CAIE Further Paper 3 2024 June — Question 5 7 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2024
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 centre of mass calculation and equilibrium analysis. Part (a) involves routine application of the formula for removing a triangular section (3 marks suggests straightforward calculation). Part (b) requires setting up equilibrium conditions with the given angle, involving geometric reasoning and algebraic manipulation. While it requires multiple techniques and careful coordinate geometry, it follows predictable patterns for this topic without requiring novel insight.
Spec6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces
\includegraphics{figure_5} A uniform lamina is in the form of a triangle \(OBC\), with \(OC = 18a\), \(OB = 24a\) and angle \(COB = 90°\). The point \(A\) on \(OB\) is such that \(OA = x\) (see diagram). The triangle \(OAC\) is removed from the lamina.
  1. Find, in terms of \(a\) and \(x\), the distance of the centre of mass of the remaining object \(ABC\) from \(OC\). [3]
The object \(ABC\) is suspended from \(C\). In its equilibrium position, the side \(AB\) makes an angle \(\theta\) with the vertical, where \(\tan\theta = \frac{8}{5}\).
  1. Find \(x\) in terms of \(a\). [4]
Question 5:
AnswerMarks
5(a)OBC OAC ABC
1
Area 24a18a 9ax 216a2 9ax
2
Centre of mass 8a 1x x
from OC 3
Moments about OC
 216a2 9ax  x 216a28a9ax 1 x
AnswerMarks Guidance
3M1 A1 Note that moments about OB is M0 (y 6a).
Moments equation about OC with all terms
present, allow sign error, dimensionally correct.
All correct for A1.
576a2 x2 x24a
x  or
AnswerMarks Guidance
72a3x 3A1 Accept any equivalent form.
Alternative solution to question 5(a)
AnswerMarks
Consider system as equivalent to particles at (0, 18a), (x, 0) and (24a, 0)B1
Then the x-coordinate of the centre of mass is at 1x24a
AnswerMarks
3M1 A1
3
AnswerMarks Guidance
OBCOAC ABC
Area1
24a18a
AnswerMarks Guidance
29ax 216a2 9ax
Centre of mass
AnswerMarks Guidance
from OC8a 1x
3x
QuestionAnswer Marks
AnswerMarks
5(b)18a6a 18a y y 18a
tan or or or
x x 1 5ax 1 2a 1x
AnswerMarks Guidance
2M1 A1 Either way up (their value for x may be
substituted in).
x2 30ax144a2 0
AnswerMarks Guidance
Or, with simplified form, 30ax24aM1 Obtain homogeneous (quadratic) equation
Note that if simplified form of x is used, equation
will be linear.
AnswerMarks Guidance
x6aA1 Single correct answer only.
4
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
--- 5(a) ---
5(a) | OBC OAC ABC
1
Area 24a18a 9ax 216a2 9ax
2
Centre of mass 8a 1x x
from OC 3
Moments about OC
 216a2 9ax  x 216a28a9ax 1 x
3 | M1 A1 | Note that moments about OB is M0 (y 6a).
Moments equation about OC with all terms
present, allow sign error, dimensionally correct.
All correct for A1.
576a2 x2 x24a
x  or
72a3x 3 | A1 | Accept any equivalent form.
Alternative solution to question 5(a)
Consider system as equivalent to particles at (0, 18a), (x, 0) and (24a, 0) | B1
Then the x-coordinate of the centre of mass is at 1x24a
3 | M1 A1
3
OBC | OAC | ABC
Area | 1
24a18a
2 | 9ax | 216a2 9ax
Centre of mass
from OC | 8a | 1x
3 | x
Question | Answer | Marks | Guidance
--- 5(b) ---
5(b) | 18a6a 18a y y 18a
tan or or or
x x 1 5ax 1 2a 1x
2 | M1 A1 | Either way up (their value for x may be
substituted in).
x2 30ax144a2 0
Or, with simplified form, 30ax24a | M1 | Obtain homogeneous (quadratic) equation
Note that if simplified form of x is used, equation
will be linear.
x6a | A1 | Single correct answer only.
4
Question | Answer | Marks | Guidance
\includegraphics{figure_5}

A uniform lamina is in the form of a triangle $OBC$, with $OC = 18a$, $OB = 24a$ and angle $COB = 90°$. The point $A$ on $OB$ is such that $OA = x$ (see diagram). The triangle $OAC$ 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 $ABC$ from $OC$. [3]
\end{enumerate}

The object $ABC$ is suspended from $C$. In its equilibrium position, the side $AB$ makes an angle $\theta$ with the vertical, where $\tan\theta = \frac{8}{5}$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find $x$ in terms of $a$. [4]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q5 [7]}}
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