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CIE 9231 2022 November Paper 32 Q3

A Level / CIE / FM

CIE 9231 2022 Nov Paper 32 Paper · Question 3

题目

Problem

One end of a light elastic string, of natural length aa and modulus of elasticity 163Mg\frac{16}3Mg, is attached to a fixed point OO. A particle PP of mass 4M4M is attached to the other end of the string and hangs vertically in equilibrium. Another particle of mass 2M2M is attached to PP and the combined particle is then released from rest. The speed of the combined particle when it has descended a distance 14a\frac14a is vv.

Find an expression for vv in terms of gg and aa.

(6)
题目中文翻译

一根自然长度为 aa、弹性模量为 163Mg\frac{16}3Mg 的轻质弹性细绳的一端连接在固定点 OO。一个质量为 4M4M 的质点 PP 连接在细绳的另一端,并竖直悬挂,处于平衡状态。另一个质量为 2M2M 的质点连接到 PP 上,然后将合成质点从静止释放。合成质点下降 14a\frac14a 的距离后,其速率为 vv。

求 vv 用 gg 和 aa 表示的表达式。

(6)
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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先用原来悬挂质量的平衡求伸长。加上第二个粒子后,下降 a/4a/4 使伸长增加到 aa,用重力势能的减少扣去弹性势能的增加。

答题过程

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Let the initial extension be ee. Equilibrium before adding the second mass gives

16Mg3ae= 4Mg,e= 4Mg(3a)16Mg=34a.\begin{align*} \frac{16Mg}{3a}e=&\,4Mg,\\[4mm] e=&\,\frac{4Mg(3a)}{16Mg}=\frac34a. \end{align*}

After descending a/4a/4 the extension is aa. Conservation of energy for the combined mass 6M6M gives

6Mga4=16Mg6a(a2−(34a)2)+12(6M)v2.6Mg\frac a4 =\frac{16Mg}{6a}\bigg(a^2-\bigg(\frac34a\bigg)^2\bigg) +\frac12(6M)v^2.

The increase in elastic energy is

ΔE= 8Mg3a(a2−916a2)= 8Mg3a⋅716a2=76Mga.\begin{align*} \Delta E=&\,\frac{8Mg}{3a}\bigg(a^2-\frac9{16}a^2\bigg)\\[4mm] =&\,\frac{8Mg}{3a}\cdot\frac7{16}a^2=\frac76Mga. \end{align*}

Hence

3Mv2= (32−76)Mga=13Mga,v2= ag9.\begin{align*} 3Mv^2=&\,\bigg(\frac32-\frac76\bigg)Mga=\frac13Mga,\\[4mm] v^2=&\,\frac{ag}{9}. \end{align*}

Taking the positive square root gives v=13agv=\frac13\sqrt{ag}.