📚 Year 9 Cambridge Physics: Case Study Practical Exercises | 案例分析实战演练
Welcome to this practical guide on case study exercises for Year 9 Cambridge Physics. In this article, we will walk through several real‑world scenarios where physics principles are applied, helping you develop problem‑solving skills crucial for your exams and beyond. Each case study presents a situation, highlights the relevant concepts, and demonstrates a step‑by‑step approach to analysis. You will find opportunities to test your understanding with quick practice tasks embedded in every section.
欢迎来到这份 Year 9 剑桥物理案例分析实战演练指南。我们将一起走过多个真实情境，在这些情境中物理原理被灵活应用，从而帮助你培养对考试及未来学习都至关重要的解题能力。每个案例都会给出一种场景，点明相关的物理概念，并展示一步步分析的方法。每个小节中还嵌入了快速练习任务，供你检验自己的理解。
1. Braking Distance of a Car | 汽车刹车距离案例分析
A car travels at 20 m/s on a dry road. The driver spots a hazard and takes 0.7 s to react before pressing the brake. The brakes then provide a constant deceleration of 6 m/s². We need to determine the total stopping distance – the sum of the distance covered during the driver's reaction time and the distance covered while the brakes are applied.
一辆汽车在干燥路面上以 20 m/s 的速度行驶。驾驶员发现危险，在踩下刹车之前用了 0.7 秒的反应时间。刹车随后提供 6 m/s² 的恒定减速度。我们需要求出总停车距离——即驾驶员反应时间内行驶的距离与刹车过程中行驶的距离之和。
The reaction distance is calculated from uniform motion: distance = speed × time. Braking distance comes from the equation of motion v² = u² + 2as, where final velocity v = 0, initial velocity u = 20 m/s, and acceleration a = -6 m/s².
反应距离由匀速运动计算：距离 = 速度 × 时间。刹车距离则来自运动方程 v² = u² + 2as，其中末速度 v = 0，初速度 u = 20 m/s，加速度 a = -6 m/s²。
Let us work through the solution step by step.
我们来一步步解出结果。
- Reaction distance = 20 m/s × 0.7 s = 14 m.
- 反应距离 = 20 m/s × 0.7 s = 14 米。
- For braking: 0² = 20² + 2 × (–6) × s → 0 = 400 – 12s → s = 400/12 ≈ 33.3 m.
- 刹车过程：0² = 20² + 2 × (–6) × s → 0 = 400 – 12s → s = 400/12 ≈ 33.3 米。
- Total stopping distance = 14 m + 33.3 m = 47.3 m.
- 总停车距离 = 14 米 + 33.3 米 = 47.3 米。
This analysis shows that even a short reaction time contributes significantly to the overall stopping distance. The concept is essential for understanding road safety and the physics of motion. As a practice task, recalculate the stopping distance if the same car travels on a wet road where the deceleration drops to 4 m/s² (assume reaction time unchanged).
这个分析表明，即使很短暂的反应时间也会对总停车距离产生显著影响。这一概念对理解道路安全及运动物理学至关重要。作为练习任务，请重新计算同一辆车在湿滑路面行驶、减速度降至 4 m/s² 时的停车距离（假设反应时间不变）。
2. Skier’s Energy Transformation | 滑雪者的能量转换
A skier starts from rest at the top of a frictionless slope 50 m high. We want to find her speed at the bottom using energy conservation. The key idea is that gravitational potential energy (GPE) at the top is transformed entirely into kinetic energy (KE) at the bottom, since no energy is lost to friction.
一位滑雪者从 50 米高的无摩擦斜坡顶端由静止出发。我们想利用能量守恒求出她到达坡底时的速度。核心思路是：顶端的重力势能（GPE）在底部完全转化为动能（KE），因为没有摩擦损耗。
The conservation equation is: mgh = ½mv². Notice that the mass m cancels out. Solving for v gives v = √2gh. With g = 9.8 m/s² and h = 50 m, we obtain the theoretical maximum speed.
