📚 Year 10 WJEC Physics: Case Study Practice | 案例分析实战演练
Case studies bring physics to life by showing how concepts are used to solve everyday problems. This resource walks you through ten worked examples, each targeting a key topic from the Year 10 WJEC Physics specification. Every step is explained in English and Chinese to help you build both your physics understanding and your bilingual terminology.
案例分析让物理活起来，展示如何用概念解决日常问题。本文带领你演练十个从十年级 WJEC 物理课程中精选的实战例子，每个步骤均用中英双语讲解，帮助你同时提升物理思维和双语术语能力。
1. Case Study: Braking Distance | 案例一：刹车距离
A car travels at 20 m/s on a wet road. The driver sees a hazard and takes 0.7 s to react. The brakes then produce a deceleration of 6.0 m/s². Let’s find the total stopping distance.
一辆汽车在湿滑路面上以 20 m/s 的速度行驶。驾驶员发现险情，经过 0.7 s 才做出反应。随后刹车产生 6.0 m/s² 的减速度。我们来计算总刹停距离。
Thinking distance = speed × reaction time = 20 m/s × 0.7 s = 14 m.
反应距离 = 速度 × 反应时间 = 20 m/s × 0.7 s = 14 m。
v² = u² + 2as
Since the car stops, v = 0; rearrange to find braking distance s = u²/(2a).
由于汽车最终停下，v = 0；整理得制动距离 s = u²/(2a)。
Substitute u = 20 m/s and a = 6.0 m/s²: s = (20)² / (2×6) = 400/12 ≈ 33.3 m.
代入 u = 20 m/s，a = 6.0 m/s²：s = (20)² / (2×6) = 400/12 ≈ 33.3 m。
Total stopping distance = thinking distance + braking distance = 14 m + 33.3 m = 47.3 m.
总刹停距离 = 反应距离 + 制动距离 = 14 m + 33.3 m = 47.3 m。
2. Case Study: Newton’s Second Law in a Car | 案例二：汽车的牛顿第二定律
A car of mass 1200 kg starts from rest. The engine supplies a driving force of 3000 N, while resistive forces total 500 N. Calculate the acceleration.
一辆质量为 1200 kg 的汽车从静止出发。引擎提供 3000 N 的驱动力，阻力共计 500 N。计算加速度。
Resultant force = driving force − resistive force = 3000 N − 500 N = 2500 N.
净力 = 驱动力 − 阻力 = 3000 N − 500 N = 2500 N。
F = m a
Rearrange to a = F/m = 2500 N / 1200 kg ≈ 2.08 m/s².
整理得 a = F/m = 2500 N / 1200 kg ≈ 2.08 m/s²。
The car accelerates at about 2.08 m/s² in the direction of the resultant force.
汽车沿净力方向产生约 2.08 m/s² 的加速度。
3. Case Study: Object on an Inclined Plane | 案例三：斜面上的物体
A 20 kg box rests on a frictionless slope that makes a 30° angle with the horizontal. Find the component of the weight acting down the slope and explain why it tends to slide.
一个 20 kg 的箱子静止在无摩擦的斜面上，斜面与水平面的夹角为 30°。求重力沿斜面的分量，并解释箱体为什么会下滑。
F_parallel = m g sin θ
Here m = 20 kg, g = 10 N/kg, θ = 30°. sin 30° = 0.5.
已知 m = 20 kg, g = 10 N/kg, θ = 30°。sin 30° = 0.5。
F_parallel = 20 × 10 × 0.5 = 100 N.
F_parallel = 20 × 10 × 0.5 = 100 N。
The weight component pulling the box down the slope is 100 N. With no friction, this unbalanced force produces an acceleration.
沿斜面向下拉箱子的重力分量为 100 N。由于没有摩擦，这个非平衡力会使箱子产生加速度。
4. Case Study: Work and Power – Lifting a Load | 案例四：功和功率——提升重物
A lift raises a 500 kg load vertically at a constant speed of 2 m/s for a height of 10 m in 5 seconds. Calculate the work done and the power output of the lift motor.
一部电梯以 2 m/s 的恒定速度将 500 kg 的重物垂直提升 10 m，用时 5 秒。计算电梯电动机所做的功和输出功率。
The force needed to lift the load equals its weight: F = m g = 500 kg × 10 N/kg = 5000 N.
