📚 Case Study Practical Exercises for AQA Year 10 Physics | AQA 十年级物理案例分析实战演练
Applying physics concepts to real-world scenarios is a vital skill in the AQA Year 10 course. This article walks you through a series of case studies that mirror the style of exam problems, helping you practise energy transfers, electricity, forces, waves, and more. Work through each example to build confidence in tackling unfamiliar contexts.
将物理概念应用于现实场景是 AQA 十年级课程的一项关键技能。本文将带你分析一系列模拟考试题型的案例，帮助你练习能量转换、电学、力、波等知识点。仔细研读每个实例，培养应对陌生情境的信心。
1. Case 1: Roller Coaster Energy Transfer | 案例一：过山车的能量转换
A roller coaster car with a total mass of 500 kg is lifted to the top of a 30 m high hill and released from rest. Assuming that friction and air resistance are negligible, determine the speed of the car when it reaches the bottom of the hill.
一辆总质量为 500 kg 的过山车被提升到 30 m 高的坡顶并从静止释放。假设摩擦力和空气阻力可以忽略不计，求过山车到达坡底时的速度。
At the top, the car has gravitational potential energy relative to the bottom: Ep = mgh. As it descends, this energy is converted into kinetic energy. Using the principle of conservation of energy, we can write mgh = ½mv², where v is the speed at the bottom.
在坡顶，过山车相对于坡底具有重力势能：Ep = mgh。下滑过程中，这部分能量转化为动能。根据能量守恒定律，我们可以写出 mgh = ½mv²，其中 v 是坡底的速度。
v = √(2gh) = √(2 × 9.8 m/s² × 30 m) = √588 ≈ 24.2 m/s
Notice that the mass cancels out, so the speed depends only on the height. In reality, some energy would be transferred to thermal stores because of friction, and the actual speed would be slightly lower.
注意质量被消去，因此速度只取决于高度。在现实中，由于摩擦，部分能量会转移到热能储存，实际速度会略低一些。
2. Case 2: Choosing the Correct Fuse | 案例二：选择合适的保险丝
A portable heater is rated at 2.2 kW and operates on the UK mains supply of 230 V. The appliance is fitted with a plug that contains a fuse. The available fuse ratings are 3 A, 5 A and 13 A. Determine which fuse should be used and explain why.
一台便携式加热器额定功率为 2.2 kW，使用英国 230 V 市电供电。该电器插头内置保险丝，可选的保险丝额定值有 3 A、5 A 和 13 A。判断应选用哪一个保险丝，并解释原因。
First, calculate the normal operating current using the equation P = IV. Rearranging gives I = P / V. Note: 2.2 kW = 2200 W.
首先，用公式 P = I V 计算正常工作电流。变形得 I = P / V。注意：2.2 kW = 2200 W。
I = 2200 W ÷ 230 V ≈ 9.57 A
The current drawn during normal operation is about 9.6 A. The fuse rating must be slightly higher than the normal current, so the 13 A fuse is the appropriate choice. If a 5 A fuse were used, it would blow as soon as the heater was switched on; a 3 A fuse would also fail. The 13 A fuse allows the heater to work safely while still providing protection against excessive current.
正常工作时的电流约为 9.6 A。保险丝的额定值必须略高于正常工作电流，因此 13 A 保险丝是合适的选择。如果使用 5 A 保险丝，一通电就会熔断；3 A 的也会失效。13 A 保险丝既能保证加热器正常工作，又能在出现过流时提供保护。
3. Case 3: Thinking and Braking Distance | 案例三：思考距离与制动距离
A car travels at a steady speed of 20 m/s. The driver spots a hazard and reacts after 0.7 seconds. The brakes then produce a constant deceleration of 6 m/s². Calculate the total stopping distance.
一辆汽车以 20 m/s 的恒定速度行驶。驾驶员发现危险，经过 0.7 秒后作出反应。随后刹车产生恒定的 6 m/s² 减速度。计算总停车距离。
The overall stopping distance is the sum of the thinking distance and the braking distance. Thinking distance is the distance travelled during the reaction time before the brakes are applied.
