📚 Year 13 Edexcel Physics: Case Study Hands-on Practice | Year 13 Edexcel 物理：案例分析实战演练
Mastering A-level Physics requires more than memorising formulas — you must analyse unfamiliar scenarios, select correct principles, and execute multi-step solutions under time pressure. This article guides you through eight extended case studies common in Edexcel Year 13 exams. Each case dissects the underlying physics and demonstrates rigorous problem-solving aligned with the specification: mechanics, fields, nuclear physics, oscillations, and thermal physics. We will work through numerical examples, highlight common pitfalls, and reinforce exam technique. Let's dive into real-context practice.
掌握 A-level 物理不仅仅是记忆公式——你必须在陌生情境中分析问题，选择正确的原理，并在时间压力下执行多步骤求解。本文带你深入 Edexcel Year 13 考试中常见的八个拓展案例分析。每个案例剖析背后的物理本质，并展示与考纲严格一致的解题过程：力学、场、核物理、振动和热物理。我们将演练数值案例，指出常见陷阱，并强化应试技巧。现在进入真实情景的实战训练。
1. Projectile Motion Case Study | 抛体运动案例研究
A cricket ball is struck from ground level with a speed of 28 m s⁻¹ at an angle of 35° to the horizontal. The outfield is flat and air resistance negligible. We need the time of flight, maximum height, and horizontal range. This tests SUVAT in two dimensions and independence of vertical and horizontal motions.
一个板球从地面以 28 m s⁻¹ 的速度、与水平面成 35° 角击出。外场平坦且忽略空气阻力。需求飞行时间、最大高度和水平射程。这考查二维运动学中的 SUVAT 以及竖直和水平运动的独立性。
pex: final vertical velocity = 0. Using v = u + at with a = -9.81 m s⁻², 0 = 16.1 - 9.81 tup, giving tup ≈ 1.64 s. Total time of flight = 2 × 1.64 = 3.28 s.
分解初速度：uₓ = 28 cos 35° ≈ 22.9 m s⁻¹，uᵧ = 28 sin 35° ≈ 16.1 m s⁻¹。上升至最高点竖直运动：末速度竖直分量为零。用 v = u + at，a = -9.81 m s⁻²，0 = 16.1 - 9.81 tup，得 tup ≈ 1.64 s。总飞行时间 = 2 × 1.64 = 3.28 s。
get to double the time for symmetrical flight or misuse sine/cosine. In exam, show clear resolution and state the assumption of no air drag.
最大高度：s = uᵧ t + ½ a t² = 16.1×1.64 + 0.5×(-9.81)×(1.64)² ≈ 26.4 - 13.2 = 13.2 m。水平射程 = uₓ × 总时间 = 22.9 × 3.28 ≈ 75.1 m。考生常忘记对称飞行时间加倍或错用正弦/余弦。考试中要清晰分解并声明忽略空气阻力。
2. Simple Harmonic Motion Analysis | 简谐运动分析
A 0.500 kg mass attached to a spring of stiffness 80 N m⁻¹ oscillates on a frictionless surface. It is displaced 0.040 m from equilibrium and released. Determine the angular frequency, period, maximum speed, and maximum acceleration. This links to SHM equations and energy.
一个 0.500 kg 的质量连接在劲度系数为 80 N m⁻¹ 的弹簧上，在光滑水平面上振动。它偏离平衡位置 0.040 m 后释放。求角频率、周期、最大速度和最大加速度。这联系简谐运动方程和能量。
Angular frequency ω = √k/m = √80 / 0.500 = √160 ≈ 12.65 rad s⁻¹. Period T = 2π/ω = 2π/12.65 ≈ 0.497 s. Maximum speed vmax = ωA = 12.65 × 0.040 = 0.506 m s⁻¹. Maximum acceleration amax = ω²A = (12.65)² × 0.040 = 160 × 0.040 = 6.4 m s⁻².
角频率 ω = √k/m = √80 / 0.500 = √160 ≈ 12.65 rad s⁻¹。周期 T = 2π/ω = 2π/12.65 ≈ 0.497 s。最大速度 vmax = ωA = 12.65 × 0.040 = 0.506 m s⁻¹。最大加速度 amax = ω²A = (12.65)² × 0.040 = 160 × 0.040 = 6.4 m s⁻²。
818) = -0.0327 m. The negative indicates direction opposite to initial displacement. Always confirm the initial condition to choose sine or cosine form.
