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CIE 9709 2022 June Paper 41 Q5

A Level / CIE / M1

CIE 9709 2022 June Paper 41 Paper · Question 5

题目

Problem

Two racing cars AA and BB are at rest alongside each other at a point OO on a straight horizontal test track. The mass of AA is 1200 kg1200\,\mathrm{kg}. The engine of AA produces a constant driving force of 4500 N4500\,\mathrm{N}. When AA arrives at a point PP its speed is 25 m s−125\,\mathrm{m\,s^{-1}}. The distance OPOP is d md\,\mathrm{m}. The work done against the resistance force experienced by AA between OO and PP is 75 000 J75\,000\,\mathrm{J}.

(a) Show that d=100d=100.

(3)

Car BB starts off at the same instant as car AA. The two cars arrive at PP simultaneously and with the same speed. The engine of BB produces a driving force of 3200 N3200\,\mathrm{N} and the car experiences a constant resistance to motion of 1200 N1200\,\mathrm{N}.

(b) Find the mass of BB.

(3)

(c) Find the steady speed which BB can maintain when its engine is working at the same rate as it is at PP.

(3)

题目中文翻译

两辆赛车 AA 和 BB 并排静止在水平直线测试跑道上的 OO 点。AA 的质量为 1200 kg1200\,\mathrm{kg},其发动机产生 4500 N4500\,\mathrm{N} 的恒定驱动力。AA 到达 PP 点时,速率为 25 m s−125\,\mathrm{m\,s^{-1}}。距离 OPOP 为 d md\,\mathrm{m}。AA 在 OO 与 PP 之间克服所受阻力做的功为 75 000 J75\,000\,\mathrm{J}。

(a) 证明 d=100d=100。

(3)

赛车 BB 与赛车 AA 同时出发。两辆车同时到达 PP,且到达时速率相同。BB 的发动机产生 3200 N3200\,\mathrm{N} 的驱动力,赛车受到 1200 N1200\,\mathrm{N} 的恒定运动阻力。

(b) 求 BB 的质量。

(3)

(c) 当 BB 的发动机以其在 PP 点时相同的功率工作时,求 BB 能维持的稳定速率。

(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

阻力只给全程做功,不能假设恒加速度。用驱动力功减阻力功等于末动能,实际推出d。

答题过程

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The car starts from rest. By the work-energy equation,

4500d−75000= 12(1200)(252)= 375000.\begin{align*} 4500d-75000=&\,\frac12(1200)(25^2)\\[4mm] =&\,375000. \end{align*}

Thus 4500d=4500004500d=450000, giving

d=100.\boxed{d=100}.

(b)

解法一

思路

展开

B驱动力与阻力都恒定,所以B匀加速;已知路程100及末速25,可先求加速度再求质量。

答题过程

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For car BB, constant acceleration gives

252=2a(100),a=3.125 m s−2.25^2=2a(100),\qquad a=3.125\,\mathrm{m\,s^{-2}}.

Newton’s second law gives

3200−1200=ma.3200-1200=ma.

Thus

m=20003.125=640 kg.\boxed{m=\frac{2000}{3.125}=640\,\mathrm{kg}}.

解法二

思路

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也可直接对B列能量方程,净功转为从静止增加的动能。

答题过程

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For car BB, the work-energy equation is

(3200−1200)(100)=12m(252).(3200-1200)(100)=\frac12m(25^2).

Hence

m=400000625=640 kg.\boxed{m=\frac{400000}{625}=640\,\mathrm{kg}}.

(c)

解法一

思路

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先求B在P处的功率,再令同功率下稳速驱动力等于阻力。

答题过程

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At PP, the engine power of BB is

P=3200(25)=80000 W.P=3200(25)=80000\,\mathrm{W}.

At steady speed vv with this power,

80000v=1200.\frac{80000}{v}=1200.

Therefore

v=2003 m s−1≈66.7 m s−1.\boxed{v=\frac{200}{3}\,\mathrm{m\,s^{-1}}} \approx66.7\,\mathrm{m\,s^{-1}}.