📚 A-Level Further Mathematics 9665 Mechanics Topic Test: Key Concepts Explained | A-Level 进阶数学 9665 力学专题测试：知识点精讲
Welcome to this comprehensive revision guide for the Oxford AQA International A-Level Further Mathematics 9665 Mechanics topic test. Mechanics in Further Mathematics extends beyond the standard A-Level syllabus, introducing more sophisticated modelling techniques and deeper mathematical rigour. This article systematically unpacks the essential knowledge areas you need to master, including advanced kinematics, dynamics of rigid bodies, work-energy principles, impulse and momentum, circular motion, and centres of mass. Each section is carefully structured to align with the 9665 specification requirements, ensuring you build both conceptual understanding and problem-solving fluency.
欢迎阅读这篇针对 Oxford AQA International A-Level 进阶数学 9665 力学专题测试的全面复习指南。进阶数学中的力学超越了标准 A-Level 大纲，引入了更复杂的建模技巧和更深的数学严谨性。本文系统地梳理了你需要掌握的核心知识领域，包括高阶运动学、刚体动力学、功与能原理、冲量与动量、圆周运动以及质心。每个部分都精心编排，与 9665 大纲要求保持一致，确保你既能建立概念理解，又能提升解题的流畅度。
1. Advanced Kinematics in One and Two Dimensions | 一维与二维高阶运动学
In Further Mechanics, kinematics extends into vector notation and parametric forms. Displacement, velocity, and acceleration are treated as functions of time, often requiring differentiation and integration of vector quantities. For a particle moving in a plane, its position vector is given by r(t) = x(t)i + y(t)j. Velocity is the first derivative v = dr/dt, and acceleration is the second derivative a = d²r/dt². You must be comfortable switching between Cartesian components and magnitude-direction representations.
在进阶力学中，运动学拓展到了向量表示和参数形式。位移、速度和加速度被视作时间的函数，通常需要对向量进行微分和积分运算。对于一个在平面内运动的质点，其位置向量表示为 r(t) = x(t)i + y(t)j。速度是一阶导数 v = dr/dt，加速度是二阶导数 a = d²r/dt²。你必须熟练地在笛卡尔分量表示法和模长-方向表示法之间进行切换。
Projectile motion under uniform gravity is a core application. Assuming initial speed U and launch angle θ, the horizontal and vertical displacements are modelled parametrically. The Cartesian equation of the trajectory eliminates t, yielding a parabolic path. Key derived results include time of flight T = 2U sinθ/g, maximum height H = (U sinθ)²/(2g), and horizontal range R = (U² sin2θ)/g. Air resistance is neglected in these standard models.
均匀重力场下的抛体运动是一个核心应用。假设初速度为 U、发射角为 θ，水平和垂直位移可以用参数形式建模。轨迹的笛卡尔方程通过消去 t 得到，表现为一条抛物线路径。关键推导结果包括飞行时间 T = 2U sinθ/g、最大高度 H = (U sinθ)²/(2g) 以及水平射程 R = (U² sin2θ)/g。在这些标准模型中，空气阻力被忽略。
Variable acceleration problems demand fluency with calculus. If acceleration is given as a function of time, velocity and displacement are found by direct integration, with initial conditions determining the constants. When acceleration is a function of displacement, use the alternative form a = v dv/dx. This leads to separable differential equations, a technique tested frequently in the 9665 topic test.
变加速问题要求熟练运用微积分。如果加速度以时间函数的形式给出，速度和位移通过直接积分求得，初始条件决定积分常数。当加速度是位移的函数时，需要使用替代形式 a = v dv/dx。这会导向可分离的微分方程，这一技巧在 9665 专题测试中经常考到。
2. Newton’s Laws and Connected Particles | 牛顿定律与连接体系统
Newton’s three laws form the axiomatic foundation of classical mechanics. The first law defines inertial frames; the second law expresses resultant force as F = ma; the third law asserts action-reaction pairs. In Further Mathematics, these laws are applied to systems of connected particles, pulleys, and bodies on inclined planes with friction. Resolving forces into components along and perpendicular to the plane of motion is a fundamental skill.
