📚 Year 9 Edexcel Physics: Case Study Practical Exercises | Year 9 Edexcel 物理：案例分析实战演练
Case studies in Year 9 Edexcel Physics sharpen your ability to apply scientific principles to real-world situations. This article presents a collection of practical exercises, each designed to test your understanding of key topics – from motion and energy to electricity and waves. Work through the scenarios, check the step-by-step analysis, and strengthen your problem-solving skills.
Year 9 Edexcel 物理中的案例分析能锻炼你将科学原理应用于真实情境的能力。本文汇编了一系列实战练习，每个练习都旨在检验你对运动、能量、电学、波动等重点专题的理解。跟着情景走，查看逐步分析，夯实你的解题技能。
1. Case Study 1: The Moving Car | 案例一：运动的汽车
A car starts from rest and accelerates uniformly to a velocity of 20 m/s in 10 seconds. The driver then maintains this speed for a further 15 seconds. Determine the acceleration during the first phase and the total distance travelled.
一辆汽车从静止开始匀加速，10秒后速度达到20 m/s。之后司机保持这个速度继续行驶15秒。求第一阶段的加速度以及汽车行驶的总距离。
= (v - u) ÷ t. Since the car starts from rest, u = 0 m/s, v = 20 m/s, t = 10 s. Therefore:
计算加速度，用公式 a = (v - u) ÷ t。汽车从静止起步，u = 0 m/s，v = 20 m/s，t = 10 s。因此：
a = (20 - 0) ÷ 10 = 2 m/s²
peed phase, distance = speed × time = 20 × 15 = 300 m. Total distance = 100 + 300 = 400 m.
加速阶段的行驶距离用 s = u t + ½ a t² 计算。代入得 s₁ = 0 × 10 + ½ × 2 × 10² = 100 m。匀速阶段，距离 = 速度 × 时间 = 20 × 15 = 300 m。总距离 = 100 + 300 = 400 m。
2. Case Study 2: Energy in a Roller Coaster | 案例二：过山车中的能量
l potential energy at the top and the maximum speed the car could reach at the bottom. (Take g = 10 N/kg)
一辆质量为500 kg的过山车被拉到第一座小山的顶部，离地40 m。假设没有能量损失，计算车在顶部的重力势能以及到达底部时可能达到的最大速度。（取 g = 10 N/kg）
00 000 J. At the bottom, all GPE is converted into kinetic energy (KE), so KE = 200 000 J.
重力势能 (GPE) = m g h。代入：GPE = 500 × 10 × 40 = 200 000 J。到达底部时，所有重力势能转化为动能 (KE)，所以 KE = 200 000 J。
KE = ½ m v² → 200 000 = ½ × 500 × v² → v² = 800 → v = √800 ≈ 28.3 m/s
The maximum speed at the bottom is therefore about 28.3 m/s. In practice, friction and air resistance would reduce this value.
因此底部最大速度约为28.3 m/s。实际中，摩擦和空气阻力会使这一数值减小。
3. Case Study 3: Sound Waves in a Concert Hall | 案例三：音乐厅中的声波
A sound wave with a frequency of 440 Hz travels through air at a speed of 340 m/s. The wave reflects off a wall 17 m away from the source. Calculate the wavelength and the time delay before the echo is heard.
一个频率为440 Hz的声波在空气中以340 m/s的速度传播。声波从距离声源17 m的墙壁反射回来。计算波长以及听到回声的延迟时间。
440 ≈ 0.773 m. The sound travels to the wall and back, a total distance of 2 × 17 = 34 m.
波速公式：v = f λ。变形得 λ = v ÷ f = 340 ÷ 440 ≈ 0.773 m。声音传播到墙壁再返回，总距离为 2 × 17 = 34 m。
time = distance ÷ speed = 34 ÷ 340 = 0.1 s
Thus the echo is heard 0.1 seconds after the original sound. This brief gap explains why echoes are distinct only when the reflecting surface is far enough away.
