📚 Year 10 CAIE Physics: Case Study Practical Exercises | Year 10 CAIE 物理：案例分析实战演练
Welcome to this comprehensive set of case studies designed to strengthen your problem-solving skills in Year 10 CAIE Physics. Each scenario challenges you to apply physical principles to real-world situations, covering kinematics, forces, energy, waves, electricity, and more. Carefully work through the explanations and calculations to master the concepts required for your IGCSE examinations.
欢迎来到这组综合性案例分析，旨在强化你在Year 10 CAIE物理中的解题能力。每个情景都要求你将物理原理应用于实际问题，涵盖运动学、力、能量、波、电学等内容。仔细学习这些解释和计算，以掌握IGCSE考试所需的各个概念。
1. Kinematics – Car Braking Distance | 运动学 – 汽车刹车距离
A car travels at a constant speed of 20 m/s on a dry road. The driver sees a hazard and applies the brakes after a reaction time of 0.70 s. The car then decelerates uniformly at 6.0 m/s² until it stops. Calculate the total stopping distance. First, find the thinking distance: distance = speed × reaction time = 20 m/s × 0.70 s = 14 m. Next, use v² = u² + 2as to find the braking distance. Here u = 20 m/s, v = 0, a = -6.0 m/s². Rearranging: 0 = (20)² + 2 × (-6.0) × s, giving s = 400/12 = 33.3 m. Total stopping distance = 14 m + 33.3 m = 47.3 m.
一辆汽车在干燥路面上以 20 m/s 的恒定速度行驶。驾驶员发现危险，在 0.70 s 的反应时间后踩下刹车。随后汽车以 6.0 m/s² 的匀减速一直减速至停止。计算总停车距离。首先求思考距离：距离 = 速度 × 反应时间 = 20 m/s × 0.70 s = 14 m。接着用 v² = u² + 2as 计算制动距离。这里 u = 20 m/s，v = 0，a = -6.0 m/s²。整理得：0 = (20)² + 2 × (-6.0) × s，得 s = 400/12 = 33.3 m。总停车距离 = 14 m + 33.3 m = 47.3 m。
The key insight is that the total stopping distance is the sum of the thinking distance (during which the car travels at constant speed) and the braking distance (during which the car slows down). Factors such as tiredness, alcohol, or mobile phone use increase reaction time, while wet or icy roads reduce the deceleration. Both effects can dramatically increase the stopping distance, often leading to collisions.
关键在于总停车距离是思考距离（汽车在此期间匀速行驶）与制动距离（汽车减速期间）之和。疲劳、酒精或使用手机等因素会增加反应时间，而潮湿或结冰路面会降低减速性能。这两种效应都会显著增加停车距离，常常导致碰撞事故。
2. Velocity–Time Graph Analysis | 速度–时间图分析
A motorcycle accelerates uniformly from rest to 30 m/s in 10 s. It then travels at constant speed for 35 s, and finally decelerates uniformly to rest in 10 s. Sketch the velocity–time graph and find the total distance travelled. The graph consists of three sections: a straight sloping line from (0,0) to (10,30), a horizontal line from (10,30) to (45,30), and a straight sloping line down to (55,0). The area under a velocity–time graph gives the distance. Area 1 (triangle): ½ × 10 s × 30 m/s = 150 m. Area 2 (rectangle): 35 s × 30 m/s = 1050 m. Area 3 (triangle): ½ × 10 s × 30 m/s = 150 m. Total distance = 150 + 1050 + 150 = 1350 m.
一辆摩托车从静止开始匀加速，10 s 内达到 30 m/s。随后匀速行驶 35 s，最后在 10 s 内匀减速至静止。画出速度–时间图并求行驶的总距离。图像由三部分组成：从 (0,0) 到 (10,30) 的倾斜直线，从 (10,30) 到 (45,30) 的水平直线，以及向下倾斜到 (55,0) 的直线。速度–时间图下的面积表示行驶距离。面积 1（三角形）：½ × 10 s × 30 m/s = 150 m。面积 2（矩形）：35 s × 30 m/s = 1050 m。面积 3（三角形）：½ × 10 s × 30 m/s = 150 m。总距离 = 150 + 1050 + 150 = 1350 m。
We can also calculate the acceleration in the first phase: gradient = (30 – 0) / 10 = 3.0 m/s². The deceleration phase gives a negative gradient: (0 – 30) / 10 = -3.0 m/s². This type of graph is common in IGCSE questions, and the method of finding distance by area is essential. Always check the units and whether the motion is uniform.
