📚 IB & Edexcel Mathematics: Mechanics Key Concepts | IB & Edexcel 数学：力学考点精讲
Whether you are tackling the IB Mathematics HL Mechanics option or the Edexcel A Level Mechanics modules, a strong grasp of fundamental principles is essential. This guide distils the key topics, equations and problem-solving strategies that appear in both curricula, helping you revise efficiently and secure top marks.
无论你正在攻克 IB 数学 HL 力学选项，还是 Edexcel A Level 力学模块，扎实掌握基本原理至关重要。本文梳理了两个课程体系中共同的重点主题、核心方程和解题策略，助你高效复习，争取高分。
1. Equations of Motion (SUVAT) | 匀加速运动方程
For an object moving in a straight line with constant acceleration, the five SUVAT equations connect displacement s, initial velocity u, final velocity v, acceleration a and time t. They are only valid when acceleration is constant.
对于在直线方向上做匀加速运动的物体，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。这些方程仅在加速度恒定时成立。
v = u + at gives the velocity after a time interval t.
v = u + at 给出了经过时间 t 后的速度。
s = ut + ½at² calculates the displacement from initial velocity and acceleration.
s = ut + ½at² 利用初速度和加速度计算位移。
s = ½(u + v)t relates displacement to the average velocity.
s = ½(u + v)t 将位移与平均速度联系起来。
v² = u² + 2as links velocities to displacement without mentioning time.
v² = u² + 2as 给出了不涉及时间时速度与位移的关系。
s = vt − ½at² is the displacement formula using final velocity.
s = vt − ½at² 是利用末速度的位移公式。
Always state the positive direction, list the known u, v, a, s, t values, and select the equation that contains only one unknown.
始终明确正方向，列出已知量 u、v、a、s、t，并选择只含一个未知量的方程。
2. Newton’s Laws of Motion | 牛顿运动定律
Newton’s three laws govern the relationship between force and motion. They apply in inertial reference frames and form the foundation of classical mechanics.
牛顿三大定律规定了力与运动的关系。它们适用于惯性参考系，是经典力学的基础。
First Law: An object remains at rest or moves with constant velocity unless acted upon by a resultant external force.
第一定律：物体将保持静止或匀速直线运动状态，除非受到外力的作用。
Second Law: Resultant force F = ma, where m is the mass and a is the acceleration in the direction of the force.
第二定律：合力 F = ma，其中 m 为质量，a 为沿力方向的加速度。
Third Law: If body A exerts a force on body B, then body B exerts an equal and opposite force on body A. These forces act on different bodies.
第三定律：若物体 A 对物体 B 施加力，则物体 B 同时对物体 A 施加大小相等、方向相反的力。这两个力作用在不同物体上。
Always consider the entire system or individual components when applying Newton’s second law to connected particles.
在处理连接体问题时，应用牛顿第二定律时要区分整体系统还是单个部分进行分析。
3. Free Body Diagrams and Resolving Forces | 受力分析与力的分解
Drawing a clear free body diagram showing all forces acting on an object is the vital first step in any mechanics problem. Forces must be resolved into perpendicular components, usually horizontal and vertical.
在解任何力学问题时，清晰画出显示物体所有受力情况的隔离受力图是至关重要的第一步。力必须分解为相互垂直的分量，通常是水平与竖直方向。
If a force F makes an angle θ with the horizontal, its components are F cos θ (horizontal) and F sin θ (vertical).
如果一个力 F 与水平方向成 θ 角，其分量为 F cos θ（水平）和 F sin θ（竖直）。
When an object is in equilibrium, the vector sum of all forces is zero: ΣFx = 0 and ΣFy = 0.
当物体处于平衡状态时，所有力的矢量和为零：ΣFx = 0 且 ΣFy = 0。
Use the sign convention consistently. Tension, weight, normal reaction, friction, and applied forces must all be accounted for.
要始终如一地使用正负号约定。必须将张力、重力、法向反作用力、摩擦力及其他外力都考虑在内。
4. Inclined Planes and Friction | 斜面与摩擦力
On a smooth or rough inclined plane, weight is resolved parallel and perpendicular to the plane. The perpendicular component is mg cos θ; the parallel component down the slope is mg sin θ.
在光滑或粗糙斜面上，重力需分解为平行和垂直于斜面的分量。垂直分量为 mg cos θ，沿斜面下滑的分量为 mg sin θ。
Friction F opposes relative motion or the tendency to move. Static friction F ≤ μsR, while kinetic friction is F = μkR, where R is the normal reaction.
摩擦力 F 阻碍相对运动或运动趋势。静摩擦力 F ≤ μsR，而动摩擦力为 F = μkR，其中 R 是法向反作用力。
When an object is on the point of sliding, use the limiting friction F = μR (often written simply as μ).
当物体即将滑动时，使用极限摩擦力 F = μR（通常简写为 μ）。
Always check whether the plane is smooth (no friction) or rough, and whether equilibrium or acceleration is involved.
务必确认斜面是光滑（无摩擦）还是粗糙的，以及问题是涉及平衡状态还是加速运动。
5. Connected Particles and Pulleys | 连接体与滑轮系统
For two particles connected by a light inextensible string over a smooth pulley, the tension is the same on both sides and the accelerations are equal in magnitude.
对于用轻质且不可伸长的绳子跨过光滑滑轮连接的两个物体，绳子两端的张力相等，且加速度大小相同。
Write separate equations of motion for each particle using Newton’s second law, then solve simultaneously.
应用牛顿第二定律分别列出每个物体的运动方程，然后联立求解。
If masses are m₁ and m₂ with m₂ > m₁, the acceleration is often a = (m₂ − m₁)g / (m₁ + m₂) and tension T = (2m₁m₂)g / (m₁ + m₂) for a simple pulley system.
