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CIE 9231 2024 November Paper 32 Q1

A Level / CIE / FM

CIE 9231 2024 Nov Paper 32 Paper · Question 1

题目

Problem

A particle of mass 2 kg2\text{ kg} is attached to one end of a light elastic string of natural length 0.8 m0.8\text{ m} and modulus of elasticity 100 N100\text{ N}. The other end of the string is attached to a fixed point OO on a smooth horizontal surface. The particle is moving in a horizontal circle about OO with the string taut and with constant angular speed 55 radians per second.

Find the extension of the string.

(3)
题目中文翻译

一个质量为 2 kg2\text{ kg} 的质点连接在一根原长为 0.8 m0.8\text{ m}、弹性模量为 100 N100\text{ N} 的轻弹性绳的一端。绳的另一端连接在光滑水平面上的固定点 OO。质点绕 OO 在水平圆周上运动,绳保持绷紧,角速率恒为每秒 55 弧度。

求绳的伸长量。

(3)
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 (gg) is needed, use 10 m s−210\,\mathrm{m\,s^{-2}}.

需要使用重力加速度 gg 的数值时,取 10 m s−210\,\mathrm{m\,s^{-2}}。

解答

解法一

思路

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绳的张力提供向心力。圆周半径是自然长度与伸长之和,把胡克定律代入径向方程。

答题过程

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Let xx be the extension in metres. The radius is 0.8+x0.8+x, and the tension is 100x/0.8100x/0.8 by Hooke’s law. Hence

125x= 2(0.8+x)(52)=40+50x,75x= 40.\begin{align*} 125x=&\,2(0.8+x)(5^2)=40+50x,\\[4mm] 75x=&\,40. \end{align*}

This gives x=8/15 m≈0.533 mx=8/15\,\mathrm{m}\approx0.533\,\mathrm{m}.