我们也可以计算第一阶段的加速度：斜率 = (30 – 0) / 10 = 3.0 m/s²。减速阶段的斜率为负：(0 – 30) / 10 = -3.0 m/s²。这类图形在 IGCSE 考题中很常见，通过面积求距离的方法是必不可少的。始终要检查单位以及运动是否均匀。
3. Newton’s Second Law – Trolley on an Incline | 牛顿第二定律 – 倾斜轨道上的小车
A trolley of mass 0.80 kg is placed on a frictionless slope inclined at 25° to the horizontal. A constant force of 4.5 N parallel to the slope pulls the trolley upwards. Determine the net force and hence the acceleration. First, calculate the component of weight down the slope: W_parallel = m g sin θ = 0.80 × 10 × sin 25°. Using sin 25° ≈ 0.423, W_parallel ≈ 0.80 × 10 × 0.423 = 3.38 N. The applied force is 4.5 N up the slope, so the net force = 4.5 N – 3.38 N = 1.12 N up the slope. Applying F = m a, we get a = F_net / m = 1.12 / 0.80 = 1.4 m/s².
一辆质量为 0.80 kg 的小车放在与水平面成 25° 的无摩擦斜面上。一个平行于斜面的 4.5 N 恒力将小车向上拉。求合力和加速度。首先计算重力沿斜面向下的分量：W_parallel = m g sin θ = 0.80 × 10 × sin 25°。sin 25° ≈ 0.423，得 W_parallel ≈ 0.80 × 10 × 0.423 = 3.38 N。施加的力为 4.5 N 沿斜面向上，因此合力 = 4.5 N – 3.38 N = 1.12 N 沿斜面向上。应用 F = m a，得 a = F_net / m = 1.12 / 0.80 = 1.4 m/s²。
If friction were present, we would subtract the frictional force from the applied force as well. The direction of the net force determines the direction of acceleration. This case study reinforces that you must resolve forces into components along the line of motion and always consider all forces acting on the body.
如果存在摩擦力，我们还应从施加的力中减去摩擦力。合力的方向决定加速度的方向。本案例强化了这样一个要点：必须将力沿运动方向分解，并始终考虑作用在物体上的所有力。
4. Terminal Velocity – Parachutist in Free Fall | 终极速度 – 自由落体中的跳伞员
A skydiver of mass 70 kg jumps from a plane and initially accelerates at 9.8 m/s². As he falls, air resistance increases with speed. Explain why he eventually reaches a terminal velocity of about 55 m/s before opening his parachute. The skydiver’s weight is W = mg = 70 × 9.8 ≈ 686 N. Initially, the drag force is negligible, so net force ≈ 686 N downward. As speed builds up, the drag force, D, increases (proportional to v² at high speeds). When D equals the weight, net force becomes zero, and by Newton’s first law the skydiver continues at a constant velocity. This is terminal velocity. Given that the drag force D = ½ C ρ A v², for the skydiver the relevant parameters result in D = 686 N when v ≈ 55 m/s.
一名质量为 70 kg 的跳伞员从飞机上跳下，最初以 9.8 m/s² 加速。在下落过程中，空气阻力随速度增大而增大。请解释为什么他在打开降落伞前最终会达到约 55 m/s 的终极速度。跳伞员的体重为 W = mg = 70 × 9.8 ≈ 686 N。最初，阻力可忽略不计，因此合力 ≈ 686 N 向下。随着速度增加，阻力 D 增大（高速时正比于 v²）。当 D 等于体重时，合力为零，根据牛顿第一定律，跳伞员将以恒定速度继续下落，此即终极速度。假设阻力 D = ½ C ρ A v²，对于该跳伞员，相关参数使得当 v ≈ 55 m/s 时 D = 686 N。
After the parachute opens, the cross-sectional area A increases dramatically, so the drag force becomes much larger than weight immediately, causing a rapid deceleration until a new, much lower terminal velocity is reached (about 5 m/s). This sequence of events is crucial for understanding force balance and motion graphs in IGCSE.
打开降落伞后，横截面积 A 急剧增大，因此阻力立即远大于体重，导致急剧减速，直至达到一个新的、低得多的终极速度（约 5 m/s）。这一系列事件对于理解 IGCSE 中力的平衡和运动图线至关重要。
5. Energy Conservation – Roller Coaster Top to Bottom | 能量守恒 – 过山车从顶到底
A roller coaster car of mass 500 kg starts from rest at the top of a hill 40 m above the ground. Assuming no energy losses, determine its speed at the bottom. Use the principle of conservation of energy: gravitational potential energy at the top is converted into kinetic energy at the bottom. At the top: GPE = m g h = 500 × 10 × 40 = 200,000 J. At the bottom: KE = ½ m v². Equating: ½ × 500 × v² = 200,000. This gives v² = (200,000 × 2) / 500 = 800, so v = √800 ≈ 28.3 m/s.
