📚 International A-Level Physics Unit 4: Key Formula Derivations from the Jan 19 Mark Scheme | 国际 A-Level 物理第四单元：2019 年 1 月评分方案核心公式推导
The January 2019 Unit 4 (Further Mechanics, Fields and Particles) mark scheme rewards students who can not only recall equations but also explain their origins. Understanding derivations deepens your grasp of physical principles and helps secure the highest marks on ‘show that’ questions. This article walks through eight central derivations that frequently appear, from circular motion dynamics to exponential capacitor discharge.
2019 年 1 月的 Unit 4（进阶力学、场与粒子）评分方案奖励那些不仅能背诵方程、更能解释其由来的学生。理解推导过程可以加深你对物理原理的掌握，并有助于在“证明”类题目中拿下最高分。本文将逐一讲解八个常见核心推导，涵盖从圆周运动动力学到电容指数放电等内容。
1. Centripetal Acceleration: a = v²/r = rω² | 向心加速度：a = v²/r = rω²
Consider an object moving at constant speed v in a circle of radius r. In a short time Δt, the object moves through an angle Δθ and its velocity vector turns by the same angle. The change in velocity Δv is directed towards the centre and has magnitude v Δθ. Since Δθ = v Δt / r, the magnitude of the average acceleration is a = Δv / Δt = v² / r. In the limit Δt → 0, this becomes the instantaneous centripetal acceleration.
考虑一个物体以恒定速度 v 在半径为 r 的圆周上运动。在很短的时间 Δt 内，物体转过角度 Δθ，其速度矢量也转过同样的角度。速度变化量 Δv 指向圆心，大小为 v Δθ。因为 Δθ = v Δt / r，平均加速度的大小为 a = Δv / Δt = v² / r。取极限 Δt → 0，即得到瞬时向心加速度。
Substituting v = rω gives the alternative form a = rω². Both expressions are identical and are used depending on whether angular velocity ω or linear speed v is given.
代入 v = rω 可得另一种形式 a = rω²。两种表达式完全相同，可根据已知量是角速度 ω 还是线速度 v 来选用。
2. Centripetal Force: F = mv²/r = mrω² | 向心力：F = mv²/r = mrω²
Newton’s second law states that F = ma. Taking the centripetal acceleration derived above and multiplying by mass m immediately yields the centripetal force required to keep an object in uniform circular motion: F = mv²/r = mrω². This force is always directed towards the centre of the circle.
牛顿第二定律指出 F = ma。将上面推导出的向心加速度乘以质量 m，直接得到维持物体做匀速圆周运动所需的向心力：F = mv²/r = mrω²。这个力始终指向圆心。
In the exam, you may need to identify which physical force (tension, gravity, friction, magnetic) provides this centripetal force and equate it to mrω² or mv²/r to solve problems such as banked tracks, conical pendulums or charged particles in magnetic fields.
在考试中，你可能需要识别哪个实际力（张力、重力、摩擦力、磁力）提供了这个向心力，并将其与 mrω² 或 mv²/r 联立，以解决如倾斜轨道、圆锥摆或带电粒子在磁场中运动等问题。
3. Derivation of T² ∝ r³ from Newton’s Law of Gravitation | 由牛顿万有引力定律推导 T² ∝ r³
For a planet of mass m orbiting a star of mass M, the gravitational force provides the centripetal force: GMm/r² = mv²/r. Cancel m and multiply by r to get GM/r = v². Substituting v = 2πr/T gives GM/r = (4π²r²)/T². Rearranging leads to T² = (4π²/GM) r³, which is Kepler’s third law for circular orbits.
对于一颗质量为 m 的行星绕质量为 M 的恒星运行，万有引力提供向心力：GMm/r² = mv²/r。消去 m 并两边乘以 r 得到 GM/r = v²。代入 v = 2πr/T 得 GM/r = (4π²r²)/T²。整理后得到 T² = (4π²/GM) r³，这就是圆轨道下的开普勒第三定律。
This derivation is frequently required in Unit 4 as a ‘show that’ exercise. The constant of proportionality depends only on the central mass, which allows us to calculate the mass of the Sun or other stars from orbital data.
