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CIE 9231 2024 June Paper 33 Q2

A Level / CIE / FM

CIE 9231 2024 June Paper 33 Paper · Question 2

题目

Problem

A particle PP of mass mm is attached to one end of a light elastic string of natural length aa and modulus of elasticity 2mg2mg. A particle QQ of mass kmkm is attached to the other end of the string. Particle PP lies on a smooth horizontal table. The string has part of its length in contact with the table and then passes through a small smooth hole HH in the table.

Particle PP moves in a horizontal circle on the surface of the table with constant speed 12ga\sqrt{\frac12ga}. Particle QQ hangs in equilibrium vertically below the hole with HQ=14aHQ=\frac14a.

(a) Find, in terms of aa, the extension in the string.

(4)

(b) Find the value of kk.

(2)
题目中文翻译

质量为 mm 的质点 PP 连接在一根原长为 aa、弹性模量为 2mg2mg 的轻弹性绳的一端。质量为 kmkm 的质点 QQ 连接在绳的另一端。质点 PP 位于光滑水平桌面上。绳的一部分与桌面接触,然后穿过桌面上的光滑小孔 HH。

质点 PP 以恒定速率 12ga\sqrt{\frac12ga} 在桌面上的水平圆周上运动。质点 QQ 悬挂在小孔的正下方并处于平衡状态,且 HQ=14aHQ=\frac14a。

(a) 用 aa 表示绳的伸长量。

(4)

(b) 求 kk 的值。

(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

展开

圆周半径是桌面上的绳长,等于总长减去 HQHQ。用胡克定律和向心力方程表示同一个张力。

答题过程

展开

Let xx be the extension. The radius is (a+x)−a/4=3a/4+x(a+x)-a/4=3a/4+x. Hooke’s law and the radial equation give

T= 2mgxa,T= m(ga/2)3a/4+x.\begin{align*} T=&\,\frac{2mgx}{a},\\[4mm] T=&\,\frac{m(ga/2)}{3a/4+x}. \end{align*}

Equating these expressions gives

2x(3a4+x)= a22,4x2+3ax−a2= 0,(4x−a)(x+a)= 0.\begin{align*} 2x\bigg(\frac{3a}{4}+x\bigg)=&\,\frac{a^2}{2},\\[4mm] 4x^2+3ax-a^2=&\,0,\\[4mm] (4x-a)(x+a)=&\,0. \end{align*}

The roots are a/4a/4 and −a-a. The latter is impossible for a taut string, so x=a/4x=a/4.

(b)

解法一

思路

展开

对悬挂的 QQ 做竖直平衡。小孔光滑,两侧张力相同。

答题过程

展开

The tension is 2mg(a/4)/a=mg/22mg(a/4)/a=mg/2. Since QQ is in equilibrium, T=kmgT=kmg, so k=1/2k=1/2.