OCR数学 力学 SUVAT 摩擦力 力矩 抛体运动
OCR A Level Mathematics A 的第三张卷 H240/03 叫 Pure Mathematics and Mechanics，2 小时 100 分，官方 spec 写明这张卷分成两个部分、每部分约 50 分，力学独占大约一半。很多学生把力学当成套公式的一块，真正拉开差距的是三件事：正方向写没写、受力图画没画、答案有没有按卷面要求收成 3 位有效数字。试卷第 2、3 页印着公式表，suvat 四式和二维的向量形式都给了，背公式救不了这一块。下面按 OCR 力学考纲的四块内容（量与单位、运动学、力与牛顿定律、力矩）逐块拆开，每块配一道按官方口径算过的例题。很多学生以为力学只占 50 分可以先放着，按 H240/03 的结构，力学部分丢 12 分等于整个 A Level 少掉大约 4%。
Paper 3 of OCR A Level Mathematics A (H240/03) is Pure Mathematics and Mechanics: two hours, 100 marks, and by the official specification two sections of approximately 50 marks each, so mechanics is worth about a third of the whole qualification. In practice examiners see the same handful of failures year after year, and none of them comes from a missing formula, because the suvat equations are printed on pages 2 and 3 of the question paper. What separates a comfortable pass from a high grade on this paper is process: a declared positive direction, a labelled force diagram, and final answers rounded to three significant figures unless the question says otherwise.
量与单位：从 kg 到 N 和矢量记号
力学的第一块内容是量与单位，OCR 把它单列出来，因为这一块决定了后面所有计算能不能自洽。质量用 kg，加速度用 m s^-2，力的单位 N 是 F = ma 推出来的复合单位，1 N = 1 kg m s^-2，写答案时单位错一个量纲整题作废。很多学生以为单位只是最后补上去的装饰，实际上把 60 kg 直接当成 60 N 是选择题里最常见的失分点，重量必须写成 W = mg = 60 x 9.8 = 588 N 才能进方程。位移、速度、加速度都是矢量，OCR 在二维问题里允许 i、j 记号，公式表给出的正是向量形式的 suvat。做题时先写一句取向右为正，或者取竖直向上为正，后面所有符号才有依据，阅卷时看不到方向约定就很难给方法分。凡是问大小 magnitude 的题目，答案必须是正数，负号属于方向信息，不能塞进大小里。
Every mechanics question starts with units, because OCR lists quantities and units as content in its own right; force is a derived unit and one newton is exactly one kilogram metre per second squared, so a wrong unit can invalidate an otherwise correct calculation. Many students lose easy marks by treating a mass of 60 kg as a force, so weight has to be written as W = mg = 60 x 9.8 = 588 N before it goes into any equation. Displacement, velocity and acceleration are vectors, so state a positive direction in words at the start; when a question asks for a magnitude, give the positive value, and keep the sign for the direction.
匀变速直线运动：先列已知量再挑 suvat
suvat 的五件套是 s、u、v、a、t，四个公式每一个都少一个量，选式的动作应该是先列出题目给了什么、要求什么，再看哪个公式用不上多余的量。u = 4 m/s、a = 2 m s^-2、t = 5 s 时，v = u + at = 4 + 2 x 5 = 14 m/s，再用 s = ut + 0.5at^2 = 4 x 5 + 0.5 x 2 x 25 = 45 m。竖直方向把 g 取 9.8，向上为正时 a = -9.8，21 m/s 竖直上抛，最高点 v = 0，用 v^2 = u^2 + 2as 得 0 = 441 - 19.6s，s = 22.5 m，到顶时间 21 / 9.8 = 2.14 s。很多学生以为最高点速度为零、加速度也为零，所以球停在了半空，这个理解会让后面的受力题全部写错，最高点的加速度仍然是 9.8 m s^-2 向下。常见的错误是漏掉分段，先匀减速后反向匀加速的题目要分两段列式，位移的正负要跟着所设正方向走。回代检验很便宜，把算出的 v 和 t 代回另一个公式，两边对不上就说明符号用反了。
The five suvat quantities are s, u, v, a and t, and each of the four equations leaves one of them out, so list what you know and what you want before choosing; a worked example with u = 4 m/s, a = 2 m s^-2 and t = 5 s gives v = 14 m/s and s = 45 m. Vertically, taking g = 9.8 and upwards as positive gives a = -9.8, so a ball thrown up at 21 m/s has v = 0 at the top, 0 = 441 - 19.6s, s = 22.5 m and a time to the top of 2.14 s. Many students believe the velocity and the acceleration are both zero at the highest point, which then breaks every force question that follows, because the acceleration there is still 9.8 m s^-2 downwards.
