📚 Year 8 Cambridge Physics: Case Study Practice | Year 8 剑桥物理：案例分析实战演练
In Year 8 Cambridge Physics, case studies bridge the gap between theory and real-world application. This article presents a series of practical scenarios designed to strengthen your problem-solving skills, covering topics from speed to moments. Each case includes a clear description, key physics principles, and step-by-step analysis with answers. Use these examples to revise for end-of-unit tests and to build confidence in explaining physical phenomena using correct scientific terms.
在 Year 8 剑桥物理中，案例研究弥合了理论与现实应用之间的鸿沟。本文提供了一系列实战场景，旨在强化你的问题解决技能，涵盖从速度到力矩的多个主题。每个案例包含清晰的描述、关键物理原理以及逐步分析并给出答案。利用这些例子为单元测试复习，并建立用正确科学术语解释物理现象的信心。
1. Case Study: Skateboard Speed | 案例一：滑板速度
A student skateboards along a straight path in the park. She travels 150 metres in 30 seconds, stops for 10 seconds to tie her shoelace, then continues another 90 metres in 20 seconds. What is her average speed for the entire journey?
一名学生在公园的直道上滑滑板。她在30秒内滑行150米，停下来10秒系鞋带，然后在20秒内继续滑行90米。她整个行程的平均速度是多少？
0 s + 10 s + 20 s = 60 s. Therefore, average speed = 240 m ÷ 60 s = 4 m/s. The rest period is included because average speed considers the whole journey time.
要计算平均速度，使用公式：平均速度 = 总距离 ÷ 总时间。总距离 = 150 m + 90 m = 240 m。总时间包括休息时间：30 s + 10 s + 20 s = 60 s。因此，平均速度 = 240 m ÷ 60 s = 4 m/s。休息时间被纳入是因为平均速度考虑整个行程时间。
Key takeaway: Average speed is not simply the mean of the moving speeds; it is total distance divided by total time, including any stops.
关键要点：平均速度不仅仅是移动速度的平均值；它是总距离除以总时间，包括任何停顿。
2. Case Study: Snowshoes and Pressure | 案例二：雪鞋与压强
An explorer weighing 800 N walks on fresh snow. When wearing normal boots with a total sole area of 0.025 m², she sinks into the snow. With snowshoes of total area 0.20 m², she stays on the surface. Explain why, using pressure calculations.
一名重800 N的探险家在新鲜雪地上行走。穿着总鞋底面积为0.025 m²的普通靴子时，她会陷入雪中。穿上总面积为0.20 m²的雪鞋时，她能停留在雪面上。请用压强计算解释原因。
ly increase the area, reducing the pressure on the snow to one-eighth. A smaller pressure means the snow is less likely to compress and collapse, so the explorer does not sink.
压强 = 力 ÷ 面积。首先，靴子：压强 = 800 N ÷ 0.025 m² = 32 000 Pa。其次，雪鞋：压强 = 800 N ÷ 0.20 m² = 4000 Pa。雪鞋显著增大了面积，使雪面上的压强降至八分之一。压强越小，雪就不容易被压缩和塌陷，因此探险家不会下沉。
This illustrates why vehicles with wide tyres or tracks perform better on soft ground – they spread the weight over a larger area.
这说明了为什么宽轮胎或履带车辆在松软地面上表现更好——它们将重量分散到了更大的面积上。
3. Case Study: Series Circuit Lamp Failure | 案例三：串联电路灯泡故障
A circuit contains two lamps connected in series with a 6 V battery. When both lamps are working, each glows normally. Suddenly, one lamp goes out and the other also turns off. Suggest a reason and describe how you would test the circuit with a voltmeter.
一个电路包含两只灯泡串联连接到6 V电池。当两只灯泡都工作时，每只正常发光。突然，一只灯泡熄灭，另一只也灭了。提出一个原因，并描述你将如何用电压表测试电路。
In a series circuit, if one lamp fails (filament breaks), the circuit is broken and current stops everywhere. So both lamps turn off. To test, place a voltmeter across the terminals of the suspicious lamp. If it reads the full battery voltage (6 V), that lamp is faulty – the broken filament acts like an infinite resistance, with all the voltage dropped across it. If the voltmeter reads 0 V, the fault may be elsewhere, but with a simple series loop it is almost certainly that lamp.
在串联电路中，如果一只灯泡损坏（灯丝断开），电路断开，各处电流停止。因此两只灯泡都熄灭。为了测试，将电压表跨接在可疑灯泡的两端。如果读数为满电池电压（6 V），那只灯泡就有故障——断开的灯丝如同无穷大电阻，所有电压都降在其上。如果电压表读数为0 V，故障可能在别处，但在简单的串联回路中几乎可以肯定就是该灯泡故障。
Remember: voltmeter must be connected in parallel and, in a healthy circuit, would show the lamp's share of voltage, not the total.