许多问题要求 t = 0.2 s 时的位移。用 x = A cos(ωt) 因为从振幅开始。x = 0.040 cos(12.65×0.2) = 0.040 cos(2.53 rad) ≈ 0.040 × (-0.818) = -0.0327 m。负号表示与初始位移方向相反。务必确认初始条件以选择正弦或余弦形式。
3. Capacitor Discharge in a DC Circuit | 直流电路中的电容器放电
A 220 μF capacitor is charged to 12.0 V and then discharged through a 47 kΩ resistor. We require the time constant, the time for voltage to drop to 3.0 V, and the half-life. This quantitative case exercises exponential decay and logarithmic manipulation.
一个 220 μF 电容器充电至 12.0 V，然后通过一个 47 kΩ 电阻放电。求时间常数、电压降至 3.0 V 所需时间以及半衰期。这个定量案例练习指数衰减和对数运算。
Time constant τ = RC = 47×10³ × 220×10⁻⁶ = 10.34 s. For discharge V = V₀ e-t/RC. Setting V = 3.0 V: 3.0 = 12.0 e-t/10.34 → e-t/10.34 Verify: after 7.17 s, V = 12 × 0.5 = 6 V; after another half-life, 3 V, matching 14.34 s.
时间常数 τ = RC = 47×10³ × 220×10⁻⁶ = 10.34 s。放电时 V = V₀ e-t/RC。设 V = 3.0 V：3.0 = 12.0 e-t/10.34 → e-t/10.34 = 0.25 → -t/10.34 = ln(0.25) = -1.386 → t = 14.3 s。半衰期 T₁/₂ = τ ln 2 = 10.34 × 0.693 = 7.17 s。验证：7.17 s 后 V = 12 × 0.5 = 6 V；再过一个半衰期得 3 V，与 14.34 s 吻合。
Examiners may ask you to linearise the data. Plotting ln V against t yields a straight line with gradient -1/RC and intercept ln V₀. This is a core practical skill. Also be prepared to discuss effect of reducing R or C on the discharge rate.
考官可能要求将数据线性化。绘 ln V 对 t 图得一条直线，斜率为 -1/RC，截距为 ln V₀。这是核心实验技能。还需准备讨论减小 R 或 C 对放电速率的影响。
4. Radioactive Decay Calculations | 放射性衰变计算
10⁵ Bq. Calculate the decay constant, the number of nuclei present initially, and the activity after 365 days. This case links activity, number of nuclei, and exponential decay law.
一个纯 ²¹⁰Po 样品（半衰期 138 天）初始活度为 2.5 × 10⁵ Bq。计算衰变常数、初始核的数量以及 365 天后的活度。此案例联系活度、核的数量与指数衰减规律。
itial number N₀ = A₀ / λ. Using SI: N₀ = 2.5×10⁵ / 5.81×10⁻⁸ ≈ 4.30×10¹² nuclei. Activity after 365 days: A = A₀ e-λt with t = 365 days, λ in day⁻¹: A = 2.5×10⁵ × e-0.00502×365 = 2.5×10⁵ × e-1.832 ≈ 2.5×10⁵ × 0.160 = 4.00×10⁴ Bq.
衰变常数 λ = ln 2 / T₁/₂ = 0.693 / (138 × 24 × 3600) s = 0.693 / (1.192×10⁷ s) = 5.81×10⁻⁸ s⁻¹。也可用天：λ = 0.693/138 = 0.00502 day⁻¹。初始核数 N₀ = A₀ / λ。用国际单位制：N₀ = 2.5×10⁵ / 5.81×10⁻⁸ ≈ 4.30×10¹² 个核。365 天后的活度：A = A₀ e-λt，t = 365 天，λ 用天⁻¹：A = 2.5×10⁵ × e-0.00502×365 = 2.5×10⁵ × e-1.832 ≈ 2.5×10⁵ × 0.160 = 4.00×10⁴ Bq。
can also use number of half-lives: 365/138 ≈ 2.645 half-lives. Fraction remaining = (1/2)2.645 lculation. This method is quicker for estimation but ensure accurate log work for precise values.
当活度降至初始的某个分数，也可用经历的半衰期数量：365/138 ≈ 2.645 个半衰期。剩余分数 = (1/2)2.645 = 0.160，与指数计算结果一致。此法估算更快，但有精确要求时仍需准确的对数运算。
5. Gravitational Field and Satellite Orbits | 引力场与卫星轨道
A satellite of mass 450 kg orbits Earth at an altitude of 600 km above the surface. Earth' 6.67×10⁻¹¹ N m² kg⁻². Find the orbital speed, period, and total energy. This tests Newton's law of gravitation and circular motion.
