质点 $P$ 从水平面上的点 $O$ 投出，并仅在重力作用下运动。$P$ 的初速度为 $25\,\mathrm{m\,s^{-1}}$，与水平方向成角 $\theta$，其中 $\tan\theta=\frac43$。在点 $A$，$P$ 的运动方向与经过 $A$ 的竖直向下方向成 $45^\circ$ 角。

(a) 通过对轨迹方程求导或用其他方法，求 $A$ 的坐标。

在点 $A$，质点撞上一个固定光滑挡板、反弹并落在水平面上。该挡板与水平方向成 $45^\circ$ 角。

(b) 求 $P$ 与挡板碰撞前瞬间的速率。

(c) 已知挡板与质点之间的恢复系数为 $\frac19$，求 $P$ 撞上挡板后运动的水平距离。

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.

Where a numerical value for the acceleration due to gravity ($g$) is needed, use $10\,\mathrm{m\,s^{-2}}$.

需要重力加速度的数值时，取 $g=10\,\mathrm{m\,s^{-2}}$。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。

轨迹切线斜率就是竖直速度除以水平速度。到 $A$ 时向下倾斜 $45^\circ$，因此斜率为 $-1$。

Since $\tan\theta=4/3$, the initial horizontal speed is $15$ and $\cos\theta=3/5$. Using $y=x\tan\theta-gx^2/(2u^2\cos^2\theta)$ with $u=25$ and $g=10$, the trajectory is

At $A$ the direction is $45^\circ$ from the downward vertical, so the tangent slope is $-1$. Thus $\frac43-\frac{2x}{45}=-1$, giving $x_A=105/2$. Substitution gives $y_A=\frac43(105/2)-(105/2)^2/45=70-245/4=35/4$. Hence $\boxed{A=(105/2,\ 35/4)}$ in metres relative to $O$.

也可先用速度分量确定到达 $A$ 的时间。此时竖直向下的分量大小等于水平分量。

The initial velocity components are $15$ horizontally and $20$ vertically upward. At time $t$ they are $15$ and $20-10t$. The direction at $A$ gives $20-10t=-15$, so $t=7/2$. The displacements are

碰撞前，水平与竖直速度分量分别为 $15$ 和 $-15$。速度大小用勾股定理求出。

The velocity at $A$ before impact is $(15,-15)\,\mathrm{m\,s^{-1}}$. Its magnitude is $\sqrt{15^2+(-15)^2}=\sqrt{450}$, giving $\boxed{15\sqrt2\,\mathrm{m\,s^{-1}}}$.

粒子确实撞上障碍物，所以障碍物不是与入射速度平行的那条斜线。另一条倾角为 $45^\circ$ 的斜线恰好与速度垂直，碰后速度反向并缩为原来的九分之一。再由竖直位移求落地时间。

The incoming direction is $45^\circ$ below the horizontal. A barrier with that same slope would be parallel to the incoming velocity. Since the particle strikes the barrier, its slope is the other one, and the impact is normal. The smooth barrier and $e=1/9$ give the outgoing velocity components $(-5/3,5/3)$.

If $\tau$ is the time from impact to landing, the vertical displacement is $-35/4$, giving

Multiplying by $12$ gives $60\tau^2-20\tau-105=5(2\tau-3)(6\tau+7)=0$. The roots are $3/2$ and $-7/6$; time after impact is positive, so $\tau=3/2$. The horizontal distance travelled is therefore $(5/3)(3/2)=\boxed{\frac52\,\mathrm m}$.

也可从碰撞点起，沿反弹的水平运动方向取正。用碰后的轨迹方程，直接求落地时的水平距离。

The impact is normal, so the outgoing horizontal speed and upward speed are each $5/3$. Let $D$ be the horizontal distance along the outgoing direction from $A$. The vertical displacement from $A$ is

Setting this equal to $-35/4$ gives $36D^2-20D-175=0$. Factoring gives $(2D-5)(18D+35)=0$. Since $D$ is a positive distance, reject $-35/18$ and take $\boxed{D=5/2\,\mathrm m}$.