| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Mechanics (Further AS Paper 2 Mechanics) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Rotating disc with friction |
| Difficulty | Standard +0.3 This is a straightforward circular motion problem requiring application of F=mrω² with friction providing centripetal force. The calculation is direct with given values, requiring only unit conversion (grams to kg, cm to m) and rearrangement of a standard formula. While it's Further Maths content, the problem-solving is routine with no novel insight needed, making it slightly easier than average overall. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks |
|---|---|
| 9(a) | Draws a correct force diagram |
| Answer | Marks | Guidance |
|---|---|---|
| and friction and no other forces. | 1.1b | B1 |
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 9(b) | Recalls a correct expression for |
| Answer | Marks | Guidance |
|---|---|---|
| towards O | 1.2 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| r | 3.3 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| or v | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| AWRT 3.7 | 1.1b | A1 |
| Subtotal | 4 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 9(c) | States an appropriate modelling |
| Answer | Marks | Guidance |
|---|---|---|
| No air resistance forces. | 3.5a | E1 |
| Subtotal | 1 | |
| Question total | 6 | |
| Question Paper total | 40 |
Question 9:
--- 9(a) ---
9(a) | Draws a correct force diagram
showing weight, normal reaction
and friction and no other forces. | 1.1b | B1
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 9(b) ---
9(b) | Recalls a correct expression for
the acceleration or force
towards O | 1.2 | B1 | a = rω2
F = mrω2
0.01 = (0.0036)(0.2)ω2
0.01
ω =
0.0036×0.2
ω = 3.7
Forms the equation F = mrω2 or
mv2
F =
r | 3.3 | M1
Obtains a correct equation in ω
or v | 1.1b | A1
Obtains ω = 3.7
AWRT 3.7 | 1.1b | A1
Subtotal | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 9(c) ---
9(c) | States an appropriate modelling
assumption.
Eg
The coin is modelled as a
particle.
The radius is the distance to the
centre of mass of the coin from
O
No air resistance forces. | 3.5a | E1 | The coin is modelled as a particle.
Subtotal | 1
Question total | 6
Question Paper total | 40A small coin is placed at a point $C$ on a rough horizontal turntable, with centre $O$, as shown in the diagram below.
\includegraphics{figure_9}
The mass of the coin is 3.6 grams.
The distance $OC$ is 20 cm
The turntable rotates about a vertical axis through $O$, with constant angular speed $\omega$ radians per second.
\begin{enumerate}[label=(\alph*)]
\item Draw a diagram to show all the forces acting on the coin. [1 mark]
\item The maximum value of friction is 0.01 newtons and the coin does not slip during the motion.
Find the maximum value of $\omega$
Give your answer to two significant figures. [4 marks]
\item State one modelling assumption you have made to answer part (b). [1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2024 Q9 [6]}}