📚 A-Level Physics Paper 2 (June 2019): Exam Report on Formula Derivations | A-Level物理试卷2（2019年6月）：公式推导考试报告
Every year, the A-Level Physics Paper 2 exam demands more than just rote learning; it tests whether students can derive fundamental equations from first principles. The June 2019 sitting was no exception, with several high-mark questions dedicated to step-by-step derivations across mechanics, fields and nuclear physics. This article examines the key formula derivations that appeared, draws on the official examiner report to highlight common mistakes, and provides clear, annotatable derivation pathways to help future candidates succeed.
每年的A-Level物理试卷2考试都不仅仅要求死记硬背，它考察学生是否能从基本原理推导出基本方程。2019年6月的考试也不例外，试卷中好几道高分题专门考查了力学、场和核物理中的逐步推导。本文梳理了考查的关键公式推导，依据官方考官报告指出常见错误，并提供清晰、可注解的推导路径，帮助未来的考生取得成功。
1. Overview of the 2019 Paper 2 Exam | 2019年试卷2概述
The June 2019 Paper 2 covered Further Mechanics, Thermal Physics, Fields, and Nuclear Physics. The structure included multiple-choice questions and a longer written section. Examiner feedback consistently emphasised that many candidates lost marks on derivations because they could not link physical concepts to algebraic manipulation. Questions on centripetal acceleration, gravitational fields, capacitor energy, and radioactive decay featured prominently in the derivation category.
2019年6月的试卷2涵盖了进阶力学、热物理、场和核物理。试卷结构包含选择题和较长的简答部分。考官反馈一再强调，许多考生在推导题上失分，因为他们无法将物理概念与代数运算联系起来。向心加速度、引力场、电容器能量以及放射性衰变等推导题在试卷中占据突出位置。
The examiner report explicitly stated that students who presented clear, logical sequences with labelled diagrams and correct vector geometry scored highest. In contrast, those who jumped straight to the final formula without justifying intermediate steps rarely attained full marks.
考官报告明确指出，那些展示出清晰、逻辑的推导顺序，并配有标注图和正确矢量几何的学生得分最高。相反，那些跳过中间步骤直接给出最终公式的学生很少能拿到满分。
2. Deriving Centripetal Acceleration | 向心加速度推导
One of the most challenging derivation questions asked candidates to derive a = v²/r for an object moving in a circle at constant speed. The examiner noted that only about 40% of responses contained a fully correct vector-based argument.
最具挑战性的推导题之一要求考生推导出物体以恒定速率做圆周运动的向心加速度a = v²/r。考官指出，只有约40%的答卷包含了完全正确的矢量论证。
Step 1: Consider a particle moving with speed v around a circle of radius r. In a short time Δt, it sweeps out an angle Δθ = vΔt / r. The velocity vectors at the start and end of the interval have equal magnitude v but different directions.
步骤1：考虑一个粒子以速率v沿半径为r的圆周运动。在短时间Δt内，它扫过的角度Δθ = vΔt / r。起止时刻的速度矢量大小均为v，但方向不同。
Step 2: The change in velocity Δv is found by vector subtraction. For small Δθ, the magnitude of Δv is approximately vΔθ. Crucially, the direction of Δv points exactly towards the centre of the circle, as revealed by the similarity between the velocity triangle and the triangle formed by two radii.
步骤2：速度变化Δv通过矢量减法求得。对于很小的Δθ，Δv的大小近似为vΔθ。关键的是，Δv的方向正指向圆心，这可通过速度矢量三角形与两条半径构成的三角形相似来揭示。
Step 3: Acceleration magnitude a = Δv / Δt ≈ vΔθ / Δt = v × (v/r) = v²/r. Since Δv is directed radially inward, the acceleration is centripetal. Many candidates lost marks by treating Δv as a scalar or failing to discuss why Δv points towards the centre.
步骤3：加速度大小a = Δv / Δt ≈ vΔθ / Δt = v × (v/r) = v²/r。因为Δv指向径向向内，所以加速度是向心的。许多考生因将Δv视为标量，或未能解释为何Δv指向圆心而失分。
The report advised drawing separate velocity vectors and explicitly marking the angle between them as Δθ. Stronger candidates also included a small-angle approximation statement.
