📚 A-Level WJEC Maths: Mechanics Revision Guide | A-Level WJEC 数学：力学考点精讲
Mechanics is a core component of the WJEC A-Level Mathematics specification, requiring students to model real-world situations using mathematical principles. From kinematics and forces to energy and momentum, this guide breaks down the essential topics, key formulas, and common exam pitfalls to help you build confidence and accuracy in Mechanics. Each section is presented with clear explanations, worked examples, and crucial tips for success in both the AS and A2 units.
力学是 WJEC A-Level 数学大纲的核心组成部分，要求学生运用数学原理对现实情境进行建模。从运动学和力，到能量和动量，本指南梳理了必考知识点、关键公式和常见考试陷阱，帮助你在力学部分建立信心，提高解题准确率。每个章节都配有清晰的解释、典型示例，以及针对 AS 和 A2 单元的重要应试技巧。
1. Kinematics and SUVAT Equations | 运动学与 SUVAT 方程
Kinematics describes motion without considering its causes. In WJEC Mechanics, constant acceleration (uniform acceleration) problems are solved using the SUVAT equations. The five standard formulas, which link displacement s, initial velocity u, final velocity v, acceleration a, and time t, are essential tools.
运动学描述物体的运动而不考虑引起运动的原因。在 WJEC 力学中，匀加速运动问题使用 SUVAT 方程求解。这五个标准公式联系了位移 s、初速度 u、末速度 v、加速度 a 和时间 t，是必备工具。
v = u + at s = ut + ½at² v² = u² + 2as s = ½(u + v)t s = vt − ½at²
The choice of which equation to use depends on the known and unknown quantities. Always define a positive direction before substituting values. For vertical motion under gravity, a is usually taken as ±9.8 m s⁻², with signs carefully assigned. Many candidates lose marks by neglecting direction when dealing with upward and downward movements.
选择哪个方程取决于已知量和未知量。代入数值前务必先规定正方向。对于重力作用下的竖直运动，加速度 a 通常取 ±9.8 m s⁻²，并需谨慎规定符号。许多考生在处理上抛和下落时因忽略方向而丢分。
When an object is moving freely under gravity, you may need to treat upward motion as positive (a = −9.8) and recognise that velocity at the highest point is zero. Graphs of displacement–time and velocity–time can also be analysed to find gradients and areas, reinforcing the link between the SUVAT equations and calculus.
物体只在重力下运动时，可将向上定为正方向（a = −9.8），并注意最高点速度为零。也可利用位移–时间图和速度–时间图分析斜率和面积，这加强了 SUVAT 方程与微积分的联系。
2. Newton’s Laws and Free-Body Diagrams | 牛顿定律与受力图
Dynamics explains how forces affect motion. Newton’s three laws are fundamental: the first law (constant velocity if no resultant force), the second law (F = ma), and the third law (action–reaction pairs). In WJEC Mechanics, you must apply F = ma in the direction of motion, using a clear free‑body diagram to identify all forces.
动力学解释力如何影响运动。牛顿三定律是基础：第一定律（若合力为零则速度恒定），第二定律（F = ma），第三定律（作用力与反作用力）。在 WJEC 力学中，必须使用清晰的受力图识别所有力，并沿运动方向应用 F = ma。
Common forces include weight (mg), normal reaction, tension, driving force, and resistive forces such as friction or air resistance. When multiple particles are connected, you should either treat the whole system or set up equations for each particle separately, using the fact that tension is the same on both sides of a light inextensible string passing over a smooth pulley.
常见力包括重力（mg）、法向反力、张力、驱动力以及摩擦力或空气阻力等阻碍力。当多个物体连接在一起时，既可对整体系统应用牛顿第二定律，也可对各物体单独列方程，利用轻绳绕过光滑滑轮时张力处处相等这一条件。
A typical exam question involves a block being pulled along a rough horizontal surface. You will need to resolve vertically to find the normal reaction, then use Ffriction = μR to find the frictional force, and finally apply F = ma horizontally. Always check that your sign convention is consistent.
典型的考题是粗糙水平面上拉物块。需先竖直方向分解求出法向反力，再用 Ffriction = μR 计算摩擦力，最后水平方向应用 F = ma。切记符号规定要一致。
3. Resolving Forces and Equilibrium | 力的分解与平衡
When a particle or rigid body is in equilibrium, the resultant force in any direction is zero and the total moment about any point is zero. For concurrent forces, we often resolve into perpendicular components (usually horizontal and vertical) and set the sum of components to zero in each direction.
