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Pearson Edexcel · International A-Level · Further Mathematics หัวข้อ 6 4:58 การบรรยายภาษาอังกฤษ · คำบรรยายภาษาอังกฤษ + 中文 ลอยตัวบนภาพ

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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。

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