提升重物需要的力等于其重量：F = m g = 500 kg × 10 N/kg = 5000 N。
Work done = force × distance moved in the direction of the force = 5000 N × 10 m = 50 000 J (50 kJ).
做功 = 力 × 沿力方向移动的距离 = 5000 N × 10 m = 50 000 J（50 kJ）。
P = W / t
Power = 50 000 J / 5 s = 10 000 W (10 kW).
功率 = 50 000 J / 5 s = 10 000 W（10 kW）。
The motor delivers 10 kW of power. Because the speed is constant, the lifting force equals the weight.
电动机输出 10 kW 的功率。因为速度不变，提升力等于重力。
5. Case Study: Series Circuit Analysis | 案例五：串联电路分析
Two resistors, R₁ = 4 Ω and R₂ = 6 Ω, are connected in series across a 12 V battery. Determine the total resistance, the circuit current, and the voltage across each resistor.
两个电阻 R₁ = 4 Ω 和 R₂ = 6 Ω 串联在 12 V 电池两端。求总电阻、电路电流以及每个电阻两端的电压。
For a series circuit, total resistance R_total = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω.
对于串联电路，总电阻 R_total = R₁ + R₂ = 4 Ω + 6 Ω = 10 Ω。
I = V / R
Current I = 12 V / 10 Ω = 1.2 A.
电流 I = 12 V / 10 Ω = 1.2 A。
Voltage across R₁: V₁ = I × R₁ = 1.2 A × 4 Ω = 4.8 V.
R₁ 两端电压：V₁ = I × R₁ = 1.2 A × 4 Ω = 4.8 V。
Voltage across R₂: V₂ = I × R₂ = 1.2 A × 6 Ω = 7.2 V. (The voltages add up to 12 V.)
R₂ 两端电压：V₂ = I × R₂ = 1.2 A × 6 Ω = 7.2 V。（两个电压之和为 12 V。）
6. Case Study: Choosing a Fuse for a Kettle | 案例六：为电热水壶选择保险丝
An electric kettle is marked 230 V, 2000 W. It connects to the UK mains supply (230 V). What is the normal operating current? Which standard fuse should be used: 3 A, 5 A, or 13 A?
某电热水壶标有 230 V、2000 W，连接到 230 V 的英国市电。其正常工作电流是多少？应选用哪种标准保险丝：3 A、5 A 还是 13 A？
P = I V
Rearranging gives I = P / V = 2000 W / 230 V ≈ 8.7 A.
整理得 I = P / V = 2000 W / 230 V ≈ 8.7 A。
The normal current is about 8.7 A. A 3 A fuse would blow, and a 5 A fuse is very close to the normal current so may blow during small surges. Therefore, the safest choice is the 13 A fuse.
正常工作电流约 8.7 A。3 A 保险丝会熔断，5 A 保险丝非常接近正常电流，稍有浪涌便可能熔断。最安全的选择是 13 A 保险丝。
The fuse protects the cable, so its rating should be slightly above the normal operating current.
保险丝用于保护导线，其额定值应略高于正常工作电流。
7. Case Study: Seismic Waves – P-waves and S-waves | 案例七：地震波——P波与S波
P-waves travel at 8 km/s and S-waves at 5 km/s through the Earth’s crust. A seismograph records the P-wave 15 seconds before the S-wave from the same earthquake. Calculate the distance to the epicentre.
P 波在地壳中的速度为 8 km/s，S 波为 5 km/s。某地震仪记录到同一地震的 P 波比 S 波早到 15 秒。计算震中距离。
Let d be the distance. Time for P-wave: t_P = d / 8. Time for S-wave: t_S = d / 5.
设距离为 d。P 波时间：t_P = d / 8，S 波时间：t_S = d / 5。
The time difference is t_S − t_P = 15 s: d/5 − d/8 = 15.
时间差 t_S − t_P = 15 s：d/5 − d/8 = 15。
Combine the fractions: (8d − 5d) / 40 = 3d / 40 = 15. Therefore, d = 15 × 40 / 3 = 200 km.
通分：(8d − 5d) / 40 = 3d / 40 = 15。因此，d = 15 × 40 / 3 = 200 km。
The epicentre is 200 km away from the seismograph station.
震中距离地震仪台站 200 km。
8. Case Study: Radio
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