Year 10 WJEC Physics: Case Study Practice in Action | 案例分析实战演练

📚 Year 10 WJEC Physics: Case Study Practice in Action | 案例分析实战演练

Physics is not just about recalling formulas; it is about applying them to real-world situations. In WJEC Year 10, you will frequently encounter case studies that ask you to analyse a scenario, extract data, choose the right equations, and present a logical solution. This article provides a structured approach and a series of worked examples to build your confidence.

物理学习不仅仅是背诵公式，更在于将其应用于实际情境。WJEC 十年级考试中经常会遇到案例分析题，要求你分析一个场景、提取数据、选择合适的方程，并给出有逻辑的解答。本文将提供一个结构化的解题思路，并通过一系列典型例题来帮助建立信心。

1. How to Approach a Physics Case Study | 如何解答物理案例分析

Read the entire question twice. Underline the quantities given and the unknown you need to find. Pay attention to units – are they in metres, seconds, kilograms? Convert everything into SI base units (metres, kilograms, seconds, amperes) before starting your calculation.

将题目完整阅读两遍。划出已知量和待求的未知量。注意单位 – 是米、秒还是千克？在开始计算之前，先把所有数据转换到国际单位制的基本单位（米、千克、秒、安培）。

Decide which area of physics the problem belongs to: mechanics, electricity, energy, waves, thermal physics, or nuclear physics. Then list the relevant equations from the formula sheet. For example, if the case involves constant acceleration, you may need v = u + at or v² = u² + 2as.

确定该问题属于物理学的哪一板块：力学、电学、能量、波、热学还是核物理。然后从公式表中列出相关的方程。例如，如果案例涉及匀加速运动，你可能需要用到 v = u + at 或 v² = u² + 2as。

Substitute the values carefully and perform the calculation. Always include units in your working. Finally, check whether your answer makes sense. Is the speed of a falling apple reasonable? Is the current too high for a domestic circuit? This reflection is a key part of case study analysis.

仔细代入数值并完成计算。计算过程中始终要带着单位。最后，检查答案是否合理。苹果下落的速度是否在合理范围内？电流是否对家用电路来说过高？这种反思是案例分析的关键环节。

2. Free Fall of an Apple | 苹果的自由落体

Scenario: An apple of mass 0.15 kg falls from a branch 5.0 m above the ground. Assume g = 9.8 m/s² and ignore air resistance. Calculate the speed of the apple just before it hits the ground and the time taken to fall.

情景：一个质量为 0.15 kg 的苹果从离地面 5.0 m 高的树枝上落下。取 g = 9.8 m/s²，忽略空气阻力。求苹果落地前瞬间的速度以及下落所用时间。

We know: initial velocity u = 0 m/s, displacement s = 5.0 m, acceleration a = 9.8 m/s² downwards. To find the final velocity v, we use the equation that does not involve time:

v² = u² + 2as

已知：初速度 u = 0 m/s，位移 s = 5.0 m，加速度 a = 9.8 m/s² 向下。为求末速度 v，我们使用不包含时间的公式：

v² = u² + 2as

Substituting: v² = 0 + 2 × 9.8 × 5.0 = 98. So v = √98 ≈ 9.90 m/s. The speed just before impact is about 9.9 m/s.

代入得：v² = 0 + 2 × 9.8 × 5.0 = 98。因此 v = √98 ≈ 9.90 m/s。落地前的速度约为 9.9 m/s。

To find time t, use: v = u + at. 9.90 = 0 + 9.8 × t, so t = 9.90 ÷ 9.8 ≈ 1.01 s. The apple takes about 1.0 second to reach the ground.

为求时间 t，使用 v = u + at。9.90 = 0 + 9.8 × t，得 t = 9.90 ÷ 9.8 ≈ 1.01 s。苹果大约需要 1.0 秒落到地面。

Notice that the mass of the apple is not needed for this calculation. In the absence of air resistance, all objects fall with the same acceleration. This is a core idea in mechanics.

注意，苹果的质量在此计算中并无用武之地。在没有空气阻力的情况下，所有物体都以相同的加速度下落。这是力学中的一个核心概念。

3. Pushing a Box up an Inclined Plane | 沿斜面推箱子

Scenario: A 20 kg box is pushed up a slope inclined at 30° to the horizontal, with the push force F applied parallel to the slope. The box moves at a steady speed. The coefficient of dynamic friction between the box and the slope is 0.30. Take g = 9.8 N/kg. Calculate the push force required.

