CAIE Further Paper 3 2023 June — Question 4 8 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2023
SessionJune
Marks8
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Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeComposite solid with hemisphere and cylinder/cone
DifficultyChallenging +1.2 This is a standard Further Maths centre of mass problem requiring systematic application of formulas for composite bodies (hemisphere and cylinder), followed by an equilibrium condition on an inclined plane. Part (a) involves routine calculations with known COM positions and masses, while part (b) requires setting up a toppling condition. The multi-step nature and Further Maths context place it above average difficulty, but the techniques are well-practiced and straightforward to apply.
Spec6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces
\includegraphics{figure_4} An object is formed from a solid hemisphere, of radius \(2a\), and a solid cylinder, of radius \(a\) and height \(d\). The hemisphere and the cylinder are made of the same material. The cylinder is attached to the plane face of the hemisphere. The line \(OC\) forms a diameter of the base of the cylinder, where \(C\) is the centre of the plane face of the hemisphere and \(O\) is common to both circumferences (see diagram). Relative to axes through \(O\), parallel and perpendicular to \(OC\) as shown, the centre of mass of the object is \((\bar{x}, \bar{y})\).
  1. Show that \(\bar{x} = \frac{32a^2 + 3ad}{16a + 3d}\) and find an expression, in terms of \(a\) and \(d\), for \(\bar{y}\). [5]
The object is placed on a rough plane which is inclined to the horizontal at an angle \(\theta\) where \(\sin\theta = \frac{1}{6}\). The object is in equilibrium with \(CO\) horizontal, where \(CO\) lies in a vertical plane through a line of greatest slope.
  1. Find \(d\) in terms of \(a\). [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)[Mass is proportional to volume]
Distance of Distance of
Volume centre of mass centre of mass
from vertical axis from OC
2 3
2a3
Hemisphere 2a  2a
3 8
1
Cylinder a2d a d
2
2
Object 2a3 a2d x y
3
  2 2a3 a2d  x  16 a32a a2da
AnswerMarks Guidance
3  3M1 A1 Moments equation, dimensionally correct, correct
number of terms. Allow sign errors.
32a2 3ad
Simplify to x 
AnswerMarks Guidance
16a3dA1 AG. At least one line of intermediate working.
  2 2a3 πa2d  y  16 πa3   3 2a   πa2d 1 d
AnswerMarks Guidance
3  3  8  2M1 Moments equation, dimensionally correct, correct
number of terms. Allow sign errors.
3  d2 8a2
y 
AnswerMarks Guidance
216a3dA1 AEF
5
AnswerMarks
VolumeDistance of
centre of mass
AnswerMarks
from vertical axisDistance of
centre of mass
from OC
AnswerMarks
Hemisphere2
2a3
AnswerMarks Guidance
32a 3
 2a
8
AnswerMarks Guidance
Cylindera2d a
d
2
AnswerMarks
Object2
2a3 a2d
AnswerMarks Guidance
3x y
QuestionAnswer Marks
AnswerMarks
4(b)2ax
sin
AnswerMarks
2aB1
1 32a2 3ad
2a 2a
6 16a3d
5
16a3d32a3d
AnswerMarks Guidance
3M1 Remove fractions
8
d  a
AnswerMarks
3A1
3
AnswerMarks Guidance
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) | [Mass is proportional to volume]
Distance of Distance of
Volume centre of mass centre of mass
from vertical axis from OC
2 3
2a3
Hemisphere 2a  2a
3 8
1
Cylinder a2d a d
2
2
Object 2a3 a2d x y
3
  2 2a3 a2d  x  16 a32a a2da
3  3 | M1 A1 | Moments equation, dimensionally correct, correct
number of terms. Allow sign errors.
32a2 3ad
Simplify to x 
16a3d | A1 | AG. At least one line of intermediate working.
  2 2a3 πa2d  y  16 πa3   3 2a   πa2d 1 d
3  3  8  2 | M1 | Moments equation, dimensionally correct, correct
number of terms. Allow sign errors.
3  d2 8a2
y 
216a3d | A1 | AEF
5
Volume | Distance of
centre of mass
from vertical axis | Distance of
centre of mass
from OC
Hemisphere | 2
2a3
3 | 2a | 3
 2a
8
Cylinder | a2d | a | 1
d
2
Object | 2
2a3 a2d
3 | x | y
Question | Answer | Marks | Guidance
--- 4(b) ---
4(b) | 2ax
sin
2a | B1
1 32a2 3ad
2a 2a
6 16a3d
5
16a3d32a3d
3 | M1 | Remove fractions
8
d  a
3 | A1
3
Question | Answer | Marks | Guidance
\includegraphics{figure_4}

An object is formed from a solid hemisphere, of radius $2a$, and a solid cylinder, of radius $a$ and height $d$. The hemisphere and the cylinder are made of the same material. The cylinder is attached to the plane face of the hemisphere. The line $OC$ forms a diameter of the base of the cylinder, where $C$ is the centre of the plane face of the hemisphere and $O$ is common to both circumferences (see diagram). Relative to axes through $O$, parallel and perpendicular to $OC$ as shown, the centre of mass of the object is $(\bar{x}, \bar{y})$.

\begin{enumerate}[label=(\alph*)]
\item Show that $\bar{x} = \frac{32a^2 + 3ad}{16a + 3d}$ and find an expression, in terms of $a$ and $d$, for $\bar{y}$. [5]
\end{enumerate}

The object is placed on a rough plane which is inclined to the horizontal at an angle $\theta$ where $\sin\theta = \frac{1}{6}$. The object is in equilibrium with $CO$ horizontal, where $CO$ lies in a vertical plane through a line of greatest slope.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find $d$ in terms of $a$. [3]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 3 2023 Q4 [8]}}
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