A-Level Edexcel Physics: Detailed Solutions to Typical Problems | A-Level Edexcel 物理：典型例题详解

📚 A-Level Edexcel Physics: Detailed Solutions to Typical Problems | A-Level Edexcel 物理：典型例题详解

Mastering A-Level Edexcel Physics requires more than memorising formulas; it demands the ability to apply concepts to unfamiliar situations. This article walks you through ten carefully selected typical problems, each broken down into clear, logical steps. You will see how to identify key quantities, choose the right equations, and carry out calculations just as you would in an exam. Every section pairs an English explanation with a Chinese translation so that bilingual learners can reinforce their understanding in both languages.

掌握 A-Level Edexcel 物理不仅需要记住公式，更需要将概念应用到未知情境中。本文带你逐一攻克十道精心挑选的典型例题，每一步都拆解成清晰的逻辑环节。你将学会如何识别关键量、选取正确方程并完成计算，如同在真正的考试中一样。每个部分都采用中英对照讲解，帮助双语学习者在两种语言中同步巩固理解。

1. Uniform Acceleration: Applying SUVAT Equations | 匀加速运动：应用SUVAT方程

A car accelerates uniformly from rest at 2.0 m s⁻² for 5.0 s. Calculate the distance it travels and its final velocity.

一辆汽车从静止开始以 2.0 m s⁻² 的加速度匀加速运动了 5.0 s。求汽车行驶的距离和末速度。

First, extract the known quantities and assign standard symbols: initial velocity u = 0, acceleration a = 2.0 m s⁻², time t = 5.0 s. The unknown variables are displacement s and final velocity v. We choose SUVAT equations that contain only one unknown at a time.

首先，提取已知量并赋予标准符号：初速度 u = 0，加速度 a = 2.0 m s⁻²，时间 t = 5.0 s。未知量是位移 s 和末速度 v。我们选择每次只含一个未知量的SUVAT方程。

s = ut + ½at²

Substituting: s = 0 × 5.0 + ½ × 2.0 × (5.0)² = 0 + ½ × 2.0 × 25 = 25 m.

代入计算：s = 0 × 5.0 + ½ × 2.0 × (5.0)² = 0 + ½ × 2.0 × 25 = 25 m。

v = u + at

Substituting: v = 0 + 2.0 × 5.0 = 10.0 m s⁻¹. Hence the car travels 25 m and reaches 10 m s⁻¹.

代入计算：v = 0 + 2.0 × 5.0 = 10.0 m s⁻¹。因此汽车行驶了25 m，末速度为10 m s⁻¹。

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2. Projectile Motion: Maximum Height and Range | 抛体运动：最大高度与射程

A ball is projected from level ground with an initial speed of 20 m s⁻¹ at an angle of 30° above the horizontal. Determine the maximum height reached and the total horizontal range. Take g = 9.8 m s⁻².

一球从水平地面以初速度 20 m s⁻¹、仰角 30° 抛出。求到达的最大高度和水平射程。取 g = 9.8 m s⁻²。

Resolve the initial velocity: vertical component u_y = 20 sin30° = 20 × 0.5 = 10 m s⁻¹; horizontal component u_x = 20 cos30° ≈ 17.3 m s⁻¹. At the highest point the vertical velocity becomes zero.

分解初速度：竖直分量 u_y = 20 sin30° = 20 × 0.5 = 10 m s⁻¹；水平分量 u_x = 20 cos30° ≈ 17.3 m s⁻¹。在最高点，竖直速度为零。

v_y² = u_y² – 2 g H

Set v_y = 0 → 0 = (10)² – 2 × 9.8 × H → H = 100 / (2 × 9.8) ≈ 5.1 m. That is the maximum height.

令 v_y = 0 → 0 = (10)² – 2 × 9.8 × H → H = 100 / (2 × 9.8) ≈ 5.1 m。这就是最大高度。

For total range, find the total time of flight. Using s_y = u_y t – ½ g t², the vertical displacement returns to zero: 0 = 10 t – ½ × 9.8 t² → t(10 – 4.9 t) = 0, giving t = 0 or t ≈ 2.04 s. Range R = u_x × t ≈ 17.3 × 2.04 ≈ 35.3 m.

