📚 A-Level Maths Unit 5 Jan20: Common Pitfalls Summary | A-Level 数学单元5 2020年1月卷 易错点总结
The January 2020 Mechanics 1 (WME01) paper, often referred to as Unit 5 in many schools, revealed a series of typical mistakes that can cost students valuable marks. This article highlights the most common pitfalls, explains why they occur, and shows you how to avoid them through careful analysis and structured working.
2020年1月的力学1（WME01）试卷，在许多学校被称为单元5，暴露了一系列学生容易丢分的典型错误。本文总结最常见的易错点，分析错误原因，并通过细致的分析和规范化的解题步骤告诉你如何避免这些陷阱。
1. Directional sign errors in SUVAT equations | SUVAT方程中的方向符号错误
Many candidates lost marks by failing to assign a consistent positive direction when using SUVAT equations. In a vertical motion problem, if upward is taken as positive, the acceleration due to gravity must be written as -9.8 m s⁻². Forgetting the negative sign leads to wrong values for displacement and time.
许多考生在使用SUVAT方程时没有设定一致的正方向，从而丢分。在竖直运动问题中，如果取向上为正，重力加速度必须写成-9.8 m s⁻²。漏掉负号会导致位移和时间的值全部错误。
A reliable approach is to sketch a quick arrow showing the chosen positive direction and write the values of s, u, v, a, t with their signs clearly indicated. When a ball thrown upwards reaches its highest point, its velocity v is zero, not the acceleration.
一个可靠的方法是画出一个简短箭头标明所选的正方向，并清晰标出s, u, v, a, t各量及其符号。当向上抛出的球达到最高点时，速度v为零，而加速度不为零。
2. Misplacing components on an inclined plane | 斜面上分力的错误放置
A very frequent error was confusing the parallel and perpendicular components of weight on a slope. The weight mg is always vertically downward. For a plane inclined at angle θ to the horizontal, the component parallel to the slope is mg sin θ, and the component perpendicular to the slope is mg cos θ. Reversing these two is a classic mistake.
一个非常常见的错误是在斜面上混淆重力的平行分量和垂直分量。重力mg总是竖直向下的。对于与水平面夹角θ的斜面，平行于斜面的分量为mg sin θ，垂直于斜面的分量为mg cos θ。把这两个分量弄反是经典的错误。
Use a right-angled triangle with mg as the hypotenuse. The angle between the vertical and the perpendicular to the plane is θ. Therefore, the side opposite θ is mg sin θ, which lies along the slope. In friction problems, the normal reaction R equals the perpendicular component only if no other forces act vertically to the plane, so always resolve systematically.
画一个以mg为斜边的直角三角形。竖直线与平面垂线之间的夹角是θ，因此θ的对边是mg sin θ，方向沿斜面。在摩擦力问题中，只有当没有其他垂直于斜面的力作用时，法向反力R才等于该垂直分量，因此始终要系统地分解。
3. Tension and thrust in connected particles | 连接质点的张力和推力
In connected particle questions, students frequently misapply Newton’s second law by writing separate equations for each particle without accounting for the tension correctly. A common oversight is treating the tension as acting in the same direction on both particles; in a light inextensible string, tension pulls each body toward the string.
在连接体问题中，学生经常错误地应用牛顿第二定律，为每个质点单独列方程时没有正确处理张力。一个常见的疏忽是把张力当作对两个质点都沿同一方向作用；在轻质不可伸长的绳中，张力把每个物体拉向绳子方向。
When treating the whole system, the internal tension cancels out. However, when isolating a single particle, the tension must be included with the correct direction. In pulleys, if one particle is accelerating upward, the tension pulling it up is greater than its weight, while for the descending particle the weight exceeds the tension. Writing the equation of motion with consistent acceleration signs is crucial.
当考虑到整个系统时，内部的张力会相互抵消。但单独分析一个质点时，张力必须以正确方向纳入。在滑轮问题中，如果一个质点加速向上，则向上拉它的张力大于其重力，而下落的质点重力大于张力。用一致的加速度符号写出运动方程至关重要。
4. Conservation of momentum: ignoring sign for direction | 动量守恒：忽略方向的符号
Collision problems in the January 2020 paper required careful handling of velocity signs. Many scripts showed the equation total momentum before = total momentum after, but failed to make velocities in opposite directions have opposite signs. Writing every velocity to the right as positive and those to the left as negative is essential.
