| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2023 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Composite solid with hemisphere and cylinder/cone |
| Difficulty | Challenging +1.2 This is a standard two-part Further Maths mechanics question. Part (a) uses energy conservation and circular motion equations (routine for FM students). Part (b) applies Newton's second law at point A. Both parts follow well-established solution methods with no novel insights required, though the algebra requires care. Typical of Further Maths Paper 3 standard questions. |
| Spec | 6.02j Conservation with elastics: springs and strings6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At \(B\), \(mg\sin\theta = \dfrac{m4ag}{5a}\) | M1 | Allow cos instead of sin for M1. Do not award until tension \(= 0\) used. Mass must be seen. No sign error. |
| \(\sin\theta = \dfrac{4}{5}\) | A1 | |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At \(A\), \(T - mg\cos\theta = \dfrac{mu^2}{a}\) | B1 | |
| Energy: \(\frac{1}{2}mu^2 - \frac{1}{2}m\times\dfrac{4ag}{5} = mga(\cos\theta+\sin\theta)\) | M1 A1 | Energy equation with 4 terms, dimensionally correct. Mass must be present, allow sign errors. Must see \(\frac{1}{2}\) in the KE terms. |
| Solve to find \(T\) | M1 | Complete method leading to an expression in \(mg\) for \(T\). |
| \(T = \dfrac{21}{5}mg\) | A1 | CWO |
| 5 |
## Question 3:
**Part (a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| At $B$, $mg\sin\theta = \dfrac{m4ag}{5a}$ | M1 | Allow cos instead of sin for M1. Do not award until tension $= 0$ used. Mass must be seen. No sign error. |
| $\sin\theta = \dfrac{4}{5}$ | A1 | |
| | **2** | |
**Part (b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| At $A$, $T - mg\cos\theta = \dfrac{mu^2}{a}$ | B1 | |
| Energy: $\frac{1}{2}mu^2 - \frac{1}{2}m\times\dfrac{4ag}{5} = mga(\cos\theta+\sin\theta)$ | M1 A1 | Energy equation with 4 terms, dimensionally correct. Mass must be present, allow sign errors. Must see $\frac{1}{2}$ in the KE terms. |
| Solve to find $T$ | M1 | Complete method leading to an expression in $mg$ for $T$. |
| $T = \dfrac{21}{5}mg$ | A1 | CWO |
| | **5** | |
---3 A 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 $P$ is held at the point $A$, where $O A$ makes an angle $\theta$ with the downward vertical through $O$, and with the string taut. The particle $P$ is projected perpendicular to $O A$ in an upwards direction with speed $u$. It then starts to move along a circular path in a vertical plane. The string goes slack when $P$ is at $B$, where angle $A O B$ is $90 ^ { \circ }$ and the speed of $P$ is $\sqrt { \frac { 4 } { 5 } \mathrm { ag } }$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $\sin \theta$.
\item Find, in terms of $m$ and $g$, the tension in the string when $P$ is at $A$.\\
\includegraphics[max width=\textwidth, alt={}, center]{454be64a-204f-4fa4-a5fc-72fd88e1289f-06_846_767_258_689}
An object is formed from a solid hemisphere, of radius $2 a$, and a solid cylinder, of radius $a$ and height $d$. The hemisphere and the cylinder are made of the same material. The cylinder is attached to the plane face of the hemisphere. The line $O C$ forms a diameter of the base of the cylinder, where $C$ is the centre of the plane face of the hemisphere and $O$ is common to both circumferences (see diagram). Relative to axes through $O$, parallel and perpendicular to $O C$ as shown, the centre of mass of the object is ( $\mathrm { x } , \mathrm { y }$ ).
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2023 Q3 [7]}}