AS AQA Further Maths Unit 2 Paper Walkthrough (Jan 2019) | AS AQA 进阶数学第二单元试卷详解（2019年1月）

📚 AS AQA Further Maths Unit 2 Paper Walkthrough (Jan 2019) | AS AQA 进阶数学第二单元试卷详解（2019年1月）

The January 2019 AQA AS Further Mathematics Unit 2 paper (7366/2) is a 1½-hour examination worth 75 marks. The paper is divided into three optional sections: Section A (Mechanics), Section B (Statistics) and Section C (Discrete). You must answer all questions from exactly one section. This article breaks down every topic area, provides worked examples in the style of the paper, and highlights the traps that cost students marks.

2019年1月AQA AS进阶数学第二单元试卷（7366/2）考试时长1.5小时，满分75分。试卷分为三个选做部分：A部分（力学）、B部分（统计学）和C部分（离散数学）。考生必须从其中一个部分作答全部题目。本文将逐一剖析每个知识领域，提供与试卷风格一致的例题详解，并指出学生失分的常见陷阱。

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1. Paper Structure and Mark Allocation | 试卷结构与分值分布

Each section contains a mixture of short and extended-response questions. Mechanics focuses on momentum, impulse, collisions, work, energy and power. Statistics centres on the Poisson distribution, hypothesis testing, and continuous distributions such as exponential and uniform. Discrete covers graph algorithms (Prim, Kruskal, Dijkstra) and critical path analysis. You should plan roughly 1 minute per mark, leaving the final 10 minutes to check units, significant figures and algebraic signs.

每个部分均包含简答题和扩展回答题。力学部分重点考查动量、冲量、碰撞、功、能量和功率；统计学部分围绕泊松分布、假设检验以及指数分布和均匀分布等连续分布；离散部分涵盖图论算法（普里姆、克鲁斯卡尔、迪杰斯特拉）和关键路径分析。建议按每题分值分配时间，约每分钟1分，最后留出10分钟检查单位、有效数字和代数符号。

Typical mark allocations observed in this series:

该系列试卷中常见的分值分配如下：

• Momentum and impulse questions: 6–9 marks | 动量与冲量题：6–9分

• Collision problems with restitution: 8–11 marks | 含恢复系数的碰撞问题：8–11分

• Work-energy-power chains: 7–10 marks | 功能关系与功率题：7–10分

• Poisson calculations and hypothesis tests: 9–13 marks | 泊松计算与假设检验：9–13分

• Exponential and uniform distribution: 6–9 marks | 指数分布与均匀分布：6–9分

• Graph algorithms (Dijkstra/Prim/Kruskal): 10–14 marks | 图论算法题：10–14分

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2. Mechanics: Momentum and Impulse | 力学：动量与冲量

Momentum is the product of mass and velocity, p = mv, measured in kg·m/s. Impulse is the change in momentum and equals the average force multiplied by the time of contact: Impulse = Ft = mv − mu. These definitions are the foundation of Section A.

动量是质量与速度的乘积，p = mv，单位为kg·m/s。冲量是动量的变化量，等于平均作用力乘以接触时间：冲量 = Ft = mv − mu。这些定义是A部分的基础。

Impulse = Ft = m(v − u) | 冲量 = Ft = m(v − u)

Worked example (Jan 19 style): A particle of mass 0.5 kg, moving at 6 m/s, is struck by a constant force of 4 N in the direction of motion for 0.3 s. Find the new speed.

例题（2019年1月风格）：质量为0.5 kg的质点以6 m/s运动，受到一个与其运动方向相同的4 N恒定力作用0.3 s。求其新速度。

Impulse = Ft = 4 × 0.3 = 1.2 N·s = 0.5(v − 6)

Solving: v − 6 = 2.4 so v = 8.4 m/s. Note that impulse is a vector; if the force opposed motion, we would write Ft = 0.5(v − 6) with F = −4 N. Sign errors here are the single most common cause of lost marks.

解得：v − 6 = 2.4，故v = 8.4 m/s。注意冲量是矢量；若力与运动方向相反，则应以F = −4 N代入Ft = 0.5(v − 6)。符号错误是此处最常见的失分原因。

In the Jan 2019 paper, momentum questions typically link to the conservation principle: for two bodies in an isolated system, total momentum before equals total momentum after. Always define a positive direction and write a clear equation.

