Edexcel数学 力学 SUVAT 抛体 力矩 摩擦
Edexcel A Level 数学的力学内容全部放在 Paper 3，试卷代码 9MA0/03，两小时 100 分，Section A 统计 50 分，Section B 力学 50 分，整卷占最终成绩的三分之一。很多学生把复习时间几乎全投给两卷纯数，到春天才发现在这 50 分里有十几分是标准化流程题，画对受力图、选对公式就能拿到。力学真正的门槛不在公式数量，Edexcel 公式簿只印了 F = ma 与 v = dr/dt 两行，其余推导都得自己写。拉开差距的是三件事：受力图是否画全，判断该用哪条公式，以及答案是否取成阅卷能接受的有效数字。2026 年 6 月这一卷的评分口径没有变化，Pearson 官方在 Principal Examiner Feedback（2018 夏季，9MA0/03）里写明 g 一律取 9.8，最终答案给 2 到 3 位有效数字，比这更精确会被扣分。下面按八个考点拆开讲，每个考点配一个可以直接抄进错题本的例题。
Mechanics fills Section B of Paper 3 (9MA0/03), a two-hour paper worth 100 marks and one third of the A Level, 50 marks of which are mechanics. Most candidates spend the year on the two pure papers and only in the spring notice how much of that section is standard routine, where a complete force diagram and the right equation carry the marks. The Edexcel formula booklet hands you F = ma and v = dr/dt and little else; scores move when your diagram is complete, your chosen equation matches the missing quantity, and your final answer is rounded to 2 or 3 significant figures as the Pearson Principal Examiner Feedback (Summer 2018, 9MA0/03) requires.
SUVAT 四式与已知量判断
匀加速直线运动只有四条公式：v = u + at、s = ut + (1/2)at^2、v^2 = u^2 + 2as、s = (u + v)t/2。选公式的方法很简单，看题目没有给你哪个量：没给位移就用 v = u + at，没给末速度用 s = ut + (1/2)at^2，没给时间用 v^2 = u^2 + 2as，没给加速度用 s = (u + v)t/2。真正容易翻车的是方向符号，做题前先定一个正方向并写在草稿纸最上面，这样减速时 a 取负值，位移为负说明物体落在出发点另一侧。竖直上抛的球以 12 m/s 出手，g 取 9.8，到最高点用 t = 12 / 9.8 = 1.22 s，最高点离手 s = 144 / 19.6 = 7.35 m，回到手上总共 2.45 s。学生最常把上抛阶段的 g 也写成正值，于是算出一个永远落不下来的球。阅卷时会分别核对位移与路程，速度与速率都必须取正值，距离与位移混用是这一节最常见的扣分点。
Constant acceleration runs on four equations: v = u + at, s = ut + (1/2)at^2, v^2 = u^2 + 2as and s = (u + v)t/2. Choose the one that does not contain the quantity the question withholds, which removes almost all hesitation under time pressure; then fix a positive direction at the top of your page before substituting anything, so a deceleration enters as a negative value and a negative displacement simply means the other side of the start. A ball thrown straight up at 12 m/s stops after 1.22 s and rises 7.35 m at g = 9.8, returning at 2.45 s; examiners check displacement against distance and expect speed and velocity to stay positive.
