📚 Practical Case Study Drill for Pre-U CCEA Physics | Pre-U CCEA 物理：案例分析实战演练
Multi-stage mechanics problems are a staple of Pre-U and CCEA A-level Physics exams. They test your ability to link concepts such as energy conservation, work done by friction, and projectile motion in a single coherent scenario. This case study walks you through a complete worked example, highlights common pitfalls, and provides a structured method you can apply in your own revision and exams.
多阶段力学问题是 Pre-U 和 CCEA A-level 物理考试中的常见题型。它们考查你能否在一个连贯的场景中串联能量守恒、摩擦力做功和抛体运动等概念。本篇案例分析将带你完整演练一道例题，指出常见错误，并提供一套可在复习和考试中直接使用的结构化解题方法。
1. Case Scenario: From Ramp to Free Fall | 案例情景：从斜面到自由落体
A small ball of mass 0.50 kg is released from rest at the top of a smooth incline. The incline makes an angle of 30° with the horizontal and has a vertical height of 2.0 m. At the bottom, the ball slides onto a rough horizontal surface of length 1.5 m, where the coefficient of kinetic friction is 0.20. The far end of the rough surface is the edge of a table 1.225 m above the floor. The ball leaves the table horizontally and falls to the ground. Take g = 9.8 m s⁻² and ignore air resistance.
一个质量为 0.50 kg 的小球从光滑斜面的顶端由静止释放。斜面与水平面的夹角为 30°，竖直高度为 2.0 m。小球滑到底部后进入一段长度为 1.5 m 的粗糙水平面，动摩擦因数为 0.20。粗糙面的末端是距离地面 1.225 m 的桌子边缘。小球以水平速度离开桌面，掉落到地面上。取 g = 9.8 m s⁻²，忽略空气阻力。
Find the horizontal distance from the table edge to the point where the ball strikes the floor.
求小球落地点到桌缘的水平距离。
2. Stage 1 – Acceleration Down the Smooth Incline | 阶段一——沿光滑斜面加速
Since the incline is smooth, no energy is lost to friction during the descent. We can apply conservation of mechanical energy. The ball loses gravitational potential energy (mgh) and gains kinetic energy (½mv²). Setting the initial kinetic energy to zero gives v₁ = √(2gh) at the bottom of the incline.
因为斜面是光滑的，下滑过程中没有因摩擦造成的能量损失。我们可以运用机械能守恒定律。小球减少的重力势能 (mgh) 转化为动能 (½mv²)。设初动能为零，到达斜面底端的速度 v₁ = √(2gh)。
v₁ = √(2 × 9.8 × 2.0) = √39.2 ≈ 6.26 m s⁻¹
Notice the incline angle is not needed for the speed calculation because only the vertical height matters for energy conservation. The angle would be required if we were finding acceleration or time down the slope.
注意，计算速度时不需要斜面倾角，因为能量守恒只与竖直高度有关。若要计算沿斜面的加速度或下滑时间，才需要角度。
3. Stage 2 – Work Done by Friction on the Rough Surface | 阶段二——粗糙水平面上的摩擦力做功
On the horizontal rough section, the kinetic friction force fₖ = μₖN acts opposite to the motion. Since the surface is horizontal, the normal reaction N equals the weight mg. Thus fₖ = μₖmg. The work done by friction over the distance d removes kinetic energy: W = fₖ × d = μₖmgd.
在水平粗糙段上，动摩擦力 fₖ = μₖN 的方向与运动方向相反。由于水平面是水平的，支持力 N 等于重力 mg，因此 fₖ = μₖmg。摩擦力在距离 d 上做的功消耗了动能：W = fₖ × d = μₖmgd。
W = 0.20 × 0.50 × 9.8 × 1.5 = 1.47 J
This positive value of work done by friction represents the energy dissipated as heat. We will subtract it from the kinetic energy the ball had at the start of the rough section.
这个摩擦力的正功值代表以热能形式耗散的能量。我们将从进入粗糙面时的动能中减掉它。
4. Applying the Work–Energy Theorem | 运用动能定理
The kinetic energy at the beginning of the rough surface is KE₁ = ½mv₁². After sliding across the rough patch, the remaining kinetic energy KE₂ = KE₁ − W. From KE₂ we can find the speed v₂ the ball has just as it leaves the table edge.
