📚 Year 8 SQA Physics: Case Study Practical Exercises | Year 8 SQA 物理：案例分析实战演练
In Year 8 SQA Physics, learning is not just about remembering facts – it is about applying your knowledge to new situations. Case studies help you develop problem-solving skills and see how physics explains the world around us. This article provides a series of practical case studies, each with a real-world scenario, guiding you through the analysis step by step. By working through these exercises, you will strengthen your understanding of motion, forces, energy, electricity, matter, and waves.
在Year 8 SQA物理课程中，学习不仅仅是记忆知识点，更是将所学知识应用到新的情境中。案例分析能帮助你培养解决问题的能力，并理解物理是如何解释我们周围世界的。本文提供了一系列实用的案例分析，每一个都包含真实场景，并逐步引导你进行分析。通过练习这些题目，你将加深对运动、力、能量、电、物质和波的理解。
1. Analysing a Cyclist's Motion | 1. 分析自行车运动员的运动
A cyclist recorded the distance travelled at different times during a short race. The data is shown in the table below. We can use this data to calculate average speeds for each segment and describe the motion.
一位自行车运动员记录了短距离比赛中不同时刻行驶的距离。数据如下表所示。我们可以利用这些数据计算每一段的平均速度，并描述运动情况。
| Time (s) / 时间 (秒) | Distance (m) / 距离 (米) |
|---|---|
| 0 | 0 |
| 3 | 18 |
| 6 | 45 |
| 9 | 45 |
| 12 | 75 |
Average speed is calculated using the formula:
平均速度用以下公式计算：
v = Δd / Δt
where Δd is the change in distance and Δt is the change in time. Let us find the speed for each interval.
其中 Δd 是距离的变化量，Δt 是时间的变化量。现在我们来计算每个区间的速度。
From 0 to 3 s: Δd = 18 – 0 = 18 m, Δt = 3 s, so v = 18 / 3 = 6 m/s.
从 0 到 3 秒：Δd = 18 – 0 = 18 m，Δt = 3 s，因此 v = 18 / 3 = 6 m/s。
From 3 to 6 s: Δd = 45 – 18 = 27 m, Δt = 3 s, so v = 27 / 3 = 9 m/s.
从 3 到 6 秒：Δd = 45 – 18 = 27 m，Δt = 3 s，因此 v = 27 / 3 = 9 m/s。
From 6 to 9 s: Δd = 45 – 45 = 0 m, Δt = 3 s, so v = 0 / 3 = 0 m/s. The cyclist was stationary.
从 6 到 9 秒：Δd = 45 – 45 = 0 m，Δt = 3 s，因此 v = 0 / 3 = 0 m/s。自行车运动员处于静止状态。
From 9 to 12 s: Δd = 75 – 45 = 30 m, Δt = 3 s, so v = 30 / 3 = 10 m/s. This is the fastest segment.
从 9 到 12 秒：Δd = 75 – 45 = 30 m，Δt = 3 s，因此 v = 30 / 3 = 10 m/s。这是最快的区间。
The analysis shows that the cyclist accelerated at first, then stopped briefly, and finally sprinted at the highest speed. Interpreting graphs and data tables in this way is a key skill in physics.
分析表明，自行车运动员起初加速，然后短暂停下，最后以最高速度冲刺。用这种方式解读图表和数据表格是物理学中的一项关键技能。
2. Weight on Earth and the Moon | 2. 地球和月球上的重量
and depends on the gravitational field strength, g. On Earth, g ≈ 10 N/kg; on the Moon, g ≈ 1.6 N/kg.
一位宇航员的质量为 80 kg。质量是物体所含物质的量，不会随位置而改变。然而，重量是由于重力而产生的力，取决于引力场强度 g。在地球上，g ≈ 10 N/kg；在月球上，g ≈ 1.6 N/kg。
Weight is calculated using:
重量用下式计算：
W = m × g
ight on the Moon is much smaller, even though the mass stays exactly the same. This explains why astronauts can jump higher and lift heavy objects more easily on the lunar surface. It also reminds us that weight is a force, measured in newtons (N), while mass is measured in kilograms (kg).
在地球上：W = 80 kg × 10 N/kg = 800 N。在月球上：W = 80 kg × 1.6 N/kg = 128 N。尽管质量完全相同，但月球上的重量小得多。这就解释了为什么宇航员在月球表面能跳得更高，更容易举起重物。这也提醒我们，重量是一种力，单位是牛顿 (N)，而质量则以千克 (kg) 为单位。
3. Energy on a Roller Coaster | 3. 过山车上的能量
A roller coaster car of mass 500 kg sits at the top of the first hill, 30 m above the ground. We take the ground as the reference level where gravitational potential energy (Eₚ) is zero.
一辆质量为 500 kg 的过山车停在了最高点的第一个山坡上，离地 30 m。我们以地面作为重力势能 (Eₚ) 为零的参考面。
The gravitational potential energy is given by:
重力势能由下式给出：
Eₚ = mgh
000 J. If we ignore friction and air resistance, all of this potential energy will be converted into kinetic energy (Eₖ) at the bottom of the hill.
