Mechanics Core Concepts: Cambridge AS & A Level Revision | 力学核心概念：剑桥 AS & A Level 复习

📚 Mechanics Core Concepts: Cambridge AS & A Level Revision | 力学核心概念：剑桥 AS & A Level 复习

This revision guide covers the essential mechanics topics in the Cambridge International AS & A Level Mathematics Mechanics coursebook by Jan Dangerfield, Stuart Haring and colleagues. It is designed for CIE candidates who need clear definitions, key formulas and exam-ready methods.

本复习指南涵盖剑桥国际 AS & A Level 数学力学教材（Jan Dangerfield、Stuart Haring 等）的核心力学主题，帮助 CIE 考生掌握清晰定义、关键公式与考试方法。

1. Scalars, Vectors and Units | 标量、矢量与单位

In mechanics, every quantity is either a scalar or a vector. Scalar quantities such as mass, time, speed and energy have magnitude only. Vector quantities such as displacement, velocity, acceleration and force have both magnitude and direction. You must separate distance from displacement and speed from velocity. Distance is the total length travelled, while displacement is the straight-line change in position from the starting point. Speed is the rate at which distance changes; velocity is the rate at which displacement changes. In Cambridge AS & A Level Mechanics, vector answers should include direction unless the question states otherwise.

在力学中，每个物理量要么是标量，要么是矢量。质量、时间、速率和能量等标量只有大小；位移、速度、加速度和力等矢量既有大小又有方向。必须区分路程与位移、速率与速度。路程是运动轨迹的总长度，而位移是从起点到终点的直线位置变化。速率是路程变化的速率；速度是位移变化的速率。在剑桥 AS & A Level 力学中，矢量答案应包含方向，除非题目另有说明。

• Use SI units: metre (m), second (s), kilogram (kg), newton (N), joule (J), watt (W).
• 使用国际单位：米（m）、秒（s）、千克（kg）、牛顿（N）、焦耳（J）、瓦特（W）。

---

2. SUVAT Equations of Constant Acceleration | 匀加速运动的 SUVAT 方程

The equations of uniformly accelerated motion, often called the SUVAT equations, apply only when acceleration is constant. They link five quantities: displacement s, initial velocity u, final velocity v, acceleration a and time t. Always list the three known values and the quantity you need before choosing an equation.

匀加速运动方程常被称为 SUVAT 方程，仅在加速度恒定时使用。它们联系五个量：位移 s、初速度 u、末速度 v、加速度 a 和时间 t。在选择方程前，先列出三个已知量和待求量。

v = u + at

s = ½(u + v)t

s = ut + ½at²

v² = u² + 2as

Quantity	Symbol	SI unit
Displacement	s	m
Initial velocity	u	m s⁻¹
Final velocity	v	m s⁻¹
Acceleration	a	m s⁻²
Time	t	s

---

3. Reading Motion Graphs | 解读运动图像

Displacement-time, velocity-time and acceleration-time graphs give a rapid visual summary of motion. The gradient of a displacement-time graph gives velocity. The gradient of a velocity-time graph gives acceleration, and the area under a velocity-time graph gives displacement. The area under an acceleration-time graph gives the change in velocity.

位移-时间图、速度-时间图和加速度-时间图可以快速直观地描述运动。位移-时间图的斜率给出速度。速度-时间图的斜率给出加速度，速度-时间图下的面积给出位移。加速度-时间图下的面积给出速度变化量。

When a graph is not a straight line, you may still estimate the required quantity using a tangent for the gradient or counting squares for the area. Exam questions often ask you to describe the stages of motion: constant velocity, constant acceleration, or rest.

当图像不是直线时，仍可用切线求斜率或用数方格的方法估算面积。考试题常要求描述运动的各个阶段：匀速、匀加速或静止。

---

4. Newton’s Laws and Resultant Force | 牛顿定律与合力

Newton’s second law states that the resultant force acting on a particle is equal to the product of its mass and acceleration: F = ma. This is a vector equation. If several forces act, first find the resultant force in each direction, then apply the equation separately along each axis. For connected particles such as two masses linked by a light inextensible string passing over a smooth pulley, draw separate free-body force diagrams. Because the string is inextensible, the particles have the same acceleration; because it is light, the tension is the same throughout.

牛顿第二定律指出，作用在质点上的合力等于质量与加速度的乘积：F = ma。这是一个矢量方程。如果有多个力作用，先求每个方向的合力，再沿各轴分别应用该方程。对于通过光滑滑轮用轻质不可伸长的绳子连接的两个物体等连接体问题，应分别画受力图。因为绳子不可伸长，两个物体的加速度相同；因为绳子轻质，绳中张力处处相同。

When a particle is on a smooth horizontal surface, the normal reaction balances its weight, so the vertical resultant is zero. On a rough surface or an inclined plane, always include friction and weight components in your force diagram.

当质点位于光滑水平面上时，法向反作用力与其重量平衡，因此竖直方向的合力为零。在粗糙表面或斜面上，务必在受力图中包含摩擦力和重力的分量。

---

5. Resolving Forces and Equilibrium | 力的分解与平衡

A force can be resolved into perpendicular components, usually horizontal and vertical or parallel and perpendicular to a slope. On an inclined plane at angle θ to the horizontal, the weight mg is usually resolved into mg sinθ down the slope and mg cosθ perpendicular to the slope. A particle is in equilibrium when the vector sum of all forces is zero. In component form, the sum of horizontal force components is zero and the sum of vertical force components is zero.

一个力可以分解为互相垂直的分量，通常是水平与竖直分量，或沿斜面与垂直斜面的分量。在与水平面成角 θ 的斜面上，重力 mg 通常分解为沿斜面向下的 mg sinθ 和垂直斜面的 mg cosθ。当所有力的矢量和为零时，质点处于平衡状态。用分量表示，即水平方向合力为零，竖直方向合力为零。

If only three non-parallel forces act on a particle in equilibrium, they can be represented by a closed triangle. This gives a quick geometric method for finding an unknown force, but resolving components is usually more reliable in exam solutions.

如果只有三个互不平行的力作用在处于平衡状态的质点上，它们可以构成闭合三角形。这提供了一种快速求解未知力的几何方法，但在考试解答中，分解分量通常更可靠。

---

6. Friction: Static and Kinetic | 摩擦力：静摩擦与动摩擦

Friction opposes relative motion or attempted motion between two surfaces in contact. If the normal contact force is R and the coefficient of friction is μ, the maximum static friction is μR. If the object is sliding, the kinetic friction is usually modelled as F = μR. If the object is not moving and the required friction is less than μR, it remains in equilibrium. The inequality F ≤ μR is used for limiting equilibrium.

摩擦力阻碍接触面之间的相对运动或相对运动趋势。若法向接触力为 R，摩擦系数为 μ，则最大静摩擦力为 μR。若物体正在滑动，动摩擦力通常建模为 F = μR。若物体静止且所需摩擦力小于 μR，则物体保持平衡。处于极限平衡时使用不等式 F ≤ μR。

On an inclined plane, the normal reaction is usually mg cosθ, so the maximum friction is μmg cosθ. The particle is just about to slip down the plane when mg sinθ = μmg cosθ, giving tanθ = μ.

在斜面上，法向反作用力通常为 mg cosθ，因此最大摩擦力为 μmg cosθ。当质点刚好

Published by TutorHao | IB Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）