CAIE Further Paper 3 2021 November — Question 4 7 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2021
SessionNovember
Marks7
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Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeSolid with removed cylinder or hemisphere from solid
DifficultyChallenging +1.2 This is a standard centre of mass problem requiring subtraction of volumes and moments, followed by applying equilibrium conditions for suspension. Part (a) involves routine application of standard formulae for hemisphere and cylinder centres of mass with algebraic manipulation to reach the given result. Part (b) requires geometric reasoning about the suspension angle and solving a resulting equation. While it involves multiple steps and careful bookkeeping, the techniques are all standard for Further Maths mechanics with no novel insights required.
Spec6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces
\includegraphics{figure_4} An object is formed by removing a solid cylinder, of height \(ka\) and radius \(\frac{1}{2}a\), from a uniform solid hemisphere of radius \(a\). The axes of symmetry of the hemisphere and the cylinder coincide and one circular face of the cylinder coincides with the plane face of the hemisphere. \(AB\) is a diameter of the circular face of the hemisphere (see diagram).
  1. Show that the distance of the centre of mass of the object from \(AB\) is \(\frac{3a(2-k^2)}{2(8-3k)}\). [4] When the object is freely suspended from the point \(A\), the line \(AB\) makes an angle \(\theta\) with the downward vertical, where \(\tan\theta = \frac{7}{18}\).
  2. Find the possible values of \(k\). [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)Volume Centre of mass from AB
2 3
Hemisphere πa3 a
3 8
a ka
Cylinder πka( )2
2 2
2
2 a
Remainder πa3 −πka   x
AnswerMarks Guidance
3 2M1 Attempt at moments, 3 terms.
Taking moments about AB:
2 a 2 2 3   a 2 ka
 πa3 −πka   ×x =  πa3× a  − πka   × 
3 2  3 8   2 2 
AnswerMarks
   A1
A1Any 2 terms correct.
All correct.
( )
3a 2−k2
x =
AnswerMarks Guidance
2 ( 8−3k )A1 Shown convincingly, AG.
4
AnswerMarks
4(b)x
tanθ=
a
( )
3 2−k2
7
=
AnswerMarks Guidance
2 ( 8−3k ) 18B1
27k2 −21k+2=0M1 Rearrange to form quadratic.
2 1
k = and k =
AnswerMarks Guidance
3 9A1 Both answers correct.
3
AnswerMarks
VolumeCentre of mass from AB
Hemisphere2
πa3
AnswerMarks
33
a
8
AnswerMarks
Cylindera
πka( )2
AnswerMarks
2ka
2
AnswerMarks
Remainder2
2 a
πa3 −πka
 
AnswerMarks Guidance
3 2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) | Volume Centre of mass from AB
2 3
Hemisphere πa3 a
3 8
a ka
Cylinder πka( )2
2 2
2
2 a
Remainder πa3 −πka   x
3 2 | M1 | Attempt at moments, 3 terms.
Taking moments about AB:
2 a 2 2 3   a 2 ka
 πa3 −πka   ×x =  πa3× a  − πka   × 
3 2  3 8   2 2 
    | A1
A1 | Any 2 terms correct.
All correct.
( )
3a 2−k2
x =
2 ( 8−3k ) | A1 | Shown convincingly, AG.
4
--- 4(b) ---
4(b) | x
tanθ=
a
( )
3 2−k2
7
=
2 ( 8−3k ) 18 | B1
27k2 −21k+2=0 | M1 | Rearrange to form quadratic.
2 1
k = and k =
3 9 | A1 | Both answers correct.
3
Volume | Centre of mass from AB
Hemisphere | 2
πa3
3 | 3
a
8
Cylinder | a
πka( )2
2 | ka
2
Remainder | 2
2 a
πa3 −πka
 
3 2 | x
Question | Answer | Marks | Guidance
\includegraphics{figure_4}

An object is formed by removing a solid cylinder, of height $ka$ and radius $\frac{1}{2}a$, from a uniform solid hemisphere of radius $a$. The axes of symmetry of the hemisphere and the cylinder coincide and one circular face of the cylinder coincides with the plane face of the hemisphere. $AB$ is a diameter of the circular face of the hemisphere (see diagram).

\begin{enumerate}[label=(\alph*)]
\item Show that the distance of the centre of mass of the object from $AB$ is $\frac{3a(2-k^2)}{2(8-3k)}$. [4]

When the object is freely suspended from the point $A$, the line $AB$ makes an angle $\theta$ with the downward vertical, where $\tan\theta = \frac{7}{18}$.

\item Find the possible values of $k$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q4 [7]}}
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