📚 IB & CIE Physics: Worked Examples Explained | IB 与 CIE 物理：典型例题详解
Mastering physics at IB and CIE A-Level requires more than memorising formulas – you must learn to apply concepts to unfamiliar situations. This article presents ten typical worked examples spanning mechanics, fields, circuits, thermodynamics and waves, with step-by-step reasoning in both English and Chinese. Each example targets a core topic common to both syllabuses, helping you bridge the gap between theory and examination performance.
在 IB 和 CIE A-Level 物理中取得高分，不仅需要熟记公式，还必须学会将概念应用到陌生情境中。本文精选了十个典型例题，涵盖力学、场、电路、热力学和波动，用中英双语逐步解析。每个例题都针对两个大纲共同的核心知识点，帮助你弥合理论与应试之间的差距。
1. Kinematics – Projectile Motion | 运动学 – 抛体运动
and (c) the horizontal range. Assume air resistance is negligible and use g = 9.81 m s⁻².
题目：一个高尔夫球从地面以 28.0 m s⁻¹ 的初速度、与水平面成 35.0° 角击出。计算 (a) 飞行时间，(b) 最大高度，(c) 水平射程。忽略空气阻力，取 g = 9.81 m s⁻²。
28 s. (b) Maximum height occurs when vertical velocity becomes zero: h = vᵧ₀² / (2 g) = (16.06)² / (2 × 9.81) ≈ 13.14 m. (c) Horizontal range R = vₓ₀ × t = 22.94 × 3.28 ≈ 75.3 m. Note that the trajectory equation could also be used, but the component method is safer.
解答：将初速度分解为水平和竖直分量。水平分量 vₓ₀ = 28.0 cos 35.0° ≈ 22.94 m s⁻¹；竖直分量 vᵧ₀ = 28.0 sin 35.0° ≈ 16.06 m s⁻¹。(a) 返回同一水平面的飞行时间 t = 2 vᵧ₀ / g = 2 × 16.06 / 9.81 ≈ 3.28 s。(b) 最大高度出现在竖直速度为零时：h = vᵧ₀² / (2 g) = (16.06)² / (2 × 9.81) ≈ 13.14 m。(c) 水平射程 R = vₓ₀ × t = 22.94 × 3.28 ≈ 75.3 m。也可使用轨迹方程，但分量法更为稳妥。
2. Newton’s Laws – Connected Bodies and Friction | 牛顿定律 – 连接体与摩擦力
(a) the acceleration of the system and (b) the tension in the string. Take g = 9.8 m s⁻².
题目：两木块质量分别为 m₁ = 3.0 kg 和 m₂ = 2.0 kg，用轻质不可伸长的细绳跨过无摩擦滑轮连接。m₁ 放置在粗糙水平桌面上，动摩擦因数 μₖ = 0.25，m₂ 自由悬挂。系统由静止释放。求 (a) 系统的加速度，(b) 绳的张力。取 g = 9.8 m s⁻²。
left. For m₂: weight m₂ g downward, tension T upward. Apply Newton’s second law. For m₁: T – fₖ = m₁ a → T – 0.25 × 3.0 × 9.8 = 3.0 a → T – 7.35 = 3.0 a. For m₂: m₂ g – T = m₂ a → 2.0 × 9.8 – T = 2.0 a → 19.6 – T = 2.0 a. Add the two equations to eliminate T: (19.6 – 7.35) = (3.0 + 2.0) a → 12.25 = 5.0 a → a = 2.45 m s⁻². Substitute back: T = 19.6 – 2.0 × 2.45 = 19.6 – 4.9 = 14.7 N. The system accelerates at 2.45 m s⁻² and the string tension is 14.7 N.
