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M3 2026 June A Q1

A Level / Edexcel / M3

IAL 2026 June A Paper · Question 1

题目

Problem

1.

A car moves on a horizontal circular path round a banked bend on a race track.

The radius of the path is 60 m60\,\mathrm{m}.

The bend is banked at 20∘20^\circ to the horizontal, as shown in Figure 1.

The coefficient of friction between the car and the track is 0.250.25

The maximum speed at which the car can be driven round the bend without slipping sideways is V m s−1V\,\mathrm{m\,s^{-1}}

The car and driver are modelled as a single particle of mass 700 kg700\,\mathrm{kg}.

Using the model,

(a) find the magnitude of the normal reaction between the car and the track when the speed of the car is V m s−1V\,\mathrm{m\,s^{-1}}

(4)

(b) find the value of VV.

(4)
题目中文翻译

一辆汽车沿赛道上有横向倾角的弯道，在水平圆周路径上行驶。

路径的半径为 60 m60\,\mathrm{m}。

如图 1 所示，弯道与水平面之间的横向倾角为 20∘20^\circ。

汽车与赛道之间的摩擦系数为 0.250.25。

汽车沿弯道行驶而不发生侧滑的最大速率为 V m s−1V\,\mathrm{m\,s^{-1}}。

将汽车与驾驶员视为质量为 700 kg700\,\mathrm{kg} 的单一质点。

利用此模型，

(a) 求汽车速率为 V m s−1V\,\mathrm{m\,s^{-1}} 时，汽车与赛道之间的法向反作用力大小；

(4)

(b) 求 VV 的值。

(4)

解答

(a)

解法一

思路

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最大速率时，摩擦力沿斜面向下并达到极限值 μR\mu R。竖直方向没有加速度，因此可先单独求支持力。

答题过程

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At the maximum speed, the car tends to slip up the bank, so friction acts down the bank at its limiting value

F=μR=0.25R.F=\mu R=0.25R.

The forces are the normal reaction perpendicular to the bank, friction down the bank, and the weight vertically downwards.

Resolving vertically,

Rcos⁡20∘−0.25Rsin⁡20∘=700g.R\cos20^\circ-0.25R\sin20^\circ=700g.

Therefore

R=700(9.8)cos⁡20∘−0.25sin⁡20∘=8031.02… N,R=\frac{700(9.8)}{\cos20^\circ-0.25\sin20^\circ} =8031.02\ldots\ \mathrm N,

so

R=8.03×103 N(3 s.f.).\boxed{R=8.03\times10^3\ \mathrm N\quad\text{(3 s.f.)}.}

(b)

解法一

思路

展开

水平合力提供向心加速度。摩擦沿斜面向下，它的水平分量也朝向圆心。使用未舍入的支持力求速率。

答题过程

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Resolving horizontally towards the centre of the circular path,

Rsin⁡20∘+0.25Rcos⁡20∘=700V260.R\sin20^\circ+0.25R\cos20^\circ=\frac{700V^2}{60}.

Substitute the unrounded expression for RR from (a):

V2=60gsin⁡20∘+0.25cos⁡20∘cos⁡20∘−0.25sin⁡20∘.V^2=60g\frac{\sin20^\circ+0.25\cos20^\circ} {\cos20^\circ-0.25\sin20^\circ}.

Hence

V=19.9 m s−1(3 s.f.).\boxed{V=19.9\ \mathrm{m\,s^{-1}}\quad\text{(3 s.f.)}.}