📚 Case Study Practical Exercises for Year 9 OCR Physics | 九年级OCR物理案例分析实战演练
This article presents a series of real-world case studies designed to help Year 9 students master key concepts in OCR Physics. Each scenario is broken down with step-by-step analysis, linking observations to the underlying principles of forces, energy, waves, electricity and matter. By working through these practical investigations, you will strengthen your ability to apply scientific knowledge to unfamiliar contexts – a vital skill for end-of-topic tests and future GCSE success.
本文提供一系列真实世界的案例分析，帮助九年级学生掌握OCR物理的核心概念。每个情景都通过分步剖析，将观察现象与力、能量、波、电和物质的基本原理联系起来。通过完成这些实践探究，你将强化将科学知识应用于陌生情境的能力，这对单元测验和未来的GCSE考试至关重要。
1. Case Study: Braking Distance and Speed | 案例：刹车距离与速度
A driver applies the brakes on a dry road at 13 m/s. The car skids to a stop, leaving 18 m of skid marks. When the same car brakes on the same surface at 20 m/s, the skid marks measure 42 m. Use energy ideas to explain why doubling the speed more than doubles the braking distance.
一辆汽车在干燥路面上以13 m/s的速度刹车，滑行18 m后停下。同一辆车在同一路面以20 m/s刹车时，滑痕长达42 m。用能量观点解释为什么速度增加一倍，刹车距离增加不止一倍。
he speed, the kinetic energy becomes four times larger. Since the friction force is roughly constant on the same surface, the work done (force × distance) must also be four times greater, so the braking distance increases by a factor of about four, not just two. In the data, 20 m/s is about 1.5 times 13 m/s, so we expect distance to increase by a factor of roughly (20/13)² ≈ 2.37, and 42 m is indeed about 2.33 times 18 m, confirming the squared relationship.
摩擦力使车辆停止所做的功等于汽车的初始动能。动能与速度的平方成正比（KE = ½ m v²）。如果速度加倍，动能变为四倍。由于同一路面的摩擦力大致恒定，摩擦力做功（力 × 距离）也需要变为四倍，因此刹车距离大约增加至原来的四倍，而不仅仅是两倍。在给出的数据中，20 m/s 约为13 m/s 的1.5倍，因此距离预期增加约 (20/13)² ≈ 2.37 倍，而42 m 恰好约为18 m 的2.33倍，证实了平方关系。
2. Case Study: Bungee Jump Energy Transformation | 案例：蹦极能量转换
A bungee jumper of mass 75 kg leaps from a bridge. The rope stretches, and the jumper bounces up and down several times before coming to rest. Describe the main energy transfers from the moment of jumping until the jumper temporarily stops at the lowest point for the first time.
一名质量为75 kg的蹦极者从桥上跳下。绳索伸展，跳跃者上下弹跳数次后逐渐静止。描述从起跳瞬间到跳跃者第一次在最低点暂时停止期间的主要能量转换。
At the moment of jumping, the jumper has gravitational potential energy relative to the lowest point. As they fall, this energy changes into kinetic energy. When the rope begins to stretch, kinetic energy and additional gravitational potential energy are converted into elastic potential energy stored in the rope. At the lowest point, the jumper has minimum gravitational potential energy, zero kinetic energy, and maximum elastic potential energy in the rope. Some energy is also transferred to the surroundings as thermal energy due to air resistance and internal friction in the rope, which is why the bounce height decreases over time.
起跳瞬间，跳跃者相对于最低点具有重力势能。下落时，重力势能转化为动能。当绳索开始伸展，动能和进一步损失的重力势能转化为储存在绳索中的弹性势能。在最低点，跳跃者的重力势能最小、动能为零，绳索中的弹性势能达到最大。同时，由于空气阻力和绳索内摩擦，部分能量会以热能形式传递到周围环境，这就是弹跳高度随时间降低的原因。
3. Case Study: Series and Parallel Fairy Lights | 案例：串联与并联装饰灯
A set of 20 identical bulbs is connected in series to a 230 V supply. One bulb blows, and the whole string goes dark. The owner replaces the blown bulb with a wire link, and the remaining bulbs light up but appear brighter. Later, a second string designed with 20 bulbs in parallel is examined – when one bulb fails, the others stay on. Explain these observations using circuit rules.
