📚 Edexcel Maths: Mechanics Key Points | Edexcel 数学：力学 考点精讲
In Edexcel A Level Mathematics, the Mechanics component challenges you to apply mathematical models to real-world physical situations. This revision guide covers the essential topics, key formulas, and typical problem types that appear in the exam. By mastering these core concepts, you will be well prepared to tackle kinematics, forces, momentum, moments, energy, and more.
在 Edexcel A Level 数学中，力学部分考察你将数学模型应用于真实物理情境的能力。本考点精讲涵盖了考试中出现的核心主题、关键公式和典型题型。掌握这些基础概念后，你将能从容应对运动学、力、动量、力矩、能量等内容。
1. SUVAT Equations | 匀加速运动方程
The five SUVAT equations connect displacement (s), initial velocity (u), final velocity (v), acceleration (a) and time (t) for motion in a straight line with constant acceleration. They are the foundation of kinematics.
五个 SUVAT 方程将位移 (s)、初速度 (u)、末速度 (v)、加速度 (a) 和时间 (t) 联系起来，适用于匀加速直线运动，是运动学的基础。
Always start by listing the known quantities and the one you need to find. Then choose the equation that contains only those four variables.
解题时先列出已知量和所求量，然后选择只含这四个变量的方程。
v = u + at
s = ut + ½at²
s = ½(u + v)t
v² = u² + 2as
s = vt - ½at²
Remember to define a positive direction. Displacement, velocity and acceleration are vectors; their signs depend on that direction.
切记要规定正方向。位移、速度和加速度都是矢量，其正负取决于所选正方向。
2. Forces and Newton's Laws | 力与牛顿定律
Newton's three laws govern all force interactions. The second law, F = ma, is the most frequently used equation in mechanics problems.
牛顿三定律支配着所有力的相互作用。第二定律 F = ma 是力学问题中最常用的方程。
Always draw a clear force diagram showing weight (mg), normal reaction (R), tension (T), friction (Fr) and any applied forces. Resolve forces parallel and perpendicular to the slope when an inclined plane is present.
始终画出清晰的受力图，标出重力 (mg)、法向反力 (R)、张力 (T)、摩擦力 (Fr) 以及施加的外力。有斜面时，沿平行和垂直斜面方向分解力。
For a particle on a smooth inclined plane of angle θ, the component of weight down the slope is mg sin θ and the perpendicular component is mg cos θ.
对于放在光滑斜面上倾角为 θ 的质点，重力沿斜面向下的分量为 mg sin θ，垂直斜面的分量为 mg cos θ。
Friction opposes motion and its maximum value is FMAX = μR, where μ is the coefficient of friction.
摩擦力阻碍运动，其最大值满足 FMAX = μR，其中 μ 为摩擦系数。
3. Connected Particles | 连接体问题
same tension and acceleration magnitude. Treat each particle separately, applying F = ma.
当物体通过轻质且不可伸长的绳子跨过光滑滑轮连接时，绳上张力处处相等，加速度大小相同。分别对每个物体应用 F = ma。
stem with masses m₁ and m₂, the acceleration is a = (m₂ - m₁)g/(m₁ + m₂) assuming m₂ > m₁.
对每个质量列出运动方程，然后联立求解。对于典型的滑轮系统，如果 m₂ > m₁，加速度为 a = (m₂ - m₁)g/(m₁ + m₂)。
inside a lift accelerating upward or downward. The apparent weight is R = m(g ± a), where '+' is for upward acceleration.
电梯问题常涉及物体在加速上升或下降的电梯中。视重为 R = m(g ± a)，加速上升时取 '+' 号。
4. Momentum and Impulse | 动量与冲量
Momentum is defined as p = mv. Impulse is the change in momentum, given by I = mv - mu, and also equals the average force multiplied by time: I = Ft.
动量定义为 p = mv。冲量是动量的变化量，即 I = mv - mu，也等于平均力乘以时间：I = Ft。
: total momentum before = total momentum after, provided no external resultant force acts.
在碰撞和爆炸问题中，如果系统不受合外力，则动量守恒：碰撞前总动量 = 碰撞后总动量。
sistently for velocities. For two particles colliding directly, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.
始终规定正方向并一致使用速度的正负号。两物体直接碰撞时，m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂。
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: Moment = F × d.
力对一点的力矩等于力的大小乘以该点到力作用线的垂直距离：力矩 = F × d。
For a rigid body in equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments. This is the principle of moments.
对于处于平衡的刚体，对任意点的顺时针力矩之和等于逆时针力矩之和，这就是力矩原理。
When a uniform rod is involved, its weight acts at the midpoint. Use the concept of a resultant moment to find unknown forces or distances.
