Formula Derivations in 9630 International A Level Physics | 9630 国际 A-Level 物理公式推导

📚 Formula Derivations in 9630 International A Level Physics | 9630 国际 A-Level 物理公式推导

Understanding the derivation of key physics formulas is essential for success in the 9630 International A Level Physics specification. This article walks through the most important derivations, showing clear step‑by‑step reasoning in both English and Chinese. Mastering these will deepen your conceptual understanding and equip you to handle unfamiliar problems in the exam.

掌握关键物理公式的推导是 9630 国际 A‑Level 物理考试成功的基础。本文梳理了最重要的公式推导，以中英双语清晰展示逐步推理。熟练掌握这些内容将加深你对概念的理解，并让你有能力应对考试中的陌生问题。

1. Kinematic Equations for Uniform Acceleration | 匀加速运动的运动学方程

The kinematic equations relate displacement s, initial velocity u, final velocity v, acceleration a, and time t for motion in a straight line with constant acceleration. Starting from the definition of acceleration:

运动学方程描述了匀变速直线运动中位移 s、初速度 u、末速度 v、加速度 a 和时间 t 之间的关系。我们从加速度的定义出发：

a = (v − u) / t

Rearranging gives the first equation: v = u + at.

整理得到第一个方程：v = u + at。

Average velocity for uniform acceleration is (u + v)/2, so displacement is average velocity times time:

匀加速运动的平均速度为 (u + v)/2，因此位移为平均速度乘以时间：

s = ((u + v) / 2) × t

Substitute v = u + at into the displacement equation to obtain s = ut + ½at²:

将 v = u + at 代入位移方程，得到 s = ut + ½at²：

s = ut + ½at²

To eliminate time, solve the first equation for t: t = (v − u)/a, and substitute into s = (u + v)t/2:

为消去时间，将第一个方程解出 t：t = (v − u)/a，并代入 s = (u + v)t/2：

s = (u + v)/2 × (v − u)/a = (v² − u²) / (2a)

Rearranging gives the third kinematic equation: v² = u² + 2as.

整理得到第三个运动学方程：v² = u² + 2as。

These three equations form the foundation for solving constant‑acceleration problems in one dimension.

这三个方程构成了解决一维匀加速问题的基础。

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2. Newton’s Second Law and Momentum | 牛顿第二定律与动量

Newton’s second law is commonly expressed as F = ma, but its most general form involves the rate of change of momentum. Momentum p is defined as mv. For constant mass, the derivative of momentum gives:

牛顿第二定律通常表示为 F = ma，但其最普遍的形式涉及动量对时间的变化率。动量 p 定义为 mv。对于质量不变的情况，动量的导数给出：

F = dp/dt = d(mv)/dt = m(dv/dt) = ma

Thus F = ma is a special case. The more powerful statement is that net force equals the rate of change of momentum: F = Δp / Δt. This is used for variable‑mass systems (e.g. rockets) and in deriving the impulse–momentum relationship:

因此 F = ma 是一个特例。更强大的表述是合外力等于动量变化率：F = Δp / Δt。这用于变质量系统（如火箭）以及推导冲量‑动量关系：

FΔt = Δp

The area under a force–time graph gives impulse, which equals the change in momentum. This derivation is vital for understanding collisions and explosions.

力‑时间图下的面积代表冲量，等于动量的变化。这一推导对于理解碰撞和爆炸至关重要。

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3. Work and the Kinetic Energy Theorem | 功与动能定理

Work done by a constant force F acting over a displacement s in the direction of the force is W = Fs. Using F = ma and the kinematic equation v² = u² + 2as, we can derive the kinetic energy formula:

恒力 F 在力方向上的位移 s 上做的功为 W = Fs。利用 F = ma 和运动学方程 v² = u² + 2as，可以推导出动能公式：

W = Fs = ma × (v² − u²)/(2a) = ½m(v² − u²) = ½mv² − ½mu²

This shows that the net work done on an object equals its change in kinetic energy. The kinetic energy of an object of mass m moving at speed v is therefore defined as Eₖ = ½mv².

这表明物体所受的净功等于其动能的变化。因此质量为 m、速度为 v 的物体的动能定义为 Eₖ = ½mv²。

This derivation is fundamental to solving mechanics problems involving energy conservation without explicit force analysis.

这一推导对于解决涉及能量守恒的力学问题而无需显式受力分析具有基础性意义。

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4. Gravitational Potential Energy | 重力势能

When an object of mass m is lifted through a height Δh near the Earth’s surface, the work done against gravity is mgΔh. This work is stored as gravitational potential energy (GPE).

当质量为 m 的物体在地球表面附近被提升高度 Δh 时，克服重力做的功为 mgΔh。这份功以重力势能 (GPE) 的形式储存起来。

ΔEₚ = mgΔh

If we set a reference level where GPE is zero, we have Eₚ = mgh. For larger distances from Earth, the universal gravitational potential energy formula must be used: U = −GMm/r, where M is Earth’s mass and r the distance from the centre. The negative sign indicates a bound system.

