Edexcel数学 力学 匀加速 速度图 受力 力矩
Edexcel 的 A-Level 数学把力学放在 Paper 3 的后半段，和统计各占 50 分。内容不算多，考法却相当固定：读题、画受力图、选公式、代数字、按有效数字写答案。很多学生把五条匀加速公式背得滚瓜烂熟，进了考场仍卡在第一步的图上。下面按 Edexcel 9MA0 考纲里力学的四个专题顺序，把这 50 分拆成十个具体动作讲。
Mechanics occupies Section B of Paper 3 and is worth 50 of the 100 marks; Pearson prints it as a separate booklet with the paper code 9MA0/32, so the section is effectively a one-hour paper of its own. The content list is short: quantities and units, kinematics, forces and Newton's laws, moments. What varies is the packaging, since almost every question arrives as a described situation rather than an equation. Marks are lost in the setup far more often than in the algebra, and the ten habits below address exactly that stage.
Paper 3 的力学是 50 分一小时
Edexcel 的 9MA0 一共三份卷子，每份 2 小时、100 分、各占 33.33%。第三份卷号 9MA0/03，标题是 Statistics and Mechanics，卷面分成 Section A 统计和 Section B 力学。两段各 50 分，考试时也各给一小时。Pearson 在考务上把力学单独印成一份小册子，页首写着 Paper 32: Mechanics。这份卷子的封面上还有一条计算器限制：不能带符号代数运算、微积分功能，也不能存有可检索的公式。拿到卷子先看分值分布，再决定每道题花多久。50 分对 60 分钟，一分大约一分钟出头，12 分的题给到 15 分钟就该收手。先做会做的部分，剩下的回头补，比在某一问上死磕划算。力学的题目大致按专题排布，运动学、受力与摩擦、连接体、力矩与向量都会出现。读题时先归到某一类，就知道该翻公式册的哪一部分，也知道该画哪种图。
Each of the three papers is worth 100 marks and 33.33%, and the third one splits cleanly into statistics and mechanics; Pearson prints the mechanics half as Paper 32 with its own cover, its own 50 marks and its own instruction that calculators must not offer symbolic algebra, calculus or stored formulae. That cover tells you what kind of thinking is expected: arithmetic is cheap here, so the marks sit in the modelling. A sensible plan divides the hour by marks rather than by questions and leaves a stubborn part for later. Sorting each question into kinematics, forces and friction, connected bodies, or moments also tells you which page of the booklet you will need, which saves time before the writing starts.
匀加速五条公式与缺量选式
五条匀加速公式都印在公式册 Mathematical Formulae and Statistical Tables 的 Mechanics 部分，这一册是 2017 年 7 月发布的 Issue 1。它们分别是 v = u + at、s = ut + ½at²、s = vt - ½at²、v² = u² + 2as、s = ½(u + v)t。考试不考记忆，考的是选哪一条。稳妥的做法是先写下 s、u、v、a、t 五个符号，把题给的三项填进去，把要求的一项圈出来，再看剩下那个不需要的量。找一条既不含多余量、又能直接解出答案的式子，一步到位。常见的错误是看见加速度就写 v = u + at，结果题目问的是位移，白算一遍还丢了分。手算一遍就明白：一辆车初速 12 m s⁻¹、加速度 -2 m s⁻²，问四秒内的位移，格子里 s 与 a 未知，u 与 t 已知，缺的是 v，选 s = ut + ½at²，得 48 减 16 等于 32 m。若先求 v 再算位移，多一步，也多一个可能算错的地方。
All five formulae are printed in the booklet, which means recall is not being tested; selection is. Listing s, u, v, a and t on the page, filling in the three values the question supplies and circling the one it asks for turns the choice into a mechanical step. The equation that omits the unwanted quantity is the one to use, and it usually yields the answer without any simultaneous work. Reaching for v = u + at whenever an acceleration appears is the classic slip, because it answers a question about speed when the paper asked about distance. A car at 12 m s⁻¹ decelerating at 2 m s⁻² covers 32 m in four seconds, which s = ut + ½at² delivers in one line and the two-step route only reproduces with extra room for error.
速度时间图 斜率是加速度 面积是位移
速度时间图是力学里最省时间的一件工具。斜率读出加速度，图线与时间轴围出的面积读出位移。轴下方的面积取负值，表示这段时间的位移与选定正方向相反。题目给出三段折线时，分段算面积再相加，比套公式快得多。求总路程时把负面积取绝对值再加,求净位移时保留符号。向上的运动一般定为正方向，减速就对应负的加速度。很多学生以为图线下的面积总是正的，遇到折返运动就把位移写成了路程。折返的情况要分段处理：先算速度减到零的时刻，再分别算前后两段的面积。刹车类题目里，给的时间可能比实际停下所需的时间长，多出来的那段车已经静止，照着公式硬算会得出负位移。
A velocity-time sketch converts a word problem into two readings: gradient gives acceleration, enclosed area gives displacement, and area below the axis counts as negative displacement. Splitting a broken line into trapezoids and triangles is usually quicker than feeding three intervals through the formulae, and it keeps the direction of travel visible. Total distance takes the absolute value of every region, whereas net displacement keeps the signs. Fixing upward as positive at the start of the question makes a deceleration appear as a negative gradient instead of a surprise. Reversal needs splitting at the instant the velocity reaches zero, and braking questions often quote a duration longer than the vehicle actually takes to stop, so the formula applied blindly returns a negative displacement.