涉及均匀杆时，其重力作用在杆的中点。利用合力矩的概念可求解未知力或距离。
6. Equilibrium and Statics | 平衡与静力学
A particle is in equilibrium when the resultant force in any direction is zero. This means the vector sum of all forces is zero, and components in perpendicular directions balance individually.
当质点在任何方向上合力均为零时，质点处于平衡。这意味着所有力的矢量和为零，且相互垂直方向上的分力分别平衡。
about to slip down. Resolve parallel and perpendicular to find limiting angle: tan θ = μ.
对于粗糙斜面上处于极限平衡的质点，摩擦力达到 μR 并沿斜面向上（若质点即将下滑）。分解后可得极限角满足 tan θ = μ。
For a rigid body in equilibrium, both resultant force and resultant moment must be zero. This gives up to three independent equations: two for forces and one for moments.
对于平衡的刚体，合外力与合力矩都必须为零。这最多提供三个独立方程：两个力方程和一个力矩方程。
7. Work, Energy and Power | 功、能量与功率
Work done by a constant force is W = Fs cos θ, where s is displacement and θ is the angle between force and displacement. Energy is the capacity to do work.
恒力做的功为 W = Fs cos θ，其中 s 是位移，θ 是力与位移的夹角。能量是做功的能力。
The work–energy principle states that the total work done by all forces acting on a particle equals the change in kinetic energy: Work done = ½mv² - ½mu².
功能原理指出，合外力对质点做的总功等于动能的变化量：总功 = ½mv² - ½mu²。
Gravitational potential energy is mgh, where h is the vertical height relative to a chosen datum. Power is the rate of doing work, P = Fv for a force moving at speed v in its direction.
重力势能为 mgh，h 是相对于所选参考面的垂直高度。功率是做功的快慢，当一个力沿其方向以速度 v 运动时，P = Fv。
8. Variable Acceleration | 变加速度运动
cement x as a function of time t, velocity v = dx/dt and acceleration a = dv/dt = d²x/dt².
当加速度不恒定时，运动学需要微积分。若位移 x 表示为时间 t 的函数，则速度 v = dx/dt，加速度 a = dv/dt = d²x/dt²。
d by integration: v = ∫ a dt, x = ∫ v dt, using initial conditions to determine constants.
反之，若加速度表示为时间的函数，可以通过积分求速度和位移：v = ∫ a dt，x = ∫ v dt，并用初始条件确定积分常数。
de the constant of integration and evaluate using given conditions, e.g. v = 5 when t = 0.
常见表达式如 a = 6t - 2 或 v = 3t² + 4。记得加上积分常数，并利用给定条件（如 t = 0 时 v = 5）求值。
9. Projectiles | 抛体运动
acceleration due to gravity (g = 9.8 m/s²). The two directions are treated independently.
抛体运动可分解为水平匀速运动和竖直方向重力加速度恒定的运动 (g = 9.8 m/s²)。两个方向独立处理。
d with speed u at an angle θ to the horizontal, the initial components are uₓ = u cos θ, uy = u sin θ. Time of flight, maximum height and range are derived using SUVAT vertically.
一质点以速率 u 与水平成 θ 角抛出，初始速度分量为 uₓ = u cos θ，uy = u sin θ。飞行时间、最大高度和射程可通过对竖直方向应用 SUVAT 求得。
ntal velocity remains u cos θ. Range = (u² sin 2θ)/g and maximum height = (u² sin²θ)/(2g).
在最高点，竖直速度为零，但水平速度仍保持 u cos θ。射程 = (u² sin 2θ)/g，最大高度 = (u² sin²θ)/(2g)。
10. Resolving Forces and Friction | 力的分解与摩擦力
Resolving forces into perpendicular components simplifies equilibrium and motion problems. On an inclined plane, resolve weight into mg sin θ (down the slope) and mg cos θ (perpendicular).
将力分解为互相垂直的分量可以简化平衡和运动问题。在斜面上，将重力分解为 mg sin θ（沿斜面向下）和 mg cos θ（垂直斜面）。
friction = μR; if in limiting equilibrium, friction reaches μR but can be less otherwise.
涉及摩擦力时，务必判断物体是正在运动还是处于极限平衡。若在运动，摩擦力 = μR；若处于极限平衡，摩擦力达到 μR，其他情形可能小于该值。
ntal surface pulled by an inclined force, resolve horizontally and vertically to find the normal reaction and condition for motion.
利用摩擦不等式条件：F ≤ μR。对于粗糙水平面上受倾斜拉力作用的物体，分别沿水平和竖直分解，求出法向反力和运动条件。
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