📚 International A-level Unit M2 (Mechanics) Complete Revision Guide | 国际A-level M2（力学）完整复习指南
The M2 (Mechanics 2) unit forms a core component of the International A-level Mathematics qualification. Building directly upon the foundations established in M1, it introduces more sophisticated mathematical tools for modelling physical systems: projectile motion, variable acceleration, energy methods, impulse and momentum, elastic strings, centre of mass, moments, and motion in a circle. This guide provides a systematic revision of every major topic in the M2 specification, with the essential theory, worked strategies, and exam-focused tips you need to succeed.
M2（力学2）单元是国际A-level数学资格认证的核心组成部分。它在M1所建立的根基之上，引入了更精密的数学工具来模拟物理系统：抛体运动、变加速度、能量方法、冲量与动量、弹性绳、质心、力矩以及圆周运动。本指南系统性地复习M2考纲中的每一个主要专题，为你提供考试成功所需的核心理论、解题策略和应试要点。
1. Projectile Motion | 抛体运动
Projectile motion describes the flight of a particle moving freely under gravity after being given an initial velocity. The crucial insight is that the horizontal and vertical motions are independent. The horizontal acceleration is zero (assuming no air resistance), so the horizontal velocity remains constant throughout the flight. The vertical acceleration is constant at g = 9.8 m s⁻² downward, making the vertical motion identical to motion in a straight line.
抛体运动描述的是粒子在获得初速度后仅在重力作用下自由飞行的过程。关键洞察在于水平与垂直方向的运动是相互独立的。水平方向加速度为零（假设没有空气阻力），因此水平速度在整个飞行过程中保持不变。垂直方向的加速度恒为 g = 9.8 m s⁻² 向下，使垂直运动等同于直线运动。
For a particle launched with speed U at an angle θ above the horizontal, the velocity components at time t are:
vₓ = U cos θ (constant)
vᵧ = U sin θ − g t
The displacement components follow from integrating these velocities. Taking the launch point as the origin:
x = (U cos θ) t
y = (U sin θ) t − ½ g t²
- Time of flight is found by setting y = 0 (returning to launch height): t = 2U sin θ / g.
- Maximum height occurs when vᵧ = 0, giving H = U² sin² θ / (2g).
- The range on level ground is R = U² sin 2θ / g.
When the landing point is at a different height to the launch point, solve the vertical displacement equation using the quadratic formula, and then substitute the resulting time into the horizontal displacement equation. Use the positive root of the quadratic that corresponds to physically meaningful motion.
当落点与发射点的高度不同时，先用二次方程求根公式解垂直位移方程，然后将所得时间代入水平位移方程。取二次方程中具有物理意义正根。
2. Variable Acceleration | 变加速度运动
In M2, acceleration is no longer assumed constant. Instead, acceleration may be given as a function of time a(t), or occasionally as a function of displacement. Calculus becomes the central tool: differentiating displacement gives velocity, differentiating velocity gives acceleration; conversely, integrating acceleration gives velocity, and integrating velocity gives displacement.
在M2中，加速度不再被假定为恒定值。相反，加速度可能被表示为时间的函数 a(t)，或者偶尔表示为位移的函数。微积分成为核心工具：对位移求导得到速度，对速度求导得到加速度；反过来，对加速度积分得到速度，对速度积分得到位移。
v = ∫ a dt s = ∫ v dt
For example, if a = 3t² − 2t m s⁻², then integrating once yields:
v = t³ − t² + C
The constant C is determined by the initial condition, e.g. if v = 4 m s⁻¹ when t = 0, then C = 4. Integrating v again produces displacement, and the second integration constant is likewise found from the initial displacement. When acceleration is given as a function of displacement, use the chain rule relation:
a = v (dv/dx)
which allows you to integrate ∫ a dx = ½v² and thus relate speed to displacement directly. In examinations, always write down the integration constant explicitly before substituting initial conditions; omitting this step is a common and costly error.
