Mastering Formula Derivations in AS Physics Unit 2: Insights from the Jan 20 Exam Report | 掌握AS物理Unit 2公式推导：2020年1月考卷深度剖析

📚 Mastering Formula Derivations in AS Physics Unit 2: Insights from the Jan 20 Exam Report | 掌握AS物理Unit 2公式推导：2020年1月考卷深度剖析

Mastering formula derivations is the key to truly understanding AS Physics Unit 2 topics such as mechanics, materials, waves, and electricity. In the January 2020 examination, many students lost marks because they could not derive key equations from first principles. This article walks you through the essential derivations that appear most often in exams, helping you move beyond rote memorisation and develop a deep conceptual understanding.

掌握公式推导是真正理解AS物理Unit 2中力学、材料、波和电学等主题的关键。在2020年1月的考试中，许多学生因为无法从基本原理推导出关键方程而失分。本文将带你逐一掌握考试中最常出现的核心推导，帮助你超越死记硬背，建立深刻的概念理解。

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1. Understanding Acceleration to Get the First SUVAT Equation | 理解加速度并导出第一个运动学方程

Acceleration is defined as the rate of change of velocity. For uniform (constant) acceleration, the change in velocity is simply final velocity v minus initial velocity u over time t. The definition a = (v – u) / t is your starting point.

加速度定义为速度的变化率。对于匀加速度，速度变化量即为末速度v减初速度u除以时间t。定义式 a = (v – u) / t 是整个推导的起点。

By multiplying both sides by t and adding u, you obtain v = u + at, which is the first of the SUVAT equations. This simple rearrangement is often overlooked by students who try to memorise the result without appreciating its origin.

等式两边同乘以t并加上u，即可得到 v = u + at，这是匀加速直线运动方程组的第一个。这个简单的代数变换常被忽略，学生往往直接记忆结果却不知其来龙去脉。

v = u + at

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2. Displacement via Average Velocity | 通过平均速度求位移

When acceleration is constant, velocity changes linearly, so the average velocity is the arithmetic mean of the initial and final speeds: v_avg = (u + v) / 2. Displacement s is then just average velocity multiplied by time t.

当加速度恒定时，速度呈线性变化，因此平均速度等于初末速度的算术平均值：v_avg = (u + v) / 2。位移 s 就等于平均速度与时间 t 的乘积。

Substituting v = u + at from the first equation into the average velocity formula, we get s = ((u + [u + at])/2) × t = (2u + at) × t / 2 = ut + ½at². This derivation links the two fundamental relationships neatly.

将第一式 v = u + at 代入平均速度公式，可得 s = ((u + [u + at])/2) × t = ut + ½at²。这个推导自然地将两个基本关系连接起来，而非凭空出现。

s = ut + ½at²

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3. Eliminating Time: The Timeless Equation | 消去时间：与时间无关的运动方程

Sometimes we know initial and final speeds along with displacement, but not the time taken. To derive v² = u² + 2as, start with v = u + at and solve for t = (v – u)/a. Insert this t into s = ut + ½at² and simplify: s = u((v-u)/a) + ½a((v-u)/a)².

有时我们已知初末速度和位移，却不知道所用时间。要推导 v² = u² + 2as，可从 v = u + at 解出 t = (v – u)/a，然后代入 s = ut + ½at² 并化简：s = u((v-u)/a) + ½a((v-u)/a)²。

After careful algebraic manipulation, the terms combine to give v² – u² = 2as, or v² = u² + 2as. This equation is particularly useful in energy-related problems.

经过仔细的代数运算，项合并后得到 v² – u² = 2as，即 v² = u² + 2as。这个方程在处理与能量相关的问题时尤为有用。

v² = u² + 2as

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4. Kinetic Energy and the Work-Energy Theorem | 动能及功能定理

Consider a constant force F acting on an object to displace it by s. The work done is W = F s. Using Newton’s second law F = ma and the equation v² = u² + 2as, we can express acceleration a as (v² – u²)/(2s). Substituting into the work equation yields W = m × (v² – u²)/(2s) × s = ½mv² – ½mu².

设一个恒定力F作用于物体，使之产生位移s，做功为 W = F s。利用牛顿第二定律 F = ma 以及 v² = u² + 2as，可将加速度 a 表示成 (v² – u²)/(2s)。代入做功式子得到 W = m × (v² – u²)/(2s) × s = ½mv² – ½mu²。

This shows that the net work transfers energy that appears as a change in kinetic energy. Hence, kinetic energy is defined as KE = ½mv². Understanding this derivation prevents the common mistake of forgetting the factor ½.

这表明净功转移的能量表现为动能的变化。由此定义动能 KE = ½mv²。理解这个推导能避免常见错误的出现，比如漏掉系数 ½。

KE = ½mv²

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5. Conservation of Linear Momentum from Newton’s Third Law | 由牛顿第三定律推导动量守恒

Imagine two objects colliding. Object 1 exerts an impulse FΔt on object 2, and by Newton’s third law, object 2 exerts an equal and opposite impulse -FΔt on object 1. The impulse equals the change in momentum: for object 1, m₁v₁ – m₁u₁ = FΔt, and for object 2, m₂v₂ – m₂u₂ = -FΔt.

