CAIE Year 13 Physics: Case Study Practical Exercises | CAIE Year 13 物理：案例分析实战演练

📚 CAIE Year 13 Physics: Case Study Practical Exercises | CAIE Year 13 物理：案例分析实战演练

Welcome to this focused revision session. The CAIE Year 13 Physics syllabus demands not only conceptual understanding but also the ability to dissect and solve complex numerical and theoretical problems. This article brings together ten carefully crafted case studies that span Mechanics, Gravitational and Electric Fields, Electromagnetism, Oscillations, Quantum and Nuclear Physics. Each example is broken down step by step, highlighting typical errors and providing the thinking process you need to score top marks in Paper 4 and beyond.

欢迎来到本次重点复习。CAIE Year 13 物理大纲不仅要求概念理解，更要求拆解并求解复杂计算与理论题的能力。本文精选十个案例，涵盖力学、引力与电场、电磁学、振动、量子与核物理。每个例子逐步拆解，突显典型错误，并展示拿下 Paper 4 及更高分所需的解题思维。

1. Projectile Motion: Range and Maximum Height | 抛体运动：射程与最大高度

A projectile is launched from ground level with an initial speed of 20 m s⁻¹ at an angle of 30° to the horizontal. Calculate the time of flight, the horizontal range, and the maximum height. Take g = 9.81 m s⁻² and neglect air resistance.

一抛体从地面以 20 m s⁻¹ 的初速度、与水平成 30° 角发射。计算飞行时间、水平射程和最大高度。取 g = 9.81 m s⁻²，忽略空气阻力。

Resolve the initial velocity: horizontal component = 20 cos30° ≈ 17.32 m s⁻¹; vertical component = 20 sin30° = 10 m s⁻¹. Always keep the horizontal and vertical motions independent.

分解初速度：水平分量 20 cos30° ≈ 17.32 m s⁻¹；竖直分量 20 sin30° = 10 m s⁻¹。务必保持水平与竖直运动独立。

Choose upward as positive. The vertical displacement is zero when the projectile returns to the ground. Using s = ut + ½at² gives 0 = 10t − ½×9.81t². Solving yields the non‑zero root t = 2.04 s. Alternatively, use the direct formula t = (2u sinθ)/g.

取向上为正。落地时竖直位移为零。由 s = ut + ½at² 得 0 = 10t − ½×9.81t²，解得非零根 t = 2.04 s。也可直接用 t = (2u sinθ)/g。

The horizontal range is found from constant horizontal velocity: R = uₓ t = 17.32 × 2.04 ≈ 35.3 m.

水平射程利用匀速：R = 水平分量 × t = 17.32 × 2.04 ≈ 35.3 m。

At maximum height, the vertical velocity is momentarily zero. Using v² = u² + 2as with v = 0, u = 10 m s⁻¹, a = −g: 0 = 10² − 2×9.81×H ⇒ H = 100/(2×9.81) ≈ 5.10 m.

最大高度时竖直速度瞬间为零。用 v² = u² + 2as，取 v = 0, u = 10, a = −g 得 H ≈ 5.10 m。

t = (2u sinθ)/g ,  R = u cosθ × t ,  H = (u sinθ)²/(2g)

Common pitfalls: forgetting the negative acceleration when upward is positive, mixing horizontal and vertical components, or using the full initial speed in the vertical equations. Always draw a clear resolution diagram.

常见错误：向上为正时忘记加速度为负；混淆水平与竖直分量；或在竖直方程中误用合初速度。务必画出分解示意图。

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2. Circular Motion: Banked Curve without Friction | 圆周运动：无摩擦倾斜弯道

A car rounds a circular banked curve of radius 100 m. The road is banked at 20° to the horizontal, and there is no reliance on sideways friction. Determine the ideal speed of the car so that it stays in the centre of the lane.

一辆汽车驶过半径 100 m 的圆形倾斜弯道，路面与水平成 20° 角，不依赖侧向摩擦力。求使车辆保持在车道中央的理想速度。

The horizontal component of the normal reaction provides the centripetal force: N sinθ = mv²/r. The vertical component balances weight: N cosθ = mg. Dividing the two equations eliminates the unknown normal force N, giving tanθ = v²/(rg).