一组20个相同的小灯泡串联在230 V电源上。一只灯泡烧坏后整串熄灭。主人用一段导线代替烧坏的灯泡，其余灯泡亮起但明显更亮。后来查看另一串采用20个灯泡并联的设计——其中一个灯泡损坏后，其他灯泡依然亮着。用电路规律解释这些现象。
with a wire of negligible resistance, the total resistance of the circuit decreases. With the same 230 V supply, current increases (I = V / R). Since the remaining 19 bulbs have a fixed resistance, the larger current makes each bulb brighter. However, the higher current may cause these bulbs to blow sooner. In the parallel circuit, each bulb has its own separate loop connected directly to the supply. If one loop breaks, current can still flow through the other branches, so the remaining bulbs continue to glow at their normal brightness because each receives the full supply voltage.
串联电路中电流只有一条通路。如果一个灯泡烧坏，电路断开，各处电流都停止。当用一根电阻可忽略的导线替代烧坏的灯泡后，电路的总电阻减小。在相同的230 V电源下，电流增大（I = V / R）。剩下的19个灯泡电阻不变，更大的电流使每个灯泡更亮，但高电流可能导致它们更快烧毁。在并联电路中，每个灯泡都有自己独立的回路，直接连接电源。其中一个回路断开时，电流仍可流过其他支路，因此其余灯泡继续正常发光，且每个都获得完整的电源电压。
4. Case Study: House Insulation and Heat Loss | 案例：房屋隔热与热量散失
A homeowner measures the temperature loss in different rooms. The loft loses heat at 120 W through the ceiling, the windows lose 85 W, and the walls lose 200 W. The homeowner installs 200 mm of loft insulation, double glazing, and cavity wall insulation. Explain how each measure reduces heat transfer, referring to conduction, convection and radiation.
某房主测量了不同房间的热量损失。阁楼天花板散热120 W，窗户散热85 W，墙壁散热200 W。房主安装了200 mm厚的阁楼保温层、双层玻璃以及空心墙隔热层。解释每种措施如何通过传导、对流和辐射来减少热传递。
Loft insulation, commonly mineral wool, traps air in tiny pockets. Still air is a very poor conductor, so it reduces heat loss by conduction. It also stops warm air in the room from rising and setting up convection currents in the loft space. Double glazing consists of two glass panes with a narrow gap of dry air or argon. The gas layer greatly reduces conduction because gases conduct much less than solids. The narrow gap also minimises convection currents. Cavity wall insulation fills the air gap between inner and outer brick layers with foam or mineral wool. This eliminates the main convection path that would otherwise carry heat across the cavity, and the insulating material itself has low thermal conductivity. Radiation losses are relatively small at room temperatures, but reflective foil on insulation can further reduce radiation.
阁楼保温层通常为矿棉，能将空气困在微小空隙中。静止的空气是极差的导热体，因此它通过减少传导来降低热损失。同时它还能阻止室内暖空气上升并在阁楼空间形成对流。双层玻璃由两片玻璃及其间的干燥空气或氩气窄间隙构成。气体层大幅降低传导，因为气体的导热能力远低于固体。窄间隙也最大限度抑制了对流。空心墙隔热层用泡沫或矿棉填充内外砖层之间的空隙，这消除了原本会在空心层中传递热量的主要对流途径，而隔热材料本身导热系数低。室温下辐射散热相对较小，但隔热层上的反射箔可以进一步减少辐射。
5. Case Study: Sound Waves on an Oscilloscope | 案例：用示波器显示声波
A student connects a microphone to an oscilloscope. When a tuning fork of 440 Hz is struck near the microphone, the trace shows a smooth, regular wave. Another tuning fork of 880 Hz produces a wave with twice the number of complete cycles on the screen for the same time base setting. A third, louder sound gives a taller trace with unchanged frequency. Interpret these observations.
一名学生将麦克风连接到示波器。当440 Hz的音叉在麦克风附近被敲击时，屏幕上显示出平滑的规则波形。另一个880 Hz的音叉在相同的时间基准设定下，产生的波形周期数是前者的两倍。第三个更响亮的声音给出更高的波形，但频率不变。解释这些观察结果。
The oscilloscope displays a voltage-time graph of the sound signal. The number of complete wave cycles across the screen for a fixed time base indicates frequency; more cycles in the same time interval means a higher frequency. The 880 Hz wave shows exactly twice as many cycles because its frequency is double. The height (amplitude) of the trace represents the loudness or volume of the sound. A louder sound transfers more energy and produces a pressure wave with larger amplitude, but the number of vibrations per second stays the same, so the trace is taller but the horizontal spacing of the wave pattern is identical.
