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

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

Mastering numerical problems is crucial for success in CCEA A-Level Physics. This drill focuses on the most common calculation types across the AS and A2 units, reinforcing equation recall, unit conversion and a systematic problem-solving routine. Work through each worked example carefully, then cover the solution and repeat it from memory.

掌握数值计算题对于 CCEA A-Level 物理考试成功至关重要。本专项训练聚焦 AS 和 A2 单元中最常见的计算题型，强化公式记忆、单位转换和系统的解题流程。请认真完成每个典型例题，随后遮住解答，凭记忆重做一遍。

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1. Why Calculation Skills Decide Your Grade | 计算能力为何决定你的成绩

In CCEA A-Level Physics, a large share of the marks in every written paper depends on producing a correct numerical answer under time pressure. A candidate who understands the physics but slips in algebra, prefixes or significant figures loses marks that were entirely within reach.

在 CCEA A-Level 物理中，每份笔试试卷都有相当大比例的分数取决于能否在时间压力下写出正确的数值答案。理解了物理却因代数、词头或有效数字出错而丢分的考生，失去的其实是完全可以拿到的分数。

The mark schemes reward method as well as the final answer. Substituting correctly into a named equation, showing the rearrangement and stating the unit can earn most of the marks even when the arithmetic goes wrong, so never hide your working.

评分方案既奖励最终答案，也奖励解题方法。正确代入已写出的公式、展示移项过程并写出单位，即使算术出错也能拿到大部分步骤分，因此绝不要隐藏你的解题过程。

Every calculation in this article follows the same five-step routine: identify the known quantities, select the equation, convert to SI units, substitute and evaluate, then sanity-check the magnitude and the unit of the result.

本文中每道计算题都遵循同样的五步流程：确定已知量、选择公式、换算为国际单位制、代入并求值，最后检查结果的数量级与单位是否合理。

Typical CCEA calculation topics include kinematics, Newton's laws, momentum, work and energy, DC circuits, waves and photons, thermal physics, and nuclear decay. The drills below cover each of these in turn.

CCEA 常见的计算主题包括运动学、牛顿定律、动量、功与能量、直流电路、波与光子、热学以及核衰变。下文依次覆盖这些内容。

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2. The Core Toolkit: Units, Prefixes and Significant Figures | 核心工具：单位、词头与有效数字

Almost every lost mark in a calculation can be traced back to a unit error. Before substituting, convert mass to kg, length to m, time to s, current to A and temperature to K. A speed given in km h⁻¹ must be divided by 3.6 to become m s⁻¹.

计算题中几乎每一处失分都可以追溯到单位错误。代入之前，先把质量换算为 kg、长度换算为 m、时间换算为 s、电流换算为 A、温度换算为 K。以 km h⁻¹ 给出的速度必须除以 3.6 才能变成 m s⁻¹。

s. Converting 3.5 μC to 3.5 × 10⁻⁶ C, or 2.4 GJ to 2.4 × 10⁹ J, should take only a second.

下表列出了 CCEA 试卷中要求你一眼就能认出的词头。把 3.5 μC 写成 3.5 × 10⁻⁶ C，或把 2.4 GJ 写成 2.4 × 10⁹ J，应该只需一秒钟。

Prefix	Symbol	Factor
pico	p	10⁻¹²
nano	n	10⁻⁹
micro	μ	10⁻⁶
milli	m	10⁻³
kilo	k	10³
mega	M	10⁶
giga	G	10⁹

For significant figures, quote the final answer to the same number of significant figures as the least precise value in the question, normally two or three. Intermediate values should be kept to full calculator accuracy and only rounded at the end.

关于有效数字，最终答案应与题目中精度最低的数据保持相同的有效数字位数，通常取两到三位。中间过程应保留计算器的完整精度，只在最后一步取整。

W h = 3.6 × 10⁶ J. Writing them on the first line of your rough work prevents panic later.

值得背下来的常用换算包括 1 eV = 1.60 × 10⁻¹⁹ J、1 u = 1.66 × 10⁻²⁷ kg、0 °C = 273 K 以及 1 kW h = 3.6 × 10⁶ J。在草稿纸第一行写下它们，可以避免后面慌乱。

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3. Kinematics: SUVAT Equations Worked Drill | 运动学：匀加速直线运动方程实战

Question: a cyclist accelerates uniformly from 3 m s⁻¹ to 9 m s⁻¹ while covering a distance of 30 m. Determine the acceleration and the time taken.

题目：自行车手从 3 m s⁻¹ 匀加速至 9 m s⁻¹，位移为 30 m。求加速度和所用时间。

Step 1 – ty v = 9 m s⁻¹, displacement s = 30 m. The unknowns are the acceleration a and the time t.

步骤 1 – 确定已知量：初速度 u = 3 m s⁻¹，末速度 v = 9 m s⁻¹，位移 s = 30 m。未知量为加速度 a 和时间 t。

Step 2 – Choose the equation that contains the three knowns and the first unknown, and that does not contain time:

步骤 2 – 选用包含三个已知量和第一个未知量、且不含时间的方程：

v² = u² + 2as

Step 3 – 2 m s⁻². The positive sign confirms that the acceleration acts in the direction of motion.

步骤 3 – 移项并代入：a = (v² − u²) / (2s) = (9² − 3²) / (2 × 30) = (81 − 9) / 60 = 72 / 60 = 1.2 m s⁻²。正号表示加速度方向与运动方向相同。

Step 4 – Find the time using v = u + at, so t = (v − u) / a = (9 − 3) / 1.2 = 6 / 1.2 = 5.0 s.

