WJEC物理 AS 矢量 运动学 动量 杨氏模量
WJEC 的 AS 物理第一张卷叫 Motion, Energy and Matter，1 小时 30 分钟里要考基本物理、运动学、动力学、能量概念、固体受应力、用辐射研究恒星、粒子与核结构七个主题。这篇按 WJEC 官方 spec（版本 3，2023 年 10 月）的条目顺序，把最常考的公式、官方样卷里出现过的数字与最常见的扣分点排在一起。所有卷面信息都取自 WJEC 官方 spec 与官方样卷 Specimen Assessment Materials，具体分值可以自己回查核对。选题池里写的 Motion Energy Charge 是 2008 年之前旧版 PH1 单元的标题，现行 spec 用的名字是 Motion, Energy and Matter。很多学生以为力学部分只要背公式，实际上样卷第 1 题第 (a) 问只给单位，让你证明 F = mv^2/2D 成立，考的是量纲而不是代数。下面按七个主题展开，每一块都给出一到两个可以直接套用的计算例子。
Unit 1 of WJEC AS Physics is called Motion, Energy and Matter, and its single 90-minute paper covers seven topics: basic physics, kinematics, dynamics, energy concepts, solids under stress, using radiation to investigate stars, and particles and nuclear structure. Everything quoted below comes from the WJEC specification (version 3, October 2023) and the official Specimen Assessment Materials, so every mark tariff can be checked against the published papers. The older label Motion Energy Charge belongs to the pre-2008 PH1 unit, and the current paper rewards unit checking and graph reading as heavily as algebra.
官方卷面结构与七个主题 / Paper Structure and the Seven Topics
Unit 1 与 Unit 2 共同构成 AS，两张卷各 1 小时 30 分钟、各 80 分，各占整个 A Level 的 20%，也就是各占 AS 的一半。官方样卷把 80 分拆成七道题，分值依次是 12、11、14、9、15、11、8，其中第 5 题 (b) 是作答质量（QER）的评分点。考生须知写明除了试卷还要带计算器和 Data Booklet，数据手册随卷提供常用公式与常数。A Level 阶段再加 Unit 3（2 小时 15 分钟、100 分、25%）、Unit 4（2 小时、100 分、25%）与 Unit 5 实验考试（50 分、10%，含 25 分实验任务与 25 分数据分析任务）。很多学生以为 Unit 1 只是力学入门，实际上它的 80 分里有相当一部分落在固体受应力与粒子、核结构这两块偏材料的内容上。样卷第 1 题就用一道量纲题开场，第一分起就要求你把单位写清楚。
Unit 1 and Unit 2 together make up the AS award, each running 1 hour 30 minutes for 80 marks, which is 20% of the A Level and half of the AS. The official specimen paper splits those 80 marks across seven questions worth 12, 11, 14, 9, 15, 11 and 8, with the quality of extended response assessed in question 5(b); candidates are told to bring a calculator and the Data Booklet that carries the equations and constants. The A Level adds Unit 3 (2 hours 15 minutes, 100 marks, 25%), Unit 4 (2 hours, 100 marks, 25%) and the Unit 5 practical examination (50 marks, 10%, split into a 25-mark experimental task and a 25-mark analysis task).
六个基本单位与量纲校验 / Base SI Units and Homogeneity
基本物理在 spec 里的条目 (a) 到 (k) 覆盖六个基本 SI 单位 kg、m、s、A、mol、K，用基本单位改写导出单位，以及用量纲判断方程是否成立。样卷第 1 题 (a) 给了一个待检验的关系 F = mv^2/2D，要求只用单位说明它成立：左边牛顿写成 kg m s^-2，右边是 kg 乘以 (m s^-1)^2 再除以 m，展开同样是 kg m s^-2。焦耳与瓦特要背成 J = kg m^2 s^-2 与 W = kg m^2 s^-3，考试时没有时间现场拆。学生最常把量纲校验做成两边数值相除，阅卷时这类答案不给分，因为检验只看单位指数是否相等，数字本身不参与判断。密度 ρ = m/V 这一条也挂在基本物理里，规则固体的体积先用游标卡尺或千分尺量边长再算，量具读数要估读到最小分度的下一位。官方指定的实验任务里，测量固体密度与用杠杆原理测未知质量就是从这里派生的。
Basic physics runs from item (a) to item (k) in the specification: the six base SI units are kg, m, s, A, mol and K, and derived units such as the newton, joule and watt have to be rewritten in terms of them. In question 1(a) of the specimen paper candidates show that F = mv^2/2D is homogeneous, with the newton on the left becoming kg m s^-2 and the right-hand side reducing to the same combination. Density belongs to the same topic, and the volume of a regular solid comes from vernier calipers or a micrometer read one place beyond the smallest division.
