两个半径相等、质量均为 $m$ 的均匀光滑球 $A$ 和 $B$ 在光滑水平面上分别以速率 $u_A$ 和 $u_B$ 运动并发生碰撞。碰撞前一瞬间，$A$ 的运动方向与球心连线成 $\alpha$ 角，$B$ 的运动方向与球心连线成 $\beta$ 角（见图）。碰撞后紧接着，$A$ 的运动方向与球心连线成 $\theta$ 角，$B$ 的运动方向与球心连线成 $\varphi$ 角。已知 $\cos\alpha=\cos\varphi=\frac45$，$\cos\beta=\frac{24}{25}$，且 $\cos\theta=\frac35$。

(a) 求 $A$ 的动能损失百分比。

(b) 求两球之间的恢复系数。

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 $10\,\mathrm{m\,s^{-2}}$。

光滑球间的冲量沿球心连线，所以垂直于球心连线的速度分量不变。用这一条件求 $A$ 的速率比。

Let the speeds after impact be $v_A$ and $v_B$. Since $\sin\alpha=3/5$ and $\sin\theta=4/5$,

The percentage loss of kinetic energy for $A$ is

先由垂直分量不变得到 $B$ 的碰后速率，再沿球心连线列动量守恒。用分离速率与接近速率之比求恢复系数。

For $B$, $\sin\beta=\sqrt{1-(24/25)^2}=7/25$ and $\sin\varphi=3/5$. The unchanged perpendicular component gives

Take the direction from $A$ to $B$ along the line of centres as positive. On the separating branch where $A$ rebounds and $B$ moves forwards, conservation of momentum along this line gives

Dividing by $m$ and substituting the speed ratios gives

Newton’s law of restitution then gives

To check the directions, the positive magnitudes of the final parallel components are $\frac9{20}u_A$ and $\frac{28}{75}u_B$. If both components were positive, the same momentum equation with a positive $A$ term would give $u_B=\frac{21}{80}u_A$, making $B$ slower than $A$ along the line of centres. If both were negative, momentum would give $u_B=\frac{375}{176}u_A$, again making the signed component of $B$ smaller than that of $A$. A positive component for $A$ and a negative one for $B$ also gives approach. Thus the chosen branch is the one that allows separation.