Formula Derivations in Cambridge IGCSE Physics | 剑桥 IGCSE 物理公式推导

📚 Formula Derivations in Cambridge IGCSE Physics | 剑桥 IGCSE 物理公式推导

Understanding how key formulas are derived not only deepens your comprehension but also helps you apply them correctly and remember them more easily during exams. This article walks you through the essential formula derivations in Cambridge IGCSE Physics, showing step-by-step reasoning from definitions to the final equations.

理解关键公式的推导过程不仅能加深你的理解，还能帮助你在考试中正确应用并更轻松地记住它们。本文带你一步步梳理剑桥 IGCSE 物理中的核心公式推导，展示从定义到最终方程的逻辑推演。

1. Defining Speed and Acceleration | 定义速度与加速度

Speed is the rate at which distance is covered. If an object travels a distance s in a time t, its average speed v is given by v = s ÷ t.

速度是距离随时间的变化率。若物体在时间 t 内通过距离 s，其平均速度 v 为 v = s ÷ t。

Acceleration is defined as the rate of change of velocity. When an object’s velocity changes from an initial value u to a final value v over a time interval t, the acceleration a is calculated as a = (v – u) ÷ t.

加速度定义为速度的变化率。当物体的速度在时间 t 内从初速度 u 变化到末速度 v 时，加速度 a 计算为 a = (v – u) ÷ t。

Both quantities are vector quantities in more advanced settings, but in IGCSE we mainly handle motion along a straight line, allowing us to use signed scalars for direction.

这两个量在更高阶的物理中是矢量，但在 IGCSE 中我们主要处理直线运动，可以用带符号的标量表示方向。

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2. The Uniform Acceleration Equations (suvat) | 匀加速运动方程 (suvat)

From the definition of acceleration a = (v – u) / t, we can rearrange directly to obtain the first suvat equation: v = u + at.

由加速度定义 a = (v – u) / t 可直接整理得到第一个匀加速方程：v = u + at。

v = u + at

When acceleration is constant, the average velocity is simply the mean of u and v, i.e. (u + v) ÷ 2. The displacement s equals average velocity multiplied by time: s = ((u + v) ÷ 2) × t.

加速度恒定时，平均速度为 u 和 v 的平均值，即 (u + v) ÷ 2。位移 s 等于平均速度乘以时间：s = ((u + v) ÷ 2) × t。

Substituting v = u + at into this displacement expression gives s = ((u + u + at) ÷ 2) t = ((2u + at) ÷ 2) t = ut + ½at².

将 v = u + at 代入位移表达式得 s = ((u + u + at) ÷ 2) t = ((2u + at) ÷ 2) t = ut + ½at²。

s = ut + ½at²

To derive the third equation, start with v = u + at and square both sides: v² = (u + at)² = u² + 2uat + a²t². Factor out 2a: v² = u² + 2a(ut + ½at²). Recognise the bracket as s, so we obtain v² = u² + 2as.

为推导第三个方程，从 v = u + at 两边平方开始：v² = (u + at)² = u² + 2uat + a²t²。提出因子 2a：v² = u² + 2a(ut + ½at²)。括号内正是 s，于是得到 v² = u² + 2as。

v² = u² + 2as

These three equations, known as the suvat equations, are the foundation of kinematics in IGCSE Physics.

这三个方程是 IGCSE 物理运动学的基础，常被称为 suvat 方程。

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

Newton’s second law states that the resultant force on an object is proportional to the rate of change of its momentum. If we express momentum as p = mv (where m is mass and v is velocity), then force F = Δp / t.

牛顿第二定律表明，物体所受合外力与其动量变化率成正比。若动量 p = mv（m 为质量，v 为速度），则力 F = Δp / t。

For a constant mass, the change in momentum is Δp = m(v – u). Substituting this into the expression for force gives F = m(v – u) / t. Since acceleration a = (v – u) / t, we arrive at F = ma.