这个推导在 Unit 4 中经常作为“证明”练习出现。比例常数仅取决于中心天体的质量，这使我们能够通过轨道数据计算太阳或其他恒星的质量。
4. Electric Field Strength of a Point Charge: E = kQ/r² | 点电荷的电场强度：E = kQ/r²
From Coulomb’s law, the force on a test charge q due to a point charge Q is F = kQq/r², where k = 1/(4πε₀). Since electric field strength is defined as E = F/q, substituting gives E = kQ/r². The direction is radial: outward from a positive charge and inward towards a negative charge.
由库仑定律可知，点电荷 Q 对试探电荷 q 的作用力为 F = kQq/r²，其中 k = 1/(4πε₀)。因为电场强度定义为 E = F/q，代入后得到 E = kQ/r²。其方向沿径向：正电荷向外，负电荷向内。
This inverse-square relationship mirrors the gravitational field g = GM/r². Understanding the mathematical symmetry helps students recall and apply both field equations confidently in Unit 4 problems.
这种平方反比关系与引力场 g = GM/r² 类似。理解数学上的对称性有助于学生在 Unit 4 的问题中自信地回忆和应用这两个场方程。
5. Capacitor Discharge Equation: Q = Q₀ e^(−t/RC) | 电容放电方程：Q = Q₀ e^(−t/RC)
When a capacitor of capacitance C discharges through a resistor R, the current I is the rate of flow of charge: I = −dQ/dt. By definition, Q = CV and V = IR, so Q = C × IR, giving I = Q/(RC). Combining, −dQ/dt = Q/(RC). Separating variables: dQ/Q = −dt/(RC). Integrating from Q₀ at t = 0 to Q at time t yields ln(Q/Q₀) = −t/(RC), hence Q = Q₀ e^(−t/RC).
当电容 C 通过电阻 R 放电时，电流 I 是电荷流动的速率：I = −dQ/dt。根据定义，Q = CV 且 V = IR，因此 Q = C × IR，得到 I = Q/(RC)。联立得 −dQ/dt = Q/(RC)。分离变量：dQ/Q = −dt/(RC)。从 t = 0 时电荷为 Q₀ 积分到 t 时刻电荷为 Q，得到 ln(Q/Q₀) = −t/(RC)，因此 Q = Q₀ e^(−t/RC)。
Voltage and current follow the same exponential decay pattern: V = V₀ e^(−t/RC) and I = I₀ e^(−t/RC). The time constant τ = RC is the time for the quantity to fall to about 37% of its initial value.
电压和电流遵循相同的指数衰减规律：V = V₀ e^(−t/RC) 和 I = I₀ e^(−t/RC)。时间常数 τ = RC 是使相关量降至初始值约 37% 所需的时间。
6. Magnetic Force on a Moving Charge: F = Bqv sin θ | 运动电荷在磁场中的受力：F = Bqv sin θ
The magnetic force on a single charged particle is derived from the more general force on a current-carrying wire. For a wire, F = BIL sin θ. Current I = nAqv, where n is the number of charge carriers per unit volume, A is cross-sectional area, q is the charge on each carrier, and v is drift velocity. The number of carriers in length L is nAL, so total force on them is B (nAqv) L sin θ. Dividing by nAL gives the force on one charge: F = Bqv sin θ.
单个带电粒子所受的磁力可以从更一般的载流导线受力公式推导出来。对于一段导线，F = BIL sin θ。电流 I = nAqv，其中 n 是单位体积内的载流子数目，A 是横截面积，q 是每个载流子的电荷，v 是漂移速度。长度为 L 的导线中载流子总数为 nAL，因此它们所受的总力为 B (nAqv) L sin θ。除以 nAL 即得一个电荷所受的力：F = Bqv sin θ。
When the velocity is perpendicular to the magnetic field (θ = 90°), F = Bqv. This force acts as a centripetal force, leading to r = mv/(Bq), a favourite exam derivation for determining the radius of a particle’s circular path in a magnetic field.
当速度与磁场垂直时（θ = 90°），F = Bqv。这个力充当向心力，从而导出 r = mv/(Bq)，这是考试中常用于确定粒子在磁场中圆周运动半径的一个常见推导。
7. Momentum Conservation and Impulse | 动量守恒与冲量
Newton’s second law in its most general form is F = dp/dt. If the net external force on a system is zero, then dp/dt = 0, so the total momentum p is constant. This is the law of conservation of momentum. For two colliding objects, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.
