📚 Case Study Practical Drills for Year 10 Eduqas Physics | Year 10 Eduqas 物理：案例分析实战演练
Mastering physics is not just about memorising formulas; it is about learning to think like a scientist. In the Year 10 Eduqas Physics course, you are expected to apply your knowledge to real-world situations, analyse data, and explain phenomena using scientific principles. This article presents a series of case study drills that will help you sharpen these skills. Each section focuses on a key topic, provides a realistic scenario, and walks you through the step-by-step reasoning you need to succeed in your exams.
掌握物理不仅仅是记住公式，而是要学会像科学家一样思考。在 Year 10 Eduqas 物理课程中，你需要将知识应用于现实世界的情境，分析数据，并用科学原理解释现象。本文提供了一系列案例分析训练，帮助你打磨这些技能。每个小节聚焦一个关键主题，给出一个贴近实际的情境，并带你一步步走过考试成功所需的推理过程。
1. Analysing Speed and Acceleration | 分析速度与加速度
A police officer uses a speed gun to track a car travelling along a straight road. The car passes point A at 12 m/s and reaches point B, 200 m away, 10 seconds later. We need to determine the car's acceleration, assuming it is uniform, and its speed at point B.
一名警察使用测速枪追踪一辆在笔直道路上行驶的汽车。汽车通过 A 点时速度为 12 m/s，10 秒后到达 200 米外的 B 点。我们需要求出汽车的加速度（假设为匀加速）以及它在 B 点的速度。
the SUVAT equations. We can use s = ut + ½at² to find a. Substituting: 200 = 12×10 + ½×a×10², giving 200 = 120 + 50a. So 50a = 80, a = 1.6 m/s². Then use v = u + at: v = 12 + 1.6×10 = 28 m/s. This example shows how to select the correct equation based on given data.
首先列出已知量：初速度 u = 12 m/s，位移 s = 200 m，时间 t = 10 s。需要求出加速度 a 和末速度 v。回想 SUVAT 方程。我们可以用 s = ut + ½at² 求 a。代入：200 = 12×10 + ½×a×10²，得出 200 = 120 + 50a。所以 50a = 80，a = 1.6 m/s²。然后用 v = u + at：v = 12 + 1.6×10 = 28 m/s。这个例子展示了如何根据已知数据选择正确的方程。
2. Newton's Second Law in Action | 牛顿第二定律的应用
Consider a delivery van of mass 2500 kg accelerating at 0.8 m/s² when the driving force from the engine is 3200 N. What is the total resistive force acting on the van? This case tests your understanding of resultant force.
考虑一辆质量为 2500 kg 的厢式货车，发动机驱动力为 3200 N，加速度为 0.8 m/s²。作用在货车上的总阻力是多少？这个案例考查你对合力的理解。
rce − resistive force. Rearranging: resistive force = driving force − resultant force = 3200 − 2000 = 1200 N. This demonstrates how to apply Newton' when multiple forces are present. Always draw a free-body diagram to keep direction clear.
用 F = ma，合力 = 2500 × 0.8 = 2000 N。驱动力与阻力方向相反。所以，合力 = 驱动力 − 阻力。移项得：阻力 = 驱动力 − 合力 = 3200 − 2000 = 1200 N。这展示了在存在多个力时如何应用牛顿第二定律。务必画出受力示意图以明辨方向。
3. Energy Transfers in a Roller Coaster | 过山车中的能量转移
A roller coaster car of mass 500 kg starts from rest at the top of a hill 40 m high. Assuming no energy losses, calculate its speed at the bottom. Also, explain the energy changes if friction is considered.
一辆质量为 500 kg 的过山车从 40 米高的山顶静止出发。假设无能量损失，计算它在山底的速度。若考虑摩擦力，解释能量如何变化。
ssuming no losses. So ½mv² = 196 000. Solving: v² = (2 × 196 000) / 500 = 784, v = √784 = 28 m/s. With friction, some GPE would be dissipated as thermal energy in the wheels and track, so KE and speed would be lower. This case reinforces energy conservation and dissipation.
山顶的重力势能 (GPE) = mgh = 500 × 9.8 × 40 = 196 000 J。在山底，若无损失，所有 GPE 都转化为动能 (KE)。所以 ½mv² = 196 000。求解：v² = (2 × 196 000) / 500 = 784，v = √784 = 28 m/s。有摩擦时，部分 GPE 会以热能的形式散逸到车轮和轨道上，因此 KE 和速度都会变小。这个案例强化了能量守恒和耗散的概念。
4. Circuit Analysis: Current, Voltage and Resistance | 电路分析：电流、电压与电阻
A student builds a circuit with a 12 V battery and two resistors in series: 4 Ω and 8 Ω. Calculate the total resistance, the current in the circuit, and the voltage across each resistor. Then, the resistors are rearranged in parallel—find the total resistance and current from the battery.
