📚 Year 11 Edexcel Physics: Case Study Practice | Edexcel 物理 Year 11：案例分析实战演练
This article presents a series of real-world case studies to help Year 11 students master Edexcel GCSE Physics. Each case integrates key concepts such as forces, energy, waves, electricity, and radioactivity, showing how theory is applied to solve practical problems. Work through each example to strengthen your analytical skills and exam technique.
本文通过一系列真实情境的案例分析，帮助Year 11学生掌握Edexcel GCSE物理核心内容。每个案例融合了力、能量、波、电和放射性等关键概念，展示如何将理论应用于解决实际问题。请逐一研读，提升你的分析能力和应试技巧。
1. Free Fall and Motion Graphs | 案例一：自由落体与运动图像
A steel ball is dropped from rest at the top of a cliff 45 m high. Ignore air resistance. Take g = 9.8 m/s².
一个钢球从45米高的悬崖顶部由静止释放。忽略空气阻力，取g = 9.8 m/s²。
We first find the time to hit the ground using s = ut + ½at². With u = 0, s = 45 m, a = 9.8 m/s², so 45 = ½ × 9.8 × t², giving t² = 9.18, t ≈ 3.03 s.
我们先用 s = ut + ½at² 求落地时间。初速 u = 0，位移 s = 45 m，加速度 a = 9.8 m/s²，因此 45 = ½ × 9.8 × t²，得 t² = 9.18，t ≈ 3.03 秒。
The final velocity is v = u + at = 0 + 9.8 × 3.03 ≈ 29.7 m/s. The velocity-time graph is a straight line starting at the origin with slope 9.8. The area under the graph equals the displacement.
末速度 v = u + at = 0 + 9.8 × 3.03 ≈ 29.7 m/s。速度–时间图像是一条过原点的直线，斜率为9.8。图像下的面积等于位移。
This case highlights uniform acceleration equations and graphical analysis of motion.
此案例突出了匀加速运动方程和运动的图像分析。
2. Resultant Force and Acceleration | 案例二：合力与加速度
A 1200 kg car accelerates from rest to 27 m/s in 9.0 s on a straight road. The driving force is 4800 N. Calculate the resistive force acting on the car.
一辆1200 kg的汽车在平直道路上从静止加速到27 m/s，用时9.0秒。驱动力为4800 N。求作用在汽车上的阻力。
Acceleration a = (v – u)/t = (27 – 0)/9.0 = 3.0 m/s². Using F = ma, the net force required is 1200 × 3.0 = 3600 N. The net force is the driving force minus resistive force. So 4800 – Fresist = 3600, giving resistive force = 1200 N.
加速度 a = (v – u)/t = (27 – 0)/9.0 = 3.0 m/s²。由 F = ma，所需合力为 1200 × 3.0 = 3600 N。合力等于驱动力减阻力，因此 4800 – F阻力 = 3600，得阻力为1200 N。
This demonstrates the link between resultant force, mass, and acceleration, and how to isolate unknown opposing forces.
本例展示了合力、质量和加速度的联系，以及如何隔离未知的反向力。
3. Energy Transfers in a Roller Coaster | 案例三：过山车的能量转换
A roller coaster car of mass 500 kg starts from rest at point A, 30 m above the ground. It descends to point B at ground level. Assume no energy losses. Find the speed at B, and explain the energy changes.
一辆质量500 kg的过山车在离地30 m的A点由静止出发，下滑至地面B点。假设无能量损失。求B点的速度，并解释能量变化。
At A, all energy is gravitational potential energy: Ep = mgh = 500 × 9.8 × 30 = 147 000 J. At B, all this has converted to kinetic energy: Ek = ½mv². So ½ × 500 × v² = 147 000, v² = 588, v ≈ 24.2 m/s.
在A点，所有能量为重力势能：Ep = mgh = 500 × 9.8 × 30 = 147 000 J。在B点，全部转化为动能：Ek = ½mv²。因此 ½ × 500 × v² = 147 000，v² = 588，v ≈ 24.2 m/s。
The energy converts from gravitational potential to kinetic. If friction were present, some would be transferred to thermal energy, reducing the final speed.
能量由重力势能转化为动能。若存在摩擦，一部分会转化为热能，使末速度减小。
4. Series and Parallel Circuits | 案例四：串联和并联电路
A circuit contains a 12 V battery and three resistors: R₁ = 4 Ω, R₂ = 6 Ω, R₃ = 12 Ω. First, all three are connected in series. Then, R₂ and R₃ are connected in parallel, and this combination is in series with R₁. For each arrangement, calculate the total resistance, the current from the battery, and the voltage across R₁.
