📚 Case Study Practice for Year 11 CAIE Physics | Year 11 CAIE 物理：案例分析实战演练
Physics is not just about memorising equations—it is about applying them to real-world scenarios. This article presents a series of case studies carefully selected for Year 11 CAIE Physics students. Each case develops problem-solving skills by guiding you through analysis, formula selection, and step-by-step calculations. By practising these examples, you will strengthen your ability to interpret exam-style questions and demonstrate deep understanding in your answers.
物理不仅仅是背诵公式——更重要的是将它们应用于真实情境。本文为Year 11 CAIE物理学生精心挑选了一系列案例分析。每个案例通过引导你进行分析、选择和逐步计算，来锻炼解题能力。通过这些练习，你将增强解读考试题型的能力，并在答案中展示深刻的理解。
1. Motion of a Sprinting Athlete | 短跑运动员的运动
A sprinter accelerates from rest at 2.5 m/s² for 4.0 seconds and then maintains a constant speed for the remaining 6.0 seconds of a 100 m race. Does she finish the race within this time, and what is her average speed?
一名短跑运动员从静止开始以2.5 m/s²加速4.0秒，然后在100米比赛的剩余6.0秒内保持匀速。她能在这段时间内完成比赛吗？她的平均速度是多少？
Step 1 – Find the maximum speed: Use v = u + at. Here u = 0, a = 2.5 m/s², t = 4.0 s, so v = 0 + 2.5 × 4.0 = 10 m/s.
步骤1 – 求最大速度：使用 v = u + at。这里 u = 0，a = 2.5 m/s²，t = 4.0 s，因此 v = 0 + 2.5 × 4.0 = 10 m/s。
Step 2 – Distance during acceleration: s₁ = ut + ½at² = 0 + 0.5 × 2.5 × 4.0² = 20 m.
步骤2 – 加速阶段的距离：s₁ = ut + ½at² = 0 + 0.5 × 2.5 × 4.0² = 20 m。
Step 3 – Distance at constant speed: Remaining time = 6.0 s, so s₂ = v × t = 10 × 6.0 = 60 m. Total distance = 20 + 60 = 80 m. She does not finish 100 m; she covers only 80 m.
步骤3 – 匀速阶段的距离：剩余时间 = 6.0 s，所以 s₂ = v × t = 10 × 6.0 = 60 m。总距离 = 20 + 60 = 80 m。她未能完成100米，只跑了80米。
Step 4 – Average speed: v_avg = total distance / total time = 80 / 10 = 8.0 m/s.
步骤4 – 平均速度：v_avg = 总距离 / 总时间 = 80 / 10 = 8.0 m/s。
This case highlights the importance of dividing motion into distinct phases. Even if the acceleration phase is powerful, the total distance must be checked against the required race length.
这个案例强调了将运动分为不同阶段的重要性。即使加速阶段很有力，也必须核对总距离是否达到比赛要求的长度。
2. Forces in an Accelerating Lift | 加速电梯中的力
A person of mass 60 kg stands on a weighing scale inside a lift. The lift accelerates upward at 2.0 m/s². What reading does the scale show? Take g = 9.8 m/s².
一个质量为60公斤的人站在电梯内的体重秤上。电梯以2.0 m/s²向上加速。体重秤的读数是多少？取 g = 9.8 m/s²。
The scale reads the normal reaction force N. Apply Newton’s second law: resultant force = ma. Upward direction positive: N − mg = ma, so N = m(g + a).
体重秤读数是支持力 N。应用牛顿第二定律：合力 = ma。取向上为正方向：N − mg = ma，所以 N = m(g + a)。
N = 60 × (9.8 + 2.0) = 60 × 11.8 = 708 N. In kg-wt equivalent, divide by 9.8 → about 72.2 kg. The scale shows a higher reading because the lift accelerates upward, making the person feel heavier.
N = 60 × (9.8 + 2.0) = 60 × 11.8 = 708 N。转换为公斤力，除以9.8 → 约72.2 kg。因为电梯向上加速，人感觉更重，体重秤读数变大。
If the lift were accelerating downward at 2.0 m/s², then N = m(g − a) = 60 × 7.8 = 468 N, appearing lighter. This principle is crucial for understanding apparent weight in non-inertial frames.
