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M2 2023 June Q6

A Level / Edexcel / M2

IAL 2023 June Paper · Question 6

题目

Problem

6.

A van of mass 900 kg900\,\mathrm{kg} is moving along a straight horizontal road.

The resistance to the motion of the van is modelled as a constant force of magnitude 600 N600\,\mathrm{N}.

The engine of the van is working at a constant rate of 24 kW24\,\mathrm{kW}.

At the instant when the speed of the van is V m s−1V\,\mathrm{m\,s^{-1}}, the acceleration of the van is 2 m s−22\,\mathrm{m\,s^{-2}}

(a) Find the value of VV

(4)

Later on, the van is towing a trailer of mass 700 kg700\,\mathrm{kg} up a straight road inclined at an angle θ\theta to the horizontal, where sin⁡θ=114\sin\theta=\frac1{14}

The trailer is attached to the van by a towbar, as shown in Figure 3.

The towbar is parallel to the direction of motion of the van and the trailer.

The resistance to the motion of the van from non-gravitational forces is modelled as a constant force of magnitude 600 N600\,\mathrm{N}.

The resistance to the motion of the trailer from non-gravitational forces is modelled as a constant force of magnitude 550 N550\,\mathrm{N}.

The towbar is modelled as a light rod.

The engine of the van is working at a constant rate of 24 kW24\,\mathrm{kW}.

(b) Find the acceleration of the van at the instant when the van and the trailer are moving with speed 8 m s−18\,\mathrm{m\,s^{-1}}

(4)

At the instant when the van and the trailer are moving up the road at 9 m s−19\,\mathrm{m\,s^{-1}}, the towbar breaks. The trailer continues to move in a straight line up the road until it comes to instantaneous rest.

The distance moved by the trailer as it slows from a speed of 9 m s−19\,\mathrm{m\,s^{-1}} to instantaneous rest is dd metres.

(c) Use the work-energy principle to find the value of dd.

(4)
题目中文翻译

质量为 900 kg900\,\mathrm{kg} 的厢式货车沿水平直路行驶。

将货车所受的运动阻力视为大小为 600 N600\,\mathrm{N} 的恒力。

货车发动机以恒定功率 24 kW24\,\mathrm{kW} 工作。

在货车速率为 V m s−1V\,\mathrm{m\,s^{-1}} 的瞬间，其加速度为 2 m s−22\,\mathrm{m\,s^{-2}}。

(a) 求 VV 的值。

(4)

后来，货车牵引质量为 700 kg700\,\mathrm{kg} 的拖车，沿一条与水平面成 θ\theta 角的直路上坡，其中 sin⁡θ=114\sin\theta=\frac1{14}。

如图 3 所示，拖车通过牵引杆与货车连接。

牵引杆平行于货车与拖车的运动方向。

将货车所受的非重力运动阻力视为大小为 600 N600\,\mathrm{N} 的恒力。

将拖车所受的非重力运动阻力视为大小为 550 N550\,\mathrm{N} 的恒力。

将牵引杆视为轻杆。

货车发动机以恒定功率 24 kW24\,\mathrm{kW} 工作。

(b) 求货车与拖车以速率 8 m s−18\,\mathrm{m\,s^{-1}} 行驶的瞬间，货车的加速度。

(4)

在货车与拖车以 9 m s−19\,\mathrm{m\,s^{-1}} 的速率上坡的瞬间，牵引杆断裂。拖车继续沿直线上坡，直至瞬时静止。

拖车从速率 9 m s−19\,\mathrm{m\,s^{-1}} 减至瞬时静止所运动的距离为 dd 米。

(c) 利用功—能原理求 dd 的值。

(4)

解答

(a)

解法一

思路

展开

水平路段的合力等于牵引力减阻力。功率先换成瓦，再用 F=P/VF=P/V。

答题过程

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Let the driving force be FF N. Resolving in the direction of motion,

F−600=900(2),F-600=900(2),

so F=2400F=2400 N. Since the engine power is 24000 W,

24000=FV=2400V.24000=FV=2400V.

Therefore

V=10 m s−1.\boxed{V=10\text{ m s}^{-1}}.

(b)

解法一

思路

展开

视两车为整体以消去杆力。合质量为 16001600，两项阻力都要计入；重力沿坡向下，不能只计拖车或只计货车。

答题过程

展开

At speed 8 m s−1^{-1}, the driving force is

F=240008=3000 N.F=\frac{24000}{8}=3000\text{ N}.

For the van and trailer together, resolving up the road gives

3000−(900+700)gsin⁡θ−(600+550)=(900+700)a.3000-(900+700)g\sin\theta-(600+550)=(900+700)a.

Using sin⁡θ=1/14\sin\theta=1/14,

3000−1600(9.8)(114)−1150=1600a.3000-1600(9.8)\left(\frac1{14}\right)-1150=1600a.

Thus

3000−1120−1150=1600a,3000-1120-1150=1600a,

and hence

a=0.456 m s−2 up the road.\boxed{a=0.456\text{ m s}^{-2}\text{ up the road}}.

(c)

解法一

思路

展开

断杆后只研究拖车，初速是 99。按题目要求用功—能：初动能变成势能增加和对拖车阻力所做的功，不再计货车质量或发动机功。

答题过程

展开

After the towbar breaks, the trailer has no driving force. Its initial kinetic energy is used to gain gravitational potential energy and overcome its own resistance of 550 N.

By the work-energy principle,

12(700)(9)2=700g(dsin⁡θ)+550d.\frac12(700)(9)^2=700g(d\sin\theta)+550d.

Therefore

28350=(700(9.8)114+550)d=1040d.28350=\left(700(9.8)\frac1{14}+550\right)d=1040d.

Hence

d=27.3 m.\boxed{d=27.3\text{ m}}.