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CIE 9709 2024 June Paper 41 Q5

A Level / CIE / M1

CIE 9709 2024 June Paper 41 Paper · Question 5

题目

Problem

A straight slope of length 60 m60\,\mathrm{m} is inclined at an angle of 12∘12^\circ to the horizontal. A bobsled starts at the top of the slope with a speed of 5 m s−15\,\mathrm{m\,s^{-1}}. The bobsled slides directly down the slope.

(a) It is given that there is no resistance to the bobsled’s motion.

Find its speed when it reaches the bottom of the slope.

(3)

(b) It is given instead that the coefficient of friction between the bobsled and the slope is 0.030.03.

Find the time that it takes for the bobsled to reach the bottom of the slope.

(5)

题目中文翻译

一个长 60 m60\,\mathrm{m} 的笔直斜坡与水平面成 12∘12^\circ 角。一架雪橇从坡顶以 5 m s−15\,\mathrm{m\,s^{-1}} 的速率开始运动，沿斜坡直接向下滑行。

(a) 已知雪橇的运动不受阻力。

求雪橇到达坡底时的速率。

(3)

(b) 改为已知雪橇与斜坡之间的摩擦系数为 0.030.03。

求雪橇到达坡底所需的时间。

(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

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无阻力时沿坡只有重力分量，得到恒加速度后用速度与路程公式。

答题过程

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If the bobsled has mass mm, resolving down the slope gives

ma=mgsin⁡12∘,a=10sin⁡12∘.ma=mg\sin12^\circ,\qquad a=10\sin12^\circ.

Hence

v2=52+2(10sin⁡12∘)(60).v^2=5^2+2(10\sin12^\circ)(60).

Taking the positive root,

v≈16.6 m s−1.v\approx\boxed{16.6\,\mathrm{m\,s^{-1}}}.

解法二

思路

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也可用机械能守恒，竖直下降高度为60乘斜角正弦。

答题过程

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The loss of potential energy equals the gain in kinetic energy:

mg(60sin⁡12∘)=12m(v2−52).mg(60\sin12^\circ)=\frac12m(v^2-5^2).

Cancelling mm and using g=10g=10,

v=25+1200sin⁡12∘≈16.6 m s−1.v=\sqrt{25+1200\sin12^\circ} \approx\boxed{16.6\,\mathrm{m\,s^{-1}}}.

(b)

解法一

思路

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支持力为重力法向分量，摩擦向上。求出净加速度后解位移二次方程，只取正时间。

答题过程

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The normal reaction and friction are

R= mgcos⁡12∘,F= 0.03mgcos⁡12∘.\begin{align*} R=&\,mg\cos12^\circ,\\[4mm] F=&\,0.03mg\cos12^\circ. \end{align*}

Resolving down the slope,

ma=mgsin⁡12∘−F,ma=mg\sin12^\circ-F,

so

a=10(sin⁡12∘−0.03cos⁡12∘)=1.7856726… m s−2.a=10(\sin12^\circ-0.03\cos12^\circ) =1.7856726\ldots\,\mathrm{m\,s^{-2}}.

The time satisfies

60=5t+12at2.60=5t+\frac12at^2.

The positive root is

t=−5+25+120aa≈5.86 s.t=\frac{-5+\sqrt{25+120a}}{a} \approx\boxed{5.86\,\mathrm{s}}.

解法二

思路

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能量法先求底端速率，再利用恒加速度运动的平均速率求用时。摩擦力恒定，因而加速度恒定。

答题过程

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The normal reaction is R=mgcos⁡12∘R=mg\cos12^\circ, so friction is F=0.03mgcos⁡12∘F=0.03mg\cos12^\circ. The work-energy equation is

12m(v2−25)=mg(60sin⁡12∘)−60F.\frac12m(v^2-25)=mg(60\sin12^\circ)-60F.

Thus

v2=25+1200(sin⁡12∘−0.03cos⁡12∘),v^2=25+1200(\sin12^\circ-0.03\cos12^\circ),

giving v=15.468701… m s−1v=15.468701\ldots\,\mathrm{m\,s^{-1}}. The forces, and hence acceleration, are constant. Therefore

60=5+v2t.60=\frac{5+v}{2}t. t=1205+v≈5.86 s.t=\frac{120}{5+v}\approx\boxed{5.86\,\mathrm{s}}.