Year 9 AQA Physics: Case Study Practical Drills | AQA Year 9 物理：案例分析实战演练

📚 Year 9 AQA Physics: Case Study Practical Drills | AQA Year 9 物理：案例分析实战演练

Case study questions are an excellent way to sharpen your problem‑solving skills in AQA Year 9 Physics. By working through real‑world scenarios step by step, you learn to identify the relevant formulas, substitute values correctly, and interpret your results. This article presents eight carefully chosen case studies covering electricity, forces, energy, pressure, motion, density, work & power, and moments. Each case is unpacked with full working, helping you master the mathematical and conceptual demands of the course.

案例分析是磨练 AQA 物理 Year 9 解题能力的绝佳方式。通过逐步完成真实情境问题，你将学会识别相关公式、正确代入数值并解释结果。本文精选了八个案例，涵盖电路、力、能量、压强、运动、密度、功与功率以及力距，每个案例都附有完整推导过程，帮助你掌握课程所需的数学与概念要求。

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1. Series Circuit Analysis | 串联电路分析

t’s find the total resistance, current, and the potential difference across each resistor.

一个学生用一节 12 V 电池和三个电阻 R₁ = 2 Ω, R₂ = 4 Ω, R₃ = 6 Ω 搭建了一个串联电路。我们来求总电阻、电流以及每个电阻两端的电压。

Step 1: Total resistance. In series, resistances add up directly. So Rₜₒₜₐₗ = R₁ + R₂ + R₃ = 2 + 4 + 6 = 12 Ω.

步骤 1：总电阻。 在串联中，电阻直接相加。因此 Rₜₒₜₐₗ = R₁ + R₂ + R₃ = 2 + 4 + 6 = 12 Ω。

Step 2: Current in the circuit. g Ohm’s law V = IR, rearrange to I = V / R. Here V = 12 V, R = 12 Ω, so I = 12 / 12 = 1 A.

步骤 2：电路中的电流。 使用欧姆定律 V = IR，变形为 I = V / R。这里 V = 12 V，R = 12 Ω，因此 I = 12 / 12 = 1 A。

Step 3: Individual voltages. = 2 V; V₂ = 1 × 4 = 4 V; V₃ = 1 × 6 = 6 V. Check: 2 + 4 + 6 = 12 V, matching the battery.

步骤 3：各电阻电压。 V₁ = I × R₁ = 1 × 2 = 2 V；V₂ = 1 × 4 = 4 V；V₃ = 1 × 6 = 6 V。验证：2 + 4 + 6 = 12 V，等于电池电压。

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2. Parallel Circuit Branch Currents | 并联电路分支电流

Now the same two resistors, 3 Ω and 6 Ω, are connected in parallel across a 12 V supply. Determine the total resistance, main‑line current, and the current in each branch.

现在将同样的两个电阻 3 Ω 和 6 Ω 并联，接在 12 V 电源上。求总电阻、干路电流以及每条支路的电流。

Step 1: Total resistance. = 6/3 = 2 Ω. A quicker check: product/sum = (3×6)/(3+6)=18/9=2 Ω, valid for two resistors.

步骤 1：总电阻。 并联时，1/Rₜₒₜₐₗ = 1/R₁ + 1/R₂。因此 1/Rₜₒₜₐₗ = 1/3 + 1/6 = 2/6 + 1/6 = 3/6。所以 Rₜₒₜₐₗ = 6/3 = 2 Ω。快速验证：积/和 = (3×6)/(3+6)=18/9=2 Ω，适用于两个电阻。

Step 2: Main current. I_total = V / R_total = 12 V / 2 Ω = 6 A.

步骤 2：干路电流。 Iₜₒₜₐₗ = V / Rₜₒₜₐₗ = 12 V / 2 Ω = 6 A。

Step 3: Branch currents. 12 V. I₁ = 12 / 3 = 4 A; I₂ = 12 / 6 = 2 A. These add to 6 A, confirming the calculation.

步骤 3：支路电流。 每条支路电压都是 12 V。I₁ = 12 / 3 = 4 A；I₂ = 12 / 6 = 2 A。两者之和为 6 A，验证无误。

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3. Energy Conversion in Free Fall | 自由落体中的能量转换

Ignore air resistance. Calculate the speed at which it hits the ground. Take g = 10 N/kg.

