📚 A-Level WJEC Physics: Worked Examples Explained | A-Level WJEC 物理：典型例题详解
Mastering A-Level WJEC Physics requires more than just knowing the formulas—it demands the ability to apply concepts to unfamiliar problems. This article presents a collection of carefully selected worked examples, each unpacking a key topic in the WJEC specification. By following the step-by-step solutions, you will build problem-solving confidence and deepen your understanding of core principles like mechanics, electricity, waves, and modern physics. Every example is paired with a clear English explanation and an equivalent Chinese translation, making it ideal for bilingual learners.
掌握 A-Level WJEC 物理不仅需要记住公式，更要求能够将概念应用于陌生问题。本文精选了一系列典型例题，逐一拆解 WJEC 考纲中的重点主题。通过跟随逐步的解题过程，你将培养解题信心，并加深对力学、电学、波和现代物理等核心原理的理解。每个例题都配有清晰的英文解释和对应的中文翻译，非常适合双语学习者。
1. Kinematics: Constant Acceleration | 运动学：匀加速直线运动
A car accelerates uniformly from rest to 25 m s⁻¹ in 8.0 seconds. Calculate the acceleration and the distance travelled during this time.
一辆汽车从静止开始匀加速，8.0 秒后速度达到 25 m s⁻¹。计算加速度和这段时间内的位移。
5 × (0 + 25) × 8.0 = 100 m. Alternatively, s = ut + ½ a t² = 0 + 0.5 × 3.125 × 64 = 100 m.
使用匀变速运动公式。首先加速度 a = (v - u) / t = (25 - 0) / 8.0 = 3.125 m s⁻² ≈ 3.1 m s⁻²。距离 s = ½ (u + v) t = 0.5 × (0 + 25) × 8.0 = 100 m。也可用 s = ut + ½ a t² = 0 + 0.5 × 3.125 × 64 = 100 m。
2. Newton's Second Law and Friction | 牛顿第二定律与摩擦力
efficient of kinetic friction is 0.20. Find the acceleration of the block. (g = 9.8 m s⁻²)
一个质量为 4.0 kg 的木块被一个与水平方向成 30° 斜向上的 18 N 力拉着在水平面上运动。动摩擦因数为 0.20。求木块的加速度。（g = 9.8 m s⁻²）
N = 0.20 × 30.2 = 6.04 N. Net horizontal force = 15.6 - 6.04 = 9.56 N. Acceleration a = Fnet / m = 9.56 / 4.0 = 2.39 m s⁻².
分解拉力：水平分量 Fx = 18 cos30° = 15.6 N，竖直分量 Fy = 18 sin30° = 9.0 N。支持力 N = mg - Fy = (4.0×9.8) - 9.0 = 30.2 N。摩擦力 f = μ N = 0.20 × 30.2 = 6.04 N。合力水平方向 = 15.6 - 6.04 = 9.56 N。加速度 a = Fnet / m = 9.56 / 4.0 = 2.39 m s⁻²。
3. Work, Energy and Power | 功、能与功率
rovides a power of 4.5 kW. Calculate the efficiency of the lifting system. (g = 9.8 m s⁻²)
一台起重机以 0.80 m s⁻¹ 的恒定速度垂直吊起 500 kg 的重物。电动机提供的功率为 4.5 kW。计算起吊系统的效率。（g = 9.8 m s⁻²）
ed to lift the load at constant speed equals its weight: F = mg = 500 × 9.8 = 4900 N. So Pout = 4900 × 0.80 = 3920 W = 3.92 kW. Efficiency η = (Pout / Pin) × 100% = (3.92 / 4.5) × 100% = 87.1%.
有用输出功率 = 力 × 速度。匀速提升所需的力等于重力：F = mg = 500 × 9.8 = 4900 N。因此 Pout = 4900 × 0.80 = 3920 W = 3.92 kW。效率 η = (Pout / Pin) × 100% = (3.92 / 4.5) × 100% = 87.1%。
4. Momentum and Impulse | 动量与冲量
A tennis ball of mass 0.058 kg strikes a racket at 35 m s⁻¹ and rebounds at 45 m s⁻¹ in the opposite direction. Contact time is 0.040 s. Calculate the average force exerted on the ball.
