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M1 2025 Oct A Q5

A Level / Edexcel / M1

IAL 2025 Oct A Paper · Question 5

题目

Problem

5.

Two particles PP and QQ have masses 0.1 kg0.1\,\mathrm{kg} and 0.5 kg0.5\,\mathrm{kg} respectively.

The particles are attached to the ends of a light inextensible string.

Particle PP is held at rest on a rough horizontal table.

The string lies along the table and passes over a small smooth pulley which is fixed to the edge of the table.

Particle QQ is at rest on a smooth plane which is inclined to the horizontal at an angle θ\theta, where tan⁡θ=43\tan\theta=\frac43

The string lies in the vertical plane which contains the pulley and a line of greatest slope of the inclined plane, as shown in Figure 2.

Particle PP is released from rest with the string taut.

During the first 0.5 s0.5\,\mathrm{s} of the motion PP does not reach the pulley and QQ moves 0.75 m0.75\,\mathrm{m} down the plane.

(a) Find the tension in the string during the first 0.5 s0.5\,\mathrm{s} of the motion.

(6)

(b) Find the coefficient of friction between PP and the table.

(5)
题目中文翻译

两质点 PP 和 QQ 的质量分别为 0.1 kg0.1\,\mathrm{kg} 和 0.5 kg0.5\,\mathrm{kg}。

两质点连接在一根不可伸长的轻绳的两端。

质点 PP 被保持静止在粗糙水平桌面上。

绳沿桌面放置,并跨过一个固定在桌边的小光滑滑轮。

质点 QQ 静止在与水平面成 θ\theta 角的光滑斜面上,其中 tan⁡θ=43\tan\theta=\frac43。

如图 2 所示,绳位于包含滑轮和斜面一条最大坡度线的竖直平面内。

质点 PP 由静止释放,绳绷紧。

运动的前 0.5 s0.5\,\mathrm{s} 内,PP 未到达滑轮,QQ 沿斜面向下运动 0.75 m0.75\,\mathrm{m}。

(a) 求运动的前 0.5 s0.5\,\mathrm{s} 内绳中的张力。

(6)

(b) 求 PP 与桌面之间的摩擦系数。

(5)

解答

(a)

解法一

思路

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恒定外力使加速度恒定。由 QQ 的位移和运动时间求加速度,再沿光滑斜面向下列运动方程。

答题过程

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Since tan⁡θ=4/3\tan\theta=4/3,

sin⁡θ=45.\sin\theta=\frac45.

Using s=ut+12at2s=ut+\frac12at^2 for QQ,

0.75=12a(0.5)2⟹a=6 m s−2.0.75=\frac12a(0.5)^2 \quad\Longrightarrow\quad a=6\text{ m s}^{-2}.

The string is light and the pulley smooth, so its tension is TT throughout. Resolving down the smooth plane for QQ,

0.5gsin⁡θ−T= 0.5a,0.5(9.8)45−T= 0.5(6),T= 0.92 N.\begin{align*} 0.5g\sin\theta-T=&\,0.5a,\\ 0.5(9.8)\frac45-T=&\,0.5(6),\\ T=&\,\boxed{0.92\text{ N}}. \end{align*}

(b)

解法一

思路

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不可伸长的绳使 PP 的加速度大小也为 66。桌面上支持力等于 PP 的重量,摩擦力反向于运动;斜面光滑,不应给 QQ 加摩擦力。

答题过程

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For PP, vertical equilibrium gives

R=0.1g=0.98 N.R=0.1g=0.98\text{ N}.

Taking the direction towards the pulley as positive, Newton’s second law gives

T−F=0.1a.T-F=0.1a.

Thus

F=0.92−0.1(6)=0.32 N.F=0.92-0.1(6)=0.32\text{ N}.

Using F=μRF=\mu R,

μ=0.320.98=1649=0.327(3 s.f.).\mu=\frac{0.32}{0.98}=\frac{16}{49} =\boxed{0.327}\quad(3\text{ s.f.}).