📚 Year 9 WJEC Physics: Case Study Practical Drills | Year 9 WJEC 物理：案例分析实战演练
In Year 9 WJEC Physics, real-world problem solving is not just an add-on — it is the heart of how you learn to think like a scientist. Case studies pull together forces, energy, electricity and waves and ask you to apply them in unfamiliar situations. This article walks through carefully chosen practical drills designed to sharpen your analytical skills and prepare you for the style of questions that appear in WJEC assessments.
在 Year 9 WJEC 物理中，解决实际问题不只是附加内容——它是学习像科学家一样思考的核心。案例研究将力、能量、电学与波动整合起来，要求你将它们应用于陌生情境。本文精选了一系列实战演练，旨在磨练你的分析能力，帮助你熟悉 WJEC 考试中常见的问题风格。
1. Braking Distance Investigation | 刹车距离探究
Imagine a car travelling at 12 m/s on a dry road. The driver sees a hazard and applies the brakes. The graph of velocity against time shows a straight line dropping to zero in 3 seconds. Use the graph to calculate the braking distance.
设想一辆汽车在干燥路面上以 12 m/s 的速度行驶。驾驶员发现危险并踩下刹车。速度‑时间图显示一条直线在 3 秒内降至零。利用该图计算刹车距离。
The area under a velocity– ou should also discuss how wet roads or worn tyres would reduce the deceleration and increase stopping distance beyond thinking distance plus braking distance.
速度‑时间图下方的面积代表位移。因为线是直线，面积是一个三角形：½ × 底 × 高 = ½ × 3 秒 × 12 米/秒 = 18 米。这就是刹车距离。你还应讨论潮湿路面或磨损轮胎会如何减小减速度，并使总停车距离超过反应距离与刹车距离之和。
2. Energy Transfers in a Roller Coaster | 过山车中的能量转移
timate its speed at the bottom of the hill using conservation of energy. Take g = 10 N/kg.
一辆 500 kg 的过山车被拉到 30 米高的坡顶。在顶部速度极低。忽略摩擦，利用能量守恒估算它到达底部时的速度。取 g = 10 N/kg。
250 v² = 150 000 → v² = 600 → v = √600 ≈ 24.5 m/s. In reality, some energy is transferred as heat due to friction, so the speed would be slightly lower.
顶部的重力势能 = mgh = 500 × 10 × 30 = 150 000 J。在底部，这些能量全部转化为动能：½mv²。因此 ½ × 500 × v² = 150 000 → 250 v² = 150 000 → v² = 600 → v = √600 ≈ 24.5 m/s。实际上，由于摩擦会有部分能量转化为热能，因此速度会略低。
3. Electrical Fault in a Parallel Circuit | 并联电路中的故障
Two identical bulbs are connected in parallel to a 6 V battery. Each bulb has a resistance of 12 Ω. Suddenly, one bulb goes out but the other stays lit. The ammeter reading changes from 1.0 A to 0.5 A. Explain what happened and support your answer with calculations.
两个相同的灯泡并联接到 6 V 电池上。每个灯泡的电阻为 12 Ω。突然，一个灯泡熄灭而另一个仍亮着。电流表读数从 1.0 A 变为 0.5 A。解释发生了什么并用计算佐证。
circuit has only one 12 Ω bulb. Current = 6/12 = 0.5 A. The drop in current matches the reading. The most likely fault is a broken filament in the blown bulb, which creates an open circuit in that branch, so all current flows through the working bulb.
两个灯泡正常时，并联总电阻 = (12 × 12)/(12 + 12) = 6 Ω。电池输出电流 = V/R = 6/6 = 1 A。当一个灯泡损坏后，电路中只剩一个 12 Ω 灯泡。电流 = 6/12 = 0.5 A。电流下降与读数吻合。最可能的故障是坏灯泡灯丝断裂，在该支路形成断路，因此所有电流都流过完好的灯泡。
4. Speed of Sound Using Echoes | 利用回声测声速
A student stands 85 m from a large wall and claps two wooden blocks together. She hears the echo 0.5 s later. Calculate the speed of sound in air. Discuss why the experiment might give a slightly different value on a cold day.
