| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | June |
| Marks | 18 |
| 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 standard M3/Further Mechanics question on centres of mass requiring integration techniques. Part (i) uses standard lamina formulas, part (ii) applies solid of revolution formulas with a straightforward substitution (x² = y - 5), and part (iii) cleverly uses subtraction of solids. While it requires multiple techniques and careful setup, these are well-practiced methods for Further Maths students with no novel conceptual leaps required. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area is \(\int_0^3 (x^2 + 5)\,\mathrm{d}x\) | M1 | For \(\int(x^2+5)\,\mathrm{d}x\) |
| \(= \left[\frac{1}{3}x^3 + 5x\right]_0^3\ (= 24)\) | A1 | For \(\frac{1}{3}x^3 + 5x\) |
| \(\int xy\,\mathrm{d}x = \int_0^3 (x^3 + 5x)\,\mathrm{d}x\) | M1 | For \(\int xy\,\mathrm{d}x\) |
| \(= \left[\frac{1}{4}x^4 + \frac{5}{2}x^2\right]_0^3\ \left(= \frac{171}{4}\right)\) | A1 | For \(\frac{1}{4}x^4 + \frac{5}{2}x^2\) |
| \(\bar{x} = \frac{42.75}{24} = \frac{57}{32} = 1.78125\) | A1 | |
| \(\int\frac{1}{2}y^2\,\mathrm{d}x = \int_0^3 \frac{1}{2}(x^4 + 10x^2 + 25)\,\mathrm{d}x\) | M1 | For \(\int y^2\,\mathrm{d}x\) |
| \(= \left[\frac{1}{10}x^5 + \frac{5}{3}x^3 + \frac{25}{2}x\right]_0^3\ (= 106.8)\) | A2 | For \(\frac{1}{10}x^5 + \frac{5}{3}x^3 + \frac{25}{2}x\); Give A1 for two terms correct |
| \(\bar{y} = \frac{106.8}{24} = \frac{89}{20} = 4.45\) | A1 | |
| Total | 9 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Volume is \(\int\pi x^2\,\mathrm{d}y = \int_5^{14}\pi(y-5)\,\mathrm{d}y\) | M1 | For \(\int(y-5)\,\mathrm{d}y\) |
| \(= \pi\left[\frac{1}{2}y^2 - 5y\right]_5^{14}\ (= 40.5\pi)\) | A1 | For \(\left[\frac{1}{2}y^2 - 5y\right]_5^{14}\) |
| \(\int\pi x^2 y\,\mathrm{d}y = \int_5^{14}\pi(y^2 - 5y)\,\mathrm{d}y\) | M1 | For \(\int x^2 y\,\mathrm{d}x\) |
| \(= \pi\left[\frac{1}{3}y^3 - \frac{5}{2}y^2\right]_5^{14}\ (= 445.5\pi)\) | A1 | For \(\frac{1}{3}y^3 - \frac{5}{2}y^2\) |
| \(\bar{y} = \frac{445.5\pi}{40.5\pi}\) | M1 | *Dependent on previous M1M1* |
| \(= 11\) | A1 | |
| Total | 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Volume of whole cylinder is \(\pi\times 3^2\times 14 = 126\pi\) | ||
| \(126\pi\times 7 = 40.5\pi\times 11 + (126\pi - 40.5\pi)\times\bar{y}_A\) | M1 A1 | Using formula for composite body |
| \(\bar{y}_A = \frac{126\pi\times 7 - 40.5\pi\times 11}{126\pi - 40.5\pi}\) | ||
| \(= \frac{97}{19} = 5.105\) (4 sf) | A1 cao | |
| Total | 3 |
# Question 4:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area is $\int_0^3 (x^2 + 5)\,\mathrm{d}x$ | M1 | For $\int(x^2+5)\,\mathrm{d}x$ |
| $= \left[\frac{1}{3}x^3 + 5x\right]_0^3\ (= 24)$ | A1 | For $\frac{1}{3}x^3 + 5x$ |
