题目

Problem

One end of a light inextensible string of length $a$ is attached to a fixed point $O$ . A particle of mass $m$ is attached to the other end of the string. The particle is held at the point $A$ with the string taut. The angle between $OA$ and the downward vertical is equal to $\alpha$ , where $\cos\alpha=\frac45$ . The particle is projected from $A$ , perpendicular to the string in an upwards direction, with a speed $\sqrt{3ga}$ . It then moves along a circular path in a vertical plane. The string first goes slack when it makes an angle $\theta$ with the upward vertical through $O$ .

Find the value of $\cos\theta$ .

(5)

题目中文翻译

一根长为 $a$ 的轻质不可伸长细绳的一端连接在固定点 $O$ 。一个质量为 $m$ 的质点连接在细绳的另一端。将质点保持在点 $A$ ，使细绳绷紧。 $OA$ 与竖直向下方向的夹角等于 $\alpha$ ，其中 $\cos\alpha=\frac45$ 。质点从 $A$ 点以速率 $\sqrt{3ga}$ 沿垂直于细绳的向上方向射出。随后，它在竖直平面内沿圆周运动。当细绳与经过 $O$ 的竖直向上方向成 $\theta$ 角时，细绳第一次松弛。

求 $\cos\theta$ 的值。

(5)

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

解法一

思路

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在绳刚松弛处，重力的径向分量提供全部向心力。把该处速率平方代入从 $A$ 开始的能量方程。

答题过程

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Let $v$ be the speed when the string becomes slack. At that point,

mg\cos\theta=\frac{mv^2}{a}.

The rise from $A$ is $a(\cos\alpha+\cos\theta)$ , so conservation of energy gives

\begin{align*} \frac12m(3ag)-\frac12mv^2=&\,mga(\cos\alpha+\cos\theta),\\[4mm] v^2=&\,3ag-2ag(\cos\alpha+\cos\theta). \end{align*}

Substituting $v^2=ag\cos\theta$ and $\cos\alpha=4/5$ , and dividing by $ag$ , gives

\begin{align*} \cos\theta=&\,3-2\bigg(\frac45+\cos\theta\bigg),\\[4mm] 3\cos\theta=&\,3-\frac85=\frac75,\\[4mm] \cos\theta=&\,\frac7{15}. \end{align*}

CIE 9231 2022 June Paper 31 Q2

题目

解答

解法一

思路

答题过程