取 g = 10 N/kg：Eₚ = 500 × 10 × 30 = 150,000 J。如果忽略摩擦和空气阻力，所有这些势能将在山坡底部转化为动能 (Eₖ)。
Kinetic energy is expressed as:
动能表示为：
Eₖ = ½mv²
150,000, which simplifies to 250 v² = 150,000, so v² = 600. Taking the square root gives:
令 Eₖ = 150,000 J，得到 ½ × 500 × v² = 150,000，简化得 250 v² = 150,000，因此 v² = 600。开平方得：
v = √600 ≈ 24.5 m/s
This is the speed at the lowest point. In reality, some energy is transferred to the surroundings due to friction, so the actual speed would be slightly lower. This case study illustrates the principle of conservation of energy and the interchange between Eₚ and Eₖ.
这是最低点的速度。在现实中，由于摩擦，部分能量会转移到周围环境中，因此实际速度会略低。本案例展示了能量守恒原理以及 Eₚ 与 Eₖ 之间的相互转化。
4. Troubleshooting a Series Circuit | 4. 排障串联电路
A student builds a series circuit with a 6 V battery and two identical lamps, Lamp A and Lamp B. When the switch is closed, Lamp A lights up brightly, but Lamp B remains off. The student needs to identify the possible fault.
一名学生用一节 6 V 电池和两个相同的灯泡（灯泡 A 和灯泡 B）搭建了一个串联电路。当开关闭合时，灯泡 A 亮了，但灯泡 B 一直不亮。这名学生需要找出可能的故障。
In a series circuit, the same current flows through all components. If one lamp is working and the other is not, the fault could be:
在串联电路中，所有元件都流过相同的电流。如果一个灯泡亮而另一个不亮，故障可能是：
A broken filament in Lamp B – this creates an open circuit, but in a true series circuit, this would stop current everywhere, so Lamp A would also go out. Therefore, this is unlikely if Lamp A is lit.
灯泡 B 的灯丝断了——这会形成断路，但在真正的串联电路中，这将导致电路中没有任何电流，因此灯泡 A 也会熄灭。既然灯泡 A 亮着，这个可能性不大。
A short circuit across Lamp B – if a wire or a fault creates a path of very low resistance in parallel with Lamp B, current will bypass Lamp B and flow through the short circuit. Lamp A still receives the full current, so it glows. This is the most likely explanation.
灯泡 B 被短路——如果一根导线或故障在灯泡 B 两端形成了一条电阻极低的路径，电流便会绕过灯泡 B 流过短路处。灯泡 A 仍能获得全部电流，因此发光。这是最有可能的解释。
To confirm, a voltmeter could be connected across Lamp B. If the reading is 0 V, a short circuit is present. If the reading is 6 V, the lamp is open-circuited (but then Lamp A would be off). This fault-finding approach helps develop logical diagnostic skills.
为确认故障，可将电压表并联在灯泡 B 两端。如果读数为 0 V，说明存在短路；如果读数为 6 V，说明灯泡断路（但那样灯泡 A 也会熄灭）。这种排障方法有助于培养逻辑诊断能力。
5. Density and Floating | 5. 密度与浮沉
A student wants to predict whether an object will float or sink in water. The rule is simple: if the object's density is less than the density of water (1.0 g/cm³ or 1000 kg/m³), it floats; otherwise it sinks.
一名学生想预测一个物体在水中会浮起还是下沉。规则很简单：如果物体的密度小于水的密度（1.0 g/cm³ 或 1000 kg/m³），它就会浮起；否则就会下沉。
Object 1 has a mass of 200 g and a volume of 250 cm³. Its density is:
物体 1 的质量为 200 g，体积为 250 cm³。其密度为：
ρ = m / V = 200 g / 250 cm³ = 0.8 g/cm³
Since 0.8 g/cm³ < 1.0 g/cm³, Object 1 will float. In fact, exactly 80% of its volume will be submerged, with the remaining 20% above the surface, because its density is 0.8 times that of water.
由于 0.8 g/cm³ < 1.0 g/cm³，物体 1 将浮起。事实上，因其密度是水的 0.8 倍，它将有 80% 的体积浸没在水中，20% 露出水面。
Object 2 has a mass of 500 kg and a volume of 0.4 m³. Its density in SI units is:
物体 2 的质量为 500 kg，体积为 0.4 m³。用国际单位制计算其密度：
ρ = 500 kg / 0.4 m³ = 1250 kg/m³
This is greater than 1000 kg/m³, so Object 2 will sink. Ships are built with large hollow hulls to reduce their average density, allowing them to float despite being made of steel. Understanding density is essential for designing boats, submarines, and even hot-air balloons.
这个值大于 1000 kg/m³，因此物体 2 会下沉。船舶通过建造大型空心船体来降低平均密度，使得它们即使用钢铁打造也能浮在水面上。理解密度对于设计船只、潜水艇甚至热气球都至关重要。
6. Designing a Vacuum Flask | 6. 设计保温瓶
A vacuum flask (or thermos) keeps liquids hot or cold by minimising heat transfer. The design tackles all three mechanisms: conduction, convection, and radiation.
