📚 OxfordAQA PH04 June 2023 Formula Derivations | OxfordAQA PH04 2023年6月公式推导
In the OxfordAQA A-level Physics Unit 4 (PH04) examination, a significant number of marks are awarded for the ability to derive key equations from first principles. The June 2023 marking scheme (Final MS v1.0) explicitly tests these derivations, rewarding logical steps, clear diagrams and accurate algebraic manipulation. This article revisits the most important formula derivations that frequently appear in PH04, presenting them in a step‑by‑step bilingual format to help you secure full marks on those high‑value questions.
在 OxfordAQA A-level 物理单元四（PH04）考试中，相当一部分分数是对从基本原理推导关键方程式的能力的考查。2023年6月的评分方案（最终版 v1.0）明确地考查这些推导，按逻辑步骤、清晰图示和准确的代数运算给分。本文重新梳理 PH04 中最常出现的重要公式推导，以逐步双语讲解的形式呈现，帮助你在那些高分值题目上拿到满分。
1. Derivation of Centripetal Acceleration | 向心加速度公式推导
Consider an object moving with constant speed v in a circle of radius r. Although the speed is constant, the direction changes, so there is an acceleration towards the centre. To derive a = v²/r, we draw a vector diagram of velocities at two close positions separated by a small time Δt.
考虑一个物体以恒定速率 v 在半径为 r 的圆周上运动。尽管速率不变，但方向不断改变，因此存在指向圆心的加速度。为推导 a = v²/r，我们画出两个靠近位置处速度的矢量图，时间间隔为 Δt。
The angle Δθ between the two velocity vectors is the same as the angle between the two radius lines, because velocity is always tangential. The change in velocity Δv points approximately towards the centre. For a small angle, the magnitude Δv ≈ v Δθ, and the arc length Δs = r Δθ, so Δθ = v Δt / r. Substituting gives Δv = v (v Δt / r) = v² Δt / r.
两个速度矢量之间的夹角 Δθ 与两条半径线之间的夹角相同，因为速度总是沿切线方向。速度变化量 Δv 近似指向圆心。对于小角度，其大小 Δv ≈ v Δθ，弧长 Δs = r Δθ，故 Δθ = v Δt / r。代入得 Δv = v (v Δt / r) = v² Δt / r。
By definition, acceleration a = Δv / Δt, so a = (v² Δt / r) / Δt = v²/r. This is the required centripetal acceleration directed towards the centre.
根据定义，加速度 a = Δv / Δt，因此 a = (v² Δt / r) / Δt = v²/r。这就是指向圆心的向心加速度。
a = v² / r
The June 2023 mark scheme expects the velocity triangle to be drawn and labelled, with explicit mention that for small Δθ, sin(Δθ) ≈ Δθ, or that the triangle is isosceles and the resultant points to the centre. Also, the expression ω = v / r may be used to obtain a = ω² r.
2023年6月的评分方案要求画出并标注速度三角形，明确提到对于小角度Δθ，sin(Δθ) ≈ Δθ，或者三角形为等腰三角形且合矢量指向圆心。此外，也可用 ω = v / r 得出 a = ω² r。
2. Combining Circular Motion and Newton’s Law of Gravitation | 圆周运动与万有引力定律的结合
For a planet or satellite in a stable circular orbit, the centripetal force is provided by the gravitational force. Setting the centripetal force equal to the gravitational force yields the orbital speed equation.
对于在稳定圆轨道上运行的行星或卫星，向心力由万有引力提供。令向心力等于万有引力，即可得到轨道速率方程。
Centripetal force needed: F = m v² / r. Gravitational force: F = G M m / r². Equating: m v² / r = G M m / r². Cancel m from both sides (the mass of the satellite does not affect the orbital speed). Multiply both sides by r: v² = G M / r.
所需的向心力：F = m v² / r。万有引力：F = G M m / r²。两者相等：m v² / r = G M m / r²。两边消去 m（卫星质量不影响轨道速率）。两边乘以 r：v² = G M / r。
v = √(G M / r)
This derivation is frequently required in PH04. The mark scheme insists on clear reasoning: state the forces, equate them, cancel m and rearrange. Often, a follow‑up asks for the period T: using v = 2π r / T, substitute to obtain T² = (4π² / G M) r³, which is Kepler’s third law.