一个学生用 12 V 电池和两个电阻（4 Ω 和 8 Ω）串联构建电路。计算总电阻、电路中的电流以及每个电阻两端的电压。然后将电阻改为并联——求总电阻和电池输出的电流。
For series: Rtotal = R₁ + R₂ = 4 + 8 = 12 Ω. Current I = V / Rtotal R₁ = 1 × 4 = 4 V; across 8 Ω: V₂ = 1 × 8 = 8 V. Notice they add to 12 V. For parallel: 1/Rtotal = 1/4 + 1/8 = (2 + 1)/8 = 3/8, so Rtotal × 3/8 = 4.5 A. This drill reinforces the differences between series and parallel circuits.
串联时：总电阻 Rtotal = R₁ + R₂ = 4 + 8 = 12 Ω。电流 I = V / Rtotal = 12 / 12 = 1 A。4 Ω 电阻两端电压：V₁ = I × R₁ = 1 × 4 = 4 V；8 Ω 电阻两端：V₂ = 1 × 8 = 8 V。注意它们相加为 12 V。并联时：1/Rtotal = 1/4 + 1/8 = (2+1)/8 = 3/8，故 Rtotal = 8/3 ≈ 2.67 Ω。电池输出电流：I = 12 / (8/3) = 12 × 3/8 = 4.5 A。这个训练巩固了串联与并联电路的区别。
5. Wave Speed, Frequency and Wavelength | 波速、频率与波长
A sonar system on a ship emits a sound pulse of frequency 50 kHz towards the seabed. The pulse returns after 0.8 seconds. Given the speed of sound in seawater is 1500 m/s, calculate the depth of the sea and the wavelength of the sound wave.
船上的声纳系统向海床发射频率为 50 kHz 的声脉冲。脉冲在 0.8 秒后返回。已知声音在海水中的速度为 1500 m/s，计算海的深度以及声波的波长。
wavelength, use v = fλ. λ = v / f = 1500 / 50 000 = 0.03 m (3 cm). This case illustrates a common pitfall— time covers the distance twice. It also links wave properties with practical application.
深度可由 距离 = 速度 × 时间 求得，但要记住时间是往返的。单程时间 = 0.8 / 2 = 0.4 s。深度 = 1500 × 0.4 = 600 m。求波长时使用 v = fλ。λ = v / f = 1500 / 50 000 = 0.03 m (3 cm)。这个案例展示了一个常见陷阱——忘记回声时间覆盖了双倍距离。同时将波的性质与实际应用联系起来。
6. 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. Plot a decay curve and determine the activity after 24 days. Explain why this isotope is suitable for medical tracing but must be handled carefully.
一块碘-131 样品的初始活度为 800 Bq，半衰期为 8 天。画出衰变曲线，并确定 24 天后的活度。解释为什么这种同位素适合医用示踪但必须小心处理。
Each half-life, activity halves. After 8 days: 400 Bq; after 16 days: 200 Bq; after 24 days: 100 Bq. The decay curve is exponential. Iodine-131 is suitable as a tracer because it emits detectable radiation and has a short enough half-life to minimise patient exposure, but it must be handled with care because its beta and gamma emissions can damage tissue if dose is too high. This links to risk–benefit analysis in physics.
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。衰变曲线呈指数形态。碘-131 适合作为示踪剂，因为它放出可探测的辐射，且半衰期足够短，能尽量降低对病人的照射。但它必须小心处理，因为其贝塔和伽马辐射在剂量过大时会损伤组织。这联系到了物理学中的风险-收益分析。
7. Resolving Forces on a Slope | 斜面上力的分解
A crate of weight 600 N rests on a frictionless ramp inclined at 30° to the horizontal. Find the component of the weight parallel to the slope that pulls it downwards, and the perpendicular component pressing into the surface. Then determine the tension in a rope holding the crate stationary.
一个重 600 N 的木箱静止在倾角 30° 的光滑斜面上。求重力沿斜面方向向下的分力，以及垂直于斜面压入表面的分力。然后计算保持木箱静止的绳子中的张力。
600 × cos30° = 600 × 0.866 = 520 N (approx). Since the rope holds the crate stationary, tension T must balance the parallel component, so T = 300 N. This case develops skills in resolving forces, essential for understanding motion on inclines.
