AQA国际A-Level数学力学Unit2考点精讲
一、矢量与力的分解 | Vectors and Resolving Forces
在力学中，力、速度、加速度和位移都是既有大小又有方向的矢量（vector）。考试中最重要的基本功是把一个力分解成互相垂直的两个分量：沿水平方向 Fx = F cosθ，沿竖直方向 Fy = F sinθ。几乎所有涉及斜面、悬绳、抛体的题目，第一步都是先画受力图，再选正交方向分解，把矢量问题转化为两个独立的标量问题。例如，把斜面上的重力沿斜面和垂直斜面方向分解，摩擦力的计算会立刻清晰。记住：分解的方向选取得好，题目就会简单一半。
In mechanics, force, velocity, acceleration and displacement are all vector quantities with both magnitude and direction. The most fundamental exam skill is resolving a single force into two perpendicular components: a horizontal component Fx = F cosθ and a vertical component Fy = F sinθ. For almost every question involving slopes, suspended strings or projectiles, the first step is to draw a free-body diagram, choose two perpendicular directions, and resolve, turning a vector problem into two independent scalar problems. Resolving weight along and perpendicular to a slope, for instance, immediately clarifies how friction should be handled. A well-chosen pair of directions makes any question considerably simpler.
二、共点力平衡与合力为零 | Concurrent Forces in Equilibrium
当一个物体静止或匀速直线运动时，它处于平衡状态（equilibrium），作用在它上面所有力的合力必须为零。具体解题时，把平衡条件写成两个方程：水平方向合力为零，竖直方向合力为零。若题目给出三个共点力，还可以用拉密定理（Lami's theorem）或画封闭的力三角形（triangle of forces）快速求解。例如，一盏重 20 N 的灯由两条与竖直方向各成 30° 的绳悬挂，两条绳的张力大小相等，各约为 11.5 N。常见陷阱是误把"合力为零"写成"某个力为零"，必须对每个方向分别列方程。
When an object is at rest or moving in a straight line at constant speed, it is in equilibrium, and the resultant of all forces acting on it must be zero. In practice, you write the equilibrium condition as two equations: the horizontal resultant is zero and the vertical resultant is zero. If a question gives three concurrent forces, you can apply Lami's theorem or draw a closed triangle of forces to solve quickly. For example, a lamp weighing 20 N hangs from two strings each making 30° with the vertical; the tensions are equal and each is about 11.5 N. A common trap is writing "one force equals zero" instead of "the resultant is zero"; you must set up a separate equation for each direction.
三、牛顿三大运动定律 | Newton's Three Laws of Motion
牛顿第一定律说明物体保持静止或匀速运动状态，除非受到外力，它引出了惯性（inertia）的概念。第二定律给出核心公式 F = ma，即合力等于质量乘以加速度，这是力学题目的万能钥匙。第三定律强调作用力与反作用力大小相等、方向相反，但作用在不同物体上，所以不能相互抵消。例如，质量 2 kg 的物体受合力 10 N 作用，加速度就是 a = F/m = 5 m/s^2。解题时要先确定研究对象，只考虑作用在它身上的力，再用 F = ma 沿运动方向列方程。
Newton's first law states that a body remains at rest or in uniform motion unless acted on by an external force, introducing the concept of inertia. The second law gives the core formula F = ma: the resultant force equals mass times acceleration, and it is the master key to nearly every mechanics problem. The third law emphasises that action and reaction are equal in magnitude and opposite in direction, but they act on different bodies, so they never cancel. For example, a 2 kg body acted on by a resultant force of 10 N accelerates at a = F/m = 5 m/s^2. When solving, identify your target object, consider only the forces acting on it, then apply F = ma along the direction of motion.
四、斜面、摩擦力与动摩擦因数 | Inclined Planes, Friction and the Coefficient of Friction
斜面上物体的重力要分解成平行斜面的分量 mg sinθ 和垂直斜面的分量 mg cosθ，其中 θ 是斜面与水平面的夹角。摩擦力分为两类：物体将要滑动时是最大静摩擦力，正在滑动时是滑动摩擦力 F = μR，其中 R 是法向反作用力，μ 是摩擦因数。判断一个物体是否静止，要先比较 mg sinθ 与最大静摩擦力的大小。若 mg sinθ 小于 μmg cosθ，物体保持静止，此时静摩擦力大小等于 mg sinθ，而非 μmg cosθ。方向永远是阻碍相对运动或相对运动趋势。
On an inclined plane, the weight is resolved into a component parallel to the plane, mg sinθ, and a component perpendicular to the plane, mg cosθ, where θ is the angle of the slope. Friction comes in two forms: the limiting static friction just before sliding, and kinetic friction F = μR during sliding, where R is the normal reaction and μ is the coefficient of friction. To decide whether an object stays at rest, first compare mg sinθ with the maximum static friction. If mg sinθ is less than μmg cosθ, the object stays still, and the actual friction equals mg sinθ rather than μmg cosθ. Friction always opposes relative motion or the tendency toward it.
