📚 A-Level Mathematics 9660 Mechanics Unit 2 (MA05) Key Concepts | A-Level数学9660力学单元2(MA05)知识点精讲
This comprehensive guide covers the essential topics in Mechanics Unit 2 (MA05) for the International A-Level Mathematics 9660 specification. We explore projectiles, variable acceleration, work and energy, power, Hooke’s law, moments, equilibrium of rigid bodies, centre of mass, and the application of calculus to dynamics. Every concept is explained with clear English–Chinese paired explanations, key formulas, and typical exam insights.
本精讲全面覆盖国际A-Level数学9660力学单元2（MA05）的核心知识点。内容包括抛体运动、变加速度、功与能、功率、胡克定律、力矩、刚体平衡、质心以及微积分在动力学中的应用。每个概念均以英中双语对照讲解，并配有核心公式和典型考点剖析。
1. Projectile Motion | 抛体运动
A projectile moves under constant gravity with zero horizontal acceleration. The horizontal velocity remains constant, while the vertical motion is governed by the uniform acceleration equations.
抛体在恒定重力下运动，水平方向加速度为零。水平速度保持不变，而竖直方向遵循匀加速运动规律。
We resolve the initial velocity u at angle θ: uₓ = u cos θ, uᵧ = u sin θ. The equations of motion become:
我们将初速度 u 按角度 θ 分解：uₓ = u cos θ，uᵧ = u sin θ。运动方程变为：
vₓ = u cos θ, sₓ = u cos θ × t
vᵧ = u sin θ − gt, sᵧ = u sin θ × t − ½gt²
The maximum height H occurs when vᵧ = 0, giving H = (u² sin²θ) / (2g). The total time of flight T = 2u sinθ / g, and the horizontal range R = (u² sin 2θ) / g.
当 vᵧ = 0 时达到最大高度 H，H = (u² sin²θ) / (2g)。总飞行时间 T = 2u sinθ / g，水平射程 R = (u² sin 2θ) / g。
For a given speed, the maximum range is achieved when θ = 45°. The trajectory is parabolic, described by the equation y = x tan θ − (g x²) / (2u² cos²θ).
对于给定速率，当 θ = 45° 时射程最大。轨迹为抛物线，方程为 y = x tan θ − (g x²) / (2u² cos²θ)。
2. Variable Acceleration Using Calculus | 用微积分处理变加速度
When acceleration is not constant, we use calculus to link displacement x, velocity v, and acceleration a.
当加速度不恒定时，我们借助微积分联系位移 x、速度 v 和加速度 a。
The fundamental relations are a = dv/dt, v = dx/dt, and a = v dv/dx. We find velocity by integration: v = ∫ a dt, and displacement: x = ∫ v dt. Initial conditions are used to determine constants of integration.
基本关系式为 a = dv/dt，v = dx/dt，以及 a = v dv/dx。通过对加速度积分求速度：v = ∫ a dt，位移：x = ∫ v dt，并利用初始条件确定积分常数。
v(t) = v₀ + ∫₀ᵗ a(s) ds, x(t) = x₀ + ∫₀ᵗ v(s) ds
For example, if a = 6t, then v = 3t² + C. With v(0) = 2, we get v = 3t² + 2, and then x = ∫(3t² + 2) dt = t³ + 2t + D.
例如，若 a = 6t，则 v = 3t² + C。若 v(0) = 2，得 v = 3t² + 2，进而 x = ∫(3t² + 2) dt = t³ + 2t + D。
Problems often ask for maximum velocity or when a particle changes direction. Set v = 0 to find turning points, and use a = v dv/dx to relate position and speed.
考题常要求最大速度或粒子改变方向的时刻。设 v = 0 求转向点，并可用 a = v dv/dx 联系位置与速率。
3. Work and Energy Principles | 功与能原理
Work done by a constant force is the product of the force and the displacement in its direction: W = F s cos θ. When the force varies, we use integration: W = ∫ F dx.
恒力做功等于力与沿力方向位移的乘积：W = F s cos θ。若力变化，则用积分：W = ∫ F dx。
The work–energy theorem states that the net work done on a particle equals its change in kinetic energy: W_net = ΔKE = ½mv² − ½mu².
功能定理指出，作用于质点的净功等于其动能变化量：W_net = ΔKE = ½mv² − ½mu²。
W = ∫ₓ₁ˣ² F dx = ½m v₂² − ½m v₁²
Work done against gravity increases gravitational potential energy (GPE = mgh), while work against a spring stores elastic potential energy. For systems with only conservative forces, total mechanical energy is conserved.