总停车距离等于思考距离与制动距离之和。思考距离是指在刹车起作用之前的反应时间内汽车行驶的距离。
Thinking distance = speed × reaction time = 20 m/s × 0.7 s = 14 m
For the braking phase, the car decelerates from 20 m/s to 0 m/s. Using v² = u² + 2as, with final velocity v = 0, initial velocity u = 20 m/s, and acceleration a = -6 m/s²:
制动阶段，汽车从 20 m/s 减速到 0 m/s。利用 v² = u² + 2as，其中末速度 v = 0，初速度 u = 20 m/s，加速度 a = -6 m/s²：
0 = (20)² + 2(-6)s ⇒ 0 = 400 – 12s ⇒ s = 400 ÷ 12 ≈ 33.3 m
Total stopping distance = 14 m + 33.3 m = 47.3 m (to 1 decimal place). Factors such as wet roads, worn tyres, or driver tiredness could increase this distance significantly.
总停车距离 = 14 m + 33.3 m = 47.3 m（保留一位小数）。雨天、轮胎磨损或驾驶员疲劳等因素都会显著增加这一距离。
4. Case 4: Cooling a Hot Drink | 案例四：热饮冷却
A student pours 200 g of coffee at 80 °C into a mug. The coffee cools to 50 °C before the student takes a sip. The specific heat capacity of coffee is 4200 J/kg°C. How much energy is transferred from the coffee to the surroundings?
一名学生将 200 g、80 °C 的咖啡倒入杯子中。在喝之前，咖啡冷却到 50 °C。咖啡的比热容为 4200 J/kg°C。问有多少能量从咖啡传递到了周围环境？
The energy transferred as heat can be calculated using ΔE = m c Δθ. First, convert the mass to kilograms: 200 g = 0.2 kg. The temperature change Δθ = 80 °C – 50 °C = 30 °C.
以热量形式传递的能量可用 ΔE = m c Δθ 计算。先将质量换算为千克：200 g = 0.2 kg。温度变化 Δθ = 80 °C – 50 °C = 30 °C。
ΔE = 0.2 kg × 4200 J/kg°C × 30 °C = 25 200 J
Thus, 25 200 joules leave the coffee. This energy increases the internal energy of the mug and the surrounding air. The rate of cooling could be slowed down by using an insulated mug, which reduces thermal conduction.
因此，咖啡释放了 25 200 焦耳的能量。这部分能量增加了杯子和周围空气的内能。使用保温杯可以减缓冷却速率，因为它能减少热传导。
5. Case 5: Solar Panel Efficiency | 案例五：太阳能电池板效率
A rooftop solar panel has an area of 2.5 m². On a clear day, the radiation arriving from the Sun is 1000 W/m². The panel generates an electrical power output of 375 W. Calculate the efficiency of the panel.
一块屋顶太阳能电池板的面积为 2.5 m²。晴天时，来自太阳的辐射强度为 1000 W/m²。该电池板输出的电功率为 375 W。计算电池板的效率。
The total power input from sunlight is the intensity multiplied by the area: Pin = 1000 W/m² × 2.5 m² = 2500 W. The useful power output is Pout = 375 W.
太阳光输入的总功率等于辐射强度乘以面积：Pin = 1000 W/m² × 2.5 m² = 2500 W。有用输出功率 Pout = 375 W。
Efficiency = (Pout ÷ Pin) × 100% = (375 ÷ 2500) × 100% = 15%
An efficiency of 15% is typical for many commercial photovoltaic panels. The rest of the energy is mainly transferred into thermal stores, meaning the panel heats up. Improving efficiency is a major goal in solar technology.
15% 的效率对于许多商用光伏板来说是典型值。其余的能量主要转化为内能，使得电池板升温。提高效率是太阳能技术的主要目标之一。
6. Case 6: Submarine Pressure | 案例六：潜水艇承受的压强
A submarine dives to a depth of 120 m in seawater of density 1030 kg/m³. Calculate the increase in pressure due to the water at this depth. (g = 9.8 N/kg)
一艘潜水艇下潜到海水密度为 1030 kg/m³ 的 120 m 深处。计算该深度处由于水引起的压强增加量。(g = 9.8 N/kg)
The pressure exerted by a column of liquid is given by p = ρgh, where ρ is the density, g is the gravitational field strength, and h is the depth.
液柱产生的压强由公式 p = ρgh 给出，其中 ρ 是密度，g 是重力场强度，h 是深度。
p = 1030 kg/m³ × 9.8 N/kg × 120 m = 1 211 280 Pa
This is about 1.21 × 10⁶ Pa, or roughly 12 times atmospheric pressure (≈ 100 000 Pa). The total pressure on the submarine would be this water pressure plus the atmospheric pressure at the surface, but the question asked only for the increase.
