📚 Case Analysis Practical Exercises | 案例分析实战演练
In Year 11 Cambridge IGCSE Physics, case analysis is vital for developing problem‑solving skills and connecting theoretical concepts to real‑world situations. The following exercises cover kinematics, forces, energy, electricity, waves, thermal physics, pressure, nuclear physics, moments, elasticity and circular motion. Each case is presented with a short scenario, followed by a step‑by‑step guided solution to help you practise applying equations logically and interpreting results.
在 Year 11 剑桥 IGCSE 物理中，案例分析对于培养解决问题的能力以及将理论概念与实际情况联系起来至关重要。以下练习涵盖运动学、力、能量、电学、波、热物理、压强、核物理、力矩、弹性和圆周运动。每个案例都给出了一个简短的情景，随后是分步指导解答，以帮助你练习有逻辑地应用方程和解释结果。
1. Kinematics: Stopping Distance | 运动学：刹车距离
A car travels at 20 m s⁻¹ along a straight road. The driver spots a hazard and applies the brakes, producing a constant deceleration of 5 m s⁻². Determine the distance travelled while the car comes to rest.
一辆汽车以 20 m s⁻¹ 的速度沿直线道路行驶。司机发现危险并刹车，产生恒定的减速度 5 m s⁻²。求汽车从刹车到静止所行驶的距离。
v² = u² + 2as. Here final velocity v = 0, initial velocity u = 20 m s⁻¹, acceleration a = –5 m s⁻². Substituting yields 0 = 20² + 2(–5)s, so s = 400 ÷ 10 = 40 m.
选择方程 v² = u² + 2as。这里末速度 v = 0，初速度 u = 20 m s⁻¹，加速度 a = –5 m s⁻²。代入得到 0 = 20² + 2(–5)s，因此 s = 400 ÷ 10 = 40 m。
Always note the sign of acceleration when a body slows down. The negative value reduces the term 2as, allowing s to be positive. A common mistake is forgetting the negative sign, which would incorrectly give a negative distance.
当物体减速时务必要注意加速度的符号。负号使 2as 项变小，从而让 s 为正值。常见的错误是忘记负号，这样会错误地得出负的距离。
2. Forces: Tension in a Crane Cable | 力：起重机缆绳张力
celeration of 1.2 m s⁻². Assume g = 9.8 m s⁻². Calculate the tension in the lifting cable.
一根 500 kg 的钢梁以 1.2 m s⁻² 的向上加速度被竖直吊起。取 g = 9.8 m s⁻²。计算吊索的张力。
Apply Newton’s second law to the beam: T – ion. Rearranging gives T = m(g + a). Substituting: T = 500(9.8 + 1.2) = 500 × 11 = 5500 N.
对钢梁应用牛顿第二定律：T – mg = ma，其中 T 是张力。整理得 T = m(g + a)。代入：T = 500(9.8 + 1.2) = 500 × 11 = 5500 N。
dditional force during acceleration is due to the need to overcome both weight and provide the net force for acceleration.
如果钢梁向下加速或匀速运动，张力会不同。匀速时 a = 0，因此 T = mg = 4900 N。加速时额外的力是因为既要克服重量，又要提供产生加速度的净力。
3. Energy: Roller Coaster Loop | 能量守恒：过山车环形
A roller coaster car of mass 600 kg starts from rest at height 45 m. It enters a vertical loop of radius 12 m. Neglecting friction, find the speed of the car at the top of the loop.
一辆质量为 600 kg 的过山车从 45 m 高的地方由静止出发，进入一个半径为 12 m 的竖直环。忽略摩擦，求过山车在环顶时的速度。
5 = 9.8×24 + ½v² → 441 = 235.2 + 0.5v² → v² = 2×(205.8) = 411.6 → v = √411.6 ≈ 20.3 m s⁻¹.
