OCR数学 H240力学 运动学 牛顿定律 力矩
OCR 的 A Level 数学 A（H240）把力学放在 Paper 3，和纯数合成一张两小时的卷子。整卷 100 分，力学通常占 30 分到 40 分。OCR 官方考纲把力学拆成三块：3.02 运动学、3.03 力与牛顿定律、3.04 力矩。这三块的题目几乎每年都会互相串，比如一道斜面题里既有牛顿第二定律又有摩擦力。下面按这三块的顺序走，每块配一个能直接套用的做题流程和一个带数字的例子。
Paper 3 of OCR A Level Mathematics A (H240) combines pure mathematics with mechanics in a single two-hour paper worth 100 marks, and mechanics usually accounts for 30 to 40 of them. The specification splits that content into kinematics (3.02), forces and Newton's laws (3.03) and moments (3.04), yet questions routinely mix all three inside one context such as a block on a rough slope. The walkthrough below follows that order and pairs each topic with a working routine and a numeric example.
一、Paper 3 里的力学范围
OCR 考纲把这些内容编号到 3.02 到 3.04，纯数部分先做，力学题接在后面。3.02 运动学包括匀加速公式、速度-时间图、非匀加速和抛体。3.03 力与牛顿定律包括受力图、力的分解、摩擦、斜面和连接体。3.04 力矩包括刚体平衡和力矩方程。很多学生以为力学就是背五个公式，其实失分点多在受力图漏力。卷面统一用 g = 9.8 m s^-2，题目另给数值时按题目。答案一般保留三位有效数字。力学题单题多在 5 到 14 分之间，最后一问常把两个物体或两个方向连起来。近年真题的背景换成了无人机、滑板、货箱这类场景，问法还是求加速度、求时间、求某个力。
Mechanics is examined under specification points 3.02 to 3.04, and it follows a pure section rather than opening the paper. Kinematics covers the constant acceleration formulae, velocity-time graphs, non-uniform acceleration and projectiles. Forces and Newton's laws cover force diagrams, resolving, friction, inclined planes and connected particles, while moments cover equilibrium of a rigid body and setting up a moment equation. Marks are lost far more often through a missing arrow on a force diagram than through algebra. Values are expected to three significant figures unless a question says otherwise.
二、suvat 五个公式与符号约定
匀加速的五个公式里，能直接用哪一个取决于题目给了哪三个量。先写下 s、u、v、a、t 五个符号和已知值，缺的那个往往就是题目要你求的。符号约定要在动笔前定好：我习惯取运动方向为正，于是减速时 a 是负数，向上抛时 a = -9.8。很多学生以为加速度为正就一定在加速，遇到"减速上升"就把符号写反了。位移和速度都是矢量，返回出发点时位移为零，路程却不是零。单位要跟着走，时间用秒，距离用米，速度用 m s^-1。举个数字，u = 8 m s^-1、a = 1.5 m s^-2、t = 6 s 时，v = 8 + 1.5 × 6 = 17 m s^-1，s = 8 × 6 + 0.5 × 1.5 × 36 = 75 m。题目若问还要多久停下，就把 v = 0 代进同一个式子，别另起一套符号。
Which of the five constant acceleration formulae to use is decided by the three quantities a question hands you. Write s, u, v, a and t down with the known values filled in, and the missing one is normally what is being asked. A body thrown upwards and still rising has a negative acceleration, which is the most common sign slip in this topic. Displacement and velocity are vectors, so a return to the start gives zero displacement but a non-zero distance.
三、速度-时间图读什么
速度-时间图上有两个读数：斜率是加速度，曲线下的面积是位移。图线在时间轴下方时面积算负值，回到原点那一刻正负面积正好抵消。匀加速的图线是直线，非匀加速的图线是曲线，这时候斜率要用切线求。常见的错误是把面积直接当成路程，方向一变就不再相等。求最大速度时先看加速度何时为零，也就是图线的最高点或最低点。这类题通常 4 到 6 分，画图和读图各占一半。画的时候把起点、转向点、终点标出来，分段算面积不容易漏。若给的是距离-时间图，斜率才是速度，这一点容易和速度-时间图混。
Two readings matter on a velocity-time graph: the gradient gives acceleration and the area under the line gives displacement. Area below the time axis counts as negative, so the two regions cancel when a body returns to its starting point. Constant acceleration produces a straight line and non-uniform acceleration a curve, where the gradient has to come from a tangent. Treating the area as distance only works while the direction of motion stays unchanged.
四、矢量形式的运动学
H240 的力学题会用到 i、j 两个方向的写法，位移、速度、加速度都可以写成列向量。匀加速公式照样成立：v = u + at，r = r0 + ut + 0.5at^2，只是每个分量单独算。两个方向的时间必须相同，这一点决定了题目能不能解。若加速度只有竖直分量，水平速度在整个过程中保持不变。求两点间距离时先把位移向量减出来，再取模长。求方向时用 tan 求角，注意题目要的是方位角还是与水平方向的夹角。举个数字，u = (4, 1) m s^-1、a = (0, -9.8) m s^-2，t = 2 s 时 v = (4, -18.6) m s^-1，速率约 19.0 m s^-1。
Questions on this specification use i and j notation, so displacement, velocity and acceleration can all be written as column vectors. The constant acceleration formulae still hold in vector form, with each component handled separately. Time is shared by both components, which is what usually makes a problem solvable. Where acceleration has a vertical component only, horizontal velocity stays constant throughout, and the distance between two positions is the modulus of the displacement between them.
