📚 AS Physics Paper 2 Jan 2018 Formula Derivation Walkthrough | 2018年1月AS物理第二卷公式推导详解
The January 2018 AS Physics Paper 2 exam featured several derivation questions that challenged students to show a deep understanding of fundamental principles. This article revisits those derivations, explaining every step in detail so that you can tackle similar problems with confidence.
2018年1月的AS物理第二卷考试包含多道公式推导题，要求学生展示对基本原理的深刻理解。本文回顾这些推导过程，详细解释每一步，让你能够自信应对类似问题。
1. Understanding Formula Derivation Questions | 理解公式推导题型
In AS Physics, derivation questions assess your ability to connect concepts logically. Rather than memorising results, you must show how equations follow from definitions, laws, or previously derived relationships. Examiners look for clear logical steps, stated assumptions, and correct use of symbols.
在AS物理中，推导题考查逻辑连接概念的能力。你不能只记住结果，还必须展示方程是如何从定义、定律或之前推导的关系中得出的。考官看重清晰的逻辑步骤、明确的假设以及正确的符号使用。
Typical derivation questions in Paper 2 include deriving kinematic equations, energy formulas, electrical relationships, and wave properties. The January 2018 paper specifically tested derivations of kinetic energy, parallel resistors, and wave speed on a string, among others.
第二卷常见的推导题包括推导运动学方程、能量公式、电学关系以及波的特性。2018年1月的试卷特别考查了动能公式、并联电阻以及弦上波速等推导。
2. Derivation of Kinetic Energy (Ek = ½ m v²) | 动能公式推导
A classic question on the January 2018 paper asked students to derive the kinetic energy formula from the concepts of work and Newton's second law.
2018年1月试卷中的一道经典题目，要求学生从功和牛顿第二定律出发推导动能公式。
e by a constant force: W = F s, where s is the displacement in the direction of the force.
推导从恒力做功的定义开始：W = F s，其中 s 是沿力方向的位移。
Next, apply Newton' a s. Assume the object accelerates uniformly from initial velocity u to final velocity v.
接着，应用牛顿第二定律 F = m a 替换力，得到 W = m a s。假设物体从初速度 u 匀加速到末速度 v。
acceleration, and displacement: v² = u² + 2 a s. Rearranging gives s = (v² - u²) / (2 a).
使用联系速度、加速度和位移的运动学方程：v² = u² + 2 a s，变形得 s = (v² - u²) / (2 a)。
f the object starts from rest (u = 0), the work done equals the kinetic energy gained, so:
将 s 的表达式代入功的方程：W = m a × (v² - u²) / (2 a) = ½ m (v² - u²)。若物体从静止开始（u = 0），则所做的功等于获得的动能，因此：
Ek = ½ m v²
This derivation highlights how kinetic energy is the work required to accelerate an object from rest to speed v. In the exam, stating the assumption of no energy loss and constant net force is essential.
这个推导强调动能是将物体从静止加速到速度 v 所需的功。在考试中，说明无能量损失和恒合力的假设至关重要。
3. Deriving the Parallel Resistor Formula (1/R = 1/R₁ + 1/R₂) | 并联电阻公式推导
Another frequent derivation in Paper 2 involves finding the equivalent resistance of two resistors in parallel – a question that appeared in the January 2018 exam.
第二卷中另一常见的推导是求两个并联电阻的等效电阻，该题出现在2018年1月考试中。
ss each is the same: V = V₁ = V₂. The total current I from the source splits: I = I₁ + I₂.
对于并联的两个电阻 R₁ 和 R₂，每个两端的电势差相同：V = V₁ = V₂。总电流 I 从电源流出后分流：I = I₁ + I₂。
Apply Ohm' ₁ and I₂ = V / R₂. For the equivalent single resistor R, the same total current I = V / R.
对各支路应用欧姆定律：I₁ = V / R₁，I₂ = V / R₂。对于等效电阻 R，总电流 I = V / R。
V / R₁ + V / R₂. Cancel the common factor V (since V ≠ 0) to obtain the parallel formula:
代入电流求和式：V / R = V / R₁ + V / R₂。消去公因子 V（V ≠ 0），得到并联公式：
1 / R = 1 / R₁ + 1 / R₂
This result can be extended to any number of resistors. Remember that the equivalent resistance in parallel is always less than the smallest individual resistance.
