📚 AS Physics Paper 1 Mark Scheme January 2018: Essential Formula Derivations | AS物理试卷1评分方案2018年1月：核心公式推导
Understanding how fundamental physics equations are derived is a crucial skill assessed in the January 2018 AS Physics Paper 1. The mark scheme rewards clear logical steps, correct use of definitions, and precise algebraic manipulation rather than mere recall of final results. This article revisits the key formula derivations that underpinned the paper, providing a step-by‑step guide that mirrors the reasoning expected by examiners. By mastering these derivations, students can strengthen their grasp on core mechanics, materials, waves, and electricity, turning formula memorisation into confident application.
理解基本物理方程的推导过程是2018年1月AS物理试卷1考查的一项关键能力。评分方案奖励清晰的逻辑步骤、正确定义的运用以及精确的代数操作，而不仅仅是对最终公式的回忆。本文重温了该试卷所依托的核心公式推导，逐步呈现考卷期望的推理过程。通过掌握这些推导，学生可以加深对力学、材料、波和电学核心内容的理解，将公式记忆转化为自信的应用。
1. Deriving the SUVAT Equations from Definitions of Velocity and Acceleration | 从速度与加速度的定义推导匀变速运动方程组
Start with constant acceleration defined as a = (v – u)/t. Rearranging gives the first SUVAT equation: v = u + at. To find displacement, use average velocity: for uniform acceleration, average velocity = (u + v)/2. Displacement s is then s = average velocity × time = ((u + v)/2) × t. Substituting v from the first equation gives s = (u + u + at)t/2 = (2u + at)t/2 = ut + (1/2)at². Mark schemes give credit for explicitly stating that acceleration is constant and that displacement equals average velocity multiplied by time.
从恒定加速度的定义 a = (v – u)/t 出发。移项可得第一个匀变速运动方程：v = u + at。为求位移，利用平均速度：匀加速运动中平均速度 = (u + v)/2。于是位移 s = 平均速度 × 时间 = ((u + v)/2) × t。将第一个方程中的 v 代入，得 s = (u + u + at)t/2 = (2u + at)t/2 = ut + ½at²。评分方案特别奖励明确说明加速度恒定以及位移等于平均速度乘以时间这一关键步骤。
To derive v² = u² + 2as, eliminate t from the first two equations. From v = u + at we have t = (v – u)/a. Substitute into s = ((u + v)/2) × t to obtain s = ((u + v)/2) × (v – u)/a = (v² – u²)/(2a). Rearranging gives the third equation. Examiners look for correct algebraic manipulation and the clear identification of the eliminated variable.
推导 v² = u² + 2as 时，需从前面两个方程中消去 t。由 v = u + at 得 t = (v – u)/a。代入 s = ((u + v)/2) × t，得 s = ((u + v)/2) × (v – u)/a = (v² – u²)/(2a)。整理即得。阅卷人关注代数处理的正确性以及对消元变量的清晰识别。
2. Deriving an Expression for g from a Free‑Fall Experiment | 从自由落体实验推导重力加速度表达式
When an object is dropped from rest (u = 0) and falls through a height h in time t, the equation s = ut + ½at² becomes h = ½gt². Rearranging gives g = 2h/t². In the January 2018 context, candidates were often asked to explain how a graph of h against t² yields g. The gradient of such a graph is g/2, so g = 2 × gradient. A mark is awarded for explaining that the graph is a straight line through the origin, validating the relationship h ∝ t².
物体从静止（u = 0）自由下落，经过时间 t 位移为 h，则方程 s = ut + ½at² 变为 h = ½gt²。整理得 g = 2h/t²。在2018年1月试题中，常要求学生解释如何利用 h-t² 图线求得 g。该图线的斜率为 g/2，因此 g = 2 × 斜率。评分点包括说明图线为过原点的直线，从而验证 h ∝ t² 的结论。
To minimise uncertainty, multiple timings are taken for each height and repeated readings reduce random error. The derived formula shows that g can be obtained even without knowledge of the initial velocity, provided it is consistently zero. This derivation is a staple of the AS practical assessment objectives.
