| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Mechanics (Further AS Paper 2 Mechanics) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Horizontal circular track – friction only (no banking) |
| Difficulty | Standard +0.3 This is a straightforward application of circular motion with friction providing centripetal force (F = mv²/r). Part (a) requires identifying that the safe speed uses the minimum friction value, then converting units. Part (b) tests understanding that wet conditions reduce maximum friction. Standard Further Maths mechanics with clear setup and routine calculations, slightly above average due to the reasoning/interpretation elements. |
| Spec | 6.01a Dimensions: M, L, T notation6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Force towards centre of circle \(= \frac{990v^2}{48}\) | M1 | Uses correct formula to obtain expression for magnitude of resultant force or acceleration |
| Friction \(= \frac{990v^2}{48}\) | A1 | Forms correct equation involving friction |
| \(v = \sqrt{\frac{48(7300)}{990}}\) | M1 | Deduces value of \(F\) to be used and substitutes to find maximum safe speed in m s\(^{-1}\) |
| \(v = 18.8\ldots\) m s\(^{-1}\), so \(v = 42\) mph | A1 | Obtains correct maximum safe speed converting to mph; AWRT 40 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Wet conditions reduce friction | E1 | Infers that on a wet day friction would be reduced |
| \(10000\text{ N} > 9200\text{ N}\), Gary's assumption is wrong | E1 | Infers that 10000 N is incorrect and concludes Gary's revised assumption is wrong |
## Question 5(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Force towards centre of circle $= \frac{990v^2}{48}$ | M1 | Uses correct formula to obtain expression for magnitude of resultant force or acceleration |
| Friction $= \frac{990v^2}{48}$ | A1 | Forms correct equation involving friction |
| $v = \sqrt{\frac{48(7300)}{990}}$ | M1 | Deduces value of $F$ to be used and substitutes to find maximum safe speed in m s$^{-1}$ |
| $v = 18.8\ldots$ m s$^{-1}$, so $v = 42$ mph | A1 | Obtains correct maximum safe speed converting to mph; AWRT 40 |
## Question 5(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Wet conditions reduce friction | E1 | Infers that on a wet day friction would be reduced |
| $10000\text{ N} > 9200\text{ N}$, Gary's assumption is wrong | E1 | Infers that 10000 N is incorrect and concludes Gary's revised assumption is wrong |
---5 A car travels around a roundabout at a constant speed. The surface of the roundabout is horizontal.
The car has mass 990 kg and the path of the car is a circular arc of radius 48 metres.\\
A simple model assumes that the car is a particle and the only horizontal force acting on it as it travels around the roundabout is friction.
On a dry day typical values of friction, $F$, between the surface of the roundabout and the tyres of the car are
$$7300 \mathrm {~N} \leq F \leq 9200 \mathrm {~N}$$
5
\begin{enumerate}[label=(\alph*)]
\item Using this model calculate a safe speed limit, in miles per hour, for the car as it travels around the roundabout.
Explain your reasoning fully.\\
Note that there are 1600 metres in one mile.\\
5
\item Gary assumes that on a wet day typical values for friction, $F$, are
$$5400 \mathrm {~N} \leq F \leq 10000 \mathrm {~N}$$
Comment on the validity of Gary's revised assumption.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2018 Q5 [6]}}