五、变加速用微积分
加速度不是常数时，五个公式全部失效，只能回到定义。a = dv/dt，v = ds/dt，链式法则还给出 a = v dv/ds。题目通常给 v 关于 t 的表达式，求 a 就求导，求 s 就积分。积分常数靠初始条件定，t = 0 时 s = 0 就把常数写成零。举个具体例子，v = 3t^2 - 2t，则 a = 6t - 2，t = 2 时 a = 10 m s^-2。s = t^3 - t^2，t = 2 时 s = 4 m，t = 3 时 s = 18 m，这一秒内走了 14 m。很多学生在这里忘记加积分常数，或者把 t 直接代进导数式子求位移。反过来，若题目给 a 关于 t 的表达式，求 v 要积分一次，求 s 要积分两次，两个常数各由一个初始条件定。
Once acceleration varies, none of the five formulae apply and the definitions take over: acceleration is the derivative of velocity, velocity is the derivative of displacement, and the chain rule adds acceleration as velocity times the gradient of velocity with respect to displacement. A question normally supplies velocity as a function of time, so differentiating gives acceleration and integrating gives displacement. The constant of integration comes from an initial condition such as zero displacement at zero time. For v = 3t^2 - 2t the acceleration is 6t - 2, giving 10 m s^-2 at two seconds, while displacement runs 4 m at two seconds and 18 m at three seconds.
六、抛体运动拆成两个一维问题
抛体题唯一的技巧是拆：水平方向匀速，竖直方向匀加速，两个方向只共用时间。竖直方向取向上为正，则 a = -9.8，最高点时竖直速度为 0。水平方向的初速度一般不变，射程就是水平速度乘总时间。从高处水平抛出时竖直初速度为 0，落体时间由高度决定：20 m 高、g = 9.8 时 t = 根号(2 × 20 / 9.8) = 2.02 s。水平初速 12 m s^-1 时射程为 24.2 m，落地速度由两个分量合成，约 23.1 m s^-1。学生最常把重力也加到水平方向上，结果整道题的射程都算错。斜抛时初速度要先分解成水平分量 u cos θ 和竖直分量 u sin θ，再各走各的。
The one technique in projectile work is separation: horizontal motion stays uniform, vertical motion is uniformly accelerated, and only the time is shared. Taking upward as positive makes gravity -9.8, so vertical velocity is zero at the highest point. A projectile launched horizontally has zero initial vertical velocity, and the fall time follows from the height alone. From 20 m the fall takes 2.02 seconds, so a horizontal speed of 12 m s^-1 gives a range of 24.2 m and an impact speed near 23.1 m s^-1. Adding gravity to the horizontal component is the error that wrecks the rest of the working.
七、牛顿第二定律与受力图
F = ma 里的 F 是合力，不是某一个力，这一点决定了受力图画得对不对。画图时把物体单独框出来，只画它受到的力：重力、支持力、绳子拉力、摩擦力和外加推力。然后按运动方向分解，沿运动方向写一条 F = ma，垂直方向写合力为零。连接体里绳子的拉力对两端物体方向相反、大小相同，这是解题的关键约束。平衡题里 a = 0，方程退化成合力为零。很多学生以为支持力永远等于重力，在斜面或有竖直加速度时这里就出错了。投影时把所有的力都投到两个轴上，别只投重力。答案的符号要回到你定的正方向去解释，负号只表示方向与正方向相反。
The force in Newton's second law is the resultant, not any single force, which is why the diagram decides whether the algebra works. Isolate the body and draw only what acts on it: weight, normal reaction, tension, friction and any applied push or pull. Resolve along the direction of motion and set the resultant equal to mass times acceleration there. In a connected system the tension pulls the two bodies in opposite directions with equal magnitude, and that shared value links the equations. Normal reaction is not automatically equal to weight, which is where slope and lift problems go wrong.
八、斜面、摩擦与极限平衡
斜面上要把重力分解成沿斜面和垂直斜面两个分量：沿斜面向下是 mg sin θ，垂直斜面是 mg cos θ。摩擦力的上限是 μR，其中 R 是垂直斜面的支持力，静止时 F ≤ μR，滑动时 F = μR。方向的判断只看趋势，物体想往下滑，摩擦就向上。极限平衡就是摩擦力刚好达到 μR，再加一点力就动。举个数字，5 kg 的物体放在 30 度斜面上、μ = 0.2：R = 5 × 9.8 × cos30 = 42.4 N，最大摩擦 8.49 N，沿斜面向下的分量 24.5 N，合力 16.0 N，加速度 3.2 m s^-2。常见的错误是斜面上也写 R = mg。若物体沿斜面向上运动，摩擦改为沿斜面向下，符号必须重判。
On a slope the weight splits into a component along the plane, mg sin θ downwards, and one perpendicular to it, mg cos θ. Friction is capped by μR with R the normal reaction, so a body at rest satisfies friction less than or equal to μR while a sliding body sits at the limit. Direction follows the tendency of motion only, and limiting equilibrium means friction has just reached its maximum. A 5 kg block on a 30 degree slope with a coefficient of 0.2 has a normal reaction of 42.4 N, a friction limit of 8.49 N and a downslope component of 24.5 N, leaving 3.2 m s^-2 of acceleration. Writing the normal reaction as weight is the mistake this setup is built to catch.