这大约是 1.21 × 10⁶ Pa，约为大气压（≈ 100 000 Pa）的 12 倍。潜水艇受到的总压强是此水压加上海面的大气压，但题目只要求计算增加量。
7. Case 7: Bat Echolocation | 案例七：蝙蝠的回声定位
A bat emits a high-frequency sound pulse while flying towards a cave wall. The echo is detected 0.015 seconds after emission. Given the speed of sound in air is 340 m/s, how far away is the wall?
一只蝙蝠在朝洞穴墙壁飞行时发出高频声脉冲，0.015 秒后检测到回声。已知空气中声速为 340 m/s，问墙壁距离多远？
The sound travels from the bat to the wall and back. Therefore, the total distance travelled by the sound is twice the distance from the bat to the wall: 2d = v × t.
声音从蝙蝠传到墙壁再返回。因此，声音传播的总路程是蝙蝠到墙壁距离的两倍：2d = v × t。
d = (v × t) ÷ 2 = (340 m/s × 0.015 s) ÷ 2 = 2.55 m
The wall is 2.55 metres away. Bats use this principle of echolocation to navigate and hunt in the dark. Ultrasound pulses are also used in medicine for foetal imaging because they are non-ionising.
墙壁距离为 2.55 米。蝙蝠利用回声定位的原理在黑暗中导航和捕食。医学上也用超声波脉冲进行胎儿成像，因为超声波是非电离的。
8. Case 8: Stretching a Spring | 案例八：拉伸弹簧
A student hangs a spring from a clamp and adds weights. With a load of 6.0 N, the spring extends by 0.030 m. Assuming the spring obeys Hooke’s law, calculate the spring constant. Then predict the extension when a 10.0 N load is applied.
一名学生将弹簧悬挂在铁架台上，并添加砝码。当负载为 6.0 N 时，弹簧的伸长量为 0.030 m。假设弹簧遵循胡克定律，计算弹簧常数，然后预测施加 10.0 N 负载时的伸长量。
Hooke’s law states that the extension is directly proportional to the force, provided the limit of proportionality is not exceeded: F = kx, where k is the spring constant.
胡克定律指出，只要不超过比例极限，伸长量与力成正比：F = kx，其中 k 是弹簧常数。
k = F ÷ x = 6.0 N ÷ 0.030 m = 200 N/m
For a 10.0 N load, the extension is found by rearranging: x = F ÷ k = 10.0 N ÷ 200 N/m = 0.050 m (5.0 cm). If the spring were loaded beyond its elastic limit, permanent deformation would occur, and Hooke’s law would no longer apply.
对于 10.0 N 的负载，伸长量由变形公式求得：x = F ÷ k = 10.0 N ÷ 200 N/m = 0.050 m (5.0 cm)。如果弹簧加载超过其弹性极限，就会发生永久变形，胡克定律也不再适用。
9. Case 9: Carbon Dating a Fossil | 案例九：碳定年法测定化石
An archaeologist discovers a wooden tool in an excavation. A fresh sample of the same wood gives a carbon-14 count rate of 32 counts per minute. The ancient tool gives a count rate of 8 counts per minute. The half-life of carbon-14 is 5730 years. Estimate the age of the tool.
一位考古学家在发掘中发现了一件木质工具。同种木材的新鲜样本碳-14 计数率为每分钟 32 次。古代工具的计数率为每分钟 8 次。碳-14 的半衰期为 5730 年。估算该工具的年代。
Each half-life reduces the number of radioactive carbon-14 nuclei by half. The count rate has fallen from 32 to 8, which is a factor of ¼. To go from 32 to 16 is one half-life; from 16 to 8 is a second half-life.
每经过一个半衰期，放射性碳-14 原子核的数量减少一半。计数率从 32 降至 8，这是原始值的 ¼。从 32 降到 16 经历一个半衰期，从 16 降到 8 经历第二个半衰期。
Therefore, two half-lives have passed. Age = 2 × 5730 years = 11 460 years.
因此，已经过了两个半衰期。年龄 = 2 × 5730 年 = 11 460 年。
Carbon dating is useful for organic materials up to about 50 000 years old. It relies on the assumption that the concentration of carbon-14 in the atmosphere has been constant, which can be calibrated using tree rings.
碳定年法适用于约 5 万年内的有机物质。它基于假设大气中碳-14 的浓度是恒定的，这可以通过树木年轮进行校准。
Published by TutorHao | Physics Revision Series | aleveler.com
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