📚 Case Study Practical Exercises | 案例分析实战演练
In Year 8 SQA Physics, applying theory to real-world scenarios is just as important as learning formulas. This article presents a series of case studies designed to sharpen your problem-solving skills, covering electricity, forces, energy, density, pressure, and more. Each case is broken down step by step, helping you to think like a scientist and tackle exam-style application questions with confidence.
在八年级苏格兰资格认证局物理课程中，将理论应用于实际场景与学习公式同样重要。本文提供了一系列案例研究，旨在提高你的问题解决能力，涵盖电学、力、能量、密度、压强等内容。每个案例都逐步分解，帮助你像科学家一样思考，自信地应对考试风格的应用题。
1. Diagnosing a Simple Circuit Fault | 简单电路故障诊断
Case: A student builds a series circuit with a battery, a switch, and two bulbs. When the switch is closed, neither bulb lights up. The battery is new, and all connections appear secure. What could be the fault, and how would you test each possibility?
案例：一名学生用电池、开关和两个灯泡搭建了一个串联电路。闭合开关后，两个灯泡都不亮。电池是新的，所有连接看起来都牢固。可能是什么故障？你会如何测试每种可能性？
First, recall that for a series circuit to work, there must be a complete conducting path. An open circuit anywhere will stop the current. Possible faults: a blown bulb (broken filament), a faulty switch, or an incorrect connection that bypasses a component. To diagnose, we could use a known working bulb to test each component individually, or use a voltmeter to check for voltage drops across components. If the switch is faulty, bridging it with a wire should light the bulbs. If one bulb is blown, replacing it will fix the circuit. Always test one variable at a time.
首先，回想一下，要使串联电路工作，必须有一个完整的导通路径。任何地方的断路都会阻止电流。可能的故障包括：灯泡烧坏（灯丝断裂）、开关故障，或者错误的连接导致某个部件被旁路。诊断时，我们可以用一个已知完好的灯泡单独测试每个元件，或者用电压表检查元件两端的电压降。如果开关有故障，用导线将它短接应该能使灯泡点亮。如果其中一个灯泡烧坏了，更换它就能修复电路。每次只测试一个变量。
2. Forces in a Tug-of-War | 拔河中的力
Case: Team A pulls a rope to the left with a force of 450 N. Team B pulls to the right with 520 N. There is also a frictional force of 30 N acting on the rope against the ground. Calculate the resultant force and determine the direction of motion.
案例：A队用450牛的力向左拉绳子，B队用520牛的力向右拉。绳子与地面之间还有30牛的摩擦力。计算合力并确定运动方向。
tion is to the right (+70 N), friction acts to the left (-30 N). Resultant force = 520 − 450 − 30 = +40 N. The rope accelerates to the right with a net force of 40 N. This shows how to combine horizontal forces, remembering that friction always opposes the direction of sliding or intended motion.
我们将向右设为正方向。B队的力：+520牛；A队的力：-450牛。摩擦力阻碍运动；由于没有摩擦力时的净力向右（+70牛），所以摩擦力向左（-30牛）。合力 = 520 − 450 − 30 = +40牛。绳子以40牛的合力向右加速。这展示了如何合并水平力，谨记摩擦力总是与滑动方向或即将运动的方向相反。
3. Energy Conversion in a Roller Coaster | 过山车中的能量转换
gligible friction, calculate its kinetic energy and speed at the bottom. (Use g = 10 N/kg)
案例：一辆质量为500千克的过山车位于30米高的山顶。假设摩擦力可忽略不计，计算它到达底部时的动能和速度。（g取10牛/千克）
Gravitational potential energy at the top: Ep times; 30 = 150,000 J. At the bottom, all this energy has converted into kinetic energy (Ek). So Ek = 150,000 J. Using Ek = ½ mv²;, we rearrange: v² = 2Ek case illustrates the principle of conservation of mechanical energy and the interchange between potential and kinetic energy.
