题目

Problem

The points $A$ and $B$ are at the same horizontal level a distance $4a$ apart. The ends of a light elastic string, of natural length $4a$ and modulus of elasticity $\lambda$ , are attached to $A$ and $B$ . A particle $P$ of mass $m$ is attached to the midpoint of the string. The system is in equilibrium with $P$ at a distance $\frac32a$ below $M$ , the midpoint of $AB$ .

(a) Find $\lambda$ in terms of $m$ and $g$ .

(3)

The particle $P$ is pulled down vertically and released from rest at a distance $\frac83a$ below $M$ .

(b) Find, in terms of $a$ and $g$ , the speed of $P$ as it passes through $M$ in the subsequent motion.

(4)

题目中文翻译

点 $A$ 和 $B$ 位于同一水平高度，相距 $4a$ 。一根原长为 $4a$ 、弹性模量为 $\lambda$ 的轻弹性绳的两端分别连接在 $A$ 和 $B$ 。质量为 $m$ 的质点 $P$ 连接在绳的中点。系统处于平衡状态，此时 $P$ 位于 $AB$ 的中点 $M$ 下方 $\frac32a$ 处。

(a) 用 $m$ 和 $g$ 表示 $\lambda$ 。

(3)

将质点 $P$ 竖直向下拉动，在 $M$ 下方 $\frac83a$ 处由静止释放。

(b) 用 $a$ 和 $g$ 表示 $P$ 在随后的运动中经过 $M$ 时的速率。

(4)

原卷通用作答要求

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 $10\,\mathrm{m\,s^{-2}}$ 。

解答

(a)

解法一

思路

展开

平衡时两半段绳对称。由直角三角形求每段的长度与张力的竖直分量，再用胡克定律。

答题过程

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Each half of the string has natural length $2a$ . At equilibrium each half has length

\sqrt{(2a)^2+(3a/2)^2}=\frac52a,

so its extension is $a/2$ and its tension is $\lambda(a/2)/(2a)=\lambda/4$ . The vertical component of each tension is its magnitude multiplied by $(3a/2)/(5a/2)=3/5$ . Thus

\begin{align*} 2\frac{\lambda}{4}\frac35=&\,mg,\\[4mm] \frac{3\lambda}{10}=&\,mg,\\[4mm] \lambda=&\,\frac{10}{3}mg. \end{align*}

(b)

解法一

思路

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释放时先由几何求两段绳的伸长。在 $M$ ，两段恰好恢复自然长度，用弹性势能的减少量减去重力势能的增加量。

答题过程

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At release each half has length

\sqrt{(2a)^2+(8a/3)^2}=\frac{10}{3}a

and extension $4a/3$ . The total initial elastic potential energy is

\begin{align*} E=&\,2\frac{\lambda}{2(2a)}\bigg(\frac43a\bigg)^2\\[4mm] =&\,\frac89\lambda a\\[4mm] =&\,\frac89\cdot\frac{10}{3}mga=\frac{80}{27}mga. \end{align*}

At $M$ the string is at its natural length and the particle has risen by $8a/3$ . Conservation of energy gives

\begin{align*} \frac12mv^2=&\,\frac{80}{27}mga-\frac83mga\\[4mm] =&\,\frac{80-72}{27}mga=\frac8{27}mga,\\[4mm] v^2=&\,\frac{16}{27}ag. \end{align*}

Therefore $v=\frac{4}{3\sqrt3}\sqrt{ag}$ .

CIE 9231 2024 June Paper 31 Q2

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程