| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Solid on inclined plane - toppling |
| Difficulty | Challenging +1.3 This is a multi-part M3 question requiring solid of revolution centre of mass (integration with Pappus), algebraic manipulation to reach a given form, toppling analysis on an inclined plane, and lamina centre of mass. While technically demanding with several steps, these are standard M3 techniques applied systematically rather than requiring novel insight. The toppling proof is elegant but follows directly from comparing the centre of mass position to the critical angle. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Volume is \(\displaystyle\int_0^k \pi(e^{-x})^2\,dx\) | M1 | \(\pi\) may be omitted throughout |
| \(= \pi\left[-\frac{1}{2}e^{-2x}\right]_0^k \left\{= \frac{1}{2}\pi(1-e^{-2k})\right\}\) | A1 | For \(-\frac{1}{2}e^{-2x}\) |
| \(\displaystyle\int \pi xy^2\,dx\) | M1 | |
| \(= \displaystyle\int_0^k \pi x e^{-2x}\,dx = \pi\left[-\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x}\right]_0^k\) | A1A1 | For \(-\frac{1}{2}xe^{-2x}\) and \(-\frac{1}{4}e^{-2x}\) |
| \(= \frac{1}{4}\pi(1 - 2ke^{-2k} - e^{-2k})\) | ||
| \(\bar{x} = \dfrac{1 - 2ke^{-2k} - e^{-2k}}{2(1-e^{-2k})}\) | A1 | Any correct form |
| \(= \dfrac{1-e^{-2k}}{2(1-e^{-2k})} - \dfrac{2ke^{-2k}}{2(1-e^{-2k})} = \dfrac{1}{2} - \dfrac{k}{e^{2k}-1}\) | E1 [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(OG < \frac{1}{2}\) for all values of \(k\) | B1 | OR \(\dfrac{k}{e^{2k}-1} > 0\) o.e. stated or implied; allow \(\bar{x} \to \frac{1}{2}\) as \(k \to \infty\) for B1 |
| \(OP = (1)\tan 30° = \dfrac{1}{\sqrt{3}}\) \((= 0.577)\) | M1, A1 | Trigonometry in OAP or OAG; or \(O\hat{A}G < \tan^{-1}\frac{1}{2}\) \((= 26.6°)\) |
| \(OG < OP\) (or \(O\hat{A}G < 30°\)) so G is to the right of AP and solid will not topple | E1 [4] | Fully correct explanation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Area is \(\displaystyle\int_0^k e^{-x}\,dx = \left[-e^{-x}\right]_0^k\) \((= 1-e^{-k})\) | B1 | |
| \(\displaystyle\int xy\,dx\) | M1 | |
| \(= \displaystyle\int_0^k xe^{-x}\,dx = \left[-xe^{-x} - e^{-x}\right]_0^k\) | A1 | |
| \(\bar{x} = \dfrac{1-ke^{-k}-e^{-k}}{1-e^{-k}}\) | A1 | Any correct form; e.g. \(1 - \dfrac{k}{e^k-1}\) |
| \(\displaystyle\int \frac{1}{2}y^2\,dx\) | M1 | For \(\displaystyle\int \ldots y^2\,dx\) |
| \(= \displaystyle\int_0^k \frac{1}{2}e^{-2x}\,dx = \left[-\frac{1}{4}e^{-2x}\right]_0^k\) | A1 | |
| \(\bar{y} = \dfrac{1-e^{-2k}}{4(1-e^{-k})}\) | A1 [7] | Any correct form; e.g. \(\frac{1}{4}(1+e^{-k})\) |
# Question 4:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Volume is $\displaystyle\int_0^k \pi(e^{-x})^2\,dx$ | M1 | $\pi$ may be omitted throughout |
| $= \pi\left[-\frac{1}{2}e^{-2x}\right]_0^k \left\{= \frac{1}{2}\pi(1-e^{-2k})\right\}$ | A1 | For $-\frac{1}{2}e^{-2x}$ |
| $\displaystyle\int \pi xy^2\,dx$ | M1 | |
| $= \displaystyle\int_0^k \pi x e^{-2x}\,dx = \pi\left[-\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x}\right]_0^k$ | A1A1 | For $-\frac{1}{2}xe^{-2x}$ and $-\frac{1}{4}e^{-2x}$ |
| $= \frac{1}{4}\pi(1 - 2ke^{-2k} - e^{-2k})$ | | |
