📚 Year 9 CAIE Physics: Case Study Practical Drills | Year 9 CAIE 物理：案例分析实战演练
In the CAIE Year 9 Physics course, case study questions are designed to test your ability to apply scientific knowledge to unfamiliar, real-world situations. These questions often present data in tables, graphs, or descriptive paragraphs, and ask you to perform calculations, explain phenomena, or make predictions. This article will guide you through a series of practical case studies, helping you develop the skills needed to analyse data, choose the correct formulas, and communicate your reasoning clearly. Each case reflects topics you have covered, such as motion, forces, energy, electricity, density, pressure, and waves.
在 CAIE 九年级物理课程中，案例分析题旨在检验你将科学知识应用于陌生、真实情境的能力。这些题目通常以表格、图表或描述段落的形式提供数据，并要求你进行计算、解释现象或作出预测。本文将带领你完成一系列实战案例，帮助你培养分析数据、选择正确公式和清晰表达推理过程的技能。每个案例都涵盖了你学过的主题，如运动、力、能量、电、密度、压强和波。
1. Understanding Case Study Questions | 理解案例分析题
A case study in physics is not just about recalling facts; it is about using concepts like a detective. You will be given a scenario, such as a car accelerating or a diver jumping, along with some measurements. Your task is to identify which physical principles apply, extract the relevant data, and apply equations accurately. Always read the entire case before attempting any calculation, and pay close attention to units and significant figures. The best approach is to break the problem into small steps.
物理的案例分析不仅仅是回忆事实，而是像侦探一样运用概念。你会得到一个场景，比如一辆汽车加速或一名跳水运动员起跳，并附有一些测量数据。你的任务是识别适用的物理原理，提取相关数据，并准确应用方程。在任何计算之前，始终通读整个案例，并密切关注单位和有效数字。最好的方法是把问题分解成小步骤。
Many case study questions will ask you to explain trends or make comparisons. For example, you might need to describe why acceleration decreases as speed increases for a falling object. Such explanations require linking observations to key ideas like air resistance and resultant force. Practising with structured answers now will boost your confidence in exams.
许多案例分析题会要求你解释趋势或进行比较。例如，你可能需要描述为什么下落物体的加速度随着速度增加而减小。这样的解释需要将观察结果与空气阻力和合力等关键概念联系起来。现在用结构化的答案进行练习，将在考试中增强你的信心。
2. Key Skills for Tackling Case Studies | 攻克案例分析的关键技能
Before diving into specific examples, let us highlight the essential skills you need. First, data extraction: pick out numbers and units from the text or graph carefully. Second, formula selection ange in velocity / time, force = mass × acceleration, pressure = force / area, and density = mass / volume. Third, unit conversion: always work in SI units (metres, seconds, kilograms) unless the question states otherwise. Fourth, graph interpretation: recognise that the gradient of a distance–time graph gives speed, and the gradient of a speed–time graph gives acceleration. The area under a speed–time graph represents distance travelled.
在进入具体示例之前，让我们先强调你需要的关键技能。第一，数据提取：仔细从文字或图表中找出数字和单位。第二，公式选择：决定哪个方程将已知量与未知量联系起来。对于九年级，常见方程包括 速度 = 路程 / 时间、加速度 = 速度变化量 / 时间、力 = 质量 × 加速度、压强 = 力 / 面积、密度 = 质量 / 体积。第三，单位换算：始终使用国际单位制（米、秒、千克），除非题目另有说明。第四，图表解读：认识到路程–时间图的斜率表示速度，速度–时间图的斜率表示加速度。速度–时间图下的面积代表移动的距离。
Another crucial skill is explaining using physics vocabulary. Avoid vague phrases like 'it gets bigger'. Instead, state 'the resultant force decreases, so the acceleration decreases'. Wherever possible, back up your reasoning with an equation. For instance, in a question about pressure, you can say 'pressure is inversely proportional to area, so a smaller area gives a larger pressure'. This shows the examiner you understand the underlying principle, not just the outcome.