速度时间图像的斜率与面积
v-t 图像上斜率就是加速度，图线与时间轴之间的面积就是位移。图像落在时间轴下方时面积记为负值，如果题目问路程，就要把轴下方的面积取绝对值再加起来。梯形面积用 (上底 + 下底) 乘高再除以 2，三角形面积用底乘高除以 2，不必背公式，把图形切成梯形与三角形更快。一辆车在 8 秒内速度从 0 均匀升到 20 m/s，加速度是 2.5 m/s^2，位移就是 80 m；接着 6 秒匀减速到停下，再走 60 m，前后共 140 m。常见的错误是拿 v-t 图算位移时直接读最后速度乘总时间，忽略了前半段还没加速到那个速度。a-t 图像同理，面积给的是速度变化量；题目给出 x-t 图要求画 v-t 图时，把每段的斜率一步步转过去就不会错。
On a velocity-time graph the gradient is acceleration and the area between the line and the time axis is displacement, with areas below the axis counted as negative. Ask yourself whether the question wants displacement or distance travelled, because only the second one sums the absolute areas; trapezium and triangle splits beat memorised formulas. A car reaching 20 m/s from rest over 8 s has a = 2.5 m/s^2 and covers 80 m, then 60 m more while braking to a stop in the following 6 s, giving 140 m in total. The same reasoning runs backwards: an acceleration-time graph gives a change in velocity as its area, and a distance-time graph hands you the gradients.
抛体运动的水平与竖直独立
抛体题的第一步是分解初速度：水平分量 U cos θ 全程不变，竖直分量 U sin θ 受重力影响。两个方向分别列式，再用同一个时间 t 把两式连起来。竖直方向是匀加速运动，SUVAT 四条公式全部可用；水平方向只有匀速一条 x = U cos θ 乘 t。最高点竖直速度为 0，水平速度却一点没减，很多学生以为最高点速度整体归零，这一步想错后面全错。以 20 m/s、仰角 30 度斜抛，飞行时间 t = 2 乘 20 乘 sin 30 / 9.8 = 2.04 s，射程为 20 cos 30 乘 2.04 = 35.3 m。若题目要求评估模型局限，只写 the model is not accurate 拿不到分，Pearson 2018 年 6 月 9MA0/03 的 mark scheme 给出的参考答案是目标有实际尺寸、空气阻力、风与球的旋转，要写出会改变结果的具体物理因素。
For projectiles, split the initial velocity first: the horizontal component U cos θ never changes, while the vertical component U sin θ behaves like any falling body under gravity. Keep the two directions independent and join them only through the shared time t; at the highest point the vertical velocity is zero but the horizontal velocity is exactly what it was at launch, and treating the whole velocity as zero is the error that ruins the rest of the question. A ball launched at 20 m/s and 30 degrees is airborne for 2.04 s and lands 35.3 m away, and a model-criticism part expects named physical factors such as air resistance, wind, spin or the size of the target rather than a general comment.
牛顿第二定律与电梯连接体
F = ma 的用法可以固定成四步：画受力图，选正方向，写出合力，解方程。电梯题的关键是法向力不等于重力，人站在秤上，秤读的是法向力而不是体重。电梯以 2 m/s^2 向上加速时，5 kg 物体的秤读数是 m(g + a) = 5 乘 11.8 = 59 N，比静止时的 49 N 大；电梯以同样大小向下加速时读数变成 5 乘 7.8 = 39 N，这一对数字就是超重与失重的全部内容。连接体先整体求加速度，再隔离其中一个物体求绳张力。4 kg 与 6 kg 用轻绳连在光滑桌面上，20 N 拉力作用在 6 kg 上，整体加速度 a = 20 / 10 = 2 m/s^2，绳张力只负责拉动 4 kg，所以 T = 4 乘 2 = 8 N。学生最常怀疑滑轮两侧张力不相等，其实在轻绳与光滑滑轮的前提下两段张力始终相同，差别只出现在有质量或有摩擦的滑轮上。
Newton's second law is a routine rather than an idea: draw the forces, fix a positive direction, write the resultant, then solve the equation. In a lift the scale reading is the normal reaction, so a 5 kg mass accelerating upwards at 2 m/s^2 reads 59 N while the same mass accelerating downwards reads 39 N, which is the whole of the apparent weight story. For connected particles find the acceleration of the whole system first and then isolate one body for the tension; a 20 N pull on 6 kg joined to 4 kg on a smooth table gives a = 2 m/s^2 and T = 8 N, and the tension is the same on both sides of a light string over a smooth pulley.