山顶的重力势能：Ep = mgh = 500 × 10 × 30 = 150,000焦。到达底部时，所有能量都转化为动能（Ek）。因此Ek = 150,000焦。利用Ek = ½ mv²;，整理得：v² = 2Ek / m = 2 × 150,000 / 500 = 600。所以v = √600 ≈ 24.5米/秒。该案例展示了机械能守恒原理以及势能和动能之间的相互转化。
4. Calculating Density from Irregular Shape | 通过不规则形状计算密度
Case: A stone of mass 240 g is placed into a measuring cylinder containing 80 cm³ of water. The water level rises to 140 cm³. Calculate the density of the stone in g/cm³ and in kg/m³.
案例：将一块质量为240克的石头放入装有80立方厘米水的量筒中，水面上升到140立方厘米。计算石头的密度，以克/立方厘米和千克/立方米表示。
nvert to kg/m³, multiply by 1000: 4.0 × 1000 = 4000 kg/m³. This method of water displacement is essential for finding the volume of irregular objects, and linking units helps reinforce the relationship between g/cm³ and kg/m³.
石头的体积 = 最终体积 − 初始体积 = 140 − 80 = 60立方厘米。密度 = 质量 / 体积 = 240克 / 60立方厘米 = 4.0克/立方厘米。换算成千克/立方米需乘以1000：4.0 × 1000 = 4000千克/立方米。这种排水法对于测量不规则物体的体积至关重要，单位换算有助于巩固克/立方厘米与千克/立方米之间的关系。
5. Pressure Exerted by a Stiletto Heel | 高跟鞋跟产生的压强
Case: A person weighing 600 N stands on one stiletto heel with a circular area of 0.5 cm². Calculate the pressure on the floor in N/cm² and Pa. Compare this to the pressure when standing flat-footed on an area of 120 cm².
案例：一个重600牛的人用一只高跟鞋跟站立，鞋跟为圆形，面积为0.5平方厘米。计算地板受到的压强，以牛/平方厘米和帕斯卡表示。将此与平足站立时120平方厘米的面积所产生的压强进行比较。
00 = 12,000,000 Pa (12 MPa). For flat foot: P = 600 / 120 = 5 N/cm², which is 50,000 Pa. The enormous pressure from the heel explains why stilettos can damage wooden floors. This demonstrates the inverse relationship between pressure and area.
压强 = 力 / 面积。对于鞋跟：P = 600 / 0.5 = 1200牛/平方厘米。将牛/平方厘米转换为帕斯卡，注意1牛/平方厘米 = 10,000帕，所以P = 1200 × 10,000 = 12,000,000帕（12兆帕）。对于平足：P = 600 / 120 = 5牛/平方厘米，即50,000帕。鞋跟产生的巨大压强解释了为什么高跟鞋会损坏木地板。这说明了压强与面积之间的反比关系。
6. Lever as a Simple Machine | 作为简单机械的杠杆
Case: A crowbar is used to lift a heavy rock. The fulcrum is placed 0.2 m from the rock, and the effort is applied 1.2 m from the fulcrum. If the rock has a weight of 800 N, what minimum effort is needed? What is the mechanical advantage?
案例：用一根撬棍撬起一块重石头。支点离石头0.2米，动力作用点离支点1.2米。如果石头重800牛，至少需要多大的动力？机械效益是多少？
800 × 0.2) / 1.2 = 160 / 1.2 ≈ 133.3 N. Mechanical advantage (MA) = load / effort = 800 / 133.3 ≈ 6.0. Alternatively, MA = effort arm / load arm = 1.2 / 0.2 = 6. This shows how a lever multiplies force, making heavy tasks easier.
对于处于平衡状态的杠杆：动力 × 动力臂 = 阻力 × 阻力臂。阻力臂 = 0.2米，动力臂 = 1.2米。因此 E × 1.2 = 800 × 0.2。E = (800 × 0.2) / 1.2 = 160 / 1.2 ≈ 133.3牛。机械效益 (MA) = 负载 / 动力 = 800 / 133.3 ≈ 6.0。或者，MA = 动力臂 / 阻力臂 = 1.2 / 0.2 = 6。这显示了杠杆如何放大力量，使重活变得轻松。
7. Conduction, Convection, and Radiation in a Room | 房间中的热传导、热对流和热辐射
Case: A radiator is placed under a window. A person sitting 2 metres away feels warm even though air is a poor conductor. Explain the mechanisms of heat transfer involved.
