📚 Year 8 Edexcel Physics: Case Study Practical Drills | Year 8 Edexcel 物理：案例分析实战演练
Case studies are a powerful way to connect classroom physics with real-world situations. This article provides ten practical drills covering key topics from the Year 8 Edexcel physics curriculum: forces, motion, energy, electricity, waves, and pressure. Each case presents a scenario, followed by step-by-step analysis using the principles you have learned. Work through them to strengthen your problem-solving skills and exam confidence.
案例研究是将课堂物理知识与现实世界联系起来的有效方式。本文提供了十个实战练习，涵盖Year 8 Edexcel物理课程的关键主题：力、运动、能量、电学、波和压强。每个案例都呈现一个情景，然后运用所学原理进行逐步分析。通过练习，可以加强你的解题能力和考试信心。
1. Case Study 1: The 100 m Sprint | 案例分析1：百米短跑
A runner completes the 100-metre dash in 12.5 seconds. (a) Calculate his average speed in metres per second (m/s). (b) Convert your answer to kilometres per hour (km/h). (c) Sketch a distance‑time graph for his motion, assuming he accelerates at the start and then holds a steady speed.
一名短跑运动员用12.5秒跑完100米。(a) 计算他的平均速度，单位米/秒。(b) 将答案转换为千米/小时。(c) 画出他运动的距离‑时间图草图，假设他从起跑加速后保持匀速。
formula, v = s / t = 100 m / 12.5 s = 8.0 m/s. His average speed is 8.0 metres per second.
解答：(a) 平均速度 = 总距离 ÷ 总时间。利用公式 v = s / t = 100 m ÷ 12.5 s = 8.0 m/s。他的平均速度为8.0米/秒。
So 8.0 × 3.6 = 28.8 km/h. At this average pace he would cover 28.8 kilometres in one hour.
(b) 将米/秒转换为千米/小时，乘以3.6。因此8.0 × 3.6 = 28.8 km/h。以这个平均速度，他一小时可以跑完28.8公里。
(c) The distance‑time graph starts with a curve that becomes steeper (acceleration phase), then changes to a straight line when max speed is reached. Label the x‑axis ‘time (s)’ and y‑axis ‘distance (m)’. Mark the point (12.5 s, 100 m).
(c) 距离‑时间图起初为弯曲上升（斜率变陡），表示加速阶段，之后达到最大速度时变为斜率不变的直线。标注x轴“时间（秒）”，y轴“距离（米）”。标记点(12.5 s, 100 m)。
2. Case Study 2: Interpreting a Distance‑Time Graph | 案例分析2：解读距离‑时间图
The graph below describes a cyclist’s journey. From 0 to 2 s the line is flat. From 2 s to 5 s the line is straight and sloping upward. From 5 s to 7 s the line curves upward, becoming steeper. From 7 s to 10 s the line curves in the opposite direction and flattens. Identify the motion in each interval and calculate the speed between 2 s and 5 s if the distance changes from 0 m to 15 m.
下图描述了一位自行车手的旅程。0到2秒线段水平；2到5秒为向上倾斜的直线；5到7秒向上弯曲且越来越陡；7到10秒反向弯曲并趋于水平。识别每一段的运动状态，并计算2秒到5秒间的速度，已知距离从0米变为15米。
0–2 s: horizontal line → the cyclist is stationary (stopped). 2–5 s: straight sloping line → constant speed. Speed = Δdistance / Δtime = (15 m – 0 m) / (5 s – 2 s) = 15 m / 3 s = 5 m/s. 5–7 s: steepening curve → acceleration (speeding up). 7–10 s: levelling‑off curve → deceleration (slowing down) until stationary.
0–2秒：水平线 → 自行车手静止（停止）。2–5秒：倾斜直线 → 匀速运动。速度 = 距离变化量 / 时间变化量 = (15 m – 0 m) ÷ (5 s – 2 s) = 15 m / 3 s = 5 m/s。5–7秒：变陡的曲线 → 加速（速度增加）。7–10秒：变平的曲线 → 减速（速度减小）直至静止。
3. Case Study 3: Forces on a Cyclist | 案例分析3：自行车手的受力
A girl starts cycling from rest on a flat road. She pushes hard on the pedals, and the bike accelerates. Draw the forces acting on the bike and the girl. Explain how the forces change when she reaches a steady speed, and what happens when she stops pedalling.
