A-Level Physics Core Formula Review & Typical Problem Applications | A-Level 物理核心公式梳理与典型题型应用

📚 A-Level Physics Core Formula Review & Typical Problem Applications | A-Level 物理核心公式梳理与典型题型应用

Physics at A-Level is not about memorising every equation in isolation — it is about understanding the fundamental relationships between physical quantities and knowing when and how to apply them under exam conditions. This revision guide consolidates the most frequently tested formulas across all major topics and pairs each with a classic exam-style application to help you move from recall to problem-solving.

A-Level 物理考试并非要求孤立地背下所有公式，而是要理解物理量之间的本质联系，并清楚在什么条件下、如何应用这些公式来解题。本篇复习指南将各核心板块的高频公式集中梳理，并配合典型考题应用，帮助你从”记住公式”真正过渡到”会做题目”。

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1. Kinematics & Linear Motion | 运动学与直线运动

The SUVAT equations describe motion with constant acceleration. You must be able to choose the correct equation based on which variables are given and which are required. A common mistake is applying these equations to situations with changing acceleration, such as vertical motion with air resistance.

SUVAT 方程组描述的是匀加速直线运动。解题时必须依据已知量和待求量选择合适的方程。常见的错误是将这些公式套用于加速度变化的场景，例如考虑空气阻力时的竖直运动。

v = u + at
s = ut + ½at²
v² = u² + 2as

Typical problem: A ball is thrown vertically upward with an initial speed of 15 m s⁻¹. Calculate the maximum height reached (take g = 9.81 m s⁻²).

典型题目：小球以初速度 15 m s⁻¹ 竖直上抛，求能达到的最大高度（取 g = 9.81 m s⁻²）。

The most efficient approach is to note that at maximum height, v = 0. Using v² = u² + 2as with upward as positive: 0 = 15² + 2(−9.81)s, giving s = 225 ÷ 19.62 ≈ 11.5 m.

最高点处速度 v = 0，因此选用 v² = u² + 2as，取向上为正方向：0 = 15² + 2(−9.81)s，解得 s = 225 ÷ 19.62 ≈ 11.5 m。

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2. Newton’s Laws & Forces | 牛顿定律与力

Newton’s second law, F = ma, is the bridge between dynamics and kinematics. In inclined-plane problems, always resolve forces parallel and perpendicular to the plane. Remember that the normal reaction is not always equal to the weight — on an incline of angle θ, R = mg cosθ.

牛顿第二定律 F = ma 是连接动力学与运动学的桥梁。在斜面问题中，务必沿平行于斜面和垂直于斜面的方向分解力。注意支持力未必等于重力——当斜面倾角为 θ 时，R = mg cosθ。

F = ma
W = mg
F ≤ μR

Typical problem: A 3.0 kg block slides down a 30° incline with a coefficient of kinetic friction of 0.20. Find the acceleration.

典型题目：质量 3.0 kg 的木块沿倾角 30° 的斜面下滑，动摩擦因数 μ = 0.20，求加速度。

Parallel to the slope: mg sinθ − μR = ma, where R = mg cosθ. Substituting: 3.0 × 9.81 × 0.5 − 0.20 × 3.0 × 9.81 × 0.866 = 3.0a. This gives 14.715 − 5.097 ≈ 3.0a, so a ≈ 3.21 m s⁻².

沿斜面方向：mg sinθ − μR = ma，其中 R = mg cosθ。代入：3.0 × 9.81 × 0.5 − 0.20 × 3.0 × 9.81 × 0.866 = 3.0a，得 14.715 − 5.097 ≈ 3.0a，因此 a ≈ 3.21 m s⁻²。

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3. Work, Energy & Power | 功、能与功率

The work-energy theorem states that the net work done on a body equals its change in kinetic energy. In A-Level questions, energy conservation is often the fastest route, provided you correctly account for all energy stores and transfers, including heat from friction.

动能定理指出，合外力对物体所做的功等于物体动能的变化。在 A-Level 题目中，能量守恒往往是最快的解法，前提是正确列出所有能量储库与转移，包括摩擦产生的热量。

W = Fs cosθ
KE = ½mv²
PE = mgh
P = W/t = Fv

Typical problem: A car of mass 1200 kg accelerates from rest to 20 m s⁻¹ over a distance of 200 m on a horizontal road. The average resistive force is 400 N. Calculate the engine power output, assuming constant acceleration and that power is measured at the final instant.

