📚 Year 12 SQA Physics Case Study Drills | 12年级 SQA 物理案例分析实战演练
Case study questions are a significant part of the SQA Higher Physics assessment. They test your ability to apply physics principles to unfamiliar contexts, analyse real experimental data, and evaluate procedures critically. This article offers a practical walk-through of typical case studies, step‑by‑step analytical techniques, and examiner insights to help you tackle these challenges confidently.
案例分析题是 SQA 高等物理考试的重要组成部分。它考察你将物理原理应用于不熟悉情境、分析真实实验数据以及批判性评价实验过程的能力。本文通过讲解典型案例、逐步分析方法和考官评分视角，帮助你自信地应对这些挑战。
1. Understanding SQA Physics Case Study Questions | 理解 SQA 物理案例分析题
SQA case studies present a scenario, experimental procedure, data table, or graph, followed by a series of linked questions. You will often be asked to process data, plot a graph, calculate a quantity with its uncertainty, and comment on the reliability of the experiment.
SQA 案例分析题会给出一个情境、实验步骤、数据表或图表，然后提出一连串相互关联的问题。你通常需要处理数据、绘制图像、计算某个物理量及其不确定度，并评价实验的可靠性。
Marks are awarded for demonstrating knowledge of relevant physics, selecting appropriate equations, performing correct substitutions, giving the final answer with a suitable unit and significant figures, and providing reasoned evaluative comments.
分数针对展示相关物理知识、选择正确公式、准确代入、给出带有合适单位和有效数字的最终答案，以及提供有依据的评价性评论来授予。
Common contexts include mechanics, electricity, waves, and nuclear radiation. You must be able to handle real data with random and systematic uncertainties and recognise the limitations of the apparatus.
常见的考查背景包括力学、电学、波动和核辐射。你必须能处理带有随机和系统不确定度的真实数据，并认识仪器的局限性。
2. Case Study 1: Measuring g Using Free Fall | 案例1：用自由落体测量重力加速度 g
A student investigates the acceleration due to gravity by dropping a steel ball from various heights and recording the fall time with an electronic timer. The ball is released from rest. The data are shown in the table below.
一名学生通过从不同高度释放钢球并用电子计时器记录下落时间来研究重力加速度。小球从静止释放。数据如下表所示。
| h / m | t₁ / s | t₂ / s | t₃ / s | tavg / s | (tavg)² / s² |
|---|---|---|---|---|---|
| 0.500 | 0.32 | 0.33 | 0.32 | 0.323 | 0.104 |
| 0.800 | 0.40 | 0.41 | 0.40 | 0.403 | 0.162 |
| 1.200 | 0.49 | 0.50 | 0.49 | 0.493 | 0.243 |
| 1.600 | 0.57 | 0.58 | 0.57 | 0.573 | 0.328 |
plotting h against t² should yield a straight line through the origin with a slope equal to g/2.
从静止开始自由下落时，高度与时间的关系为 h = ½ g t²。因此，画出 h 对 t² 的图像应得到一条过原点的直线，其斜率为 g/2。
To determine g, calculate the gradient of the best‑fit line on an h vs t² graph. Choose two well‑parated points on the line, not data points, and compute (Δh)/(Δ(t²)). Then g = 2 × slope.
为了求 g，在 h–t² 图中计算最佳拟合线的斜率。选取线上两个相距较远的点（而非数据点），计算 (Δh)/(Δ(t²))。然后 g = 2 × 斜率。
Mark allocation typically requires: correct calculation of average times and t², labelled axes with units, sensible scales, accurately plotted points, a best‑fit straight line, a large gradient triangle, a final g value with units, and an estimate of the uncertainty in g using max‑min gradient lines or error bars.
评分通常要求：正确计算平均时间和 t²、坐标轴标注单位和标签、合理分度、准确描点、最佳拟合直线、大三角形的斜率计算、带单位的最终 g 值，以及用最大/最小斜率或误差棒估算 g 的不确定度。
For the given data, the slope is approximately (0.800 – 0)/(0.162 – accepted value is 9.8 m/s², but the line should be drawn using all points and the uncertainty evaluated. Typical sources of uncertainty are the timer's resolution and human reaction time for release.