设初始高度 h₁ = 45 m，环顶高度 h₂ = 2×12 = 24 m。机械能守恒给出 mgh₁ = mgh₂ + ½mv²。消去 m：8×45 = 9.8×24 + ½v² → 441 = 235.2 + 0.5v² → v² = 2×205.8 = 411.6 → v = √411.6 ≈ 20.3 m s⁻¹。
At the top of the loop the car must also have a centripetal force requirement. If the speed were too low, the car would fall. Checking: required minimum speed vmin = √gr = √9.8×12 ≈ 10.8 m s⁻¹. Our calculated speed is well above this, so the car safely passes the loop.
在环顶，过山车还必须满足向心力要求。如果速度太低，车就会坠落。检查：所需的最低速度 vmin = √gr = √9.8×12 ≈ 10.8 m s⁻¹。我们算出的速度远高于此，因此过山车可以安全通过环顶。
4. Momentum: Two‑Car Collision | 动量守恒：两车碰撞
Car A (mass 1200 kg) moves at 15 m s⁻¹ and collides with stationary car B (mass 900 kg). The cars lock together on impact. Calculate the common velocity immediately after the collision.
A 车（质量 1200 kg）以 15 m s⁻¹ 行驶，与静止的 B 车（质量 900 kg）碰撞。碰撞后两车锁在一起。求碰撞后瞬间的共同速度。
By conservation of momentum: mA uA + mB uB = (mA + mB)v. Initially uB = 0. So 1200×15 = (1200+900)v → 18000 = 2100v → v = 18000 ÷ 2100 ≈ 8.57 m s⁻¹.
根据动量守恒：mA uA + mB uB = (mA + mB)v。初始 uB = 0。因此 1200×15 = (1200+900)v → 18000 = 2100v → v = 18000 ÷ 2100 ≈ 8.57 m s⁻¹。
½×2100×(8.57)² ≈ 77100 J. The lost energy is converted mainly to thermal and sound energy.
由于是非弹性碰撞，动能不守恒。计算初始动能 = ½×1200×15² = 135000 J。末动能 = ½×2100×(8.57)² ≈ 77100 J。损失的能量主要转化为热能和声能。
5. Waves: Two‑Source Interference in Ripple Tank | 波：水波槽双源干涉
Two coherent sources in a ripple tank oscillate in phase at frequency 15 Hz. Wave speed in water is 30 cm s⁻¹. At a point P, the path difference to the sources is 4.0 cm. State whether constructive or destructive interference occurs at P.
水波槽中两个相干振源以 15 Hz 的频率同相振动。水中的波速为 30 cm s⁻¹。在 P 点，到两个振源的波程差为 4.0 cm。说明 P 点发生的是相长干涉还是相消干涉。
ultiple of the wavelength, the waves arrive in phase and constructive interference occurs.
首先求波长 λ = v/f = 30 cm s⁻¹ ÷ 15 Hz = 2.0 cm。波程差 = 4.0 cm = 2λ。因为这是波长的整数倍，两列波同相到达，发生相长干涉。
If the path difference had been 1.5λ or any odd multiple of half‑wavelength, destructive interference would result. In the double‑slit experiment with light, the same principle governs bright and dark fringes.
如果波程差是 1.5λ 或任何半波长的奇数倍，就会发生相消干涉。在光的双缝实验中，同样的原理决定了明条纹和暗条纹。
6. Electricity: Thermistor Potential Divider | 电学：热敏电阻分压电路
A thermistor is connected in series with a fixed 2.0 kΩ resistor across a 6.0 V battery. At 25 °C the thermistor resistance is 4.0 kΩ. Calculate the output voltage across the fixed resistor at this temperature.
一个热敏电阻与一个 2.0 kΩ 的固定电阻串联，接在 6.0 V 的电池上。在 25 °C 时热敏电阻为 4.0 kΩ。计算此温度下固定电阻两端的输出电压。
Use the potential divider formula: Vout = Vsupply × Rfixed / (Rtherm + Rfixed). Vout = 6.0 × 2000 / (4000+2000) = 6.0 × (2000/6000) = 6.0 × ⅓ = 2.0 V.
