OCR MEI Further Mechanics B AS Specimen — Question 6 12 marks

Exam BoardOCR MEI
ModuleFurther Mechanics B AS (Further Mechanics B AS)
SessionSpecimen
Marks12
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Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeToppling on inclined plane
DifficultyStandard +0.8 This is a multi-part Further Maths mechanics question requiring volume of revolution (standard integral), centre of mass calculation (more involved integral with moment), and toppling analysis. While the techniques are standard for FM students, the question demands careful setup of multiple integrals, algebraic manipulation to reach given answers, and application of equilibrium conditions. The 'show that' format and three-part structure with increasing complexity places it moderately above average difficulty.
Spec4.08d Volumes of revolution: about x and y axes6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces
6 In this question you must show detailed reasoning. As shown in Fig. 6.1, the region R is bounded by the lines \(x = 1 , x = 2 , y = 0\) and the curve \(y = 2 x ^ { 2 }\) for \(1 \leq x \leq 2\). A uniform solid of revolution, S , is formed when R is rotated through \(360 ^ { \circ }\) about the \(x\)-axis. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a01b2e46-e213-4f20-bc2e-5852061d8b91-6_725_449_539_751} \captionsetup{labelformat=empty} \caption{Fig. 6.1}
\end{figure}
  1. Show that the volume of S is \(\frac { 124 \pi } { 5 }\).
  2. Show that the distance of the centre of mass of S from the centre of its smaller circular plane surface is \(\frac { 43 } { 62 }\). Fig. 6.2 shows S placed so that its smaller circular plane surface is in contact with a slope inclined at \(\alpha ^ { \circ }\) to the horizontal. S does not slip but is on the point of tipping. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{a01b2e46-e213-4f20-bc2e-5852061d8b91-6_458_565_2014_694} \captionsetup{labelformat=empty} \caption{Fig. 6.2}
    \end{figure}
  3. Find the value of \(\alpha\), giving your answer in degrees correct to 3 significant figures.
Question 6:
AnswerMarks Guidance
6(i) DR
2(cid:11) 2x2(cid:12) 2
V (cid:32)(cid:83)(cid:179) dx
1
(cid:32)(cid:83)(cid:179) 2(cid:11) 4x4(cid:12) dx= (cid:83) (cid:170)4 x5 (cid:186) 2
(cid:171) (cid:187)
1 (cid:172)5 (cid:188)
1
4
(cid:83)(32(cid:16)1)
5
124(cid:83)
(cid:32) AG
AnswerMarks
5M1
A1
C
A1
AnswerMarks
[3]1.1a
1.1
I
AnswerMarks
1.1E
Allow use of surface density = 1 without
comment
M
At least 2 terms correct. Allow no limits.
AnswerMarks Guidance
6(ii) 2 (cid:11) 2x2(cid:12) 2
Vx (cid:32)(cid:83)(cid:179) x dx
1
2
(cid:170)2x6(cid:186)
(cid:32)(cid:83)(cid:179) 2(cid:11) 4x5(cid:12) dx= (cid:83)(cid:171) (cid:187)
1 (cid:172) 3 (cid:188)
1
(cid:32)42(cid:83)
42(cid:83)
Need x (cid:16)1 = (cid:16)1
(cid:167)124(cid:83)(cid:183)
(cid:168) (cid:184)
(cid:169) 5 (cid:185)
210
= (cid:16)1
124
43
AG
AnswerMarks
62M1
A1
A1
M1
E1
AnswerMarks
[5]3.1b
1.1
1.1
2.2a
AnswerMarks
2.1Limits not required
At least 2 terms correct. Allow no limits.
N
Any form
Must involve expression with their x
E
MClearly established
AnswerMarks Guidance
6(iii) DR
G
α
2
C
α
C is the centre of the base of S and G its CoM
Radius of the base is 2
G is vertically above a point on the circumference of
the base of S
2
tan(cid:68)(cid:32)
CG
43
where CG is from (i) E
62
AnswerMarks
so angle is 70.87477… so 70.9° (to 3 s. f.)B1
B1
C
M1
A1
AnswerMarks
[4]3.1b
2.2a
I
1.1
AnswerMarks
1.1N
E
M
May be implicit e.g. from obtaining the
correct angle
Question 6:
6 | (i) | DR
2(cid:11) 2x2(cid:12) 2
V (cid:32)(cid:83)(cid:179) dx
1
(cid:32)(cid:83)(cid:179) 2(cid:11) 4x4(cid:12) dx= (cid:83) (cid:170)4 x5 (cid:186) 2
(cid:171) (cid:187)
1 (cid:172)5 (cid:188)
1
4
(cid:83)(32(cid:16)1)
5
124(cid:83)
(cid:32) AG
5 | M1
A1
C
A1
[3] | 1.1a
1.1
I
1.1 | E
Allow use of surface density = 1 without
comment
M
At least 2 terms correct. Allow no limits.
6 | (ii) | 2 (cid:11) 2x2(cid:12) 2
Vx (cid:32)(cid:83)(cid:179) x dx
1
2
(cid:170)2x6(cid:186)
(cid:32)(cid:83)(cid:179) 2(cid:11) 4x5(cid:12) dx= (cid:83)(cid:171) (cid:187)
1 (cid:172) 3 (cid:188)
1
(cid:32)42(cid:83)
42(cid:83)
Need x (cid:16)1 = (cid:16)1
(cid:167)124(cid:83)(cid:183)
(cid:168) (cid:184)
(cid:169) 5 (cid:185)
210
= (cid:16)1
124
43
AG
62 | M1
A1
A1
M1
E1
[5] | 3.1b
1.1
1.1
2.2a
2.1 | Limits not required
At least 2 terms correct. Allow no limits.
N
Any form
Must involve expression with their x
E
MClearly established
6 | (iii) | DR
G
α
2
C
α
C is the centre of the base of S and G its CoM
Radius of the base is 2
G is vertically above a point on the circumference of
the base of S
2
tan(cid:68)(cid:32)
CG
43
where CG is from (i) E
62
so angle is 70.87477… so 70.9° (to 3 s. f.) | B1
B1
C
M1
A1
[4] | 3.1b
2.2a
I
1.1
1.1 | N
E
M
May be implicit e.g. from obtaining the
correct angle
6 In this question you must show detailed reasoning.
As shown in Fig. 6.1, the region R is bounded by the lines $x = 1 , x = 2 , y = 0$ and the curve $y = 2 x ^ { 2 }$ for $1 \leq x \leq 2$. A uniform solid of revolution, S , is formed when R is rotated through $360 ^ { \circ }$ about the $x$-axis.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{a01b2e46-e213-4f20-bc2e-5852061d8b91-6_725_449_539_751}
\captionsetup{labelformat=empty}
\caption{Fig. 6.1}
\end{center}
\end{figure}

(i) Show that the volume of S is $\frac { 124 \pi } { 5 }$.\\
(ii) Show that the distance of the centre of mass of S from the centre of its smaller circular plane surface is $\frac { 43 } { 62 }$.

Fig. 6.2 shows S placed so that its smaller circular plane surface is in contact with a slope inclined at $\alpha ^ { \circ }$ to the horizontal. S does not slip but is on the point of tipping.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{a01b2e46-e213-4f20-bc2e-5852061d8b91-6_458_565_2014_694}
\captionsetup{labelformat=empty}
\caption{Fig. 6.2}
\end{center}
\end{figure}

(iii) Find the value of $\alpha$, giving your answer in degrees correct to 3 significant figures.

\hfill \mbox{\textit{OCR MEI Further Mechanics B AS  Q6 [12]}}
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