Case Study Practical Exercises | 案例分析实战演练

📚 Case Study Practical Exercises | 案例分析实战演练

Case studies in physics help you apply what you’ve learned to real-life situations. They develop your problem‑solving skills and show you how the laws of physics work in everyday contexts. This article presents ten hands‑on case studies designed for Year 7 SQA Physics. Each case asks you to analyse data, carry out calculations, and draw conclusions from the evidence. Read each study carefully, try to work out the answers before reading the solutions, and strengthen your understanding of key topics such as speed, forces, circuits, energy, density, floating and sinking, sound, light and simple machines.

物理案例分析能帮助你运用所学知识解决真实情境中的问题。它们能提升你的解题能力，并让你看到物理定律在日常生活中的作用。本文为七年级SQA物理设计了十个动手案例。每个案例都要求你分析数据、进行计算并从证据中得出结论。仔细阅读每个案例，试着在看答案之前解出问题，从而加深你对速度、力、电路、能量、密度、沉浮、声音、光和简单机械等关键主题的理解。

1. Case Study 1: Calculating Speed | 案例1：计算速度

A cyclist travels 600 metres in 120 seconds. Calculate her average speed in metres per second.

一名自行车骑手在120秒内骑行了600米。计算她的平均速度，单位为米每秒。

Use the formula: speed = distance ÷ time. So, speed = 600 m ÷ 120 s = 5 m/s. The cyclist’s average speed is 5 metres per second.

使用公式：速度 = 距离 ÷ 时间。因此，速度 = 600 米 ÷ 120 秒 = 5 米/秒。这名骑手的平均速度为5米每秒。

Now a car moves at a constant speed of 20 m/s for 15 seconds. How far does it travel? Rearranging the equation: distance = speed × time. Distance = 20 m/s × 15 s = 300 m.

现在一辆汽车以20米/秒的恒定速度行驶了15秒。它行驶了多远？整理方程：距离 = 速度 × 时间。距离 = 20 米/秒 × 15 秒 = 300 米。

This case shows that once you understand the relationship between distance, speed and time, you can solve for any unknown quantity. Whether you are analysing a school sports day or planning a journey, the speed equation is a fundamental tool in physics.

这一案例表明，一旦你理解了距离、速度和时间之间的关系，你就能求出任何一个未知量。无论是分析学校运动会还是规划一次旅程，速度方程都是物理学中的一个基本工具。

2. Case Study 2: Forces in Action | 案例2：力的作用

Two teams engage in a tug‑of‑war. Team A pulls with a force of 450 N to the right, and Team B pulls with 380 N to the left. Determine the resultant force and its direction.

两队进行拔河比赛。A队向右施加450牛的拉力，B队向左施加380牛的拉力。确定合力及其方向。

The forces are opposite in direction. Subtract the smaller force from the larger: 450 N − 380 N = 70 N. The resultant force is 70 N towards the direction of Team A (to the right).

这两个力方向相反。用较大的力减去较小的力：450牛 − 380牛 = 70牛。合力为70牛，方向朝向A队（向右）。

If the forces were both acting in the same direction, say two people pushing a car with 200 N and 150 N forward, the resultant would be 200 N + 150 N = 350 N forward. Recognising when forces add up or cancel out is essential for understanding how objects start, stop or change direction.

如果两个力方向相同，比如两个人一起向前推车，力分别为200牛和150牛，那么合力就是200牛 + 150牛 = 350牛向前。识别力是相加还是抵消对于理解物体如何起动、停止或改变方向至关重要。

3. Case Study 3: Friction Investigation | 案例3：摩擦力研究

A pupil slides a wooden block across different surfaces and measures the force needed to keep it moving at a steady speed. The results are: wooden bench – 2.2 N, carpet – 5.8 N, and ice‑like sheet – 0.9 N. Explain the pattern.

一名学生让木块在不同表面上滑动，并测量使其保持匀速运动所需的力。结果如下：木制桌面—2.2牛，地毯—5.8牛，冰面状薄板—0.9牛。解释这一规律。

The force required equals the frictional force because the block moves at constant speed (balanced forces). Carpet has the roughest texture and produces the greatest friction, so more force is needed. The ice‑like sheet is very smooth, giving very low friction, so only a small force is required.

由于木块匀速运动（平衡力），所需的力等于摩擦力。地毯表面最粗糙，产生的摩擦力最大，因此需要的力也最大。冰面状薄板非常光滑，摩擦力很小，所以只需很小的力。

To reduce friction, engineers use lubricants or ball bearings. To increase friction, we use rougher materials or apply greater pressure. Carefully controlled experiments like this one reveal the quantitative link between surface roughness and friction.

为减小摩擦，工程师使用润滑剂或滚珠轴承。为增大摩擦，我们使用更粗糙的材料或施加更大的压力。像这样的受控实验揭示了表面粗糙度与摩擦力之间的定量联系。

4. Case Study 4: Simple Circuits | 案例4：简单电路

A series circuit contains a battery, a switch, and two identical bulbs. When the switch is closed, both bulbs light up. One bulb is then unscrewed. Predict what happens and explain why.

