📚 Year 9 CAIE Physics: Mastering Case Studies | Year 9 CAIE物理：案例分析实战演练
Case studies bring physics to life by applying concepts to real-world situations. In this article we work through ten practical scenarios that test your understanding of forces, energy, electricity, waves, pressure and thermal physics. Each case is examined step by step in the way a CAIE examiner expects, so you learn both the physics and the exam technique needed for Year 9 CAIE Physics.
案例分析把物理概念应用到真实情境中，让物理"活"起来。本文将通过十个实际场景，检测你对力、能量、电学、波动、压强和热物理的理解。每个案例都按 CAIE 考官的期待逐步拆解，让你同时掌握物理知识和 Year 9 CAIE 物理所需的答题技巧。
1. What Case Studies Test and How to Answer Them | 案例分析考什么、怎么答
A case study is not a memory test. The examiner gives you an unfamiliar situation - a car braking, a ship floating, a kettle boiling - and checks whether you can select the right physics, use the right equation and comment on the answer. Marks are usually split between calculation (up to 3 marks), explanation using correct terminology (2-3 marks) and a final judgement or evaluation (1-2 marks).
案例分析不是背诵测试。考官给出一个陌生情境——汽车刹车、轮船漂浮、水壶烧水——考查你能否选出正确的物理知识、使用正确的公式，并对结果作出评价。分数通常分为三部分：计算（最多 3 分）、用规范术语解释（2-3 分）以及最后的判断或评估（1-2 分）。
Before you write anything, use this four-step routine. It works for every case in this article and for every Checkpoint or IGCSE-style question you will meet later.
动笔之前，请使用下面这套四步流程。它对本文每个案例、以及你今后遇到的所有 Checkpoint 或 IGCSE 风格题目都适用。
- Step 1: Read the scenario twice and underline every number together with its unit.
- Step 2: Name the physics topic and write down the equation you need before substituting.
- Step 3: Substitute the values, keep the units in every line, and round to a sensible number of significant figures.
- Step 4: Check that the answer is physically reasonable and add one sentence of judgement.
- 第一步：把情境读两遍，划出每一个数字及其单位。
- 第二步：先说出物理主题，写出所需公式，再代入数值。
- 第三步：代入计算，每一行都带单位，并按合理的有效数字取整。
- 第四步：检查答案在物理上是否合理，并补一句判断性结论。
2. Case Study 1: Braking Distances and Road Safety | 案例一：刹车距离与道路安全
d from speed × time: 20 m/s × 0.7 s = 14 m. No braking has happened yet - the car is still moving at full speed.
想象你正以 20 m/s 的速度行驶，突然一个小孩跑到路上。你的反应时间是 0.7 秒。在这段时间内，汽车行驶的思考距离由"速度 × 时间"求得：20 m/s × 0.7 s = 14 m。此时还没有开始刹车，汽车仍以原速前进。
ime comes from change in speed ÷ deceleration: 20 ÷ 8 = 2.5 s. The braking distance is therefore 10 m/s × 2.5 s = 25 m.
接着你踩下刹车，汽车以 8 m/s² 减速。用平均速度法——速度从 20 m/s 降到 0，平均速度为 10 m/s——刹车时间由"速度变化量 ÷ 减速度"求得：20 ÷ 8 = 2.5 s。因此刹车距离为 10 m/s × 2.5 s = 25 m。
is roughly ten car lengths, which is why the Highway Code tells drivers to leave a two-second gap in dry conditions and more when the road is wet.
总停车距离是两部分之和：14 m + 25 m = 39 m。这大约相当于十个车身长度，所以交通规则要求在干燥路面上保持两秒车距，湿滑路面上还要更长。
Now double the speed to 40 m/s. The thinking distance doubles to 28 m, but the braking distance does not double - it quadruples to about 100 m, because kinetic energy depends on speed squared. The new total is about 128 m, more than three times the original. This is why speed limits save lives: a small increase in speed causes a large increase in stopping distance.
现在把速度加倍到 40 m/s。思考距离加倍至 28 m，但刹车距离不是加倍，而是约变为四倍，约 100 m，因为动能与速度的平方成正比。新的总停车距离约为 128 m，是原来的三倍多。这就是限速能救命的原因：速度小幅上升，停车距离却大幅增加。
peed × reaction time stopping distance = thinking distance + braking distance
it: 20² ÷ (2 × 8) = 400 ÷ 16 = 25 m, exactly the same answer as the average speed method.
