מכניקה 2
Pearson Edexcel · International A-Level · Further Mathematics נושא 5 5:46 קריאת קול באנגלית · תרגום אנגלי + סינית שרוף בתוך הסרטון
פרקים
תרגום טקסט
Paper WME02: 1 hour 30 minutes, 75 marks, with P1 to P4 and M1 assumed.
WME02 试卷：1 小时 30 分钟，75 分，默认已学 P1 到 P4 和 M1。
Memorise kinetic and potential energy, work and power; v and s as integrals; and restitution, separation speed over approach speed, between 0 and 1.
记住动能和势能、功和功率；v 和 s 的积分形式；以及恢复系数，即分离速度除以接近速度，在 0 到 1 之间。
Acceleration k times 3 minus t squared, from rest at O.
加速度 a = k(3 − t²)，从 O 静止出发。
Integrate: v is k times 3t minus t cubed over 3, with no constant because it starts at rest.
积分：v = k(3t − t³/3)，因为从静止开始，没有常数。
v at 1.5 is 13.5, so k is 4.
v(1.5) = 13.5，所以 k = 4。
Integrate again for s, then find where it turns.
再积分得 s，然后找它掉头的位置。
The displacement s equals 6 t squared minus t to the 4 over 3 rises to 27, where v is zero at t equals 3, then falls back to O.
位移 s = 6t² − t⁴/3 上升到 27，此时 t = 3、v 为零，然后回到 O。
The total distance is 27 out plus 27 back: 54 metres.
总路程是去程 27 加回程 27：54 米。
Differentiate the position component by component: v is 2 over root t i minus 3j, which is i minus 3j at t equals 4.
对位置逐分量求导：v = (2/√t)i − 3j，t = 4 时是 i − 3j。
Then apply the impulse, 6i minus 3j, to the 3 kilogram mass: the new velocity is 3i minus 4j, speed 5.
然后对 3 千克的质量施加冲量 6i − 3j：新速度是 3i − 4j，速率 5。
Weight each piece by its length and take moments about AB.
每段按长度加权，对 AB 取矩。
The big arc gives 8a squared; the small arc, on the other side, minus 2a squared; the straights, nothing.
大弧给出 8a²；小弧在另一侧，给出 −2a²；直线部分为零。
The total length is 3 pi plus 2, times a.
总长度是 (3π + 2)a。
Arc AB of radius 2a sits above the straight wire, arc CD of radius a below it.
半径 2a 的弧 AB 在直线上方，半径 a 的弧 CD 在下方。
The pull of the big arc wins: the centre of mass is 6a over 3 pi plus 2 above AB, on the line of symmetry.
大弧占优：质心在 AB 上方 6a/(3π + 2)，位于对称轴上。
Hung from A, the centre of mass hangs vertically below A.
从 A 悬挂时，质心在 A 的正下方。
At A, the tangent of angle OAG is 3 over 3 pi plus 2.
在 A 处，tan∠OAG = 3/(3π + 2)。
OA equals OE, so angle OAE is 45 degrees, and the angle of AE to the vertical is the difference.
OA = OE，所以 ∠OAE = 45°，AE 与竖直方向的夹角就是两者之差。
The work-energy principle: total work equals the change in kinetic energy, with resistance entering as minus R s.
功能原理：总功等于动能的变化，阻力以 −Rs 计入。
Power is F v: at constant speed the driving force balances the resistances, and any excess force accelerates.
功率 P = Fv：匀速时牵引力与阻力平衡，多余的力产生加速度。
A 750 kilogram car climbs with 20 kilowatts.
750 千克的汽车以 20 千瓦爬坡。
Gravity along the slope is 750 g over 7, which is 1050, plus 200 resistance: 1250 newtons.
重力沿斜面的分量是 750g/7，即 1050，加上阻力 200：共 1250 牛顿。
The driving force is 20000 over v, which balances 1250 at 16 metres per second.
