📚 A-Level Physics Unit 3 Mark Scheme Jun22 Formula Derivation | A-Level物理单元3评分方案2022年6月公式推导
In A-Level Physics Unit 3, one of the most frequently examined skills is taking a non-linear physical relationship and transforming it into a straight-line form. This enables students to determine unknown quantities from the gradient or intercept, exactly as required by the June 2022 mark scheme. This article breaks down the derivation process step by step, using a classic free-fall experiment to determine the acceleration due to gravity, g, and highlights how examiners award marks for logical reasoning, correct graph plotting, and uncertainty analysis.
在A-Level物理单元3中，最常见的考查技能之一就是把非线性的物理关系转化为直线形式。这使学生能够从斜率或截距中求出未知量，这正是2022年6月评分方案所要求的。本文通过一个经典的测定重力加速度g的自由落体实验，逐步解析公式推导过程，并重点说明考官如何对逻辑推理、正确作图以及不确定度分析给予评分。
1. Understanding the Role of Formula Derivation in Unit 3 | 理解单元3中公式推导的作用
The June 2022 mark scheme for Unit 3 makes it clear that simply memorising final results is insufficient. Candidates must demonstrate the ability to manipulate an equation into the form y = mx + c, correctly identify which quantities to plot on the vertical and horizontal axes, and explain how the gradient or intercept relates to the target physical constant. This process is not just algebraic work; it also involves judging the reliability of data and propagating uncertainties using the extremes of the best-fit line.
2022年6月的单元3评分方案明确指出，仅靠记忆最终结果是不够的。考生必须展示将方程变形为 y = mx + c 形式的能力，正确识别纵轴和横轴应绘制的物理量，并解释斜率或截距如何与目标物理常数相联系。这一过程不仅是代数推导，还包括判断数据的可靠性以及利用最佳拟合线的极值传递不确定度。
Examiners often allocate separate marks for the derivation, for labelling axes with appropriate units, for calculating the gradient correctly, for stating the final value of the required quantity, and for the uncertainty estimate. Familiarity with the mark scheme structure allows you to pick up these marks systematically.
考官通常会为推导过程、坐标轴标注（含单位）、正确计算斜率、给出最终所求量的数值以及不确定度估计分别评分。熟悉评分方案的结构能帮助你系统性地拿到这些分数。
2. The Free Fall Experiment: A Classic Example | 自由落体实验：经典范例
Consider an experiment in which a dense object is dropped from rest and the time t taken to fall through a measured vertical distance s is recorded. The accepted kinematic equation, assuming negligible air resistance, is s = ½ g t², where g is the acceleration of free fall. This relationship is clearly non-linear because s is proportional to t², not to t.
考虑这样一个实验：让一个密度较大的物体从静止下落，记录其通过一段测量好的竖直距离 s 所需的时间 t。在忽略空气阻力的情况下，公认的运动学方程为 s = ½ g t²，其中 g 是自由落体加速度。这个关系显然是非线性的，因为 s 与 t² 成正比，而不是与 t 成正比。
In Unit 3, you might be provided with a set of (s, t) data and asked to determine g. The mark scheme expects you to recognise that plotting s against t gives a parabola, from which extracting a reliable gradient is difficult. Therefore, you must linearise the relationship first.
在单元3中，你可能会得到一组 (s, t) 数据并被要求测定 g。评分方案期望你认识到，绘制 s–t 图像得到的是抛物线，很难从中提取可靠的斜率。因此，你必须先将关系线性化。
3. Step-by-Step Derivation of the Linear Form | 逐步推导线性形式
Start with the original equation:
从原始方程出发：
s = ½ g t²
To obtain a straight-line graph, we need to identify what to plot so that the equation resembles y = m x + c. Here there is no constant term, so the intercept will ideally be zero. Compare:
为了得到直线图，我们需要确定绘制什么物理量，使方程与 y = m x + c 形式一致。此例中没有常数项，因此截距理论上应为零。比较：
y = m x → s = (½ g) × t²
This suggests that if we plot s on the vertical axis and t² on the horizontal axis, the graph should be a straight line passing through the origin, with gradient m = ½ g. The required quantity, g, can then be found from g = 2 × gradient.
这表明，如果我们将 s 绘制在纵轴，t² 绘制在横轴，图像应该是一条过原点的直线，斜率为 m = ½ g。然后可由 g = 2 × 斜率 求出所需量 g。
It is essential to write the derivation in full:
完整写出推导过程至关重要：
s = ½ g t² → s = (g/2) t² → plot s vs t², gradient = g/2, therefore g = 2 × gradient
Examiners award a mark for clearly showing this rearrangement and stating what the gradient represents.