电路包含一个12 V电池和三个电阻：R₁ = 4 Ω，R₂ = 6 Ω，R₃ = 12 Ω。首先，三个电阻串联；然后，将R₂和R₃并联，再与R₁串联。对于每种连接，计算总电阻、电池输出电流及R₁两端电压。
Series: Rtotal = 4 + 6 + 12 = 22 Ω. Current I = V/R = 12/22 ≈ 0.545 A. Voltage across R₁ = I × 4 = 2.18 V.
串联：总电阻 = 4 + 6 + 12 = 22 Ω。电流 I = V/R = 12/22 ≈ 0.545 A。R₁两端电压 = I × 4 = 2.18 V。
Parallel combination: 1/Rp = 1/6 + 1/12 = 2/12 + 1/12 = 3/12, so Rp = 4 Ω. Total resistance = 4 + 4 = 8 Ω. Current = 12/8 = 1.5 A. Voltage across R₁ = 1.5 × 4 = 6 V. The remaining 6 V is across the parallel pair.
并联组合：1/Rp = 1/6 + 1/12 = 2/12 + 1/12 = 3/12，得 Rp = 4 Ω。总电阻 = 4 + 4 = 8 Ω。电流 = 12/8 = 1.5 A。R₁电压 = 1.5 × 4 = 6 V。并联部分电压为6 V。
This case reinforces the rules for combining resistances and applying Ohm’s law to complex circuits.
此案例强化了电阻组合规则以及欧姆定律在复杂电路中的应用。
5. Wave Speed, Frequency, and Wavelength | 案例五：波速、频率和波长
A water wave travels at 2.4 m/s. Its wavelength is 0.80 m. Determine the frequency and period of the wave. If the wave enters a shallower region where the speed reduces to 1.8 m/s, what happens to the frequency and wavelength?
一个水波以2.4 m/s的速度传播，波长为0.80 m。求波的频率和周期。如果波进入较浅区域，波速降至1.8 m/s，频率和波长如何变化？
Using v = fλ, frequency f = v/λ = 2.4 / 0.80 = 3.0 Hz. Period T = 1/f = 1/3.0 ≈ 0.333 s. When the wave enters shallow water, frequency remains unchanged (determined by the source). Therefore new wavelength λ’ = v’/f = 1.8 / 3.0 = 0.60 m. The wavelength decreases.
由 v = fλ，频率 f = v/λ = 2.4 / 0.80 = 3.0 Hz。周期 T = 1/f = 1/3.0 ≈ 0.333 s。当波进入浅水区，频率不变（由波源决定），因此新波长 λ’ = v’/f = 1.8 / 3.0 = 0.60 m。波长变短。
This illustrates the wave equation and the fact that frequency remains constant when a wave crosses a boundary.
这说明了波动方程以及波越过边界时频率保持不变的性质。
6. Radioactive Decay and Half-life | 案例六：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. Plot a decay curve and determine the activity after 24 days, and the time taken for the activity to drop to 100 Bq.
一份碘-131样品初始活度为800 Bq，半衰期为8天。画出衰变曲线，求24天后的活度，以及活度降至100 Bq所需的时间。
After each half-life, the activity halves. After 8 days: 400 Bq; 16 days: 200 Bq; 24 days: 100 Bq. So after 24 days, activity is 100 Bq. To drop from 800 Bq to 100 Bq requires three half-lives, which is 24 days. The graph shows exponential decay.
每经过一个半衰期，活度减半。8天后：400 Bq；16天后：200 Bq；24天后：100 Bq。因此24天后活度为100 Bq。要从800 Bq降至100 Bq需3个半衰期，即24天。图像呈指数衰减。
Understanding half-life is crucial for nuclear physics, medical tracers, and radioactive dating.
理解半衰期对核物理、医学示踪和放射性测年至关重要。
7. Hooke’s Law and Elastic Potential Energy | 案例七：胡克定律与弹性势能
A spring extends by 0.15 m when a force of 12 N is applied. Calculate the spring constant and the elastic potential energy stored.
一根弹簧在被施加12 N的力时伸长0.15 m。计算弹簧常数和储存的弹性势能。
By Hooke’s law F = kx, so k = F/x = 12 / 0.15 = 80 N/m. The elastic potential energy E = ½kx² = ½ × 80 × (0.15)² = 0.9 J. If the extension doubled, the energy stored would be four times greater, because E ∝ x².
