📚 Key Derivations in Edexcel IAL Physics Unit 4 (PH04) | 爱德思国际A-Level物理单元4(PH04)公式推导
Unit 4 of the Edexcel International A-Level Physics syllabus – Further Mechanics, Fields and Particles – demands a solid understanding of how fundamental equations are derived. Mastering these derivations not only helps you answer long-answer questions but also deepens your grasp of physical principles. This article walks you through the key derivations you are expected to know, using clear step-by-step reasoning and examiner-friendly formats.
爱德思国际A-Level物理课程单元4（进阶力学、场与粒子）要求学生不仅会使用公式，还必须理解公式的推导过程。掌握这些推导不仅能帮你在长答题中拿分，更能加深对物理原理的理解。本文将通过清晰的逐步推理，带你梳理考试中必须掌握的核心公式推导。
1. Impulse and Change in Momentum | 冲量与动量变化
aw in the form F = ma and the definition of acceleration, we can derive this relationship.
冲量-动量定理指出，物体受到的冲量等于其动量的变化量。从牛顿第二定律 F = ma 和加速度的定义出发，可以推导出这一关系。
e impulse J, and the right side is the change in momentum Δp. Therefore, J = Δp = mv - mu.
由于加速度 a = Δv/Δt，代入 F = ma 得到 F = m(Δv/Δt)。两边同乘以 Δt 得到 F Δt = mΔv。左边是冲量 J，右边是动量变化量 Δp。因此 J = Δp = mv – mu。
F Δt = m(v – u)
In force–time graphs, the area under the curve represents the impulse, which equals the change in momentum even if the force is not constant.
在力–时间图中，曲线下的面积代表冲量，即使力不是恒定的，面积仍等于动量的变化量。
2. Conservation of Momentum in Two Dimensions | 二维动量守恒
plane, momentum is conserved independently along each axis. Consider two bodies colliding – l final momentum vector. Resolving into perpendicular directions (e.g., x and y) gives two scalar equations.
当物体在平面内发生碰撞时，动量在每一个坐标轴方向上独立守恒。考虑两个物体碰撞——总初动量矢量等于总末动量矢量。将矢量沿垂直方向（如x和y方向）分解，可得到两个标量方程。
For a collision between a moving object A and a stationary object B: before collision, total momentum mAuA is along the x‑axis. After collision, mAvA cos θA + mBvB cos θB = mAuA, and perpendicularly, mAvA sin θA – mBvB sin θB = 0. These equations allow us to find unknown velocities or angles.
对于运动物体A与静止物体B的碰撞：碰撞前，总动量 mAuA 沿x轴方向。碰撞后，mAvA cos θA + mBvB cos θB = mAuA，在垂直方向，mAvA sin θA – mBvB sin θB = 0。这两个方程可用来求解未知的速度大小或角度。
Σpx before = Σpx after, Σpy before = Σpy after
The derivation relies on the vector nature of momentum; it is extremely common in mark schemes for PH04.
这一推导依赖于动量的矢量性；这是PH04评分标准中极其常见的考点。
3. Centripetal Acceleration Derivation | 向心加速度的推导
To derive a = v²/r, consider the small change in velocity vector Δv over a short time Δt.
物体以恒定速率v在半径为r的圆周上运动时，具有指向圆心的加速度。为了推导 a = v²/r，考虑在短时间 Δt 内速度矢量的微小变化 Δv。
gle swept out. The distance travelled is v Δt = r Δθ, so Δθ = v Δt / r. Substituting gives |Δv| Because Δv is directed towards the centre, the instantaneous acceleration is centripetal.
在 Δt 趋近于零时，Δv 的大小近似为 v Δθ，其中 Δθ 是转过的微小角度。走过的弧长 v Δt = r Δθ，因此 Δθ = v Δt / r。代入得 |Δv| = v (v Δt / r) = v² Δt / r。平均加速度 a = Δv/Δt = v²/r。由于 Δv 指向圆心，瞬时加速度即为向心加速度。
a = v²/r = r ω²
n be expressed as a = rω². This vector derivation is a classic piece of further mechanics.