摩擦系数与斜面受力
Edexcel 公式簿把摩擦写成 F ≤ μR，等号只在临界滑动时成立。水平面上推 8 kg 的箱子，μ 取 0.3，临界推力是 μR = 0.3 乘 8 乘 9.8 = 23.5 N；推力小于这个值时物体不动，此时摩擦力等于推力而不是 μR，这一步最容易丢分。斜面上要先把重力分成沿斜面与垂直斜面两个分量：R = mg cos θ，沿斜面向下的分量是 mg sin θ。仰角 20 度、μ 取 0.3 时，R = 73.7 N，最大摩擦 22.1 N，下滑分量 26.8 N 更大，物体必然下滑，加速度由 (mg sin θ - μR) / m 得到 0.59 m/s^2。斜面静摩擦的临界角满足 tan θ = μ，这个关系能把选择题从两分钟压缩到二十秒。摩擦力方向永远与相对运动或相对运动趋势相反，画错箭头会让整道题的符号全反。
Edexcel prints friction as F ≤ μR, with equality only at limiting equilibrium, so a push below 23.5 N on an 8 kg box leaves friction equal to the push rather than to μR. On a slope resolve the weight first to get R = mg cos θ and a downslope component mg sin θ; at 20 degrees with μ = 0.3 the friction tops out at 22.1 N against a 26.8 N pull, so the box accelerates at 0.59 m/s^2. Limiting equilibrium on an incline also gives the compact relation tan θ = μ, and friction always opposes relative motion or the tendency to move, which sets the sign of every term.
力矩平衡与梯子问题
力矩的定义是力乘上它到支点的垂直距离，斜着拉的力要先分解出垂直分量再乘。均匀杆的重心在中点，均匀平板的重心在对角线交点，这两个结论在力矩题里几乎年年用到。刚性体平衡有两个条件：合力为零，以及绕任意一点取矩的合力矩为零。6 m 均匀木板重 90 N，支点在离左端 1 m 处，重心距支点 2 m，顺时针矩是 90 乘 2 = 180 N m，右端挂重物 W 距支点 5 m，由 5W = 180 得 W = 36 N。梯子靠墙的题有固定套路：把支点取在梯子与地面的接触点，一次消掉摩擦力与法向力两个未知量。学生最常把斜边的长度当成力的作用线距离，斜杆与斜面的题会因此算错一大截，正确做法是先画垂线再量长度。
A moment is force times the perpendicular distance to the pivot, so a force acting at an angle must be resolved before you multiply, and choosing the pivot where a ladder meets rough ground removes friction and the normal reaction in one step. Uniform rods balance at their midpoint and uniform laminae at their centre of mass, while equilibrium of a rigid body needs both zero resultant force and zero resultant moment about any point. Taking moments about the base is the standard route through ladder questions, and the common failure is using the sloping length instead of the perpendicular distance, so draw the perpendicular first and measure that.
变加速的微分与积分
变加速题里 v = dx/dt 与 a = dv/dt 是主公式，公式簿写作 v = dr/dt 与 a = dv/dt = d^2r/dt^2，符号换成 r 是为了把位置与位移分开。给定 v 求位移要积分再代入初始条件定常数，给定 a 求速度同理。求最大速度令 a = 0，求再次静止令 v = 0，这两个动作覆盖了绝大多数小问。速度 v = 6t - t^2 时 a = 6 - 2t，t = 3 s 速度最大为 9 m/s；位移 x = 3t^2 - t^3/3 + C，若 t = 0 时物体在原点，则 t = 6 s 时位移为 36 m，而且此刻 v 回到 0，物体正好停下。常见的错误是把积分常数丢掉，第二次求位移就全错。判断运动方向变没变，只需看 v 在区间内是否跨过 0，跨过就要分段积分。
With variable acceleration the toolkit is calculus: v = dx/dt and a = dv/dt = d^2x/dt^2, both printed in the Edexcel formula booklet. Integrate v to get displacement and fix the constant of integration with the initial condition, set a = 0 for maximum speed and v = 0 for the instant the particle comes to rest. For v = 6t - t^2 the maximum speed is 9 m/s at t = 3 s, and starting from the origin the displacement at t = 6 s is 36 m; if v crosses zero inside the interval the integration has to be split into stages.