因此回声在原始声音之后0.1秒听到。这一短暂的间隔解释了为什么只有当反射面足够远时回声才清晰可辨。
4. Case Study 4: Electric Circuits at Home | 案例四：家庭电路
A 12 V battery is connected to two resistors in series: a 4 Ω resistor and an 8 Ω resistor. Calculate the total resistance, the current flowing, and the potential difference across each resistor.
一个12 V电池与两个电阻串联：一个4 Ω电阻和一个8 Ω电阻。计算总电阻、流过的电流以及每个电阻两端的电势差。
In series, total resistance Rtotal = R₁ + R₂ = 4 + 8 = 12 Ω. Using Ohm's law, I = V ÷ Rtotal = 12 ÷ 12 = 1 A.
串联电路中，总电阻 Rtotal = R₁ + R₂ = 4 + 8 = 12 Ω。用欧姆定律，I = V ÷ Rtotal = 12 ÷ 12 = 1 A。
8 Ω resistor: V₂ = I × R₂ = 1 × 8 = 8 V. Notice that 4 V + 8 V = 12 V, the supply voltage.
4 Ω电阻两端的电势差：V₁ = I × R₁ = 1 × 4 = 4 V。8 Ω电阻两端的电势差：V₂ = I × R₂ = 1 × 8 = 8 V。注意到 4 V + 8 V = 12 V，正是电源电压。
5. Case Study 5: Magnetic Forces and Compasses | 案例五：磁力与指南针
A bar magnet is placed beneath a sheet of paper, and iron filings are sprinkled on top. The filings form a pattern of lines that curve from the north pole to the south pole. Explain why this happens and predict the direction a plotting compass would point at three different positions.
将一根条形磁铁放在一张纸下面，撒上铁屑。铁屑形成从北极指向南极的曲线图案。解释原因，并预测在三个不同位置处指向罗盘会指向什么方向。
Each iron filing becomes a tiny temporary magnet in the magnetic field, aligning with the field lines. The lines emerge from the north pole and enter the south pole. A plotting compass placed near the north pole will point away from it; placed at the side, it will align tangentially to the field line; placed near the south pole, it will point towards it.
每一粒铁屑在磁场中变成了微小的临时磁体，顺着磁场线排列。磁感线从北极发出，进入南极。放在北极附近的指向罗盘会背离北极；放在侧面时，会沿磁感线切线方向；放在南极附近时，会指向南极。
- Above the north pole: compass needle's north pole points away from the magnet.
- Mid-way: needle lines up parallel to the bar magnet, north seeking pole pointing towards the south pole of the magnet.
- Above the south pole: needle's north pole points directly towards the magnet's south pole.
- 北极上方：罗盘的北极背离磁铁。
- 中间位置：罗盘与条形磁铁平行，指北极指向磁铁的南极。
- 南极上方：罗盘的北极直接指向磁铁的南极。
6. Case Study 6: Renewable Energy Project | 案例六：可再生能源项目
A small island requires 500 kWh of electricity per day. It plans to install solar panels that produce 150 W per panel on average during daylight hours (8 hours). How many panels are needed to meet the daily demand, and what is one major advantage and one disadvantage of this energy source?
一个小岛每天需要500 kWh的电力。计划安装太阳能电池板，每块在白天（8小时）平均产生150 W。需要多少块电池板才能满足日常需求？这种能源的一个主要优点和一个缺点是什么？
er day. Number of panels = 500 000 ÷ 1200 ≈ 416.7, so 417 panels are needed (rounding up).
首先将日需求转换为瓦时：500 kWh = 500 000 Wh。每块电池板每日供能 150 W × 8 h = 1200 Wh。所需板数 = 500 000 ÷ 1200 ≈ 416.7，因此需要417块（向上取整）。
Advantage: Solar energy is renewable and produces no greenhouse gases during operation. Disadvantage: Output depends on weather and daylight, so energy storage or backup is required for nights and cloudy periods.
优点：太阳能是可再生能源，运行时不产生温室气体。缺点：发电量取决于天气和日照，因此夜间和阴天需要储能或备用电源。
7. Case Study 7: Cooling Curve of Water | 案例七：水的冷却曲线
A beaker of hot water at 80 °C is left to cool in a room at 20 °C. The temperature is recorded every minute. After about 15 minutes, the temperature reaches 30 °C. Describe the shape of the cooling curve and explain the rate of heat transfer.
