Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2023 June Paper 43 Q2

A Level / CIE / M1

CIE 9709 2023 June Paper 43 Paper · Question 2

题目

Problem

A car of mass 1500 kg1500\,\mathrm{kg} is towing a trailer of mass m kgm\,\mathrm{kg} along a straight horizontal road. The car and the trailer are connected by a tow-bar which is horizontal, light and rigid. There is a resistance force of F NF\,\mathrm{N} on the car and a resistance force of 200 N200\,\mathrm{N} on the trailer. The driving force of the car’s engine is 3200 N3200\,\mathrm{N}, the acceleration of the car is 1.25 m s−21.25\,\mathrm{m\,s^{-2}} and the tension in the tow-bar is 300 N300\,\mathrm{N}.

Find the value of mm and the value of FF.

(4)

题目中文翻译

一辆质量为 1500 kg1500\,\mathrm{kg} 的汽车沿笔直的水平道路牵引一辆质量为 m kgm\,\mathrm{kg} 的拖车。汽车与拖车由一根水平、轻质且刚性的牵引杆连接。汽车受到 F NF\,\mathrm{N} 的阻力,拖车受到 200 N200\,\mathrm{N} 的阻力。汽车发动机的驱动力为 3200 N3200\,\mathrm{N},汽车的加速度为 1.25 m s−21.25\,\mathrm{m\,s^{-2}},牵引杆中的张力为 300 N300\,\mathrm{N}。

求 mm 和 FF 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

展开

分别对拖车和汽车列牛顿第二定律,直接求m和F,不必额外联立整个系统。

答题过程

展开

For the trailer, in the direction of motion,

300−200=1.25m,m=80.300-200=1.25m,\qquad \boxed{m=80}.

For the car,

3200−F−300=1500(1.25).3200-F-300=1500(1.25).

Thus

F=2900−1875=1025.\boxed{F=2900-1875=1025}.