为减小不确定度，每个高度多次计时并取平均值以减少随机误差。推导出的公式表明，只要初速度始终为零，即可求得 g，无需其他假设。该推导是AS实验评估目标的经典内容。
3. Deriving Centripetal Acceleration a = v²/r | 推导向心加速度公式 a = v²/r
Consider an object moving at constant speed v around a circle of radius r. In a short time Δt, the object moves from point A to B, covering an arc length vΔt. The change in velocity Δv is directed towards the centre. From the geometry of similar triangles (velocity triangle and displacement triangle), we have |Δv|/v = chord AB / r. For small Δt, chord AB ≈ arc length vΔt. Thus |Δv|/v = vΔt/r, giving |Δv|/Δt = v²/r. As Δt→0, |Δv|/Δt is the magnitude of acceleration a. Hence a = v²/r. The mark scheme rewards the use of vector diagrams and the small‑angle approximation.
考虑物体以恒定速率 v 沿半径 r 的圆周运动。在很短的 Δt 内，物体从 A 运动到 B，经过弧长 vΔt。速度变化量 Δv 指向圆心。由相似三角形（速度三角形与位移三角形）的几何关系得 |Δv|/v = 弦 AB / r。对于很小的 Δt，弦 AB ≈ 弧长 vΔt。因此 |Δv|/v = vΔt/r，得 |Δv|/Δt = v²/r。当 Δt→0 时，|Δv|/Δt 即加速度 a 的大小。故 a = v²/r。评分方案鼓励使用矢量图和微小角度近似。
Substituting v = ωr gives the alternative form a = ω²r. Examiners expect students to be able to switch between these forms and to state that the acceleration is always perpendicular to the velocity, changing direction but not speed.
代入 v = ωr 可得另一形式 a = ω²r。阅卷人期望学生能在两种形式间转换，并指出加速度始终垂直于速度，只改变方向而不改变速率。
4. Principle of Moments: Deriving the Equilibrium Condition | 力矩原理：推导平衡条件
For an object in rotational equilibrium, the sum of clockwise moments about any pivot must equal the sum of anticlockwise moments. This can be derived by considering a rigid body free to rotate about a fixed point. Each force produces a turning effect given by moment = force × perpendicular distance from pivot. If the object is stationary (or rotating at constant angular velocity), the resultant moment is zero. Taking moments about the pivot and setting Σ (F × d) = 0 yields the equilibrium condition. In a typical 2018 question, candidates had to derive an unknown distance or mass using this principle, showing clearly which moments are clockwise and which are anticlockwise.
对于绕任一支点处于转动平衡的物体，顺时针力矩之和必等于逆时针力矩之和。这可通过考虑一个绕固定点自由旋转的刚体来推导。每个力产生的转动效应用力矩 = 力 × 力到支点的垂直距离表示。若物体静止（或匀角速转动），合力矩为零。对支点取矩并令 Σ (F × d) = 0 即得平衡条件。在典型的2018年试题中，考生需运用该原理推导未知距离或质量，并清晰标出哪些力矩是顺时针、哪些是逆时针。
Mark scheme annotations often include ‘identifies pivot’, ‘states perpendicular distance’, and ‘equates clockwise to anticlockwise’ as key scoring points. The derivation is valid regardless of the pivot chosen, a fact occasionally tested in the paper.
评分方案常将“明确支点”、“标注垂直距离”以及“令顺时针力矩等于逆时针力矩”列为得分要点。无论选择哪个点为支点，推导均有效，这一点在试卷中偶有考查。
5. Deriving Young Modulus E = (F/A) / (ΔL/L) | 推导杨氏模量公式
The Young modulus characterises a material’s stiffness when under tensile stress. Stress is defined as force per unit cross‑sectional area: stress = F/A. Strain is the extension per unit original length: strain = ΔL/L. By definition, Young modulus E is the ratio of stress to strain within the proportional limit. Therefore
E = stress/strain = (F/A) / (ΔL/L)
which can be rearranged to F = (EA/L) ΔL. In the January 2018 paper, candidates were required to derive this expression from first‑principle measurements of a wire: recording F, A, original L, and extension ΔL. Marks were allocated for correctly identifying the independent and dependent variables and for plotting a suitable graph to determine E. A graph of F against ΔL yields a straight line with gradient EA/L, so E = gradient × (L/A). This derivation links experimental procedure directly to material property.
