📚 IB AQA Physics: Typical Example Problems Explained | IB AQA 物理：典型例题详解
Physics at the IB and AQA level challenges students to apply fundamental principles to a variety of real-world and theoretical situations. Both curricula emphasise problem-solving, mathematical reasoning and conceptual clarity. In this article, we work through carefully selected example problems that mirror the style and difficulty of IB and AQA physics examinations. Each problem is broken down step by step, with explanations in both English and Chinese, to help you build confidence and exam technique. From kinematics and forces to electricity and quantum phenomena, these examples cover the core topics that appear year after year.
IB 与 AQA 物理课程要求学生将基本原理运用于各种现实和理论场景，两个体系都强调解题能力、数学推理和概念清晰。本文精选了贴近 IB 和 AQA 物理考试风格与难度的典型例题，逐一详解。每道题都分步呈现，提供中英双语的解答说明，帮助你建立信心并掌握应试方法。从运动学、力到电学、量子现象，这些案例覆盖了年年必考的核心主题。
1. Kinematics with Constant Acceleration | 匀加速运动学
Problem: A car accelerates uniformly from rest at 3.0 m/s² for 8.0 s. It then maintains constant velocity for 12.0 s before decelerating uniformly to rest in 5.0 s. Find the total distance travelled.
题目：一辆汽车从静止开始以 3.0 m/s² 匀加速 8.0 s，然后匀速行驶 12.0 s，最后匀减速至静止，用时 5.0 s。求总位移。
Solution: We split the motion into three phases. Phase 1: using s = ut + ½at² with u=0, a=3.0, t=8.0 gives s₁ = 0 + ½×3.0×8.0² = 96 m. Final speed after phase 1 is v = u + at = 0 + 3.0×8.0 = 24 m/s. Phase 2: constant speed 24 m/s for 12.0 s, s₂ = v×t = 24×12.0 = 288 m. Phase 3: deceleration from 24 m/s to 0 in 5.0 s, acceleration a = (0−24)/5.0 = −4.8 m/s². Distance s₃ = ut + ½at² = 24×5.0 + ½×(−4.8)×5.0² = 120 − 60 = 60 m. Total distance = 96 + 288 + 60 = 444 m.
解析：将运动分为三个阶段。阶段一：由 s = ut + ½at²，u=0，a=3.0，t=8.0，得 s₁ = ½×3.0×8.0² = 96 m。末速度 v = u + at = 24 m/s。阶段二：匀速 24 m/s 持续 12.0 s，s₂ = 24×12.0 = 288 m。阶段三：从 24 m/s 减速到 0 用时 5.0 s，加速度 a = (0−24)/5.0 = −4.8 m/s²。位移 s₃ = 24×5.0 + ½×(−4.8)×5.0² = 60 m。总位移 = 444 m。
2. Projectile Motion | 抛体运动
Problem: A ball is kicked from ground level with an initial speed of 20 m/s at an angle of 30° above the horizontal. Air resistance is negligible. Calculate the time of flight, maximum height and horizontal range. (g = 9.8 m/s²)
题目：一球以 20 m/s 的初速率、与水平方向成 30° 发射，忽略空气阻力。求飞行时间、最大高度和水平射程。(g = 9.8 m/s²)
Solution: Resolve initial velocity: vₓ = 20 cos30° = 17.32 m/s, vₙ = 20 sin30° = 10 m/s. Time of flight: total time t = 2vₙ/g = 2×10/9.8 = 2.04 s. Maximum height: H = vₙ²/(2g) = 10²/(2×9.8) = 5.10 m. Range: R = vₓ × t = 17.32×2.04 = 35.3 m. Alternatively, R = (v² sin2θ)/g = (400×sin60°)/9.8 = 35.3 m.