牛顿三大定律构成了经典力学的公理基础。第一定律定义了惯性参考系；第二定律将合力表示为 F = ma；第三定律确立了作用力与反作用力对。在进阶数学中，这些定律被应用于连接体系统、滑轮以及带摩擦的斜面物体。将力沿运动平面方向和垂直方向进行分解，是一项基本技能。
For a particle on a rough inclined plane at angle θ, the normal reaction is R = mg cosθ. The maximum static friction is F_max = μR, where μ is the coefficient of friction. Motion occurs down the plane when the component of weight mg sinθ exceeds μR. When solving connected particle problems, draw clear force diagrams, label tension T in inextensible strings, and write equations of motion for each particle separately before solving simultaneously.
对于一个位于粗糙斜面、倾角为 θ 的质点，法向反作用力为 R = mg cosθ。最大静摩擦力为 F_max = μR，其中 μ 是摩擦系数。当重力分量 mg sinθ 超过 μR 时，物体沿斜面下滑。在解决连接体问题时，要画出清晰的受力图，标出不可伸长的绳子中的张力 T，并分别为每个质点写出运动方程，然后联立求解。
Pulley systems often involve two masses connected by a light inextensible string passing over a smooth pulley. The string transmits tension uniformly, and the magnitudes of acceleration for both masses are equal. Deriving the acceleration a = (m₁ – m₂)g/(m₁ + m₂) for a simple Atwood machine, and understanding how to modify this when masses rest on surfaces or experience friction, is essential for success in the topic test.
滑轮系统通常涉及两个通过轻质不可伸长绳子连接、绳子跨过光滑滑轮的物体。绳子均匀传递张力，两个物体加速度的大小相等。对于简单的阿特伍德机，推导出加速度 a = (m₁ – m₂)g/(m₁ + m₂)，并理解当物体放置在表面上或受到摩擦力时如何对公式进行修改，对于在专题测试中取得成功至关重要。
3. Work, Energy, and Power | 功、能与功率
The work done by a constant force F moving its point of application a displacement s in the direction of the force is W = Fs. When force and displacement are not aligned, the scalar product is used: W = F · s = Fs cosθ. For variable forces, work is the integral of force with respect to displacement: W = ∫F · dr. The work-energy principle states that the net work done on a particle equals its change in kinetic energy.
一个恒力 F 使其作用点沿力的方向发生位移 s 所做的功为 W = Fs。当力与位移不在同一方向时，需要使用标量积：W = F · s = Fs cosθ。对于变力，功是力对位移的积分：W = ∫F · dr。功与能原理指出，作用在一个质点上的净功等于其动能的变化量。
Kinetic energy is defined as KE = ½mv². Gravitational potential energy is GPE = mgh, measured from a chosen reference level. Elastic potential energy stored in a spring obeying Hooke’s law is EPE = ½kx², where k is the spring constant and x is the extension or compression. The principle of conservation of mechanical energy applies when only conservative forces (gravity, elastic spring forces) do work.
动能定义为 KE = ½mv²。重力势能为 GPE = mgh，从选定的参考水平面开始测量。储存在遵循胡克定律的弹簧中的弹性势能为 EPE = ½kx²，其中 k 是弹簧常量，x 是伸长量或压缩量。当只有保守力（重力、弹簧弹力）做功时，机械能守恒原理适用。
Power is the rate of doing work, defined as P = dW/dt. For a force moving its point of application at velocity v, instantaneous power is P = F · v. In problems involving vehicles moving against resistance, the tractive force exerted by the engine relates to power by F = P/v. At maximum speed, the tractive force equals the total resistance, enabling calculation of terminal velocity.