示波器显示的是声音信号的电压-时间图。在固定时间基准下，屏幕上完整波形的个数表示频率；相同时间间隔内周期数越多，频率越高。880 Hz的波形恰好显示两倍的周期数，因为其频率是两倍。波形的高度（振幅）代表声音的响度或音量。更响亮的声音传递更多能量，产生的压力波振幅更大，但每秒振动次数不变，因此波形更高，但波形在水平方向的间距完全相同。
6. Case Study: Investigating Floating and Sinking | 案例：探究浮沉现象
A block of wood floats with 40% erged in water. Use Archimedes’ principle to find the density of aluminium in g/cm³. (Take density of water as 1.0 g/cm³, g = 10 N/kg.)
一块木头漂浮时有40%的体积露出水面。同等大小的铝块则完全沉没。铝块在空气中重2.7 N，完全浸没在水中时重1.7 N。利用阿基米德原理求出铝的密度，以g/cm³为单位。（水的密度取1.0 g/cm³，g取10 N/kg。）
The apparent loss in weight of the aluminium block is the upthrust. Upthrust = 2.7 N – ed = 1.0 N ÷ 10 N/kg = 0.1 kg = 100 g. Since the density of water is 1.0 g/cm³, the volume of water displaced (and hence the volume of the aluminium block) is 100 cm³. The mass of the aluminium block in air is 2.7 N ÷ 10 N/kg = 0.27 kg = 270 g. Therefore density = mass / volume = 270 g / 100 cm³ = 2.7 g/cm³. The wooden block floats because its density is less than water; 40% of its volume is above water, meaning 60% is submerged, so its average density is 0.6 g/cm³.
铝块表观重量减少的量即为浮力。浮力 = 2.7 N – 1.7 N = 1.0 N。根据阿基米德原理，浮力等于排开水的重量。排开水重1.0 N，因此排开水的质量 = 1.0 N ÷ 10 N/kg = 0.1 kg = 100 g。水的密度为1.0 g/cm³，所以排开水的体积（即铝块的体积）为100 cm³。铝块在空气中的质量为 2.7 N ÷ 10 N/kg = 0.27 kg = 270 g。因此密度 = 质量/体积 = 270 g / 100 cm³ = 2.7 g/cm³。木块之所以漂浮是因为其密度小于水；40%的体积在水面以上，意味着60%浸没，所以其平均密度为0.6 g/cm³。
7. Case Study: Electromagnet Strength Investigation | 案例：电磁铁强度探究
A student builds an electromagnet by wrapping insulated wire around an iron nail. She changes the number of turns and measures how many paper clips the magnet can pick up. The table shows results: 10 turns – 4 clips; 20 turns – 7 clips; 30 turns – 11 clips; 40 turns – 14 clips. She then repeats with a thicker steel core and picks up 20 clips with 40 turns. Explain the trends and identify the key variables.
一名学生用绝缘导线缠绕铁钉制作电磁铁。她改变线圈匝数，测量磁铁能吸起多少回形针。数据如表：10匝—4枚；20匝—7枚；30匝—11枚；40匝—14枚。然后她用一个更粗的钢芯重复实验，40匝时吸起了20枚。解释趋势并识别关键变量。
The independent variable is the number of coil turns (and later the core type). The dependent variable is the number of paper clips picked up, which indicates magnetic strength. Control variables include the current in the coil, the type of wire, and the distance to the clips. As the number of turns increases, the magnetic field becomes stronger because each turn adds to the total magnetising effect. This is why more clips are attracted. The increase is not perfectly linear due to limits of the nail’s magnetisation. When a thicker steel core is used, the electromagnet becomes even stronger for the same number of turns. Steel, like iron, is ferromagnetic and concentrates the magnetic field, but a thicker core can sustain a larger magnetic flux, boosting the strength. However, steel may retain magnetism when current stops, which is a key difference from soft iron.
自变量是线圈匝数（以及后来的铁芯类型）。因变量是吸起的回形针数量，它指示了磁性强度。控制变量包括线圈中的电流、导线类型以及与回形针的距离。随着匝数增加，磁场更强，因为每匝线圈都增强总磁化效果，所以能吸引更多回形针。增长并非完全线性，因为铁钉的磁化有极限。使用更粗的钢芯后，相同匝数下电磁铁变得更强。钢同铁一样是铁磁性的，可以集中磁场，但更粗的铁芯能承载更大的磁通量，提升强度。不过，电流断开后钢可能保留磁性，这是与软铁的重要区别。
8. Case Study: Light Refraction Through a Glass Block | 案例：光通过玻璃砖的折射
t bends away from the normal and emerges parallel to the incident ray. If the glass has a refractive index of 1.5, explain why the emergent ray is parallel and calculate the angle of refraction inside the glass using Snell’s law. (sin 30° = 0.5)
一束光以与法线成30°的角度射入矩形玻璃砖。它在玻璃中向法线靠拢。当射出玻璃砖时，它偏离法线并平行于入射光线射出。如果玻璃的折射率为1.5，解释为什么出射光线平行，并用斯涅尔定律计算玻璃内部的折射角。（sin 30° = 0.5）
= 0.3333, giving r ≈ 19.5°. Inside the glass, the ray strikes the opposite face at the same angle r. Applying Snell’s law at the glass-air boundary: 1.5 × sin 19.5° = 1 × sin e. Since sin 19.5° = 0.3333, sin e = 1.5 × 0.3333 = 0.5, so e = 30°. The emergent angle equals the incident angle because the two refractions reverse the deviation. The emergent ray is shifted sideways but remains parallel to the original ray, a result of the identical entering and leaving angles.