另一个关键技能是用物理词汇进行解释。避免模糊的短语，如“它变大了”。相反，要说明“合力减小，因此加速度减小”。尽可能用方程来支持你的推理。例如，在一个关于压强的问题中，你可以说“压强与面积成反比，因此面积越小，压强越大”。这向考官表明你理解了背后的原理，而不仅仅是结果。
3. Case 1: The Accelerating Bicycle | 案例1：加速的自行车
; at 2 s, velocity = 4 m/s; at 4 s, velocity = 8 m/s; at 6 s, velocity = 12 m/s; at 8 s, velocity = 16 m/s. The cyclist and bicycle together have a mass of 80 kg. The total forward driving force from the pedalling is 100 N, and a constant frictional force of 36 N opposes the motion. Your task is to analyse the motion and the forces involved.
一名自行车手从静止开始沿直路运动。传感器每隔2秒记录一次速度，如下表所示：0秒时速度为0 m/s；2秒时为4 m/s；4秒时为8 m/s；6秒时为12 m/s；8秒时为16 m/s。车手和自行车总质量为80 kg。踩踏产生的总前进驱动力为100 N，恒定摩擦力36 N阻碍运动。你的任务是分析运动和涉及的力。
= (16 m/s - 0 m/s) / 8 s = 2 m/s². You can also use any two points; the motion is uniformly accelerated. The value is 2 m/s².
步骤1：计算加速度。使用数据，速度每2秒增加4 m/s。加速度 a = 速度变化量 / 所用时间 = (16 m/s - 0 m/s) / 8 s = 2 m/s²。你也可以使用任意两点；运动是匀加速的。数值为2 m/s²。
bicycle system. Resultant force = driving force - friction = 100 N - 36 N = 64 N forwards.
步骤2：计算作用在自行车系统上的合力。合力 = 驱动力 - 摩擦力 = 100 N - 36 N = 64 N，方向向前。
Step 3: Verify acceleration using Newton' F / m = 64 N / 80 kg = 0.8 m/s². This does not match the kinematic acceleration of 2 m/s²! g force is not constant. In an exam, you would be expected to notice the mismatch and suggest the driving force must be larger, or friction smaller, or there is an additional force. This sharpens your critical thinking.
步骤3：用牛顿第二定律验证加速度。F = m a，因此预测加速度 = F / m = 64 N / 80 kg = 0.8 m/s²。这与运动学得到的加速度2 m/s²不匹配！这告诉我们需检查数据或假设。也许给出的质量不是有效的总质量，或者驱动力不是恒定的。考试中，你应注意到不匹配，并提出驱动力必须更大，或摩擦更小，或者存在额外的力。这能锻炼你的批判性思维。
Step 4: Calculate the total distance travelled in 8 seconds using the velocity– × height = ½ × 8 s × 16 m/s = 64 m. Alternatively, use s = ½ (u+v) t = ½ (0+16)×8 = 64 m.
步骤4：使用速度–时间图方法计算8秒内移动的总距离。由于图形是从(0,0)到(8,16)的直线，图下面积是一个三角形：距离 = ½ × 底 × 高 = ½ × 8 s × 16 m/s = 64 m。或者用 s = ½ (u+v) t = ½ (0+16)×8 = 64 m。
4. Case 2: Forces on a Diver | 案例2：跳水运动员的受力
A 50 kg diver jumps off a 10 m high platform. For the first second after leaving the platform, she is in free fall, ignoring air resistance. After that, she enters the water and experiences an upward water resistance force. At the instant she hits the water, her velocity is 14 m/s vertically downwards. While underwater, the water exerts a constant upward force of 2000 N, and her weight still acts downwards. Analyse her motion before and after hitting the water.