一辆质量为 500 kg 的过山车从离地 40 m 高的山顶静止出发。假设无能量损失，求其在底部的速度。运用能量守恒原理：山顶的重力势能转化为底部的动能。山顶：GPE = m g h = 500 × 10 × 40 = 200,000 J。底部：KE = ½ m v²。等式：½ × 500 × v² = 200,000。得 v² = (200,000 × 2) / 500 = 800，故 v = √800 ≈ 28.3 m/s。
In reality, some energy is lost to friction and air resistance, so the actual speed would be slightly lower. This case study also illustrates that the mass cancels out in the ideal situation: mgh = ½ m v² ⇒ v = √(2gh). Thus, any roller coaster car, regardless of mass, would achieve the same speed at the bottom from the same height. Energy problems often require you to ignore non-conservative forces for IGCSE calculations.
现实中会因摩擦和空气阻力损失部分能量，因此实际速度略低。该案例还表明，在理想情况下质量可以消掉：mgh = ½ m v² ⇒ v = √(2gh)。因此无论质量如何，任何过山车从相同高度下落都会达到相同的底部速度。IGCSE 计算中通常需要忽略非保守力。
6. Pressure and Hydraulic Systems – Car Brake | 压强与液压系统 – 汽车刹车
A hydraulic car brake system has a small master piston of area 2.0 cm² and a larger brake piston of area 45 cm². The driver applies a force of 30 N on the brake pedal, which is transmitted to the master piston. Calculate the force exerted by the brake piston on the brake disc. The pressure in the hydraulic fluid is the same everywhere (Pascal’s principle): P = F₁ / A₁. P = 30 N / 2.0 cm² = 15 N/cm². Note: keep units consistent; we can work in cm². Then the force on the larger piston: F₂ = P × A₂ = 15 N/cm² × 45 cm² = 675 N.
某汽车液压刹车系统中，小主活塞的面积为 2.0 cm²，较大的制动活塞面积为 45 cm²。驾驶员施加 30 N 的力在刹车踏板上，并传递给主活塞。求制动活塞施加在刹车片上的力。根据帕斯卡原理，液压油中各处的压强相等：P = F₁ / A₁ = 30 N / 2.0 cm² = 15 N/cm²。注意保持单位一致；可用 cm² 计算。那么大活塞上的力：F₂ = P × A₂ = 15 N/cm² × 45 cm² = 675 N。
The hydraulic system multiplies the force 22.5 times (675/30). However, the distance moved by the small piston is larger than that moved by the larger piston because the work done (force × distance) is conserved ideally. This system demonstrates how pressure is transmitted through a fluid and is used in car brakes, hydraulic jacks, and industrial presses.
该液压系统将力放大了 22.5 倍（675/30）。然而，小活塞移动的距离大于大活塞，因为理想情况下做功（力 × 距离）是守恒的。该系统展示了压强在液体中如何传递，并应用于汽车刹车、液压千斤顶和工业压力机中。
7. Thermal Physics – Cooling a Cup of Tea | 热物理 – 冷却一杯茶
A student pours 0.30 kg of tea at 85 °C into a cup. The tea cools to 55 °C in a room at 20 °C. The specific heat capacity of tea is approximately 4200 J/kg°C. Calculate the heat energy lost by the tea. Q = m c Δθ = 0.30 × 4200 × (85 – 55) = 0.30 × 4200 × 30 = 37,800 J. This energy is transferred to the surroundings mainly by conduction, convection, and radiation.
一名学生把 0.30 kg 85 °C 的茶水倒入杯中。茶水在 20 °C 的房间内冷却到 55 °C。茶水的比热容约为 4200 J/kg°C。计算茶水损失的热量。Q = m c Δθ = 0.30 × 4200 × (85 – 55) = 0.30 × 4200 × 30 = 37,800 J。这部分能量主要通过传导、对流和辐射传递到了周围环境中。
The rate of cooling depends on the temperature difference between the tea and the room. Initially, the difference is 65 °C, so cooling is rapid. As the tea approaches room temperature, the rate slows down. To keep tea hot longer, one could use a lid to reduce evaporation and convection, or a double-walled cup to reduce conduction. Understanding heat transfer mechanisms is vital for thermal physics case studies.
冷却速率取决于茶水和房间的温差。初始温差为 65 °C，因此冷却迅速。随着茶水接近室温，冷却速率减慢。为了让茶保温更久，可以使用杯盖以减少蒸发和对流，或使用双层杯以减少传导。理解热传递机制对热物理案例分析至关重要。
8. Waves – Echo Sounding to Measure Distance | 波 – 回声测距
A ship sends a sound pulse directly downwards to the seabed. The echo is received 0.48 s later. The speed of sound in seawater is 1500 m/s. Calculate the depth of the sea at this location. The total distance travelled by the sound wave is twice the depth (down and back). Distance = speed × time, so total distance = 1500 m/s × 0.48 s = 720 m. Therefore, depth = total distance / 2 = 360 m.
