📚 IGCSE WJEC Maths: Mechanics Essentials | IGCSE WJEC 数学：力学 考点精讲
Mechanics in the IGCSE WJEC Mathematics specification focuses on modelling the motion of objects and the forces that cause this motion. This revision guide covers the key concepts you must master, from kinematics and Newton’s laws to momentum and energy. Each section pairs concise English explanations with their Chinese equivalents, ensuring you build a bilingual understanding while tackling typical WJEC-style problems.
在 IGCSE WJEC 数学大纲中，力学部分侧重于对物体运动及其受力进行建模。本考点精讲涵盖了从运动学、牛顿定律到动量与能量等必须掌握的核心概念。每一节都提供了简明的中英文对照解释，帮助你在构建双语思维的同时，攻克 WJEC 风格的典型考题。
1. Kinematics: Displacement, Velocity and Acceleration | 运动学：位移、速度和加速度
Displacement (s) is a vector quantity describing the change in position of an object in a given direction. Velocity (v) is the rate of change of displacement, also a vector. Acceleration (a) is the rate of change of velocity, measured in m/s2. In contrast, distance and speed are scalar quantities and lack directional information.
位移（s）是一个矢量，描述物体在给定方向上的位置变化。速度（v）是位移的变化率，同样为矢量。加速度（a）是速度的变化率，单位为 m/s2。与之相对，距离和速率是标量，不包含方向信息。
When acceleration is constant, we can use the five fundamental quantities: u (initial velocity), v (final velocity), a (acceleration), t (time), and s (displacement). The four SUVAT equations link these. Always ensure that the signs you assign to vectors are consistent with a chosen positive direction, especially when dealing with opposing forces or directions of motion.
当加速度恒定时，我们可以使用五个基本量：u（初速度）、v（末速度）、a（加速度）、t（时间）和 s（位移）。这四个 SUVAT 方程将它们联系起来。务必确保为矢量赋予的正负号与你选定的正方向一致，尤其是在处理相反的作用力或运动方向时。
| Quantity | Type | Unit |
|---|---|---|
| Displacement (s) | Vector | m |
| Velocity (u, v) | Vector | m/s |
| Acceleration (a) | Vector | m/s2 |
| Time (t) | Scalar | s |
2. SUVAT Equations for Constant Acceleration | 匀加速运动的 SUVAT 方程
The four equations of motion for constant acceleration are derived from the definitions of velocity and acceleration. They must only be applied when the acceleration is uniform. The equations are:
这四个匀加速运动方程由速度和加速度的定义推导而来。它们仅在加速度恒定时才能使用。方程如下：
v = u + at
s = ut + ½at2
v2 = u2 + 2as
s = ½(u + v)t
To solve a problem, list the known quantities and the one you need to find. Choose the equation that includes all the knowns and the unknown. For example, if a train accelerates from 10 m/s to 25 m/s over 150 m, you would use v2 = u2 + 2as to find a.
解题时，列出已知量和待求量，选择包含所有已知量和未知量的方程。例如，一列火车从 10 m/s 加速到 25 m/s 通过 150 m，此时应使用 v2 = u2 + 2as 求加速度 a。
Remember: if an object is slowing down, acceleration is negative relative to the direction of motion. Always state the positive direction clearly before substituting values. The unit for displacement may be km in some questions, so convert to metres if necessary.
记住：若物体在减速，则加速度相对于运动方向为负值。务必在代入数值前明确指定正方向。有些题目中位移单位可能为 km，必要时需转换为米。
3. Interpreting Velocity-Time Graphs | 速度-时间图解读
A velocity-time (v-t) graph provides rich information about motion. The gradient of the line represents acceleration; a horizontal line implies constant velocity (zero acceleration). The area between the graph and the time axis gives the displacement, while areas below the axis indicate negative displacement.
速度-时间（v-t）图提供了有关运动的丰富信息。图线的斜率代表加速度；水平线意味着速度恒定（加速度为零）。图线与时间轴之间的面积表示位移，而时间轴以下的面积表示负位移。
When the motion has multiple stages, compute the displacement for each region separately. For a non-uniform acceleration, the area must be estimated, but WJEC questions typically feature straight-line segments, making area calculations straightforward. Always check whether the question asks for displacement or total distance travelled — total distance is the sum of the absolute values of all areas.
