📚 A-Level AQA Physics: Calculation Questions Intensive Practice | A-Level AQA 物理：计算题专项训练
Calculation questions form the backbone of AQA A-Level Physics examinations, often contributing over 40% of the total marks. Mastering these questions requires not only memorising equations but also developing a systematic approach to identifying variables, selecting the correct formula, and carrying out unit conversions. This intensive practice guide walks you through the most frequently tested problem types, pairing clear explanations with worked examples to build both confidence and precision.
计算题是 AQA A-Level 物理考试的核心，通常占总分的 40% 以上。要熟练解答这类题目，不仅要记住公式，还要培养系统的解题思路：识别变量、选择正确的方程以及进行单位换算。这份专项训练指南带你走过最常考的问题类型，通过清晰的解释和练习题配对，帮助你建立信心和准确度。
1. Kinematics Equations | 运动学方程
The four SUVAT equations model uniformly accelerated motion along a straight line. They link displacement (s), initial velocity (u), final velocity (v), acceleration (a), and time (t). The equations are: v = u + at, s = ut + ½at², v² = u² + 2as, and s = ½(u + v)t. In AQA problems, always define a positive direction and remember that acceleration due to gravity is g = 9.81 m s⁻² near Earth’s surface.
四个 SUVAT 方程描述了匀加速直线运动，关联位移(s)、初速度(u)、末速度(v)、加速度(a)和时间(t)。方程分别为：v = u + at、s = ut + ½at²、v² = u² + 2as 和 s = ½(u + v)t。在 AQA 试题中，务必先规定正方向，并记住地球表面附近的重力加速度 g = 9.81 m s⁻²。
A stone is thrown vertically upwards with an initial speed of 14 m s⁻¹ from ground level. Calculate the maximum height reached and the time taken to return to the ground. Taking up as positive, u = 14 m s⁻¹, a = -9.81 m s⁻², v = 0 at maximum height. Using v² = u² + 2as gives 0 = 14² + 2(-9.81)s, so s = 196 / 19.62 = 9.99 m. For the total time of flight, we can use s = ut + ½at² with s = 0: 0 = 14t – 4.905t², factorising gives t(14 – 4.905t) = 0, hence t = 0 or t = 14 / 4.905 = 2.85 s.
一块石头以 14 m s⁻¹ 的初速度从地面竖直向上抛出。计算到达的最大高度和返回地面所需的时间。取向上为正，u = 14 m s⁻¹，a = -9.81 m s⁻²，在最高点 v = 0。利用 v² = u² + 2as 得到 0 = 14² + 2(-9.81)s，解得 s = 196 / 19.62 = 9.99 m。对于全程飞行时间，使用 s = ut + ½at² 并令 s = 0：0 = 14t – 4.905t²，提取公因式得 t(14 – 4.905t) = 0，因此 t = 0 或 t = 14 / 4.905 = 2.85 s。
2. Newton’s Laws and Friction | 牛顿定律与摩擦力
Newton’s second law, F = ma, is the foundation of force problems. On inclined planes, the weight component along the slope is mg sinθ, and the normal reaction is mg cosθ. Friction is given by f ≤ μR, where μ is the coefficient of friction and R is the normal contact force. Many AQA questions combine these to find acceleration or limiting angles.
牛顿第二定律 F = ma 是力学问题的基础。在斜面上，沿斜面的重力分量为 mg sinθ，法向反作用力为 mg cosθ。摩擦力由 f ≤ μR 给出，其中 μ 为摩擦系数，R 为法向接触力。许多 AQA 题目会结合这些关系求解加速度或临界角度。
A 5.0 kg block is placed on a rough slope inclined at 25° to the horizontal. The coefficient of static friction between the block and the surface is 0.45. Determine whether the block will slide, and if it does, find its acceleration down the slope when μ becomes kinetic and equals 0.35. The downhill component of weight = mg sin25° = 5.0 × 9.81 × sin25° = 20.7 N. Maximum static friction = μₛR = 0.45 × (5.0 × 9.81 × cos25°) = 0.45 × 44.5 N = 20.0 N. Since the downhill force (20.7 N) exceeds 20.0 N, the block slides. Kinetic friction = 0.35 × 44.5 = 15.6 N. Resultant force = 20.7 – 15.6 = 5.1 N, giving a = F/m = 5.1 / 5.0 = 1.02 m s⁻².
