📚 A-Level AQA Physics: Worked Example Problems | A-Level AQA 物理：典型例题详解
Welcome to this focused revision guide covering typical worked examples for AQA A Level Physics. Each problem has been selected to target key skills from the specification, including mechanics, fields, circuits, waves and modern physics. Detailed step-by-step solutions are provided, with each solution step explained in both English and Chinese to help you master the logic and calculation methods required in the exam.
欢迎阅读这份针对 AQA A Level 物理的典型例题详解精讲。每道题都选自考纲核心内容，涵盖力学、场、电路、波动和近代物理。解答逐步展开，每个步骤均提供中英双语解析，帮助你掌握考试所需的逻辑与计算方法。
1. Projectile Motion: Range and Time of Flight | 抛体运动：射程与飞行时间
Problem: A ball is projected from ground level with a speed of 20 m s⁻¹ at an angle of 30° to the horizontal. Assume g = 9.81 m s⁻². Calculate the time of flight and the horizontal range.
题目：一球以 20 m s⁻¹ 的初速度从地面与水平面成 30° 角抛出。取 g = 9.81 m s⁻²，求飞行时间和水平射程。
Step 1: Resolve the initial velocity into horizontal and vertical components. u_x = 20 cos30° ≈ 17.32 m s⁻¹, u_y = 20 sin30° = 10.0 m s⁻¹.
步骤1：将初速度分解为水平和竖直分量。u_x = 20 cos30° ≈ 17.32 m s⁻¹，u_y = 20 sin30° = 10.0 m s⁻¹。
Step 2: Use the vertical motion equation s = u_y t − ½gt². When the ball returns to the ground, vertical displacement s = 0. Thus 0 = 10t − ½ × 9.81 × t². Factorising gives t(10 − 4.905t) = 0, so non-zero time of flight t = 10 / 4.905 ≈ 2.04 s.
步骤2：利用竖直运动方程 s = u_y t − ½gt²。球落回地面时竖直位移 s = 0，得 0 = 10t − ½ × 9.81 × t²。因式分解得 t(10 − 4.905t) = 0，故非零解飞行时间 t = 10 / 4.905 ≈ 2.04 s。
Step 3: Horizontal range = u_x × time of flight = 17.32 × 2.04 ≈ 35.3 m.
步骤3：水平射程 = u_x × 飞行时间 = 17.32 × 2.04 ≈ 35.3 m。
2. Newton’s Laws: Connected Particles | 牛顿定律：连接体问题
Problem: Two blocks, A (5.0 kg) and B (3.0 kg), are connected by a light inextensible string over a smooth pulley. Block A rests on a smooth horizontal table, while block B hangs freely. Find the acceleration of the system and the tension in the string. (g = 9.81 m s⁻²)
题目：两物块 A（5.0 kg）和 B（3.0 kg）用轻质不可伸长的细绳跨过光滑滑轮连接。A 置于光滑水平桌面，B 自由悬挂。求系统加速度和绳中张力。（g = 9.81 m s⁻²）
Step 1: Draw free-body diagrams. For A on the table: the only horizontal force is tension T, so T = m_A a = 5a. For B hanging: weight acts downwards, tension upwards, so 3g − T = 3a.
步骤1：画受力图。桌面上的 A：仅受水平方向张力 T，故 T = m_A a = 5a。悬挂的 B：重力向下，张力向上，得 3g − T = 3a。
Step 2: Substitute T = 5a into the second equation: 3g − 5a = 3a → 3g = 8a → a = 3g / 8 = (3 × 9.81) / 8 ≈ 3.68 m s⁻².
步骤2：将 T = 5a 代入第二式：3g − 5a = 3a → 3g = 8a → a = 3g / 8 = (3 × 9.81) / 8 ≈ 3.68 m s⁻²。
Step 3: Calculate tension: T = 5a = 5 × 3.68 ≈ 18.4 N.
步骤3：计算张力：T = 5a = 5 × 3.68 ≈ 18.4 N。
3. Work, Energy and Power: Spring and Incline | 功、能量与功率：弹簧与斜面
Problem: A spring of stiffness k = 200 N m⁻¹ is compressed by 0.10 m and used to launch a 0.50 kg block up a smooth incline of 30°. Determine the maximum distance the block travels along the incline before momentarily stopping.