解答：画受力图。对 m₁：拉力 T 向右，动摩擦力 fₖ = μₖ N = μₖ m₁ g 向左。对 m₂：重力 m₂ g 向下，拉力 T 向上。应用牛顿第二定律。对 m₁：T – fₖ = m₁ a → T – 0.25 × 3.0 × 9.8 = 3.0 a → T – 7.35 = 3.0 a。对 m₂：m₂ g – T = m₂ a → 2.0 × 9.8 – T = 2.0 a → 19.6 – T = 2.0 a。两式相加消去 T：(19.6 – 7.35) = (3.0 + 2.0) a → 12.25 = 5.0 a → a = 2.45 m s⁻²。回代得 T = 19.6 – 2.0 × 2.45 = 19.6 – 4.9 = 14.7 N。系统加速度为 2.45 m s⁻²，绳中张力为 14.7 N。
3. Work, Energy and Power – Conservation of Mechanical Energy | 功、能量与功率 – 机械能守恒
d for the car to just make it over a loop of radius 10 m (assuming point C is at the top of the loop). Use g = 9.8 m s⁻².
题目：一辆过山车质量为 500 kg，从离地 40 m 的 A 点静止释放。它沿无摩擦轨道下滑到地面的 B 点，然后上升到高度 25 m 的 C 点。计算 (a) 在 B 点的速率，(b) 在 C 点的速率，(c) 为使过山车刚好通过半径为 10 m 的圆环顶部（假设 C 点为圆环最高点），A 点的最小高度。取 g = 9.8 m s⁻²。
n of mechanical energy, loss in gravitational PE = gain in KE. From A to B: m g hₐ = ½ m vB² → vB = √2 g hₐ = √2 × 9.8 × 40 = √784 ≈ 28.0 m s⁻¹. (b) From A to C: mg(hₐ – hC) = ½ m vC² → vC = √(2 g (40 – 25)) = √2 × 9.8 × 15 of a loop of radius r = 10 m, at the top the centripetal force is provided by weight: m vtop² / r = m g → vtop = √g r = √9.8 × 10 = √98 0 m s⁻¹. Using energy conservation from A to the top of the loop (height 2r = 20 m): m g hmin = m g (2r) + ½ m vtop² → hmin = 2r + vtop²/(2g) = 20 + 98/(2×9.8) = 20 + 5 = 25 m. Thus the minimum release height is 25 m.
解答：(a) 利用机械能守恒，重力势能减少量等于动能增加量。从 A 到 B：m g hₐ = ½ m vB² → vB = √2 g hₐ = √2 × 9.8 × 40 = √784 ≈ 28.0 m s⁻¹。(b) 从 A 到 C：mg(hₐ – hC) = ½ m vC² → vC = √(2 g (40 – 25)) = √2 × 9.8 × 15 = √294 ≈ 17.1 m s⁻¹。(c) 过山车刚好通过半径为 10 m 的圆环顶部时，在最高点向心力由重力提供：m vtop² / r = m g → vtop = √g r = √9.8 × 10 ≈ 9.90 m s⁻¹。利用从 A 到圆环顶部（高 20 m）的能量守恒：m g hmin = m g (2r) + ½ m vtop² → hmin = 20 + (98)/(2×9.8) = 25 m。因此最小释放高度为 25 m。
4. Momentum and Impulse – Collisions | 动量与冲量 – 碰撞
Problem: A 0.50 kg ball moving at 6.0 m s⁻¹ overtakes a 0.80 kg ball moving at 2.0 m s⁻¹ in the same direction on a smooth surface. After a head-on elastic collision, find the final velocities of both balls.
