AQA A-Level物理 力学 SUVAT 力矩 动量 功与能
力学在 AQA A-Level Physics 7408 里不是独立一张卷，它属于 Section 3.4 Mechanics and materials，和材料部分一起被放进 Paper 1。按 AQA 官方 spec 的 scheme of assessment，Paper 1 考 Section 1 到 5 再加 6.1 Periodic motion，时长 2 小时，满分 85 分，占整个 A-Level 的 34%，其中 60 分是简答与长答，25 分是选择题。同一份官方文件写明，全科至少 40% 的分数需要数学技能（Level 2 及以上），至少 15% 的分数考实验能力。力学这一块几乎每道计算题都同时用到这两条。下面把标量矢量、匀加速运动、运动图像、牛顿定律、力矩、动量、功与能、材料八块按考点顺序拆开，每块后面跟一段英文，方便国际学校的学生把内容与课堂上的英文术语对上。
Mechanics sits inside Section 3.4 of the AQA A-Level Physics 7408 specification, and it is examined on Paper 1 together with materials. Paper 1 covers Sections 1 to 5 plus 6.1 Periodic motion: two hours, 85 marks, 34 per cent of the whole A-Level, split into 60 marks of written answers and 25 multiple choice questions. The same specification fixes two numbers worth remembering, because they explain why mechanics questions look the way they do. At least 40 per cent of the marks across the qualification need mathematical skills at Level 2 or above, and at least 15 per cent assess practical work, so almost every mechanics calculation earns marks for method and for handling numbers properly.
标量与矢量的分解与合成
标量只有大小，矢量同时有大小和方向，AQA 在 3.4.1.1 要求能用平行四边形法则与三角形法则把矢量合成一个合力。稳妥的做法是先把每个矢量拆到互相垂直的两个方向上，再用勾股定理求合力大小、用反三角函数求它与某条边的夹角。真正决定对错的是角度基准。题目写拉力与水平方向成 35 度，横向分量就是 F cos 35 度、纵向是 F sin 35 度；如果那句改成与竖直方向成 35 度，两个函数必须互换。很多学生以为背下 sin 和 cos 的写法就够了，考场上错的常是没看清角度从哪条边量起。负号也要一次定好，声明向右为正之后，所有向左的力与反方向的位移都带负号进方程，中途换约定是丢分最多的写法。
Scalars carry magnitude only, while vectors carry magnitude and direction, and the specification asks you to combine them with the parallelogram or triangle rule. The reliable method is to resolve each vector into two perpendicular components first, then use Pythagoras for the magnitude and an inverse trigonometric function for the direction. A common mistake is misreading the angle: a force at 35 degrees to the horizontal gives cos 35 for the horizontal component and sin 35 for the vertical one, and those swap the moment the angle is measured from the vertical instead. Examiners also see marks lost through sign errors, so fix a positive direction before the first equation and keep it to the end.
匀加速运动的公式挑选与自由落体
沿直线匀加速这一节，AQA 在 3.4.1.3 要求把运动学公式用在直线运动与自由落体上。能直接用的式子只有四个：末速度等于初速度加加速度乘时间，位移等于初速度乘时间加二分之一加速度乘时间平方，末速度平方等于初速度平方加二倍加速度乘位移，位移还等于初末速度和的一半乘时间。挑公式的判据只有一条，看题目给的三个量与所求的量，剩下的字母就是你要回避的。自由落体把加速度换成 g，取 9.81 米每二次方秒，向上为正时 g 记为负值。举例，物体从静止下落 3 秒，位移是二分之一乘 9.81 乘 9，得 44.1 米，末速度是 9.81 乘 3，得 29.4 米每秒。常见的错误是不换单位就代数字，题目里出现千米每小时、厘米或毫秒，都要先换成国际单位。
The equations of motion are used for motion in a straight line and for free fall, and only four of them matter; pick the equation by listing the three quantities the question gives and the one it asks for, then ignore the variable that never appears. In free fall the acceleration becomes g, taken as 9.81 metres per second squared, and it enters with a negative sign if you have called upwards positive. Worked example: an object released from rest falls for 3 seconds, so its displacement is half of 9.81 times 9, which is 44.1 metres, and its final speed is 9.81 times 3, which is 29.4 metres per second. Units are where most marks disappear, so convert kilometres per hour, centimetres and milliseconds before substituting anything.