标量与矢量、分解与共面合成 / Scalars, Vectors and Resolution
spec 条目 (d) 要求分清标量与矢量，位移、速度、加速度、力属于矢量，速率、时间、密度、压强属于标量。条目 (e) 把运算限制在两个互相垂直的矢量上，所以考场上最常见的两条路是勾股定理与三角函数，而不是复杂的矢量多边形。把一个力分解成两个垂直分量时，斜面是最典型的场景：重量 mg 沿斜面方向的分量是 mg sinθ，垂直斜面的分量是 mg cosθ。以 2.0 kg 的物块放在 30 度斜面上算，沿斜面向下的分量是 2.0 乘以 9.8 再乘以 0.50，等于 9.8 N。多个方向做功时用 W = Fx cosθ，4.0 N 的力与位移成 60 度、移动 3.0 m，功就是 4.0 乘以 3.0 再乘以 0.50，等于 6.0 J。常见的错误是角度取错，题目给的是力与竖直方向的夹角，却直接代进 cosθ 当成了与位移的夹角。
Items (d) to (f) of the first topic ask for the scalar and vector distinction, the addition and subtraction of coplanar vectors limited to two perpendicular directions, and the resolution of a vector into two perpendicular components. On an incline the weight mg splits into mg sinθ along the slope and mg cosθ perpendicular to it, so a 2.0 kg block on a 30 degree slope has a component of 9.8 N pulling it downhill. Work done by a force at an angle uses W = Fx cosθ, which gives 6.0 J for 4.0 N acting at 60 degrees over 3.0 m.
密度、力矩平衡与重心 / Density, Moments and Centre of Gravity
力矩等于力乘以力臂，spec 条目 (i) 与 (j) 要求用杠杆原理求未知质量，并能说出圆柱、球与长方体均匀体的重心位置。平衡条件是合力为零且净力矩为零，两句话都要出现在过程里，只写力矩相等往往会丢一分。样卷第 4 题用米尺、一个 0.2 N 的砝码和一只试管测滚珠质量：砝码挂在离左端 4.0 cm 处，试管挂在离支点 38.4 cm 处，取支点力矩相等，0.2 乘以 4.0 等于 W 乘以 38.4，得到重量约 0.021 N，质量约 2.1 g。密度的公式是 ρ = m/V，同一个物体分成几块时密度不变，体积与质量按比例分配。错得最多的是把力臂当成物体长度，而不是支点到力的作用线的垂直距离。阅卷时看到先写平衡条件再代入数据的卷面，即使算术出错也能拿到方法分。
The turning effect of a force is measured by the moment, force times perpendicular distance from the pivot, and equilibrium needs both the resultant force and the net moment to be zero. In the specimen paper a metre ruler, a 0.2 N weight and a test tube are used to find the mass of a ball bearing: balance at 4.0 cm and 38.4 cm gives a weight near 0.021 N and a mass of about 2.1 g. Density is mass over volume, and splitting an object leaves the density unchanged while the mass and volume stay in proportion.
