📚 Year 13 AQA Physics: In-Depth Analysis of Past Paper Questions | Year 13 AQA 物理：历年真题深度解析
Mastering AQA A-Level Physics requires more than memorising formulas – it demands the ability to apply concepts to unfamiliar contexts under time pressure. This article takes you through a selection of typical past-paper problems from the Year 13 specification, breaking down the reasoning, mark-scheme expectations and common pitfalls. Whether you are revisiting circular motion, fields, nuclear physics or thermodynamics, these worked examples will sharpen your exam technique and deepen your conceptual understanding.
掌握 AQA A-Level 物理不仅仅需要记忆公式，更需要你在时间压力下将概念灵活应用于陌生情境。本文精选了 Year 13 考纲中的典型历年真题，一步步拆解解题思路、评分标准要求和常见失分点。无论你正在复习圆周运动、场、核物理还是热力学，这些范例讲解都能帮你打磨应试技巧，深化对物理本质的理解。
1. Circular Motion & Centripetal Force | 圆周运动与向心力
A car of mass 1200 kg travels over a humpback bridge with a radius of curvature 40 m at a constant speed of 12 m s⁻¹. Calculate the normal reaction force exerted by the road on the car when it is at the top of the bridge.
一辆质量为 1200 kg 的汽车以 12 m s⁻¹ 的恒定速度通过一座曲率半径为 40 m 的拱桥。计算汽车在桥顶时路面对它的法向支持力。
Fnet = mg − N = m v² / r
es: 1200 × 9.81 − N = 1200 × (12)² / 40. This gives N = 11772 − 4320 = 7452 N (≈7.5 × 10³ N to 2 s.f.).
在桥顶，指向曲率中心的合力由重力减去支持力提供。代入数值：1200 × 9.81 − N = 1200 × (12)² / 40。解得 N = 11772 − 4320 = 7452 N（保留两位有效数字约 7.5 × 10³ N）。
Common mistake: ents write N − mg = m v² / r, which would imply the normal force is larger than the weight – . Always draw a free-body diagram and check the direction of the centripetal acceleration.
常见错误：许多学生写 N − mg = m v² / r，这意味着支持力大于重力——如果汽车刚好能留在路面上，这在物理上是不可能的。务必画出受力分析图，并核对向心加速度的方向。
2. Simple Harmonic Motion & Energy | 简谐运动与能量
A 0.50 kg mass on a spring oscillates with amplitude 0.080 m and angular frequency 12 rad s⁻¹. Determine the maximum speed and the total energy of the system.
一个 0.50 kg 的物体在弹簧上以振幅 0.080 m、角频率 12 rad s⁻¹ 振动。求最大速度和系统的总能量。
vmax = ωA, Etotal = ½ m vmax² = ½ k A²
vmax atively, using k = m ω², k = 0.50 × 144 = 72 N m⁻¹, then E = ½ × 72 × (0.080)² = 0.2304 J.
vmax = 12 × 0.080 = 0.96 m s⁻¹。总能量 E = ½ × 0.50 × (0.96)² = 0.2304 J ≈ 0.23 J。也可以先用 k = m ω² 得 k = 0.50 × 144 = 72 N m⁻¹，然后 E = ½ × 72 × (0.080)² = 0.2304 J。
Mark-scheme tip: You can earn marks for stating the relationship between total energy and maximum kinetic energy. Ensure you square the angular frequency correctly when finding k. Also, always give the final answer to an appropriate number of significant figures – here 2 s.f. is suitable.
评分标准提示：只要你写出总能量与最大动能的关系式就可以得分。计算 k 时注意角频率的平方要算对。此外，最终答案务必按有效数字给出——本题保留两位有效数字比较合适。
3. Gravitational Fields & Orbital Mechanics | 引力场与轨道力学
Use the data for the Moon' e Earth (period 27.3 days, orbital radius 3.84 × 10⁸ m) to estimate the mass of the Earth.
