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CIE 9709 2025 November Paper 43 Q2

A Level / CIE / M1

CIE 9709 2025 Nov Paper 43 Paper · Question 2

题目

Problem

The engine of a motorcycle can generate a maximum power of 105 kW105\,\mathrm{kW}. The mass of the motorcycle and its rider is 400 kg400\,\mathrm{kg}. The total resistance to the motion of the motorcycle and its rider is cv Ncv\,\mathrm{N}, where v m s−1v\,\mathrm{m\,s^{-1}} is the motorcyclist’s speed and cc is a constant.

The motorcyclist travels along a straight horizontal road under maximum engine power. When the motorcyclist’s speed is 35 m s−135\,\mathrm{m\,s^{-1}}, his acceleration is 0.5 m s−20.5\,\mathrm{m\,s^{-2}}.

(a) Show that c=80c=80.

(3)

The motorcyclist now travels up a straight hill under maximum engine power. The hill makes an angle of sin⁡−1(320)\sin^{-1}\big(\frac{3}{20}\big) with the horizontal.

(b) Find the steady speed at which the motorcyclist travels up the hill.

(3)
题目中文翻译

一辆摩托车的发动机能够产生的最大功率为 105 kW105\,\mathrm{kW}。摩托车与骑手的总质量为 400 kg400\,\mathrm{kg}。摩托车与骑手运动所受的总阻力为 cv Ncv\,\mathrm{N},其中 v m s−1v\,\mathrm{m\,s^{-1}} 是骑手的速率,cc 为常数。

骑手在水平直路上以发动机的最大功率行驶。当骑手的速率为 35 m s−135\,\mathrm{m\,s^{-1}} 时,他的加速度为 0.5 m s−20.5\,\mathrm{m\,s^{-2}}。

(a) 证明 c=80c=80。

(3)

骑手现在以发动机的最大功率沿直山坡上行。山坡与水平面的夹角为 sin⁡−1(320)\sin^{-1}\big(\frac{3}{20}\big)。

(b) 求骑手沿山坡上行时的稳定速率。

(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)

解法一

思路

展开

105kW换成W,用P=Fv求牵引力,再把速度35代入阻力cv。

答题过程

展开

At 35 m s−135\,\mathrm{m\,s^{-1}}, the driving force is

D=10500035=3000 N.D=\frac{105000}{35}=3000\text{ N}.

Newton’s second law along the horizontal road gives

3000−35c=400(0.5),3000-35c=400(0.5), 35c=2800,c=80,35c=2800, \qquad \boxed{c=80},

as required.

(b)

解法一

思路

展开

稳速上坡合力零,但阻力仍随速度变。乘正速率v后解二次方程并舍负根。

答题过程

展开

At steady speed v>0v>0, the driving force balances the resistance and downslope weight component:

105000v=80v+400(10)320.\frac{105000}{v}=80v+400(10)\frac3{20}.

Thus

80v2+600v−105000= 0,4v2+30v−5250= 0.\begin{align*} 80v^2+600v-105000=&\,0,\\[4mm] 4v^2+30v-5250=&\,0. \end{align*}

The roots are

v=−15±58494.v=\frac{-15\pm5\sqrt{849}}4.

Rejecting the negative root, the speed is

v=32.672005711…≈32.7 m s−1.v=32.672005711\ldots\approx\boxed{32.7\text{ m s}^{-1}}.