考官会为清楚展示这一变形并说明斜率所代表的含义而给予分数。
4. Identifying Variables for a Straight-Line Graph | 识别直线图的变量
A common mistake is to plot t² against s, which reverses the axes and leads to a gradient equal to 2/g. The mark scheme insists that the independent variable (the one you control or that is set first) should be plotted on the horizontal axis. Since t is measured as a result of changing s, t is the dependent variable. However, after squaring, t² becomes the quantity we calculate. In many mark schemes, the horizontal axis is the quantity that is manipulated to achieve a linear relationship; here it is t². Meanwhile, the vertical axis is the directly measured distance s. Always ensure that axes are labelled as ‘s / m’ and ‘t² / s²’ with units.
常见的错误是绘制 t²–s 图像，这会颠倒坐标轴，导致斜率等于 2/g。评分方案强调，自变量（你控制的或最先设定的量）应绘制在横轴。因为 t 是因改变 s 而测得的结果，t 是因变量。但平方后，t² 成了我们计算的量。在许多评分方案中，横轴是为了获得线性关系而被操作的那个量，此处是 t²。同时，纵轴是直接测得的距离 s。务必确保坐标轴标注为 ‘s / m’ 和 ‘t² / s²’，并包含单位。
In some experiments, such as those involving an initial velocity, the linearised equation may contain an intercept. For instance, if s = u t + ½ g t², dividing by t gives s/t = u + (g/2) t. Then plotting s/t against t yields a straight line with gradient g/2 and intercept u. Recognising the correct variables for different scenarios earns high marks.
在某些含有初速度的实验中，线性化后的方程可能包含截距。例如，若 s = u t + ½ g t²，两边除以 t 得到 s/t = u + (g/2) t。此时绘制 s/t–t 图会得到一条直线，斜率为 g/2，截距为 u。能识别不同情形下的正确变量，可以赢得高分。
5. Plotting the Graph and Calculating the Gradient | 绘制图像并计算斜率
Once the suitable variables are chosen, plot the data points on graph paper or using appropriate software. The mark scheme expects the points to occupy at least half the grid in both directions, the axes to be scaled evenly, and error bars to be included where uncertainties are given. Always draw a single best-fit straight line, balancing points on either side.
一旦选择了合适的变量，就在坐标纸或使用合适软件绘制数据点。评分方案要求数据点在两个方向上至少占据网格的一半，坐标轴均匀分度，并在给定不确定度时包含误差棒。务必画一条单一的最佳拟合直线，使两侧的数据点大致平衡。
To determine the gradient, select two points that lie on the line of best fit, not necessarily data points, and that are far apart to minimise percentage error. Record the coordinates (x₁, y₁) and (x₂, y₂), then compute:
为求斜率，选取最佳拟合直线上的两点（未必是原始数据点），且两点应相隔较远以减小百分误差。记录坐标 (x₁, y₁) 和 (x₂, y₂)，然后计算：
gradient = (y₂ − y₁) / (x₂ − x₁)
Always include the units of the gradient, which will be m s⁻² (from m divided by s²) in the s vs t² graph. Reading the triangle directly from the graph must be clearly indicated with a large triangle. These steps are explicitly rewarded in the June 22 mark scheme.
务必注明斜率的单位，在 s–t² 图中，斜率单位是 m s⁻²（米除以秒平方）。在图上读三角形时，必须用大三角形清晰标出。这些步骤在2022年6月的评分方案中都有明确的分数奖励。
6. Determining g from the Gradient | 从斜率求g
With the gradient m obtained, the value of g follows directly:
得到斜率 m 后，g 的值可直接求出：
g = 2 × m
For example, if the gradient of the s vs t² graph is 4.90 m s⁻², then g = 9.80 m s⁻². The mark scheme expects you to give the final value to an appropriate number of significant figures, usually matching the precision of the original data. A typical requirement is 3 significant figures, but this depends on the data provided.
例如，若 s–t² 图的斜率为 4.90 m s⁻²，则 g = 9.80 m s⁻²。评分方案希望你以适当的有效数字给出最终值，通常与原始数据的精度保持一致。典型的有效数字为3位，但视具体数据而定。
You must also express the final result with its absolute uncertainty, derived from the uncertainty in the gradient. A statement such as ‘g = 9.80 ± 0.12 m s⁻²’ is often required for the highest marks.
