A-Level Physics: Calculation Problem Drill | A-Level 物理：计算题专项训练

📚 A-Level Physics: Calculation Problem Drill | A-Level 物理：计算题专项训练

Calculation problems form the backbone of A-Level Physics assessments. Mastering them requires not only memorising formulae but also developing a systematic approach to problem-solving. This article drills into key topic areas, providing step-by-step methods, worked examples, and essential tips.

计算题是 A-Level 物理考试的核心。掌握它们不仅需要记住公式，还需要培养系统的解题方法。本文针对重点专题，提供分步方法、典型例题和关键技巧。

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1. Kinematics and Projectile Motion | 运动学与抛体运动

Start by listing the known quantities with their signs, e.g. u = 20 m s⁻¹, a = -9.8 m s⁻². Select the SUVAT equation that omits the unwanted unknown.

首先列出已知量及其符号，例如 u = 20 m s⁻¹, a = -9.8 m s⁻²。然后选择不包含多余未知量的 SUVAT 方程。

v = u + at,   s = ut + ½at²,   v² = u² + 2as,   s = ½(u+v)t

For projectile motion, resolve initial velocity into horizontal u_x = u cosθ and vertical u_y = u sinθ. Horizontal motion is uniform; vertical motion uses a = -g. Combine them to find height, range, and time of flight.

对于抛体运动，将初速度分解为水平分量 u_x = u cosθ 和竖直分量 u_y = u sinθ。水平运动是匀速的，竖直方向用 a = -g 的匀加速方程。两者结合可求最大高度、射程和飞行时间。

Worked example: A ball is projected at 20 m s⁻¹ at 30° above horizontal; g = 9.8 m s⁻². Find maximum height and range.

典型例题：一球以 20 m s⁻¹ 的速度、与水平成 30° 角抛出，g = 9.8 m s⁻²。求最大高度和射程。

Step 1 – resolve: u_y = 20 sin30° = 10 m s⁻¹. At the highest point v_y = 0, so use v² = u_y² + 2a s_y → 0 = 10² + 2(-9.8)s_y → s_y = 100 / 19.6 ≈ 5.1 m.

步骤 1 – 分解：u_y = 20 sin30° = 10 m s⁻¹。最高点 v_y = 0，使用 v² = u_y² + 2a s_y → 0 = 100 – 19.6 s_y → s_y ≈ 5.1 m。

Step 2 – time of flight: vertical displacement s_y = 0, so s = u_y t + ½ a t² → 0 = 10t – 4.9t² → t = 0 or t = 20/4.9 ≈ 2.04 s. Horizontal range: u_x = 20 cos30° ≈ 17.32 m s⁻¹, R = u_x × t ≈ 35.3 m.

步骤 2 – 飞行时间：竖直位移 s_y = 0，由 s = u_y t + ½ a t² 得 0 = 10t – 4.9t²，解得 t ≈ 2.04 s。水平射程：u_x = 20 cos30° ≈ 17.32 m s⁻¹，R = u_x × t ≈ 35.3 m。

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2. Forces and Newton’s Laws | 力与牛顿定律

Draw a free-body diagram and resolve forces along perpendicular axes. Write ΣF = ma for each direction. Friction is f = μR, where R is the normal reaction.

画出受力图，将力沿垂直方向分解。对每个方向列出 ΣF = ma。摩擦力 f = μR，其中 R 是法向接触力。

Example: A 5.0 kg block slides down a 30° incline with μ = 0.20. Determine the acceleration.

例题：一个 5.0 kg 的滑块沿 30° 斜面下滑，μ = 0.20。求加速度。

Down-slope component of weight = mg sin30° = 5 × 9.8 × 0.5 = 24.5 N. Normal reaction R = mg cos30° = 5 × 9.8 × 0.866 = 42.4 N. Friction f = 0.20 × 42.4 = 8.48 N up the slope. Net force F = 24.5 – 8.48 = 16.02 N. Hence a = F/m = 16.02 / 5.0 ≈ 3.2 m s⁻².