利用月球绕地球运行的数据（周期 27.3 天、轨道半径 3.84 × 10⁸ m）估算地球质量。
G M m / r² = m r (2π/T)² → M = (4π² r³) / (G T²)
[6.67 × 10⁻¹¹ × (2.359 × 10⁶)²] ≈ 6.0 × 10²⁴ kg. The calculation is sensitive to rounding – keep all intermediate values in the calculator.
将 T = 27.3 × 24 × 3600 = 2.359 × 10⁶ s。然后 M = [4π² × (3.84 × 10⁸)³] / [6.67 × 10⁻¹¹ × (2.359 × 10⁶)²] ≈ 6.0 × 10²⁴ kg。该计算对四舍五入很敏感——在计算器中保留所有中间值。
Pitfall: Students often forget to convert the period into seconds or mistakenly use the radius of the Earth instead of the orbital radius. Also, note that the derived mass is the central body – the Earth, not the Moon.
易错点：学生经常忘记把周期换算成秒，或者误用了地球半径而不是轨道半径。还要注意，推导出的质量是中心天体——地球，而不是月球。
4. Electric Fields & Charged Particle Deflection | 电场与带电粒子偏转
horizontally. The plates are 0.040 m apart and have a potential difference of 200 V across them. Calculate the vertical deflection after the electron travels 0.10 m between the plates.
一个电子以 1.5 × 10⁷ m s⁻¹ 的水平速度进入两平行板之间的区域。板间距 0.040 m，板间电势差为 200 V。计算电子在板间行进 0.10 m 后的竖直偏转量。
E = V / d, F = e E, a = F / m, t = x / u, y = ½ a t²
0⁷) = 6.67 × 10⁻⁹ s. Deflection y = ½ × 8.78 × 10¹⁴ × (6.67 × 10⁻⁹)² ≈ 0.0195 m (19.5 mm).
E = 200 / 0.040 = 5000 V m⁻¹。加速度 a = (1.60 × 10⁻¹⁹ × 5000) / (9.11 × 10⁻³¹) = 8.78 × 10¹⁴ m s⁻²。时间 t = 0.10 / (1.5 × 10⁷) = 6.67 × 10⁻⁹ s。偏转 y = ½ × 8.78 × 10¹⁴ × (6.67 × 10⁻⁹)² ≈ 0.0195 m（19.5 mm）。
Exam commentary: A significant number of candidates treat the electron's motion as if it is launched at an angle. Always separate horizontal (constant velocity) and vertical (constant acceleration) components. The vertical deflection formula is derived from kinematics, not from a pre-memorised result, so show the steps clearly.
考试点评：相当一部分考生把电子的运动当作斜抛来处理。一定要把水平方向（匀速）和竖直方向（匀加速）分开处理。竖直偏转公式要从运动学推导，而不能直接套用某个记忆公式，因此请清晰展示步骤。
5. Capacitance & Exponential Decay | 电容与指数衰减
A 470 μF capacitor is charged to 6.0 V and then discharged through a resistor. The potential difference falls to 2.0 V in 12 s. Find the resistance of the resistor.
一个 470 μF 的电容充电至 6.0 V，然后通过一个电阻放电。经过 12 s 后，电势差降至 2.0 V。求电阻的阻值。
V = V₀ e−t/RC ⇒ ln(V₀/V) = t / (RC)
= 10.92 / (470 × 10⁻⁶) ≈ 23 200 Ω (23 kΩ). The answer should be given to 2 s.f. as 23 kΩ.
变形：RC = t / ln(V₀/V) = 12 / ln(6.0/2.0) = 12 / ln 3 ≈ 12 / 1.0986 = 10.92 s。因此 R = 10.92 / (470 × 10⁻⁶) ≈ 23 200 Ω（23 kΩ）。答案保留两位有效数字为 23 kΩ。
Common slip: Forgetting to convert microfarads to farads leads to an answer wrong by a factor of 10⁶. Also, some students use the half-life formula incorrectly; the exponential decay method is safer when the voltage does not exactly halve. Always check the unit of your final resistance – it must be in ohms.
