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CIE 9709 2024 November Paper 41 Q7

A Level / CIE / M1

CIE 9709 2024 Nov Paper 41 Paper · Question 7

题目

Problem

A car has mass 1200 kg1200\,\mathrm{kg}. When the car is travelling at a speed of v m s−1v\,\mathrm{m\,s^{-1}}, there is a resistive force of magnitude kv Nkv\,\mathrm{N}. The maximum power of the car’s engine is 92.16 kW92.16\,\mathrm{kW}.

(a) The car travels along a straight level road.

(i) The car has a greatest possible constant speed of 48 m s−148\,\mathrm{m\,s^{-1}}.

Show that k=40k=40.

(1)

(ii) At an instant when its speed is 45 m s−145\,\mathrm{m\,s^{-1}}, find the greatest possible acceleration of the car.

(3)

(b) The car now travels at a constant speed up a hill inclined at an angle of sin⁡−10.15\sin^{-1}0.15 to the horizontal.

Find the greatest possible speed of the car going up the hill.

(4)
题目中文翻译

一辆汽车的质量为 1200 kg1200\,\mathrm{kg}。当汽车以 v m s−1v\,\mathrm{m\,s^{-1}} 的速率行驶时,受到大小为 kv Nkv\,\mathrm{N} 的阻力。汽车发动机的最大功率为 92.16 kW92.16\,\mathrm{kW}。

(a) 汽车沿水平直路行驶。

(i) 汽车能够维持的最大恒定速率为 48 m s−148\,\mathrm{m\,s^{-1}}。

证明 k=40k=40。

(1)

(ii) 在汽车速率为 45 m s−145\,\mathrm{m\,s^{-1}} 的某一瞬间,求汽车的最大可能加速度。

(3)

(b) 现在汽车以恒定速率沿一条与水平面成 sin⁡−10.15\sin^{-1}0.15 角的山坡上行。

求汽车上坡时的最大可能速率。

(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

(a)(i)

解法一

思路

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最高恒速时合力为零,推力等于kv,发动机已达到最大功率。

答题过程

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At the greatest constant speed, the driving force equals the resistance. Therefore

92160=(48k)(48),k=92160482=40.92160=(48k)(48),\qquad \boxed{k=\frac{92160}{48^2}=40}.

(a)(ii)

解法一

思路

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给定速率下最大推力等于最大功率除以速率,再扣除阻力求合力。

答题过程

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At speed 45 m s−145\,\mathrm{m\,s^{-1}}, the greatest driving force is

F=9216045=2048 N.F=\frac{92160}{45}=2048\,\mathrm{N}.

Newton’s second law gives

1200a=2048−40(45)=248.1200a=2048-40(45)=248.

Hence

a=31150 m s−2≈0.207 m s−2.\boxed{a=\frac{31}{150}\,\mathrm{m\,s^{-2}}} \approx0.207\,\mathrm{m\,s^{-2}}.

(b)

解法一

思路

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匀速上坡时推力需平衡阻力及重力沿坡分量。用最大功率建立速度的二次方程,只保留正根。

答题过程

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Let the greatest constant speed be v>0v>0. The driving force is

F=40v+1200(10)(0.15)=40v+1800.F=40v+1200(10)(0.15)=40v+1800.

At maximum power,

(40v+1800)v=92160.(40v+1800)v=92160.

Thus

40v2+1800v−92160=0,40v^2+1800v-92160=0,

and the positive root is

v= −1800+18002+160(92160)80≈ 30.5 m s−1.\begin{align*} v=&\,\frac{-1800+\sqrt{1800^2+160(92160)}}{80}\\[4mm] \approx&\,\boxed{30.5\,\mathrm{m\,s^{-1}}}. \end{align*}

The other root is negative and cannot represent the speed.