牛顿第二定律的最普遍形式是 F = dp/dt。如果系统所受合外力为零，则 dp/dt = 0，因此总动量 p 守恒。这就是动量守恒定律。对于两个碰撞物体，有 m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂。
Impulse is defined as the integral of force over time, which equals the change in momentum: J = ∫F dt = Δp. The area under a force–time graph gives the impulse, a key exam skill for solving collision and rebound problems.
冲量定义为力对时间的积分，等于动量的变化：J = ∫F dt = Δp。力–时间图像下的面积表示冲量，这是解决碰撞和反弹问题时的一个重要考试技能。
8. Relating Electric Potential and Field: E = −dV/dr | 电势与电场的关系：E = −dV/dr
For a radial electric field, the potential at a distance r from a point charge Q is V = kQ/r. The electric field strength E is the negative gradient of potential: E = −dV/dr. Differentiating V with respect to r gives dV/dr = −kQ/r², so E = −(−kQ/r²) = kQ/r², which recovers the earlier point charge field formula. The minus sign indicates that field lines point in the direction of decreasing potential.
对于径向电场，距离点电荷 Q 为 r 处的电势为 V = kQ/r。电场强度 E 是电势的负梯度：E = −dV/dr。将 V 对 r 求导得 dV/dr = −kQ/r²，因此 E = −(−kQ/r²) = kQ/r²，这就回到了之前的点电荷场强公式。负号表明电场线指向电势降低的方向。
In a uniform field, E = V/d, which is the special case where potential decreases linearly with distance. Understanding this link is crucial for tackling questions about equipotential surfaces and field strength in capacitor problems.
在匀强电场中，E = V/d，这是电势随距离线性下降的特殊情形。理解这种联系对于解决电容问题中的等势面和场强问题至关重要。
9. Radioactive Decay Law and Half-life | 放射性衰变定律与半衰期
The activity A of a radioactive sample is the rate of decay: A = −dN/dt. Experiment shows activity is proportional to the number of undecayed nuclei N: A = λN, where λ is the decay constant. Hence −dN/dt = λN. Solving the differential equation yields N = N₀ e^(−λt). The half-life t₁/₂ is the time for N to halve, so ½N₀ = N₀ e^(−λ t₁/₂), giving t₁/₂ = ln 2 / λ.
放射性样品的活度 A 是衰变速率：A = −dN/dt。实验表明活度与未衰变原子核数目 N 成正比：A = λN，其中 λ 是衰变常量。因此 −dN/dt = λN。解这个微分方程得到 N = N₀ e^(−λt)。半衰期 t₁/₂ 是 N 减半所需的时间，故 ½N₀ = N₀ e^(−λ t₁/₂)，解得 t₁/₂ = ln 2 / λ。
Students should be able to derive the relationship N = N₀ (1/2)^(n) for n half-lives as well as use the exponential form to calculate decay-corrected activities, a skill tested in the Unit 4 particle physics section.
学生应能推导出 n 个半衰期后 N = N₀ (1/2)^(n) 的关系式，并能使用指数形式计算经衰变修正的活度，这是 Unit 4 粒子物理部分考查的一项技能。
10. Energy Stored in a Capacitor: E = ½QV = ½CV² = ½Q²/C | 电容器储存的能量：E = ½QV = ½CV² = ½Q²/C
When a capacitor is charged, the p.d. across it builds up from 0 to V. The work done dW to add a small increment of charge dq when the p.d. is v is dW = v dq. Since q = Cv, dW = (q/C) dq. Integrating from 0 to Q gives total work W = ∫₀^Q (q/C) dq = Q²/(2C) = ½QV. Substituting Q = CV gives ½CV².
电容器充电时，其两端电势差从 0 增加到 V。当电势差为 v 时，增加微小电荷 dq 所做的功为 dW = v dq。由于 q = Cv，有 dW = (q/C) dq。从 0 到 Q 积分得到总功 W = ∫₀^Q (q/C) dq = Q²/(2C) = ½QV。代入 Q = CV 得 ½CV²。
This energy is stored in the electric field between the plates. The derivation by integration is a standard ‘show that’ task and highlights why the factor ½ appears, compared to the simple product QV.
这些能量储存在两极板之间的电场中。通过积分进行推导是标准的“证明”任务，并揭示了为何会出现因子 ½，而非简单的乘积 QV。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导