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CIE 9709 2023 March Paper 42 Q4

A Level / CIE / M1

CIE 9709 2023 March Paper 42 Paper · Question 4

题目

Problem

A toy railway locomotive of mass 0.8 kg0.8\,\mathrm{kg} is towing a truck of mass 0.4 kg0.4\,\mathrm{kg} on a straight horizontal track at a constant speed of 2 m s−12\,\mathrm{m\,s^{-1}}. There is a constant resistance force of magnitude 0.2 N0.2\,\mathrm{N} on the locomotive, but no resistance force on the truck. There is a light rigid horizontal coupling connecting the locomotive and the truck.

(a) State the tension in the coupling.

(1)

(b) Find the power produced by the locomotive’s engine.

(1)

The power produced by the locomotive’s engine is now changed to 1.2 W1.2\,\mathrm{W}.

(c) Find the magnitude of the tension in the coupling at the instant that the locomotive begins to accelerate.

(5)

题目中文翻译

一辆质量为 0.8 kg0.8\,\mathrm{kg} 的玩具火车机车牵引着一节质量为 0.4 kg0.4\,\mathrm{kg} 的货车,沿笔直的水平轨道以 2 m s−12\,\mathrm{m\,s^{-1}} 的恒定速率行驶。机车受到大小恒为 0.2 N0.2\,\mathrm{N} 的阻力,但货车不受阻力。机车与货车由一根轻质、刚性的水平连接杆连接。

(a) 写出连接杆中的张力。

(1)

(b) 求机车发动机输出的功率。

(1)

机车发动机输出的功率现在改为 1.2 W1.2\,\mathrm{W}。

(c) 求机车开始加速的瞬间,连接杆中张力的大小。

(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

展开

拖车无阻力又匀速,水平方向唯一可能的连接器力只能为零。

答题过程

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The truck has no resistance and zero acceleration. Hence its horizontal resultant is zero, so

T=0 N.\boxed{T=0\,\mathrm{N}}.

(b)

解法一

思路

展开

整体匀速时驱动力只需平衡机车阻力。

答题过程

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The driving force is 0.2 N0.2\,\mathrm{N}. Thus

P=0.2(2)=0.4 W.\boxed{P=0.2(2)=0.4\,\mathrm{W}}.

(c)

解法一

思路

展开

功率改变的瞬间速度仍为2,先求新驱动力,再用整体加速度与拖车方程求张力。

答题过程

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At the instant the power changes, the speed remains 2 m s−12\,\mathrm{m\,s^{-1}}. The new driving force is

F=1.22=0.6 N.F=\frac{1.2}{2}=0.6\,\mathrm{N}.

For the whole system,

0.6−0.2=(0.8+0.4)a,a=13 m s−2.0.6-0.2=(0.8+0.4)a, \qquad a=\frac13\,\mathrm{m\,s^{-2}}.

For the truck,

T=0.4a=215 N.\boxed{T=0.4a=\frac2{15}\,\mathrm{N}}.