📚 Mastering Mechanics Unit 2 (MA05): Moments, Projectiles, Energy and Collisions | 掌握力学单元 2 (MA05)：力矩、抛体运动、能量与碰撞
This revision guide targets AQA International A-level Mathematics 9660, Mechanics Unit 2 (MA05). It brings together the core principles you must be able to use with precision: moments, centres of mass, projectile motion, work-energy methods, impulse and collisions, and uniform circular motion. The unit often combines two or more of these topics in a single question, so the ability to switch between models is essential.
本复习指南针对 AQA 国际 A-level 数学 9660 力学单元 2 (MA05)。它汇集了你必须能够准确使用的核心原理：力矩、质心、抛体运动、功与能量方法、冲量与碰撞，以及匀速圆周运动。该单元经常在一个题目中组合两个或更多主题，因此在模型之间切换的能力至关重要。
1. Unit Overview and Assessment Approach | 单元概述与考试方法
MA05 builds directly on Mechanics Unit 1 by extending Newton’s laws to rigid bodies, two-dimensional motion and systems of particles. You are expected to model rods as uniform, strings as light and inextensible, pulleys as smooth, and air resistance as negligible unless a question states otherwise. These modelling assumptions simplify real situations into solvable equations.
MA05 在力学单元 1 的基础上，将牛顿定律扩展到刚体、二维运动和质点系。你应把杆建模为均匀杆，绳子为轻且不可伸长，滑轮为光滑，空气阻力除非题目另有说明否则忽略不计。这些建模假设将真实情境简化为可求解的方程。
Examination questions test three main skills: selecting the correct mechanical principle, applying algebra and calculus accurately, and interpreting results in context. Common command words include ‘find’, ‘show that’, ‘hence’ and ‘state the direction’. ‘Show that’ questions require a clear algebraic route, not just a final formula.
考试题目考查三项主要能力：选择正确的力学原理，准确运用代数与微积分，以及在具体情境中解释结果。常见指令词包括 ‘find’、’show that’、’hence’ 和 ‘state the direction’。’Show that’ 类题目要求清晰的代数推导过程，而不只是写出最终公式。
2. Moments and Equilibrium of Rigid Bodies | 力矩与刚体平衡
The moment of a force about a pivot is a measure of its turning effect. It is defined by M = F × d⊥, where d⊥ is the perpendicular distance from the pivot to the line of action of the force. When the force acts at an angle θ to the line joining pivot and point of application, the moment is M = Fd sin θ. The SI unit of moment is the newton metre, N m.
力对支点的力矩是其转动效应的量度。它定义为 M = F × d⊥，其中 d⊥ 是从支点到力作用线的垂直距离。当力与支点和作用点连线成 θ 角时，力矩为 M = Fd sin θ。力矩的国际单位是牛顿米，N m。
For a body in static equilibrium, two conditions must hold: the resultant force in any direction is zero, and the resultant moment about any chosen point is zero. Choosing the pivot to lie on an unknown force eliminates that unknown from the moment equation. A couple consists of two equal, opposite parallel forces and has moment equal to one force times the perpendicular distance between the forces.
对于处于静态平衡的物体，必须满足两个条件：任意方向的合力为零，且对任意选定点的合力矩为零。选择支点位于某个未知力上，可以从力矩方程中消去该未知量。力偶由两个大小相等、方向相反的平行力组成，其力矩等于其中一个力乘以两力之间的垂直距离。
M = Fd sin θ | C = F × d
3. Centre of Mass | 质心
The centre of mass is the point at which the whole mass of a body can be considered to act. For a system of particles, the coordinates are given by x̄ = Σmᵢxᵢ / Σmᵢ and ȳ = Σmᵢyᵢ / Σmᵢ. For a composite body, split the body into standard shapes and use the same principle in a table.
质心是物体全部质量可以看作集中于该点的点。对于质点系，其坐标为 x̄ = Σmᵢxᵢ / Σmᵢ 和 ȳ = Σmᵢyᵢ / Σmᵢ。对于组合体，将物体拆分为标准形状，并在表格中使用相同原理计算。
Standard results you must know include: a uniform rod has its centre at its midpoint; a rectangle at the intersections of its diagonals; a triangular lamina at one third of the way along each median from the base; and a uniform semicircular lamina at 4r/(3π) from its diameter. Always take moments about the same axis when finding ȳ or x̄.
