AQA物理 力学 运动学 suvat 牛顿定律 动量
AQA 的 A-Level Physics 7408 把力学与材料放在 Section 3.4，Paper 1 覆盖 Section 3.1 到 3.5，时长两小时、满分 85，占总分 34%（AQA 7408 specification）。力学题的计算难度不高，但过程分给得细，公式、代入、结果与单位各占一点。下面按考纲顺序把运动学、抛体、牛顿定律、力矩、动量、功与能、弹簧、杨氏模量、终端速度与必做实验拆成十节，每节给出考纲要求、常见错法和可以照着写的答题格式。
Section 3.4 of the AQA A-Level Physics 7408 specification covers mechanics and materials, and Paper 1, which assesses Sections 3.1 to 3.5 over two hours and 85 marks for 34 per cent of the qualification, is where most of it is examined. The calculations are rarely hard; the marks are lost in method, because the equation, the substitution and the answer with its unit are each worth something. The ten sections below follow the specification order and pair each requirement with the mistakes that show up in scripts and the layout that earns the marks.
一、v-t 图与 suvat 的适用条件
AQA 7408 的运动学题大多从图像入手：v-t 图的斜率是加速度，图线与时间轴围出的面积是位移，落在轴下方的部分要记负值。四个 suvat 公式只在加速度恒定时成立，题目出现「速度逐渐减小」这类描述时，要先判断是不是匀减速。选公式的办法是列出已知的三个量和要求的量，再找不含多余字母的那一条，比硬代数省时间。常见的错误是把厘米直接代进以米为单位的公式，或者方向符号处理反了，算出的数值差一个量级。位移是矢量，往返一次位移可以为零，而路程一直在增加，选择题常在这里设陷阱。AQA 的评分细则一般给公式 1 分、代入 1 分、结果与单位 1 分，写出公式再代入比直接写答案稳。
Most kinematics questions in 7408 start from a graph, where the gradient of a velocity-time plot is acceleration and the enclosed area is displacement, negative below the time axis; the four suvat equations apply only to constant acceleration, so establish that condition before substituting anything. Choose an equation by listing the three known quantities and the unknown, then take the one that omits the unused symbol. Converting centimetres to metres and keeping a consistent positive direction are the two steps where marks disappear most often. Displacement is a vector, so a there-and-back trip can end at zero displacement while distance keeps growing, and AQA mark schemes normally give one mark for the equation, one for substitution and one for the answer with its unit.
二、抛体的水平与竖直分运动
抛体题的固定做法是把运动拆成水平与竖直两条线：忽略空气阻力时水平方向没有力，速度保持不变；竖直方向只有重力，加速度取 9.81 m/s^2 向下。两条线共享同一个时间，所以先用能算的方向求时间，再代到另一边求位移或速度。很多学生以为越重的物体飞得越远，实际上质量在无阻力抛体里被约掉，射程由初速度和发射角决定。常见的错误是记住最高点竖直速度为零，却忘了那里的水平速度仍然存在，结果整道题卡住。题目给出射程或最大高度时，用竖直方向的对称性最快，上升时间和下落时间相等。AQA 的数据表把 g 写成 9.81 m/s^2，答题时只用 9.8 而答案要求三位有效数字会被判不等。
Split a projectile into a horizontal line with no force once drag is ignored and a vertical line where gravity alone gives 9.81 m/s^2 downwards; the two lines share one time, so solve whichever direction is determined first and carry that time across. Mass cancels in a drag-free projectile, which is why range depends on launch speed and angle rather than weight. At the top of the path the vertical velocity is zero while the horizontal velocity is untouched, and using the symmetry of rise and fall is the fastest route when a question gives maximum height or range. Data booklets in AQA papers quote g as 9.81 m/s^2, so a value rounded to 9.8 can fail a three-significant-figure mark.
三、牛顿第二定律与受力图
力学题的起手动作是画受力图，把重力、支持力、摩擦力、张力逐个标清楚，再沿运动方向分解。牛顿第二定律里的 F 是合力，不是题目提到的某一个力，这一点在阅卷时错得最多。第三定律的成对力作用在两个不同物体上，所以不会互相抵消，书压桌面的支持力与书的重力并不是一对。平衡状态只表示合力为零，与速度大小无关，匀速上升的电梯同样处于平衡。斜面题把重力分解成 mg sin(theta) 沿斜面和 mg cos(theta) 垂直斜面，摩擦力用后者乘摩擦系数。连接体题先用整体求加速度，再隔离其中一个物体求张力，两步顺序反了会多出未知数。
Start every mechanics question with a free-body diagram naming weight, normal contact force, friction and tension before resolving along the direction of motion. The F in F = ma is the resultant force rather than any single labelled force, and examiner reports on 7408 papers show this is the most common error in the section. Third-law pairs act on two different bodies and therefore never cancel; the normal force of a table on a book and the weight of the book are not a pair. Equilibrium means zero resultant force and says nothing about speed, so a lift rising steadily is in equilibrium. On a slope, resolve weight into mg sin(theta) down the incline and mg cos(theta) into the surface, using the second component for friction.
