AS Mathematics: Mechanics Key Points Review | AS 数学：力学考点精讲

📚 AS Mathematics: Mechanics Key Points Review | AS 数学：力学考点精讲

This comprehensive guide covers essential mechanics topics at AS Level, including kinematics, Newton’s laws, moments, momentum, work and energy, and friction. Each section breaks down key concepts and common exam applications, pairing clear English explanations with equivalent Chinese translations to strengthen bilingual understanding and exam technique.

本指南全面覆盖了AS阶段力学的核心考点，包括运动学、牛顿定律、力矩、动量、功与能以及摩擦力。每个小节都分解了关键概念与常见考题应用，通过中英文配对讲解，强化双语理解与解题技巧。

1. Kinematics and SUVAT Equations | 运动学与匀加速方程

For straight‑line motion with constant acceleration, the five SUVAT variables are displacement (s), initial velocity (u), final velocity (v), acceleration (a) and time (t). The four standard equations each omit one variable: v = u + at (no s), s = ut + ½at² (no v), s = ½(u + v)t (no a), and v² = u² + 2as (no t).

在匀加速直线运动中，五个SUVAT变量为位移（s）、初速度（u）、末速度（v）、加速度（a）和时间（t）。四个标准方程各自缺一个变量：v = u + at（无s），s = ut + ½at²（无v），s = ½(u + v)t（无a），以及 v² = u² + 2as（无t）。

Always define a positive direction before substituting values. For example, if upward is positive, the acceleration due to gravity becomes a = −9.8 m s⁻². A stone thrown downwards from rest would have u = 0, a = +9.8 m s⁻² if downward is chosen as positive.

代入数值前必须明确正方向。例如，若取向上为正，重力加速度 a = −9.8 m s⁻²。而一个由静止向下抛出的石块，若取向下方为正，则 u = 0，a = +9.8 m s⁻²。

Do not mix units; convert everything to SI (metres, seconds) before solving. The equation s = ½(u + v)t is particularly useful when acceleration is not given but average velocity is easy to find.

切勿混用单位；解题前先将所有量转换为国际单位制（米、秒）。当未给出加速度但平均速度容易求得时，方程 s = ½(u + v)t 尤其好用。

2. Displacement–Time and Velocity–Time Graphs | 位移–时间与速度–时间图像

The gradient of a displacement–time (s–t) graph gives velocity. A straight line indicates constant velocity; a curve shows acceleration or deceleration. The gradient of a velocity–time (v–t) graph gives acceleration, and the area under the graph represents displacement.

位移–时间（s–t）图像的斜率表示速度。直线代表匀速，曲线则表明有加速度或减速度。速度–时间（v–t）图像的斜率表示加速度，图像下的面积表示位移。

When a v–t graph lies below the time axis, the area counts as negative displacement, indicating motion in the opposite direction. For non‑linear graphs, estimate the area by counting squares or using trapezium methods if the shape allows.

当 v–t 图位于时间轴下方时，面积计为负位移，表明质点向反方向运动。对于非线性图像，可通过数格或利用梯形法估算面积。

Always label axes and show key values such as initial velocity, time of zero velocity, and total time. Converting between s–t, v–t and a–t graphs is a common exam skill.

务必标注坐标轴并标明关键值，如初速度、速度为零的时刻和总时间。在 s–t、v–t 与 a–t 图像之间相互转换是常见的考查技能。

3. Projectile Motion | 抛体运动

Projectile motion is analysed by resolving initial velocity into horizontal and vertical components. Horizontal motion has constant speed: vₓ = u cos θ, x = (u cos θ)t. Vertical motion has constant acceleration due to gravity: vᵧ = u sin θ − gt, y = (u sin θ)t − ½gt², with upward taken as positive.

抛体运动通过将初速度分解为水平和竖直分量来分析。水平方向为匀速：vₓ = u cos θ，x = (u cos θ)t。竖直方向受重力加速度影响：vᵧ = u sin θ − gt，y = (u sin θ)t − ½gt²，取向上为正。

The time of flight is found by setting y = 0 (assuming launch and landing on same horizontal level): t = (2u sin θ)/g. Maximum height occurs when vᵧ = 0: h = (u² sin² θ)/(2g). The horizontal range is R = (u² sin 2θ)/g.

飞行时间由设 y = 0 求得（假设起落点在同一水平面）：t = (2u sin θ)/g。最大高度出现在 vᵧ = 0 时：h = (u² sin² θ)/(2g)。水平射程为 R = (u² sin 2θ)/g。

For projectiles launched from a height or landing on inclined planes, you must set up separate equations for horizontal and vertical displacements and link them through time t. Always state the direction of the acceleration due to gravity clearly.