例如，若 a = 3t² − 2t m s⁻²，则积分一次得：
v = t³ − t² + C
常数 C 由初始条件确定，例如当 t = 0 时 v = 4 m s⁻¹，则 C = 4。对 v 再次积分得到位移，第二个积分常数同样由初始位移决定。当加速度表示为位移的函数时，利用链式法则关系：
a = v (dv/dx)
从而可以直接对 ∫ a dx = ½v² 积分，将速度与位移直接联系起来。在考试中，务必在代入初始条件之前明确写出积分常数；省略这一步是常见且代价高昂的错误。
3. Work and Energy | 功与能量
Work is the energy transferred to a particle by a force acting over a displacement. For a constant force F acting in the direction of motion over distance d, the work done is W = F d. In general, the work done by a variable force is the integral of force with respect to displacement:
功是力在位移上作用所传递给粒子的能量。对于沿运动方向作用的恒力 F 在距离 d 上所做的功为 W = F d。一般来说，变力所做的功是力对位移的积分：
W = ∫ F dx
The kinetic energy of a particle of mass m moving with speed v is:
KE = ½ m v²
Gravitational potential energy relative to a reference level is:
PE = m g h
The work–energy principle states that the total work done on a particle equals its change in kinetic energy:
W_total = ΔKE = ½ m v² − ½ m u²
This principle is especially powerful when dealing with motion along curved paths or under variable forces, where SUVAT equations do not apply. Always identify all forces doing work: gravity, applied forces, friction, and resistance. Forces perpendicular to the displacement (such as the normal reaction on a horizontal surface) do no work.
粒子的质量为 m、速度为 v 时，其动能为：
KE = ½ m v²
相对于参考平面，重力势能为：
PE = m g h
功–能原理指出：对粒子所做的总功等于其动能的变化量：
W_total = ΔKE = ½ m v² − ½ m u²
该原理在处理曲线运动或变力作用下的运动时尤为强大，因为此时SUVAT方程不再适用。务必识别所有做功的力：重力、施加的力、摩擦力和阻力。垂直于位移的力（如水平面上法向反作用力）不做功。
4. Power | 功率
Power is the rate at which work is done, measured in watts (W), where one watt equals one joule per second. For a force F acting on a particle moving with velocity v, the power developed is:
功率是做功的速率，单位为瓦特（W），一瓦特等于每秒一焦耳。对于作用在速度为 v 的粒子上的力 F，其产生的功率为：
P = F v
If a vehicle’s engine produces constant power P, then the driving force is F = P/v, which decreases as speed increases. When modelling a vehicle accelerating from rest, the driving force tends to infinity as speed approaches zero, so a constant-power model only applies above some small initial speed. In practice, problems often specify that the maximum power is developed when the vehicle reaches a certain speed.
如果车辆发动机产生恒定功率 P，则驱动力为 F = P/v，它随速度增大而减小。当模拟车辆从静止开始加速时，驱动力在速度趋近零时趋向无穷，因此恒定功率模型仅适用于某个微小初速度之上。实际问题中，通常会指定在车辆达到某速度时产生最大功率。
The average power over a time interval is the total work done divided by the time taken: P_average = W / t. When a variable force acts, use P = dW/dt. A common exam question involves a cyclist or car climbing a slope: the power output must overcome both the resistance and the component of weight down the slope. At constant speed up a slope of angle θ, the equation is:
在一段时间间隔内的平均功率是总功除以所用时间：P_average = W / t。当变力作用时，使用 P = dW/dt。一个常见的考题涉及骑自行车者或汽车爬坡：功率输出必须同时克服阻力和重力沿坡道的分量。在沿倾角为 θ 的斜坡匀速上升时，方程为：
P = (R + m g sin θ) v
where R is the resistance force. Note that on level ground, θ = 0, so the weight component vanishes.
其中 R 是阻力。注意在水平地面上，θ = 0，因此重力分量消失。
5. Impulse and Momentum | 冲量与动量
The linear momentum of a particle of mass m and velocity v is m v, a vector quantity measured in kg m s⁻¹. Newton’s second law can be expressed in terms of momentum: the resultant force equals the rate of change of momentum. Consequently, when a constant force F acts for time t:
质量为 m、速度为 v 的粒子的线性动量为 m v，这是一个矢量量，单位为 kg m s⁻¹。牛顿第二定律可以用动量来表述：合力等于动量变化率。因此，当恒力 F 作用时间 t 时：
Impulse = F t = m v − m u
The impulse–momentum equation links the impulse delivered by a force to the change in momentum. For variable forces, the impulse is ∫ F dt, which equals the area under the force–time graph. In such problems, graphing the force–time curve and calculating the area is often the most efficient route.
冲量–动量方程将力传递的冲量与动量变化联系起来。对于变力，冲量为 ∫ F dt，即力–时间图像下的面积。在这类问题中，绘制力–时间曲线并计算面积通常是最有效的方法。
The principle of conservation of momentum states that in an isolated system (no external forces), the total momentum before an event equals the total momentum after the event. This applies to all collisions and explosions:
动量守恒原理指出：在孤立系统中（无外力），事件前后的总动量相等。这适用于所有碰撞和爆炸：
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
When solving collision problems, always define a positive direction first, and treat opposite directions as negative velocities. Draw a clear before–after diagram, labelling all masses and velocities. This reduces sign errors, which are the most common cause of incorrect answers in momentum questions.
