小球 $A$、$B$ 在光滑水平地面上沿同一直线相向运动。$A$ 的质量为 $2m$，$B$ 的质量为 $3m$。 两球发生正碰。碰撞前瞬间，$A$ 的速率为 $2u$，$B$ 的速率为 $u$。$A$ 与 $B$ 之间的恢复系数为 $e$，其中 $e>0$。

将小球视为质点，

(a) 证明碰撞后瞬间 $B$ 的速率为 $\frac15u(1+6e)$。

与球 $A$ 碰撞后，球 $B$ 撞上与其运动方向垂直的光滑固定竖直墙。

$B$ 与墙之间的恢复系数为 $\frac57$。

球 $B$ 从墙反弹，随后 $A$、$B$ 发生第二次正碰。

(b) 求 $e$ 的可能取值范围。

统一取 $A$ 碰前方向为正，把两球碰后速度先当作有符号量。这样不必预先假设 $A$ 是否反向。

Take the initial direction of $A$ as positive. Let the signed velocities immediately after collision be $p$ for $A$ and $q$ for $B$.

Conservation of momentum gives

so

The approach speed is $2u-(-u)=3u$. Newton’s law of restitution gives

Substituting into the momentum equation,

Hence

Therefore the speed of $B$ is

$A$ 碰后可能向右、静止或向左，不宜把某个表达式无条件当作它向左的速率。沿用有符号速度：$B$ 从墙反弹后位于 $A$ 右侧，两者间距须逐渐减小才会再次碰撞。

From (a), the signed velocity of $A$ after the first collision is

After rebounding from the fixed wall, $B$ has signed velocity

At this stage $A$ is to the left of $B$. Their separation decreases precisely when $p>r$. Thus a second collision requires

Since $u>0$,

Combining with $e>0$,

Equality is excluded: the velocities would be equal, so the positive gap would not close. This signed-velocity argument also includes the cases in which $A$ continues forwards or is momentarily at rest after the first collision.