📚 Formula Derivations in 9630-PH03 International A-Level Physics Specimen Paper v4.2 | 9630-PH03国际A-Level物理样卷v4.2公式推导
The 9630-PH03 International A-Level Physics specimen paper (version 4.2) is a Unit 3 practical skills paper from the Edexcel IAL specification. Many of its questions require candidates to handle data, interpret graphs, and justify experimental procedures by deriving key relationships from first principles. This article walks through the most important formula derivations that appear in the context of such practical assessments, emphasising the physical reasoning and the algebraic steps needed to linearise equations and extract meaningful quantities.
9630-PH03 国际 A-Level 物理样卷（版本 4.2）是 Edexcel IAL 考试中单元 3 实验技能卷。它的许多题目要求考生处理数据、解释图表，并通过从基本原理推导关键关系来证明实验步骤的合理性。本文逐条讲解在此类实验评估中出现的最重要的公式推导，重点说明物理推理以及将方程线性化并提取有意义物理量所需的代数步骤。
1. Simple Pendulum: Deriving the T² vs L Relationship | 单摆：推导 T² 与 L 的关系
The period T of a simple pendulum of length L for small oscillations is given by T = 2π√(L/g), where g is the acceleration due to gravity. Squaring both sides removes the square root and gives a direct proportionality between T² and L.
小角度下单摆的周期 T 与摆长 L 的关系为 T = 2π√(L/g)，其中 g 是重力加速度。两边平方消去根号，得到 T² 与 L 之间的正比关系。
T² = (4π²/g) × L
Hence, a graph of T² against L yields a straight line through the origin. The gradient is 4π²/g, so g can be calculated as g = 4π² / slope. This linearisation is a core skill tested very frequently in Unit 3 papers.
因此，绘制 T²-L 图将得到一条过原点的直线。斜率为 4π²/g，由此可计算出 g = 4π² / 斜率。这种线性化方法是单元 3 试卷中极为常考的核心技能。
2. Resistivity of a Wire: From Ohm’s Law and Geometry | 导线电阻率：从欧姆定律和几何推导
The resistance R of a uniform wire is related to its resistivity ρ, length L and cross-sectional area A by R = ρL / A. For a cylindrical wire of diameter d, the area is A = πd²/4.
均匀导线的电阻 R 与其电阻率 ρ、长度 L 和横截面积 A 之间的关系为 R = ρL / A。对于直径为 d 的圆柱形导线，截面积为 A = πd²/4。
Substituting A gives R = (4ρL) / (πd²), which can be rearranged to the explicit form:
代入面积后得 R = (4ρL) / (πd²)，可整理为显式形式：
ρ = (πd²R) / (4L)
In a typical experiment, d is measured with a micrometer, and R is obtained from V–I measurements. If R is plotted against L, the graph has slope = 4ρ/(πd²), making it possible to determine ρ from the slope. If V is plotted against I for a fixed length, the resistance is the gradient, and the resistivity can then be calculated.
在典型实验中，d 用千分尺测量，R 由 V–I 测量获得。如果绘制 R-L 图，斜率 = 4ρ/(πd²)，因此可由斜率求出 ρ。若对固定长度导线绘制 V-I 图，其斜率即为电阻，进而可计算电阻率。
3. Young Modulus of a Wire: Extension and Diameter | 金属丝的杨氏模量：伸长量与直径
Young modulus E is defined as tensile stress divided by tensile strain: E = (F/A) / (ΔL/L) = FL / (AΔL). For a wire of diameter d, A = πd²/4, so
杨氏模量 E 定义为拉伸应力除以拉伸应变：E = (F/A) / (ΔL/L) = FL / (AΔL)。对于直径为 d 的金属丝，A = πd²/4，因此
E = (4FL) / (πd²ΔL)
Here F is the applied force (often mg from hanging masses), L is the original length, and ΔL is the extension. In an experiment, a graph of F against ΔL is linear; its gradient is (Eπd²)/(4L), so E = (4L × gradient) / (πd²). The formula also highlights why accurate measurement of d is critical — the diameter appears squared, magnifying uncertainties.
其中 F 是施加的力（常由悬挂砝码的重力 mg 提供），L 是原长，ΔL 是伸长量。实验中绘制 F-ΔL 图是一条直线；其斜率为 (Eπd²)/(4L)，因此 E = (4L × 斜率) / (πd²)。该公式也说明了精确测量直径至关重要——直径以平方形式出现，会放大不确定度。
4. Spring Combinations: Series and Parallel Effective Constants | 弹簧串并联：等效劲度系数
When two springs with stiffness constants k₁ and k₂ are connected in series, the same force F acts through both, and the total extension is the sum of individual extensions: Δx_total = Δx₁ + Δx₂ = F/k₁ + F/k₂.
两根劲度系数分别为 k₁ 和 k₂ 的弹簧串联时，相同的力 F 作用于两者，总伸长量为各自伸长量之和：Δx_total = Δx₁ + Δx₂ = F/k₁ + F/k₂。
Since the effective spring constant k_eff = F / Δx_total, we obtain 1/k_eff = 1/k₁ + 1/k₂, giving
由等效劲度系数 k_eff = F / Δx_total，得到 1/k_eff = 1/k₁ + 1/k₂，即
k_eff = k₁k₂ / (k₁ + k₂)
For parallel arrangement, the extensions are equal (Δx) and the total force is shared: F_total = k₁Δx + k₂Δx = (k₁ + k₂)Δx. Therefore, the effective constant is simply the sum:
并联时，两根弹簧的伸长量相同（Δx），总力为两者之和：F_total = k₁Δx + k₂Δx = (k₁ + k₂)Δx。因此，等效劲度系数就是两者之和：
k_eff = k₁ + k₂
These derivations are often needed when analysing energy storage or when designing experiments to measure an unknown spring constant using combinations.