功率是做功的速率，定义为 P = dW/dt。对于一个以速度 v 移动其作用点的力，瞬时功率为 P = F · v。在涉及车辆克服阻力运动的问题中，发动机施加的牵引力与功率的关系为 F = P/v。在最大速度下，牵引力等于总阻力，从而可以计算出极限速度。
4. Impulse and Momentum in One Dimension | 一维冲量与动量
Momentum is a vector quantity defined as p = mv. Impulse, the effect of a force acting over a time interval, is the change in momentum: J = Δp = ∫F dt. For constant force, impulse simplifies to J = FΔt. The impulse-momentum theorem provides a powerful alternative to Newton’s second law, particularly useful when forces act for very short durations, such as in collisions or impacts.
动量是一个矢量，定义为 p = mv。冲量是力在一段时间间隔内作用所产生的效果，等于动量的变化量：J = Δp = ∫F dt。对于恒力，冲量简化为 J = FΔt。冲量-动量定理为牛顿第二定律提供了一个有力的替代方案，在力作用时间非常短的情况下（例如碰撞或冲击中）尤为有用。
The principle of conservation of linear momentum states that if no external forces act on a system, the total momentum before and after an interaction remains constant. For two colliding bodies along a straight line: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. This principle is fundamental to solving collision problems, including those where particles coalesce (perfectly inelastic collisions) or rebound.
动量守恒定律指出，如果没有外力作用于系统，相互作用前后的总动量保持不变。对于沿同一直线碰撞的两个物体：m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂。这一原理是解决碰撞问题的基础，包括质点合为一体（完全非弹性碰撞）或发生反弹的情况。
Newton’s experimental law of restitution defines the coefficient of restitution e, which measures the elasticity of a collision. For a direct impact between two bodies, e = (v₂ – v₁) / (u₁ – u₂), where u and v are velocities before and after impact. The value e = 1 corresponds to perfectly elastic collisions (kinetic energy conserved); e = 0 corresponds to perfectly inelastic collisions. Most real collisions have 0 < e < 1.
牛顿实验性恢复定律定义了恢复系数 e，用以衡量碰撞的弹性程度。对于两个物体之间的正碰，e = (v₂ – v₁) / (u₁ – u₂)，其中 u 和 v 分别是碰撞前后的速度。e = 1 对应完全弹性碰撞（动能守恒）；e = 0 对应完全非弹性碰撞。大多数真实的碰撞满足 0 < e < 1。
5. Oblique Impacts and Momentum in Two Dimensions | 斜碰与二维动量
Oblique impacts require vector treatment of momentum and impulse. The key insight is to decompose velocities into components parallel and perpendicular to the line of impact (the common normal at the point of contact). The component parallel to the line of impact obeys the law of restitution and conservation of momentum along that direction. The component perpendicular to the line of impact remains unchanged for smooth bodies because no impulse acts in that direction.
斜碰需要对动量和冲量进行向量处理。关键思路是将速度分解为平行于和垂直于碰撞线（接触点处的公法线）的分量。平行于碰撞线的分量遵循恢复定律和该方向上的动量守恒。垂直于碰撞线的分量对于光滑物体保持不变，因为在该方向上没有冲量作用。
For two smooth spheres colliding obliquely, let the line of centres be the x-axis. Before impact, velocities have components u₁ₓ, u₁ᵧ, u₂ₓ, u₂ᵧ. After impact, the y-components are unchanged: v₁ᵧ = u₁ᵧ, v₂ᵧ = u₂ᵧ. The x-components satisfy the usual one-dimensional impact equations incorporating the coefficient of restitution e. Solving these simultaneous equations yields all post-collision velocity components.
对于两个光滑球体发生斜碰的情况，设球心连线为 x 轴。碰撞前，速度分量为 u₁ₓ、u₁ᵧ、u₂ₓ、u₂ᵧ。碰撞后，y 方向分量不变：v₁ᵧ = u₁ᵧ，v₂ᵧ = u₂ᵧ。x 方向分量则满足包含恢复系数 e 的常规一维碰撞方程。解这些联立方程即可得到所有碰撞后的速度分量。
Vector impulse on each particle equals its change in momentum. The impulse exerted by the first particle on the second acts along the line of centres. This vector approach is crucial for solving problems where particles strike fixed planes obliquely or where the line of impact is not aligned with the initial direction of motion. Always draw clear vector diagrams and label pre- and post-collision velocities.
