📚 Year 10 Cambridge Physics: Case Study Practice | 剑桥10年级物理：案例分析实战演练
Case studies are an excellent way to connect physics theory to real-world situations. In Year 10 Cambridge Physics, you are expected to apply equations, interpret data, and explain phenomena using scientific reasoning. This article presents eight carefully selected case studies covering key topics: motion, waves, electricity, energy, pressure, radioactivity, optics, and momentum. Each case includes a scenario, analysis, and solution steps — just like in an exam. By working through these examples, you will build the confidence and skills needed to tackle unfamiliar problems.
案例分析是将物理理论与现实情境联系起来的极佳方式。在剑桥10年级物理中，你需要运用公式、解读数据，并用科学推理来解释现象。本文精选了八个案例，涵盖运动、波、电学、能量、压强、放射现象、光学和动量等关键主题。每个案例都包含情境、分析与解题步骤，就像考试一样。通过这些练习，你将培养自信与技能，从容应对陌生题型。
1. Case Study 1: Stopping Distance of a Car | 案例一：汽车的刹车距离
A driver travelling at 20 m/s sees a hazard and applies the brakes. The car decelerates uniformly at 5 m/s². The driver's reaction time is 0.8 s. Determine the total stopping distance. This problem separates stopping distance into thinking distance and braking distance. The thinking distance is the distance travelled during the reaction time at constant speed. Braking distance is the distance covered while decelerating to rest.
一辆汽车以 20 m/s 的速度行驶，司机发现危险后踩下刹车。汽车以 5 m/s² 的加速度均匀减速。司机的反应时间为 0.8 s。求总刹车距离。该问题将刹车距离分为反应距离和制动距离。反应距离是在反应时间内以恒定速度行驶的距离。制动距离则是从开始减速到停止所经过的距离。
10s → s = 40 m. Total stopping distance = 16 + 40 = 56 m. This case highlights the importance of reaction time and deceleration on road safety.
反应距离 = 速度 × 反应时间 = 20 × 0.8 = 16 m。制动距离运用公式 v² = u² + 2as，其中 v = 0, u = 20 m/s, a = -5 m/s²。代入得 0 = (20)² + 2(-5)s → 0 = 400 - 10s → s = 40 m。总刹车距离 = 16 + 40 = 56 m。此案例凸显了反应时间和减速度对道路安全的重要性。
2. Case Study 2: Echo Sounding to Measure Depth | 案例二：利用回声测量深度
A ship sends a sound pulse vertically downwards. The echo is received after 0.6 s. The speed of sound in water is 1500 m/s. Calculate the depth of the sea. Echo sounding uses the principle that sound travels at a known speed, reflects off the seabed, and returns. The time measured is the round-trip time. Therefore, the depth is half the distance travelled by the sound.
一艘船垂直向下发射声脉冲，0.6 s 后接收到回声。声音在水中的速度为 1500 m/s。计算海水深度。回声测深利用声音以已知速度传播，遇到海床反射并返回的原理。测量的时间是往返时间。因此，水深是声音传播距离的一半。
ctice, sonar systems use this method for mapping the ocean floor and detecting submarines.
声音传播的距离 = 速度 × 时间 = 1500 × 0.6 = 900 m。由于这是来回距离，水深 = 900 / 2 = 450 m。实际应用中，声纳系统利用此方法绘制海底地图和探测潜艇。
3. Case Study 3: Electrical Circuit Fault Diagnosis | 案例三：电路故障诊断
A student builds a series circuit with a 12 V battery, a switch, a lamp, and a fixed resistor. The lamp does not light when the switch is closed. The student measures the voltage across the lamp and gets 12 V, while the voltage across the resistor is 0 V. Diagnose the fault. In a series circuit, if a component breaks (open circuit), current stops. The PD across the break equals the supply voltage. Here, the lamp might have a broken filament, causing an open circuit.
