📚 Year 7 Edexcel Physics: Case Study Practice | 7年级Edexcel物理：案例分析实战演练
In Year 7 Edexcel Physics, being able to apply your knowledge to real-world situations is a vital skill. Case study questions test your understanding of core concepts like forces, energy, electricity, waves, and matter by presenting everyday scenarios. This article guides you through eight carefully selected case studies, each followed by a step-by-step analysis in both English and Chinese. By practising these examples, you will strengthen your problem-solving abilities and gain confidence for exams.
在7年级Edexcel物理中，能够将知识应用到真实情境是一项关键技能。案例分析题通过呈现日常场景，考察你对力、能量、电、波和物质等核心概念的理解。本文将通过八个精心挑选的案例，每个案例都用中英双语逐步分析。练习这些例子，你将提高解决问题的能力，并在考试中更有信心。
1. Sprint Speed Calculation | 百米赛跑速度计算
A student runs 100 metres in 14 seconds during a school sports day. Calculate her average speed in metres per second (m/s).
一名学生在学校运动会上用14秒跑完100米。计算她的平均速度，单位是米每秒（m/s）。
We use the formula that links distance, time and speed:
我们使用联系距离、时间和速度的公式：
average speed = total distance ÷ total time
Substituting the given values: speed = 100 m ÷ 14 s = 7.14 m/s (to 2 decimal places). The runner’s average speed is about 7.14 m/s.
代入数值：速度 = 100 m ÷ 14 s = 7.14 m/s（保留两位小数）。该跑步者的平均速度约为7.14 m/s。
Always check that the units are consistent: distance in metres and time in seconds gives speed in m/s. If the distance was given in km, you would need to convert to metres first.
一定要检查单位是否一致：距离用米，时间用秒，得出的速度单位就是米每秒。如果距离是以千米给出的，你需要先将其转换成米。
2. Spring Extension and Hooke’s Law | 弹簧伸长与胡克定律
A spring has an unstretched length of 8 cm. When a weight of 3 N is hung from it, the length becomes 11 cm. Assuming the spring does not exceed its elastic limit, what will its length be when a 5 N weight is attached?
一根弹簧原长为8 cm。挂上3 N的重物后，长度变成11 cm。假设弹簧没有超出弹性限度，那么挂上5 N重物时，它的长度将是多少？
First, find the extension caused by the 3 N force: extension = new length − original length = 11 cm − 8 cm = 3 cm.
首先，求出3 N力产生的伸长量：伸长量 = 新长度 − 原长 = 11 cm − 8 cm = 3 cm。
Hooke’s Law tells us that extension is proportional to force (F ∝ e) for small loads. So doubling the force doubles the extension. To find the extension for a 5 N force, we use the ratio: extension for 5 N = (5 N ÷ 3 N) × 3 cm = 5 cm.
胡克定律告诉我们，在载荷较小时伸长量与力成正比（F ∝ e）。因此力加倍，伸长量也加倍。要计算5 N力对应的伸长量，我们采用比例法：5 N对应的伸长量 = (5 N ÷ 3 N) × 3 cm = 5 cm。
The total length with the 5 N weight = original length + extension = 8 cm + 5 cm = 13 cm.
挂上5 N重物时的总长度 = 原长 + 伸长量 = 8 cm + 5 cm = 13 cm。
Key point: Extension is the increase in length, not the total length. Always subtract the original length before applying the proportion.
要点：伸长量是长度的增加量，不是总长度。在用比例计算前，务必先减去原长。
3. Faulty Bulb in a Series Circuit | 串联电路中的故障灯泡
A circuit contains a battery, a switch and two identical bulbs connected in series. When the switch is closed, bulb A glows brightly but bulb B remains completely dark. Suggest a possible reason for this observation.
一个电路包含电池、开关和两个完全相同的灯泡串联。闭合开关后，灯泡A发光很亮，但灯泡B完全不亮。对这一现象提出一种可能的解释。
If bulb B had a broken filament, the circuit would be open (broken) and neither bulb would light up. Since bulb A is on, the circuit must be complete. Therefore, bulb B is not receiving current because it is likely being bypassed.
