📚 Edexcel Physics: Detailed Worked Examples | Edexcel 物理：典型例题详解
This article presents a selection of typical Edexcel A Level Physics problems with step-by-step solutions. Each example is chosen to reinforce key concepts across mechanics, electricity, waves, quantum physics and nuclear physics. The worked solutions demonstrate effective problem-solving strategies and clear application of formulae, helping you master the skills required for the exam.
本文精选了 Edexcel A Level 物理中的典型例题，并给出逐步详解。每个例子都旨在巩固力学、电学、波动、量子物理和核物理的核心概念。这些解题过程展示了高效的问题解决策略和公式的清晰应用，帮助同学们掌握考试必备的技能。
1. Kinematics: Two-Stage Motion Problem | 运动学：两阶段运动问题
A car starts from rest, accelerates uniformly for 10 s and then travels at constant speed for a further 20 s. The total distance covered is 400 m. Calculate the acceleration of the car.
一辆汽车从静止开始，先匀加速运动 10 秒，然后以恒定速度行驶 20 秒。总路程为 400 m。求汽车的加速度。
= ut₁ + ½at₁² = ½a(10)² = 50a. The final velocity after acceleration is v = u + at₁ = 10a.
设加速度为 a。在第一阶段，初速 u = 0，时间 t₁ = 10 s。位移 s₁ = ut₁ + ½at₁² = ½a(10)² = 50a。加速后的末速度 v = u + at₁ = 10a。
he car moves at constant velocity v for t₂ = 20 s. Distance s₂ = v t₂ = (10a) × 20 = 200a.
在第二阶段，汽车以恒定速度 v 运动 t₂ = 20 s。位移 s₂ = v t₂ = (10a) × 20 = 200a。
= 50a + 200a = 250a. Set equal to 400 m: 250a = 400, therefore a = 400 / 250 = 1.6 m s⁻².
总位移 s = s₁ + s₂ = 50a + 200a = 250a。等于 400 m：250a = 400，因此加速度 a = 400 / 250 = 1.6 m s⁻²。
2. Newton's Laws: Force at an Angle with Friction | 牛顿定律：斜向拉力与摩擦力
A block of mass 5.0 kg rests on a rough horizontal surface. A force of 30 N is applied at 30° above the horizontal. The coefficient of kinetic friction between the block and the surface is 0.40. Find the acceleration of the block.
一个质量为 5.0 kg 的木块放在粗糙水平面上。一个大小为 30 N 的力以与水平方向成 30° 向上拉木块。木块与地面间的动摩擦因数为 0.40。求木块的加速度。
Resolve the applied force: horizontal component Fₓ = 30 cos 30° = 30 × (√3/2) ≈ 25.98 N. Vertical component Fᵧ = 30 sin 30° = 15 N upwards.
分解拉力：水平分量 Fₓ = 30 cos 30° = 30 × (√3/2) ≈ 25.98 N。竖直分量 Fᵧ = 30 sin 30° = 15 N（向上）。
mg = 5.0 × 9.81 = 49.05 N downward. The normal reaction N = W - Fᵧ = 49.05 - 15 = 34.05 N.
木块重力 W = mg = 5.0 × 9.81 = 49.05 N（向下）。支持力 N = W - Fᵧ = 49.05 - 15 = 34.05 N。
Kinetic friction f = μ N = 0.40 × 34.05 = 13.62 N opposing motion.
动摩擦力 f = μ N = 0.40 × 34.05 = 13.62 N，方向与运动相反。
Net horizontal force Fnet = Fₓ - f = 25.98 - 13.62 = 12.36 N. Acceleration a = Fnet / m = 12.36 / 5.0 ≈ 2.47 m s⁻².
水平方向合力 Fnet = Fₓ - f = 25.98 - 13.62 = 12.36 N。加速度 a = Fnet / m = 12.36 / 5.0 ≈ 2.47 m s⁻²。
3. Energy Conservation: Incline and Spring | 能量守恒：斜面与弹簧
tal spring with spring constant k = 200 N m⁻¹. Find the maximum compression of the spring.
一个 2.0 kg 的物体从静止沿光滑斜面下滑，斜面高度 5.0 m。在底端物体撞击一个水平放置的弹簧，弹簧劲度系数 k = 200 N m⁻¹。求弹簧的最大压缩量。
gy is conserved. Loss in gravitational potential energy = mgh = 2.0 × 9.81 × 5.0 = 98.1 J.
由于无摩擦，机械能守恒。减少的重力势能 = mgh = 2.0 × 9.81 × 5.0 = 98.1 J。
This energy is converted entirely into elastic potential energy of the spring: ½ k xmax² = 98.1 J.
这部分能量全部转化为弹簧的弹性势能：½ k xmax² = 98.1 J。
Rearrange: xmax² = (2 × 98.1) / 200 = 196.2 / 200 = 0.981. Therefore xmax = √0.981 ≈ 0.990 m (about 99 cm).