矩形玻璃砖的两面平行。当光线射入时，由斯涅尔定律 n₁ sin θ₁ = n₂ sin θ₂，从空气到玻璃：1 × sin 30° = 1.5 × sin r，解得 sin r = 0.5 / 1.5 = 0.3333，r ≈ 19.5°。在玻璃内部，光线以相同角度 r 射到对面。在玻璃-空气界面应用斯涅尔定律：1.5 × sin 19.5° = 1 × sin e。因为 sin 19.5° = 0.3333，sin e = 1.5 × 0.3333 = 0.5，故出射角 e = 30°。出射角等于入射角，因为两次折射恰好逆转了偏折。出射光线发生侧移但仍与入射光线平行，这由入射角和出射角相等所决定。
9. Case Study: Liquid Pressure in a Dam | 案例：水坝中的液体压强
base, given density of water = 1000 kg/m³ and g = 10 N/kg. Explain why pressure increases with depth and how this shapes the dam design.
一名工程师设计水坝。坝底水深25 m。坝壁底部比顶部更厚。已知水的密度为1000 kg/m³，g = 10 N/kg，计算坝底的水压，并解释为什么压强随深度增加及这如何影响坝体设计。
This pressure acts in all directions at a given depth. The deeper you go, the greater the weight of water above that level, so pressure increases linearly with depth. At the top of the dam, the depth and pressure are small; at the base, both are large. To withstand the higher force, the wall must be thickest at the bottom where the turning moment about the base is greatest. This is why dam walls have a triangular cross-section, tapering from a wide base to a narrower crest.
液体中的压强 = h ρ g = 25 m × 1000 kg/m³ × 10 N/kg = 250 000 Pa（即250 kPa）。该压强在给定深度的各个方向都有作用。深度越深，上方水的重量越大，所以压强随深度线性增加。大坝顶部，水深小，压强低；底部水深大，压强大。为了承受更大的力，坝壁在底部必须最厚，因为底部所受倾覆力矩最大。这就是坝壁截面呈三角形、由宽阔底部向较窄坝顶收窄的原因。
10. Case Study: Electrical Power and the Cost of Running Appliances | 案例：电功率与电器运行成本
A household uses a 2 kW heater for 5 hours, six 20 W LED bulbs for 8 hours, and a 0.5 kW fridge that runs for 24 hours in one day. Electricity costs 15 pence per kWh. Calculate the total energy used in kWh, the total cost for that day, and discuss which appliance contributes most to the bill.
某家庭一天内使用一台2 kW暖风机5小时、六只20 W LED灯泡8小时，以及一台0.5 kW冰箱运行24小时。电价为15便士每千瓦时。计算总用电量（kWh）和当日总电费，并讨论哪一电器对电费贡献最大。
Fridge: 0.5 kW × 24 h = 12 kWh. Total energy = 10 + 0.96 + 12 = 22.96 kWh. Cost = 22.96 × £0.15 = £3.444 (about £ s much but runs for only 5 hours. The bulbs, despite the long duration, consume very little due to low power. This shows that both power and time determine energy transferred, so efficient appliances with low power ratings can significantly reduce bills if they have long running times.
暖风机用电量 = 2 kW × 5 h = 10 kWh。LED灯泡：每只20 W = 0.02 kW，六只合计0.12 kW，用电量 = 0.12 kW × 8 h = 0.96 kWh。冰箱：0.5 kW × 24 h = 12 kWh。总用电量 = 10 + 0.96 + 12 = 22.96 kWh。费用 = 22.96 × £0.15 = £3.444（约 £3.44）。冰箱用电最多（12 kWh），因为它全天运行，尽管额定功率比暖风机低。暖风机用电量接近冰箱，但仅运行5小时。灯泡虽然使用时间长，但因功率低，用电量很小。这表明功率和时间共同决定传递的能量，因此即使长时间运行，选择功率低的高效电器也能显著减少电费。
Published by TutorHao | Physics Revision Series | aleveler.com
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