2020年1月试卷中的碰撞问题要求仔细处理速度符号。很多答卷写下了总动量前=总动量后的方程，但没有让相反方向的速度带上相反的符号。必须规定向右的速度为正，向左的为负。
For a particle of mass m₁ moving at 4 m s⁻¹ to the right colliding with m₂ at rest, the total momentum is 4m₁. If after impact m₁ rebounds at 1 m s⁻¹, its velocity is -1, not 1. The sign error leads to an incorrect mass ratio or speed. A momentum table helps: write each mass, initial velocity with sign, final velocity with sign, then sum the products.
质量为m₁的质点以4 m s⁻¹向右运动，与静止的m₂发生碰撞，总动量为4m₁。如果碰后m₁以1 m s⁻¹弹回，其速度应为-1，而不是1。符号错误会导致质量比或速度计算错误。可以借助动量表：列出每个质量、带符号的初速度、带符号的末速度，然后求和。
5. Impulse-momentum equation with vectors | 矢量冲量-动量方程
When impulses are given in vector form, such as (3i – 2j) N s, students often forget to multiply the final velocity vector by the mass or subtract initial momentum vector correctly. The formula I = mv – mu is a vector equation; every term must be expressed in i and j components.
当冲量以矢量形式给出，例如(3i – 2j) N s，学生经常忘记用质量乘上末速度矢量，或错误地减去初动量矢量。公式I = mv – mu是矢量方程；每一项都必须用i和j分量表示。
Set up two separate scalar equations from the i and j components. Write the initial velocity vector, the final velocity vector, then equate the impulse vector to m(v – u). A common mistake is using speed instead of velocity, losing the direction information. Always keep the i and j components paired until the final answer.
根据i和j分量列出两个独立的标量方程。写出初速度矢量、末速度矢量，然后令冲量矢量等于m(v – u)。一个常见错误是用速率代替速度，丢失了方向信息。在得出最终答案之前，始终将i和j分量成对处理。
6. Friction direction and inequality | 摩擦力方向与不等式
Determining the direction of friction is another major stumbling block. Friction opposes the relative motion or the tendency to move. In equilibrium problems, if the direction of an applied force is uncertain, students may guess the friction direction incorrectly, leading to a sign reversal in the equilibrium equations.
判断摩擦力的方向是另一个主要障碍。摩擦力与相对运动或运动趋势方向相反。在平衡问题中，如果作用力的方向不确定，学生可能猜错摩擦方向，导致平衡方程中符号反转。
When using F ≤ μR, remember that static friction is a range, not a fixed value. In limiting equilibrium, F = μR; otherwise, the friction is less than μR. Many candidates mistakenly set F = μR without checking whether the system is about to slip. Always state the assumption clearly before substitution.
在使用F ≤ μR时，记住静摩擦力是一个范围，而不是固定值。在极限平衡状态下，F = μR；否则摩擦力小于μR。很多考生在没有判断系统是否即将滑动的情况下就将F设为μR。代入前一定要清楚地陈述假设。
7. Velocity-time graph: displacement vs distance | 速度-时间图：位移与路程
Interpreting velocity-time graphs caused confusion. The area under the graph gives displacement, but if the graph dips below the time axis, that area is negative. To find the total distance travelled, you must sum the absolute values of the areas above and below the axis.
对速度-时间图的解读引起了混淆。图线下的面积表示位移，但如果图线落到时间轴下方，那块面积为负值。要求总路程，必须把轴上方面积和下方面积的绝对值加起来。
Many students simply integrated the function without considering where the velocity changes sign. When a particle reverses direction, the integral of velocity from start to finish gives the net change in position, not the total path length. Splitting the time interval at the roots of v(t) is necessary for distance.
许多学生只是对函数进行积分而没有考虑速度在何处变号。当质点反向运动时，从起点到终点的速度积分给出的是位置的净变化，而非运动路径总长。要计算路程，必须在v(t)的零点处分割时间区间。
8. Unit conversion fails (km/h to m/s) | 单位换算失败
A surprisingly high number of candidates forgot to convert kilometres per hour to metres per second before using SUVAT. The conversion factor is dividing by 3.6: 72 km/h = 20 m/s. Leaving speeds in km/h while using time in seconds produces nonsense accelerations.
出人意料的是，很多考生在使用SUVAT之前忘记将千米每小时换算为米每秒。换算因子是除以3.6：72 km/h = 20 m/s。保留km/h的速度而时间用秒，会得出毫无意义的加速度。
It is good practice to write all quantities in SI units at the start: metres, seconds, m s⁻¹, m s⁻². In a braking distance problem, if the initial speed is 90 km/h, immediately convert to 25 m s⁻¹. A quick unit check before calculation can prevent losing several marks in an otherwise straightforward question.
