📚 AQA MM04 January 2021 Markscheme Common Mistakes Summary | AQA 数学：aqa-ma04-w-ms-jan21 易错点总结
This article is based on the examiner’s markscheme and common pitfalls observed in the AQA MM04 (Mechanics 4) January 2021 paper. It identifies the most frequent errors made by candidates and shows you how to avoid them. Whether you are revising for a mock or the final exam, studying these mistakes will sharpen your problem‑solving skills and boost your final grade.
本文基于 AQA MM04（力学 4）2021 年 1 月试卷的评分方案和常见错误编写而成，总结了考生最高频的失分点，并告诉你如何规避这些陷阱。无论你是准备模拟考还是最终大考，认真研读这些易错点都能有效提升解题能力，帮助你在考试中拿到更高的分数。
1. Misinterpreting Relative Velocity | 相对速度的误解
Many candidates incorrectly apply the relative velocity formula when two particles are approaching each other. The key is to write vA – vB or vB – vA in a single consistent direction. In the January 2021 markscheme, marks were consistently lost because students wrote expressions like v1 + v2 for an approaching case without drawing a proper vector diagram.
许多考生在处理两个物体相互靠近的情况时错误地使用了相对速度公式。关键在于沿同一个参考方向写出vA – vB 或 vB – vA。在 2021 年 1 月的评分方案中，学生因未画矢量图就直接写出类似于 v1 + v2 的表达式而反复丢分。
Always sketch a small arrow diagram. If particle A is moving to the right at 3 m s⁻¹ and B to the left at 2 m s⁻¹, taking right as positive gives vA = +3, vB = –2. The relative velocity vA – vB = 3 – (–2) = 5 m s⁻¹. Many candidates simply added the speeds to get 5, but when the formula required a formal subtraction, they lost method marks.
一定要画一个简单的箭头示意图。假设物体 A 以 3 m s⁻¹ 速度向右运动，物体 B 以 2 m s⁻¹ 速度向左运动，取向右为正方向则 vA = +3，vB = –2。相对速度vA – vB = 3 – (–2) = 5 m s⁻¹。很多考生直接把速率相加得出 5，但当公式要求严格做减法时，他们会因步骤不清而丢失方法分。
vAB = vA – vB (taking directions into account)
vAB = vA – vB （务必考虑方向）
2. Incorrect Sign Conventions in Momentum Equations | 动量方程中正负号的错误使用
In the January 2021 paper, a common mistake was writing the conservation of momentum without attaching a clear sign to each velocity. For a collision, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ only works if the same positive direction is used for all velocities. Students often assigned positive values to all speeds, which then gave incorrect final velocities when substituted into Newton’s experimental law.
在 2021 年 1 月的试卷中，一个常见错误是写动量守恒时没有为每个速度明确指定正负号。对于碰撞问题，只有在所有速度使用相同正方向的前提下，m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ 才成立。学生经常把所有速率都带正号，导致代入牛顿实验定律时算出的末速度错误。
The markscheme penalised forgetting to convert a left‑moving velocity from a diagram into a negative value. Even after a diagram was drawn, candidates sometimes wrote v₂ = +1.5 from the diagram, then substituted –1.5 into the restitution equation, creating inconsistency.
评分方案对以下情况会扣分：画了图，却忘了把向左运动的速度转化为负值。有些考生虽然画了图，从图上读出 v₂ = +1.5，随后在恢复系数方程中又用 –1.5，造成了前后不一致。
Checklist:
- Draw a clear “positive” arrow.
- Assign + and – to every velocity.
- Use the same signs in both momentum and restitution equations.
自查清单：
- 画出一个清晰的“正方向”箭头。
- 为每一个速度标上 + 或 –。
- 在动量方程和恢复系数方程中使用相同的符号。
3. Forgetting to Consider Energy Changes | 忽略能量变化
MM04 often includes problems where a particle moves against a variable force or on a rough incline. Many candidates in January 2021 lost marks by applying the work–energy principle incorrectly, counting only the gravitational potential energy change but omitting the work done against friction or air resistance.
MM04 经常包含物体在变力或粗糙斜面上运动的题目。2021 年 1 月的很多考生因错误应用动能定理而丢分：只计算了重力势能的变化，却忽略了克服摩擦力或空气阻力所做的功。
The standard formula is:
Total work done = ΔK.E. + ΔG.P.E. + work against friction
标准公式为：
合力做功 = 动能变化 + 重力势能变化 + 克服摩擦做功
If a particle slides from rest down a rough slope, the loss in G.P.E. does not all become K.E. Some energy is dissipated as heat. The examiners’ report stressed that candidates must set out the energy equation clearly and state work done against friction = µR × d. R is the normal reaction, not the weight, and d is the distance along the slope.
