📚 IB CCEA Maths: Mechanics Key Points | IB CCEA 数学：力学考点精讲
Mechanics lies at the heart of applied mathematics, bridging pure theory and physical reality. In both IB Mathematics (especially the Calculus and Vectors topics) and CCEA Mechanics modules, you will need to master kinematics, Newtonian dynamics, energy, momentum, and circular motion. This article consolidates the essential concepts, formulas, and problem-solving strategies you must know for examinations.
力学是应用数学的核心，连接着纯理论与物理现实。在 IB 数学（尤其是微积分和向量部分）以及 CCEA 力学模块中，你需要掌握运动学、牛顿动力学、能量、动量和圆周运动。本文整合了考试必须掌握的核心概念、公式与解题策略。
1. Displacement, Velocity, Acceleration | 位移、速度与加速度
Displacement (s) is a vector quantity describing the change in position from a reference point. Velocity (v) is the rate of change of displacement with respect to time, and acceleration (a) is the rate of change of velocity. All are vector quantities, meaning direction matters. In one-dimensional motion, a sign convention (e.g., rightwards positive) determines whether these quantities are positive or negative.
位移 (s) 是一个矢量，描述从参考点出发的位置变化。速度 (v) 是位移对时间的变化率，加速度 (a) 是速度对时间的变化率。它们都是矢量，方向至关重要。在一维运动中，需要先规定正方向（例如向右为正），以决定它们的正负。
Instantaneous velocity v = ds/dt, and instantaneous acceleration a = dv/dt = d²s/dt². Understanding these derivative relationships is fundamental for calculus-based kinematics problems.
瞬时速度 v = ds/dt，瞬时加速度 a = dv/dt = d²s/dt²。理解这些导数关系是解决基于微积分的运动学问题的基础。
2. Calculus in Kinematics | 运动学中的微积分
When displacement is given as a function of time, s(t), differentiate to obtain velocity and acceleration. Conversely, if acceleration is expressed as a(t), integrate with respect to time to find velocity, and integrate velocity to find displacement. Always determine the constant of integration using initial conditions (e.g., at t = 0, v = u, s = 0).
当位移表示为时间的函数 s(t) 时，求导可得速度和加速度。反过来，若加速度为 a(t)，对其关于时间积分得到速度，再积分一次得到位移。务必利用初始条件（如 t = 0 时 v = u，s = 0）确定积分常数。
For instance, if a = 6t – 2, integrate to get v = 3t² – 2t + C. With v(0) = 5, C = 5, so v = 3t² – 2t + 5. Integrate again for s. This technique appears frequently in IB Paper 2 and CCEA mechanics questions.
例如，若 a = 6t – 2，积分得 v = 3t² – 2t + C。由 v(0) = 5 得 C = 5，故 v = 3t² – 2t + 5。再积分一次即得位移。这种技巧常见于 IB 试卷二和 CCEA 力学试题。
3. Equations of Motion (SUVAT) | 匀变速直线运动方程
For constant acceleration in a straight line, five key quantities relate to each other: u (initial velocity), v (final velocity), a (acceleration), s (displacement), and t (time). The four SUVAT equations are:
在匀加速直线运动中，五个关键量相互关联：u（初速度）、v（末速度）、a（加速度）、s（位移）和 t（时间）。四个 SUVAT 方程如下：
v = u + at
s = ut + ½at²
v² = u² + 2as
s = ½(u + v)t
| v = u + at | 缺少 s |
| s = ut + ½at² | 缺少 v |
| v² = u² + 2as | 缺少 t |
| s = ½(u + v)t | 缺少 a |
Choose the equation that contains the three known quantities and the one unknown you need. Remember to assign a consistent positive direction; acceleration due to gravity, g = 9.8 m s⁻², acts downwards, so be careful with signs in vertical motion.
选择包含三个已知量和待求未知量的方程。务必设定一致的正方向；重力加速度 g = 9.8 m s⁻² 方向向下，因此处理竖直运动时需注意正负号。
4. Projectile Motion | 抛体运动
Projectile motion is analysed by resolving the initial velocity into horizontal and vertical components. Horizontally, there is no acceleration (ignoring air resistance), so horizontal velocity uₓ = u cos θ remains constant. Vertically, the motion experiences constant acceleration a = -g (if upwards is positive), so SUVAT equations apply to the vertical component.
分析抛体运动时，先将初速度分解为水平与竖直分量。水平方向无加速度（忽略空气阻力），因此水平速度 uₓ = u cos θ 保持不变。竖直方向受恒定加速度 a = -g（若以向上为正），故可将 SUVAT 方程应用于竖直分量。
The time of flight, maximum height, and range are derived from these two independent motions. The trajectory equation y = x tan θ – (g x²)/(2u² cos² θ) combines them, which is useful for finding the path equation in parametric form, often examined in IB vectors and calculus topics.
飞行时间、最大高度和水平射程均由这两个独立运动导出。轨迹方程 y = x tan θ – (g x²)/(2u² cos² θ) 将两者结合，在参数形式下可用来求路径方程，这在 IB 向量和微积分专题中常出现。
5. 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, where F is the resultant force, m is mass, a is acceleration. Third Law: Forces appear in equal and opposite action-reaction pairs. In mechanics problems, always draw a free-body diagram showing all forces acting on each body.
牛顿第一定律：若不受合外力，物体保持静止或匀速直线运动。第二定律：F = ma，F 为合外力，m 为质量，a 为加速度。第三定律：力以大小相等、方向相反的作用力与反作用力成对出现。在力学问题中，务必画出每个物体的受力示意图。
Common forces include weight (mg), normal reaction, tension, friction (F ≤ μR), and thrust. When objects are connected by a light inextensible string, they share the same acceleration and tension magnitude. Tackle such problems by writing equations of motion for each mass separately and solving simultaneously.
