📚 A-Level Mathematics: Mechanics Essentials | A-Level 数学：力学考点精讲
Mechanics in A-Level Mathematics brings physical intuition into mathematical modelling. You will learn to describe motion using kinematic equations, apply Newton’s laws to connected objects, and use concepts of energy and momentum to solve real-world problems. This article recaps the key topics you must master for the exam, with paired English and Chinese explanations for every point.
A-Level 数学中的力学将物理直觉引入数学建模。你将学会用运动学方程描述运动，将牛顿定律应用于连接物体，并使用能量和动量概念解决现实问题。本文梳理了考试必须掌握的核心考点，每个要点均提供中英双语对照讲解。
1. Constant Acceleration Formulae (SUVAT) | 匀加速运动公式 (SUVAT)
The five SUVAT equations link displacement (s), initial velocity (u), final velocity (v), acceleration (a) and time (t) for motion in a straight line with constant acceleration. You must be able to select the right equation based on the given and unknown variables.
五个运动学方程将匀变速直线运动中的位移 (s)、初速度 (u)、末速度 (v)、加速度 (a) 和时间 (t) 联系起来。你必须能够根据已知量和未知量选择合适的方程。
The equations are: v = u + at; s = ut + ½at²; s = ½(u+v)t; v² = u² + 2as; and s = vt − ½at². Remember that displacement, velocity and acceleration are vectors – signs indicate direction. Always define a positive direction first.
方程包括：v = u + at；s = ut + ½at²；s = ½(u+v)t；v² = u² + 2as；以及 s = vt − ½at²。请记住位移、速度和加速度均为矢量——符号代表方向。务必先规定正方向。
v = u + at
s = ½(u + v)t
v² = u² + 2as
Vertical motion under gravity uses a = ±9.8 m s⁻². For a particle projected upwards, velocity becomes zero at the highest point, and time of flight is twice the time to the peak.
重力作用下的竖直运动使用 a = ±9.8 m s⁻²。物体上抛时，最高点速度为零，总飞行时间是到达最高点时间的两倍。
2. Motion Graphs and Their Interpretations | 运动图像及其解读
Displacement–time, velocity–time and acceleration–time graphs give a visual summary of motion. The gradient of a displacement–time graph gives velocity; the gradient of a velocity–time graph gives acceleration. The area under a velocity–time graph gives displacement.
位移–时间图、速度–时间图和加速度–时间图是运动的直观总结。位移–时间图的斜率表示速度；速度–时间图的斜率表示加速度。速度–时间图下的面积表示位移。
A straight line on a velocity–time graph implies constant acceleration. A horizontal line means constant velocity (zero acceleration). The area between the curve and the time axis can be found using integration when acceleration is not constant.
速度–时间图上的直线表示匀加速。水平线表示匀速（加速度为零）。当加速度不恒定时，可通过积分求曲线与时间轴之间的面积。
In exam questions, you may need to move between graphs, write equations from given graphs, or calculate total distance by summing absolute areas under a v–t graph.
考试题可能要求你在不同图像之间转换、根据图像写出方程，或通过对 v–t 图的面积取绝对值求和来计算总路程。
3. Newton’s Three Laws of Motion | 牛顿运动三定律
Newton’s First Law: An object remains at rest or moves with constant velocity unless acted upon by an external resultant force. This introduces the concept of inertia.
牛顿第一定律：物体将保持静止或匀速直线运动状态，除非受到外力的合力作用。由此引入惯性的概念。
Newton’s Second Law: F = ma. The resultant force on an object equals its mass multiplied by its acceleration. Force and acceleration are vectors in the same direction.
牛顿第二定律：F = ma。作用于物体的合力等于其质量乘以加速度。力与加速度是方向相同的矢量。
Newton’s Third Law: For every action, there is an equal and opposite reaction. These forces act on different bodies and are of the same type.
牛顿第三定律：每一个作用力都有一个大小相等、方向相反的反作用力。这两个力作用在不同物体上且属于同一类型。
Apply these laws to any mechanics problem by first drawing a clear force diagram, resolving forces along convenient directions, and writing F = ma for each particle.
应用这些定律时，首先要画出清晰的受力图，沿合适的方向分解力，并对每个质点写出 F = ma。
4. Resolving Forces and Equilibrium | 力的分解与平衡
When a particle is in equilibrium, the vector sum of all forces acting on it is zero. This means the horizontal components sum to zero and the vertical components sum to zero simultaneously.
