| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2020 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Composite solid with hemisphere and cylinder/cone |
| Difficulty | Standard +0.8 This is a standard Further Maths centre of mass problem requiring calculation of individual centres of mass for cone and cylinder, then using the composite formula. Part (b) requires checking the equilibrium condition (vertical line through COM falls within base of support), which is a routine application once part (a) is complete. The multi-step nature and Further Maths context place it above average, but it follows a well-established template without requiring novel geometric insight. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Cone volume \(= \frac{1}{3}\pi(3r)^2 \cdot 4r\), centre of mass from base \(= 4r+r\); Cylinder volume \(= \pi(3r)^2 \cdot 4r\), centre of mass \(= 2r\); Combined volume \(= \frac{4}{3}\pi(3r)^2 \cdot 4r\), centre \(= \bar{x}\) | B1 | Distances correct |
| Taking moments about base of cylinder: \(\bar{x} \cdot \frac{4}{3}\pi(3r)^2 \cdot 4r = \frac{1}{3}\pi(3r)^2 \cdot 4r \cdot 5r + \pi(3r)^2 \cdot 4r \cdot 2r\) | M1 A1 | Moments equation |
| \(\bar{x} = \frac{11}{4}r\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Condition: \(OG\cos\theta < OA\) (where \(O\) is vertex of cone and \(OA\) is slant height of cone) | B1 | Correct condition for equilibrium |
| \(\left(4r + \frac{5r}{4}\right) \times \frac{4}{5} < 5r\) | M1 | Expression in terms of \(r\) |
| \(21 < 25\) True | A1 | Correct conclusion, with correct working |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Cone volume $= \frac{1}{3}\pi(3r)^2 \cdot 4r$, centre of mass from base $= 4r+r$; Cylinder volume $= \pi(3r)^2 \cdot 4r$, centre of mass $= 2r$; Combined volume $= \frac{4}{3}\pi(3r)^2 \cdot 4r$, centre $= \bar{x}$ | B1 | Distances correct |
| Taking moments about base of cylinder: $\bar{x} \cdot \frac{4}{3}\pi(3r)^2 \cdot 4r = \frac{1}{3}\pi(3r)^2 \cdot 4r \cdot 5r + \pi(3r)^2 \cdot 4r \cdot 2r$ | M1 A1 | Moments equation |
| $\bar{x} = \frac{11}{4}r$ | A1 | |
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## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Condition: $OG\cos\theta < OA$ (where $O$ is vertex of cone and $OA$ is slant height of cone) | B1 | Correct condition for equilibrium |
| $\left(4r + \frac{5r}{4}\right) \times \frac{4}{5} < 5r$ | M1 | Expression in terms of $r$ |
| $21 < 25$ True | A1 | Correct conclusion, with correct working |
---3 An object consists of a uniform solid circular cone, of vertical height $4 r$ and radius $3 r$, and a uniform solid cylinder, of height $4 r$ and radius $3 r$. The circular base of the cone and one of the circular faces of the cylinder are joined together so that they coincide. The cone and the cylinder are made of the same material.
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of the object from the end of the cylinder that is not attached to the cone.
\item Show that the object can rest in equilibrium with the curved surface of the cone in contact with a horizontal surface.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q3 [7]}}