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CIE 9231 2022 November Paper 31 Q2

A Level / CIE / FM

CIE 9231 2022 Nov Paper 31 Paper · Question 2

题目

Problem

A light elastic string has natural length aa and modulus of elasticity 4mg4mg. One end of the string is fixed to a point OO 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 along the surface in the direction OPOP. When the length of the string is 54a\frac54a, the speed of PP is vv. When the length of the string is 32a\frac32a, the speed of PP is 12v\frac12v.

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

(4)

(b) Find, in terms of gg, the acceleration of PP when the stretched length of the string is 32a\frac32a.

(2)
题目中文翻译

一根轻质弹性细绳的自然长度为 aa,弹性模量为 4mg4mg。细绳的一端固定在光滑水平面上的点 OO。一个质量为 mm 的质点 PP 连接在细绳的另一端。质点 PP 沿该平面向 OPOP 方向射出。当细绳的长度为 54a\frac54a 时,PP 的速率为 vv。当细绳的长度为 32a\frac32a 时,PP 的速率为 12v\frac12v。

(a) 求 vv 用 aa 和 gg 表示的表达式。

(4)

(b) 当细绳拉伸后的长度为 32a\frac32a 时,求 PP 的加速度,用 gg 表示。

(2)
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

(a)

解法一

思路

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动能的减少量等于弹性势能的增加量。两处绳长分别减去自然长度 aa,得到 a/4a/4 与 a/2a/2 的伸长。

答题过程

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Conservation of energy gives

 12m(v2−v24)= 4mg2a[(a2)2−(a4)2]= 2mga⋅3a216= 38mga.\begin{align*} &\,\frac12m\bigg(v^2-\frac{v^2}{4}\bigg)\\[4mm] =&\,\frac{4mg}{2a}\bigg[\bigg(\frac a2\bigg)^2-\bigg(\frac a4\bigg)^2\bigg]\\[4mm] =&\,\frac{2mg}{a}\cdot\frac{3a^2}{16}\\[4mm] =&\,\frac38mga. \end{align*}

Thus 38mv2=38mga\frac38mv^2=\frac38mga, giving v=agv=\sqrt{ag}.

(b)

解法一

思路

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此时沿桌面的力只有绳张力,方向指向 OO。先按胡克定律求其大小,再除以质量。

答题过程

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The extension is a/2a/2, so the tension is (4mg/a)(a/2)=2mg(4mg/a)(a/2)=2mg. Newton’s second law gives mA=2mgmA=2mg, so the acceleration is A=2gA=2g towards OO.