一名50 kg的跳水运动员从10米高的跳台跳下。离开跳台后的第一秒内，她处于自由落体状态，忽略空气阻力。之后她入水，并受到向上的水阻力。在她接触水面的瞬间，速度为垂直向下14 m/s。水下时，水施加2000 N的恒定向上力，她的体重仍然向下作用。分析她入水前后的运动。
g h = 2×10×10 = 200, v ≈ 14.1 m/s, which matches the given 14 m/s. So the data is consistent.
首先，用能量或运动学验证自由落体速度。重力场强 g = 10 N/kg。如果她从静止下落10 m，用 v² = 2 g h = 2×10×10 = 200，v ≈ 14.1 m/s，与给出的14 m/s吻合。数据一致。
pward as positive) = 2000 N - 500 N = 1500 N upwards. So acceleration a = F / m = 1500 N / 50 kg = 30 m/s² upwards. That means she decelerates rapidly, which makes sense for stopping in water.
现在，计算她在水下的加速度。她的体重 = m g = 50 kg × 10 N/kg = 500 N 向下。合力 = 向上水阻力 - 体重（取向上为正）= 2000 N - 500 N = 1500 N 向上。所以加速度 a = F / m = 1500 N / 50 kg = 30 m/s² 向上。这意味着她迅速减速，这符合在水中停下的情况。
ful with sign: let downward be negative, upward positive. Initial velocity u = -14 m/s, final v = 0, a = +30 m/s². Then t = (0 - (-14)) / 30 = 14/30 ≈ 0.47 s. She comes to rest in less than half a second; the stopping distance can be found with v² = u² + 2as: 0 = (-14)² + 2×30×s → 196 = -60 s? 6/60 = -3.27 m. The negative sign indicates the displacement is downward, meaning she travels 3.27 m down while stopping. This is a good check on water depth required.
如果她以14 m/s入水，并以30 m/s²减速，停止所需时间 t = (v - u)/a = (0 - (-14))/30，取向下为负。但注意符号：设向下为负，向上为正。初速度 u = -14 m/s，末 v = 0，a = +30 m/s²。则 t = (0 - (-14)) / 30 = 14/30 ≈ 0.47 s。她在不到半秒内停下；用 v² = u² + 2as 可求出停止距离：0 = (-14)² + 2×30×s → 196 = -60 s？等一下，a 向上，位移 s 向上，需注意：使用 u = -14 m/s（向下），a = +30 m/s²（向上），v = 0，那么 0² = (-14)² + 2(30)s → 0 = 196 + 60s → s = -196/60 = -3.27 m。负号表示位移向下，意味着她在停下时下行了3.27 m。这对要求的水深是个很好的检验。
5. Case 3: Energy Changes in a Roller Coaster | 案例3：过山车的能量变化
A roller coaster car of mass 600 kg starts from rest at the top of a hill 25 m above the ground. Assume no energy is lost to friction or air resistance. The car descends to the bottom and then rises to a second hill 10 m high. Analyse the energy transfers and find the speed at the bottom and at the top of the second hill.
一辆质量600 kg的过山车从地面以上25 m的山顶由静止出发。假设没有因摩擦或空气阻力损失能量。车子下滑到底部，然后上升到10 m高的第二个山顶。分析能量转移，并计算在底部和第二个山顶时的速度。
verted to kinetic energy (KE) (since the height is zero). So KE = 150,000 J. Using KE = ½ m v², we have 150,000 = ½ × 600 × v² → 150,000 = 300 v² → v² = 500 → v = √500 ≈ 22.4 m/s. The speed at the bottom is about 22.4 m/s.
首先，计算起点处的重力势能 (GPE)：GPE = m g h = 600 kg × 10 N/kg × 25 m = 150,000 J。到底部时，所有这些 GPE 都转化为动能 (KE)（因为高度为零）。所以 KE = 150,000 J。利用 KE = ½ m v²，得到 150,000 = ½ × 600 × v² → 150,000 = 300 v² → v² = 500 → v = √500 ≈ 22.4 m/s。底部速度约为22.4 m/s。
maining energy is KE: total energy 150,000 J - 60,000 J = 90,000 J. So KE at the top of the second hill is 90,000 J. Then ½ m v² = 90,000 → ½ × 600 × v² = 90,000 → 300 v² = 90,000 → v² = 300 → v = √300 ≈ 17.3 m/s. The car therefore slows down but still has considerable speed.