常见失误：忘记把微法换算成法会导致答案差 10⁶ 倍。此外，有些学生错误地使用半衰期公式；当电压没有正好减半时，指数衰减方法更稳妥。务必检查最终电阻的单位——它必须是欧姆。
6. Electromagnetic Induction & Alternating EMF | 电磁感应与交变电动势
A rectangular coil of 80 turns, area 0.030 m², rotates at 50 rev s⁻¹ in a uniform magnetic field of 0.25 T. Derive an expression for the induced emf and calculate its peak value.
一个 80 匝、面积 0.030 m² 的矩形线圈在 0.25 T 的匀强磁场中以 50 rev s⁻¹ 转动。推导感应电动势的表达式并计算其峰值。
Flux linkage NΦ = B A N cos(θ), θ = ω t ⇒ ε = −d(NΦ)/dt = B A N ω sin(ω t)
V (≈ 1.9 × 10² V). The sinusoidal variation comes from the rate of change of flux linkage.
ω = 2π × 50 = 100π rad s⁻¹。峰值电动势 ε₀ = B A N ω = 0.25 × 0.030 × 80 × 100π ≈ 188.5 V（≈ 1.9 × 10² V）。正弦变化源于磁链变化率的变化。
Exam focus: Many marks are awarded for clearly stating Faraday's law and for recognising that the peak emf occurs when the plane of the coil is parallel to the field (maximum rate of flux cutting). Use radians for angular frequency when evaluating trigonometric functions, though here only the peak value is requested.
考试重点：明确写出法拉第定律并认识到峰值电动势出现在线圈平面与磁场平行时（磁通量切割率最大），这两点能拿到很多分数。虽然本题只求峰值，但在涉及三角函数计算时务必使用弧度。
7. Radioactive Decay & Half-Life | 放射性衰变与半衰期
The activity of a sample of ¹⁴C is 0.23 Bq. Fresh carbon of the same mass has an activity of 0.30 Bq. The half-life of ¹⁴C is 5730 years. Estimate the age of the sample.
一份 ¹⁴C 样品的活度为 0.23 Bq。同质量的新鲜碳的活度为 0.30 Bq。¹⁴C 的半衰期为 5730 年。估算该样品的年代。
A = A₀ e−λ t, λ = ln 2 / T₁/₂
λ = ln 2 / 5730 ≈ 1.21 × 10⁻⁴ yr⁻¹. Then 0.23 = 0.30 e−λ t → e−λ t ) = ln(1.304) ≈ 0.2655 → t ≈ 0.2655 / (1.21 × 10⁻⁴) ≈ 2190 years (≈ 2200 years to 2 s.f.).
λ = ln 2 / 5730 ≈ 1.21 × 10⁻⁴ yr⁻¹。然后 0.23 = 0.30 e−λ t → e−λ t .30 ≈ 0.7667 → λ t = ln(1/0.7667) = ln(1.304) ≈ 0.2655 → t ≈ 0.2655 / (1.21 × 10⁻⁴) ≈ 2190 年（保留两位有效数字约 2200 年）。
Typical error: Using the half-life directly without converting to decay constant, or taking the average of the two activities to find time. The exponential relationship is non-linear, so you must use logarithms. Also ensure the final age is expressed in a consistent unit (years).
典型错误：不转换成衰变常量而直接套用半衰期，或者用两个活度的平均值去求时间。指数关系是非线性的，因此必须使用对数。还要确保最终年代的单位统一（年）。
8. Nuclear Binding Energy & Fission | 核结合能与裂变
rest masses: ²³⁵U = 235.0439 u, ¹⁴¹Ba = 140.9144 u, ⁹²Kr = 91.9262 u, neutron = 1.00867 u.
计算以下裂变反应释放的能量：²³⁵U + n → ¹⁴¹Ba + ⁹²Kr + 3n。使用下列静止质量：²³⁵U = 235.0439 u，¹⁴¹Ba = 140.9144 u，⁹²Kr = 91.9262 u，中子 = 1.00867 u。
Δm = m_reactants − m_products, E = Δm c², 1 u = 931.5 MeV
× 1.00867 = 235.86661 u. Mass defect Δm = 0.18596 u. Energy = 0.18596 × 931.5 ≈ 173.2 MeV.
