Newton's Laws: A-Level CCEA Mathematics Revision | A-Level CCEA 数学：牛顿定律考点精讲

📚 Newton’s Laws: A-Level CCEA Mathematics Revision | A-Level CCEA 数学：牛顿定律考点精讲

Newton’s laws form the foundation of classical mechanics and are a core part of the CCEA A-Level Mathematics syllabus, appearing in Mechanics 1 and Mechanics 2. Understanding these laws is essential for solving problems involving forces, motion, and connected particles. This revision guide provides a thorough breakdown of each law, demonstrates common applications, and highlights exam techniques to help you secure top marks.

牛顿定律是经典力学的基石，也是 CCEA A-Level 数学大纲中力学1和力学2的核心内容。理解这些定律对于解决涉及力、运动和连接质点的问题至关重要。本复习指南详细解析了每条定律，演示了常见应用，并强调了考试技巧，助你获得高分。

1. Introduction to Newton’s Laws | 牛顿定律简介

Newton’s three laws of motion are a fundamental part of the CCEA A-Level Mathematics Mechanics modules. They provide the framework for analysing the forces acting on particles and rigid bodies, and for predicting the resulting motion. In CCEA examinations, you will be expected to state these laws, apply them in a variety of contexts, and combine them with kinematic equations.

牛顿三定律是 CCEA A-Level 数学力学模块的基础内容。它们为分析作用于质点和刚体上的力以及预测其运动提供了框架。在 CCEA 考试中，你不仅需要陈述这些定律，还要在各种情境中应用它们，并将其与运动学方程相结合。

First Law (Law of Inertia): A body remains at rest or continues to move at a constant velocity unless acted upon by a net external force.

第一定律（惯性定律）：如果物体不受净外力作用，它将保持静止或匀速直线运动状态。

Second Law: The resultant force acting on a body is equal to the rate of change of its momentum. For constant mass, this simplifies to F = m a, where F is the net force, m is the mass, and a is the acceleration.

第二定律：作用在物体上的合力等于其动量的变化率。若质量恒定，可简化为 F = m a，其中 F 为净力，m 为质量，a 为加速度。

Third Law: For every action, there is an equal and opposite reaction. If body A exerts a force on body B, then body B exerts a force of equal size but opposite direction on body A.

第三定律：对于每一个作用力，都有一个大小相等、方向相反的反作用力。如果物体 A 对物体 B 施加一个力，那么物体 B 也对物体 A 施加一个大小相等但方向相反的力。

2. Newton’s First Law and Equilibrium | 牛顿第一定律与平衡

A particle is said to be in equilibrium if the vector sum of all forces acting on it is zero. According to Newton’s first law, this implies the particle either remains at rest or moves with constant velocity in a straight line. In exam problems, equilibrium conditions often appear when forces are balanced, such as an object on a rough slope about to slip, or a system of connected particles moving uniformly.

如果作用在质点上的所有力的矢量和为零，则称该质点处于平衡状态。根据牛顿第一定律，这意味着质点要么保持静止，要么沿直线以恒定速度运动。在考试题中，平衡条件常出现在力平衡的情形下，例如放在粗糙斜面上即将滑动的物体，或者作匀速运动的连接质点系统。

To solve equilibrium problems, we resolve forces into horizontal and vertical components (or parallel and perpendicular to an inclined plane) and set the net force in each direction to zero:

解决平衡问题时，我们将力分解为水平和竖直分量（或沿斜面及垂直于斜面方向），并令每个方向上的净力为零：

ΣF_x = 0, ΣF_y = 0

Always draw a clear force diagram. Mark all forces—weight (mg), normal reaction (R), tension (T), friction (f), and any applied forces. A correct force diagram is half the solution.

一定要画清晰的受力图。标出所有力——重力（mg）、法向反作用力（R）、张力（T）、摩擦力（f）以及任何外力。画对受力图就等于完成了一半的解答。

3. Newton’s Second Law: F = m a | 牛顿第二定律：F = m a

Newton’s second law states that the resultant force acting on a body is proportional to the rate of change of its momentum. If the mass of the body remains constant, the law simplifies to the well-known equation:

牛顿第二定律指出，作用在物体上的合力与其动量变化率成正比。如果物体的质量保持不变，该定律简化为大家熟知的公式：

F = m a

Here, F is the net force (in newtons, N), m is the mass (in kilograms, kg), and a is the acceleration (in m s⁻²). The equation is a vector equation—the acceleration is always in the same direction as the resultant force.