报告建议分别画出速度矢量，并明确标注它们之间的夹角为Δθ。表现较好的考生还会写出小角度近似说明。
3. Escape Velocity from Gravitational Potential | 从引力势推导逃逸速度
A multi-step question required candidates to derive the escape speed vₑ = √(2GM/R) from a planet’s surface. The examiners observed that students often confused gravitational potential V = -GM/r with gravitational potential energy, or mishandled the sign.
一道多步题要求考生从行星表面推导出逃逸速度vₑ = √(2GM/R)。考官发现，学生经常混淆引力势V = -GM/r与引力势能，或在符号上出错。
Start with the gravitational potential energy at the surface: U = -GMm/R. At infinite distance, both kinetic energy and potential energy approach zero. By conservation of energy, ½mvₑ² + ( -GMm/R ) = 0.
开始时，行星表面的引力势能为U = -GMm/R。在无限远处，动能和势能均趋近于零。根据能量守恒，½mvₑ² + ( -GMm/R ) = 0。
Rearrange to obtain ½mvₑ² = GMm/R, cancel m, and solve for vₑ = √(2GM/R). The examiner report stressed that candidates must write the general energy conservation equation first and substitute the correct signs. A frequent error was writing +GMm/R instead of -GMm/R, leading to an imaginary escape speed under further manipulation.
整理得½mvₑ² = GMm/R，约去m，解出vₑ = √(2GM/R)。考官报告强调，考生必须先写出一般的能量守恒方程，再代入正确的符号。常见的错误是将势能写作+GMm/R，而不是-GMm/R，导致后续运算会出现虚数逃逸速度。
Several answers incorrectly tried to use circular orbit centripetal force equations; the report clarified that escape speed derivation is strictly an energy argument, not a force balance.
一些答案错误地试图使用圆周运动的向心力方程；报告澄清，逃逸速度的推导严格采用能量论证，而非力平衡。
4. Energy Stored in a Capacitor | 电容器储存的能量
Deriving the formula for energy stored in a capacitor, W = ½QV, from the work done to charge it featured in the fields section. The exam report indicated that many students could quote the formula but could not show the integration step that yields the factor ½.
从充电做功推导电容器储存能量公式W = ½QV，出现在场部分。考试报告显示，许多学生能说出公式，却无法展示得出系数½的积分步骤。
Consider a capacitor being charged. At an intermediate stage when charge is q and p.d. is v, the infinitesimal work done to add an extra charge dq is dW = v dq. Using v = q/C, we have dW = (q/C) dq.
考虑电容器正在充电。在中间阶段，当电荷为q、电压为v时，添加额外电荷dq所做的元功为dW = v dq。利用v = q/C，得dW = (q/C) dq。
Integrate from 0 to Q: W = ∫₀^Q (q/C) dq = (1/C)[½q²]₀^Q = ½ Q²/C. Since Q = CV, this simplifies to W = ½CV² = ½QV. Examiners noted that candidates who attempted to just multiply final Q and V missed the important fact that p.d. rises linearly during charging, so average p.d. is V/2.
从0积分至Q：W = ∫₀^Q (q/C) dq = (1/C)[½q²]₀^Q = ½ Q²/C。由于Q = CV，简化为W = ½CV² = ½QV。考官指出，那些试图直接用最终Q和V相乘的考生忽略了重要事实——充电过程中电压是线性上升的，因此平均电压为V/2。
The report recommended including a graph of V against q and shading the area under the line to support the derivation, a method that was awarded marks.
报告建议在推导时附上V-q图，并对图线下方面积进行阴影标注，这一方法可获得分数。
5. Electric Potential of a Point Charge | 点电荷的电势
Another frequently examined derivation was the electric potential V at a distance r from a point charge Q, starting from Coulomb’s law. The 2019 report commented that students were often unable to set up the integral correctly with limits from infinity to r.
