📚 AS Further Maths Unit 2 January 2020: Mechanics and Statistics Question Breakdown | AS 进阶数学单元2 2020年1月真题题型解析
This article provides a detailed walkthrough of question types from the AQA AS Further Mathematics Unit 2 paper of January 2020. Unit 2 covers mechanics and statistics topics that extend the pure maths core. By dissecting each section, you will see how to approach typical exam problems, where common pitfalls lie, and how to structure your answers to gain full marks. The paper is designed to test both your conceptual understanding and your ability to apply formulas accurately in unfamiliar contexts.
本文深入解析了 AQA 考试局 AS 进阶数学单元2 2020年1月真题中的典型题型。单元2 涵盖力学与统计，是对纯数核心内容的拓展。我们将逐节剖析考题，帮助你掌握典型问题的解题思路，识别常见陷阱，并学习如何组织答案以获得满分。这份试卷旨在同时考查你对概念的理解以及在不熟悉情境中准确应用公式的能力。
1. Projectile Motion with Horizontal Launch | 水平抛射运动
In this question, a particle is projected horizontally from a cliff top with speed 15 m s⁻¹. The cliff is 20 m high. You are asked to calculate the time of flight and the horizontal distance travelled before hitting the ground. The key is to split the motion into vertical and horizontal components. Vertically, the initial velocity is 0 m s⁻¹, acceleration is g = 9.8 m s⁻², and displacement is 20 m downwards. Using s = ut + ½at² gives 20 = ½ × 9.8 × t², so t = √(40/9.8) ≈ 2.02 s.
题目中，一个质点从悬崖顶以 15 m s⁻¹ 的水平速度抛出，悬崖高度为 20 m。要求计算飞行时间和落地前的水平距离。解题关键是将运动分解为垂直和水平两个方向。垂直方向上，初速度为 0 m s⁻¹，加速度 g = 9.8 m s⁻²，位移向下 20 m。利用 s = ut + ½at² 得到 20 = ½ × 9.8 × t²，因此 t = √(40/9.8) ≈ 2.02 s。
Horizontally, there is no acceleration, so distance = speed × time = 15 × 2.02 ≈ 30.3 m. Many students forget that the vertical displacement is positive when taken as the direction of gravity, but the sign convention must be consistent. This question often appears early in the paper to ease you into mechanics.
水平方向没有加速度，因此水平距离 = 速度 × 时间 = 15 × 2.02 ≈ 30.3 m。许多学生忽略垂直位移在取重力方向为正时是正值，但符号约定必须前后一致。这种题型通常出现在试卷开头，帮助你平稳进入力学部分。
2. Resolving Forces on an Inclined Plane | 斜面上的力分解
A block of mass 5 kg rests on a smooth plane inclined at 30° to the horizontal. It is held in equilibrium by a force P acting parallel to the plane, directed up the slope. You need to find P and the normal reaction R. The weight component down the plane is 5g sin 30°. Since g is usually taken as 9.8 m s⁻², this equals 5 × 9.8 × 0.5 = 24.5 N. Thus P = 24.5 N.
一个质量为 5 kg 的物块静止在倾角为 30° 的光滑斜面上，通过一个沿斜面向上、平行于斜面的力 P 保持平衡。需要求 P 和法向反作用力 R。重力沿斜面方向的分量为 5g sin 30°。通常 g 取 9.8 m s⁻²，因此该分量为 5 × 9.8 × 0.5 = 24.5 N，所以 P = 24.5 N。
The normal reaction is perpendicular to the plane: R = 5g cos 30° = 5 × 9.8 × (√3/2) ≈ 42.4 N. A common mistake is confusing sine and cosine for the perpendicular component. Always draw a clear force diagram and label the angle correctly. This type of equilibrium question tests fundamental resolving skills.
