Edexcel物理 力学 SUVAT 受力分析 动量 力矩
力学是 Edexcel A Level 物理（9PH0）里最早讲、也最容易被低估的一块。它在 Paper 1 占一个完整板块，真正扣分的地方却常常不在公式本身，而在符号、方向和单位。
下面按卷面顺序把运动、受力、功与能、动量、实验与数据处理拆开，每个小节给出可以直接代入的算式和一个具体数字。
Mechanics arrives early in the Edexcel course and is usually judged as the easy part, which is exactly why weak habits survive for two years. The sections below follow the order of the specification and keep an eye on where marks actually disappear.
力学在 9PH0 卷面的位置与分值
Paper 1 的代码是 9PH0/01，时长 1 小时 45 分钟，满分 90 分，占整个 A Level 的 30%。这一卷把 Working as a Physicist、Mechanics 与 Electric Circuits 放在一起考，力学通常拿到三分之一左右的题量。Paper 3 是 General and Practical Principles，2 小时 30 分钟 120 分，占 40%，会从实验与数据处理的角度回头考力学的读数与作图。Edexcel 在每份卷子开头都给一份 List of Data, Formulae and Relationships，运动学四条公式、F = ma、W = Fs cosθ 都在里面。公式册给出来，选哪一条仍旧要自己判断。很多学生以为公式册等于不用记，翻书翻掉的时间比背公式更多。考试局在官方 Specification 里把力学拆成运动、力、功与能三组，条文编号一直排到 2.3 节之后。
Paper 1 carries the code 9PH0/01, runs for 1 hour 45 minutes and is worth 90 marks, or 30 per cent of the qualification, combining Working as a Physicist, Mechanics and Electric Circuits with mechanics taking roughly a third of the paper. Paper 3, at 2 hours 30 minutes and 120 marks, returns to mechanics through experimental design and data handling. A formula booklet sits at the front of every paper, yet deciding which line applies remains the candidate's own job. Examiners watch students leaf through that booklet while the clock runs, and the hesitation costs more marks than the algebra ever does.
位移时间图与速度时间图的斜率面积
位移时间图上，割线的斜率给平均速度，切线的斜率给瞬时速度。速度时间图的斜率是加速度，图线下的面积是位移。Edexcel 的条文把这三件事写在同一条要求里，并且明确要求处理加速度不均匀的情形。曲线向下弯说明速度在减小，速度本身仍然是正的，物体继续朝同一方向走。面积带符号，时间轴下方的部分算负位移，问路程时要把各段取绝对值后相加。常见的错误是把 v-t 图当路线图读，看见线往下走就以为物体在倒退。阅卷时这类方向判断一错，后面的位移和平均速度会连着丢分。
A displacement-time graph gives average velocity from a chord and instantaneous velocity from a tangent, while on a velocity-time graph the gradient is acceleration and the area beneath the line is displacement. Edexcel groups these readings into one requirement and expects non-uniform acceleration to be handled as well. A falling line on a velocity-time graph means slowing down rather than turning back, and signed areas below the axis have to be subtracted when total displacement is asked for. Candidates who read the graph as a map of the journey tend to lose the direction mark, and usually the average velocity mark that follows it.
四条匀加速公式与选式条件
匀加速直线运动的四条公式都在公式册里：s = (u + v)t/2、v = u + at、s = ut + at²/2、v² = u² + 2as。五个量 s、u、v、a、t 之中，每条公式恰好缺一个，所以先看题目没给哪个量、也没问哪个量，那个量就决定了选哪一条。一颗 1.8 m 高处的钢球自由下落，用 v² = u² + 2as 得 v² = 2 × 9.81 × 1.8 = 35.3，落地速度 5.9 m/s，全程 0.61 s。若改用 s = ut + at²/2，解 t 要开根号，多一步运算也多一个出错点。符号一次设定好：取向上为正，g 就写 -9.81 m/s²，位移向上为正。很多学生以为加速度为负就是减速，其实符号只跟选定的正方向有关，与快慢无关。
All four constant-acceleration equations appear in the booklet and each one omits exactly one of the five quantities, so the variable that is neither given nor asked for picks the equation for you. A steel ball released from 1.8 m reaches the floor at about 5.9 m/s after 0.61 s, straight from v squared equals u squared plus 2as. Working through the form containing t squared forces a square root and adds another place to slip. Fix the positive direction once at the start and let g carry the minus sign, then the algebra looks after the rest.