当质点或刚体处于平衡状态时，任意方向上的合力为零，且对任意点的合力矩为零。对于共点力系，通常分解为互相垂直的分量（一般为水平和竖直两个方向），并令每个方向的分量之和为零。
On an inclined plane, it is most efficient to resolve parallel and perpendicular to the slope. The weight mg is replaced by components mg sin θ (down the slope) and mg cos θ (perpendicular to the slope). Careful choice of resolution axes can greatly simplify the algebra.
在斜面上，沿斜面方向和垂直于斜面方向分解效率最高。重力 mg 可分解为 mg sin θ（沿斜面向下）和 mg cos θ（垂直于斜面）。恰当选择坐标轴可以极大简化代数运算。
Equilibrium problems often combine resolution with taking moments. For a rod resting against a wall, the forces include weight, normal reactions, and friction. Writing two perpendicular resolution equations plus one moment equation is the standard method; do not forget that friction always acts to oppose relative motion.
平衡问题常将分解和取矩结合起来。对于靠墙的杆，力包括重力、法向反力和摩擦力。标准方法为两个垂直方向的分解方程加上一个力矩方程；切记摩擦力总是与相对运动趋势方向相反。
4. Friction | 摩擦力
Friction is a resistive force that opposes motion or the tendency to move. The WJEC specification requires you to distinguish between limiting equilibrium (static friction) and dynamic friction. The law Fmax = μR applies when the object is on the point of sliding, but when sliding occurs, friction is taken as F = μR with a constant coefficient μ.
摩擦力是一种阻碍运动或运动趋势的阻力。WJEC 大纲要求区分极限平衡（静摩擦）与动摩擦。当物体即将滑动时满足 Fmax = μR；而滑动一旦发生，摩擦力按 F = μR 计算，μ 为常数。
On a rough inclined plane, the condition for a block to remain at rest is that the angle of inclination θ is less than or equal to the angle of friction λ, where μ = tan λ. This concept appears regularly in questions about a block on a slope that is gradually tilted.
在粗糙斜面上，物块保持静止的条件是倾角 θ 不大于摩擦角 λ，满足 μ = tan λ。这一概念经常出现在逐渐倾斜斜面的题目中。
Always calculate the normal reaction correctly before applying the friction law. When a horizontal force acts on a particle on a rough plane, the normal reaction is not simply mg but is modified by the vertical component of the applied force. Missing this modification is one of the most common errors.
应用摩擦定律前务必先正确计算出法向反力。当一个水平力作用在粗糙平面上的质点时，法向反力并非简单地等于 mg，而是会受到外力竖直分量的影响。忽略这一调整是最常见的错误之一。
5. Moments | 力矩
The moment of a force about a point is the product of the force and the perpendicular distance from the point to the line of action of the force. In WJEC Moments questions, you must be able to calculate clockwise and anticlockwise moments and use the principle of moments: for a system in equilibrium, total clockwise moments equal total anticlockwise moments.
力对某点的力矩等于力的大小乘以该点到力作用线的垂直距离。在 WJEC 力矩题中，需要能够计算顺时针力矩和逆时针力矩，并运用力矩原理：平衡系统下，总顺时针力矩等于总逆时针力矩。
Many problems involve a uniform rod, where weight acts at the centre. Tension, reactions, and applied forces can all produce moments. The key is to select a pivot that eliminates unknown forces from the moment equation, typically choosing a point through which a reaction or unknown force passes.
许多问题涉及均质杆，其重力作用在杆的中点。张力、反力和外加力都能产生力矩。关键技巧是选择一个可以消去未知力的支点来列力矩方程，通常选择某个反力或未知力通过的点。
When a rod is on the point of tilting about a pivot, the reaction at any other support becomes zero. This limiting condition allows you to find the maximum load that can be placed before tilting. Combined moments and resolution problems, such as a ladder resting against a wall, are a staple of WJEC exams.
当杆即将绕某个支点转动时，其他支撑处的反力变为零。利用这一极限条件可求出在即将倾斜前能承载的最大负荷。力矩与分解相结合的问题，例如靠墙的梯子，是 WJEC 考试中的必考内容。
6. Vectors in Mechanics | 力学中的向量
Mechanical quantities such as displacement, velocity, acceleration, and force are vectors and can be expressed in component form using i and j unit vectors. For instance, a position vector might be given as r = (2t³ i + 5t j) m, from which velocity is obtained by differentiation.
位移、速度、加速度和力等力学量都是向量，可以用 i 和 j 单位向量以分量形式表示。例如，位置向量可能为 r = (2t³ i + 5t j) m，对其求导可得速度。
The magnitude of a vector ai + bj is √(a² + b²), and its direction can be given as the angle from the positive i-direction, found using θ = tan⁻¹(b/a). Vector notation simplifies many problems, especially those involving relative motion or motion in two dimensions, such as projectiles.