题目：一根劲度系数 k = 200 N m⁻¹ 的弹簧被压缩 0.10 m，用于将 0.50 kg 的物块沿光滑 30° 斜面向上发射。求物块在斜面上滑行的最大距离（瞬间停止前）。
Step 1: Elastic potential energy stored = ½kx² = ½ × 200 × (0.10)² = 1.0 J. This energy converts entirely into gravitational potential energy as the block rises (no friction).
步骤1：弹性势能 = ½kx² = ½ × 200 × (0.10)² = 1.0 J。该能量完全转化为物块上升的重力势能（无摩擦）。
Step 2: Gain in GPE = mgh, where h is the vertical height. h = s sin30°, with s being the distance along the slope. So 1.0 = 0.50 × 9.81 × s × sin30°.
步骤2：增加的重力势能 = mgh，h 为竖直高度。h = s sin30°，s 为沿斜面的距离。因此 1.0 = 0.50 × 9.81 × s × sin30°。
Step 3: Solve for s: s = 1.0 / (0.50 × 9.81 × 0.5) = 1.0 / 2.4525 ≈ 0.408 m.
步骤3：求解 s：s = 1.0 / (0.50 × 9.81 × 0.5) = 1.0 / 2.4525 ≈ 0.408 m。
4. Circular Motion: Bridge Problem | 圆周运动：拱桥问题
Problem: A car travels over a convex bridge of radius 50 m. At what speed will the car just lose contact with the road at the top of the bridge? (g = 9.81 m s⁻²)
题目：一辆汽车驶过半径 50 m 的凸形桥。车在桥顶刚好离开桥面的速度是多少？（g = 9.81 m s⁻²）
Step 1: At the point of losing contact, the normal reaction N = 0. The centripetal force is provided entirely by the weight: mg = mv²/r.
步骤1：在即将离开桥面的瞬间，支持力 N = 0。向心力全部由重力提供：mg = mv²/r。
Step 2: Cancel m and rearrange: v² = g r → v = √(g r) = √(9.81 × 50) ≈ √490.5 ≈ 22.1 m s⁻¹.
步骤2：约去 m 并整理：v² = g r → v = √(g r) = √(9.81 × 50) ≈ √490.5 ≈ 22.1 m s⁻¹。
Thus the speed must be about 22.1 m s⁻¹ for the car to feel weightless at the top.
因此，当车速约为 22.1 m s⁻¹ 时，在桥顶会感到失重。
5. Simple Harmonic Motion (SHM): Maximum Values | 简谐运动：最大值计算
Problem: A particle performs SHM with amplitude 0.050 m and period 2.0 s. Determine its maximum speed and maximum acceleration.
题目：一质点做振幅 0.050 m、周期 2.0 s 的简谐运动。求其最大速度和最大加速度。
Step 1: Calculate angular frequency ω = 2π / T = 2π / 2.0 = π ≈ 3.14 rad s⁻¹.
步骤1：计算角频率 ω = 2π / T = 2π / 2.0 = π ≈ 3.14 rad s⁻¹。
Step 2: Maximum speed v_max = ωA = π × 0.050 ≈ 0.157 m s⁻¹.
步骤2：最大速度 v_max = ωA = π × 0.050 ≈ 0.157 m s⁻¹。
Step 3: Maximum acceleration a_max = ω²A = π² × 0.050 ≈ 9.87 × 0.050 ≈ 0.494 m s⁻².
步骤3：最大加速度 a_max = ω²A = π² × 0.050 ≈ 9.87 × 0.050 ≈ 0.494 m s⁻²。
6. Electric Fields: Zero Field Point | 电场：电场零点位置
Problem: Two point charges, +2.0 μC and −3.0 μC, are placed 0.10 m apart in a vacuum. Find the position along the line joining them where the resultant electric field is zero.
题目：两点电荷 +2.0 μC 和 −3.0 μC 在真空中相距 0.10 m。求连线上合电场为零的位置。
Step 1: Zero field cannot lie between opposite charges because their fields point in the same direction there. The zero point must be on the side of the smaller magnitude charge, i.e., beyond the +2.0 μC charge. Let distance from +2.0 μC be x.