suvat 公式与运动图像 / suvat Equations and Motion Graphs
运动学要求掌握位移、平均速度与瞬时速度、平均加速度与瞬时加速度，并能从位移-时间图读到速度、从速度-时间图读到加速度与位移。匀加速公式的结论都印在 Data Booklet 上，样卷第 3 题 (a) 却要求用速度-时间图自己推出 x = ½at^2，思路是图线下方三角形的面积等于 ½ 乘以底乘以高。数值练习可以这样配：一辆车 8.0 s 内从静止加到 20 m/s，加速度是 20 除以 8.0 等于 2.5 m/s^2，这 8.0 s 内走过的距离是 ½ 乘以 2.5 乘以 8.0 的平方，等于 80 m。速度-时间图里曲线下方的面积代表位移，斜率代表加速度，负斜率不等于减速，只代表加速度方向与所取正方向相反。样卷第 3 题 (b) 给的是竖直上抛：小球在释放后 0.3 s 与 1.9 s 各经过同一窗台一次，对称中点 1.1 s 就是最高点，初速度是 9.8 乘以 1.1 约 10.8 m/s，窗台高度约 2.8 m。很多学生以为图像题只靠看图，评分点却要求写出读图依据，例如斜率等于加速度、面积等于位移。
Kinematics covers displacement, mean and instantaneous values of speed, velocity and acceleration, and the graphical representation of each one. The uniformly accelerated equations are printed in the Data Booklet, yet question 3(a) of the specimen paper still asks candidates to derive x = ½at^2 from the area of a velocity-time triangle, and a car reaching 20 m/s in 8.0 s accelerates at 2.5 m/s^2 over 80 m. The same paper throws a ball past a window ledge at 0.3 s and again at 1.9 s, so the midpoint at 1.1 s marks the top of the flight and gives an initial speed near 10.8 m/s with a ledge height near 2.8 m.
自由落体、终端速度与抛体 / Free Fall, Terminal Velocity and Projectiles
spec 条目 (e) 要求描述有空气阻力和没有空气阻力时物体在重力场中的运动，终端速度是其中的关键词：阻力随速度增大，直到与重量相等，加速度变成零。条目 (f) 与 (g) 考的是竖直方向与水平方向的独立性，抛体可以当成一个匀速直线运动加一个自由落体运动来处理。给一个具体数字：从 45 m 高的悬崖以 20 m/s 水平抛出，落地时间只看竖直方向，t 等于根号下 2 乘以 45 除以 9.8，约 3.0 s，水平距离是 20 乘以 3.0，等于 60 m。落到地面时的竖直速度是 9.8 乘以 3.0 约 29 m/s，合成速度要用勾股定理，方向由反正切给出。很多学生以为抛得越快落得越晚，其实竖直方向的运动与水平速度无关，只是水平距离更远。指定实验里测量重力加速度用自由落体配合光电门或电磁铁计时，g 由 2h 除以 t^2 得到，这一步最容易丢在高度与时间的单位换算上。
Item (e) describes motion in a gravitational field with and without air resistance, where drag grows with speed until it balances the weight and the acceleration falls to zero. Vertical and horizontal motion are independent, so a projectile can be treated as constant velocity across plus free fall down; a ball thrown horizontally at 20 m/s from 45 m takes about 3.0 s to land and travels 60 m horizontally. The specified practical for this topic measures g by freefall, using light gates or an electromagnet release and the relationship g = 2h/t^2.
受力分析图与牛顿第二定律 / Free Body Diagrams and F = ma
动力学从力的概念与牛顿第三定律开始，spec 条目 (a) 到 (c) 要求会画受力分析图，并在质量不变时用 F = ma 求解。受力分析图要把每个力画在作用点上并标注名称，常见的四个力是重量、支持力、摩擦力与外力，画漏一个就会在后续计算里连锁出错。指定实验是验证牛顿第二定律，用小车、配重片和数据记录仪测出不同拉力下的加速度，再把 a 对 F 作图，斜率是质量的倒数。样卷第 1 题 (b) 给了同一类数据：拉力小于 0.5 N 时加速度为零，原因是静摩擦力与外力平衡，只有超过这个限度物体才开始动。以 0.50 kg 的小车为例，加 5.0 N 的拉力，忽略摩擦时 a = 10 m/s^2，摩擦为 1.0 N 时实际加速度变成 8.0 m/s^2。学生最常把摩擦力方向画反，或者把支持力当成重量的等值力写进方程，只有在水平面上两者大小才相等。
Dynamics starts from the concept of force and Newton's third law, then moves on to free body diagrams and F = ma for constant mass. A free body diagram needs every force labelled at its point of action, and the specified practical investigates Newton's second law with a trolley, masses and a data logger, plotting acceleration against force to get a gradient of 1/m. The specimen paper shows zero acceleration below 0.5 N because static friction balances the applied force, and a 0.50 kg trolley pulled by 5.0 N accelerates at 10 m/s^2, or 8.0 m/s^2 once 1.0 N of friction is included.