一名学生站在离一堵大墙 85 米远处，拍击两木块。她 0.5 秒后听到回声。计算空气中声音的传播速度，并讨论为何在寒冷天气下实验值会略有不同。
cold day, air particles move more slowly, so sound travels slower; the measured speed would be lower than 340 m/s. The student could improve accuracy by timing multiple claps (e.g., 10 echoes) and averaging.
声音传播到墙壁再返回，因此距离 = 2 × 85 = 170 m。速度 = 距离/时间 = 170 m / 0.5 s = 340 m/s。在寒冷天气下，空气粒子运动较慢，声音传播更慢；测得的声速会低于 340 m/s。学生可以通过对多次拍击（例如 10 次回声）计时并取平均值来提高精度。
5. Forces on a Skydiver: Terminal Velocity | 跳伞者的受力：终端速度
ydiver jumps from a plane. After a few seconds, she reaches terminal velocity. Draw a free‑ rating. Calculate the size of the air resistance force at terminal velocity. (g = 10 N/kg)
一名 70 kg 的跳伞者从飞机跳下。几秒后，她达到终端速度。画出受力示意图，解释为什么跳伞者停止加速，并计算她在终端速度时所受空气阻力的大小。(g = 10 N/kg)
This means air resistance (drag) must equal weight, i.e., 700 N upward. The two forces are balanced. As she opens her parachute, surface area increases dramatically, drag becomes larger than weight, causing deceleration until a new, lower terminal velocity is reached.
重力 = mg = 70 × 10 = 700 N 向下。在终端速度时，跳伞者匀速下落，因此合力为零。这意味着空气阻力必须等于重力，即 700 N 向上。两个力平衡。当她打开降落伞时，表面积急剧增大，阻力超过重力，导致减速，直到达到一个新的、更低的终端速度。
6. Density of an Irregular Object | 不规则物体的密度
An irregular stone has a mass of 180 g. A measuring cylinder contains 50 cm³ of water. When the stone is fully submerged, the water level rises to 125 cm³. Calculate the density of the stone and state whether it will float or sink in water.
一块不规则石块的质量为 180 g。量筒中装有 50 cm³ 水。当石块完全浸没后，水面升至 125 cm³。计算石块的密度，并说明它在水中会漂浮还是下沉。
Volume of stone = 125 – cement method works for any object that does not dissolve in water. To improve reliability, repeat and take the mean.
石块体积 = 125 – 50 = 75 cm³。密度 = 质量/体积 = 180 g / 75 cm³ = 2.4 g/cm³。水的密度为 1.0 g/cm³，石块密度更大，因此会下沉。这种排水法适用于任何不溶于水的物体。为提高可靠性，可重复测量并取平均值。
7. Moments and Equilibrium: Seesaw Problem | 力矩与平衡：跷跷板问题
A uniform plank of length 4 m is pivoted at its centre. A 300 N child sits 1.5 m to the left of the pivot. Where must a 500 N adult sit on the right side to balance the seesaw? Check your answer by calculating the total clockwise and anticlockwise moments.
一根 4 m 长的均匀木板在其中点支起。一个重 300 N 的孩子坐在支点左侧 1.5 m 处。一个重 500 N 的成人应坐在右侧何处才能使跷跷板平衡？通过计算总顺时针力矩和逆时针力矩验证答案。
he right. Check: clockwise moment = 500 × 0.9 = 450 Nm. They are equal, so the seesaw is in equilibrium. If the pivot were not at the centre, you would also need to account for the plank's weight.