典型题目：质量 1200 kg 的汽车在水平路面上从静止加速到 20 m s⁻¹，行驶距离 200 m，平均阻力 400 N。求最终时刻发动机的输出功率（设加速度恒定）。

Final instantaneous power P = F_total × v. Net force from kinematics: a = v²/(2s) = 400/400 = 1 m s⁻², so F_net = ma = 1200 N. Traction force F = F_net + resistance = 1600 N. Hence P = 1600 × 20 = 32,000 W = 32 kW.

末时刻瞬时功率 P = F_合 × v。由运动学求加速度：a = v²/(2s) = 400/400 = 1 m s⁻²，F_合 = ma = 1200 N。牵引力 F = 1200 + 400 = 1600 N，因此 P = 1600 × 20 = 32,000 W = 32 kW。

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4. Circular Motion & Gravitation | 圆周运动与万有引力

For uniform circular motion, the resultant force acts toward the centre and equals mv²/r. Gravitational problems often combine this with Newton’s law of gravitation to derive orbital speed or period. A frequent pitfall is forgetting that the centripetal force is the gravitational force in orbital motion — there is no separate “balancing” force.

在匀速圆周运动中，合力指向圆心，大小等于 mv²/r。万有引力题目通常将向心力公式与牛顿万有引力定律结合，推导轨道速度或周期。常见误区是忘记在轨道运动中，万有引力本身就是向心力，并不存在另一个”平衡”力。

a = v²/r = ω²r
F = mv²/r = mω²r
F = GMm/r²

Typical problem: A satellite orbits the Earth at an altitude of 400 km. The radius of Earth is 6370 km and g at the surface is 9.81 m s⁻². Estimate the orbital speed.

典型题目：一颗卫星在距地面 400 km 的高度绕地球运动，地球半径 6370 km，地面重力加速度 9.81 m s⁻²，估算卫星的轨道速度。

Equate gravitational force to centripetal force: GMm/r² = mv²/r, so v = √(GM/r). Since GM = gR² at the surface, v = √(gR²/r) = √(9.81 × (6.37 × 10⁶)² / 6.77 × 10⁶) ≈ 7.67 × 10³ m s⁻¹.

令万有引力等于向心力：GMm/r² = mv²/r，得 v = √(GM/r)。由地面的 GM = gR²，所以 v = √(gR²/r) = √(9.81 × (6.37 × 10⁶)² / 6.77 × 10⁶) ≈ 7.67 × 10³ m s⁻¹。

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5. Simple Harmonic Motion | 简谐运动

Simple harmonic motion (SHM) is defined by the condition that acceleration is proportional to displacement and directed toward the equilibrium position: a = −ω²x. In SHM problems, identify the equilibrium position first, then determine whether the system behaves like a mass–spring or pendulum. Time period equations are essential for solving oscillator questions.

简谐运动的定义条件是加速度与位移成正比且方向始终指向平衡位置：a = −ω²x。解 SHM 题目时，先找出平衡位置，再判断系统属于弹簧振子还是单摆。周期公式是解题的关键工具。

a = −ω²x
x = A cos(ωt)
T = 2π√(m/k)
T = 2π√(L/g)

Typical problem: A 0.50 kg mass on a spring oscillates with a period of 0.80 s. Calculate the spring constant k. What is the maximum acceleration if the amplitude is 0.10 m?

典型题目：质量 0.50 kg 的物体在弹簧上做周期为 0.80 s 的简谐振动，求劲度系数 k。若振幅为 0.10 m，最大加速度是多少？

T = 2π√(m/k) ⇒ k = 4π²m/T² = 4π² × 0.50 / 0.64 ≈ 30.8 N m⁻¹. Maximum acceleration occurs at maximum displacement: a_max = ω²A = (2π/T)²A = (2π/0.80)² × 0.10 ≈ 6.17 m s⁻².

T = 2π√(m/k) ⇒ k = 4π²m/T² = 4π² × 0.50 / 0.64 ≈ 30.8 N m⁻¹。最大加速度出现在最大位移处：a_max = ω²A = (2π/T)²A = (2π/0.80)² × 0.10 ≈ 6.17 m s⁻²。

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6. Electric Circuits | 直流电路

DC circuit analysis in A-Level requires mastery of Ohm’s law, Kirchhoff’s laws, and the equations for power. When dealing with non-ideal cells, remember that the internal resistance r is in series with the external load. The terminal voltage is less than the EMF when current flows. Kirchhoff’s laws are indispensable in multi-loop circuits, where you must assign consistent direction conventions.

