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M1 2026 Jan A Q6

A Level / Edexcel / M1

IAL 2026 Jan A Paper · Question 6

题目

Problem

6.

A car pulls a trailer along a straight horizontal road using a light inextensible towbar. The mass of the car is M kgM\,\mathrm{kg}, the mass of the trailer is 600 kg600\,\mathrm{kg} and the towbar is horizontal and parallel to the direction of motion. There is a resistance to motion of magnitude 200 N200\,\mathrm{N} acting on the car and a resistance to motion of magnitude 100 N100\,\mathrm{N} acting on the trailer.

The driver of the car spots a hazard ahead. Instantly he reduces the force produced by the engine of the car to zero and applies the brakes of the car. The brakes produce a braking force on the car of magnitude 6500 N6500\,\mathrm{N} and the car and the trailer have a constant deceleration of magnitude 4 m s−24\,\mathrm{m\,s^{-2}}

Given that the resistances to motion on the car and trailer are unchanged and that the car comes to rest after travelling 40.5 m40.5\,\mathrm{m} from the point where the brakes were applied,

find

(a) the thrust in the towbar while the car is braking,

(3)

(b) the value of MM,

(3)

(c) the time it takes for the car to stop after the brakes are applied.

(3)
题目中文翻译

汽车通过一根不可伸长的轻质牵引杆，沿水平直路拖动一辆拖车。 汽车的质量为 M kgM\,\mathrm{kg}，拖车的质量为 600 kg600\,\mathrm{kg}，牵引杆水平且平行于运动方向。 作用在汽车上的运动阻力大小为 200 N200\,\mathrm{N}，作用在拖车上的运动阻力大小为 100 N100\,\mathrm{N}。

汽车司机发现前方有危险，立即将汽车发动机产生的力减至零，并对汽车施加制动。 制动对汽车产生大小为 6500 N6500\,\mathrm{N} 的制动力，汽车和拖车的恒定减速度大小为 4 m s−24\,\mathrm{m\,s^{-2}}。

已知汽车和拖车所受的运动阻力保持不变，汽车从开始制动的位置再行驶 40.5 m40.5\,\mathrm{m} 后静止，

求

(a) 汽车制动时牵引杆中的推力，

(3)

(b) MM 的值，

(3)

(c) 从开始制动到汽车停止所用的时间。

(3)

解答

(a)

解法一

思路

展开

制动时拖车受到牵引杆向后的推力，杆处于受压状态。取原运动方向为正，加速度为 −4-4。

答题过程

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Let the thrust be T NT\text{ N} and take the direction of motion as positive. For the trailer,

−100−T=600(−4).-100-T=600(-4).

Therefore

T=2300 N.\boxed{T=2300\text{ N}}.

(b)

解法一

思路

展开

对汽车和拖车整体列方程，牵引杆内力相消。制动力与两部分阻力均与运动方向相反。

答题过程

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For the car and trailer together,

−(6500+200+100)=(M+600)(−4).-(6500+200+100)=(M+600)(-4).

Hence

6800=4(M+600),6800=4(M+600), M+600=1700⟹M=1100.M+600=1700\quad\Longrightarrow\quad\boxed{M=1100}.

解法二

思路

展开

也可单独考虑汽车。拖车通过受压的杆对汽车施加向前的推力，所以这一项与制动力方向相反。

答题过程

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Taking forwards as positive for the car,

2300−6500−200=−4M.2300-6500-200=-4M.

Thus

−4400=−4M⟹M=1100.-4400=-4M\quad\Longrightarrow\quad\boxed{M=1100}.

(c)

解法一

思路

展开

先由停止距离求制动初速度，再由速度变化求时间。不能把物体开始制动误当作从静止开始。

答题过程

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Let the speed when the brakes are applied be uu. Using v2=u2+2asv^2=u^2+2as,

0=u2−2(4)(40.5)⟹u=18 m s−1.0=u^2-2(4)(40.5) \quad\Longrightarrow\quad u=18\text{ m s}^{-1}.

Now use v=u+atv=u+at:

0=18−4t⟹t=4.5 s.0=18-4t \quad\Longrightarrow\quad\boxed{t=4.5\text{ s}}.