如果电梯以2.0 m/s²向下加速，那么 N = m(g − a) = 60 × 7.8 = 468 N，人会感觉变轻。这一原理对于理解非惯性系中的视重至关重要。
3. Energy Conservation in a Roller Coaster | 过山车中的能量守恒
A roller coaster car of mass 500 kg starts from rest at point A, 30 m above the ground. It travels down a frictionless track to point B at ground level, then rises to point C, 20 m high. Find the speed at B and C. (g = 9.8 m/s²)
一辆质量为500 kg的过山车从A点静止出发，A点离地30米。它沿着无摩擦轨道滑到地面B点，然后上升到20米高的C点。求B点和C点的速度。（g = 9.8 m/s²）
At A: kinetic energy KE = 0, potential energy PE = mgh = 500 × 9.8 × 30 = 147 000 J. By conservation, at B: PE = 0, KE = 147 000 J, so ½mv² = 147 000 → v² = 2 × 147 000 / 500 = 588 → v ≈ 24.2 m/s.
在A点：动能 KE = 0，势能 PE = mgh = 500 × 9.8 × 30 = 147 000 J。根据能量守恒，在B点：PE = 0，KE = 147 000 J，所以 ½mv² = 147 000 → v² = 2 × 147 000 / 500 = 588 → v ≈ 24.2 m/s。
At C: height 20 m, PE = 500 × 9.8 × 20 = 98 000 J. Remaining energy becomes KE: 147 000 − 98 000 = 49 000 J. Then ½mv² = 49 000 → v² = 196 → v = 14.0 m/s.
在C点：高度20 m，PE = 500 × 9.8 × 20 = 98 000 J。剩余能量转化为动能：147 000 − 98 000 = 49 000 J。因此 ½mv² = 49 000 → v² = 196 → v = 14.0 m/s。
This classic problem demonstrates the swap between gravitational potential and kinetic energy. Note that the mass cancels out in such idealised cases, so speed depends only on height change.
这个经典问题展示了重力势能与动能之间的转换。注意在这种理想情况下质量会被约掉，速度只取决于高度变化。
4. Momentum in a Collision | 碰撞中的动量
A trolley of mass 2.0 kg moves at 3.0 m/s and collides with a stationary trolley of mass 1.0 kg. After the collision, the first trolley slows to 1.0 m/s. Find the velocity of the second trolley. Is the collision elastic?
一辆质量为2.0 kg的小车以3.0 m/s的速度运动，与一辆静止的质量为1.0 kg的小车碰撞。碰撞后第一辆小车的速度减为1.0 m/s。求第二辆小车的速度。碰撞是弹性的吗？
Momentum before = 2.0 × 3.0 + 1.0 × 0 = 6.0 kg·m/s. After collision: 2.0 × 1.0 + 1.0 × v = 2.0 + v. By conservation, 2.0 + v = 6.0 → v = 4.0 m/s.
碰撞前动量 = 2.0 × 3.0 + 1.0 × 0 = 6.0 kg·m/s。碰撞后：2.0 × 1.0 + 1.0 × v = 2.0 + v。由动量守恒得 2.0 + v = 6.0 → v = 4.0 m/s。
Check kinetic energy: before KE = 0.5 × 2.0 × 3.0² = 9.0 J. After KE = 0.5 × 2.0 × 1.0² + 0.5 × 1.0 × 4.0² = 1.0 + 8.0 = 9.0 J. KE is conserved, so the collision is elastic.
检查动能：碰撞前 KE = 0.5 × 2.0 × 3.0² = 9.0 J。碰撞后 KE = 0.5 × 2.0 × 1.0² + 0.5 × 1.0 × 4.0² = 1.0 + 8.0 = 9.0 J。动能守恒，所以碰撞是弹性的。
If the collision were inelastic, some kinetic energy would be converted to other forms. Always verify using both momentum and energy where relevant.
如果碰撞是非弹性的，部分动能将转化为其他形式。在相关情况下，务必同时用动量和能量进行验证。
5. Pressure in a Hydraulic Lift | 液压升降机中的压强
A hydraulic jack has a small piston of area 0.02 m² and a large piston of area 0.50 m². A force of 100 N is applied to the small piston. What load can the large piston lift? How far must the small piston move to raise the load by 0.10 m?
一台液压千斤顶的小活塞面积为0.02 m²，大活塞面积为0.50 m²。对小活塞施加100 N的力。大活塞能举起多大的重物？为了将重物提升0.10 m，小活塞需要移动多远？
Pressure is transmitted equally: P = F₁/A₁ = F₂/A₂. So F₂ = F₁ × (A₂/A₁) = 100 × (0.50 / 0.02) = 100 × 25 = 2500 N. That is a load mass of about 255 kg (using g = 9.8).
压强等值传递：P = F₁/A₁ = F₂/A₂。所以 F₂ = F₁ × (A₂/A₁) = 100 × (0.50 / 0.02) = 100 × 25 = 2500 N。这相当于约255 kg的重物（取 g = 9.8）。
Work input = work output (ignoring friction): F₁d₁ = F₂d₂. Therefore d₁ = (F₂ / F₁) × d₂ = (2500 / 100) × 0.10 = 25 × 0.10 = 2.5 m. The small piston must move a much larger distance to gain force.