能量守恒方程为：mgh = ½mv²。注意质量 m 会约掉。解出 v 得到 v = √2gh。取 g = 9.8 m/s²，h = 50 米，即可算出理论最大速度。
v = √2 × 9.8 × 50 = √980 ≈ 31.3 m/s
In reality, friction and air resistance convert some mechanical energy into thermal energy, so the actual speed will be lower. This case study highlights the power of energy analysis: we do not need to know the slope’s shape or the skier’s mass. Now consider what happens if 20% of the initial GPE is converted to thermal energy – what would the speed be?
现实中，摩擦和空气阻力会将部分机械能转化为热能，因此实际速度会更低。这个案例突显了能量分析的优势：我们不需要知道斜坡形状或滑雪者的质量。现在思考：如果 20% 的初始重力势能转化为热能，速度会变成多少？
3. Hydraulic Lift: Pascal’s Principle | 液压升降机：帕斯卡原理
A hydraulic lift has a small piston of area 2 cm² and a large piston of area 200 cm². A force of 100 N is applied to the small piston. Pascal's principle tells us that the pressure applied to an enclosed fluid is transmitted undiminished throughout the fluid. Hence, the pressure on both pistons is equal.
一台液压升降机的小活塞面积为 2 cm²，大活塞面积为 200 cm²。在小活塞上施加 100 N 的力。帕斯卡原理告诉我们：施加在封闭流体上的压强会大小不变地向流体各处传递。因此，两个活塞上的压强相等。
We calculate the pressure at the small piston: p = F₁ / A₁ = 100 N / 2 cm² = 50 N/cm². The same pressure acts on the large piston, so the output force is F₂ = p × A₂ = 50 N/cm² × 200 cm² = 10 000 N. This shows how a small input force can lift a heavy load.
我们计算小活塞上的压强：p = F₁ / A₁ = 100 N / 2 cm² = 50 N/cm²。相同的压强作用在大活塞上，因此输出力为 F₂ = p × A₂ = 50 N/cm² × 200 cm² = 10 000 N。这展示了一个小的输入力如何举起重物。
What if the small piston moves 10 cm downward? By volume conservation, the fluid displaced from the small cylinder equals the fluid gained by the large cylinder: A₁d₁ = A₂d₂ ) / 200 cm² = 0.1 cm. The large piston rises only a tiny distance, but with enormous force – a classic force–distance trade‑off.
如果小活塞向下移动 10 厘米会怎样？根据体积守恒，从小液压缸排出的液体体积等于大液压缸获得的体积：A₁d₁ = A₂d₂。因此 d₂ = (2 cm² × 10 cm) / 200 cm² = 0.1 cm。大活塞只上升极小的距离，但获得了巨大的力——这是经典的力–距离权衡。
4. Faulty Circuit Detective | 故障电路侦探
Imagine a simple series circuit containing a battery, a switch, and two identical bulbs. When the switch is closed, one bulb glows brightly but the other remains completely dark. As a physics detective, you suspect either a broken filament in the dark bulb or a loose connection in the wire leading to it. A voltmeter is your tool to locate the fault.
设想一个简单的串联电路，包含电池、开关和两只相同的灯泡。当开关闭合时，一只灯泡明亮发光，另一只却完全黑暗。作为物理侦探，你怀疑要么是暗灯泡的灯丝断了，要么是通向它的导线有松脱。万用表（电压档）就是你定位故障的工具。
Connect the voltmeter across various points: across the battery shows normal voltage (e.g. 6 V). Across the glowing bulb reads nearly 0 V, because it has very low resistance when other parts are fine. Across the dark bulb, however, the voltmeter reads 6 V – the full battery voltage. This indicates an open circuit inside that bulb: the broken filament creates an infinite resistance, so all the voltage drops across the gap, and no current flows.
