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CIE 9709 2024 November Paper 42 Q3

A Level / CIE / M1

CIE 9709 2024 Nov Paper 42 Paper · Question 3

题目

Problem

A cyclist is riding along a straight horizontal road. The total mass of the cyclist and his bicycle is 90 kg90\,\mathrm{kg}. The power exerted by the cyclist is 250 W250\,\mathrm{W}. At an instant when the cyclist’s speed is 5 m s−15\,\mathrm{m\,s^{-1}}, his acceleration is 0.1 m s−20.1\,\mathrm{m\,s^{-2}}.

(a) Find the value of the constant resistance to motion acting on the cyclist.

(3)

The cyclist comes to the bottom of a hill inclined at 2∘2^\circ to the horizontal.

(b) Given that the power and resistance to motion are unchanged, find the steady speed which the cyclist could maintain when riding up the hill.

(2)

题目中文翻译

一名骑车人沿一条笔直的水平道路骑行。骑车人和自行车的总质量为 90 kg90\,\mathrm{kg}。骑车人输出的功率为 250 W250\,\mathrm{W}。在骑车人的速率为 5 m s−15\,\mathrm{m\,s^{-1}} 的某一瞬间,他的加速度为 0.1 m s−20.1\,\mathrm{m\,s^{-2}}。

(a) 求作用于骑车人的恒定运动阻力的大小。

(3)

骑车人来到一个与水平面成 2∘2^\circ 角的山坡底部。

(b) 已知功率和运动阻力保持不变,求骑车人上坡时能够保持的稳定速率。

(2)

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

先由功率求驱动力,再用牛顿第二定律扣出阻力。

答题过程

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The driving force at speed 5 m s−15\,\mathrm{m\,s^{-1}} is

F=2505=50 N.F=\frac{250}{5}=50\,\mathrm{N}.

If the resistance is RR, Newton’s second law gives

50−R=90(0.1),R=41 N.50-R=90(0.1),\qquad \boxed{R=41\,\mathrm{N}}.

(b)

解法一

思路

展开

稳速上坡时推力需同时平衡阻力和重力沿坡分量,再以功率除以该推力求速率。

答题过程

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At steady speed vv, resolving up the hill gives

250v−41−90(10)sin⁡2∘=0.\frac{250}{v}-41-90(10)\sin2^\circ=0.

Therefore

v=25041+900sin⁡2∘≈3.45 m s−1.v=\frac{250}{41+900\sin2^\circ} \approx\boxed{3.45\,\mathrm{m\,s^{-1}}}.