向量 ai + bj 的模为 √(a² + b²)，方向可用与正 i 方向的夹角表示，由 θ = tan⁻¹(b/a) 求得。向量记号能简化许多问题，尤其是相对运动或二维运动（如抛体运动）问题。
When using vectors along with calculus, remember that differentiating position gives velocity, and differentiating velocity gives acceleration; integration performs the reverse operation. Be prepared to find the time at which two moving objects are parallel, perpendicular, or closest to each other by analysing their velocity vectors.
将向量与微积分结合使用时，记住位置求导得速度，速度求导得加速度；积分则进行逆运算。学会通过分析速度向量求解两个运动物体平行、垂直或距离最近的时刻是常见考点。
7. Projectile Motion | 抛体运动
Projectile motion is analysed by treating the horizontal and vertical components separately. Horizontally, velocity is constant (assuming no air resistance); vertically, the motion is uniformly accelerated with acceleration −g (or +g depending on sign convention). The initial velocity U at angle θ to the horizontal is resolved into U cos θ (horizontal) and U sin θ (vertical).
抛体运动通过将水平和竖直分量分开来处理。水平方向速度恒定（假设无空气阻力）；竖直方向为加速度为 −g（或 +g，取决于符号规定）的匀加速运动。初速度 U 与水平方向夹角为 θ，分解为 U cos θ（水平）和 U sin θ（竖直）。
The key equations for a projectile launched from ground level are: time of flight = (2U sin θ)/g, maximum height = (U² sin² θ)/(2g), and horizontal range = (U² sin 2θ)/g. These derived formulas can save time, but you must be able to derive them using the SUVAT equations when the launch or landing height differs.
从地面发射的抛体运动关键公式为：飞行时间 = (2U sin θ)/g，最大高度 = (U² sin² θ)/(2g)，水平射程 = (U² sin 2θ)/g。记住这些导出公式可以节省时间，但当起抛点与落地点高度不同时，必须学会用 SUVAT 方程自行推导。
In WJEC questions, you may need to find the velocity (speed and direction) at a given time, or the position vector of the particle. Vector methods are particularly neat for projectile problems; writing the position as r = (U cos θ) t i + [(U sin θ) t − ½gt²] j allows you to find the height for any horizontal distance.
在 WJEC 考题中，可能需要求某个时刻的速度（大小和方向）或位置向量。用向量方法处理抛体问题特别简洁；位置写为 r = (U cos θ) t i + [(U sin θ) t − ½gt²] j，即可求出任意水平距离对应的高度。
8. Work, Energy and Power | 功、能与功率
The work done by a constant force is Work = F d cos θ, where d is the displacement and θ the angle between force and displacement. In Mechanics problems, work is often calculated when a force moves its point of application along a line. The work–energy principle states that the total work done by external forces equals the change in kinetic energy.
恒力所做的功为 Work = F d cos θ，其中 d 为位移，θ 为力与位移的夹角。在力学问题中，常需计算力沿直线移动其作用点时所做的功。功能原理指出，外力所作的总功等于动能的变化量。
Kinetic energy (KE) is given by ½mv², and gravitational potential energy (GPE) by mgh. When only gravity and normal forces are acting (no friction), mechanical energy is conserved: initial KE + initial GPE = final KE + final GPE. With friction, the work done against friction is equal to the energy dissipated.
动能（KE）由 ½mv² 给出，重力势能（GPE）为 mgh。当只有重力和法向力作用（无摩擦）时，机械能守恒：初动能 + 初重力势能 = 末动能 + 末重力势能。若有摩擦力，克服摩擦所做的功等于耗散的能量。
Power is the rate of doing work and is defined as P = Fv for a force moving its point at velocity v in the direction of the force. Typical exam questions ask for the maximum speed of a car given its engine power and the total resistance, or the power required to maintain a constant speed up a slope.
功率是做功的速率，当力以速度 v 沿力的方向移动其作用点时，功率定义为 P = Fv。常见考题包括根据发动机功率和总阻力求汽车的最大速度，或维持上坡匀速所需的功率。
9. Momentum and Impulse | 动量与冲量
Momentum is a vector quantity given by p = mv. The impulse of a constant force F acting for time t is Ft, and it equals the change in momentum: Ft = mv − mu. This impulse–momentum theorem is especially useful for collisions and sudden changes in motion.