单独的刹车距离公式来自 v² = 2 a s，即 s = v² ÷ (2a)。验证一下：20² ÷ (2 × 8) = 400 ÷ 16 = 25 m，与平均速度法的结果完全一致。
3. Case Study 2: Forces in a Tug-of-War | 案例二：拔河比赛中的力
Two teams pull on a rope. A common mistake is to say that the stronger team pulls harder on the rope. In fact the tension in the rope is the same at both ends - that is Newton's third law in action - so the tension alone cannot decide the winner. What decides it is the resultant force on each team, which depends mainly on the friction between their shoes and the ground.
两队拉一根绳子。常见错误是说"力气大的一队拉绳更用力"。实际上绳子两端的拉力大小相等——这正是牛顿第三定律——所以单看绳的张力无法判断胜负。真正决定胜负的是每队所受的合力，而合力主要取决于队员鞋底与地面之间的摩擦力。
When the forces on a body are balanced, the resultant force is zero and the body either stays still or keeps moving at a constant velocity. When they are unbalanced, the body accelerates in the direction of the resultant force. The rope in a tug-of-war is only still when the two pulls are equal in size and opposite in direction.
当物体受力平衡时，合力为零，物体要么保持静止，要么保持匀速运动；当受力不平衡时，物体沿合力方向加速。拔河中绳子只有在两侧拉力大小相等、方向相反时才是静止的。
to the left. Using resultant force = mass × acceleration, a 200 kg total mass accelerates at 100 ÷ 200 = 0.5 m/s².
假设左队拉力为 800 N，右队拉力为 700 N。合力为向左 100 N，因此绳子和两队一起向左加速。由"合力 = 质量 × 加速度"，总质量为 200 kg 时加速度为 100 ÷ 200 = 0.5 m/s²。
- Always list the forces: weight, normal contact force, tension, friction and any push or pull.
- Draw a free body diagram with arrows starting from the object and labelled with sizes in newtons.
- Decide balanced or unbalanced before doing any calculation.
- State the direction of the resultant force and therefore the direction of acceleration.
- 先列出所有力：重力、法向支持力、张力、摩擦力以及任何推力或拉力。
- 画受力图，箭头从物体出发，并用牛顿标出大小。
- 计算前先判断平衡还是不平衡。
- 说明合力方向，从而说明加速度方向。
4. Case Study 3: Energy Transfers and Efficiency on a Roller Coaster | 案例三：过山车中的能量转移与效率
A roller coaster car at the top of a hill stores gravitational potential energy. As it falls, that store is transferred to the kinetic energy store of the car, and some is transferred by heating to the track and the air because of friction and air resistance. Energy is never destroyed - it is only transferred between stores.
过山车位于坡顶时具有重力势能。下落过程中，这些能量转移到小车的动能，同时由于摩擦和空气阻力，一部分能量通过做功转移到轨道和空气的内能。能量不会被消灭，只是在各种能量形式之间转移。
estions do, we get 100 000 J. Always check which value of g the question tells you to use.
假设小车质量为 500 kg，高度为 20 m。重力势能为 m g h = 500 × 9.8 × 20 = 98 000 J。如果题目要求用 g = 10 N/kg（很多 Year 9 题目如此），则为 100 000 J。务必看清题目给定的 g 值。
e that the mass cancels out, so a heavier car reaches the same speed from the same height.
如果没有任何能量损失，这 98 000 J 会全部变成动能。由 ½ m v² = 98 000 J 得 v² = 392，所以 v ≈ 19.8 m/s。注意质量被消去，因此从同一高度下落，更重的车也会达到同样的速度。
In reality about 20% 00 J becomes kinetic energy. This gives v ≈ 17.7 m/s. The efficiency of the transfer is 80%, calculated as useful output energy ÷ total input energy × 100%. The lost energy has warmed the track and the surrounding air.
现实中约有 20% 的能量通过做功转化为内能，只有 78 400 J 变成动能，此时 v ≈ 17.7 m/s。该能量转移的效率为 80%，计算方法是有用输出能量 ÷ 总输入能量 × 100%。损失的能量使轨道和周围空气温度升高。
m g h KE = ½ m v² efficiency = useful output ÷ total input × 100%
A strong answer also mentions the law of conservation of energy and states clearly that the "missing" energy has not disappeared but has been transferred to the thermal energy store of the surroundings.
高分答案还会提到能量守恒定律，并明确指出"消失"的能量并未消失，而是转移到了周围环境的内能中。
5. Case Study 4: Electricity in a Household Circuit | 案例四：家庭电路中的电学
n a fuse slightly larger than this, so a 13 A fuse is the correct choice; a 3 A fuse would melt immediately.
一只电水壶标有 2200 W、230 V。由"功率 = 电压 × 电流"得 I = P ÷ V = 2200 ÷ 230 ≈ 9.6 A。插头中所配保险丝应略大于该电流，因此应选 13 A；若用 3 A 保险丝会立刻熔断
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