如果物体从静止沿粗糙斜面下滑，减少的重力势能并不会全部转化为动能，部分能量以热的形式耗散了。考官报告强调，考生必须清晰列出能量方程，并写出克服摩擦做功 = µR × d。R 是法向反力，不是重力，d 是沿斜面的距离。
A table can help you organise the terms:
| Energy term | Expression | Sign convention |
|---|---|---|
| Initial K.E. | ½ m u² | + (if moving) |
| Final K.E. | ½ m v² | – (moves to the other side) |
| Δ G.P.E. | mgΔh | Loss: +; Gain: – |
| Work vs friction | µR d | Always subtract from total |
一个表格可以帮助你整理各项：
| 能量项 | 表达式 | 符号规则 |
|---|---|---|
| 初动能 | ½ m u² | + （若在运动） |
| 末动能 | ½ m v² | – （移项处理） |
| 重力势能变化 | mgΔh | 减少：+；增加：– |
| 克服摩擦做功 | µR d | 总是从总能量中减去 |
4. Errors in Resolving Forces on Inclined Planes | 斜面上力的分解错误
When dealing with a particle on a rough inclined plane, many candidates in the January 2021 exam resolved perpendicular to the plane incorrectly. They often wrote R = mg cos θ and then used F = µR but forgot to check whether there was an additional applied force with a perpendicular component.
在处理粗糙斜面上的物体时，2021 年 1 月考卷中的许多考生在垂直于斜面的分解上犯了错误。他们经常写成R = mg cos θ，然后直接使用 F = µR，却忘记检查是否存在带有垂直分量的额外外力。
The correct approach is to sum all force components perpendicular to the plane:
R = mg cos θ ± (any other force’s perpendicular component)
正确的做法是把所有垂直于斜面的分力求和：
R = mg cos θ ± （其他外力的垂直分量）
For example, if an extra horizontal force P is pushing the particle into the slope, then R = mg cos θ + P sin θ (depending on geometry). This R then feeds into the friction calculation, and a mistake in R directly affects the acceleration and equilibrium conditions.
例如，若有一个额外的水平力 P 把物体压向斜面，那么R = mg cos θ + P sin θ（取决于几何关系）。这个 R 会进入摩擦力的计算，而 R 的错误将直接影响加速度和平衡条件的求解。
5. Miscalculation of Centre of Mass | 质心计算错误
In the MM04 paper, composite shapes often appear — such as a lamina with a circle cut out. A typical mistake is to treat the cut‑out as a positive mass instead of subtracting its contribution. The January 2021 markscheme consistently awarded method marks only when the mass of the removed piece was taken as negative.
在 MM04 试卷中，经常出现组合图形——例如带圆形切孔的薄片。一个典型的错误是把切掉的孔视为正质量，而不是减去它的贡献。2021 年 1 月的评分方案规定，只有当移除部分的质量被视为负值时才能获得方法分。
The general formula is:
x̄ = (m₁x₁ – m₂x₂) / (m₁ – m₂)
通用公式为：
x̄ = (m₁x₁ – m₂x₂) / (m₁ – m₂)
Where m₁ is the mass of the original complete shape, and m₂ is the mass that has been removed. Many candidates wrote a plus sign between the two terms, which would be correct only if they had already defined m₂ as negative, but they rarely showed that assumption.
其中 m₁ 是原完整图形的质量，m₂ 是被移除部分的质量。很多考生在两项之间使用了加号，这只有在已经将 m₂ 定义为负数时才成立，但他们几乎从未说明这一假设。
Another frequent error was confusing the distance x in the formula with the coordinate of the centroid of the removed part. The distance must be measured from the same reference point (usually one corner of the lamina).
另一个常见错误是把公式中的距离 x 与移除部分的形心坐标混淆。距离必须从同一个参考点（通常是薄板的一个角）量起。
6. Using the Wrong Coefficient of Restitution Formula | 恢复系数公式的误用
Newton’s experimental law is e = (speed of separation) / (speed of approach). In many January 2021 responses, students reversed the numerator and denominator or omitted the absolute value. The markscheme accepted both e = (v₂ – v₁) / (u₁ – u₂) along a defined positive line, but only if the signs were consistent.
