Mécanique 3
Pearson Edexcel · International A-Level · Further Mathematics Sujet 6 4:58 Narration en anglais · Sous-titres anglais + 中文 incrustés
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Paper WME03: 1 hour 30 minutes, 75 marks, with everything before it assumed, and answers kept in symbols.
WME03 试卷：1 小时 30 分钟，75 分，默认之前所有内容都已学，答案用符号表示。
Memorise Hooke's law and elastic energy, the SHM equation and its period, acceleration towards the centre, the centre of mass of a solid, and v d v by d x.
记住胡克定律和弹性势能、简谐运动方程及其周期、指向圆心的加速度、实心体的质心，以及 v dv/dx。
With a equals 4x minus 2, write a as v d v by d x and integrate against x.
a = 4x − 2，把 a 写成 v dv/dx 并对 x 积分。
The condition gives a half v squared equals 2 x squared minus 2x plus a half, a perfect square: v is 1 minus 2x.
由条件得 v²/2 = 2x² − 2x + 1/2，是完全平方：v = 1 − 2x。
Then d v by d t is minus 2v, and v equals e to the 2 minus 2t.
于是 dv/dt = −2v，v = e^{2 − 2t}。
A string of natural length L and modulus lambda, stretched by e, has tension lambda e over L.
原长为 L、弹性模量为 λ 的绳子伸长 e 时，张力是 λe/L。
The energy stored is lambda e squared over 2L, a third store beside kinetic and gravitational energy.
储存的能量是 λe²/(2L)，是动能和重力势能之外的第三种能量。
A string of natural length 3a is fixed between A and B, 4a apart, and P is projected down from C, the midpoint.
原长 3a 的绳子固定在相距 4a 的 A、B 之间，P 从中点 C 向下抛出。
Each half has natural length 3a over 2.
每半段原长 3a/2。
At C each half is 2a, an extension of a half a.
在 C 处每半段长 2a，伸长 a/2。
At D, 3a over 2 below, each half is 5a over 2, an extension of a.
在 C 下方 3a/2 的 D 处，每半段长 5a/2，伸长 a。
Falling 3a over 2 releases 3 halves m g a of gravitational energy.
下落 3a/2 释放 (3/2)mga 的重力势能。
It pays for the change in kinetic energy, the half m g a stored in the string, and the work against resistance.
它支付动能的变化、绳中储存的 (1/2)mga，以及克服阻力做的功。
Solving, V squared is 29 a g over 10.
解得 V² = 29ag/10。
A rod of length 2a is hinged at A, held by a vertical string from B of length 1.2a.
长 2a 的杆在 A 处铰接，由 B 处一根长 1.2a 的竖直绳子拉住。
The 3-4-5 triangle puts B 1.6a across, and the midpoint 0.8a across.
3-4-5 三角形使 B 水平距离为 1.6a，中点为 0.8a。
Moments about A give T equals half m g, and Hooke's law gives the natural length: 24 over 25 a.
对 A 取矩得 T = mg/2，胡克定律给出原长：(24/25)a。
P sits between two strings, A to P and P to B.
P 位于两根绳子之间，A 到 P 和 P 到 B。
At E, 4.5L from A, both tensions are half m g: E is genuine.
在距 A 4.5L 的 E 处，两个张力都是 mg/2：E 确实是平衡点。
Displace P by x towards B: one string shortens, the other lengthens, and the net force is minus 4 m g over 3L times x.
把 P 向 B 移动 x：一根绳缩短，另一根伸长，合力是 −(4mg/3L)x。
That is SHM, with period pi root 3L over g.
这就是简谐运动，周期 π√(3L/g)。
Released from an end, the speed follows a sine.
从端点释放后，速度按正弦变化。
It stays below half the maximum for a third of every oscillation, so the slow time is a third of the period: pi root L over 3g.
它在每次振动的三分之一时间里低于最大值的一半，所以慢速时间是周期的三分之一：π√(L/3g)。
P moves in a circle on a smooth table, held by a string of length a to O above, at 30 degrees to the table.
P 在光滑桌面上做圆周运动，由一根长 a 的绳子连到上方的 O，绳与桌面成 30°。
The radius is a cos 30.
半径是 a cos 30°。
Towards the centre, T cos 30 equals m r omega squared, so T is half m g.
指向圆心：T cos 30° = mrω²，所以 T = mg/2。
Vertically, the reaction R is m g minus T sine 30: three quarters m g.
竖直方向，反力 R = mg − T sin 30°，即 (3/4)mg。
P is released at A on a smooth sphere and leaves at B.
P 在光滑球面上的 A 点释放，在 B 点离开。
Energy gives v squared equals 2 g r times cos alpha minus sine alpha.
能量给出 v² = 2gr(cos α − sin α)。
P leaves at B, where R is zero, so the weight's component towards O provides all of m v squared over r.
P 在 B 点离开，此时 R 为零，所以重力指向 O 的分量提供全部 mv²/r。
After leaving, P is a projectile and its horizontal velocity, v sine alpha, is conserved.
离开后 P 是抛体，水平速度 v sin α 守恒。
Level with O, energy gives its speed W.
到与 O 同高时，由能量得速率 W。
The angle to the horizontal has cos squared theta equal to 4 over 39.
与水平方向的夹角满足 cos²θ = 4/39。
Slice the solid hemisphere into thin discs parallel to the plane face.
把实心半球切成平行于平面的薄圆盘。
With y squared equal to a squared minus x squared, the top integral is a to the 4 over 4 and the bottom 2 a cubed over 3: the centre of mass is 3a over 8 from the plane face.
y² = a² − x²，分子积分是 a⁴/4，分母是 2a³/3：质心距平面 3a/8。
A hemisphere of radius a has a hemisphere of radius a over 2 removed.
半径 a 的半球挖去一个半径 a/2 的半球。
Masses go as radius cubed: 8, 1 and 7.
质量与半径立方成正比：8、1 和 7。
Along the axis the centre is 45a over 112; across it, a over 14.
沿轴方向质心在 45a/112；横向在 a/14。
Resting on its curved surface, the centre of mass sits above the contact point, so tan theta is 8 over 45.
曲面着地静止时，质心在接触点正上方，所以 tan θ = 8/45。
Check yourself.
自我检测。
When a depends on x, write it as v d v by d x.
a 依赖于 x 时，写成 v dv/dx。
Elastic energy is lambda e squared over 2L.
弹性势能是 λe²/(2L)。
x double dot equals minus omega squared x proves SHM.
ẍ = −ω²x 证明简谐运动。
A particle leaves a surface when R is zero.
R = 0 时质点离开表面。
And a solid hemisphere balances 3a over 8 from its face.
实心半球的质心距平面 3a/8。