OCR MEI Further Mechanics B AS 2021 November — Question 3 8 marks

Exam BoardOCR MEI
ModuleFurther Mechanics B AS (Further Mechanics B AS)
Year2021
SessionNovember
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeComposite solid with hemisphere and cylinder/cone
DifficultyStandard +0.8 This is a multi-part centre of mass problem requiring (a) proving the standard cone COM result using integration or formula, and (b) solving for a density ratio using the condition that the combined COM lies on the boundary. Part (a) is routine for Further Maths students; part (b) requires setting up and solving an equation involving volumes, densities, and COM positions, which is moderately challenging but follows standard techniques for composite bodies.
Spec6.04a Centre of mass: gravitational effect6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces

3 In this question you must show detailed reasoning. [In this question you may use the formula: Volume of cone \(= \frac { 1 } { 3 } \times\) base area × height.]
The region between the line \(\mathrm { y } = - 3 \mathrm { x } + 3 \mathrm { a }\), where \(a > 0\), the \(x\)-axis and the \(y\)-axis is rotated about the \(y\)-axis to form a uniform right circular cone C with base radius \(a\).
  1. Show that the centre of mass of C is \(\frac { 3 } { 4 } a\) from its base. The cone C is fixed on top of a uniform cube, B , of edge length \(2 a\), so that there is no overlap. Fig. 3.1 shows a side view of C and B fixed together; Fig. 3.2 shows a view of C and B from above. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{37798594-8cb0-48aa-8401-090f09e25dff-3_570_323_785_246} \captionsetup{labelformat=empty} \caption{Fig. 3.1}
    \end{figure} \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{37798594-8cb0-48aa-8401-090f09e25dff-3_309_319_982_753} \captionsetup{labelformat=empty} \caption{Fig. 3.2}
    \end{figure} The centre of mass of the combined shape lies on the boundary of C and B .
    The density of \(B\) is not equal to the density of \(C\).
  2. Determine the exact value of \(\frac { \text { density of } \mathrm { C } } { \text { density of } \mathrm { B } }\).
    [0pt] [3]

Question 3:
AnswerMarks Guidance
3(a) Vol of C is 𝜋𝑎3
3𝑎 3𝑎−𝑦 2
(𝜋𝑎3×𝑦̅) = 𝜋∫ ( ) 𝑦d𝑦
3
AnswerMarks
0M1
A12.3
1.1Allow missing π or limits or bracket
wrong, for M1 but y essential. Left
side can be implied, or wrong.
𝜋 9 6 1 3𝑎
(𝜋𝑎3×𝑦̅ =) [ 𝑎2𝑦2− 𝑎𝑦3+ 𝑦4]
9 2 3 4
AnswerMarks Guidance
0A1FT 1.1
Must be a follow through from their
attempt at rearrangement for x2
3
𝑦̅ = 𝑎
AnswerMarks Guidance
4A1 1.1
[5]
AnswerMarks
(b)3
𝜋𝑎3𝐷 × 𝑎 = 8𝑎3𝐷 ×𝑎
AnswerMarks
𝐶 4 𝐵M1
A1FT3.1b
1.1Allow ft from (a) for M1A1; allow
errors in cube for M1 onlyDensity of cone is 𝐷 and of
𝐶
cube 𝐷
𝐵
32
Fraction is
AnswerMarks Guidance
3𝜋A1FT 1.1
[3]
Question 3:
3 | (a) | Vol of C is 𝜋𝑎3 | B1 | 2.1
3𝑎 3𝑎−𝑦 2
(𝜋𝑎3×𝑦̅) = 𝜋∫ ( ) 𝑦d𝑦
3
0 | M1
A1 | 2.3
1.1 | Allow missing π or limits or bracket
wrong, for M1 but y essential. Left
side can be implied, or wrong.
𝜋 9 6 1 3𝑎
(𝜋𝑎3×𝑦̅ =) [ 𝑎2𝑦2− 𝑎𝑦3+ 𝑦4]
9 2 3 4
0 | A1FT | 1.1 | Or equivalent
Must be a follow through from their
attempt at rearrangement for x2
3
𝑦̅ = 𝑎
4 | A1 | 1.1 | AG
[5]
(b) | 3
𝜋𝑎3𝐷 × 𝑎 = 8𝑎3𝐷 ×𝑎
𝐶 4 𝐵 | M1
A1FT | 3.1b
1.1 | Allow ft from (a) for M1A1; allow
errors in cube for M1 only | Density of cone is 𝐷 and of
𝐶
cube 𝐷
𝐵
32
Fraction is
3𝜋 | A1FT | 1.1
[3]
3 In this question you must show detailed reasoning.
[In this question you may use the formula: Volume of cone $= \frac { 1 } { 3 } \times$ base area × height.]\\
The region between the line $\mathrm { y } = - 3 \mathrm { x } + 3 \mathrm { a }$, where $a > 0$, the $x$-axis and the $y$-axis is rotated about the $y$-axis to form a uniform right circular cone C with base radius $a$.
\begin{enumerate}[label=(\alph*)]
\item Show that the centre of mass of C is $\frac { 3 } { 4 } a$ from its base.

The cone C is fixed on top of a uniform cube, B , of edge length $2 a$, so that there is no overlap. Fig. 3.1 shows a side view of C and B fixed together; Fig. 3.2 shows a view of C and B from above.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{37798594-8cb0-48aa-8401-090f09e25dff-3_570_323_785_246}
\captionsetup{labelformat=empty}
\caption{Fig. 3.1}
\end{center}
\end{figure}

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{37798594-8cb0-48aa-8401-090f09e25dff-3_309_319_982_753}
\captionsetup{labelformat=empty}
\caption{Fig. 3.2}
\end{center}
\end{figure}

The centre of mass of the combined shape lies on the boundary of C and B .\\
The density of $B$ is not equal to the density of $C$.
\item Determine the exact value of $\frac { \text { density of } \mathrm { C } } { \text { density of } \mathrm { B } }$.\\[0pt]
[3]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics B AS 2021 Q3 [8]}}