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 ↗
TopicCentre of Mass 1
TypeCentre of mass with variable parameter
DifficultyStandard +0.8 This is a continuation problem requiring students to use equilibrium conditions (moments about suspension point) with the given angle to find unknown parameters. It involves setting up and solving an equation from the tangent condition, requiring understanding of centre of mass location and moment equilibrium. The algebraic manipulation is moderate but the conceptual setup and connection between angle and mass distribution requires solid understanding of the topic.
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)
y h 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)
y h 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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