一个女孩在平路上从静止开始骑行。她用力踩脚踏板，自行车加速。画出作用在自行车和女孩上的力。解释当她达到匀速时力如何变化，以及当她停止踩踏时会怎样。
d weight, and upward normal reaction from the ground. While accelerating, thrust is larger than the total resistive forces → unbalanced force → acceleration. At steady speed, thrust equals total resistance → forces are balanced → zero resultant force → constant velocity. When she stops pedalling, thrust disappears; resistive forces slow her down (deceleration) until she stops.
受力包括：后轮推地产生的前进推力（摩擦力）、向后的空气阻力和滚动摩擦阻力、向下的重力，以及地面向上的支持力。加速时，推力大于总阻力 → 不平衡力 → 加速。匀速时，推力等于总阻力 → 力平衡 → 合力为零 → 速度恒定。当她停止踩踏，推力消失；阻力使她减速，直到停下来。
4. Case Study 4: Energy Changes on a Bungee Jump | 案例分析4：蹦极跳中的能量变化
A bungee jumper leaps from a bridge. Describe the main energy transfers from the moment she jumps until she reaches the lowest point. At what point is her kinetic energy greatest, and why?
一名蹦极者从桥上跳下。描述从她跳下直至最低点的主要能量转换。她的动能最大值出现在哪里，为什么？
s maximal just before the bungee cord starts to stretch significantly, because after that some energy starts storing as elastic potential energy (EPE). As the cord stretches, KE and remaining GPE gradually convert to EPE. At the lowest point, all the original GPE is stored as EPE (and a little thermal energy due to air resistance).
一开始，她具有最大的重力势能（GPE），动能为零。下落时，GPE 转化为动能（KE）→ 她加速。动能最大值出现在蹦极绳开始明显拉伸之前，因为之后部分能量开始储存为弹性势能（EPE）。随着绳索伸长，动能和剩余的重力势能逐渐转化为弹性势能。在最低点，所有初始重力势能都转化为弹性势能储存（以及少量因空气阻力产生的内能）。
5. Case Study 5: Lever and Load | 案例分析5：杠杆与负载
A worker uses a 2‑metre crowbar to lift a heavy rock. The pivot is placed 0.5 m from the rock. The rock weighs 500 N. What force must be applied at the other end to just lift the rock? Treat the crowbar as a weightless lever.
一名工人用一根2米长的撬棍抬起一块重石。支点距离石块0.5米。石块重500 N。另一端需要施加多大的力才能刚好撬起石块？将撬棍视为无重杠杆。
Moment of a force = force × perpendicular distance from pivot
力矩 = 力 × 到支点的垂直距离
Distance from pivot to rock = 0.5 m. Distance from pivot to applied force = 2 m – 0 → effort = 250 / 1.5 ≈ 167 N (to two significant figures). The worker must apply about 167 N.
支点到石块距离 = 0.5 m。支点到施力点距离 = 2 m – 0.5 m = 1.5 m。平衡时：逆时针力矩 = 顺时针力矩。负载力矩 = 500 N × 0.5 m = 250 N·m。施力力矩 = 施力 × 1.5 m。所以施力 × 1.5 = 250 → 施力 = 250 / 1.5 ≈ 167 N（取两位有效数字）。工人需要施加约167 N的力。
6. Case Study 6: Series Circuit Troubleshooting | 案例分析6：串联电路故障排查
A series circuit contains a 6 V battery and two identical bulbs, both lit. Suddenly one bulb’s filament breaks. Explain what happens to the other bulb and why. Draw the circuit before and after the fault.
一个串联电路包含一个6 V电池和两个相同灯泡，两个灯泡都亮着。突然一个灯泡的灯丝烧断了。解释另一个灯泡会怎样，为什么。画出故障前后的电路图。
. No current can flow anywhere in the circuit. Therefore, the other bulb also turns off immediately. Before the fault, the battery voltage was shared between the two bulbs, giving each about 3 V. After the break, the voltage across the broken bulb becomes 6 V, but no current flows through either bulb.
