CAIE Further Paper 3 2021 November — Question 4 8 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2021
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeLamina with removed triangle/rectangle/square
DifficultyStandard +0.8 This is a multi-step centre of mass problem requiring coordinate geometry, composite body techniques, and equilibrium analysis. Part (a) involves finding the centroid of an irregular lamina by subtraction (5 marks), requiring careful coordinate work. Part (b) requires applying the equilibrium condition that the centre of mass must lie above the base, leading to an inequality. While the techniques are standard for Further Maths mechanics, the problem requires sustained reasoning across multiple steps and careful algebraic manipulation, placing it moderately above average difficulty.
Spec6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces
\includegraphics{figure_4} A uniform lamina \(AECF\) is formed by removing two identical triangles \(BCE\) and \(CDF\) from a square lamina \(ABCD\). The square has side \(3a\) and \(EB = DF = h\) (see diagram).
  1. Find the distance of the centre of mass of the lamina \(AECF\) from \(AD\) and from \(AB\), giving your answers in terms of \(a\) and \(h\). [5]
The lamina \(AECF\) is placed vertically on its edge \(AE\) on a horizontal plane.
  1. Find, in terms of \(a\), the set of values of \(h\) for which the lamina remains in equilibrium. [3]
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)Area Centre of mass from AD
3
Square 9a2 a
2
3
CDF ah a
2
3 1
BEC ah 3a− h
2 3
AnswerMarks Guidance
Resulting AEFC 9a2 −3ah xM1 Attempt at moments with three terms.
Taking moments about AD:
( )  3  3  3  1 
9a2 −3ah x =  9a2× a  −  ah×a  −  ah×  3a− h 
AnswerMarks
 2  2  2  3 A1
A1Two terms correct.
All correct.
27a2 −12ah+h2 9a−h
x = =
AnswerMarks Guidance
6 ( 3a−h )   6  A1 AEF
y =xB1 By symmetry or equal to their x.
5
AnswerMarks Guidance
4(b)For equilibrium, x ⩽ 3a – h
27a2 – 12ah + h2 ⩽ 6(3a – h)2B1 Accept strict inequality.
27a2– 24ah + 5h2 ⩾ 0M1 Homogeneous 3-term quadratic inequality.
9
h ⩽ a
AnswerMarks Guidance
5A1 CAO.
3
AnswerMarks Guidance
AreaCentre of mass from AD
Square9a2 3
a
2
AnswerMarks
CDF3
ah
AnswerMarks
2a
BEC3
ah
AnswerMarks
21
3a− h
3
AnswerMarks Guidance
Resulting AEFC9a2 −3ah x
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) | Area Centre of mass from AD
3
Square 9a2 a
2
3
CDF ah a
2
3 1
BEC ah 3a− h
2 3
Resulting AEFC 9a2 −3ah x | M1 | Attempt at moments with three terms.
Taking moments about AD:
( )  3  3  3  1 
9a2 −3ah x =  9a2× a  −  ah×a  −  ah×  3a− h 
 2  2  2  3  | A1
A1 | Two terms correct.
All correct.
27a2 −12ah+h2 9a−h

x = =
6 ( 3a−h )   6   | A1 | AEF
y =x | B1 | By symmetry or equal to their x.
5
--- 4(b) ---
4(b) | For equilibrium, x ⩽ 3a – h
27a2 – 12ah + h2 ⩽ 6(3a – h)2 | B1 | Accept strict inequality.
27a2– 24ah + 5h2 ⩾ 0 | M1 | Homogeneous 3-term quadratic inequality.
9
h ⩽ a
5 | A1 | CAO.
3
Area | Centre of mass from AD
Square | 9a2 | 3
a
2
CDF | 3
ah
2 | a
BEC | 3
ah
2 | 1
3a− h
3
Resulting AEFC | 9a2 −3ah | x
Question | Answer | Marks | Guidance
\includegraphics{figure_4}

A uniform lamina $AECF$ is formed by removing two identical triangles $BCE$ and $CDF$ from a square lamina $ABCD$. The square has side $3a$ and $EB = DF = h$ (see diagram).

\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of the lamina $AECF$ from $AD$ and from $AB$, giving your answers in terms of $a$ and $h$. [5]
\end{enumerate}

The lamina $AECF$ is placed vertically on its edge $AE$ on a horizontal plane.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find, in terms of $a$, the set of values of $h$ for which the lamina remains in equilibrium. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q4 [8]}}
This paper (7 questions)
View full paper
Q1 4 Q2 6 Q3 6 Q4 8 Q5 7 Q6 8 Q7 11