一根轻弹性绳的自然长度为 $8a$，弹性模量为 $5mg$。一个质量为 $m$ 的质点 $P$ 连接在绳的中点。绳的两端分别连接在光滑水平桌面上相距 $12a$ 的点 $A$ 和 $B$ 处。将质点 $P$ 保持在桌面上，使得 $AP=BP=L$（见图）。质点 $P$ 从静止释放。当 $P$ 位于 $AB$ 的中点时，其速率为 $\sqrt{80ag}$。

(a) 用 $a$ 表示 $L$。

(b) 用 $g$ 表示 $P$ 的初始加速度。

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.

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。

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

需要使用重力加速度 $g$ 的数值时，取 $10\,\mathrm{m\,s^{-2}}$。

桌面光滑且水平，动能的增加完全来自绳的弹性势能减少。注意到达 $AB$ 中点时绳仍有伸长。

At release the whole string has extension $2L-8a$. At the midpoint of $AB$, the total length is $12a$, giving extension $4a$. Since $AP=BP$ throughout this motion, both halves have equal extension and the total elastic energy is given by the whole-string expression. Conservation of energy gives

Cancelling $mg$ and multiplying by $16a/5$ gives

Since $L\ge6a$, we have $2L-8a>0$, so $2L-8a=12a$ and $L=10a$.

释放时两段张力相同，垂直于对称轴的分量互相抵消。由 $6a$、$10a$ 构成的直角三角形求沿对称轴的分量。

The initial tension is $5mg(2L-8a)/(8a)=15mg/2$. If $M$ is the midpoint of $AB$, then

Each tension has a component $T(8a/10a)=4T/5$ towards $M$. If the initial acceleration has magnitude $A_0$, Newton’s second law gives

Thus $A_0=12g$ towards $M$.