一块 2 kg 的砖块从 5 m 高处自由落下，忽略空气阻力。计算它撞击地面时的速率。取 g = 10 N/kg。

Step 1: Gravitational potential energy lost. 0 × 5 = 100 J. This energy is completely converted into kinetic energy just before impact.

步骤 1：失去的重力势能。 GPE = mgh = 2 × 10 × 5 = 100 J。这些能量在落地前完全转化为动能。

mgh = ½mv²

Step 2: Cancel mass and solve for v. v = √200 ≈ 14.1 m/s. (Using g=9.8 gives v ≈ 9.9 m/s, but the principle remains the same.)

步骤 2：消去质量求 v。 2 × 10 × 5 = ½ × v² → 100 = 0.5 v² → v² = 200，因此 v = √200 ≈ 14.1 m/s。（用 g=9.8 时 v≈9.9 m/s，原理一致。）

Step 3: Interpretation. Notice the mass cancels out; in the absence of air resistance, all objects fall at the same rate regardless of mass. The speed depends only on the drop height.

步骤 3：解释。 注意质量被消去了；在没有空气阻力时，所有物体无论质量大小下落速率相同。速率只取决于下落高度。

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4. Pressure Under Foot – Elephant vs Human | 足下压强——大象与人类

ea 0.1 m²) with that of a 70 kg person standing on one foot (area 0.02 m²). Use g = 10 N/kg.

压强表示力在某一面积上的集中程度。比较一头质量为 5000 kg 的大象（四足站立，每只脚掌面积 0.1 m²）与一个 70 kg 的人单脚站立（面积 0.02 m²）所产生的压强。取 g = 10 N/kg。

Step 1: Elephants force and total area. N. Total contact area = 4 × 0.1 = 0.4 m². Pressure P = F / A = 50 000 / 0.4 = 125 000 Pa.

步骤 1：大象重力和总面积。 重力 = mg = 5000 × 10 = 50 000 N。总接触面积 = 4 × 0.1 = 0.4 m²。压强 P = F / A = 50 000 / 0.4 = 125 000 Pa。

Step 2: Human when on one foot. ephant is much heavier, its pressure is only about 3.6 times greater because the force spreads over a much larger area. Standing on one foot, the person experiences a remarkably high pressure under that small sole.

步骤 2：人类单脚站立时。 重力 = 70 × 10 = 700 N。面积 = 0.02 m²。P = 700 / 0.02 = 35 000 Pa。尽管大象重得多，但由于力分布在更大的面积上，其压强仅为人类的约 3.6 倍。人单脚站立时，脚掌小的面积下产生了相当高的压强。

Case	Force (N)	Area (m²)	Pressure (Pa)
Elephant (4 feet)	50 000	0.4	125 000
Human (1 foot)	700	0.02	35 000

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5. Speed, Distance and Time – A Car Journey | 速率、距离与时间——一段汽车行程

A car travels at a constant speed of 15 m/s for 5 minutes. How far does it go? This seems straightforward, but converting units correctly is a key exam skill.

一辆汽车以 15 m/s 的恒定速率行驶了 5 分钟。它走了多远？这看似简单，但正确转换单位是关键的考试技能。

Step 1: Convert time to seconds. 5 minutes = 5 × 60 = 300 s.

步骤 1：将时间转换为秒。 5 分钟 = 5 × 60 = 300 s。

Step 2: Apply s = vt. 15 m/s × 300 s = 4500 m, or 4.5 km. Always check that units cancel: m/s × s gives metres.

步骤 2：应用 s = vt。 s = 15 m/s × 300 s = 4500 m，即 4.5 km。务必检查单位抵消：m/s × s 得到米。

Step 3: Reverse thinking. ce and speed, you can find time by t = s/v. For example, how long to cover 900 m at 15 m/s? t = 900 / 15 = 60 s.

步骤 3：逆向思考。 如果已知距离和速率，可用 t = s/v 求时间。例如，以 15 m/s 行驶 900 m 需要多久？t = 900 / 15 = 60 s。

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6. Density Determination – Identify the Metal | 测定密度——辨别金属

ngular metal block has a mass of 540 g and dimensions 5 cm × 6 cm × 10 cm. Is it aluminium? The density of aluminium is about 2.7 g/cm³.

一块长方体金属质量为 540 g，尺寸为 5 cm × 6 cm × 10 cm。它是铝吗？铝的密度约为 2.7 g/cm³。

Step 1: Volume of the block. = length × width × height = 5 × 6 × 10 = 300 cm³. Always write the unit after the number.