题目：一个 0.50 kg 的小球以 6.0 m s⁻¹ 的速度追上前方同向以 2.0 m s⁻¹ 运动的 0.80 kg 小球，在光滑水平面上发生正面弹性碰撞。求碰撞后两球的末速度。
Solution: For a one-dimensional elastic collision, both momentum and kinetic energy are conserved. Let mA = 0.50 kg, uA = 6.0 m s⁻¹; mB = 0.80 kg, uB = 2.0 m s⁻¹. The relative velocity of approach equals relative velocity of separation: uA – uB = vB – vA. So 6.0 – 2.0 = 4.0 = vB – vA. Momentum conservation: mA uA + mB uB = mA vA + mB vB → 0.50×6.0 + 0.80×2.0 = 0.50 vA + 0.80 vB → 3.0 + 1.6 = 4.6 = 0.50 vA + 0.80 vB. Multiply by 10: 46 = 5 vA + 8 vB. Substitute vB = vA + 4.0: 46 = 5 vA + 8(vA + 4) = 5v_A + 8v_A + 32 = 13 vA + 32 → 14 = 13 vA → vA ≈ 1.08 m s⁻¹. Then vB . The 0.50 kg ball slows down to 1.08 m s⁻¹, and the 0.80 kg ball speeds up to 5.08 m s⁻¹.
解答：一维弹性碰撞同时遵守动量守恒和动能守恒。设 mA = 0.50 kg, uA = 6.0 m s⁻¹；mB = 0.80 kg, uB = 2.0 m s⁻¹。相对接近速率等于相对分离速率：uA – uB = vB – vA，即 6.0 – 2.0 = 4.0 = vB – vA。动量守恒：mA uA + mB uB = mA vA + mB vB → 3.0 + 1.6 = 4.6 = 0.50 vA + 0.80 vB，两边乘 10 得 46 = 5 vA + 8 vB。代入 vB = vA + 4.0：46 = 5 vA + 8(vA + 4) = 13 vA + 32 → 14 = 13 vA → vA ≈ 1.08 m s⁻¹，vB ≈ 5.08 m s⁻¹。0.50 kg 球减速至 1.08 m s⁻¹，0.80 kg 球加速至 5.08 m s⁻¹。
5. Circular Motion and Gravitation | 圆周运动与万有引力
.8 m s⁻². Determine (a) the orbital radius, (b) the satellite’s orbital speed, and (c) the period of revolution.
题目：一颗人造卫星在某高度绕地球做圆周运动，该处的重力加速度为 4.9 m s⁻²。地球半径 R = 6.37 × 10⁶ m，地表重力加速度 g = 9.8 m s⁻²。求 (a) 轨道半径，(b) 卫星的轨道速率，(c) 公转周期。
Solution: (a) Gravitational acceleration at distance r is given by g' = g (R / r)². Therefore 4.9 = 9.8 (6.37×10⁶ / r)² → (6.37×10⁶ / r)² = 0.5 → r = 6.37×10⁶ / √0 r motion, centripetal acceleration equals the local gravitational acceleration: v² / r = g' → v = √r g' = √9.01×10⁶ × 4.9 ≈ √4.415×10⁷ / v = 2π × 9.01×10⁶ / 6.64×10³ ≈ (5.66×10⁷) / 6.64×10³ ≈ 8.52×10³ s, or about 142 minutes.
解答：(a) 距离 r 处的重力加速度为 g' = g (R / r)²。由 4.9 = 9.8 (6.37×10⁶ / r)² 得 (6.37×10⁶ / r)² = 0.5 → r ≈ 9.01×10⁶ m。(b) 圆周运动中向心加速度等于当地重力加速度：v² / r = g' → v = √r g' = √9.01×10⁶ × 4.9 ≈ 6.64×10³ m s⁻¹。(c) 周期 T = 2π r / v ≈ 8.52×10³ s，约 142 分钟。
6. Electric Fields and Potential | 电场与电势
Problem: Two point charges q₁ = +2.0 μC and q₂ = – b) Calculate the electric potential at that midpoint. Take 1/(4π ε₀) = 9.0 × 10⁹ N m² C⁻².