法向反力的水平分量提供向心力：N sinθ = mv²/r；竖直分量平衡重力：N cosθ = mg。两式相除消去 N，得 tanθ = v²/(rg)。

Rearrange to find the ideal speed: v = √(rg tanθ).

整理得理想速度：v = √(rg tanθ)。

v = √(rg tanθ)

Substitute r = 100 m, g = 9.81 m s⁻², θ = 20°: tan20° ≈ 0.3640, so v = √(100 × 9.81 × 0.3640) = √(357.3) ≈ 18.9 m s⁻¹ (about 68 km h⁻¹).

代入 r = 100 m, g = 9.81 m s⁻², θ = 20°：tan20° ≈ 0.3640，得 v = √(100 × 9.81 × 0.3640) = √(357.3) ≈ 18.9 m s⁻¹（约 68 km h⁻¹）。

If the speed is lower than this, the car would slide down the slope; if higher, it would slide up. The derivation assumes the friction force is zero — a common exam condition. Remember that the normal reaction is not equal to mg here; it is N = mg/cosθ, which is larger than mg.

速度低于此值车会向弯道内侧滑，高于则会外滑。推导假设摩擦力为零，这是常见考题条件。注意此处的法向反力不等于 mg，而是 N = mg/cosθ，大于 mg。

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3. Gravitational Field: Escape Velocity from Earth | 引力场：地球逃逸速度

Use the following data to calculate the escape velocity from the Earth’s surface: Mass of Earth M = 5.97×10²⁴ kg, radius R = 6.37×10⁶ m, universal gravitational constant G = 6.67×10⁻¹¹ N m² kg⁻².

利用以下数据计算地球表面的逃逸速度：地球质量 M = 5.97×10²⁴ kg，半径 R = 6.37×10⁶ m，万有引力常量 G = 6.67×10⁻¹¹ N m² kg⁻²。

Escape velocity is the minimum speed needed for an object to just reach infinity with zero remaining kinetic energy. Equate initial kinetic energy to the magnitude of gravitational potential energy: ½mv² = GMm / R.

逃逸速度是物体刚好能到达无穷远处且剩余动能为零的最小速度。初动能等于引力势能的大小：½mv² = GMm / R。

v_esc = √(2GM / R)

Notice m cancels — escape velocity is independent of the object’s mass. Plug in the numbers: v_esc = √(2 × 6.67×10⁻¹¹ × 5.97×10²⁴ / 6.37×10⁶). Compute the numerator: 2GM = 2 × 3.982×10¹⁴ ≈ 7.964×10¹⁴. Dividing by R gives ≈ 1.250×10⁸. Taking the square root yields v_esc ≈ 1.12×10⁴ m s⁻¹ = 11.2 km s⁻¹.

注意 m 消去，逃逸速度与物体质量无关。代入数值：分子 2GM ≈ 7.964×10¹⁴，除以 R 约 1.250×10⁸，开平方得 v_esc ≈ 1.12×10⁴ m s⁻¹ = 11.2 km s⁻¹。

In exams, you may be asked to derive the formula or to compare escape velocities between different planets. Always express the final answer with correct units and to 2 or 3 significant figures. Remember that the surface gravitational field strength g = GM/R² is often used to rewrite the escape velocity as v_esc = √(2gR).

考试中可能要求推导公式或比较不同行星的逃逸速度。答案务必带单位并保留 2 到 3 位有效数字。注意表面重力场强 g = GM/R²，常可将逃逸速度改写为 v_esc = √(2gR)。

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4. Electric Field: Electron Deflection in a Parallel Plate | 电场：平行板中电子偏转

An electron enters the region between two horizontal parallel plates with a horizontal speed of 2.0×10⁷ m s⁻¹. The plates are 2.0 cm apart and have a potential difference of 200 V. The electron travels a horizontal distance of 5.0 cm while between the plates. Determine the vertical deflection of the electron as it leaves the plates

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