| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2022 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring simultaneous analysis of two equilibrium states with circular motion. Students must set up force equations for both particles in two different configurations, eliminate the tension variable, and solve a system involving trigonometric relationships. The multi-step reasoning, need to handle two coupled scenarios, and algebraic manipulation of the resulting equations place this significantly above average difficulty, though the individual techniques (resolving forces, circular motion equations) are standard for Further Maths. |
| Spec | 3.03k Connected particles: pulleys and equilibrium6.05a Angular velocity: definitions |
| Answer | Marks |
|---|---|
| 6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 6(a) | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | B1 | May be implied. |
| Tsin=mr2 =mhtan2 | B1 | Allow r for radius. |
| Answer | Marks |
|---|---|
| 7h | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 4 7h | M1 | Second scenario, equivalent result . |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | A1 | Combine convincingly to obtain given result. |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 6(b) | First scenario: N+Tcos=mg |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | B1 | Both. |
| Answer | Marks | Guidance |
|---|---|---|
| 7 7 | M1 | 12cos−6cos=7 |
| Answer | Marks |
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| 4 | A1 |
| Answer | Marks |
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| 7 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 6:
6 | Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear.
PMT
9231/32 Cambridge International AS & A Level – Mark Scheme October/November 2022
PUBLISHED
Abbreviations
AEF/OE Any Equivalent Form (of answer is equally acceptable) / Or Equivalent
AG Answer Given on the question paper (so extra checking is needed to ensure that the detailed working leading to the result is valid)
CAO Correct Answer Only (emphasising that no ‘follow through’ from a previous error is allowed)
CWO Correct Working Only
ISW Ignore Subsequent Working
SOI Seen Or Implied
SC Special Case (detailing the mark to be given for a specific wrong solution, or a case where some standard marking practice is to be varied in the
light of a particular circumstance)
WWW Without Wrong Working
AWRT Answer Which Rounds To
© UCLES 2022 Page 5 of 15
Question | Answer | Marks | Guidance
--- 6(a) ---
6(a) | 6
T = mg
7 | B1 | May be implied.
Tsin=mr2 =mhtan2 | B1 | Allow r for radius.
Radius of circle = htan
6g
[So 2 = cos]
7h | B1
9 6g
In second scenario, 2 = cos
4 7h | M1 | Second scenario, equivalent result .
6g 4 6g
Equate, cos= cos giving
7h 9 7h
4
cos= cos
9 | A1 | Combine convincingly to obtain given result.
AG
5
Question | Answer | Marks | Guidance
--- 6(b) ---
6(b) | First scenario: N+Tcos=mg
1
Second scenario, N +Tcos=mg
2 | B1 | Both.
6 12
Equate: mg− mgcos=2mg− mgcos
7 7 | M1 | 12cos−6cos=7
3
cos=
4 | A1
5
N = mg
7 | A1
4
Question | Answer | Marks | Guidance\includegraphics{figure_6}
A light inextensible string is threaded through a fixed smooth ring $R$ which is at a height $h$ above a smooth horizontal surface. One end of the string is attached to a particle $A$ of mass $m$. The other end of the string is attached to a particle $B$ of mass $\frac{1}{2}m$. The particle $A$ moves in a horizontal circle on the surface. The particle $B$ hangs in equilibrium below the ring and above the surface (see diagram).
When $A$ has constant angular speed $\omega$, the angle between $AR$ and $BR$ is $\theta$ and the normal reaction between $A$ and the surface is $N$.
When $A$ has constant angular speed $\frac{3}{2}\omega$, the angle between $AR$ and $BR$ is $\alpha$ and the normal reaction between $A$ and the surface is $\frac{1}{2}N$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\cos \theta = \frac{4}{9}\cos \alpha$. [5]
\item Find $N$ in terms of $m$ and $g$ and find the value of $\cos \alpha$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2022 Q6 [9]}}