📚 Year 10 Eduqas Physics: Case Study Practice | Year 10 Eduqas 物理：案例分析实战演练
Physics case studies test your ability to connect theory with real-world situations. They appear throughout the Eduqas Year 10 course and require a logical, step-by-step approach. This article presents practical case studies covering motion, forces, energy, electricity, waves and nuclear physics, with worked solutions and exam tips.
物理案例分析考查你将理论与现实情境联系起来的能力。它们贯穿Eduqas十年级课程，需要有逻辑的逐步解题方法。本文呈现涵盖运动、力、能量、电学、波和核物理的实用案例，附带详细解答和考试技巧。
1. Understanding Case Studies in Physics | 理解物理案例分析
A case study typically describes a scenario and poses several questions. Your first job is to extract all the given data: known values, constants and the target quantity. Always write them down before choosing an equation.
一个案例分析通常会描述一个场景并提出若干问题。你的首要工作是提取所有已知数据：已知量、常量和所求量。在选择公式之前，务必先把它们写下来。
Read the problem twice and underline key numbers and units. Draw a simple diagram if it helps, especially for forces or circuit layouts. This reduces careless mistakes.
仔细阅读问题两遍，划出关键数字和单位。如果有帮助，画一个简图，特别是在受力或电路布局时。这能减少粗心错误。
Check that all units are consistent (e.g. convert cm to m, minutes to seconds). Use the correct significant figures based on the data given. This habit is highly valued in Eduqas mark schemes.
检查所有单位是否统一（例如将厘米转为米，分钟转为秒）。根据所给数据使用合适的有效数字。这种习惯在Eduqas评分方案中很受重视。
2. Case Study 1: Motion of a Car | 案例一：汽车的运动
A car accelerates uniformly from rest and reaches a velocity of 28 m/s in 7.0 seconds. Calculate its acceleration and the distance it travels during this time.
一辆汽车从静止开始匀加速，在7.0秒内速度达到28 m/s。计算它的加速度和在这段时间内行驶的距离。
s, final velocity v = 28 m/s, time t = 7.0 s. Unknowns: acceleration a and displacement s.
列出已知量：初速度 u = 0 m/s，末速度 v = 28 m/s，时间 t = 7.0 s。未知量：加速度 a 和位移 s。
Use v = u + at to find a. Rearranging gives a = (v – u) / t = (28 – 0) / 7.0 = 4.0 m/s². The acceleration is 4.0 m/s².
使用 v = u + at 求 a。整理得 a = (v – u) / t = (28 – 0) / 7.0 = 4.0 m/s²。加速度为4.0 m/s²。
e s = (u + v)t / 2 = (0 + 28) × 7.0 / 2 = 98 m. Both methods agree, which is a good check.
现在使用 s = ut + ½at² 求 s = 0 + ½ × 4.0 × (7.0)² = 0 + 2.0 × 49 = 98 m。你也可以使用 s = (u + v)t / 2 = (0 + 28) × 7.0 / 2 = 98 m。两种方法一致，这是很好的验证。
3. Applying SUVAT Equations | 应用SUVAT方程
The five key equations of uniformly accelerated motion are often called SUVAT. They link displacement (s), initial velocity (u), final velocity (v), acceleration (a) and time (t). Choosing the right equation depends on which quantities you know and which you need.
匀加速运动的五个关键方程常被称为SUVAT。它们联系了位移(s)、初速度(u)、末速度(v)、加速度(a)和时间(t)。选择哪个方程取决于你已知哪些量以及需要求哪个量。
ut + ½at². The third is v² = u² + 2as. The fourth is s = (u + v)t / 2. The fifth is s = vt – ½at², though the first four are used most often at Year 10 level.
第一个方程是 v = u + at。第二个是 s = ut + ½at²。第三个是 v² = u² + 2as。第四个是 s = (u + v)t / 2。第五个是 s = vt – ½at²，但在十年级最常用的是前四个。
| Equation | Quantities involved |
| v = u + at | v, u, a, t |
| s = ut + ½at² | s, u, a, t |
| v² = u² + 2as | v, u, a, s |
| s = (u + v)t / 2 | s, u, v, t |
When using SUVAT, acceleration must be constant. Practise by writing down the knowns and the unknown, then ticking off which equations contain those symbols. This prevents time-wasting in the exam.
使用SUVAT时，加速度必须恒定。通过写下已知量和未知量，然后勾选出包含这些符号的方程进行练习。这能避免在考试中浪费时间。
4. Case Study 2: Energy Transfers in a Roller Coaster | 案例二：过山车中的能量转移
on and air resistance, calculate the speed of the carriage at the bottom. Use g = 10 N/kg.
一辆总质量为800 kg的过山车从25米高的山顶由静止出发。假设摩擦和空气阻力可忽略不计，计算它到达底部时的速度。使用 g = 10 N/kg。
25 = 200,000 J. At the bottom, all GPE has been converted into kinetic energy (KE) = ½mv².
在山顶，过山车具有重力势能 (GPE) = mgh = 800 × 10 × 25 = 200,000 J。在底部，所有GPE已转化为动能(KE) = ½mv²。
= ½mv². Cancel m (mass cancels out, so speed does not depend on mass) and solve for v: v = √2gh = √2 × 10 × 25 = √500 ≈ 22.4 m/s. So the speed at the bottom is about 22.4 m/s.
根据能量守恒，mgh = ½mv²。消去m（质量被约掉，因此速度与质量无关）并求解v：v = √2gh = √2 × 10 × 25 = √500 ≈ 22.4 m/s。因此底部速度约为22.4 m/s。
If friction were present, some energy would be transferred to thermal energy, and the final speed would be lower. Case studies often ask you to state this assumption.
如果存在摩擦，部分能量会转化为热能，最终速度会更低。案例分析常要求你陈述这一假设。
5. Analyzing Forces and Free-Body Diagrams | 分析力与受力图
A wooden box of mass 12 kg is pushed across a rough floor with a horizontal force of 50 N. The friction force opposing the motion is 26 N. Find the resultant force and the acceleration of the box.
一个质量为12 kg的木箱被一个50 N的水平力推着在粗糙地板上滑动。阻碍运动的摩擦力为26 N。求合力和木箱的加速度。
tion to the left (26 N). The vertical forces cancel. The resultant horizontal force = 50 N – 26 N = 24 N to the right.
画出受力图：重力向下 (120 N)，法向接触力向上 (120 N)，施加力向右 (50 N)，摩擦力向左 (26 N)。竖直方向力抵消。水平方向合力 = 50 N – 26 N = 24 N，方向向右。
Apply Newton' F / m = 24 / 12 = 2.0 m/s². The box accelerates at 2.0 m/s² in the direction of the push.
应用牛顿第二定律：F = ma，因此 a = F / m = 24 / 12 = 2.0 m/s²。木箱以2.0 m/s²的加速度
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