| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Banked track – with friction (find maximum/minimum speed or friction coefficient) |
| Difficulty | Standard +0.8 This M3 banked track problem requires resolving forces in two directions, applying circular motion principles, and handling friction in limiting equilibrium cases. Part (a) is standard, but parts (b) and (c) require careful force resolution with friction acting in opposite directions and algebraic manipulation to find μ and maximum speed—more demanding than typical M3 questions but still within standard syllabus scope. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| (a) resolve ↑: \(K \cos 35 - mg = 0\), \(K = \frac{mg}{\cos 35}\) | M1 | |
| resolve ←: \(K \sin 35 = \frac{m v^2}{r} = \frac{m v^2}{25}\) | M1 A1 | |
| combining, \(v^2 = 25g \tan 35\) giving \(v = 13.1 \text{ m s}^{-1}\) (3sf) | M1; A1 | |
| (b) resolve ↑: \(K \cos 35 + \mu R \sin 35 - mg = 0\) | M1 A1 | |
| \(R = \frac{mg}{\cos 35 + \mu \sin 35}\) | M1 | |
| resolve ←: \(K \sin 35 - \mu R \cos 35 = \frac{m v^2}{r} = \frac{100m}{25}\) | M1 A1 | |
| combining, \(\frac{mg(\sin 35 - \mu \cos 35)}{\cos 35 + \mu \sin 35} = 4m\) | M1 A1 | |
| giving \(g(\sin 35 - \mu \cos 35) = 4(\cos 35 + \mu \sin 35)\) \(\therefore \mu = \frac{4\sin 35 - 4\cos 35}{4\sin 35 + g\cos 35} = 0.227\) (3sf) | M1; A1 | |
| (c) resolve ↑: \(K \cos 35 - \mu R \sin 35 - mg = 0\) | M1 | |
| \(R = \frac{mg}{\cos 35 - \mu \sin 35}\) | M1 | |
| resolve ←: \(K \sin 35 + \mu R \cos 35 = \frac{m v^2}{r} = \frac{m v^2}{25}\) | M1 | |
| combining, \(\frac{mg(\sin 35 + \mu \cos 35)}{\cos 35 - \mu \sin 35} = \frac{m v^2}{25}\) | M1 | |
| giving \(v^2 = \frac{25g(\sin 35 + \mu \cos 35)}{\cos 35 - \mu \sin 35}\) \(\therefore v = 16 \text{ m s}^{-1}\) (2sf) | M1; A1 | (18) |
(a) resolve ↑: $K \cos 35 - mg = 0$, $K = \frac{mg}{\cos 35}$ | M1 |
resolve ←: $K \sin 35 = \frac{m v^2}{r} = \frac{m v^2}{25}$ | M1 A1 |
combining, $v^2 = 25g \tan 35$ giving $v = 13.1 \text{ m s}^{-1}$ (3sf) | M1; A1 |
(b) resolve ↑: $K \cos 35 + \mu R \sin 35 - mg = 0$ | M1 A1 |
$R = \frac{mg}{\cos 35 + \mu \sin 35}$ | M1 |
resolve ←: $K \sin 35 - \mu R \cos 35 = \frac{m v^2}{r} = \frac{100m}{25}$ | M1 A1 |
combining, $\frac{mg(\sin 35 - \mu \cos 35)}{\cos 35 + \mu \sin 35} = 4m$ | M1 A1 |
giving $g(\sin 35 - \mu \cos 35) = 4(\cos 35 + \mu \sin 35)$ $\therefore \mu = \frac{4\sin 35 - 4\cos 35}{4\sin 35 + g\cos 35} = 0.227$ (3sf) | M1; A1 |
(c) resolve ↑: $K \cos 35 - \mu R \sin 35 - mg = 0$ | M1 |
$R = \frac{mg}{\cos 35 - \mu \sin 35}$ | M1 |
resolve ←: $K \sin 35 + \mu R \cos 35 = \frac{m v^2}{r} = \frac{m v^2}{25}$ | M1 |
combining, $\frac{mg(\sin 35 + \mu \cos 35)}{\cos 35 - \mu \sin 35} = \frac{m v^2}{25}$ | M1 |
giving $v^2 = \frac{25g(\sin 35 + \mu \cos 35)}{\cos 35 - \mu \sin 35}$ $\therefore v = 16 \text{ m s}^{-1}$ (2sf) | M1; A1 | (18)
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**Total: (75)**7. A cyclist is travelling round a circular bend of radius 25 m on a track which is banked at an angle of $35 ^ { \circ }$ to the horizontal.
In a model of the situation, the cyclist and her bicycle are represented by a particle of mass 60 kg and air resistance and friction are ignored.
Using this model and assuming that the cyclist is not slipping,
\begin{enumerate}[label=(\alph*)]
\item find, correct to 3 significant figures, the speed at which she is travelling.
In tests it is found that the cyclist must travel at a minimum speed of $10 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to prevent the bicycle from slipping down the slope. A more refined model is now used with a coefficient of friction between the bicycle and the track of $\mu$.
Using this model,
\item show that $\mu = 0.227$, correct to 3 significant figures,
\item find, correct to 2 significant figures, the maximum speed at which the cyclist can travel without slipping up the slope.
END
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q7 [18]}}