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M3 2026 Jan A Q6

A Level / Edexcel / M3

IAL 2026 Jan A Paper · Question 6

题目

Problem

6.

The path followed by a motorcycle round a circular racetrack is modelled as a horizontal circle of radius 50 m50\,\mathrm{m}. The track is banked at an angle θ\theta to the horizontal, where sin⁡θ=35\sin\theta=\dfrac35. The motorcycle travels round the track at constant speed.

The motorcycle is modelled as a particle and air resistance can be ignored.

In an initial model it is assumed that there is no sideways friction between the motorcycle tyres and the track.

Using this model,

(a) find the speed, in m s−1\mathrm{m\,s^{-1}}, of the motorcycle.

(5)

In a refined model it is assumed that there is sideways friction.

The coefficient of friction between the motorcycle tyres and the track is 14\dfrac14

It is still assumed that air resistance can be ignored and that the motorcycle is modelled as a particle.

The motorcycle’s path is unchanged.

Using this model,

(b) find the maximum speed, in m s−1\mathrm{m\,s^{-1}}, at which the motorcycle can travel without slipping sideways.

(8)
题目中文翻译

将摩托车沿圆形赛道行驶的路径视为半径为 50 m50\,\mathrm{m} 的水平圆周。赛道与水平面之间的横向倾角为 θ\theta，其中 sin⁡θ=35\sin\theta=\dfrac35。摩托车以恒定速率沿赛道行驶。

将摩托车视为质点，忽略空气阻力。

最初的模型假设摩托车轮胎与赛道之间没有侧向摩擦力。

利用此模型，

(a) 求摩托车的速率，单位为 m s−1\mathrm{m\,s^{-1}}。

(5)

改进后的模型假设存在侧向摩擦力。

摩托车轮胎与赛道之间的摩擦系数为 14\dfrac14。

仍假设可忽略空气阻力，并将摩托车视为质点。

摩托车的路径保持不变。

利用此模型，

(b) 求摩托车不发生侧滑时的最大速率，单位为 m s−1\mathrm{m\,s^{-1}}。

(8)

解答

(a)

解法一

思路

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向心加速度沿水平方向。将支持力分解成竖直和水平分量，再相除消去支持力；它不等于 mgcos⁡θmg\cos\theta。

答题过程

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Since sin⁡θ=3/5\sin\theta=3/5 and θ\theta is acute,

cos⁡θ=45,tan⁡θ=34.\cos\theta=\frac45,\qquad \tan\theta=\frac34.

Let mm be the mass and RR the normal reaction. With no sideways friction, vertical resolution and horizontal radial motion give

Rcos⁡θ=mg,Rsin⁡θ=mv250.R\cos\theta=mg,\qquad R\sin\theta=\frac{mv^2}{50}.

Dividing,

v250g=tan⁡θ=34,v2=75g2=367.5.\frac{v^2}{50g}=\tan\theta=\frac34, \qquad v^2=\frac{75g}{2}=367.5.

Thus

v=19.2 m s−1(3 s.f.).\boxed{v=19.2\ \mathrm{m\,s^{-1}}\quad\text{(3 s.f.)}.}

解法二

思路

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沿斜面方向分解可以直接消去支持力，但水平向心加速度在这个方向也有分量，不能误写成沿斜面平衡。

答题过程

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Resolve down the bank, towards the inner edge of the track. The normal reaction has no component in this direction, while the horizontal inward acceleration has component (v2/50)cos⁡θ(v^2/50)\cos\theta. Hence

mgsin⁡θ=mv250cos⁡θ.mg\sin\theta=m\frac{v^2}{50}\cos\theta.

Therefore

v2=50gtan⁡θ=50(9.8)34=367.5,v^2=50g\tan\theta=50(9.8)\frac34=367.5,

and

v=19.2 m s−1(3 s.f.).\boxed{v=19.2\ \mathrm{m\,s^{-1}}\quad\text{(3 s.f.)}.}

(b)

解法一

思路

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最大速率时，车有沿坡向上侧滑的趋势，所以摩擦沿坡向下，并达到极限值。这里的支持力与第(a)问不同，必须重新设未知量。

答题过程

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At the maximum speed VV, friction acts down the bank. Let the new normal reaction be SS, and let the limiting friction be FF. Then

F=μS=S4.F=\mu S=\frac S4.

There is no vertical acceleration, so

Scos⁡θ−Fsin⁡θ=mg.S\cos\theta-F\sin\theta=mg.

The horizontal resultant acts towards the centre:

Ssin⁡θ+Fcos⁡θ=mV250.S\sin\theta+F\cos\theta=\frac{mV^2}{50}.

Substitute sin⁡θ=3/5\sin\theta=3/5, cos⁡θ=4/5\cos\theta=4/5 and F=S/4F=S/4:

S(45−320)=mg⟹S=20mg13,S\left(\frac45-\frac3{20}\right)=mg \quad\Longrightarrow\quad S=\frac{20mg}{13}, S(35+15)=mV250.S\left(\frac35+\frac15\right)=\frac{mV^2}{50}.

Thus

mV250=4520mg13=16mg13,V2=800g13.\frac{mV^2}{50}=\frac45\frac{20mg}{13}=\frac{16mg}{13}, \qquad V^2=\frac{800g}{13}.

Consequently,

V=800(9.8)13=24.6 m s−1(3 s.f.).\boxed{V=\sqrt{\frac{800(9.8)}{13}}=24.6\ \mathrm{m\,s^{-1}}\quad\text{(3 s.f.)}.}