输入功 = 输出功（忽略摩擦）：F₁d₁ = F₂d₂。因此 d₁ = (F₂ / F₁) × d₂ = (2500 / 100) × 0.10 = 25 × 0.10 = 2.5 m。小活塞必须移动更长的距离以获得力的增益。
Hydraulic systems perfectly illustrate the force–distance trade-off inherent in simple machines, grounded in Pascal’s principle.
液压系统完美地展示了基于帕斯卡原理的简单机械中力与距离的权衡。
6. Thermal Energy and Specific Heat Capacity | 热能与此热容
An aluminium block of mass 0.80 kg is heated by an electric heater supplying 50 W for 5.0 minutes. The temperature rises from 22 °C to 47 °C. Calculate the specific heat capacity of aluminium and compare it with the accepted value of 900 J/(kg·°C). Account for any discrepancy.
一个质量为0.80 kg的铝块用50 W的电热器加热5.0分钟。温度从22 °C升至47 °C。计算铝的比热容，并与公认值900 J/(kg·°C)比较。解释任何差异。
Energy supplied: E = P × t = 50 W × (5.0 × 60) s = 15 000 J. Temperature change Δθ = 47 − 22 = 25 °C. Using E = mcΔθ, we get c = E / (mΔθ) = 15 000 / (0.80 × 25) = 15 000 / 20 = 750 J/(kg·°C).
供应的能量：E = P × t = 50 W × (5.0 × 60) s = 15 000 J。温度变化 Δθ = 47 − 22 = 25 °C。利用 E = mcΔθ，得出 c = E / (mΔθ) = 15 000 / (0.80 × 25) = 15 000 / 20 = 750 J/(kg·°C)。
The experimental value 750 J/(kg·°C) is lower than 900. This suggests energy losses to the surroundings—the block radiated heat, or the heater was not perfectly efficient in transferring energy. To reduce losses, insulation would be used.
实验值750 J/(kg·°C)低于900。这表明有能量散失到周围环境中——铝块辐射了热量，或者加热器传递能量的效率不完美。为减少损失，需要使用隔热措施。
Recognising systematic errors is a key skill in practical physics. Always compare results with standard values and discuss possible reasons for deviations.
识别系统误差是实验物理中的关键技能。始终将结果与标准值进行比较，并讨论产生偏差的可能原因。
7. Wave Speed from Ripple Tank Observations | 从水波槽观测求波速
In a ripple tank, a dipper produces waves of frequency 12 Hz. The distance between the first and sixth bright crests is 25 cm. Calculate the wavelength and the speed of the waves.
在水波槽中，一个振源产生频率为12 Hz的波。第一和第六个亮波峰之间的距离为25 cm。计算波长和波速。
The distance from the 1st to the 6th crest spans 5 full wavelengths (count: 1→2, 2→3, 3→4, 4→5, 5→6). So 5λ = 25 cm → λ = 5.0 cm = 0.050 m.
从第一个波峰到第六个波峰跨越了5个完整波长（计数：1→2, 2→3, 3→4, 4→5, 5→6）。所以 5λ = 25 cm → λ = 5.0 cm = 0.050 m。
Wave speed v = f × λ = 12 Hz × 0.050 m = 0.60 m/s.
波速 v = f × λ = 12 Hz × 0.050 m = 0.60 m/s。
Care must be taken when counting wavelengths between numbered crests: for N crests, the number of wavelengths is N−1. This simple counting error can cost marks in exams.
在数两个编号波峰之间的波长时要小心：对于N个波峰，波长数为 N−1。这个简单的计数错误会在考试中失分。
8. Electrical Circuit Analysis – Series and Parallel | 电路分析 – 串联与并联
A 12 V battery is connected to three resistors: R₁ = 4 Ω and R₂ = 6 Ω are in parallel, and this combination is in series with R₃ = 3 Ω. Calculate the total resistance, the current drawn from the battery, and the potential difference across each resistor.
一个12 V的电池连接到三个电阻：R₁ = 4 Ω 和 R₂ = 6 Ω 并联，然后这个组合与 R₃ = 3 Ω 串联。计算总电阻、电池供给的电流和每个电阻两端的电压。
Parallel resistance: 1/R_par = 1/4 + 1/6 = 3/12 + 2/12 = 5/12, so R_par = 12/5 = 2.4 Ω. Total R = R_par + R₃ = 2.4 + 3.0 = 5.4 Ω.