利用关系式 v = rω，向心加速度也可写成 a = rω²。这一矢量推导是进阶力学的经典内容。
4. Electric Field Strength and Potential Gradient | 电场强度与电势梯度
d is derived from the definition of potential difference as the work done per unit charge.
对于相距为d、电势差为V的两平行板间的匀强电场，场强E恒定。公式E = V/d 可通过电势差的定义——单位电荷所做的功——推导出来。
o W = qEd. By definition, V = W/q, therefore V = Ed, and rearranging gives E = V/d. In non‑uniform fields, the general relation is E = – where the negative sign indicates that E points in the direction of decreasing potential.
对检验电荷q从一板移动到另一板所做的功 W = F d，而 F = qE，所以 W = qEd。根据定义 V = W/q，因此 V = Ed，整理得 E = V/d。在非匀强电场中，一般关系式为 E = –dV/dr，负号表示场强指向电势降低的方向。
Uniform field: E = V/d General: E = – dV/dr
This derivation elegantly links the vector field strength with the scalar potential, a concept frequently tested in the fields section of PH04.
这一推导巧妙地将矢量场强与标量电势联系起来，是PH04中场部分频繁考查的概念。
5. Capacitance and the Energy Stored | 电容与储能公式的推导
citor can be found by considering the work done to move charge onto the plates against the rising potential.
电容C定义为储存的电荷量与电势差之比：C = Q/V。电容器储存的能量可以通过考虑将电荷移动到极板上克服逐渐升高的电势所做的功来推导。
rgy W = ∫ (q/C) dq = ½ Q²/C. Substituting Q = CV yields the other forms: E = ½ QV = ½ CV².
当小量电荷 dq 在电势 V 下被添加到极板时，做功 dW = V dq。利用 V = q/C，得 dW = (q/C) dq。从0积分到Q得到总能量 W = ∫ (q/C) dq = ½ Q²/C。代入 Q = CV 可得其他形式：E = ½ QV = ½ CV²。
E = ½ Q V = ½ C V² = ½ Q² / C
This derivation shows why the factor ½ appears – it arises from the average potential experienced by the accumulating charge.
这个推导解释了为什么会出现½因子——它源于电荷积累过程中经历的平均电势。
6. Charging and Discharging of Capacitors | 电容器的充放电指数公式
tarting from Kirchhoff’s voltage law and the definition of current, we can derive Q = Q₀ e–t/RC.
在RC串联电路中，放电时电容器上的电荷量按指数规律衰减。从基尔霍夫电压定律和电流的定义出发，可以推导出 Q = Q₀ e–t/RC。
cross the capacitor V = Q/C equals the p.d. across the resistor IR, so Q/C = IR. Since I = –dQ/dt (current reduces charge), we obtain dQ/dt = –Q/(RC). Solving this differential equation yields Q = Q₀ e–t/RC. The time constant τ = RC is the time for the charge to fall to 37% of its initial value.
放电过程中，电容器两端电压 V = Q/C 等于电阻两端电压 IR，因此 Q/C = IR。因为电流 I = –dQ/dt（电流使电荷减少），得到 dQ/dt = –Q/(RC)。解这个微分方程得到 Q = Q₀ e–t/RC。时间常数 τ = RC，是电荷量降至初始值37%所需的时间。
Q = Q₀ e–t/RC, τ = RC
Similar equations apply for current and voltage, and the derivations are a perfect illustration of exponential behaviour examined in Unit 4.
电流和电压也有类似的指数公式，这些推导完美地展示了单元4所考查的指数行为。
7. Magnetic Flux and Faraday’s Law | 磁通量与法拉第电磁感应定律
ates that the magnitude of the induced emf is equal to the rate of change of flux linkage.
磁通量Φ定义为磁通密度B、面积A以及磁场方向与面积法线夹角余弦的乘积：Φ = BA cos θ。法拉第电磁感应定律指出，感应电动势的大小等于磁通链的变化率。
For a coil of N turns, the flux linkage is NΦ. The induced emf is ε = – esents Lenz’s law, indicating the direction of the induced emf opposes the change in flux.