动量守恒与弹性非弹性碰撞 / Momentum Conservation and Collisions
动量是质量与速度的乘积，spec 条目 (e) 把冲量写成动量变化率，即 F 等于 Δp 除以 Δt，这条式子在解释缓冲装置时最好用。条目 (f) 要求用动量守恒处理一维的弹性与非弹性碰撞，弹性碰撞动能不损失，非弹性碰撞动能有损失，粘在一起的碰撞一定属于后者。算一个例子：2.0 kg 的台车以 3.0 m/s 撞上静止的 1.0 kg 台车并粘在一起，碰撞后速度是 6.0 除以 3.0 等于 2.0 m/s，碰前动能 9.0 J，碰后 6.0 J，损失 3.0 J。这 3.0 J 的去向要写清楚，转化成了内能与声能，只写损耗了会被判概念不清。安全气囊与溃缩区是官方给定的应用背景，延长作用时间可以减小平均力，动量变化不变，力与时间成反比。阅卷时常见的失分是漏掉方向符号，定好向右为正之后向左的速度必须带负号，两个物体反向运动的题尤其明显。
Momentum is mass times velocity, and the rate of change of momentum gives the force in F = Δp/Δt, which is the cleanest way to explain crumple zones and air bags. Conservation of momentum handles one-dimensional collisions, with kinetic energy unchanged in an elastic collision and reduced in an inelastic one, as when two trolleys stick together. A 2.0 kg trolley at 3.0 m/s hitting a stationary 1.0 kg trolley gives 2.0 m/s afterwards, dropping the kinetic energy from 9.0 J to 6.0 J and transferring 3.0 J to internal and sound energy.
功、动能、势能与功能关系 / Work, Kinetic Energy and the Work-Energy Relationship
能量概念的核心公式是功 W = Fx cosθ、动能 ½mv^2、重力势能 mgΔh 与弹性势能 ½kx^2，spec 条目 (a) 到 (d) 要求把它们与功能关系 Fx = ½mv^2 减去 ½mu^2 连起来用。样卷第 2 题给了最典型的一道：弓弦的水平拉力 800 N，弦中张力约 1170 N，储存的能量就是力-伸长曲线下方的面积。用 ½Fx 代入 1170 N 与约 0.43 m 的拉伸量，得到约 250 J，一支 50 g 的箭以 100 m/s 离弦，动能是 ½ 乘以 0.05 乘以 10000，同样是 250 J，两条结果对上说明题目里的能量是自洽的。常见的错误是把功写成 Fx 而忽略角度，力垂直于位移时 cos90 等于零，做功为零，例如水平提着箱子走路，重力不做功。功能关系最常用来避开加速度，尤其是变力或路径不明确时，直接从能量差得到速度比用运动学方程省一步。
The energy topic links work, kinetic energy, gravitational potential energy and elastic potential energy through the work-energy relationship Fx = ½mv^2 - ½mu^2. In the specimen paper a longbow with a draw force of 800 N and a string tension near 1170 N stores roughly 250 J, and a 50 g arrow leaving at 100 m/s carries the same 250 J, so the two routes agree and the answer checks itself. Work is Fx cosθ, which vanishes when the force is perpendicular to the displacement, and the energy route often avoids finding an acceleration that the question never asks for.