一颗 450 kg 的卫星在距地球表面 600 km 的高度绕地运行。地球质量 M = 5.97×10²⁴ kg，半径 R = 6.37×10⁶ m，G = 6.67×10⁻¹¹ N m² kg⁻²。求轨道速率、周期和总能量。这考查牛顿万有引力定律和圆周运动。
×10⁶ m. For circular orbit: gravitational force provides centripetal force: GMm/r² = mv²/r ⇒ v = √GM/r = √6.67×10⁻¹¹ × 5.97×10²⁴ / 6.97×10⁶ = √3.98×10¹⁴ / 6.97×10⁶ = √5.71×10⁷ ≈ 7.56×10³ m s⁻¹. Period T = 2πr/v = 2π×6.97×10⁶ / 7.56×10³ ≈ 5.79×10³ s (about 96.5 min).
轨道半径 r = 6.37×10⁶ + 600×10³ = 6.97×10⁶ m。圆周轨道：万有引力提供向心力：GMm/r² = mv²/r ⇒ v = √GM/r = √6.67×10⁻¹¹ × 5.97×10²⁴ / 6.97×10⁶ = √3.98×10¹⁴ / 6.97×10⁶ = √5.71×10⁷ ≈ 7.56×10³ m s⁻¹。周期 T = 2πr/v = 2π×6.97×10⁶ / 7.56×10³ ≈ 5.79×10³ s（约 96.5 分钟）。
alue indicates a bound orbit. Knowledge of escape velocity and energy graphs often appears. Always convert altitude to radius.
总机械能 = - (6.67×10⁻¹¹×5.97×10²⁴×450) / (2×6.97×10⁶) = - (1.79×10¹⁷) / (1.394×10⁷) ≈ -1.28×10¹⁰ J。负值表示束缚轨道。常涉及逃逸速度和能量图像的知识。务必把高度转换为半径。
6. Linear Momentum and Collisions | 线动量与碰撞
A 1200 kg car moving at 18 m s⁻¹ collides with a stationary 800 kg car. After the impact, they lock together. Determine the velocity after collision and the kinetic energy lost. If the collision time is 0.3 s, find the average force on each car. This demonstrates conservation of momentum and impulse.
一辆 1200 kg 的汽车以 18 m s⁻¹ 行驶，与一辆静止的 800 kg 汽车碰撞。碰撞后两车锁在一起。求碰撞后的速度和动能损失。若碰撞时长为 0.3 s，求每辆车受到的平均力。这展示动量守恒和冲量。
fore: ½ ×1200×18² = 194400 J. After: ½ ×2000×10.8² = 116640 J. Loss = 77760 J. This energy becomes heat and deformation.
碰撞前总动量 = 1200×18 + 0 = 21600 kg m s⁻¹。碰撞后总质量 = 2000 kg，速度 v = 21600/2000 = 10.8 m s⁻¹。碰撞前动能：½ ×1200×18² = 194400 J；碰撞后：½ ×2000×10.8² = 116640 J。损失 77760 J。这些能量转化为热和形变。
(10.8-18) = -8640 kg m s⁻¹, so force magnitude also 28800 N but opposite direction (Newton' third law). Exam trick: a graph of force vs time with area representing impulse is common.
平均力：对静止车，Δp = 800×10.8 = 8640 kg m s⁻¹，F = Δp/Δt = 8640/0.3 = 28800 N。对运动车，Δp = 1200×(10.8-18) = -8640 kg m s⁻¹，故力的大小也是 28800 N 但方向相反（牛顿第三定律）。常见考题：力-时间图，面积代表冲量。
7. Electromagnetic Induction Application | 电磁感应应用
A conducting rod of length 0.25 m moves perpendicular to a uniform magnetic field of 0.80 T at a constant speed of 6.0 m s⁻¹. The rod is part of a circuit with total resistance 4.0 Ω. Calculate induced emf, current, electrical power, and the force needed to keep the rod moving. This links Faraday's law, Lenz's law, and energy.
一根长 0.25 m 的导体杆垂直于 0.80 T 的匀强磁场以 6.0 m s⁻¹ 的恒定速率运动。杆是一个总电阻为 4.0 Ω 的回路的一部分。计算感应电动势、电流、电功率以及维持杆运动所需的力。这联系法拉第定律、楞次定律和能量。
or P = ε I = 1.2 × 0.30 = 0.36 W). The magnetic force on the rod due to induced current: FB opposes motion, so an external force of 0.060 N must be applied to maintain constant speed.