常见力包括重力 (mg)、法向反力、张力、摩擦力 (F ≤ μR) 和推力。当物体由轻质且不可伸长的绳子连接时，它们具有相同的加速度和张力大小。处理这类问题时，分别对每个物体列运动方程，并联立求解。
6. Resolving Forces and Equilibrium | 力的分解与平衡
When multiple forces act at a point, the resultant can be found by resolving each force into perpendicular components (usually horizontal and vertical). For equilibrium, the vector sum of forces must be zero: ΣFₓ = 0 and ΣFᵧ = 0. This principle is used to find unknown tensions, reactions, or angles in static configurations.
当多个力作用于同一点时，可将每个力分解为相互垂直的分量（通常沿水平和竖直方向）。对于平衡状态，力的矢量和必须为零：ΣFₓ = 0 且 ΣFᵧ = 0。这一原理可用来求解静力配置中的未知张力、反力或角度。
Friction often plays a crucial role in equilibrium. The limiting friction F_max = μR, where μ is the coefficient of static friction and R is the normal reaction. If the required friction to maintain equilibrium is less than μR, the system remains stationary.
摩擦力在平衡中经常起关键作用。最大静摩擦力 F_max = μR，其中 μ 为静摩擦系数，R 为法向反力。若维持平衡所需的摩擦力小于 μR，则系统保持静止。
7. Moments and Torque | 力矩与扭矩
The moment of a force about a point is the product of the force and the perpendicular distance from the point to its line of action: Moment = F × d. Moments that tend to cause clockwise or anticlockwise rotation are assigned opposite signs. For an object in rotational equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments about any pivot.
力对某点的力矩等于力的大小乘以该点到力作用线的垂直距离：力矩 = F × d。使物体产生顺时针或逆时针转动趋势的力矩用正负号加以区分。对于处于转动平衡的物体，关于任意支点，顺时针力矩之和等于逆时针力矩之和。
In IB and CCEA, you may encounter rigid body equilibrium problems requiring both force and moment equations. When a rod is supported at two points, taking moments about one support eliminates the reaction at that point, simplifying calculations.
在 IB 和 CCEA 中，你会遇到需要同时列力平衡方程和力矩平衡方程的刚体平衡问题。当一根杆在两点被支撑时，对其中一个支点取矩可以消去该点的反力，从而简化计算。
8. Momentum and Impulse | 动量与冲量
Linear momentum p = mv, a vector quantity. Impulse J is the change in momentum, equal to the average force multiplied by the time interval: J = F Δt = Δp. In collisions and explosions, the principle of conservation of momentum applies provided no external resultant force acts: total momentum before = total momentum after.
线动量 p = mv，是矢量。冲量 J 是动量的变化量，等于平均力乘以作用时间：J = F Δt = Δp。在碰撞和爆炸过程中，只要没有合外力作用，动量守恒定律适用：碰撞前总动量 = 碰撞后总动量。
For collisions, the coefficient of restitution e = (relative speed after separation)/(relative speed before approach), where 0 ≤ e ≤ 1. An e = 1 indicates a perfectly elastic collision; e = 0 a perfectly inelastic collision where bodies coalesce. Use the conservation of momentum together with the restitution equation to find final velocities.
对于碰撞，恢复系数 e = (分离后的相对速度)/(接近前的相对速度)，其中 0 ≤ e ≤ 1。 e = 1 表示完全弹性碰撞；e = 0 表示完全非弹性碰撞，此时物体合为一体。结合动量守恒与恢复系数方程即可求出末速度。
9. Work, Energy and Power | 功、能与功率
Work done by a constant force is W = F s cos θ, where θ is the angle between force and displacement. Kinetic energy KE = ½mv². Gravitational potential energy GPE = mgh. The work-energy principle states that the net work done on an object equals its change in kinetic energy: W_net = ΔKE.
恒力做功 W = F s cos θ，θ 为力与位移的夹角。动能 KE = ½mv²。重力势能 GPE = mgh。功能原理指出，物体所受合外力的功等于其动能的变化量：W_合 = ΔKE。
Power is the rate of doing work: P = W/t for average power; instantaneous power P = Fv, where v is velocity in the direction of the force. In problems involving vehicles moving up slopes, the driving force produced by the engine is often calculated from power and speed, then resolved against resistive forces.
功率是做功的快慢：平均功率 P = W/t；瞬时功率 P = Fv，v 为沿力方向的速度。在车辆爬坡问题中，发动机提供的驱动力常通过功率与速度求得，再与阻力等进行分析。
10. Circular Motion | 圆周运动
An object moving in a circle of radius r at constant speed v still experiences acceleration towards the centre: centripetal acceleration a = v²/r = rω², where ω is angular speed (ω = θ/t). The centripetal force required is F = mv²/r = mrω², provided by tension, friction, normal reaction, or gravity depending on context.
以恒速率 v 在半径为 r 的圆周上运动的物体仍具有指向圆心的加速度：向心加速度 a = v²/r = rω²，其中 ω 为角速度 (ω = θ/t)。所需向心力 F = mv²/r = mrω²，由张力、摩擦力、法向反力或重力等提供。
For motion in a vertical circle, the speed is not constant; energy conservation often links the speed at different points. At the top of a loop, the minimum speed for the object to stay on the track is given by mg = mv²/r, so v_min = √(gr). This is a classic application combining circular dynamics and energy.
对于竖直面内的圆周运动，速率并非恒定；往往用能量守恒将不同位置的速度联系起来。在圆环最高点，物体不脱离轨道的最小速率满足 mg = mv²/r，故 v_min = √(gr)。这是综合圆周动力学与能量的经典应用。
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