五、连接体与滑轮系统 | Connected Particles and Pulleys
两个物体用轻绳（light inextensible string）连接、绕过光滑滑轮时，绳中张力（tension）处处相等，两个物体加速度大小相同。解题方法有两种：一是分别对每个物体用 F = ma 列方程，二是把整个系统看作一个整体列整体方程，再回头求张力。例如，质量 3 kg 和 5 kg 的物体用轻绳绕过光滑定滑轮连接，整体加速度为 a = (5-3)g/(5+3) = g/4，方向朝重的一侧。轻绳意味着绳子质量可忽略，因此两端张力相等；不可伸长意味着两端速度、加速度大小相同。这是连接体题目的两个核心假设。
When two objects are joined by a light inextensible string passing over a smooth pulley, the tension is the same throughout the string and both objects share the same magnitude of acceleration. There are two solution methods: apply F = ma to each object separately, or treat the whole system as one body to write a single equation, then return to find the tension. For example, with masses of 3 kg and 5 kg joined over a smooth pulley, the system accelerates at a = (5-3)g/(5+3) = g/4 toward the heavier side. A light string means its mass is negligible, so the tension is equal at both ends; inextensibility means the two ends move with the same speed and acceleration. These are the two core assumptions behind connected-particle questions.
六、匀加速运动与SUVAT方程 | Constant Acceleration and the SUVAT Equations
当加速度恒定时，五个量 s（位移）、u（初速度）、v（末速度）、a（加速度）、t（时间）之间由四个方程联系：v = u + at，s = ut + 1/2 at^2，v^2 = u^2 + 2as，s = (u + v)/2 × t。每个方程包含五个量中的四个，解题时先列出已知量和所求量，选择包含这四个量的那个方程。例如，汽车从静止以 2 m/s^2 加速 5 秒，位移 s = 1/2 × 2 × 5^2 = 25 m。务必先规定正方向，把负方向的速度和加速度带上负号代入。
When acceleration is constant, the five quantities s (displacement), u (initial velocity), v (final velocity), a (acceleration) and t (time) are linked by four equations: v = u + at, s = ut + 1/2 at^2, v^2 = u^2 + 2as and s = (u + v)/2 × t. Each equation involves four of the five quantities, so first list what you know and what you want, then choose the equation containing those four. For example, a car accelerating from rest at 2 m/s^2 for 5 seconds travels s = 1/2 × 2 × 5^2 = 25 m. Always define a positive direction and substitute negative signs for velocities and accelerations pointing the other way.
七、抛体运动的水平与竖直分解 | Projectile Motion: Horizontal and Vertical Components
抛体运动可以分解为两个独立的运动：水平方向不受力，做匀速直线运动，x = u cosθ × t；竖直方向只受重力，做匀加速运动，用 SUVAT 方程求解。关键结论：飞到最高点时竖直速度为 0，但水平速度保持不变；飞行时间完全由竖直运动决定；轨迹是抛物线。飞行时间 T = 2u sinθ / g，最大高度 H = u^2 sin^2θ / 2g，这两个公式在答题时可以直接引用。典型考题包括求最大高度、飞行时间和水平射程，其中水平射程 R = u^2 sin2θ / g 在 θ = 45° 时达到最大。
Projectile motion splits into two independent motions: horizontally there is no force, so the object moves at constant velocity with x = u cosθ × t; vertically it is accelerated by gravity alone, so the SUVAT equations apply. The key results are: at the highest point the vertical velocity is zero but the horizontal velocity is unchanged; the time of flight is decided entirely by the vertical motion; and the path is a parabola. The time of flight is T = 2u sinθ / g and the maximum height is H = u^2 sin^2θ / 2g, which you may quote directly in an answer. Typical questions ask for the maximum height, time of flight and horizontal range, where the range R = u^2 sin2θ / g reaches a maximum when θ = 45°.
八、动量守恒与碰撞 | Conservation of Momentum and Collisions
动量（momentum）定义为 p = mv，是矢量。当系统不受外力或外力合力为零时，总动量守恒，碰撞前后 m1u1 + m2u2 = m1v1 + m2v2。碰撞分两类：弹性碰撞（elastic collision）中动能也守恒，可以用两个守恒方程联立求解；非弹性碰撞（inelastic collision）中动能不守恒，只有动量守恒。完全非弹性碰撞中两物体碰撞后粘在一起以共同速度运动。注意动量守恒方程要按矢量处理，先规定正方向，方向相反的动量必须带负号。
Momentum is defined as p = mv and is a vector. When no external force acts, or the resultant external force is zero, the total momentum is conserved, so before and after a collision m1u1 + m2u2 = m1v1 + m2v2. Collisions fall into two types: in an elastic collision kinetic energy is also conserved, so you solve the two conservation equations together; in an inelastic collision kinetic energy is not conserved and momentum alone survives. In a perfectly inelastic collision the two bodies stick together and move with a common velocity. Remember that the momentum equation is a vector statement, so fix a positive direction first and give momenta in the opposite direction a negative sign.