牵引力是 20000/v，在 16 米每秒时与 1250 平衡。
Slower, at 10 it is 2000, and the excess 750 newtons gives 1 metre per second squared.
速度较慢、为 10 时牵引力是 2000，多出的 750 牛顿产生 1 米每二次方秒的加速度。
The engine is cut at A at 10 metres per second.
在 A 点以 10 米每秒关闭引擎。
Its 37500 joules of kinetic energy go into height, 750 g d sine alpha, and into resistance, 200 d, over the slope distance.
37500 焦耳的动能转化为高度势能 750g·d sin α，以及沿斜面距离做的阻力功 200d。
That is 1250 d, so d is 30 metres.
合计 1250d，所以 d = 30 米。
Per unit mass the projectile starts with 200.
按单位质量，抛体初始能量为 200。
At 11 metres up its components are V and 2V, so 200 equals five halves V squared plus 110, and V is 6.
在 11 米高处，分量是 V 和 2V，所以 200 = (5/2)V² + 110，V = 6。
It meets a 60 degree plane at right angles, so it moves 30 degrees below horizontal at 8 root 3, and energy gives the height: about 10 metres.
它垂直撞上 60° 的斜面，所以速度方向在水平线下 30°，速率 8√3，由能量得高度：约 10 米。
P, of mass 2m at speed u, strikes Q, of mass m, at rest.
质量 2m、速度 u 的 P 撞击静止的质量为 m 的 Q。
Momentum gives 2u equals 2 v P plus v Q; restitution gives e u equals v Q minus v P.
动量守恒：2u = 2v_P + v_Q；恢复系数：eu = v_Q − v_P。
Solving gives both speeds, and the kinetic energy lost is m u squared over 3 times 1 minus e squared.
解得两个速度，损失的动能是 (mu²/3)(1 − e²)。
The impulse on Q is 8 over 9 m u, so e is a third.
Q 受到的冲量是 (8/9)mu，所以 e = 1/3。
P strikes Q: afterwards P keeps 5u over 9 and Q leaves at 8u over 9.
P 撞击 Q：之后 P 保持 5u/9，Q 以 8u/9 离开。
Q then rebounds from a wall to match P's speed, so the wall's restitution is 5 over 8.
Q 随后从墙上反弹，速率与 P 相同，所以墙的恢复系数是 5/8。
Successive impacts are bookkeeping.
连续碰撞是记账。
Solve the first collision completely, carry the surviving speeds into the next impact, and at a wall reverse the direction and scale the speed by e.
先完整解出第一次碰撞，把剩下的速度带入下一次碰撞，碰墙时反向并乘以 e。
Equate only what the question ties together.
只把题目联系起来的量设为相等。
A uniform rod AB rests on rough ground at A and against a smooth rail at C, whose reaction S is perpendicular to the rod.
均匀杆 AB 一端 A 在粗糙地面上，并靠在 C 处的光滑栏杆上，栏杆的反力 S 垂直于杆。
Moments about A remove both ground forces at once.
对 A 取矩可同时消去地面的两个力。
Vertically, V is m g minus S cos theta, and at the limit F equals S sine theta.
竖直方向 V = mg − S cos θ，极限时 F = S sin θ。
At the limit with mu four sevenths, F equals mu V.
μ = 4/7 的极限状态下，F = μV。
Divide by cos squared theta and replace sec squared with 1 plus tan squared: 6 tan squared theta minus 7 tan theta plus 2 equals 0.
两边除以 cos²θ，并把 sec²θ 换成 1 + tan²θ：6tan²θ − 7tan θ + 2 = 0。
Check yourself.
自我检测。
When a particle turns back, add both legs.
质点掉头时，把两段路程相加。
Wire pieces are weighted by length.
线框按长度加权。
At constant speed, v is P over F.
匀速时 v = P/F。
The kinetic energy loss carries 1 minus e squared.
动能损失带有 1 − e² 因子。
And for a rod on a rail, take moments about the ground contact.
杆靠在栏杆上时，对地面接触点取矩。