克服重力做功增加重力势能 (GPE = mgh)，克服弹力做功储存弹性势能。对只有保守力的系统，总机械能守恒。
4. Kinetic and Potential Energy | 动能与势能
Kinetic energy (KE) of a particle of mass m moving with speed v is given by KE = ½mv². It is always non‑negative and is measured in joules (J).
质量为 m、速率为 v 的质点的动能 (KE) 为 KE = ½mv²，始终非负，单位为焦耳 (J)。
Gravitational potential energy near the Earth’s surface is GPE = mgh, where h is the vertical height above an arbitrary reference level. Changes in GPE depend only on height difference.
近地表重力势能为 GPE = mgh，h 为相对于任意参考面的竖直高度。GPE 的变化仅取决于高度差。
Elastic potential energy stored in a stretched spring or string obeys Hooke’s law: EPE = ½kx² or, for a light elastic string, EPE = λx²/(2l), where λ is the modulus of elasticity and l the natural length.
拉伸弹簧或弹性绳中储存的弹性势能遵循胡克定律：EPE = ½kx²，或对轻质弹性绳，EPE = λx²/(2l)，其中 λ 为弹性模量，l 为原长。
Total mechanical energy = KE + GPE + EPE
When only gravity and elastic forces do work, the sum remains constant. This principle simplifies many dynamical problems.
当只有重力和弹力做功时，三者之和保持不变，可简化许多动力学问题。
5. Power and Efficiency | 功率与效率
Power is the rate of doing work: P = dW/dt. For a constant force moving at velocity v, the instantaneous power is P = F v. The SI unit is the watt (W).
功率是做功的快慢：P = dW/dt。对以速度 v 运动的恒力，瞬时功率为 P = F v。国际单位是瓦特 (W)。
P = F v cos θ
When a vehicle moves at constant speed against resistance, the driving force equals the resistance, and the power developed is P = R v. Maximum power and maximum tractive effort are key concepts in vehicle dynamics.
当车辆匀速行驶抵抗阻力时，驱动力等于阻力，产生功率为 P = R v。最大功率与最大牵引力是车辆动力学的重要概念。
Efficiency is defined as useful output power divided by total input power, often expressed as a percentage. In mechanics, energy losses arise from friction and air resistance.
效率定义为有用输出功率除以总输入功率，通常以百分数表示。力学中的能量损失来自摩擦和空气阻力。
Problems may involve static resistance, variable slope, and limits on power, requiring careful use of P = Fv and Newton’s second law.
考题可能涉及恒定阻力、可变坡度和功率限制，需综合运用 P = Fv 和牛顿第二定律。
6. Hooke’s Law and Elastic Potential Energy | 胡克定律与弹性势能
Hooke’s law states that the tension T in an elastic string or spring is proportional to its extension x from its natural length l: T = λx/l, where λ is the modulus of elasticity.
胡克定律指出，弹性绳或弹簧中的张力 T 与其相对于原长 l 的伸长量 x 成正比：T = λx/l，λ 为弹性模量。
The energy stored is the work done in stretching: EPE = ∫₀ˣ T dx = λx²/(2l) or equivalently ½kx² with k = λ/l.
储存的能量为拉伸过程中所做的功：EPE = ∫₀ˣ T dx = λx²/(2l)，或写成 ½kx²，其中 k = λ/l。
EPE = λx²/(2l) = ½kx²
In dynamics, when a particle is attached to an elastic string moving vertically or on a smooth incline, energy conservation often gives the quickest solution for speed at a given extension.
在动力学中，当质点连接弹性绳在竖直方向或光滑斜面上运动时，能量守恒常可快速求得给定伸长量下的速率。
Note that a string is only taut when extended; if the string becomes slack, the tension is zero and EPE is zero.
注意，弹性绳仅在伸长时绷紧；若绳子松弛，张力为零，弹性势能亦为零。
7. Moments of Forces | 力矩
The moment of a force about a point is a measure of its turning effect. It is defined as the product of the force and the perpendicular distance from the point to the line of action: M = F d.
力对一点的力矩衡量其转动效应，定义为力与点到力作用线的垂直距离的乘积：M = F d。
Moments are usually taken as positive in one sense (e.g., clockwise) and negative in the opposite. The net moment about any point for a body in rotational equilibrium is zero.
力矩通常规定某一转向（如顺时针）为正，反向为负。刚体处于转动平衡时，对任一点的合力矩为零。
When a rod is acted upon by several forces, taking moments about a support eliminates unknown reaction forces, making it easier to solve for tensions or other reactions.