该结果可推广至任意数量电阻。记住，并联等效电阻总是小于最小的单个电阻。
4. Deriving Drift Velocity from I = n q v A | 从微观电流表达式推导漂移速度
d asked to derive the drift velocity v or to explain its derivation from first principles.
在2018年1月试卷的某些版本中，题目给出微观电流方程 I = n q v A，要求推导漂移速度 v，或从基本原理解释其推导。
Consider a conductor with cross-sectional area A. The number of charge carriers per unit volume (number density) is n, each carrying charge q. In a time t, the carriers move a distance v t, sweeping out a volume A v t.
考虑横截面积为 A 的导体。单位体积内的载流子数目（数密度）为 n，每个载流子所带电荷为 q。在时间 t 内，载流子移动距离 v t，扫过的体积为 A v t。
iers in that volume multiplied by the charge per carrier: Q = (n × A v t) × q = n q v A t.
在时间 t 内通过某点的总电荷 Q 等于该体积内的载流子数乘以每个载流子的电荷：Q = (n × A v t) × q = n q v A t。
s the rate of flow of charge: I = Q / t. Substituting Q gives I = n q v A t / t = n q v A.
电流 I 定义为电荷流动的速率：I = Q / t。代入 Q 得 I = n q v A t / t = n q v A。
q A). This derivation assumes that all carriers move with the same average drift velocity.
整理得到漂移速度 v = I / (n q A)。该推导假设所有载流子以相同平均漂移速度运动。
5. Wave Speed on a Stretched String (v = √T/μ) | 弦上波速与张力关系推导
A challenging derivation question on waves in the January 2018 paper involved showing that the speed of a transverse wave on a taut string depends on tension T and linear mass density μ (mass per unit length).
2018年1月试卷中一道有挑战性的波动物理推导题，需要证明张紧弦上横波的波速取决于张力 T 和线密度 μ（单位长度质量）。
g along a curved pulse. The segment can be treated as an arc of a circle of radius R, so its centripetal acceleration is v² / R.
考虑一小段长度为 Δl 的弦，其质量 m = μ Δl，随弯曲脉冲运动。该小段可视为半径为 R 的圆弧，因此其向心加速度为 v² / R。
θ, and the inward resultant force is approximately 2T sin(Δθ/2) ≈ T Δθ (for small angles).
向心力由张力的分量提供。若该段对圆心张角为小角 Δθ，则 Δl = R Δθ，向内的合力约为 2T sin(Δθ/2) ≈ T Δθ（小角近似）。
Apply Newton' / R = μ (R Δθ) v² / R = μ Δθ v². Cancel Δθ (which is non-zero) to obtain T = μ v², hence:
对向心运动应用牛顿第二定律：T Δθ = (μ Δl) × v² / R = μ (R Δθ) v² / R = μ Δθ v²。消去非零的 Δθ，得到 T = μ v²，因此：
v = √T / μ
This derivation illustrates the powerful use of small-angle approximations and the concept of centripetal force. Examiners expect you to state the key assumptions: the pulse is small, tension constant, and the string is perfectly flexible.
这个推导展示了小角近似和向心力概念的巧妙运用。考官希望你能说明关键假设：脉冲微小、张力恒定且弦完全柔软。
6. Deriving Power Dissipated in a Resistor (P = I² R = V² / R) | 电阻功率公式推导
The relationship between power, current, voltage, and resistance is a staple of AS electricity questions. The Jan 2018 paper required students to derive the alternative forms of electrical power.
功率、电流、电压和电阻之间的关系是AS电学题目的基本内容。2018年1月试卷要求学生推导电功率的其它形式。
electrical power: P = V I, where P is power, V is potential difference, and I is current.
从电功率的定义出发：P = V I，其中 P 为功率，V 为电势差，I 为电流。
Apply Ohm' that power dissipated is proportional to the square of the current at constant resistance.
应用欧姆定律 V = I R 替换 V：P = (I R) × I = I² R。此形式表明，电阻恒定时，耗散功率与电流的平方成正比。
Alternatively, using I = V / R from Ohm' R) = V² / R. Both forms are equivalent and useful in different circuit analysis scenarios.
