CAIE M2 2018 November — Question 2 6 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2018
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeCentre of mass with variable parameter
DifficultyStandard +0.8 This is a two-part centre of mass problem requiring geometric visualization, application of toppling conditions (vertical line through CM passes through pivot point), and composite body calculations. Part (i) involves standard CM formulas with given dimensions, while part (ii) requires setting up and solving an equation for an unknown dimension. The conceptual understanding of toppling equilibrium and 3D geometry makes this moderately challenging, though the mathematical techniques are straightforward once the setup is clear.
Spec6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces

\includegraphics{figure_2} A uniform object is made by attaching the base of a solid hemisphere to the base of a solid cone so that the object has an axis of symmetry. The base of the cone has radius \(0.3\text{ m}\), and the hemisphere has radius \(0.2\text{ m}\). The object is placed on a horizontal plane with a point \(A\) on the curved surface of the hemisphere and a point \(B\) on the circumference of the cone in contact with the plane (see diagram).
  1. Given that the object is on the point of toppling about \(B\), find the distance of the centre of mass of the object from the base of the cone. [3]
  2. Given instead that the object is on the point of toppling about \(A\), calculate the height of the cone. [3]
[The volume of a cone is \(\frac{1}{3}\pi r^2 h\). The volume of a hemisphere is \(\frac{2}{3}\pi r^3\).]

Question 2:

AnswerMarks Guidance
2(i)cosθ = 0.2/0.3 B1
tanθ = x/0.3M1
x = 0.335(41..)A1
3

AnswerMarks Guidance
2(ii)M1 Attempt to take moments about A
(π0.32h/3)×(h/4) = (2π0.23/3)(3×0.2/8)A1
h = 0.231A1
3
AnswerMarks Guidance
QuestionAnswer Marks
Question 2:
--- 2(i) ---
2(i) | cosθ = 0.2/0.3 | B1 | Axis makes an angle θ with the horizontal
tanθ = x/0.3 | M1
x = 0.335(41..) | A1
3
--- 2(ii) ---
2(ii) | M1 | Attempt to take moments about A
(π0.32h/3)×(h/4) = (2π0.23/3)(3×0.2/8) | A1
h = 0.231 | A1
3
Question | Answer | Marks | Guidance
\includegraphics{figure_2}

A uniform object is made by attaching the base of a solid hemisphere to the base of a solid cone so that the object has an axis of symmetry. The base of the cone has radius $0.3\text{ m}$, and the hemisphere has radius $0.2\text{ m}$. The object is placed on a horizontal plane with a point $A$ on the curved surface of the hemisphere and a point $B$ on the circumference of the cone in contact with the plane (see diagram).

\begin{enumerate}[label=(\roman*)]
\item Given that the object is on the point of toppling about $B$, find the distance of the centre of mass of the object from the base of the cone. [3]

\item Given instead that the object is on the point of toppling about $A$, calculate the height of the cone. [3]
\end{enumerate}

[The volume of a cone is $\frac{1}{3}\pi r^2 h$. The volume of a hemisphere is $\frac{2}{3}\pi r^3$.]

\hfill \mbox{\textit{CAIE M2 2018 Q2 [6]}}