A-Level 直流电路分析需要掌握欧姆定律、基尔霍夫定律和功率公式。在处理非理想电源时，注意内阻 r 与外部负载串联。当电路有电流时，路端电压小于电动势。对于多回路电路，基尔霍夫定律是必备工具，使用时要保持一致的方向约定。

V = IR
P = IV = I²R = V²/R
R = ρL/A
ε = I(R + r)

Typical problem: A battery of EMF 6.0 V and internal resistance 0.5 Ω is connected to a 2.5 Ω resistor. Find the current and the terminal voltage.

典型题目：电动势 6.0 V、内阻 0.5 Ω 的电池与 2.5 Ω 的电阻串联，求电路中的电流和路端电压。

The total resistance is 2.5 + 0.5 = 3.0 Ω, so I = 6.0 / 3.0 = 2.0 A. Terminal voltage V = ε − Ir = 6.0 − 2.0 × 0.5 = 5.0 V. This equals I × R = 2.0 × 2.5 = 5.0 V, confirming consistency.

总电阻为 2.5 + 0.5 = 3.0 Ω，故 I = 6.0 / 3.0 = 2.0 A。路端电压 V = ε − Ir = 6.0 − 2.0 × 0.5 = 5.0 V。该值也等于 I × R = 2.0 × 2.5 = 5.0 V，验证了计算一致。

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7. Electric Fields & Capacitance | 电场与电容

Electric field questions typically fall into two categories: uniform fields between parallel plates (E = V/d) and radial fields around point charges (E = kQ/r²). Capacitor problems often involve charging/discharging curves or energy storage. In series, capacitors combine like resistors in parallel; in parallel, they combine like resistors in series. The exponential nature of capacitor discharge appears frequently in data-analysis questions.

电场题目通常分为两类：平行板间的匀强电场（E = V/d）和点电荷周围的径向电场（E = kQ/r²）。电容器问题常涉及充放电曲线或储能计算。电容串联时等效电容的求法类似于电阻并联；电容并联时则类似于电阻串联。电容放电的指数衰减特性常出现在数据分析题中。

E = F/q
E = V/d
C = Q/V
E_cap = ½QV = ½CV² = ½Q²/C

Typical problem: A parallel-plate capacitor of capacitance 20 μF is charged to a potential difference of 100 V. It is then disconnected from the supply and the plate separation is doubled. What happens to the stored energy?

典型题目：一个电容为 20 μF 的平行板电容器充电到 100 V 后与电源断开，然后将极板距离加倍，储存的电场能如何变化？

Initially Q = CV = 20 × 10⁻⁶ × 100 = 2.0 × 10⁻³ C and E = ½QV = ½ × 2.0 × 10⁻³ × 100 = 0.10 J. When disconnected, Q is constant. Doubling d halves C (since C ∝ 1/d), so V = Q/C doubles to 200 V. Final energy = ½QV = ½ × 2.0 × 10⁻³ × 200 = 0.20 J — the energy doubles, and this extra energy comes from the work done in separating the plates.

初始 Q = CV = 20 × 10⁻⁶ × 100 = 2.0 × 10⁻³ C，E = ½QV = ½ × 2.0 × 10⁻³ × 100 = 0.10 J。断开电源后 Q 保持不变。极板距离加倍使 C 减半（C ∝ 1/d），所以 V = Q/C 加倍到 200 V。末能量 E = ½QV = ½ × 2.0 × 10⁻³ × 200 = 0.20 J——能量加倍，多出的能量来源于拉开极板时外力所做的功。

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8. Magnetic Fields & Electromagnetic Induction | 磁场与电磁感应

Magnetic force on a current-carrying conductor F = BIL and on a moving charge F = Bqv are the cornerstones of magnetism questions. For electromagnetic induction, Faraday’s law relates induced EMF to the rate of change of flux linkage. Lenz’s law determines direction and is a direct consequence of energy conservation. Transformer and motor questions often combine these ideas with mechanical power.

载流导体在磁场中所受的安培力 F = BIL 以及运动电荷所受的洛伦兹力 F = Bqv 是磁场问题的基石。对于电磁感应，法拉第定律将感应电动势与磁链变化率联系起来；楞次定律决定感应方向，是能量守恒的直接推论。变压器和电动机题目常将这些概念与机械功率结合考查。

F = BIL sinθ
F = Bqv
Φ = BA cosθ
ε = −N ΔΦ/Δt

Typical problem: A coil of 200 turns and area 4.0 × 10⁻³ m² is placed perpendicular to a magnetic field that decreases uniformly from 0.50 T to 0 T in 0.20 s. Calculate the magnitude of the induced EMF.