一个 5.0 kg 的物块放在与水平面成 25° 的粗糙斜面上。物块与表面之间的静摩擦系数为 0.45。判断物块是否会滑动，若会滑动，当动摩擦系数变为 0.35 时，求其沿斜面下滑的加速度。沿斜面的重力分量 = mg sin25° = 5.0 × 9.81 × sin25° = 20.7 N。最大静摩擦力 = μₛR = 0.45 × (5.0 × 9.81 × cos25°) = 0.45 × 44.5 N = 20.0 N。因为下滑力(20.7 N)超过 20.0 N，物块会滑动。动摩擦力 = 0.35 × 44.5 = 15.6 N。合力 = 20.7 – 15.6 = 5.1 N，加速度 a = F/m = 5.1 / 5.0 = 1.02 m s⁻²。
3. Work, Energy and Power | 功、能量和功率
Energy principles often provide a quicker route than resolving forces. Kinetic energy Eₖ = ½mv², gravitational potential energy Eₚ = mgh, and work done W = Fs cosθ. Power is the rate of doing work: P = W/t = Fv. In AQA papers, you are expected to apply conservation of energy or the work-energy theorem to solve problems involving dissipative forces.
能量方法通常比力的分解更快捷。动能 Eₖ = ½mv²，重力势能 Eₚ = mgh，做功 W = Fs cosθ。功率是做功的速率：P = W/t = Fv。在 AQA 试卷中，要求考生运用能量守恒或功能定理解决涉及耗散力的问题。
A crane lifts a 250 kg load vertically at a constant speed of 0.80 m s⁻¹. Calculate the power output of the crane’s motor and the energy transferred in lifting the load through 12 m. Since speed is constant, the tension in the cable equals the weight: T = mg = 250 × 9.81 = 2452.5 N. Power P = Fv = 2452.5 × 0.80 = 1962 W (≈ 1.96 kW). The work done lifting through 12 m is W = Fs = 2452.5 × 12 = 29 430 J. Alternatively, using potential energy gain: mgh = 250 × 9.81 × 12 = 29 430 J, confirming the answer.
一台起重机以 0.80 m s⁻¹ 的恒定速度垂直起吊 250 kg 的重物。计算起重机电机的输出功率以及将重物提升 12 m 所传递的能量。由于速度恒定，缆绳张力等于重力：T = mg = 250 × 9.81 = 2452.5 N。功率 P = Fv = 2452.5 × 0.80 = 1962 W（约 1.96 kW）。提升 12 m 所做的功 W = Fs = 2452.5 × 12 = 29 430 J。另一种方法，重力势能的增加：mgh = 250 × 9.81 × 12 = 29 430 J，两者一致。
4. Conservation of Momentum | 动量守恒
In a closed system with no external forces, total momentum before a collision equals total momentum after. Momentum p = mv is a vector quantity, so direction must be accounted for. Collisions may be elastic (kinetic energy conserved) or inelastic (some kinetic energy dissipated). Typical AQA calculations involve recoil speeds, explosions, or bullet-block interactions.
在没有外力的封闭系统中，碰撞前的总动量等于碰撞后的总动量。动量 p = mv 是矢量，因此必须考虑方向。碰撞可分为弹性碰撞（动能守恒）和非弹性碰撞（部分动能耗散）。AQA 典型计算题涉及反冲速度、爆炸或子弹与物块作用。
A 0.60 kg trolley moving at 2.5 m s⁻¹ to the right collides with a stationary 0.90 kg trolley. After the inelastic collision they move off together. Find the common speed and the kinetic energy lost. Using momentum conservation: (0.60 × 2.5) + (0.90 × 0) = (0.60 + 0.90)v, so v = 1.5 / 1.5 = 1.0 m s⁻¹. Initial kinetic energy = ½ × 0.60 × 2.5² = 1.875 J. Final kinetic energy = ½ × 1.50 × 1.0² = 0.75 J. Energy lost = 1.875 – 0.75 = 1.125 J.
