp and the marker position on the wire when just taut. The diameter d was measured at several points and averaged.
实验装置包括将一根长金属丝的一端牢固地夹紧，另一端绕过滑轮，并挂上砝码挂钩。每次逐步增加砝码，记录金属丝的伸长量。原长L是固定夹具到金属丝刚好绷直时标记位置之间的距离。直径d在多个位置测量后取平均值。
The principle uses Hooke' x). Rearranging, the relationship between force F and extension x is F = (EA/L) x. This indicates that a graph of F (y‑axis) against x (x‑ by measuring the gradient, the Young modulus can be calculated as E = (gradient × L) / A.
实验原理基于胡克定律和杨氏模量E的定义。对于处于弹性形变的金属丝，应力σ = F/A，应变ε = x/L，因此E = σ/ε = (F/A) / (x/L) = FL / (Ax)。整理后，力F与伸长量x的关系为 F = (EA/L) x。这种形式表明，以F为y轴，x为x轴绘制的图像应得到一条通过原点的直线，其斜率k = EA/L。因此，通过测量斜率，杨氏模量可由E = (斜率 × L) / A 计算得出。
The Jun19 insert contained a table with force readings (in N) and corresponding extensions (in mm), with some gaps to be filled using the given data. The values were chosen so that the wire remained within its elastic limit, and the straight‑line fit validates this.
2019年6月的插入材料包含了一张表格，其中有力的读数（以N为单位）和相应的伸长量（以mm为单位），部分空缺需要根据给定数据来填写。所选的数值确保金属丝保持在弹性限度内，直线拟合可以验证这一点。
4. Data Recording and Tables | 数据记录与表格
e 9.81 or 10). The extension x has been corrected for any zero error in the scale (if the marker did not start exactly at zero). The original length L = 1.500 m with uncertainty ±2 mm, and the mean diameter after zero correction is d = 0.33 mm (3.3 × 10⁻⁴ m). The cross‑sectional area is A = π × (0.33 × 10⁻³)² / 4 ≈ 8.55 × 10⁻⁸ m².
下面是一个与插入材料类似的再造表格。所有质量值都已转换为力F（使用g = 9.81 N kg⁻¹，尽管有时插入材料会用9.81或10）。伸长量x已针对标尺的零点误差（如果标记初始未精确对准零点）进行了修正。原长L = 1.500 m，不确定度±2 mm；零点修正后的平均直径d = 0.33 mm（3.3 × 10⁻⁴ m）。横截面积A = π × (0.33 × 10⁻³)² / 4 ≈ 8.55 × 10⁻⁸ m²。
| Mass added / g | Force F / N | Extension x / mm |
|---|---|---|
| 0 | 0.00 | 0.00 |
| 200 | 1.96 | 0.60 |
| 400 | 3.92 | 1.20 |
| 600 | 5.89 | 1.80 |
| 800 | 7.85 | 2.40 |
| 1000 | 9.81 | 3.00 |
In the exam, you might be asked to complete a column for the mass added or to calculate forces. The insert often gives one or two extra rows to allow for a reading with the load decreasing to check for hysteresis, but the above suffices for analysis. Notice that the extensions are small (millimetres) – this reinforces how vital precise instruments are.
在考试中，你可能被要求填充增加的质量一列或者计算力。插入材料通常会额外提供一两行数据，用于在卸载过程中读取，以检查迟滞现象，但上述数据已足够用于分析。请注意伸长量很小（毫米级），这更加说明了精确仪器的重要性。
5. Plotting the Graph: Force vs Extension | 绘制图表：力与伸长量
cupy more than half the grid, so scales must be chosen carefully. For the vertical axis (0–10 N), a scale of 1 cm = 1 N works well. For the horizontal axis (0– quired by the question. In the Jun19 insert, the uncertainty in x was given as ±0.2 mm, so each plotted point carries a horizontal error bar of ±0.2 mm. The uncertainty in F is often taken as ±0.1 N.
插入材料要求将力F绘制在纵轴，伸长量x绘制在横轴。绘制的图表应占据坐标纸一半以上的面积，因此必须谨慎选择刻度。对于纵轴（0–10 N），1 cm = 1 N的刻度很合适。对于横轴（0–3.0 mm），1 cm = 0.25 mm或2 cm = 0.5 mm的刻度较为合理，因为它能使点分散开来。每个数据点用整齐的十字（×）标出，如果题目需要还要加上误差棒。在2019年6月的插入材料中，x的不确定度给定为±0.2 mm，因此每个数据点都带有±0.2 mm的水平误差棒。F的不确定度通常取±0.1 N。
A line of best fit (LBF) should be drawn through the origin (since when no force is applied, extension is zero). Check if all points lie close to the line; if one is obviously anomalous, identify it and leave it out of the fit. In our ideal data, all points lie perfectly on a straight line, so the LBF passes exactly through them. For a more realistic scenario, you would draw a LBF that has an equal distribution of points above and below the line.
应绘制一条通过原点的最佳拟合线（LBF）（因为不施加力时伸长量为零）。检查所有点是否紧靠直线；如果有某个点明显异常，应将其识别出来并在拟合中舍弃。在我们的理想数据中，所有点都完美地落在一条直线上，因此最佳拟合线精确穿过它们。在更真实的情况中，你需要画一条LBF，使得直线两侧点的分布均匀。
To determine the uncertainty in the gradient, draw two additional lines: the maximum gradient line (steepest) and the minimum gradient line (shallowest) that still pass through the error bars of most points. The gradient uncertainty Δm is then (mmax − mmin) / 2.