四、力矩、力偶与刚体平衡
力矩等于力乘以力臂，力臂是从支点到力作用线的垂直距离，直接拿斜边长会算大。力偶是两个大小相等、方向相反但不共线的力，它的力矩等于其中一个力乘以两力之间的垂直间距，与支点位置无关。刚体平衡要同时满足合力为零和合力矩为零，题目给三个力时通常连续取矩，把支点选在未知力最多的位置。均匀杆的重心在中点，非均匀杆要先求重心位置再列方程。很多学生以为挂在墙上的梯子只受重力和地面支持力，漏掉墙面的法向力和地面摩擦力，方程数就不够。AQA 的力矩题常带人体或工具情境，受力点画错一位，后面的数全错。
A moment is force times the perpendicular distance from the pivot to the line of action, and using the sloping length instead of that perpendicular distance inflates the answer; a couple consists of two equal, opposite and non-collinear forces whose moment is one force times their perpendicular separation, independent of where the pivot sits. A rigid body in equilibrium needs both zero resultant force and zero resultant moment, so with three forces take moments about the point carrying the most unknowns to eliminate them. The centre of mass of a uniform rod is at its midpoint, while a non-uniform rod needs its centre found before any equation is written. Ladder problems lose marks when the normal force from the wall or friction at the ground is omitted.
五、动量守恒与弹性碰撞
动量 p = mv 是矢量，一维问题里先规定正方向，再给每个速度带符号。冲量等于动量变化，F = Δp/Δt 解释了为什么安全气囊能减小受力：动量的变化量不变，但作用时间拉长，力就变小。系统在某方向不受外力时，该方向动量守恒，爆炸和反冲题都用这一条。弹性碰撞同时守恒动能与动量，完全非弹性碰撞后两物体共速，动能损失最多。常见的错误是只对其中一个物体用动量守恒，或者忘了把碰撞后的共同速度当成一个未知量。AQA 的动量题常和冲量图像结合，F-t 图下的面积就是冲量。
Momentum p = mv is a vector, so fix a positive direction and carry signs through every velocity in one-dimensional problems; impulse equals change in momentum, and F = Δp/Δt explains why a crumple zone reduces the force while leaving the momentum change unchanged, since contact time is stretched. A system with no external force along a direction conserves momentum in that direction, which covers explosions and recoil as well as collisions. Elastic collisions conserve kinetic energy as well as momentum, while a perfectly inelastic collision leaves both bodies moving together with the largest kinetic energy loss. Applying the conservation law to only one of the two bodies is the standard mistake.
六、功、能量与功率
功的定义是 W = Fs cos(theta)，只有当力的方向与位移有夹角时才有分量做功，垂直于运动的支持力不做功。动能定理把合力做的功与动能变化连起来，用它比全程追加速度更省步骤，尤其是速度未知的中间状态。重力势能的变化只依赖高度差，与路径无关，所以沿斜面下滑与自由落体到达同一高度的动能相同。功率 P = W/t 也等于 Fv，汽车在恒定功率下加速时牵引力随速度下降，加速度越来越小。效率等于有用功除以总输入能量，题目给的百分比要乘回到能量上再列方程。常见的错误是把摩擦力做的功记成正功，导致能量守恒式符号出错。
Work is W = Fs cos(theta), so a force perpendicular to the displacement transfers no energy and the normal contact force on a sliding block does no work. The work-energy theorem links the resultant work to the change in kinetic energy and usually shortens a solution compared with tracking acceleration, especially when an intermediate speed is unknown. Gravitational potential energy depends only on height difference, so sliding down a slope and falling freely to the same level give equal kinetic energy at the bottom; power is P = W/t as well as Fv, which is why a car on constant power accelerates less and less as speed rises. Friction does negative work, and losing that minus sign breaks the energy equation.