一个串联电路包含一个电池、一个开关和两个相同的灯泡。闭合开关时，两个灯泡都发光。然后旋下一个灯泡。预测会发生什么并解释原因。

In a series circuit, there is only one path for the current. If one bulb is removed or fails, the circuit is broken and current stops flowing. The other bulb will go out immediately.

在串联电路中，电流只有一条通路。如果一个灯泡被取下或损坏，电路断开，电流停止流动。另一个灯泡会立即熄灭。

Now imagine the bulbs are connected in parallel. Each bulb has its own separate loop. Removing one bulb does not break the complete circuit for the other. The second bulb will remain lit. This case highlights why household wiring uses parallel circuits—so that if one appliance fails, the others keep working.

现在假设灯泡是并联的。每个灯泡有自己独立的回路。取下一个灯泡不会断开另一个灯泡的完整电路，因此第二个灯泡仍会发光。这个案例说明了为什么家庭布线使用并联电路——这样即使一个电器出故障，其他电器也能继续工作。

5. Case Study 5: Energy Transfers | 案例5：能量转换

A battery‑operated fan is switched on. Describe the main energy changes that occur from the battery to the spinning blades.

打开一个电池驱动的小风扇。描述从电池到扇叶转动过程中发生的主要能量变化。

The chemical energy stored in the battery is converted into electrical energy in the wires. The electrical energy is then changed into kinetic (movement) energy of the fan blades. Some energy is also transferred to the surroundings as thermal energy (heat) and sound. The overall transfer can be summarised as: chemical → electrical → kinetic + thermal + sound.

电池中储存的化学能转化为导线中的电能。电能随后转化为风扇叶片的动能（运动能）。还有一部分能量以热能（热量）和声能的形式散逸到周围环境中。总的能量转移可概括为：化学能 → 电能 → 动能 + 热能 + 声能。

Energy is never created or destroyed, only transferred. Identifying the starting and ending stores of energy helps you understand efficiency. In a real fan, not all the electrical energy ends up as useful kinetic energy—some is wasted as heat and sound.

能量既不会凭空产生也不会消失，只会发生转移。辨识能量的起点和终点存储形式有助于你理解效率。在实际的风扇中，并非所有的电能都转化为了有用的动能——一部分以热和声的形式被浪费了。

6. Case Study 6: Density Dilemma | 案例6：密度难题

A metal cube has a mass of 540 g and measures 2 cm × 3 cm × 10 cm. Calculate its density in g/cm³. Is it likely to be aluminium (density about 2.7 g/cm³)?

一个金属立方体质量为540克，尺寸为2厘米 × 3厘米 × 10厘米。计算其密度，单位是克/立方厘米。它可能是铝吗（密度大约为2.7克/立方厘米）？

First, find the volume: 2 cm × 3 cm × 10 cm = 60 cm³. Density = mass ÷ volume = 540 g ÷ 60 cm³ = 9 g/cm³. This density is much higher than aluminium’s, so it is not aluminium. Pure copper, for example, has a density close to 8.96 g/cm³, so the cube could be made of a metal like copper or brass.

首先，计算体积：2厘米 × 3厘米 × 10厘米 = 60立方厘米。密度 = 质量 ÷ 体积 = 540克 ÷ 60立方厘米 = 9克/立方厘米。这一密度远高于铝的密度，因此它不是铝。例如，纯铜的密度接近8.96克/立方厘米，所以这个立方体可能是由铜或黄铜等金属制成的。

Using density to identify materials is a common technique in forensic science and engineering. By measuring mass and volume, you can match an unknown sample to a list of known densities and make a reliable identification.

利用密度来鉴别材料是法医学和工程学中常用的技术。通过测量质量和体积，你可以将未知样品与一系列已知密度进行比对，并作出可靠的鉴定。

7. Case Study 7: Sinking and Floating | 案例7：沉浮问题

An object weighs 8 N in air and 5 N when fully submerged in water. Calculate the upthrust (buoyant force) and determine whether the object floats or sinks if placed gently on the water surface.

一个物体在空气中重8牛，完全浸没在水中时称重为5牛。计算它所受到的浮力，并判断如果将其轻轻放在水面上，它会漂浮还是下沉。

Upthrust = weight in air − apparent weight in water = 8 N − 5 N = 3 N. Since the upthrust when fully submerged is only 3 N, while the object’s weight is 8 N, the weight is greater than the maximum possible upthrust. Therefore, the object will sink. If the upthrust had been greater than or equal to the weight, the object would float.

浮力 = 在空气中的重量 − 在水中的视重 = 8牛 − 5牛 = 3牛。由于完全浸没时的浮力仅为3牛，而物体的重量为8牛，重力大于可能产生的最大浮力，因此物体会下沉。若浮力大于或等于重量，物体就会漂浮。

Designing ships and submarines relies on manipulating buoyancy. By changing the shape to displace more water, a steel ship can float even though steel itself is denser than water. This case demonstrates why volume and shape matter just as much as density.

