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CIE 9709 2025 March Paper 42 Q4

A Level / CIE / M1

CIE 9709 2025 March Paper 42 Paper · Question 4

题目

Problem

Two particles AA and BB have masses 0.3 kg0.3\,\mathrm{kg} and 0.1 kg0.1\,\mathrm{kg} respectively. The particles are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley, and the particles hang vertically below the pulley. Both particles are initially at a height of x mx\,\mathrm{m} above horizontal ground (see diagram). The system is released from rest.

(a) Find the tension in the string and the acceleration of the particles.

(4)

During the subsequent motion, BB does not reach the pulley. When AA reaches the ground, it comes to rest.

(b) Given that the greatest height of BB above the ground is 1.2 m1.2\,\mathrm{m}, find the value of xx.

(3)
题目中文翻译

两个质点 AA 和 BB 的质量分别为 0.3 kg0.3\,\mathrm{kg} 和 0.1 kg0.1\,\mathrm{kg},连接在一根轻质不可伸长细绳的两端。细绳跨过一个光滑定滑轮,质点竖直悬挂在滑轮下方。两个质点最初距水平地面的高度均为 x mx\,\mathrm{m}(见图)。系统从静止释放。

(a) 求细绳中的张力及质点的加速度。

(4)

在随后的运动中,BB 不会到达滑轮。当 AA 到达地面时,它停止运动。

(b) 已知 BB 离地的最大高度为 1.2 m1.2\,\mathrm{m},求 xx 的值。

(3)
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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轻绳、光滑滑轮保证两边张力相同,两粒子的加速度大小相同。分别沿各自运动方向列牛顿第二定律。

答题过程

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Let aa be the common acceleration magnitude and TT the tension. AA accelerates downwards and BB upwards. Newton’s second law gives

3−T=0.3a,T−1=0.1a.3-T=0.3a,\qquad T-1=0.1a.

Adding,

2=0.4a,a=5 m s−2.2=0.4a,\qquad \boxed{a=5\,\mathrm{m\,s^{-2}}}.

Substituting into the equation for BB,

T=1+0.1(5)=1.5 N.T=1+0.1(5)=\boxed{1.5\,\mathrm{N}}.

(b)

解法一

思路

展开

分开两段:落地前绳绷紧,B上升x;落地后绳松弛,B仅受重力继续上升。

答题过程

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When AA reaches the ground, both particles have moved a distance xx. Their speed vv satisfies

v2=2(5)x=10x.v^2=2(5)x=10x.

At that instant BB is at height 2x2x. The string then becomes slack, and BB continues upwards freely. If its additional rise is hh,

0=10x−2(10)h,h=x2.0=10x-2(10)h,\qquad h=\frac{x}{2}.

Thus its greatest height is

2x+x2=1.2,x=0.48.2x+\frac{x}{2}=1.2, \qquad \boxed{x=0.48}.