| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2017 |
| Session | June |
| Marks | 13 |
| 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 +0.3 This is a standard M3 banked track problem with two parts: (a) finding speed without friction using standard circular motion equations, and (b) incorporating friction to find maximum speed. Both parts follow well-established methods taught in M3, requiring resolution of forces and application of F=mv²/r, but the friction component adds modest complexity beyond the basic case. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(R(\uparrow)\ R\cos\theta = mg\) | B1 | Resolve vertically, equation must be fully correct |
| \(R(\rightarrow)\ R\sin\theta = m\dfrac{v^2}{50}\) | M1A1 | Form equation of motion horizontally. \(R\) must be resolved; acceleration can be in either form |
| \(\tan\theta = \dfrac{3}{4} = \dfrac{v^2}{50g}\) | dM1 | Use \(\tan\theta = 3/4\) with equations to reach \(v^2 = \ldots\) or \(v = \ldots\) Depends on M1 above |
| \(v^2 = \dfrac{150g}{4}\), \(v = 19.17\ldots = 19\) or \(19.2\ \text{ms}^{-1}\) | A1 (5) | Correct value of \(v\). Must be 2 or 3 sf |
| ALT: Equation parallel to track: \(mg\sin\theta = m\dfrac{v^2}{50}\cos\theta\) | M1 weight and acceleration both resolved, A1A1 one mark each term (50 or \(r\)). No B mark. |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(R(\uparrow)\ R\cos\theta - F\sin\theta = mg\) | M1A1 | Resolve vertically, \(R\) and \(F\) both resolved. Treat \(\mu mg\) as \(F\) for first 5 marks |
| \(R(\rightarrow)\ R\sin\theta + F\cos\theta = m\dfrac{v^2}{50}\) | M1A1A1 | Resolve horizontally, \(R\) and \(F\) both resolved, acceleration in either form. Treat \(\mu mg\) as \(F\) |
| \(F = \dfrac{1}{4}R\) | B1 | (\(\mu mg\) scores B0) |
| \(\dfrac{4}{5}R - \dfrac{1}{4}R \times \dfrac{3}{5} = mg\) | ||
| \(\dfrac{3}{5}R + \dfrac{1}{4}R \times \dfrac{4}{5} = m\dfrac{v^2}{50}\) | ||
| \(\dfrac{v^2}{50g} = \dfrac{4}{5} \div \dfrac{13}{20}\), \(v = 24.55\ldots = 25\) or \(24.6\ \text{ms}^{-1}\) | dM1,A1 (8) | Eliminate \(R\) and \(F\) to obtain equation for \(v^2\). Depends on two previous M marks. Correct value of \(v\), must be 2 or 3 sf |
| ALT: Parallel to track: \(F + mg\sin\theta = m(v^2/50)\cos\theta\) | M1A1, 50 or \(r\) | |
| Perpendicular to track: \(R - mg\cos\theta = m(v^2/50)\sin\theta\) | M1A1A1, 50 or \(r\) |
# Question 6:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $R(\uparrow)\ R\cos\theta = mg$ | B1 | Resolve vertically, equation must be fully correct |
| $R(\rightarrow)\ R\sin\theta = m\dfrac{v^2}{50}$ | M1A1 | Form equation of motion horizontally. $R$ must be resolved; acceleration can be in either form |
| $\tan\theta = \dfrac{3}{4} = \dfrac{v^2}{50g}$ | dM1 | Use $\tan\theta = 3/4$ with equations to reach $v^2 = \ldots$ or $v = \ldots$ Depends on M1 above |
| $v^2 = \dfrac{150g}{4}$, $v = 19.17\ldots = 19$ or $19.2\ \text{ms}^{-1}$ | A1 (5) | Correct value of $v$. Must be 2 or 3 sf |
**ALT:** Equation parallel to track: $mg\sin\theta = m\dfrac{v^2}{50}\cos\theta$ | M1 weight and acceleration both resolved, A1A1 one mark each term (50 or $r$). No B mark.
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $R(\uparrow)\ R\cos\theta - F\sin\theta = mg$ | M1A1 | Resolve vertically, $R$ and $F$ both resolved. Treat $\mu mg$ as $F$ for first 5 marks |
| $R(\rightarrow)\ R\sin\theta + F\cos\theta = m\dfrac{v^2}{50}$ | M1A1A1 | Resolve horizontally, $R$ and $F$ both resolved, acceleration in either form. Treat $\mu mg$ as $F$ |
| $F = \dfrac{1}{4}R$ | B1 | ($\mu mg$ scores B0) |
| $\dfrac{4}{5}R - \dfrac{1}{4}R \times \dfrac{3}{5} = mg$ | | |
| $\dfrac{3}{5}R + \dfrac{1}{4}R \times \dfrac{4}{5} = m\dfrac{v^2}{50}$ | | |
| $\dfrac{v^2}{50g} = \dfrac{4}{5} \div \dfrac{13}{20}$, $v = 24.55\ldots = 25$ or $24.6\ \text{ms}^{-1}$ | dM1,A1 (8) | Eliminate $R$ and $F$ to obtain equation for $v^2$. Depends on two previous M marks. Correct value of $v$, must be 2 or 3 sf |
**ALT:** Parallel to track: $F + mg\sin\theta = m(v^2/50)\cos\theta$ | M1A1, 50 or $r$
Perpendicular to track: $R - mg\cos\theta = m(v^2/50)\sin\theta$ | M1A1A1, 50 or $r$
---6. The path followed by a motorcycle round a circular race track is modelled as a horizontal circle of radius 50 m . The track is banked at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 3 } { 5 }$. 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.
\begin{enumerate}[label=(\alph*)]
\item Find the speed, in $\mathrm { m } \mathrm { s } ^ { - 1 }$, of the motorcycle.
In a refined model it is assumed that there is sideways friction. The coefficient of friction between the motorcycle tyres and the track is $\frac { 1 } { 4 }$. 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,
\item find the maximum speed, in $\mathrm { m } \mathrm { s } ^ { - 1 }$, at which the motorcycle can travel without slipping sideways.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2017 Q6 [13]}}