Year 8 WJEC Physics: Interdisciplinary Comprehensive Question Training | 八年级 WJEC 物理：跨学科综合题型训练

📚 Year 8 WJEC Physics: Interdisciplinary Comprehensive Question Training | 八年级 WJEC 物理：跨学科综合题型训练

Physics does not exist in isolation – it connects deeply with mathematics, biology, geography, technology and even art. This article presents ten carefully designed interdisciplinary questions that mirror the WJEC Year 8 style, helping you build confidence in applying physics concepts to solve real-world problems. Each question is followed by a step-by-step solution designed to strengthen your analytical skills and subject integration.

物理学并非孤立存在——它与数学、生物学、地理、技术甚至艺术紧密相连。本文精心设计了十道跨学科综合题，模仿 WJEC 八年级考试风格，帮助你建立信心运用物理概念解决现实问题。每道题后都附有分步解答，旨在强化你的分析能力和学科融合思维。

---

1. Calculating Wave Speed in a Ripple Tank | 计算波纹槽中的波速（数学）

A student sets up a ripple tank and generates waves with a frequency of 5 Hz. The distance between five consecutive crests is 12 cm. Calculate the wave speed. (Hint: Remember what 'distance between five crests' means for wavelength.)

一名学生设置了一个波纹槽，产生频率为 5 Hz 的波。五个连续波峰之间的距离为 12 cm。计算波速。（提示：记住“五个波峰之间的距离”对波长的含义。）

Step 1: Determine the wavelength. wavelengths (since there are 4 gaps between 5 crests). So 4λ = 12 cm → λ = 3 cm = 0.03 m.

步骤 1：确定波长。 5 个波峰之间的距离等于 4 个波长（因为 5 个波峰之间有 4 个间隔）。所以 4λ = 12 cm → λ = 3 cm = 0.03 m。

Step 2: Apply the wave equation. wave speed v = frequency × wavelength. v = 5 Hz × 0.03 m = 0.15 m/s.

步骤 2：应用波动方程。 波速 v = 频率 × 波长。v = 5 Hz × 0.03 m = 0.15 m/s。

This question blends physics with arithmetic and spatial reasoning, a common cross-curricular skill in WJEC assessments.

此题将物理与算术和空间推理相结合，是 WJEC 评估中常见的跨学科技能。

---

2. Interpreting a Distance-Time Graph of a Cyclist | 解释自行车手的距离-时间图（数学/体育）

A cyclist travels along a straight road. Her journey is described as follows: from 0 to 2 seconds she covers 10 meters; from 2 to 5 seconds she stops; from 5 to 8 seconds she returns to the start at a constant speed. (a) Describe how you would plot the distance-time graph. (b) Calculate her speed during the return journey.

一名自行车手沿直路行驶。她的行程描述如下：第 0 至 2 秒共前进 10 米；第 2 至 5 秒停止；第 5 至 8 秒以恒定速度返回起点。(a) 描述如何绘制距离-时间图。(b) 计算她在返回过程中的速度。

Part (a): The graph has three stages: a sloping line up from (0,0) to (2,10), a flat horizontal line from (2,10) to (5,10), and then a sloping line down from (5,10) to (8,0). This interpretation links motion to graphical representation, a key maths skill.

第 (a) 部分： 图表分为三个阶段：一条从 (0,0) 到 (2,10) 的上斜线，一段从 (2,10) 到 (5,10) 的水平线，然后是一条从 (5,10) 到 (8,0) 的下斜线。这种解读将运动与图形表示联系起来，是一项关键的数学技能。

Part (b) Calculation: ÷ time = 10 ÷ 3 ≈ 3.33 m/s. The negative slope confirms constant speed back to the start.

第 (b) 部分计算： 返回路程距离 = 10 m，时间 = 3 s。速度 = 距离 ÷ 时间 = 10 ÷ 3 ≈ 3.33 m/s。负斜率证实了返回起点的恒定速度。

---

3. Designing a Series Circuit for a Science Game | 设计科学游戏中的串联电路（设计与技术）

Your task is to design a buzzer game: a lamp must light up whenever a main switch is closed, but a buzzer should only sound when an additional 'trigger' switch is pressed. Describe in words how you would connect these components, and draw a description of the circuit.

你的任务是设计一个蜂鸣器游戏：只要闭合一个主开关，一盏灯就必须亮起；但蜂鸣器仅当另一个“触发”开关被按下时才发声。描述你将如何连接这些元件，并画出电路的描述。

Circuit description: Connect the lamp in series with the main switch and the battery – this loop is always complete when the main switch is closed. For the buzzer, connect it in parallel with the lamp, but place the trigger switch in series with the buzzer alone. Therefore, the buzzer only operates when both the main switch is closed (providing power) and the trigger switch is closed.

电路描述： 将电灯与主开关和电池串联——当主开关闭合时，该回路始终接通。对于蜂鸣器，将其与电灯并联，但将触发开关仅与蜂鸣器串联。因此，蜂鸣器仅在主开关闭合（供电）且触发开关闭合时才会工作。

This task integrates physics with design technology and logical thinking, requiring you to apply knowledge of parallel and series circuits in a practical context.

