| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Minor (Further Mechanics Minor) |
| Year | 2020 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Toppling on inclined plane |
| Difficulty | Standard +0.3 This is a straightforward centre of mass problem requiring standard toppling conditions (vertical line through COM passes through pivot edge) and basic COM calculations for composite bodies. Part (a) uses simple geometry of toppling, part (b) applies tan α = base/2height, and part (c) uses the standard weighted average formula for composite COM. All techniques are routine for Further Mechanics students with no novel insight required. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass |
| Answer | Marks |
|---|---|
| 1(a) | 12= 1h |
| Answer | Marks |
|---|---|
| h=36 | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | Uses the fact that the height of the |
| Answer | Marks |
|---|---|
| solids | This mark may be implied |
| Answer | Marks |
|---|---|
| 1(b) | tanα= 6 |
| Answer | Marks |
|---|---|
| α=26.6 | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | 1 w |
| Answer | Marks |
|---|---|
| reciprocal) | or α+tan−112 =90 |
| Answer | Marks |
|---|---|
| 1(c) | 288×12+( 24+1h )( 6h )=( 288+6h ) y |
| Answer | Marks |
|---|---|
| 7 | M1 |
| Answer | Marks |
|---|---|
| [2] | 2.1 |
| 2.2a | Takes moments about base of |
| Answer | Marks |
|---|---|
| oe | Allow with their h from (a) |
Question 1:
--- 1(a) ---
1(a) | 12= 1h
3
h=36 | M1
A1
[2] | 1.1
1.1 | Uses the fact that the height of the
centre of mass is the same for both
solids | This mark may be implied
by sight of correct answer
--- 1(b) ---
1(b) | tanα= 6
12
α=26.6 | M1
A1
[2] | 1.1
1.1 | 1 w
For using tanα= 2 (allow
12
reciprocal) | or α+tan−112 =90
6
For reference:
26.5650511…
--- 1(c) ---
1(c) | 288×12+( 24+1h )( 6h )=( 288+6h ) y
3
y =156
7 | M1
A1
[2] | 2.1
2.2a | Takes moments about base of
compound shape – correct number of
terms and dimensionally consistent
oe | Allow with their h from (a)
22.28571429…
1 A uniform solid rectangular prism has cross-section with width $w \mathrm {~cm}$ and height 24 cm . Another uniform solid prism has cross-section in the shape of an isosceles triangle with width $w \mathrm {~cm}$ and height $h \mathrm {~cm}$. The prisms are both placed with their axes vertical on a rough horizontal plane (see Fig. 1.1, which shows the cross-sections through the centres of mass of both solids).
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6418c1b7-092a-4747-bc88-1b57815c6ad9-2_520_1123_520_246}
\captionsetup{labelformat=empty}
\caption{Fig. 1.1}
\end{center}
\end{figure}
The plane is slowly tilted and both solids remain in equilibrium until the angle of inclination of the plane reaches $\alpha$, when both solids topple simultaneously.
\begin{enumerate}[label=(\alph*)]
\item Determine the value of $h$.
It is given that $w = 12$.
\item Determine the value of $\alpha$.
Both prisms are made from the same material and are of uniform density. The triangular prism is now placed on top of the rectangular prism to form a composite body C such that the base of the triangular prism coincides with the top of the rectangular prism. A cross-section of C is shown in Fig. 1.2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6418c1b7-092a-4747-bc88-1b57815c6ad9-2_777_439_1784_258}
\captionsetup{labelformat=empty}
\caption{Fig. 1.2}
\end{center}
\end{figure}
\item Determine the height of the centre of mass of C from its base.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2020 Q1 [6]}}