AS AQA Further Mathematics Paper 2 (June 2022) | AS AQA 进阶数学卷2（2022年6月）

📚 AS AQA Further Mathematics Paper 2 (June 2022) | AS AQA 进阶数学卷2（2022年6月）

The AQA AS Further Mathematics Paper 2 (component code 7366/2) for June 2022 is the applied paper of the AS qualification. You answer the questions from one of three options: Further Mechanics, Further Statistics or Discrete Mathematics. This guide reviews the paper structure, the most frequently examined topics and the common errors students made, so you can walk into the exam room fully prepared.

AQA AS 进阶数学第二卷（试卷代码 7366/2）于 2022 年 6 月开考，是本资格的应用数学试卷。考生需要从进阶力学、进阶统计或离散数学三个选项中选做其中一组。本文将系统梳理试卷结构、高频考点以及学生在考场上的常见失误，帮助你做好充分准备。

1. Paper Structure | 试卷结构

The paper lasts 1 hour 30 minutes and is worth 80 marks. You choose one optional topic and answer all questions in that section. A formula booklet is provided, but you must decide quickly which option you have practised most.

本卷考试时长为 1 小时 30 分钟，满分 80 分。你需要在三个选修专题中选择一个，并完成该部分全部题目。考试会提供公式册，但你需要快速判断自己最熟悉、练得最多的选项。

Component	Length	Marks	Options
7366/2	1 h 30 min	80	Further Mechanics / Further Statistics / Discrete

The June 2022 paper kept the same structure as previous series. Many candidates lost marks because they spent too long on the first part of their chosen option and then rushed the final, sometimes easier, questions.

2022 年 6 月试卷的结构与往年保持一致。许多考生在前半部分题目上耗时过多，导致最后较容易的题目来不及完成，这是十分可惜的失分原因。

2. Assessment Objectives | 考核目标

Three assessment objectives are tested across the paper. AO1 rewards correct use of standard methods, AO2 tests reasoning in proofs and justifications, and AO3 focuses on modelling real-world situations with mathematics.

本卷考查三个考核目标：AO1 考查标准方法与公式的准确运用，AO2 考查证明与论证中的逻辑推理，AO3 考查将现实情境转化为数学模型的能力。

AO1: about 50% of the marks – routine calculations, using formulae correctly.

AO1：约占 50% 分值——常规计算，正确套用公式。
AO2: about 25% – constructing arguments, e.g. proof by induction or verifying conditions.

AO2：约占 25% 分值——构建论证，例如数学归纳法或验证使用条件。
AO3: about 25% – interpreting a problem, choosing a model, and explaining results.

AO3：约占 25% 分值——理解问题、选择模型并解释结果。

In the June 2022 series, several structured questions began with a routine calculation and then asked for a comment on the context. For example, a mechanics question might finish with “state whether your answer is sensible and why”. Do not ignore these final lines.

2022 年 6 月考试中，许多分层设问的题目先进行常规计算，再要求结合背景作出解释。例如力学题最后可能要求“说明你的结果是否合理并解释原因”。千万不要忽略最后这几句问句。

3. Further Mechanics: Core Topics | 进阶力学：核心考点

If you chose Further Mechanics, the essential topics are momentum, impulse, collisions, work, energy and power. You must be fluent with vector notation and sign conventions.

如果你选择了进阶力学，核心考点包括动量、冲量、碰撞、功、能与功率。你必须熟练使用向量记号并严格遵循正方向约定。

The two fundamental equations you will use again and again are the conservation of linear momentum and Newton’s experimental law of restitution:

你会反复使用两个基本方程：动量守恒定律和牛顿碰撞恢复系数公式：

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ , v₂ − v₁ = −e(u₂ − u₁)

Here e is the coefficient of restitution, with 0 ≤ e ≤ 1. For perfectly elastic collisions e = 1; for perfectly inelastic ones e = 0, so the particles move together.

其中 e 为恢复系数，满足 0 ≤ e ≤ 1。完全弹性碰撞中 e = 1；完全非弹性碰撞中 e = 0，此时两物体粘合在一起运动。

Impulse on a particle is defined as the change in momentum: I = mv − mu. When two particles are involved, remember Newton’s third law: the impulses are equal in magnitude but opposite in direction, which is why total momentum is conserved.