记住：电压表必须并联连接，在完好电路中，它只会显示该灯泡分得的电压，而非总电压。
4. Case Study: Roller Coaster Energy Changes | 案例四：过山车能量变化
A roller coaster car starts from rest at the top of a 40 m hill. Ignoring friction, describe the energy transformations as it moves down the first drop and up the next smaller hill. Use the terms gravitational potential energy (GPE) and kinetic energy (KE).
一辆过山车从40米高的山顶静止启动。忽略摩擦，描述它冲下第一个坡道并冲上下一个较矮山坡时的能量转换。使用重力势能和动能的术语。
At the top, the car has maximum GPE and zero KE. As it descends, GPE decreases and KE increases – the car speeds up. At the bottom of the drop, GPE is at a minimum (we can set it to zero) and KE is maximum. When climbing the next hill, KE is converted back into GPE: the car slows down. Since no energy is lost, the total mechanical energy remains constant. The car will not reach the original 40 m height on the next hill unless that hill is shorter; some KE remains.
在顶部，车具有最大重力势能和零动能。下降时，重力势能减少，动能增加——车加速。在坡底，重力势能最小（可设定为零），动能最大。当爬上下一个山坡时，动能重新转化为重力势能：车减速。由于没有能量损失，总机械能守恒。车不会在下一个山坡达到原有的40米高度，除非该山坡更矮；一部分动能会保留。
ght (mgh). KE = ½ × mass × speed². These formulas can calculate exact values at any point.
重力势能 = 质量 × 重力场强 × 高度 (mgh)。动能 = ½ × 质量 × 速度²。这些公式可以计算任何点的精确值。
5. Case Study: Measuring Density of an Odd Rock | 案例五：测量不规则石块的密度
A student wants to find the density of a small, irregularly shaped stone. She measures its mass as 120 g using a balance. To find volume, she uses a displacement can and a measuring cylinder. The water level in the cylinder rises from 50 cm³ to 70 cm³. Calculate the density in g/cm³ and kg/m³. Explain whether the stone will float in water (density 1 g/cm³).
一名学生想测量一块不规则小石头的密度。她用天平测得质量为120 g。为求体积，她使用排水罐和量筒。量筒中水位从50 cm³上升至70 cm³。计算密度（以g/cm³和kg/m³表示）。解释该石头是否会浮在水（密度1 g/cm³）上。
Volume of stone = 70 cm³ – ÷ 20 cm³ = 6 g/cm³. To convert to kg/m³, multiply by 1000: 6 g/cm³ = 6000 kg/m³. The stone' 1 g/cm³, so it will sink in water. Objects sink if their density is greater than the fluid's density.
石头体积 = 70 cm³ – 50 cm³ = 20 cm³。密度 = 质量 ÷ 体积 = 120 g ÷ 20 cm³ = 6 g/cm³。转换为kg/m³，乘以1000：6 g/cm³ = 6000 kg/m³。该石头的密度大于1 g/cm³，因此会沉入水中。物体密度大于流体密度时会下沉。
This method works for any irregular solid that does not absorb water and sinks. For floating objects, a sinker is used.
此方法适用于任何不吸水且下沉的不规则固体。对于漂浮物体，需使用沉块。
6. Case Study: Echo in a Canyon | 案例六：峡谷回声
A hiker shouts towards a vertical cliff and hears the echo 2.4 seconds later. The speed of sound in air that day is 340 m/s. How far away is the cliff? Explain why the time must be divided by two.
一名徒步者朝着垂直悬崖喊叫，2.4秒后听到回声。当天空气中的声速为340 m/s。悬崖有多远？解释为什么时间必须除以2。
e = 340 m/s × 2.4 s = 816 m. Therefore, distance to cliff = 816 m ÷ 2 = 408 m. Always halve the total distance when using echo time to find one-way distance.
声音传播到悬崖再返回，所以声音传播的距离是到悬崖距离的两倍。声音传播距离 = 速度 × 时间 = 340 m/s × 2.4 s = 816 m。因此，到悬崖的距离 = 816 m ÷ 2 = 408 m。利用回声时间求单向距离时，总是要将总距离减半。
Echoes are used by bats and ships (sonar) to navigate and find food or depth.
回声被蝙蝠和船只（声纳）用于导航和寻找食物或测量深度。
7. Case Study: Friction and Braking | 案例七：摩擦力与刹车
A cyclist is moving at 8 m/s on a level road. She applies the brakes and comes to a stop in 4 seconds. The total mass of bike and rider is 75 kg. Assuming a constant braking force, calculate the deceleration and the size of the frictional force between tyres and road.
一名骑行者在平路上以8 m/s的速度行驶。她捏下刹车，4秒后停下。自行车和骑行者的总质量为75 kg。假设刹车力恒定，计算减速度以及轮胎与路面间摩擦力的大小。
First, deceleration a = change in speed ÷ time = (0 – 8 m/s) ÷ 4 s = –2 m/s²; the magnitude is 2 m/s². Using Newton' N. This force acts in the direction opposite to motion, provided by the road on the tyres.