沿斜面的重力分量 = mg sin30° = 5 × 9.8 × 0.5 = 24.5 N。法向力 R = mg cos30° = 5 × 9.8 × 0.866 = 42.4 N。摩擦力 f = 0.20 × 42.4 = 8.48 N 沿斜面向上。合力 F = 24.5 – 8.48 = 16.02 N。因此加速度 a = F/m = 16.02 / 5.0 ≈ 3.2 m s⁻²。

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3. Work, Energy, and Power | 功、能量与功率

Use the work–energy theorem: net work = ΔEₖ. Kinetic energy Eₖ = ½mv², gravitational potential energy Eₚ = mgh. In conservative systems, mechanical energy is conserved. Power P = Fv or P = ΔW/Δt.

运用功能原理：合外力的功 = 动能变化量 ΔEₖ。动能 Eₖ = ½mv²，重力势能 Eₚ = mgh。在保守系统中机械能守恒。功率 P = Fv 或 P = ΔW/Δt。

Eₖ = ½mv²,   Eₚ = mgh,   P = Fv

Example: A car of mass 1200 kg accelerates from rest to 25 m s⁻¹ in 8.0 s on a flat road. Calculate the average power delivered to the car.

例题：一辆 1200 kg 的汽车从静止加速到 25 m s⁻¹，耗时 8.0 s，平路。求汽车获得的平均功率。

Work done = ΔEₖ = ½ × 1200 × (25)² – 0 = 375 000 J. Average power = work / time = 375 000 / 8.0 = 46 875 W ≈ 47 kW.

做功 = 动能增量 = ½ × 1200 × 25² = 375 000 J。平均功率 = 功 / 时间 = 375 000 / 8.0 ≈ 47 kW。

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4. Momentum and Collisions | 动量与碰撞

Momentum is conserved in any collision provided no external force acts. For two bodies: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. In a perfectly elastic collision, kinetic energy is also conserved.

只要没有外力作用，碰撞前后动量守恒。对于两个物体：m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂。在完全弹性碰撞中，动能也守恒。

Example: A 2.0 kg trolley moving at 3.0 m s⁻¹ collides elastically with a stationary 1.0 kg trolley. Find their velocities after collision.

例题：一辆 2.0 kg 的小车以 3.0 m s⁻¹ 的速度与一辆静止的 1.0 kg 小车发生弹性碰撞。求碰后各自速度。

Using conservation of momentum: 2.0 × 3.0 + 0 = 2.0 v₁ + 1.0 v₂ → 6 = 2v₁ + v₂. Kinetic energy: ½×2.0×9 = ½×2.0 v₁² + ½×1.0 v₂² → 9 = v₁² + ½ v₂². Solve: from first, v₂ = 6 – 2v₁. Substitute: 9 = v₁² + ½(6 – 2v₁)² → 9 = v₁² + ½(36 – 24v₁ + 4v₁²) = v₁² + 18 – 12v₁ + 2v₁² = 3v₁² – 12v₁ + 18. Thus 3v₁² – 12v₁ + 9 = 0 → v₁² – 4v₁ + 3 = 0 → (v₁ – 1)(v₁ – 3) = 0. v₁ = 1 m s⁻¹ (reject v₁ = 3 as no change). Then v₂ = 6 – 2 = 4 m s⁻¹.

动量守恒：2.0×3.0 + 0 = 2.0 v₁ + 1.0 v₂ → 6 = 2v₁ + v₂。动能守恒：9 = v₁² + ½ v₂²。由第一式得 v₂ = 6 – 2v₁，代入第二式得 9 = v₁² + ½(6 – 2v₁)²，化简得 v₁² – 4v₁ + 3 = 0，解得 v₁ = 1 m s⁻¹，v₂ = 4 m s⁻¹。

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5. Circular Motion and Gravitation | 圆周运动与万有引力

Centripetal acceleration a = v²/r = rω², and centripetal force F = mv²/r = mrω². For gravitational orbits, equate centripetal force to GMm/r². Use Kepler’s law T² ∝ r³ for satellites.

