📚 GCSE WJEC Physics: Calculation Practice Drills | GCSE WJEC 物理：计算题专项训练
Strong calculation skills are essential for GCSE WJEC Physics. This workbook-style guide targets the most common numeric topics: motion, forces, energy, electricity, waves and matter. Each section breaks down a key formula, provides a worked example and sets a practice problem, so you can move from "I recognise the equation" to "I can score full marks under timed conditions". Work through all 12 drills in order, and always write the formula, the substitution and the unit on every line of working.
扎实的计算能力是 GCSE WJEC 物理成功的关键。这本练习指南针对最常见的数值题型：运动、力、能量、电学、波和物质。每个小节解析一个核心公式，给出例题并设置练习题，帮助你把"认得公式"变成"限时状态下拿满分"。请按顺序完成全部 12 个训练，并且每一步都写出公式、代入过程和单位。
1. Speed, Distance and Time | 速度、距离和时间
The average speed of an object is the distance travelled divided by the time taken. This is the most frequently examined equation in the whole WJEC specification, and it also appears inside wave, momentum and energy questions, so it must be automatic.
物体的平均速率等于它移动的距离除以所用的时间。这是整个 WJEC 考纲中出现频率最高的公式，它还会出现在波、动量和能量题目里，所以必须做到条件反射般的熟练。
v = s ÷ t
tes to seconds (× 60). A common trap is mixing km with seconds, which produces a meaningless answer.
其中 v 为速率（米/秒），s 为距离（米），t 为时间（秒）。代入前请务必统一单位：公里换米（× 1000），分钟换秒（× 60）。常见的失分点是"公里配秒"，这样算出的结果毫无意义。
Example: A runner completes 400 m in 50 s. Calculate the average speed.
例题：一名跑步者用 50 秒跑完 400 米，求平均速率。
Answer: v = s ÷ t = 400 ÷ 50 = 8 m/s.
解答：v = s ÷ t = 400 ÷ 50 = 8 m/s。
Practice: A cyclist travels 9 km in 30 minutes. Find the speed in m/s. (Convert first!)
练习：一骑自行车者 30 分钟骑行 9 公里，求速率（单位：m/s）。（先转换单位！）
Answer: s = 9 × 1000 = 9000 m; t = 30 × 60 = 1800 s; v = 9000 ÷ 1800 = 5 m/s.
答案：s = 9 × 1000 = 9000 m；t = 30 × 60 = 1800 s；v = 9000 ÷ 1800 = 5 m/s。
Remember that average speed is total distance divided by total time for the whole journey. Instantaneous speed is the speed at one particular moment, and it is what a speedometer shows. A car that drives 60 km/h for one hour and then stops for an hour has an average speed of 30 km/h, not 60 km/h.
请记住：平均速率是"全程总距离 ÷ 总时间"；瞬时速率是某一瞬间的速率，也就是速度表显示的数值。一辆车以 60 km/h 开一小时、再停一小时，它的平均速率是 30 km/h，而不是 60 km/h。
2. Acceleration | 加速度
Acceleration is the rate of change of velocity. The equation linking initial velocity (u), final velocity (v), time (t) and acceleration (a) is one of the highest-value equations in the motion topic.
加速度是速度的变化率。联系初速度（u）、末速度（v）、时间（t）和加速度（a）的公式是运动学部分分值最高的公式之一。
a = (v − u) ÷ t
where a is in m/s² (metres per second squared). Note the order of subtraction: it is always final minus initial. If the object slows down, the answer is negative and we call it a deceleration (or negative acceleration).
其中 a 的单位为 m/s²（米每二次方秒）。注意减法顺序：永远是"末速度减初速度"。如果物体在减速，结果为负值，我们称之为减速度（或负加速度）。
Example: A car accelerates from 5 m/s to 25 m/s in 4 s. Find the acceleration.
例题：一辆汽车在 4 秒内从 5 m/s 加速到 25 m/s，求加速度。
Answer: a = (25 − 5) ÷ 4 = 20 ÷ 4 = 5 m/s².
解答：a = (25 − 5) ÷ 4 = 20 ÷ 4 = 5 m/s²。
Practice: A train slows from 30 m/s to 8 m/s in 10 s. Calculate its acceleration and state whether it is accelerating or decelerating.