每个质点受到的冲量向量等于其动量的变化量。第一个质点施加给第二个质点的冲量沿球心连线方向。这种向量方法对于质点斜向撞击固定平面或碰撞线与初始运动方向不一致的问题至关重要。始终要画出清晰的向量图，并标出碰撞前后的速度。
6. Circular Motion with Constant Speed | 匀速圆周运动
A particle moving in a circle of radius r with constant angular speed ω has speed v = rω. Though the speed is constant, the velocity is continuously changing direction, so the particle experiences centripetal acceleration directed towards the centre of the circle. The magnitude of this acceleration is a = v²/r = rω². According to Newton’s second law, the net force towards the centre is the centripetal force F = mv²/r = mrω².
一个以恒定角速度 ω 在半径为 r 的圆周上运动的质点，其速率为 v = rω。尽管速率恒定，但速度方向不断变化，因此质点受到一个指向圆心的向心加速度。该加速度的大小为 a = v²/r = rω²。根据牛顿第二定律，指向圆心的净力即为向心力 F = mv²/r = mrω²。
The centripetal force is not a distinct type of force but rather the resultant of actual forces acting radially inwards. Common sources include tension in a string (conical pendulum, horizontal circle), the normal reaction from a track (vertical circular motion, looping), friction (car rounding a banked curve), and gravitational attraction (orbital motion). Analysing circular motion requires resolving forces along the radial direction and equating the net inward force to mv²/r or mrω².
向心力并非一种独特的力，而是实际作用力沿径向向内的合力。常见的来源包括绳子中的张力（圆锥摆、水平圆周）、轨道的法向反作用力（竖直圆周运动、过山车翻滚）、摩擦力（汽车转弯）和万有引力（轨道运动）。分析圆周运动需要沿径向分解力，并将向内的净力等于 mv²/r 或 mrω²。
The conical pendulum consists of a mass attached to a string, moving in a horizontal circle with the string tracing out a cone. Resolving vertically, T cosθ = mg. Resolving radially, T sinθ = mrω². The geometry relates radius r to string length L by r = L sinθ. Eliminating T and r yields the relationship between angular speed and the angle: ω² = g/(L cosθ). The period of revolution is T_period = 2π√(L cosθ/g).
圆锥摆由一个系在绳子上的质量块组成，它在水平面内做圆周运动，绳子划出一个圆锥面。竖直方向分解得 T cosθ = mg。径向分解得 T sinθ = mrω²。几何关系将半径 r 与绳长 L 关联起来：r = L sinθ。消去 T 和 r 得到角速度与角度之间的关系：ω² = g/(L cosθ)。旋转周期为 T_period = 2π√(L cosθ/g)。
7. Vertical Circular Motion and Energy | 竖直平面圆周运动与能量
When a particle moves in a vertical circle, gravity does work, so speed varies with position. Conservation of mechanical energy, combined with radial force resolution, is the standard analytical approach. For a particle attached to a light rod or string of length r, at an angular position θ measured from the downward vertical, the speed v at any point can be found from the initial conditions using energy conservation.
当质点在竖直面内做圆周运动时，重力做功，因此速率随位置变化。机械能守恒，结合径向力的分解，是标准的分析方法。对于一个系在长度为 r 的轻杆或绳子上的质点，在从竖直向下方向起量度为 θ 的角位置处，任意点的速率 v 都可以通过初始条件和能量守恒求得。
A critical condition for completing a full vertical circle on a string is that the string must remain taut at the highest point. At the top, the tension T and weight mg both act downwards, providing the centripetal force: T + mg = mv²/r. For the string to be just taut, T ≥ 0, so the minimum speed at the top is v_min = √(gr). Using energy conservation, the required minimum speed at the lowest point is then √(5gr).
