OCR MEI Further Mechanics B AS 2022 June — Question 3 10 marks

Exam BoardOCR MEI
ModuleFurther Mechanics B AS (Further Mechanics B AS)
Year2022
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeComposite solid with hemisphere and cylinder/cone
DifficultyStandard +0.8 This is a multi-part Further Maths mechanics question requiring volume of revolution integration, centre of mass calculations for composite solids, and toppling analysis. Part (a) requires detailed proof using integration by parts or similar techniques. While the individual techniques are standard for FM students, the combination of calculus-based COM, composite body analysis, and toppling condition makes this moderately challenging, though still within expected FM scope.
Spec4.08d Volumes of revolution: about x and y axes6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces

3 Fig. 3.1 shows the curve with equation \(y = x ^ { 2 } + 3\). The region \(R\), shown shaded, is bounded by the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = 2\). A uniform solid of revolution S is formed by rotating the region R through \(2 \pi\) about the \(x\)-axis. The volume of \(S\) is \(\frac { 202 } { 5 } \pi\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{feb9a438-26b0-41d3-b044-6acd6efccde0-3_392_547_511_246} \captionsetup{labelformat=empty} \caption{Fig. 3.1}
\end{figure} \section*{(a) In this question you must show detailed reasoning.} Show that the \(x\)-coordinate of the centre of mass of S is \(\frac { 395 } { 303 }\). S is fixed to a cylinder of base radius 3 units and height 2 units to form the uniform solid D . The smaller circular face of S is joined to the top circular face of the cylinder, as shown in Fig. 3.2. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{feb9a438-26b0-41d3-b044-6acd6efccde0-3_394_556_1491_244} \captionsetup{labelformat=empty} \caption{Fig. 3.2}
\end{figure} (b) Find the distance of the centre of mass of D from its smaller circular face. D is placed with its smaller circular face in contact with a rough plane which is inclined at an angle of \(30 ^ { \circ }\) to the horizontal. It is given that D does not slip.
(c) Determine whether D topples.

Question 3:
AnswerMarks Guidance
3(a) [ 202 πœ‹π‘₯Μ… =] πœ‹βˆ«π‘₯(π‘₯2+3)2dπ‘₯
5M1 3.1b
= πœ‹βˆ«(π‘₯5+6π‘₯3+9π‘₯)A1 1.1
1 6 9 2
= πœ‹[ π‘₯6+ π‘₯4+ π‘₯2]
6 4 2
AnswerMarks Guidance
0M1 1.1
correct on ft
202 64 96 36 158
πœ‹π‘₯Μ… = πœ‹( + + )[= ]
AnswerMarks
5 6 4 2 3Must be seen – or equivalent
395
π‘₯Μ… =
AnswerMarks Guidance
303A1 1.1
[4]
AnswerMarks
(b)202 1001
(58.4πœ‹Γ—π‘₯Μ…) = 18πœ‹Γ—1+ πœ‹Γ—
AnswerMarks Guidance
5 303M1 3.1b
1 2Do not need LHS here
A11.1 For both masses correct
A11.1 For both distances correct
π‘₯Μ… = 2.59A1 1.1
on LHS
[4]
AnswerMarks Guidance
(c)2.59tan30Β° = 1.495… M1
Or by finding tanπœƒ = or
π‘‘β„Žπ‘’π‘–π‘Ÿ 2.59
AnswerMarks
inverseβ€¦π‘‘β„Žπ‘’π‘–π‘Ÿ 2.59
Allow tanπœƒ = for
3
M1
AnswerMarks Guidance
This is less than 3 , so does not toppleA1FT 2.2a
Note: angle of toppling is
AnswerMarks
arctan(3/2.59…)=49.155…oFT correct method to
incorrect CoM
[2]
Question 3:
3 | (a) | [ 202 πœ‹π‘₯Μ… =] πœ‹βˆ«π‘₯(π‘₯2+3)2dπ‘₯
5 | M1 | 3.1b | DR in part (a)
= πœ‹βˆ«(π‘₯5+6π‘₯3+9π‘₯) | A1 | 1.1
1 6 9 2
= πœ‹[ π‘₯6+ π‘₯4+ π‘₯2]
6 4 2
0 | M1 | 1.1 | Use of limits and at least one term
correct on ft
202 64 96 36 158
πœ‹π‘₯Μ… = πœ‹( + + )[= ]
5 6 4 2 3 | Must be seen – or equivalent
395
π‘₯Μ… =
303 | A1 | 1.1 | AG
[4]
(b) | 202 1001
(58.4πœ‹Γ—π‘₯Μ…) = 18πœ‹Γ—1+ πœ‹Γ—
5 303 | M1 | 3.1b | For attempting 𝑀π‘₯ +π‘šπ‘₯
1 2 | Do not need LHS here
A1 | 1.1 | For both masses correct | Do not need LHS here
A1 | 1.1 | For both distances correct | Do not need LHS here
π‘₯Μ… = 2.59 | A1 | 1.1 | 2.5936 Accept 2.6 | Must come from 58.4πœ‹Γ—π‘₯Μ…
on LHS
[4]
(c) | 2.59tan30Β° = 1.495… | M1 | 3.1b | 3
Or by finding tanπœƒ = or
π‘‘β„Žπ‘’π‘–π‘Ÿ 2.59
inverse… | π‘‘β„Žπ‘’π‘–π‘Ÿ 2.59
Allow tanπœƒ = for
3
M1
This is less than 3 , so does not topple | A1FT | 2.2a | …and correct deduction
Note: angle of toppling is
arctan(3/2.59…)=49.155…o | FT correct method to
incorrect CoM
[2]
3 Fig. 3.1 shows the curve with equation $y = x ^ { 2 } + 3$. The region $R$, shown shaded, is bounded by the curve, the $x$-axis, the $y$-axis and the line $x = 2$. A uniform solid of revolution S is formed by rotating the region R through $2 \pi$ about the $x$-axis.

The volume of $S$ is $\frac { 202 } { 5 } \pi$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{feb9a438-26b0-41d3-b044-6acd6efccde0-3_392_547_511_246}
\captionsetup{labelformat=empty}
\caption{Fig. 3.1}
\end{center}
\end{figure}

\section*{(a) In this question you must show detailed reasoning.}
Show that the $x$-coordinate of the centre of mass of S is $\frac { 395 } { 303 }$.

S is fixed to a cylinder of base radius 3 units and height 2 units to form the uniform solid D . The smaller circular face of S is joined to the top circular face of the cylinder, as shown in Fig. 3.2.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{feb9a438-26b0-41d3-b044-6acd6efccde0-3_394_556_1491_244}
\captionsetup{labelformat=empty}
\caption{Fig. 3.2}
\end{center}
\end{figure}

(b) Find the distance of the centre of mass of D from its smaller circular face.

D is placed with its smaller circular face in contact with a rough plane which is inclined at an angle of $30 ^ { \circ }$ to the horizontal. It is given that D does not slip.\\
(c) Determine whether D topples.

\hfill \mbox{\textit{OCR MEI Further Mechanics B AS 2022 Q3 [10]}}