AS June 2022 Q3
3. A cyclist is travelling around a circular track which is banked at an angle \(\alpha\) to the horizontal, where \(\tan\alpha = \dfrac{3}{4}\)
The cyclist moves with constant speed in a horizontal circle of radius \(r\).
In an initial model,
- the cyclist and her cycle are modelled as a particle
- the track is modelled as being rough so that there is sideways friction between the tyres of the cycle and the track, with coefficient of friction \(\mu\),
where \(\mu \lt \dfrac{4}{3}\)
Using this model, the maximum speed that the cyclist can travel around the track in a horizontal circle of radius \(r\), without slipping sideways, is \(V\).
In a new simplified model,
- the cyclist and her cycle are modelled as a particle
- the motion is now modelled so that there is no sideways friction between the tyres of the cycle and the track
Using this new model, the speed that the cyclist can travel around the track in a horizontal circle of radius \(r\), without slipping sideways, is \(U\).
| Scheme | Marks | AO |
|---|---|---|
| Resolving vertically | M1 | 3.4 |
| \(R\cos\alpha - F\sin\alpha = mg\) | A1 | 1.1b |
| Equation of motion horizontally | M1 | 3.4 |
| \(R\sin\alpha + F\cos\alpha = \dfrac{mV^2}{r}\) | A1 | 1.1b |
| Use of \(F = \mu R\) | M1 | 3.4 |
| Solve for \(V\) | M1 | 3.1b |
| \(V = \sqrt{\dfrac{(3+4\mu)rg}{4-3\mu}}\) * | A1* | 1.1b |
| (7) |
Notes
M1: Correct no. of terms, dim correct, condone sin/cos confusion and sign errors
A1: Correct equation
M1: Correct no. of terms, dim correct, condone sin/cos confusion and sign errors
A1: Correct equation
M1: Independent but must be used in an equation
M1: Substitute for trig and solve for \(V\). Dependent on preceding M marks.
A1*: Correct given answer correctly obtained
| Scheme | Marks | AO |
|---|---|---|
| Use of \(\mu = 0\) oe | M1 | 2.1 |
| \(U = \sqrt{\dfrac{3rg}{4}}\) | A1 | 1.1b |
| (2) |
Notes
M1: If they don’t use \(\mu = 0\), we need to see the first 6 marks from (a), without friction
A1: cao
| Scheme | Marks | AO |
|---|---|---|
| Since \(3 + 4\mu \gt 3\) and \(4 - 3\mu \lt 4\) oe | M1 | 2.1 |
| \(\dfrac{3}{4} \lt \dfrac{3+4\mu}{4-3\mu}\) and hence \(U \lt V\) * | A1* | 2.2a |
| (2) | ||
| (11 marks) |
Notes
M1: Any convincing argument
A1*: Given answer correctly obtained
SC: Allow M1A0 if they work in reverse to show that if \(U \lt V\) then \(\mu \gt 0\) and make an appropriate comment