📚 Case Study Practical Drill | 案例分析实战演练
Case studies are a powerful way to apply your physics knowledge to real-world situations. By breaking down a problem step by step, you learn how to identify the relevant concepts, choose the correct equations, and interpret your results. This practical drill will guide you through nine carefully selected scenarios, each designed to strengthen your understanding of key Year 8 OCR physics topics such as motion, forces, energy, electricity, waves, and density. Read each case carefully, attempt the calculations yourself, and then compare your reasoning with the worked solutions provided.
案例分析是将物理知识应用于实际情况的有效方式。通过逐步分解问题，你可以学会如何识别相关概念、选择正确的方程并解释结果。这次实战演练将带你分析九个精选情景，每个情景都旨在巩固你对八年级 OCR 物理关键主题的理解，包括运动、力、能量、电学、波动和密度。请仔细阅读每个案例，先自己尝试计算，再与提供的解题过程进行对比。
1. Case Study 1: The Sprinter’s Sprint | 案例一：短跑运动员的冲刺
A sprinter runs a 100-metre race and crosses the finish line in 12.5 seconds. We assume her speed is nearly constant throughout the race, so we can calculate her average speed.
一名短跑运动员跑完 100 米比赛，用时 12.5 秒。我们假设她在整个赛程中速度几乎恒定，这样便能计算她的平均速率。
Question: What was the sprinter’s average speed in metres per second?
问题：这名运动员的平均速率是多少米每秒？
Analysis: Average speed is defined as the total distance travelled divided by the total time taken. Since the motion is in a straight line, speed and velocity have the same magnitude here.
分析：平均速率定义为总路程除以总时间。因为这里是直线运动，速率和速度的大小相等。
average speed = total distance ÷ total time
Substitute the given values: distance = 100 m, time = 12.5 s.
代入已知数值：路程 = 100 m，时间 = 12.5 s。
v = 100 m ÷ 12.5 s = 8.0 m/s
The sprinter’s average speed was 8.0 metres per second. This tells us how fast she ran, but it does not tell us anything about the direction, because speed is a scalar quantity.
这名短跑运动员的平均速率为 8.0 米每秒。这告诉我们她跑得有多快，但并没有给出方向信息，因为速率是标量。
Key point: Always write down the formula first, then plug in numbers, and finally state the answer with units. Checking that the units match (metres and seconds) is an essential habit.
关键点：务必先写下公式，再代入数据，最后写明带单位的答案。检查单位（米和秒）是否匹配是必须养成的习惯。
2. Case Study 2: The Cyclist’s Push | 案例二：自行车手的加速
A cyclist on a flat road accelerates from rest to 10 m/s in 5 seconds. The total mass of the cyclist and bicycle is 70 kg. We want to find the acceleration and the average forward force acting during this time.
一位自行车手在平路上从静止开始加速，5 秒内速度达到 10 m/s。车手和自行车的总质量为 70 kg。我们需要求出这段时间内的加速度和平均前进力。
Question: What is the cyclist’s acceleration, and what force produces this acceleration?
问题：车手的加速度是多少？产生该加速度的力是多大？
Analysis: Acceleration is the rate of change of velocity. Because the cyclist starts from rest, the initial velocity v₀ = 0 m/s, and the final velocity v = 10 m/s. The change in velocity Δv = v – v₀ = 10 m/s.
分析：加速度是速度的变化率。车手从静止开始，初速度 v₀ = 0 m/s，末速度 v = 10 m/s，速度变化量 Δv = v – v₀ = 10 m/s。
a = Δv ÷ t = (10 m/s) ÷ 5 s = 2 m/s²
Now, using Newton’s second law, the resultant force equals mass times acceleration.
现在，利用牛顿第二定律，合力等于质量乘以加速度。
F = m × a = 70 kg × 2 m/s² = 140 N
The acceleration is 2 metres per second squared, and the average driving force is 140 newtons. Note that in reality, some of this force would also have to overcome friction and air resistance, so the forward force exerted by the cyclist would be larger than 140 N if those forces are present.
