CAIE Further Paper 3 2024 November — Question 4 3 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2024
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeNon-uniform beam on supports
DifficultyStandard +0.8 This is part (b) of a multi-part moments question requiring students to apply equilibrium conditions with a specific geometric constraint (tan θ = 1/2) and given parameter h = 8a/3 to find k. It involves resolving forces/moments in two directions, using trigonometry, and solving resulting equations—standard Further Maths mechanics but requiring careful algebraic manipulation across multiple steps.
Spec6.04e Rigid body equilibrium: coplanar forces
When the object is suspended from \(A\), the angle between \(AB\) and the vertical is \(\theta\), where \(\tan \theta = \frac{1}{2}\).
  1. Given that \(h = \frac{8}{3}a\), find the possible values of \(k\). [3]
Question 4:
AnswerMarks
4(a)Large Small Object
2  2  4 
Volume a2h   a  kh a2h1− k
3   9 
Centre of 1 1
mass from h kh x
AB 2 2
Moments about AB:
2
 4  1 2  1
a2h1− k y = a2h h− a kh kh
 
AnswerMarks Guidance
 9  2 3  2B1 Correct volumes and distances for large and small.
M1Moments equation with 3 terms, dimensionally
correct.
AnswerMarks
A1Correct, unsimplified.
( 9−4k2)
h
y =
AnswerMarks
2(9−4k)A1
4
AnswerMarks
4(b)( 9−4k2)
h
y 3
tan= : =
AnswerMarks Guidance
a 2(9−4k)a 2B1 FT FT their part (a)
8
Use h= a and simplify to quadratic in k: 32k2 −36k+9=0
AnswerMarks
3M1
3 3
k = ,
AnswerMarks
8 4A1
3
AnswerMarks Guidance
LargeSmall Object
Volumea2h 2
2 
 a kh
 
AnswerMarks
3  4 
a2h1− k
 9 
Centre of
mass from
AnswerMarks
AB1
h
AnswerMarks
21
kh
AnswerMarks Guidance
2x
QuestionAnswer Marks 
Question 4:
--- 4(a) ---
4(a) | Large Small Object
2  2  4 
Volume a2h   a  kh a2h1− k
3   9 
Centre of 1 1
mass from h kh x
AB 2 2
Moments about AB:
2
 4  1 2  1
a2h1− k y = a2h h− a kh kh
 
 9  2 3  2 | B1 | Correct volumes and distances for large and small.
M1 | Moments equation with 3 terms, dimensionally
correct.
A1 | Correct, unsimplified.
( 9−4k2)
h
y =
2(9−4k) | A1
4
--- 4(b) ---
4(b) | ( 9−4k2)
h
y 3
tan= : =
a 2(9−4k)a 2 | B1 FT | FT their part (a)
8
Use h= a and simplify to quadratic in k: 32k2 −36k+9=0
3 | M1
3 3
k = ,
8 4 | A1
3
Large | Small | Object
Volume | a2h | 2
2 
 a kh
 
3  |  4 
a2h1− k
 9 
Centre of
mass from
AB | 1
h
2 | 1
kh
2 | x
Question | Answer | Marks | Guidance
When the object is suspended from $A$, the angle between $AB$ and the vertical is $\theta$, where $\tan \theta = \frac{1}{2}$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Given that $h = \frac{8}{3}a$, find the possible values of $k$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2024 Q4 [3]}}
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Q1 5 Q2 5 Q3 6 Q3 2 Q4 3 Q5 4 Q6 3 Q7 4