一辆 0.60 kg 的小车以 2.5 m s⁻¹ 的速度向右运动，与一辆静止的 0.90 kg 小车发生非弹性碰撞后粘连在一起。求它们的共同速度和损失的动能。根据动量守恒：(0.60 × 2.5) + (0.90 × 0) = (0.60 + 0.90)v，解得 v = 1.5 / 1.5 = 1.0 m s⁻¹。初始动能 = ½ × 0.60 × 2.5² = 1.875 J。末动能 = ½ × 1.50 × 1.0² = 0.75 J。能量损失 = 1.875 – 0.75 = 1.125 J。
5. Circular Motion | 圆周运动
An object moving in a circle at constant speed experiences an acceleration towards the centre: centripetal acceleration a = v²/r = ω²r, where v is linear speed, r is radius, and ω is angular speed in rad s⁻¹. Centripetal force F = mv²/r = mω²r is provided by tension, friction, gravity, or the normal reaction. Exam problems often ask for the minimum speed to complete a vertical loop.
物体以恒定速率做圆周运动时具有指向圆心的向心加速度：a = v²/r = ω²r，其中 v 是线速率，r 是半径，ω 是角速度（单位 rad s⁻¹）。向心力 F = mv²/r = mω²r 可由张力、摩擦力、重力或法向反作用力提供。考试常要求计算完成竖直圆周运动的最小速率。
A 0.40 kg mass is whirled in a horizontal circle of radius 0.80 m on a frictionless table, attached to a string that passes through a hole and supports a hanging 0.60 kg mass. Find the speed required to keep the hanging mass stationary. Tension in the string must equal the weight of the hanging mass: T = 0.60 × 9.81 = 5.886 N. This tension provides the centripetal force for the whirling mass: T = mv²/r → 5.886 = 0.40 × v² / 0.80 → v² = (5.886 × 0.80) / 0.40 = 11.772 → v = 3.43 m s⁻¹.
一个 0.40 kg 的物体在光滑桌面上做水平圆周运动，半径为 0.80 m，通过绳子穿过小孔悬挂一个 0.60 kg 的重物。求使悬挂重物保持静止所需的速率。绳中张力必须等于悬挂重物的重量：T = 0.60 × 9.81 = 5.886 N。该张力为旋转物体提供向心力：T = mv²/r → 5.886 = 0.40 × v² / 0.80 → v² = (5.886 × 0.80) / 0.40 = 11.772 → v = 3.43 m s⁻¹。
6. Simple Harmonic Motion | 简谐运动
Simple harmonic motion (SHM) occurs when acceleration is directly proportional to displacement from equilibrium and directed towards it: a = -ω²x. The time period for a mass-spring system is T = 2π√(m/k) and for a simple pendulum T = 2π√(l/g). Maximum speed v_max = ωA and maximum acceleration a_max = ω²A, where A is amplitude. AQA questions frequently require energy calculations using E_total = ½mω²A².
简谐运动（SHM）是指加速度与离开平衡位置的位移成正比且方向相反的运动：a = -ω²x。弹簧振子的周期 T = 2π√(m/k)，单摆的周期 T = 2π√(l/g)。最大速率 v_max = ωA，最大加速度 a_max = ω²A，其中 A 为振幅。AQA 题目常要求用总能量 E_total = ½mω²A² 进行能量计算。
A 0.25 kg mass on a spring oscillates with amplitude 0.030 m and period 0.50 s. Determine the spring constant and the maximum kinetic energy of the mass. From T = 2π√(m/k), we solve for k: T² = 4π² m/k, so k = 4π² m / T² = 4π² × 0.25 / 0.50² = (π² × 1.0) / 0.25 = 39.5 N m⁻¹. Angular frequency ω = 2π/T = 2π/0.50 = 4π rad s⁻¹. Maximum kinetic energy equals total energy: E = ½mω²A² = 0.5 × 0.25 × (4π)² × 0.030² = 0.125 × 16π² × 0.0009 = 0.125 × 0.142 = 0.0178 J.
