📚 Typical Example Problems in IB CCEA Physics | IB CCEA 物理典型例题详解
This article provides a carefully curated selection of worked examples spanning the core topics of IB and CCEA A-Level Physics. Each problem is broken down step by step, with English and Chinese explanations running side by side. The goal is to strengthen conceptual understanding and problem-solving technique for typical examination questions.
本文精选了涵盖 IB 与 CCEA 物理核心主题的典型例题，并逐步拆解分析。每个步骤均配有中英文对照解释，旨在强化对典型考题的概念理解与解题技巧。
1. Projectile Motion | 抛体运动例题
A ball is kicked 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 reached. Assume negligible air resistance and g = 9.8 m s⁻².
一个球从地面以 20 m s⁻¹ 的初速度与水平方向成 30° 角踢出。计算飞行时间、水平射程和最大高度。忽略空气阻力，取 g = 9.8 m s⁻²。
Resolve the initial velocity into horizontal and vertical components. The horizontal component vᵪ = u cosθ = 20 cos30° = 20 × (√3/2) ≈ 17.3 m s⁻¹. The vertical component vᵧ = u sinθ = 20 sin30° = 20 × 0.5 = 10.0 m s⁻¹.
将初速度分解为水平和竖直分量。水平分量 vᵪ = u cosθ = 20 cos30° = 20 × (√3/2) ≈ 17.3 m s⁻¹。竖直分量 vᵧ = u sinθ = 20 sin30° = 20 × 0.5 = 10.0 m s⁻¹。
The time of flight depends only on vertical motion. Using s = uᵧ t + ½ a t², with s = 0 (returns to ground), 0 = 10 t – 4.9 t². Factoring gives t(10 – 4.9t) = 0, so t = 0 or t = 10/4.9 ≈ 2.04 s. The flight time is about 2.04 s.
飞行时间仅取决于竖直运动。由 s = uᵧ t + ½ a t²，其中 s = 0（落回地面），得 0 = 10 t – 4.9 t²。因式分解得 t(10 – 4.9t) = 0，故 t = 0 或 t = 10/4.9 ≈ 2.04 s，飞行时间约为 2.04 s。
The horizontal range is found from constant horizontal velocity: R = vᵪ × t = 17.3 × 2.04 ≈ 35.3 m. The maximum height occurs when vᵧ = 0. Using vᵧ² = uᵧ² + 2a s, 0 = 10² – 2×9.8×h, giving h = 100/19.6 ≈ 5.10 m.
水平射程由匀速水平运动求得：R = vᵪ × t = 17.3 × 2.04 ≈ 35.3 m。最大高度发生在 vᵧ = 0 时，由 vᵧ² = uᵧ² + 2a s，0 = 10² – 2×9.8×h，得 h = 100/19.6 ≈ 5.10 m。
2. Connected Masses on an Incline | 斜面上的连接体问题
Two blocks are connected by a light inextensible string over a frictionless pulley. Block A of mass 4.0 kg rests on a smooth slope inclined at 30° to the horizontal. Block B of mass 3.0 kg hangs vertically. Determine the acceleration of the system and the tension in the string. Take g = 9.8 m s⁻².
两个物块由一根轻质不可伸长的绳子跨过光滑滑轮连接。物块 A 质量 4.0 kg 静置于倾角 30° 的光滑斜面上，物块 B 质量 3.0 kg 竖直悬挂。求系统的加速度和绳中张力。取 g = 9.8 m s⁻²。
For block A on the slope, the component of weight down the slope is mₐ g sinθ = 4.0 × 9.8 × sin30° = 4.0 × 9.8 × 0.5 = 19.6 N. The equation of motion for A is: T – 19.6 = 4.0 a, assuming acceleration down the slope for B pulls A up the slope. Here we must choose a consistent direction; let’s assume B falls so A moves up the slope. Then for A: T – mₐ g sinθ = mₐ a.
对于斜面上的物块 A，沿斜面的重力分量为 mₐ g sinθ = 4.0 × 9.8 × sin30° = 19.6 N。A 的运动方程为：T – 19.6 = 4.0 a，这里假设 B 下落使 A 沿斜面向上运动，故对于 A：T – mₐ g sinθ = mₐ a。
For hanging block B, weight m_b g = 3.0 × 9.8 = 29.4 N acts downward, tension T acts upward. The equation: 29.4 – T = 3.0 a. Solving the two equations simultaneously: T = 19.6 + 4.0a and T = 29.4 – 3.0a. Equating: 19.6 + 4.0a = 29.4 – 3.0a → 7.0a = 9.8 → a = 1.4 m s⁻². Then T = 19.6 + 4.0×1.4 = 25.2 N (or 29.4 – 3.0×1.4 = 25.2 N).