你还必须结合斜率的不确定度，给出带有绝对不确定度的最终结果。最高分通常要求写出如 ‘g = 9.80 ± 0.12 m s⁻²’ 这样的陈述。
7. Uncertainty Analysis: Maximum and Minimum Gradients | 不确定性分析：最大和最小斜率
The June 2022 mark scheme consistently rewards the method of determining uncertainty by drawing two additional lines: the steepest (maximum) and the shallowest (minimum) lines that are still consistent with the error bars. This ‘worst-fit’ approach gives a range for the gradient, which then translates to an uncertainty in g.
2022年6月的评分方案一贯奖励通过绘制两条附加直线来确定不确定度的方法：即仍与误差棒一致的最陡（最大）线和最浅（最小）线。这种“最差拟合”方法给出了斜率的取值范围，进而转化为 g 的不确定度。
The process is:
具体步骤如下：
-
Draw the best-fit line and calculate its gradient m_best.
绘制最佳拟合线，计算其斜率 m_best。
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Draw the maximum gradient line that passes through the extreme ends of the majority of error bars and calculate m_max.
绘制通过大多数误差棒极端端的最大斜率线，计算 m_max。
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Draw the minimum gradient line similarly and calculate m_min.
类似地绘制最小斜率线，计算 m_min。
The uncertainty in the gradient ∆m is then half the range:
斜率的不确定度 ∆m 取范围为一半：
∆m = (m_max − m_min) / 2
If g = 2m, the absolute uncertainty in g is ∆g = 2 × ∆m. Always express ∆g to 1 significant figure and round g to the same decimal place. Many marks are lost by candidates who forget to double the uncertainty in m when finding ∆g.
若 g = 2m，则 g 的绝对不确定度为 ∆g = 2 × ∆m。始终将 ∆g 保留1位有效数字，并将 g 舍入到相同的小数位。很多考生在求 ∆g 时忘记将 m 的不确定度乘以2，因而丢分。
8. Worked Example with Hypothetical Data | 假设数据计算示例
Suppose the following data were collected for a steel ball dropped from various heights:
假设对从不同高度释放的钢球采集了以下数据：
| s / m | t / s | t² / s² |
|---|---|---|
| 0.50 | 0.319 | 0.102 |
| 1.00 | 0.452 | 0.204 |
| 1.50 | 0.553 | 0.306 |
| 2.00 | 0.639 | 0.408 |
| 2.50 | 0.714 | 0.510 |
Plot s vs t². The best-fit gradient calculated from two points (0.100, 0.49) and (0.500, 2.45) gives:
绘制 s–t² 图。从直线上的两点 (0.100, 0.49) 和 (0.500, 2.45) 计算最佳斜率：
m_best = (2.45 − 0.49) / (0.500 − 0.100) = 1.96 / 0.400 = 4.90 m s⁻²
Then g_best = 2 × 4.90 = 9.80 m s⁻². If the maximum gradient is 5.05 m s⁻² and the minimum is 4.77 m s⁻², then ∆m = (5.05 − 4.77)/2 = 0.14 m s⁻². Hence ∆g = 0.28 m s⁻². Report g = 9.80 ± 0.28 m s⁻², or more appropriately rounded to g = 9.8 ± 0.3 m s⁻² (uncertainty to 1 s.f.). This aligns with the precision expected in the mark scheme.
因此 g_best = 2 × 4.90 = 9.80 m s⁻²。若最大斜率为 5.05 m s⁻²，最小斜率为 4.77 m s⁻²，则 ∆m = (5.05 − 4.77)/2 = 0.14 m s⁻²。于是 ∆g = 0.28 m s⁻²。可报告 g = 9.80 ± 0.28 m s⁻²，或更恰当地四舍五入为 g = 9.8 ± 0.3 m s⁻²（不确定度取1位有效数字）。这与评分方案要求的精度一致。
9. Mark Scheme Insights: What Examiners Look For | 评分方案洞察：考官期望
The June 2022 Unit 3 mark scheme for this type of question is highly structured. Marks are typically allocated as: (1) correct rearrangement of equation and identification of gradient; (2) correct axis labels with units; (3) sensible scales and accurate plotting; (4) line of best fit; (5) large triangle for gradient calculation; (6) correct gradient value with units; (7) substitution to find g; (8) plotting of max/min lines; (9) calculation of uncertainty in gradient; (10) correct uncertainty in g and proper presentation of final result. Missing any of these steps results in lost marks, even if the final value is close to the true value.