速度时间图：斜率是加速度、面积是位移
图像题在 OCR 力学部分每年都出现，读图的规则只有两条：直线的斜率是加速度，图线与时间轴围出的面积是位移，轴下方的面积算负值。一辆车 0 到 4 s 保持 12 m/s，斜率是 0，这段时间合力为零；接下来 6 s 匀减速到停止，斜率 -2 m s^-2，说明合力方向与运动方向相反。位移把两块面积加起来，12 x 4 = 48 m，加上 0.5 x 12 x 6 = 36 m，共 84 m，平均速度 84 / 10 = 8.4 m/s，与 12 的一半加权结果一致。很多学生把图线围出的面积当成速度，或者把时间轴上下的面积直接相加，前者读错量纲，后者在往返运动的题目里必然丢分。曲线上的切线斜率只能给出某一时刻的加速度，这时要转去用微分，图上读到的信息不足以直接算。
Velocity-time graphs carry two examinable pieces of information in OCR mechanics: the gradient of a straight line is the acceleration, and the area between the line and the time axis is the displacement, with anything below the axis counted as negative. A car holding 12 m/s for 4 s and then decelerating uniformly to rest over 6 s covers 48 m plus 36 m, so 84 m in total, and its average speed is 8.4 m/s. Many students add areas above and below the axis together, which ruins any question involving a change of direction, and on a curved graph a gradient only gives the acceleration at one instant, so calculus takes over.
变加速运动：v = ds/dt 与 a = dv/dt
加速度不再是常数时 suvat 全部失效，只能回到微积分的定义：速度是位移对时间的导数，加速度是速度对时间的导数，反向积分就是从加速度回到速度、再回到位移，每一层都要用初始条件定常数。设 a = 6t - 2，单位 m s^-2，t = 0 时 v = 3 m/s、s = 0，积分得 v = 3t^2 - 2t + 3，t = 2 时 v = 12 - 4 + 3 = 11 m/s；再积分得 s = t^3 - t^2 + 3t，t = 2 时 s = 8 - 4 + 6 = 10 m。速度取极值的时刻出现在 a = 0，由 6t - 2 = 0 得 t = 0.33 s，代回速度式得 v = 2.67 m/s。学生最常把积分常数丢掉，或者把 t = 0 代进了错误的那一层式子。带 detailed reasoning 要求的题目必须把积分步骤写在卷面上，只交一个数字拿不到过程分。
When acceleration varies, the suvat equations no longer hold and the definitions take over: velocity is the derivative of displacement, acceleration is the derivative of velocity, and integrating in reverse needs an initial condition at every step to fix the constant. With a = 6t - 2, v = 3 at t = 0 and s = 0 at t = 0, the first integration gives v = 3t^2 - 2t + 3, so v = 11 m/s at t = 2, and the second gives s = t^3 - t^2 + 3t = 10 m at the same instant; the speed is stationary where a = 0, at t = 0.33 s. A common mistake is dropping the constant of integration, and questions labelled detailed reasoning expect the integration steps on the page rather than a bare answer.