自由落体与 g 取 9.8 的答案习惯
卷首说明写明 g 取 9.8 m s⁻²，除非题目另给数值，答案按需要给两位或三位有效数字。这条规则看着琐碎，扣分却很实在：算出 19.6 写成 19.613 会被判精度不符，写成 20 又可能被判有效数字不足。竖直上抛的题先定正方向，再决定 g 前面的符号。在最高点速度为零，加速度仍是 9.8，两个量不要混。落地时间用 s = ut + ½at² 解二次方程，取正根，负根没有物理意义。阅卷时最常见的一种失分，是全程把 g 当成 10 来算，前后的数字全部对不上标准答案。一颗石子从静止自由下落三秒，位移是 ½ 乘 9.8 乘 9，得 44.1 m，三秒末的速度是 9.8 乘 3，得 29.4 m s⁻¹，两个答案都留三位有效数字。
The rubric at the top of the paper fixes g at 9.8 m s⁻² unless a question says otherwise, and asks for answers to two or three significant figures as appropriate. That rule matters because a correct 19.6 written as 19.613 loses the accuracy mark while 20 can lose it for the opposite reason. Choose a positive direction before substituting, since the sign in front of g follows from that choice rather than from the motion itself. At the highest point the velocity is zero and the acceleration is still 9.8, a distinction that examiners test every year. A stone released from rest falls 44.1 m in three seconds and reaches 29.4 m s⁻¹, both quoted to three significant figures rather than to whatever the display shows.
变加速用 a = dv/dt 积分求位移
匀加速公式只管加速度恒定的情况，题目一旦把速度写成时间的函数，就得回到微积分。速度对时间求导得加速度，加速度对时间积分得速度变化，速度对时间积分得位移。积分后出现的常数用初始条件定：t 等于零时的位置或速度代进去解。Paper 3 的所有纯数学内容都算已知，所以求导与积分在这里不算超纲。看到 v = 3t² - 8t 这类表达式，先判断题目问的是某一时刻的加速度，还是一段时间内的位移。这一步判断错了，后面写得再整齐也拿不到方法分。定积分的上下限就是时间区间的两端，常数在相减时自然抵消，所以求位移变化量可以直接算定积分。要求某一时刻的位置，才需要把初始条件代进去定常数。
Once the velocity is given as a function of time, the constant-acceleration formulae are out and calculus takes over: differentiate to find acceleration, integrate to recover velocity or displacement, and fix the constant of integration with the initial condition. Pure mathematics is assumed knowledge for Paper 3, so nothing here is off-syllabus. The decision that matters is reading whether a question wants the acceleration at one instant or the displacement over an interval, because the two routes diverge immediately. A definite integral between the two times cancels the constant on its own, and only a position at a named instant needs the initial condition worked out.
受力图先画重力 再补反力与张力
受力图画对了，F = ma 只是填空。顺序建议固定：先画重力 mg，竖直向下；再画接触面给的法向反力 R，垂直于接触面；绳子只拉不推，张力沿绳、方向背离物体；最后补摩擦力与题目给的推力。图上只画作用在这个物体上的力，别把它施给别人的力也画进来。光滑表示没有摩擦，轻绳表示绳子的质量可以忽略，不可伸长的绳两端加速度大小相同。很多学生以为反力总等于重力，斜面上一算就发现不是。连接体问题里，两个物体各画一张受力图，绳两端的张力大小相同，加速度大小相同而方向不同。整体法把两件当成一件，只看外力，求加速度更快；隔离法能解出内部张力。两种方法的结果应当一致，对不上就说明某张图漏了力。
Draw the weight first, vertically down, then add the normal reaction perpendicular to the surface, then any tension acting away from the body along a light inextensible string, then friction and applied forces. Only forces acting on the chosen body belong on the diagram, which is why the force it exerts on something else stays off. Smooth means no friction, light means the string's mass is ignored, and inextensible means both ends share the same magnitude of acceleration. Assuming the normal reaction equals the weight is safe on level ground and wrong on every incline. Connected bodies need one diagram each, with the same tension at both ends of the string and equal magnitudes of acceleration, and a whole-system check should reproduce whatever the separate diagrams produce.