当车子爬上10 m的山顶时，部分 KE 重新转化为 GPE。10 m高度处的 GPE = m g h = 600 × 10 × 10 = 60,000 J。根据能量守恒，剩余能量为 KE：总能量150,000 J - 60,000 J = 90,000 J。所以第二个山顶处的 KE 为90,000 J。然后 ½ m v² = 90,000 → ½ × 600 × v² = 90,000 → 300 v² = 90,000 → v² = 300 → v = √300 ≈ 17.3 m/s。因此车子减速，但仍有相当快的速度。
Notice how energy methods avoid complicated kinematics, which would be difficult on a curved track. This case illustrates the power of the energy conservation principle. A common error is to forget that mass cancels out when finding speed if you set equations equal, but it is safer to calculate step by step.
注意能量方法如何避开了在弯曲轨道上复杂的运动学计算。这个案例说明了能量守恒原理的强大。常见错误是当通过等式相等求速度时忘记质量可消去，但更稳妥的做法是逐步计算。
6. Case 4: Electrical Circuit in a Torch | 案例4：手电筒中的电路
A torch contains two 1.5 V cells in series, providing a total of 3.0 V. The bulb has a resistance of 6 Ω when lit. The switch and connecting wires have negligible resistance. A student also finds that the bulb can be modelled as a pure resistor. Determine the current flowing in the circuit and the power dissipated by the bulb. Later, the student replaces the bulb with a resistor of 12 Ω and connects it in parallel with the original 6 Ω bulb. Calculate the total current drawn from the battery and the current through each branch.
一个手电筒包含两节串联的1.5 V电池，提供总计3.0 V电压。灯泡点亮时电阻为6 Ω。开关和连接导线电阻可忽略。一学生发现灯泡可视为纯电阻。确定电路中的电流和灯泡消耗的功率。后来，该学生用一个12 Ω电阻替换灯泡，并将其与原来的6 Ω灯泡并联。计算电池提供的总电流和流过每个支路的电流。
Part 1: With a single 6 Ω bulb across 3.0 V, using Ohm' natively, P = I² R = (0.5)² × 6 = 0.25 × 6 = 1.5 W. So the bulb glows with 1.5 W of power.
第一部分：单个6 Ω灯泡接在3.0 V两端，用欧姆定律：I = V / R = 3.0 V / 6 Ω = 0.5 A。功率 P = V × I = 3.0 V × 0.5 A = 1.5 W。或者 P = I² R = (0.5)² × 6 = 0.25 × 6 = 1.5 W。因此灯泡以1.5 W功率发光。
5 A. Current through the 12 Ω branch: I₂ = 3.0 V / 12 Ω = 0.25 A. Total current from the battery = I₁ + I₂ = 0.5 + 0.25 = 0.75 A. The effective resistance of the parallel combination is R = (1/6 + 1/12)⁻¹ = (3/12)⁻¹ = 12/3 = 4 Ω, which matches V / Itotal = 3.0 / 0.75 = 4 Ω. This confirms the calculation.
第二部分：现在6 Ω灯泡和12 Ω电阻并联。每个支路的电压仍为3.0 V。流过6 Ω支路的电流：I₁ = 3.0 V / 6 Ω = 0.5 A。流过12 Ω支路的电流：I₂ = 3.0 V / 12 Ω = 0.25 A。电池提供的总电流 = I₁ + I₂ = 0.5 + 0.25 = 0.75 A。并联组合的有效电阻 R = (1/6 + 1/12)⁻¹ = (3/12)⁻¹ = 12/3 = 4 Ω，这与 V / Itotal = 3.0 / 0.75 = 4 Ω 相符。这验证了计算。
Important: In parallel circuits, the current divides but the voltage stays the same. Many students mistakenly think the current is the same everywhere. Remember, the battery provides more total current when resistance is lower, which is why adding a parallel path increases total current.