将电压表接在不同测点：接在电池两端显示正常电压（如 6 V）。并联在亮灯两端读数接近 0 V，因为其他部分正常时它本身电阻很小。然而，并联在暗灯两端时，电压表读数为 6 V——全电池电压。这表明该灯泡内部断路：断裂的灯丝造成无限大电阻，全部电压降在断开处，电路中没有电流。
If the voltmeter had read 0 V across the dark bulb and the battery voltage appeared across a section of wire, the fault would be a broken wire. Use this logical approach to diagnose circuits. Challenge: draw the circuit and indicate where you would place the voltmeter to confirm a short circuit in one bulb.
如果电压表在暗灯两端读数为 0 V，而在一段导线上出现电池电压，那么故障就是导线断了。用这种逻辑方法去诊断电路。挑战：画出电路并标出为确认某只灯泡短路而应放置电压表的位置。
5. Submarine Buoyancy | 潜水艇的浮力
A submarine controls its vertical position by adjusting its average density. When the ballast tanks are filled with water, the submarine’s overall density becomes greater than that of the surrounding seawater, and it sinks. To rise, compressed air forces water out of the tanks, lowering the average density until it is less than the seawater’s density.
潜水艇通过调节自身平均密度来控制垂直位置。当压载水箱注满水时，潜水艇的整体密度大于周围海水密度，它便下潜。要上浮时，压缩空气将水排出水箱，使平均密度降低，直至小于海水密度。
Archimedes' principle states that the upthrust equals the weight of fluid displaced. If a submarine has a volume of 200 m³ and the seawater density is 1025 kg/m³, the maximum upthrust is upthrust = ρ × V × g = 1025 × 200 × 9.8 ≈ 2 009 000 N 00 × 9.8 = 1 764 000 N. The weight is less than upthrust, so the submarine would rise unless it takes on more water.
阿基米德原理指出，浮力等于排开液体的重量。若潜水艇体积为 200 m³，海水密度为 1025 kg/m³，最大浮力为 浮力 = ρ × V × g = 1025 × 200 × 9.8 ≈ 2 009 000 N。要使潜水艇上浮，其总重量必须小于此浮力。若总质量（含压载水）为 180 000 kg，重量 = 180 000 × 9.8 = 1 764 000 N。重量小于浮力，因此潜水艇会上浮，除非再注入更多水。
This case study links density, pressure, and forces. You can explore what mass of water must be taken in to achieve neutral buoyancy (weight exactly equals upthrust). Calculate the required total mass and compare it with the mass of the empty submarine.
本案例将密度、压强和力联系在一起。你可以探索必须吸入多少质量的水才能实现中性浮力（重量恰好等于浮力）。计算出所需的总质量，并与空艇质量作比较。
6. Keeping Drinks Cool: Thermal Insulation | 饮料保温：热绝缘
A vacuum flask keeps a cold drink chilled for hours by minimising heat transfer. The double‑wall construction with a vacuum between the walls virtually eliminates conduction and convection. Additionally, the inner surfaces are silvered to reflect infrared radiation, reducing heat gain from the surroundings.
保温瓶通过最大限度地减少热传递来让冷饮保持冰凉数小时。双层壁结构以及壁间的真空几乎消除了传导和对流。此外，内壁镀银可以反射红外辐射，减少从周围环境吸收的热量。
In a simpler demonstration, you can wrap a chilled can in a layer of foam insulation 1 cm thick. The rate of conductive heat transfer through the foam is given by Q/t = k A (T₂ – T₁) / d, where k is the thermal conductivity, A is surface area, (T₂ – T₁) is temperature difference, and d is thickness. Foam has a low k (~0.03 W/m·K), so even a modest thickness greatly slows warming. By calculating the time for the drink to rise by 5 °C, students appreciate the role of insulating materials.
在一个更简单的演示中，你可以用 1 厘米厚的泡沫隔热层包裹一罐冷饮。通过泡沫的导热速率由 Q/t = k A (T₂ – T₁) / d 给出，其中 k 是导热系数，A 是表面积，(T₂ – T₁) 是温差，d 是厚度。泡沫的 k 值很低（约 0.03 W/m·K），因此即使厚度不大，也能大大减缓升温。通过计算饮料升高 5 °C 所需的时间，学生可以体会隔热材料的作用。
Thermal case studies highlight the everyday importance of conduction, convection and radiation. For your own investigation, design an experiment to compare the effectiveness of different insulating layers – aluminium foil, cotton wool, and bubble wrap – using identical containers of warm water.