对于N匝线圈，磁通链为 NΦ。感应电动势 ε = –N dΦ/dt。在实际问题中，平均感应电动势常用有限变化量表示：ε_avg = N (ΔΦ/Δt)。负号代表楞次定律，表明感应电动势的方向总是反抗磁通量的变化。
ε = –N ΔΦ / Δt
v) can also be done directly from Faraday’s law by considering the area swept out per unit time.
通过考虑单位时间内扫过的面积，也可以直接从法拉第定律推导出导体在匀强磁场中运动产生的感应电动势公式 ε = BLv。
8. Motion of a Charged Particle in a Magnetic Field | 带电粒子在磁场中的偏转
es a magnetic force F = BQv, which acts as the centripetal force, causing circular motion.
带电粒子垂直于匀强磁场运动时，受到磁力 F = BQv，该力充当向心力，使粒子做圆周运动。
eriod T = 2πr/v = 2πm/(BQ) is independent of speed, a key result used in cyclotron design.
令磁力等于向心力：BQv = mv²/r，解得半径 r = mv/(BQ)。该式表明半径与动量成正比，与磁感应强度和电荷量成反比。周期 T = 2πr/v = 2πm/(BQ) 与速度无关，这是回旋加速器设计中的重要结论。
r = mv / (BQ), T = 2πm / (BQ)
This derivation elegantly combines mechanics with magnetic fields and is a regular feature in PH04 mark schemes for particle physics topics.
这一推导将力学与磁场巧妙结合，是PH04评分标准中粒子物理部分的常见内容。
9. Mass–Energy Equivalence in Particle Annihilation | 粒子湮灭中的质能等价
Einstein’s mass– l in particle physics. When a particle and its antiparticle annihilate, their combined rest mass is converted entirely into energy carried by photons.
爱因斯坦的质能关系式 E = mc² 是粒子物理学的基石。当一个粒子与它的反粒子湮灭时，它们的总静止质量会完全转化为由光子携带的能量。
For an electron–positron annihilation, the total rest mass 2me (where me lectron rest mass) is converted into two photons. Conservation of energy gives 2 × hf = 2mec², so each photon has energy Ephoton = mec² and frequency f = mec²/h. The minimum energy required for pair production is the reverse process, requiring a photon of at least 2mec².
对于电子–正电子湮灭，总静止质量 2me（me为电子静质量）转化为两个光子。根据能量守恒，2 × hf = 2mec²，因此每个光子的能量 Ephoton = mec²，频率 f = mec²/h。电子对产生是逆过程，需要光子能量至少为 2mec²。
E = mc², Annihilation: 2m₀c² → 2hf
These derivations using energy conservation appear frequently in Unit 4 particle physics questions, often requiring students to calculate photon energies or frequencies.
这些利用能量守恒的推导频繁出现在单元4的粒子物理考题中，通常要求学生计算光子能量或频率。
10. Kinetic Energy and Relativistic Momentum in Accelerators | 加速器中的动能与相对论动量
In particle accelerators, charged particles are accelerated to high energies. At low speeds the classical kinetic energy formula ½ mv² applies, but as speeds approach the speed of light, relativistic effects become significant.
在粒子加速器中，带电粒子被加速到高能量。低速时经典动能公式 ½ mv² 适用，但当速度接近光速时，相对论效应变得明显。
The total energy of a particle is E = γm₀c², where γ = 1/√1 – v²/c². The kinetic energy is Ek = (γ – uency and recognise that for high speeds, the simple cyclotron fails because the particle’s mass increases, altering the time period.
粒子的总能量为 E = γm₀c²，其中 γ = 1/√1 – v²/c²。动能为 Ek = (γ – 1)m₀c²，在低速下可化简为 ½ m₀v²。动量变为 p = γm₀v。在单元4中，学生可能需要推导回旋加速器的工作频率，并认识到高速时简单的回旋加速器会失效，因为粒子质量增加，改变了周期。
Ek = (γ – 1)m₀c², p = γm₀v, γ = 1/√1 – v²/c²
While full relativistic derivations are not required, being able to quote and use these relationships to explain accelerator physics is vital for top marks in PH04.
尽管不要求进行完整的相对论推导，但能够引用并运用这些关系解释加速器物理，对于在PH04取得高分至关重要。
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