质点 $A$、$B$ 和 $C$ 的质量分别为 $2m$、$m$ 和 $3m$，沿同一直线静止在光滑水平面上，$B$ 位于 $A$ 和 $C$ 之间。质点 $A$ 以速率 $2u$ 向质点 $B$ 投射，并与 $B$ 发生正碰。

每对质点之间的恢复系数均为 $e$。

(a) (i) 证明 $B$ 与 $A$ 碰撞后瞬间的速率为 $\frac43u(1+e)$。

(ii) 求 $A$ 与 $B$ 碰撞后瞬间的速率。

当 $A$ 与 $B$ 碰撞的瞬间，质点 $C$ 以速率 $u$ 向 $B$ 投射，使 $B$ 和 $C$ 发生正碰。

(b) 证明 $A$ 和 $B$ 将发生第二次碰撞。

统一取从 $A$ 指向 $C$ 的方向为正。先写动量守恒和恢复系数方程，再联立消元；恢复系数使用分离速率除以接近速率。

Take the direction from $A$ towards $C$ as positive. Let the velocities of $A$ and $B$ immediately after their first collision be $v$ and $w$ respectively.

Conservation of momentum gives

Newton’s law of restitution gives

Thus $v=w-2eu$, and substitution gives

Therefore

which is positive and is therefore the speed of $B$.

把上一问的 $B$ 速度代回恢复系数方程。还要检查所得有符号速度为正，才可以直接把它写成速率。

Using $v=w-2eu$,

Since $0\le e\le1$, this is positive. Thus the speed of $A$ is

$C$ 在第二次碰撞前向左运动，因此初速度为 $-u$。求出 $B$ 碰后有符号速度为负，再结合 $A$ 仍在其左方并向右运动，才能说明两者必定再次碰撞。

Before the collision between $B$ and $C$, their velocities are

Let their velocities afterwards be $x$ and $y$, respectively, using the same positive direction as before.

Conservation of momentum gives

so

The restitution equation is

Subtracting three times the restitution equation from the momentum equation,

Therefore

Since $u>0$ and $0\le e\le1$, $x<0$: $B$ moves back towards $A$.

Meanwhile, $A$ is still moving to the right with positive velocity

At the collision between $B$ and $C$, $A$ is to the left of $B$. Afterwards they move towards each other, with closing speed $v-x>0$. Hence there will be a second collision between $A$ and $B$.