在2019年1月试卷中，动量题通常与守恒原理结合：对于孤立系统中的两个物体，碰撞前后的总动量相等。务必先规定正方向，再写出清晰的方程。

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3. Mechanics: Collisions and Newton’s Law of Restitution | 力学：碰撞与牛顿恢复系数定律

For direct collisions between two smooth spheres, two equations apply simultaneously. First, conservation of momentum:

对于两个光滑球体之间的对心碰撞，需同时使用两个方程。首先是动量守恒方程：

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Second, Newton’s law of restitution defines the coefficient e (0 ≤ e ≤ 1):

其次，牛顿恢复系数定律定义系数e（0 ≤ e ≤ 1）：

e = (v₂ − v₁) / (u₁ − u₂)

Note the ordering: the numerator subtracts the final velocities in the same order as the denominator subtracts the initial velocities. For a perfectly elastic collision e = 1; for an inelastic collision that coalesces, e = 0 and v₁ = v₂.

注意顺序：分子中末速度相减的顺序与分母中初速度相减的顺序一致。完全弹性碰撞e = 1；完全非弹性碰撞（粘合）时e = 0且v₁ = v₂。

Worked example: A 2 kg sphere A moving at 5 m/s strikes a stationary 4 kg sphere B directly. Given e = 0.5, find the velocities after impact.

例题：质量为2 kg的球A以5 m/s的速度与静止的4 kg球B发生正碰。已知e = 0.5，求碰撞后的速度。

10 = 2v_A + 4v_B → 5 = v_A + 2v_B

0.5 = (v_B − v_A) / 5 → v_B − v_A = 2.5

Substituting v_A = v_B − 2.5 into the momentum equation: (v_B − 2.5) + 2v_B = 5, giving 3v_B = 7.5, so v_B = 2.5 m/s and v_A = 0 m/s. Sphere A stops completely and B moves forward at 2.5 m/s. Always check that the result is physically plausible — a negative v_A with e < 1 would indicate a direction reversal, which is possible but must be consistent with both equations.

将v_A = v_B − 2.5代入动量方程：(v_B − 2.5) + 2v_B = 5，得到3v_B = 7.5，故v_B = 2.5 m/s，v_A = 0 m/s。球A完全停止，球B以2.5 m/s前进。务必检查结果是否物理上合理——当e < 1时v_A为负值表示方向反转，这虽可能，但必须与两个方程一致。

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4. Mechanics: Work, Energy and Power | 力学：功、能量与功率

The work-energy principle states that the total work done on a particle equals its change in kinetic energy. The key equations are:

功能原理指出：对质点所做的总功等于其动能的变化量。关键公式如下：

Work done = Fd | Kinetic energy = ½mv² | Potential energy = mgh | Power = Fv

功 = Fd | 动能 = ½mv² | 势能 = mgh | 功率 = Fv

Worked example: A particle of mass 2 kg is projected up a rough plane inclined at 20° with initial speed 12 m/s. The coefficient of friction is 0.15. Find the distance travelled before the particle comes to rest.

例题：质量为2 kg的质点以12 m/s的初速度沿倾角为20°的粗糙斜面向上运动。摩擦系数为0.15。求质点停止前通过的距离。

Resolving perpendicular to the plane: R = 2g cos20° = 2 × 9.8 × 0.9397 ≈ 18.42 N. Friction = μR = 0.15 × 18.42 ≈ 2.76 N.

垂直于斜面方向分解：R = 2g cos20° = 2 × 9.8 × 0.9397 ≈ 18.42 N。摩擦力 = μR = 0.15 × 18.42 ≈ 2.76 N。

Along the plane, friction and the weight component both oppose motion. Total resisting force = 2.76 + 2g sin20° = 2.76 + 6.70 = 9.46 N. By the work-energy principle: work done = ½mv², so 9.46 × d = ½ × 2 × 144 = 144, giving d ≈ 15.2 m.

沿斜面方向，摩擦力和重力分量都阻碍运动。总阻力 = 2.76 + 2g sin20° = 2.76 + 6.70 = 9.46 N。由功能原理：功 = ½mv²，故9.46 × d = ½ × 2 × 144 = 144，解得d ≈ 15.2 m。

For power questions, remember that when speed is constant up a slope, engine force equals resistance plus the component of weight down the slope; then power = F × v. Use consistent units — forces in N, distances in m, times in s.

对于功率问题，记住：当物体沿斜面匀速上升时，发动机牵引力等于阻力加上重力沿斜面向下的分量；然后功率 = F × v。注意使用一致的单位——力的单位为N，距离单位为m，时间单位为s。

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5. Statistics: The Poisson Distribution | 统计学：泊松分布

The Poisson distribution models the number of rare events occurring in a fixed interval of time or space, with mean rate λ. The probability function is:

泊松分布用于建模在固定时间或空间间隔内稀有事件发生的次数，平均发生

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