练习：一列火车在 10 秒内从 30 m/s 减速到 8 m/s，求它的加速度，并说明是加速还是减速。
Answer: a = (8 − 30) ÷ 10 = −22 ÷ 10 = −2.2 m/s², so it is decelerating.
答案：a = (8 − 30) ÷ 10 = −22 ÷ 10 = −2.2 m/s²，因此它在减速。
A negative sign alone does not lose marks, but the written statement "it is decelerating" often carries a separate mark. Also check the unit: acceleration is m/s², never m/s. If you write m/s, an examiner will treat the final answer as wrong even when the number is correct.
只写负号通常不扣分，但"它在减速"这句文字表述往往单独占一分。同时要检查单位：加速度单位是 m/s²，绝不是 m/s。如果写成 m/s，即使数字正确，考官也会判最终答案错误。
3. Motion Graphs: Gradient and Area | 运动图像：斜率与面积
Graph questions reward three simple rules. On a distance-time graph, the gradient equals speed. On a velocity-time graph, the gradient equals acceleration and the area under the line equals the distance travelled.
图像题只需掌握三条规则：在距离—时间图上，斜率等于速率；在速度—时间图上，斜率等于加速度，而图线与横轴之间的面积等于移动的距离。
gradient = Δy ÷ Δx area of triangle = ½ × base × height
Break a complicated velocity-time graph into rectangles and triangles, calculate each area separately, then add them. Take areas below the time axis as negative if the question asks for displacement rather than distance.
遇到复杂的速度—时间图，把它拆成矩形和三角形，分别求面积后相加。如果题目问的是位移而非路程，横轴下方的面积要取负值。
Example: A cyclist accelerates steadily from rest to 20 m/s in 5 s, then travels at a constant 20 m/s for 10 s. Find the total distance.
例题：一名骑行者从静止匀加速，5 秒后达到 20 m/s，随后以 20 m/s 匀速行驶 10 秒。求总距离。
Answer: triangle = ½ × 5 × 20 = 50 m; rectangle = 20 × 10 = 200 m; total = 250 m.
解答：三角形 = ½ × 5 × 20 = 50 m；矩形 = 20 × 10 = 200 m；总计 = 250 m。
Practice: A car brakes from 24 m/s to rest in 6 s. Use the graph idea (area of a triangle) to find the braking distance.
练习：一辆汽车在 6 秒内从 24 m/s 刹车至停止。用图像思路（三角形面积）求刹车距离。
Answer: ½ × 6 × 24 = 72 m.
答案：½ × 6 × 24 = 72 m。
Always label the axes with quantity and unit before you start, and check whether the axis says distance or displacement. Flat sections mean constant speed on a distance-time graph and constant velocity on a velocity-time graph, not "stationary".
动笔前先给坐标轴标上物理量和单位，并确认横轴写的是"距离"还是"位移"。在距离—时间图上，水平线段表示匀速；在速度—时间图上，水平线段表示匀速运动，而不是"静止"。
4. Weight and Newton's Second Law | 重力与牛顿第二定律
Weight is the force of gravity on a mass, and Newton's second law links force, mass and acceleration. Both appear every year in WJEC papers, usually in a context such as a lift, a rocket or a car.
重力是作用在质量上的引力，而牛顿第二定律把力、质量和加速度联系起来。这两个公式每年都会出现在 WJEC 试卷中，常见情境包括电梯、火箭或汽车。
W = m g F = m a
F is the resultant force in newtons, m is mass in kilograms and a is acceleration in m/s².
在 W = m g 中，g 是重力场强度，地球表面取 9.8 N/kg（部分 WJEC 试卷允许取 10 N/kg，请先阅读卷首说明）。在 F = m a 中，F 是合力，单位牛顿；m 是质量，单位千克；a 是加速度，单位 m/s²。
Example: A pupil of mass 60 kg stands on the floor. Calculate her weight. Then find the force needed to accelerate a 1500 kg car at 2 m/s².
例题：一名质量为 60 kg 的学生站在地面上，求她的重力；再求让一辆 1500 kg 的汽车以 2 m/s² 加速所需的力。
Answer: W = 60 × 9.8 = 588 N. F = 1500 × 2 = 3000 N.
解答：W = 60 × 9.8 = 588 N。F = 1500 × 2 = 3000 N。
Practice: A 0.5 kg ball is kicked with a resultant force of 6 N. Find its acceleration.