牛顿实验定律为 e = （分离速率）/（靠近速率）。在 2021 年 1 月的许多答卷中，学生颠倒了分子和分母，或者漏掉了绝对值。评分方案接受沿定义正方向写出的 e = (v₂ – v₁) / (u₁ – u₂)，但前提是符号一致。
The biggest trap occurs in oblique collisions, where students apply the coefficient normally for the component along the line of centres but forget to set the tangential component unchanged. For smooth spheres, the velocity component perpendicular to the line of centres remains the same. Candidates frequently changed both components, which lost all accuracy marks.
最大的陷阱出现在斜碰问题中：学生通常能正确对连心线方向的分量使用恢复系数，却忘记切向速度分量保持不变。对于光滑球体，与连心线垂直的速度分量碰撞前后不变。考生经常把两个分量都改变，导致准确性分全部丢失。
Use this step‑by‑step method:
- Choose axes: one along the line of centres, one perpendicular.
- Apply e = (v₂′ – v₁′) / (u₁ – u₂) along the line of centres.
- Conserve tangential component: v₁⊥ = u₁⊥, v₂⊥ = u₂⊥.
- Conserve momentum along the line of centres.
使用以下分步方法：
- 建立坐标轴：一个沿连心线，一个与之垂直。
- 沿连心线方向应用 e = (v₂′ – v₁′) / (u₁ – u₂)。
- 切向分量守恒：v₁⊥ = u₁⊥，v₂⊥ = u₂⊥。
- 沿连心线方向动量守恒。
7. Confusing Impulse and Momentum | 冲量与动量混淆
The vector impulse exerted on a particle equals the change in its momentum: I = m v – m u. In the January 2021 markscheme, candidates frequently wrote the impulse on A as m(v₁ + u₁), adding the speeds, or they reversed the direction. Remember: the impulse is in the direction of the change in velocity.
作用在物体上的冲量矢量等于其动量的变化：I = m v – m u。在 2021 年 1 月的评分方案中，考生频繁地将作用在 A 上的冲量写成 m(v₁ + u₁)，进行了速率相加，或者弄反了方向。请记住：冲量的方向是速度变化的方向。
In a collision between A and B, the impulse on A is equal and opposite to the impulse on B. This is a direct consequence of Newton’s third law. Mistakes occurred when candidates tried to find the impulse on A by using B’s momentum change but got the sign wrong.
在 A 与 B 的碰撞中，作用在 A 上的冲量与作用在 B 上的冲量大小相等、方向相反，这是牛顿第三定律的直接结果。当考生试图用 B 的动量变化求 A 上的冲量时，常因正负号搞错而出错。
Ion A = mA (vA – uA) = – Ion B
I作用在A上 = mA (vA – uA) = – I作用在B上
Always perform the subtraction v – u vectorially. If u is negative and v is positive, the change might be larger than the initial speed, which is fine — do not artificially change the sign to make the impulse look “neater”.
一定要按矢量方式进行减法 v – u。如果 u 为负，v 为正，变化量可能比初速率更大，这完全正常——不要人为地去修改符号来让冲量显得更“简洁”。
8. Projectile Motion with Air Resistance | 带空气阻力的抛体运动
Questions involving air resistance usually require setting up a differential equation. The January 2021 paper asked for the terminal velocity of a particle falling under gravity against a resistive force kv or kv². A frequent mistake was to write mg – kv = ma but then forget that at terminal velocity a = 0, so vterm = mg / k.
涉及空气阻力的题目通常需要建立微分方程。2021 年 1 月的试卷要求计算物体在重力下受到阻力 kv 或 kv² 作用时的终极速度。一个常见错误是写出了 mg – kv = ma，却忘记在终极速度时 a = 0，因此vterm = mg / k。
If the resistive force is kv², then at terminal speed:
mg = k vterm² ⇒ vterm = √(mg / k)
如果阻力为 kv²，则在终极速度时：
mg = k vterm² ⇒ vterm = √(mg / k)
Many candidates also made errors when integrating to find velocity as a function of time. They failed to separate variables correctly, especially when the equation was m(dv/dt) = mg – kv. The correct steps involve rewriting as:
∫ dv / (mg – kv) = ∫ (1/m) dt
很多考生在通过积分求速度与时间的关系时也犯了错误，尤其是当方程为 m(dv/dt) = mg – kv 时，他们未能正确分离变量。正确的步骤为：
∫ dv / (mg – kv) = ∫ (1/m) dt
Then use ln|mg – kv|. Dropping the absolute value or forgetting the constant of integration were common reasons for losing the final accuracy marks.