根据给定数据，斜率大约为 (0.800 – 0)/(0.162 – 0) ≈ 4.94 m/s²，因此 g ≈ 9.9 m/s²。公认值是 9.8 m/s²，但图像需基于所有点绘制并评估不确定度。典型的不确定度来源是计时器的分辨率和人为释放时的反应时间。
3. Case Study 2: Ohm's Law and Resistance | 案例2：欧姆定律与电阻
In an electrical experiment, a student connects a fixed resistor to a variable power supply and measures the potential difference V across the resistor and the current I through it. The data collected are:
在一个电学实验中，学生将定值电阻连接到可调电源，并测量电阻两端的电势差 V 和通过它的电流 I。收集的数据如下：
| V / V | I / A |
|---|---|
| 0.50 | 0.05 |
| 1.00 | 0.10 |
| 1.50 | 0.15 |
| 2.00 | 0.19 |
| 2.50 | 0.25 |
The student is required to plot an I‑V characteristic graph and decide whether the resistor obeys Ohm's law. Since the resistor is ohmic, the graph should show a straight line through the origin; the resistance R is the reciprocal of the slope of the I‑V graph, or more conveniently, the slope of a V‑I graph.
要求绘制 I‑V 特性曲线并判断该电阻是否遵循欧姆定律。由于电阻是欧姆导体，图像应是一条过原点的直线；电阻 R 是 I‑V 图斜率的倒数，或者更简便地，是 V‑I 图的斜率。
From the data, you can plot V on the y‑axis and I on the x‑axis. A careful best‑ Ω. The manufacturer’s value is 10.0 Ω, but tolerance and heating effects need discussion.
根据数据，你可以将 V 标在 y 轴，I 标在 x 轴。小心绘制最佳拟合线，其斜率近似为 R。使用点 (0,0) 和 (2.50, 0.25)，R = 2.50 V / 0.25 A = 10.0 Ω。标称值是 10.0 Ω，但需要讨论容差和发热效应。
Notice the slight curvature at higher currents (0.19 A for 2.00 V), which hints at an increase in resistance due to heating. In your evaluation, comment on how warming can cause a deviation from pure ohmic behaviour and how using a short duty cycle could reduce this effect.
请注意较高电流时（2.00 V 对应 0.19 A）的轻微弯曲，这暗示电阻因发热而增大。在评价中，要评论升温如何导致偏离纯粹的欧姆行为，以及使用短时通电方式可以减小这种影响。
4. Case Study 3: Conservation of Momentum | 案例3：动量守恒
0.250 kg) which is initially stationary. The velocities before and after the collision are:
在气垫导轨上，两个滑块发生碰撞，用光电门记录速度。滑块 A（质量 m₁ = 0.250 kg）向初始静止的滑块 B（质量 m₂ = 0.250 kg）运动。碰撞前后的速度如下：
Before: v₁ = 0.45 m s⁻¹, v₂ = 0 m s⁻¹
碰撞前：v₁ = 0.45 m s⁻¹，v₂ = 0 m s⁻¹
After: v₁' = −0.03 m s⁻¹, v₂' = 0.42 m s⁻¹
碰撞后：v₁' = −0.03 m s⁻¹，v₂' = 0.42 m s⁻¹
The calculation of total momentum before gives pbefore = m₁v₁ + m₂v₂ = 0.250 × 0.45 + 0 = 0.1125 kg m s⁻¹. After the collision, pafter .250 × 0.42 = −0.0075 + 0.105 = 0.0975 kg m s⁻¹. The apparent loss in momentum is about 13%, which exceeds typical uncertainty limits.
计算碰撞前的总动量：pbefore = m₁v₁ + m₂v₂ = 0.250 × 0.45 + 0 = 0.1125 kg m s⁻¹。碰撞后，pafter = 0.250 × (−0.03) + 0.250 × 0.42 = −0.0075 + 0.105 = 0.0975 kg m s⁻¹。动量的表观损失约为 13%，超出了典型的不确定度范围。
In your analysis, you must identify possible reasons: the collision is not perfectly elastic, friction from the track, misalignment of light gates, or the small recoil of glider A being difficult to measure accurately. You should also calculate the kinetic energy before and after to classify the type of collision.
在分析中，你必须找出可能的原因：碰撞不是完全弹性的、导轨存在摩擦、光电门未对准、或滑块 A 的微小反弹难以精确测量。你还应该计算碰撞前后的动能，以对碰撞类型进行分类。
5. Case Study 4: Young's Double‑Slit Interference | 案例4：杨氏双缝干涉
he student measures the fringe spacing Δx by taking the distance across 10 fringes and dividing by 10. Five trials give: 2.1 mm, 2.0 mm, 2.1 mm, 2.2 mm, 2.0 mm.
一束未知波长的激光照射到间距 d = 0.50 mm 的双缝上。在距离 D = 1.50 m 的屏幕上观察干涉图样。学生测量 10 个条纹的间距并除以 10 得到条纹间距 Δx。五次测量结果为：2.1 mm、2.0 mm、2.1 mm、2.2 mm、2.0 mm。
The mean fringe spacing is (2.1+2.0+2.1+2.2+2.0)/5
Published by TutorHao | Year 12 Physics Revision Series | aleveler.com
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