一个 0.25 kg 的物体连接在弹簧上，以振幅 0.030 m、周期 0.50 s 振动。求弹簧的劲度系数和物体的最大动能。由 T = 2π√(m/k) 解出 k：T² = 4π² m/k，因此 k = 4π² m / T² = 4π² × 0.25 / 0.50² = 39.5 N m⁻¹。角频率 ω = 2π/T = 2π/0.50 = 4π rad s⁻¹。最大动能等于总能量：E = ½mω²A² = 0.5 × 0.25 × (4π)² × 0.030² = 0.125 × 16π² × 0.0009 ≈ 0.0178 J。
7. Electric Fields and Coulomb’s Law | 电场与库仑定律
Coulomb’s law describes the force between two point charges: F = kQq/r², where k = 8.99×10⁹ N m² C⁻². Electric field strength is force per unit charge, E = F/q. In a uniform field between parallel plates, E = V/d, where V is the potential difference and d is plate separation. Work done on a charge moving through a potential difference is W = QV.
库仑定律描述两个点电荷之间的作用力：F = kQq/r²，其中 k = 8.99×10⁹ N m² C⁻²。电场强度是单位电荷所受的力，E = F/q。在平行板间的匀强电场中，E = V/d，其中 V 是电势差，d 是板间距。电荷在电势差中移动所做的功为 W = QV。
Two small charged spheres, one of +4.0 μC and the other of -6.0 μC, are separated by 0.25 m in a vacuum. Calculate the magnitude of the electrostatic force between them and state whether it is attractive or repulsive. Using Coulomb’s law: F = (8.99×10⁹) × (4.0×10⁻⁶) × (6.0×10⁻⁶) / (0.25)² = (8.99×10⁹) × (24×10⁻¹²) / 0.0625 = (215.76×10⁻³) / 0.0625 = 3.45 N. The charges have opposite signs, so the force is attractive.
两个带电小球，一个带 +4.0 μC，另一个带 -6.0 μC，在真空中相距 0.25 m。计算它们之间静电力的大小，并说明是吸引力还是排斥力。根据库仑定律：F = (8.99×10⁹) × (4.0×10⁻⁶) × (6.0×10⁻⁶) / (0.25)² = (8.99×10⁹) × (24×10⁻¹²) / 0.0625 = 3.45 N。两电荷电性相反，因此是吸引力。
8. Circuit Analysis and Resistive Networks | 电路分析与电阻网络
Series and parallel resistor combinations form the basis of many circuit calculations. For series: R_total = R₁ + R₂ + …; for parallel: 1/R_total = 1/R₁ + 1/R₂ + … . Ohm’s law V = IR and Kirchhoff’s voltage and current laws are essential tools. Potential divider networks are particularly common in AQA problems, where the output voltage V_out = V_in × (R₂ / (R₁ + R₂)).
串并联电阻组合是许多电路计算的基础。串联：R_total = R₁ + R₂ + …；并联：1/R_total = 1/R₁ + 1/R₂ + …。欧姆定律 V = IR 以及基尔霍夫电压和电流定律是必不可少的工具。分压器网络在 AQA 题目中尤为常见，其输出电压 V_out = V_in × (R₂ / (R₁ + R₂))。
A 12.0 V battery of negligible internal resistance is connected to a network consisting of a 10 Ω resistor in series with a parallel combination of a 15 Ω and a 30 Ω resistor. Find the total current drawn from the battery and the potential difference across the 15 Ω resistor. The parallel section has resistance 1/R_p = 1/15 + 1/30 = 3/30 = 1/10, so R_p = 10 Ω. Total circuit resistance = 10 Ω + 10 Ω = 20 Ω. Total current I = V / R_total = 12.0 / 20 = 0.60 A. The potential difference across the parallel section is V_p = I × R_p = 0.60 × 10 = 6.0 V, and this is the voltage across the 15 Ω resistor.