加速度为 2 米每二次方秒，平均驱动力为 140 牛顿。注意，在实际情况中，该力的一部分还需用来克服摩擦和空气阻力，因此若存在这些阻力，车手施加的前进力会大于 140 N。
3. Case Study 3: The Diver’s Energy Transformation | 案例三：跳水运动员的能量转换
A 50 kg diver stands on a 10-metre-high platform. She steps off and falls vertically into the water below. We ignore air resistance and use g = 10 N/kg.
一位 50 kg 的跳水运动员站在 10 米高的跳台上。她迈步离开跳台，竖直落入下方的水中。我们忽略空气阻力，并取 g = 10 N/kg。
Question: How much gravitational potential energy does she lose during the fall, and what is her speed just before hitting the water?
问题：她下落过程中损失了多少重力势能？她入水前的速度是多少？
Analysis: Gravitational potential energy (Eₚ) is given by mass × gravitational field strength × height. As she falls, this energy is converted into kinetic energy (Eₖ). At the top, Eₖ = 0; just before hitting the water, all the lost Eₚ has become kinetic energy.
分析：重力势能 Eₚ 由质量 × 重力场强度 × 高度计算。下落过程中，该能量转化为动能 Eₖ。在最高点，Eₖ = 0；入水前瞬间，所有损失的重力势能都变成了动能。
Eₚ = m × g × h = 50 kg × 10 N/kg × 10 m = 5000 J
So the kinetic energy at the bottom is also 5000 J. The kinetic energy formula is Eₖ = ½ m v². We rearrange to find v.
因此，底部的动能也为 5000 J。动能公式为 Eₖ = ½ m v²。我们变形求 v。
5000 J = ½ × 50 kg × v² → 5000 = 25 × v² → v² = 200 → v = √200 ≈ 14.1 m/s
The diver loses 5000 joules of gravitational potential energy and enters the water at about 14.1 m/s. This example beautifully demonstrates the principle of conservation of energy: energy changes form but is never destroyed.
她损失了 5000 焦耳的重力势能，并以约 14.1 m/s 的速度入水。这个例子完美地展示了能量守恒原理：能量改变形式，但永不消失。
4. Case Study 4: The Light Bulb’s Brightness | 案例四：灯泡的亮度
A filament lamp is labelled “60 W, 240 V”. It is connected to a 240 V mains supply and lights at full brightness.
一只白炽灯标有“60 W，240 V”。它连接到 240 V 的市电上，正常发光。
Question: Calculate the current flowing through the bulb and its resistance when operating at full brightness.
问题：计算灯泡正常发光时流过的电流及其电阻。
Analysis: Electrical power P, potential difference V, and current I are linked by P = I × V. For a resistor, Ohm’s law states V = I × R, so once we know the current, we can find the resistance R.
分析：电功率 P、电势差 V 和电流 I 之间的关系为 P = I × V。对于电阻元件，欧姆定律指出 V = I × R，因此知道电流后即可求出电阻 R。
I = P ÷ V = 60 W ÷ 240 V = 0.25 A
Now use Ohm’s law to find resistance.
现在用欧姆定律求电阻。
R = V ÷ I = 240 V ÷ 0.25 A = 960 Ω
The current is 0.25 amperes and the resistance is 960 ohms. Note that the filament’s resistance changes with temperature, so this is the resistance only at its operating temperature; when the bulb is cold, the resistance is much lower.
电流为 0.25 安培，电阻为 960 欧姆。注意，灯丝的电阻随温度变化，因此该电阻仅为其工作温度下的数值；灯泡冷却时电阻要低得多。
5. Case Study 5: Seeing Around Corners with a Mirror | 案例五：用镜子看拐角
A student stands 2 metres in front of a flat mirror mounted on the wall. She uses the mirror to observe a clock that is on the side wall, 1.5 metres behind her and 1 metre to her right.
一名学生站在离墙上平面镜 2 米处。她利用镜子观察身后 1.5 米、右侧 1 米处侧墙上的时钟。
Question: Explain how the mirror forms an image that allows her to see the clock. How far behind the mirror does the image of the clock appear?