然后使用 ln|mg – kv|。漏掉绝对值符号或忘记积分常数是丢失最终准确性分的常见原因。
9. Variable Force Integration Limits | 变力做功的积分限设置错误
Work done by a variable force F(x) over a displacement from x = a to x = b is given by W = ∫ab F(x) dx. In the January 2021 paper, some candidates integrated over the entire range but used negative signs incorrectly when the particle reversed direction.
变力 F(x) 在从 x = a 到 x = b 的位移上所做的功由W = ∫ab F(x) dx 给出。在 2021 年 1 月的试卷中，部分考生对全程进行了积分，但当物体反转方向时，他们没有正确处理正负号。
If the particle changes direction, you must split the integral into separate intervals where the velocity has a constant sign. The work is scalar, but the direction of F relative to the displacement matters. When F opposes the motion, the integrand becomes negative, or you can calculate work against the force and record it as positive work done against the system.
如果物体改变运动方向，则必须将积分分段，在每一段内速度的符号保持不变。功是标量，但力 F 相对于位移的方向至关重要。当 F 与运动方向相反时，被积函数为负值，或者你可以计算克服该力所做的功，并将其记录为对外部系统做的正功。
Leave a small table in your answer:
| Segment | x from | to | Direction of F vs motion | Work integral |
|---|---|---|---|---|
| 1 | 0 | d | Same | ∫₀ᵈ F(x) dx |
| 2 (reverse) | d | 0 | Opposite | –∫₀ᵈ F(x) dx (or ∫ᵈ₀ (–F) dx) |
在答案中列出一个小表格：
| 分段 | x 从 | 到 | F 与运动的方向关系 | 功的积分 |
|---|---|---|---|---|
| 1 | 0 | d | 相同 | ∫₀ᵈ F(x) dx |
| 2（返回） | d | 0 | 相反 | –∫₀ᵈ F(x) dx（或 ∫ᵈ₀ (–F) dx） |
This method ensures you do not double‑count or cancel work incorrectly.
这种方法可确保你不会重复计算或错误地抵消功。
10. Equilibrium of Rigid Bodies: Moments | 刚体平衡：力矩错误
When a ladder or a beam rests against a rough wall and floor, candidates often miss one of the normal reactions or friction forces. In the January 2021 MM04 paper, a classic mistake was to take moments about a point and forget that the friction at the floor produces a moment when the line of action does not pass through the pivot point.
当梯子或横梁靠在粗糙的墙和地面上时，考生常常会漏掉一个法向反力或摩擦力。在 2021 年 1 月的 MM04 试卷中，一个经典错误是：对某点取矩时，忘记了地面摩擦力会由于作用线不通过矩心而产生力矩。
To avoid this, sketch a large free‑body diagram, label every force (weight, normal reactions, friction), and clearly mark the perpendicular distance from the pivot to each line of action. Practice writing the moment equation in full before solving.
避免这一错误的方法是画一个较大的受力图，标出每一个力（重力、法向反力、摩擦力），并清楚地标明从矩心到每条力作用线的垂直距离。在求解前，务必先完整地写出力矩方程。
Anticlockwise moments = Clockwise moments (about any chosen point)
逆时针力矩之和 = 顺时针力矩之和（关于任意选定的点）
The January 2021 markscheme also penalised candidates who wrote moment = force × distance along the beam rather than perpendicular distance. For a sloping beam, you must use d = (length) × sin(angle) or cos(angle).
2021 年 1 月的评分方案也对那些写成“力矩 = 力 × 沿梁的距离”而非垂直距离的考生进行了扣分。对于倾斜梁，必须使用 d = （梁长）× sin(θ) 或 cos(θ)。
11. Errors in Vector Notation for Collisions | 碰撞中的向量符号错误
The January 2021 paper contained a question where velocities were given in i–j notation. Candidates frequently forgot that when applying conservation of momentum in vector form, they must work component‑wise. Trying to combine i and j components in a single equation without separating them was a common cause of error.
2021 年 1 月的试卷中有一道题用 i–j 向量表示速度。考生常常忘记，在应用向量形式的动量守恒时，必须对每个分量分别运算。试图在不分离分量的情况下将 i 和 j 分量混合在同一个方程中是常见的错误原因。
Write:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Then equate the i‑components and the j‑components separately. The restitution equation also applies only along the line of centres, which may be at an angle to the i‑direction. To use it correctly, you must first resolve velocities parallel and perpendicular to the line of centres. Many candidates lost marks by assuming the coefficient of restitution could be applied to the i‑component directly.
写出上式后，分别令 i 分量相等，j 分量相等。恢复系数方程也仅适用于沿连心线方向，
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