两个半径相等的均匀光滑球 $A$ 和 $B$ 的质量分别为 $4m$ 和 $m$。球 $B$ 静止在光滑水平面上。球 $A$ 以速度 $u$ 在该平面上运动，并与球 $B$ 发生正碰。碰撞后，$A$ 的动量是 $B$ 的动量的三倍。

求恢复系数 $e$ 的值。

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.

Where a numerical value for the acceleration due to gravity ($g$) is needed, use $10\,\mathrm{m\,s^{-2}}$.

需要重力加速度的数值时，取 $g=10\,\mathrm{m\,s^{-2}}$。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。

碰撞后两球的总动量不变，并按 $3:1$ 分配。先求两个速度，再用分离速度除以接近速度。

Take the initial direction of $A$ as positive. The total momentum is $4mu$. Since the final momentum of $A$ is three times that of $B$, the final momenta are $3mu$ and $mu$, respectively. Thus the final velocities are $v_A=3u/4$ and $v_B=u$.

Newton’s law of restitution gives

先联立动量守恒和恢复系数方程，把碰后速度写成含 $e$ 的式子，再使用给定动量比。

Taking the initial direction as positive, momentum conservation and restitution give

Substituting $v_B=v_A+eu$ into the first equation gives $5v_A=(4-e)u$. Thus

The stated momentum ratio is $4mv_A=3mv_B$. Hence

so $\boxed{e=1/4}$.