| 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 +0.8 This is a multi-step centre of mass problem requiring similar triangles to find the truncated height, volume calculations for composite solids (cone minus cylinder), and application of the centre of mass formula for removed volumes. While the techniques are standard for M2, the problem requires careful geometric reasoning and algebraic manipulation across multiple steps, making it moderately challenging but not requiring novel insight. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Height of conical tip \(= 1.2 \times \frac{0.2}{0.5} = 0.48\) | M1 | Use ratio of corresponding sides, similar figures |
| Cylindrical height \(= 1.2 - 0.48 = 0.72\) | A1 | AG |
| Volume removed \(= \pi 0.2^2 \times \frac{0.48}{3} + \pi 0.2^2 \times 0.72\) \((= 0.0064\pi + 0.0288\pi)\) | M1 | |
| Volume removed \(= 0.0352\pi\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Moment of cone removed about the base \(= 0.0064\pi(0.72 + \frac{0.48}{4}) = 0.0064\pi \times 0.84\) | B1 | |
| Moment of cylinder removed about the base \(= 0.0288\pi \times \frac{0.72}{2} = 0.0288\pi \times 0.36\) | B1 | |
| Moment of the original cone about the base \(= \pi \times 0.5^2 \times \frac{1.2}{3} \times 0.3 = 0.1\pi \times 0.3\) | B1 | |
| M1 | Attempt to take moments about the base | |
| \(0.1\pi \times 0.3 = 0.0064\pi \times 0.84 + 0.0288\pi \times 0.36 + 0.0648\pi x\) | A1 | Note \(0.0648\pi\) is the volume of the object |
| \(x = 0.22\) m | A1 |
## Question 7(i):
Height of conical tip $= 1.2 \times \frac{0.2}{0.5} = 0.48$ | M1 | Use ratio of corresponding sides, similar figures
Cylindrical height $= 1.2 - 0.48 = 0.72$ | A1 | AG
Volume removed $= \pi 0.2^2 \times \frac{0.48}{3} + \pi 0.2^2 \times 0.72$ $(= 0.0064\pi + 0.0288\pi)$ | M1 |
Volume removed $= 0.0352\pi$ | A1 | AG
**Total: 4 marks**
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## Question 7(ii):
Moment of cone removed about the base $= 0.0064\pi(0.72 + \frac{0.48}{4}) = 0.0064\pi \times 0.84$ | B1 |
Moment of cylinder removed about the base $= 0.0288\pi \times \frac{0.72}{2} = 0.0288\pi \times 0.36$ | B1 |
Moment of the original cone about the base $= \pi \times 0.5^2 \times \frac{1.2}{3} \times 0.3 = 0.1\pi \times 0.3$ | B1 |
| M1 | Attempt to take moments about the base
$0.1\pi \times 0.3 = 0.0064\pi \times 0.84 + 0.0288\pi \times 0.36 + 0.0648\pi x$ | A1 | Note $0.0648\pi$ is the volume of the object
$x = 0.22$ m | A1 |
**Total: 6 marks**7\\
\includegraphics[max width=\textwidth, alt={}, center]{8dda6c21-7cb5-43b6-9a34-485bdf4042c4-12_732_581_260_774}
A uniform solid cone has height 1.2 m and base radius 0.5 m . A uniform object is made by drilling a cylindrical hole of radius 0.2 m through the cone along the axis of symmetry (see diagram).\\
(i) Show that the height of the object is 0.72 m and that the volume of the cone removed by the drilling is $0.0352 \pi \mathrm {~m} ^ { 3 }$.\\[0pt]
[The volume of a cone is $\frac { 1 } { 3 } \pi r ^ { 2 } h$.]\\
(ii) Find the distance of the centre of mass of the object from its base.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE M2 2018 Q7 [10]}}