| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2021 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Difficulty | Standard +0.8 This is a 3D circular motion problem requiring resolution of forces in multiple directions, application of Newton's second law for circular motion, and careful geometric reasoning with the string constraint. While the setup is standard for Further Maths mechanics, coordinating the vertical equilibrium of B, the circular motion of A, and the normal reaction condition requires systematic multi-step analysis beyond routine exercises. |
| Spec | 3.03n Equilibrium in 2D: particle under forces6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| 3(a) | For B: T =mg | B1 |
| For A: R+Tcosθ=3mg | M1 | All 3 terms, allow sign errors, allow sin/cos mix. |
| Answer | Marks |
|---|---|
| 5 | A1 |
| Answer | Marks |
|---|---|
| 3(b) | 3mv2 |
| Answer | Marks | Guidance |
|---|---|---|
| r | M1 | May be seen in part (a), allow sin/cos mix. |
| r= ARsinθ | B1 | Or equivalent. |
| Answer | Marks |
|---|---|
| 4 4 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 3:
--- 3(a) ---
3(a) | For B: T =mg | B1 | May be embedded.
For A: R+Tcosθ=3mg | M1 | All 3 terms, allow sign errors, allow sin/cos mix.
3
Use given R to obtain cosθ=
5 | A1
3
--- 3(b) ---
3(b) | 3mv2
Tsinθ=
r | M1 | May be seen in part (a), allow sin/cos mix.
r= ARsinθ | B1 | Or equivalent.
3a 1
[Combine to give AR= , so] BR= a
4 4 | A1
3
Question | Answer | Marks | Guidance\includegraphics{figure_3}
Particles $A$ and $B$, of masses $3m$ and $m$ respectively, are connected by a light inextensible string of length $a$ that passes through a fixed smooth ring $R$. Particle $B$ hangs in equilibrium vertically below the ring. Particle $A$ moves in horizontal circles on a smooth horizontal surface with speed $\frac{2}{3}\sqrt{ga}$. The angle between $AR$ and $BR$ is $\theta$ (see diagram). The normal reaction between $A$ and the surface is $\frac{15}{2}mg$.
\begin{enumerate}[label=(\alph*)]
\item Find $\cos \theta$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q3 [3]}}