使用分压公式：Vout = Vsupply × Rfixed / (Rtherm + Rfixed)。Vout = 6.0 × 2000 / (4000+2000) = 6.0 × (2000/6000) = 6.0 × ⅓ = 2.0 V。
As temperature increases, the thermistor’s resistance decreases (assuming an NTC thermistor), so more voltage drops across the fixed resistor. This circuit is commonly used in temperature sensors and fire alarms.
随着温度升高，热敏电阻的阻值减小（假设是负温度系数热敏电阻），因此固定电阻上的压降增大。这种电路常用于温度传感器和火灾报警器。
7. Thermal Physics: Measuring Specific Heat Capacity of a Metal | 热物理：测量金属的比热容
A 0.500 kg metal block is heated with an electric heater delivering 50 W for 600 s. The temperature rises from 20.0 °C to 45.0 °C. Assuming no heat losses, calculate the specific heat capacity of the metal.
一个 0.500 kg 的金属块用 50 W 的电热器加热 600 s，温度从 20.0 °C 升至 45.0 °C。假设没有热损失，计算该金属的比热容。
Using Q = mcΔθ, c = Q / (mΔθ) = 30000 / (0.500 × 25.0) = 30000 / 12.5 = 2400 J kg⁻¹ °C⁻¹.
提供的能量 Q = Pt = 50 × 600 = 30 000 J。利用 Q = mcΔθ，c = Q / (mΔθ) = 30000 / (0.500 × 25.0) = 30000 / 12.5 = 2400 J kg⁻¹ °C⁻¹。
In real experiments, heat losses to the surroundings are inevitable. Using insulation and a lid can reduce these losses. A more accurate method is the method of mixtures, where thermal equilibrium is achieved.
在真实实验中，向环境的热损失不可避免。使用隔热材料和盖子可减少损失。一种更准确的方法是混合法，通过达到热平衡来测量。
8. Pressure: Hydraulic Car Jack | 压强：液压千斤顶
A hydraulic jack has a small piston of radius 2.0 cm and a large piston of radius 12.0 cm. An effort force of 80 N is applied on the small piston. Find the maximum load the large piston can lift.
一台液压千斤顶的小活塞半径为 2.0 cm，大活塞半径为 12.0 cm。在小活塞上施加 80 N 的作用力。求大活塞能举起的最大负载。
Pressure is transmitted equally: P = Fsmall / Asmall = Flarge / Alarge. A = πr². Asmall = π(0.020)², Alarge = π(0.12)². Ratio of areas = (0.12/0.020)² = 36. So Flarge = 80 × 36 = 2880 N. This corresponds to a mass of about 2880/9.8 ≈ 294 kg.
压强等值传递：P = Fsmall / Asmall = Flarge / Alarge。A = πr²。Asmall = π(0.020)²，Alarge = π(0.12)²。面积比 = (0.12/0.020)² = 36。因此 Flarge = 80 × 36 = 2880 N。这对应的质量约为 2880/9.8 ≈ 294 kg。
The hydraulic jack is a classic application of Pascal’s principle. The small force moves a larger distance while the large force moves a smaller distance, conserving work (neglecting friction).
液压千斤顶是帕斯卡原理的经典应用。小力移动较大的距离，而大力移动较小的距离，从而功守恒（忽略摩擦）。
9. Nuclear Physics: Carbon‑14 Dating | 核物理：碳‑14 测年
An archaeologist finds a wooden artefact. Measurements show its carbon‑14 activity is 3.2 counts per minute, while a living sample of the same mass gives 12.8 counts per minute. The half‑life of carbon‑14 is 5730 years. Calculate the age of the artefact.
一位考古学家发现一件木质文物。测量显示其碳‑14 的活度为每分钟 3.2 个计数，而相同质量的现代样品活度为每分钟 12.8 个计数。碳‑14 的半衰期为 5730 年。计算该文物的年代。
Activity ratio = 3.2 / 12.8 = ¼. This means two half‑lives have passed because (½)² = ¼. Therefore age = 2 × 5730 = 11460 years.