另一个常考的推导是利用库仑定律推出点电荷Q在距离r处的电势V。2019年报告评价说，学生往往无法正确设定从无穷远到r的积分上下限。
Begin by stating the radial electric field strength E = Q / (4πε₀r²). The potential at r is the work done per unit charge in bringing a small positive test charge from infinity to r: V = -∫_∞^r E·dr. Because E and dr are in opposite directions (E outward, dr inward), the dot product introduces a negative sign, giving V = -∫_∞^r (Q/(4πε₀r²)) dr.
首先写出径向电场强度E = Q / (4πε₀r²)。在r处的电势是将微小正检验电荷从无穷远移至r时每单位电荷所做的功：V = -∫_∞^r E·dr。由于E和dr方向相反（E向外，dr向内），点积引入负号，得到V = -∫_∞^r (Q/(4πε₀r²)) dr。
Evaluate the integral: V = – [ -Q/(4πε₀r) ]_∞^r = Q/(4πε₀r) – 0 = Q/(4πε₀r). Common errors included omitting the negative sign in the work integral, treating E as a constant, or forgetting to convert the integration variable r properly.
计算积分：V = – [ -Q/(4πε₀r) ]_∞^r = Q/(4πε₀r) – 0 = Q/(4πε₀r)。常见错误包括漏写功积分中的负号、将E视为常数，或者未正确转换积分变量r。
Examiners advised stating clearly that ∫ 1/r² dr = -1/r, and showing the substitution step. A small diagram with field lines and path direction also helped excellent candidates secure full marks.
考官建议明确写出∫ 1/r² dr = -1/r，并展示代入步骤。一幅标注有电场线和路径方向的小图也有助于优秀考生获得满分。
6. Charged Particle in a Magnetic Field | 磁场中的带电粒子
The radius of curvature for a charged particle moving perpendicular to a uniform magnetic field, r = mv/(BQ), was another required derivation. According to the examiner report, many students recognised the centripetal force condition but failed to explain why the magnetic force provides that centripetal force.
对于垂直进入匀强磁场的带电粒子，其曲率半径r = mv/(BQ)是另一个需要推导的公式。据考官报告，许多学生认出了向心力条件，但未能解释为何磁力充当了向心力。
Start from the magnetic force on a moving charge: F = BQv sinθ. When θ = 90°, F = BQv. This force is always perpendicular to the velocity, so it does no work and causes circular motion. Therefore, BQv = mv²/r.
从运动电荷受到的磁力出发：F = BQv sinθ。当θ = 90°时，F = BQv。这个力始终垂直于速度，因此不做功，并导致圆周运动。因此，BQv = mv²/r。
Rearrange to r = mv/(BQ). Candidates frequently lost marks by omitting the justification that the magnetic force is the centripetal force and by not clearly stating that the particle’s path is circular. Additionally, the examiner report warned against confusing this formula with the electric field case.
整理得r = mv/(BQ)。考生常常因没有说明磁力就是向心力，也没有明确陈述粒子轨迹为圆周而失分。此外，考官报告警告不要将此公式与电场情形混淆。
In more demanding variations, students had to combine this with the kinetic energy ½mv² = qV to express r in terms of accelerating potential. This multi-step linkage was handled well by only the top third of the candidates.
在要求更高的变式中，学生需要将此与动能½mv² = qV结合，用加速电压表示r。只有位于前三分之一的高分考生能很好地完成这种多步联动。
7. Simple Harmonic Motion Displacement Equation | 简谐运动的位移方程
A common derivation for the displacement equation x = A cos(ωt) or x = A sin(ωt) appeared in the context of an oscillator. The report stated that many candidates lost marks by confusing angular frequency ω with angular velocity and by not linking the solution to the defining differential equation a = -ω²x.
在振动情境中，出现了常见的位移方程 x = A cos(ωt) 或 x = A sin(ωt) 的推导。报告指出，许多考生因混淆角频率ω与角速度，以及没有将解与定义微分方程 a = -ω²x 联系起来而失分。
In SHM, acceleration is proportional to negative displacement: d²x/dt² = -ω²x. A general solution is x = A cos(ωt + φ). Substituting back verifies the solution: dx/dt = -Aω sin(ωt + φ), d²x/dt² = -Aω² cos(ωt + φ) = -ω²x.