整理得：xmax² = (2 × 98.1) / 200 = 196.2 / 200 = 0.981。因此最大压缩量 xmax = √0.981 ≈ 0.990 m（约 99 cm）。
4. Elastic Collision in One Dimension | 一维弹性碰撞
A 0.50 kg ball moving at 4.0 m s⁻¹ collides elastically with a stationary 0.30 kg ball on a frictionless track. Determine the velocities of both balls after the collision.
一个 0.50 kg 的小球以 4.0 m s⁻¹ 的速度在光滑轨道上运动，与一个静止的 0.30 kg 小球发生弹性碰撞。求碰撞后两球的速度。
For a perfectly elastic head-on collision, both momentum and kinetic energy are conserved. Using the standard formulae for one stationary object:
对于完全弹性正碰，动量和动能均守恒。使用经典公式（一球静止）：
v₁' = (m₁ - m₂) / (m₁ + m₂) * u₁, v₂' = (2m₁) / (m₁ + m₂) * u₁.
v₁' = (m₁ - m₂) / (m₁ + m₂) · u₁, v₂' = (2m₁) / (m₁ + m₂) · u₁。
Substitute m₁ = 0.50 kg, m₂ = 0.30 kg, u₁ = 4.0 m s⁻¹:
代入 m₁ = 0.50 kg，m₂ = 0.30 kg，u₁ = 4.0 m s⁻¹：
v₁' = (0.50 - 0.30) / (0.50 + 0.30) × 4.0 = (0.20 / 0.80) × 4.0 = 0.25 × 4.0 = 1.0 m s⁻¹.
v₁' = (0.50 - 0.30) / (0.50 + 0.30) × 4.0 = (0.20 / 0.80) × 4.0 = 0.25 × 4.0 = 1.0 m s⁻¹。
v₂' = (2 × 0.50) / (0.80) × 4.0 = (1.0 / 0.80) × 4.0 = 1.25 × 4.0 = 5.0 m s⁻¹.
v₂' = (2 × 0.50) / 0.80 × 4.0 = (1.0 / 0.80) × 4.0 = 1.25 × 4.0 = 5.0 m s⁻¹。
The 0.50 kg ball continues forward at 1.0 m s⁻¹; the 0.30 kg ball moves forward at 5.0 m s⁻¹.
0.50 kg 的球以 1.0 m s⁻¹ 向前运动；0.30 kg 的球以 5.0 m s⁻¹ 向前运动。
5. DC Circuits: Kirchhoff's Laws | 直流电路：基尔霍夫定律
In the circuit below, E₁ = 10 V, E₂ = 5 V, R₁ = 2 Ω, R₂ = 3 Ω, R₃ = 4 Ω. Use Kirchhoff' left loop contains E₁, R₁ and R₂; the right loop contains R₂, R₃ and E₂, with R₂ shared.)
在下图电路中，E₁ = 10 V，E₂ = 5 V，R₁ = 2 Ω，R₂ = 3 Ω，R₃ = 4 Ω。用基尔霍夫定律求通过每个电阻的电流。（假设电路有两个回路：左回路包含 E₁、R₁ 和 R₂；右回路包含 R₂、R₃ 和 E₂，R₂ 为共用电阻。）
Assign loop currents: I₁ flows clockwise in the left loop, I₂ flows clockwise in the right loop. Then current through R₂ is I₁ - I₂ (downward).
设定回路电流：左回路顺时针方向电流 I₁，右回路顺时针方向电流 I₂。则通过 R₂ 的电流（向下）为 I₁ - I₂。
Apply Kirchhoff' ft loop: E₁ - I₁R₁ - (I₁ - I₂)R₂ = 0 → 10 - 2I₁ - 3(I₁ - I₂) = 0 → 10 - 5I₁ + 3I₂ = 0. (1)
对左回路应用基尔霍夫电压定律：E₁ - I₁R₁ - (I₁ - I₂)R₂ = 0 → 10 - 2I₁ - 3(I₁ - I₂) = 0 → 10 - 5I₁ + 3I₂ = 0。 (1)
Right loop: - (I₂ - I₁)R₂ - I₂R₃ - E₂ = 0? o I₂ is I₂R₂, but direction of I₁ also contributes. Write as: -I₂R₃ - E₂ + (I₁ - I₂)R₂ = 0? ent through R₂ is I₁ - I₂ downward, so the drop in the clockwise direction is +(I₁ - I₂)R₂? Let' ugh R₃ with current I₂, drop = -4I₂, then through R₂ with current I₂ relative to this loop? ith E₂, resistor R₃ and shared R₂, the sum: -5 + 4I₂ + 3(I₂ - I₁) = 0 → 7I₂ - 3I₁ = 5. Let's use that.