你必须掌握的标准结果包括：均匀杆的质心在其中点；矩形在对角线交点；三角形薄片位于每条中线从底边起三分之一处；均匀半圆薄片在距直径 4r/(3π) 处。求 ȳ 或 x̄ 时，始终对同一轴取矩。
4. Kinematics with Variable Acceleration | 变加速度运动学
When acceleration is not constant, the SUVAT equations do not apply. Instead, use calculus: velocity is the derivative of displacement, and acceleration is the derivative of velocity. Conversely, velocity is the integral of acceleration with respect to time, and displacement is the integral of velocity with respect to time. Initial conditions determine the constants of integration.
当加速度不是常量时，SUVAT 方程不适用。此时应使用微积分：速度是位移的导数，加速度是速度的导数。反过来，速度是加速度对时间的积分，位移是速度对时间的积分。初始条件决定积分常数。
v = ∫a dt | s = ∫v dt | a = dv/dt
For two-dimensional motion, write displacement as r = x i + y j, where i and j are unit vectors. Differentiate each component separately. For example, if v = 6t i + (4 − 3t²) j, then a = 6 i − 6t j. Always include units in final answers and check that initial conditions are used exactly.
对于二维运动，将位移写作 r = x i + y j，其中 i 和 j 为单位向量。对每个分量分别求导。例如，若 v = 6t i + (4 − 3t²) j，则 a = 6 i − 6t j。最终答案中始终包含单位，并准确使用初始条件。
5. Projectile Motion | 抛体运动
A projectile is modelled as a particle moving under gravity only. The horizontal acceleration is zero, so horizontal velocity is constant. The vertical acceleration is −g, usually taken as −9.8 m s⁻². If the initial speed is u at angle θ to the horizontal, resolve into uₓ = u cos θ and u_y = u sin θ.
抛体被建模为仅在重力作用下运动的质点。水平加速度为零，因此水平速度恒定。竖直加速度为 −g，通常取 −9.8 m s⁻²。如果初速度为 u，与水平方向成 θ 角，则分解为 uₓ = u cos θ 和 u_y = u sin θ。
x = u cos θ × t | y = u sin θ × t − ½ g t²
The time of flight for a projectile returning to the same horizontal level is T = 2u sin θ / g, the horizontal range is R = u² sin 2θ / g, and the greatest height is H = u² sin²θ / (2g). The maximum range for a given speed occurs at θ = 45°. When the landing height differs, solve y = H_final instead of y = 0.
抛体返回同一水平高度时的飞行时间为 T = 2u sin θ / g，水平射程为 R = u² sin 2θ / g，最大高度为 H = u² sin²θ / (2g)。给定速度下的最大射程出现在 θ = 45°。当落地高度不同时，应对 y = H_final 求解，而不是 y = 0。
6. Work, Energy and Power | 功、能与功率
Work done by a constant force is W = F s cos θ, where s is the displacement and θ is the angle between force and displacement. Kinetic energy is KE = ½ m v², and gravitational potential energy is GPE = mgh. The work-energy principle states that the total work done by all external forces equals the change in kinetic energy.
恒力做功为 W = F s cos θ，其中 s 是位移，θ 是力与位移之间的夹角。动能为 KE = ½ m v²，重力势能为 GPE = mgh。功—能原理指出，所有外力所做的总功等于动能的变化量。
Power is the rate of doing work: P = W / t for average power, and P = Fv for a force moving at speed v along its line of action. When a car or particle moves up a rough incline, include work done against friction and changes in GPE in the energy equation. Quoting energy values in joules avoids sign errors from vector forces.
功率是做功的速率：平均功率为 P = W / t，力沿其作用线以速度 v 移动时的功率为 P = Fv。当汽车或质点沿粗糙斜面向上运动时，应在能量方程中包含克服摩擦力所做的功和重力势能的变化。以焦耳为单位表示能量值可避免矢量力带来的符号错误。
7. Momentum and Impulse | 动量与冲量
Momentum is a vector quantity defined by p = mv. Impulse is the change in momentum caused by a force acting over time. For a constant force, I = F Δt, but the impulse-momentum principle is always true: impulse equals final momentum minus initial momentum, I = mv − mu.