情景：一个 20 kg 的箱子被沿斜面向上推动，斜面与水平面的夹角为 30°，推力 F 与斜面平行。箱子匀速运动。箱与斜面之间的动摩擦系数为 0.30，取 g = 9.8 N/kg。求所需推力的大小。

Since the box moves at constant speed, the resultant force parallel to the slope is zero. The forces acting along the slope are: the component of weight down the slope (mg sinθ), the friction force f down the slope, and the push force F up the slope. For equilibrium: F = mg sinθ + f.

因为箱子匀速运动，平行斜面方向的合力为零。沿斜面方向的力有：向下滑的重力分量（mg sinθ）、向下的摩擦力 f 以及向上的推力 F。由平衡条件：F = mg sinθ + f。

The normal reaction R balances the perpendicular component of weight: R = mg cosθ. The dynamic friction is f = μ R = μ mg cosθ. Combining: F = mg sinθ + μ mg cosθ = mg (sinθ + μ cosθ).

法向反力 R 与重力的垂直斜面分量平衡：R = mg cosθ。动摩擦力为 f = μ R = μ mg cosθ。合并可得：F = mg sinθ + μ mg cosθ = mg (sinθ + μ cosθ)。

Substitute values: m = 20 kg, g = 9.8 N/kg, θ = 30°, μ = 0.30. sin30° = 0.5, cos30° ≈ 0.866. F = 20 × 9.8 × (0.5 + 0.30 × 0.866) = 196 × (0.5 + 0.2598) = 196 × 0.7598 ≈ 149 N. The required push force is about 149 N.

代入数值：m = 20 kg, g = 9.8 N/kg, θ = 30°, μ = 0.30。sin30° = 0.5, cos30° ≈ 0.866。F = 20 × 9.8 × (0.5 + 0.30 × 0.866) = 196 × (0.5 + 0.2598) = 196 × 0.7598 ≈ 149 N。所需推力约为 149 N。

4. Christmas Light Circuits | 圣诞灯串电路

Scenario: A string of 12 identical filament bulbs is connected in series to a 36 V supply. Each bulb has a resistance of 5.0 Ω when lit. Find the voltage across each bulb and the current flowing in the circuit. One bulb blows (its filament breaks). Explain what happens to the remaining bulbs. The same 12 bulbs are then rewired in parallel across the same 36 V supply. Comment on the brightness of each bulb and any safety concerns.

情景：一串由 12 个相同灯丝灯泡串联而成的灯串连接在 36 V 电源上。每个灯泡点亮时的电阻为 5.0 Ω。求每个灯泡两端的电压和电路中的电流。如果其中一个灯泡烧毁（灯丝断裂），请解释其他灯泡会如何。随后，这 12 个灯泡被改为并联连接到同一个 36 V 电源上。请分析每个灯泡的亮度以及可能的安全问题。

In series, the total resistance R_total = 12 × 5.0 Ω = 60 Ω. The current I = V_supply / R_total = 36 V / 60 Ω = 0.60 A. The voltage across each bulb is V_bulb = I × R_bulb = 0.60 A × 5.0 Ω = 3.0 V. Alternatively, since the bulbs are identical, they share the total voltage equally: 36 V / 12 = 3.0 V per bulb.

串联时，总电阻 R_total = 12 × 5.0 Ω = 60 Ω。电流 I = V_supply / R_total = 36 V / 60 Ω = 0.60 A。每个灯泡两端的电压为 V_bulb = I × R_bulb = 0.60 A × 5.0 Ω = 3.0 V。或者，由于灯泡相同，它们平均分担总电压：36 V / 12 = 3.0 V 每灯。

If one bulb blows, the series circuit is broken. Current stops flowing, and all the other bulbs go out. This is a classic disadvantage of series wiring for decorative lights.

如果一个灯泡烧毁，串联电路就被切断。电流停止流动，所有其他灯泡都会熄灭。这是装饰灯串采用串联连接的一个典型缺点。

When the same 12 bulbs are connected in parallel to 36 V, each bulb receives the full 36 V. The current through each bulb becomes I_bulb = 36 V / 5.0 Ω = 7.2 A. The total current drawn from the supply is 12 × 7.2 A = 86.4 A. This is extremely high and would blow a standard fuse or cause overheating of the wires. The bulbs would be exceedingly bright but would likely burn out immediately or pose a fire hazard. Parallel connection at full voltage is dangerous without additional protective components.