案例：散热器安装在窗户下方。一个坐在两米外的人即使空气是热的不良导体也会感到温暖。解释所涉及的热传递机制。
The radiator heats the air in contact with it by conduction. Warm air expands, becomes less dense, and rises — this is convection, setting up a circulation current that distributes heat around the room. Additionally, the radiator emits infrared radiation, which travels in straight lines, directly warming objects and people without heating the air itself. The person feels warmth from both the warm air convection and the radiant heat. The window placement helps counteract cold drafts descending from the glass.
散热器通过传导加热与之接触的空气。暖空气膨胀，密度变小并上升——这是对流，形成循环气流，将热量分布到整个房间。此外，散热器发出红外辐射，以直线传播，直接加热物体和人，而不加热空气本身。人从温暖的空气对流和辐射热两方面感到温暖。将散热器安装在窗下有助于抵消从玻璃下沉的冷气流。
8. Reflection of Light from a Plane Mirror | 平面镜对光的反射
Case: A ray of light strikes a plane mirror at an angle of incidence of 35°. Draw the reflected ray and state the angle of reflection. If the mirror is rotated by 10° clockwise, what happens to the reflected ray?
案例：一束光线以35°的入射角射到平面镜上。画出反射光线并说明反射角。如果平面镜顺时针旋转10°，反射光线会怎样？
ted ray makes a 35° angle with the normal, on the opposite side. If the mirror rotates by 10° clockwise, the normal also rotates by 10°, so the incident angle changes by 10°. The new angle of incidence becomes 45° (if rotation increases it) or 25° (if decreases), depending on direction. Consequently, the reflected ray changes by twice the rotation angle, i.e., 20°. This demonstrates how mirror rotation affects beam deflection.
反射定律：入射角等于反射角。因此反射光线与法线成35°角，位于法线另一侧。如果镜面顺时针旋转10°，法线也旋转10°，因此入射角改变10°。新的入射角变为45°（如果旋转增大了角度）或25°（如果减小了角度），具体取决于方向。因此，反射光线改变的角度是镜面旋转角度的两倍，即20°。这展示了镜面旋转如何影响光束偏转。
9. Speed of Sound vs. Speed of Light in a Thunderstorm | 雷暴中声速与光速的比较
Case: During a thunderstorm, you see a lightning flash and hear the thunder 4 seconds later. If the speed of sound is 340 m/s, how far away was the lightning? Why can we assume the light travel time is negligible?
案例：在雷暴中，你看到闪电后4秒听到雷声。如果声速为340米/秒，闪电离你有多远？为什么我们可以假设光的传播时间可以忽略不计？
d of light is so great (300,000,000 m/s) that the light from the flash reaches you almost instantaneously over this short distance. Thus, the time delay is solely due to the slower speed of sound. This calculation is a classic application of the wave speed equation, linking real-world observation to physical principles.
距离 = 速度 × 时间。声音用了4秒钟到达你，所以距离 = 340 × 4 = 1360米。光速极快（300,000,000米/秒），在如此短的距离内，闪光几乎瞬间到达你。因此，时间延迟完全是由于声速较慢造成的。这个计算是波速方程的经典应用，将现实世界观察与物理原理联系起来。
10. Investigating Springs and Hooke's Law | 探究弹簧与胡克定律
Case: A spring is 10 cm long. When a 2 N weight is hung on it, the length becomes 14 cm. When a 5 N weight is added, the length becomes 22 cm. Is the spring obeying Hooke's Law? Find the spring constant.
案例：一根弹簧原长10厘米。挂上2牛的砝码后，长度变为14厘米。再挂上5牛砝码后，长度变为22厘米。弹簧是否遵循胡克定律？求出弹簧的劲度系数。
.14 m = 14 cm, so total length = 10 + 14 = 24 cm. But observed length is 22 cm, extension = 12 cm. This is less than expected, so the spring does not obey Hooke's Law for larger forces— c limit. The case highlights how to test proportionality and the concept of elastic limit.