小球刚进入粗糙面时的动能为 KE₁ = ½mv₁²。滑过粗糙段后，剩余的动能 KE₂ = KE₁ − W。由 KE₂ 可求出小球离开桌缘瞬间的速度 v₂。
- English: KE₁ = 0.5 × 0.50 × (6.26)² ≈ 9.80 J
- 中文：KE₁ = 0.5 × 0.50 × (6.26)² ≈ 9.80 J
- English: KE₂ = 9.80 − 1.47 = 8.33 J
- 中文：KE₂ = 9.80 − 1.47 = 8.33 J
v₂ = √(2 × 8.33 / 0.50) = √33.32 ≈ 5.77 m s⁻¹
This v₂ is entirely horizontal because the rough track is horizontal. It serves as the launch speed for the projectile stage.
这个 v₂ 完全是水平方向的速度，因为粗糙轨道是水平的。它将作为抛体运动阶段的发射初速度。
5. Stage 3 – Projectile Motion Setup | 阶段三——抛体运动设置
Once the ball leaves the table, it behaves as a projectile launched horizontally from a height H = 1.225 m. The initial vertical velocity is zero, and the only acceleration is g = 9.8 m s⁻² downward. We analyse the vertical and horizontal motions independently.
小球离开桌面后，可视为从高度 H = 1.225 m 处水平抛出的抛体。竖直方向的初速度为零，只有向下的加速度 g = 9.8 m s⁻²。我们分别分析竖直和水平两个方向的运动。
The vertical motion determines the time of flight. The horizontal motion determines the range, since horizontal velocity remains constant (air resistance neglected).
竖直分运动决定飞行时间；水平分运动决定射程，因为水平速度保持不变（忽略空气阻力）。
6. Calculating the Time of Flight | 计算飞行时间
Use the kinematic equation for displacement under constant acceleration: s = ut + ½at². Taking downward as positive, u_y = 0, s = H = 1.225 m, a = g = 9.8 m s⁻². Thus H = ½gt².
运用匀加速直线运动的位移公式：s = ut + ½at²。取向下为正方向，u_y = 0，s = H = 1.225 m，a = g = 9.8 m s⁻²，于是 H = ½gt²。
t = √(2H/g) = √(2 × 1.225 / 9.8) = √(2.45 / 9.8) = √0.25 = 0.50 s
The numbers are chosen here to give a clean result: the ball takes exactly 0.50 s to reach the floor. In exam problems, the time of flight may not be a round number, but the method remains identical.
这里的数据特意选取了整数结果：小球到达地面的时间恰好为 0.50 s。在考试题中，飞行时间可能不是整数，但解题方法完全相同。
7. Horizontal Range Calculation | 水平射程计算
Horizontally, the ball moves with constant speed v₂ = 5.77 m s⁻¹ for the duration of the flight. Thus the range R = v₂ × t.
水平方向上，小球以恒定速率 v₂ = 5.77 m s⁻¹ 运动整个飞行时间，故水平射程 R = v₂ × t。
R = 5.77 × 0.50 = 2.885 m ≈ 2.89 m
The ball lands approximately 2.89 m from the edge of the table. This is the final answer to the multi-stage problem.
小球落在距离桌缘约 2.89 m 的地方。这就是这道多阶段问题的最终答案。
8. Assembling the Complete Solution | 汇总完整解题过程
A well-structured answer in an exam should show clear logical steps: (i) energy conservation on the incline, (ii) work–energy principle on the rough surface, (iii) projectile motion split into vertical free fall and horizontal constant velocity. Present each step with the relevant equation, substitution, and final value.
