| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Major (Further Mechanics Major) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of lamina by integration |
| Difficulty | Standard +0.8 This is a multi-part Further Mechanics question requiring center of mass calculation via integration (finding moments and area of a cubic curve region), followed by toppling analysis in two orientations. While the integration is moderately challenging with a cubic function, the toppling conditions are standard applications of comparing the vertical line through the center of mass with the pivot edge. The question requires solid technique across multiple concepts but follows predictable patterns for this topic. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks |
|---|---|
| 7(a) | Meets x-axis at (2, 0) |
| Answer | Marks |
|---|---|
| 35 | B1 |
| Answer | Marks |
|---|---|
| [5] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | soi (e.g. intercepts axis when x = 2 or as the top limit of an |
| Answer | Marks |
|---|---|
| 7(b) | 0.6 21 |
| Answer | Marks |
|---|---|
| 30>21.9... or 22 so will topple | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1b |
| 2.2a | x |
| Answer | Marks | Guidance |
|---|---|---|
| 0.6 21 | M1 | y |
| Answer | Marks | Guidance |
|---|---|---|
| 68.0...+30(=98.0...)>90 so will topple | A1 | Correct angle (to at least 2 sf rot), comparison (oe) or |
| Answer | Marks |
|---|---|
| 7(c) | 52 |
| Answer | Marks |
|---|---|
| 68.0...>30 so will not topple | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1b |
| 2.2a | y |
| Answer | Marks | Guidance |
|---|---|---|
| 35 | M1 | x |
| Answer | Marks | Guidance |
|---|---|---|
| 21.9...+30(=51.9... or 52)<90 so will not topple | A1 | Correct angle (to at least 2 sf rot), comparison (oe) or |
Question 7:
--- 7(a) ---
7(a) | Meets x-axis at (2, 0)
Area of L is 4
Use of 4x =∫xydx or 4y = 1∫y2dx
2
x =0.6
y = 52
35 | B1
B1
M1
A1
A1
[5] | 1.1
1.1
1.1a
1.1
1.1 | soi (e.g. intercepts axis when x = 2 or as the top limit of an
integral)
2
BC or from ∫ (x3 −3x2 +4)dx
0
soi – either correct coordinate implies this mark. Must
substitute correct equation for y which is y = x3−3x2 +4-
integral limits not required for this mark and allow their
area of L (which if correct is 4) but their area must be
numerical
BC
BC (1.485714…) allow awrt 1.49 www (but not just 1.5)
--- 7(b) ---
7(b) | 0.6 21
tanα= = ⇒α=21.9...
52 52
35
30>21.9... or 22 so will topple | M1
A1
[2] | 3.1b
2.2a | x
Use of tanα= (or reciprocal) with their values from (a)
y
and find corresponding value of α(to at least 2 sf rot) oe
Correct angle (to at least 2 sf rot), comparison (or a correct
argument) and ‘will topple’ or compare 0.57… with 21
52
Alternative Solution
52
35 52
tanα= = ⇒α=68.0...
0.6 21 | M1 | y
Use of tanα= (or reciprocal) and find corresponding
x
value of α(to at least 2 sf rot) oe
68.0...+30(=98.0...)>90 so will topple | A1 | Correct angle (to at least 2 sf rot), comparison (oe) or
could mention ‘vertical’ (oe), and ‘will topple’
--- 7(c) ---
7(c) | 52
35 52
tanα= = ⇒α=68.0...
0.6 21
68.0...>30 so will not topple | M1
A1
[2] | 3.1b
2.2a | y
Use of tanα= (or reciprocal) with their values from
x
part (a) and find corresponding value of α(to at least 2 sf
rot) oe
Correct angle (to at least 2 sf rot), comparison (or a correct
argument) and ‘will not topple’ or compare 0.57… with 52
21
Alternative Solution
0.6 21
tanα= = ⇒α=21.9...
52 52
35 | M1 | x
Use of tanα= (or reciprocal) with their values from (a)
y
and find corresponding value of α(to at least 2 sf rot) oe
21.9...+30(=51.9... or 52)<90 so will not topple | A1 | Correct angle (to at least 2 sf rot), comparison (oe) or
could mention ‘vertical’ (oe), and ‘will not topple’The region bounded by the curve $y = x^3 - 3x^2 + 4$, the positive $x$-axis and the positive $y$-axis is occupied by a uniform lamina L. The vertices of L are O, A and B, where O is the origin, A is a point on the positive $x$-axis and B is a point on the positive $y$-axis (see diagram).
\includegraphics{figure_7}
\begin{enumerate}[label=(\alph*)]
\item Determine the coordinates of the centre of mass of L. [5]
\end{enumerate}
The lamina L is the cross-section through the centre of mass of a uniform solid prism M.
The prism M is placed on an inclined plane, which makes an angle of $30°$ with the horizontal, so that OA lies along a line of greatest slope of the plane with O lower down the plane than A.
It is given that M does \textbf{not} slip on the plane.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Determine whether M will topple in this case. Give a reason to support your answer. [2]
\end{enumerate}
The prism M is now placed on the same inclined plane so that OB lies along a line of greatest slope of the plane with O lower down the plane than B.
It is given that M still does \textbf{not} slip on the plane.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Determine whether M will topple in this case. Give a reason to support your answer. [2]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2024 Q7 [9]}}