感应电动势 ε = B l v = 0.80 × 0.25 × 6.0 = 1.2 V。电流 I = ε / R = 1.2 / 4.0 = 0.30 A。电功率 P = I² R = 0.30² × 4.0 = 0.36 W（或 P = ε I = 1.2 × 0.30 = 0.36 W）。由于感应电流，杆所受磁力：FB = B I l = 0.80 × 0.30 × 0.25 = 0.060 N。此力阻碍运动，故需施加 0.060 N 的外力以维持匀速。
Mechanical power input = Fext rical power, demonstrating energy conservation. For AC generator scenarios, expect sinusoidal emf variation. In Edexcel, often pair this with a plot of flux linkage change.
输入的机械功率 = Fext × v = 0.060 × 6.0 = 0.36 W，等于电功率，体现能量守恒。在交流发电机情境中，会出现正弦变化的电动势。Edexcel 考试中常结合磁通链变化图像。
8. Nuclear Binding Energy Problem | 核结合能问题
ding energy, and binding energy per nucleon. This highlights E = Δm c² and nuclear stability.
原子核 ⁵⁶Fe 的质量为 55.934937 u。已知质子质量 = 1.007276 u，中子质量 = 1.008665 u，1 u = 931.5 MeV/c²。计算质量亏损、结合能和每核子的结合能。这突出 E = Δm c² 和核稳定性。
25995 = 56.449126 u. Mass defect Δm = 56.449126 - 55.934937 = 0.514189 u. Binding energy Eb /nucleon. This lies near the peak of the curve of binding energy per nucleon, showing iron's stability.
⁵⁶Fe 有 26 个质子和 30 个中子。组成部分总质量 = 26×1.007276 + 30×1.008665 = 26.189176 + 30.25995 = 56.449126 u。质量亏损 Δm = 56.449126 - 55.934937 = 0.514189 u。结合能 Eb = Δm × 931.5 = 0.514189 × 931.5 ≈ 479.0 MeV。每核子结合能 = 479.0 / 56 ≈ 8.55 MeV/核子。该值接近每核子结合能曲线的峰值，表明铁的稳定性。
o joules when needed using 1 eV = 1.6×10⁻¹⁹ J. Show clear steps separating mass defect and energy conversion.
在裂变或聚变计算中，比较产物与反应物质量以求得释放的能量。记得使用一致单位：需要时将 MeV 转换为焦耳，1 eV = 1.6×10⁻¹⁹ J。清晰展示质量亏损和能量转换的步骤。
9. Damping and Resonance Scenario | 阻尼与共振情景
A lightly damped mass-spring system has natural frequency 2.5 Hz. A driver applies a sinusoidal force of variable frequency. At resonance, the amplitude is 8.0 cm. When frequency is 3.0 Hz, amplitude drops to 2.0 cm. Discuss the relationship between damping, Q-factor, and the amplitude-frequency graph. Calculate the Q-factor approximately using the half-power bandwidth method.
一个轻度阻尼的质量-弹簧系统固有频率为 2.5 Hz。驱动装置施加可变频率的正弦力。共振时振幅为 8.0 cm；频率为 3.0 Hz 时，振幅降至 2.0 cm。讨论阻尼、品质因子与振幅-频率图像的关系。用半功率带宽法近似计算 Q 因子。
Q-factor quantifies the sharpness of resonance. Bandwidth Δf is defined between frequencies where power is half the resonant power, which corresponds to amplitude of Ares/√2 ≈ 0.707 Ares y compute bandwidth without data points around resonance, but we can estimate if the 5.66 cm level occurs at, say, f₁ = 2.3 Hz and f₂ = 2.7 Hz, then Δf = 0.4 Hz, Q = f₀/Δf = 2.5/0.4 = 6.25. In practice, you would be given data.
品质因子 Q 表征共振的锐度。带宽 Δf 定义为功率降至共振功率一半所对应的频率范围，此时振幅为 Ares/√2 ≈ 0.707 Ares。这里共振振幅 8.0 cm，半功率振幅 ≈ 5.66 cm。题目给出 3.0 Hz 处的振幅为 2.0 cm（远低于此值）。没有共振两侧的频率点不能直接计算带宽，但假若 5.66 cm 发生在 f₁ = 2.3 Hz 和 f₂ = 2.7 Hz，则 Δf = 0.4 Hz，Q = f₀/Δf = 2.5/0.4 = 6.25。实际考试会给出数据点。
Light damping gives high Q and narrow resonance; heavy damping broadens the curve. In an exam, you might plot amplitude vs frequency and identify natural frequency, phase difference changes from 0 to π, and discuss effects of damping on forced oscillations. Real examples: tuned mass dampers in buildings or suspension bridges.
