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CIE 9709 2025 June Paper 42 Q5

A Level / CIE / M1

CIE 9709 2025 June Paper 42 Paper · Question 5

题目

Problem

One end of a light inextensible string is attached to a particle AA of mass 3 kg3\,\mathrm{kg}. The other end of the string is attached to a particle BB of mass 4 kg4\,\mathrm{kg}. Particle AA is in contact with a rough plane inclined at 30∘30^\circ to the horizontal, and particle BB is in contact with a smooth horizontal plane. A second light inextensible string is attached to BB. The other end of this second string is attached to a particle CC of mass 5 kg5\,\mathrm{kg} which hangs vertically.

Both strings are taut and pass over small smooth pulleys that are fixed at the ends of the horizontal plane. The part of the string from AA to the pulley is parallel to a line of greatest slope of the inclined plane, and AA, BB and CC are in the same vertical plane (see diagram).

The system is released from rest. In the subsequent motion, CC moves vertically downwards with acceleration 2 m s−22\,\mathrm{m\,s^{-2}}, and neither AA nor BB reach a pulley.

(a) Find the tensions in each of the strings.

(3)

(b) Find the coefficient of friction between AA and the inclined plane.

(4)

When the system has been in motion for 1.5 s1.5\,\mathrm{s}, the string attached to AA breaks.

(c) Find the total distance that AA travels up the plane from the instant that the system is released from rest to the instant that AA comes to instantaneous rest.

(5)
题目中文翻译

一根轻质不可伸长细绳的一端连接质量为 3 kg3\,\mathrm{kg} 的质点 AA,另一端连接质量为 4 kg4\,\mathrm{kg} 的质点 BB。质点 AA 与一个和水平面成 30∘30^\circ 角的粗糙斜面接触,质点 BB 与一个光滑水平面接触。第二根轻质不可伸长细绳连接在 BB 上,其另一端连接一个质量为 5 kg5\,\mathrm{kg}、竖直悬挂的质点 CC。

两根细绳均绷紧,并跨过固定在水平面两端的小光滑滑轮。从 AA 到滑轮的那段细绳平行于斜面的一条最大坡度线,且 AA、BB、CC 位于同一竖直平面内(见图)。

系统从静止释放。在随后的运动中,CC 以 2 m s−22\,\mathrm{m\,s^{-2}} 的加速度竖直向下运动,且 AA 和 BB 均不会到达滑轮。

(a) 求每根细绳中的张力。

(3)

(b) 求 AA 与斜面之间的摩擦系数。

(4)

系统运动 1.5 s1.5\,\mathrm{s} 后,连接 AA 的细绳断裂。

(c) 求从系统由静止释放的瞬间至 AA 瞬时静止的瞬间,AA 沿斜面向上运动的总路程。

(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

(a)

解法一

思路

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两绳张力不同,先由悬挂C求BC绳张力,再由水平B求AB绳张力。

答题过程

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Let TABT_{AB} and TBCT_{BC} be the tensions in the strings joining those particles. For CC, taking downwards as positive,

5(10)−TBC=5(2),TBC=40 N.5(10)-T_{BC}=5(2), \qquad \boxed{T_{BC}=40\text{ N}}.

For BB, taking its direction of motion as positive,

TBC−TAB=4(2).T_{BC}-T_{AB}=4(2).

Therefore

TAB=40−8=32 N.T_{AB}=40-8=\boxed{32\text{ N}}.

(b)

解法一

思路

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A上行时摩擦向下,用已求32N张力求摩擦力11,再除正压力。

答题过程

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For AA, the normal contact force is

R=3(10)cos⁡30∘=153 N.R=3(10)\cos30^\circ=15\sqrt3\text{ N}.

Resolving up the plane,

32−F−3(10)sin⁡30∘=3(2),F=11 N.32-F-3(10)\sin30^\circ=3(2), \qquad F=11\text{ N}.

Hence

μ= FR=11153= 11345≈0.423 (3 s.f.).\begin{align*} \mu=&\,\frac F R=\frac{11}{15\sqrt3}\\[4mm] =&\,\frac{11\sqrt3}{45}\approx\boxed{0.423}\text{ (3 s.f.)}. \end{align*}

(c)

解法一

思路

展开

断绳前以2加速1.5秒,已走2.25、速率3;断绳后重力与11N摩擦使A减速,再把两段距离相加。

答题过程

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Before the string breaks, AA travels

s1=12(2)(1.5)2=2.25 ms_1=\frac12(2)(1.5)^2=2.25\text{ m}

and has speed u=2(1.5)=3 m s−1u=2(1.5)=3\,\mathrm{m\,s^{-1}}.

After the break, the tension on AA is zero. Taking up the plane as positive,

3a= −11−3(10)sin⁡30∘=−26,a= −263 m s−2.\begin{align*} 3a=&\,-11-3(10)\sin30^\circ=-26,\\[4mm] a=&\,-\frac{26}{3}\text{ m s}^{-2}. \end{align*}

If the additional distance to rest is s2s_2,

0=32+2(−263)s2,s2=2752 m.0=3^2+2\biggl(-\frac{26}{3}\biggr)s_2, \qquad s_2=\frac{27}{52}\text{ m}.

Thus the total distance up the plane is

s1+s2= 94+2752=3613≈ 2.77 m (3 s.f.).\begin{align*} s_1+s_2=&\,\frac94+\frac{27}{52}=\frac{36}{13}\\[4mm] \approx&\,\boxed{2.77\text{ m}}\text{ (3 s.f.)}. \end{align*}