一艘船向正下方的海底发射一个声音脉冲。0.48 s 后接收到回声。海水中声速为 1500 m/s。计算此处海水深度。声波传播的总距离是深度的两倍（去和回）。距离 = 速度 × 时间，因此总距离 = 1500 m/s × 0.48 s = 720 m。所以深度 = 总距离 / 2 = 360 m。
This technique, called sonar, is widely used for mapping the ocean floor and by fishing vessels to locate shoals of fish. The underlying principle is the wave speed equation v = f λ and the relationship distance = v × t. Remember that for echo problems, the time measured includes the return journey. Always halve the product to obtain the one-way distance.
这项技术称为声呐，广泛用于绘制海底地图和渔船定位鱼群。其基本原理是波速方程 v = f λ 以及距离 = v × t 的关系。务必记住，对于回声问题，测得的时间包含了往返行程，因此需要将乘积减半才能得到单程距离。
9. Electricity – Series Circuit with Two Resistors | 电学 – 两个电阻的串联电路
A 12 V battery is connected in series to a 6 Ω resistor and an 18 Ω resistor. Determine the total resistance, the circuit current, and the potential difference across each resistor. For resistors in series: R_total = R₁ + R₂ = 6 + 18 = 24 Ω. Using Ohm’s law: I = V / R_total = 12 V / 24 Ω = 0.50 A. The potential difference across the 6 Ω resistor: V₁ = I × R₁ = 0.50 × 6 = 3.0 V. Across the 18 Ω resistor: V₂ = I × R₂ = 0.50 × 18 = 9.0 V. Check: 3.0 V + 9.0 V = 12 V, which matches the supply.
一个 12 V 的电池与一个 6 Ω 电阻和一个 18 Ω 电阻串联。求总电阻、电路中的电流以及每个电阻两端的电压。串联电阻：R_total = R₁ + R₂ = 6 + 18 = 24 Ω。使用欧姆定律：I = V / R_total = 12 V / 24 Ω = 0.50 A。6 Ω 电阻两端电压：V₁ = I × R₁ = 0.50 × 6 = 3.0 V。18 Ω 电阻两端电压：V₂ = I × R₂ = 0.50 × 18 = 9.0 V。验证：3.0 V + 9.0 V = 12 V，与电源电压相符。
In a series circuit, the current is the same everywhere, and the sum of the p.d.s equals the source voltage. The larger resistor takes a larger share of the voltage. This principle is used in potential divider circuits. If one of the resistors were replaced by a variable resistor, you could control the output voltage for sensors.
在串联电路中，电流处处相等，各部分电压之和等于电源电压。电阻越大，分得的电压越多。这一原理应用于分压电路中。如果其中一个电阻换为可变电阻，你就可以控制传感器等的输出电压。
10. Moment of a Force – Using a Crowbar to Lift a Heavy Load | 力矩 – 用撬棍抬起重物
A worker uses a crowbar to lift a heavy crate. The crowbar is 1.5 m long and the pivot is placed 0.15 m from the end under the crate. The worker applies a downward force of 200 N at the opposite end. Calculate the maximum weight of the crate that can be lifted. Assume the crowbar is uniform and its weight is negligible. The principle of moments: clockwise moment = anticlockwise moment. Taking moments about the pivot: Effort force moment = Load force moment. The effort is 200 N, and its distance from the pivot is (1.5 – 0.15) = 1.35 m. So, 200 N × 1.35 m = Load × 0.15 m. Load = (200 × 1.35) / 0.15 = 270 / 0.15 = 1800 N.
一名工人用撬棍抬起一个沉重的板条箱。撬棍长 1.5 m，支点位于距箱底一端 0.15 m 处。工人在另一端向下施加 200 N 的力。求可以抬起的最大箱体重量。假设撬棍均匀且其自重可忽略不计。力矩原理：顺时针力矩 = 逆时针力矩。以支点为中心取矩：动力矩 = 阻力矩。动力为 200 N，其至支点的距离为 (1.5 – 0.15) = 1.35 m。故 200 N × 1.35 m = Load × 0.15 m。Load = (200 × 1.35) / 0.15 = 270 / 0.15 = 1800 N。
The crowbar acts as a force multiplier, giving a mechanical advantage of Load / Effort = 1800 / 200 = 9. This high advantage is due to the large ratio of distances (1.35 / 0.15). Levers are simple machines that can reduce the effort needed to overcome a load. In problems, always identify the pivot and ensure that the moments are balanced.
撬棍起到了力放大作用，机械效益为 Load / Effort = 1800 / 200 = 9。如此高的效益源于距离之比（1.35 / 0.15）很大。杠杆是一种简单机械，能够减小克服负荷所需要的力。解题时，一定要先找出支点并确保力矩平衡。
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