一名学生搭建了一个包含 12 V 电池、开关、灯泡和固定电阻的串联电路。闭合开关后，灯泡不亮。学生测得灯泡两端电压为 12 V，电阻两端电压为 0 V。诊断故障。在串联电路中，如果某个元件断路，电流停止。断路处的电势差等于电源电压。这里，灯泡的灯丝可能断裂，导致断路。
e broken lamp acts as an infinite resistance, so all 12 V appears across it. A working circuit would share the voltage. Replacing the lamp should fix the fault. This case tests understanding of potential divider behaviour and fault analysis.
没有电流时，电阻无电势差（V = IR = 0）。断路的灯泡相当于无限大电阻，因此 12 V 全部落在其两端。正常电路会分压。更换灯泡即可排除故障。此案例考察对分压行为和故障分析的理解。
4. Case Study 4: Energy Transfer in a Roller Coaster | 案例四：过山车中的能量转换
gy changes. At the top, the car has gravitational potential energy (GPE). As it descends, GPE is converted to kinetic energy (KE). By conservation of energy, mgh = ½mv², so v = √2gh. This ignores air resistance and friction, but gives a theoretical maximum.
一辆质量为 500 kg 的过山车从 30 m 高处静止出发。假设无摩擦，求它在坡底的速度，并解释能量变化。在顶部，过山车具有重力势能（GPE）。下滑过程中，GPE 转换为动能（KE）。根据能量守恒，mgh = ½mv²，因此 v = √2gh。计算忽略了空气阻力和摩擦，但给出理论最大值。
v = √2 × 9.8 × 30 rgy is transferred to thermal energy due to friction, so the actual speed would be lower. This principle underpins the design of all amusement park rides.
v = √2 × 9.8 × 30 = √588 ≈ 24.2 m/s。现实中，部分能量因摩擦转化为热能，因此实际速度会低一些。这一原理支撑着所有游乐园设施的设计。
5. Case Study 5: Pressure and Hydraulic Lifts | 案例五：压强与液压升降机
piston. Calculate the force exerted by the large piston. Hydraulic systems work on Pascal' ly in all directions. Pressure at small piston = F₁/A₁ = Pressure at large piston = F₂/A₂.
一台液压升降机的小活塞面积为 0.02 m²，大活塞面积为 0.5 m²。对小活塞施加 200 N 的力。计算大活塞输出的力。液压系统基于帕斯卡原理：施加在密闭液体上的压强会等大地向各个方向传递。小活塞压强 = F₁/A₁ = 大活塞压强 = F₂/A₂。
hus a small input force can lift a much heavier load by using area ratios. Hydraulic brakes in cars operate on the same principle.
压强 = 200 / 0.02 = 10 000 Pa。则 F₂ = 压强 × A₂ = 10 000 × 0.5 = 5000 N。由此可见，借助面积比例，较小的输入力可举起重得多的负载。汽车液压刹车也是基于同一原理。
6. Case Study 6: Half-life and Radioactive Dating | 案例六：半衰期与放射性定年
A sample of ancient wood contains 25% onential pattern, but simple halving sequences work for whole numbers of half-lives. If 25% remains, that means two half-lives have passed (100% → 50% → 25%). Therefore, age = 2 × 5730 = 11 460 years.
一块古代木材含有活树中碳-14 含量的 25%。碳-14 的半衰期为 5730 年。估算该木材的年代。放射性衰变遵循指数规律，但在半衰期取整数时可用简单的减半序列。若剩余 25%，意味着经过了两个半衰期（100% → 50% → 25%）。因此，年代 = 2 × 5730 = 11 460 年。
Radiocarbon dating is widely used in archaeology to determine the age of organic materials up to about 50 000 years. Understanding half-life also helps with radioactive waste management and medical isotopes.
放射性碳定年法广泛应用于考古学，可测定约五万年以内的有机材料年代。理解半衰期还有助于放射性废物管理和医用同位素应用。
7. Case Study 7: Refraction and Lens Power | 案例七：折射与透镜焦度
ry distant object, the rays are parallel and converge at the focal point. Thus, the focal length f equals the image distance for distant objects: f = 15 cm = 0.15 m. Lens power (in dioptres) is 1 / f (metres).