对于悬挂的物块 B，重力 m_b g = 3.0 × 9.8 = 29.4 N 向下，绳张力 T 向上。方程：29.4 – T = 3.0 a。联立两式：T = 19.6 + 4.0a 且 T = 29.4 – 3.0a，令其相等得 19.6 + 4.0a = 29.4 – 3.0a → 7.0a = 9.8 → a = 1.4 m s⁻²。于是 T = 19.6 + 4.0×1.4 = 25.2 N（或 29.4 – 3.0×1.4 = 25.2 N）。
3. Critical Speed in Vertical Circular Motion | 竖直圆周运动的临界速度
A roller coaster car of mass 500 kg goes over the top of a circular loop of radius 15 m. What is the minimum speed at the top so that the car does not lose contact with the track? What is the normal reaction force when the speed at the top is 20 m s⁻¹?
一辆质量为 500 kg 的过山车通过半径为 15 m 的圆形环轨顶部。车在顶部不掉落的最小速度是多少？若顶部速度为 20 m s⁻¹，轨道对车的支持力为多大？
At the top, the centripetal force is provided by weight plus normal reaction: mg + N = mv²/r. For the minimum speed to just maintain contact, the normal reaction N = 0. Thus mg = mv²/r, giving v = √(gr) = √(9.8 × 15) = √147 ≈ 12.1 m s⁻¹.
在顶部，向心力由重力和支持力共同提供：mg + N = mv²/r。为恰好保持接触，支持力 N = 0，于是 mg = mv²/r，得 v = √(gr) = √(9.8 × 15) = √147 ≈ 12.1 m s⁻¹。
When the speed at the top is 20 m s⁻¹, we use the full equation: mg + N = mv²/r. Therefore N = m(v²/r – g) = 500 × (20²/15 – 9.8) = 500 × (400/15 – 9.8) = 500 × (26.67 – 9.8) = 500 × 16.87 ≈ 8435 N. The reaction force is about 8400 N upward (pushing the car toward the centre).
当顶部速度为 20 m s⁻¹ 时，用完整方程：mg + N = mv²/r。可得 N = m(v²/r – g) = 500 × (20²/15 – 9.8) = 500 × (400/15 – 9.8) = 500 × (26.67 – 9.8) = 500 × 16.87 ≈ 8435 N。支持力约为 8400 N，方向向上（指向圆心）。
4. Satellite Orbital Velocity and Period | 卫星的轨道速度与周期
A satellite orbits Earth at an altitude of 300 km above the surface. Earth’s radius is 6400 km and its mass is 6.0 × 10²⁴ kg. Determine the orbital speed and the period of the satellite. G = 6.67 × 10⁻¹¹ N m² kg⁻².
一颗卫星在距地球表面 300 km 高度处绕地球运行。地球半径为 6400 km，质量为 6.0 × 10²⁴ kg。计算卫星的轨道速度和周期。G = 6.67 × 10⁻¹¹ N m² kg⁻²。
The orbital radius r = (6400 + 300) km = 6700 km = 6.7 × 10⁶ m. Gravitational force provides centripetal force: GMm/r² = mv²/r. Thus v² = GM/r, v = √(GM/r).
轨道半径 r = (6400 + 300) km = 6700 km = 6.7 × 10⁶ m。万有引力提供向心力：GMm/r² = mv²/r，因此 v² = GM/r，v = √(GM/r)。
Calculate v = √(6.67×10⁻¹¹ × 6.0×10²⁴ / 6.7×10⁶) = √(4.002×10¹⁴ / 6.7×10⁶) = √(5.973×10⁷) ≈ √(5.97×10⁷) ≈ 7.73×10³ m s⁻¹, about 7.73 km s⁻¹. The period T = 2πr / v = 2×π×6.7×10⁶ / 7.73×10³ ≈ (4.21×10⁷) / 7.73×10³ ≈ 5.45×10³ s, or about 91 minutes.
计算 v = √(6.67×10⁻¹¹ × 6.0×10²⁴ / 6.7×10⁶) = √(4.002×10¹⁴ / 6.7×10⁶) = √(5.973×10⁷) ≈ √(5.97×10⁷) ≈ 7.73×10³ m s⁻¹，约为 7.73 km s⁻¹。周期 T = 2πr / v = 2×π×6.7×10⁶ / 7.73×10³ ≈ (4.21×10⁷) / 7.73×10³ ≈ 5.45×10³ s，约 91 分钟。
5. Energy in Simple Harmonic Motion | 简谐运动中的能量
A mass of 0.50 kg hangs from a spring with spring constant 200 N m⁻¹. It is pulled down 0.040 m from equilibrium and released. Find the angular frequency, the maximum speed, and the total mechanical energy of the system.