| $\bar{x} = \dfrac{1 - 2ke^{-2k} - e^{-2k}}{2(1-e^{-2k})}$ | A1 | Any correct form |
| $= \dfrac{1-e^{-2k}}{2(1-e^{-2k})} - \dfrac{2ke^{-2k}}{2(1-e^{-2k})} = \dfrac{1}{2} - \dfrac{k}{e^{2k}-1}$ | E1 [7] | |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $OG < \frac{1}{2}$ for all values of $k$ | B1 | OR $\dfrac{k}{e^{2k}-1} > 0$ o.e. stated or implied; allow $\bar{x} \to \frac{1}{2}$ as $k \to \infty$ for B1 |
| $OP = (1)\tan 30° = \dfrac{1}{\sqrt{3}}$ $(= 0.577)$ | M1, A1 | Trigonometry in OAP or OAG; or $O\hat{A}G < \tan^{-1}\frac{1}{2}$ $(= 26.6°)$ |
| $OG < OP$ (or $O\hat{A}G < 30°$) so G is to the right of AP and solid will not topple | E1 [4] | Fully correct explanation |
## Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Area is $\displaystyle\int_0^k e^{-x}\,dx = \left[-e^{-x}\right]_0^k$ $(= 1-e^{-k})$ | B1 | |
| $\displaystyle\int xy\,dx$ | M1 | |
| $= \displaystyle\int_0^k xe^{-x}\,dx = \left[-xe^{-x} - e^{-x}\right]_0^k$ | A1 | |
| $\bar{x} = \dfrac{1-ke^{-k}-e^{-k}}{1-e^{-k}}$ | A1 | Any correct form; e.g. $1 - \dfrac{k}{e^k-1}$ |
| $\displaystyle\int \frac{1}{2}y^2\,dx$ | M1 | For $\displaystyle\int \ldots y^2\,dx$ |
| $= \displaystyle\int_0^k \frac{1}{2}e^{-2x}\,dx = \left[-\frac{1}{4}e^{-2x}\right]_0^k$ | A1 | |
| $\bar{y} = \dfrac{1-e^{-2k}}{4(1-e^{-k})}$ | A1 [7] | Any correct form; e.g. $\frac{1}{4}(1+e^{-k})$ |4 The region $R$ is bounded by the $x$-axis, the $y$-axis, the curve $y = \mathrm { e } ^ { - x }$ and the line $x = k$, where $k$ is a positive constant.\\
(i) The region $R$ is rotated through $2 \pi$ radians about the $x$-axis to form a uniform solid of revolution. Find the $x$-coordinate of the centre of mass of this solid, and show that it can be written in the form
$$\frac { 1 } { 2 } - \frac { k } { \mathrm { e } ^ { 2 k } - 1 } .$$
(ii) The solid in part (i) is placed with its larger circular face in contact with a rough plane inclined at $60 ^ { \circ }$ to the horizontal, as shown in Fig. 4, and you are given that no slipping occurs.
\begin{figure}[h]
\begin{center}
\begin{tikzpicture}[line width=0.8pt, line cap=round, line join=round]
% Horizontal dashed line
\draw[densely dashed] (0,0) -- (3.5,0);
% Angle arc and text
\draw (1.2,0) arc (0:60:1.2);
\node at (30:0.75) {$60^\circ$};
% Inclined plane
\draw (0,0) -- (60:6.5);
% The symmetric object
% - Top and bottom are flat and parallel to the slope
% - Sides are convex (bulging outward slightly)
% - Perfectly symmetric
\begin{scope}[shift={(60:3.5)}, rotate=60]
\draw[fill=white, thick]
(-1.5, 0) -- (1.5, 0) % Flat bottom on the incline
to[bend left=30] (0.8, 3) % Convex curve for the right side
-- (-0.8, 3) % Flat top edge, parallel to base/slope
to[bend left=30] cycle; % Convex curve for the left side
\end{scope}
\end{tikzpicture}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
Show that, whatever the value of $k$, the solid will not topple.\\
(iii) A uniform lamina occupies the region $R$. Find, in terms of $k$, the coordinates of the centre of mass of this lamina.
\section*{END OF QUESTION PAPER}
\hfill \mbox{\textit{OCR MEI M3 2014 Q4 [18]}}