质量分别为 $4m\,\mathrm{kg}$ 和 $3m\,\mathrm{kg}$ 的两个质点 $P$ 和 $Q$ 静止在光滑水平地面上。$P$ 以 $6.5\,\mathrm{m\,s^{-1}}$ 的速率直接朝 $Q$ 发射。两质点碰撞后沿同一方向运动，$P$ 和 $Q$ 的速率分别为 $2\,\mathrm{m\,s^{-1}}$ 和 $v\,\mathrm{m\,s^{-1}}$。

(a) (i) 证明 $v=6$。

(ii) 已知 $P$ 与 $Q$ 碰撞时损失的能量为 $45\,\mathrm{J}$，求 $m$ 的值。

$Q$ 从 $P$ 与 $Q$ 第一次碰撞的位置运动了 $9\,\mathrm{m}$ 后，撞上一面垂直于 $Q$ 运动方向的竖直墙并反弹。$Q$ 撞墙后的速率为 $0.5\,\mathrm{m\,s^{-1}}$。

(b) 求墙到 $P$ 与 $Q$ 第二次碰撞位置的距离。

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 $10\,\mathrm{m\,s^{-2}}$。

碰撞前Q静止；保留正质量参数m列动量守恒，再约去。

Taking $P$‘s original direction as positive, conservation of momentum gives

Since $m>0$,

as required.

初动能减去两粒子末动能等于45J，使用(a)(i)的速度6。

Thus

Q先到墙时P已经前进3m，剩6m间距；反弹后相向而行，闭合速率2.5。

After the first collision, $Q$ takes

to reach the wall. During this time $P$ moves $2(1.5)=3\,\mathrm{m}$, so it is then $9-3=6\,\mathrm{m}$ from the wall.

After $Q$ rebounds, the particles approach each other at relative speed $2+0.5=2.5\,\mathrm{m\,s^{-1}}$. Their time to meet is

The second collision occurs at distance

from the wall.

也可设第二次碰撞点距墙x，用反弹之后两粒子所用时间相等直接解x。

The time for $Q$ to reach the wall is $9/6=1.5$ seconds. By then $P$ has moved $3\,\mathrm{m}$ and is $6\,\mathrm{m}$ from the wall.

Let the next collision point be $x$ metres from the wall. After the rebound, $Q$ travels $x$ metres at speed $0.5$, while $P$ travels $6-x$ metres at speed 2 in the same time. Hence