并联电阻：1/R_par = 1/4 + 1/6 = 3/12 + 2/12 = 5/12，所以 R_par = 12/5 = 2.4 Ω。总电阻 R = R_par + R₃ = 2.4 + 3.0 = 5.4 Ω。
Current from battery: I = V / R_total = 12 / 5.4 ≈ 2.22 A. This current flows entirely through R₃, so V_R3 = I × R₃ = 2.22 × 3.0 ≈ 6.67 V. Then voltage across parallel pair = 12 − 6.67 = 5.33 V. Across R₁ and R₂ individually, V is 5.33 V, so I₁ = 5.33 / 4 = 1.33 A, I₂ = 5.33 / 6 = 0.89 A (check sum ≈ 2.22 A).
电池供给电流：I = V / R_total = 12 / 5.4 ≈ 2.22 A。该电流全部流过 R₃，所以 V_R3 = I × R₃ = 2.22 × 3.0 ≈ 6.67 V。然后并联组合两端的电压 = 12 − 6.67 = 5.33 V。R₁ 和 R₂ 各自的电压都是 5.33 V，因此 I₁ = 5.33 / 4 = 1.33 A，I₂ = 5.33 / 6 = 0.89 A（验证：两者之和 ≈ 2.22 A）。
Mastering circuit reduction and voltage/current division rules is essential for tackling more complex networks in CAIE examinations.
掌握电路的简化和电压/电流分配规则对于应对CAIE考试中更复杂的电路网络至关重要。
9. Electromagnetic Induction in a Simple Generator | 简单发电机中的电磁感应
A rectangular coil of 200 turns, each of area 0.0025 m², rotates in a magnetic field of strength 0.40 T. The coil makes 50 revolutions per second. Estimate the maximum induced e.m.f. and state two ways to increase it.
一个200匝的矩形线圈，每匝面积为0.0025 m²，在磁感应强度为0.40 T的磁场中旋转。线圈每秒转动50转。估算最大感应电动势，并说出两种增大电动势的方法。
Flux linkage per turn = BA = 0.40 × 0.0025 = 0.0010 Wb. For N turns, total flux linkage = NΦ = 200 × 0.0010 = 0.20 Wb. Angular frequency ω = 2πf = 2π × 50 = 100π ≈ 314 rad/s.
每匝的磁链 = BA = 0.40 × 0.0025 = 0.0010 Wb。N匝总磁链 = NΦ = 200 × 0.0010 = 0.20 Wb。角频率 ω = 2πf = 2π × 50 = 100π ≈ 314 rad/s。
Maximum induced e.m.f. |ε_max| = NΦω = 0.20 × 100π = 20π ≈ 62.8 V.
最大感应电动势 |ε_max| = NΦω = 0.20 × 100π = 20π ≈ 62.8 V。
To increase the output: (1) use stronger magnets to increase B; (2) increase the rotation speed f; (3) use more turns or a larger coil area. Induction is the working principle behind most power generation.
提高输出可以： (1) 使用更强的磁铁以增大B； (2) 提高转速f； (3) 增加匝数或线圈面积。电磁感应是大多数发电方式的原理。
10. Radioactive Decay and Half-Life | 放射性衰变与半衰期
A sample of iodine-131 has an initial activity of 800 Bq. Its half-life is 8 days. What will be the activity after 24 days? How long would it take for the activity to drop below 50 Bq?
一个碘-131样品的初始活度为800 Bq。它的半衰期为8天。24天后活度是多少？活度降到低于50 Bq需要多长时间？
After n half-lives, activity A = A₀ × (½)^n. In 24 days, number of half-lives n = 24 / 8 = 3. So A = 800 × (½)³ = 800 × 1/8 = 100 Bq.
经过n个半衰期后，活度 A = A₀ × (½)^n。24天内，半衰期个数 n = 24 / 8 = 3。所以 A = 800 × (½)³ = 800 × 1/8 = 100 Bq。
For A < 50 Bq, we need A₀/2^n < 50 → 800/2^n < 50 → 2^n > 800/50 = 16. The smallest integer n such that 2^n > 16 is n = 5 (2^4=16 is not strictly greater, but activity would be exactly 50 Bq; to drop below 50 we need n=5, activity = 800/32 = 25 Bq). Time = 5 × 8 = 40 days.
要求 A < 50 Bq，需要 800/2^n < 50 → 2^n > 800/50 = 16。满足 2^n > 16 的最小整数 n 是5（因为 2^4=16 并不严格大于，活度恰好为50 Bq；要低于50，需 n=5，活度 = 800/32 = 25 Bq）。时间 = 5 × 8 = 40天。
Understanding exponential decay and half-life calculations is fundamental for nuclear physics applications, including medical isotopes and carbon dating.
理解指数衰变和半衰期计算对于核物理应用（包括医用同位素和碳定年）至关重要。
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