📚 AQA A Level Maths MA05 Unit M2: Worked Example Responses | AQA A Level 数学 MA05 单元 M2 范例作答解析
AQA A Level Maths Unit M2 (Mechanics 2) tests your ability to model motion and forces in more advanced situations: projectiles, energy methods, impulse, moments, centres of mass and circular motion. The examiner is not only checking your final answer but also how clearly you communicate the physical principles, diagrams, and algebraic steps. This article collects example responses for the main question types in MA05, showing what a high-scoring answer looks like and where candidates often lose marks.
AQA A Level 数学单元 M2（力学 2）考查你在更复杂情境中建立运动与力模型的能力：抛体运动、能量方法、冲量、力矩、质心和圆周运动。考官不仅看最终答案，还看你如何清晰地表达物理原理、图示和代数步骤。本文汇集 MA05 主要题型的范例作答，展示高分答案的样子以及考生常在哪里丢分。
1. Understanding MA05 Unit M2: Command Words and Mark Schemes | 理解 MA05 单元 M2：指令词与评分方案
In M2, questions often use command words such as find, show that, determine, and state. A show that question requires a complete derivation with no hidden jumps; you must give the given result exactly, usually by quoting a principle and substituting values carefully.
在 M2 中，题目常使用 find、show that、determine 和 state 等指令词。Show that 题要求完整的推导，不能跳步；你必须引用某一原理并仔细代入数值，精确得到给定结果。
Examiners allocate marks for method (M), accuracy (A), and sometimes explanation or diagram (E). For example, in a work-energy question, quoting the work-energy principle and identifying the correct energy terms can earn method marks even if a numerical slip occurs later.
考官将分数分为方法分（M）、准确分（A），有时还有解释或图示分（E）。例如，在功能关系题中，引用功能原理并识别正确的能量项，即使后面计算出现数值小错，也能获得方法分。
Therefore, a strong example response should show the principle, clear working, and a final answer stated to the required accuracy, usually 2 or 3 significant figures unless the question says otherwise.
因此，一份好的范例作答应展示原理、清晰的过程，并以题目要求的精度给出最终答案，除非题目另有说明，通常保留 2 到 3 位有效数字。
2. Projectile Motion: Model Answer Structure | 抛体运动：标准答案结构
, uᵧ = u sin θ, and write the SUVAT equations with the convention that upward is positive.
对于抛体运动，最高分作答从一开始就区分水平与竖直运动。写出 uₓ = u cos θ、uᵧ = u sin θ，并规定向上为正，列出 SUVAT 方程。
Example response for Find the greatest height reached by a particle projected with speed 24 m s⁻¹ at an angle 30° above the horizontal where uᵧ = 24 sin 30° = 12 m s⁻¹, gives 0 = 12² − 2 × 9.8 × H, so H = 144 ÷ 19.6 = 7.35 m.
例如 求一质点以 24 m s⁻¹ 与水平成 30° 角抛出所达到的最大高度 的范例作答：在最高点，竖直速度为零。利用 vᵧ² = uᵧ² − 2gH，其中 uᵧ = 24 sin 30° = 12 m s⁻¹，得 0 = 12² − 2 × 9.8 × H，因此 H = 144 ÷ 19.6 = 7.35 m。
. If you use upward as positive, the acceleration is −9.8 m s⁻² in the vertical equations.
考官提示：在使用 g = 9.8 之前，一定要先规定正方向。如果取向上为正，竖直方程中的加速度应写作 −9.8 m s⁻²。
zontal level, or use the given height if the particle lands above or below the projection point.
对于射程和飞行时间，写出竖直位移方程，若落点在相同水平面上，则令 sᵧ = 0；若落点高于或低于抛射点，则使用题目给出的高度。
3. Work, Energy and Power: Showing the Principle Clearly | 功、能与功率：清晰展示原理
PE. A model response always states this principle and then lists each energy change separately.
功能原理指出，外力所做的总功等于机械能的变化量，常写作 W = ΔKE + ΔGPE。范例作答总是先陈述该原理，然后分别列出每个能量变化。
kg climbs a 200 m slope at constant speed 15 m s⁻¹ with resistance 400 N. Find the engine' n solves F × 200 = 400 × 200 + 900 × 9.8 × h, after determining h from the slope geometry.
例如：一辆质量 900 kg 的汽车以 15 m s⁻¹ 的恒定速度爬上一段 200 m 的斜坡，阻力为 400 N。求发动机的驱动力。由于速度恒定，ΔKE = 0。发动机的有用功为 F × 200。克服阻力做的功为 400 × 200，重力势能增加量为 900 × 9.8 × h，其中 h 为竖直高度。作答随后在由斜面几何关系求出 h 后，解方程 F × 200 = 400 × 200 + 900 × 9.8 × h。
ower you are finding and quote P = Fv only when the force acts in the direction of motion.
功率题需要区分平均功率 P = 功 ÷ 时间 和瞬时功率 P = Fv。在作答中，要指出你求的是哪一种功率，并且只有在力沿运动方向时才引用 P = Fv。
4. Impulse and Momentum: Vector Form and Sign Conventions | 冲量与动量：向量形式与符号约定
s the change in momentum. In M2, this is often applied in one dimension first, then extended to two-dimensional vector form.
冲量-动量原理为 I = mv − mu，其中 I 是冲量，mv − mu 是动量变化量
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