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CIE 9709 2023 November Paper 42 Q2

A Level / CIE / M1

CIE 9709 2023 Nov Paper 42 Paper · Question 2

题目

Problem

The diagram shows a smooth ring RR, of mass m kgm\,\mathrm{kg}, threaded on a light inextensible string. A horizontal force of magnitude 2 N2\,\mathrm{N} acts on RR. The ends of the string are attached to fixed points AA and BB on a vertical wall. The part ARAR of the string makes an angle of 30∘30^\circ with the vertical, the part BRBR makes an angle of 40∘40^\circ with the vertical and the string is taut. The ring is in equilibrium.

Find the tension in the string and find the value of mm.

(5)

题目中文翻译

如图所示,一个质量为 m kgm\,\mathrm{kg} 的光滑圆环 RR 套在一根不可伸长的轻绳上。一个大小为 2 N2\,\mathrm{N} 的水平力作用于 RR。绳的两端连接到竖直墙壁上的固定点 AA 和 BB。绳的 ARAR 段与竖直方向成 30∘30^\circ 角,BRBR 段与竖直方向成 40∘40^\circ 角,且绳处于绷紧状态。圆环处于平衡状态。

求绳的张力及 mm 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 (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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光滑环上同一根轻绳两侧张力相同。两个张力都向左,但上、下分量方向相反,分别列平衡式。

答题过程

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Since the ring is smooth, the tension is the same, TT, in both parts of the string. Horizontal equilibrium gives

Tsin⁡30∘+Tsin⁡40∘=2.T\sin30^\circ+T\sin40^\circ=2.

Thus

T=2sin⁡30∘+sin⁡40∘≈1.75 N(3 s.f.).T=\frac{2}{\sin30^\circ+\sin40^\circ} \approx\boxed{1.75\,\mathrm{N}}\quad\text{(3 s.f.)}.

Vertical equilibrium gives

Tcos⁡30∘−Tcos⁡40∘−mg=0.T\cos30^\circ-T\cos40^\circ-mg=0.

Using the unrounded tension and g=10g=10,

m=T(cos⁡30∘−cos⁡40∘)10≈0.0175.m=\frac{T(\cos30^\circ-\cos40^\circ)}{10} \approx\boxed{0.0175}.