问题：解释镜子如何成像，使她能看到时钟。时钟的像出现在镜子后方多远？
Analysis: A flat mirror produces a virtual image that appears to be as far behind the mirror as the object is in front. The law of reflection (angle of incidence = angle of reflection) ensures that light rays from the clock can be reflected into her eyes. We must calculate the perpendicular distance from the clock to the mirror.
分析：平面镜产生虚像，像在镜后的距离等于物体在镜前的距离。反射定律（入射角 = 反射角）确保来自时钟的光线能反射进入她的眼睛。我们需要计算时钟到镜子的垂直距离。
The clock is located 1.5 m behind the student, and the student is 2 m in front of the mirror. The perpendicular distance from the clock to the mirror is therefore 1.5 m + 2 m = 3.5 m. The image of the clock will appear 3.5 m behind the mirror, directly along the line of sight after reflection.
时钟位于学生身后 1.5 米，而学生站在镜前 2 米，因此时钟到镜子的垂直距离为 1.5 m + 2 m = 3.5 m。时钟的像将出现在镜子后方 3.5 米处，恰好位于反射后的视线方向上。
The student does not need to turn around; she simply looks at the mirror and sees the virtual image of the clock as if it were behind the wall. This is a useful application of the law of reflection.
该学生无需转身，只需观看镜子，就能看到时钟的虚像，仿佛它在墙壁后方。这是反射定律的一项实际应用。
6. Case Study 6: Echo from a Cliff | 案例六：悬崖的回声
On a school trip, a student shouts loudly towards a vertical cliff face and hears the echo 2.6 seconds later. The temperature is about 20 °C, so the speed of sound in air is approximately 340 m/s.
在学校旅行中，一名学生对着一面竖直悬崖大声呼喊，2.6 秒后听到回声。当时气温约为 20 °C，空气中的声速约为 340 m/s。
Question: Estimate the distance between the student and the cliff.
问题：估算学生与悬崖之间的距离。
Analysis: The sound travels from the student to the cliff and back again, so the total distance covered by the sound is twice the distance to the cliff. Time for the round trip is 2.6 s. Therefore, the distance to the cliff is half of the total distance travelled by the sound.
分析：声音从学生传到悬崖再反射回来，因此声音走过的总路程是到悬崖距离的两倍。往返时间为 2.6 秒。所以到悬崖的距离是声音总路程的一半。
total distance = speed × time = 340 m/s × 2.6 s = 884 m
distance to cliff = 884 m ÷ 2 = 442 m
The student is roughly 442 metres from the cliff. This method, called echolocation, is used by bats and by ships’ sonar systems, though with ultrasound rather than audible sound.
学生与悬崖的距离大约为 442 米。这种方法称为回声定位，蝙蝠和船舶声呐系统都使用类似原理，不过它们使用的是超声波而非可听声。
7. Case Study 7: Finding the Density of a Lump of Metal | 案例七：测量金属块的密度
An irregular lump of metal is placed on a top-pan balance and its mass is recorded as 180 g. A measuring cylinder is filled to the 50 cm³ mark with water. When the metal is carefully lowered into the water, the water level rises to 74 cm³.
一块不规则金属放在上盘天平上，称得质量为 180 g。量筒中盛水至 50 cm³ 标记处。将金属小心放入水中后，水面上升到 74 cm³。
Question: Calculate the density of the metal in g/cm³. Suggest what metal it might be.
问题：计算该金属的密度（单位 g/cm³），并推测可能是什么金属。
Analysis: Density is mass per unit volume. The volume of the metal is equal to the volume of water displaced, which is the rise in the water level. We must convert units appropriately if needed, but here g/cm³ is directly convenient.