杨氏模量描述材料在拉伸应力下的刚度。应力定义为力与横截面积之比：stress = F/A。应变定义为伸长量与原长之比：strain = ΔL/L。按定义，杨氏模量 E 是在比例极限内应力与应变之比。因此 E = 应力/应变 = (F/A) / (ΔL/L)，可整理为 F = (EA/L) ΔL。在2018年1月试卷中，要求考生通过金属丝的测量数据推导该式：记录 F、A、原长 L 和伸长量 ΔL。得分点包括正确识别自变量和因变量，并绘制合适图线以确定 E。F-ΔL 图线为一条斜率为 EA/L 的直线，故 E = 斜率 × (L/A)。该推导将实验步骤直接与材料性质联系起来。
6. Deriving Resistivity ρ = RA/L | 推导电阻率公式
For a uniform conductor of length L and cross‑sectional area A, resistance R is directly proportional to L and inversely proportional to A. Introducing the constant of proportionality, resistivity ρ, gives
R = ρL/A
Rearranging yields ρ = RA/L. The derivation can be approached from the microscopic view: the drift velocity of charge carriers is affected by the length (more collisions) and cross‑sectional area (more pathways). The 2018 mark scheme expected students to recognise that this equation holds only for ohmic materials at constant temperature and that resistivity is an intrinsic property. Questions often require analysis of experimental data where resistance is measured for wires of different lengths but the same cross‑section and material. Plotting R against L gives a straight line through the origin with gradient ρ/A, so ρ = gradient × A.
对于长 L、横截面积 A 的均匀导体，电阻 R 与 L 成正比，与 A 成反比。引入比例常数电阻率 ρ，得 R = ρL/A。移项得 ρ = RA/L。该推导可从微观角度理解：载流子的漂移速度受长度（更多碰撞）和横截面积（更多通路）影响。2018年评分方案期望学生认识到此式仅在恒温欧姆材料中成立，且电阻率是固有属性。试题常要求分析实验数据，即测量相同材料、相同截面积但长度不同的导线电阻。作 R-L 图可得一条过原点、斜率为 ρ/A 的直线，故 ρ = 斜率 × A。
7. Deriving Total Resistance for Series and Parallel Circuits | 推导串联与并联电路的总电阻
Series: The same current I flows through each resistor. The total p.d. V across the combination is the sum of individual p.d.s: V = V₁ + V₂ + … By Ohm’s law V₁ = IR₁, V₂ = IR₂, so V = I(R₁ + R₂ + …). The total resistance Rtotal = V/I = R₁ + R₂ + … This derivation appears in mark schemes as ‘p.d.s add, current is common’.
串联：每个电阻流过相同的电流 I。组合两端的总电压 V 等于各分电压之和：V = V₁ + V₂ + …。由欧姆定律 V₁ = IR₁, V₂ = IR₂，故 V = I(R₁ + R₂ + …)。总电阻 Rtotal = V/I = R₁ + R₂ + …。评分方案中此项推导的关键点为“电压相加，电流相同”。
Parallel: The p.d. V across each resistor is the same. The total current I is the sum of branch currents: I = I₁ + I₂ + … Using I₁ = V/R₁, I₂ = V/R₂ gives I = V(1/R₁ + 1/R₂ + …). Hence total resistance is given by 1/Rtotal = I/V = 1/R₁ + 1/R₂ + … The mark scheme credits stating that ‘current splits, voltage is common’. In the January 2018 exam, candidates often had to derive an expression for two parallel resistors, Rtotal = (R₁R₂)/(R₁+R₂), and explain why the total resistance is always less than the smallest individual resistance.
并联：各电阻两端电压 V 相同。总电流 I 为各支路电流之和：I = I₁ + I₂ + …。利用 I₁ = V/R₁, I₂ = V/R₂ 得 I = V(1/R₁ + 1/R₂ + …)。故总电阻满足 1/Rtotal = I/V = 1/R₁ + 1/R₂ + …。评分中奖励明确指出“电流分流，电压相同”。在2018年1月考试中，考生常需推导两个并联电阻的公式 Rtotal = (R₁R₂)/(R₁+R₂)，并解释为何总电阻总是小于最小的那个分电阻。
8. Deriving Electrical Power Equations P = VI, P = I²R, P = V²/R | 推导电功率公式
Power is defined as energy transferred per unit time. When a charge Q moves through a potential difference V, the electrical work done is W = QV. In a circuit carrying steady current I, charge Q = It. Substituting gives W = VIt. Therefore power
P = W/t = VI
Using Ohm’s law V = IR for an ohmic component, we obtain P = I²R. Alternatively, substituting I = V/R gives P = V²/R. The 2018 paper rewarded showing the derivation from first principles and specifying the conditions under which each form is used. For instance, P = I²R is convenient for resistive heating in series circuits (current constant), while P = V²/R is useful for parallel branches (voltage constant).
