| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Mechanics (Further Paper 3 Mechanics) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: string becomes slack |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring energy conservation and circular motion analysis with string tension becoming zero at a specific angle. Part (a) involves multiple steps: applying energy conservation from bottom to point C, using the condition that tension becomes zero (centripetal force equals weight component), and algebraic manipulation to reach the given result. Part (b) tests understanding of modelling assumptions. While conceptually demanding and requiring careful coordinate geometry with the 30° angle, it follows a standard framework for vertical circular motion problems that Further Maths students practice extensively. The 'show that' format provides a target, reducing difficulty slightly. |
| Spec | 6.05e Radial/tangential acceleration6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| 9(a) | Resolves radially at the point C. | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| point C.. | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| U | 1.1a | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| two points and change in GPE. | 3.4 | M1 |
| Obtains a correct energy equation. | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| AG | 2.1 | R1 |
| Subtotal | 6 | |
| Q | Marking Instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 9(b) | States that air resistance has been | |
| ignored. | 3.5a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | 2.2b | R1 |
| Subtotal | 2 | |
| Question Total | 8 |
Question 9:
--- 9(a) ---
9(a) | Resolves radially at the point C. | 3.3 | M1 | mv2
=mgcos30
a
ag 3
v2 = agcos30=
2
1 1
mU2 = mv2 +mga(1+cos30)
2 2
U2 =v2 +2ga(1+cos30)
ag 3 3
U2 = +2ga1+
2 2
3
U2 = ag +2+ 3
2
3
U2 = ag2+3
2
ag ( )
U2 = 4+3 3
2
Obtains a correct expression for
the speed or speed squared at the
point C.. | 1.1b | A1
Recognises that the initial speed is
U | 1.1a | B1
Uses conservation of energy to
form a three-term equation to find
the speed or speed squared of the
particle at the point C.
Must see consideration of KE at
two points and change in GPE. | 3.4 | M1
Obtains a correct energy equation. | 1.1b | A1
Completes a reasoned argument to
a g ( )
obtain U 2 = 4 + 3 3
2
AG | 2.1 | R1
Subtotal | 6
Q | Marking Instructions | AO | Marks | Typical Solution
--- 9(b) ---
9(b) | States that air resistance has been
ignored. | 3.5a | M1 | As air resistance has been ignored,
some energy would be lost so the
value of U2 must be greater than
the value calculated.
ag( )
U2 > 4+3 3
2
a g ( )
States U 2 > 4 + 3 3 .
2
ag( )
Condone U2 4+3 3
2 | 2.2b | R1
Subtotal | 2
Question Total | 8A small sphere, of mass $m$, is attached to one end of a light inextensible string of length $a$
The other end of the string is attached to a fixed point $O$
The sphere is at rest in equilibrium directly below $O$ when it is struck, giving it a horizontal impulse of magnitude $mU$
After the impulse, the sphere follows a circular path in a vertical plane containing the point $O$ until the string becomes slack at the point $C$
At $C$ the string makes an angle of 30° with the upward vertical through $O$, as shown in the diagram below.
\includegraphics{figure_9}
\begin{enumerate}[label=(\alph*)]
\item Show that
$$U^2 = \frac{ag}{2}\left(4 + 3\sqrt{3}\right)$$
where $g$ is the acceleration due to gravity.
[6 marks]
\item With reference to any modelling assumptions that you have made, explain why giving your answer as an inequality would be more appropriate, and state this inequality.
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2024 Q9 [8]}}