这个推导在 PH04 中频繁出现。评分方案要求清晰的推理：陈述作用力、令其相等、消去 m 并重新整理。通常，后续问题会要求求解周期 T：利用 v = 2π r / T，代入可得 T² = (4π² / G M) r³，即开普勒第三定律。
3. Simple Harmonic Motion (SHM): a = – ω² x | 简谐运动基本方程推导
SHM is defined as motion in which the acceleration is directly proportional to the displacement from equilibrium and always directed towards that equilibrium. The defining equation a = – ω² x can be derived from the projection of uniform circular motion onto a diameter.
简谐运动定义为加速度与离开平衡位置的位移成正比且始终指向平衡位置的运动。定义方程 a = – ω² x 可以由匀速圆周运动在直径上的投影推导得出。
Consider a point P moving in a circle of radius A (amplitude) with constant angular speed ω. Its displacement x along the horizontal diameter is x = A cos(ω t). The velocity in SHM is the horizontal component of the circular velocity: v = – ω A sin(ω t). Differentiating velocity with respect to time gives acceleration: a = – ω² A cos(ω t) = – ω² x.
考虑一个点 P 以半径 A（振幅）和恒定角速度 ω 做圆周运动。它在水平直径上的位移为 x = A cos(ω t)。简谐运动的速度是圆周速度的水平分量：v = – ω A sin(ω t)。将速度对时间求导得到加速度：a = – ω² A cos(ω t) = – ω² x。
a = – ω² x
In the June 2023 PH04 mark scheme, candidates can also start from Newton’s second law applied to a mass‑spring system, where restoring force F = – k x, leading to a = – (k/m) x. Then identify ω² = k/m. Either path is acceptable provided the steps are clearly communicated.
在2023年6月 PH04 评分方案中，考生也可以从牛顿第二定律应用于质量‑弹簧系统出发，回复力 F = – k x，得到 a = – (k/m) x，并确定 ω² = k/m。只要步骤表述清楚，任何一种路线均可。
4. Period of a Mass–Spring System | 弹簧谐振子周期推导
For a mass m attached to a spring of spring constant k, the restoring force when displaced by x is F = – k x. Using Newton’s second law, m a = – k x, so a = – (k / m) x.
对于一个连接在劲度系数为 k 的弹簧上的质量块 m，当位移为 x 时，回复力为 F = – k x。利用牛顿第二定律，m a = – k x，故 a = – (k / m) x。
Comparing with the standard SHM equation a = – ω² x, we identify ω² = k / m. Since ω = 2π / T, it follows that (2π / T)² = k / m, giving T² = 4π² m / k.
与标准简谐运动方程 a = – ω² x 对比，可知 ω² = k / m。因为 ω = 2π / T，可得 (2π / T)² = k / m，从而 T² = 4π² m / k。
T = 2π √(m / k)
This derivation is straightforward but requires stating the equivalence between the two forms of acceleration. The mark scheme often looks for the explicit comparison step and the correct handling of the square root when solving for T.
这个推导很直接，但要求说明两种加速度形式的等价性。评分方案常常注意比较步骤的明确性以及解出 T 时平方根处理的正确性。
5. Derivation of the Simple Pendulum Period | 单摆周期公式推导
For a simple pendulum of length L, the restoring force along the arc is caused by the component of weight: F = – m g sin θ, where θ is the angular displacement. For small angles (θ < 10°), sin θ ≈ θ in radians.
对于长度为 L 的单摆，沿圆弧方向的回复力由重力的分量引起：F = – m g sin θ，其中 θ 为角位移。对于小角度（θ < 10°），sin θ ≈ θ（弧度制）。
Since the linear displacement x = L θ, the restoring acceleration a = – g θ = – g (x / L). Thus a = – (g / L) x. Comparing with a = – ω² x gives ω² = g / L. Using ω = 2π / T, we obtain (2π / T)² = g / L, so T² = 4π² L / g.
由于线位移 x = L θ，回复加速度 a = – g θ = – g (x / L)。因此 a = – (g / L) x。与 a = – ω² x 对比，得 ω² = g / L。代入 ω = 2π / T，可得 (2π / T)² = g / L，即 T² = 4π² L / g。
T = 2π √(L / g)
The PH04 mark scheme awards marks for stating the small‑angle approximation, converting the angular displacement to linear displacement, and correctly identifying the constant of proportionality. Diagrams showing the restoring component –mg sin θ are also valued.