当杆受到多个力作用时，对某支点取矩可消去未知反力，便于求解张力或其他约束反力。
Levers, ladders, and bridges are typical exam contexts. Always draw a clear force diagram and indicate the perpendicular distances.
杠杆、梯子和桥梁是典型的考试情境。务必绘制清晰的受力图并标出垂直距离。
8. Equilibrium of Rigid Bodies | 刚体平衡
A rigid body is in static equilibrium when both the resultant force and the resultant moment are zero. This gives two vector conditions, or three scalar equations in two dimensions.
当合外力与合力矩均为零时，刚体处于静平衡。在二维情况下表示为三个标量方程。
Σ F_x = 0, Σ F_y = 0, Σ M_A = 0
Choosing the point A wisely, e.g., at a hinge or where two unknown forces meet, simplifies the moment equation.
恰当选择矩心 A，如铰链或两未知力交点，可简化力矩方程。
Typical problems involve uniform rods, ladders leaning against rough walls, and systems with strings and pulleys. Friction introduces a limiting condition: F ≤ μR, where μ is the coefficient of friction.
典型问题包括均质杆、倚靠粗糙墙面的梯子以及带有绳和滑轮的体系。摩擦引入极限条件：F ≤ μR，其中 μ 为摩擦系数。
The possibility of sliding or toppling must be checked. For a body on an inclined plane, resolve weight parallel and perpendicular to the slope and apply equilibrium or impending motion conditions.
需检查滑动或倾覆的可能性。对于斜面上的物体，沿斜面与垂直斜面分解重力，应用平衡或即将运动条件。
9. Centre of Mass | 质心
The centre of mass of a system of particles is the weighted average of their positions: r_cm = ( Σ m_i r_i ) / Σ m_i . In coordinates: x_cm = Σ(m_i x_i)/Σm_i, y_cm = Σ(m_i y_i)/Σm_i.
质点系的质心是位置的加权平均：r_cm = ( Σ m_i r_i ) / Σ m_i。在坐标中：x_cm = Σ(m_i x_i)/Σm_i，y_cm = Σ(m_i y_i)/Σm_i。
For uniform laminas, the centre of mass lies on any axis of symmetry. Standard results exist: for a uniform triangle, the centre of mass is at the intersection of the medians, ⅓ of the way from the base along the median.
对于均质薄片，质心位于对称轴上。标准结论有：均质三角形的质心在中线交点，距底边中线的 ⅓ 处。
A composite body can be treated by splitting into simple shapes and using the formulas for centres of mass. For a uniform sector of angle 2α and radius r, the centre of mass lies on the axis of symmetry at distance (2r sin α)/(3α) from the centre.
组合体可拆分为简单形状并使用质心公式处理。对于圆心角 2α、半径 r 的均质扇形，质心在对称轴上，距圆心 (2r sin α)/(3α)。
Hanging a lamina freely from a point means the centre of mass is vertically below the pivot. This fact helps determine unknown angles or lengths.
将薄片从某点自由悬挂，质心竖直位于悬挂点下方。利用此性质可求未知角度或长度。
10. Motion with Variable Forces | 变力作用下的运动
When a force depends on displacement, such as a spring force, we can integrate to find work done and use the work–energy theorem. Alternatively, write a = v dv/dx and integrate.
当力依赖于位移时（如弹簧力），可通过积分求功并应用功能原理，或利用 a = v dv/dx 进行积分。
For a particle of mass m moving under a force F(x), we have: ∫ F(x) dx = ½m v² − ½m u². This bypasses the need to solve the differential equation of motion directly.
对于在力 F(x) 作用下的质量为 m 的质点，有 ∫ F(x) dx = ½m v² − ½m u²。这避免直接求解运动微分方程。
A classic example: a particle attached to a spring moves from rest at extension a to extension b; the change in kinetic energy equals the difference in EPE: ½m v² = (λ/(2l))(a² − b²).
经典例题：连接弹簧的质点从伸长量 a 处由静止运动到 b 处，动能变化等于弹性势能之差：½m v² = (λ/(2l))(a² − b²)。
When drag forces depend on velocity, the equation of motion becomes a first‑order differential equation, often solved by separation of variables or integrating factors.
当阻力依赖于速度时，运动方程化为一阶微分方程，常通过分离变量或积分因子求解。
Problems integrating up thrust, variable resistance, or power require careful handling of initial conditions and sign conventions. Always check for maxima and minima using dv/dt = 0.
涉及变推力、变阻力或变功率的积分问题，须谨慎处理初始条件与符号约定。通常利用 dv/dt = 0 检查最大或最小值。
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