一杯80 °C的热水在20 °C的房间中自然冷却。每分钟记录一次温度。大约15分钟后温度降至30 °C。描述冷却曲线的形状并解释传热速率。
The cooling curve is not linear. Initially, the temperature drops rapidly because the temperature difference between the water and the room is large, leading to a high rate of heat transfer. As the water cools, the temperature difference decreases, so the rate of cooling slows down. The curve flattens as it approaches room temperature.
冷却曲线不是线性的。起初温度下降很快，因为水与室温的温差大，传热速率高。随着水温下降，温差减小，冷却速率变慢。曲线在接近室温时变得平缓。
Newton's law of cooling states that the rate of heat loss is proportional to the temperature difference between the object and its surroundings. This explains the exponential-like decay observed.
牛顿冷却定律指出，热量散失的速率与物体和环境的温差成正比。这就解释了观察到的类似指数衰减的趋势。
8. Case Study 8: Pressure in Liquids | 案例八：液体压强
econd tank holds mercury (density 13 600 kg/m³) to the same depth, how does the pressure compare?
一个圆柱形水箱水深2.5 m。计算水在底部产生的压强。（水密度 = 1000 kg/m³，g = 10 N/kg）。如果另一个水箱装有相同深度的水银（密度13 600 kg/m³），压强会怎样变化？
Pressure p = ρ g h
reater due to the higher density, illustrating why a small depth of mercury can balance a tall column of water in a barometer.
对于水：p = 1000 × 10 × 2.5 = 25 000 Pa（即25 kPa）。对于水银：p = 13 600 × 10 × 2.5 = 340 000 Pa。由于密度更大，压强显著增加，这解释了为什么气压计中一小段水银可以平衡较高的水柱。
9. Case Study 9: Moments and Levers | 案例九：力矩与杠杆
A uniform plank 4 m long is pivoted at its centre. A 30 N weight is placed 1.2 m to the left of the pivot. Where must a 20 N weight be placed on the right-hand side to balance the plank?
一块均匀的长木板长4 m，在中心处支起。一个30 N的重物放在支点左侧1.2 m处。需在支点右侧何处放置一个20 N的重物才能使木板平衡？
For rotational equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments. Taking moments about the pivot:
为使转动平衡，顺时针力矩之和等于逆时针力矩之和。以支点为轴取力矩：
Anticlockwise moment = 30 N × 1.2 m = 36 N m
e d. Clockwise moment = 20 N × d. Set 20 d = 36, so d = 1.8 m from the pivot on the right.
设右侧20 N重物距支点为 d。顺时针力矩 = 20 N × d。令 20 d = 36，得 d = 1.8 m，放在支点右侧1.8 m处。
The plank itself is uniform and pivoted at its centre, so its weight does not produce a net moment.
木板本身均匀且支在中心，因此其重力不产生净力矩。
10. Case Study 10: Light and Reflection | 案例十：光的反射
A ray of light strikes a plane mirror at an angle of incidence of 35°. Draw a labelled diagram showing the incident ray, reflected ray, normal, and angles. State the law of reflection and calculate the angle between the incident ray and the reflected ray.
一束光线以35°的入射角射向平面镜。画出标注入射光线、反射光线、法线和角度的示意图。陈述反射定律，并计算入射光线与反射光线之间的夹角。
from the normal. Here the angle of incidence i = 35°, so the angle of reflection r = 35°.
反射定律：入射角等于反射角，两者都从法线量起。此处入射角 i = 35°，因此反射角 r = 35°。
d show the normal as a dashed line perpendicular to the mirror, with the two rays symmetrical around it.
入射光线与反射光线之间的夹角是法线同一侧两角之和，i + r = 35° + 35° = 70°。图中应用虚线表示垂直于镜面的法线，两条光线对称分布在法线两侧。
Published by TutorHao | Physics Revision Series | aleveler.com
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