在绳子上完成完整竖直圆周的一个关键条件是，绳子在最高点必须保持紧绷。在顶部，张力 T 和重力 mg 都向下作用，提供向心力：T + mg = mv²/r。为了使绳子刚好绷直，T ≥ 0，因此顶部的最小速率为 v_min = √(gr)。利用能量守恒，最低点所需的最小速率则为 √(5gr)。
For a bead threaded on a smooth circular wire or a particle on the outside of a smooth sphere, the normal reaction may drop to zero at some point, causing the particle to leave the circular path. The condition for loss of contact is that the normal reaction N = 0. Solving N = 0 simultaneously with energy and radial force equations determines the angle and speed at which the particle detaches from the surface.
对于穿在光滑圆形金属丝上的珠子，或位于光滑球体外表面上的质点，法向反作用力可能在某个点降为零，导致质点脱离圆形路径。脱离接触的条件是法向反作用力 N = 0。将 N = 0 与能量方程和径向力方程联立求解，可以确定质点脱离表面时的角度和速率。
8. Rigid Body Statics and Moments | 刚体静力学与力矩
A rigid body is an idealised system where the distances between all constituent particles remain fixed. For a rigid body in static equilibrium, both the resultant force and the resultant moment about any point must be zero. The moment of a force about a point is the product of the force’s magnitude and the perpendicular distance from the point to the line of action: M = Fd. Moments tending to cause rotation in opposite directions are assigned opposite signs.
刚体是一个理想化系统，其中所有组成质点之间的距离保持固定。对于处于静力平衡的刚体，合力和对任意点的合力矩都必须为零。力对某点的力矩是力的大小乘以该点到力作用线的垂直距离：M = Fd。倾向于引起相反方向旋转的力矩被赋予相反的符号。
Solving rigid body equilibrium problems typically involves resolving forces in two perpendicular directions and taking moments about a strategically chosen point. Choosing a point through which unknown forces pass eliminates those forces from the moment equation, simplifying the algebra. Common scenarios include ladders leaning against rough walls, uniform rods hinged or supported, and loaded beams.
解决刚体平衡问题通常涉及沿两个互相垂直的方向分解力，并选取一个策略性位置作为力矩中心。选择一个未知力穿过的点作为矩心，可以使这些力不出现在力矩方程中，从而简化代数运算。常见的情形包括斜靠在粗糙墙壁上的梯子、铰接或支撑的均质杆以及受载梁。
The coefficient of friction and limiting equilibrium are central to ladder problems. At limiting equilibrium, the friction force attains its maximum value F = μR. Writing equilibrium equations for horizontal and vertical forces, together with a moment equation, yields conditions on the angle of inclination. The ladder will slip if the required friction exceeds μR at either contact point.
摩擦系数和极限平衡是梯子问题的核心。在极限平衡状态下，摩擦力达到其最大值 F = μR。写出水平和竖直方向的力平衡方程，再加上力矩方程，就能得到倾角的条件。如果在任一接触点所需的摩擦力超过 μR，梯子就会滑倒。
9. Centre of Mass of Discrete and Continuous Bodies | 离散与连续物体的质心
The centre of mass of a system of particles is the weighted average of their positions, with each particle’s mass as the weighting factor. For n particles in a plane, the coordinates are: x̄ = Σmᵢxᵢ / Σmᵢ, ȳ = Σmᵢyᵢ / Σmᵢ. For uniform bodies, the centre of mass coincides with the geometric centroid. Standard results for uniform laminas and solids should be memorised, including triangles, sectors, arcs, hemispheres, and cones.