步骤 1：金属块体积。 V = 长 × 宽 × 高 = 5 × 6 × 10 = 300 cm³。始终记得写出单位。

Step 2: Density. ρ (density) = mass / volume = 540 g / 300 cm³ = 1.8 g/cm³.

步骤 2：密度。 ρ = 质量 / 体积 = 540 g / 300 cm³ = 1.8 g/cm³。

Step 3: Conclusion. 1.8 g/cm³ is far lower than 2.7 g/cm³, so the block is not pure aluminium. It could be an alloy or a different material altogether, such as magnesium (density around 1.74 g/cm³). This shows how density can help identify substances.

步骤 3：结论。 1.8 g/cm³ 远低于 2.7 g/cm³，因此这块金属不是纯铝。它可能是一种合金或其他材料，例如镁（密度约 1.74 g/cm³）。这表明密度可用来鉴别物质。

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7. Work Done and Power of a Crane | 起重机的做功与功率

10 m in 20 s. Calculate the work done by the crane and its power output. Take g = 10 N/kg.

一台起重机在 20 s 内将一根 500 kg 的钢梁垂直吊起 10 m。计算起重机所做的功及其输出功率。取 g = 10 N/kg。

Step 1: Work done. t the beam equals its weight: F = mg = 500 × 10 = 5000 N. So W = 5000 N × 10 m = 50 000 J.

步骤 1：做功。 功 = 力 × 距离。吊起钢梁所需的力等于其重力：F = mg = 500 × 10 = 5000 N。因此 W = 5000 N × 10 m = 50 000 J。

Step 2: Power. 500 W (or 2.5 kW). The same work done in half the time would double the power requirement.

步骤 2：功率。 功率是做功的速率：P = W / t = 50 000 J / 20 s = 2500 W（即 2.5 kW）。若时间减半做同样的功，功率需求将翻倍。

Alternative approach: me = 10/20 = 0.5 m/s, so P = 5000 N × 0.5 m/s = 2500 W. Both methods give the same result.

替代方法： 有学生喜欢用 P = Fv。平均速率 v = 距离/时间 = 10/20 = 0.5 m/s，因此 P = 5000 N × 0.5 m/s = 2500 W。两种方法结果一致。

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8. Lever Moments – Lifting a Heavy Load | 杠杆力距——撬起重物

A crowbar is used to lift an 800 N boulder. The fulcrum is placed 0.3 m from the load end, and the total bar length from fulcrum to effort end is 1.5 m. Find the minimum effort force needed.

用一根撬棍抬升一块 800 N 的巨石。支点距重物端 0.3 m，支点到施力端的撬棍总长为 1.5 m。求所需的最小施力。

Step 1: Identify arms. Load arm d_load = 0.3 m. Effort arm d_effort arm may be the full length from fulcrum to handle if the force is applied at the very end.

步骤 1：确定力臂。 负载力臂 d_load = 0.3 m。施力力臂 d_effort = 1.5 m。注意，如果力施加在手柄最末端，施力力臂可取支点至手柄的全长。

Step 2: Principle of moments. ibrium, clockwise moment = anticlockwise moment. Taking the fulcrum as the pivot: Load × d_load = Effort × d_effort. So 800 × 0.3 = Effort × 1.5.

步骤 2：力距原理。 要平衡，顺时针力距 = 逆时针力距。以支点为支点：负载 × 负载力臂 = 施力 × 施力力臂。因此 800 × 0.3 = 施力 × 1.5。

Step 3: Solve for effort. Effort = (800 × 0.3) / 1.5 = 240 / 1.5 = 160 N. The effort is only one‑ imes longer than the load arm. This illustrates how a lever provides mechanical advantage.

步骤 3：求施力。 施力 = (800 × 0.3) / 1.5 = 240 / 1.5 = 160 N。施力仅为负载重量的五分之一，因为施力力臂是负载力臂的五倍。这说明杠杆如何提供机械优势。

Checking units: The moment is measured in newton‑metres (Nm) here, but the calculation is consistent because we compare forces directly using the ratio of arm lengths.

检查单位： 此处力距单位为牛·米 (Nm)，但利用力臂长度比直接比较力时计算是自洽的。

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Published by TutorHao | Physics Revision Series | aleveler.com

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