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CIE 9231 2023 June Paper 33 Q2

A Level / CIE / FM

CIE 9231 2023 June Paper 33 Paper · Question 2

题目

Problem

One end of a light elastic string, of natural length aa and modulus of elasticity λmg\lambda mg, is attached to a fixed point OO. The string lies on a smooth horizontal surface. A particle PP of mass mm is attached to the other end of the string. The particle PP is projected in the direction OPOP. When the length of the string is 43a\frac43a, the speed of PP is 2ag\sqrt{2ag}. When the length of the string is 53a\frac53a, the speed of PP is 122ag\frac12\sqrt{2ag}.

Find the value of λ\lambda.

(4)
题目中文翻译

一根自然长度为 aa、弹性模量为 λmg\lambda mg 的轻弹性绳的一端连接在固定点 OO 处。绳位于光滑水平面上。一个质量为 mm 的质点 PP 连接在绳的另一端。质点 PP 沿 OPOP 方向抛出。当绳长为 43a\frac43a 时,PP 的速率为 2ag\sqrt{2ag};当绳长为 53a\frac53a 时,PP 的速率为 122ag\frac12\sqrt{2ag}。

求 λ\lambda 的值。

(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 (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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桌面光滑且水平,动能的减少量等于绳的弹性势能增加量。计算两处伸长时都要减去自然长度 aa。

答题过程

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The extensions at lengths 4a/34a/3 and 5a/35a/3 are a/3a/3 and 2a/32a/3. Between those points the kinetic energy decreases by

12m(2ag−2ag4)=34mga.\frac12m\bigg(2ag-\frac{2ag}{4}\bigg)=\frac34mga.

The elastic potential energy increases by

 λmg2a[(2a3)2−(a3)2]= λmg2a⋅3a29= λ6mga.\begin{align*} &\,\frac{\lambda mg}{2a}\bigg[\bigg(\frac{2a}{3}\bigg)^2-\bigg(\frac a3\bigg)^2\bigg]\\[4mm] =&\,\frac{\lambda mg}{2a}\cdot\frac{3a^2}{9}\\[4mm] =&\,\frac{\lambda}{6}mga. \end{align*}

Equating the energy changes and dividing by mga>0mga>0 gives λ/6=3/4\lambda/6=3/4, so λ=9/2\lambda=9/2.