| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Mechanics (Further Paper 3 Mechanics) |
| Year | 2024 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Conical pendulum – particle on horizontal surface |
| Difficulty | Moderate -0.8 This is a straightforward circular motion question requiring direct application of standard formulas: a = v²/r, T = ma, and ω = v/r with unit conversion. All three parts are routine calculations with no problem-solving or conceptual challenges, making it easier than average even for Further Maths students. |
| Spec | 6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks |
|---|---|
| 4(a) | States or uses the formula for the |
| Answer | Marks | Guidance |
|---|---|---|
| PI by correct answer. | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Condone missing units. | 1.1b | A1 |
| Subtotal | 2 | |
| Q | Marking Instructions | AO |
| Answer | Marks |
|---|---|
| 4(b) | Obtains 60 |
| Answer | Marks | Guidance |
|---|---|---|
| Condone missing units. | 1.1b | B1F |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Q | Marking Instructions | AO |
| Answer | Marks |
|---|---|
| 4(c) | Selects and uses a method to find |
| Answer | Marks | Guidance |
|---|---|---|
| second) or 300 (rad per minute). | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| AG | 2.1 | R1 |
| Subtotal | 2 | |
| Question total | 5 | |
| Q | Marking Instructions | AO |
Question 4:
--- 4(a) ---
4(a) | States or uses the formula for the
acceleration under circular motion.
PI by correct answer. | 1.1a | M1 | 4 2
a = = 2 0 m s – 2
0 . 8
Obtains 20
Condone missing units. | 1.1b | A1
Subtotal | 2
Q | Marking Instructions | AO | Marks | Typical Solution
--- 4(b) ---
4(b) | Obtains 60
Follow through their acceleration
Condone missing units. | 1.1b | B1F | T = 3 × 20
= 60 N
Subtotal | 1
Q | Marking Instructions | AO | Marks | Typical Solution
--- 4(c) ---
4(c) | Selects and uses a method to find
the angular speed for example
seeing 5 (rad s-1) 0.79 (revs per
second) or 300 (rad per minute). | 1.1a | M1 | 4×60
=47.746....
2π×0.8
=48 rpm
Completes a reasoned argument to
obtain 48.
Condone missing units.
AG | 2.1 | R1
Subtotal | 2
Question total | 5
Q | Marking Instructions | AO | Marks | Typical SolutionA particle of mass 3 kg is attached to one end of a light inextensible string.
The other end of the string is attached to a fixed point on a smooth horizontal surface.
The particle is set into motion so that it moves with a constant speed 4 m s$^{-1}$ in a circular path with radius 0.8 metres on the horizontal surface.
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of the particle.
[2 marks]
\item Find the tension in the string.
[1 mark]
\item Show that the angular speed of the particle is 48 revolutions per minute correct to two significant figures.
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2024 Q4 [5]}}