三个质点 $A$、$B$ 和 $C$ 的质量分别为 $0.3\,\mathrm{kg}$、$0.4\,\mathrm{kg}$ 和 $m\,\mathrm{kg}$，它们在光滑水平面上沿同一直线静止放置。$B$ 与 $C$ 之间的距离为 $2.1\,\mathrm{m}$。$A$ 以 $2\,\mathrm{m\,s^{-1}}$ 的速率直接朝 $B$ 发射。$A$ 与 $B$ 碰撞后，$A$ 的速率减小为 $0.6\,\mathrm{m\,s^{-1}}$，仍沿原方向运动。

(a) 证明碰撞后 $B$ 的速率为 $1.05\,\mathrm{m\,s^{-1}}$。

$A$ 与 $B$ 碰撞后，$B$ 直接朝 $C$ 运动。随后质点 $B$ 与 $C$ 碰撞。此次碰撞后，两质点合为一体，以 $0.5\,\mathrm{m\,s^{-1}}$ 的共同速率运动。

(b) 求 $m$。

(c) 求从 $B$ 与 $C$ 碰撞的瞬间起，直到 $A$ 与合并后的质点碰撞为止所需的时间。

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 $10\,\mathrm{m\,s^{-2}}$。

沿A初始方向守恒第一次碰撞动量，逐步推出B的目标速率。

Conservation of momentum gives

Thus $0.4v_B=0.6-0.18=0.42$, so

第二次碰撞B、C粘合，末动量使用合质量。

Conservation of momentum at the second collision gives

Thus $0.4+m=0.84$, giving

第二次碰撞前B走2.1米用2秒，此时A已走1.2米，尚落后0.9米。第二次碰撞后用同向追及速率0.1求时间。

Between the first and second collisions, $B$ takes

During this time $A$ moves $0.6(2)=1.2\,\mathrm{m}$. At the second collision, the combined particle is therefore $2.1-1.2=0.9\,\mathrm{m}$ ahead of $A$.

Their speeds remain constant on the smooth plane, so the closing speed is $0.6-0.5=0.1\,\mathrm{m\,s^{-1}}$. The required time after the second collision is

也可从第一次碰撞起统一计时，写A与合体的位置相等，最后扣除前两秒。

Take the position of the first collision as origin and let $T$ be time since that collision. The second collision occurs at $T=2$, at position $2.1\,\mathrm{m}$.

For $T\ge2$, the positions are

At the next collision,

Thus the time measured from the second collision is

另设第二次碰撞后合体走过的距离d，A需多走0.9米。两者用时相等，先求距离再求时间。

The time between the first two collisions is $2.1/1.05=2\,\mathrm{s}$, so at the second collision $A$ is behind by $2.1-0.6(2)=0.9\,\mathrm{m}$.

If the combined particle subsequently travels $d\,\mathrm{m}$ before $A$ catches it, $A$ travels $d+0.9\,\mathrm{m}$. Equal times give

Thus $0.5d+0.45=0.6d$, giving $d=4.5$. Hence