向量表示与二维受力
力学里的向量统一用 i 与 j 写，位置 r = x i + y j，速度与加速度逐项微分。求合力时把各力的 i 分量与 j 分量分别相加，再让 F = ma 拆成两个标量方程。二维抛体也能用向量写，此时加速度恒为 -9.8 j，速度随时间线性变化。两物体相遇的条件是位置向量相等，把两个 r 写成同一个 t 的表达式再联立，这是连接体之外向量最常见的考法。学生最常把位置向量当成位移向量用，忘记位移要减掉起点坐标。速度向量的大小用勾股定理算，方向用 tan θ = j 分量除以 i 分量给出，角度必须写清是相对哪条轴量出来的。
Vectors in mechanics are written with i and j, so position, velocity and acceleration are differentiated component by component. Add the i-components and j-components of the forces separately to turn F = ma into two scalar equations, and remember that a projectile in vector form simply carries a constant acceleration of -9.8 j. Two particles meet when their position vectors are equal, which becomes one equation in t; magnitudes come from Pythagoras, directions from tan θ = j / i, and displacement is always a position minus the starting position.
有效数字与阅卷口径
Pearson 官方 Principal Examiner Feedback（2018 夏季，9MA0/03）写明力学计算里 g 取 9.8，除非题面另给，最终答案给 2 到 3 位有效数字，比这更精确会被扣分，但 g 的精确倍数通常可以接受。中间步骤要多留两位，只在最后结果取整，中途四舍五入再代入是失分的主要原因。题面出现 show that 时必须写出足够细节，只有答案没有过程可能一分都拿不到。单位别漏，力矩的单位是 N m 而不是 J，两者混写在阅卷里会直接判错。分数与百分数的精度同理，四分之三写成 0.75 是精确值，写成 0.8 就不再相等。
Pearson's Principal Examiner Feedback for 9MA0/03 in Summer 2018 states that g is taken as 9.8 unless the question says otherwise, and that final answers should be given to 2 or 3 significant figures, since more accurate answers are penalised. Keep the extra digits through the working and round once at the end. A show that part needs enough visible method to earn every mark, and moments are measured in N m rather than J, so mixing the two units costs accuracy marks even when the number is right.
每周真题切分与错题归因
复习节奏可以按 Section B 的 50 分来切：每周一次限时 60 分钟做完一整份力学部分，再用 40 分钟对着官方 mark scheme 逐题归因。归因分四类就够用：受力图漏力，符号方向取错，公式选错，结果精度不合口径。这四类里精度问题往往能在两周内修好，受力图需要反复重画才有手感。错题本上不要只抄正确答案，把当时自己写的那一步也抄下来，对照 mark scheme 的 M 分与 A 分看分丢在哪一步。练习素材用近年 9MA0/03 的 Section B 与 8MA0 的应用卷，题目难度接近而题量更小，适合月考前的短练。每周固定同一时间做，比考前突击两周有效得多，力学的手感来自重复而不是顿悟。
A workable weekly routine is one timed 60-minute run through a mechanics section followed by 40 minutes of marking your own work against the official mark scheme. Sort every lost mark into one of four buckets: a missing force on the diagram, a wrong sign or direction, the wrong equation, or an answer rounded outside the accepted tolerance. Copy the step you actually wrote into your error log, because the M and A marks in the scheme show exactly where the method broke down; recent 9MA0/03 Section B papers and 8MA0 applied papers make good short practice.
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