九、连接体的整体法与隔离法
两个物体用绳子连在一起时，加速度大小相同，方向各自沿自己的运动方向。求加速度用整体法：把两个物体当成一个质量为 m1 + m2 的整体，外力只剩重力和摩擦。求绳子拉力必须隔离其中一个物体，因为整体法里拉力是内力，已经抵消掉了。挂在桌边的物体贡献的是它的重力，桌面上的物体贡献的是摩擦和它自己的质量。举个数字，4 kg 放在桌面、μ = 0.25，2 kg 悬挂：摩擦 = 0.25 × 4 × 9.8 = 9.8 N，加速度 = (2 × 9.8 - 9.8) / 6 = 1.63 m s^-2，拉力 = 2 × (9.8 - 1.63) = 16.3 N。这一步最容易漏掉桌面的摩擦，或者把悬挂物体的质量也塞进分母。滑轮光滑时绳两侧拉力相同，滑轮有质量时题目会另给条件。
Two bodies joined by a string share the same magnitude of acceleration, each along its own line of motion. Treating them as one body of combined mass gives the acceleration, because the tension is then internal and cancels. The tension itself needs one body isolated, since the combined treatment removes it. A hanging mass contributes its weight while a mass on a table contributes friction and its own inertia. With 4 kg on a table at a coefficient of 0.25 and 2 kg hanging, friction is 9.8 N, acceleration is 1.63 m s^-2 and tension is 16.3 N.
十、力矩与刚体平衡
力矩等于力乘以支点到力作用线的垂直距离，单位是 N m。刚体平衡要同时满足两件事：合力为零，对任意一点的力矩和为零。均匀杆的重心在中点，重量作用在那里；非均匀杆的重心位置未知，要设成 x 再用平衡条件求。选支点有技巧，把支点选在未知力的作用线上，那个力的力矩为零，方程里就少一个未知数。举个数字，4 m 长、6 kg 的均匀杆 AB 在 A 点和距 A 三米的 C 点各有支撑，B 端挂 3 kg：杆重 58.8 N，挂重 29.4 N，对 A 取矩得 R_C = 78.4 N，再由竖直方向平衡得 R_A = 9.8 N。物体将要倾倒的临界状态是某一端支持力刚好降为零。力矩的方向用顺时针和逆时针区分，方程写成顺时针之和等于逆时针之和最不容易乱。
A moment is the force multiplied by the perpendicular distance from the pivot to its line of action, measured in newton metres. Equilibrium of a rigid body needs two statements at once: zero resultant force and zero total moment about any point. A uniform rod carries its weight at the midpoint, while a non-uniform one has its centre of mass at an unknown position that the equilibrium conditions can locate. Choosing the pivot on the line of an unknown force removes that force from the equation. For a 4 m rod of 6 kg supported at one end and at 3 m, with 3 kg hung at the far end, moments about the first support give 78.4 N there and vertical equilibrium leaves 9.8 N at the other. Tipping begins when one of those reactions falls to zero.
十一、考场上的写法与时间
Paper 3 两小时 100 分，力学部分一般 30 到 40 分，留给它大约 40 分钟。拿到题先读关键词，判断属于 3.02、3.03 还是 3.04，再画图，最后代公式。受力图要让阅卷人一眼看懂，箭头起点在物体上，标出每个力的大小或符号。中间步骤要写出来，方法分往往比最终答案分多，跳步会丢 M 分。数值答案带单位，方向写清楚是向上还是沿斜面向下。最后留两分钟检查符号和有效数字，一个符号错常常连带后面几问全错。最后一问若一时没有思路，先把能写的方程写上去，方法分照样给。
The paper gives two hours for 100 marks and the mechanics questions usually take about 40 minutes of that. Read the wording first to decide whether a question is kinematics, forces or moments, then draw, then substitute. A force diagram should let a marker see at a glance which arrows act and where they start, and the intermediate line is worth writing because method marks outweigh the final value. Answers need units, and any direction should be stated as upwards or down the slope rather than left implicit.
海外直订Higher Mathematics for AQA GCSE 高等数学AQA GCSE¥645.0📱 去京东购买京东精选
需要一对一辅导？Need One-on-One Tutoring?
A-Level / GCSE / IB / KS2-KS3 全科辅导，OCR、Edexcel、AQA、CAIE 考纲逐条对照
16621398022 同微信
Visit aleveler.com
或截图后用「扫一扫 → 相册」选择本图