图中所示为两个相同的光滑均匀球 $A$ 和 $B$，它们半径相等，质量均为 $m$。两球在光滑水平面上运动，并分别以速率 $2u$ 和 $3u$ 发生碰撞。碰撞前的瞬间，$A$ 的运动方向与两球连心线成角 $\theta$，$B$ 的运动方向垂直于 $A$ 的运动方向。碰撞后，$B$ 沿垂直于连心线的方向运动。两球之间的恢复系数为 $\frac13$。

(a) 求 $\tan\theta$ 的值。

(b) 求此次碰撞造成的总动能损失。

(c) 求此次碰撞使 $A$ 的运动方向偏转的角度，以度为单位。

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 $10\,\mathrm{m\,s^{-2}}$。

取 $A$ 指向 $B$ 为正。由两球碰前速度互相垂直，从图上得到两球沿球心连线的分量。$B$ 碰后该分量为零。

Let $v$ be the signed component of the final velocity of $A$ along the line from $A$ to $B$. Initially the components of $A$ and $B$ on this line are $2u\cos\theta$ and $-3u\sin\theta$. After collision the component of $B$ is zero. Conservation of momentum and restitution give

Eliminating $v$ yields $8\cos\theta=6\sin\theta$, so $\tan\theta=4/3$.

光滑球的碰撞只改变沿球心连线的分量。只比较这些分量对应的碰前、碰后动能。

Since $\cos\theta=3/5$ and $\sin\theta=4/5$, the initial parallel components are $6u/5$ and $-12u/5$. The final ones are $-6u/5$ and zero. The perpendicular components are unchanged. Thus the total loss is

也可分别计算碰撞前后的总动能。碰后用平行和垂直分量的平方和求各球速率的平方。

The initial total kinetic energy is $\frac12m((2u)^2+(3u)^2)=13mu^2/2$. After collision the perpendicular components have magnitudes $2u\sin\theta=8u/5$ for $A$ and $3u\cos\theta=9u/5$ for $B$. The parallel component magnitudes are $6u/5$ and zero. Therefore the final total kinetic energy is

Subtracting gives a loss of $\big(\frac{13}{2}-\frac{181}{50}\big)mu^2=\frac{72}{25}mu^2$.

对 $A$，垂直分量不变，而沿球心连线的分量大小不变、方向反转。根据这两个向量的位置求转角。

The initial and final velocities of $A$ have parallel components $6u/5$ and $-6u/5$ and the same perpendicular component. Each makes the acute angle $\theta$ with the corresponding horizontal direction along the line of centres. The deflection is therefore