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步骤 4 – 用 v = u + at 求时间，t = (v − u) / a = (9 − 3) / 1.2 = 6 / 1.2 = 5.0 s。

Step 5 – 0 m, which matches the data given. Always finish a kinematics question with an independent check like this.

步骤 5 – 用平均速度验证：s = ½(u + v)t = ½(3 + 9) × 5.0 = 30 m，与题给位移一致。运动学题目务必用这种独立方法作最终检验。

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4. SUVAT Variable Reference Table | SUVAT 变量速查表

The five SUVAT variables form a closed set. If you can list what you know and what you need, the correct equation is usually determined automatically, because each equation omits exactly one variable.

五个 SUVAT 变量构成一个完整的集合。只要能列出已知量和所求量，正确的公式通常就自动确定了，因为每个方程恰好缺少一个变量。

Variable	Meaning	SI unit
u	initial velocity 初速度	m s⁻¹
v	final velocity 末速度	m s⁻¹
a	acceleration 加速度	m s⁻²
s	displacement 位移	m
t	time 时间	s

A second table shows which variable is missing from each of the five standard equations. This is the fastest way to select the right equation in an exam.

第二张表显示五个标准方程各自缺少哪个变量。这是考场上选择正确方程最快的方法。

Equation	Missing variable
v = u + at	s
s = ut + ½at²	v
v² = u² + 2as	t
s = ½(u + v)t	a
s = vt − ½at²	u

l components separately, using a = 0 horizontally and a = 9.81 m s⁻² vertically downwards.

请记住，这些方程只适用于加速度恒定的情形。处理抛体运动时，把水平与竖直分量分开处理：水平方向 a = 0，竖直向下方向 a = 9.81 m s⁻²。

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5. Newton's Laws and Connected Bodies | 牛顿定律与连接体

Newton' force, W = mg, and must be calculated before applying any vertical equilibrium condition.

牛顿第二定律是力学的核心工具：合力等于质量乘以加速度。重力是一种力，W = mg，在应用任何竖直方向平衡条件之前必须先算出它。

F = ma    W = mg    F = μN

s⁻². If a friction force of 5.0 N also acts, the resultant becomes 15 N and a = 3.0 m s⁻².

例题：光滑水平面上质量为 5.0 kg 的物块受到 20 N 的水平拉力。加速度 a = F / m = 20 / 5.0 = 4.0 m s⁻²。若同时存在 5.0 N 的摩擦力，合力变为 15 N，则 a = 3.0 m s⁻²。

θ, so the acceleration down the slope is g sin θ. At θ = 30°, a = 9.81 × 0.5 = 4.9 m s⁻².

对于倾角为 θ 的光滑斜面，重力沿斜面方向的分量为 mg sin θ，法向反力为 mg cos θ，因此沿斜面下滑的加速度为 g sin θ。当 θ = 30° 时，a = 9.81 × 0.5 = 4.9 m s⁻²。

ne body to find the tension. With masses 3.0 kg and 1.0 kg hanging freely, a = (3.0 − 1.0) × 9.81 / (3.0 + 1.0) = 4.9 m s⁻², and the tension is T = 1.0 × (9.81 + 4.9) = 14.7 N.

对于用轻质不可伸长细绳跨过光滑滑轮连接的物体，先把整体看作一个系统求加速度，再隔离其中一个物体求绳中张力。若两端悬挂质量分别为 3.0 kg 与 1.0 kg，则 a = (3.0 − 1.0) × 9.81 / (3.0 + 1.0) = 4.9 m s⁻²，张力 T = 1.0 × (9.81 + 4.9) = 14.7 N。

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6. Momentum, Impulse and Collisions | 动量、冲量与碰撞

t time, Ft = Δp, which explains why a longer collision time reduces the force experienced.

线动量是质量与速度的乘积，p = mv，它是一个矢量。冲量是力与作用时间的乘积，Ft = Δp，这解释了为什么碰撞时间越长、物体所受的力越小。

p = mv    Ft = Δp    Σp_before = Σp_after

−8.0 − 12) = −3.0 kg m s⁻¹, so the magnitude of the impulse is 3.0 N s and it acts away from the wall.

例题：质量为 0.15 kg 的球以 12 m s⁻¹ 撞向墙壁，再以 8.0 m s⁻¹ 反弹。取初速度方向为正方向，Δp = m(v − u) = 0.15 × (−8.0 − 12) = −3.0 kg m s⁻¹，故冲量大小为 3.0 N s，方向背离墙壁。

. If safety padding increases the contact time to 0.30 s, the force falls to 10 N, which is exactly why crumple zones and helmets work.

若接触时间为 0.060 s，平均力为 F = Δp / t = 3.0 / 0.060 = 50 N。若安全衬垫把接触时间延长到 0.30 s，力就降到 10 N，这正是溃缩吸能区和头盔的设计原理。

r. Total momentum before is 8.0 kg m s⁻¹, so the common velocity is 8.0 / 5.0 = 1.6 m s⁻¹.

动量守恒例题：质量为 2.0 kg 的小车以 4.0 m s⁻¹ 运动，与静止的 3.0 kg 小车碰撞后粘在一起运动。碰前总动量为 8.0 kg m s⁻¹，故共同速度 v = 8.0 / 5.0 = 1.6 m s⁻¹。

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