当分析能量储存或设计通过组合测量未知弹簧劲度系数的实验时，经常需要这些推导。
5. Capacitor Discharge: Linearising the Exponential Decay | 电容器放电：指数衰减线性化
The voltage V across a discharging capacitor of capacitance C through a resistor R follows the exponential law V = V₀ e^(−t/RC), where V₀ is the initial voltage at t = 0.
容量为 C 的电容器通过电阻 R 放电时，其两端电压 V 遵循指数规律 V = V₀ e^(−t/RC)，其中 V₀ 是 t = 0 时的初始电压。
Taking the natural logarithm of both sides linearises the equation:
对两边取自然对数可使方程线性化：
ln V = ln V₀ − (1/RC) t
Thus, a graph of ln V against t produces a straight line with gradient −1/RC and y-intercept ln V₀. From the gradient, the time constant RC can be found; if R is known, C can be determined. This logarithmic transformation is a standard requirement in practical skills assessments.
因此，绘制 ln V-t 图将得到一条斜率为 −1/RC、截距为 ln V₀ 的直线。由斜率可求出时间常数 RC；若已知 R，可求出 C。这种对数变换是实验技能评估中的标准要求。
6. Diffraction Grating: The Grating Equation d sin θ = nλ | 衍射光栅：光栅方程 d sin θ = nλ
A diffraction grating with slit spacing d (where d = 1/N, N lines per metre) produces constructive interference when the path difference between light from adjacent slits equals an integer multiple of the wavelength λ.
光栅常数为 d（d = 1/N，N 为每米刻线数）的衍射光栅，当相邻狭缝出射光的光程差等于波长 λ 的整数倍时，产生相长干涉。
From the geometry shown in the specimen paper, the path difference is d sin θ, so for the n‑th order maximum:
根据样卷中展示的几何关系，光程差为 d sin θ，因此对于第 n 级明纹：
d sin θ = nλ
For small angles, sin θ ≈ θ (in radians), allowing a simple linear treatment. In an experiment, θ is measured for known n, and λ is calculated. The derivation also underpins the use of the grating to obtain a more accurate wavelength than a double slit.
小角度时 sin θ ≈ θ（以弧度计），可进行简单的线性处理。实验中，对已知级次 n 测量角度 θ，即可计算 λ。该推导也是用光栅获得比双缝更精确波长的依据。
7. Viscosity by Falling Sphere: Stokes’ Law and Terminal Velocity | 落球法测粘度：斯托克斯定律与终端速度
When a small sphere of radius r and density ρ_s falls through a fluid of density ρ_f, it reaches a terminal velocity v when the resultant force is zero. The forces are weight mg downward, upthrust U upward, and viscous drag F_d = 6πηrv (Stokes’ law) upward.
半径为 r、密度为 ρ_s 的小球在密度为 ρ_f 的流体中下落时，当合力为零时达到终端速度 v。受力情况为：重力 mg 向下，浮力 U 向上，粘滞阻力 F_d = 6πηrv（斯托克斯定律）向上。
The mass of the sphere is m = (4/3)πr³ρ_s and the upthrust is U = (4/3)πr³ρ_f g. At terminal velocity, mg = U + F_d, giving:
小球质量 m = (4/3)πr³ρ_s，浮力 U = (4/3)πr³ρ_f g。在终端速度时，mg = U + F_d，得：
(4/3)πr³(ρ_s − ρ_f)g = 6πηrv
Solving for v yields the working formula often used in the laboratory:
解出 v 得到实验室常用的计算公式：
v = [2r²g(ρ_s − ρ_f)] / (9η)
Thus, the coefficient of viscosity η can be extracted by measuring v, r and the densities. This derivation demonstrates the balance of forces and highlights the dependence on r², which makes the sphere’s radius the most critical measurement.
因此，通过测量 v、r 及密度，可求出粘滞系数 η。该推导体现了力的平衡关系，并突出了 v 对 r² 的依赖，使得小球半径成为最关键的测量量。
8. Newton’s Second Law on an Air Track: a = F/m | 气垫导轨验证牛顿第二定律：a = F/m
On a friction‑compensated air track, a glider of mass M is accelerated by a weight of mass m hanging over a pulley. The tension T in the string provides the net force on the glider, but the system can be analysed as a whole.
在已补偿摩擦的气垫导轨上，一质量为 M 的滑行器由跨过滑轮的砝码质量 m 驱动加速。细绳中的张力 T 提供滑行器的净力，但可将系统视为整体进行分析。
Applying Newton’s second law to the entire accelerating mass (M + m): the driving force is the weight mg, so
对整个加速质量 (M + m) 应用牛顿第二定律：驱动力为重力 mg，因此
mg = (M + m) a ⇒ a = [m / (M + m)] g
If M is kept constant and m is varied, a graph of a against the force mg (or simply a against m) can be plotted. More commonly, the specimen paper may expect a linearisation by keeping total mass constant: transfer masses from the glider to the hanger so that (M + m) is constant, giving a ∝ F. The derived relationship a = F / M_total confirms the proportionality between acceleration and net force, with the slope yielding 1/(M + m).
若 M 保持不变而改变 m，可绘制 a 随重力 mg（或直接随 m）变化的图。更常见的是，样卷可能期望通过保持总质量不变来实现线性化：将质量从滑行器转移到挂钩上，使 (M + m) 为定值，从而得到 a ∝ F。导出的关系式 a = F / M_total 确认了加速度与合外力成正比，斜率即为 1/(M + m)。
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