一个凸透镜将远处物体的清晰像投射在距透镜 15 cm 的屏幕上。求焦距和透镜焦度。对于很远的物体，入射光线可视为平行光，会聚于焦点。因此，焦距 f 就等于此时像距：f = 15 cm = 0.15 m。透镜焦度（屈光度）= 1 / f（米）。
ositive power indicates a converging lens. This simple method demonstrates how you can measure focal length outdoors using sunlight or a distant window.
焦度 = 1 / 0.15 = +6.67 D（屈光度）。正焦度表明是会聚透镜。这个简单方法展示了如何利用太阳光或远处的窗户在室外测量焦距。
8. Case Study 8: Momentum and Collisions | 案例八：动量与碰撞
. In the absence of external forces, total momentum is conserved. Momentum before = m₁u₁ + m₂u₂ = (0.5 × 4) + (1.5 × 0) = 2 kg m/s. Total mass after = 0.5 + 1.5 = 2 kg. Common velocity v = total momentum / total mass = 2 / 2 = 1 m/s.
一辆 0.5 kg 的小车以 4 m/s 的速度运动，与一辆静止的 1.5 kg 小车碰撞后黏在一起。求它们碰撞后的共同速度。在没有外力作用时，总动量守恒。碰前动量 = m₁u₁ + m₂u₂ = (0.5 × 4) + (1.5 × 0) = 2 kg m/s。碰后总质量 = 0.5 + 1.5 = 2 kg。共同速度 v = 总动量 / 总质量 = 2 / 2 = 1 m/s。
This demonstrates that the heavier stationary object reduces the speed of the moving one. Inelastic collisions are common; kinetic energy is not conserved here, but momentum always is.
这表明较重的静止物体会降低运动物体的速度。非弹性碰撞十分常见；此时动能不守恒，但动量总是守恒的。
9. Case Study 9: Specific Heat Capacity in Cooking | 案例九：烹饪中的比热容
ired. Energy = mass × specific heat capacity × temperature change = 2 × 4200 × (100 - 20) = 2 × 4200 × 80 = 672 000 J. This is 672 kJ. The case illustrates why water takes a long time to boil — pecific heat capacity means it can absorb a lot of energy without a huge temperature rise.
一位厨师用电热水壶将 2 kg 水从 20 °C 加热到 100 °C。水的比热容为 4200 J/(kg °C)。计算所需能量。能量 = 质量 × 比热容 × 温度变化 = 2 × 4200 × (100 - 20) = 2 × 4200 × 80 = 672 000 J，即 672 kJ。此案例说明水为什么需要较长时间煮沸——其高比热容意味着可以吸收大量能量而温度不会急剧升高。
er = 672 000 / 2000 = 336 s, about 5.6 minutes. In practice, heat losses extend this time.
如果水壶功率为 2000 W，理论加热时间（忽略热损失）为能量 / 功率 = 672 000 / 2000 = 336 s，约 5.6 分钟。实际中，热损失会延长这一时间。
10. Case Study 10: Wave Speed on a String | 案例十：弦上的波速
ave speed. For a standing wave with three antinodes on a string fixed at both ends, the pattern corresponds to the third harmonic. The wavelength λ = 2L / n, where n = number of antinodes. So λ = 2 × 0.8 / 3 = 1.6 / 3 ≈ 0.533 m. Wave speed v = f λ = 120 × 0.533 ≈ 64 m/s.
一名学生振动一根 0.8 m 长的弦，观察到具有三个波腹的驻波。振动器频率为 120 Hz。求波速。对于两端固定的弦，三个波腹对应第三谐频。波长 λ = 2L / n，其中 n 为波腹数。因此 λ = 2 × 0.8 / 3 = 1.6 / 3 ≈ 0.533 m。波速 v = f λ = 120 × 0.533 ≈ 64 m/s。
ase connects frequency, wavelength, and boundary conditions, a favourite in Cambridge exams.
或者，利用 n 次谐频公式：f = nv/(2L)，则 v = 2L f / n = 2 × 0.8 × 120 / 3 = 192 / 3 = 64 m/s。此案例将频率、波长和边界条件联系起来，是剑桥考试中的热门题型。
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