| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2021 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Ratio of tensions/forces |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring application of circular motion principles at arbitrary positions, energy conservation, and algebraic manipulation of tension ratios. While the techniques are standard (Newton's second law in circular motion, conservation of energy), the problem requires careful coordinate geometry with the angle θ, setting up two tension equations at symmetric positions, and solving the resulting system. The 6-mark allocation for part (a) indicates substantial working, and the need to find an arbitrary angle rather than standard positions (top/bottom) elevates difficulty above typical A-level questions. |
| Spec | 6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| 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) | 5ag |
| Answer | Marks | Guidance |
|---|---|---|
| A a | B1 | N2L |
| Answer | Marks | Guidance |
|---|---|---|
| B a | B1 | N2L |
| Answer | Marks | Guidance |
|---|---|---|
| 2 2 | M1 | Energy equation with correct number of terms. |
| v2 =5ag−4gacosθ | A1 | Accept multiplied by m and/or divided by a. |
| Use ratio of tensions = 9 : 5 | M1 | Use ratio and simplify to an expression in cosθ. |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | A1 | CAO |
| Answer | Marks |
|---|---|
| 6(b) | Greatest speed at lowest point |
| Answer | Marks | Guidance |
|---|---|---|
| 2 2 | M1 | Energy equation including lowest point, correct number |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | A1 FT | Ft their cosθ from part (a). |
| 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
9231/31 Cambridge International AS & A Level – Mark Scheme October/November 2021
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 2021 Page 5 of 12
Question | Answer | Marks | Guidance
--- 6(a) ---
6(a) | 5ag
At A T −mgcosθ=m×
A a | B1 | N2L
v2
At B T +mgcosθ=m×
B a | B1 | N2L
1 1
m×5ag− mv2 =mga×2cosθ
2 2 | M1 | Energy equation with correct number of terms.
v2 =5ag−4gacosθ | A1 | Accept multiplied by m and/or divided by a.
Use ratio of tensions = 9 : 5 | M1 | Use ratio and simplify to an expression in cosθ.
2
cosθ=
5 | A1 | CAO
6
--- 6(b) ---
6(b) | Greatest speed at lowest point
− 1 m×5ag+ 1 mV2 =mga×( 1−cosθ)
2 2 | M1 | Energy equation including lowest point, correct number
of terms.
31ag
V =
5 | A1 FT | Ft their cosθ from part (a).
2
Question | Answer | Marks | GuidanceA particle $P$, 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 particle $P$ moves in complete vertical circles about $O$ with the string taut. The points $A$ and $B$ are on the path of $P$ with $AB$ a diameter of the circle. $OA$ makes an angle $\theta$ with the downward vertical through $O$ and $OB$ makes an angle $\theta$ with the upward vertical through $O$. The speed of $P$ when it is at $A$ is $\sqrt{5ag}$.
The ratio of the tension in the string when $P$ is at $A$ to the tension in the string when $P$ is at $B$ is $9 : 5$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\cos \theta$. [6]
\item Find, in terms of $a$ and $g$, the greatest speed of $P$ during its motion. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q6 [8]}}