步骤1：异种电荷之间电场同向，不可能为零。零点必在较小电荷的外侧，即超出 +2.0 μC 的位置。设离 +2.0 μC 距离为 x。
Step 2: Magnitudes of fields must be equal: k × 2.0×10⁻⁶ / x² = k × 3.0×10⁻⁶ / (0.10 + x)². Cancel k and 10⁻⁶: 2/x² = 3/(0.10+x)².
步骤2：电场大小相等：k × 2.0×10⁻⁶ / x² = k × 3.0×10⁻⁶ / (0.10 + x)²。消去 k 和 10⁻⁶ 得 2/x² = 3/(0.10+x)²。
Step 3: Cross-multiply and take square roots: √2 / x = √3 / (0.10+x) → (0.10+x) = x√(3/2) ≈ 1.225x → 0.10 = 0.225x → x ≈ 0.444 m. (Alternatively, solving gives x ≈ 0.178 m? Let’s carefully re-evaluate: Actually, 2(0.1+x)²=3x² → 2(0.01+0.2x+x²)=3x² → 0.02+0.4x+2x²=3x² → 0=x²−0.4x−0.02. Solve: x = [0.4 ± √(0.16+0.08)]/2 = [0.4 ± √0.24]/2 ≈ [0.4 ± 0.4899]/2. Positive root: (0.8899)/2 ≈ 0.445 m. Yes, approx 0.44 m. I’ll use 0.44 m for simplicity.)
步骤3：交叉相乘并开平方：(0.10+x) = x√(3/2) ≈ 1.225x → 0.10 = 0.225x → x ≈ 0.444 m。精确解二次方程得 x ≈ 0.44 m。
Therefore the field is zero at a distance of about 0.44 m from the +2.0 μC charge, on the side opposite the −3.0 μC charge.
因此电场为零的点在离 +2.0 μC 约 0.44 m 的外侧。
7. DC Circuits: Internal Resistance and Terminal p.d. | 直流电路：内阻与端电压
Problem: A battery of e.m.f. 12.0 V is connected to a 4.0 Ω external resistor. The terminal p.d. across the battery is measured as 10.0 V. Calculate the internal resistance of the battery and the short-circuit current.
题目：一电动势为 12.0 V 的电池连接 4.0 Ω 外电阻，测得电池端电压为 10.0 V。求电池内阻和短路电流。
Step 1: Current in the circuit I = V_R / R = 10.0 / 4.0 = 2.5 A.
步骤1：电路中的电流 I = V_R / R = 10.0 / 4.0 = 2.5 A。
Step 2: Lost volts across internal resistance = e.m.f. − terminal p.d. = 12.0 − 10.0 = 2.0 V. So internal resistance r = lost volts / I = 2.0 / 2.5 = 0.80 Ω.
步骤2：内阻上损失的电压 = 电动势 − 端电压 = 12.0 − 10.0 = 2.0 V。故内阻 r = 损失电压 / I = 2.0 / 2.5 = 0.80 Ω。
Step 3: Short-circuit current I_sc = e.m.f. / r = 12.0 / 0.80 = 15 A.
步骤3：短路电流 I_sc = 电动势 / r = 12.0 / 0.80 = 15 A。
8. Magnetic Fields: Force on a Current-Carrying Wire | 磁场：载流导线安培力
Problem: A straight wire of length 0.50 m carries a current of 3.0 A perpendicular to a uniform magnetic field of flux density 0.20 T. Calculate the magnetic force on the wire.
题目：一根长 0.50 m 的直导线通有 3.0 A 电流，与 0.20 T 的匀强磁场垂直。求导线所受的磁力。
Step 1: Use F = B I l sinθ. Since the wire is perpendicular to the field, θ = 90°, sinθ = 1. So F = B I l = 0.20 × 3.0 × 0.50 = 0.30 N.
步骤1：用公式 F = B I l sinθ。由于导线与磁场垂直，θ = 90°，sinθ = 1。故 F = 0.20 × 3.0 × 0.50 = 0.30 N。
Step 2: Determine direction using Fleming’s left-hand rule: The force is perpendicular to both current and field directions.
步骤2：用左手定则判断方向：力同时垂直于电流和磁场方向。
9. Electromagnetic Induction: Motional EMF | 电磁感应：动生电动势
Problem: A conducting rod of length 0.40 m moves at a constant velocity of 5.0 m s⁻¹ perpendicular to a uniform magnetic field of 0.35 T. What is the magnitude of the induced e.m.f. across the rod?