法向反作用力垂直于斜面：R = 5g cos 30° = 5 × 9.8 × (√3/2) ≈ 42.4 N。常见错误是混淆垂直分量对应的正弦和余弦。务必画出清晰的受力分析图，并正确标注角度。此类平衡问题考查的是基本的分解技巧。
3. Coefficient of Friction from Motion on a Rough Slope | 粗糙斜面上的动摩擦系数
A particle slides down a rough plane inclined at angle α, where tan α = 3/4. The acceleration is measured as 2 m s⁻². You are to find the coefficient of friction μ. The resultant force down the slope is mg sin α − μR, where R = mg cos α. Using F = ma gives mg sin α − μ mg cos α = m × 2, so g sin α − μ g cos α = 2. Substitute sin α = 3/5 and cos α = 4/5 from the tangent ratio. With g = 9.8, we get 9.8×(3/5) − μ × 9.8×(4/5) = 2, leading to 5.88 − 7.84μ = 2, and thus μ = (5.88 − 2) / 7.84 = 0.494… ≈ 0.49 (2 s.f.).
一个质点沿粗糙斜面下滑，斜面倾角 α 满足 tan α = 3/4，测得加速度为 2 m s⁻²。要求计算动摩擦系数 μ。沿斜面方向的合力为 mg sin α − μR，其中 R = mg cos α。由 F = ma 得 mg sin α − μ mg cos α = m × 2，化简为 g sin α − μ g cos α = 2。根据正切值可得 sin α = 3/5，cos α = 4/5。取 g = 9.8，代入得 9.8×(3/5) − μ × 9.8×(4/5) = 2，即 5.88 − 7.84μ = 2，因此 μ = (5.88 − 2) / 7.84 = 0.494… ≈ 0.49（保留两位有效数字）。
This question requires careful algebra and exact trigonometric conversions. Students often incorrectly use tan α directly in the force equation. Remember that tan α = 3/4 implies a 3-4-5 triangle, so sine and cosine follow immediately. Always simplify the equation by cancelling m before substituting values.
这道题要求细致的代数运算和精确的三角转换。学生常错误地将 tan α 直接代入力的方程。要记住 tan α = 3/4 蕴含 3-4-5 三角形，因此正弦和余弦可直接得出。始终在代入数值前约去质量 m，以简化方程。
4. Moments and Equilibrium of a Rod | 力矩与杆的平衡
A uniform rod AB of length 4 m and weight 30 N is freely hinged at A to a vertical wall. It is held horizontally by a light string attached at B, making an angle of 40° with the rod. You must find the tension T in the string and the horizontal and vertical components of the reaction at the hinge. Taking moments about A eliminates the hinge reaction: clockwise moment of weight = 30 × 2 (acting at the centre) equals anticlockwise moment of tension: T sin 40° × 4. So 60 = 4T sin 40°, giving T = 60 / (4 sin 40°) ≈ 23.3 N.
一根均匀杆 AB 长 4 m，重 30 N，在 A 端通过光滑铰链连接在竖直墙上。杆由一根系于 B 端的轻绳保持水平，绳与杆的夹角为 40°。需求出绳中张力 T 以及铰链处反作用力的水平和竖直分量。对 A 点取力矩以消除铰链反力：重力的顺时针力矩 = 30 × 2（作用在中心）等于张力的逆时针力矩：T sin 40° × 4。因此 60 = 4T sin 40°，解得 T = 60 / (4 sin 40°) ≈ 23.3 N。
Then resolve horizontally: R_H = T cos 40° ≈ 23.3 × cos 40° ≈ 17.8 N. Vertically: R_V + T sin 40° = 30, so R_V = 30 − 23.3 sin 40° ≈ 30 − 15.0 = 15.0 N. When a rod is in equilibrium, always check that the sum of vertical forces and the sum of horizontal forces are both zero. The hinge provides both components to balance the external forces.
