两个半径相同的均匀光滑球体 $A$ 和 $B$ 的质量分别为 $m$ 和 $km$。两球位于水平面上。球体 $A$ 以速率 $u$ 向静止的球体 $B$ 运动。两球发生碰撞。在即将碰撞前，$A$ 的运动方向与球心连线成 $\alpha$ 角。两球之间的恢复系数为 $\frac12$。

(a) 证明碰撞后 $B$ 的速率为 $\frac{3u\cos\alpha}{2(1+k)}$，并求碰撞后 $A$ 沿球心连线方向的速率的表达式，用 $k$、$u$ 和 $\alpha$ 表示。

碰撞后，$A$ 的动能等于 $B$ 的动能。

(b) 已知 $\tan\alpha=\frac23$，求 $k$ 的可能值。

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.

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 $10\,\mathrm{m\,s^{-2}}$。

沿球心连线写动量守恒和恢复定律。先求 $B$ 的碰后速率，再求 $A$ 的沿线有符号分量。

Take the direction from $A$ to $B$ as positive along the line of centres. Let $v$ and $w$ be the signed final components for $A$ and $B$. Conservation of momentum and restitution give

Adding gives $(1+k)w=\frac32u\cos\alpha$. Since $B$ has no perpendicular component, its speed is

The signed component of $A$ is

Its speed along the line is the magnitude $|v|$.

$A$ 的垂直分量仍是 $u\sin\alpha$，而 $B$ 只有沿球心连线的分量。用动能相等建立关于 $k$ 的二次方程。

The kinetic energy equality is

Since $\cos\alpha>0$ and $u>0$, dividing by $mu^2\cos^2\alpha/2$ and using part (a) gives

Using $\tan^2\alpha=4/9$ and multiplying by the positive quantity $36(1+k)^2$ gives

Expanding gives $16(1+2k+k^2)+9(4-4k+k^2)=81k$. Collecting terms and factoring,

Both masses are positive in the resulting cases, so $k=4/5$ or $13/5$.