📚 Cambridge KS3 Physics: Case Study Practice | 剑桥KS3物理：案例分析实战演练
Case studies are an excellent way to deepen your understanding of physics concepts and prepare for assessments. In this article, we will walk through several real-world style problems, showing you how to break them down, apply the correct formulas, and explain your reasoning clearly. Each case is designed to mirror the type of questions you may encounter in the Cambridge KS3 Physics curriculum, focusing on energy, electricity, forces, materials, and simple machines.
案例分析是加深物理概念理解、应对评估的绝佳方式。在这篇文章中，我们将走进几个贴近现实的问题，教你如何分解问题、应用正确的公式并清晰地解释推理过程。每个案例都模拟了剑桥KS3物理课程中可能出现的题型，重点涵盖能量、电学、力、材料和简单机械。
1. Understanding Case Studies in Physics | 理解物理案例分析
A case study in physics is not just about getting the right numerical answer — it requires you to show your working, state the principles involved, and often evaluate the data or design of an experiment. When tackling a case study, always begin by identifying the relevant physics topics, listing the known quantities, and noting what you need to find. Then, select the appropriate equation, substitute values carefully, and interpret the result in the context of the problem.
物理中的案例分析远不止于得出正确的数值答案——它要求你展示解题步骤、阐明涉及的原理，还常常需要评估数据或实验设计。处理案例分析时，首先确定相关的物理主题，列出已知量，并标注需要求解的是什么。然后选择合适的方程，仔细代入数值，并在问题背景下解释结果。
2. Case 1: Energy Changes on a Roller Coaster | 案例1：过山车的能量变化
reaches maximum speed. Explain the energy transfers that take place and calculate the maximum kinetic energy (KE) of the car. Use g = 10 m/s².
一辆质量为 500 kg 的过山车在 30 m 高的山顶从静止出发。忽略摩擦和空气阻力。在山脚，过山车达到最大速度。请解释发生的能量转换，并计算过山车的最大动能（KE）。取 g = 10 m/s²。
nverted to kinetic energy. At the bottom, all GPE has become KE (assuming no energy losses). Therefore, KEmax s for maximum KE. The energy transfer is: gravitational potential energy → kinetic energy.
在顶部，过山车具有重力势能 GPE = mgh = 500 × 10 × 30 = 150000 J。下滑过程中，GPE转化为动能。在底部，所有GPE都变成了KE（假设无能量损失）。因此，KEₘₐₓ = 150000 J。通过 KE = ½mv² 可以求出速度，但题目只要求计算最大动能。能量转换过程为：重力势能 → 动能。
Key point: In an ideal system, the total mechanical energy is conserved, and the energy simply changes form.
要点：在理想系统中，总机械能守恒，能量仅改变了形式。
3. Case 2: Electrical Circuit with a Variable Resistor | 案例2：带可变电阻的电路
A student builds a series circuit containing a 6.0 V battery, a fixed 10 Ω resistor, a variable resistor (rheostat), and an ammeter. She adjusts the variable resistor until the ammeter reads 0.20 A. Calculate the total resistance of the circuit and the resistance setting of the variable resistor. Draw the circuit diagram and explain what happens to the current if the variable resistor’s resistance is decreased.
一名学生搭建了一个串联电路，包含一个 6.0 V 电池、一个固定的 10 Ω 电阻、一个可变电阻（变阻器）和一个电流表。她调节可变电阻，直到电流表读数为 0.20 A。计算电路的总电阻以及可变电阻的阻值。画出电路图，并解释若减小可变电阻的阻值，电流将如何变化。
Using Ohm’s law: V = IR, so total resistance Rtotal = V / I = 6.0 / 0.20 = 30 Ω. In a series circuit, Rtotal = Rfixed + Rvariable. Hence, Rvariable sistor is decreased, total resistance drops, causing the current to increase, as I = V / Rtotal with constant battery voltage. The circuit diagram should show all components in a single loop.
应用欧姆定律：V = IR，总电阻 Rₜₒₜₐₗ = V / I = 6.0 / 0.20 = 30 Ω。串联电路中，Rₜₒₜₐₗ = Rₓₓ + Rᵥₐᵣₐₑₗₑ。因此，Rᵥₐᵣₐₑₗₑ = 30 − 10 = 20 Ω。若减小可变电阻，总电阻下降，导致电流增大，因为电池电压恒定时 I = V / Rₜₒₜₐₗ。电路图应显示所有元件在单一回路中。
This case emphasises the inverse relationship between resistance and current in a circuit with a fixed voltage.
此案例强调了在电压固定时，电阻与电流的反比关系。
4. Case 3: Investigating Density of Irregular Objects | 案例3：不规则物体密度的探究
measuring cylinder containing 50 cm³ of water; the water level rises to 68 cm³. Calculate the density of the stone and state whether it will float in water (density of water = 1.0 g/cm³). Explain the method used.
一位考古学家发现了一个形状奇特的小型石质文物。她用天平测得质量为 48.0 g。为测量体积，她将石块放入装有 50 cm³ 水的量筒中，水面升至 68 cm³。计算石块的密度，并说明它是否会在水中漂浮（水的密度 = 1.0 g/cm³）。解释所用的方法。
m³. Since the density is greater than that of water, the stone will sink. The method applied is water displacement, which allows the volume of an irregular solid to be measured. The object must be fully submerged and not soluble in water.
石块体积 = 68 − 50 = 18 cm³。密度 ρ = 质量 / 体积 = 48.0 / 18 ≈ 2.67 g/cm³。由于密度大于水的密度，石块会下沉。所用方法是排水法，它可用来测量不规则固体的体积。物体必须完全浸没且不溶于水。
Remember: floating or sinking depends on whether an object’s density is less or greater than the fluid’s density.
