CAIE Further Paper 3 2023 June — Question 3 7 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2023
SessionJune
Marks7
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Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeLamina with removed triangle/rectangle/square
DifficultyStandard +0.8 This is a standard Further Maths mechanics problem requiring composite body centre of mass calculation using the removal method, followed by applying the toppling condition. While it involves multiple steps (finding centroids of triangles, using the composite formula, then applying equilibrium), the techniques are routine for FM students and the geometry is straightforward with a right-angled triangle. The algebraic manipulation is moderate but not particularly challenging.
Spec6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces
\includegraphics{figure_3} A uniform lamina is in the form of a triangle \(ABC\), with \(AC = 8a\), \(BC = 6a\) and angle \(ACB = 90°\). The point \(D\) on \(AC\) is such that \(AD = 3a\). The point \(E\) on \(CB\) is such that \(CE = x\) (see diagram). The triangle \(CDE\) is removed from the lamina.
  1. Find, in terms of \(a\) and \(x\), the distance of the centre of mass of the remaining object \(ADEB\) from \(AC\). [4]
The object \(ADEB\) is on the point of toppling about the point \(E\) when the object is in the vertical plane with its edge \(EB\) on a smooth horizontal surface.
  1. Find \(x\) in terms of \(a\). [3]
Question 3:
AnswerMarks
3(a)[Mass is proportional to area]
Area Centre of mass
from AC
ABC 1 2a
.6a.8a (24a2)
2
DEC 1 1
x.5a x
2 3
ADEB 5 x
24a2  xa
AnswerMarks Guidance
2B1 All correct for ABC and DEC.
 5  1 5
Moments [about AC] x  24a2  xa  24a22a x ax
AnswerMarks Guidance
 2  3 2M1 All moment terms present, dimensionally correct,
allow sign error.
AnswerMarks
A1All correct moments about AC.
288a2 5x2
x 
AnswerMarks Guidance
348a5xA1 AEF
4
AnswerMarks
AreaCentre of mass
from AC
AnswerMarks
ABC1
.6a.8a (24a2)
AnswerMarks
22a
DEC1
x.5a
AnswerMarks
21
x
3
AnswerMarks
ADEB5
24a2  xa
AnswerMarks Guidance
2x
QuestionAnswer Marks
AnswerMarks
3(b)288a2 5x2
On the point of toppling about E: x x, x
AnswerMarks Guidance
348a5xB1 FT FT their expression for x from part (a).
Rearrange to 3-term quadratic: 10x2 144ax288a2 0M1 Allow 3-term inequality.
25x12ax12a0,
12
x a
AnswerMarks Guidance
5A1 Single correct answer, no inequality, CWO.
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 3:
--- 3(a) ---
3(a) | [Mass is proportional to area]
Area Centre of mass
from AC
ABC 1 2a
.6a.8a (24a2)
2
DEC 1 1
x.5a x
2 3
ADEB 5 x
24a2  xa
2 | B1 | All correct for ABC and DEC.
 5  1 5
Moments [about AC] x  24a2  xa  24a22a x ax
 2  3 2 | M1 | All moment terms present, dimensionally correct,
allow sign error.
A1 | All correct moments about AC.
288a2 5x2
x 
348a5x | A1 | AEF
4
Area | Centre of mass
from AC
ABC | 1
.6a.8a (24a2)
2 | 2a
DEC | 1
x.5a
2 | 1
x
3
ADEB | 5
24a2  xa
2 | x
Question | Answer | Marks | Guidance
--- 3(b) ---
3(b) | 288a2 5x2
On the point of toppling about E: x x, x
348a5x | B1 FT | FT their expression for x from part (a).
Rearrange to 3-term quadratic: 10x2 144ax288a2 0 | M1 | Allow 3-term inequality.
25x12ax12a0,
12
x a
5 | A1 | Single correct answer, no inequality, CWO.
3
Question | Answer | Marks | Guidance
\includegraphics{figure_3}

A uniform lamina is in the form of a triangle $ABC$, with $AC = 8a$, $BC = 6a$ and angle $ACB = 90°$. The point $D$ on $AC$ is such that $AD = 3a$. The point $E$ on $CB$ is such that $CE = x$ (see diagram). The triangle $CDE$ is removed from the lamina.

\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $a$ and $x$, the distance of the centre of mass of the remaining object $ADEB$ from $AC$. [4]
\end{enumerate}

The object $ADEB$ is on the point of toppling about the point $E$ when the object is in the vertical plane with its edge $EB$ on a smooth horizontal surface.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find $x$ in terms of $a$. [3]
\end{enumerate}

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