一个质量为 $0.2\,\mathrm{kg}$ 的质点 $P$ 从水平地面以 $25\,\mathrm{m\,s^{-1}}$ 的速率竖直向上抛出。

(a) 证明 $P$ 到达地面上方 $20\,\mathrm{m}$ 处时的速率为 $15\,\mathrm{m\,s^{-1}}$。

当 $P$ 到达地面上方 $20\,\mathrm{m}$ 处时，它与另一个质量为 $0.1\,\mathrm{kg}$、正以 $20\,\mathrm{m\,s^{-1}}$ 的速率向下运动的质点 $Q$ 碰撞。碰撞使 $P$ 瞬时静止。

(b) 求碰撞后一瞬间 $Q$ 的速度。

当 $P$ 到达地面时，它沿竖直方向向上反弹，反弹速率为撞地前一瞬间速率的一半。

(c) 求 $P$ 和 $Q$ 下一次碰撞时离地面的高度。

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 $10\,\mathrm{m\,s^{-2}}$。

用竖直运动的速度与位移关系，把高度20和初速25代入后实际推出目标速率。

Taking upwards as positive,

Therefore the speed at height $20\,\mathrm{m}$ is

也可用能量守恒，将上升时增加的势能等同于动能减少。

Conservation of mechanical energy gives

Hence $25^2-v^2=400$, so $v^2=225$. Thus

P第一次到达20米时向上，Q向下。以向上为正列动量守恒，求得的正负号说明碰后方向。

Take upwards as positive. At the first arrival of $P$ at height $20\,\mathrm{m}$, its velocity is $+15\,\mathrm{m\,s^{-1}}$. If the velocity of $Q$ after collision is $v$,

Hence

碰撞后P从20米高处自由落体，用时2秒，落地速率20、反弹速率10。以反弹为新计时起点，求两者高度相等时刻。

After the first collision, $P$ falls from rest through $20\,\mathrm{m}$. Thus

and its impact speed is $10t_0=20\,\mathrm{m\,s^{-1}}$. It rebounds upwards at $10\,\mathrm{m\,s^{-1}}$.

At this instant, $Q$ has height and velocity

Let $\tau$ be the time since the rebound. The heights are

Equating them gives $20\tau=20$, so $\tau=1$. Hence the next collision height is

也可从第一次碰撞统一计时，把P反弹后的时间写成t减二，避免另设计时变量。

After the first collision, $P$ falls from rest through $20\,\mathrm{m}$ in $2\,\mathrm{s}$ and reaches the ground at speed $20\,\mathrm{m\,s^{-1}}$. Its rebound speed is therefore $10\,\mathrm{m\,s^{-1}}$.

Let $t$ be the time after the first collision. For $t\ge2$, their heights above the ground are

Equating and expanding gives

so $t=3$, which lies after the rebound. Therefore