该任务将物理与设计技术和逻辑思维相结合，要求你在实际情境中应用串并联电路的知识。

---

4. Analysing Thermal Energy: Heating a Hot Chocolate | 分析热能：加热一杯热巧克力（化学/营养）

You heat 200 g of milk from 10°C to 60°C using a 1000 W heater running for 60 seconds. The specific heat capacity of milk is approximately 4000 J/kg°C. (a) Calculate the electrical energy supplied by the heater. (b) Calculate the thermal energy absorbed by the milk. (c) Determine the efficiency of the heating process.

你用一台 1000 W 的加热器加热 200 g 牛奶，从 10°C 升温至 60°C，加热时间为 60 秒。牛奶的比热容约为 4000 J/kg°C。(a) 计算加热器提供的电能。(b) 计算牛奶吸收的热能。(c) 确定加热过程的效率。

Part (a): energy supplied = power × time = 1000 W × 60 s = 60 000 J.

第 (a) 部分： 提供的能量 = 功率 × 时间 = 1000 W × 60 s = 60 000 J。

Part (b): bsorbed = mass × specific heat capacity × temperature change = 0.2 × 4000 × 50 = 40 000 J.

第 (b) 部分： 质量单位转换为 kg：200 g = 0.2 kg。温度变化 = 50°C。吸收的能量 = 质量 × 比热容 × 温度变化 = 0.2 × 4000 × 50 = 40 000 J。

Part (c): efficiency = (useful output) ÷ (input) × 100% = 40 000 ÷ 60 000 × 100% ≈ 66.7% nks physics to everyday life and chemistry, emphasising energy transfers and conservation.

第 (c) 部分： 效率 = (有用输出) ÷ (输入) × 100% = 40 000 ÷ 60 000 × 100% ≈ 66.7%。这连接了物理与日常生活和化学，强调了能量转移与守恒。

---

5. Investigating Hooke's Law with a Spring | 探究胡克定律——弹簧实验（数学/数据处理）

A spring stretches 4 cm when a 2 N weight is hung from it. (a) Calculate the spring constant in N/cm. (b) Predict the extension if a 5 N weight is hung, assuming the elastic limit is not exceeded. (c) State a real-world application of Hooke's Law outside of physics class.

一根弹簧悬挂 2 N 的重物时，伸长 4 cm。(a) 计算弹簧常量，单位用 N/cm。(b) 假设未超过弹性极限，预测悬挂 5 N 重物时的伸长量。(c) 说出胡克定律在物理课堂之外的一个实际应用。

Part (a): Hooke's Law: F = kx. k = F ÷ x = 2 N ÷ 4 cm = 0.5 N/cm.

第 (a) 部分： 胡克定律：F = kx。k = F ÷ x = 2 N ÷ 4 cm = 0.5 N/cm。

Part (b): N ÷ 0.5 N/cm = 10 cm. This shows direct proportionality, a core mathematical relationship.

第 (b) 部分： x = F ÷ k = 5 N ÷ 0.5 N/cm = 10 cm。这体现了正比例关系，是核心的数学关系。

Part (c): Applications include spring scales in markets, suspension systems in vehicles, or even the mechanism inside a pen. The question encourages linking physics to engineering and commerce.

第 (c) 部分： 应用包括市场中的弹簧秤、车辆的悬挂系统，甚至是圆珠笔内的机械结构。该问题鼓励将物理学与工程和商业联系起来。

---

广告 / Sponsored

6. Building a Pinhole Camera and Image Formation | 制作针孔相机与成像（艺术/几何）

A pinhole camera has a box length of 20 cm. A 30 cm tall candle is placed 50 cm in front of the pinhole. (a) Calculate the image height. (b) State whether the image is upright or inverted, and explain why. (c) Suggest how the image would change if the pinhole were made larger.

一个针孔相机的盒子长度为 20 cm。一支 30 cm 高的蜡烛放置在针孔前 50 cm 处。(a) 计算像的高度。(b) 说明像是正立还是倒立，并解释原因。(c) 如果针孔变大，像会如何变化？

Part (a): ct distance = 20 cm ÷ 50 cm = 0.4. Image height = m × object height = 0.4 × 30 cm = 12 cm.

第 (a) 部分： 利用相似三角形，放大率 m = 像距 ÷ 物距 = 20 cm ÷ 50 cm = 0.4。像的高度 = m × 物体高度 = 0.4 × 30 cm = 12 cm。

Part (b): The image is inverted because light travels in straight lines, crossing at the pinhole. This geometry is a direct link between physics and art (photography).

第 (b) 部分： 像是倒立的，因为光沿直线传播，在针孔处交叉。这种几何关系直接联系了物理与艺术（摄影）。

Part (c): A larger pinhole would make the image brighter but blurrier, as multiple rays from the same point on the object reach different spots on the screen. This touches on the physics of light and cameras.

