📚 Year 7 CAIE Physics: Case Study Practice | 七年级 CAIE 物理：案例分析实战演练
Case study practice is one of the most effective ways to learn physics. By working through real-world scenarios, you can connect abstract ideas to everyday experiences. In this article, we will examine ten carefully selected case studies that cover the essential topics of the Year 7 CAIE Physics syllabus. Each case study will guide you through the problem, the scientific principles involved, and a clear step-by-step solution.
案例分析练习是学习物理最有效的方法之一。通过研究真实世界的情景，你可以把抽象的概念和日常经验联系起来。在这篇文章中，我们将分析十个精心挑选的案例，它们覆盖了七年级 CAIE 物理课程的核心主题。每个案例都会带你逐步了解问题、其中涉及的科学原理以及清晰的解答步骤。
1. Case Study: Calculating Average Speed | 案例分析：计算平均速度
A student runs 200 metres in 25 seconds. What is her average speed?
一名学生用 25 秒跑了 200 米。她的平均速度是多少？
t moves over a certain distance. The formula is speed = distance ÷ time, often written as:
在物理中，平均速度告诉我们物体在一定距离内移动的快慢。公式是 速度 = 路程 ÷ 时间，通常写为：
v = d ÷ t
= 200 m, t = 25 s. So v = 200 m ÷ 25 s = 8 m/s. Her average speed is 8 metres per second.
这里，路程 d = 200 m，时间 t = 25 s。所以 v = 200 m ÷ 25 s = 8 m/s。她的平均速度是 8 米每秒。
Remember that the result only gives the average speed; the student might have run faster at the start and slower near the end.
记住，这个结果只给出平均速度；这名学生可能在开始跑得较快，接近终点时较慢。
2. Case Study: Balanced and Unbalanced Forces | 案例分析：平衡力与非平衡力
A wooden box is pushed to the right with a force of 10 N. Friction acts to the left with a force of 4 N. Determine the net force and explain whether the forces are balanced.
一个木箱被向右的 10 N 力推动。摩擦力向左，大小为 4 N。求合力，并说明这些力是否平衡。
e difference between them. Here, net force = 10 N (right) − 4 N (left) = 6 N to the right.
当两个力沿相反方向作用时，合力是它们之间的差。这里，合力 = 10 N (右) − 4 N (左) = 6 N 向右。
Because the net force is not zero, the forces are unbalanced. An unbalanced force causes the box to accelerate to the right.
因为合力不为零，所以这些力是非平衡力。非平衡力会使木箱向右加速。
If friction had been 10 N as well, the net force would be 0 N and the box would either stay still or move at a constant speed – that is a balanced force situation.
如果摩擦力也是 10 N，那么合力将为 0 N，箱子就会保持静止或匀速直线运动——那就是平衡力的情况。
3. Case Study: Density and Floating | 案例分析：密度与浮沉
A small object has a mass of 40 g and a volume of 50 cm³. Will it float or sink in water? (Density of water = 1.0 g/cm³)
一个小物体的质量为 40 g，体积为 50 cm³。它在水中是漂浮还是下沉？（水的密度 = 1.0 g/cm³）
Density measures how much mass is packed into a given volume. The formula is:
密度衡量的是单位体积内所含质量的多少。公式为：
ρ = m ÷ V
e ρ is density, m is mass and V is volume. For this object, ρ = 40 g ÷ 50 cm³ = 0.8 g/cm³.
其中 ρ 为密度，m 为质量，V 为体积。对这个物体，ρ = 40 g ÷ 50 cm³ = 0.8 g/cm³。
Since 0.8 g/cm³ is less than the density of water (1.0 g/cm³), the object will float. An object floats when its density is lower than that of the liquid.
由于 0.8 g/cm³ 小于水的密度 (1.0 g/cm³)，该物体将会漂浮。当物体的密度小于液体的密度时，物体上浮。
4. Case Study: Series Circuit Current | 案例分析：串联电路中的电流
Two identical lamps are connected in series to a cell. An ammeter placed next to the cell reads 0.5 A. What is the current through each lamp?
两个相同的灯泡串联连接到电池。连接在电池旁的电流表读数为 0.5 A。流过每个灯泡的电流是多少？
In a series circuit, the current is the same at every point. There is only one path for the charges to flow.
在串联电路中，各处电流都相同。电荷只有一条路可以流动。
Therefore, the current through the first lamp is 0.5 A, and the current through the second lamp is also 0.5 A. The ammeter reading confirms the current pushing out of the cell.
因此，流过第一个灯泡的电流是 0.5 A，流过第二个灯泡的电流也是 0.5 A。电流表读数证实了从电池出来的电流大小。
If one lamp were removed, the circuit would break and current would stop everywhere – a key property of series connections.
如果取下其中一个灯泡，电路就会断开，各处电流都将停止——这是串联连接的一个关键特性。
5. Case Study: Conductors, Convection and Radiation | 案例分析：传导、对流与辐射
A metal spoon is placed in a cup of hot water. The handle soon feels warm. Why does this happen, and what type of heat transfer is involved?
一把金属勺放在一杯热水中，不一会儿勺柄就变热了。为什么会这样？这里涉及哪种热传递方式？
The heat from the hot water travels along the spoon because metal is a good conductor. The particles at the hot end vibrate more and pass the energy to neighbouring particles. This is conduction.
热量从热水沿着勺子传递，因为金属是热的良导体。热端的粒子振动更剧烈，并将能量传递给相邻的粒子。这就是传导。
Meanwhile, the water itself moves in currents: hot water rises, cooler water sinks. This circulation is convection. Conduction and convection require particles, but radiation can travel through a vacuum, like the heat we feel from the Sun.