在简谐运动中，加速度与位移成正比且反向：d²x/dt² = -ω²x。一般解为x = A cos(ωt + φ)。代入验证：dx/dt = -Aω sin(ωt + φ)，d²x/dt² = -Aω² cos(ωt + φ) = -ω²x。
To determine A and φ, boundary conditions are used, e.g. at t=0, x = x₀ and v = 0 leads to φ = 0 and A = x₀. The examiner noted that students often could not differentiate cosine and sine correctly, thereby breaking the logical chain. They also advised writing the derivative steps explicitly rather than quoting the final result.
要确定A和φ，需使用边界条件，例如在t=0时，x = x₀ 且 v = 0，可得到φ = 0，A = x₀。考官指出，学生经常不能正确进行余弦和正弦的微分，从而打断了逻辑链。他们还建议明确写出求导步骤，而不是直接引用最终结果。
8. Radioactive Decay Law | 放射性衰变定律
Deriving the exponential decay law N = N₀e⁻λt from the activity equation dN/dt = -λN was tested. The 2019 exam report highlighted that while most students could write the differential equation, many failed to separate variables and integrate correctly, or they omitted the constant of integration.
从活度方程dN/dt = -λN推导指数衰变律 N = N₀e⁻λt 也进行了考查。2019年考试报告强调，虽然大多数学生能写出微分方程，但许多人未能正确分离变量并积分，或漏掉了积分常数。
Separate variables: (1/N) dN = -λ dt. Integrate both sides: ∫ (1/N) dN = ∫ -λ dt, giving ln N = -λt + C. At t = 0, N = N₀, so ln N₀ = C. Therefore, ln N – ln N₀ = -λt ⇒ ln(N/N₀) = -λt ⇒ N/N₀ = e⁻λt ⇒ N = N₀e⁻λt.
分离变量：(1/N) dN = -λ dt。两边积分：∫ (1/N) dN = ∫ -λ dt，得到 ln N = -λt + C。在t = 0时，N = N₀，所以ln N₀ = C。因此，ln N – ln N₀ = -λt ⇒ ln(N/N₀) = -λt ⇒ N/N₀ = e⁻λt ⇒ N = N₀e⁻λt。
Common mistakes included forgetting to integrate 1/N as ln N, incorrectly handling the minus sign, and failing to convert logarithmic forms to exponentials. The examiners recommended always writing the limits of integration to avoid sign errors.
常见错误包括忘记将1/N积分为ln N，错误处理负号，以及未能将对数形式转换为指数形式。考官建议始终写上积分上下限以避免符号错误。
9. Transformer Turn-Ratio Equation | 变压器匝数比方程
Although not a calculus-based derivation, the relation Vₚ/Vₛ = Nₚ/Nₛ for an ideal transformer required a physical justification. The examiner report observed that students often quoted the formula without linking it to Faraday’s law of electromagnetic induction.
尽管不是基于微积分的推导，理想变压器的关系式Vₚ/Vₛ = Nₚ/Nₛ仍需要物理论证。考官报告发现，学生常引用公式，却未将其与法拉第电磁感应定律联系起来。
For an ideal transformer, the same magnetic flux Φ passes through both primary and secondary coils. By Faraday’s law, the induced emf per turn is equal, so Vₚ / Nₚ = Vₛ / Nₛ, which rearranges to Vₚ/Vₛ = Nₚ/Nₛ. Candidates who simply stated the formula without mention of flux linkage lost marks.
对于理想变压器，相同的磁通量Φ穿过初级和次级线圈。根据法拉第定律，每匝感应电动势相等，因此Vₚ / Nₚ = Vₛ / Nₛ，整理得Vₚ/Vₛ = Nₚ/Nₛ。仅仅陈述公式而不提及磁链的考生失了分。
The report specifically noted that a diagram showing the core and flux lines, along with a clear statement that dΦ/dt is identical in both coils, constituted a complete answer.
报告特别指出，一幅显示铁
Published by TutorHao | A-Level 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Cancel reply