接着水平方向分解：R_H = T cos 40° ≈ 23.3 × cos 40° ≈ 17.8 N。竖直方向：R_V + T sin 40° = 30，因此 R_V = 30 − 23.3 sin 40° ≈ 30 − 15.0 = 15.0 N。杆处于平衡时，务必验证竖直方向合力与水平方向合力均为零。铰链提供两个方向的分量来平衡外力。
5. Centre of Mass of a Composite Lamina | 组合薄板的重心
A uniform lamina is formed by removing a square of side 0.5 m from a larger square of side 1 m, as shown. You are asked to find the distance of the centre of mass from the left edge AD. The original square has mass proportional to area 1 m², and its centre of mass is at (0.5, 0.5) taking A as the origin. The removed square of area 0.25 m² has centre at (0.75, 0.75). The remaining mass is 1 − 0.25 = 0.75 units. Using the formula for centre of mass of a composite body: x̄ = (m₁x₁ − m₂x₂) / (m₁ − m₂) = (1 × 0.5 − 0.25 × 0.75) / 0.75 = (0.5 − 0.1875) / 0.75 = 0.3125 / 0.75 = 0.4167 m. So x̄ = 0.417 m (3 s.f.).
一块均匀薄板是由边长为 1 m 的正方形剪去一个边长为 0.5 m 的正方形制成。要求计算重心距离左边线 AD 的距离。以 A 为原点，原正方形质量与面积 1 m² 成正比，其重心在 (0.5, 0.5)。剪去的正方形面积为 0.25 m²，中心在 (0.75, 0.75)。剩余质量为 1 − 0.25 = 0.75 单位。利用组合体重心公式：x̄ = (m₁x₁ − m₂x₂) / (m₁ − m₂) = (1 × 0.5 − 0.25 × 0.75) / 0.75 = (0.5 − 0.1875) / 0.75 = 0.3125 / 0.75 = 0.4167 m。即 x̄ = 0.417 m（三位有效数字）。
Students often misplace the centre of the removed piece. Ensure you draw the shape and label coordinates clearly. The y-coordinate can be found similarly: ȳ = (1 × 0.5 − 0.25 × 0.75) / 0.75 = same value, so the centre of mass of the L-shaped lamina lies at (0.417, 0.417). This concept is fundamental for toppling and stability problems.
学生常会弄错被挖去部分的重心位置。务必画出图形并清晰标注坐标。y 坐标可用同样方法求得：ȳ = (1 × 0.5 − 0.25 × 0.75) / 0.75，数值相同，因此 L 形薄板的重心位于 (0.417, 0.417)。这一概念对于倾倒和稳定性问题至关重要。
6. Discrete Random Variables and Expected Value | 离散型随机变量与期望值
A biased dice has probability distribution: P(X=1)=0.1, P(X=2)=0.2, P(X=3)=0.15, P(X=4)=0.25, P(X=5)=0.2, and P(X=6)=0.1. Find E(X) and Var(X). E(X) = Σ x·P(X=x) = 1×0.1 + 2×0.2 + 3×0.15 + 4×0.25 + 5×0.2 + 6×0.1 = 0.1 + 0.4 + 0.45 + 1.0 + 1.0 + 0.6 = 3.55.
一枚不均匀的骰子概率分布为：P(X=1)=0.1，P(X=2)=0.2，P(X=3)=0.15，P(X=4)=0.25，P(X=5)=0.2，P(X=6)=0.1。求 E(X) 和 Var(X)。E(X) = Σ x·P(X=x) = 1×0.1 + 2×0.2 + 3×0.15 + 4×0.25 + 5×0.2 + 6×0.1 = 0.1 + 0.4 + 0.45 + 1.0 + 1.0 + 0.6 = 3.55。
Var(X) = E(X²) − [E(X)]². First compute E(X²) = 1²×0.1 + 4×0.2 + 9×0.15 + 16×0.25 + 25×0.2 + 36×0.1 = 0.1 + 0.8 + 1.35 + 4.0 + 5.0 + 3.6 = 14.85. Then Var(X) = 14.85 − 3.55² = 14.85 − 12.6025 = 2.2475. This question is straightforward but many candidates forget to square the mean at the end. Always check the sum of probabilities equals 1.
