Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2022 November Paper 43 Q6

A Level / CIE / M1

CIE 9709 2022 Nov Paper 43 Paper · Question 6

题目

Problem

A car of mass 1750 kg1750\,\mathrm{kg} is pulling a caravan of mass 500 kg500\,\mathrm{kg}. The car and the caravan are connected by a light rigid tow-bar. The resistances to the motion of the car and caravan are 650 N650\,\mathrm{N} and 150 N150\,\mathrm{N} respectively.

(a) The car and caravan are moving along a straight horizontal road at a constant speed of 24 m s−124\,\mathrm{m\,s^{-1}}.

(i) Find the power of the car’s engine.

(2)

(ii) The engine’s power is now suddenly increased to 40 kW40\,\mathrm{kW}.

Find the instantaneous acceleration of the car and caravan and find the tension in the tow-bar.

(5)

(b) The car and caravan now travel up a straight hill, inclined at an angle sin⁡−10.14\sin^{-1}0.14 to the horizontal, at a constant speed of v m s−1v\,\mathrm{m\,s^{-1}}. The car’s engine is working at 31 kW31\,\mathrm{kW}. The resistances to the motion of the car and caravan are unchanged.

Find vv.

(3)

题目中文翻译

一辆质量为 1750 kg1750\,\mathrm{kg} 的汽车牵引一辆质量为 500 kg500\,\mathrm{kg} 的旅居拖车。汽车与拖车由一根轻质刚性牵引杆连接。汽车与拖车所受的运动阻力分别为 650 N650\,\mathrm{N} 和 150 N150\,\mathrm{N}。

(a) 汽车与拖车沿水平直路以 24 m s−124\,\mathrm{m\,s^{-1}} 的恒定速率运动。

(i) 求汽车发动机的功率。

(2)

(ii) 现在发动机的功率突然增加到 40 kW40\,\mathrm{kW}。

求汽车与拖车的瞬时加速度及牵引杆中的张力。

(5)

(b) 现在汽车与拖车沿一条与水平面夹角为 sin⁡−10.14\sin^{-1}0.14 的直山坡,以 v m s−1v\,\mathrm{m\,s^{-1}} 的恒定速率上行。汽车发动机的功率为 31 kW31\,\mathrm{kW}。汽车与拖车所受的运动阻力不变。

求 vv。

(3)

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)(i)

解法一

思路

展开

匀速时驱动力平衡两个阻力之和,再乘速率求功率。

答题过程

展开

At constant speed, the driving force is 650+150=800 N650+150=800\,\mathrm{N}. Hence

P=800(24)=19200 W.\boxed{P=800(24)=19200\,\mathrm{W}}.

(a)(ii)

解法一

思路

展开

功率突变瞬间速度仍为24,求新驱动力后用整体和拖车的牛顿方程。

答题过程

展开

The new driving force at the unchanged speed is

F=4000024=50003 N.F=\frac{40000}{24}=\frac{5000}{3}\,\mathrm{N}.

For the whole system,

50003−800=2250a,\frac{5000}{3}-800=2250a,

giving

a=52135 m s−2≈0.385 m s−2.\boxed{a=\frac{52}{135}\,\mathrm{m\,s^{-2}}} \approx0.385\,\mathrm{m\,s^{-2}}.

For the caravan,

T−150=500a.T-150=500a.

Thus

T= 150+500(52135)= 925027 N≈ 343 N.\begin{align*} T=&\,150+500\bigg(\frac{52}{135}\bigg)\\[4mm] =&\,\boxed{\frac{9250}{27}\,\mathrm{N}}\\[4mm] \approx&\,343\,\mathrm{N}. \end{align*}

(b)

解法一

思路

展开

上坡匀速,驱动力除平衡原有阻力外还需平衡整个系统的重力沿坡分量。

答题过程

展开

The driving force at steady speed is

F=800+2250(10)(0.14)=3950 N.F=800+2250(10)(0.14)=3950\,\mathrm{N}.

Hence

31000=3950v,v=62079≈7.85.31000=3950v, \qquad\boxed{v=\frac{620}{79}\approx7.85}.