求水平射程，先求总飞行时间。由竖直位移 s_y = u_y t – ½ g t²，回到地面时 s_y = 0：0 = 10 t – ½ × 9.8 t² → t(10 – 4.9 t) = 0，得 t = 0 或 t ≈ 2.04 s。射程 R = u_x × t ≈ 17.3 × 2.04 ≈ 35.3 m。

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3. Newton’s Laws and Connected Bodies | 牛顿定律与连接体

Two blocks of mass m₁ = 3.0 kg and m₂ = 2.0 kg are connected by a light inextensible string passing over a smooth pulley at the edge of a frictionless table. m₁ hangs vertically while m₂ rests on the horizontal table. Find the acceleration of the system and the tension in the string. Take g = 9.8 m s⁻².

质量 m₁ = 3.0 kg 和 m₂ = 2.0 kg 的两物块通过轻质不可伸长的绳子绕过光滑桌面边缘的光滑滑轮相连。m₁ 竖直悬挂，m₂ 静置于水平桌面上。求系统的加速度和绳中张力。取 g = 9.8 m s⁻²。

For m₁ (hanging): the forces are weight m₁g downward and tension T upward. Newton’s second law: m₁g – T = m₁a. For m₂ (on table): only horizontal force is T, so T = m₂a.

对 m₁（悬挂）：受向下的重力 m₁g 和向上的张力 T。牛顿第二定律：m₁g – T = m₁a。对 m₂（桌上）：只受水平张力 T，故 T = m₂a。

Substitute T into the first equation: m₁g – m₂a = m₁a → m₁g = (m₁ + m₂)a → a = m₁g / (m₁ + m₂) = (3.0 × 9.8) / (3.0 + 2.0) = 29.4 / 5.0 = 5.88 m s⁻².

将 T 代入第一式：m₁g – m₂a = m₁a → m₁g = (m₁ + m₂)a → a = m₁g / (m₁ + m₂) = (3.0 × 9.8) / (3.0 + 2.0) = 29.4 / 5.0 = 5.88 m s⁻²。

Then tension T = m₂a = 2.0 × 5.88 = 11.76 N (or about 11.8 N). Always check that the acceleration is less than g, as expected.

因此张力 T = m₂a = 2.0 × 5.88 = 11.76 N（约 11.8 N）。务必检查加速度小于 g，符合预期。

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4. Work, Energy and Power: Inclined Plane | 功、能与功率：斜面问题

A pulling force of 50 N parallel to a 30° incline moves a 10 kg crate up the slope at a constant speed of 0.5 m s⁻¹ over a distance of 4.0 m. Calculate the work done by the pulling force, the gain in gravitational potential energy, and the power developed. Take g = 9.8 m s⁻².

一个大小为 50 N、方向平行于 30° 斜面的拉力，把一个 10 kg 的木箱以 0.5 m s⁻¹ 的恒定速度沿斜面向上拉动了 4.0 m。计算拉力所做的功、增加的重力势能以及产生的功率。取 g = 9.8 m s⁻²。

Work done by the pulling force: W_F = F × d = 50 N × 4.0 m = 200 J. Gain in height: h = d sin30° = 4.0 × 0.5 = 2.0 m. Gain in GPE = mgh = 10 × 9.8 × 2.0 = 196 J. The difference is due to work against friction (196 J < 200 J).

拉力所做的功：W_F = F × d = 50 N × 4.0 m = 200 J。上升高度：h = d sin30° = 4.0 × 0.5 = 2.0 m。增加的重力势能 = mgh = 10 × 9.8 × 2.0 = 196 J。差值反映了克服摩擦力做的功（196 J < 200 J）。

Power developed by the pulling force: P = F v = 50 N × 0.5 m s⁻¹ = 25 W. Alternatively, P = work / time. Time = d / v = 4.0 / 0.5 = 8.0 s, so P = 200 J / 8.0 s = 25 W. Both methods agree.