质量恒定时，动量变化 Δp = m(v – u)。将其代入力的表达式得 F = m(v – u) / t。由于加速度 a = (v – u) / t，我们得到 F = ma。

F = ma

This derivation shows that F = ma is a special case of the more general law relating force and momentum. It applies whenever mass remains unchanged.

该推导表明 F = ma 是力与动量普遍关系的特例，适用于质量不变的情形。

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4. Momentum and Impulse | 动量与冲量

Momentum is defined as the product of mass and velocity: p = mv. The SI unit of momentum is kg m s⁻¹.

动量定义为质量与速度的乘积：p = mv，国际单位是 kg m s⁻¹。

The impulse imparted by a force is the product of the force and the time for which it acts: impulse = F t. According to Newton’s second law, impulse equals the change in momentum: F t = Δp = mv – mu.

力产生的冲量等于力与作用时间的乘积：冲量 = F t。由牛顿第二定律，冲量等于动量的变化量：F t = Δp = mv – mu。

This relationship is particularly useful in safety applications such as crumple zones, where increasing collision time reduces the average force.

这一关系在安全应用中十分有用，例如碰撞吸能区通过延长碰撞时间来减小平均作用力。

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5. Work Done and Mechanical Energy | 做功与机械能

Work is done when a force causes displacement in the direction of the force. If a constant force F acts over a distance d parallel to the force, the work done W is W = F d.

当力使物体沿力的方向发生位移时，力做功。若恒力 F 沿平行于力的方向移动距离 d，做功 W = F d。

W = F d

Energy is defined as the capacity to do work. Mechanical energy can exist as kinetic energy or potential energy. The work done on an object transfers energy to or from that object.

能量是做功的本领。机械能可以表现为动能或势能。对物体做功意味着能量向物体转移或从物体转出。

In the absence of friction or air resistance, the total mechanical energy (kinetic + potential) remains constant, illustrating the principle of conservation of energy.

在没有摩擦或空气阻力时，总机械能（动能加势能）保持不变，这体现了能量守恒原理。

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6. Deriving Kinetic Energy: ½mv² | 动能公式推导：½mv²

Consider an object of mass m initially at rest that is accelerated by a constant force F. The work done by the force is W = F s, where s is the displacement.

考虑一个质量为 m 的物体从静止开始受恒力 F 加速。该力做功 W = F s，s 为位移。

Using Newton’s second law, F = ma. From the suvat equation for initial velocity u = 0, we have s = (v² – 0) / (2a) = v²/(2a). Substitute these into the work done: W = ma × v²/(2a) = ½mv².

由牛顿第二定律 F = ma。初速 u = 0 时的 suvat 方程给出 s = (v² – 0)/(2a) = v²/(2a)。代入做功表达式：W = ma × v²/(2a) = ½mv²。

This work becomes the kinetic energy of the object, so kinetic energy Eₖ = ½mv².

这部分功转化为物体的动能，因此动能 Eₖ = ½mv²。

Eₖ = ½mv²

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7. Gravitational Potential Energy: mgh | 重力势能：mgh

To raise an object of mass m through a vertical height h near the Earth’s surface, you must do work against its weight mg. The minimum force required is equal to mg, applied vertically upwards.

在地球表面附近将质量为 m 的 object 竖直提升高度 h，必须克服其重力 mg 做功。所需最小力等于 mg，方向竖直向上。

If the object is lifted slowly at constant speed, the upward force equals mg and the work done is W = force × distance = mg × h = mgh.

若缓慢匀速提升，向上的力等于 mg，做功 W = 力 × 距离 = mg × h = mgh。

This work is stored as gravitational potential energy (GPE). Thus, the change in GPE is ΔEₚ = mgh. We usually set a reference level where h = 0 and write Eₚ = mgh.

这部分功储存为重力势能（GPE）。因此重力势能的变化量 ΔEₚ = mgh。通常取 h = 0 为参考面，写作 Eₚ = mgh。

Eₚ = mgh

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8. Density and Pressure in Fluids | 密度与流体压强

Density ρ is mass per unit volume: ρ = m ÷ V, where m is mass and V is volume. The SI unit is kg m⁻³.

