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CIE 9709 2025 June Paper 43 Q5

A Level / CIE / M1

CIE 9709 2025 June Paper 43 Paper · Question 5

题目

Problem

A van of mass 2500 kg2500\,\mathrm{kg} travelling at speed v m s−1v\,\mathrm{m\,s^{-1}} experiences a resistance force of kv2 Nkv^2\,\mathrm{N}. The constant power of the van’s engine is 62.5 kW62.5\,\mathrm{kW}.

(a) The steady speed that the van could maintain when moving along a straight horizontal road is 50 m s−150\,\mathrm{m\,s^{-1}}.

Show that k=0.5k=0.5, and find the acceleration of the van when its speed is 25 m s−125\,\mathrm{m\,s^{-1}} on this straight horizontal road.

(4)

The van begins to ascend a hill inclined at an angle θ∘\theta^\circ to the horizontal. The van travels along a line of greatest slope of the hill. The speed of the van at the start of the hill is 20 m s−120\,\mathrm{m\,s^{-1}}, and its acceleration is 5a m s−25a\,\mathrm{m\,s^{-2}}. Later, on the same hill, the speed of the van is 30 m s−130\,\mathrm{m\,s^{-1}} and its acceleration is a m s−2a\,\mathrm{m\,s^{-2}}. The power of the van’s engine remains at 62.5 kW62.5\,\mathrm{kW}, and the resistance force remains at 0.5v2 N0.5v^2\,\mathrm{N}.

(b) Find the value of aa and the value of θ\theta.

(5)
题目中文翻译

一辆质量为 2500 kg2500\,\mathrm{kg} 的厢式货车以 v m s−1v\,\mathrm{m\,s^{-1}} 的速率行驶时,受到 kv2 Nkv^2\,\mathrm{N} 的阻力。货车发动机的功率恒为 62.5 kW62.5\,\mathrm{kW}。

(a) 货车沿水平直路行驶时能够维持的稳定速率为 50 m s−150\,\mathrm{m\,s^{-1}}。

证明 k=0.5k=0.5,并求货车在这条水平直路上以 25 m s−125\,\mathrm{m\,s^{-1}} 的速率行驶时的加速度。

(4)

货车开始沿一条与水平面成 θ∘\theta^\circ 角的山坡上行,运动方向沿山坡的一条最大坡度线。货车在坡底的速率为 20 m s−120\,\mathrm{m\,s^{-1}},加速度为 5a m s−25a\,\mathrm{m\,s^{-2}}。之后,在同一山坡上,货车的速率为 30 m s−130\,\mathrm{m\,s^{-1}},加速度为 a m s−2a\,\mathrm{m\,s^{-2}}。货车发动机的功率仍为 62.5 kW62.5\,\mathrm{kW},阻力仍为 0.5v2 N0.5v^2\,\mathrm{N}。

(b) 求 aa 和 θ\theta 的值。

(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

展开

稳速时牵引力等于k50²,由功率关系求k;到25时牵引力和阻力均需重新计算。

答题过程

展开

At steady speed 50 m s−150\,\mathrm{m\,s^{-1}}, the driving force equals the resistance. Hence

62500=(k⋅502)(50),k=0.5,62500=(k\cdot50^2)(50), \qquad \boxed{k=0.5},

as required.

At 25 m s−125\,\mathrm{m\,s^{-1}}, Newton’s second law gives

6250025−0.5(25)2=2500a25.\frac{62500}{25}-0.5(25)^2=2500a_{25}.

Therefore

a25=2500−312.52500=0.875 m s−2.a_{25}=\frac{2500-312.5}{2500} =\boxed{0.875\text{ m s}^{-2}}.

(b)

解法一

思路

展开

同一坡的重力分量相同,列20与30速率的两条方程相减消去角度,再回代。

答题过程

展开

Resolving up the hill at the two instants gives

6250020−0.5(20)2−25000sin⁡θ=12500a,(1)\frac{62500}{20}-0.5(20)^2-25000\sin\theta=12500a,\tag{1} 6250030−0.5(30)2−25000sin⁡θ=2500a.(2)\frac{62500}{30}-0.5(30)^2-25000\sin\theta=2500a.\tag{2}

Subtracting (2) from (1),

10000a= (3125−200)−(6250030−450)= 38753.\begin{align*} 10000a=&\,(3125-200)-\biggl(\frac{62500}{30}-450\biggr)\\[4mm] =&\,\frac{3875}{3}. \end{align*}

Thus

a=31240=0.129166667…≈0.129.a=\frac{31}{240}=0.129166667\ldots\approx\boxed{0.129}.

Substituting into (2),

25000sin⁡θ= 6250030−450−250031240= 1572512.\begin{align*} 25000\sin\theta=&\,\frac{62500}{30}-450-2500\frac{31}{240}\\[4mm] =&\,\frac{15725}{12}. \end{align*}

Hence

θ=3.004630723…∘≈3.0∘ (1 d.p.).\theta=3.004630723\ldots^\circ\approx\boxed{3.0^\circ}\text{ (1 d.p.)}.