题目：一根长 0.40 m 的导体棒以 5.0 m s⁻¹ 的速度垂直于 0.35 T 的匀强磁场运动。求棒两端的感应电动势大小。
Step 1: For a moving rod cutting magnetic flux, induced e.m.f. ε = B l v (when v is perpendicular to B).
步骤1：对于切割磁力线的运动导体棒，感应电动势 ε = B l v（v 垂直于 B）。
Step 2: ε = 0.35 × 0.40 × 5.0 = 0.70 V.
步骤2：ε = 0.35 × 0.40 × 5.0 = 0.70 V。
10. Wave Superposition: Young’s Double-Slit Fringe Spacing | 波的叠加：杨氏双缝条纹间距
Problem: In a Young’s double-slit experiment, light of wavelength 600 nm illuminates two slits separated by 0.50 mm. A screen is placed 1.5 m from the slits. Calculate the fringe spacing Δy.
题目：在杨氏双缝实验中，波长为 600 nm 的光照射相距 0.50 mm 的双缝。屏幕距缝 1.5 m。求条纹间距 Δy。
Step 1: The formula for fringe separation is Δy = λD / d, where d is slit separation and D is screen distance.
步骤1：条纹间距公式为 Δy = λD / d，d 为缝间距，D 为到屏幕的距离。
Step 2: Convert all to metres: λ = 600 × 10⁻⁹ m, d = 0.50 × 10⁻³ m, D = 1.5 m. Then Δy = (600×10⁻⁹ × 1.5) / (0.50×10⁻³) = (9.0×10⁻⁷) / (5.0×10⁻⁴) = 1.8×10⁻³ m = 1.8 mm.
步骤2：单位化为米：λ = 600 × 10⁻⁹ m，d = 0.50 × 10⁻³ m，D = 1.5 m。计算 Δy = (600×10⁻⁹ × 1.5) / (0.50×10⁻³) = 1.8 mm。
11. Photoelectric Effect: Maximum Kinetic Energy | 光电效应：最大动能
Problem: Ultraviolet light of wavelength 200 nm is incident on a metal surface with a work function φ = 4.5 eV. Determine the maximum kinetic energy of emitted photoelectrons in eV and the stopping potential. (Use hc = 1240 eV nm)
题目：波长为 200 nm 的紫外光照射在逸出功 φ = 4.5 eV 的金属表面上。求发射光电子的最大动能（eV）和遏止电压。（取 hc = 1240 eV nm）
Step 1: Photon energy E = hc / λ = 1240 / 200 = 6.2 eV.
步骤1：光子能量 E = hc / λ = 1240 / 200 = 6.2 eV。
Step 2: Maximum kinetic energy K_max = E − φ = 6.2 − 4.5 = 1.7 eV.
步骤2：最大动能 K_max = E − φ = 6.2 − 4.5 = 1.7 eV。
Step 3: Stopping potential V_s = K_max / e = 1.7 V (since electron charge e). Thus a retarding potential of 1.7 V will stop the fastest electrons.
步骤3：遏止电压 V_s = K_max / e = 1.7 V。因此加上 1.7 V 的反向电压即可阻止最快的电子。
12. Nuclear Physics: Radioactive Decay Calculation | 核物理：放射性衰变计算
Problem: A radioactive source has an initial activity of 800 Bq and a half-life of 5.0 days. What will its activity be after 20 days?
题目：某放射源初始活度为 800 Bq，半衰期为 5.0 天。求 20 天后的活度。
Step 1: Number of half-lives n = total time / half-life = 20 / 5.0 = 4.
步骤1：半衰期个数 n = 总时间 / 半衰期 = 20 / 5.0 = 4。
Step 2: After each half-life, activity halves. Activity A = A₀ × (1/2)^n = 800 × (1/2)⁴ = 800 / 16 = 50 Bq.
步骤2：每经过一个半衰期活度减半。活度 A = A₀ × (1/2)^n = 800 × (1/2)⁴ = 800 / 16 = 50 Bq。
The activity drops to 50 Bq after 20 days, demonstrating exponential decay.
20 天后活度降至 50 Bq，体现了指数衰减规律。
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