设计船舶和潜艇依赖于对浮力的操控。通过改变形状以排开更多的水，一艘钢制轮船能够漂浮，即使钢铁本身的密度大于水。这个案例证明了体积和形状与密度同样重要。

8. Case Study 8: Sound Waves in Action | 案例8：声波应用

A student claps her hands near a large, flat wall and hears an echo 0.3 seconds later. Given that the speed of sound in air is about 340 m/s, estimate the distance to the wall.

一名学生在一面平整的大墙附近拍手，0.3秒后听到了回声。已知空气中的声速大约为340米/秒，估算她到墙的距离。

The sound travels to the wall and back, so the total distance it covers is speed × time = 340 m/s × 0.3 s = 102 m. This is the round‑trip distance. The one‑way distance to the wall is half of that: 102 m ÷ 2 = 51 m. Therefore, the wall is about 51 metres away.

声音传播到墙再反射回来，因此它所经过的总距离为速度 × 时间 = 340 米/秒 × 0.3 秒 = 102 米。这是往返距离。到墙的单程距离为一半：102 米 ÷ 2 = 51 米。因此，墙大约在51米外。

Echolocation works on the same principle. Bats and dolphins send out high‑frequency sounds and listen for echoes to gauge distances and locate prey. This case connects basic wave physics to how animals and technology (such as sonar) navigate the world.

回声定位依据的是同样的原理。蝙蝠和海豚发出高频声音，并倾听回声来估计距离和定位猎物。这个案例将基础的波动物理学与动物和技术（例如声纳）如何感知世界联系起来。

9. Case Study 9: Reflection of Light | 案例9：光的反射

A ray of light strikes a plane mirror with an angle of incidence of 35°. Draw a diagram in your mind and state the angle of reflection. Then determine the angle between the incident ray and the reflected ray.

一束光线以35°的入射角射向平面镜。在脑海中画出简图，并说出反射角的大小。然后确定入射光线与反射光线之间的夹角。

By the law of reflection, the angle of incidence equals the angle of reflection. So the angle of reflection is also 35°. The incident ray strikes the mirror, and the reflected ray bounces off. The total angle between them is the sum of the angle of incidence and the angle of reflection: 35° + 35° = 70°.

根据反射定律，入射角等于反射角。因此反射角也为35°。入射光线射向镜面，反射光线从镜面反弹。它们之间的总夹角是入射角与反射角之和：35° + 35° = 70°。

If you rotate the mirror by 10°, the reflected ray turns by 20°. This simple geometry is used in periscopes, kaleidoscopes and many optical instruments. Mastering the law of reflection allows you to predict how mirrors will redirect light in a huge range of practical setups.

如果你将镜子旋转10°，反射光线会转动20°。这种简单的几何关系被应用于潜望镜、万花筒和许多光学仪器中。掌握反射定律能使你预测在各种实际装置中镜子如何重新定向光线。

10. Case Study 10: Levers and Effort | 案例10：杠杆与力

A lever is used to lift a heavy rock. The effort is applied 2.0 m from the pivot, and the rock sits 0.5 m from the pivot on the other side. If the rock has a weight of 600 N, calculate the minimum effort required to just lift it (ignore the weight of the lever).

用一根杠杆来抬起一块大石头。动力施加在距离支点2.0米处，石头位于支点另一侧0.5米处。如果石头的重量为600牛，计算刚好能抬起它所需的最小动力（忽略杠杆自身重量）。

For a lever in equilibrium, the clockwise moment equals the anticlockwise moment. Moment = force × distance from pivot. Let the effort be F. Then F × 2.0 m = 600 N × 0.5 m. So F = (600 N × 0.5 m) ÷ 2.0 m = 300 Nm ÷ 2.0 m = 150 N. An effort of only 150 N can lift a 600 N rock.

对于平衡的杠杆，顺时针力矩等于逆时针力矩。力矩 = 力 × 到支点的距离。设需要的动力为 F。则 F × 2.0 米 = 600 牛 × 0.5 米。所以 F = (600 牛 × 0.5 米) ÷ 2.0 米 = 300 牛·米 ÷ 2.0 米 = 150 牛。仅仅150牛的力就能抬起600牛的石头。

Levers multiply effort by increasing the distance at which the effort acts. This is why crowbars, scissors and wheelbarrows make tasks easier. Identifying pivot, load and effort positions in real‑life levers helps you calculate the mechanical advantage and design more efficient tools.

杠杆通过增大动力作用的距离来放大动力。这就是撬棍、剪刀和手推车能省力的原因。在现实生活中的杠杆上辨识支点、负载和动力的位置，可以帮助你计算机械效益并设计出更高效的工具。

Published by TutorHao | Physics Revision Series | aleveler.com

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Case Study Practical Exercises | 案例分析实战演练