| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | L-shaped or composite rectangular lamina |
| Difficulty | Challenging +1.2 This is a multi-part centre of mass question requiring solid integration techniques and composite body reasoning. Part (a) involves volume integrals for a solid of revolution (M3 standard but requires careful setup). Part (b) involves area/moment integrals for planar laminas with a curve boundary, plus using symmetry/complementary regions in (ii). While technically demanding with multiple integration steps, these are standard M3 techniques without requiring novel insight—slightly above average difficulty due to the computational load and multi-step nature. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| \(V = \int_0^h \pi(y^4)^{\frac{1}{3}} dy\) | M1 | For \(\int \ldots (y^4)^{\frac{1}{3}} dy\) |
| \(= \pi\left[\frac{2}{3}y^{\frac{1}{3}}\right]_0^h \, \left(= \frac{2}{3}\pi h^{\frac{3}{2}}\right)\) | A1 | For \(\frac{2}{3}y^{\frac{1}{3}}\) |
| \(V\bar{y} = \int \pi x^2 y \, dy = \int_0^h \pi y^2 y \, dy\) | M1 | For \(\int x^2 y \, dy\) |
| \(= \pi\left[\frac{2}{5}y^{\frac{5}{3}}\right]_0^h \, \left(= \frac{2}{5}\pi h^{\frac{5}{3}}\right)\) | A1 | For \(\frac{2}{5}y^{\frac{5}{3}}\) |
| \(\bar{y} = \frac{\frac{2}{5}\pi h^{\frac{5}{3}}}{\frac{2}{3}\pi h^{\frac{1}{3}}} = \frac{3}{5}h\) | A1 | |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| \(A = \int_0^4 (x + \sqrt{x}) \, dx\) | M1 | For \(\int (x + \sqrt{x}) \, dx\) |
| \(= \left[\frac{1}{2}x^2 + \frac{2}{3}x^{\frac{3}{2}}\right]_0^4 \, \left(= \frac{40}{3}\right)\) | A1 | For \(\frac{1}{2}x^2 + \frac{2}{3}x^{\frac{3}{2}}\) |
| \(A\bar{x} = \int xy \, dx = \int_0^4 x(x + \sqrt{x}) \, dx\) | M1 | For \(\int xy \, dx\) |
| \(= \left[\frac{1}{3}x^3 + \frac{2}{5}x^{\frac{5}{2}}\right]_0^4 \, \left(= \frac{512}{15}\right)\) | A1 | For \(\frac{1}{3}x^3 + \frac{2}{5}x^{\frac{5}{2}}\) |
| \(\bar{x} = \frac{\frac{512}{15}}{\frac{40}{3}} = \frac{64}{25} = 2.56\) | E1 | |
| \(A\bar{y} = \int \frac{1}{2}y^2 \, dx = \int_0^4 \frac{1}{2}(x + \sqrt{x})^2 \, dx\) | M1 | For \(\ldots y^2 \, dx\) |
| \(= \left[\frac{1}{6}x^3 + \frac{2}{5}x^{\frac{5}{2}} + \frac{1}{4}x^2\right]_0^4 \, \left(= \frac{412}{15}\right)\) | A1A1 | For \(\frac{1}{6}x^3 + \frac{2}{5}x^{\frac{5}{2}} + \frac{1}{4}x^2\). Give A1 for two correct terms |
| \(\bar{y} = \frac{\frac{412}{15}}{\frac{40}{3}} = \frac{103}{50} = 2.06\) | A1 | |
| [9] |
| Answer | Marks | Guidance |
|---|---|---|
| Area of \(B\) is \(24 - \frac{40}{3} = \frac{32}{3}\) | ||
| \(\frac{32}{3}\left(\bar{x}\right) + \frac{40}{3}(2.56) = 24\left(\frac{2}{3}\right)\) | M1, M1 | CM of composite body. Correct strategy. (One coordinate sufficient) |
| \(\left(\bar{x}, \bar{y}\right) = \left(\frac{1.3}{4.175}\right)\) | A1, A1 | CAO. FT requires \(0 < \bar{y} < 6\). FT is 6.75 – 1.25\(\bar{y}_A\). No FT from wrong area |
| Answer | Marks | Guidance |
|---|---|---|
| OR | ||
| \(\int \frac{1}{4}\left(\sqrt{1 + 4y} - 1\right)^2 y \, dy\) or \(\int x(6 - x - \sqrt{x}) \, dx\) | M1 | For any one of these |
| or \(\int \frac{1}{32}\left(\sqrt{1 + 4y} - 1\right)^4 dy\) | ||
| or \(\int \frac{1}{2}(6 - x - \sqrt{x})(6 + x + \sqrt{x}) \, dx\) | ||
| \(\bar{x} = 1.3, \bar{y} = 4.175\) | M1, A1A1 | For one successful integration. For one successful integration |
| [4] |
### Part (a)
| $V = \int_0^h \pi(y^4)^{\frac{1}{3}} dy$ | M1 | For $\int \ldots (y^4)^{\frac{1}{3}} dy$ |
|---|---|---|
| $= \pi\left[\frac{2}{3}y^{\frac{1}{3}}\right]_0^h \, \left(= \frac{2}{3}\pi h^{\frac{3}{2}}\right)$ | A1 | For $\frac{2}{3}y^{\frac{1}{3}}$ |
|---|---|---|
