📚 Common Mistakes in Edexcel IAL Unit 5 (WME01) January 2021 | 爱德思IAL单元5 2021年1月真题易错点总结
This article reviews the most frequent errors made by candidates in the Edexcel International A‑Level Mathematics Unit 5 (WME01 Mechanics 1) exam from January 2021. Understanding these pitfalls can help you avoid losing marks on similar question types.
本文回顾了考生在 2021 年 1 月爱德思国际 A‑Level 数学第五单元（WME01 力学 1）考试中最常见的错误。了解这些陷阱有助于你在类似题型中避免失分。
1. Sign Conventions in SUVAT Equations | SUVAT方程中的符号约定
rticle was projected upwards from a height of 1.2 m with initial speed 14.7 m s⁻¹, taking upward as positive means a = −9.8 m s⁻². A common mistake was to use a positive value for a when the particle was moving downwards, leading to incorrect times and final velocities.
很多学生在使用 v = u + at、s = ut + ½ at² 等式子时，错误地给位移、速度和加速度赋予符号。在第二题中，一质点从 1.2 m 高处向上以 14.7 m s⁻¹ 抛出，若向上为正，则 a = −9.8 m s⁻²。常见错误是在质点向下运动时仍使用正的加速度，导致时间和末速度计算错误。
ed the physical context and accepted both mathematical solutions without checking which one corresponds to the required event (e.g. passing a point twice).
在由 s = ut + ½ at² 解二次方程时，部分考生忽略物理背景，直接取两个数学解而未验证哪一解符合题意（例如经过某点两次）。
2. Resolving Forces on an Inclined Plane | 斜面上力的分解
ncorrectly, either swapping sin and cos or missing the perpendicular component when writing the equilibrium conditions parallel and perpendicular to the plane. Always draw a clear free-body diagram and check: component down the plane = mg sin θ, normal reaction = mg cos θ + (horizontal force's perpendicular component).
在斜面问题（第四题）中，质量为 5 kg 的质点在一水平力作用下保持平衡。很多学生分解重力时出错，分不清正弦和余弦，或者在写沿斜面和垂直斜面的平衡条件时遗漏分量。务必画清晰的受力图并检查：沿斜面向下的分力 = mg sin θ，法向反力 = mg cos θ + 水平力的垂直斜面分量。
The horizontal force had both a component parallel to the plane and a component perpendicular to the plane. Forgetting to include the perpendicular component when calculating the normal reaction led to an incorrect friction or reaction value.
水平力既有沿斜面的分力，也有垂直斜面的分力。计算法向反力时忘记计入垂直分量会导致摩擦力或反力值错误。
3. Connected Particles and Tension | 连接体与绳中张力
Question 5 involved two particles connected by a light inextensible string passing over a smooth pulley, one on a rough horizontal table and the other hanging freely. A typical error was assuming that the tension is equal to the weight of the hanging particle. In connected-body problems, tension is not equal to the hanging weight because the system accelerates; it must be found by solving simultaneous equations.
第五题涉及两个由轻质不可伸长绳相连的质点，绕过光滑滑轮，一个在粗糙水平桌上，另一个自由悬挂。典型错误是假设绳中张力等于悬挂质点的重量。在连接体问题中，由于系统加速，张力不等于悬挂重量；必须通过解联立方程求出。
Another mistake was to use the same sign for acceleration for both particles without correlating their directions. If particle A moves right, particle B moves down; establishing a consistent positive direction is crucial.
另一个错误是为两个质点使用相同的加速度符号而未关联它们的运动方向。若质点 A 向右运动，质点 B 向下运动；设定一致的正方向至关重要。
4. Moments and the Choice of Pivot | 力矩与支点选择
Question 6 required finding the tension in a string holding a uniform rod, with one end against a rough wall and the string attached to the wall at a point above the rod. Many candidates took moments about the wrong point or missed one of the forces. Choosing a pivot and consistently writing the perpendicular distance is essential. The most efficient pivot is often the point where unknown forces (such as the reaction at the wall) intersect, to eliminate them from the moment equation.
第六题要求计算一根均匀杆的绳中张力，杆一端靠粗糙墙，绳系在墙上杆上方某点。很多考生取错力矩点或遗漏某个力。选择合适的支点并写出正确的垂直距离至关重要。最高效的支点通常是未知力（例如墙的反力）的交点，以便在力矩方程中消去它们。
A frequent slip was using the length of the rod rather than the perpendicular distance from the line of action to the pivot. In an inclined rod, the perpendicular distance from the pivot to the weight is (L/2) cos θ, not simply L/2.
常见失误是使用杆的长度而非力作用线到支点的垂直距离。在倾斜杆问题中，支点到重力的垂直距离是 (L/2) cos θ，而非简单的 L/2。
5. Pulley Systems and Constraint Equations | 滑轮系统与约束方程
In the two-particle pulley problem (Q5), the string length is constant, so the magnitudes of the displacements, velocities and accelerations of the two particles are equal as long as the string remains taut. Candidates often forgot to apply this constraint when linking the equations of motion, leading to inconsistent variables and unsolvable systems.