PH04 评分方案对陈述小角度近似、将角位移转换为线位移以及正确识别比例常数给予分值。绘制出回复力分量 –mg sin θ 的示意图同样会得到认可。
6. Work–Energy Principle and Kinetic Energy Formula | 功‑能原理与动能公式
The work–energy principle states that the net work done on an object equals its change in kinetic energy. To derive K.E. = ½ m v², consider a constant resultant force F acting on a mass m, causing an acceleration a over a displacement s.
功‑能原理指出，对物体所做的净功等于其动能的变化。要推导 K.E. = ½ m v²，考虑一个恒定的合力 F 作用在质量 m 上，在位移 s 上产生加速度 a。
Using the kinematic equation v² = u² + 2 a s, set initial velocity u = 0 for a body starting from rest. Then v² = 2 a s, so a s = v² / 2. Work done W = F s = m a s = m (v² / 2) = ½ m v². Hence the kinetic energy gained equals the work done, leading to the definition of kinetic energy.
利用运动学方程 v² = u² + 2 a s，设物体由静止出发，初速度 u = 0，则 v² = 2 a s，因此 a s = v² / 2。所做的功 W = F s = m a s = m (v² / 2) = ½ m v²。所以获得的动能等于做功，从而引出动能的定义。
K.E. = ½ m v²
Alternatively, the PH04 mark scheme accepts derivation using integration: W = ∫ F dx = ∫ m (dv/dt) dx = m ∫ v dv = ½ m v². This calculus‑based approach is expected for A2 candidates, and it often appears in structured questions on work done by variable forces.
此外，PH04 评分方案也接受使用积分方法：W = ∫ F dx = ∫ m (dv/dt) dx = m ∫ v dv = ½ m v²。这种基于微积分的方法适合 A2 阶段的考生，并常出现在变力做功的结构题中。
7. Conservation of Momentum from Newton’s Third Law | 由牛顿第三定律推导动量守恒
Consider two bodies, A and B, interacting with each other. According to Newton’s third law, the force exerted by A on B, F_AB, is equal in magnitude and opposite in direction to the force exerted by B on A, F_BA: F_AB = – F_BA.
考虑两个相互作用的物体 A 和 B。根据牛顿第三定律，A 对 B 的作用力 F_AB 与 B 对 A 的作用力 F_BA 大小相等、方向相反：F_AB = – F_BA。
If the forces act for the same time interval Δt, the impulses are equal and opposite: F_AB Δt = – F_BA Δt. Impulse equals change in momentum, so Δp_B = – Δp_A, meaning the total change in momentum of the system Δp_A + Δp_B = 0. Hence total momentum is conserved provided no external forces act.
如果力作用的时间间隔 Δt 相同，则冲量大小相等、方向相反：F_AB Δt = – F_BA Δt。冲量等于动量的变化，所以 Δp_B = – Δp_A，即系统总动量的变化 Δp_A + Δp_B = 0。因此在无外力作用的条件下，总动量守恒。
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂
The June 2023 scheme expects explicit mention that the derivation relies on Newton’s third law and the definition of impulse as F Δt = Δ(m v). Diagrams showing the forces during collision can also be useful auxiliary marks.
2023年6月的评分方案要求明确指出该推导依赖牛顿第三定律以及冲量定义 F Δt = Δ(m v)。展示碰撞过程中作用力的示意图也有助于获得辅助分。
8. Capacitor Discharge Equation | 电容器放电方程
For a capacitor of capacitance C discharging through a resistor R, the charge q on the plates decreases over time. At any instant, the p.d. V = q / C, and by Kirchhoff’s voltage law, V = I R. Since the current I = – dq / dt (negative because charge decreases), we have q / C = – R dq / dt.
对于一个电容为 C 的电容器通过电阻 R 放电，极板上的电荷量 q 随时间减少。在任一时刻，电势差 V = q / C，根据基尔霍夫电压定律，V = I R。由于电流 I = – dq / dt（负号表示电荷减少），可得 q / C = – R dq / dt。
Separating variables: (1/q) dq = – 1/(RC) dt. Integrating both sides from initial charge Q₀ to q and from 0 to t gives ln(q / Q₀) = – t / (RC). Thus q = Q₀ e^(– t / RC).