在考试中，一份结构清晰的答案应当展示明确的逻辑步骤：(i) 斜面上的能量守恒；(ii) 粗糙面上的动能定理；(iii) 将抛体运动分解为竖直自由落体和水平匀速直线运动。每一步都写出相关方程、代入数据和最终数值。
Using a summary table can help the examiner follow your reasoning:
使用汇总表格可以帮助考官追踪你的推理过程：
| Stage | Key Equation | Result |
|---|---|---|
| Incline | v₁ = √(2gh) | 6.26 m s⁻¹ |
| Rough surface | ½mv₂² = ½mv₁² − μₖmgd | v₂ = 5.77 m s⁻¹ |
| Free fall | t = √(2H/g) | 0.50 s |
| Horizontal range | R = v₂t | 2.89 m |
9. What if the Incline Were Not Smooth? | 假如斜面不光滑？
If friction were also present on the incline, the mechanical energy approach would need modification. The work done by friction along the slope would have to be calculated using fₖ = μₖN = μₖmg cos θ over the slant distance. This work would further reduce the kinetic energy at the bottom.
如果斜面上也存在摩擦，那么机械能守恒的方法就需要修改。需用 fₖ = μₖN = μₖmg cos θ 计算沿斜面摩擦力在整个斜面长度上做的功。这部分功将进一步减少到达底部的动能。
slant distance L = h / sin θ = 2.0 / sin30° = 4.0 m
You would then use the work–energy theorem from the very start: KE at bottom = mgh − fₖ(incline) × L. This extra complexity is a common extension in exam questions.
此时就需要从一开始就使用动能定理：到达底部的动能 = mgh − fₖ(斜面) × L。这一复杂化是考试中常见的拓展问法。
10. Common Mistakes and How to Avoid Them | 常见错误与避错技巧
- Mixing energy and force methods incorrectly. Some students try to find the speed after the rough patch by using v² = u² + 2as with a = μg, which is correct for deceleration on a horizontal rough plane when only friction acts, but they often forget to check if the object stops before the end. Always verify that the distance is enough to stop the object; if not, use work–energy to find remaining speed.
- 错误混用能量法与动力学公式。 有些学生想用 v² = u² + 2as 来求粗糙段后的速度，其中 a = μg，这只有在仅有摩擦力作用时才对，但他们常常忘记检验物体会不会在到达末端前停下。务必验证距离是否足以使物体停下；如果不够，就用功能关系求剩余速度。
- Forgetting the horizontal launch condition. Students sometimes add a vertical component to the initial velocity at the table edge. The problem states the ball leaves the table horizontally, so initial vertical velocity is zero.
- 忘了水平发射条件。 有的学生会在桌缘处给初速度添加竖直分量。题目明确小球以水平速度离开，因此竖直初速为零。
- Confusing g values. Stick to the given value (9.8 or 9.81) consistently. Mixing g = 10 for one part and g = 9.8 for another will ruin precision.
- g 值混淆。 始终使用题目给定的 g 值 (9.8 或 9.81)。一部分用 10、另一部分用 9.8 会破坏精度。
11. Experimental Verification Ideas | 实验验证思路
In a practical context, you could verify this analysis using a ramp, a rough mat, and a table. By measuring the actual range with a metre rule and filming the motion, you could compare the predicted range with the experimental mean. The difference would help estimate energy losses due to rolling (if a ball is used) or air resistance.
在实验情境中，你可以用斜面、粗糙垫和桌子来验证上述分析。通过用米尺测量实际射程并拍摄运动过程，可以对比理论预测射程与实验均值。两者之差有助于估算因滚动（如果使用小球）或空气阻力造成的能量损失。
You might also use light gates to measure v₂ directly and check whether the theoretical value matches the measurement within uncertainty. This links closely to the practical skills assessed in CCEA Physics.
你还可以使用光电门直接测量 v₂，检验理论值是否在不确定度范围内与实测值吻合。这与 CCEA 物理所评估的实验技能密切相关。
12. Summary and Key Takeaways | 总结与关键要点
This case study demonstrates how to break a complex multi-stage problem into manageable blocks: energy conversion, work against friction, and two-dimensional kinematics. Always identify the start and end points for each energy transfer, and choose the most efficient principle (conservation of energy or work–energy theorem) for that stage. Practising such integrated problems will sharpen your analytical skills and prepare you for the highest-mark questions on the CCEA specification.
本案例分析展示了如何将复杂的多阶段问题拆解为易于处理的模块：能量转化、克服摩擦力做功以及二维运动学。始终明确每段能量转移的初末状态，并为该阶段选择效率最高的原理（能量守恒或动能定理）。多练习这类综合性问题能提升你的分析能力，为 CCEA 考纲中分值最高的题目做好准备。
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