一个质量为 0.058 kg 的网球以 35 m s⁻¹ 的速度撞击球拍后，以 45 m s⁻¹ 的速度反向弹回。接触时间为 0.040 s。计算球拍对球的平均作用力。
5 - (-35)) = 0.058 × 80 = 4.64 kg m s⁻¹. Average force F = Δp / Δt = 4.64 / 0.040 = 116 N.
设反弹方向为正。初速度 u = -35 m s⁻¹，末速度 v = +45 m s⁻¹。动量变化量 Δp = m(v - u) = 0.058 × (45 - (-35)) = 0.058 × 80 = 4.64 kg m s⁻¹。平均作用力 F = Δp / Δt = 4.64 / 0.040 = 116 N。
5. Circular Motion: Horizontal Circle | 圆周运动：水平圆周
A mass of 0.20 kg is whirled in a horizontal circle of radius 0.80 m on a frictionless table. The string can withstand a maximum tension of 50 N. Determine the maximum angular speed before the string breaks.
一个 0.20 kg 的物体在光滑水平面上做半径为 0.80 m 的水平圆周运动，绳子能承受的最大拉力为 50 N。求绳子断裂前的最大角速度。
The centripetal force is provided by tension: T = m ω² r. Rearranging, ω_max = √(Tmax / (m r)) = √(50 / (0.20 × 0.80)) = √50 / 0.16 = √312.5 ≈ 17.7 rad s⁻¹.
向心力由绳子拉力提供：T = m ω² r。变形得 ω_max = √(Tmax / (m r)) = √(50 / (0.20 × 0.80)) = √50 / 0.16 = √312.5 ≈ 17.7 rad s⁻¹。
6. Electric Fields: Coulomb's Law | 电场：库仑定律
Calculate the magnitude of the electrostatic force between them. (ε₀ = 8.85 × 10⁻¹² F m⁻¹)
两个点电荷 +3.0 μC 和 -4.0 μC 在真空中相距 0.50 m。计算它们之间静电力的大小。(ε₀ = 8.85 × 10⁻¹² F m⁻¹)
Using Coulomb's law: F = (1/(4πε₀)) × |q₁ q₂| 4.0×10⁻⁶) / (0.50)² = (8.99×10⁹ × 12×10⁻¹²) / 0.25 = (0.10788) / 0.25 = 0.4315 N ≈ 0.43 N.
使用库仑定律：F = (1/(4πε₀)) × |q₁ q₂| / r²。常数 1/(4πε₀) = 8.99 × 10⁹ N m² C⁻²。 × 4.0×10⁻⁶) / (0.50)² = (8.99×10⁹ × 12×10⁻¹²) / 0.25 = 0.10788 / 0.25 = 0.4315 N ≈ 0.43 N。
7. DC Circuits: Kirchhoff's Laws | 直流电路：基尔霍夫定律
Two resistors, 4.0 Ω and 6.0 Ω, are connected in parallel; this combination is in series with a 2.0 Ω resistor. A 12 V battery of negligible internal resistance is connected across the whole network. Find the current through the 6.0 Ω resistor.
两个电阻 4.0 Ω 和 6.0 Ω 并联后，再与一个 2.0 Ω 电阻串联。一个内阻可忽略的 12 V 电池接在整个网络两端。求通过 6.0 Ω 电阻的电流。
First, equivalent resistance of the parallel section: 1/Rp = 1/4 + 1/6 = 5/12, so Rp = 12/5 = 2.4 Ω. Total circuit resistance Rtotal = 2.4 + 2.0 = 4.4 Ω. Total current Itotal = V / Rtotal = 12 / 4.4 = 2.727 A. The voltage across the parallel branch is Vp = Itotal × Rp = 2.727 × 2.4 = 6.545 V. Current through the 6.0 Ω resistor: I₆ = Vp / 6.0 = 6.545 / 6.0 = 1.09 A.