对于从一定高度抛出或落在斜面上的抛体，需分别列出水平与竖直位移方程，并通过时间 t 建立联系。必须明确重力加速度的方向。

4. Newton’s Laws of Motion | 牛顿运动定律

Newton’s First Law states that an object remains at rest or moves with constant velocity unless acted upon by a resultant external force. Newton’s Second Law gives the equation of motion: F = ma, where F is the resultant force in newtons, m is mass in kg, and a is acceleration in m s⁻².

牛顿第一定律指出，除非受到合外力的作用，物体将保持静止或匀速直线运动。牛顿第二定律给出运动方程：F = ma，其中 F 为合外力（牛顿），m 为质量（千克），a 为加速度（米每二次方秒）。

Newton’s Third Law reminds us that if body A exerts a force on body B, body B exerts an equal and opposite force on body A. These forces act on different bodies, so they do not cancel in a free‑body diagram of a single object.

牛顿第三定律指出，若物体A给物体B一个力，则物体B给物体A一个大小相等、方向相反的力。这两个力作用在不同物体上，因此在对单个物体作受力分析时不会抵消。

When drawing force diagrams, include weight, normal reaction, tension, friction and any applied forces. Resolve forces along the direction of acceleration and perpendicular to it. The resultant in the direction of motion equals ma.

画受力图时，应标出重力、法向反作用力、拉力、摩擦力以及任何外加力。沿加速度方向及其垂直方向分解力。运动方向的合外力等于 ma。

5. Connected Particles and Pulleys | 连接体与滑轮

For light, inextensible strings passing over smooth pulleys, the tension T is the same throughout the string. Each particle is treated separately with its own equation of motion: T − mg = ma or mg − T = ma depending on direction of motion, where the acceleration a has the same magnitude for both masses.

对于跨过光滑滑轮的轻质不可伸长绳，绳上张力 T 处处相等。每个质点单独列运动方程：T − mg = ma 或 mg − T = ma，视运动方向而定，且两个物体的加速度大小 a 相同。

An alternative approach is to treat the whole system as one, applying F = ma to the total mass using the net driving force, e.g. (m₁ − m₂)g = (m₁ + m₂)a for a simple Atwood machine. This quickly gives a, after which T can be found by substituting into one particle’s equation.

另一种方法是把系统视为整体，用总质量与净驱动力列 F = ma，例如简单阿特伍德机中 (m₁ − m₂)g = (m₁ + m₂)a。这样可快速求出 a，再代回任一物体的方程即可求得 T。

When particles are connected on a horizontal table with one hanging, the tension pulls the table‑bound mass, and friction may oppose motion. Write equations for each mass and solve simultaneously, ensuring signs agree with the chosen positive direction.

当两个物体一个在水平桌面上、另一个悬吊时，绳的张力拉动桌面上的物体，摩擦力可能阻碍运动。为每个物体列出方程并联立求解，确保符号与选定的正方向一致。

6. Forces in Equilibrium | 力的平衡

A body is in equilibrium when the resultant force in any direction is zero. For two‑dimensional problems, resolve all forces into perpendicular components (usually horizontally and vertically) and set ΣFₓ = 0, ΣFᵧ = 0. The vector triangle of forces closes.

物体处于平衡状态时，任意方向上的合外力为零。对于二维问题，将所有的力分解为相互垂直的分量（通常为水平与竖直），并令 ΣFₓ = 0，ΣFᵧ = 0。力的矢量三角形闭合。

For three non‑parallel forces in equilibrium, they must be concurrent (their lines of action intersect at a single point). Lami’s theorem can be used: F₁/sin α = F₂/sin β = F₃/sin γ, where α is the angle opposite F₁, etc.

对于三个不平行力作用下的平衡，它们必须共点（作用线交于一点）。此时可使用拉密定理：F₁/sin α = F₂/sin β = F₃/sin γ，其中 α 为 F₁ 所对的角，依此类推。

On inclined planes, the weight component down the plane is mg sin θ, and the normal reaction is mg cos θ. Equilibrium requires that the sum of forces parallel to the plane is zero, usually balanced by friction or tension.

在斜面上，重力沿斜面的分量为 mg sin θ，法向反作用力为 mg cos θ。平衡要求沿斜面方向的合力为零，通常由摩擦力或拉力平衡。

7. Moments | 力矩

The moment of a force about a point is defined as the product of the force and the perpendicular distance from the point to its line of action: moment = F × d. Moments are measured in newton‑metres (N m) and can be clockwise or anticlockwise.