摩擦力 F ≤ μR 与极限平衡
摩擦分两种状态，写法不能混。物体还在滑动或刚好要滑动时，摩擦力取到最大值 F = μR。物体静止且没有滑动趋势的极限时，摩擦力只满足 F ≤ μR，具体数值由平衡条件解出来。题目里出现「即将滑动」「limiting equilibrium」这类字眼，就是在提示用等号。摩擦力的方向永远与相对运动或相对运动趋势相反，与外力方向无关。斜面上的临界角用 tanθ = μ 求，这个结果从 F = μR 与沿面平衡两条式子联立得到。常见的错误是静止状态下也硬套 F = μR，把实际摩擦算大了。五千克的箱子放在水平地面上，μ 取 0.4，反力等于 49 N，最大静摩擦约 19.6 N。推力小于这个数时箱子不动，此时摩擦力等于推力本身，而不是 19.6 N。
Friction has two regimes and they are not interchangeable: at the point of sliding the value reaches its maximum F = μR, while a body in equilibrium with no tendency to move satisfies only the inequality F ≤ μR and takes whatever value the balance equations demand. Phrases such as about to slip or limiting equilibrium are the signal to replace the inequality with an equality. Its direction always opposes relative motion or the tendency to move, which is not necessarily the direction of the applied force. Applying F = μR to a body that is sitting still is the commonest error in this topic, because it invents friction that is not there. A 5 kg box with μ = 0.4 has a normal reaction of 49 N and a limiting friction near 19.6 N, yet a push of 10 N meets a friction of 10 N and no movement at all.
斜面分解中 sin 与 cos 的归属
斜面问题的第一动作是把重力分解成两个分量。沿斜面向下的分量是 mg sinθ，垂直斜面的分量是 mg cosθ。θ 取斜面与水平面的夹角，不是斜面与竖直方向的夹角，两者互余，弄反了整道题的方程都反。沿斜面方向的方程写成 mg sinθ 减去摩擦力等于质量乘加速度，垂直方向写成 R 等于 mg cosθ。粗糙斜面上还要判断滑动方向，摩擦力与它相反。光滑斜面只剩沿面的 mg sinθ，加速度直接等于 g sinθ，与质量无关。这个结论可以用来检查答案是否合理。θ 等于 30° 时沿面的加速度是 4.9 m s⁻²，θ 趋于 90° 时它趋于 9.8，θ 趋于零时它趋于零。算出超过 9.8 的加速度，一定在分解或代入时出了错。
Resolving starts from one split: the component down the slope is mg sinθ and the component into the surface is mg cosθ, where θ is measured from the horizontal. Taking the complement by mistake flips every equation that follows, so write θ on the diagram before writing any component. Along the plane the balance reads mg sinθ minus friction equals ma, while perpendicular to it the normal reaction carries mg cosθ. On a smooth slope the mass cancels and the acceleration is simply g sinθ, which is a quick sanity check on any numerical answer. At 30 degrees that gives 4.9 m s⁻², tending to 9.8 as the slope steepens towards vertical and to zero as it flattens, so anything above 9.8 signals a mistake in the resolution.
力矩平衡与梯子问题的取矩点
力矩等于力乘上力臂，力臂是转轴到力的作用线的垂直距离。刚体平衡要同时满足两个条件：合力为零，绕任意一点的合力矩为零。解题时挑一个未知力穿过的点作转轴，那个力的力臂为零，方程里就少一个未知量。梯子靠墙是考纲里点名的模型，墙光滑、地面粗糙，两端各有一个反力，中间还有重力作用在梯子的中点。均匀梯子的重力作用在几何中心，这一点决定了力臂的大小。力偶是两个大小相等、方向相反、不共线的力，它的合力为零，合力矩不为零。很多学生以为平衡只要求合力为零，忘了力矩那一条，梯子题就写不下去。
A moment is force times the perpendicular distance from the pivot to the line of action, and a rigid body in equilibrium needs both zero resultant force and zero resultant moment about any point. Choosing a pivot that an unknown force passes through removes that force from the equation entirely, which is the single most useful trick in this topic. The ladder problem named in the specification combines a smooth wall, a rough floor and the weight of a uniform ladder acting at its midpoint. A couple has zero resultant force but a non-zero moment, so it can only be balanced by another couple.
二维向量运动学 r = ut + ½at²
二维运动的最省事写法是把位移、速度、加速度都写成 i 与 j 两个分量，然后让两个方向各用一套匀加速公式。r = ut + ½at² 在向量形式下依然成立，逐分量代入即可，单位向量不参与运算。题目给出两个时刻的位置，先相减得到位移向量，再代公式求加速度，最后取模长得到大小。位移大小用勾股定理算，方向用 atan 求与水平方向的夹角，别把分量当成答案。2018 年的 9MA0/03 力学部分就有一道这样的题，先积分速度求位置向量，再求两点间距离。分量算完不合成，是这类题最常见的失分方式。
Two-dimensional motion is easiest in components: write displacement, velocity and acceleration as i and j parts, then run the constant-acceleration formulae along each axis separately. The vector form r = ut + ½at² holds unchanged, and the unit vectors simply carry through the arithmetic. When two positions are given, subtract them to get a displacement vector, substitute for the acceleration, then take the magnitude for the answer. Leaving the result as components is the usual way to lose the final accuracy mark, because the question asked for a distance or a bearing, not a pair of numbers.
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