一个内阻可忽略的 12.0 V 电池连接到一个网络，该网络由一个 10 Ω 电阻与一个 15 Ω 和 30 Ω 的并联组合串联而成。求电池提供的总电流以及 15 Ω 电阻两端的电势差。并联部分电阻为 1/R_p = 1/15 + 1/30 = 3/30 = 1/10，故 R_p = 10 Ω。电路总电阻 = 10 Ω + 10 Ω = 20 Ω。总电流 I = V / R_total = 12.0 / 20 = 0.60 A。并联部分的电势差 V_p = I × R_p = 0.60 × 10 = 6.0 V，这也是 15 Ω 电阻两端的电压。
9. Capacitors and Time Constant | 电容器与时间常数
A capacitor stores charge Q = CV, and the energy stored is E = ½CV². The time constant for an RC circuit is τ = RC, representing the time for the charge or voltage to fall to 37% of its initial value during discharge. Exponential decay is modelled by Q = Q₀ e^{-t/RC} and V = V₀ e^{-t/RC}. AQA questions frequently ask for calculations of half-life or the time to reach a specific voltage.
电容器储存电荷 Q = CV，储存的能量为 E = ½CV²。RC 电路的时间常数 τ = RC，代表放电过程中电荷或电压降至初始值 37% 所需的时间。指数衰减由 Q = Q₀ e^{-t/RC} 和 V = V₀ e^{-t/RC} 描述。AQA 试题经常要求计算半衰期或达到特定电压的时间。
A 220 μF capacitor is charged to 9.0 V and then discharged through a 47 kΩ resistor. Calculate the time constant and the time required for the voltage to drop to 1.5 V. Time constant τ = RC = (220×10⁻⁶) × (47×10³) = 10.34 s. Using V = V₀ e^{-t/τ}, we set 1.5 = 9.0 e^{-t/10.34} → e^{-t/10.34} = 1.5/9.0 = 1/6. Taking natural logs: -t/10.34 = ln(1/6) = -ln6, so t = 10.34 × ln6 ≈ 10.34 × 1.7918 = 18.5 s.
一个 220 μF 的电容器被充电至 9.0 V，然后通过一个 47 kΩ 的电阻放电。计算时间常数以及电压降至 1.5 V 所需的时间。时间常数 τ = RC = (220×10⁻⁶) × (47×10³) = 10.34 s。利用 V = V₀ e^{-t/τ}，令 1.5 = 9.0 e^{-t/10.34} → e^{-t/10.34} = 1.5/9.0 = 1/6。取自然对数：-t/10.34 = ln(1/6) = -ln6，因此 t = 10.34 × ln6 ≈ 10.34 × 1.7918 = 18.5 s。
10. Nuclear Physics and Radioactive Decay | 核物理与放射性衰变
Radioactive decay follows an exponential law: N = N₀ e^{-λt}, where λ is the decay constant. Activity A = λN, and the half-life T₁/₂ = ln2 / λ. In AQA papers, you may be asked to convert between half-life and decay constant, or to find the age of a sample using carbon-dating principles. Always pay attention to units of time.
放射性衰变遵循指数规律：N = N₀ e^{-λt}，其中 λ 是衰变常量。活度 A = λN，半衰期 T₁/₂ = ln2 / λ。在 AQA 试卷中，可能要求在半衰期和衰变常量之间转换，或利用碳定年法原则求出样本的年代。要始终注意时间单位。
A radioactive isotope has a half-life of 5.0 days. A sample initially contains 8.0×10¹⁵ nuclei. Calculate the decay constant in s⁻¹ and the number of nuclei remaining after 10 days. First convert half-life to seconds: T₁/₂ = 5.0 × 24 × 3600 = 4.32×10⁵ s. Decay constant λ = ln2 / T₁/₂ = 0.693 / (4.32×10⁵) = 1.60×10⁻⁶ s⁻¹. After 10 days (two half-lives), the fraction remaining is (1/2)² = 1/4, so nuclei remaining = 8.0×10¹⁵ × 1/4 = 2.0×10¹⁵.
某种放射性同位素的半衰期为 5.0 天。一个样本最初含有 8.0×10¹⁵ 个原子核。计算以 s⁻¹ 为单位的衰变常量以及 10 天后剩余的原子核数。首先将半衰期转换为秒：T₁/₂ = 5.0 × 24 × 3600 = 4.32×10⁵ s。衰变常量 λ = ln2 / T₁/₂ = 0.693 / (4.32×10⁵) = 1.60×10⁻⁶ s⁻¹。10 天后（两个半衰期），剩余比例为 (1/2)² = 1/4，因此剩余原子核数 = 8.0×10¹⁵ × 1/4 = 2.0×10¹⁵。
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