解析：分解初速度：水平 vₓ = 20 cos30° = 17.32 m/s，竖直 vₙ = 20 sin30° = 10 m/s。飞行时间 t = 2vₙ/g = 2.04 s。最大高度 H = vₙ²/(2g) = 5.10 m。水平射程 R = vₓ×t = 35.3 m，亦可用公式 R = (v² sin2θ)/g 验证。
3. Newton’s Laws and Connected Bodies | 牛顿定律与连接体
Problem: Two blocks of masses 4.0 kg and 2.0 kg are connected by a light string over a frictionless pulley. The 4.0 kg mass rests on a smooth horizontal table, while the 2.0 kg mass hangs vertically. Find the acceleration of the system and the tension in the string. (g = 9.8 m/s²)
题目：两个质量分别为 4.0 kg 和 2.0 kg 的物块用轻绳跨过无摩擦定滑轮连接，4.0 kg 块置于光滑水平桌面，2.0 kg 块竖直悬挂。求系统的加速度和绳中张力。(g = 9.8 m/s²)
Solution: Apply Newton’s second law to each mass. For hanging mass m₂ = 2.0 kg: m₂g − T = m₂a. For table mass m₁ = 4.0 kg: T = m₁a. Adding equations: m₂g = (m₁+m₂)a => a = (2.0×9.8)/(4.0+2.0) = 19.6/6.0 = 3.27 m/s². Tension T = m₁a = 4.0×3.27 = 13.1 N.
解析：对悬挂物 m₂：m₂g − T = m₂a。对桌面物 m₁：T = m₁a。两式相加得 m₂g = (m₁+m₂)a，a = 3.27 m/s²。张力 T = 4.0×3.27 = 13.1 N。
4. Work, Energy and Power | 功、能与功率
Problem: A 1500 kg car accelerates from 10 m/s to 25 m/s on a level road. If the average driving force is 6000 N and frictional force is 500 N, calculate the work done by the engine and the distance over which this occurs. Also find the average power output.
题目：一辆 1500 kg 的汽车在水平路面由 10 m/s 加速到 25 m/s，平均驱动力 6000 N，摩擦力 500 N。计算发动机做功、加速距离以及平均输出功率。
Solution: Net force F_net = 6000 − 500 = 5500 N. Work-energy theorem: W_net = ΔKE = ½m(v² − u²) = ½×1500×(25² − 10²) = 750×(625 − 100) = 393,750 J. This net work is also F_net × s, so s = W_net / F_net = 393,750 / 5500 = 71.6 m. Work done by engine = 6000×71.6 = 429,600 J. Average power = engine work / time. Find time using a = F_net/m = 5500/1500 = 3.667 m/s², t = (v−u)/a = 15/3.667 = 4.09 s. Average power = 429,600/4.09 = 105,000 W ≈ 105 kW.
解析：合力 F_net = 6000 − 500 = 5500 N。由动能定理：W_net = ΔKE = ½×1500×(25² − 10²) = 393,750 J。又 W_net = F_net × s，得 s = 71.6 m。发动机做功 = 6000×71.6 = 429,600 J。加速度 a = F_net/m = 3.667 m/s²，时间 t = (25−10)/3.667 = 4.09 s。平均功率 = 429,600/4.09 ≈ 105 kW。
5. Momentum and Collisions | 动量与碰撞
Problem: A 0.200 kg clay ball moving at 15 m/s strikes a stationary 0.800 kg block on a frictionless surface and sticks to it. Find the final velocity of the combined object and the kinetic energy lost in the collision.
题目：一个 0.200 kg 的粘土球以 15 m/s 的速度撞击静止在光滑表面上的 0.800 kg 物块并粘在一起。求复合体的末速度和碰撞中损失的动能。
Solution: Momentum conservation: m₁u₁ + m₂u₂ = (m₁+m₂)v. Here u₂=0. v = (0.200×15)/(0.200+0.800) = 3.0/1.0 = 3.0 m/s. Initial KE = ½×0.200×15² = 22.5 J. Final KE = ½×1.00×3.0² = 4.5 J. Loss = 22.5 − 4.5 = 18 J.
解析：动量守恒：(0.200)×15 = (0.200+0.800)×v，v = 3.0 m/s。初动能 = ½×0.200×15² = 22.5 J。末动能 = ½×1.00×3.0² = 4.5 J。损失动能 18 J，转化为内能。
6. Circular Motion | 圆周运动
Problem: A 0.500 kg mass is attached to a string of length 1.20 m and swung in a horizontal circle at 2.0 revolutions per second. Calculate the angular velocity, linear speed, centripetal acceleration and tension in the string.
题目：0.500 kg 的小球系于长 1.20 m 的绳，在水平面内每秒转 2.0 圈。求角速度、线速度、向心加速度和绳的张力。
Solution: Frequency f = 2.0 Hz. Angular velocity ω = 2πf = 2π×2.0 = 12.57 rad/s. Linear speed v = ωr = 12.57×1.20 = 15.1 m/s. Centripetal acceleration a = v²/r = (15.1)²/1.20 = 190 m/s² (or ω²r). Tension provides centripetal force: T = m a = 0.500×190 = 95.0 N.
