CAIE M2 Specimen — Question 6 9 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
SessionSpecimen
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeConical or hemispherical shell composite
DifficultyChallenging +1.2 This is a standard composite centre of mass problem requiring knowledge of standard results (hemisphere shell CM at r/2, cone CM at h/4 from base), application of the toppling condition (vertical line through G passes through contact point), and algebraic manipulation. Part (i) involves geometric reasoning and combining CM formulas; part (ii) requires using moments about a point. While multi-step, it follows a predictable pattern for M2 composite body questions with no novel insight required.
Spec6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces

6 \includegraphics[max width=\textwidth, alt={}, center]{add3948c-3b45-4e67-ac84-e2ca935afd64-08_442_953_237_596} An object is formed by joining a hemispherical shell of radius 0.2 m and a solid cone with base radius 0.2 m and height \(h \mathrm {~m}\) along their circumferences. The centre of mass, \(G\), of the object is \(d \mathrm {~m}\) from the vertex of the cone on the axis of symmetry of the object. The object rests in equilibrium on a horizontal plane, with the curved surface of the cone in contact with the plane (see diagram). The object is on the point of toppling.
  1. Show that \(d = h + \frac { 0.04 } { h }\).
  2. It is given that the cone is uniform and of weight 4 N , and that the hemispherical shell is uniform and of weight \(W \mathrm {~N}\). Given also that \(h = 0.8\), find \(W\).

Question 6(i):
AnswerMarks Guidance
AnswerMarks Guidance
\(d\cos\theta = h/\cos\theta\)1 M1
\(\cos\theta = \dfrac{h}{\sqrt{0.2^2 + h^2}}\), \(d = \dfrac{h}{(h^2/(0.04 + h^2))}\)1 M1
\(d = h + \dfrac{0.04}{h}\) AG1 A1
Total3
Question 6(ii):
AnswerMarks Guidance
AnswerMarks Guidance
\(0.6 \times 4 + 0.9W = d(4 + W)\)2 M1, A1
\(d = 0.8 + \dfrac{0.2^2}{0.8}\)1 B1
\(2.4 + 0.9W = 0.85(4 + W)\)1 M1
\(0.05W = 1\)1 A1
\(W = 20\)1 A1
Total6
## Question 6(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $d\cos\theta = h/\cos\theta$ | 1 | M1 | $\theta$ = semi-vertical angle |
| $\cos\theta = \dfrac{h}{\sqrt{0.2^2 + h^2}}$, $d = \dfrac{h}{(h^2/(0.04 + h^2))}$ | 1 | M1 | |
| $d = h + \dfrac{0.04}{h}$ AG | 1 | A1 | |
| **Total** | **3** | | |

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## Question 6(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.6 \times 4 + 0.9W = d(4 + W)$ | 2 | M1, A1 | Table of moments idea |
| $d = 0.8 + \dfrac{0.2^2}{0.8}$ | 1 | B1 | 0.85 |
| $2.4 + 0.9W = 0.85(4 + W)$ | 1 | M1 | |
| $0.05W = 1$ | 1 | A1 | |
| $W = 20$ | 1 | A1 | |
| **Total** | **6** | | |

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6\\
\includegraphics[max width=\textwidth, alt={}, center]{add3948c-3b45-4e67-ac84-e2ca935afd64-08_442_953_237_596}

An object is formed by joining a hemispherical shell of radius 0.2 m and a solid cone with base radius 0.2 m and height $h \mathrm {~m}$ along their circumferences. The centre of mass, $G$, of the object is $d \mathrm {~m}$ from the vertex of the cone on the axis of symmetry of the object. The object rests in equilibrium on a horizontal plane, with the curved surface of the cone in contact with the plane (see diagram). The object is on the point of toppling.\\
(i) Show that $d = h + \frac { 0.04 } { h }$.\\

(ii) It is given that the cone is uniform and of weight 4 N , and that the hemispherical shell is uniform and of weight $W \mathrm {~N}$. Given also that $h = 0.8$, find $W$.\\

\hfill \mbox{\textit{CAIE M2  Q6 [9]}}