当运动包含多个阶段时，需要分别计算每个区域的位移。对于非匀加速运动，面积需要估算，但 WJEC 考题通常由直线段构成，因此面积计算相对简单。一定要辨认题目要求的是位移还是总路程——总路程是所有面积的绝对值之和。
4. Newton’s First and Second Laws | 牛顿第一和第二定律
Newton’s First Law states that an object remains at rest or in uniform motion in a straight line unless acted upon by a resultant external force. This is the principle of inertia. Newton’s Second Law relates the resultant force F, mass m, and acceleration a: F = ma.
牛顿第一定律指出，除非受到合外力作用，否则物体将保持静止或匀速直线运动状态。这就是惯性原理。牛顿第二定律将合外力 F、质量 m 和加速度 a 联系起来：F = ma。
The resultant force is the vector sum of all forces acting on a particle. When several forces act, you must resolve them into components or combine parallel forces to find the net force. The acceleration produced is always in the direction of the resultant force. Units: force is measured in newtons (N), where 1 N = 1 kg m/s2.
合外力是作用在质点上所有力的矢量和。当多个力同时作用时，必须将其分解为分量，或合并平行力以求得净力。产生的加速度方向始终与合外力方向一致。单位：力以牛顿（N）为单位，1 N = 1 kg m/s2。
5. Free-Body Diagrams and Equilibrium | 受力图与平衡
A free-body diagram isolates a particle and shows all the forces acting on it with labelled arrows. Common forces include weight (mg, acting downwards), normal reaction (perpendicular to the surface), tension (along a string or rod), and friction (parallel to the surface, opposing motion or tendency).
受力图将质点隔离出来，用标记好的箭头表示作用在其上的所有力。常见的力包括：重力（mg，竖直向下）、法向反作用力（垂直于表面）、拉力（沿绳或杆的方向）和摩擦力（平行于表面，与运动或运动趋势方向相反）。
A particle is in equilibrium if the resultant force in any direction is zero. This means the vector sum of all forces is zero. For two-dimensional equilibrium, you can check that the sum of horizontal components is zero and the sum of vertical components is zero. Solving equilibrium problems often involves setting up two independent equations.
如果质点在任意方向上的合外力为零，则它处于平衡状态。这意味着所有力的矢量和为零。对于二维平衡问题，可以检查水平分量的代数和为零，且竖直分量的代数和为零。求解平衡问题通常需要建立两个独立的方程。
6. Resolving Forces into Components | 力的分解
When a force acts at an angle to the coordinate axes, we resolve it into perpendicular components. If a force F makes an angle θ with the horizontal, its horizontal component is F cosθ and its vertical component is F sinθ. This technique is essential for analysing inclined planes and connected particles.
当一个力与坐标轴成一定角度时，我们需要将其分解为互相垂直的分量。若力 F 与水平方向的夹角为 θ，则它的水平分量为 F cosθ，竖直分量为 F sinθ。该技巧在分析斜面问题和连接体问题时至关重要。
For a particle on a smooth inclined plane, the weight is resolved parallel and perpendicular to the slope. The parallel component (mg sinθ) causes acceleration down the slope if friction is negligible. The perpendicular component (mg cosθ) is balanced by the normal reaction R.
对于光滑斜面上的质点，重力被分解为平行和垂直于斜面的两个分量。若摩擦力可忽略，则平行分量（mg sinθ）使质点沿斜面向下加速。垂直分量（mg cosθ）由法向反力 R 平衡。
R = mg cosθ
Resultant force down slope = mg sinθ
7. Friction | 摩擦力
Friction is a resistive force that opposes the motion or potential motion between two surfaces. The maximum static friction is given by Fmax = μR, where μ is the coefficient of friction and R is the normal reaction. For a particle at rest, the actual friction can be less than or equal to Fmax.
摩擦力是一种阻碍两个表面之间相对运动或相对运动趋势的阻力。最大静摩擦力由 Fmax = μR 给出，其中 μ 是摩擦系数，R 是法向反力。对于静止的质点，实际摩擦力可以小于或等于 Fmax。
When an object is sliding, the kinetic friction is constant and equals μkR, where μk is typically slightly smaller than the static coefficient. In many WJEC problems, the friction is described as ‘limiting’, meaning the object is on the point of moving and friction has just reached its maximum value.
