质量为 $m\text{ kg}$ 的质点 $P$ 连接在一根原长为 $2\text{ m}$、弹性模量为 $2mg\text{ N}$ 的轻弹性绳的一端。绳的另一端连接在固定点 $O$。质点 $P$ 悬挂在 $O$ 的正下方并处于平衡状态。将质点 $P$ 从平衡位置竖直向下拉动 $d\text{ m}$，然后由静止释放。

(a) 已知质点在随后的运动中恰好到达 $O$，求 $d$ 的值。

(b) 由此求 $P$ 在 $O$ 下方 $2\text{ m}$ 处时的速率。

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 $10\,\mathrm{m\,s^{-2}}$。

先用悬挂平衡求弹性绳的伸长。释放时有弹性势能，恰好到达 $O$ 时绳已松弛且速率为零。

If the equilibrium extension is $e$, Hooke’s law gives $(2mg/2)e=mg$, so $e=1\,\mathrm{m}$. At release the extension is $1+d$ and the distance below $O$ is $3+d$, in metres. At $O$ both the elastic potential energy and kinetic energy are zero. Equating the initial elastic energy to the gain in gravitational potential energy gives

Cancelling $mg$ and multiplying by $2$ gives $1+2d+d^2=6+2d$, so $d^2=5$. The downward pull distance is positive, so $d=\sqrt5$.

从释放位置到 $O$ 下方 $2\,\mathrm{m}$ 使用能量守恒。这个位置恰为绳长等于自然长度的位置。

At $2\,\mathrm{m}$ below $O$ the string has no elastic potential energy. The particle has risen $1+d$ metres from release. Hence, if its speed is $V$,

With $d^2=5$ and $g=10$, we get $V^2=40$ and $V=2\sqrt{10}\,\mathrm{m\,s^{-1}}$.

也可从绳刚松弛的位置推到最高点。绳松弛后，质点只受重力，继续上升 $2\,\mathrm{m}$ 后停下。

From the point $2\,\mathrm{m}$ below $O$ to $O$, the tension is zero and the particle moves freely under gravity. Its kinetic energy at the start of this interval is the gain in gravitational potential energy:

Thus $V=2\sqrt{10}\,\mathrm{m\,s^{-1}}$.