拉力产生的功率：P = F v = 50 N × 0.5 m s⁻¹ = 25 W。或用 P = 功 / 时间，时间 = d / v = 4.0 / 0.5 = 8.0 s，故 P = 200 J / 8.0 s = 25 W。两种方法结果一致。

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5. Stress, Strain and the Young Modulus | 应力、应变与杨氏模量

A copper wire of length 2.500 m and cross-sectional area 1.2 × 10⁻⁷ m² stretches by 3.8 mm when a 4.0 kg mass hangs from it. Calculate the stress, strain, and Young modulus of copper. Take g = 9.8 m s⁻².

一根长 2.500 m、横截面积为 1.2 × 10⁻⁷ m² 的铜丝，在悬挂 4.0 kg 的重物时伸长了 3.8 mm。计算应力、应变和铜的杨氏模量。取 g = 9.8 m s⁻²。

Force F = mg = 4.0 × 9.8 = 39.2 N. Stress σ = F / A = 39.2 / (1.2 × 10⁻⁷) ≈ 3.27 × 10⁸ Pa. Strain ε = extension / original length = (3.8 × 10⁻³ m) / 2.500 m = 1.52 × 10⁻³. Young modulus E = σ / ε = 3.27 × 10⁸ / 1.52 × 10⁻³ ≈ 2.15 × 10¹¹ Pa. This is close to the accepted value for copper (≈ 1.1–1.3 × 10¹¹ Pa), though exact values depend on the specimen.

力 F = mg = 4.0 × 9.8 = 39.2 N。应力 σ = F / A = 39.2 / (1.2 × 10⁻⁷) ≈ 3.27 × 10⁸ Pa。应变 ε = 伸长量 / 原长 = (3.8 × 10⁻³ m) / 2.500 m = 1.52 × 10⁻³。杨氏模量 E = σ / ε = 3.27 × 10⁸ / 1.52 × 10⁻³ ≈ 2.15 × 10¹¹ Pa。与铜的公认值（约 1.1–1.3 × 10¹¹ Pa）接近，具体数值取决于样品。

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6. Kirchhoff’s Laws in DC Circuits | 直流电路中的基尔霍夫定律

A circuit consists of a 12 V battery and two resistors in parallel: R₁ = 6.0 Ω and R₂ = 3.0 Ω. This combination is connected in series with a third resistor R₃ = 4.0 Ω. Find the total current supplied by the battery and the current through R₁.

一个电路由一个 12 V 电池与两个并联电阻 R₁ = 6.0 Ω、R₂ = 3.0 Ω 组合后再与第三个电阻 R₃ = 4.0 Ω 串联构成。求电池提供的总电流以及流过 R₁ 的电流。

First, find the equivalent resistance of the parallel pair: 1/R_par = 1/6.0 + 1/3.0 = 1/6 + 2/6 = 3/6, so R_par = 2.0 Ω. Total circuit resistance R_total = R_par + R₃ = 2.0 + 4.0 = 6.0 Ω. Total current I_total = V / R_total = 12 V / 6.0 Ω = 2.0 A.

首先求并联部分的等效电阻：1/R_par = 1/6.0 + 1/3.0 = 1/6 + 2/6 = 3/6，故 R_par = 2.0 Ω。电路总电阻 R_total = R_par + R₃ = 2.0 + 4.0 = 6.0 Ω。总电流 I_total = V / R_total = 12 V / 6.0 Ω = 2.0 A。

The voltage across the parallel section is V_par = I_total × R_par = 2.0 × 2.0 = 4.0 V. Then the current through R₁ is I₁ = V_par / R₁ = 4.0 V / 6.0 Ω = 0.667 A (approximately). Likewise, I₂ = 4.0 / 3.0 = 1.333 A, and the sum matches the total 2.0 A.