当质点处于平衡状态时，作用在其上的所有力的矢量和为零。这意味着水平分力和竖直分力各自加起来均为零。
Resolve forces into two perpendicular directions, usually horizontal and vertical, or along an inclined plane and normal to it. Use sine and cosine to split a force F at angle θ: component along the direction = F cos θ; component perpendicular = F sin θ.
将力分解为两个互相垂直的方向，通常为水平和竖直方向，或沿斜面及其垂直方向。将一个与方向成 θ 角的力 F 分解：沿该方向的分量为 F cos θ，垂直分量为 F sin θ。
If three forces act at a point and the system is in equilibrium, the forces can be represented by a closed triangle. Lami’s theorem is a special case for three coplanar, concurrent forces in equilibrium.
若三个力作用在一点且系统平衡，则这三个力可构成一个闭合三角形。拉米定理是三力共面共点平衡的一种特殊情况。
5. Friction: Static and Kinetic | 摩擦力：静摩擦与动摩擦
Friction opposes relative motion between two surfaces in contact. Static friction acts when there is no movement, up to a maximum value Fmax = μsR, where μs is the coefficient of static friction and R is the normal reaction.
摩擦力阻碍接触面间的相对运动。静摩擦在没有相对运动时起作用，最大可达 Fmax = μsR，其中 μs 为静摩擦系数，R 为法向反力。
Once motion begins, kinetic (or dynamic) friction applies: F = μkR, where μk is the coefficient of kinetic friction, usually slightly less than μs.
一旦开始运动，便适用动摩擦：F = μkR，其中 μk 为动摩擦系数，通常略小于 μs。
On a rough inclined plane at angle θ, the condition for a particle about to slide down is tan θ = μ. For general motion up or down the slope, include friction opposite to the direction of motion and write F = ma parallel to the plane.
在粗糙斜面上，当质点即将下滑时，满足 tan θ = μ。对于沿斜面向上或向下的运动，摩擦力方向与运动方向相反，需沿斜面平行方向写出 F = ma。
6. Connected Particles and Pulleys | 连接体与滑轮模型
When two particles are connected by a light, inextensible string passing over a smooth pulley, their accelerations are equal in magnitude but may be in different directions. The tension in the string is the same on both sides of the pulley.
当两个质点用一根轻质且不可伸长的绳子跨过光滑滑轮连接时，两者的加速度大小相等但方向可能不同。滑轮两侧绳子的张力大小相等。
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Treat each particle separately: draw a force diagram and write F = ma.
分别处理每个质点：画受力图并写出 F = ma。
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For a particle on a horizontal table connected to a hanging particle, the table may be smooth (no friction) or rough (include friction).
对于水平桌面上的质点与悬挂质点相连的情形，桌面可能是光滑的（无摩擦）或粗糙的（需计入摩擦力）。
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If the string breaks or becomes slack, tension becomes zero and particles move under gravity and/or friction independently.
如果绳子断开或松弛，张力变为零，各质点将在重力和/或摩擦力作用下独立运动。
Lift problems: a person or object inside a lift experiences apparent weight = m(g ± a) depending on whether the lift accelerates upwards (+) or downwards (−).
电梯问题：电梯内的人或物体感受到的表现重量取决于电梯加速度方向：向上加速时为 m(g + a)，向下加速时为 m(g − a)。
7. Momentum and Impulse | 动量与冲量
Momentum is defined as p = mv, a vector quantity with unit kg m s⁻¹. Impulse equals the change in momentum: I = mv − mu, and also equals average force × time: I = Ft.
动量定义为 p = mv，是矢量，单位为 kg m s⁻¹。冲量等于动量的变化：I = mv − mu，也等于平均力乘以时间：I = Ft。
The principle of conservation of linear momentum applies when no external resultant force acts on a system. For a collision or explosion, total momentum before = total momentum after.
当系统不受外力作用时，动量守恒定律成立。对于碰撞或爆炸，系统碰撞前（或爆炸前）的总动量等于碰撞后（或爆炸后）的总动量。
Collisions may be elastic (kinetic energy conserved) or inelastic. In a perfectly inelastic collision, particles stick together. The coefficient of restitution e, which appears in Further Mathematics, is not required for standard A-Level Mechanics.
碰撞可以是弹性的（动能守恒）或非弹性的。在完全非弹性碰撞中，物体粘在一起。恢复系数 e 的知识出现在进阶数学中，标准 A-Level 力学不作要求。
Outcome:
结论：
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
8. Moments and Centre of Mass | 力矩与质心
The moment of a force about a point measures its turning effect: moment = force × perpendicular distance from the point. The SI unit is newton-metre (N m).
