| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2020 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Solid with removed cone from cone or cylinder |
| Difficulty | Standard +0.8 This is a standard Further Maths centre of mass problem requiring composite body techniques (subtracting volumes and moments) and equilibrium analysis. Part (a) involves routine application of the formula for centre of mass of composite bodies with known cone COM positions, requiring algebraic manipulation over 4 marks. Part (b) applies equilibrium geometry with a specific numerical value. While methodical and requiring careful bookkeeping, it follows standard procedures without novel insight, placing it moderately above average difficulty. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks |
|---|---|
| 4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw). |
| Answer | Marks |
|---|---|
| 4(a) | Volume Centre of mass from AB |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | B1 | For 9h/8 or 3h/8 (unsimplified) |
| Answer | Marks | Guidance |
|---|---|---|
| 3 6 8 6 8 | M1 A1 | Moments equation: Allow use of relative masses 1, 26, 27 |
| Answer | Marks |
|---|---|
| 26 | A1 |
| Answer | Marks |
|---|---|
| 4(b) | x |
| Answer | Marks |
|---|---|
| 3r | M1 |
| Answer | Marks |
|---|---|
| 8 | A1 |
| Answer | Marks |
|---|---|
| Volume | Centre of mass from AB |
| Small cone | 1 h |
| Answer | Marks |
|---|---|
| 3 2 | 1 h 9h |
| Answer | Marks |
|---|---|
| Large cone | 1 π( )2 3h |
| Answer | Marks |
|---|---|
| 3 2 | 1 3h 3h |
| Answer | Marks |
|---|---|
| Object | 26 π( )2 |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | x | |
| Question | Answer | Marks |
Question 4:
4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
--- 4(a) ---
4(a) | Volume Centre of mass from AB
1 h 1 h 9h
Small cone πr2. h+ . =
3 2 4 2 8
Large cone 1 π( 3r )2 . 3h 1 . 3h = 3h
3 2 4 2 8
26 π( )2
Object r h x
6 | B1 | For 9h/8 or 3h/8 (unsimplified)
Take moments about AB
13 27 3h 1 9h
πr2h.x = πr2h. − πr2h.
3 6 8 6 8 | M1 A1 | Moments equation: Allow use of relative masses 1, 26, 27
9h
x =
26 | A1
4
--- 4(b) ---
4(b) | x
tanθ=
3r | M1
3h 13
(= ) Use h= r
26r 4
3
tanθ=
8 | A1
2
Volume | Centre of mass from AB
Small cone | 1 h
πr2.
3 2 | 1 h 9h
h+ . =
4 2 8
Large cone | 1 π( )2 3h
3r .
3 2 | 1 3h 3h
. =
4 2 8
Object | 26 π( )2
r h
6 | x
Question | Answer | Marks | Guidance\includegraphics{figure_4}
The diagram shows the cross-section $ABCD$ of a uniform solid object which is formed by removing a cone with cross-section $DCE$ from the top of a larger cone with cross-section $ABE$. The perpendicular distance between $AB$ and $DC$ is $h$, the diameter $AB$ is $6r$ and the diameter $DC$ is $2r$.
\begin{enumerate}[label=(\alph*)]
\item Find an expression, in terms of $h$, for the distance of the centre of mass of the solid object from $AB$. [4]
\end{enumerate}
The object is freely suspended from the point $B$ and hangs in equilibrium. The angle between $AB$ and the downward vertical through $B$ is $\theta$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Given that $h = \frac{13}{4}r$, find the value of $\tan\theta$. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q4 [6]}}