Var(X) = E(X²) − [E(X)]²。先计算 E(X²) = 1²×0.1 + 4×0.2 + 9×0.15 + 16×0.25 + 25×0.2 + 36×0.1 = 0.1 + 0.8 + 1.35 + 4.0 + 5.0 + 3.6 = 14.85。则 Var(X) = 14.85 − 3.55² = 14.85 − 12.6025 = 2.2475。此题简单明了，但许多考生会在最后忘记将均值平方。务必检查概率之和等于 1。
7. Binomial Distribution and Cumulative Probability | 二项分布及累积概率
A manufacturer claims that 8% of its light bulbs are defective. A sample of 20 bulbs is taken. Find the probability that exactly 2 are defective, and the probability that at most 2 are defective. Let X ~ B(20, 0.08). P(X=2) = ²⁰C₂ × (0.08)² × (0.92)¹⁸. Using calculator: ²⁰C₂ = 190, (0.08)² = 0.0064, (0.92)¹⁸ ≈ 0.2089, so P(X=2) ≈ 190 × 0.0064 × 0.2089 = 0.253 (3 s.f.). For P(X ≤ 2), sum P(X=0) + P(X=1) + P(X=2). P(X=0) = (0.92)²⁰ ≈ 0.1887, P(X=1) = 20 × 0.08 × (0.92)¹⁹ ≈ 20 × 0.08 × 0.2051 ≈ 0.3282. So cumulative probability ≈ 0.1887 + 0.3282 + 0.253 = 0.7699 ≈ 0.770.
某制造商声称其灯泡中有 8% 是次品。现抽取 20 个灯泡作为样本。求恰好有 2 个次品的概率，以及至多 2 个次品的概率。设 X ~ B(20, 0.08)。P(X=2) = ²⁰C₂ × (0.08)² × (0.92)¹⁸。使用计算器：²⁰C₂ = 190，(0.08)² = 0.0064，(0.92)¹⁸ ≈ 0.2089，因此 P(X=2) ≈ 190 × 0.0064 × 0.2089 = 0.253（三位有效数字）。对于 P(X ≤ 2)，求 P(X=0) + P(X=1) + P(X=2) 之和。P(X=0) = (0.92)²⁰ ≈ 0.1887，P(X=1) = 20 × 0.08 × (0.92)¹⁹ ≈ 20 × 0.08 × 0.2051 ≈ 0.3282。因此累积概率 ≈ 0.1887 + 0.3282 + 0.253 = 0.7699 ≈ 0.770。
This is a standard binomial calculation. Be careful with reading ‘at most’ vs ‘at least’. Some students mistakenly find P(X ≥ 2) by subtracting the lower tail from 1. In hypothesis testing, this tail probability is crucial for finding p-values.
这是标准的二项分布计算。要注意区分“至多”与“至少”。有些学生会错误地用 1 减去下尾概率来计算 P(X ≥ 2)。在假设检验中，这种尾部概率对于求 p 值至关重要。
8. Poisson Distribution as an Approximation | 泊松分布近似
A rare disease occurs in 0.5% of the population. Find the probability that in a random sample of 400 people, exactly 3 have the disease. Since n is large and p is small, we use Poisson approximation with λ = np = 400 × 0.005 = 2. Then P(X=3) = e⁻² × 2³ / 3! = e⁻² × 8 / 6 ≈ (0.1353 × 8) / 6 = 1.0824 / 6 = 0.1804 ≈ 0.180. The binomial calculation would give ⁴⁰⁰C₃ × (0.005)³ × (0.995)³⁹⁷, which is numerically close. This approximation saves time and reduces calculator errors.
某罕见病在人群中的发病率为 0.5%。求在一个由 400 人组成的随机样本中，恰好有 3 人患病的概率。由于 n 很大而 p 很小，可使用泊松近似，取 λ = np = 400 × 0.005 = 2。则 P(X=3) = e⁻² × 2³ / 3! = e⁻² × 8 / 6 ≈ (0.1353 × 8) / 6 = 1.0824 / 6 = 0.1804 ≈ 0.180。若用二项式计算，则为 ⁴⁰⁰C₃ × (0.005)³ × (0.995)³⁹⁷，结果在数值上相近。这种近似可节省时间并减少计算器输入错误。
Always check that the conditions for Poisson approximation are met: n > 50 and np < 5 usually. Also note that the Poisson distribution is useful for modelling events occurring independently at a constant average rate, but here it serves as an approximation to the binomial.
