质量为 $m$ 的质点 $P$ 沿一条水平直线朝该直线上的固定点 $A$ 运动。质点受到的阻力与 $P$ 和 $A$ 之间距离的平方成反比。初始时，$P$ 和 $A$ 之间的距离为 $d$，$P$ 的速率为 $u$。当 $P$ 和 $A$ 之间的距离为 $\frac12d$ 时，质点已损失其动能的 $\frac34$。

求 $P$ 到 $A$ 的最小距离，用 $d$ 表示。

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 $10\,\mathrm{m\,s^{-2}}$。

以到 $A$ 的距离为坐标，把阻力方向和速度方向分清。对速度的微分方程积分，用已知的动能损失确定比例常数。

Let $r$ be the distance from $A$, and let $v=-\mathrm{d}r/\mathrm{d}t$ be the speed towards $A$ while $P$ is moving. Write the magnitude of the resistance as $k/r^2$, with $k>0$. Taking the direction towards $A$ as positive in Newton’s second law gives

Using $\mathrm{d}r/\mathrm{d}t=-v$ accounts for the sign in the second equation. Integrating from the initial state gives

Hence

At $r=d/2$, the loss is $\frac34\cdot\frac12mu^2$, so

The kinetic energy decreases as $r$ decreases. It first reaches zero when

Thus $1/r=1/d+4/(3d)=7/(3d)$, so the minimum distance is $\boxed{\frac37d}$.

也可以从起点沿运动方向量距离。此时若质点已走了 $x$，它到 $A$ 的距离就是 $d-x$。

Let $x$ be the distance travelled from the initial point. Newton’s second law gives

Since $\int(d-x)^{-2}\,\mathrm{d}x=(d-x)^{-1}$, integrating gives $\frac12mv^2=C-\frac{k}{d-x}$. At $x=0$ and $x=d/2$, respectively,

Subtracting gives $3mu^2/8=k/d$, so $k=\frac38mdu^2$. Then $C=mu^2/2+k/d=\frac78mu^2$. At the first stop, $v=0$, so