CAIE Further Paper 3 2024 November — Question 4 4 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2024
SessionNovember
Marks4
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Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeSolid with removed cylinder or hemisphere from solid
DifficultyStandard +0.3 This is a standard centre of mass problem using the removal method with cylinders. Students need to apply the formula for composite bodies (large cylinder minus small cylinder), use the standard result that a cylinder's centre of mass is at its geometric centre, and perform straightforward algebraic manipulation. The setup is clear and the method is routine for Further Maths students, making it slightly easier than average.
Spec6.04c Composite bodies: centre of mass
\includegraphics{figure_4} An object is formed by removing a cylinder of radius \(\frac{2}{3}a\) and height \(kh\) (\(k < 1\)) from a uniform solid cylinder of radius \(a\) and height \(h\). The vertical axes of symmetry of the two cylinders coincide. The upper faces of the two cylinders are in the same plane as each other. The points \(A\) and \(B\) are the opposite ends of a diameter of the upper face of the object (see diagram).
  1. Find, in terms of \(h\) and \(k\), the distance of the centre of mass of the object from \(AB\). [4]
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
\includegraphics{figure_4}

An object is formed by removing a cylinder of radius $\frac{2}{3}a$ and height $kh$ ($k < 1$) from a uniform solid cylinder of radius $a$ and height $h$. The vertical axes of symmetry of the two cylinders coincide. The upper faces of the two cylinders are in the same plane as each other. The points $A$ and $B$ are the opposite ends of a diameter of the upper face of the object (see diagram).

\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $h$ and $k$, the distance of the centre of mass of the object from $AB$. [4]
\end{enumerate}

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