如果设定零势能参考面，则有 Eₚ = mgh。对于距离地球较远的情况，必须使用万有引力势能公式：U = −GMm/r，其中 M 为地球质量，r 为到地心的距离。负号表示束缚系统。

These derivations link work, force, and energy in a consistent framework required by the 9630 syllabus.

这些推导将功、力和能量统一在一个连贯的框架中，符合 9630 课程标准的要求。

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5. Centripetal Acceleration | 向心加速度

An object moving in a circle of radius r with constant speed v undergoes acceleration directed towards the centre. The magnitude of this centripetal acceleration is a = v²/r. This can be derived using vector geometry.

在半径为 r 的圆周上以恒定速率 v 运动的物体具有指向圆心的加速度。该向心加速度的大小为 a = v²/r，可通过矢量几何推导。

Consider a short time interval Δt. The change in velocity Δv points towards the centre. The triangle formed by the velocity vectors is similar to the triangle formed by the radii and the arc length. Since arc length ≈ vΔt, we have:

考虑一个短时间间隔 Δt。速度变化 Δv 指向圆心。速度矢量构成的三角形与半径和弧长构成的三角形相似。由于弧长 ≈ vΔt，有：

Δv / v ≈ arc length / r = vΔt / r → Δv / Δt = v²/r

Taking the limit Δt → 0 gives the instantaneous acceleration: a = v²/r. Using v = ωr, it can also be written as a = ω²r.

取 Δt → 0 的极限，得到瞬时加速度：a = v²/r。利用 v = ωr，也可写成 a = ω²r。

This derivation is essential for explaining circular motion and applications such as banked tracks and satellites.

该推导对于解释圆周运动以及倾斜轨道和卫星等应用至关重要。

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6. Newton’s Law of Gravitation | 牛顿万有引力定律

Newton’s law of gravitation states that the force between two point masses is proportional to the product of their masses and inversely proportional to the square of the distance between them:

牛顿万有引力定律指出，两个质点之间的引力与它们的质量乘积成正比，与它们之间距离的平方成反比：

F = Gm₁m₂ / r²

Combining this with Newton’s second law and centripetal acceleration for a satellite in circular orbit allows derivation of Kepler’s third law and the gravitational field strength g = GM/r². Near the Earth’s surface, r ≈ R (Earth’s radius), so g = GM/R².

将此式与牛顿第二定律以及圆周运动卫星的向心加速度相结合，可以推导开普勒第三定律和引力场强度 g = GM/r²。在地球表面附近，r ≈ R（地球半径），因此 g = GM/R²。

This shows that the acceleration due to gravity is the same for all objects and is determined by the Earth’s mass and radius.

这表明重力加速度对所有物体都相同，并由地球的质量和半径决定。

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7. Simple Harmonic Motion | 简谐运动

Simple harmonic motion (SHM) occurs when the restoring force is proportional to the displacement from equilibrium and acts towards it: F = −kx. Using Newton’s second law gives m(d²x/dt²) = −kx, which leads to the differential equation:

简谐运动 (SHM) 发生时，回复力与偏离平衡位置的位移成正比且方向相反：F = −kx。运用牛顿第二定律得到 m(d²x/dt²) = −kx，导出微分方程：

d²x/dt² = −(k/m)x

Defining ω² = k/m, the solution is a sinusoidal function: x = A cos(ωt + φ). The angular frequency is ω = √(k/m), and the period T = 2π/ω = 2π√(m/k). For a simple pendulum, k/m is replaced by g/L, giving T = 2π√(L/g).

定义 ω² = k/m，解为正弦函数：x = A cos(ωt + φ)。角频率为 ω = √(k/m)，周期 T = 2π/ω = 2π√(m/k)。对于单摆，k/m 替换为 g/L，得到 T = 2π√(L/g)。

SHM derivations connect kinematics and dynamics, and they are fundamental for understanding oscillations in springs and pendulums.

简谐运动的推导将运动学与动力学联系起来，是理解弹簧振动和摆运动的基础。

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8. Capacitor Discharge | 电容器放电

When a capacitor discharges through a resistor, the rate at which charge leaves the capacitor is related to the current: I = −dQ/dt. Since V = IR and Q = CV, we have:

当电容器通过电阻器放电时，电荷离开电容器的速率与电流相关：I = −dQ/dt。利用 V = IR 和 Q = CV，我们得到：

−dQ/dt = Q / (RC)

This first‑order differential equation solves to Q = Q₀ e^−t/RC. Differentiating gives the current: I = I₀ e^−t/RC, where I₀ = Q₀/RC. Similarly, voltage decays as V = V₀ e^−t/RC.

这个一阶微分方程的解为 Q = Q₀ e^−t/RC。对其求导得到电流：I = I₀ e^−t/RC，其中 I₀ = Q₀/RC。同理，电压按照 V = V₀ e^−t/RC 衰减。

The time constant τ = RC is the time for the charge to fall to 1/e ≈ 37% of its initial value. This exponential decay pattern is a key result in RC circuits.

