| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Major (Further Mechanics Major) |
| Session | Specimen |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of lamina by integration |
| Difficulty | Challenging +1.2 This is a multi-part centre of mass question requiring integration to find centroids, composite body calculations, and moment equilibrium. While it involves several steps and careful bookkeeping across three parts, the techniques are standard Further Maths mechanics: integrating to find centre of mass (part i), combining centres of mass using the formula for composite bodies (part ii), and applying moment equilibrium with tension conditions (part iii). The 7+4+5=16 marks reflect length rather than conceptual difficulty—each part follows established procedures without requiring novel insight or particularly challenging integration. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | (i) | P |
| Answer | Marks |
|---|---|
| 3 15 5 | B1 |
| Answer | Marks |
|---|---|
| [7] | 2.4 |
| Answer | Marks |
|---|---|
| 1.1 | Symmetry needs to be stated |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | (ii) | Let X ( = XG) be distance of com from CD |
| Answer | Marks |
|---|---|
| 3 | M1 |
| Answer | Marks |
|---|---|
| [4] | 1.1 |
| Answer | Marks |
|---|---|
| 2.1 | Moments |
| Answer | Marks | Guidance |
|---|---|---|
| 11 | (iii) | (cid:167) 2k(cid:183) |
| Answer | Marks |
|---|---|
| 6 | B1 |
| Answer | Marks |
|---|---|
| A1 | 3.4 |
| Answer | Marks |
|---|---|
| 1.1 | Take moments: all terms present |
| Answer | Marks | Guidance |
|---|---|---|
| D A | M1 | M |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | A1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 10k(cid:14)30 3 | M1 | |
| E | C | Solving 3-term quadratic, |
| Answer | Marks |
|---|---|
| Comment on taking positive root | B1 |
| Answer | Marks |
|---|---|
| P | [5] |
Question 11:
11 | (i) | P
DR
x (cid:32)0 by symmetry
S
y(cid:86) (cid:180) (cid:181) 1 1 k (cid:11) 1(cid:16)x2(cid:12) dx(cid:32)(cid:86) (cid:180) (cid:181) 1 1 k2(cid:11) 1(cid:16)2x2 (cid:14)x4(cid:12) dx
(cid:182) 2 (cid:182) 8
(cid:16)1 (cid:16)1
1 1
(cid:170)1 (cid:167) 1 (cid:183)(cid:186) (cid:170)1 (cid:167) 2 1 (cid:183)(cid:186)
y (cid:171) k(cid:168)x(cid:16) x3 (cid:184)(cid:187) (cid:32) (cid:171) k2 (cid:168)x(cid:16) x3(cid:14) x5 (cid:184)(cid:187)
(cid:172)2 (cid:169) 3 (cid:185)(cid:188) (cid:172)8 (cid:169) 3 5 (cid:185)(cid:188)
(cid:16)1 (cid:16)1
2 2 1
ky (cid:32) k2: y (cid:32) k AG
3 15 5 | B1
M1
M1
A1
A1
M1
A1
[7] | 2.4
2.5
1.1
2.1
1.1
1.1
1.1 | Symmetry needs to be stated
Use of correct formula
Reasonable attempt at integration
LHS
RHS
Substitute limits
11 | (ii) | Let X ( = XG) be distance of com from CD
(cid:167)1(cid:183) 2 (cid:167) 1 (cid:183)
2(cid:168) (cid:184)(cid:14) k(cid:168)1(cid:14) k(cid:184)
(cid:169)2(cid:185) 3 (cid:169) 5 (cid:185)
X (cid:32)
2
2(cid:14) k
3
1
1(cid:14) (10k(cid:14)2k2)
15 2k2 (cid:14)10k(cid:14)15
(cid:32) (cid:32) AG
1 10k(cid:14)30
(6(cid:14)2k)
3 | M1
A1
A1
A1
[4] | 1.1
1.1
1.1
2.1 | Moments
Numerator
DenominaNtor
E
11 | (iii) | (cid:167) 2k(cid:183)
Consider forces vertically: 3T (cid:32)(cid:168) 2(cid:14) (cid:184)g
(cid:169) 3 (cid:185)
Take moments about D:
(cid:167) 2k(cid:183) (cid:11) 2k2 (cid:14)10k(cid:14)15 (cid:12)
2T (cid:32)(cid:168)2(cid:14) (cid:184)g(cid:117)
(cid:169) 3 (cid:185) (10k(cid:14)30)
Comment on taking positive root
1(cid:11) (cid:12)
k (cid:32) (cid:16)5(cid:14) 115 (cid:32)0.9539...
6 | B1
M1
M1
B1
A1 | 3.4
3.1b
1.1
2.3
1.1 | Take moments: all terms present
N
SoElving 3-term quadratic,
dependent on previous M1
BC
Alternative Methods
T :T (cid:32)1:2(cid:159)DG:GA(cid:32)2:1
D A | M1 | M
Use of ratio of moments
2
(cid:159)DG(cid:32)
3 | A1 | A1 | I
2k2 (cid:14)10k(cid:14)15 2
(cid:32)
10k(cid:14)30 3 | M1
E | C | Solving 3-term quadratic,
dependent on previous M1
1(cid:11) (cid:12)
(cid:159)k (cid:32) (cid:16)5(cid:14) 115 (cid:32)0.9539...
6
Comment on taking positive root | B1
A1
P | [5]
Solving 3-term quadratic,
dependent on previous M1The region bounded by the $x$-axis and the curve $y = \frac{1}{2}k(1-x^2)$ for $-1 \leq x \leq 1$ is occupied by a uniform lamina, as shown in Fig. 11.1.
\includegraphics{figure_11_1}
\begin{enumerate}[label=(\roman*)]
\item In this question you must show detailed reasoning.
Show that the centre of mass of the lamina is at $\left(0, \frac{1}{5}k\right)$. [7]
\end{enumerate}
A shop sign is modelled as a uniform lamina in the form of the lamina in part (i) attached to a rectangle ABCD, where AB = 2 and BC = 1. The sign is suspended by two vertical wires attached at A and D, as shown in Fig. 11.2.
\includegraphics{figure_11_2}
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Show that the centre of mass of the sign is at a distance
$$\frac{2k^2 + 10k + 15}{10k + 30}$$
from the midpoint of CD. [4]
\end{enumerate}
The tension in the wire at A is twice the tension in the wire at D.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find the value of $k$. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Major Q11 [16]}}