力对某点的力矩衡量其转动效应：力矩 = 力 × 力的作用线到该点的垂直距离。国际单位为牛顿·米 (N m)。
For a rigid body in equilibrium, the resultant force in any direction is zero and the sum of clockwise moments about any pivot equals the sum of anticlockwise moments.
对于刚体平衡，任意方向的合力为零，且对任意支点，顺时针力矩之和等于逆时针力矩之和。
The centre of mass is the point at which the entire mass of a body can be considered to act. For uniform rods and symmetrical laminas, it is at the geometric centre. Non-uniform objects may require calculation using discrete masses or integration.
质心是物体全部质量可视为集中的点。对于均匀杆和对称薄板，质心在几何中心。非均匀物体可能需要通过离散质量或积分进行计算。
Tilting and toppling conditions: If a body rests on a surface, it will topple if the line of action of the weight falls outside its base.
倾斜与倾倒条件：如果物体放置在表面上，当重力的作用线超出底座范围时，物体将倾倒。
9. Work, Energy and Power | 功、能与功率
Work done by a constant force is force × distance moved in the direction of the force. When the force is at an angle, use F d cos θ. Work is measured in joules (J).
恒力做的功等于力 × 沿力的方向移动的距离。当力与运动方向有一个夹角时，用 F d cos θ。功的单位是焦耳 (J)。
Gravitational potential energy: GPE = mgh, where h is vertical height. Kinetic energy: KE = ½mv². The work–energy principle states that total work done by all external forces equals the change in kinetic energy.
重力势能：GPE = mgh，其中 h 为竖直高度。动能：KE = ½mv²。功–能原理指出：所有外力做的总功等于动能的变化量。
Power is the rate of doing work: P = Fv for a constant force acting on a particle moving with velocity v. Power is measured in watts (W).
功率是做功的快慢：对于作用在速度为 v 的物体上的恒力，有 P = Fv。功率的单位是瓦特 (W)。
In problems involving vehicles moving on slopes, power output often equals driving force × velocity, with resistance forces opposing motion.
在涉及车辆在斜坡上运动的问题中，输出功率通常等于驱动力 × 速度，同时阻力阻碍运动。
10. Projectile Motion | 抛体运动
A projectile moves under constant vertical acceleration due to gravity and zero horizontal acceleration (assuming no air resistance). Treat horizontal and vertical motions independently.
抛体在竖直方向受恒定重力加速度影响，水平方向加速度为零（假设无空气阻力）。处理时需将水平与竖直运动分开考虑。
Horizontal: velocity u cos θ is constant, so displacement x = (u cos θ) t. Vertical: initial velocity u sin θ, acceleration −g; use SUVAT equations with signs.
水平方向：速度 u cos θ 恒定，因此位移 x = (u cos θ) t。竖直方向：初速度 u sin θ，加速度 −g；使用带有正负号的 SUVAT 方程。
Key results: maximum height H = (u² sin² θ) / (2g); range R = (u² sin 2θ) / g for a projectile on level ground. The equation of trajectory (parabola) can be derived by eliminating t.
关键结果：最大高度 H = (u² sin² θ) / (2g)；在水平地面上的射程 R = (u² sin 2θ) / g。通过消去 t 可得轨迹方程（抛物线）。
When the starting point is not at the same height as the landing point, use the full vertical equation with the final vertical displacement taken as −h or +h accordingly.
当起抛点与落地点不在同一高度时，需要使用完整的竖直运动方程，将竖直位移设为 −h 或 +h。
11. Variable Acceleration Using Calculus | 变加速运动与微积分
When acceleration is not constant, you must apply calculus. If displacement is given as a function of time x(t), velocity v = dx/dt and acceleration a = dv/dt = d²x/dt².
当加速度不是常数时，需要运用微积分。若位移表示为时间函数 x(t)，则速度 v = dx/dt，加速度 a = dv/dt = d²x/dt²。
Conversely, given acceleration as a function of time a(t), velocity is found by integration: v = ∫ a dt + C, and displacement by another integration: x = ∫ v dt + D. The constants C and D are found using initial conditions.
反之，若已知加速度作为时间的函数 a(t)，则通过积分求速度：v = ∫ a dt + C；再次积分求位移：x = ∫ v dt + D。常数 C 和 D 利用初始条件确定。
You may also be given acceleration as a function of displacement, a(x). In this case, use the relation a = v dv/dx to form a differential equation and solve.
也可能给出加速度作为位移的函数 a(x)。此时使用关系 a = v dv/dx 建立微分方程并求解。
Definite integrals can be used to find distance moved when velocity changes sign. Distance = ∫ |v| dt over the time interval.
可利用定积分求速度变号时的运动路程。路程 = ∫ |v| dt 在时间段内积分。
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