质点系的质心是其位置的加权平均值，以每个质点的质量为权重因子。对于平面内的 n 个质点，坐标为：x̄ = Σmᵢxᵢ / Σmᵢ，ȳ = Σmᵢyᵢ / Σmᵢ。对于均质物体，质心与几何形心重合。应该熟记均质薄片和固体的标准结果，包括三角形、扇形、圆弧、半球和圆锥。
For a uniform triangular lamina, the centre of mass lies at the intersection of the medians, one-third of the way from each side to the opposite vertex. For a uniform sector of a circle of radius r and angle 2α (in radians), the centre of mass is at distance 2r sinα/(3α) from the centre along the axis of symmetry. For a uniform solid hemisphere, the centre of mass is at distance 3r/8 from the centre along the axis of symmetry.
对于均质三角形薄片，质心位于中线的交点处，在每条边上从该边到对顶点的三分之一处。对于半径为 r、圆心角为 2α（以弧度计）的均质扇形，质心位于对称轴上距中心 2r sinα/(3α) 处。对于均质实心半球，质心位于对称轴上距中心 3r/8 处。
Composite bodies can be analysed by treating them as combinations of simple shapes, or by subtraction when a shape has a removed portion. The overall centre of mass is the weighted average of the centres of mass of the constituent parts. For bodies in equilibrium, the centre of mass lies vertically below the point of suspension when freely suspended, a fact used to locate centres of mass experimentally.
复合体可以通过将其视为简单形状的组合来分析，或者当一个形状有孔洞切除部分时，使用减法处理。整体的质心是各组成部分质心的加权平均值。对于处于平衡的物体，当自由悬挂时，质心位于悬挂点的竖直下方，这一事实可用于实验确定质心位置。
10. Tilting and Stability of Rigid Bodies | 刚体的倾覆与稳定性
A rigid body resting on a surface is in equilibrium provided the vertical line through its centre of mass falls within its base of support. Tilting or toppling occurs when this line moves beyond the edge of the base. Solving tilting problems involves finding the position at which the normal reaction at the pivot edge becomes the sole support, and taking moments about that edge to determine the limiting condition.
放置于表面上的刚体，只要通过其质心的竖直线落在支撑底面内，就处于平衡状态。当这条线移出底面的边缘时，就会发生倾斜或倾覆。解决倾覆问题需要找出枢轴边缘处的法向反作用力成为唯一支撑力时的位置，并以该边缘为矩心计算力矩，以确定极限条件。
For a uniform rod leaning against a wall or placed on an inclined plane, the critical angle at which tilting occurs is independent of mass but depends on geometry and friction. When a body is on the point of tilting about an edge, the normal reaction and any friction at other contact points become zero. This simplification allows direct calculation of the force or angle causing instability.
对于斜靠在墙上或放置在斜面上的均质杆，发生倾覆的临界角与质量无关，但取决于几何形状和摩擦。当物体即将绕某条边缘倾覆时，其他接触点处的法向反作用力和摩擦力变为零。这种简化使得我们能够直接计算出导致不稳定的力或角度。
Stability and equilibrium are classified as stable, unstable, or neutral. A body is in stable equilibrium if a small displacement raises its centre of mass; it is unstable if a displacement lowers the centre of mass; neutral equilibrium occurs when small displacements do not change the height of the centre of mass. These principles apply to floating bodies, rocking toys, and design of structures.
稳定性和平衡可分为稳定平衡、不稳定平衡和随遇平衡。如果一个微小的位移使物体的质心升高，则物体处于稳定平衡状态；如果位移使质心降低，则为不稳定平衡；当微小位移不改变质心高度时，出现随遇平衡。这些原理适用于浮体、摇摆玩具以及结构设计。
11. Further Topics in Dynamics: Variable Mass and Rockets | 动力学进阶：变质量与火箭
Variable mass systems, such as rockets ejecting fuel or conveyor belts collecting material, require a modification of Newton’s second law. The mass is no longer constant, so the product rule must be applied when differentiating momentum. For a rocket expelling exhaust gases at constant velocity u relative to the rocket, the thrust is given by F_thrust = u |dm/dt|, directed opposite to the exhaust direction.