务必检查泊松近似的条件是否满足：通常要求 n > 50 且 np < 5。还需注意，泊松分布适用于对以恒定平均发生率独立发生的事件进行建模，但在这里它是对二项分布的近似。
9. Normal Distribution and Inverse Normal | 正态分布与逆正态
The masses of apples are normally distributed with mean 120 g and standard deviation 15 g. A supermarket rejects apples weighing less than 100 g. What proportion is rejected? First, standardise: Z = (100 − 120) / 15 = −20/15 = −1.333… We need P(Z < −1.333). Using tables or calculator, Φ(−1.333) = 1 − Φ(1.333) ≈ 1 − 0.9088 = 0.0912. So about 9.1% are rejected. Next, find the weight exceeded by the top 10% heaviest apples. For the top 10%, we need the 90th percentile. Inverse normal: Φ⁻¹(0.9) ≈ 1.2816. Then weight = 120 + 1.2816 × 15 = 120 + 19.224 = 139.224 g ≈ 139 g.
苹果的重量服从正态分布，均值为 120 g，标准差为 15 g。某超市拒收重量低于 100 g 的苹果。求被拒收的比例。首先标准化：Z = (100 − 120) / 15 = −20/15 = −1.333… 需要求 P(Z < −1.333)。查表或使用计算器可得 Φ(−1.333) = 1 − Φ(1.333) ≈ 1 − 0.9088 = 0.0912。因此约 9.1% 的苹果被拒收。接下来，求最重的 10% 苹果所超过的重量值。对于前 10%，我们需要第 90 百分位数。逆正态：Φ⁻¹(0.9) ≈ 1.2816，然后重量 = 120 + 1.2816 × 15 = 120 + 19.224 = 139.224 g ≈ 139 g。
Many students forget to convert the tail probability correctly when using inverse normal. If the question asks for the weight that 10% exceed, the area to the left is 0.9, not 0.1. Drawing a sketch normal curve and shading the required area prevents such errors.
许多学生在使用逆正态时忘记正确转换尾部概率。如果题目问的是 10% 超过的重量，左侧面积应为 0.9 而非 0.1。画出正态曲线草图并标出要求的面积区域可以避免此类错误。
10. Hypothesis Testing with Binomial (Critical Region) | 二项分布假设检验（临界区域）
A company claims that at least 75% of customers are satisfied. A consumer group suspects the proportion is lower and surveys 30 customers, of which 18 are satisfied. Test at the 5% significance level. Let p be the true proportion satisfied. H₀: p = 0.75; H₁: p < 0.75. Under H₀, X ~ B(30, 0.75). We need the probability of observing 18 or fewer satisfied customers when p=0.75. Using calculator: P(X ≤ 18) = 0.0219 (or from tables). Since 0.0219 < 0.05, we reject H₀. There is sufficient evidence to suggest the proportion is less than 75%.
某公司声称至少 75% 的顾客是满意的。一个消费者团体怀疑该比例偏低，于是调查了 30 名顾客，其中 18 人满意。在 5% 显著性水平下进行检验。设 p 为真实的满意比例。H₀: p = 0.75；H₁: p < 0.75。在 H₀ 下，X ~ B(30, 0.75)。我们需要计算当 p=0.75 时，观察到 18 个或更少满意顾客的概率。使用计算器：P(X ≤ 18) = 0.0219（或查表）。由于 0.0219 < 0.05，我们拒绝 H₀。有充分证据表明满意比例小于 75%。
Alternatively, one can find the critical region. For a one-tailed test with n=30 and p=0.75, the lower critical value is the largest x such that P(X ≤ x) ≤ 0.05. From tables, P(X ≤ 18) = 0.0219, and P(X ≤ 19) = 0.0519, so the critical region is X ≤ 18. Our observed value 18 falls in the critical region, leading to the same conclusion. This structured approach is essential to secure all method marks.