解析：频率 f = 2.0 Hz。角速度 ω = 2πf = 12.57 rad/s。线速度 v = ωr = 15.1 m/s。向心加速度 a = v²/r = 190 m/s²。张力 T = m a = 95.0 N。
7. Simple Harmonic Motion | 简谐运动
Problem: A 0.200 kg mass oscillates on a spring with spring constant 50 N/m. The amplitude is 0.080 m. Determine the angular frequency, maximum speed, maximum acceleration, and total energy of the system. Ignore damping.
题目：0.200 kg 的物块在劲度系数 50 N/m 的弹簧上振动，振幅 0.080 m。求角频率、最大速度、最大加速度和系统总能量（忽略阻尼）。
Solution: Angular frequency ω = √(k/m) = √(50/0.200) = √250 = 15.81 rad/s. v_max = ωA = 15.81×0.080 = 1.265 m/s. a_max = ω²A = 250×0.080 = 20.0 m/s². Total energy E = ½kA² = ½×50×(0.080)² = 0.16 J.
解析：角频率 ω = √(k/m) = 15.81 rad/s。v_max = ωA = 1.265 m/s。a_max = ω²A = 20.0 m/s²。总能量 E = ½kA² = 0.16 J。
8. Electric Fields and Potential | 电场与电势
Problem: Two point charges, +4.0 μC and +9.0 μC, are separated by 0.50 m. Determine the position along the line joining them where the net electric field is zero. Also calculate the electric potential at that point. (k = 8.99×10⁹ N·m²/C²)
题目：两点电荷 +4.0 μC 和 +9.0 μC 相距 0.50 m。求连线上合场强为零的位置，并计算该点电势。(k = 8.99×10⁹ N·m²/C²)
Solution: Let distance from 4.0 μC charge to zero-field point be x m. Then distance to 9.0 μC is (0.50 − x). Field magnitudes equal: k×4.0/x² = k×9.0/(0.50−x)² → 4/x² = 9/(0.5−x)² → taking square root: 2/x = 3/(0.5−x) → 2(0.5−x) = 3x → 1 − 2x = 3x → 5x = 1 → x = 0.20 m. So position is 0.20 m from the 4.0 μC charge. Potential V = k×4.0×10⁻⁶/0.20 + k×9.0×10⁻⁶/0.30 = 8.99×10⁹×(20×10⁻⁶ + 30×10⁻⁶) = 8.99×10⁹×50×10⁻⁶ = 4.495×10⁵ V ≈ 4.5×10⁵ V.
解析：设离 4.0 μC 距离为 x，则离 9.0 μC 为 (0.50−x)。场强相等：k×4.0/x² = k×9.0/(0.50−x)²，开方得 2/x = 3/(0.5−x)，解得 x = 0.20 m。该点电势 V = k(4.0×10⁻⁶/0.20 + 9.0×10⁻⁶/0.30) ≈ 4.5×10⁵ V。
9. Circuit Analysis | 电路分析
Problem: A 12.0 V battery with internal resistance 0.50 Ω is connected to a parallel combination of a 6.0 Ω and a 3.0 Ω resistor. Find the terminal voltage of the battery and the current through each resistor.
题目：12.0 V 电池内阻 0.50 Ω，连接到一个 6.0 Ω 与 3.0 Ω 的并联组合上。求路端电压及各电阻电流。
Solution: Equivalent resistance of parallel: 1/R_parallel = 1/6 + 1/3 = 0.5 → R_parallel = 2.0 Ω. Total circuit resistance R_total = internal r + R_parallel = 0.50 + 2.0 = 2.50 Ω. Total current I = emf / R_total = 12.0/2.5 = 4.8 A. Terminal voltage V = emf − I×r = 12.0 − 4.8×0.50 = 12.0 − 2.4 = 9.6 V. Voltage across parallel branch is 9.6 V. I₆ = 9.6/6.0 = 1.6 A; I₃ = 9.6/3.0 = 3.2 A (sum = 4.8 A).