在双质点滑轮问题（第五题）中，绳长恒定，因此只要绳子保持张紧，两质点的位移、速度和加速度大小相等。考生在连接运动方程时常忘记应用这一约束，导致变量不一致、方程组无法求解。
Additionally, when the string passes over a smooth pulley, the tension on both sides is equal — but only if the pulley is smooth and light. This condition was sometimes misapplied in non-ideal pulley setups.
此外，当绳子绕过光滑滑轮时，两侧张力相等——但前提是滑轮光滑且轻质。这一条件在非理想滑轮情境下常被误用。
6. Velocity–Time Graphs and Distance | 速度-时间图像与路程
Question 3 presented a velocity–time graph for a car journey. A common mistake was to equate distance with displacement and simply find the area under the graph including signs. When the velocity is negative, the area still contributes to distance (total area), but some students subtracted the negative area, thus computing displacement instead. Distance is the sum of the absolute areas.
第三题给出了汽车行驶的速度-时间图像。常见错误是将路程等同于位移，简单求带符号的面积。速度为负时，该面积仍计入路程（总面积），但部分学生减去负面积，从而算成了位移。路程是各块面积的绝对值之和。
Many also struggled to convert the trapezium areas correctly or misread the time axis, especially when the graph included a period of uniform deceleration.
很多学生也无法正确计算梯形面积或误读时间轴，尤其是当图像包含匀减速阶段时。
7. Vector Addition and Bearing | 矢量加法与方位角
Question 4 required finding the resultant of two forces given as vectors, then stating its magnitude and bearing. Students often gave the angle to the horizontal correctly but misinterpreted the bearing. Bearing is measured clockwise from north, whereas the angle from arctan(j/i) is measured from the positive i‑direction (east). Conversion is required: if the resultant direction is 30° above the eastward horizontal, the bearing is 060°. Getting this wrong was extremely common.
第四题要求计算以矢量形式给出的两个力的合力，并说明其大小和方位角。学生常能正确给出与水平方向的夹角，却对方位角的定义理解有误。方位角是从正北顺时针量取，而用 arctan(j/i) 算出的角度是从正 i 方向（东）量起的。需要换算：若合力方向为东偏北 30°，则方位角为 060°。这一错误极为普遍。
In addition, some students omitted the magnitude or gave an unsimplified surd, losing accuracy marks.
此外，一些学生遗漏大小或给出未化简的根式，损失了精度分。
8. Momentum Conservation in Collisions | 碰撞中的动量守恒
n applied correctly, but many then made a sign error when combining the terms. For two particles, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂, with all velocities sharing a consistent positive direction. A number of candidates incorrectly assumed both final velocities were in the same direction, gave both as positive without checking, and derived impossible results.
第七题描述了沿同一直线运动的两个质点的直接碰撞。动量守恒定律通常能正确列出，但很多人在合并各项时出现符号错误。对于两质点，m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂，所有速度应统一正方向。不少考生错误假设两末速度同向，未经验证便均取正，导致不合理结果。
v − mu but calculated u − v. Ensure that impulse is final momentum minus initial momentum for the body receiving the impulse.
冲量常与动量变化混淆；有人写出冲量 = mv − mu 但实际算了 u − v。需确保冲量是受力物体的末动量减初动量。
9. Friction Direction and Limiting Equilibrium | 摩擦力的方向与极限平衡
In Question 5, the particle on the rough table was subject to friction. Students often drew the friction force in the wrong direction. Friction opposes relative motion or the tendency of motion. On a horizontal table with the connecting string pulling the particle to the right, friction acts to the left. However, if the system is about to move in the opposite direction under some external influence, friction must be reversed. Many lost marks by not checking the direction of the resultant force correctly.
第五题中，粗糙桌面上的质点受摩擦力作用。学生常画错摩擦力的方向。摩擦力总是阻碍相对运动或相对运动趋势。在水平桌上，若连接绳向右拉质点，摩擦力向左。但若系统将在某种外力作用下朝相反方向运动，则摩擦力方向需反过来。许多考生因未正确检验合力方向而失分。
ften used F = μR regardless of whether the friction had reached its maximum. The condition 'on the point of slipping' must be present to apply F = μR. Otherwise, friction can take any value up to μR.
在极限平衡问题（第八题，斜面上的方块）中，考生常不论是否达到最大静摩擦均直接套用 F = μR。只有出现“即将滑动”的条件时才能用 F = μR。否则摩擦力可为不超过 μR 的任意值。
10. Applying the Whole-System Approach | 整体法应用
iminating internal tension. In January 2021, Q5 could be solved more quickly by applying Newton' udents missed this shortcut and made algebraic errors when solving simultaneous equations.
对于连接体问题，考虑整个系统（例如两质点和绳）可通过消去内张力来简化加速度的求解。在 2021 年 1 月的试卷中，第五题可通过整体法快速解决：驱动力（悬挂物重）减去总阻力 = 总质量 × a。许多学生未想到这一捷径，在解联立方程时出现代数错误。
The whole-system approach must only be used in the direction of motion that is common to the system. If particles move in different directions, you must still handle signs carefully, but this method often reduces algebraic mistakes and saves time.
整体法只能用于系统共同的运动方向。若各质点运动方向不同，仍需谨慎处理符号，但此方法常可减少代数错误并节省时间。
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