密度 ρ 是单位体积的质量：ρ = m ÷ V，其中 m 为质量，V 为体积，国际单位是 kg m⁻³。

Pressure p is defined as the force acting normally per unit area: p = F ÷ A. Its unit is the pascal (Pa), where 1 Pa = 1 N m⁻².

压强 p 定义为垂直作用于单位面积的力：p = F ÷ A，单位是帕斯卡 (Pa)，1 Pa = 1 N m⁻²。

For a liquid at rest, consider a column of liquid of density ρ, height h and cross-sectional area A. The mass of the column is m = ρ × A × h. Its weight is W = mg = ρ A h g. This weight exerts a downward force on the base area A, so the pressure due to the liquid alone is p = W ÷ A = ρ A h g ÷ A = ρgh.

对于静止液体，考虑一个密度为 ρ、高为 h、底面积为 A 的液柱。柱的质量 m = ρ × A × h，重量 W = mg = ρ A h g。该重量施加在底面积 A 上，因此仅由液体产生的压强 p = W ÷ A = ρ A h g ÷ A = ρgh。

p = ρgh

This pressure increases linearly with depth and is independent of the area of the container.

这一压强随深度线性增加，与容器底面积无关。

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9. Ohm’s Law and Resistance | 欧姆定律与电阻

Ohm’s law states that, for a metallic conductor kept at constant temperature, the current I through the conductor is directly proportional to the potential difference V across it: V ∝ I.

欧姆定律指出，对于保持恒温的金属导体，通过导体的电流 I 与导体两端的电势差 V 成正比：V ∝ I。

Introducing a proportionality constant, the resistance R, gives V = I R. Therefore, resistance is defined as R = V ÷ I and is measured in ohms (Ω).

引入比例常数——电阻 R，可得 V = I R。因此电阻定义为 R = V ÷ I，单位为欧姆 (Ω)。

V = IR

The derivation rests on experimental evidence that the ratio V/I remains constant when physical conditions, particularly temperature, are unchanged.

该推导基于实验证据：当物理条件（特别是温度）不变时，V/I 的比值保持恒定。

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10. Resistors in Series and Parallel | 串联与并联电阻

When resistors are joined in series, the same current I flows through each resistor. The total potential difference V_total is the sum of the individual p.d.s: V_total = V₁ + V₂ + … . Applying Ohm’s law to each resistor: V₁ = I R₁, V₂ = I R₂, and so on. Hence V_total = I R₁ + I R₂ + … = I (R₁ + R₂ + …). Comparing with V_total = I R_total yields R_total = R₁ + R₂ + … .

电阻串联时，通过每个电阻的电流 I 相同。总电势差 V_total 等于各电阻两端电势差之和：V_total = V₁ + V₂ + … 。对每个电阻应用欧姆定律：V₁ = I R₁，V₂ = I R₂，依此类推。于是 V_total = I R₁ + I R₂ + … = I (R₁ + R₂ + …)。与 V_total = I R_total 对比，得 R_total = R₁ + R₂ + … 。

For resistors in parallel, the potential difference across each branch is the same (V). The total current I_total is the sum of the branch currents: I_total = I₁ + I₂ + … . Using I = V/R for each branch: I₁ = V / R₁, I₂ = V / R₂, … . Therefore I_total = V/R₁ + V/R₂ + … = V (1/R₁ + 1/R₂ + …). Writing I_total = V / R_total gives 1/R_total = 1/R₁ + 1/R₂ + … .

电阻并联时，各支路两端电势差相等（均为 V）。总电流 I_total 等于各支路电流之和：I_total = I₁ + I₂ + … 。对每个支路使用 I = V/R：I₁ = V / R₁，I₂ = V / R₂，…。因此 I_total = V/R₁ + V/R₂ + … = V (1/R₁ + 1/R₂ + …)。代入 I_total = V / R_total，得 1/R_total = 1/R₁ + 1/R₂ + … 。

1/R_total = 1/R₁ + 1/R₂ + …

These derivations are crucial for analysing circuits and calculating effective resistance.

这些推导对于分析电路

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