其中 F 为净力（单位为牛顿 N），m 为质量（单位为千克 kg），a 为加速度（单位为 m s⁻²）。该方程为矢量方程——加速度方向始终与合力方向一致。

One newton is defined as the force required to give a mass of 1 kg an acceleration of 1 m s⁻²: 1 N = 1 kg m s⁻². In CCEA problems, you may need to combine F = m a with constant acceleration (suvat) equations to find velocity, time, or displacement.

1 牛顿的定义是使 1 kg 质量的物体产生 1 m s⁻² 加速度所需的力：1 N = 1 kg m s⁻²。在 CCEA 题目中，你可能需要将 F = m a 与匀加速运动（suvat）方程结合起来，求解速度、时间或位移。

4. Applying F = m a in Linear Motion | 直线运动中的 F = m a 应用

Consider a car of mass 800 kg moving along a straight horizontal road. The driving force produced by the engine is 3000 N, and the total resistive force (air resistance and friction) is 500 N. The resultant force in the direction of motion is:

考虑一辆质量为 800 kg 的汽车沿平直水平公路行驶。发动机产生的驱动力为 3000 N，总阻力（空气阻力和摩擦）为 500 N。沿运动方向的合力为：

Resultant force = 3000 N – 500 N = 2500 N.

Using F = m a, the acceleration a = 2500 N ÷ 800 kg = 3.125 m s⁻².

应用 F = m a，可得加速度 a = 2500 N ÷ 800 kg = 3.125 m s⁻²。

If the car starts from rest, its velocity after 4 seconds can be found using v = u + a t: v = 0 + 3.125 × 4 = 12.5 m s⁻¹. Such questions require you to identify all forces, compute the net force, and then apply both F = m a and the suvat equations.

若汽车从静止开始运动，4 秒后的速度可用 v = u + a t 求得：v = 0 + 3.125 × 4 = 12.5 m s⁻¹。这类题目要求你找出所有力，计算净力，然后同时运用 F = m a 和 suvat 方程。

Always choose a positive direction and stick to it. Forces and accelerations acting opposite to the chosen positive direction must be given negative signs.

始终选定一个正方向并保持一致。与所选正方向相反的力和加速度必须加负号。

5. Connected Particles and Tension | 连接质点与张力

When two bodies are connected by a light, inextensible string, they experience the same magnitude of acceleration. A ‘light’ string means its mass is negligible, so the tension is the same at both ends. These simplifications allow us to write equations of motion for each particle and solve simultaneously.

当两个物体通过轻质且不可伸长的绳子连接时，它们的加速度大小相同。“轻”绳意味着其质量可忽略，因此绳子两端的张力大小相等。利用这些简化假设，我们可以对每个质点列出运动方程并联立求解。

For example, a particle of mass m on a smooth horizontal table is connected by a string passing over a frictionless pulley at the edge to a hanging particle of mass M. For the particle on the table (assuming no friction): T = m a. For the hanging particle: M g – T = M a. Solving gives a = M g / (m + M) and T = M m g / (m + M).

例如，一个质量为 m 的质点放在光滑水平桌面上，通过一根绕过桌边无摩擦滑轮的绳子与一个悬挂的质量为 M 的质点相连。对于桌面上的质点（假设无摩擦力）：T = m a。对于悬挂质点：M g – T = M a。解方程得 a = M g / (m + M)，T = M m g / (m + M)。

In CCEA exams, you may also encounter situations where a particle hangs vertically from another on a horizontal surface, or where the string passes over a pulley and the particles move in different directions. Always label tensions clearly and assign a consistent acceleration direction, usually taken as the direction of motion of the heavier particle.

在 CCEA 考试中，你可能会遇到一个质点悬挂在水平面上的另一个质点之下的情形，或者绳子绕过滑轮且各质点朝不同方向运动的情况。务必清楚地标出张力，并设定一个一致的加速度方向，通常取较重质点的运动方向为正。

6. Pulleys and String Problems | 滑轮与绳子问题

Pulleys in CCEA Mechanics are assumed to be smooth and light, meaning they do not affect the tension—the tension remains uniform throughout the string. The string is light and inextensible, so the acceleration of both particles is equal in magnitude. Typically, one particle hangs vertically while the other lies on a horizontal plane or an inclined plane.

CCEA 力学中的滑轮假设光滑且质轻，这意味着它们不影响绳子张力——整根绳子的张力保持不变。绳子轻质且不可伸长，因此两个质点的加速度大小相等。通常一个质点竖直悬挂，另一个置于水平面或斜面上。

To solve such problems, draw two separate force diagrams, one for each particle. Write an equation of motion using F = m a for each, taking care of the direction of acceleration. If one particle moves downwards, treat downward as positive for that particle. For the other particle moving horizontally or up the slope, choose its positive direction consistently with the string’s movement.

