CAIE Further Paper 3 2023 November — Question 3 7 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2023
SessionNovember
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeLamina with removed triangle/rectangle/square
DifficultyChallenging +1.2 This is a standard Further Maths centre of mass problem requiring coordinate setup, use of composite body formula (removing a triangle), and applying equilibrium conditions for suspension. The geometry is straightforward with similar triangles, and the method is well-practiced in FM syllabi. The algebra is routine though multi-step, making it moderately above average difficulty but not requiring novel insight.
Spec6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces
\includegraphics{figure_3} A uniform lamina is in the form of an isosceles triangle \(ABC\) in which \(AC = 2a\) and angle \(ABC = 90°\). The point \(D\) on \(AB\) is such that the ratio \(DB : AB = 1 : k\). The point \(E\) on \(CB\) is such that \(DE\) is parallel to \(AC\). The triangle \(DBE\) is removed from the lamina (see diagram).
  1. Find, in terms of \(k\), the distance of the centre of mass of the remaining lamina \(ADEC\) from the midpoint of \(AC\). [4]
When the lamina \(ADEC\) is freely suspended from the vertex \(A\), the edge \(AC\) makes an angle \(\theta\) with the downward vertical, where \(\tan \theta = \frac{2}{15}\).
  1. Find the value of \(k\). [3]
Question 3:
AnswerMarks
3(a)Mass is proportional to area
Centre of mass
Area
from AC
1 1
ABC 2aa a
2 3
1 a 2 2a
BDE 2  a−  
2 k 3k
1 a 2
ADEC − 2  +a2 x
AnswerMarks Guidance
2 kB1 At least two areas correct, at least one distance correct
 1 a 2  1  2a a 2
x− 2  +a2= aa2 −a−   
AnswerMarks Guidance
  2 k   3  3k kM1A1 Moments equation, dimensionally correct, correct number
of terms.
a ( k2 +k−2 )
x =
AnswerMarks Guidance
3k(k+1)A1 a ( k3−3k+2 )
Allow unsimplified single fraction .
3k ( k2 −1 )
4
AnswerMarks
AreaCentre of mass
from AC
AnswerMarks
ABC1
2aa
AnswerMarks
21
a
3
AnswerMarks
BDE1 a 2
2 
AnswerMarks
2 k2a
a−
 
3k
AnswerMarks
ADEC1 a 2
− 2  +a2
AnswerMarks Guidance
2 kx
QuestionAnswer Marks
AnswerMarks
3(b)x  5 
tan= =
 
AnswerMarks Guidance
a  18B1 With their x from part (a).
So 18 ( k2 +k−2 ) =5 ( 3k2 +3k )
AnswerMarks Guidance
k2 +k−12=0M1 Obtain a polynomial in k only, e.g. k3−13k+12=0,
may be implied.
(k+4)(k−3)=0,
AnswerMarks Guidance
k=3 onlyA1 CWO
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 3:
--- 3(a) ---
3(a) | Mass is proportional to area
Centre of mass
Area
from AC
1 1
ABC 2aa a
2 3
1 a 2 2a
BDE 2  a−  
2 k 3k
1 a 2
ADEC − 2  +a2 x
2 k | B1 | At least two areas correct, at least one distance correct
 1 a 2  1  2a a 2
x− 2  +a2= aa2 −a−   
  2 k   3  3k k | M1A1 | Moments equation, dimensionally correct, correct number
of terms.
a ( k2 +k−2 )
x =
3k(k+1) | A1 | a ( k3−3k+2 )
Allow unsimplified single fraction .
3k ( k2 −1 )
4
Area | Centre of mass
from AC
ABC | 1
2aa
2 | 1
a
3
BDE | 1 a 2
2 
2 k | 2a
a−
 
3k
ADEC | 1 a 2
− 2  +a2
2 k | x
Question | Answer | Marks | Guidance
--- 3(b) ---
3(b) | x  5 
tan= =
 
a  18 | B1 | With their x from part (a).
So 18 ( k2 +k−2 ) =5 ( 3k2 +3k )
k2 +k−12=0 | M1 | Obtain a polynomial in k only, e.g. k3−13k+12=0,
may be implied.
(k+4)(k−3)=0,
k=3 only | A1 | CWO
3
Question | Answer | Marks | Guidance
\includegraphics{figure_3}

A uniform lamina is in the form of an isosceles triangle $ABC$ in which $AC = 2a$ and angle $ABC = 90°$. The point $D$ on $AB$ is such that the ratio $DB : AB = 1 : k$. The point $E$ on $CB$ is such that $DE$ is parallel to $AC$. The triangle $DBE$ is removed from the lamina (see diagram).

\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $k$, the distance of the centre of mass of the remaining lamina $ADEC$ from the midpoint of $AC$. [4]
\end{enumerate}

When the lamina $ADEC$ is freely suspended from the vertex $A$, the edge $AC$ makes an angle $\theta$ with the downward vertical, where $\tan \theta = \frac{2}{15}$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the value of $k$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2023 Q3 [7]}}
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