一质量为 0.50 kg 的物块悬挂在劲度系数为 200 N m⁻¹ 的弹簧上。将其从平衡位置向下拉 0.040 m 后释放。求角频率、最大速度和系统的总机械能。
Angular frequency ω = √(k/m) = √(200 / 0.50) = √400 = 20 rad s⁻¹. The amplitude A = 0.040 m. In SHM, maximum speed v_max = ωA = 20 × 0.040 = 0.80 m s⁻¹.
角频率 ω = √(k/m) = √(200 / 0.50) = √400 = 20 rad s⁻¹。振幅 A = 0.040 m。在简谐运动中，最大速度 v_max = ωA = 20 × 0.040 = 0.80 m s⁻¹。
Total mechanical energy E = ½ k A² = ½ × 200 × (0.040)² = 100 × 0.0016 = 0.16 J. This energy remains constant, transforming between kinetic and potential.
总机械能 E = ½ k A² = ½ × 200 × (0.040)² = 100 × 0.0016 = 0.16 J。该能量守恒，在动能和势能之间转化。
6. Kirchhoff’s Laws in a Multi-loop Circuit | 基尔霍夫定律解多回路电路
Consider a circuit with two batteries and three resistors. Battery 1: 12 V, internal resistance 0.5 Ω; Battery 2: 6 V, internal resistance 0.3 Ω. Resistor R₁ = 4 Ω, R₂ = 2 Ω, R₃ = 10 Ω arranged such that R₁ and Battery 1 are in series in the left branch, R₂ and Battery 2 in the right branch, and R₃ connects the midpoints of the two branches. Find the current through each resistor.
考虑一个包含两节电池和三个电阻的电路。电池 1：12 V，内阻 0.5 Ω；电池 2：6 V，内阻 0.3 Ω。电阻 R₁ = 4 Ω，R₂ = 2 Ω，R₃ = 10 Ω，连接方式为：左支路串联 R₁ 和电池 1，右支路串联 R₂ 和电池 2，R₃ 跨接在两支路的中点之间。求各电阻中的电流。
Assign loop currents: let I₁ be current in left loop (clockwise), I₂ in right loop (clockwise), and I₃ = I₁ – I₂ flowing downward through R₃. Write Kirchhoff’s voltage law for left loop: –12 + 0.5I₁ + 4I₁ + 10(I₁ – I₂) = 0 → 12 = (0.5+4+10)I₁ – 10I₂ → 12 = 14.5I₁ – 10I₂. (Equation 1)
设定回路电流：设左回路电流为 I₁（顺时针），右回路电流为 I₂（顺时针），则通过 R₃ 向下的电流为 I₃ = I₁ – I₂。对左回路列基尔霍夫电压方程：–12 + 0.5I₁ + 4I₁ + 10(I₁ – I₂) = 0 → 12 = (0.5+4+10)I₁ – 10I₂ → 12 = 14.5I₁ – 10I₂。（式 1）
For the right loop: –6 + 0.3I₂ + 2I₂ + 10(I₂ – I₁) = 0 → 6 = –10I₁ + (0.3+2+10)I₂ → 6 = –10I₁ + 12.3I₂. (Equation 2) Solving simultaneously: multiply Eq1 by 10: 120 = 145I₁ – 100I₂. Multiply Eq2 by 14.5: 87 = –145I₁ + 178.35I₂. Adding gives 207 = 78.35I₂ → I₂ = 2.64 A. Substitute back: 12 = 14.5I₁ – 10×2.64 → 12 = 14.5I₁ – 26.4 → 14.5I₁ = 38.4 → I₁ = 2.65 A. Then I₃ = I₁ – I₂ = 0.01 A (negligible). So current through R₁ is 2.65 A, through R₂ is 2.64 A, through R₃ is ~0.01 A.