轻阻尼导致高 Q 值和窄共振峰；重阻尼使曲线变宽。考试中可能要绘制振幅-频率图，识别固有频率，说明相位差从 0 变到 π，讨论阻尼对受迫振动的影响。实际例子：建筑中的调谐质量阻尼器或悬索桥。
10. Thermal Physics and Ideal Gas Application | 热物理与理想气体应用
t allows volume changes. The gas is heated at constant pressure until volume doubles, then compressed isothermally back to original volume. Find initial volume, final temperature after heating, work done in isothermal compression, and sketch the p–V diagram. This combines gas laws and first law of thermodynamics.
一个密封气缸装有 0.20 mol 理想气体，温度 300 K，压强 1.0 × 10⁵ Pa。气缸有一个可移动的活塞以改变体积。气体在恒压下加热至体积加倍，然后等温压缩回原始体积。求初始体积、加热后的最终温度、等温过程中做的功，并画出 p–V 图。这结合气体定律和热力学第一定律。
= 600 K. Isothermal compression at T₂ = 600 K from V₂ to V₁. Work done on gas W = nRT ln(V₁/V₂) = 0.20 × 8.31 × 600 × ln(0.5) = 997.2 × (-0.693) = -691 J. Negative means work is done on the gas, so energy is transferred to the surroundings.
初始 ₁ = nRT/P = 0.20 × 8.31 × 300 / 1.0×10⁵ = 498.6 / 100000 = 4.986 × 10⁻³ m³ ≈ 5.0 × 10⁻³ m³。等压过程 V/T 恒定，当 V₂ = 2V₁，T₂ = 2T₁ = 600 K。等温压缩过程温度 T₂ = 600 K，从 V₂ 压缩至 V₁。对气体做功 W = nRT ln(V₁/V₂) = 0.20 × 8.31 × 600 × ln(0.5) = 997.2 × (-0.693) = -691 J。负值表示气体被做功，能量传向外界。
On the p– s heat exchanges. Many candidates mix up sign conventions; stick to the convention that work done BY the gas is positive in W = pΔV.
在 p–V 图上：等压膨胀（水平线向右），然后等温压缩（曲线向左）。用热力学第一定律 ΔU = Q - W 讨论热交换。许多考生搞混正负号约定；坚持气体对外做功时 W 为正（W = pΔV）的惯例。
11. Electric Fields and Equipotentials | 电场与等势面
A +3.0 nC point charge is fixed at the origin. A second charge of +2.0 nC is moved from a point 0.10 m away to a point 0.20 m away along a radial line. Determine the change in electric potential energy of the system and the work done by the electric field. Interpret the sign and relate to forces.
一个 +3.0 nC 点电荷固定于原点。将另一个 +2.0 nC 的电荷沿径向从 0.10 m 处移至 0.20 m 处。求系统电势能的变化和电场力所做的功。解释符号并联系力。
Electric potential energy U = k Q q / r, with k = 8.99×10⁹ N m² C⁻². Initial Ui = (8.99×10⁹)×(3.0×10⁻⁹)×(2.0×10⁻⁹) / 0.10 = (53.94×10⁻⁹) / 0.10 = 5.394×10⁻⁷ J. Final Uf = 53.94×10⁻⁹ / 0.20 = 2.697×10⁻⁷ J. Change ΔU = Uf - Ui itive work by field means the field does work pushing the charges apart (like charges repel), so kinetic energy would increase if free.
电势能 U = k Q q / r，k = 8.99×10⁹ N m² C⁻²。初始 Ui = (8.99×10⁹)×(3.0×10⁻⁹)×(2.0×10⁻⁹) / 0.10 = (53.94×10⁻⁹) / 0.10 = 5.394×10⁻⁷ J。末 Uf = 53.94×10⁻⁹ / 0.20 = 2.697×10⁻⁷ J。变化 ΔU = Uf - Ui = -2.697×10⁻⁷ J。电场力做功 W = -ΔU = +2.697×10⁻⁷ J。电场做正功意味着电场力推动电荷分开（同号相斥），若自由运动动能将增加。
Plotting V against r and E against r helps visualise the 1/r and 1/r² dependencies. Combining potential due to multiple charges requires scalar addition. For equipotentials, remember they are perpendicular to field lines. This topic bridges to capacitors and uniform fields.
绘制 V 随 r 及 E 随 r
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