一根原长为 $a$、模量为 $5mg$ 的轻弹性绳，一端连接于固定点 $O$。质量分别为 $m$ 和 $4m$ 的两个质点 $P$ 和 $Q$ 连接于绳的另一端，并竖直悬挂且处于平衡。随后将质点 $Q$ 从绳上拆下，因此质点 $P$ 从静止开始运动。

用 $a$ 表示当质点 $P$ 的速率首次等于 $\sqrt{\frac75ag}$ 时细绳的长度。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

Where a numerical value for the acceleration due to gravity ($g$) is needed, use $10\,\mathrm{m\,s^{-2}}$.

需要重力加速度的数值时，取 $g=10\,\mathrm{m\,s^{-2}}$。请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

先由两粒子悬挂时的总重量求初始伸长量。移去 $Q$ 后只对 $P$ 和弹性绳用机械能守恒，注意首次达到所求速度时绳仍伸长。

Initially the tension is $5mg$, so Hooke’s law $5mg\,e/a=5mg$ gives the initial extension $e=a$. After $Q$ is removed, let $x$ be the extension when $P$ has the stated speed. Taking gravitational potential energy zero at the initial position, the rise of $P$ is $a-x$. Conservation of energy gives

Multiplying by $10a/(mg)$ gives $25a^2=10a(a-x)+25x^2+7a^2$. Bringing the terms together gives

The particle initially accelerates upward, so the extension decreases from $a$. It reaches $x=4a/5$ before the string becomes slack. The required length is $\boxed{9a/5}$.

也可在绳仍绷紧时，以移去 $Q$ 后的新平衡位置为原点。合力与位移成正比，用简谐运动的速度与位移关系。

With only $P$ attached, the equilibrium extension is $a/5$. Let $s$ be the downward displacement from this equilibrium. While the string is taut, the downward resultant is $mg-\frac{5mg}{a}(a/5+s)=-5mgs/a$. Thus $ms''=-5mgs/a$, giving $s''=-(5g/a)s$. The initial displacement is $a-a/5=4a/5$ and the particle is released from rest. Therefore

With $v^2=7ag/5$, this gives $s^2=9a^2/25$. The first such position is $s=3a/5$ as $s$ decreases from $4a/5$. Thus the extension is $a/5+3a/5=4a/5$ and the length is $9a/5$.