两个质点 $A$ 和 $B$ 的质量分别为 $km$ 和 $m$，其中 $k$ 和 $m$ 为常数，它们可在光滑水平面上沿直线自由运动。质点 $A$ 以速率 $2u$ 向 $B$ 发射，同时 $B$ 以速率 $u$ 向 $A$ 发射。

两质点发生碰撞。碰撞后，$A$ 的速率为 $u$，且两质点均沿 $A$ 原来的运动方向运动。

已知碰撞损失了总动能的 $35\%$。用 $u$ 表示碰撞后 $B$ 的速率。

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 $10\,\mathrm{m\,s^{-2}}$。

先用带正负号的动量守恒把B的末速表示成k；再用保留65%动能建立k的二次方程，并按末运动方向舍根。

Take the initial direction of $A$ as positive. Let the velocity of $B$ after the collision be $v$. Conservation of momentum gives

Thus

The initial and final kinetic energies are

Since $35\%$ is lost, $E_f=0.65E_i$. Dividing by $mu^2/2>0$,

Both particles move in the original direction of $A$, so $v=(k-1)u>0$ requires $k>1$. Reject $k=0.1$, leaving $k=3.5$.

Therefore the speed of $B$ after the collision is $\boxed{2.5u}$.

也可先用动量守恒把质量比k消去，直接建立末速与u的关系。

Take the initial direction of $A$ as positive and write the final velocity of $B$ as $v$. Conservation of momentum gives

Thus

The energy condition is

Dividing by $m/2$ and substituting $k=1+v/u$,

The root $v=-0.9u$ contradicts the stated direction of motion, so the required speed is $\boxed{v=2.5u}$.