两个质点 $P$ 和 $Q$ 的质量分别为 $2m\,\mathrm{kg}$ 和 $m\,\mathrm{kg}$，被保持静止在同一条竖直线上。$P$ 和 $Q$ 离水平地面的高度分别为 $1\,\mathrm{m}$ 和 $2\,\mathrm{m}$。$P$ 以 $2\,\mathrm{m\,s^{-1}}$ 的速率竖直向上抛出。同时，$Q$ 从静止释放。

(a) 求两质点碰撞前一瞬间各自的速率。

(b) 已知碰撞后一瞬间 $Q$ 向下的速率为 $3.5\,\mathrm{m\,s^{-1}}$。

求 $P$ 到达地面瞬间的速率。

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 $10\,\mathrm{m\,s^{-2}}$。

两者重力加速度相同，列高度式相等时平方项会消去。速度要保留方向，最后才转成速率。

Taking upwards as positive, their heights above the ground at time $t$ are

At collision,

Their velocities immediately before collision are

Both are moving downwards, with speeds

沿向下取正方向守恒碰撞动量。再由碰撞高度到地面的下降距离求末速。

Take downwards as positive. If the velocity of $P$ immediately after collision is $u$, conservation of momentum gives

Hence $2u=7.5$, so $u=3.75\,\mathrm{m\,s^{-1}}$ downwards.

The collision height is

For the subsequent fall of $P$,

Therefore its speed at the ground is