位移时间图与速度时间图的斜率面积
运动图像是 AQA 3.4.1.3 里被反复考的一小块，判断规则只有三条。位移时间图上某点的切线斜率是那一时刻的速度，曲线向上弯说明速度在增大。速度时间图的斜率是加速度，图线与时间轴围出的面积是位移，时间轴下方的面积算负位移。加速阶段与减速阶段的面积要分开算，容易错的地方是把梯形当三角形，或者忘了用平均速度。很多学生以为位移时间图的斜率就是加速度，这个混淆几乎每次都能在解析里看到。遇到折线图时先分段，每段判断是匀速还是匀加速，再动笔算面积。
Motion graphs come up on almost every paper, and only three rules decide the marks: on a displacement-time graph the gradient of the tangent at a point is the velocity at that instant, so a curve bending upwards means the speed is rising. On a velocity-time graph the gradient gives acceleration and the area between the line and the time axis gives displacement, with anything below the axis counting as negative displacement. A frequent slip is treating a trapezium as a triangle, or forgetting to use average velocity for the uniform section. Read a piecewise graph by splitting it into sections, decide whether each one is uniform or accelerating, and only then start calculating areas.
牛顿三定律与受力分析图
牛顿三定律在 3.4.1.5，考点落在受力分析上。平衡的判据是合力为零，也就是两个方向上各自的分量之和都为零，缺一个方向就会漏解。质量与重量不是一回事，重量是质量乘 g，用牛顿表示；物体在斜面上时重量要分解成沿斜面与垂直斜面两个分量。第三定律的成对力必须作用在两个不同物体上，重力和桌面对物体的支持力作用在同一个物体上，它们不是第三定律的一对，这一条是选择题里最常见的设计。终端速度出现在合力为零的时刻，此时阻力与重力大小相等，速度不再变化。画受力图时把每个力都画成从物体中心出发的箭头，标注符号，之后列方程不容易漏项。
Newton's laws are examined through free body diagrams; equilibrium means the resultant force is zero, which in practice means the components in each of the two chosen directions add to zero, and leaving out one direction loses marks. Mass and weight are different quantities: weight is mass times g and is measured in newtons, and on a slope the weight must be resolved along and perpendicular to the surface. Newton's third law pairs act on two different bodies, so the weight of a book and the normal contact force from the table are not such a pair, because both act on the book; multiple choice questions use this trap constantly. Terminal velocity arrives when the resultant force falls to zero, the drag force matching the weight, and the speed then stays constant.
力矩平衡与力偶
力矩这一块在 3.4.1.2，定义是力乘力偶臂，力偶臂指支点到力的作用线的垂直距离，不是支点到着力点的距离。力偶是两个大小相等、方向相反且不共线的平行力，它们的合力为零但会让物体转动，力矩等于其中一个力乘两力之间垂直距离。平衡物体要同时满足合力为零与合力矩为零，取哪一点作为转动中心可以自由选择，选在未知力通过的位置可以让方程少一个未知量。举例，一根轻杆支点在左侧，右侧 2.0 米处施加向下的力，同时在 0.8 米处有 250 牛的重量，取支点求力矩平衡，力等于 250 乘 0.8 除以 2.0，得 100 牛。常见的错误是拿斜着量出的长度当力偶臂，垂直距离才是定义里的距离。
A moment is a force multiplied by the perpendicular distance from the pivot to the line of action of that force, which is not the distance to the point where the force is applied; A couple is a pair of equal, opposite and parallel forces that are not in line, so the resultant force is zero while the turning effect is not, and its moment is one force times the perpendicular separation. A body in equilibrium needs both the resultant force and the resultant moment to be zero, and the pivot may be chosen freely, so picking the point where an unknown force acts removes that unknown from the equation. Worked example: a light rod is pivoted at one end, a downward force acts at 2.0 metres and a weight of 250 newtons hangs at 0.8 metres, so taking moments about the pivot gives 250 times 0.8 divided by 2.0, that is 100 newtons.