重要：在并联电路中，电流分流而电压相同。许多学生错误地认为各处电流相同。请记住，当电阻较低时，电池提供的总电流更大，这就是为什么增加并联路径会使总电流增加。
7. Case 5: Density and Floating Bottle | 案例5：密度与漂浮的瓶子
A glass bottle has a mass of 200 g and an internal volume of 500 cm³ when empty. It is sealed and then placed in a large tank of fresh water (density 1.0 g/cm³). Will the empty bottle float or sink? Explain using density. Next, the bottle is filled with sand so that its total mass becomes 600 g, and its external volume remains approximately 510 cm³. Determine whether it will float now, and if it floats, what volume of the bottle is submerged?
一个玻璃瓶的质量为200 g，空瓶时内部容积为500 cm³。将其密封后放入一大水箱淡水中（密度1.0 g/cm³）。空瓶会漂浮还是下沉？用密度进行解释。接下来，向瓶中装入沙子，使其总质量变成600 g，其外部体积仍约为510 cm³。确定此时它是否会漂浮，如果漂浮，瓶子的浸没体积是多少？
density = mass / volume = 200 g / 500 cm³ = 0.4 g/cm³. Since 0.4 g/cm³ is less than water' ts own weight, so the floating position will have only a fraction of its volume submerged.
对于空瓶：质量 = 200 g，体积 = 500 cm³。平均密度 = 质量 / 体积 = 200 g / 500 cm³ = 0.4 g/cm³。由于0.4 g/cm³小于水的密度1.0 g/cm³，瓶子会漂浮。它将排开与其自身重量相等的水重，因此漂浮时只有一部分体积浸没。
will sink. In the floating case, we would calculate submerged volume from the upthrust: weight of bottle = weight of displaced water. For the 200 g bottle, weight = 0.2 kg × 10 N/kg = 2.0 N, but using g/cm³ and mass, we can say mass of displaced water = 200 g, so volume of displaced water = mass / density = 200 g / (1 g/cm³) = 200 cm³. Therefore, 200 cm³ of the bottle is submerged, which is 40% of its volume.
装沙后的瓶子：总质量 = 600 g，外部体积 = 510 cm³。平均密度 = 600 g / 510 cm³ ≈ 1.18 g/cm³。这大于1.0 g/cm³，因此瓶子会下沉。对于漂浮情况，我们可以通过浮力计算浸没体积：瓶子重量 = 排开水的重量。对于200 g的瓶子，重量 = 0.2 kg × 10 N/kg = 2.0 N，但使用 g/cm³ 和质量，可以说排开水的质量 = 200 g，因此排开水的体积 = 质量 / 密度 = 200 g / (1 g/cm³) = 200 cm³。因此，200 cm³的瓶子浸没，占其体积的40%。
8. Case 6: Pressure Under a Stool | 案例6：凳子下面的压强
A wooden stool has four legs, each with a circular flat bottom of radius 1.5 cm. The total weight of the stool is 60 N. A person sits on the stool, exerting an additional downward force of 500 N. Calculate the pressure exerted by each leg on the floor, assuming the weight is evenly distributed. Then, after the person leans, only two legs remain in contact. How does the pressure change? What might happen to the floor if it is made of soft wood?
一个木凳有四条腿，每条腿的底部为圆形平面，半径1.5 cm。凳子总重量为60 N。一个人坐在凳子上，施加额外向下的力500 N。假设重量均匀分布，计算每条腿对地板的压强。随后，人斜靠后只有两条腿接触地面。压强如何变化？如果地板是软木做的，可能会发生什么？
Step 1: Find the area of one leg base. Area = π r² = π × (1.5
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