| $V\bar{y} = \int \pi x^2 y \, dy = \int_0^h \pi y^2 y \, dy$ | M1 | For $\int x^2 y \, dy$ |
|---|---|---|
| $= \pi\left[\frac{2}{5}y^{\frac{5}{3}}\right]_0^h \, \left(= \frac{2}{5}\pi h^{\frac{5}{3}}\right)$ | A1 | For $\frac{2}{5}y^{\frac{5}{3}}$ |
|---|---|---|
| $\bar{y} = \frac{\frac{2}{5}\pi h^{\frac{5}{3}}}{\frac{2}{3}\pi h^{\frac{1}{3}}} = \frac{3}{5}h$ | A1 | |
|---|---|---|
| | [5] | |
|---|---|---|
### Part (b)(i)
| $A = \int_0^4 (x + \sqrt{x}) \, dx$ | M1 | For $\int (x + \sqrt{x}) \, dx$ |
|---|---|---|
| $= \left[\frac{1}{2}x^2 + \frac{2}{3}x^{\frac{3}{2}}\right]_0^4 \, \left(= \frac{40}{3}\right)$ | A1 | For $\frac{1}{2}x^2 + \frac{2}{3}x^{\frac{3}{2}}$ |
|---|---|---|
| $A\bar{x} = \int xy \, dx = \int_0^4 x(x + \sqrt{x}) \, dx$ | M1 | For $\int xy \, dx$ |
|---|---|---|
| $= \left[\frac{1}{3}x^3 + \frac{2}{5}x^{\frac{5}{2}}\right]_0^4 \, \left(= \frac{512}{15}\right)$ | A1 | For $\frac{1}{3}x^3 + \frac{2}{5}x^{\frac{5}{2}}$ |
|---|---|---|
| $\bar{x} = \frac{\frac{512}{15}}{\frac{40}{3}} = \frac{64}{25} = 2.56$ | E1 | |
|---|---|---|
| $A\bar{y} = \int \frac{1}{2}y^2 \, dx = \int_0^4 \frac{1}{2}(x + \sqrt{x})^2 \, dx$ | M1 | For $\ldots y^2 \, dx$ |
|---|---|---|
| $= \left[\frac{1}{6}x^3 + \frac{2}{5}x^{\frac{5}{2}} + \frac{1}{4}x^2\right]_0^4 \, \left(= \frac{412}{15}\right)$ | A1A1 | For $\frac{1}{6}x^3 + \frac{2}{5}x^{\frac{5}{2}} + \frac{1}{4}x^2$. Give A1 for two correct terms |
|---|---|---|
| $\bar{y} = \frac{\frac{412}{15}}{\frac{40}{3}} = \frac{103}{50} = 2.06$ | A1 | |
|---|---|---|
| | [9] | |
|---|---|---|
### Part (b)(ii)
| Area of $B$ is $24 - \frac{40}{3} = \frac{32}{3}$ | | |
|---|---|---|
| $\frac{32}{3}\left(\bar{x}\right) + \frac{40}{3}(2.56) = 24\left(\frac{2}{3}\right)$ | M1, M1 | CM of composite body. Correct strategy. (One coordinate sufficient) |
|---|---|---|
| $\left(\bar{x}, \bar{y}\right) = \left(\frac{1.3}{4.175}\right)$ | A1, A1 | CAO. FT requires $0 < \bar{y} < 6$. FT is 6.75 – 1.25$\bar{y}_A$. No FT from wrong area |
|---|---|---|
| | | |
|---|---|---|
| **OR** | | |
|---|---|---|
| $\int \frac{1}{4}\left(\sqrt{1 + 4y} - 1\right)^2 y \, dy$ or $\int x(6 - x - \sqrt{x}) \, dx$ | M1 | For any one of these |
|---|---|---|
| or $\int \frac{1}{32}\left(\sqrt{1 + 4y} - 1\right)^4 dy$ | | |
|---|---|---|
| or $\int \frac{1}{2}(6 - x - \sqrt{x})(6 + x + \sqrt{x}) \, dx$ | | |
| $\bar{x} = 1.3, \bar{y} = 4.175$ | M1, A1A1 | For one successful integration. For one successful integration |
|---|---|---|
| | [4] | |
|---|---|---|4
\begin{enumerate}[label=(\alph*)]
\item The region enclosed between the curve $y = x ^ { 4 }$ and the line $y = h$ (where $h$ is positive) is rotated about the $y$-axis to form a uniform solid of revolution. Find the $y$-coordinate of the centre of mass of this solid.
\item The region $A$ is bounded by the $x$-axis, the curve $y = x + \sqrt { x }$ for $0 \leqslant x \leqslant 4$, and the line $x = 4$. The region $B$ is bounded by the $y$-axis, the curve $y = x + \sqrt { x }$ for $0 \leqslant x \leqslant 4$, and the line $y = 6$. These regions are shown in Fig. 4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3f674569-7e99-4ba8-84f1-a1eb438e30ed-3_572_513_1779_778}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item A uniform lamina occupies the region $A$. Show that the $x$-coordinate of the centre of mass of this lamina is 2.56 , and find the $y$-coordinate.
\item Using your answer to part (i), or otherwise, find the coordinates of the centre of mass of a uniform lamina occupying the region $B$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2013 Q4 [18]}}