动量守恒与冲量求平均力
动量等于质量乘速度，是矢量，守恒的条件是系统所受外力可以忽略，或相互作用时间极短。一维碰撞与爆炸都可以用初动量之和等于末动量之和处理，方向靠符号区分。牛顿第二定律的另一种写法是力等于动量的变化率，力乘作用时间等于动量变化，也就是冲量。举例，0.15 千克的小球以 12 米每秒撞墙，以 8 米每秒反弹，接触时间 0.05 秒，动量变化是 0.15 乘 20 得 3.0 千克米每秒，平均力是 3.0 除以 0.05，得 60 牛。反冲的例子同样常见，4.0 千克的枪水平射出 0.02 千克的子弹，子弹速度 300 米每秒，枪的后退速度是 0.02 乘 300 除以 4.0，得 1.5 米每秒，方向与子弹相反。学生最常把反弹当成同方向，忘记速度变号，动量变化就少算了一倍。
Momentum is mass times velocity, it is a vector, and it is conserved when the external forces on a system are negligible or the interaction is very brief. Collisions and explosions in one dimension are handled by setting the total momentum before equal to the total momentum after, with direction carried by the signs. Newton's second law in its momentum form says the force equals the rate of change of momentum, so force times contact time equals the change in momentum. Worked example: a ball of mass 0.15 kilograms hits a wall at 12 metres per second and rebounds at 8 metres per second over 0.05 seconds, so the change in momentum is 0.15 times 20, which is 3.0 kilogram metres per second, and the average force is 3.0 divided by 0.05, which is 60 newtons.
功 能量守恒与功率
功等于力乘位移再乘两者夹角的余弦，力与位移垂直时做功为零。动能是二分之一质量乘速度平方，重力势能是质量乘 g 乘高度，弹性势能可以用二分之一力乘伸长量计算，也就是力与伸长量图像下方的面积。功率是功除以时间，也可以写成力乘速度，这两个式子在不同题目里各占优势。举例，60 千克的学生爬上 10 米高的楼层，重力势能增加 60 乘 9.81 乘 10，约 5.9 千焦；1000 千克的汽车以 20 米每秒行驶，动能是二分之一乘 1000 乘 400，得 200 千焦。常见的错误是能量守恒里漏掉摩擦生热，斜面下滑题的末速度往往比理想值小，差额就是摩擦力做的功。
Work is force times displacement times the cosine of the angle between them, so a force acting at right angles to the motion does no work at all; kinetic energy is half the mass times the speed squared, gravitational potential energy is mass times g times height, and elastic strain energy is half the force times the extension, which is the area under a force-extension graph. Power is work divided by time, and it also equals force times velocity, a form that saves work in problems about constant speed; worked example: a student of mass 60 kilograms climbs 10 metres, gaining 60 times 9.81 times 10, which is about 5.9 kilojoules, while a car of mass 1000 kilograms moving at 20 metres per second carries half of 1000 times 400, which is 200 kilojoules. Energy questions lose marks when friction is left out of the bookkeeping, since the final speed down a slope is always lower than the ideal value.