2022年6月单元3对此类题目的评分方案结构非常清晰。分数通常分配如下：(1) 正确变形方程并识别斜率含义；(2) 坐标轴标注正确并带单位；(3) 合理的标度和精确描点；(4) 最佳拟合线；(5) 用大三角形计算斜率；(6) 斜率值及其单位正确；(7) 代入求 g；(8) 绘制最大/最小线；(9) 计算斜率不确定度；(10) 正确算出 g 的不确定度并规范表述最终结果。缺少其中任一步骤都会丢分，即使最终值接近真值也没用。
Additionally, the mark scheme often includes a specific note that the final uncertainty should be given to 1 significant figure, and the calculated value of g should be rounded to the same number of decimal places. This attention to detail is what separates A* candidates from the rest.
此外，评分方案通常还会特别注明，最终不确定度应给1位有效数字，且 g 的计算值应舍入到相同的小数位。这种对细节的关注正是区分 A* 考生和其他人的关键。
10. Common Errors and How to Avoid Them | 常见错误及避免方法
One frequent error is forcing the best-fit line through the origin when the data do not support it. Even though theory predicts a zero intercept, systematic errors (like a delay in timing) might produce a small intercept. The mark scheme requires the line to be drawn through the centroid of the points unless there is a clear instruction or evidence for a fixed intercept. Always check if the question says ‘the equation suggests the line should pass through the origin’ before forcing it.
一个常见错误是，在数据并不支持的情况下，强行让最佳拟合线通过原点。即使理论预测截距为零，系统误差（如计时延迟）也可能产生一个不大的截距。评分方案要求直线通过数据点的中心，除非有明确指示或证据要求固定截距。在强制过原点前，务必核查题目是否表明‘方程表明直线应通过原点’。
Another pitfall is using data points to calculate the gradient instead of points taken from the line of best fit. The mark scheme explicitly instructs examiners to penalise this, because the line smooths out random errors. Similarly, using too small a triangle or reading coordinates incorrectly leads to inaccurate gradients.
另一个陷阱是使用数据点而非最佳拟合线上的点来计算斜率。评分方案明确要求考官对此进行扣分，因为拟合线已经平滑了随机误差。同样，使用过小的三角形或错误读取坐标也会导致斜率不准确。
Many candidates also confuse the uncertainty in t². If the raw uncertainty in t is δt, then the uncertainty in t² is 2t δt, which must be used to draw error bars on the horizontal axis. Neglecting to propagate the uncertainty correctly can cause the max/min lines to be drawn incorrectly.
许多考生还会混淆 t² 的不确定度。若 t 的原始不确定度为 δt，则 t² 的不确定度为 2t δt，这必须用来绘制横轴的误差棒。忽视对不确定度的正确传递会导致最大/最小线绘制错误。
11. Summary and Key Takeaways | 总结与关键要点
Mastering formula derivation for Unit 3 is not about memorising dozens of separate equations but about developing a systematic approach: (i) Write down the given physical law. (ii) Rearrange it to match y = mx + c. (iii) Identify which quantities to plot. (iv) Draw a clean graph with well-chosen scales. (v) Determine the gradient and its uncertainty using best-fit and worst-fit lines. (vi) Convert the gradient into the required physical constant, carefully propagating uncertainties. (vii) Present the final answer with correct significant figures and units.
掌握单元3的公式推导不是要记住几十个独立的方程，而是发展出一套系统的方法：(i) 写下所给的物理定律；(ii) 重新整理使之匹配 y = mx + c 形式；(iii) 确定要绘制的物理量；(iv) 选用合适的标度绘制清晰的图像；(v) 利用最佳拟合和最差拟合线求出斜率及其不确定度；(vi) 将斜率转化为所求物理常数，小心传递不确定度；(vii) 以正确的有效数字和单位呈现最终答案。
By aligning your approach with the June 2022 mark scheme expectations, you can approach any data analysis or practical skills question with confidence, knowing exactly where each mark is earned. Practice with past papers and always check the corresponding mark scheme to reinforce these steps.
让你的做法与2022年6月的评分方案期望保持一致，你就能自信地应对任何数据分析或实验技能题目，清楚地知道每一分来自哪里。通过历年真题练习，并始终对照相应的评分方案来强化这些步骤。
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