变质量系统，如喷射燃料的火箭或收集物料的传送带，需要对牛顿第二定律进行修正。质量不再恒定，因此在对动量求导时必须应用乘积法则。对于以相对于火箭的恒定速度 u 排出废气的火箭，推力为 F_thrust = u |dm/dt|，方向与排气方向相反。
The rocket equation in free space (no external forces) can be integrated to give the Tsiolkovsky equation: Δv = u ln(m₀/m), where m₀ is the initial mass, m is the final mass, and Δv is the change in velocity. In a uniform gravitational field, the net acceleration is reduced by g. Setting up and solving the differential equation m dv/dt = -mg + u dm/dt is a typical Further Mathematics problem.
在自由空间（无外力）中，火箭方程可以积分得到齐奥尔科夫斯基方程：Δv = u ln(m₀/m)，其中 m₀ 为初始质量，m 为最终质量，Δv 为速度变化量。在均匀重力场中，净加速度会减去 g。建立并求解微分方程 m dv/dt = -mg + u dm/dt 是典型的进阶数学问题。
Problems involving sand falling onto a moving conveyor belt, or chains being lifted from a pile, require careful consideration of the momentum change of the added mass. The force needed to accelerate the newly added material from rest to the belt speed is F = v dm/dt, where v is the belt speed and dm/dt is the rate of mass accumulation. This force is additional to any frictional or gravitational resistance.
涉及沙粒落到移动的传送带上，或者链条从堆中提起的问题，需要仔细考虑新增质量的动量变化。将新加入的材料从静止加速到传送带速度所需的力为 F = v dm/dt，其中 v 是传送带速度，dm/dt 是质量积累的速率。这个力是任何摩擦阻力或重力之外额外的力。
12. Differential Equations in Mechanics Problem-Solving | 力学问题中的微分方程
Many mechanics problems reduce to solving ordinary differential equations. First-order ODEs arise from F = m dv/dt or a = v dv/dx when force depends on velocity or displacement. Separable equations, integrating factor methods, and substitution techniques are all applicable. A common example is the motion of a particle under linear air resistance: m dv/dt = mg – kv, which yields an exponential approach to terminal velocity.
许多力学问题归结为求解常微分方程。当力依赖于速度或位移时，从 F = m dv/dt 或 a = v dv/dx 会产生一阶常微分方程。可分离方程、积分因子法和换元技巧都适用。一个常见的例子是质点在线性空气阻力下的运动：m dv/dt = mg – kv，它产生一个以指数方式逼近终端速度的解。
Second-order ODEs appear in simple harmonic motion (SHM) and damped oscillations. The standard equation for SHM is d²x/dt² = -ω²x, with general solution x = A cos(ωt) + B sin(ωt) or x = R cos(ωt – φ). The period is T = 2π/ω, independent of amplitude. Verifying that a given physical system satisfies this equation, and determining ω from the physical parameters, is a regular test requirement.
二阶常微分方程出现在简谐运动（SHM）和阻尼振动中。简谐运动的标准方程为 d²x/dt² = -ω²x，通解为 x = A cos(ωt) + B sin(ωt) 或 x = R cos(ωt – φ)。周期为 T = 2π/ω，与振幅无关。验证给定的物理系统满足该方程，并根据物理参数确定 ω，是常规的测试要求。
Setting up differential equations from physical descriptions is a higher-order skill. This involves translating a verbal problem statement into mathematical form by identifying the forces, applying Newton’s laws, and expressing constraints. Practice converting sentences such as ‘the resistance is proportional to the square of the speed’ into terms like -kv² in the force balance. Always check that the signs of terms correspond to the chosen positive direction.
根据物理描述建立微分方程是一项高阶技能。这涉及通过识别力、应用牛顿定律并表达约束条件，将文字问题表述转化为数学形式。练习将诸如“阻力与速度的平方成正比”这样的句子转化为力平衡中的 -kv² 这样的项。始终要检查各项的符号是否与选定的正方向一致。
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