保温瓶（或热水瓶）通过最大限度地减少热传递来保持液体热或冷。其设计同时应对了热传导、热对流和热辐射这三种传热方式。
The double-walled container with a vacuum between the walls almost eliminates conduction and convection, because both require a medium (solid or fluid). The vacuum contains very few particles to transfer energy.
双层容器壁之间为真空，几乎消除了热传导和对流，因为两者都需要介质（固体或流体）。真空中几乎没有粒子来传递能量。
The inner and outer surfaces are silvered to act like mirrors. This reduces heat transfer by radiation – the shiny surface reflects thermal radiation back into the liquid (or away from the cold contents), rather than emitting or absorbing it.
内壁和外壁表面镀银，像镜子一样。这减少了热辐射传递——光亮的表面将热辐射反射回液体中（或反射离开冷的物体），而不是发射或吸收辐射。
The stopper at the top is made from insulating material like cork or plastic, which is a poor conductor. It also seals the gap to prevent convection currents carrying heat in or out through the opening.
顶部的瓶塞由软木或塑料等隔热材料制成，是热的不良导体。它还密封了瓶口，防止对流气流通过开口带入或带出热量。
These features work together to keep tea hot for hours or to keep chilled juice cool. This case study links the physics of heat transfer to everyday design and engineering.
这些特性协同作用，使得热茶能保温数小时，或使冰镇果汁保持冰凉。本案例将热传递物理与日常设计和工程联系了起来。
7. Reflection from a Plane Mirror | 7. 平面镜反射
When you look into a flat bathroom mirror, you see a reflection that appears to be behind the glass. This is a virtual image – it cannot be projected onto a screen. The ray model of light helps explain the properties of this image.
当你照浴室的平面镜时，会看到一个似乎位于镜子后面的像。这是一个虚像——无法投射到屏幕上。光线的模型有助于解释该像的性质。
The image is the same size as the object. No magnification occurs with a plane mirror.
像与物体大小相同。平面镜不会产生放大或缩小。
The image is upright (the same way up as the object) but laterally inverted – left and right appear swapped.
像是正立的（与物体方向相同），但左右颠倒——左边和右边看起来互换了。
The distance from the object to the mirror equals the distance from the mirror to the image. If you stand 1 m away from the mirror, your image appears to be 1 m behind the mirror, making the total distance between you and your image 2 m.
物体到镜子的距离等于镜子到像的距离。如果你站在离镜子 1 m 处，你的像看起来在镜子后方 1 m 处，因此你与像的总距离为 2 m。
These observations can be explained by the law of reflection: the angle of incidence equals the angle of reflection, measured from the normal. When you draw ray diagrams, you can trace the reflected rays back behind the mirror to locate the virtual image. This knowledge is applied in periscopes, kaleidoscopes, and rear-view mirrors.
这些观察可以用反射定律解释：入射角等于反射角，均从法线量起。当你绘制光路图时，可以将反射光线反向延长到镜后，从而确定虚像的位置。这一知识被应用于潜望镜、万花筒和汽车后视镜中。
8. Pressure Under a Stiletto Heel | 8. 细高跟鞋下的压强
Pressure tells us how concentrated a force is. A stiletto heel can exert a huge pressure on a floor because the force (the person's weight) acts over a very small area.
压强告诉我们力有多集中。细高跟鞋会对地板施加巨大的压强，因为力（人的体重）作用在一个非常小的面积上。
erting to square metres: 0.5 cm² = 0.5 ÷ 10,000 = 0.00005 m². Pressure is calculated from:
假设一个人的质量为 50 kg。其重量 W = 50 kg × 10 N/kg = 500 N。当单脚鞋跟着地时，接触面积可能只有 0.5 cm²。换算成平方米：0.5 cm² = 0.5 ÷ 10,000 = 0.00005 m²。压强由下式计算：
P = F / A
ach with an area of 0.1 m². Total area = 4 × 0.1 = 0.4 m². The pressure under the elephant's feet is P = 50,000 N / 0.4 m² = 125,000 Pa.
因此 P = 500 N / 0.00005 m² = 10,000,000 Pa (10 MPa)。相比之下，一头大象质量为 5000 kg，重量为 50,000 N，四脚站立，每只脚面积 0.1 m²，总面积 = 4 × 0.1 = 0.4 m²。大象脚下的压强为 P = 50,000 N / 0.4 m² = 125,000 Pa。
The stiletto heel pressure (10,000,000 Pa) is 80 times greater than that of the elephant (125,000 Pa)! This explains why sharp heels can damage wooden floors, and why a sharp knife cuts more easily than a blunt one – the same force is concentrated over a much smaller area, producing a higher pressure.
细高跟鞋的压强（10,000,000 Pa）是大象压强（125,000 Pa）的 80 倍！这就解释了为什么尖细的鞋跟会损坏木地板，以及为什么锋利的刀比钝刀更容易切割——相同的力集中在更小的面积上，产生了更大的压强。
Published by TutorHao | Physics Revision Series | aleveler.com
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