二维运动与抛体：先分解再算时间
二维问题把水平与竖直拆成两个一维问题，水平方向没有加速度，速度保持 U cosθ 不变，竖直方向是 a = -9.8 的 suvat，两边共用的量只有时间。以 20 m/s、与水平成 30 度抛出：水平分量 20 x 0.866 = 17.32 m/s，竖直分量 20 x 0.5 = 10 m/s，落地时间由 0 = 10t - 4.9t^2 得 t = 2.04 s，射程 17.32 x 2.04 = 35.3 m，最大高度 10^2 / (2 x 9.8) = 5.10 m。很多学生把 20 m/s 直接代进竖直方向的式子，答案会大出一倍以上，这类错误在 OCR 的 examiner report 里年复一年地出现。最高点只有竖直速度为零，水平分量仍是 17.32 m/s，题目若问最高点的速度，答案不是 0。3 位有效数字的口径要在最后一步才收，中间过程保留 4 位以上精度。
Two-dimensional motion is split into two one-dimensional problems that share only the time: horizontally there is no acceleration, so the speed stays at U cos θ, while vertically the suvat equations apply with a = -9.8. For a launch at 20 m/s and 30 degrees, the components are 17.32 m/s and 10 m/s, and 0 = 10t - 4.9t^2 gives a flight time of 2.04 s, a range of 35.3 m and a maximum height of 5.10 m. Substituting the resultant 20 m/s straight into the vertical equations is the mistake examiners report most often, and at the highest point the horizontal component is still 17.32 m/s, so the speed there is not zero.
受力分析与牛顿第二定律
牛顿第二定律写成 F = ma，左边是合力，右边是质量乘加速度，合力方向必须与所设正方向一致。1500 kg 的汽车驱动力 4000 N、阻力 1000 N，合力 3000 N，a = 3000 / 1500 = 2.0 m s^-2。连接体先用整体法求加速度：桌面上 5 kg 滑块经滑轮与下垂的 3 kg 重物相连，绳张力对整体是内力，a = 3 x 9.8 / (3 + 5) = 3.68 m s^-2，再隔离滑块得张力 T = 5 x 3.68 = 18.4 N，隔离重物得到同样的 18.4 N。常见的错误是在整体法里把绳张力也当成外力算进去，那样会凭空多出一个 18.4 N。受力图上每个力都要有明确的施力物体，绳张力沿绳、法向力垂直于接触面，方向画错后面全错。整体法算加速度、隔离法求内力，这个顺序在连接体题目里能省下大半计算。
Newton's second law is written as F = ma, where the left side is the resultant force and its direction has to match the positive direction you declared. A 1500 kg car with a 4000 N driving force and 1000 N of resistance has a resultant of 3000 N and therefore a = 2.0 m s^-2. For connected particles, treat the system as a whole first, so a 5 kg block joined over a pulley to a hanging 3 kg mass accelerates at 3.68 m s^-2 and the tension follows from isolating either body, giving 18.4 N either way, because counting that tension as an external force shifts the whole answer.
摩擦力模型与斜面
摩擦力的官方模型是一个不等式 F <= μR，静止时静摩擦只取刚好够用的值，达到最大 F = μR 之后才开始滑动，对静止物体直接写 F = μR 是斜面题的典型失分点。4 kg 物块放在 25 度斜面上，μ = 0.35，重力的斜面分量 4 x 9.8 x sin25 = 16.6 N，法向力 R = 4 x 9.8 x cos25 = 35.5 N，最大静摩擦 0.35 x 35.5 = 12.4 N，小于 16.6 N，所以物块下滑，加速度 (16.6 - 12.4) / 4 = 1.0 m s^-2。判据可以化成 tanθ > μ，tan25 = 0.466 大于 0.35，与上面的结论一致，不需要具体数值的题目用这个不等式更省时间。很多学生把 R 直接当成 mg，把 39.2 N 当法向力会让摩擦力算成 13.7 N，最终加速度差出 0.1 m s^-2 以上。摩擦力方向永远与相对运动或相对运动趋势相反，物块下滑时摩擦力沿斜面向上。
Friction is modelled as an inequality, F <= μR: a body at rest uses only as much friction as it needs, and sliding starts once the limit μR is reached, so writing F = μR for a stationary block is a standard error. On a 25 degree slope with μ = 0.35, a 4 kg block feels 16.6 N down the slope, the normal reaction is 35.5 N, and the limiting friction of 12.4 N is too small, which leaves an acceleration of about 1.0 m s^-2; the test tan θ > μ gives the same verdict without any arithmetic. Treating the normal reaction as mg is the other common slip, because on a slope both components are smaller than the weight itself.