向心加速度 a = v²/r = rω²，向心力 F = mv²/r = mrω²。对于引力轨道，令向心力等于万有引力 GMm/r²。卫星问题中可用开普勒定律 T² ∝ r³。

F = Gm₁m₂ / r²,   v = √(GM/r)

Example: Calculate the height of a geostationary satellite above Earth’s surface. Earth mass M = 5.97×10²⁴ kg, G = 6.67×10⁻¹¹ N m² kg⁻², Earth radius R = 6.37×10⁶ m, period T = 24 h = 86 400 s.

例题：计算地球同步卫星离地高度。地球质量 M = 5.97×10²⁴ kg，G = 6.67×10⁻¹¹ N m² kg⁻²，地球半径 R = 6.37×10⁶ m，周期 T = 24 h = 86 400 s。

Using T² = (4π²/GM) r³ → r³ = (GMT²)/(4π²). Substitute: r³ = (6.67×10⁻¹¹ × 5.97×10²⁴ × (86 400)²) / (4π²) ≈ 7.54×10²². Hence r ≈ 4.23×10⁷ m. Height h = r – R = 4.23×10⁷ – 6.37×10⁶ ≈ 3.59×10⁷ m (≈ 35 900 km).

由 T² = (4π²/GM) r³ 得 r³ = (GMT²)/(4π²)。代入：r³ = (6.67×10⁻¹¹ × 5.97×10²⁴ × 86400²) / (4π²) ≈ 7.54×10²²，r ≈ 4.23×10⁷ m。高度 h = r – R ≈ 3.59×10⁷ m（约 35 900 km）。

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6. Simple Harmonic Motion | 简谐运动

SHM is defined by a = -ω²x. Displacement x = A cos(ωt) or A sin(ωt), velocity v = ±ω√(A² – x²), maximum speed v_max = ωA. Period of a mass–spring system T = 2π√(m/k), simple pendulum T = 2π√(l/g). Energy E = ½mω²A².

简谐运动由 a = -ω²x 定义。位移 x = A cos(ωt) 或 A sin(ωt)，速度 v = ±ω√(A² – x²)，最大速度 v_max = ωA。弹簧振子周期 T = 2π√(m/k)，单摆周期 T = 2π√(l/g)。能量 E = ½mω²A²。

Example: A mass of 0.50 kg on a spring of stiffness 200 N m⁻¹ is pulled 4.0 cm from equilibrium and released. Find the period and the speed as it passes equilibrium.

例题：一个 0.50 kg 的质量悬挂在劲度系数 200 N m⁻¹ 的弹簧上，从平衡位置拉开 4.0 cm 释放。求周期和经过平衡位置时的速度。

Period T = 2π√(m/k) = 2π√(0.50/200) = 2π√(0.0025) = 2π × 0.05 = 0.314 s. Angular frequency ω = 2π/T = 20 rad s⁻¹. Amplitude A = 0.040 m. Maximum speed v_max = ωA = 20 × 0.040 = 0.80 m s⁻¹.

周期 T = 2π√(m/k) = 2π√(0.50/200) = 0.314 s。角频率 ω = 2π/T = 20 rad s⁻¹。振幅 A = 0.040 m。最大速率 v_max = ωA = 20 × 0.040 = 0.80 m s⁻¹。

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7. Electric Fields and Potential | 电场与电势

Coulomb’s law F = kQq/r², electric field strength E = F/q = kQ/r². For uniform fields E = V/d. The work done in moving a charge is W = qΔV; kinetic energy gained by an accelerated charge: qV = ½mv².

库仑定律 F = kQq/r²，电场强度 E = F/q = kQ/r²。匀强电场中 E = V/d。移动电荷做的功 W = qΔV；电荷被电场加速获得的动能：qV = ½mv²。

E = V/d,   F = kQq / r²,   k = 1/(4πε₀)

Example: An electron is accelerated from rest through a p.d. of 500 V. Find its final speed. (e = 1.60×10⁻¹⁹ C, m_e = 9.11×10⁻³¹ kg)

例题：一个电子从静止经 500 V 电势差加速。求其末速度。（e = 1.60×10⁻¹⁹ C，m_e = 9.11×10⁻³¹ kg）

Energy conservation: eV = ½ m v

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