对右回路：–6 + 0.3I₂ + 2I₂ + 10(I₂ – I₁) = 0 → 6 = –10I₁ + (0.3+2+10)I₂ → 6 = –10I₁ + 12.3I₂。（式 2）联立求解：式 1 乘以 10：120 = 145I₁ – 100I₂；式 2 乘以 14.5：87 = –145I₁ + 178.35I₂。两式相加得 207 = 78.35I₂ → I₂ = 2.64 A。代入可得 12 = 14.5I₁ – 10×2.64 → 12 = 14.5I₁ – 26.4 → 14.5I₁ = 38.4 → I₁ = 2.65 A。于是 I₃ = I₁ – I₂ = 0.01 A（可忽略）。因此通过 R₁ 的电流为 2.65 A，通过 R₂ 的为 2.64 A，通过 R₃ 的约为 0.01 A。
7. Deflection of an Electron in an Electric Field | 电场中电子的偏转
An electron enters the region between two parallel plates at 2.0 × 10⁷ m s⁻¹ horizontally. The plates are 0.020 m long and have a uniform electric field of 5.0 × 10³ V m⁻¹ directed downward. How much vertical deflection occurs as the electron leaves the plates? Mass of electron = 9.11 × 10⁻³¹ kg, charge = –1.6 × 10⁻¹⁹ C.
一个电子以 2.0 × 10⁷ m s⁻¹ 的水平速度进入两平行板之间。板长 0.020 m，其间有向下的匀强电场 5.0 × 10³ V m⁻¹。求电子离开板时的竖直偏转量。电子质量 9.11 × 10⁻³¹ kg，电荷量 –1.6 × 10⁻¹⁹ C。
The electron experiences an upward electric force because the field is downward and the charge is negative. Magnitude of force F = eE = 1.6×10⁻¹⁹ × 5.0×10³ = 8.0×10⁻¹⁶ N. Acceleration a = F/m = 8.0×10⁻¹⁶ / 9.11×10⁻³¹ ≈ 8.78×10¹⁴ m s⁻² upward.
电子受到向上的电场力，因场强向下且电荷为负。力的大小 F = eE = 1.6×10⁻¹⁹ × 5.0×10³ = 8.0×10⁻¹⁶ N。加速度 a = F/m = 8.0×10⁻¹⁶ / 9.11×10⁻³¹ ≈ 8.78×10¹⁴ m s⁻² 向上。
Time spent between plates t = length / horizontal velocity = 0.020 / 2.0×10⁷ = 1.0×10⁻⁹ s. Vertical deflection Δy = ½ a t² = 0.5 × 8.78×10¹⁴ × (1.0×10⁻⁹)² = 0.5 × 8.78×10¹⁴ × 1.0×10⁻¹⁸ = 4.39×10⁻⁴ m ≈ 0.44 mm.
在板间运动的时间 t = 板长 / 水平速度 = 0.020 / 2.0×10⁷ = 1.0×10⁻⁹ s。竖直偏转量 Δy = ½ a t² = 0.5 × 8.78×10¹⁴ × (1.0×10⁻⁹)² = 0.5 × 8.78×10¹⁴ × 1.0×10⁻¹⁸ = 4.39×10⁻⁴ m ≈ 0.44 mm。
8. Motion of a Charge in a Magnetic Field | 电荷在磁场中的运动
A proton with kinetic energy 10 keV enters a uniform magnetic field of 0.50 T perpendicular to its velocity. Find the radius of the resulting circular path. Proton mass = 1.67 × 10⁻²⁷ kg, charge = 1.6 × 10⁻¹⁹ C. 1 eV = 1.6 × 10⁻¹⁹ J.
一个动能为 10 keV 的质子垂直射入 0.50 T 的匀强磁场中。求其圆周运动的半径。质子质量 1.67 × 10⁻²⁷ kg，电荷量 1.6 × 10⁻¹⁹ C。1 eV = 1.6 × 10⁻¹⁹ J。
First find the speed. Kinetic energy K = 10 × 10³ eV = 1.0×10⁴ × 1.6×10⁻¹⁹ = 1.6×10⁻¹⁵ J. K = ½ m v², so v = √(2K/m) = √(2 × 1.6×10⁻¹⁵ / 1.67×10⁻²⁷) = √(3.2×10⁻¹⁵ / 1.67×10⁻²⁷) = √(1.916×10¹²) ≈ 1.38×10⁶ m s⁻¹.
先求速度。动能 K = 10 keV = 1.0×10⁴ × 1.6×10⁻¹⁹ = 1.6×10⁻¹⁵ J。由 K = ½ m v² 得 v = √(2K/m) = √(2 × 1.6×10⁻¹⁵ / 1.67×10⁻²⁷) = √(3.2×10⁻¹⁵ / 1.67×10⁻²⁷) = √(1.916×10¹²) ≈ 1.38×10⁶ m s⁻¹。
Magnetic force provides centripetal force: qvB = mv²/r → r = mv / (qB). r = (1.67×10⁻²⁷ × 1.38×10⁶) / (1.6×10⁻¹⁹ × 0.50) = (2.30×10⁻²¹) / (8.0×10⁻²⁰) = 2.875×10⁻² m ≈ 2.9 cm.