或者，利用欧姆定律的变形 I = V / R 替换 I：P = V × (V / R) = V² / R。两种形式等价，在不同电路分析情境中各有用途。
Remember that these derivations assume the resistor obeys Ohm's law (constant resistance) and that all electrical energy is converted to heat.
记住，这些推导假设电阻服从欧姆定律（恒定电阻）且所有电能均转化为热量。
7. SUVAT Equation Derivation: v² = u² + 2 a s | 匀加速运动方程推导
dents to derive v² = u² + 2 a s from the definitions of acceleration and average velocity.
运动学推导是基础。2018年1月试卷有一题要求学生从加速度和平均速度的定义推导 v² = u² + 2 a s。
moving with constant acceleration a, the definition a = (v - u) / t gives t = (v - u) / a.
对于以恒定加速度 a 运动的物体，定义式 a = (v - u) / t 可得 t = (v - u) / a。
be expressed using average velocity: s = (u + v) / 2 × t. Substitute the expression for t:
位移 s 可用平均速度表示：s = (u + v) / 2 × t。代入 t 的表达式：
(v - u) / a = (v² - u²) / (2 a). Multiply both sides by 2 a to isolate the velocity terms:
s = (u + v) / 2 × (v - u) / a = (v² - u²) / (2 a)。两边同乘 2 a，分离速度项：
v² = u² + 2 a s
This equation is useful for problems where time is not given. Always note the assumption of constant acceleration and motion along a straight line.
该方程用于未给出时间的问题。务必注意恒定加速度和直线运动的假设。
8. Deriving Efficiency of Energy Transfer | 能量转换效率推导
Although conceptually simple, efficiency derivations appear regularly. In the Jan 2018 paper, students may have encountered a task to derive efficiency from energy input and output, or from power.
效率推导虽然概念简单，却经常出现。2018年1月试卷中，学生可能会遇到要求从能量输入输出或功率推导效率的题目。
(total energy input). In terms of power, η = (useful power output) / (total power input).
效率 η 定义为有用输出与总输入之比：η = (有用能量输出) / (总能量输入)。用功率表示时，η = (有用功率输出) / (总功率输入)。
m g h, and the total electrical energy supplied is V I t. Then η = (m g h) / (V I t) × 100% to express as a percentage.
例如，若电动机提升重物，有用功为 m g h，总输入电能为 V I t。则 η = (m g h) / (V I t) × 100%，以百分数表示。
Deriving such expressions requires converting all energies to the same units and clearly identifying useful versus wasted energy. The derivation step simply shows the fraction and the multiplication by 100%.
推导此类表达式需要将所有能量转换为相同单位，并清晰区分有用能量与废能。推导步骤仅是展示分数并乘以100%。
9. Key Exam Tips for Derivation Questions | 推导题应试技巧
In the January 2018 AS Physics Paper 2, many marks were lost due to missed steps or unclear logic. Follow these tips to maximise your score.
在2018年1月的AS物理第二卷中，许多失分源于步骤遗漏或逻辑不清。遵循以下技巧可最大化你的得分。
Always begin by writing down the definitions or basic laws you will use. State any assumptions, such as 'constant acceleration' or 'no energy loss', before manipulating equations.
始终从写出你将使用的定义或基本定律开始。在变换方程之前，说明所有假设，例如'恒定加速度'或'无能量损失'。
Show every algebraic manipulation step by step; do not jump to the final answer. Examiners award marks for intermediate workings.
逐步展示每一步代数变换；不要直接跳到最终答案。考官会给中间过程分。
Use correct notation and symbols exactly as they appear in the formula sheet. If you derive an expression, present it centre-aligned and boxed to make it stand out.
使用与公式表完全一致的符号和记号。推导出表达式后，将其居中并加框显示，使其醒目。
Finally, check units or limiting cases if possible. For a derived speed formula, test whether doubling the tension yields a √2 increase, as expected.
最后，若可能，检查单位或极限情况。对于推导出的速度公式，可检验加倍张力是否如预期导致√2倍的增加。
10. Conclusion: Mastering Derivations for A-Level Physics | 结语：掌握推导助力A-Level物理
The January 2018 AS Physics Paper 2 demonstrated that formula derivation is not about rote
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