九、冲量与冲量-动量定理 | Impulse and the Impulse-Momentum Theorem
冲量（impulse）定义为力与作用时间的乘积 I = F × t，单位是 N·s，它等于物体动量的变化量：I = mv - mu。这个定理特别适合处理撞击、踢球、反冲这类作用时间很短、力很大的问题。例如，质量为 0.4 kg 的足球以 10 m/s 撞向球门柱后以 8 m/s 反向弹回，冲量大小为 0.4 × (10 + 8) = 7.2 N·s。解题时规定好正方向，动量的变化是"末动量减初动量"，方向相反的动量要带负号。如果给出的是力-时间图像，冲量等于图像与时间轴围成的面积。
Impulse is defined as the product of force and the time it acts, I = F × t, with units of N·s, and it equals the change in the object's momentum: I = mv - mu. This theorem is ideal for short, intense interactions such as impacts, kicks and recoil. For example, a 0.4 kg football hits a goalpost at 10 m/s and rebounds at 8 m/s in the opposite direction, so the magnitude of the impulse is 0.4 × (10 + 8) = 7.2 N·s. Define a positive direction first; the change in momentum is final momentum minus initial momentum, and momenta in the opposite direction carry negative signs. If a force-time graph is given, the impulse equals the area between the graph and the time axis.
十、功、能量守恒与功率 | Work, Energy Conservation and Power
力对物体做功（work done）为 W = F × d，其中 d 是沿力的方向的位移，当力与位移成 θ 角时 W = Fd cosθ。动能 Ek = 1/2 mv^2，重力势能 Ep = mgh。当只有重力做功时机械能守恒，但若有摩擦，则损失的机械能转化为内能。例如，一个 1 kg 的小球从 2 m 高处自由落下，到达地面时的速度 v = √(2gh) = √40 ≈ 6.32 m/s。功率（power）是做功的快慢，P = W/t = Fv，常用于分析汽车爬坡、发动机输出等问题。
The work done by a force is W = F × d, where d is the displacement in the direction of the force; when the force acts at angle θ to the displacement, W = Fd cosθ. Kinetic energy is Ek = 1/2 mv^2 and gravitational potential energy is Ep = mgh. When gravity is the only force doing work, mechanical energy is conserved, but if friction is present the lost mechanical energy becomes internal energy. For example, a 1 kg ball dropped from a height of 2 m hits the ground with speed v = √(2gh) = √40 ≈ 6.32 m/s. Power is the rate of doing work, P = W/t = Fv, often used to analyse a car climbing a hill or an engine's output.
十一、力矩与刚体的静力平衡 | Moments and Static Equilibrium of a Rigid Body
力矩（moment）是力使物体绕某点转动的效果，大小等于力乘以力到支点的垂直距离（力臂），即 moment = F × d。刚体平衡需要同时满足两个条件：所有力的合力为零，以及绕任意点的合力矩为零。解这类题目的关键是先选一个合适的点取矩，通常是未知力的作用点，这样未知力不产生力矩，方程中少一个未知量。均匀杆的重心在几何中心。例如，一根长 4 m 的均匀杆重 60 N，在距一端 1 m 处用支点支撑，另一端的支持力可通过取矩求出。
A moment measures the turning effect of a force about a point, equal to the force multiplied by the perpendicular distance from the line of action to the pivot, moment = F × d. A rigid body is in equilibrium only when two conditions hold: the resultant force is zero and the resultant moment about any point is zero. The key to solving these problems is to take moments about a sensible point, usually where an unknown force acts, so that unknown produces no moment and disappears from the equation. The centre of gravity of a uniform rod is at its geometric centre. For example, a uniform rod 4 m long weighing 60 N is supported at a point 1 m from one end; the supporting force at the other end is found by taking moments.
十二、质心与质量分布 | Centre of Mass and Mass Distribution
质心（centre of mass）是物体质量的集中点，对于均匀物体，它与几何中心重合。计算质心坐标用加权平均公式：x = Σ(mi × xi) / Σmi，把物体分割成规则形状，分别求每个部分的质心和质量，再按公式合成。例如，两块质量分别为 m 和 2m 的小球相距 L，系统质心到质量为 m 的小球的距离为 2L/3。常见题型包括由几根杆拼成的框架、挖去一部分的板，以及求悬挂物体平衡时的倾斜角度。记住挖空部分要当作负质量来处理。
The centre of mass is the point where an object's mass is effectively concentrated, and for a uniform body it coincides with the geometric centre. Its coordinates are found from the weighted-average formula x = Σ(mi × xi) / Σmi: split the object into simple shapes, find the mass and centre of each part, then combine them. For example, with two small masses m and 2m a distance L apart, the centre of mass lies 2L/3 from the mass m. Common question types include frameworks built from rods, plates with a part cut out, and finding the tilt angle of a suspended object in equilibrium. Remember to treat a removed region as a negative mass.
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