力矩与刚体平衡
力矩等于力乘上力到转轴（取矩点）的垂直距离，M = Fd，刚体平衡要同时满足合力为零与合力矩为零两个条件。2.5 m 均匀杆质量 6 kg，重力 58.8 N 作用在中点 1.25 m 处，A、B 两端各有一个支撑，距 A 端 0.8 m 处再放 4 kg 重物，也就是 39.2 N。对 A 点取矩：39.2 x 0.8 + 58.8 x 1.25 = 31.4 + 73.5 = 104.9 N m，除以 2.5 m 得 B 端反力 41.9 N，再用合力为零得 A 端 56.1 N。取矩点的选择是这类题的核心技巧，未知力通过哪一点就对那一点取矩，方程里直接少一个未知数。很多学生以为均匀杆的自身重力可以忽略，题目写出 uniform 与质量时必须放在中点，位置放错会让反力差出十几牛。算出两个反力以后回加验证一遍，省下的时间比从头重算更多。
A moment is the force multiplied by the perpendicular distance to the point you take moments about, and a rigid body in equilibrium needs both the resultant force and the resultant moment to be zero. For a uniform 2.5 m rod of mass 6 kg resting on supports at A and B, with a 4 kg load 0.8 m from A, taking moments about A gives 104.9 N m, a reaction of 41.9 N at B and 56.1 N at A. Choosing the pivot so that an unknown force passes through it is the main technique here, and the rod is uniform, so its weight of 58.8 N acts at the midpoint, which decides whether the reactions come out right or wrong.
真题节奏与分值换算
时间分配按分值算最直接：H240/03 共 100 分、2 小时，平均每分 1.2 分钟，力学部分约 50 分对应 60 分钟左右，纯数部分同样 60 分钟，某一块超时要在另一块补回来。练习顺序建议先按知识点做真题，OCR 官网从 2018 年起每年都有 H240/03 的试卷与评分方案，运动学、力与牛顿定律、力矩三块各做 20 道，再整套限时 2 小时。评分方案里 M 分是方法分、A 分是答案分，过程写对但算错通常还能保住 M 分，这也是力学部分不要留白的理由。很多学生只在考前一周集中刷题，力学的手感需要两周以上的间隔重复，隔三天回做一遍错题比连做十道新题有效。整套卷做完要把失分按类别记账，符号错、单位错、有效数字错分三栏统计，下一周只补出现最多的那一栏。
Timing follows the marks: 100 marks in two hours means 1.2 minutes per mark, so the mechanics section is worth roughly an hour of the paper. OCR publishes H240/03 papers and mark schemes from 2018 onwards on its website, so work through them by topic first, around twenty questions each on kinematics, forces and moments, then sit a full paper under time. In the mark schemes, M marks reward a valid method and A marks the answer, which is exactly why a blank mechanics answer costs more than a wrong one.
阅卷口径与收尾检查
OCR 在 H240/03 卷首印着两条口径：非精确数值答案保留 3 位有效数字，除非题目另有要求；需要数值时 g 取 9.8，除非题目另行说明。答案的分值写在方括号里，2 分题通常只要一个结果加一句理由，7 分的 show that 题要求完整的推理链。收尾检查可以固定成四项：正方向写了吗、单位带了吗、有效数字收了吗、量级合理吗，把 0.59 m 写成 5.9 m 是手滑最常见的样子。错得最多的是把中间结果提前收成 3 位有效数字，负号与平方一连就会放大到 0.05 以上。show that 题要答够过程，评分方案里这类题的答案分只占一小部分，只写结论即使数值正确也拿不到分。
Two instructions printed at the front of every H240/03 paper settle most rounding arguments: give non-exact numerical answers correct to 3 significant figures unless a different degree of accuracy is specified, and use g = 9.8 unless instructed otherwise; the marks for each question are shown in brackets. A two-mark item usually wants a result plus a short reason, while a seven-mark show that question wants an unbroken argument. Round early and the error grows, and on show that questions a correct final value without working earns very little.
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