在解碰撞问题时，务必先定义一个正方向，并将相反方向视为负速度。画出清晰的前后示意图，标注所有质量和速度。这能减少符号错误——符号错误是动量问题中最常见的失分原因。
6. Direct Collisions and the Coefficient of Restitution | 直接碰撞与恢复系数
For two smooth spheres (or particles) colliding directly along their line of centres, the coefficient of restitution e relates the relative speeds before and after impact:
对于沿中心连线直接碰撞的两个光滑球体（或粒子），恢复系数 e 联系了碰撞前后的相对速度：
v₂ − v₁ = −e (u₂ − u₁)
The coefficient satisfies 0 ≤ e ≤ 1. When e = 1, the collision is perfectly elastic and kinetic energy is conserved. When e = 0, the particles coalesce and move together with the same velocity. In all cases, momentum is conserved, but kinetic energy is lost unless e = 1. The fraction of kinetic energy lost is proportional to (1 − e²); specifically:
恢复系数满足 0 ≤ e ≤ 1。当 e = 1 时，碰撞为完全弹性碰撞，动能守恒。当 e = 0 时，粒子粘合在一起以相同速度运动。在所有情况下动量都守恒，但除非 e = 1，否则动能会有损失。动能损失的占比正比于 (1 − e²)；具体而言：
ΔKE = ½ m₁ m₂ (u₁ − u₂)² (1 − e²) / (m₁ + m₂)
To solve a collision problem, write down the two governing equations: conservation of momentum and Newton’s law of restitution. Then solve the simultaneous equations for the final velocities. After a particle bounces off a fixed wall, its speed is multiplied by e and its direction reverses; the wall carries negligible momentum. A particle bouncing repeatedly on a floor follows a geometric sequence of bounce heights, since the speed after each bounce is e times the speed before it.
解碰撞问题时，写出两个控制方程：动量守恒方程和牛顿恢复定律方程，然后联立求解末速度。当粒子撞击固定墙壁反弹后，其速度大小乘以 e 且方向反转；墙壁携带的动量可忽略。粒子在地板上反复弹跳形成一个等比数列的弹跳高度序列，因为每次弹跳后的速度是弹跳前的 e 倍。
7. Elastic Strings and Springs | 弹性绳与弹簧
An elastic string obeys Hooke’s law up to its elastic limit. The tension in an elastic string (or spring) of natural length l and modulus of elasticity λ, when extended by length x, is:
弹性绳在其弹性限度内服从胡克定律。自然长度为 l、弹性模量为 λ 的弹性绳（或弹簧），当伸长量为 x 时，其张力为：
T = (λ / l) x = k x
where k = λ/l is the spring constant. The modulus of elasticity λ has units of force (newtons) in the A-level convention and is a property of the material and thickness of the string. The same formula describes a spring in compression with x taken as the compression distance, and the spring then pushes rather than pulls.
其中 k = λ/l 是劲度系数。弹性模量 λ 在A-level约定中具有力的单位（牛顿），是绳的材料和粗细的属性。同样的公式也描述弹簧被压缩的情况，此时 x 取压缩距离，弹簧产生推力而非拉力。
The elastic potential energy stored in a stretched string is the work done in extending it, obtained by integrating the tension:
拉伸绳子所储存的弹性势能是拉伸过程中所做的功，通过对张力积分得到：
EPE = (λ x²) / (2 l) = ½ k x²
In energy-conservation problems involving elastic strings, apply conservation of mechanical energy including the elastic potential energy term. For example, a particle attached to a string released from rest will lose gravitational potential energy while gaining kinetic energy and elastic potential energy. Set the total energy at position A equal to the total energy at position B, and solve for the unknown quantity.
在涉及弹性绳的能量守恒问题中，应用包含弹性势能项的机械能守恒。例如，连接在绳子上的粒子从静止释放时，会损失重力势能，同时获得动能和弹性势能。令A位置的总能量等于B位置的总能量，然后解出未知量。
8. Centre of Mass | 质心
The centre of mass of a system is the point at which the entire mass may be considered to act for the purposes of translational motion. For a set of particles m₁, m₂, …, mₙ located at position vectors r₁, r₂, …, rₙ, the centre of mass is:
系统的质心是就平动而言可将全部质量看作集中作用的一点。对于位于位置矢量 r₁, r₂, …, rₙ 处的一组粒子 m₁, m₂, …, mₙ，质心为：
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