时间常数 τ = RC 是电荷衰减到初始值 1/e（约 37%）所需的时间。这种指数衰减规律是 RC 电路的关键结论。

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9. Electromagnetic Induction | 电磁感应

Faraday’s law states that the induced electromotive force (emf) in a circuit is equal to the negative rate of change of magnetic flux linkage: ε = −d(NΦ)/dt. For a conductor of length L moving perpendicularly through a uniform magnetic field B with speed v, the flux cut per unit time is BLv, so ε = BLv.

法拉第定律指出，电路中的感应电动势等于磁通链变化的负率：ε = −d(NΦ)/dt。对于长度 L 的导体以速度 v 垂直切割均匀磁场 B 的情况，单位时间内切割的磁通量为 BLv，因此 ε = BLv。

This derivation is used for generators, transformers, and motional emf. Lenz’s law (indicated by the minus sign) ensures energy conservation by opposing the change that produced it.

该推导用于发电机、变压器和动生电动势。楞次定律（由负号表示）通过对抗引起感应的变化来确保能量守恒。

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10. Ideal Gas Equation | 理想气体状态方程

The ideal gas equation pV = nRT combines Boyle’s law, Charles’s law, and the pressure law. A simple kinetic theory derivation relates the pressure of a gas to the mean square speed of its molecules.

理想气体状态方程 pV = nRT 结合了玻意耳定律、查理定律和压强定律。一个简单的分子动理论推导将气体压强与其分子的均方速率联系起来。

Consider N molecules of mass m in a cubic box of side L. The force exerted on one wall from a collision is 2mv_x divided by the time between collisions 2L/v_x. Summing over all molecules and using the definition of pressure leads to:

考虑 N 个质量为 m 的分子在边长为 L 的立方盒中。一次碰撞对器壁施加的力为 2mv_x，除以两次碰撞的时间间隔 2L/v_x。对所有分子求和，并利用压强的定义，得到：

p = (1/3) (N/V) m⟨v²⟩

Here ⟨v²⟩ is the mean square speed. With the ideal gas internal energy U = (3/2)NkT = (3/2)nRT for a monatomic gas, and pV = NkT, where k is the Boltzmann constant, we obtain pV = nRT.

其中 ⟨v²⟩ 是均方速率。对于单原子理想气体内能 U = (3/2)NkT = (3/2)nRT，且 pV = NkT，k 为玻尔兹曼常数，我们得到 pV = nRT。

This derivation links the macroscopic gas laws with microscopic particle motion.

这一推导将宏观气体定律与微观粒子运动联系起来。

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11. Resistance and Resistivity | 电阻与电阻率

The resistance R of a uniform wire is directly proportional to its length L and inversely proportional to its cross‑sectional area A. The constant of proportionality is the resistivity ρ:

均匀导线的电阻 R 与其长度 L 成正比，与横截面积 A 成反比。比例常数即为电阻率 ρ：

R = ρL / A

This can be derived from the microscopic form of Ohm’s law: current density J = σE, where conductivity σ = 1/ρ. For a uniform wire, V = EL and I = JA, so R = V/I = (EL)/(σE A) = L/(σA) = ρL/A.

此式可由欧姆定律的微观形式推导：电流密度 J = σE，其中电导率 σ = 1/ρ。对于均匀导线，V = EL 且 I = JA，因此 R = V/I = (EL)/(σE A) = L/(σA) = ρL/A。

Understanding resistivity allows prediction of resistance changes with temperature and material selection in circuit design.

理解电阻率有助于预测电阻随温度的变化，并在电路设计中合理选材。

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12. Summary and Revision Tips | 总结与复习技巧

The derivations covered here are not just mathematical exercises—they reveal the logical structure of physics. For each topic, try to reproduce the derivation from memory, explaining the physical reasoning behind each step. Use flow charts or annotated diagrams to connect formulas. Practice past‑paper questions that ask you to derive or apply these equations, as the 9630 exam often tests your ability to work from first principles.

以上推导不仅是数学练习，它们揭示了物理学的逻辑结构。对于每个专题，尝试凭记忆复现推导过程，解释每一步背后的物理原理。使用流程图或注解示意图串联公式。练习往年真题中要求推导或应用这些方程的问题，因为 9630 考试常考查你从基本原理出发的能力。

Topic 主题	Key Derived Formula 推导出的关键公式	Starting Principle 起始原理
Kinematics	v² = u² + 2as	a = (v − u)/t, average velocity
Dynamics	F = Δp/Δt	Newton’s second law of motion
Energy	Eₖ = ½mv²	Work = Fs, F = ma
Circular Motion	a = v²/r	Vector subtraction, similar triangles
Gravitation	g = GM/r²	F = GMm/r², F = mg
SHM	T = 2π√(m/k)	F = −kx, a = −ω²x
Capacitors	Q = Q₀e⁻ᵗ/ᴿᶜ	I = −dQ/dt, Q = CV
Electromagnetism	ε = −d(NΦ)/dt	Faraday’s experiments, flux cutting
Thermal Physics	pV = nRT	Kinetic theory, p = ⅓ρ⟨v²⟩
Electricity	R = ρL/A	Ohm’s law, J = σE

This table serves as a quick reference to connect derived equations with their fundamental origins.

此表可作为快速参考，将推导出的方程与其基本原理联系起来。

Published by TutorHao | Physics Revision Series | aleveler.com

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