初态总质量 = 235.0439 + 1.00867 = 236.05257 u。末态总质量 = 140.9144 + 91.9262 + 3 × 1.00867 = 235.86661 u。质量亏损 Δm = 0.18596 u。释放能量 = 0.18596 × 931.5 ≈ 173.2 MeV。
Mark-scheme must: You must carefully count the neutrons on both sides. The conversion factor 931.5 MeV u⁻¹ is given in the data booklet, but using it correctly impresses examiners. Watch out for rounding – keep mass values to all given decimal places.
评分标准必含：你必须仔细计数两侧的中子数。转换因子 931.5 MeV u⁻¹ 在数据手册中给出，但正确使用它会给考官留下好印象。注意四舍五入——质量数值要保留所给的全部小数位。
9. Ideal Gas Law & Molecular Kinetic Theory | 理想气体定律与分子动理论
ms speed of the nitrogen molecules. (Molar mass N₂ = 0.028 kg mol⁻¹, R = 8.31 J K⁻¹ mol⁻¹)
一个容积为 2.0 × 10⁻³ m³ 的密封瓶内装有 300 K、1.01 × 10⁵ Pa 的氮气。计算气体的摩尔数和氮气分子的方均根速率。（N₂ 摩尔质量 = 0.028 kg mol⁻¹，R = 8.31 J K⁻¹ mol⁻¹）
p V = n R T, ½ m crms² = (3/2) k T ⇒ crms = √3 R T / M
n = pV/(RT) = (1.01 × 10⁵ × 2.0 × 10⁻³) / (8.31 × 300) ≈ 0.0810 mol. Then crms = √3 × 8.31 × 300 / 0.028 = √7479 / 0.028 ≈ √267107 ≈ 517 m s⁻¹.
n = pV/(RT) = (1.01 × 10⁵ × 2.0 × 10⁻³) / (8.31 × 300) ≈ 0.0810 mol。然后 crms = √3 × 8.31 × 300 / 0.028 = √7479 / 0.028 ≈ √267107 ≈ 517 m s⁻¹。
Common trap: Using the mass of a single molecule instead of the molar mass in the rms speed formula. The expression √3RT/M requires M in kg per mole. Also, failing to convert litres to m³ accounts for many lost marks.
常见陷阱：在方均根速率公式中误用单个分子的质量而不是摩尔质量。表达式 √3RT/M 中的 M 需以 kg/mol 为单位。此外，忘记把升换算为立方米也导致大量失分。
10. Astrophysics: Doppler Shift & Hubble's Law | 天体物理：多普勒频移与哈勃定律
ine the recession velocity of the galaxy and estimate its distance. (H₀ = 2.2 × 10⁻¹⁸ s⁻¹)
观测到一个星系的氢谱线波长为 680 nm，实验室波长为 656 nm。求该星系的退行速度，并估算其距离。（H₀ = 2.2 × 10⁻¹⁸ s⁻¹）
z = Δλ / λ₀ ≈ v / c (for v << c), v = H₀ d
× 10²⁴ m. Convert to light years or Mpc if required (1 Mpc ≈ 3.09 × 10²² m → d ≈ 162 Mpc).
Δλ = 680 − 656 = 24 nm。红移 z = 24 / 656 ≈ 0.0366。v = z c = 0.0366 × 3.00 × 10⁸ = 1.10 × 10⁷ m s⁻¹。然后 d = v / H₀ = 1.10 × 10⁷ / (2.2 × 10⁻¹⁸) ≈ 5.0 × 10²⁴ m。如需转换为光年或 Mpc（1 Mpc ≈ 3.09 × 10²² m → d ≈ 162 Mpc）。
Watch out: The Doppler formula for low speeds is an approximation; do not use the relativistic form unless the mark scheme explicitly requires it. Always check the units of H₀ – here it is in s⁻¹, so the distance naturally comes out in metres. A frequent error is to subtract the wavelengths incorrectly or to use the observed wavelength as the denominator.
注意：适用于低速的多普勒公式为近似式；除非评分标准明确要求，否则不要使用相对论形式。务必检查 H₀ 的单位——本题为 s⁻¹，因此距离自然以米为单位。常见错误包括波长相减出错，或者把观测波长当作分母。
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