A-Level AQA Maths: Mechanics Key Points Summary | A-Level AQA 数学：力学考点精讲

📚 A-Level AQA Maths: Mechanics Key Points Summary | A-Level AQA 数学：力学考点精讲

This revision guide covers the essential topics in the Mechanics component of AQA A-Level Mathematics. You will review SUVAT equations, projectile motion, Newton’s laws, moments, work-energy-power, momentum, and variable acceleration. Each section provides key formulas and explanations to help you consolidate your understanding and approach exam questions with confidence.

本复习指南涵盖 AQA A-Level 数学力学部分的核心考点，包括匀加速运动方程、抛体运动、牛顿定律、力矩、功-能-功率、动量以及变加速度等内容。每个小节提供关键公式和解释，帮助你巩固理解并自信应对考试。

1. SUVAT Equations and Motion Graphs | 匀加速运动方程与运动图像

In mechanics, constant acceleration problems are often solved using the SUVAT equations. The five kinematic variables are displacement s, initial velocity u, final velocity v, acceleration a, and time t. These equations link any four of these variables, allowing you to find the fifth.

在力学中，匀加速问题通常使用 SUVAT 方程求解。五个运动学变量为位移 s、初速度 u、末速度 v、加速度 a 和时间 t。这些方程将其中四个变量联系起来，从而求出第五个。

v = u + at s = ut + ½ at² s = ½ (u + v) t v² = u² + 2as

Each equation omits one of the five variables: v = u + at omits s; s = ut + ½ at² omits v; s = ½(u+v)t omits a; v² = u² + 2as omits t. Always list known and unknown quantities before selecting an equation.

每个方程都忽略了其中一个变量：v = u + at 没有 s；s = ut + ½ at² 没有 v；s = ½(u+v)t 没有 a；v² = u² + 2as 没有 t。在选择方程之前，先列出已知量和未知量。

Displacement-time (s-t) and velocity-time (v-t) graphs are especially important. The gradient of an s-t graph gives velocity; the gradient of a v-t graph gives acceleration. The area between the v-t graph and the time axis gives displacement. A horizontal line on a v-t graph indicates constant velocity; a sloping line indicates acceleration.

位移-时间 (s-t) 图和速度-时间 (v-t) 图尤为重要。s-t 图的斜率表示速度；v-t 图的斜率表示加速度。v-t 图与时间轴之间的面积表示位移。v-t 图中的水平线表示匀速，倾斜线表示有加速度。

2. Projectile Motion | 抛体运动

In projectile motion, treat horizontal and vertical components separately. The initial speed u is resolved into horizontal component u cosθ and vertical component u sinθ. Air resistance is assumed negligible, so horizontal velocity remains constant, while vertical motion has constant downward acceleration g.

在抛体运动中，需将水平与竖直分量分开处理。初速度 u 分解为水平分量 u cosθ 和竖直分量 u sinθ。空气阻力忽略不计，因此水平速度保持恒定，竖直方向以加速度 g 向下作匀加速运动。

Time of flight T = 2u sinθ / g, Maximum height H = u² sin²θ / (2g), Range R = u² sin 2θ / g

These formulas apply when the projectile is launched from and lands on the same horizontal level. The time of flight is twice the time to reach maximum height, reflecting the symmetry of the parabolic trajectory.

上述公式适用于抛出点与落地点在同一水平面的情况。飞行时间等于到达最高点所用时间的两倍，体现了抛物线轨迹的对称性。

For any projectile, the horizontal motion is uniform, so the horizontal displacement is x = u cosθ × t. The vertical motion is uniformly accelerated, so y = u sinθ t − ½ g t². These parametric equations describe the path.

对任意抛体，水平方向是匀速运动，水平位移 x = u cosθ × t；竖直方向是匀加速运动，y = u sinθ t − ½ g t²。这些参数方程描述了轨迹。

3. Forces and Newton’s Laws | 力与牛顿定律

Newton’s First Law: an object remains at rest or in uniform motion unless acted upon by a resultant force. Second Law: F = ma, the net force equals mass times acceleration. Third Law: every action force has an equal and opposite reaction force, acting on different bodies.

牛顿第一定律：物体保持静止或匀速直线运动状态，除非受到合外力作用。第二定律：F = ma，合外力等于质量乘以加速度。第三定律：每个作用力都有一个大小相等、方向相反的反作用力，作用在不同物体上。

Weight W = mg, where m is mass and g is the acceleration due to gravity (9.8 ms⁻² on Earth). Mass is a scalar measured in kg, while weight is a vector measured in newtons. Never confuse the two when drawing force diagrams.

重量 W = mg，其中 m 是质量，g 是重力加速度（地球表面约为 9.8 ms⁻²）。质量是标量，单位为 kg；重量是矢量，单位为 N。在受力分析图中切勿混淆二者。

When multiple forces act on a particle, the resultant force is the vector sum. Equilibrium occurs when the net force is zero, meaning the particle is either stationary or moving with constant velocity. Always resolve forces into perpendicular directions before summing.

当多个力作用在一个质点上时，合力为这些力的矢量和。当合外力为零时，质点处于平衡状态，即静止或匀速直线运动。需先将力分解到互相垂直的方向上再求和。

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

To analyse forces in different directions, use resolution. A force F at an angle θ to a given direction can be split into two perpendicular components: F cosθ along the direction and F sinθ perpendicular to it. Choose a suitable pair of axes for the problem.

分析不同方向的力时，需进行力的分解。一个与给定方向夹角为 θ 的力 F 可分解为两个相互垂直的分量：沿该方向的分量为 F cosθ，垂直分量为 F sinθ。应针对具体问题选取合适的坐标轴。

For a particle in equilibrium, the sum of force components in any direction is zero: ΣF_x = 0 and ΣF_y =

Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）