| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Mechanics (Further AS Paper 2 Mechanics) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle inside smooth hollow cylinder |
| Difficulty | Standard +0.3 This is a straightforward circular motion problem requiring standard mechanics techniques: resolving forces vertically (friction = weight), applying F=mrω² horizontally, and stating a modeling assumption. All steps are routine for Further Maths students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks |
|---|---|
| 4(a) | Draws an accurate force diagram |
| Answer | Marks | Guidance |
|---|---|---|
| reaction | 3.3 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| F = 50g = 490 N | 2.4 | E1 |
| Answer | Marks |
|---|---|
| 4(b) | Uses correct formula with either v |
| Answer | Marks | Guidance |
|---|---|---|
| the acceleration | 3.4 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Must use 980 for the reaction | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| AWRT 2.1 | 1.1b | A1 |
| Answer | Marks |
|---|---|
| 4(c) | States one appropriate modelling |
| Answer | Marks | Guidance |
|---|---|---|
| The radius is exactly 4.6 metres | 3.5a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| radius | 2.4 | A1 |
| Total | 7 | |
| Q | Marking Instructions | AO |
Question 4:
--- 4(a) ---
4(a) | Draws an accurate force diagram
with vertical forces of weight and
friction identified
Condone omission of normal
reaction | 3.3 | B1 | To stop sliding down the wall friction
must equal the weight of Stephi
F = 50g = 490 N
Explains that friction and Stephi’s
weight must be equal to remain in
equilibrium and deduces that
F = 50g = 490 N | 2.4 | E1
--- 4(b) ---
4(b) | Uses correct formula with either v
orωto obtain an expression for the
magnitude of the resultant force or
the acceleration | 3.4 | B1 | Force towards centre of circle =
mrω2
980=50(4.6)ω2
980
ω=
50×4.6
ω=2.064.....
ω=2.1
Forms a correct equation in v or
ωinvolving the normal reaction
Must use 980 for the reaction | 1.1a | M1
Obtain the correct value of ω
AWRT 2.1 | 1.1b | A1
--- 4(c) ---
4(c) | States one appropriate modelling
assumption
Accept
No air resistance
or
The radius is exactly 4.6 metres | 3.5a | M1 | Stephi is modelled as particle.
This means that 4.6 metres can be
used as the radius of the circle,
which would not be the case
otherwise.
Provides correct reasoning about
their modelling assumption
Accept
Without air resistance there is no
horizontal frictional force
or
There is no need to consider
variations in the value of the
radius | 2.4 | A1
Total | 7
Q | Marking Instructions | AO | Marks | Typical SolutionIn this question use $g = 9.8\,\text{m}\,\text{s}^{-2}$
A ride in a fairground consists of a hollow vertical cylinder of radius 4.6 metres with a horizontal floor.
Stephi, who has mass 50 kilograms, stands inside the cylinder with her back against the curved surface.
The cylinder begins to rotate about a vertical axis through the centre of the cylinder.
When the cylinder is rotating at a constant angular speed of $\omega$ radians per second, the magnitude of the normal reaction between Stephi and the curved surface is 980 newtons.
The floor is lowered and Stephi remains against the curved surface with her feet above the floor, as shown in the diagram.
\includegraphics{figure_4}
\begin{enumerate}[label=(\alph*)]
\item Explain, with the aid of a force diagram, why the magnitude of the frictional force acting on Stephi is 490 newtons.
[2 marks]
\item Find $\omega$
[3 marks]
\item State one modelling assumption that you have used in this question.
Explain the effect of this assumption.
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics 2019 Q4 [7]}}