练习：一个 0.5 kg 的球受到 6 N 的合力，求它的加速度。
Answer: a = F ÷ m = 6 ÷ 0.5 = 12 m/s².
答案：a = F ÷ m = 6 ÷ 0.5 = 12 m/s²。
o a stationary 1 kg trolley, the combined speed is (2 × 3) ÷ (2 + 1) = 2 m/s, because momentum before equals momentum after.
拓展：动量定义为 p = m v，在封闭系统中总动量守恒。若一个 2 kg 的小车以 3 m/s 撞击并粘连在一个静止的 1 kg 小车上，则结合后的速率为 (2 × 3) ÷ (2 + 1) = 2 m/s，因为碰撞前后的动量相等。
5. Work Done, Gravitational and Kinetic Energy | 功、重力势能与动能
Work is done whenever a force moves an object through a distance. Energy transferred mechanically can be stored as gravitational potential energy or as kinetic energy, and WJEC almost always asks you to move between these three equations.
只要有作用力使物体移动一段距离，就有做功。机械方式转移的能量可以储存为重力势能或动能，而 WJEC 几乎每次都会要求你在这三个公式之间转换。
W = F d GPE = m g h KE = ½ m v²
Work done and energy are both measured in joules (J). A joule is the work done by a force of one newton moving an object one metre in the direction of the force. Notice that kinetic energy depends on the square of the speed, so doubling the speed gives four times the kinetic energy.
功和能量的单位都是焦耳（J）。1 焦耳等于 1 牛顿的力沿力的方向使物体移动 1 米所做的功。注意动能与速率的平方成正比，因此速率加倍，动能变为原来的四倍。
Example: A 2 kg box is lifted 3 m onto a shelf. Calculate the gain in gravitational potential energy. Then calculate the kinetic energy of a 1000 kg car travelling at 20 m/s.
例题：把一个 2 kg 的箱子抬到 3 m 高的架子上，求重力势能的增加量；再求一辆 1000 kg 的汽车以 20 m/s 行驶时的动能。
Answer: GPE = 2 × 9.8 × 3 = 58.8 J. KE = ½ × 1000 × 20² = ½ × 1000 × 400 = 200 000 J.
解答：GPE = 2 × 9.8 × 3 = 58.8 J。KE = ½ × 1000 × 20² = ½ × 1000 × 400 = 200 000 J。
Practice: A 0.4 kg ball is dropped from a height of 5 m. Assuming no air resistance, use energy conservation to find its speed just before it lands.
练习：一个 0.4 kg 的球从 5 m 高处落下。假设不计空气阻力，用能量守恒求它落地前的速率。
Answer: m g h = ½ m v², so v² = 2 × 9.8 × 5 = 98 and v = √98 ≈ 9.9 m/s. Notice that the mass cancels, so a 4 kg ball would land at the same speed.
答案：m g h = ½ m v²，所以 v² = 2 × 9.8 × 5 = 98，v = √98 ≈ 9.9 m/s。注意质量被消掉，所以 4 kg 的球落地速率完全相同。
In exam answers, always state the assumption "no energy lost to air resistance or friction" if the question says the surfaces are smooth or the motion is ideal. Total energy in equals total energy out, and in a real system the difference becomes thermal energy.
答题时，如果题目说明表面光滑或运动理想，一定要写出"忽略空气阻力或摩擦造成的能量损失"这一假设。输入总能量等于输出总能量，而在真实系统中，差额会转化为热能。
6. Power and Efficiency | 功率与效率
Power is the rate of transferring energy, and efficiency compares the useful energy output with the total energy input. These two calculations are often combined in a single WJEC question about appliances or motors.
功率是能量转移的快慢程度，效率则比较有用能量输出与总能量输入。在 WJEC 关于电器或电动机的题目中，这两个计算常常合并出现。
P = E ÷ t efficiency = useful output ÷ total input × 100%
re 1 W = 1 J/s. Efficiency can be given as a decimal or a percentage; if the question says "as a percentage", multiply by 100 and include the % sign.
功率的单位是瓦特（W），1 W = 1 J/s。效率可以用小数或百分数表示；如果题目要求"以百分数表示"，请乘 100 并写上 % 号。
Example: A motor transfers 3000 J of energy in 5 s. Find its power. A lamp takes in 50 J and gives out 8 J of useful light energy. Find its efficiency.