如果灯泡B的灯丝断了，电路就会开路（断开），两个灯泡都不会亮。既然灯泡A亮了，说明电路是完整的。因此，灯泡B没有电流通过，很可能是被短路了。
A short circuit occurs when a conducting path with very low resistance is connected directly across the terminals of bulb B, for example by a stray wire touching both sides. Most of the current then flows through this easier path instead of through the bulb, so bulb B does not glow.
短路是指灯泡B两端被一根电阻极低的导线直接连接，例如一根散落的导线同时碰到了灯泡的两端。此时，绝大部分电流从这条更容易的路径流过，而不经过灯泡，因此灯泡B不发光。
Always rule out the simplest option first: an open circuit would affect the whole series loop. In this case, the fault is a specific short across bulb B.
总是先排除最简单的情况：断路会影响整个串联回路。在此案例中，故障是灯泡B两端的特定短路。
4. Energy Changes on a Roller Coaster | 过山车上的能量转换
A roller coaster car of mass 200 kg is pulled to the top of a 25 m high hill. It then descends without any initial push. Describe the energy transformations that take place, assuming no friction or air resistance.
一辆质量为200 kg的过山车被拉到25米高的坡顶，然后无初速地滑下。假设没有摩擦和空气阻力，描述发生的能量转换。
At the top, the car has maximum gravitational potential energy (GPE) because of its height. GPE = mass × gravitational field strength × height = 200 kg × 10 N/kg × 25 m = 50 000 J. It has negligible kinetic energy (KE).
在坡顶，由于它的高度，过山车具有最大的重力势能（GPE）。GPE = 质量 × 重力场强 × 高度 = 200 kg × 10 N/kg × 25 m = 50 000 J。它的动能（KE）几乎为零。
As the car moves downwards, the GPE decreases and is converted into kinetic energy. The car speeds up. Halfway down, half the GPE has become KE. At the bottom of the hill, all 50 000 J of GPE has been transformed into KE (ignoring friction).
当过山车向下运动时，重力势能减少并转化为动能。车子加速下滑。在半坡处，一半的重力势能已变成动能。到达坡底时，所有50 000 J的重力势能都转化成了动能（忽略摩擦）。
The speed at the bottom can be calculated using ½ m v² = 50 000 J. Solving gives v = √(2 × 50 000 ÷ 200) = √500 ≈ 22.4 m/s. This illustrates the conservation of energy.
坡底的速度可以通过½ m v² = 50 000 J求出。解得v = √(2 × 50 000 ÷ 200) = √500 ≈ 22.4 m/s。这说明了能量守恒定律。
5. Identifying a Metal by Density | 用密度鉴定金属
A student measures a block of shiny metal. The mass is 445.5 g and the volume is 55 cm³. Using a density table, identify the possible metal from the list: aluminium (2.7 g/cm³), iron (7.9 g/cm³), copper (8.9 g/cm³), lead (11.3 g/cm³).
一名学生测量了一块闪亮的金属。质量为445.5 g，体积为55 cm³。查阅密度表，从下列选项中找出可能的金属：铝（2.7 g/cm³）、铁（7.9 g/cm³）、铜（8.9 g/cm³）、铅（11.3 g/cm³）。
Density is defined as mass per unit volume. Use the formula:
密度定义为单位体积的质量。使用公式：
density = mass ÷ volume
Plug in the data: density = 445.5 g ÷ 55 cm³ = 8.1 g/cm³ (to 1 decimal place).
代入数据：密度 = 445.5 g ÷ 55 cm³ = 8.1 g/cm³（保留一位小数）。
The calculated value of 8.1 g/cm³ is closest to iron’s density of 7.9 g/cm³. There might be small measurement errors, but the metal is almost certainly iron. Copper (8.9 g/cm³) is far less likely given the result.
计算出的8.1 g/cm³最接近铁的密度7.9 g/cm³。可能存在微小的测量误差，但该金属几乎可以肯定是铁。相对于结果，铜（8.9 g/cm³）的可能性要小得多。
6. Law of Reflection in a Periscope | 潜望镜中的反射定律
A light ray strikes a plane mirror inside a periscope at an angle of 30° to the mirror surface. Determine the angle of reflection, and explain how you would draw the reflected ray.
