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CIE 9709 2025 November Paper 42 Q2

A Level / CIE / M1

CIE 9709 2025 Nov Paper 42 Paper · Question 2

题目

Problem

A railway locomotive of mass 240 000 kg240\,000\,\mathrm{kg} is towing a coach of mass 36 000 kg36\,000\,\mathrm{kg} down a hill inclined at an angle of sin⁡−10.04\sin^{-1}0.04 to the horizontal. The driving force produced by the locomotive is 450 000 N450\,000\,\mathrm{N} and there are resistances to motion of 120 000 N120\,000\,\mathrm{N} on the locomotive and 15 000 N15\,000\,\mathrm{N} on the coach. The coupling between the locomotive and the coach is light, rigid and parallel to the hill.

Find the acceleration of the locomotive and the tension in the coupling.

(5)
题目中文翻译

一台质量为 240 000 kg240\,000\,\mathrm{kg} 的铁路机车拖着一节质量为 36 000 kg36\,000\,\mathrm{kg} 的车厢,沿与水平面成 sin⁡−10.04\sin^{-1}0.04 角的山坡下行。机车产生的驱动力为 450 000 N450\,000\,\mathrm{N},机车和车厢所受的运动阻力分别为 120 000 N120\,000\,\mathrm{N} 和 15 000 N15\,000\,\mathrm{N}。机车与车厢之间的连接器轻质、刚性且平行于山坡。

求机车的加速度及连接器中的张力。

(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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对机车加车厢整体求下坡加速度,再对车厢单独求张力,重力分量均朝下坡。

答题过程

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Take down the hill as positive. The total mass is 240000+36000=276000 kg240000+36000=276000\,\mathrm{kg}. For the whole system,

276000a= 450000+276000(10)(0.04) −120000−15000= 425400.\begin{align*} 276000a=&\,450000+276000(10)(0.04)\\[4mm] &\,\hspace{2pt}-120000-15000\\[4mm] =&\,425400. \end{align*}

Hence

a=709460=1.541304348…≈1.54 m s−2 down the hill.a=\frac{709}{460}=1.541304348\ldots \approx\boxed{1.54\text{ m s}^{-2}}\text{ down the hill}.

For the coach alone,

T+36000(10)(0.04)−15000=36000a.T+36000(10)(0.04)-15000=36000a.

Using the unrounded acceleration,

T=36000709460+600=56086.956522…≈56100 N (3 s.f.).T=36000\frac{709}{460}+600 =56086.956522\ldots \approx\boxed{56100\text{ N}}\text{ (3 s.f.)}.