典型题目：一个 200 匝、面积 4.0 × 10⁻³ m² 的线圈垂直于磁场放置，磁感应强度在 0.20 s 内从 0.50 T 均匀减小到 0 T，求感应电动势的大小。

Initial flux per turn = BA = 0.50 × 4.0 × 10⁻³ = 2.0 × 10⁻³ Wb. Final flux is 0, so |ε| = N|ΔΦ/Δt| = 200 × (2.0 × 10⁻³ − 0) / 0.20 = 2.0 V. The negative sign in Faraday’s law reminds you that the induced current opposes the change causing it.

每匝初始磁通量 Φ = BA = 0.50 × 4.0 × 10⁻³ = 2.0 × 10⁻³ Wb。末磁通量为 0，因此 |ε| = N|ΔΦ/Δt| = 200 × (2.0 × 10⁻³ − 0) / 0.20 = 2.0 V。法拉第定律中的负号提示感应电流总是阻碍引起它的磁通量变化。

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9. Wave Properties & Interference | 波动性质与干涉

Wave equations link wave speed, frequency and wavelength. For interference, the double-slit formula and diffraction grating equation are frequently tested. In addition, the intensity of a wave is proportional to the square of its amplitude. For exam success, be able to distinguish between progressive and stationary waves and to identify nodes and antinodes in interference patterns.

波的方程联系了波速、频率和波长。对于干涉现象，双缝公式和衍射光栅方程是高频考点。此外，波的强度与振幅的平方成正比。考试成功的关键在于能区分行波与驻波，并能识别干涉图样中的波节与波腹。

v = fλ
d sinθ = nλ
I ∝ A²

Typical problem: Light of wavelength 580 nm illuminates a diffraction grating with 600 lines per mm. Calculate the angle of the second-order maximum.

典型题目：波长 580 nm 的光垂直照射到每毫米 600 条刻痕的衍射光栅上，求第二级亮纹的衍射角。

Grating spacing d = 1/600 mm = 1.67 × 10⁻⁶ m. For n = 2: sinθ = nλ/d = (2 × 580 × 10⁻⁹) / 1.67 × 10⁻⁶ = 0.695. Therefore θ = sin⁻¹(0.695) ≈ 44.0°. Check: sinθ would exceed 1 if the order did not exist, so always verify that nλ/d ≤ 1.

光栅常数 d = 1/600 mm = 1.67 × 10⁻⁶ m。对 n = 2：sinθ = nλ/d = (2 × 580 × 10⁻⁹) / 1.67 × 10⁻⁶ = 0.695，因此 θ = sin⁻¹(0.695) ≈ 44.0°。注意验证 nλ/d ≤ 1，若 sinθ 大于 1 则说明该级次不存在。

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10. Quantum Physics & Particle Behaviour | 量子物理与粒子行为

Quantum physics at A-Level revolves around the photoelectric effect, the wave–particle duality of matter, and energy levels. The key equation E = hf links a photon’s energy to its frequency; the work function φ represents the minimum energy needed to eject an electron. The de Broglie wavelength of a particle is obtained from its momentum. These topics also reinforce the idea that transitions between energy levels correspond to photon emission or absorption.

A-Level 量子物理围绕光电效应、物质的波粒二象性和能级展开。核心方程 E = hf 将光子能量与频率联系起来；逸出功 φ 是使电子逸出所需的最小能量。德布罗意波长由粒子动量决定。这些知识点也强化了能级跃迁与光子发射、吸收之间的对应关系。

E = hf
hf = φ + KE_max
λ = h/mv
E = mc²

Typical problem: A metal surface has a work function of 3.2 eV. What is the maximum kinetic energy of the emitted photoelectrons when light of wavelength 350 nm shines on it?

典型题目：金属表面的逸出功为 3.2 eV，当波长为 350 nm 的光照射其上时，求发射的光电子最大动能。

Photon energy E = hc/λ = (6.63 × 10⁻³⁴ × 3.0 × 10⁸) / 350 × 10⁻⁹ = 5.68 × 10⁻¹⁹ J. Convert to eV: 5.68 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ = 3.55 eV. Therefore KE_max = E − φ = 3.55 − 3.20 = 0.35 eV.