记住：浮或沉取决于物体的密度是小于还是大于流体的密度。
5. Case 4: Hooke's Law and Spring Extension | 案例4：胡克定律与弹簧伸长
th = 17.2 cm. Plot a graph of force vs. extension, determine the spring constant, and predict the length at 3.0 N. State Hooke’s law.
一名学生在弹簧上悬挂砝码并记录如下：在 0 N 时，长度为 10.0 cm；在 2.0 N 时，长度为 12.4 cm；在 4.0 N 时，长度为 14.8 cm；在 6.0 N 时，长度为 17.2 cm。绘制力与伸长量的关系图，确定弹簧常数，并预测在 3.0 N 时的长度。陈述胡克定律。
tional to extension. Hooke’s law: F = kx, where k is the spring constant. Using 4.0 N and 0.048 m (convert to metres), k = F / x = 4.0 / 0.048 ≈ 83.3 N/m. At 3.0 N, extension = F / k = 3.0 / 83.3 ≈ 0.036 m = 3.6 cm. Predicted length = 10.0 + 3.6 = 13.6 cm.
伸长量 = 拉伸后的长度 − 原长（10.0 cm）。2.0 N时：伸长量 = 2.4 cm；4.0 N时：4.8 cm；6.0 N时：7.2 cm。数据表明力与伸长量成正比。胡克定律：F = kx，其中k是弹簧常数。利用 4.0 N 和 0.048 m（换算为米），k = F / x = 4.0 / 0.048 ≈ 83.3 N/m。在 3.0 N 时，伸长量 = F / k = 3.0 / 83.3 ≈ 0.036 m = 3.6 cm。预测长度 = 10.0 + 3.6 = 13.6 cm。
Be careful to always convert to standard units and check the limit of proportionality — beyond it, Hooke’s law no longer applies.
注意始终要换算为标准单位，并检查比例极限——超出极限后，胡克定律不再适用。
6. Case 5: Using a Lever to Lift a Load | 案例5：用杠杆提起重物
A worker uses a crowbar as a lever to lift a heavy crate. The effort is applied 1.5 m from the pivot, and the load (the crate) is 0.3 m from the pivot on the opposite side. The crate’s weight is 800 N. Calculate the minimum effort required to lift the crate. What class of lever is this, and how does it provide mechanical advantage?
一名工人用撬棍作为杠杆来抬起一个沉重的板条箱。动力作用点距支点 1.5 m，负载（板条箱）在支点另一侧 0.3 m 处。板条箱重 800 N。计算抬起箱子所需的最小动力。这是什么类型的杠杆？它如何提供机械优势？
This is a first-class lever (pivot between effort and load). Mechanical advantage = load / effort = 800 / 160 = 5. The lever multiplies the force by a factor of 5, meaning the worker only needs to exert 160 N to lift 800 N.
对于平衡的杠杆：动力 × 动力臂 = 负载 × 负载臂。因此，动力 × 1.5 = 800 × 0.3。求得动力 = (800 × 0.3) / 1.5 = 240 / 1.5 = 160 N。这是一类杠杆（支点在动力和负载之间）。机械优势 = 负载 / 动力 = 800 / 160 = 5。杠杆将力放大了 5 倍，意味着工人只需施加 160 N 的力就能提起 800 N 的重物。
strates the principle of moments: clockwise moment = anticlockwise moment for equilibrium.
这展示了力矩原理：平衡状态下，顺时针力矩 = 逆时针力矩。
7. Common Mistakes and How to Avoid Them | 常见错误及避免方法
One frequent mistake is mixing up units — using centimetres instead of metres in calculations, or grams instead of kilograms when dealing with force and energy. Always convert to SI units before substituting into formulas. Another error is forgetting to state the direction of energy transfer or the assumption of no energy loss. Students also sometimes misinterpret series and parallel circuits, causing incorrect resistance calculations.
一个常见错误是弄混单位——在计算中使用厘米而非米，或在涉及力和能量时用克而非千克。代入公式前，务必换算为国际单位。另一个错误是忘记说明能量传递的方向或忽略“无能量损失”的假设。学生有时也会混淆串联和并联电路，导致电阻计算错误。
When plotting graphs, ensure the independent variable is on the x-axis, label axes with quantities and units, and draw a best-fit line rather than simply joining points. In Hooke’s law problems, always calculate extension by subtracting the original length.
绘制图表时，确保自变量在 x 轴，用物理量和单位标注坐标轴，并画出最佳拟合线，而非简单连接各点。在胡克定律问题中，始终通过减去原长来计算伸长量。
8. Tips for Success in Physics Case Studies | 物理案例分析的提分技巧
To excel in case studies, practise reading the problem carefully and highlight key numbers and terms like 'constant speed', 'neglect friction', or 'series circuit'. Show your working step-by-step and write down the formulas used. Even if the final answer is wrong, you can earn marks for correct reasoning. Always include units in your final answer and check whether your result makes sense in the real world (e.g., would a person really be able to lift that? Is the speed realistic?).
要在案例分析中脱颖而出，需练习仔细阅读题目，并圈出关键数字和术语，如“匀速”、“忽略摩擦”或“串联电路”。分步骤展示计算过程，并写出所用的公式。即便最终答案错误，正确的推理也能得分。最终答案务必包含单位，并检查结果在现实中是否合理（例如，一个人真的能抬起那个重物吗？这个速度合乎实际吗？）。
Finally, connect each case to the underlying physics principle, such as conservation of energy, Ohm’s law, Archimedes’ principle, or moments. This not only strengthens your answer but also deepens your conceptual understanding.
最后，将每个案例与背后的物理原理联系起来，如能量守恒、欧姆定律、阿基米德原理或力矩。这不仅能充实你的答案，还能加深概念理解。
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