抛体运动的水平与竖直分解
抛体在水平方向不受力，水平速度保持不变；竖直方向只受重力，按 9.81 m/s² 向下匀加速。两个方向共用同一个时间 t，这是它们之间唯一的连接点。以 20 m/s、与水平成 30 度抛出时，水平分量是 20 cos30 = 17.3 m/s，竖直分量是 20 sin30 = 10.0 m/s。到最高点竖直速度为零，用时 10.0/9.81 = 1.02 s，上升高度 5.1 m，水平射程 17.3 × 2.04 = 35.3 m。常见的错误是把 20 m/s 整段塞进竖直方向的公式，把初速度当成纯竖直分量。题目没有说明忽略空气阻力时，水平速度不变的假设就不能随便用。
A projectile keeps its horizontal velocity because nothing pushes horizontally, while gravity alone governs the vertical motion at 9.81 m/s squared downwards, and the shared time is the only link between the two. Launched at 20 m/s and 30 degrees, the components come out at 17.3 m/s and 10.0 m/s, which gives a 1.02 s climb to a 5.1 m peak and a 35.3 m range. Feeding the full 20 m/s into a vertical equation is the standard slip, and it survives all the way to the final answer. Air resistance may be dropped only when the question says it is negligible.
矢量分解与合力计算
一个力与水平成 θ 角时，水平分量是 F cosθ，竖直分量是 F sinθ，角度必须先交代清楚量的是哪一条边。两个互相垂直的力合成，大小用勾股定理，方向用 tanθ = 竖直分量/水平分量 给出。30 N 与 40 N 垂直时合力 50 N，与 30 N 的夹角约 53 度。三个力平衡时，任意两个的合力与第三个等大反向，画成三角形能直接看出比例关系。斜面问题里，重力沿斜面是 mg sinθ，垂直斜面是 mg cosθ，很多学生把这两个分量写反。作图时先标角度落在哪里，再决定用 sin 还是 cos。
Resolving a force at angle theta gives F cos theta along the horizontal and F sin theta along the vertical, provided the angle is tied to a named side before any numbers appear. Two perpendicular forces of 30 N and 40 N combine to 50 N at about 53 degrees to the smaller one, and three forces in equilibrium close into a triangle. On an incline the weight splits into mg sin theta down the slope and mg cos theta into the surface, a pair that is swapped more often than any other. Mark the angle on the diagram before reaching for a calculator.
自由体受力图与牛顿第二定律
自由体图只画作用在这个物体上的力，不要把物体施加给别人的力也画进去。重量 W = mg 永远竖直向下，法向力垂直于接触面，张力沿绳，摩擦力沿接触面并与相对运动方向相反。F = ma 里的 F 是合力，不是其中某一个力，题目给了几个力就先求矢量和。5 kg 的木块放在 30 度光滑斜面上，沿斜面合力是 5 × 9.81 × sin30 = 24.5 N，加速度 4.9 m/s²。光滑意味着摩擦为零，只有粗糙面才需要引入 μ。很多学生把法向力直接写成 mg，斜面上只要出现角度，这个写法就错了。
A free-body diagram shows only the forces acting on the object, never the forces it exerts elsewhere. Weight is mg straight down, the normal reaction is perpendicular to the surface, tension follows the string and friction opposes relative sliding along the surface. In F equals ma the F is the resultant, so several forces must be added as vectors first; a 5 kg block on a smooth 30 degree slope therefore accelerates at 4.9 m/s squared. Writing the normal reaction as mg is the error that turns up most often once a slope is involved.
力矩与转动平衡
力矩等于力乘以力臂，力臂是支点到力作用线的垂直距离，而不是支点到作用点的距离。物体既不平移也不转动，需要同时满足合力为零与合力矩为零。一对大小相等、方向相反的平行力构成力偶，力偶的力矩是力乘以两力之间的垂直间距，与支点选在哪里无关。一根 1.5 m 长的撬棍，支点距短端 0.5 m，短端受到 200 N 的向下力，要在长端加 100 N 才能平衡。重心越低、基底越宽，物体越不容易翻倒，判断是否翻倒看重心垂线有没有落在基底之外。常见的错误是把力臂量成斜边长度。
A moment is force times the perpendicular distance from the pivot to the line of action, not to the point where the force is applied. Equilibrium of an extended body needs both zero resultant force and zero resultant moment, and a couple produces a moment equal to the force times the separation of the two lines, wherever the pivot sits. On a 1.5 m crowbar pivoted 0.5 m from its short end, a 200 N push there is balanced by 100 N at the long end. Stability follows the same geometry: a low centre of gravity over a wide base keeps the vertical through it inside the base.
功、动能与功率的换算
功等于力乘以位移、再乘力与位移夹角的余弦，即 W = Fs cosθ，力垂直于位移时做功为零。动能是 mv²/2，重力势能变化是 mgΔh，只有重力做功时两者可以互相换算。1200 kg 的汽车从静止加速到 20 m/s，动能增加 240 kJ，若用 12 s 完成，平均功率 20 kW。功率的第二种写法 P = Fv 在匀速行驶时最好用，牵引力直接由阻力大小决定。效率是有用输出与总输入的比值，摩擦生热的那部分要算在总输入里。常见的错误是把 kW 与 kWh 混着用，前者是功率，后者是能量。
Work is force times displacement times the cosine of the angle between them, so a force at right angles to the motion transfers no energy at all. Kinetic energy grows with the square of speed, which is why a 1200 kg car needs 240 kJ to reach 20 m/s and delivers 20 kW when it manages that in 12 s. At steady speed the form P equals Fv is quicker, with the driving force fixed by the resistive forces. Efficiency compares useful output with total input, and heat from friction belongs on the input side of that ratio.