在串联电路中，电流只有一条路径。当灯丝烧断，该路径断开 → 电路成为断路。整个电路中没有电流流过。因此另一个灯泡也立即熄灭。故障前，电池电压由两个灯泡分担，每个约3 V。断路后，断开的灯泡两端电压变为6 V，但没有电流流过任何一个灯泡。
7. Case Study 7: Current in a Parallel Circuit | 案例分析7：并联电路中的电流
Two bulbs, A and B, are connected in parallel to a 12 V battery. The current through bulb A is measured as 0.4 A, and the current through bulb B is 0.3 A. Find the current drawn from the battery. State the rule you used.
两个灯泡A和B并联在12 V电池上。测得通过A的电流为0.4 A，通过B的电流为0.3 A。求从电池流出的总电流。并说明你使用的规律。
t law): Iₜₒₜₐₗ = I₁ + I₂. Here Iₜₒₜₐₗ = 0.4 A + 0.3 A = 0.7 A. The battery supplies 0.7 A. The voltage across each bulb is the full 12 V because they are in parallel.
在并联电路中，总电流等于各支路电流之和。这是电流守恒规律（基尔霍夫第一定律）：Iₜₒₜₐₗ = I₁ + I₂。这里 Iₜₒₜₐₗ = 0.4 A + 0.3 A = 0.7 A。电池提供0.7 A电流。由于并联，每个灯泡两端的电压都是12 V。
8. Case Study 8: Reflection in a Plane Mirror | 案例分析8：平面镜反射
A light ray strikes a flat mirror at an angle of 30° to the mirror surface. (a) What is the angle of incidence? (b) Draw a ray diagram showing the incident ray, reflected ray, normal, and all relevant angles. (c) Describe the image formed by a plane mirror.
一束光线以与镜面成30°的角度射向平面镜。(a) 入射角是多少？(b) 画出显示入射光线、反射光线、法线和所有相关角度的光路图。(c) 描述平面镜所成的像。
nt ray and the normal. The angle to the mirror surface is 30°, so angle of incidence = 90° – e of incidence = 60°. The image is virtual (cannot be projected), laterally inverted (left‑ as the object, and located the same distance behind the mirror as the object is in front.
入射角是入射光线与法线之间的夹角。光线与镜面夹角为30°，因此入射角 = 90° – 30° = 60°。根据反射定律，反射角 = 入射角 = 60°。平面镜所成的像为虚像（不能呈现在屏幕上）、左右颠倒、与物体大小相等，且位于镜后与物距相等的位置。
9. Case Study 9: Echo Sounding | 案例分析9：回声测距
A sonar device on a boat sends a sound pulse down to the seabed. The echo is detected 0.2 seconds later. The speed of sound in seawater is 1500 m/s. Calculate the depth of the water.
船上的声呐向下发射一个声脉冲，0.2秒后收到回声。声波在海水中的传播速度是1500 m/s。计算海水的深度。
. Therefore, depth = total distance / 2 = 300 m / 2 = 150 m. The water is 150 metres deep.
声波从船底传到海底再返回，因此总路程是深度的两倍。总路程 = 速度 × 时间 = 1500 m/s × 0.2 s = 300 m。所以深度 = 总路程 / 2 = 300 m / 2 = 150 m。海水深度为150米。
10. Case Study 10: Comparing Pressures | 案例分析10：比较压强
A woman of weight 600 N stands on one stiletto heel of area 1 cm². An elephant weighs 40 000 N and stands with all four feet, each foot area 0.1 m². Calculate the pressure exerted in each case and explain why the heel causes more damage to a floor.
一名重600 N的女性用面积为1 cm²的高跟鞋跟站立。一头大象重40 000 N，四脚着地，每只脚掌面积为0.1 m²。计算每种情况下的压强，并解释为何高跟鞋对地板的破坏更大。
Pressure = force ÷ area
压强 = 力 ÷ 面积
0.4 m² = 100 000 Pa (100 kPa). The heel pressure is 60 times greater despite the much smaller weight, because the force is concentrated on a tiny area, which can easily puncture or dent floor materials.
高跟鞋跟：面积 = 1 cm² = 0.0001 m²。压强 = 600 N ÷ 0.0001 m² = 6 000 000 Pa（6 MPa）。大象：总面积 = 4 × 0.1 m² = 0.4 m²。压强 = 40 000 N ÷ 0.4 m² = 100 000 Pa（100 kPa）。尽管重量小得多，鞋跟压强却是大象的60倍，因为力集中在极小的面积上，很容易压穿或损坏地板材料。
Published by TutorHao | Physics Revision Series | aleveler.com
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