质量分别为 $2.4\,\mathrm{kg}$ 和 $1.2\,\mathrm{kg}$ 的两个质点 $A$ 和 $B$ 由一根绕过固定光滑滑轮的轻质不可伸长绳子连接。$A$ 被保持在距水平面 $2.1\,\mathrm{m}$ 的上方，$B$ 距该水平面 $1.5\,\mathrm{m}$。两质点竖直悬挂，并从静止释放。在随后的运动中，$A$ 到达该平面且不反弹，$B$ 不会到达滑轮。

(a) 证明 $A$ 到达该平面前绳中的张力为 $16\,\mathrm{N}$，并求 $A$ 到达该平面前两质点的加速度大小。

(b) 求 $B$ 距该平面的最大高度。

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 $10\,\mathrm{m\,s^{-2}}$。

A较重而向下运动，两粒子加速度大小相同。分别写牛顿第二定律，先求加速度，再推出给定张力。

Let the common acceleration have magnitude $a\,\mathrm{m\,s^{-2}}$ and the tension be $T\,\mathrm{N}$. Taking downwards for $A$ and upwards for $B$,

Adding gives $12=3.6a$, so

Substitution into the equation for $B$ gives

as required.

A落地前用共同加速度求速率；落地后绳子松弛，B只受重力继续上升。最大高度还要加上初始离地高度。

When $A$ reaches the plane, both particles have speed $u$, where

At this instant $B$ is $1.5+2.1=3.6\,\mathrm{m}$ above the plane. The string then becomes slack and $B$ rises freely under gravity. If its further rise is $h$,

Its greatest height above the plane is therefore

把B从释放到最高点作为整体过程。初末均静止，张力只在绳子拉紧的前2.1米内做功；这部分功转化为B的重力势能。

Let the total rise of $B$ from release to its highest point be $H\,\mathrm{m}$. Its initial and final kinetic energies are both zero. The tension does work only while the string is taut, through a distance of $2.1\,\mathrm{m}$.

The work-energy equation for $B$ over the whole ascent is

Thus $H=33.6/12=2.8\,\mathrm{m}$, and the greatest height above the plane is