右回路：顺时针绕行，先经过 E₂ 从负极到正极电位升 +5 V，再经过 R₃ 电位降 -4I₂，然后经过 R₂：由于通过 R₂ 的电流为 I₁ - I₂（向下），顺时针方向经过 R₂ 的电位降为 +3(I₂ - I₁)。方程：-5 + 4I₂ + 3(I₂ - I₁) = 0 → 7I₂ - 3I₁ = 5。 (2)
Solve equations (1): 5I₁ - 3I₂ = 10 and (2): -3I₁ + 7I₂ = 5.
解方程组：(1) 5I₁ - 3I₂ = 10，(2) -3I₁ + 7I₂ = 5。
₁ = 10 + 3I₂ = 10 + 6.346 = 16.346 → I₁ = 3.269 A. Current through R₂ = I₁ - I₂ ≈ 1.154 A.
(1)×3: 15I₁ - 9I₂ = 30；(2)×5: -15I₁ + 35I₂ = 25。相加得 26I₂ = 55，I₂ ≈ 2.115 A。代入 (1)：5I₁ = 10 + 3I₂ = 16.346，I₁ ≈ 3.269 A。R₂ 中的电流 = I₁ - I₂ ≈ 1.154 A。
6. Young's Double-Slit Interference | 杨氏双缝干涉
In a Young's double-slit experiment, the slit separation is 0.50 mm and the screen is placed 2.0 m from the slits. The fringe separation on the screen is found to be 2.0 mm. Calculate the wavelength of the light used.
在杨氏双缝实验中，双缝间距为 0.50 mm，屏幕距双缝 2.0 m。测得条纹间距为 2.0 mm。计算所用光的波长。
g Δy is Δy = λD / d, where λ is wavelength, D is distance to screen, d is slit separation.
条纹间距公式为 Δy = λD / d，其中 λ 是波长，D 是屏距，d 是双缝间距。
nvert all lengths to metres: Δy = 2.0 × 10⁻³ m, d = 0.50 × 10⁻³ = 5.0 × 10⁻⁴ m, D = 2.0 m.
整理得：λ = Δy d / D。统一单位：Δy = 2.0 × 10⁻³ m，d = 5.0 × 10⁻⁴ m，D = 2.0 m。
λ = (2.0 × 10⁻³) × (5.0 × 10⁻⁴) / 2.0 = (10.0 × 10⁻⁷) / 2.0 = 5.0 × 10⁻⁷ m = 500 nm.
λ = (2.0 × 10⁻³) × (5.0 × 10⁻⁴) / 2.0 = (10.0 × 10⁻⁷) / 2.0 = 5.0 × 10⁻⁷ m = 500 nm。
7. Photoelectric Effect: Kinetic Energy Calculation | 光电效应：动能计算
Sodium has a work function of 2.3 eV. Light of wavelength 300 nm is incident on a sodium surface. Determine the maximum kinetic energy of the emitted photoelectrons.
钠的功函数为 2.3 eV。波长为 300 nm 的光照射到钠表面。求发射出的光电子的最大动能。
Photon energy Ephoton = hc / λ. Using hc = 1240 eV·nm (approximation), Ephoton = 1240 eV·nm / 300 nm = 4.13 eV.
光子能量 Ephoton = hc / λ。利用近似关系 hc = 1240 eV·nm，得 Ephoton = 1240 eV·nm / 300 nm = 4.13 eV。
Einstein's photoelectric equation: Kmax = Ephoton - Φ = 4.13 eV - 2.3 eV = 1.83 eV.
爱因斯坦光电方程：Kmax = Ephoton - Φ = 4.13 eV - 2.3 eV = 1.83 eV。
If required in joules: 1 eV = 1.60 × 10⁻¹⁹ J, so Kmax = 1.83 × 1.60 × 10⁻¹⁹ J ≈ 2.93 × 10⁻¹⁹ J.
如需以焦耳表示：1 eV = 1.60 × 10⁻¹⁹ J，故 Kmax = 1.83 × 1.60 × 10⁻¹⁹ J ≈ 2.93 × 10⁻¹⁹ J。
8. Mass-Energy Equivalence in Nuclear Reactions | 核反应中的质能等价
on has a mass defect of 0.10 u. Calculate the energy released in MeV. (1 u = 931.5 MeV/c²)
某核反应的质量亏损为 0.10 u。计算释放的能量（以 MeV 为单位）。(1 u = 931.5 MeV/c²)
given by ΔE = Δm c². With Δm in atomic mass units, ΔE = 0.10 u × 931.5 MeV/u = 93.15 MeV.
释放能量用质能方程 ΔE = Δm c² 计算。用原子质量单位，ΔE = 0.10 u × 931.5 MeV/u = 93.15 MeV。
The reaction releases approximately 93 MeV of energy, which is typical for nuclear processes and much larger than chemical energy changes.
该反应释放约 93 MeV 的能量，这是核过程的典型值，远大于化学反应的能量变化。
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