活度比 = 3.2 / 12.8 = ¼。这表明已经过了两个半衰期，因为 (½)² = ¼。因此年代 = 2 × 5730 = 11460 年。
For non‑ be used with decay constant λ = ln2 / T₁,₂. Here the simple ratio avoids using logarithms.
对于非整数倍的情况，可使用指数衰变方程 N = N₀ e⁻ᵏᵗ，其中衰变常数 λ = ln2 / T₁,₂。这里简单的比值避免了使用对数。
10. Moments: Balancing a See‑Saw | 力矩：跷跷板平衡
A see‑saw is 4.0 m long and pivoted at its centre. A child of weight 350 N sits 1.6 m from the pivot on one side. Where must a second child of weight 280 N sit on the opposite side to balance the see‑saw?
一个跷跷板长 4.0 m，在中心处支起。一个体重 350 N 的小孩坐在距支点 1.6 m 的一侧。体重 280 N 的第二个小孩应坐在另一侧什么位置才能使跷跷板平衡？
nts. 350 N × 1.6 m = 280 N × d → d = (350 × 1.6) / 280 = 560 / 280 = 2.0 m from the pivot.
为达到平衡，顺时针力矩之和等于逆时针力矩之和。350 N × 1.6 m = 280 N × d → d = (350 × 1.6) / 280 = 560 / 280 = 2.0 m。
If the pivot is not at the centre, the weight of the see‑saw itself must be considered. The principle of moments is a direct consequence of rotational equilibrium.
如果支点不在中心，跷跷板自身的重量就必须加以考虑。力矩原理是转动平衡的直接结果。
11. Hooke’s Law: Spring Extension and Energy | 胡克定律：弹簧伸长与能量
m⁻¹ is stretched by 0.050 m beyond its natural length. Calculate the elastic potential energy stored in the spring.
一根弹簧的劲度系数 k = 400 N m⁻¹，被拉伸到比原长长 0.050 m 的长度。计算储存在弹簧中的弹性势能。
E = ½kx² = ½ × 400 × (0.050)² = 200 × 0.0025 = 0.50 J.
E = ½kx² = ½ × 400 × (0.050)² = 200 × 0.0025 = 0.50 J。
Hooke's law applies only up to the limit of proportionality. If the spring is stretched beyond its elastic limit, it will be permanently deformed and the stored energy calculated by this formula will not be accurate.
胡克定律只在比例极限内适用。如果弹簧被拉伸超过弹性极限，它将发生永久变形，用该公式计算的储存能量将不准确。
12. Circular Motion: Car on a Flat Curve | 圆周运动：汽车过水平弯道
A 1100 kg car rounds a flat curve of radius 55 m at a speed of 18 m s⁻¹. The coefficient of static friction between the tyres and the road is 0.60. Determine whether the car can safely negotiate the curve without skidding.
一辆 1100 kg 的汽车以 18 m s⁻¹ 的速度驶过一个半径为 55 m 的水平弯道。轮胎与路面间的静摩擦系数为 0.60。判断汽车是否能安全过弯而不打滑。
Required centripetal force Fc = mv²/r = 1100 × 18² / 55 = 1100 × 324 / 55 = 6480 N. Maximum static friction Fmax 6468 N. Since the required force (6480 N) slightly exceeds the available friction, the car is likely to skid.
所需的向心力 Fc = mv²/r = 1100 × 18² / 55 = 1100 × 324 / 55 = 6480 N。最大静摩擦力 Fmax = μR = 0.60 × (1100 × 9.8) = 0.60 × 10780 = 6468 N。由于所需的力（6480 N）略大于可提供的摩擦力，汽车很可能打滑。
To improve safety, the driver should reduce speed or the road should be banked. Banking provides a component of the normal force to contribute to the centripetal force, reducing reliance on friction.
为了提高安全性，司机应降低速度，或者道路应设置倾斜弯道。倾斜路面可以使法向力的一个分力提供向心力，从而减少对摩擦的依赖。
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