AQA数学 Mechanics 力学 运动学 动量 力矩
AQA 的 A-Level 数学（7357）没有把力学单独做成一张卷，而是放进 Paper 2，与纯数学内容混合考核。很多同学复习时把时间全投给微积分和证明，到了力学大题却拿不到步骤分。这篇笔记按 AQA 官方力学板块顺序梳理：P Quantities and units（量与单位）、Q Kinematics（运动学）、R Forces and Newton's laws（力与牛顿定律）、S Moments（力矩），每个板块给出核心公式、例题和常见失分点，帮你在 Paper 2 里把力学变成稳定得分项。
In the AQA A-level Mathematics specification (7357), mechanics is not a separate paper. It sits inside Paper 2 and is mixed with pure mathematics questions. Many students put all their revision time into calculus and proof, then lose easy marks on mechanics. This guide follows the four official mechanics areas: quantities and units, kinematics, forces and Newton's laws, and moments. Each part gives the key formulas, a worked example, and the mistakes examiners see most often, so that mechanics becomes a reliable source of marks in Paper 2.
一、力学在 7357 中的位置与分值
Paper 2 满分 100 分，考试时长 2 小时，考查范围为 Paper 1 的全部内容，再加上 J Vectors（向量）、P Quantities and units、Q Kinematics、R Forces and Newton's laws、S Moments。力学板块加上向量，大约占这份卷子的三分之一以上。也就是说，力学不是附加题，而是决定你能否从 A 冲上 A* 的关键区段。题目顺序不固定，但力学题常出现在后半段，往往以实际情景包装：电梯升降、绳子拉箱子、梯子靠墙、木板搭在支点上。
Paper 2 is worth 100 marks and lasts two hours. It assesses everything from Paper 1 plus vectors, quantities and units, kinematics, forces and Newton's laws, and moments. Together with vectors, the mechanics content makes up well over a third of this paper. Mechanics is therefore not an optional extra; it decides whether you move from a grade A to an A*. The order of questions is not fixed, but mechanics questions usually appear in the later half, wrapped in real-world contexts such as a lift accelerating, a box pulled by a rope, a ladder against a wall, or a plank resting on a support.
复习节奏上，力学最怕只看不算。公式背得再熟，如果受力图画错一条线，后面全部连锁崩塌。建议每学完一个板块，立刻用 3 到 5 道真题验证：先画图，再列方程，最后核对单位与有效数字。
When revising mechanics, reading is not enough. A single wrong arrow on a force diagram destroys every later step. After each topic, test yourself with three to five past-paper questions: draw the diagram first, then set up the equations, then check units and significant figures.
二、运动学 Kinematics：公式选择是第一步
Q Kinematics 的核心是匀加速直线运动的五个量：初始速度 u、末速度 v、加速度 a、位移 s、时间 t。常用的四个公式是 v = u + at、s = ut + (1/2)at^2、v^2 = u^2 + 2as、s = ((u + v)/2)t。做题第一步不是算，而是圈出题目给了哪三个量、要求哪一个量，选出唯一覆盖这四者的公式。很多失分并非计算错误，而是选错了公式，把不含 t 的题硬套含 t 的式子。
The heart of kinematics is the five quantities of motion with constant acceleration: initial velocity u, final velocity v, acceleration a, displacement s, and time t. The four standard formulas are v = u + at, s = ut + (1/2)at^2, v^2 = u^2 + 2as, and s = ((u + v)/2)t. The first step is never arithmetic; it is listing which three quantities are given and which one is required, then choosing the single formula that connects those four. A large share of lost marks comes from picking the wrong formula rather than from calculation errors.
例题：一辆车以 20 m/s 匀速行驶，司机刹车产生 5 m/s^2 的减速度，求完全停下前行驶的距离。已知 u = 20、v = 0、a = -5，要求 s，正好对应 v^2 = u^2 + 2as，代入得 0 = 400 - 10s，s = 40 米。若题目还问刹车用时，再用 v = u + at 得 t = 4 秒。养成先写已知量清单的习惯，可以避开绝大多数公式误用。
Worked example: a car travels at 20 m/s and brakes with a deceleration of 5 m/s^2. Find the stopping distance. Here u = 20, v = 0 and a = -5, and s is required, so v^2 = u^2 + 2as gives 0 = 400 - 10s, hence s = 40 metres. If the time is also asked for, v = u + at gives t = 4 seconds. Writing a clear list of known quantities before touching the calculator removes most formula errors.
二维情形下，水平方向与竖直方向必须分开处理：水平方向通常没有加速度，竖直方向加速度为 g（AQA 使用 g = 9.8 m/s^2，除非题目另有说明）。两个方向共用的量只有时间 t，这也是连接两个方向的桥梁。计算时不要跨方向混用公式，这是抛体题最常见的错误来源。
In two dimensions, horizontal and vertical motion must be treated separately. Horizontally there is usually no acceleration, while vertically the acceleration is g, which AQA takes as 9.8 m/s^2 unless stated otherwise. The only quantity shared by both directions is time, and that is the link between them. Never mix formulas across directions; doing so is the most common source of errors in projectile questions.