当同样的 12 个灯泡并联在 36 V 电源上时，每个灯泡都承受全额 36 V 电压。通过每个灯泡的电流变为 I_bulb = 36 V / 5.0 Ω = 7.2 A。电源需要提供的总电流为 12 × 7.2 A = 86.4 A。这个电流极大，会烧断普通保险丝或导致电线过热。灯泡会异常明亮，但很可能瞬间烧毁或引发火灾危险。在没有额外保护元件的情况下，直接并联在全电压下是非常危险的。

5. Energy Efficiency of a Kettle | 电水壶的效率

Scenario: An electric kettle has a power rating of 2200 W. It is used to heat 1.0 kg of water from 20 °C to 100 °C. The water reaches boiling point in 2 minutes and 15 seconds. The specific heat capacity of water is 4200 J/(kg °C). Calculate the useful energy transferred to the water, the electrical energy supplied by the kettle, and the efficiency of the kettle.

情景：一个额定功率为 2200 W 的电水壶用来将 1.0 kg 水从 20 °C 加热到 100 °C。水在 2 分 15 秒后达到沸点。水的比热容为 4200 J/(kg °C)。计算水获得的有用能量、水壶提供的电能，以及水壶的效率。

Useful energy Q = m × c × Δθ. Here m = 1.0 kg, c = 4200 J/(kg °C), Δθ = (100 – 20) = 80 °C. So Q = 1.0 × 4200 × 80 = 336 000 J.

有用能量 Q = m × c × Δθ。其中 m = 1.0 kg，c = 4200 J/(kg °C)，Δθ = (100 – 20) = 80 °C。因此 Q = 1.0 × 4200 × 80 = 336 000 J。

Time t must be in seconds: 2 min 15 s = (2 × 60) + 15 = 135 s. Electrical energy supplied E = P × t = 2200 W × 135 s = 297 000 J.

时间 t 必须换算为秒：2 分 15 秒 = (2 × 60) + 15 = 135 s。消耗的电能 E = P × t = 2200 W × 135 s = 297 000 J。

Efficiency = (useful energy output / total energy input) × 100% = (336 000 J / 297 000 J) × 100% ≈ 113%. This result seems to exceed 100%, which is impossible. The apparent inconsistency suggests a problem: either the time was measured inaccurately, the kettle power rating is nominal and the actual power drawn was higher, or some energy was transferred from the heating element before timing started. In a real experiment, such a reading would prompt us to check the data. For this calculation, we must assume ideal values; but typically we expect an efficiency less than 100% (e.g., if the actual time were longer). The key is to apply the formula correctly and recognise that efficiency cannot truly exceed 100%.

效率 = (有用输出能量 / 总输入能量) × 100% = (336 000 J / 297 000 J) × 100% ≈ 113%。结果似乎超过了 100%，这在实际中是不可能的。这一明显的不一致表明可能存在问题：也许计时不准，水壶的标称功率不准确而实际消耗功率更高，或者有部分能量在计时开始前就已经转移给了水。在真实实验中，这样的数据会促使我们去核查测量过程。此处我们仅展示计算方法；正常情况下我们期望效率低于 100%（例如如果实际所需时间更长）。关键是要正确应用公式，并认识到效率不可能真正超过 100%。

6. Echo Sounding from a Ship | 船舶回声测深

Scenario: A research ship sends a short ultrasonic pulse vertically downward. The echo is detected 0.18 s after transmission. The speed of sound in seawater is 1520 m/s. Calculate the depth of the water beneath the ship.

情景：一艘科考船垂直向下发射一个短超声波脉冲。0.18 秒后接收到回声。海水中的声速为 1520 m/s。计算船下方海水的深度。

The pulse travels down to the seabed and back up to the ship. The total distance travelled by the sound is 2 × depth. Using the equation distance = speed × time, we have 2d = v × t, so d = (v × t) / 2.

脉冲从船上传到海底再返回船上。声音传播的总距离是水深的 2 倍。使用公式距离 = 速度 × 时间，有 2d = v × t，因此 d = (v × t) / 2。

Substitute v = 1520 m/s, t = 0.18 s. d = (1520 × 0.18) / 2 = 273.6 / 2 = 136.8 m. The depth is approximately 137 m.

代入 v = 1520 m/s，t = 0.18 s。d = (1520 × 0.18) / 2 = 273.6 / 2 = 136.8 m。水深约为 137 m。

Echo sounding is a practical application of the wave equation. It is important

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Year 10 WJEC Physics: Case Study Practice in Action | 案例分析实战演练