功率、效率与耗散力 / Power, Efficiency and Dissipative Forces
功率是能量转移的速率，公式 P = W 除以 t 与 P = Fv 都要会用，考试里更常见的是后者，因为题目往往给恒定速度与牵引力。效率的定义是有效能量转移除以总能量输入再乘以 100%，这个百分数必须落在 0 到 100 之间，算出 120% 说明分子分母选错了。算一个例子：50 kg 的学生 8.0 s 内爬高 5.0 m，克服重力做功是 50 乘以 9.8 乘以 5.0 等于 2450 J，功率约 306 W。耗散力指摩擦与阻力，它们把机械能转移到周围环境，使系统效率下降，题目常要求写出这部分能量的去向，而不是写损失了。很多学生以为效率题只要做除法，评分点却要求先分别写出有效转移与总输入两项，再相除。阅卷时看到只给一个百分数、没有中间两项的卷面，方法分通常拿不到。
Power is the rate of energy transfer, and P = Fv is the form that appears most often because questions supply a constant speed and a driving force. Efficiency is useful energy transfer divided by total energy input, expressed as a percentage, so a 50 kg student climbing 5.0 m in 8.0 s does 2450 J of work against gravity at about 306 W. Dissipative forces such as friction and drag move energy out of the system, and answers should say where that energy goes rather than calling it lost.
应力、应变与杨氏模量实验 / Stress, Strain and the Young Modulus Practical
胡克定律写成 F = kx，其中 k 是产生单位伸长所需的力，spec 条目 (b) 再往上加一层：拉伸应力 σ = F 除以 A，拉伸应变 ε = Δl 除以 l，杨氏模量 E = σ 除以 ε，条件是胡克定律成立。形变过程做的功等于力-伸长图下方的面积，胡克定律成立时是 ½Fx，这条结论把能量与材料性质接上了。算一根直径 0.40 mm 的钢丝：横截面积是 π 乘以 0.20 毫米的平方，约 1.26 x 10^-7 m^2，挂 20 N 的砝码得到应力约 1.6 x 10^8 Pa。1.500 m 的丝伸长 1.8 mm 时应变是 1.2 x 10^-3，杨氏模量约 1.3 x 10^11 Pa。官方指定的实验是测量金属丝的杨氏模量，用螺旋测微计量直径、用标尺量伸长，直径要取多点平均值，因为直径在面积里是平方项，误差会被放大一倍。这一步最容易做成只量一次直径，得到的 E 与手册值差出一个量级。
Hooke's law reads F = kx, with k the force per unit extension, and topic 5 adds tensile stress as F/A, tensile strain as Δl/l and the Young modulus as stress over strain while Hooke's law holds. The work done in deforming a solid equals the area under a force-extension graph, which is ½Fx within the elastic region; a 0.40 mm diameter wire has an area near 1.26 x 10^-7 m^2, so 20 N gives about 1.6 x 10^8 Pa and an extension of 1.8 mm over 1.500 m gives a modulus near 1.3 x 10^11 Pa. The specified practical determines that modulus for a metal wire, and the diameter has to be averaged over several readings because it enters the area as a squared term.
金属、脆性材料与橡胶的力-伸长曲线 / Force-Extension Curves for Metals, Brittle Materials and Rubber
spec 条目 (d) 把固体分成晶体、非晶体（含玻璃与陶瓷）与高分子三类，条目 (e) 到 (g) 要求分别读出金属、脆性材料与橡胶曲线的特征。铜这类金属的曲线先走线性弹性段，然后进入塑性段，位错运动带来永久形变，加外来原子、加晶界或让位错互相阻挡都能提高强度，接近断裂时出现颈缩与延性断裂。玻璃这类脆性材料一直遵守胡克定律直到断裂，裂纹从表面缺陷扩展，把表面压上预应力或用细纤维都能提高断裂应力，钢化玻璃与预应力混凝土就是这两个思路。橡胶只是近似线性，杨氏模量很低，伸长来自分子链被拉直并抵抗热运动，加载与卸载曲线不重合的现象叫滞后。复习可以排成四周：第一周六个基本单位、矢量与力矩，第二周运动学与自由落体测 g，第三周动力学与动量，第四周能量、功率与杨氏模量实验，每周配一套真题，QER 那道题单独写一遍。
Solids are classified as crystalline, amorphous or polymeric, and the force-extension curves differ sharply between a metal such as copper, a brittle material such as glass, and rubber. A metal yields plastically once dislocations move, and its strength rises when foreign atoms, grain boundaries or other dislocations block that movement, ending in necking and ductile fracture. Glass obeys Hooke's law up to brittle fracture and is strengthened by surface compression, as in toughened glass and pre-stressed concrete, while rubber only roughly follows Hooke's law and shows hysteresis.
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