一个光滑平面与水平面成 $\theta$ 角，其中 $\sin\theta=\frac45$。一根轻弹簧的自然长度为 $10a$，弹性模量为 $1.6mg$，其一端连接于斜面上的固定点 $O$。弹簧的另一端连接于质量为 $0.2m$ 的质点 $P$。点 $O$ 和质点 $P$ 位于斜面的同一条最大坡度线上，且 $P$ 位于 $O$ 上方。质点 $P$ 处于静止状态。

(a) 证明距离 $OP$ 为 $9a$。

一个质量为 $0.3m$ 的质点 $Q$ 被保持在斜面上经过 $O$ 的最大坡度线上，位于 $P$ 上方距离为 $10a$ 的位置。从静止释放质点 $Q$，随后 $Q$ 与质点 $P$ 碰撞并合为一体。

(b) 求碰撞后紧接着合成质点的速率，用 $a$ 和 $g$ 表示。

(c) 碰撞使弹簧进一步压缩。求合成质点到 $O$ 的最小距离，用 $a$ 表示。

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 $10\,\mathrm{m\,s^{-2}}$。

静止时弹簧必须向上推 $P$，因此弹簧处于压缩状态。沿斜面把弹力与重力分量平衡。

Let $c$ be the compression of the spring. The component of the weight down the plane is $0.2mg(4/5)$. At equilibrium,

This gives $c=a$, so $OP=10a-a=9a$.

先用匀加速公式求 $Q$ 碰前的速率，再用沿斜面的动量守恒求合体速率。

The acceleration of $Q$ down the plane before impact is $\frac45g$. If $u$ is its speed just before impact, then

For the speed $V$ immediately after the particles coalesce, conservation of momentum gives

Hence $V=\frac{12}{5}\sqrt{ag}$ down the plane.

也可以用重力势能的减少量求 $Q$ 的碰前动能，然后在碰撞瞬间使用动量守恒。

Before impact, $Q$ drops through a vertical distance $10a\sin\theta=8a$. By conservation of energy,

so $u=4\sqrt{ag}$. The momentum down the plane is conserved during impact, giving

从碰撞后开始用能量守恒。设合体再向下移动 $x$，弹簧压缩量从 $a$ 增为 $a+x$，在最靠近 $O$ 处速率为零。

If the combined particle first stops after travelling a further distance $x$ down the plane, the increase in elastic potential energy is

This equals the initial kinetic energy after impact plus the decrease in gravitational potential energy. Thus

The EPE coefficient is $1.6mg/(20a)=0.08mg/a$. The kinetic energy is $\frac14m(144ag/25)=1.44mga$, and the gravitational term is $0.4mgx$. Dividing by $0.08mg/a$ gives

Thus

Reject $x=-3a$ since the further downward distance is positive. The positive root is $x=6a$, before the particle can reach $O$. Its minimum distance from $O$ is $9a-6a=3a$.