好的做法是开始就把所有量换成国际单位：米、秒、m s⁻¹、m s⁻²。在刹车距离问题中，若初速为90 km/h，应立刻转换为25 m s⁻¹。计算前快速检查单位可以避免在一道原本简单的题目中丢掉好几分。
9. Equilibrium vs resultant force | 平衡与合力
In static equilibrium questions, the resultant force is zero. Students often write equations of motion as if the particle were accelerating. For a particle on a smooth inclined plane held by a horizontal force, resolving parallel and perpendicular to the plane correctly yields two simultaneous equations that can be solved for the unknown force and reaction.
在静力平衡问题中，合力为零。学生常常写出好像质点正在加速的运动方程。对于被水平力维持在光滑斜面上的质点，正确沿斜面和垂直斜面分解可以得到两个联立方程，求解未知力和反力。
The mistake is to include an ma term when the particle is stationary or moving with constant velocity. Always check the wording: “rests in equilibrium”, “moving at constant speed”, or “on the point of sliding”. Only when there is acceleration or deceleration should ma appear in Newton’s second law.
错误在于质点静止或匀速运动时加入了ma项。务必检查题目措辞：”静止平衡”，”以恒定速度运动”或”即将滑动”。只有在有加速度或减速时，牛顿第二定律中才该出现ma。
10. Using constant acceleration formulas for variable acceleration | 对变加速运动使用匀加速公式
A common serious error is automatically reaching for the SUVAT equations when the acceleration is given as a function of time or displacement. SUVAT equations only apply when acceleration is constant. If a = 3t – 2, you must use calculus: integrate acceleration to find velocity, and integrate velocity to find displacement.
一个普遍且严重的错误是，当加速度表示为时间或位移的函数时，不假思索地套用SUVAT方程。SUVAT方程仅适用于加速度恒定的情况。若a = 3t – 2，则必须用微积分：对加速度积分求速度，再对速度积分求位移。
The January 2020 paper tested this distinction with a variable acceleration context. Candidates who applied v = u + at received no credit. Always inspect the wording: “constant acceleration” versus “acceleration at time t seconds is …”. The presence of t in the expression for a signals calculus is required.
2020年1月试卷通过一个变加速情境考查了这一区别。套用v = u + at的考生没有得分。要仔细分辨表述：”匀加速度”与”在t秒时的加速度为…”。若在a的表达式中出现了t，就意味着需要微积分。
11. Misreading vector notation in kinematics | 运动学中矢量符号的误读
Position vectors, velocity vectors, and constant acceleration vectors often use i and j notation. Students who wrote the magnitude of a vector when a direction was required, or mixed up position and velocity, lost marks. If the position vector is r = (3t² – 2)i + (5t)j, then velocity v = dr/dt = (6t)i + 5j, not 6t + 5.
位置矢量、速度矢量和匀加速度矢量常用i和j表示。需要给出方向时却只给出矢量模长的学生，或混淆位置与速度的学生，都丢了分。若位置矢量为r = (3t² – 2)i + (5t)j，则速度v = dr/dt = (6t)i + 5j，而不是6t + 5。
The bearing or angle of the velocity relative to i can be found using tan θ = (j-component)/(i-component). Many simply gave the j-component as the angle, which is conceptually wrong. Practice differentiating and integrating vectors term by term, always preserving the i and j directions.
速度相对于i的方向角可用tan θ = (j分量)/(i分量)求得。许多学生直接把j分量当成角度，这在概念上是错误的。要练习对矢量逐项微分和积分，始终保留i和j的方向。
12. Overlooking the difference between a rod and a string | 忽略杆和绳的区别
In moments or connected particle problems involving rods and strings, some candidates treated a light rod as if it could only carry tension, like a string. A rod can experience both tension and thrust (compression). If an object is thrusting outward against a rod, the force is a thrust pushing away from the rod, not a tension pulling toward it. Confusing these led to reversed force arrows and incorrect moment equations.
在涉及杆和绳的力矩或连接体问题中，一些考生将轻杆视为只能承受张力，像绳子一样。杆既能承受拉力(张力)，也能承受推力(压缩)。如果一个物体向外推杆，该力为背离杆的推力，而不是朝向杆的拉力。混淆这两者导致力箭头画反和力矩方程错误。
When a uniform rod is in equilibrium, the reaction at a pivot or hinge is not necessarily vertical. Resolving horizontally and vertically, and taking moments about a point, catches the correct direction. Never assume a hinge reaction acts along the rod.
当一根均匀杆处于平衡时，铰链或转轴处的反力不一定是竖直的。通过水平、竖直分解，并对一点取矩，就能得到正确的方向。切莫假设铰链反力沿杆方向。
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