动量是向量，由 p = mv 给出。恒力 F 作用时间 t 的冲量为 Ft，它等于动量的变化量：Ft = mv − mu。这一冲量–动量定理对碰撞和运动突变问题特别有用。
In the absence of external forces, the total momentum of a system is conserved. For two particles colliding, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. WJEC often combines conservation of momentum with Newton’s experimental law, using the coefficient of restitution e, where v₂ − v₁ = e(u₁ − u₂) for a direct impact.
在没有外力的情况下，系统的总动量守恒。两质点碰撞时，m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂。WJEC 常将动量守恒与牛顿实验定律结合，引入恢复系数 e，对于正碰有 v₂ − v₁ = e(u₁ − u₂)。
You must be careful with sign conventions, especially in two‑dimensional collisions. Using vector form for impulse and momentum can help manage multiple directions. Impulse can also be found from the area under a force–time graph when the force varies.
务必注意符号规定，尤其在二维碰撞问题中。冲量和动量采用向量形式有助于处理多方向问题。当力变化时，冲量也可由力–时间图下的面积求出。
10. Variable Acceleration and Calculus | 变加速运动与微积分
When acceleration is not constant, the SUVAT equations no longer apply. Instead, we use calculus: velocity is the derivative of displacement with respect to time, v = ds/dt; acceleration is the derivative of velocity, a = dv/dt = d²s/dt². Conversely, velocity is the integral of acceleration, and displacement is the integral of velocity.
当加速度不恒定时，SUVAT 方程不再适用。此时需使用微积分：速度是位移对时间的导数，v = ds/dt；加速度是速度的导数，a = dv/dt = d²s/dt²。反之，速度是加速度的积分，位移是速度的积分。
WJEC exam questions may provide acceleration as a function of time, a(t), or as a function of displacement, a(x). In the latter case, the relation a = v dv/dx is indispensable. This allows you to link velocity and displacement directly, which is particularly useful in oscillation problems or when finding maximum speed.
WJEC 考题可能给出加速度作为时间的函数 a(t)，或作为位移的函数 a(x)。后一种情况下，关系式 a = v dv/dx 不可或缺。这使得速度和位移直接联系起来，在振动问题或求最大速度时尤其有用。
Always find the constant of integration using initial conditions. Questions often ask for the greatest height reached by a particle moving against gravity with variable air resistance; you will need to integrate and set v = 0 to find the maximum displacement. Differentiating and integrating velocity vectors is the vector equivalent for two-dimensional problems.
始终利用初始条件求出积分常数。题目常要求计算带有可变空气阻力时质点竖直上抛的最大高度；此时需积分并令 v = 0 求出最大位移。对于二维问题，对速度向量求导和积分是向量形式的等效操作。
11. Connected Particles and Pulleys | 连接体与滑轮系统
Connected particle problems, commonly featuring smooth pulleys and light inextensible strings, are a frequent WJEC topic. The fundamental assumptions are that tension is uniform along the string, and the magnitudes of acceleration of connected particles are equal (for a string that does not break). Set up equations of motion for each mass using F = ma, with careful attention to the direction of acceleration.
连接体问题，通常涉及光滑滑轮和轻质不可伸长绳，是 WJEC 的高频考点。基本假设是绳上张力处处相等，且相连各物体的加速度大小相等（对于未断开的绳子）。对每个质量分别按 F = ma 建立运动方程，并仔细注意加速度方向。
For a simple system of two particles connected by a string over a smooth pulley, let m₁ > m₂ so the system accelerates. Write one equation for the heavier mass (weight − tension = m₁a) and one for the lighter mass (tension − weight = m₂a). Solving simultaneously yields a = (m₁ − m₂)g / (m₁ + m₂) and the tension T = (2m₁m₂ g) / (m₁ + m₂). Understanding the derivation is more valuable than memorising the result.
对于两个质点通过轻绳跨过光滑滑轮的简单系统，设 m₁ > m₂，则系统加速运动。对较重质点列方程（重力 − 张力 = m₁a），对较轻质点列方程（张力 − 重力 = m₂a）。联立解得 a = (m₁ − m₂)g / (m₁ + m₂) 和张力 T = (2m₁m₂ g) / (m₁ + m₂)。理解推导过程比死记结果更有价值。
When a particle rests on a table and is attached by a string passing over a pulley at the edge to a freely hanging mass, you must include the resistance, such as friction, acting on the table particle. Resolve vertically to find the normal reaction, then use F = μR for friction. Correct application of Newton’s second law to both blocks simultaneously, or to the whole system, will yield the acceleration.
当一质点置于桌面，用绕过桌边滑轮的绳子与自由悬挂的物体相连时，必须计及桌面质点所受的阻力，如摩擦力。竖直分解求出法向反力，再用 F = μR 计算摩擦。同时对两物块或对整体系统正确应用牛顿第二定律即可求得加速度。
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