两面光滑竖直墙面相互垂直。质量为 $m$ 的光滑球体 $A$ 静止在光滑水平面上，到每面墙的距离均为 $d$。一个相同的光滑球体 $B$ 在该水平面上以速率 $u$ 运动，在两球发生碰撞时，其运动方向与两球的连心线成 $\theta$ 角（见图）。碰撞后，两球到达墙面所用的时间相同。两球之间的恢复系数为 $\frac12$。

(a) 求 $\tan\theta$ 的值。

(b) 求此次碰撞造成的两球总动能损失的百分比。

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 $10\,\mathrm{m\,s^{-2}}$。

碰后 $A$ 直向右墙，$B$ 向下墙的速度分量保持不变。图中到两面墙的相应距离相等，因此同时到达给出速度分量相等。

Let $v_A$ and $v_B$ be the post-impact components towards the right along the line of centres. Momentum conservation and restitution give

Cancelling $m$ in the first equation and adding the equations gives $2v_A=3u\cos\theta/2$. Hence

Sphere $B$ is farther from the right wall than $A$ and has the smaller rightward speed, so its first wall is the lower one. The distance it must travel towards that wall is the same as the distance $A$ travels towards the right wall. The downward component of $B$ stays $u\sin\theta$. Equal times therefore give

由 $\tan\theta$ 求正弦和余弦。把碰后的平行和垂直分量平方相加，求总动能占初始动能的比例。

Since $\theta$ is acute, $\cos\theta=4/5$ and $\sin\theta=3/5$. The speed of $A$ is $3u/5$, and the components of $B$ have magnitudes $u/5$ and $3u/5$. Therefore

Thus the percentage loss is $100(1-19/25)\%=24\%$.