分析：密度是单位体积的质量。金属的体积等于它排开水的体积，即水面上升的刻度。若需单位转换应适当处理，但这里直接用 g/cm³ 很方便。
volume of metal = 74 cm³ – 50 cm³ = 24 cm³
density = mass ÷ volume = 180 g ÷ 24 cm³ = 7.5 g/cm³
A density of 7.5 g/cm³ is close to that of cast iron or some steels, which typically range around 7.2–7.9 g/cm³. Without further tests, we cannot be certain, but the value is a good starting point for identification.
该密度 7.5 g/cm³ 接近铸铁或某些钢材的密度（通常为 7.2–7.9 g/cm³）。没有进一步检测我们无法完全确定，但该数值为识别金属提供了很好的起点。
8. Case Study 8: Balanced Forces on a Ski Slope | 案例八：滑雪斜坡上的平衡力
A skier of mass 60 kg glides down a gentle slope at a constant speed. The slope is at an angle such that the component of her weight acting parallel to the slope is 200 N. She does not use her poles to push.
一位质量为 60 kg 的滑雪者沿缓坡匀速滑下。该斜坡的角度使得她体重沿斜坡方向的分量为 200 N。她没有用雪杖进行助推。
Question: What is the size of the total frictional force acting on the skier? Explain why her speed remains constant.
问题：作用在滑雪者身上的总摩擦力是多大？解释为什么她的速度保持恒定。
Analysis: Since the skier moves at constant speed, the forces acting on her must be balanced. The two main forces parallel to the slope are the component of weight pulling her downhill (200 N) and the frictional force (including air resistance) opposing the motion. For constant velocity, the resultant force must be zero.
分析：由于滑雪者匀速运动，作用在她身上的力一定相互平衡。平行于斜坡的两个主要力是沿斜坡向下的体重分量（200 N）和阻碍运动的摩擦力（包括空气阻力）。若要速度恒定，合力必须为零。
frictional force = 200 N (up the slope)
The frictional force is 200 N directed up the slope. This exactly cancels the downhill component of weight, so the resultant force is zero. Newton’s first law tells us that an object with zero resultant force will remain at constant speed in a straight line.
摩擦力为 200 N，方向沿斜坡向上。这恰好抵消了向下的体重分量，因此合力为零。牛顿第一定律指出，合力为零的物体将保持匀速直线运动。
9. Case Study 9: The Series Circuit in a Torch | 案例九：手电筒中的串联电路
A simple torch contains two 1.5 V cells in series, a switch, and a small bulb. The bulb has a resistance of 15 Ω when lit. The switch is closed and the bulb glows steadily.
一支简单的手电筒内有两节串联的 1.5 V 电池、一个开关和一个小灯泡。灯泡点亮时的电阻为 15 Ω。闭合开关，灯泡持续发光。
Question: Calculate the total potential difference across the bulb, the current in the circuit, and the power delivered to the bulb.
问题：计算灯泡两端的总电势差、电路中的电流以及输送给灯泡的功率。
Analysis: In series, the total voltage of the cells adds up. The two 1.5 V cells give a total electromotive force of 3.0 V. Assuming the wires and switch have negligible resistance, the full 3.0 V appears across the bulb. Using Ohm’s law, we can find the current.
分析：串联时，电池的电压相加。两节 1.5 V 电池提供的总电动势为 3.0 V。假设导线和开关电阻可忽略，灯泡两端可得到全部 3.0 V 电压。利用欧姆定律即可求出电流。
V_total = 1.5 V + 1.5 V = 3.0 V
I = V ÷ R = 3.0 V ÷ 15 Ω = 0.2 A
Finally, power is the product of potential difference and current.
最后，功率为电势差与电流的乘积。
P = V × I = 3.0 V × 0.2 A = 0.6 W
The bulb receives 0.6 watts of power. If the cells were arranged in parallel instead of series, the voltage would remain at 1.5 V, and the bulb would be much dimmer. Series connections are thus used in torches to raise the voltage.
灯泡获得 0.6 瓦的功率。如果电池采用并联而非串联，电压将保持在 1.5 V，灯泡会暗得多。因此手电筒中使用串联连接来提升电压。
Published by TutorHao | Physics Revision Series | aleveler.com
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