三、力与单位：P Quantities and units 的隐藏考点
P 板块看起来最简单，却是阅卷时扣分最隐蔽的地方。你需要在牛顿（N）、千克（kg）、米每秒平方（m/s^2）之间自如换算，也要会用 g 把质量换成重量：重量 W = mg，10 kg 的物体重量约为 98 N。一旦在受力图里把质量和重量混为一谈，牛顿第二定律的方程就会整体偏差，后面的答案再准确也拿不到分。
The quantities and units topic looks simple but hides quiet mark losses. You need to move confidently between newtons, kilograms and metres per second squared, and to convert mass into weight using W = mg, so a 10 kg object weighs about 98 N. If a force diagram confuses mass with weight, the whole equation of motion is mis-scaled and later correct arithmetic still earns nothing.
建模题还要求写清楚假设。例如把物体看作质点（particle），忽略空气阻力，绳子不可伸长且质量忽略不计，斜面光滑或有摩擦。这些假设不是客套话：它们决定了你能否直接使用匀加速公式，也决定摩擦力是否要进入方程。答完题后顺手检查有效数字和单位，AQA 通常要求保留三位有效数字或者与数据精度一致。
Modelling questions also expect explicit assumptions: treat the object as a particle, ignore air resistance, assume the string is light and inextensible, and state whether the surface is smooth or rough. These assumptions are functional, not decorative. They decide whether constant-acceleration formulas apply and whether friction enters the equations. Finish by checking significant figures and units; AQA normally expects three significant figures or the precision of the given data.
四、牛顿定律与连接体问题
牛顿第一定律说明合外力为零时物体保持静止或匀速运动，因此平衡题的判据是合力为零，而不是速度为某某数值。牛顿第二定律 F = ma 处理不平衡情形，方向约定要一致：通常取运动方向为正，再据此决定摩擦力和重力的分量符号。牛顿第三定律则提醒你，绳子拉物体与物体拉绳子的力大小相等、方向相反，作用在不同物体上，永远不要写进同一个受力方程里。
Newton's first law says an object stays at rest or moves with constant velocity when the resultant force is zero, so equilibrium is judged by zero resultant force rather than by any particular speed. Newton's second law, F = ma, handles unbalanced cases, and the sign convention must be consistent: take the direction of motion as positive, then assign signs to friction and the weight component accordingly. The third law reminds you that the tension on an object and the tension on the string are equal in magnitude, opposite in direction, and act on different bodies, so they never belong in the same equation.
连接体问题（connected particles）分两步：先整体法求加速度，再隔离法求绳子张力或接触面上的作用力。若一个物体沿桌面、另一个竖直悬挂，两个物体共用同一个加速度，系统方程为合力等于总质量乘加速度，其中悬挂物的重量提供动力，摩擦力和接触面的反作用力提供阻力。滑轮两侧物体运动方向不同，受力方向也要逐一核实。
Connected-particle problems follow two steps: first treat the system as a whole to find the acceleration, then isolate one body to find the tension or the contact force. When one object slides on a table and another hangs vertically, both share the same acceleration, and the system equation sets the net driving force equal to the total mass times acceleration, with the hanging weight driving and friction resisting. On a pulley the two objects move in different directions, so every force direction must be checked separately.
五、斜面与摩擦力
斜面上处理力时，把重力分解为平行于斜面和垂直于斜面两个分量：平行分量是 mg sin(theta)，垂直分量是 mg cos(theta)。垂直方向通常处于平衡状态，因此法向反作用力 R = mg cos(theta)；平行方向则可能是平衡，也可能是加速。摩擦力大小按 F = muR 计算，方向始终与相对运动或相对运动趋势相反。判断是静摩擦还是动摩擦，决定了你是否可以直接用 muR 这个式子。
On an inclined plane, resolve weight into components parallel and perpendicular to the surface: the parallel component is mg sin(theta) and the perpendicular component is mg cos(theta). The perpendicular direction is usually in equilibrium, so the normal reaction is R = mg cos(theta), while the parallel direction may be balanced or accelerating. Friction is F = muR and always opposes relative motion or the tendency to move. Deciding whether friction is static or limiting determines whether muR can be used directly.
例题：质量 5 kg 的箱子放在倾角 30 度的粗糙斜面上，摩擦系数 0.4，判断箱子是否下滑。平行重力分量是 5 乘 9.8 乘 sin30 等于 24.5 N，最大摩擦力是 0.4 乘 5 乘 9.8 乘 cos30 约等于 17.0 N。由于平行分量大于最大摩擦力，箱子会下滑，加速度由 (24.5 - 17.0) / 5 算出，约为 1.5 m/s^2。这类题的陷阱是忘记把重力写成两个分量，直接拿 mg 去减摩擦力。
Worked example: a 5 kg box rests on a rough plane inclined at 30 degrees with coefficient of friction 0.4. Decide whether it slides. The parallel weight component is 5 times 9.8 times sin30, which is 24.5 N, and the limiting friction is 0.4 times 5 times 9.8 times cos30, about 17.0 N. Since the parallel component exceeds limiting friction, the box slides, and the acceleration comes from (24.5 - 17.0) divided by 5, roughly 1.5 m/s^2. The classic trap is forgetting to resolve weight and subtracting friction from mg directly.