题目：两个点电荷 q₁ = +2.0 μC 和 q₂ = –3.0 μC 在真空中相距 0.40 m。(a) 求它们连线中点处的电场（大小和方向）。(b) 计算该中点处的电势。取 1/(4π ε₀) = 9.0 × 10⁹ N m² C⁻²。
ution: Midpoint distance r = 0.20 m from each charge. (a) Electric field due to q₁: E₁ = k |q₁| ×10⁵ N C⁻¹, pointing away from the positive charge (to the right). Field due to q₂: E₂ = k |q₂| / r² = 9.0×10⁹ × 3.0×10⁻⁶ / 0.04 = 2.7×10⁴ / 0.04? d the negative charge. (b) Electric potential V = k q₁ / r + k q₂ / r = (9.0×10⁹ / 0.20) × (2.0×10⁻⁶ – 3.0×10⁻⁶) = 4.5×10¹⁰ × (–1.0×10⁻⁶) = –4.5×10⁴ V.
解答：中点到每个电荷的距离 r = 0.20 m。(a) q₁ 产生的电场：E₁ = k |q₁| / r² = 4.5×10⁵ N C⁻¹，方向背离正电荷（向右）。q₂ 产生的电场：E₂ = 6.75×10⁵ N C⁻¹，方向指向负电荷（也是向右，因为负电荷电场指向它）。因此合电场 E = 1.125×10⁶ N C⁻¹，方向指向负电荷。(b) 电势 V = k(q₁ + q₂) / r = 4.5×10¹⁰ × (–1.0×10⁻⁶) = –4.5×10⁴ V。
7. DC Circuits – Kirchhoff’s Laws | 直流电路 – 基尔霍夫定律
with the parallel combination of R₂ and R₃. Calculate (a) the total current drawn from the battery, (b) the current through each resistor, and (c) the power dissipated in R₂.
题目：如图所示电路，内阻可忽略的 12 V 电池连接电阻 R₁ = 4 Ω, R₂ = 6 Ω, R₃ = 3 Ω。R₁ 与 R₂ 和 R₃ 的并联组合串联。计算 (a) 电池输出的总电流，(b) 流过每个电阻的电流，(c) R₂ 消耗的功率。
Solution: (a) First find the equivalent resistance of the parallel branch: 1/Rpar = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2 → Rpar = 2 Ω. Total resistance Rtotal = R₁ + Rpar = 4 + 2 = 6 Ω. Total current Itotal = V / Rtotal nt passes entirely through R₁, so I₁ = 2.0 A. The voltage across the parallel section is Vpar = Itotal × Rpar = 2 × 2 = 4 V. Then I₂ = Vpar / R₂ = 4 / 6 ≈ 0.667 A; I₃ = Vpar 0.667 + 1.333 = 2.0 A. (c) Power in R₂ = I₂² R₂ = (0.667)² × 6 ≈ 2.67 W, or equivalently Vpar × I₂ = 4 × 0.667 ≈ 2.67 W.
解答：(a) 先求并联支路的等效电阻：1/Rpar = 1/6 + 1/3 = 1/2 → Rpar = 2 Ω。总电阻 Rtotal = 4 + 2 = 6 Ω。总电流 Itotal = 12 / 6 = 2.0 A。(b) 此电流全部流过 R₁，故 I₁ = 2.0 A。并联部分两端电压 Vpar = 2 × 2 = 4 V。于是 I₂ = 4 / 6 ≈ 0.667 A；I₃ = 4 / 3 ≈ 1.333 A。验证：0.667 + 1.333 = 2.0 A。(c) R₂ 的功率 P = I₂² R₂ = (2/3)² × 6 = 4/9 × 6 ≈ 2.67 W。
8. Electromagnetic Induction – Faraday’s Law | 电磁感应 – 法拉第定律
Problem: A rectangular coil of 200 turns, length 0.15 m and width 0.10 m, is rotated at 50 revolutions per second in a uniform magnetic field of 0.80 T. The axis of rotation is perpendicular to the field. Find (a) the maximum emf induced, and (b) the rms emf.