解决这类问题时，要为每个质点分别画受力图，然后对每个质点按 F = m a 列运动方程，并注意加速度的方向。若某质点向下运动，可设向下为正。对于另一个沿水平面或斜面上行的质点，应使其正方向与绳子的运动保持一致。

A common exam question: a mass (4 kg) on a smooth 30° incline is connected by a string over a pulley to a freely hanging mass (3 kg). Take the x-axis up the slope as positive. For the 4 kg mass: T – 4g sin 30° = 4a. For the 3 kg mass: 3g – T = 3a. Solving yields a and T.

一个常见的考题：质量为 4 kg 的物块放在 30° 光滑斜面上，通过一根绕过滑轮的绳子与一个质量为 3 kg 的自由悬挂物块连接。取沿斜面向上为正。对 4 kg 物块：T – 4g sin 30° = 4a；对 3 kg 物块：3g – T = 3a。联立可解得 a 和 T。

7. Inclined Planes | 斜面问题

When a particle rests on or moves along an inclined plane, its weight must be resolved into components parallel and perpendicular to the plane. If the plane makes an angle θ with the horizontal, then:

当质点静置于斜面上或沿斜面运动时，必须将其重力分解为平行和垂直于斜面的分量。若斜面与水平面的夹角为 θ，则有：

Component parallel to plane = m g sin θ
Component perpendicular to plane = m g cos θ

The normal reaction R acts perpendicular to the plane and balances the perpendicular weight component, provided there is no acceleration in that direction: R = m g cos θ.

法向反作用力 R 垂直于斜面，并与重力的垂直分量相平衡（假设在该方向上无加速度）：R = m g cos θ。

If the plane is smooth, the net force along the slope is simply m g sin θ, giving an acceleration a = g sin θ down the slope. If friction is present, the friction force f acts up or down the slope opposing motion. In that case, the equation of motion along the slope becomes m g sin θ – f = m a (or + f depending on direction).

若斜面光滑，沿斜面的净力仅为 m g sin θ，产生的沿斜面向下的加速度 a = g sin θ。若存在摩擦力，摩擦力 f 沿斜面向上或向下阻碍运动。此时沿斜面的运动方程为 m g sin θ – f = m a（或 + f，取决于方向）。

Inclined plane questions frequently combine kinematics; for example, you might be asked to find the time taken to travel a certain distance or the speed at the bottom.

斜面类题目常结合运动学，例如，你可能被要求计算滑过一定距离所需的时间或到达斜面底端时的速度。

8. Friction and Limiting Friction | 摩擦力与极限摩擦

Friction is a resistive force that opposes the relative motion of two surfaces in contact. In CCEA Mechanics, friction f is modelled by the inequality:

摩擦力是两个接触表面间阻碍相对运动的阻力。在 CCEA 力学中，摩擦力 f 用不等式建模：

f ≤ μ R

where μ is the coefficient of friction (a dimensionless constant) and R is the normal reaction. When the particle is in limiting equilibrium or about to move, friction reaches its maximum value f_max = μ R.

其中 μ 为摩擦系数（无量纲常数），R 为法向反作用力。当质点处于极限平衡或即将运动时，摩擦力达到最大值 f_max = μ R。

Kinetic (sliding) friction is often taken as constant, equal to μ_k R. In many exam problems μ is the same for both static and kinetic friction unless stated otherwise. Always check whether the object is moving or stationary. If stationary but not on the point of slipping, f < μ R.

动摩擦（滑动摩擦）常视为常量，等于 μ_k R。在许多考题中，除非另有说明，静摩擦和动摩擦系数 μ 相同。务必检查物体是处于运动还是静止状态。若静止且未达到即将滑动状态，则 f < μ R。

Friction can act in either direction along a surface, always opposing motion or the tendency to move. Draw a separate friction arrow on your force diagram and consider the direction carefully—incorrect friction direction is a common error.

摩擦力可以沿表面的任一方向，总是阻碍运动或运动趋势。在受力图上单独画一个摩擦力的箭头，并仔细考虑其方向——搞错摩擦力方向是常见的错误。

9. Newton’s Third Law and Normal Reaction | 牛顿第三定律与法向反作用力

Newton’s third law states that forces occur in pairs. If object A exerts a force on object B, object B exerts an equal and opposite force on A. These two forces are of the same type (e.g., both gravitational, both normal contact) and act on different bodies.

牛顿第三定律指出，力成对出现。若物体 A 对物体 B 施加一个力，物体 B 同时对物体 A 施加一个大小相等、方向相反的力。这两个力属于

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Newton’s Laws: A-Level CCEA Mathematics Revision | A-Level CCEA 数学：牛顿定律考点精讲