力对某点的力矩定义为力的大小与从该点到力作用线的垂直距离的乘积：力矩 = F × d。力矩的单位是牛顿·米（N m），可以顺时针或逆时针。

For a rigid body in equilibrium, both the resultant force and the resultant moment about any point must be zero. Take moments about a point where an unknown force acts to eliminate it from the equation.

刚体平衡时，合力必须为零，且对任一点的合力矩也必须为零。对某个未知力作用点取矩，可以消去该未知力。

In uniform rod problems, the weight acts at the centre. If a rod is supported at two points, vertical reactions can be found by taking moments about one support and using ΣFᵧ = 0. Always state the direction of each moment clearly.

在均质杆问题中，重力作用于杆的中点。若杆由两点支撑，可对其中一个支撑点取矩，并结合 ΣFᵧ = 0 求出竖直反力。必须清晰地标明每个力矩的方向。

8. Conservation of Momentum | 动量守恒

Linear momentum is p = mv. The impulse of a force equals the change in momentum: Ft = mv − mu. Momentum is conserved in collisions and explosions when no external resultant force acts: total momentum before = total momentum after.

线动量 p = mv。力的冲量等于动量的变化：Ft = mv − mu。在没有合外力的作用下，碰撞与爆炸过程中的动量守恒：碰撞前总动量 = 碰撞后总动量。

In a direct collision of two particles, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. For perfectly inelastic collisions (particles coalesce), v₁ = v₂ = v, so the common velocity can be found directly. For elastic collisions, kinetic energy is also conserved, but momentum conservation always holds.

在两个质点的一维碰撞中，m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂。对于完全非弹性碰撞（质点合为一体），v₁ = v₂ = v，可直接求出共同速度。对于弹性碰撞，动能也守恒，但动量守恒总是成立。

Sign convention is critical: define one direction as positive and assign velocities accordingly. A velocity in the opposite direction must be entered as negative. This avoids algebraic errors when solving for unknown speeds.

符号规定至关重要：定义一个方向为正，相应赋予速度符号。相反方向的速度须取负值。这样在求解未知速度时可避免代数错误。

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 the force and displacement. The work–energy principle states that the total work done by all forces equals the change in kinetic energy: W = ½mv² − ½mu².

恒力做的功为 W = F s cos θ，其中 θ 为力与位移的夹角。功能原理指出，所有力做的总功等于动能的变化量：W = ½mv² − ½mu²。

Gravitational potential energy (GPE) is mgh, where h is the vertical height. Energy is conserved when only gravity does work; then loss in GPE = gain in KE, or vice versa. If friction or external forces do work, include them as energy transfers.

重力势能（GPE）为 mgh，其中 h 为竖直高度。当只有重力做功时，能量守恒：损失的 GPE = 获得的 KE，反之亦然。若有摩擦力或外力做功，应将其作为能量传递处理。

Power is the rate of doing work: P = W/t. For a constant force moving with velocity v, the power is also P = F v. Queries often involve a car moving at constant speed up an incline, where engine power equals the rate of work against resistance and gravity.

功率是做功的快慢：P = W/t。对于恒力作用且物体以速度 v 运动的情况，功率也可表示为 P = F v。考题常涉及汽车以恒定速度爬坡的情景，此时发动机功率等于克服阻力与重力做功的速率。

10. Friction | 摩擦力

Friction acts between two surfaces in contact and opposes relative motion. The maximum static friction is F_max = μ R, where μ is the coefficient of static friction and R is the normal reaction. For moving surfaces, kinetic friction is F = μ_k R, with μ_k slightly less than μ.

摩擦力作用于两个接触面之间，阻碍相对运动。最大静摩擦力为 F_max = μ R，其中 μ 为静摩擦系数，R 为法向反力。两表面相对滑动时，动摩擦力为 F = μ_k R，μ_k 略小于 μ。

If a body is at rest and the applied force is less than or equal to the maximum static friction, the actual friction adjusts itself to balance the applied force. Only when the applied force exceeds μ R does the body begin to move.

若物体静止，且施加的力小于或等于最大静摩擦，实际摩擦力会自动调整以平衡外力。仅当外力超过 μ R 时，物体才开始运动。

On an inclined plane, the limiting equilibrium condition is tan θ = μ, where θ is the angle of inclination at which the block is just about to slide. This is derived from mg sin θ = μ mg cos θ. Always draw clear force diagrams showing weight, normal reaction, and friction.

在斜面上，极限平衡条件为 tan θ = μ，其中 θ 是滑块刚要滑动时的倾角。该条件由 mg sin θ = μ mg cos θ 导出。务必画出清晰的受力图，标明重力、法向反力与摩擦力。

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AS Mathematics: Mechanics Key Points Review | AS 数学：力学 考点精讲