质量分别为 $2\,\mathrm{kg}$ 和 $6\,\mathrm{kg}$ 的质点 $A$ 和 $B$ 连接在一根轻质不可伸长细绳的两端。细绳跨过一个光滑定滑轮，两个质点竖直悬挂在滑轮下方。两个质点最初均被保持静止，距水平地面的高度为 $3.2\,\mathrm{m}$（见图）。质点 $A$ 以 $1.2\,\mathrm{m\,s^{-1}}$ 的速率竖直向下发射。

(a) 求细绳中的张力及质点的加速度大小。

在随后的运动中，$A$ 不会碰到地面，且两个质点均不会到达滑轮。当 $B$ 碰到地面时，它不会反弹。

(b) 求 $B$ 运动的时间。

(c) 求 $A$ 所到达的最低点与最高点之间的距离。

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初向下运动但加速度向上；以A向上、B向下为共同正方向列两式。

Let $a$ be $A$‘s upward acceleration, equal in magnitude to $B$‘s downward acceleration, and let $T$ be the tension. Newton’s second law gives

Adding,

Then

B最初向上1.2，之后加速度向下5。以下为正，落地位移3.2，解正时间。

Take downwards as positive for $B$. Its initial velocity is $-1.2\,\mathrm{m\,s^{-1}}$, so

Thus

The other root is negative and is rejected. Therefore $B$ is in motion for $\boxed{1.40\text{ s}}$.

先求A初始下行的停止距离，再求B落地时A的上行速度。绳松后A自由上升，用g而非5。

Initially $A$ moves downwards at $1.2\,\mathrm{m\,s^{-1}}$ with upward acceleration 5. If $d_1$ is the distance to its lowest point,

When $B$ reaches the ground, it has moved a net distance $3.2\,\mathrm{m}$ downwards, so $A$ has moved $3.2\,\mathrm{m}$ upwards from its initial position. Its upward speed $w$ satisfies

After $B$ hits the ground, the string becomes slack and $A$ continues upwards freely under gravity. The further rise $d_2$ is given by

Thus the distance between $A$‘s lowest and highest points is

也可用整体机械能求初始下行距离及B落地前的共同速率，再对绳松后的A使用机械能守恒。最低点与最高点分别位于初始位置的下方和上方，最后把各段高度相加。

Let $d_1$ be $A$‘s initial downward distance to its lowest point. Both particles are momentarily at rest there. Their initial kinetic energy is converted into the net increase in potential energy as $A$ descends and $B$ rises:

Thus $d_1=5.76/40=0.144$ metres.

When $B$ reaches the ground, it has descended a net distance of $3.2$ metres and $A$ has risen by the same amount. The net loss of potential energy adds to their initial kinetic energy. If their common speed is $w$,

Hence $w^2=33.44$. After $B$ stops, the string becomes slack. For $A$ alone, its kinetic energy becomes gravitational potential energy during the further rise $d_2$:

The distance between the lowest and highest points is therefore