逆时针力矩 = 300 N × 1.5 m = 450 Nm。为平衡，顺时针力矩也需为 450 Nm。因此成人与支点的距离 = 450 Nm / 500 N = 0.9 m，位于右侧。验证：顺时针力矩 = 500 × 0.9 = 450 Nm。两者相等，跷跷板平衡。若支点不在中点，还需考虑木板自身的重量所产生的力矩。
8. Resistance of a Wire: Length and Hazards | 导线电阻：长度与危险
A student investigates how the length of a nichrome wire affects its resistance. Data shows that a 50 cm wire has a resistance of 4 Ω. Predict the resistance of a 150 cm wire of the same material and thickness. Explain why using a very long extension lead without unwinding it can cause a fire risk.
一名学生研究镍铬合金线长度对电阻的影响。数据显示 50 cm 长的导线电阻为 4 Ω。预测相同材料、相同粗细下 150 cm 长导线的电阻，并解释为何不展开卷线盘而使用很长的延长线会有火灾风险。
4 Ω = 12 Ω. A coiled extension lead carries current, but the coil acts like a huge single‑ are bunched together. This causes overheating, which may melt insulation and start a fire.
电阻与长度成正比。因此 150 cm / 50 cm = 3 倍长度 → 电阻 = 3 × 4 Ω = 12 Ω。卷着的延长线通电时，线圈相当于一个巨大的单匝电磁铁，且导线因缠绕在一起而难以散热。这会导致过热，可能熔化绝缘层并引发火灾。
9. Conduction, Convection and Radiation in a House | 房屋中的传导、对流与辐射
Analyse how energy is lost from a typical UK home in winter. Focus on a single‑glazed window, a loft without insulation, and a radiator. For each, name the main heat transfer process and suggest one improvement to reduce energy waste.
分析冬季典型英国家居的能量散失。聚焦单层玻璃窗户、无隔热层的阁楼以及一个暖气片。分别指出主要的传热方式，并提出一项减少能量浪费的改进措施。
Single‑glazed window: heat escapes mainly by conduction through the glass and air, and also by radiation. Improvement: double glazing with an insulating gas layer. Loft: warm air rises by convection and then conducts through the ceiling. Improvement: laying thick fibreglass insulation to trap air. Radiator: transfers heat to the room mainly by convection (air currents) and radiation. Improvement: fit reflective foil behind radiators to reflect radiant heat back into the room.
单层玻璃窗：热量主要通过玻璃及空气的传导散失，也有辐射。改进：使用带有隔热气体层的双层玻璃。阁楼：暖空气上升通过对流，然后通过天花板传导出去。改进：铺设厚玻璃纤维保温层以锁住空气。暖气片：主要通过空气对流和辐射向房间传递热量。改进：在暖气片后安装反射箔，将辐射热反射回室内。
10. Interpreting Motion Graphs from a Cycle Ride | 解读骑行中的运动图像
A velocity–time graph for a cyclist shows: 0–10 s, acceleration to 8 m/s; 10–50 s, constant speed; 50–70 s, deceleration to rest. Find the total distance travelled and describe the cyclist’s motion in terms of forces.
一名自行车骑手的速度‑时间图显示：0–10 秒匀加速至 8 m/s；10–50 秒匀速；50–70 秒减速至静止。求总行驶距离，并从力的角度描述骑手的运动。
gle = ½ × 20 × 8 = 80 m. Total = 40 + 320 + 80 = 440 m. During acceleration, driving force > friction. At constant speed, forces balanced (driving force = friction). During deceleration, friction (and braking) provide a net backward force.
距离 = 图下方面积。加速阶段：三角形面积 = ½ × 10 × 8 = 40 m。匀速：矩形 = 40 s × 8 m/s = 320 m。减速：三角形 = ½ × 20 × 8 = 80 m。总距离 = 40 + 320 + 80 = 440 m。加速时，驱动力 > 摩擦力。匀速时力平衡（驱动力 = 摩擦力）。减速时，摩擦力（及刹车）产生向后的合力。
Published by TutorHao | Physics Revision Series | aleveler.com
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