首先，减速度 a = 速度变化量 ÷ 时间 = (0 – 8 m/s) ÷ 4 s = -2 m/s²；大小为2 m/s²。使用牛顿第二定律，力 = 质量 × 加速度。摩擦力 F = 75 kg × 2 m/s² = 150 N。该力方向与运动相反，由路面对轮胎提供。
If the road were icy, the coefficient of friction would be lower, providing less force, so the stopping distance would increase. This case links motion and forces elegantly.
如果路面结冰，摩擦系数会降低，提供的力更小，因此刹车距离会增加。这个案例巧妙地将运动与力联系起来。
8. Case Study: Seesaw Balance | 案例八：跷跷板平衡
Two children play on a seesaw. The pivot is in the middle. Child A has a weight of 400 N and sits 1.5 m from the pivot. Where must child B, weighing 500 N, sit to balance the seesaw? Also calculate the moment produced by A.
两个孩子在玩跷跷板。支点在中间。孩子A体重400 N，坐在离支点1.5米处。体重500 N的孩子B应该坐在何处才能使跷跷板平衡？同时计算A产生的力矩。
moment = anticlockwise moment. Moment = force × perpendicular distance from pivot. Child A's moment = 400 N × 1.5 m = 600 Nm (assume anticlockwise). Child B' nt ÷ force = 600 Nm ÷ 500 N = 1.2 m. B must sit 1.2 m from the pivot on the opposite side.
要实现平衡，顺时针力矩 = 逆时针力矩。力矩 = 力 × 到支点的垂直距离。A的力矩 = 400 N × 1.5 m = 600 Nm（假定逆时针）。B的力矩必须等于600 Nm。因此，B的距离 = 力矩 ÷ 力 = 600 Nm ÷ 500 N = 1.2 m。B必须坐在另一侧离支点1.2米处。
The seesaw is a first-class lever. The principle of moments is used in many tools such as spanners and crowbars.
跷跷板属于第一类杠杆。力矩原理在许多工具中使用，如扳手和撬棍。
9. Case Study: Reflection from a Plane Mirror | 案例九：平面镜反射
A ray of light strikes a plane mirror at an angle of incidence of 35°. State the angle of reflection. Draw the set-up and label the incident ray, reflected ray, normal, angle i and angle r. Explain what happens if the mirror is rotated by 10° while the incident ray remains fixed.
一束光线以35°的入射角射向平面镜。写出反射角。画出装置并标出入射光线、反射光线、法线、入射角i和反射角r。解释如果镜子旋转10°而入射光线保持不变，会发生什么。
ured from the normal. If the mirror is rotated by 10°, the angle between the mirror surface and the fixed incident ray changes, effectively changing the angle of incidence by 10° as well. The reflected ray will rotate by twice that angle, i.e., 20°, relative to its original position.
反射定律：入射角(i) = 反射角(r)。因此，r = 35°。两个角都从法线量起。如果镜子旋转10°，镜面与固定入射光线之间的夹角改变，导致入射角也改变10°。反射光线将相对于原始位置旋转两倍的角度，即20°。
This principle is applied in periscopes and optical instruments to redirect light.
这一原理应用于潜望镜和光学仪器中以改变光路。
10. Case Study: Hot Air Balloon Lift | 案例十：热气球升力
kg/m³. Calculate the upthrust (buoyant force) and the net upward force if the total weight of balloon, basket and passengers is 2500 N. (Take g = 10 N/kg.)
一个热气球总体积为2800 m³。外部冷空气密度为1.3 kg/m³，气囊内热空气密度为1.0 kg/m³。若气球、吊篮和乘客总重为2500 N，计算浮力及净上升力。（取g = 10 N/kg。）
density of hot air × g = 2800 m³ × 1.0 kg/m³ × 10 N/kg = 28 000 N. The net upward force = upthrust – (weight of hot air + weight of structure) = 36 400 N – 500 N) = 5900 N upwards. The balloon rises because the buoyant force exceeds total weight.
浮力 = 排开冷空气的重量 = 体积 × 外部空气密度 × g = 2800 m³ × 1.3 kg/m³ × 10 N/kg = 36 400 N。内部热空气重量 = 体积 × 热空气密度 × g = 2800 m³ × 1.0 kg/m³ × 10 N/kg = 28 000 N。净上升力 = 浮力 – (热空气重量 + 结构重量) = 36 400 N – (28 000 N + 2500 N) = 5900 N 向上。气球上升是因为浮力超过总重量。
This demonstrates Archimedes’ principle in gases: an object submerged in a fluid experiences an upthrust equal to the weight of the fluid displaced.
这展示了阿基米德原理在气体中的应用：浸在流体中的物体受到的浮力等于排开流体的重量。
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