光子能量 E = hc/λ = (6.63 × 10⁻³⁴ × 3.0 × 10⁸) / 350 × 10⁻⁹ = 5.68 × 10⁻¹⁹ J。换算为电子伏特：5.68 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ = 3.55 eV。因此 KE_max = E − φ = 3.55 − 3.20 = 0.35 eV。

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11. Thermodynamics & Ideal Gases | 热力学与理想气体

The ideal gas equation combines Boyle’s law, Charles’s law, and Avogadro’s principle. The molecular kinetic theory expression interprets pressure in terms of molecular motion. In thermodynamics, the first law ΔU = Q + W uses a sign convention that must be applied consistently — W is the work done on the gas and Q is the heat supplied to the gas. Students often lose marks by mixing up signs in compression and expansion processes.

理想气体状态方程综合了玻意耳定律、查理定律和阿伏伽德罗定律。分子动理论公式从分子运动的角度解释了压强的微观本质。热力学第一定律 ΔU = Q + W 的符号约定必须保持一致——W 表示外界对气体做的功，Q 表示气体从外界吸收的热量。在压缩与膨胀过程中混淆符号是常见的失分点。

pV = nRT
pV = ½ Nm⟨c²⟩​/3 = ⅓Nm⟨c²⟩
ΔU = Q + W

Typical problem: An ideal gas at 27°C, occupies 5.0 × 10⁻³ m³ when its pressure is 2.0 × 10⁵ Pa. How many moles of gas are present? If the temperature is raised to 327°C at constant volume, find the new pressure.

典型题目：理想气体在 27°C 时压强为 2.0 × 10⁵ Pa，体积为 5.0 × 10⁻³ m³，求气体的物质的量。若温度升高至 327°C 且体积不变，求新的压强。

Using pV = nRT: n = pV/RT = (2.0 × 10⁵ × 5.0 × 10⁻³) / (8.31 × 300) ≈ 0.401 mol. For constant volume, p/T = constant, so p₂/p₁ = T₂/T₁ = 600/300 = 2, giving p₂ = 4.0 × 10⁵ Pa. Always work in kelvin in gas law calculations.

由 pV = nRT：n = pV/RT = (2.0 × 10⁵ × 5.0 × 10⁻³) / (8.31 × 300) ≈ 0.401 mol。等容过程满足 p/T = 常数，因此 p₂/p₁ = T₂/T₁ = 600/300 = 2，即 p₂ = 4.0 × 10⁵ Pa。使用气体定律时务必将温度转换为开尔文。

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12. Radioactive Decay & Nuclear Physics | 放射性衰变与核物理

Radioactive decay follows first-order kinetics: the rate of decay is proportional to the number of undecayed nuclei. The decay constant λ and half-life T₁/₂ are related by T₁/₂ = ln 2 / λ. Questions frequently combine the exponential decay law with activity calculations or dating problems. Remember that the units of λ are s⁻¹, min⁻¹, or year⁻¹, and that the decay constant is independent of temperature and pressure.

放射性衰变遵循一级动力学规律：衰变率与未衰变的原子核数成正比。衰变常数 λ 与半衰期 T₁/₂ 的关系为 T₁/₂ = ln 2 / λ。考题常将指数衰变定律与活度计算或年代测定问题结合。注意 λ 的单位是 s⁻¹、min⁻¹ 或 year⁻¹，且衰变常数与温度、压强无关。

A = λN
N = N₀e^(−λt)
A = A₀e^(−λt)
T₁/₂ = ln 2 / λ

Typical problem: The half-life of carbon-14 is 5730 years. A wooden artifact contains 25% of the carbon-14 found in a living tree. How old is the artifact?

典型题目：碳-14 的半衰期为 5730 年。某木制文物中碳-14 的含量仅为活体树木中的 25%，求该文物的年代。

After one half-life, 50% remains; after two half-lives, 25% remains. Therefore the artifact is two half-lives old: age = 2 × 5730 = 1.15 × 10⁴ years. Alternatively, use N/N₀ = e^(−λt) with λ = ln 2 / T₁/₂ = 1.209 × 10⁻⁴ year⁻¹; setting 0.25 = e^(−λt) gives t = ln 4 / λ = 1.146 × 10⁴ years.

经过一个半衰期剩余 50%，经过两个半衰期剩余 25%。因此该文物经历了两个半衰期：年代 = 2 × 5730 = 1.15 × 10⁴ 年。也可用 N/N₀ = e^(−λt) 计算，其中 λ = ln 2 / T₁/₂

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