若质量分别为 m₁ 和 m₂ 且 m₂ > m₁，对于简单滑轮系统，加速度通常为 a = (m₂ − m₁)g / (m₁ + m₂)，张力为 T = (2m₁m₂)g / (m₁ + m₂)。
When one particle is on a table and the other hangs vertically, remember to include friction or normal reaction for the table part.
当一个物体在桌面上而另一个悬挂在空中时，别忘了考虑桌面部分的摩擦力或法向反作用力。
6. Momentum and Impulse | 动量与冲量
Momentum p is the product of mass and velocity: p = mv. It is a vector quantity. The impulse J delivered by a constant force F over time t is J = Ft.
动量 p 是质量与速度的乘积：p = mv，为矢量。恒力 F 在时间 t 内产生的冲量 J 为 J = Ft。
The impulse–momentum theorem states: Impulse = Change in momentum, so Ft = mv − mu.
冲量-动量定理指出：冲量 = 动量的变化量，即 Ft = mv − mu。
In collisions, the principle of conservation of momentum applies when no external forces act: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.
在碰撞过程中，若无外力作用，则动量守恒：m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂。
For collisions, know the difference between perfectly elastic (keeps kinetic energy) and inelastic (kinetic energy lost).
对于碰撞，要区分完全弹性碰撞（动能守恒）和非弹性碰撞（动能损失）。
7. Work, Energy and Power | 功、能与功率
Work done W by a constant force F moving an object through displacement s in the direction of the force is W = Fs. For an angled force, W = Fs cos θ.
恒力 F 使物体沿力的方向发生位移 s 所做的功为 W = Fs。若力与位移有夹角，则 W = Fs cos θ。
Kinetic energy KE = ½mv². Gravitational potential energy GPE = mgh. The work–energy principle states that net work done equals change in kinetic energy.
动能 KE = ½mv²，重力势能 GPE = mgh。功能原理指出，合力做的功等于动能的变化量。
For a particle moving under gravity and other forces, Loss in PE + Work done by external forces = Gain in KE + Work against friction.
对于在重力和其他力作用下的物体，势能减少 + 外力做功 = 动能增加 + 克服摩擦力做的功。
Power P is the rate of doing work: P = W/t. For a vehicle moving at constant speed v against a tractive force F, P = Fv.
功率 P 是做功的速率：P = W/t。当车辆以恒定速度 v 克服牵引力 F 行驶时，P = Fv。
8. Moments and Equilibrium | 力矩与平衡
The moment of a force about a point is the product of the force and the perpendicular distance from the line of action to the point: Moment = F × d. Anticlockwise is usually taken as positive.
力对某点的力矩等于力的大小与力的作用线到该点垂直距离的乘积：力矩 = F × d。通常规定逆时针方向为正。
For a rigid body in equilibrium, both resultant force and resultant moment must be zero: ΣF = 0 and ΣM = 0 about any point.
刚体平衡时，合力与对任意点的合力矩都必须为零：ΣF = 0 且对任意点 ΣM = 0。
When taking moments to find an unknown force, choose a pivot point where an unknown force passes through to eliminate it from the equation.
在取矩求未知力时，选择通过某个未知力的点作为支点，可使该力不在力矩方程中出现。
Be comfortable with tilting and toppling problems where the normal reaction shifts to the edge of the base.
要熟悉倾斜和翻倒问题，这类问题中法向反作用力会移动到接触面的边缘。
9. Vectors in Mechanics | 力学中的向量
Displacement, velocity, acceleration and force are all vector quantities. They can be expressed in component form using unit vectors i and j (horizontal and vertical).
位移、速度、加速度和力都是矢量。它们可以用单位向量 i（水平方向）和 j（竖直方向）的分量形式表示。
If r = xi + yj, then velocity v = dr/dt and acceleration a = dv/dt. Differentiate each component separately.
若 r = xi + yj，则速度 v = dr/dt，加速度 a = dv/dt。分别对每个分量求导即可。
Magnitude of a vector v = ai + bj is given by √(a² + b²). Direction can be found using tan θ = (b/a).
矢量 v = ai + bj 的大小为 √(a² + b²)，方向可通过 tan θ = (b/a) 求得。
In relative motion, the velocity of A relative to B is vA − vB; use vector subtraction to find approach speeds or collision courses.
相对运动中，A 相对于 B 的速度为 vA − vB；利用向量减法可求接近速度或碰撞航向。
10. Projectile Motion | 抛体运动
A projectile moves under constant vertical acceleration due to gravity (g ≈ 9.8 m/s² downwards) and zero horizontal acceleration, neglecting air resistance.
忽略空气阻力时，抛体在竖直方向上受恒定的重力加速度（g ≈ 9.8 m/s² 向下），水平方向加速度为零。
Resolve initial velocity u into horizontal component u cos θ and vertical component u sin θ. The horizontal motion is uniform, while the vertical motion uses SUVAT equations with a = −g.
把初速度 u 分解为水平分量 u cos θ 和竖直分量 u sin θ。水平方向做匀速直线运动，竖直方向用 SUVAT 方程且 a = −g。
Time of flight, maximum height and horizontal range can be derived: Range = (u² sin 2θ) / g and Max height = (u² sin²θ) / (2g).
飞行时间、最大高度和水平射程的推导公式为：射程 = (u² sin 2θ) / g，最大高度 = (u² sin²θ) / (2g)。
For symmetrical projectiles (same launch and landing height), time to the highest point is half the total time; for a given speed, 45° gives maximum range.
对于对称抛体（发射和落地高度相同），到达最高点的时间为总时间的一半；在初速度大小固定的情况下，45° 射角可获得最大射程。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导Cancel reply