动量是由 p = mv 定义的矢量。冲量是力在一段时间内作用引起的动量变化。对于恒力，I = F Δt，但冲量—动量原理始终成立：冲量等于末动量减初动量，即 I = mv − mu。
In any interaction where no external force acts, total momentum is conserved. For two particles, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. Always define the positive direction before substituting signs. A negative final velocity means the particle travels in the opposite direction to the chosen positive sense.
在任何没有外力作用的相互作用中，总动量守恒。对于两个质点，m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂。代入符号前必须先定义正方向。负的末速度表示质点沿所选正方向的相反方向运动。
8. Collisions and Newton’s Law of Restitution | 碰撞与牛顿恢复系数
Newton’s law of restitution relates the relative speeds of two particles before and after a direct collision. It states that e = separation speed / approach speed. For a direct collision between two particles with initial velocities u₁ and u₂ and final velocities v₁ and v₂, the equation is e = (v₂ − v₁) / (u₁ − u₂), where u₁ > u₂ before impact.
牛顿恢复系数定律将两个质点在对心碰撞前后的相对速度联系起来。它指出 e = 分离速度 / 接近速度。对于两个质点初速度为 u₁ 和 u₂、末速度为 v₁ 和 v₂ 的对心碰撞，方程为 e = (v₂ − v₁) / (u₁ − u₂)，其中碰撞前 u₁ > u₂。
When e = 1, the collision is perfectly elastic and kinetic energy is conserved. When e = 0, the collision is perfectly inelastic and the particles move together after impact. In most AQA questions, e lies strictly between 0 and 1, so kinetic energy is lost while momentum is conserved. Solve the momentum and restitution equations simultaneously.
当 e = 1 时，碰撞为完全弹性碰撞，动能守恒。当 e = 0 时，碰撞为完全非弹性碰撞，碰撞后质点一起运动。在大多数 AQA 题目中，e 严格介于 0 和 1 之间，因此动量守恒而动能损失。应将动量方程和恢复系数方程联立求解。
9. Uniform Circular Motion | 匀速圆周运动
For a particle moving in a circle of radius r at constant speed v, the speed does not change but the velocity direction does. The acceleration is directed toward the centre of the circle with magnitude a = v² / r = rω², where ω is the angular speed in rad s⁻¹. The relation between linear and angular speed is v = rω.
对于以恒定速度 v 在半径为 r 的圆上运动的质点，速度大小不变但方向改变。加速度指向圆心，大小为 a = v² / r = rω²，其中 ω 是以 rad s⁻¹ 为单位的角速度。线速度与角速度的关系为 v = rω。
The resultant force towards the centre is F = mv² / r = mrω². This centripetal force is not an extra force; it must be provided by tension, friction, the normal reaction, or gravity. In a vertical circle, the speed is not uniform, so energy conservation is often needed alongside the radial force equation.
指向中心的合力为 F = mv² / r = mrω²。这个向心力不是额外的力；它必须由张力、摩擦力、法向反作用力或重力提供。在竖直圆周运动中，速度不是恒定的，因此除了径向力方程外，通常还需要使用能量守恒。
10. Exam Technique and Common Pitfalls | 考试技巧与常见错误
Draw a clear, labelled diagram even for simple problems. Mark the positive direction, pivot points, angles and all forces. For moments, choose the pivot to eliminate an unknown force when possible. For projectiles, keep horizontal and vertical calculations separate and decide whether time is the common link. For collisions, assign directions to velocities before writing equations.
即使是简单问题也要绘制清晰、标注完整的示意图。标出正方向、支点、角度和所有力。对于力矩问题，尽可能选择能消去未知力的支点。对于抛体问题，将水平与竖直计算分开，并判断时间是否为共同联系。对于碰撞问题，在写方程之前先规定速度的方向。
- Use g = 9.8 m s⁻² unless the question gives another value, and quote answers to an appropriate degree of accuracy.
- Write speed as positive; use negative signs only for velocity components in a chosen positive direction.
- In ‘show that’ questions, retain surds and exact expressions until the required form is reached.
- Check that every term in an energy equation has the same unit, usually joules.
- Interpret your final result: a negative answer often represents a direction, not an error.
使用 g = 9.8 m s⁻²，除非题目给出其他值，并在答案中取适当的精度。速度写为正数；只有速度分量在选定正方向相反时才使用负号。在 ‘show that’ 题目中，保留根号和精确表达式，直到得到所需形式。检查能量方程中每一项是否具有相同单位，通常为焦耳。解释最终结果：负答案通常表示方向，而不是错误。
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