或者，可以找出临界区域。对于单尾检验，n=30 且 p=0.75，下临界值是满足 P(X ≤ x) ≤ 0.05 的最大 x 值。查表得 P(X ≤ 18) = 0.0219，P(X ≤ 19) = 0.0519，因此临界区域为 X ≤ 18。我们的观测值 18 落入临界区域，得出相同结论。这种结构化的解答方式对于拿到所有过程分至关重要。
11. Interpreting Correlation and the PMCC | 相关系数与积矩相关系数的解读
A data set gives the product moment correlation coefficient (PMCC) between two variables as r = -0.823. The question asks you to interpret this value in context, saying whether it supports the claim that higher revision hours are associated with lower stress. A value of -0.823 indicates a strong negative linear correlation, meaning that as revision hours increase, stress tends to decrease. However, you must comment on the limitations: correlation does not imply causation; there may be other factors involved.
某数据集中两个变量之间的积矩相关系数（PMCC）为 r = -0.823。题目要求你在具体情境中解读该数值，判断其是否支持“复习时间越长、压力越小”的主张。数值 -0.823 表明存在强负线性相关，意味着随着复习时间增加，压力倾向于减小。但你必须评论其局限性：相关关系并不意味着因果关系；可能涉及其他因素。
A hypothesis test for correlation often follows: H₀: ρ = 0, H₁: ρ < 0. The critical value for n=10 at 5% one-tailed is -0.5494 (from tables). Since -0.823 < -0.5494, we reject H₀ and conclude there is evidence of negative correlation in the population. Students need to be careful with the sign and the correct critical value from the formula booklet.
随后通常会进行相关系数的假设检验：H₀: ρ = 0，H₁: ρ < 0。对于 n=10，5% 单尾检验的临界值为 -0.5494（查表）。由于 -0.823 < -0.5494，我们拒绝 H₀，得出总体中存在负相关的证据。学生需注意符号以及从公式表中查找正确的临界值。
12. Summary of Key Tips for Unit 2 | 单元2 备考核心技巧总结
Always show clear diagrams for mechanics problems, labelling all forces and angles. Use exact trigonometric values where possible to avoid rounding errors. In statistics, state your hypotheses precisely and give conclusions in context, using the phrase ‘sufficient evidence’ or ‘insufficient evidence’ as appropriate. Check that your calculator is in the correct mode (degree/radian) and that you are using the correct tail for probability calculations. Time management is crucial in this paper because it mixes two distinct disciplines. Work through the mechanics sections systematically, then move on to statistics, which many candidates find more straightforward.
解力学题时务必画出清晰的受力图，标注所有的力和角度。尽量使用精确的三角比以避免舍入误差。在统计部分，精确表述原假设和备择假设，并结合情境给出结论，正确使用“有充分证据”或“证据不足”等措辞。检查计算器是否处于正确模式（角度/弧度），并确保概率计算时使用了正确的尾部区域。本场考试中时间管理尤为关键，因为它融合了两门截然不同的学科。应系统性地完成力学部分，然后再转向统计部分，许多考生觉得后者更直接。
Practise past papers under timed conditions, and mark yourself strictly against the scheme. This Unit 2 paper rewards method marks, so even if you get the wrong final answer, you can still gain most of the marks by showing correct working. Revise common mistakes, such as mixing up sine/cosine in resolution, forgetting to square the standard deviation in Normal calculations, or misinterpreting the alternative hypothesis in one-tailed tests.
在限时条件下练习往年真题，并严格按照评分标准自我批改。单元2 试卷对过程分给予很多奖励，因此即使最终答案错误，只要展示出正确的步骤，你仍能获得大部分分数。复习常见错误，例如分解时混淆正弦与余弦、在正态计算中忘记标准差应平方、或在单尾检验中误解备择假设。
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