冲量定义为动量的变化量：I = mv − mu。涉及两个物体时，要记住牛顿第三定律：两个冲量大小相等、方向相反，这正是总动量守恒的原因。

4. Worked Example: Collisions | 例题：碰撞问题

Two spheres A and B of masses 2m and m move towards each other with speeds 3u and 2u respectively. The coefficient of restitution is ½. Find the speed and direction of each sphere after impact.

质量分别为 2m 和 m 的两个小球 A、B 相向运动，速度大小分别为 3u 和 2u。恢复系数为 ½。求碰撞后两球的速度与方向。

Take the direction of A’s initial motion as positive. Then uₐ = 3u and u_b = −2u. Conservation of linear momentum gives:

取 A 初始运动方向为正方向，则 uₐ = 3u，u_b = −2u。由动量守恒得：

2m(3u) + m(−2u) = 2mvₐ + mv_b

So 2vₐ + v_b = 4u. Newton’s experimental law gives v_b − vₐ = −½(−2u − 3u) = 2.5u. Solving simultaneously:

因此 2vₐ + v_b = 4u。牛顿碰撞公式给出 v_b − vₐ = −½(−2u − 3u) = 2.5u。联立解得：

vₐ = 0.5u , v_b = 3u

Both values are positive, so both spheres move in the original direction of A after the collision. Check the energy loss: total kinetic energy falls from 11mu² to 9.5mu², showing the collision is inelastic.

两个结果均为正值，说明碰撞后两球都沿 A 原来的方向运动。检验能量：总动能从 11mu² 减少为 9.5mu²，说明碰撞确实是非弹性的。

5. Further Statistics: Poisson Distribution | 进阶统计：泊松分布

If you chose Further Statistics, the Poisson distribution is central. A discrete random variable X has distribution Po(λ) if its probability function is:

如果你选择了进阶统计，泊松分布是核心内容。离散随机变量 X 服从 Po(λ)，其概率函数为：

P(X = r) = e⁻λ λʳ / r! , r = 0, 1, 2, …

You must know the conditions: events occur independently, at a constant average rate, and singly. Popular questions ask you to verify the conditions first, then calculate probabilities.

你必须牢记泊松过程的适用条件：事件相互独立、平均发生率恒定、事件不会同时发生。常见题目会先要求你验证这些条件，再计算概率。

A key extension is the additive property: if X ~ Po(λ₁) and Y ~ Po(λ₂), and X and Y are independent, then X + Y ~ Po(λ₁ + λ₂). This is often used in two-way table questions.

一个重要推广是泊松分布的可加性：若 X ~ Po(λ₁)、Y ~ Po(λ₂) 且相互独立，则 X + Y ~ Po(λ₁ + λ₂)。这常用在涉及两类事件或两张数据表的题目中。

6. Worked Example: Hypothesis Test | 例题：泊松假设检验

The number of potholes on a road follows a Poisson distribution with mean 2.5 per week. After maintenance work, 6 potholes are reported in one week. Test at the 5% level whether the mean has increased.

某路段坑洞数每周服从均值为 2.5 的泊松分布。养护施工后，某一周报告了 6 个坑洞。在 5% 显著性水平下检验均值是否增加。

Set H₀: λ = 2.5 versus H₁: λ > 2.5. The test statistic is X ~ Po(2.5). A one-tailed test is appropriate because we only care about an increase.

建立原假设 H₀：λ = 2.5，备择假设 H₁：λ > 2.5。检验统计量 X ~ Po(2.5)。由于只关心均值是否增加，应使用单尾检验。

Compute P(X ≥ 6) = 1 − P(X ≤ 5). Using tables or a calculator:

计算 P(X ≥ 6) = 1 − P(X ≤ 5)。查表或使用计算器可得：

P(X ≤ 5) ≈ 0.9580 , so P(X ≥ 6) ≈ 0.0420

Because 0.0420 < 0.05, we reject H₀. There is sufficient evidence at the 5% level that the mean number of potholes per week has increased. Many candidate errors came from using the wrong tail or forgetting to state the conclusion in context.

因为 0.0420 < 0.05，拒绝 H₀。在 5% 显著性水平下，有充分证据表明每周平均坑洞数增加了。许多考生因选错拒绝域或忘记结合题干情境下结论而失分。

7. Discrete: Algorithms and Graphs | 离散数学：算法与图论

If you chose Discrete, the June 2022 paper emphasised algorithms: sorting, path-finding and network optimisation. You must be able to execute an algorithm and trace through a table accurately.

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AS AQA Further Mathematics Paper 2 (June 2022) | AS AQA 进阶数学卷2（2022年6月）