为了确定斜率的不确定度，需要额外画两条直线：最大梯度线（最陡峭）和最小梯度线（最平缓），这两条线仍应穿过大多数点的误差棒。然后，斜率不确定度Δm = (mmax − mmin) / 2。
6. Calculating the Gradient and Its Uncertainty | 计算斜率及其不确定度
lways take readings from a large triangle on the graph, not from data table points directly, to improve precision. If the line does not pass through all points exactly, select two widely spaced points on the LBF for the calculation.
从通过原点和点(3.00 mm, 9.81 N)的最佳拟合线出发，斜率m = ΔF / Δx = 9.81 N / 3.00 × 10⁻³ m = 3270 N m⁻¹（或3.27 × 10³ N m⁻¹）。为提高精度，应始终从图上取一个大三角形来读取数据，而非直接使用表格中的点。如果直线并非精确穿过所有点，则应在LBF上选取两个相距较远的点进行计算。
Now for the uncertainty. Suppose the maximum gradient line passes through the top of the error bar of the last point (9.81 N, 3.20 mm) and through the origin, giving mmax = 9.81 / 3.20 × 10⁻³ = 3066 N m⁻¹? Wait, that gives a smaller gradient – origin (which is zero error, so still origin) and the top of the error bar of a high point? Better approach: if we assume the error bars are vertical for force? st and shallowest lines that are consistent with all error bars. For our illustration, let' al error bars give a range. The shallowest line passes through (0,0) and (3.20 mm, 9.81 N)? ta point at x = 3.00 mm has error bar to 3.20 mm, a line from origin to the rightmost end (3.20 mm, 9.81 N) has gradient 9.81/3.20e-3 = 3066 N m⁻¹. The steepest line would go through the leftmost end (2.80 mm, 9.81 N) giving 9.81/2.80e-3 = 3504 N m⁻¹. Then uncertainty Δm = (3504 - 3066)/2 = 219 N m⁻¹. This shows how to do it. You can state m = 3270 ± 220 N m⁻¹.
现在计算不确定度。假设最大梯度线穿过最后一个点的误差棒顶部（9.81 N, 3.20 mm）和原点，则mmax = 9.81 / 3.20 × 10⁻³ = 3066 N m⁻¹？等等，这样得到的是较小的梯度——我们需要最陡峭的线。要得到最陡峭的线，应该画一条穿过误差棒左端（对于高值点）的线。如果已知水平误差棒为±0.20 mm，最陡峭的线穿过(0,0)和(2.80 mm, 9.81 N)，梯度为 9.81/2.80e-3 = 3504 N m⁻¹。最平缓的线穿过(0,0)和(3.20 mm, 9.81 N)，梯度为3066 N m⁻¹。然后不确定度Δm = (3504 − 3066)/2 = 219 N m⁻¹。因此斜率可表示为 m = 3270 ± 220 N m⁻¹。
The percentage uncertainty in gradient is (220 / 3270) × 100% ≈ 6.7%. This will be carried forward to the final Young modulus calculation.
斜率的百分比不确定度为 (220 / 3270) × 100% ≈ 6.7%。这个数值将传递到最终的杨氏模量计算中。
7. Determining the Young Modulus | 测定杨氏模量
titute the best gradient m = 3270 N m⁻¹, L = 1.500 m, and A = 8.55 × 10⁻⁸ m². This yields:
E = (3270 N m⁻¹ × 1.500 m) / (8.55 × 10⁻⁸ m²) = 5.74 × 10¹⁰ Pa
使用公式E = (斜率 × L) / A。代入最佳斜率m = 3270 N m⁻¹，L = 1.500 m，以及A = 8.55 × 10⁻⁸ m²，得到：
E = (3270 N m⁻¹ × 1.500 m) / (8.55 × 10⁻⁸ m²) = 5.74 × 10¹⁰ Pa
0¹¹ Pa), but slightly low, possibly due to systematic errors like the wire not being perfectly uniform or slight slipping at the clamp. The exam usually expects a value somewhat different from the standard, prompting an evaluation.
这个数值对于钢而言量级是正确的（通常为2.0 × 10¹¹ Pa），但略微偏低，这可能是由于系统误差，例如金属丝不够均匀或夹具处有轻微滑动。考试中通常期望得到的数值与标准值有些差异，从而引发评估。
8. Uncertainty Analysis: Percentage and Absolute Uncertainties | 不确定度分析：百分比与绝对不确定度
y in E, combine the percentage uncertainties in gradient, length and area. We already have %U(m) = 6.7%. For length L, absolute uncertainty 0.002 m, so %U(L) = (0.002 / 1.500) × 100% = 0.13%. For area A, the uncertainty derives from the diameter. Let's say the diameter d = 0.33 mm, with absolute uncertainty from half‑range of repeat readings ±0.01 mm. Then %U(d) = (0.01 / 0.33) × 100% = 3.0%. Since A ∝ d², the percentage uncertainty in A is double: %U(A) = 2 × 3.0% = 6.0%.
要计算E的绝对不确定度，需要合成斜率、长度和面积的百分比不确定度。我们已有%U(m) = 6.7%。对于长度L，绝对不确定度0.002 m，因此%U(L) = (0.002 / 1.500) × 100% = 0.13%。面积A的不确定度源于直径。假设直径d = 0.33 mm，基于重复读数半范围的绝对不确定度为±0.01 mm。那么%U(d) = (0.01 / 0.33) × 100% = 3.0%。由于A ∝ d²，面积的百分比不确定度是直径的两倍：%U(A) = 2 × 3.0% = 6.0%。
dard uncertainties, but at AS level the simple addition of percentage uncertainties (worst‑case) is often accepted: %U(E) = %U(m) + %U(L) + %U(A) = 6.7% + 0.13% + 6
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