| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Solid with removed cone from cone or cylinder |
| Difficulty | Standard +0.3 This is a standard centre of mass problem requiring similar triangles to find the truncated height, volume calculation using given formulas, and then applying the composite body method (original cone minus removed cone). While it involves multiple steps and careful bookkeeping, the techniques are routine for M2 students with no novel problem-solving insight required. |
| Spec | 6.04b Find centre of mass: using symmetry6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks |
|---|---|
| 7(i) | 0.2 |
| Answer | Marks | Guidance |
|---|---|---|
| 0.5 | M1 | Use ratio of corresponding |
| Answer | Marks | Guidance |
|---|---|---|
| Cylindrical height = 1.2 – 0.48 = 0.72 | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| (=0.0064π+0.0288π) | M1 | |
| Volume removed = 0.0352π | A1 | AG |
| Answer | Marks |
|---|---|
| 7(ii) | Moment of cone removed about the base |
| Answer | Marks |
|---|---|
| 4 | B1 |
| Answer | Marks |
|---|---|
| 2 | B1 |
| Answer | Marks |
|---|---|
| 3 | B1 |
| M1 | Attempt to take moments |
| Answer | Marks | Guidance |
|---|---|---|
| 0.0648πx | A1 | Note 0.0648π is the volume |
| Answer | Marks |
|---|---|
| x = 0.22 m | A1 |
Question 7:
--- 7(i) ---
7(i) | 0.2
Height of conical tip =1.2× = 0.48
0.5 | M1 | Use ratio of corresponding
sides, similar figures
Cylindrical height = 1.2 – 0.48 = 0.72 | A1 | AG
0.48
Volume removed=π0.22× +π0.22×0.72
3
(=0.0064π+0.0288π) | M1
Volume removed = 0.0352π | A1 | AG
4
--- 7(ii) ---
7(ii) | Moment of cone removed about the base
0.48
=0.0064π(0.72+ )=0.0064π×0.84
4 | B1
Moment of cylinder removed about the base
0.72
=0.0288π× =0.0288π×0.36
2 | B1
Moment of the original cone about the base
1.2
=π0.52× ×0.3=0.1π×0.3
3 | B1
M1 | Attempt to take moments
about the base
0.1π×0.3=0.0064π×0.84+0.0288π×0.36+
0.0648πx | A1 | Note 0.0648π is the volume
of the object
x = 0.22 m | A1
6\includegraphics{figure_7}
A uniform solid cone has height $1.2 \text{ m}$ and base radius $0.5 \text{ m}$. A uniform object is made by drilling a cylindrical hole of radius $0.2 \text{ m}$ through the cone along the axis of symmetry (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Show that the height of the object is $0.72 \text{ m}$ and that the volume of the cone removed by the drilling is $0.0352\pi \text{ m}^3$. [4]
\end{enumerate}
[The volume of a cone is $\frac{1}{3}\pi r^2 h$.]
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the distance of the centre of mass of the object from its base. [6]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2018 Q7 [10]}}