应力应变与杨氏模量
材料部分属于 3.4.2。拉伸应力等于力除以横截面积，应变等于伸长量除以原长，两者都没有单位以外的量纲问题，杨氏模量就是应力除以应变，单位是帕斯卡。力与伸长量图像的斜率是劲度系数，面积是储存的弹性能量，胡克定律只在弹性限度内成立。举例，直径 0.40 毫米的钢丝，横截面积约为 1.26 乘十的负七次方平方米，挂 40 牛的力得应力约 3.2 乘十的八次方帕斯卡，若杨氏模量为 2.0 乘十的十一次方帕斯卡，应变约 1.6 乘十的负三次方，原长 1.00 米的钢丝伸长 1.6 毫米。AQA 的实验部分对应两个必做实验，Practical 3 用自由落体法测 g，Practical 4 用简单方法测杨氏模量，官方实验器材指南在 aqa.org.uk 可以查到两份装置说明。错得最多的是把应力与力混为一谈，细线承受同样的力，应力比粗线大得多。
Materials form the second half of Section 3.4; tensile stress is force divided by cross-sectional area, tensile strain is extension divided by original length, and the Young modulus is stress divided by strain, measured in pascals. On a force-extension graph the gradient is the stiffness constant and the area underneath is the energy stored elastically, with Hooke's law holding only up to the elastic limit. Worked example: a steel wire of diameter 0.40 millimetres has an area of about 1.26 times ten to the minus seven square metres, so a load of 40 newtons produces a stress near 3.2 times ten to the eighth pascals and, with a Young modulus of 2.0 times ten to the eleventh pascals, a strain of roughly 1.6 times ten to the minus three, which stretches a 1.00 metre wire by 1.6 millimetres. AQA lists two relevant required practicals, Practical 3 for measuring g by a free-fall method and Practical 4 for the Young modulus by a simple method, and both set-up guides are published by AQA.
Paper 1 的给分口径与答案写法
把官方数据记清楚，答题时的取舍就有依据。AQA 的 7408 评分方案写得很直接：Paper 1 满分 85 分，其中 60 分是简答与长答、25 分是选择题，全科至少 40% 的分数需要数学技能，至少 15% 涉及实验。有效数字有单独的规定，评分方案允许最多 2 分给到按指定有效数字写出最终答案，题目数据里有效数字最少的那个量决定答案保留几位。解答题必须把代入过程写出来，只写答案通常拿不到方法分，尤其是抛体题：AQA 2021 年 6 月的 7408/1 评分方案里，第 5 题给出 20 米每秒、仰角 40 度，得分点是先算水平分量 20 cos 40 得 15.3 米每秒，再用位移公式建立关于时间的二次方程。做题时把数据与公式手册翻到常用公式页，边写边核对单位，比最后统一检查更稳。
Knowing the official numbers makes it easier to decide how much working to show; the AQA 7408 scheme of assessment states that Paper 1 carries 85 marks, made up of 60 marks of short and long answer questions and 25 multiple choice questions, and that at least 40 per cent of the assessment uses mathematical skills while at least 15 per cent assesses practical work. Significant figures carry their own rule, since the mark scheme allows up to 2 marks for quoting a final answer to a specified number of significant figures, decided by the datum with the fewest figures. Written questions expect the substitution to be visible, and answers without working usually miss the method marks, which is exactly what the June 2021 7408/1 mark scheme shows in its projectile question.
力学错题的三种归类
复习力学的效率取决于错题怎么归。按丢分原因分成三类最实用：一类是方向与符号，包括正方向半途改变、反弹速度没变号、力矩臂取成斜距离；一类是单位与有效数字，包括千米每小时没换、毫秒没换、答案位数与题目数据不匹配；一类是模型选错，动量能解的题硬用能量守恒，或者反过来漏掉摩擦生热。每一类各准备一页例题，考前只看这三页。顺序上建议先把矢量与受力分析练到不用想就能画图，再练公式与图像，之后接动量与能量，材料部分放在最后，因为它的计算依赖前面应力与能量的概念。很多学生按章节顺序从头背公式，结果在综合题里发现两个概念本来就该连起来用。
How well mechanics revision works depends on how you sort your mistakes; three categories cover most lost marks: direction and sign errors, including changing the positive direction part way through, forgetting that a rebounding object reverses velocity, and measuring a moment arm along a slope instead of perpendicular; unit and significant figure errors, including speeds still in kilometres per hour and times in milliseconds; and choosing the wrong model, such as forcing an energy method onto a question that only momentum can answer, or ignoring friction. Keep one page of worked examples per category and read those three pages before the exam. The useful order is vectors and free body diagrams first until they are automatic, then equations and graphs, then momentum and energy, with materials last because it reuses stress and energy ideas.
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