第 (c) 部分： 针孔变大会使像更亮但更模糊，因为物体同一点发出的多束光线会到达屏幕的不同位置。这涉及光与相机的物理知识。

---

7. Pressure in Everyday Situations: High Heels vs. Snowshoes | 日常情境中的压强：高跟鞋与雪鞋（时尚/材料科学）

A person weighs 600 N. When wearing stiletto heels, the area of one heel in contact with the ground is 0.5 cm². When wearing snowshoes, the total contact area is 600 cm². (a) Calculate the pressure under a single stiletto heel. (b) Calculate the pressure under the snowshoes. (c) Explain why snowshoes prevent sinking in snow.

一个人重 600 N。穿细高跟鞋时，一个鞋跟与地面的接触面积为 0.5 cm²。穿雪鞋时，总接触面积为 600 cm²。(a) 计算单个细高跟下的压强。(b) 计算雪鞋下的压强。(c) 解释为什么雪鞋能防止陷入雪中。

Part (a): pressure = force ÷ area. The person stands on one heel (assume half body weight or whole? Normally both heels share weight, but question says 'single stiletto heel' 300 N if balanced, but typical trick: use full weight on one heel for dramatic effect. Let' e 600 N / 0.5 cm² to illustrate high pressure. So pressure = 600 N ÷ 0.5 cm² = 1200 N/cm².

第 (a) 部分： 压强 = 力 ÷ 面积。为说明高压强，常见题目假设整个人的重量短暂落在单跟上。因此压强 = 600 N ÷ 0.5 cm² = 1200 N/cm²。

Part (b): Snowshoes: pressure = 600 N ÷ 600 cm² = 1 N/cm². This is dramatically lower.

第 (b) 部分： 雪鞋：压强 = 600 N ÷ 600 cm² = 1 N/cm²。显著降低。

Part (c): Snowshoes spread the force over a large area, reducing pressure on the snow so it does not compact and sink. This combines physics with fashion and material science.

第 (c) 部分： 雪鞋将力分散到较大面积上，减小了对雪地的压强，雪不会压实下沉。这结合了物理与时尚及材料科学。

---

8. How Sound Travels Through Different Materials | 声音如何在不同材料中传播（音乐/工程）

An alarm clock rings inside a sealed jar. As a pupil pumps out the air, the sound becomes fainter. (a) Explain why this happens. (b) Predict what you would hear if the jar were filled with water instead, given that the speed of sound in water is about 1500 m/s and in air is 340 m/s. (c) Relate this to the design of ear protectors for construction workers.

一个闹钟在密封的罐子里响起。当学生抽走空气时，声音逐渐变弱。(a) 解释为什么会这样。(b) 如果罐子里换成水，且水中声速约为 1500 m/s、空气中为 340 m/s，预测你会听到什么。(c) 将此与建筑工人耳罩的设计联系起来。

Part (a): Sound needs a medium to travel; removing air particles leaves fewer particles to vibrate, so the sound energy cannot propagate well. In a near vacuum, almost no sound is heard. This links to biology (hearing) and space science.

第 (a) 部分： 声音需要介质传播；移走空气粒子后，可振动的粒子减少，声能无法良好传播。在接近真空的环境下，几乎听不到声音。这联系到生物学（听觉）和空间科学。

Part (b): With water inside, the sound would be louder and arrive slightly sooner because water couples vibrations more efficiently, even to the outer air. However, if listening from outside the jar, the sound still must transfer from water to glass to air, so it would be clearer and a bit louder than in air alone. This integrates physics with fluid dynamics and everyday experience (swimming pools).

第 (b) 部分： 罐内装水时，声音会更响亮且传播更快，因为水能更高效地耦合振动。但如果从罐外听，声音仍需从水传到玻璃再传到空气，因此会比仅有空气时更清晰、略响。这综合了物理与流体动力学及日常体验（游泳池）。

Part (c): Ear protectors use layers of dense foam and solid materials to absorb and reflect sound, preventing excessive vibrations from reaching the eardrum. This connects physics to engineering and health and safety.

第 (c) 部分： 耳罩采用多层致密泡沫和固体材料来吸收和反射声音，防止过度振动传至鼓膜。这连接了物理与工程及健康安全。

---

9. Renewable Energy: Wind Power and Weather Data | 可再生能源：风能与气象数据（地理）

A wind turbine has a power output of 500 kW when the wind speed is 12 m/s. The power in the wind can be estimated by P_wind swept area = 1000 m². (a) Calculate the power available in the wind. (b) Find the turbine's efficiency. (c) Discuss why the turbine cannot extract 100% of the wind's power.

一台风力涡轮机在风速为 12 m/s 时的输出功率为 500 kW。风中的功率可用 P_风 = ½ × (空气密度) × (扫风面积) × (风速)³ 估算。假设空气密度 = 1.2 kg/m³，扫风面积 = 1000 m²。(a) 计算风中的可用功率。(b) 求涡轮机的效率。(c) 讨论为什么涡轮机不能提取风能功率的 100%。

Part (a): Wind speed cubed = 12³ = 1728 m³/s³

Published by TutorHao | Year 8 Physics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）