洛伦兹力提供向心力：qvB = mv²/r → r = mv / (qB)。计算得 r = (1.67×10⁻²⁷ × 1.38×10⁶) / (1.6×10⁻¹⁹ × 0.50) = (2.30×10⁻²¹) / (8.0×10⁻²⁰) = 2.875×10⁻² m ≈ 2.9 cm。
9. First Law of Thermodynamics in an Isobaric Process | 等压过程中的热力学第一定律
A cylinder contains 0.10 mol of an ideal gas at 300 K. The gas expands at constant pressure of 1.0 × 10⁵ Pa until its volume doubles. Calculate the work done by the gas, the change in internal energy, and the heat supplied. Assume C_V = 12.5 J mol⁻¹ K⁻¹ and C_P = 20.8 J mol⁻¹ K⁻¹.
一汽缸装有 0.10 mol 的理想气体，初始温度 300 K。气体在 1.0 × 10⁵ Pa 的恒压下膨胀至体积加倍。计算气体做的功、内能的变化和吸收的热量。已知 C_V = 12.5 J mol⁻¹ K⁻¹，C_P = 20.8 J mol⁻¹ K⁻¹。
At constant pressure, work done W = P ΔV. Initial volume V₁ = nRT₁/P = (0.10×8.31×300) / 1.0×10⁵ = (249.3) / 1.0×10⁵ = 2.493×10⁻³ m³. Final volume V₂ = 2V₁ = 4.986×10⁻³ m³. ΔV = 2.493×10⁻³ m³. So W = 1.0×10⁵ × 2.493×10⁻³ = 249.3 J.
恒压下，气体做功 W = P ΔV。初始体积 V₁ = nRT₁/P = (0.10×8.31×300) / 1.0×10⁵ = 249.3 / 1.0×10⁵ = 2.493×10⁻³ m³。最终体积 V₂ = 2V₁ = 4.986×10⁻³ m³，ΔV = 2.493×10⁻³ m³。故 W = 1.0×10⁵ × 2.493×10⁻³ = 249.3 J。
Since it is isobaric, T₂/T₁ = V₂/V₁ = 2, so T₂ = 600 K. Change in internal energy ΔU = n C_V ΔT = 0.10 × 12.5 × (600 – 300) = 0.10 × 12.5 × 300 = 375 J. Using the first law ΔU = Q – W, we find Q = ΔU + W = 375 + 249.3 = 624.3 J. Alternatively, Q = n C_P ΔT = 0.10 × 20.8 × 300 = 624 J, showing consistency.
因过程等压，T₂/T₁ = V₂/V₁ = 2，故 T₂ = 600 K。内能变化 ΔU = n C_V ΔT = 0.10 × 12.5 × (600 – 300) = 375 J。由热力学第一定律 ΔU = Q – W，得 Q = ΔU + W = 375 + 249.3 = 624.3 J。另一方法：Q = n C_P ΔT = 0.10 × 20.8 × 300 = 624 J，两者一致。
10. Photoelectric Effect and Threshold Frequency | 光电效应与截止频率
Ultraviolet light of wavelength 200 nm shines on a clean metal surface. The work function of the metal is 4.5 eV. Find the maximum kinetic energy of the emitted electrons and the stopping potential. Determine the threshold frequency for this metal. h = 6.63 × 10⁻³⁴ J s, c = 3.0 × 10⁸ m s⁻¹.
波长为 200 nm 的紫外光照射在清洁金属表面上，金属的逸出功为 4.5 eV。求发射光电子的最大动能和遏止电势差，并确定该金属的截止频率。h = 6.63 × 10⁻³⁴ J s，c = 3.0 × 10⁸ m s⁻¹。
Photon energy E = hf = hc/λ = (6.63×10⁻³⁴ × 3.0×10⁸) / (200×10⁻⁹) = (1.989×10⁻²⁵) / (2.0×10⁻⁷) = 9.945×10⁻¹⁹ J. Convert to eV: 9.945×10⁻¹⁹ J / 1.6×10⁻¹⁹ J eV⁻¹ ≈ 6.22 eV.
光子能量 E = hf = hc/λ = (6.63×10⁻³⁴ × 3.0×10⁸) / (200×10⁻⁹) = 9.945×10⁻¹⁹ J。换算为 eV：9.945×10⁻¹⁹ J / 1.6×10⁻¹⁹ J eV⁻¹ ≈ 6.22 eV。
Maximum kinetic energy K_max = E – Φ = 6.22 eV – 4.5 eV = 1.72 eV. In joules, K_max = 1.72 × 1.6×10⁻¹⁹ = 2.75×10⁻¹⁹ J. Stopping potential V_s = K_max / e = 1
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