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CIE 9709 2025 June Paper 43 Q2

A Level / CIE / M1

CIE 9709 2025 June Paper 43 Paper · Question 2

题目

Problem

A van of mass 4500 kg4500\,\mathrm{kg} is towing a trailer of mass 350 kg350\,\mathrm{kg} along a straight horizontal road. The van and trailer are connected by a light rigid tow-bar which is parallel to the road. There are resistance forces of X NX\,\mathrm{N} on the van and 120 N120\,\mathrm{N} on the trailer. The driving force produced by the van’s engine is 2500 N2500\,\mathrm{N}. The tension in the tow-bar is T NT\,\mathrm{N}, and the acceleration of the van is 0.4 m s−20.4\,\mathrm{m\,s^{-2}}.

Find the value of XX and the value of TT.

(4)
题目中文翻译

一辆质量为 4500 kg4500\,\mathrm{kg} 的厢式货车拖着一辆质量为 350 kg350\,\mathrm{kg} 的拖车,沿水平直路行驶。货车与拖车之间由一根平行于道路的轻质刚性拖杆连接。货车和拖车受到的阻力分别为 X NX\,\mathrm{N} 和 120 N120\,\mathrm{N}。货车发动机产生的驱动力为 2500 N2500\,\mathrm{N}。拖杆中的张力为 T NT\,\mathrm{N},货车的加速度为 0.4 m s−20.4\,\mathrm{m\,s^{-2}}。

求 XX 和 TT 的值。

(4)
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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先对拖车单独求张力,再对货车求未知阻力。

答题过程

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Taking the direction of motion as positive, for the trailer,

T−120=350(0.4),T=260.T-120=350(0.4), \qquad \boxed{T=260}.

For the van,

2500−X−T=4500(0.4).2500-X-T=4500(0.4).

Thus

X=2500−260−1800=440.X=2500-260-1800=\boxed{440}.