解析：并联等效电阻 R_parallel = (1/6 + 1/3)⁻¹ = 2.0 Ω。总电阻 = 0.50 + 2.0 = 2.50 Ω。总电流 I = 12.0/2.5 = 4.8 A。端电压 V = 12.0 − 4.8×0.50 = 9.6 V。各电流 I₆ = 9.6/6.0 = 1.6 A, I₃ = 9.6/3.0 = 3.2 A。
10. Electromagnetic Induction | 电磁感应
Problem: A rectangular coil of 200 turns, dimensions 0.10 m by 0.15 m, is rotated in a uniform magnetic field of 0.80 T. The axis of rotation is perpendicular to the field and the coil rotates at 50 revolutions per second. Find the maximum emf induced in the coil.
题目：一个 200 匝的矩形线圈，面积 0.10 m × 0.15 m，在 0.80 T 的匀强磁场中旋转，转轴与磁场垂直，转速 50 rps。求最大感应电动势。
Solution: Area A = 0.10×0.15 = 0.015 m². Angular velocity ω = 2πf = 2π×50 = 100π ≈ 314.16 rad/s. Maximum emf ε_max = NBAω = 200 × 0.80 × 0.015 × 314.16 = 200 × 0.80 × 0.015 × 314.16 = 754 V (approx). Precisely: ε_max = 200×0.80×0.015×100π = 2.4×100π = 240π ≈ 754 V.
解析：线圈面积 A = 0.015 m²，角速度 ω = 2π×50 = 100π rad/s。最大电动势 ε_max = NBAω = 200×0.80×0.015×100π = 240π ≈ 754 V。
11. Quantum Physics – Photoelectric Effect | 量子物理 – 光电效应
Problem: Zinc has a work function of 4.3 eV. Ultraviolet light of wavelength 200 nm is incident on a zinc plate. Calculate the maximum kinetic energy of emitted electrons in joules and electronvolts. (h = 6.63×10⁻³⁴ J·s, c = 3.0×10⁸ m/s, 1 eV = 1.60×10⁻¹⁹ J)
题目：锌的逸出功为 4.3 eV。紫外线波长 200 nm 照射锌板。计算发射电子的最大动能，以焦耳和电子伏特表示。(h = 6.63×10⁻³⁴ J·s, c = 3.0×10⁸ m/s, 1 eV = 1.60×10⁻¹⁹ J)
Solution: Photon energy E = 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. In eV: 9.945×10⁻¹⁹ J / 1.60×10⁻¹⁹ J/eV = 6.22 eV. Maximum KE = E − φ = 6.22 − 4.3 = 1.92 eV. In joules: 1.92 × 1.60×10⁻¹⁹ = 3.07×10⁻¹⁹ J.
解析：光子能量 E = hc/λ = 9.945×10⁻¹⁹ J = 6.22 eV。最大动能 KE_max = 6.22 − 4.3 = 1.92 eV，合 3.07×10⁻¹⁹ J。
12. Nuclear Physics – Decay and Binding Energy | 核物理 – 衰变与结合能
Problem: The isotope carbon-14 decays via beta-minus emission with a half-life of 5730 years. A sample initially contains 2.0×10¹⁰ nuclei. Calculate the activity after 10,000 years. Also find the time required for 75% of the nuclei to decay. (Use λ = ln2 / T₁/₂)
题目：碳-14 通过 β⁻ 衰变，半衰期 5730 年。样品初始含 2.0×10¹⁰ 个核。求 10,000 年后的活度。并求衰变掉 75% 所需时间。(λ = ln2 / T₁/₂)
Solution: Decay constant λ = ln2 / 5730 y⁻¹ = 0.693/5730 ≈ 1.21×10⁻⁴ y⁻¹. After 10000 y, N = N₀ e^(−λt) = 2.0×10¹⁰ × e^(−1.21×10⁻⁴×10000) = 2.0×10¹⁰ × e^(−1.21) = 2.0×10¹⁰ × 0.298 = 5.96×10⁹. Activity A = λN = 1.21×10⁻⁴ × 5.96×10⁹ = 7.21×10⁵ Bq. For 75% decay, remaining fraction = 0.25 = e^(−λt) → t = ln(4)/λ = 1.386/1.21×10⁻⁴ = 11,450 y (which is two half-lives).
解析：衰变常数 λ = ln2/5730 = 1.21×10⁻⁴ y⁻¹。10000 年后剩余核数 N = 2.0×10¹⁰ e^(−1.21) = 5.96×10⁹。活度 A = λN = 7.21×10⁵ Bq。衰变 75% 即剩余 25%，由 e^(−λt) = 0.25 得 t = ln4/λ ≈ 11,450 年（两倍半衰期）。
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