七、胡克定律与弹性势能
胡克定律 F = kΔl 只在弹性限度内成立，Δl 是形变量而不是弹簧的总长度，代入前先算差值。k 是弹簧常数，单位 N/m，两根弹簧串联时等效 k 变小，并联时变大，题目若给出原长和伸长量要先判断是不是同一根弹簧。弹性势能等于 ½FΔl，也等于 ½kΔl^2，F-Δl 图下的面积就是储存的能量。超过弹性限度后撤去力，材料不能恢复原长，此时胡克定律失效，图像开始弯曲。很多学生把测力计的读数当成重力，其实读数反映的是弹簧对物体的力，方向要按牛顿第三定律判断。AQA 的必做实验里有一个就是测弹簧的力与伸长量关系，要求作图求 k。
Hooke's law F = kΔl holds only up to the elastic limit, and Δl is the extension rather than the total length, so subtract before substituting. The spring constant k has units of N/m; identical springs in series give a smaller effective k and in parallel a larger one. Elastic potential energy is ½FΔl or ½kΔl^2, and the area under a force-extension graph is the energy stored; beyond the elastic limit the material does not return to its original length and the graph curves. One of the twelve required practicals in the AQA 7408 handbook measures force against extension for a spring and asks for k from the gradient.
八、应力应变与杨氏模量
应力等于力除以横截面积，应变等于伸长量除以原长，两者都没有单位，杨氏模量 E 等于应力除以应变，单位是 Pa。E 只由材料决定，与样品的长度和粗细无关，所以同种材料的不同金属丝画出应力应变图，弹性段斜率相同。弹性段的直线部分决定 E，取点时要用图线上相距较远的两个点算斜率，不要用原点和单点。曲线出现明显弯曲处是屈服点，之后材料进入塑性变形，断裂前能承受的最大应力叫抗拉强度。常见的错误是忘记把直径换算成半径再算面积，面积差四倍会让 E 差四倍。AQA 的必做实验要求用千分尺和游标卡尺测直径，取多次读数求平均，再作图求 E。
Stress is force per unit cross-sectional area and strain is extension per unit original length, both dimensionless, so Young's modulus E equals stress divided by strain and carries the unit pascal; E depends on the material alone rather than the length or thickness of the specimen, which is why wires of the same metal give parallel elastic sections. Read E from the gradient of the straight part using two widely separated points on the line rather than one point and the origin. The yield point marks the start of plastic deformation, and the maximum stress before fracture is the tensile strength. Using the diameter as a radius is the classic error, and it changes the area, and therefore E, by a factor of four.
九、阻力与终端速度
在有阻力的下落里，阻力随速度增大，合力先向下减小到零，加速度随之变为零，速度停在终端速度不再上升。跳伞题的分段要按加速度变化来写：刚跳出时阻力小、加速度约等于 g，速度增大后阻力增加，开伞瞬间阻力远大于重力，人减速到新的、更小的终端速度。速度时间图在开伞处会出现一个向下的台阶式转折，图线下的面积仍然是下落距离。很多学生以为终端速度时人不再受力，实际上重力与阻力大小相等方向相反，合力为零。同一下落体质量越大终端速度越高，因为重力增大而阻力取决于形状和速度。AQA 的题会给阻力与速度成正比的简单情形，这时合力写成 mg - kv，令其为零即可解出终端速度。
During a fall with drag, the resistive force grows with speed until it balances weight, the resultant force falls to zero and the speed settles at terminal velocity; a parachute question is marked in stages, with small drag and an acceleration close to g at the jump, rising drag as speed increases, and drag exceeding weight when the canopy opens so the jumper decelerates to a new and lower terminal velocity. The velocity-time graph shows a downward step at the opening, and its area still gives the distance fallen. Terminal velocity is not a state with no forces, since weight and drag are equal and opposite rather than absent. A heavier falling object reaches a higher terminal speed because weight grows while drag depends on shape and speed.
十、必做实验与不确定度
AQA 7408 的十二个必做实验中，力学相关的是用自由落体测 g 和用金属丝测杨氏模量，两者都在 Paper 3 的 Section A 里以实验情境出现。自由落体测 g 的做法是让小球通过两个光电门，用 s = ½gt^2 反解 g，时间越短相对误差越大，所以要增大下落高度并重复多次。杨氏模量实验用千分尺测直径、用标尺测原长、用测微计测伸长量，取斜率而不是单点算 E。不确定度要按读数分辨率的一半与重复读数的离散程度合成，百分比不确定度用来判断结果是否与文献值一致。常见的错误是把仪器的最小分度当成不确定度，或者算斜率时忘了把单位统一到 SI。AQA 要求最终答案的有效数字与不确定度匹配，多写两位会被扣分。
Of the twelve required practicals in the AQA 7408 handbook, the mechanics pair are the free-fall measurement of g and the Young modulus of a metal wire, and both reappear as experimental contexts in Section A of Paper 3. For g, time a ball through two light gates and rearrange s = ½gt^2, increasing the drop height and repeating runs because short times carry large relative uncertainty. The Young modulus practical measures diameter with a micrometer, original length with a metre rule and extension with a vernier, then takes E from the gradient rather than a single point. Uncertainty from a scale reading is half the smallest division and percentage uncertainty decides whether a result agrees with the published value; quoting the smallest division as the uncertainty, or leaving mixed units in a gradient, costs marks here.
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