同时，水自身也在循环运动：热水上升，冷水下沉。这种循环就是对流。传导和对流都需要介质粒子，而热辐射可以在真空中传播，就像我们感受到的太阳的热量。
6. Case Study: Reflection of Light | 案例分析：光的反射
A ray of light strikes a plane mirror at an angle of incidence of 30°. What is the angle of reflection? Draw the path of the ray and label the angles.
一束光线以 30° 的入射角射到平面镜上。反射角是多少？画出光线路径并标出角度。
The law of reflection states that the angle of incidence equals the angle of reflection. Both angles are 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。两个角都从法线（一条垂直于镜面的假想线）量起。
so r = 30°. The incident ray, the reflected ray and the normal all lie in the same plane.
这里，入射角 i = 30°，所以反射角 r = 30°。入射光线、反射光线和法线都在同一平面内。
You can visualise the ray hitting the mirror at a point, bouncing off symmetrically on the other side of the normal. If the angle were 0°, the ray would reflect straight back.
你可以想象光线射到镜面一点，在法线的另一侧对称地反射出去。如果入射角为 0°，光线就会沿原路直接反射回去。
7. Case Study: Sound Needs a Medium | 案例分析：声音需要介质
On the Moon, two astronauts are standing just a few metres apart without radios. Can they hear each other speak? Explain.
在月球上，两名宇航员相距仅几米，没有使用无线电。他们能听到彼此说话吗？请解释。
Sound is a mechanical wave that requires particles to vibrate and pass the energy along. On Earth, air particles carry the vibrations from a speaker's mouth to a listener's ear.
声音是一种机械波，需要粒子振动并将能量传递出去。在地球上，空气粒子将振动从说话者口中传到听者耳中。
The Moon has almost no atmosphere – it is a vacuum. Without air particles, the vibrations cannot travel. Therefore, the astronauts cannot hear each other directly; they must use radio waves, which do not need a medium.
月球几乎没有大气层——它是真空。没有空气粒子，振动就无法传播。因此，宇航员无法直接听到对方说话；他们必须使用无线电波，因为电磁波不需要介质。
8. Case Study: Energy Transfer in a Pendulum | 案例分析：单摆中的能量转移
A pendulum bob is lifted to one side and released. Describe the energy changes as it swings from the highest point to the lowest point and back again. Ignore air resistance.
将单摆的小球拉到一侧后释放。描述它从最高点摆到最低点再摆回来时发生的能量转化。忽略空气阻力。
At the highest point, the bob has maximum gravitational potential energy (GPE) and zero kinetic energy (KE) because it is momentarily at rest.
在最高点，小球具有最大的重力势能 (GPE) 和零动能 (KE)，因为它在那瞬间静止。
As it swings down, height decreases so GPE decreases, while speed increases so KE increases. The GPE is being converted into KE.
当它向下摆动时，高度降低，因此重力势能减少；同时速度增大，动能增加。重力势能正在转化为动能。
At the lowest point, GPE is minimum and KE is maximum. Then, as it rises on the other side, KE is converted back into GPE. In an ideal system, the sum of GPE and KE remains constant.
在最低点，重力势能最小，动能最大。然后，当它摆向另一侧升高时，动能又转化回重力势能。在理想系统中，重力势能与动能的总和保持不变。
9. Case Study: Mass and Weight | 案例分析：质量与重量
ld strength g = 10 N/kg. What would its mass and weight be on the Moon, where g ≈ 1.6 N/kg?
一块岩石的质量为 5 kg。求它在地球上的重量，地球的重力场强度 g = 10 N/kg。它在月球上的质量和重量分别是多少？月球上 g ≈ 1.6 N/kg。
Weight is the force of gravity on an object and is calculated by:
重量是作用在物体上的重力，其计算公式为：
W = m × g
On Earth, W = 5 kg × 10 N/kg = 50 N. The weight is 50 newtons.
在地球上，W = 5 kg × 10 N/kg = 50 N。重量为 50 牛顿。
owever, weight on the Moon = 5 kg × 1.6 N/kg = 8 N. The rock weighs much less on the Moon.
质量是物体所含物质的多少，不随位置改变。所以月球上的质量仍然是 5 kg。然而，月球上的重量 = 5 kg × 1.6 N/kg = 8 N。岩石在月球上轻得多。
10. Case Study: Changes of State and Temperature | 案例分析：状态变化与温度
A beaker of ice is heated steadily. The temperature stays at 0 °C while the ice melts completely. Why does the temperature not rise during melting, even though heat is being supplied?
一杯冰被均匀加热。当冰完全融化时，温度保持在 0 °C。为什么融化过程中虽然不断供热，温度却不上升？
When a solid melts, the added energy is used to break the rigid bonds holding the particles in fixed positions. This energy does not increase the kinetic energy of the particles, so the temperature remains constant.
当固体融化时，加入的能量被用来打破将粒子固定在位置的刚性键。这部分能量没有增加粒子的动能，所以温度保持不变。
Once all the ice has turned into water, any further heating raises the water's temperature. The flat region on a temperature–time graph during melting shows this 'hidden' energy, sometimes called latent heat.
一旦所有冰都变成水，继续加热就会使水温升高。温度-时间图线上在融化阶段的水平区域显示了这种“隐藏的”能量，有时被称为潜热。
This principle also applies when water boils: the temperature stays at 100 °C until all the liquid has become steam.
这一原理也适用于水沸腾时：温度保持在 100 °C，直到所有液体都变成蒸汽。
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