两个质点 $A$ 和 $B$ 的质量分别为 $3\,\mathrm{kg}$ 和 $5\,\mathrm{kg}$，由一根绕过固定光滑滑轮的轻且不可伸长的绳连接。两质点被保持在绳绷紧且绳的直线部分竖直的位置。质点 $A$ 位于一个水平面上方 $1\,\mathrm{m}$ 处，质点 $B$ 位于该平面上方 $2\,\mathrm{m}$ 处（见图）。

两质点从静止释放。在随后的运动中，$A$ 不会到达滑轮，且 $B$ 到达平面后保持与平面接触。

(a) 求绳中的张力和 $B$ 到达平面所需的时间。

(b) 求 $A$ 位于该平面上方至少 $3.25\,\mathrm{m}$ 处的持续时间。

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 $10\,\mathrm{m\,s^{-2}}$。

重者B下落、A上升，分别列牛顿第二定律求张力与加速度，再用B下降2米求时间。

Let $a$ be the common acceleration magnitude and $T$ the tension. Newton’s second law gives

Adding gives $20=8a$, so $a=2.5\,\mathrm{m\,s^{-2}}$. Thus

If $t$ is the time for $B$ to fall $2\,\mathrm{m}$ from rest,

也可先由两粒子系统的机械能守恒求落地速度，再用匀加速平均速率求用时。张力仍由牛顿第二定律求。

When $B$ falls $2\,\mathrm{m}$, $A$ rises $2\,\mathrm{m}$. The net loss of potential energy equals the total gain in kinetic energy:

Thus $v^2=10$, so $v=\sqrt{10}\,\mathrm{m\,s^{-1}}$. The acceleration is constant because all forces are constant. Using the average speed,

Also

B落地时A高3米，此后绳松弛，A自由上抛。求高度3.25米的上行、下行两个时刻，相减得到持续时间。

When $B$ reaches the plane, $A$ is at height $3\,\mathrm{m}$ and its upward speed satisfies

Afterwards the string is slack. Let $\tau$ be the time after $B$ reaches the plane. The height of $A$ is

Setting $h=3.25$ gives

with roots

Before this free flight $A$ is below $3.25\,\mathrm{m}$. During free flight the downward-opening height parabola is at least $3.25$ between these roots. The required duration is

利用最高点两侧运动对称。先求最高高度3.5米，再求从3.25米到最高点的时间，乘二。

When the string becomes slack, $A$ is at height $3\,\mathrm{m}$ and has upward speed $\sqrt{10}\,\mathrm{m\,s^{-1}}$. Its further rise to the highest point is

Thus the greatest height is $3.5\,\mathrm{m}$. The level $3.25\,\mathrm{m}$ is $0.25\,\mathrm{m}$ below this. Starting from rest at the highest point, the time to fall that distance satisfies

The ascent and descent times over this height interval are equal. Hence the duration is

还可以先求3.25米处的速度大小，再求从该速度减到零的时间，利用最高点两侧对称乘二。

The speed when the string becomes slack at height $3\,\mathrm{m}$ satisfies $u^2=2(2.5)(2)=10$. At height $3.25\,\mathrm{m}$, the upward speed $w$ satisfies

The time from this level to the highest point is

By symmetry about the highest point, the total time above or at this level is

用能量计算两个关键高度的速率，再通过重力作用下的匀减速求时间，与前面的动力学方法相互校验。

Before $B$ reaches the plane, conservation of mechanical energy for the system gives

The string then becomes slack at height $3\,\mathrm{m}$. As $A$ rises another $0.25\,\mathrm{m}$, its loss in kinetic energy equals its gain in potential energy:

Thus $w^2=5$. The time from the upward crossing of $3.25\,\mathrm{m}$ to the highest point is $\sqrt5/10$. The downward part is symmetric, so the duration is