当物体滑动时，动摩擦力恒定，等于 μkR，其中 μk 通常略小于静摩擦系数。在许多 WJEC 考题中，摩擦力被描述为“极限”状态，即物体即将发生运动，摩擦力恰好达到最大值。
8. Inclined Planes | 斜面问题
An inclined plane problem combines resolving forces with friction and Newton’s Second Law. First draw a clear diagram and mark the angle of inclination. Resolve the weight into components parallel (mg sinθ) and perpendicular (mg cosθ) to the plane. Use R = mg cosθ to find the normal reaction, then compute friction if needed.
斜面问题综合了力的分解、摩擦力和牛顿第二定律。首先画出示意图，标出倾斜角。将重力分解为平行于斜面的分量（mg sinθ）和垂直于斜面的分量（mg cosθ）。利用 R = mg cosθ 求出法向反力，然后根据需要计算摩擦力。
Apply F = ma along the plane. If the particle is accelerating downwards, the resultant force is mg sinθ – Ffriction. If it is being pulled upwards by a string, include the tension T in the equation. Always check for equilibrium conditions when asked whether the particle will remain at rest.
沿斜面应用 F = ma。若质点加速下滑，则合外力为 mg sinθ – F摩擦。若被绳向上拉，则在方程中计入拉力 T。当题目问及质点是否会保持静止时，记得检查平衡条件。
9. Connected Particles and Pulleys | 连接体与滑轮
In pulley problems, two particles are connected by a light inextensible string passing over a smooth pulley. Because the string is inextensible, both particles have the same magnitude of acceleration. The tension T is uniform throughout the string. Treat each particle separately, drawing free-body diagrams and applying F = ma.
在滑轮问题中，两个质点通过一根跨过光滑滑轮的轻质不可伸长绳相连。由于绳不可伸长，两质点的加速度大小相同。绳上各处张力 T 相等。分别处理每个质点，画出受力图并应用 F=ma。
For the heavier mass m1 moving downwards, the equation is m1g – T = m1a. For the lighter mass m2 moving upwards, T – m2g = m2a. Solve simultaneously for a and T. If the particles start from rest, you can then use SUVAT to find speed or distance travelled.
对于向下运动的较重质量 m1，方程为 m1g – T = m1a。对于向上运动的较轻质量 m2，方程为 T – m2g = m2a。联立求解 a 和 T。若质点从静止出发，则可进一步使用 SUVAT 方程求解速度或运动距离。
10. Momentum and Impulse | 动量与冲量
Momentum is defined as the product of mass and velocity: p = mv. It is a vector quantity with units kg m/s. The principle of conservation of momentum states that in the absence of external forces, the total momentum of a system remains constant. This is particularly useful in collision and explosion problems.
动量定义为质量与速度的乘积：p = mv。它是一个矢量，单位为 kg m/s。动量守恒定律指出，在没有外力作用的情况下，系统的总动量保持不变。这一定律在碰撞和爆炸问题中尤为有用。
When two particles collide and coalesce, m1u1 + m2u2 = (m1 + m2)v. Impulse is the change in momentum caused by a force acting over time: Impulse = Ft = mv – mu. Impulse has the same units as momentum and can be found from the area under a force-time graph.
当两个质点碰撞并粘合在一起时，有 m1u1 + m2u2 = (m1 + m2)v。冲量是力在一段时间内作用所引起的动量变化：冲量 = Ft = mv – mu。冲量与动量单位相同，也可通过力-时间图下的面积求得。
11. Work, Energy and Power | 功、能和功率
Work done by a constant force is given by W = Fd cosθ, where d is the displacement in the direction of the force. When the force and displacement are parallel, W = Fd. The unit of work is the joule (J). Kinetic energy (KE) is ½mv2, and gravitational potential energy (GPE) is mgh.
恒力所做的功由 W = Fd cosθ 给出，其中 d 是力方向上的位移。当力与位移平行时，W = Fd。功的单位是焦耳（J）。动能（KE）为 ½mv2，重力势能（GPE）为 mgh。
The work-energy principle states that the net work done on a particle equals its change in kinetic energy: Wnet = ½mv2 – ½mu2. This often provides a simpler solution than using SUVAT and Newton’s laws, especially when forces vary along a slope. Power is the rate of doing work, measured in watts (W): P = W/t or P = Fv for constant speed.
功能原理指出，作用在质点上的净功等于其动能的变化量：W净 = ½mv2 – ½mu2。这往往比使用 SUVAT 和牛顿定律更简便，特别是当力沿斜面变化时。功率是做功的快慢，单位为瓦特（W）：P = W/t，或当物体匀速运动时 P = Fv。
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