| $\int xy\,\mathrm{d}x = \int_0^3 (x^3 + 5x)\,\mathrm{d}x$ | M1 | For $\int xy\,\mathrm{d}x$ |
| $= \left[\frac{1}{4}x^4 + \frac{5}{2}x^2\right]_0^3\ \left(= \frac{171}{4}\right)$ | A1 | For $\frac{1}{4}x^4 + \frac{5}{2}x^2$ |
| $\bar{x} = \frac{42.75}{24} = \frac{57}{32} = 1.78125$ | A1 | |
| $\int\frac{1}{2}y^2\,\mathrm{d}x = \int_0^3 \frac{1}{2}(x^4 + 10x^2 + 25)\,\mathrm{d}x$ | M1 | For $\int y^2\,\mathrm{d}x$ |
| $= \left[\frac{1}{10}x^5 + \frac{5}{3}x^3 + \frac{25}{2}x\right]_0^3\ (= 106.8)$ | A2 | For $\frac{1}{10}x^5 + \frac{5}{3}x^3 + \frac{25}{2}x$; Give A1 for two terms correct |
| $\bar{y} = \frac{106.8}{24} = \frac{89}{20} = 4.45$ | A1 | |
| **Total** | **9** | |
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Volume is $\int\pi x^2\,\mathrm{d}y = \int_5^{14}\pi(y-5)\,\mathrm{d}y$ | M1 | For $\int(y-5)\,\mathrm{d}y$ |
| $= \pi\left[\frac{1}{2}y^2 - 5y\right]_5^{14}\ (= 40.5\pi)$ | A1 | For $\left[\frac{1}{2}y^2 - 5y\right]_5^{14}$ |
| $\int\pi x^2 y\,\mathrm{d}y = \int_5^{14}\pi(y^2 - 5y)\,\mathrm{d}y$ | M1 | For $\int x^2 y\,\mathrm{d}x$ |
| $= \pi\left[\frac{1}{3}y^3 - \frac{5}{2}y^2\right]_5^{14}\ (= 445.5\pi)$ | A1 | For $\frac{1}{3}y^3 - \frac{5}{2}y^2$ |
| $\bar{y} = \frac{445.5\pi}{40.5\pi}$ | M1 | *Dependent on previous M1M1* |
| $= 11$ | A1 | |
| **Total** | **6** | |
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Volume of whole cylinder is $\pi\times 3^2\times 14 = 126\pi$ | | |
| $126\pi\times 7 = 40.5\pi\times 11 + (126\pi - 40.5\pi)\times\bar{y}_A$ | M1 A1 | Using formula for composite body |
| $\bar{y}_A = \frac{126\pi\times 7 - 40.5\pi\times 11}{126\pi - 40.5\pi}$ | | |
| $= \frac{97}{19} = 5.105$ (4 sf) | A1 cao | |
| **Total** | **3** | |4 The region $A$ is bounded by the curve $y = x ^ { 2 } + 5$ for $0 \leqslant x \leqslant 3$, the $x$-axis, the $y$-axis and the line $x = 3$. The region $B$ is bounded by the curve $y = x ^ { 2 } + 5$ for $0 \leqslant x \leqslant 3$, the $y$-axis and the line $y = 14$. These regions are shown in Fig. 4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5ecb198d-7863-4fc2-81b6-c8b6c37b1859-5_883_554_431_794}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
(i) Find the coordinates of the centre of mass of a uniform lamina occupying the region $A$.\\
(ii) The region $B$ is rotated through $2 \pi$ radians about the $y$-axis to form a uniform solid of revolution $R$. Find the $y$-coordinate of the centre of mass of the solid $R$.\\
(iii) The region $A$ is rotated through $2 \pi$ radians about the $y$-axis to form a uniform solid of revolution $S$. Using your answer to part (ii), or otherwise, find the $y$-coordinate of the centre of mass of the solid $S$.
\hfill \mbox{\textit{OCR MEI M3 2011 Q4 [18]}}