例题：一台电动机在 5 秒内转移 3000 J 能量，求其功率。一盏灯输入 50 J，输出 8 J 有用光能，求其效率。
Answer: P = 3000 ÷ 5 = 600 W. Efficiency = 8 ÷ 50 × 100% = 16%.
解答：P = 3000 ÷ 5 = 600 W。效率 = 8 ÷ 50 × 100% = 16%。
Practice: A lift raises its load with a useful energy output of 12 000 J in 20 s. The electrical input is 20 000 J. Find the power output and the efficiency.
练习：一台电梯在 20 秒内将负载提升，有用能量输出为 12 000 J，电输入为 20 000 J。求输出功率和效率。
Answer: P = 12 000 ÷ 20 = 600 W; efficiency = 12 000 ÷ 20 000 × 100% = 60%. The other 40% is transferred as thermal energy in the motor and cables.
答案：P = 12 000 ÷ 20 = 600 W；效率 = 12 000 ÷ 20 000 × 100% = 60%。其余 40% 以热能形式散失在电动机和线缆中。
Useful phrases for the "explain" part: "energy is dissipated to the surroundings", "friction between moving parts", "sound and thermal energy are wasted". WJEC marks these descriptions separately from the calculation, so never skip the sentence.
"解释"部分的实用表述包括："能量散发到周围环境"、"运动部件之间的摩擦"、"声能和热能被浪费"。WJEC 会把这类描述与计算分开给分，所以千万不要省略这句话。
7. Charge, Current and Time | 电荷、电流和时间
Current is the rate of flow of charge. This relationship underpins the whole electricity topic: if you cannot calculate charge, you will struggle with power and energy questions later.
电流是电荷流动的速率。这个关系式是整个电学部分的基础：如果不会计算电荷量，后面的功率和能量题目就会很吃力。
Q = I t
e coulomb is the charge passing a point when a current of one ampere flows for one second.
其中 Q 为电荷量，单位库仑（C）；I 为电流，单位安培（A）；t 为时间，单位秒（s）。时间必须用秒，分钟要乘 60。1 库仑等于 1 安培的电流在 1 秒内通过某点的电荷量。
Example: A current of 2 A flows through a lamp for 30 s. Calculate the charge that passes.
例题：2 A 的电流通过一盏灯持续 30 秒，求通过的电荷量。
Answer: Q = I t = 2 × 30 = 60 C.
解答：Q = I t = 2 × 30 = 60 C。
Practice: A battery delivers a charge of 720 C in 4 minutes. Find the average current.
练习：一块电池在 4 分钟内提供 720 C 电荷量，求平均电流。
Answer: t = 4 × 60 = 240 s; I = Q ÷ t = 720 ÷ 240 = 3 A.
答案：t = 4 × 60 = 240 s；I = Q ÷ t = 720 ÷ 240 = 3 A。
In a series circuit the current is the same at every point, so the same charge passes each component. In a parallel circuit the current splits, so the total charge per second at the cell equals the sum of the currents in the branches.
串联电路中各处电流相同，因此相同电荷量通过每个元件。并联电路中电流会分流，所以电源处的总电流等于各支路电流之和。
8. Ohm's Law, Series and Parallel Resistance | 欧姆定律与串并联电阻
Ohm's law describes the link between potential difference, current and resistance, and it is the single most examined electrical equation in GCSE WJEC Physics.
欧姆定律描述电压、电流和电阻之间的关系，是 GCSE WJEC 物理中考得最多的电学公式。
V = I R
nd R is resistance in ohms (Ω). Rearranged, R = V ÷ I and I = V ÷ R. A component obeys Ohm' rtional to its potential difference, which means a straight line through the origin on a V–I graph.
其中 V 为电压，单位伏特（V）；I 为电流，单位安培（A）；R 为电阻，单位欧姆（Ω）。变形可得 R = V ÷ I 和 I = V ÷ R。若某元件的电流与电压成正比，即 V–I 图像是一条过原点的直线，则该元件遵守欧姆定律。
Example: A 12 V supply drives a current of 0.5 A through a resistor. Calculate the resistance. Then find the current in a 20 Ω resistor connected to 6 V.
例题：12 V 电源通过一个电阻器产生 0.5 A 的电流，求该电阻；再求 20 Ω 电阻接在 6 V 电源上的电流。
Answer: R = 12 ÷ 0.5 = 24 Ω. I = 6 ÷ 20 = 0.3 A.p
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