热学案例突显了传导、对流和辐射在日常生活中的重要性。作为自主探究，你可以设计一个实验，使用装有等量温水的相同容器，比较铝箔、棉花和气泡膜等不同隔热层的效果。
7. Echo Sounding in Bats | 蝙蝠的回声定位
Bats navigate and hunt insects by emitting high‑frequency sound pulses and listening for the echoes. Suppose a bat sends out an ultrasonic pulse that reflects off a moth and returns after 0.2 seconds. Since the sound wave travels from the bat to the moth and back, the total distance travelled is twice the distance to the object.
蝙蝠通过发射高频声脉冲并倾听回声来导航和捕食昆虫。假设一只蝙蝠发出一个超声波脉冲，该脉冲碰到飞蛾后于 0.2 秒后返回。由于声波从蝙蝠传播到飞蛾再返回，其总路程为物体距离的两倍。
With the speed of sound in air as 340 m/s, the distance to the moth is: d = (v × t) / 2 = (340 m/s × 0.2 s) / 2 = 34 m. This is exactly how radar and sonar work, using the round‑trip time of waves to measure distances.
取空气中声速为 340 m/s，则飞蛾的距离为：d = (v × t) / 2 = (340 m/s × 0.2 s) / 2 = 34 米。这正是雷达和声呐的工作原理——利用波往返的时间来测量距离。
The frequency of the emitted ultrasound is around 40 kHz to 100 kHz, well above human hearing range. Bats can not only detect distance but also relative motion via the Doppler effect. Challenge question: if the echo frequency is slightly higher than the emitted frequency, is the moth moving toward or away from the bat? Explain.
发射的超声波频率约在 40 kHz 至 100 kHz，远高于人耳听力范围。蝙蝠不仅能探测距离，还能通过多普勒效应感知相对运动。挑战性问题：如果回声频率略高于发射频率，飞蛾是在接近还是远离蝙蝠？请加以解释。
8. Balancing a Seesaw: Moments | 跷跷板的平衡：力矩
equal the total anticlockwise moment. Moment = force × perpendicular distance from pivot.
跷跷板是力矩原理的经典应用。要使均质跷跷板平衡，以支点为参考，总顺时针力矩必须等于总逆时针力矩。力矩 = 力 × 到支点的垂直距离。
pivot on the left. Where should a 60 kg adult sit on the right side to balance the seesaw? = 60 × g × d. Equating: 40 × 1.5 = 60 × d, so d = (40 × 1.5) / 60 = 1.0 m from the pivot.
考虑一架跷跷板：一个 40 kg 的小孩坐在支点左侧 1.5 m 处。一个 60 kg 的成人应坐在右侧什么位置才能使跷跷板平衡？忽略杠杆自重，逆时针力矩 = 40 × g × 1.5，顺时针力矩 = 60 × g × d。令其相等：40 × 1.5 = 60 × d，得 d = (40 × 1.5) / 60 = 1.0 米，即距支点 1.0 米处。
If the seesaw itself has mass 20 kg and its centre of gravity is exactly at the pivot, it adds no net moment. However, if the pivot is shifted slightly from the centre of the board, the board’s weight creates an additional moment that must be included. Find the new balance point when a uniform board of length 4 m and mass 20 kg has the pivot at 1.8 m from one end, with the same two people placed appropriately. This kind of calculation develops deeper understanding of rotational equilibrium.
若跷跷板本身质量为 20 kg，且其重心恰好在支点处，则不会产生净力矩。但是，如果支点略微偏离板子中心，板子的重量就会产生一个必须纳入的附加力矩。当一根长 4 m、质量 20 kg 的均质木板的支点设在距一端 1.8 m 处，且两人物布置相当时，请找出新的平衡位置。这类计算能加深你对转动平衡的理解。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图