一束光以与镜面成30°的角度射到潜望镜内的一块平面镜上。求出反射角，并解释如何画出反射光线。
The law of reflection states that the angle of incidence is equal to the angle of reflection, both measured from the normal (an imaginary line perpendicular to the mirror surface).
反射定律指出，入射角等于反射角，两者都是从法线（一条垂直于镜面的假想线）量起的。
The ray makes an angle of 30° with the mirror surface, so the angle of incidence with the normal is 90° − 30° = 60°. Therefore, the angle of reflection is also 60°.
光线与镜面成30°角，因此与法线的入射角为90° − 30° = 60°。所以，反射角也是60°。
To draw the reflected ray, mark the normal at the point of incidence, measure a 60° angle on the opposite side of the normal, and draw the emerging ray. Label all angles clearly.
要画出反射光线，首先在入射点标出法线，在法线的另一侧量出60°角，然后画出出射光线。务必清楚地标示所有角度。
7. Sound Travel in Vacuum | 声音在真空中的传播
In many science fiction films, characters hear loud explosions in outer space. Is this scientifically correct? Justify your answer using the particle model of sound.
在许多科幻电影中，角色能听到外太空的爆炸声。这在科学上正确吗？用声音的粒子模型证明你的答案。
Sound is a mechanical wave that requires a medium (solid, liquid or gas) to travel. It propagates by making particles vibrate and pass on the vibrations. In the near-vacuum of space, there are extremely few particles, so sound cannot be transmitted.
声音是一种机械波，需要介质（固体、液体或气体）才能传播。它通过使粒子振动并传递振动来传播。在近乎真空的太空中，粒子极其稀少，因此声音无法传递。
If an explosion occurred in space, astronauts nearby would not hear any sound through their helmets. They would only hear the vibration if the shockwave directly passed through their suit or spacecraft structure. Science fiction adds sound for dramatic effect, but in reality, space is silent.
如果太空中发生爆炸，附近的宇航员通过头盔是听不到任何声音的。他们只有通过宇航服或飞船结构的直接震动才能感觉到。科幻电影为了戏剧效果添加了声音，但实际上太空是寂静的。
On the Moon (which has no atmosphere), astronauts communicated using radio waves, which are electromagnetic and can travel through a vacuum. This is a practical application of understanding sound transmission.
在月球上（没有大气层），宇航员使用无线电波通讯，因为无线电波是电磁波，可以在真空中传播。这就是理解声音传播原理的实际应用。
8. Upthrust and Sinking | 浮力与下沉
A brick weighs 12 N in air. When submerged in water, it displaces 8 N of water. Calculate the upthrust (buoyant force) on the brick, and predict whether the brick will float or sink. Explain your reasoning.
一块砖在空气中重12 N。浸没在水中时，它排开的水重8 N。计算砖块受到的浮力，并预测砖块会浮起还是下沉。解释你的推理。
Archimedes’ principle states that the upthrust on an object submerged in a fluid is equal to the weight of the fluid displaced. Therefore, the upthrust exerted on the brick is 8 N.
阿基米德原理指出，浸在流体中的物体所受的浮力等于它排开流体的重量。因此，砖块受到的浮力为8 N。
The brick’s weight (gravity pulling it down) is 12 N. The upthrust (pushing it up) is only 8 N. The net downward force is 12 N − 8 N = 4 N. Because the weight is greater than the upthrust, the brick will sink to the bottom.
砖块的重力（向下拉）为12 N。浮力（向上推）只有8 N。向下的合力为12 N − 8 N = 4 N。因为重力大于浮力，砖块将会沉到底部。
If the upthrust had been equal to the weight, the brick would float (neutral buoyancy). If upthrust exceeded weight, it would rise to the surface. This analysis helps engineers design ships and submarines.
如果浮力与重力相等，砖块就会悬浮（中性浮力）。如果浮力大于重力，它就会上浮到水面。这种分析帮助工程师设计船舶和潜艇。
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