| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Lamina suspended in equilibrium |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics question requiring integration to find centre of mass coordinates (including a 'show that' proof involving substitution), followed by applying equilibrium conditions for a suspended lamina. The integrations are non-trivial (involving √(x²+9) which requires substitution or recognition of standard forms), and part (c) requires geometric reasoning about the equilibrium configuration. While systematic, it demands multiple advanced techniques and careful execution across three connected parts. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | DR |
| Answer | Marks |
|---|---|
| 15ln3 ln3 | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | If this is not seen in (a) the mark |
| Answer | Marks |
|---|---|
| must be seen. | 1 |
| Answer | Marks |
|---|---|
| (b) | DR |
| Answer | Marks |
|---|---|
| 15ln3 | M1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Correct integral and correct limits |
| Answer | Marks |
|---|---|
| 3 | May have used substitution |
| (c) | DR |
| Answer | Marks |
|---|---|
| 𝜃 =awrt39.6° | B1 |
| Answer | Marks |
|---|---|
| [3] | 3.4 |
| Answer | Marks |
|---|---|
| 2.2a | Finding the y coord of P. |
Question 5:
5 | (a) | DR
4 15 ( ) 4
dx=15 ln x+ x2 +9
0 x2 +9 0
=15(ln9−ln3)=15ln3
41 1
15∫ ×2𝑥(𝑥2+9)− 2𝑑𝑥
2
0
1 4
=15[(𝑥2+9)2]
0
1 1
=15(42 +9)2 −(02 +9)2
=15 5−3 =30
30 2
x = =
15ln3 ln3 | B1
M1
A1
[3] | 1.1
1.1
1.1 | If this is not seen in (a) the mark
can be awarded here if integral
correctly evaluated in (b).
Integrating. Ignore limits here.
AG. Some intermediate working
must be seen. | 1
Must be in the form 𝑘[(𝑥2+9)2]
Award for fully complete proof only
(b) | DR
1 4 15 2 225 4 1
∫ ( ) 𝑑𝑥 = ∫ 𝑑𝑥
2 √𝑥2+9 2 𝑥2+9
0 0
225 1 [𝑡𝑎𝑛−1𝑥 4
= × ]
2 3 3 0
75 4
= tan −1
2 3
75 4
−1
tan
2 3
y = =awrt2.11
15ln3 | M1
M1
A1
[3] | 1.1
1.1
1.1 | Correct integral and correct limits
Integrating into the form
𝑘[𝑡𝑎𝑛−1𝑥
]
3 | May have used substitution
(c) | DR
1 5
x = 3 => y = oe
1 8
2
3− 1.17952...
𝑙𝑛3
𝑡𝑎𝑛𝜃 = =
15 1.42538...
" "−"2.1101..."
√18
=0.8275...
𝜃 =awrt39.6° | B1
M1
A1
[3] | 3.4
2.1
2.2a | Finding the y coord of P.
5 5
= = 2 or awrt 3.54
2 2
tan𝜃 = ∆𝑥 or ∆𝑦
∆𝑦 ∆𝑥
0.691rads5 In this question you must show detailed reasoning.
The region bounded by the $x$-axis, the $y$-axis, the line $x = 4$ and the curve with equation $\mathrm { y } = \frac { 15 } { \sqrt { \mathrm { x } ^ { 2 } + 9 } }$ is occupied by a uniform lamina.
The centre of mass of the lamina is at the point $G ( \bar { x } , \bar { y } )$ (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{857eca7f-c42d-49a9-ac39-a2fb5bcb9cd5-4_944_954_598_228}
\begin{enumerate}[label=(\alph*)]
\item Show that $\bar { x } = \frac { 2 } { \ln 3 }$.
\item Determine the value of $\bar { y }$. Give your answer correct to $\mathbf { 3 }$ significant figures.\\
$P$ is the point on the curved edge of the lamina where $x = 3$. The lamina is freely suspended from $P$ and hangs in equilibrium in a vertical plane.
\item Determine the acute angle that the longest straight edge of the lamina makes with the vertical.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2022 Q5 [9]}}