画出垂直向下的重力矢量。沿斜面方向的分力 = W sinθ = 600 × sin30° = 600 × 0.5 = 300 N。垂直于斜面的分力 = W cosθ = 600 × cos30° = 600 × 0.866 ≈ 520 N。由于绳子拉住木箱使其静止，张力 T 必须平衡沿斜面的分力，因此 T = 300 N。这个案例培养了分解力的技能，对于理解斜面上的运动至关重要。
8. Pressure and Hydraulic Systems | 压强与液压系统
A hydraulic car lift has a small piston of area 0.02 m² and a large piston of area 1.2 m². A force of 100 N is applied to the small piston. Calculate the pressure transmitted and the force exerted by the large piston. Explain why this system acts as a force multiplier.
一个液压汽车升降机的小活塞面积为 0.02 m²，大活塞面积为 1.2 m²。对小活塞施加 100 N 的力。计算传递的压强以及大活塞施加的力。解释为什么这个系统可以放大力量。
e force is multiplied by a factor of 60 because the area ratio is 1.2 / 0.02 = 60. This illustrates Pascal' and the trade-off: the small piston must move a much greater distance than the large one.
压强 P = F / A = 100 / 0.02 = 5000 Pa。这个压强在液体中处处等大地传递。大活塞上的力 F₂ = P × A₂ = 5000 × 1.2 = 6000 N。力被放大了 60 倍，因为面积比为 1.2 / 0.02 = 60。这说明了帕斯卡原理以及一定的代价：小活塞移动的距离必须比大活塞大得多。
9. Specific Heat Capacity and Thermal Energy | 比热容与热能
g of water at 20°C. It is left on for 3 minutes with a heater power of 2000 W. Assuming 80% the water, calculate the final temperature. Specific heat capacity of water = 4200 J/kg°C.
一个电水壶装有 1.5 kg 温度为 20°C 的水。加热器功率为 2000 W，加热 3 分钟。假设 80% 的能量用于加热水，计算最终水温。水的比热容为 4200 J/kg°C。
8 000 = 1.5 × 4200 × Δθ. Solving: Δθ = 288 000 / (1.5 × 4200) = 288 000 / 6300 = 45.7°C (approx). Final temperature = 20 + 45.7 = 65.7°C. This exercise combines energy, power, efficiency, and heat transfer.
加热器提供的能量 = 功率 × 时间 = 2000 × (3 × 60) = 2000 × 180 = 360 000 J。水的有效能量 = 0.80 × 360 000 = 288 000 J。由 ΔE = mcΔθ，得 288 000 = 1.5 × 4200 × Δθ。求解：Δθ = 288 000 / (1.5 × 4200) = 288 000 / 6300 ≈ 45.7°C。最终水温 = 20 + 45.7 = 65.7°C。这个练习将能量、功率、效率和热传递结合在一起。
10. Investigating Springs and Elastic Limit | 探究弹簧与弹性极限
N, length = 13.6 cm; with 4 N, length = 15.0 cm; with 5 N, length = 17.0 cm. Determine the spring constant and identify if the elastic limit has been exceeded.
一名学生在弹簧上悬挂砝码并记录其长度。空载长度为 10.0 cm。数据：1 N 时长度 = 11.2 cm；2 N 时长度 = 12.4 cm；3 N 时长度 = 13.6 cm；4 N 时长度 = 15.0 cm；5 N 时长度 = 17.0 cm。确定弹簧的劲度系数并判断是否超过了弹性极限。
, 5.0 cm, 7.0 cm. Notice up to 3 N, extension increases by 1.2 cm per N, giving a constant spring constant. Using Hooke' d 3 N, extension per N is larger and non-linear, indicating the elastic limit is exceeded around 3–4 N. This case teaches data analysis and understanding of proportionality.
伸长量 = 长度 − 原长。计算得：1.2 cm, 2.4 cm, 3.6 cm, 5.0 cm, 7.0 cm。注意在 3 N 以内，伸长量每牛顿增加 1.2 cm，弹簧劲度系数恒定。由胡克定律 F = kx，k = F / x。用 1 N 时，x = 0.012 m，k = 1 / 0.012 ≈ 83.3 N/m。超过 3 N 后，每牛顿的伸长量变大且不呈线性，表明大约在 3–4 N 时超过了弹性极限。这个案例教授数据分析和理解比例关系。
Published by TutorHao | Physics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图