首先，并联部分的等效电阻：1/Rp = 1/4 + 1/6 = 5/12，故 Rp = 12/5 = 2.4 Ω。电路总电阻 Rtotal = 2.4 + 2.0 = 4.4 Ω。总电流 Itotal = V / Rtotal = 12 / 4.4 = 2.727 A。并联支路两端电压 Vp = Itotal × Rp = 2.727 × 2.4 = 6.545 V。通过 6.0 Ω 电阻的电流：I₆ = Vp / 6.0 = 6.545 / 6.0 = 1.09 A。
8. Waves: Young's Double-Slit | 波：杨氏双缝干涉
In a double-slit experiment, laser light of wavelength 635 nm produces a fringe pattern on a screen 2.00 m away. The distance between the central bright fringe and the third bright fringe is 9.53 mm. Calculate the slit separation.
在双缝实验中，波长为 635 nm 的激光在 2.00 m 远处的屏幕上产生干涉条纹。中央亮纹到第三级亮纹的距离为 9.53 mm。计算双缝间距。
(3 × 635×10⁻⁹ × 2.00) / (9.53×10⁻³) = (3.81×10⁻⁶) / (9.53×10⁻³) = 4.00×10⁻⁴ m = 0.400 mm.
相邻亮纹的条纹间距 Δy = λ D / d，其中 d 为双缝间距，D 为屏幕距离。第三级亮纹到中心的距离 y₃ = 3 λ D / d。因此 9.53×10⁻³ = 3 × (635×10⁻⁹ × 2.00) / d。整理得 = (3 × 635×10⁻⁹ × 2.00) / (9.53×10⁻³) = (3.81×10⁻⁶) / (9.53×10⁻³) = 4.00×10⁻⁴ m = 0.400 mm。
9. Ideal Gases: Kinetic Theory | 理想气体：分子动理论
until the pressure becomes 2.0 × 10⁵ Pa, find the new temperature. (R = 8.31 J mol⁻¹ K⁻¹)
一个气缸装有 0.20 mol 的氦气，压强为 1.5 × 10⁵ Pa，温度为 300 K。计算气缸的容积。如果气体在定容条件下加热，直到压强变为 2.0 × 10⁵ Pa，求新的温度。（R = 8.31 J mol⁻¹ K⁻¹）
me, P₁/T₁ = P₂/T₂, so T₂ = T₁ × (P₂/P₁) = 300 × (2.0×10⁵ / 1.5×10⁵) = 300 × (4/3) = 400 K.
使用理想气体状态方程 PV = nRT：V = nRT / P = (0.20 × 8.31 × 300) / (1.5×10⁵) = 498.6 / (1.5×10⁵) = 3.324×10⁻³ m³。定容条件下，P₁/T₁ = P₂/T₂，因此 T₂ = T₁ × (P₂/P₁) = 300 × (2.0×10⁵ / 1.5×10⁵) = 300 × (4/3) = 400 K。
10. Radioactive Decay: Half-Life | 放射性衰变：半衰期
A sample of iodine-131 has an initial activity of 800 Bq. Its half-life is 8.0 days. Determine the activity after 20 days, and find the time required for the activity to drop to 100 Bq.
一个碘-131 样品的初始活度为 800 Bq，半衰期为 8.0 天。求 20 天后的活度，以及活度降至 100 Bq 所需的时间。
Number of half-lives elapsed: n = 20 / 8.0 = 2.5. Activity A = A₀ (½)n = 800 × (½)2.5 = 800 × (1/2)2.5. Since (½)2.5 = (½)2 × (½)0.5 0.1768, A = 800 × 0.1768 = 141.4 Bq ≈ 141 Bq. To find time for A = 100 Bq: 100 = 800 × (½)t/8. So (½)t/8 = 0.125 = 1/8 = (½)3, hence t/8 = 3, t = 24 days.
已过去的半衰期个数：n = 20 / 8.0 = 2.5。活度 A = A₀ (½)n = 800 × (½)2.5 = 800 × (0.5)2 × (0.5)0.5 = 800 × 0.25 × 0.7071 = 141.4 Bq ≈ 141 Bq。求降至 100 Bq 的时间：100 = 800 × (½)t/8，所以 (½)t/8 = 0.125 = 1/8 = (½)3，故 t/8 = 3，t = 24 天。
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