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 ↗
TopicMoments
TypeCentre of mass of composite shapes
DifficultyChallenging +1.2 This is a standard moments/centre of mass problem requiring taking moments about point C when suspended, using the given angle to set up an equation, and solving for k. The constraint that the centre of mass lies within the cylinder adds a minor geometric consideration. It's a typical 4-mark further mechanics question with straightforward application of equilibrium principles, slightly above average due to the 3D geometry and algebraic manipulation involved.
Spec6.04e Rigid body equilibrium: coplanar forces
The point \(C\) is on the circumference of the base of the cone. When the combined solid is freely suspended from \(C\) and hanging in equilibrium, the diameter through \(C\) makes an angle \(\alpha\) with the downward vertical, where \(\tan \alpha = \frac{1}{5}\).
  1. Given that the centre of mass of the combined solid is within the cylinder, find the value of \(k\). [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
The point $C$ is on the circumference of the base of the cone. When the combined solid is freely suspended from $C$ and hanging in equilibrium, the diameter through $C$ makes an angle $\alpha$ with the downward vertical, where $\tan \alpha = \frac{1}{5}$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Given that the centre of mass of the combined solid is within the cylinder, find the value of $k$. [4]
\end{enumerate}

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