CAIE Further Paper 3 2020 June — Question 4 4 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2020
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeComposite solid with cone and cylinder
DifficultyStandard +0.3 This is a standard centre of mass calculation for composite solids using the formula x̄ = Σ(mx)/Σm. Students need to recall the centre of mass positions for a cone (3h/4 from vertex) and cylinder (at its midpoint), calculate volumes (πr²h/3 for cone, πr²h for cylinder), then combine using the given dimensions. It's methodical bookwork with straightforward algebra, typical of Further Maths mechanics but requiring no novel insight—slightly easier than average due to its routine nature.
Spec6.04c Composite bodies: centre of mass
\includegraphics{figure_4} A uniform solid circular cone, of vertical height \(4r\) and radius \(2r\), is attached to a uniform solid cylinder, of height \(3r\) and radius \(kr\), where \(k\) is a constant less than 2. The base of the cone is joined to one of the circular faces of the cylinder so that the axes of symmetry of the two solids coincide (see diagram). The cone and the cylinder are made of the same material.
  1. Show that the distance of the centre of mass of the combined solid from the vertex of the cone is $$\frac{(99k^2 + 96)r}{18k^2 + 32}.$$ [4]
Question 4:
AnswerMarks
4Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
AnswerMarks
4(a)Centre of mass
Mass
from vertex of cone
Cylinder π ( kr )2 .3rρ 11 r
2
Cone 1 π ( 2r )2 .4rρ 3r
3
Combined  π ( kr )2 .3r+ 1 π ( 2r )2 .4r   ρ x
AnswerMarks
 3 B1
Take moments about vertex:
x×  π ( kr )2 .3r+ 1 π ( 2r )2 .4r   = 11r × π ( kr )2 .3r+3r× 1 π ( 2r )2 .4r leading to
AnswerMarks
 3  2 3M1A1
( )
99k2 +96 r
x =
AnswerMarks
18k2 +32A1
4
AnswerMarks
MassCentre of mass
from vertex of cone
AnswerMarks Guidance
Cylinderπ ( kr )2 .3rρ 11
r
2
AnswerMarks Guidance
Cone1 π ( 2r )2 .4rρ
33r
Combined π ( kr )2 .3r+ 1 π ( 2r )2 .4r   ρ
 3 x
QuestionAnswer Marks
AnswerMarks
4(b)x −4r
tanα= ( = 1/8)
AnswerMarks
2rM1
17
x = r
AnswerMarks
4A1
Equate to answer in (a) and attempt to find kM1
4
45k2 =80: k =
AnswerMarks
3A1
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 4:
4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
--- 4(a) ---
4(a) | Centre of mass
Mass
from vertex of cone
Cylinder π ( kr )2 .3rρ 11 r
2
Cone 1 π ( 2r )2 .4rρ 3r
3
Combined  π ( kr )2 .3r+ 1 π ( 2r )2 .4r   ρ x
 3  | B1
Take moments about vertex:
x×  π ( kr )2 .3r+ 1 π ( 2r )2 .4r   = 11r × π ( kr )2 .3r+3r× 1 π ( 2r )2 .4r leading to
 3  2 3 | M1A1
( )
99k2 +96 r
x =
18k2 +32 | A1
4
Mass | Centre of mass
from vertex of cone
Cylinder | π ( kr )2 .3rρ | 11
r
2
Cone | 1 π ( 2r )2 .4rρ
3 | 3r
Combined |  π ( kr )2 .3r+ 1 π ( 2r )2 .4r   ρ
 3  | x
Question | Answer | Marks
--- 4(b) ---
4(b) | x −4r
tanα= ( = 1/8)
2r | M1
17
x = r
4 | A1
Equate to answer in (a) and attempt to find k | M1
4
45k2 =80: k =
3 | A1
4
Question | Answer | Marks
\includegraphics{figure_4}

A uniform solid circular cone, of vertical height $4r$ and radius $2r$, is attached to a uniform solid cylinder, of height $3r$ and radius $kr$, where $k$ is a constant less than 2. The base of the cone is joined to one of the circular faces of the cylinder so that the axes of symmetry of the two solids coincide (see diagram). The cone and the cylinder are made of the same material.

\begin{enumerate}[label=(\alph*)]
\item Show that the distance of the centre of mass of the combined solid from the vertex of the cone is 
$$\frac{(99k^2 + 96)r}{18k^2 + 32}.$$ [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q4 [4]}}
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