对于2牛的伸长量：14 − 10 = 4厘米 = 0.04米。劲度系数k = F / x = 2 / 0.04 = 50牛/米。对于5牛，总力 = 2 + 5 = 7牛。预期伸长量：x = F / k = 7 / 50 = 0.14米 = 14厘米，因此总长 = 10 + 14 = 24厘米。但观察到的长度是22厘米，伸长量为12厘米。这比预期小，因此弹簧在较大力时不遵循胡克定律——它很可能已经超过了弹性极限。这个案例强调了如何测试正比关系以及弹性极限的概念。
11. Balancing a See-Saw with Two Different Weights | 用两个不同重量平衡跷跷板
Case: A see-saw is 4 m long, pivoted in the middle. A child weighing 300 N sits 1.5 m from the pivot on the left side. Where should a 400 N child sit on the right side to balance it? What if the pivot is moved to 1 m from the left end?
案例：一个跷跷板长4米，支点在中间。一个重300牛的孩子坐在左侧离支点1.5米处。一个重400牛的孩子应该坐在右侧什么位置才能平衡？如果支点移到离左端1米处呢？
m pivot on right. If pivot is 1 m from left end, the left child is at distance from pivot: 1 m from left end means pivot is 1 m from left, left child is at 1.5 m from original centre? m old centre, but new pivot is 1 m from left end, so distance of left child from new pivot? ld is still sitting at a point... The problem as stated is ambiguous; clarify: the see-saw' oard. Without knowing where the child is relative to the ends, we need to assume the child remains at the same physical position? Instead, we can assume the child's position relative to the pivot changes. Let' m) -> distance from pivot = 1 m. Right child at right end (4 m) -> distance = 3 m. Then 300 × 1 < 400 × 3, not balanced. To balance, right child must sit at distance d from pivot: 300 × 1 = 400 × d => d = 0.75 m from pivot, i.e., 1.75 m from left end. This problem tests the principle of moments in a new configuration, demonstrating that pivot position drastically affects balance.
当顺时针力矩等于逆时针力矩时平衡。对于支点在中间的情况：300 × 1.5 = 400 × d，所以 d = (300 × 1.5) / 400 = 1.125米（从支点向右）。如果支点移到离左端1米处，我们需要重新考虑位置。假设两个孩子坐在板的两端：左孩子在左端（0米），距离新支点1米；右孩子在右端（4米），距离支点3米。那么300 × 1 < 400 × 3，不平衡。为达到平衡，右孩子必须坐在距离支点d处：300 × 1 = 400 × d => d = 0.75米，即从支点向右0.75米，或离左端1.75米。这个问题在新结构下测试了力矩原理，表明支点位置会极大影响平衡。
12. Designing an Experiment to Test Friction on Different Surfaces | 设计实验测试不同表面的摩擦力
Case: You are given a wooden block, a spring balance, and three surfaces: sandpaper, glass, and carpet. Design an experiment to investigate how surface roughness affects friction force. Identify variables and describe a risk assessment.
案例：给你一个木块、一个弹簧秤和三种表面：砂纸、玻璃和地毯。设计一个实验来探究表面粗糙度如何影响摩擦力。确定变量并描述风险评估。
Independent variable: type of surface (roughness). Dependent variable: friction force (measured by the spring balance when pulling the block at constant speed just as it starts to move, or the reading when it moves steadily). Control variables: mass of block (add weights if needed to keep constant), surface area of contact (use same face of block), pulling speed, spring balance. Method: Place block on surface, attach spring balance, pull horizontally with increasing force until block just begins to move; record the maximum static friction. Repeat for each surface several times and average. Risk: spring balance may snap back; wear safety goggles. Trip hazard from carpet edge. The experiment shows that rougher surfaces generally produce greater friction.
自变量：表面类型（粗糙度）。因变量：摩擦力（用弹簧秤测量，当木块刚开始运动时匀速拉动时的读数，或稳定运动时的读数）。控制变量：木块质量（如有需要可加砝码保持恒定）、接触面积（使用木块的同一面）、拉动速度、弹簧秤。方法：将木块放在表面上，连接弹簧秤，水平施加递增的力直至木块刚好开始移动；记录最大静摩擦力。对每种表面重复几次并取平均值。风险：弹簧秤可能会弹回；佩戴护目镜。地毯边缘可能造成绊倒危险。实验表明，表面越粗糙通常产生的摩擦力越大。
Published by TutorHao | Physics Revision Series | aleveler.com
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