动量守恒与冲量
动量 p = mv 是矢量，方向与速度一致，符号写错整道题基本作废。冲量 FΔt 等于动量变化 Δp，这条关系解释了安全气囊与头盔为什么有用：把碰撞时间拉长就能压低平均力。0.15 kg 的球以 12 m/s 撞墙、以 8 m/s 反弹，动量变化大小是 0.15 × (12 + 8) = 3.0 kg m/s，接触时间 0.05 s 时平均力约 60 N。封闭系统里总动量守恒，碰撞与爆炸都按这一条列式。反弹类题目最容易漏方向，取撞向墙为正时，反弹速度就该写负值。很多学生把动量当标量直接相减，算出的变化量少了一半。
Momentum is a vector, so one sign error early in a collision question usually ruins the rest of it. Impulse equals the change in momentum, which is the whole argument behind crumple zones and helmets: stretching the contact time lowers the average force. A 0.15 kg ball arriving at 12 m/s and leaving at 8 m/s changes momentum by 3.0 kg m/s, and a 0.05 s contact implies about 60 N. Total momentum stays constant in a closed system, and a rebound must carry the opposite sign to the approach.
核心实验 CP1 测自由落体加速度
Edexcel 的核心实验 CP1 要求学生测定自由落体加速度。常见做法是用电磁铁释放钢球，让它穿过两个光门，或者让一块带缺口的卡片经过单个光门，用卡片长度除以遮光时间得到瞬时速度。改变下落高度 h，把 v² 对 h 作图，直线斜率就是 2g；也可以量下落时间 t，把 s 对 t² 作图，斜率是 a/2。每个高度重复三次并剔除明显异常值，光门位置每次用同一把米尺量到毫米。空气阻力与卡片长度的测量误差都会让结果偏低，g 落在 9.6 至 9.9 m/s² 之间属于正常范围。数据点偏离直线时回到步骤里找原因，不要直接把点删掉。
Core practical 1 asks for the acceleration of a freely falling object, usually by dropping a steel ball through light gates or timing a card as it passes a single gate, with card length divided by blocking time giving the speed. Plotting v squared against h yields a straight line of gradient 2g, while plotting s against t squared yields a gradient of a/2. Repeat each height three times, discard clear anomalies and measure every position with the same millimetre rule. Air resistance and an inaccurate card length both pull the answer low, so a value between 9.6 and 9.9 m/s squared is normal rather than suspicious.
有效数字与不确定度处理
Edexcel 的评分要求答案写出单位与合理的有效数字，一般两到三位，中间计算不要提前四舍五入。百分不确定度等于绝对不确定度除以读数再乘 100，0.5 m 的长度上下浮动 1 mm 时约 0.2%。米尺分辨率是 1 mm，秒表读数到 0.01 s，人的反应时间常在 0.2 s 左右，后者才是计时实验的主要误差来源。作图求斜率时，用最陡与最缓两条直线夹出的范围给出斜率的百分不确定度。多次测量取平均能把随机误差按根号 n 压下去。常见的错误是把系统误差写成可以通过重复测量消除。
Answers need units and a sensible number of significant figures, normally two or three, with rounding left to the final line. Percentage uncertainty is the absolute uncertainty divided by the reading, so 1 mm on a 0.5 m length works out at about 0.2 per cent. A millimetre rule reads to 1 mm and a stopwatch to 0.01 s, yet human reaction time near 0.2 s dominates any timing experiment. Averages shrink random error by the square root of the number of readings, but repeating a flawed method never removes a systematic error.
力学计算题的给分规则与单位方向检查
Edexcel 的计算题按方法分与答案分给分，写出所选公式并代入数字才有方法分，只写一个结果拿不到过程分。单位漏写或者把 cm 与 m 混用是最常见的失分点，答案末尾统一换成基本单位能挡掉大半。矢量题只给大小不给方向，题目问 velocity 或 force 时就会扣掉方向那一分。中间结果多留一位，最后按题目要求的有效数字收尾。重力加速度用公式册给的 9.81 m/s²，题目另有说明才换成 10。说实话，力学题做不完往往不是算得慢，而是读题时没有先圈出已知量与所求量。
Marks in mechanics split between method and answer, so the chosen equation and the substituted numbers have to be visible; a bare result earns almost nothing. Units left out, or centimetres mixed with metres, account for a large share of dropped marks every session. A vector answer without a direction loses the final mark whenever velocity or force is requested. Round only at the end, keep one extra digit while working, and take 9.81 m/s squared from the booklet unless the question states otherwise.
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