六、力矩与平衡
S Moments 的核心思想是转动的平衡条件：对任一点取矩，顺时针力矩之和等于逆时针力矩之和。力矩等于力乘力臂，力臂是支点到力作用线的垂直距离，不是支点到作用点的距离。这一点在梯子靠墙、杆子斜挂等题里最容易出错。此外，均匀杆的重力可以看作作用在它的中点，这是大部分平衡题的关键简化。
The moments topic rests on one idea: for rotational equilibrium, the sum of clockwise moments about any point equals the sum of anticlockwise moments. A moment is force times perpendicular distance from the pivot to the line of action, not the distance to the point of application. This distinction is the main trap in ladder-against-wall and tilted-rod questions. Another key simplification is that the weight of a uniform rod acts at its midpoint.
例题：一根长 4 米、重 200 N 的均匀木板架在两个支点上，支点分别在两端。若在距离左端 1 米处放一个重 300 N 的物体，求两端支点的反作用力。对左端取矩：右边支点反力乘 4 等于 200 乘 2 加 300 乘 1，得右边反力 175 N；竖直方向合力为零给出左边反力 325 N。取矩点的选择是自由的，选得好可以让未知量自动消失，这是省时间的关键技巧。
Worked example: a uniform 4 metre plank of weight 200 N rests on two supports, one at each end. A 300 N load is placed 1 metre from the left end. Find the reactions. Taking moments about the left end, the right reaction times 4 equals 200 times 2 plus 300 times 1, giving 175 N on the right. Vertical equilibrium then gives 325 N on the left. The pivot may be chosen freely, and a clever choice removes an unknown entirely, which saves a great deal of time.
七、动量的思想：与 F = ma 同源
AQA 7357 的力学板块并不单独考查动量守恒，但动量的思想仍然值得掌握，因为它是理解牛顿第二定律的另一条路径。动量定义为 p = mv，冲量是力乘作用时间，冲量动量定理写作 Ft = mv - mu。把 F = ma 与 a = (v - u)/t 联立，正好得到同一个关系式。理解这条联系之后，处理碰撞、爆炸或者短时间大作用力的情景会顺畅很多，这也是物理学科与数学力学之间的自然桥梁。
Momentum conservation is not examined as a separate topic in the AQA 7357 mechanics section, yet the idea of momentum is still worth learning because it gives a second route into Newton's second law. Momentum is p = mv, impulse is force times time, and the impulse-momentum relation is Ft = mv - mu. Combining F = ma with a = (v - u)/t reproduces exactly the same statement. Seeing this link makes problems about collisions, explosions, or large short-lived forces far more intuitive, and it forms a natural bridge between physics and the mechanics you meet in mathematics.
八、高频失分点与复习建议
第一类失分来自方向：把向上为正、向下为负的约定中途改掉，摩擦力方向画反，或者在斜面上把 sin 与 cos 互换。第二类来自假设：题目说粗糙你却当光滑，题目说绳子轻你却把绳子质量算进去。第三类来自表述：只写答案不写单位，有效数字不统一，或者缺少必要的文字说明。AQA 的阅卷标准对过程分很友好，写清受力图和方程，即使最终数字算错也能拿到大部分步骤分。
The first source of lost marks is direction: flipping a sign convention halfway through, drawing friction the wrong way, or swapping sine and cosine on an incline. The second is assumptions: treating a rough surface as smooth, or including the mass of a string the question called light. The third is presentation: answers without units, inconsistent significant figures, or missing working. AQA marking rewards clear method, so a tidy force diagram and stated equations usually secure most of the marks even if the final number slips.
复习建议按三步走：先用真题把 P、Q、R、S 四个板块各刷十题，只求方向与公式不出错；再集中处理连接体与斜面这两类高难题；最后做限时套卷，把力学题控制在每分 1.2 分钟以内。错题本不需要抄题，只记录三件事：错在哪个方向、错用哪条公式、下次看到什么线索就该想起什么。这样积累三周，力学就能从随机得分变成稳定得分。
A three-step revision plan works well. First, work ten past-paper questions from each of the four areas, focusing only on correct directions and formula choice. Second, concentrate on connected particles and inclined planes, the two hardest families. Third, sit timed papers and keep mechanics questions to about 1.2 minutes per mark. An error log needs no copied questions, just three notes per mistake: which direction was wrong, which formula was misapplied, and what clue should trigger the right approach next time. Three weeks of this turns mechanics from a lottery into a reliable source of marks.
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