分离变量：(1/q) dq = – 1/(RC) dt。两边从初始电荷 Q₀ 积分到 q，时间从 0 到 t，得到 ln(q / Q₀) = – t / (RC)。因此 q = Q₀ e^(– t / RC)。
q = Q₀ e^(– t / RC)
From this, the voltage and current decay equations follow: V = V₀ e^(– t / RC) and I = I₀ e^(– t / RC). The time constant τ = RC is the time for the charge to fall to 1/e (about 37%) of its initial value. The PH04 mark scheme often checks the limit of integration and the correct handling of the negative sign.
由此可得电压和电流的衰减方程：V = V₀ e^(– t / RC) 和 I = I₀ e^(– t / RC)。时间常数 τ = RC 是电荷降到初始值 1/e（约37%）所需的时间。PH04 评分方案常检查积分限以及负号处理的正确性。
9. Relationship Between Electric Field Strength and Potential Gradient | 电场强度与电势梯度的关系
In a uniform electric field, the relationship E = V / d can be derived from the work done in moving a positive test charge q₀ between two equipotential plates separated by distance d. The work done W = force × distance = (q₀ E) d. Also, by definition of potential difference, W = q₀ V.
在匀强电场中，关系式 E = V / d 可以由移动检验电荷 q₀ 在相距为 d 的两块等势板间所做的功推导得出。所做的功 W = 力 × 距离 = (q₀ E) d。同时，根据电势差的定义，W = q₀ V。
Equating the two expressions: q₀ E d = q₀ V, so E = V / d. The more general form for a non‑uniform field is the negative gradient of potential: E = – dV / dr, where the negative sign indicates that the electric field points in the direction of decreasing potential.
令两式相等：q₀ E d = q₀ V，因此 E = V / d。对于非匀强电场，更一般的形式是电势的负梯度：E = – dV / dr，其中负号表示电场方向指向电势降低的方向。
E = – dV / dr (general) and E = V / d (uniform field)
In the PH04 June 2023 scheme, candidates may be asked to derive the field between parallel plates or to use the gradient of a V‑r graph for a point charge. Showing the dimension check, V m⁻¹ from J C⁻¹ m⁻¹, is often rewarded.
在2023年6月 PH04 方案中，考生可能被要求推导平行板间的电场，或利用点电荷的 V‑r 图像的斜率。展示量纲检验——由 J C⁻¹ m⁻¹ 得到 V m⁻¹——通常会得到加分。
10. Faraday’s Law and Magnetic Flux Linkage | 法拉第定律与磁通匝链数
Faraday’s law of electromagnetic induction states that the induced e.m.f. is equal to the rate of change of magnetic flux linkage. For a coil of N turns, the flux linkage is N Φ, where Φ = B A cos θ. The law is expressed as ε = – N dΦ / dt. The derivation often begins with the experimental observation that a changing magnetic flux induces an e.m.f., and the negative sign reflects Lenz’s law.
法拉第电磁感应定律指出，感应电动势等于磁通匝链数的变化率。对于 N 匝线圈，磁通匝链数为 N Φ，其中 Φ = B A cos θ。该定律写作 ε = – N dΦ / dt。推导常从实验观察出发，即变化的磁通量会感应出电动势，负号反映楞次定律。
In the context of a straight conductor of length L moving with velocity v perpendicular to a magnetic field B, the motional e.m.f. can be derived by considering the force on a charge carrier. The magnetic force F = B q v. When the conductor moves, an electric field E builds up inside until equilibrium: q E = B q v, giving E = B v. Since the potential difference across length L is ε = E L, we obtain ε = B L v.
在长度为 L 的直导体以速度 v 垂直于磁场 B 运动的情况下，可通过分析电荷载子所受的力推导出动生电动势。磁力 F = B q v。当导体移动时，内部建立起电场 E 直至平衡：q E = B q v，得 E = B v。由于在长度 L 两端的电势差为 ε = E L，因此得到 ε = B L v。
ε = – N dΦ / dt and ε = B L v (motional e.m.f.)
The mark scheme rewards linking the two expressions through flux cutting: in time Δt, the area swept is L v Δt, so the change in flux is ΔΦ = B L v Δt, leading to ε = B L v for a single turn. Emphasising Lenz’s law and the conservation of energy principle is also expected.
评分方案看重通过磁通切割将两个表达式联系起来：在 Δt 时间内扫过的面积为 L v Δt，因此磁通变化量为 ΔΦ = B L v Δt，从而对于单匝线圈得到 ε = B L v。强调楞次定律和能量守恒原理也是预期的内容。
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