题目：一个 200 匝的矩形线圈，长 0.15 m、宽 0.10 m，在磁感应强度为 0.80 T 的匀强磁场中以每秒 50 转的速率转动。转动轴与磁场垂直。求 (a) 最大感应电动势，(b) 有效值电动势。
0π rad s⁻¹. The magnetic flux linkage Φ = N B A cos(ωt). By Faraday’s law, induced emf ε = – 100π = 240π ≈ 754 V. (a) Maximum emf = 754 V. (b) For a sinusoidal output, rms emf = ε₀ / √2 ≈ 754 / 1.414 = 533 V.
解答：线圈面积 A = 0.15 × 0.10 = 0.015 m²。角频率 ω = 2π × 50 = 100π rad s⁻¹。磁链 Φ = N B A cos(ωt)。根据法拉第定律，ε = N B A ω sin(ωt)。峰值 ε₀ = N B A ω = 200 × 0.80 × 0.015 × 100π = 240π ≈ 754 V。(a) 最大电动势 ≈ 754 V。(b) 对于正弦输出，有效值 ε_rms = ε₀ / √2 ≈ 533 V。
9. Thermal Physics – Ideal Gas and Kinetic Theory | 热物理 – 理想气体与分子运动论
c) the change in internal energy. Assume the molar gas constant R = 8.31 J mol⁻¹ K⁻¹ and CV = (3/2)R.
题目：一个密封气缸装有 0.25 mol 理想气体，初始压强 1.0 × 10⁵ Pa，温度 300 K。在体积不变的情况下将气体加热至 450 K。求 (a) 最终压强，(b) 气体对外做的功，(c) 内能变化。已知气体常数 R = 8.31 J mol⁻¹ K⁻¹，CV = (3/2)R。
constant, no work is done on or by the gas: W = 0. (c) Change in internal energy ΔU = n CV ΔT = 0.25 × (3/2 × 8.31) × (450 – 300) = 0.25 × 12.465 × 150 = 0.25 × 1869.75 = 467.4 J (approximately 467 J).
解答：(a) 体积不变时 P₁/T₁ = P₂/T₂ → P₂ = 1.0×10⁵ × (450/300) = 1.5×10⁵ Pa。(b) 体积不变，气体不做功：W = 0。(c) 内能变化 ΔU = n CV ΔT = 0.25 × (3/2 × 8.31) × 150 ≈ 467 J。
10. Wave Phenomena – Young’s Double-Slit | 波动现象 – 杨氏双缝干涉
Problem: In a Young’s double-slit experiment, light of wavelength 589 nm illuminates two slits separated by 0.50 mm. The interference pattern is observed on a screen 2.0 m away. Calculate (a) the fringe separation, (b) the angular position of the third-order bright fringe, and (c) the distance from the central maximum to the second dark fringe.
题目：在杨氏双缝干涉实验中，波长 589 nm 的光照射相距 0.50 mm 的双缝。观察屏距离 2.0 m。计算 (a) 条纹间距，(b) 第三级亮纹的角位置，(c) 中央明纹到第二暗纹的距离。
ll angle so θ ≈ sin θ in radians, θ = 3.534×10⁻³ rad). (c) For dark fringes, d sin θ = (m + ½) λ for m = 0, 1, 2,... Second dark fringe corresponds to m = 1, so d sin θ = (1.5) × 589×10⁻⁹ = 8.835×10⁻⁷ → sin θ = 1.767×10⁻³. Using small-angle approximation, position on screen y = D tan θ ≈ D sin θ = 2.0 × 1.767×10⁻³ = 3.534×10⁻³ m = 3.53 mm.
解答：(a) 条纹间距 Δy = λ D / d = (589×10⁻⁹ × 2.0) / 5.0×10⁻⁴ = 2.36 mm。(b) 对于第 m 级亮纹，d sin θ = m λ。m = 3 时 sin θ ≈ 3.534×10⁻³，θ ≈ 0.202°。(c) 暗纹条件 d
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