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M1 2023 June Q7

A Level / Edexcel / M1

IAL 2023 June Paper · Question 7

题目

Problem

7.

A car of mass 1200 kg1200\,\mathrm{kg} is towing a trailer of mass 600 kg600\,\mathrm{kg} up a straight road, as shown in Figure 4.

The road is inclined at an angle α\alpha to the horizontal, where sin⁡α=112\sin\alpha=\frac{1}{12}.

The driving force produced by the engine of the car is 3000 N3000\,\mathrm{N}.

The car moves with acceleration 0.75 m s−20.75\,\mathrm{m\,s^{-2}}.

The non-gravitational resistance to motion of

  • the car is modelled as a constant force of magnitude 2R2R newtons,
  • the trailer is modelled as a constant force of magnitude RR newtons.

The car and the trailer are modelled as particles.

The tow bar between the car and trailer is modelled as a light rod that is parallel to the direction of motion.

Using the model,

(a) show that the value of RR is 6060,

(4)

(b) find the tension in the tow bar.

(3)

When the car and trailer are moving at a speed of 12 m s−112\,\mathrm{m\,s^{-1}}, the tow bar breaks.

Given that the non-gravitational resistance to motion of the trailer remains unchanged,

(c) use the model to find the further distance moved by the trailer before it first comes to rest.

(4)
题目中文翻译

如图 4 所示，一辆质量为 1200 kg1200\,\mathrm{kg} 的汽车沿笔直的上坡公路拖动一辆质量为 600 kg600\,\mathrm{kg} 的拖车。

公路与水平面成 α\alpha 角，其中 sin⁡α=112\sin\alpha=\frac{1}{12}。

汽车发动机产生的驱动力为 3000 N3000\,\mathrm{N}。

汽车的加速度为 0.75 m s−20.75\,\mathrm{m\,s^{-2}}。

非重力运动阻力的模型如下：

  • 汽车所受阻力为大小为 2R2R 牛顿的恒力，
  • 拖车所受阻力为大小为 RR 牛顿的恒力。

将汽车和拖车视为质点。

将汽车与拖车之间的牵引杆视为平行于运动方向的轻杆。

利用该模型，

(a) 证明 RR 的值为 6060，

(4)

(b) 求牵引杆中的张力。

(3)

当汽车和拖车的速率为 12 m s−112\,\mathrm{m\,s^{-1}} 时，牵引杆断裂。

已知拖车所受的非重力运动阻力保持不变，

(c) 利用该模型，求拖车首次静止之前继续运动的距离。

(4)

解答

(a)

解法一

思路

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把汽车与拖车作为整体，牵引杆中的内力抵消。沿斜面向上取正，要同时包括两者的重力分量与三个 RR 的阻力总和。

答题过程

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Taking up the road as positive for the combined system,

3000−1800gsin⁡α−3R=1800(0.75).3000-1800g\sin\alpha-3R=1800(0.75).

Using sin⁡α=1/12\sin\alpha=1/12,

3000−1800(9.8)12−3R= 1350,3000−1470−3R= 1350,3R= 180,R= 60,\begin{align*} 3000-\frac{1800(9.8)}{12}-3R=&\,1350,\\ 3000-1470-3R=&\,1350,\\ 3R=&\,180,\\ R=&\,\boxed{60}, \end{align*}

as required.

(b)

解法一

思路

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单独研究拖车，牵引杆张力向上坡，阻力和重力分量均向下坡。

答题过程

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For the trailer, taking up the road as positive,

T−600gsin⁡α−60= 600(0.75),T= 450+600(9.8)12+60,T= 1000 N.\begin{align*} T-600g\sin\alpha-60=&\,600(0.75),\\ T=&\,450+\frac{600(9.8)}{12}+60,\\ T=&\,\boxed{1000\ \mathrm N}. \end{align*}

解法二

思路

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也可单独研究汽车，汽车受牵引杆向下坡的拉力。用驱动力扣除汽车阻力及重力分量，再列牛顿第二定律。

答题过程

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For the car, taking up the road as positive,

3000−1200gsin⁡α−2(60)−T= 1200(0.75),3000−980−120−T= 900,T= 1000 N.\begin{align*} 3000-1200g\sin\alpha-2(60)-T=&\,1200(0.75),\\ 3000-980-120-T=&\,900,\\ T=&\,\boxed{1000\ \mathrm N}. \end{align*}

(c)

解法一

思路

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杆断后张力消失，拖车仍向上运动，阻力与重力分量都沿下坡。先重新求加速度，再用首次停下时速率为零求额外路程。

答题过程

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After the tow bar breaks, take up the road as positive for the trailer. Its new acceleration is given by

−60−600gsin⁡α= 600a,−60−490= 600a,a= −1112 m s−2.\begin{align*} -60-600g\sin\alpha=&\,600a,\\ -60-490=&\,600a,\\ a=&\,-\frac{11}{12}\ \mathrm{m\,s^{-2}}. \end{align*}

Let the further distance before the first stop be dd. Using v2=u2+2adv^2=u^2+2ad,

0= 122+2(−1112)d,d= 86411=78.5 m(3 s.f.).\begin{align*} 0=&\,12^2+2\left(-\frac{11}{12}\right)d,\\ d=&\,\frac{864}{11}=\boxed{78.5\ \mathrm m}\quad\text{(3 s.f.)}. \end{align*}