设想两个物体发生碰撞。物体1对物体2施加冲量 FΔt，根据牛顿第三定律，物体2同时对物体1施加等大反向的冲量 -FΔt。冲量等于动量变化：对物体1有 m₁v₁ – m₁u₁ = FΔt，对物体2有 m₂v₂ – m₂u₂ = -FΔt。

Adding the two impulse equations eliminates the force term, leaving m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. This is the principle of conservation of linear momentum, applicable when no external forces act.

将两式相加，力项被抵消，得到 m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂。这就是动量守恒定律，条件是系统不受外力作用。记住这个推导过程有助于理解动量是矢量，方向必须考虑。

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

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6. Defining Young’s Modulus from Stress and Strain | 从应力与应变定义杨氏模量

Tensile stress σ is the force applied per unit cross-sectional area: σ = F/A. Tensile strain ε is the extension relative to the original length: ε = ΔL/L. Young’s modulus E quantifies a material’s stiffness and is defined within the elastic limit as the ratio of stress to strain.

拉应力 σ 是单位截面积上所受的力：σ = F/A。拉应变 ε 是伸长量相对于原始长度的比值：ε = ΔL/L。杨氏模量 E 量化材料的刚度，在弹性限度内定义为应力与应变之比。

Substituting the expressions for stress and strain, E = (F/A) / (ΔL/L) = FL / (A ΔL). This derived formula is frequently used in experiments to find E from a force-extension graph.

将应力和应变的表达式代入，E = (F/A) / (ΔL/L) = FL / (A ΔL)。这个推导出的公式常用于实验中，通过力-伸长图线来求杨氏模量。

E = FL / (A ΔL)

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7. Relating Wave Speed, Frequency, and Wavelength | 波速、频率与波长的关系推导

A wave advances by one complete wavelength λ during one period T. Since frequency f is the number of cycles per second, f = 1/T. Therefore, the speed of the wave is distance divided by time: v = λ / T = λ × (1/T) = f λ.

在一个周期 T 内，波向前传播一个完整的波长 λ。因为频率 f 是每秒的周期数，f = 1/T。因此波速等于距离除以时间：v = λ / T = λ × (1/T) = f λ。

This derivation emphasises that the wave equation v = fλ is not an empirical law but follows directly from the definitions of frequency and wavelength. Make sure you can reproduce this logic in an exam.

这个推导强调波动方程 v = fλ 并非经验公式，而是直接源于频率和波长的定义。务必确保在考试中能清晰地写出这个逻辑过程。

v = f λ

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8. Electrical Power and Ohm’s Law Combinations | 电功率与欧姆定律组合推导

Potential difference V is defined as the work done per unit charge: V = W/Q. Current I is charge per unit time: I = Q/t. Electrical power P is the rate of doing work, so P = W/t = (VQ)/t = V I. This gives the most fundamental power equation P = VI.

电势差 V 定义为每单位电荷做的功：V = W/Q。电流 I 是单位时间通过的电量：I = Q/t。电功率 P 是做功的速率，所以 P = W/t = (VQ)/t = V I。这就得到最基本的功率等式 P = VI。

Using Ohm’s law V = IR, you can substitute to obtain two other forms: P = I × (IR) = I²R, and P = (V/R) × V = V²/R. These expressions are useful for analysing energy dissipation in resistors.

结合欧姆定律 V = IR，可以代入得到另外两种形式：P = I × (IR) = I²R，以及 P = (V/R) × V = V²/R。这些表达式在分析电阻上的能量耗散时十分有用。

P = VI = I²R = V²/R

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9. Equivalent Resistance in Series and Parallel Circuits | 串联与并联电路等效电阻推导

In a series circuit, the current I is the same through all components. The total voltage V_total equals the sum of individual voltages: V_total = V₁ + V₂ = IR₁ + IR₂ = I(R₁+R₂). Hence, the equivalent series resistance R_total = R₁ + R₂.

串联电路中，通过每个元件的电流 I 相同。总电压 V_total 等于各电压之和：V_total = V₁ + V₂ = IR₁ + IR₂ = I(R₁+R₂)。因此，等效串联电阻 R_total = R₁ + R₂。

In a parallel circuit, the voltage V across each branch is the same, and the total current I_total splits: I_total = I₁ + I₂. Express each current using Ohm’s law: I_total = V/R₁ + V/R₂ = V(1/R₁+ 1/R₂). Therefore, 1/R_total = 1/R₁ + 1/R₂.

并联电路中，各支路电压 V 相同，总电流 I_total 分流：I_total = I₁ + I₂。用欧姆定律表示各支路电流：I_total = V/R₁ + V/R₂ = V(1/R₁+ 1/R₂)。因此，1/R_total = 1/R₁ + 1/R₂。

Series: R_total = R₁ + R₂; Parallel: 1/R_total = 1/R₁ + 1/R₂

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10. Deriving the Potential Divider Formula | 分压器公式推导

A potential divider usually consists of two resistors R₁ and R₂ in series across an input voltage V_in. The current in the series circuit is I = V_in / (R₁ +

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