功率定义为单位时间转移的能量。电荷 Q 通过电势差 V 时，电场力做功 W = QV。对恒定电流 I，有 Q = It。代入得 W = VIt。因此功率 P = W/t = VI。对欧姆元件利用欧姆定律 V = IR，可得 P = I²R；或代入 I = V/R 得 P = V²/R。2018年试卷奖励从基本原理推导，并说明每种形式适用的条件。例如，P = I²R 适用于串联电路（电流恒定）中的电阻发热，而 P = V²/R 适用于并联支路（电压恒定）。
Examiners often look for a clear statement that these derived equations assume ohmic behaviour; for non‑ohmic components, only P = VI is universally valid.
阅卷人常期望看到明确陈述：这些导出式假设欧姆特性；对非欧姆元件，只有 P = VI 普遍适用。
9. Deriving the Wave Speed Equation v = fλ | 推导波速公式 v = fλ
The frequency f of a wave is the number of complete oscillations per unit time. The period T is the time for one complete oscillation, so f = 1/T. The wavelength λ is the distance travelled by the wave in one period T. Therefore, wave speed v = distance/time = λ/T. Substituting T = 1/f gives
v = fλ
In the January 2018 exam, this derivation was required for both transverse (e.g. water waves) and longitudinal (e.g. sound) waves. A common follow‑up task involved using the formula with measurements from a standing wave experiment, where the wavelength is twice the distance between adjacent nodes. The mark scheme awarded points for linking the time period of the source to the wave’s temporal period, and for linking the spatial repeat distance to the wavelength.
频率 f 是波在单位时间内完成的全振动次数。周期 T 为一次全振动所需时间，故 f = 1/T。波长 λ 为波在一个周期 T 内传播的距离。因此波速 v = 距离/时间 = λ/T。代入 T = 1/f 得 v = fλ。在2018年1月考试中，横波（如水波）和纵波（如声波）均需这一推导。常见的后续任务包括用驻波实验测得数据代入此式，其中波长为相邻波节间距的两倍。评分方案奖励将波源时间周期与波的时域周期联系起来，以及将空间重复距离与波长联系起来。
10. Deriving the Critical Angle from Snell’s Law | 从斯涅尔定律推导临界角
When light travels from a medium of refractive index n₁ into a medium of lower refractive index n₂ (n₁ > n₂), total internal reflection occurs for angles of incidence greater than the critical angle c. At the critical angle, the angle of refraction is 90°. Snell’s law states n₁ sin θ₁ = n₂ sin θ₂. Substituting θ₁ = c and θ₂ = 90° (sin 90° = 1) gives
n₁ sin c = n₂ × 1 → sin c = n₂/n₁
If the second medium is air (n₂ ≈ 1), the formula simplifies to sin c = 1/n₁. In the 2018 Paper 1, marks were assigned for correctly labelling the boundary, indicating the normal, and stating the condition n₁ > n₂. Additionally, deriving the relationship between critical angle and refractive index can be linked to optical fibre applications. Some questions asked for the derivation starting from the definition of refractive index as ratio of speeds: n₁/n₂ = v₂/v₁, but the Snell’s law route was the primary approach expected.
当光从折射率 n₁ 的介质进入折射率较低 n₂ 的介质（n₁ > n₂）时，对于大于临界角 c 的入射角会发生全内反射。在临界角处，折射角为 90°。斯涅尔定律 n₁ sin θ₁ = n₂ sin θ₂，代入 θ₁ = c，θ₂ = 90°（sin 90° = 1）得 n₁ sin c = n₂ × 1 → sin c = n₂/n₁。若第二介质为空气（n₂ ≈ 1），则简化为 sin c = 1/n₁。在2018年试卷1中，得分点包含正确标注边界、指明法线，并说明条件 n₁ > n₂。此外，推导临界角与折射率的关系可联系光纤应用。有些题目要求从折射率的速度定义出发推导，但斯涅尔定律是主要期望的方法。
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