根据胡克定律 F = kx，得 k = F/x = 12 / 0.15 = 80 N/m。弹性势能 E = ½kx² = ½ × 80 × (0.15)² = 0.9 J。若伸长量加倍，储存的能量将变为四倍，因为 E ∝ x²。
This case connects force, extension, and energy storage, useful for designing suspension systems and elastic devices.
本例将力、伸长量和能量储存联系起来，在悬架系统和弹性装置设计中非常实用。
8. Momentum Conservation in Collisions | 案例八：碰撞中的动量守恒
A 0.50 kg trolley moving at 2.0 m/s collides with a stationary 1.0 kg trolley. They stick together. Find their common velocity after the collision and calculate the kinetic energy loss.
一辆0.50 kg的小车以2.0 m/s的速度运动，与一辆静止的1.0 kg小车碰撞后粘在一起。求它们碰撞后的共同速度，并计算动能损失。
By conservation of momentum: total momentum before = 0.50 × 2.0 + 1.0 × 0 = 1.0 kg m/s. After collision, combined mass = 1.50 kg. So 1.50 × v = 1.0, v ≈ 0.667 m/s. Kinetic energy before = ½ × 0.50 × (2.0)² = 1.0 J. After = ½ × 1.50 × (0.667)² ≈ 0.333 J. Loss of KE = 0.667 J, converted to heat/sound.
由动量守恒：碰前总动量 = 0.50 × 2.0 + 1.0 × 0 = 1.0 kg m/s。碰后总质量 = 1.50 kg。因此 1.50 × v = 1.0，v ≈ 0.667 m/s。碰前动能 = ½ × 0.50 × (2.0)² = 1.0 J。碰后动能 = ½ × 1.50 × (0.667)² ≈ 0.333 J。动能损失0.667 J，转化为内能和声能。
This demonstrates momentum is always conserved in collisions, while kinetic energy is only conserved in elastic collisions.
这表明动量在碰撞中总是守恒的，而动能仅在弹性碰撞中守恒。
9. Electromagnetic Induction and Generators | 案例九：电磁感应与发电机
A coil is rotated in a uniform magnetic field, inducing an alternating voltage. Explain the principle and state two ways to increase the induced voltage.
一个线圈在匀强磁场中旋转，产生交变电压。解释此原理，并说出两种增大感应电压的方法。
Faraday’s law states that induced e.m.f. is proportional to the rate of change of magnetic flux. When the coil rotates, the flux linking it changes continuously, generating an AC voltage. Peak voltage can be increased by: (1) using a stronger magnetic field, and (2) increasing the number of turns on the coil. Rotating faster also increases the rate of flux change.
法拉第定律指出，感应电动势与磁通量的变化率成正比。当线圈旋转时，交链的磁通量不断变化，产生交流电压。提高峰值电压的方法有：(1) 使用更强的磁场，(2) 增加线圈匝数。加快旋转速度也能增大磁通变化率。
This is the basis of alternators and many power generation systems.
这是交流发电机和许多发电系统的基础原理。
10. Pressure in Liquids and Hydraulic Systems | 案例十：液体压强与液压系统
A hydraulic jack has a small piston of area 2.0 cm² and a large piston of area 50 cm². A force of 40 N is applied to the small piston. Calculate the pressure transmitted and the force exerted by the large piston. State the assumption made.
一个液压千斤顶的小活塞面积为2.0 cm²，大活塞面积为50 cm²。对小活塞施加40 N的力。计算传递的压强以及大活塞输出的力。说明所作的假设。
Pressure P = F/A = 40 N / 2.0 cm² = 20 N/cm². Convert to pascals: 2.0 cm² = 2.0 × 10⁻⁴ m², so P = 40 / (2.0×10⁻⁴) = 200 000 Pa. By Pascal’s principle, this pressure acts equally on the large piston. Force on large piston F = P × A = 200 000 × (50×10⁻⁴) = 1000 N. We assume the liquid is incompressible and there is no friction.
压强 P = F/A = 40 N / 2.0 cm² = 20 N/cm²。换算成帕斯卡：2.0 cm² = 2.0 × 10⁻⁴ m²，故 P = 40 / (2.0×10⁻⁴) = 200 000 Pa。根据帕斯卡原理，该压强等值作用于大活塞。大活塞输出力 F = P × A = 200 000 × (50×10⁻⁴) = 1000 N。我们假定液体不可压缩且无摩擦。
The jack multiplies force, illustrating how pressure is transmitted through fluids.
千斤顶放大了力，体现了压强在流体中的传递方式。
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