并联部分两端的电压为 V_par = I_total × R_par = 2.0 × 2.0 = 4.0 V。流过 R₁ 的电流 I₁ = V_par / R₁ = 4.0 V / 6.0 Ω = 0.667 A（约为）。同理 I₂ = 4.0 / 3.0 = 1.333 A，二者之和等于总电流 2.0 A。

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7. Double-Slit Interference of Light | 光的双缝干涉

Light of wavelength λ = 650 nm passes through two narrow slits separated by 0.50 mm. The interference pattern is observed on a screen 2.0 m away. Calculate the fringe spacing (distance between adjacent bright fringes).

波长为 λ = 650 nm 的光通过间距为 0.50 mm 的两条窄缝，在 2.0 m 远处的屏幕上观察干涉图样。计算条纹间距（相邻亮纹之间的距离）。

The fringe width Δy is given by Δy = λD / d, where D is the distance from slits to screen and d is the slit separation. Convert all lengths to metres: λ = 650 × 10⁻⁹ m, d = 0.50 × 10⁻³ m, D = 2.0 m.

条纹宽度 Δy 由 Δy = λD / d 给出，其中 D 为双缝到屏幕的距离，d 为缝间距。将所有长度单位换算为米：λ = 650 × 10⁻⁹ m，d = 0.50 × 10⁻³ m，D = 2.0 m。

Substituting: Δy = (650 × 10⁻⁹ × 2.0) / (0.50 × 10⁻³) = (1.30 × 10⁻⁶) / (5.0 × 10⁻⁴) = 2.6 × 10⁻³ m = 2.6 mm. So bright fringes are 2.6 mm apart.

代入计算：Δy = (650 × 10⁻⁹ × 2.0) / (0.50 × 10⁻³) = (1.30 × 10⁻⁶) / (5.0 × 10⁻⁴) = 2.6 × 10⁻³ m = 2.6 mm。因此亮条纹间距为 2.6 mm。

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8. The Photoelectric Effect | 光电效应

Ultraviolet light of frequency 1.20 × 10¹⁵ Hz strikes a metal surface with a work function of 4.80 eV. Determine the maximum kinetic energy of emitted photoelectrons in joules and electronvolts, and the stopping potential. Use h = 6.63 × 10⁻³⁴ J s, e = 1.60 × 10⁻¹⁹ C.

频率为 1.20 × 10¹⁵ Hz 的紫外光照射在逸出功为 4.80 eV 的金属表面上。求发射光电子的最大动能（以焦耳和电子伏特表示）以及遏止电压。使用 h = 6.63 × 10⁻³⁴ J s，e = 1.60 × 10⁻¹⁹ C。

Photon energy E_ph = h f = 6.63 × 10⁻³⁴ × 1.20 × 10¹⁵ = 7.956 × 10⁻¹⁹ J. Convert to eV: E_ph (eV) = (7.956 × 10⁻¹⁹) / (1.60 × 10⁻¹⁹) ≈ 4.97 eV. Maximum kinetic energy K_max = E_ph – φ = 4.97 – 4.80 = 0.17 eV. In joules: 0.17 × 1.60 × 10⁻¹⁹ = 2.72 × 10⁻²⁰ J.

光子能量 E_ph = h f = 6.63 × 10⁻³⁴ × 1.20 × 10¹⁵ = 7.956 × 10⁻¹⁹ J。换算为 eV：E_ph (eV) = (7.956 × 10⁻¹⁹) / (1.60 × 10⁻¹⁹) ≈ 4.97 eV。最大动能 K_max = E_ph – φ = 4.97 – 4.80 = 0.17 eV。以焦耳计：0.17 × 1.60 × 10⁻¹⁹ = 2.72 × 10⁻²⁰ J。

The stopping potential V_s satisfies e V_s = K_max, so V_s = K_max / e. Using eV units directly gives V_s = 0.17 V. (Or in joules: V_s = 2.72 × 10⁻²⁰ / 1.60 × 10⁻¹⁹ = 0.17 V).

遏止电压 V_s 满足 e V_s = K_max，因此 V_s = K

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