| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2009 |
| 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 Further Maths M3 centre of mass question requiring integration techniques. Part (a) involves straightforward integration of exponential functions for a 2D lamina. Part (b) requires volume and moment integrals for a solid of revolution with algebraic manipulation to reach given results. While it involves multiple steps and careful algebra, the methods are direct applications of standard formulae without requiring novel insight or particularly complex reasoning. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Area is \(\int_0^{\ln 3} e^x\,dx = \left[e^x\right]_0^{\ln 3} = 2\) | M1, A1 | |
| \(\int xy\,dx = \int_0^{\ln 3} xe^x\,dx\) | M1 | |
| \(= \left[xe^x - e^x\right]_0^{\ln 3} = 3\ln 3 - 2\) | M1, A1 | Integration by parts for \(xe^x - e^x\) |
| \(\bar{x} = \frac{3\ln 3 - 2}{2} = \frac{3}{2}\ln 3 - 1\) | A1 | ww full marks (B4). Give B3 for \(0.65\) |
| \(\int \frac{1}{2}y^2\,dx = \int_0^{\ln 3} \frac{1}{2}(e^x)^2\,dx\) | M1 | For integral of \((e^x)^2\) |
| \(= \left[\frac{1}{4}e^{2x}\right]_0^{\ln 3} = 2\) | A1 | For \(\frac{1}{4}e^{2x}\) |
| \(\bar{y} = \frac{2}{2} = 1\) | A1 | If area wrong, SC1 for \(\bar{x} = \frac{3\ln 3 - 2}{\text{area}}\) and \(\bar{y} = \frac{2}{\text{area}}\) |
| Total: 9 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Volume is \(\int \pi y^2\,dx = \int_2^a \pi\frac{36}{x^4}\,dx\) | M1 | \(\pi\) may be omitted throughout |
| \(= \pi\left[-\frac{12}{x^3}\right]_2^a = \pi\left(\frac{3}{2} - \frac{12}{a^3}\right)\) | A1 | |
| \(\int \pi xy^2\,dx = \int_2^a \pi\frac{36}{x^3}\,dx\) | M1 | |
| \(= \pi\left[-\frac{18}{x^2}\right]_2^a = \pi\left(\frac{9}{2} - \frac{18}{a^2}\right)\) | A1 | |
| \(\bar{x} = \frac{\int \pi xy^2\,dx}{\int \pi y^2\,dx}\) | M1 | |
| \(= \frac{\pi\left(\frac{9}{2} - \frac{18}{a^2}\right)}{\pi\left(\frac{3}{2} - \frac{12}{a^3}\right)} = \frac{3(a^3 - 4a)}{a^3 - 8}\) | E1 | |
| Total: 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Since \(a > 2\), \(4a > 8\) | M1 | Condone \(\geq\) instead of \(>\) throughout |
| so \(a^3 - 4a < a^3 - 8\) | A1 | |
| Hence \(\bar{x} = \frac{3(a^3-4a)}{a^3-8} < 3\) i.e. CM is less than 3 units from O | E1 | Fully acceptable explanation. Dependent on M1A1 |
| OR As \(a\to\infty\), \(\bar{x} = \frac{3(1-4a^{-2})}{1-8a^{-3}} \to 3\) | M1A1 | Accept \(\bar{x} \approx \frac{3a^3}{a^3} \to 3\) etc (M1 for \(\bar{x}\to 3\) stated, but A1 requires correct justification) |
| Since \(\bar{x}\) increases as \(a\) increases, \(\bar{x}\) is less than 3 | E1 | |
| Total: 3 |
# Question 4:
## Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| Area is $\int_0^{\ln 3} e^x\,dx = \left[e^x\right]_0^{\ln 3} = 2$ | M1, A1 | |
| $\int xy\,dx = \int_0^{\ln 3} xe^x\,dx$ | M1 | |
| $= \left[xe^x - e^x\right]_0^{\ln 3} = 3\ln 3 - 2$ | M1, A1 | Integration by parts for $xe^x - e^x$ |
| $\bar{x} = \frac{3\ln 3 - 2}{2} = \frac{3}{2}\ln 3 - 1$ | A1 | ww full marks (B4). Give B3 for $0.65$ |
| $\int \frac{1}{2}y^2\,dx = \int_0^{\ln 3} \frac{1}{2}(e^x)^2\,dx$ | M1 | For integral of $(e^x)^2$ |
| $= \left[\frac{1}{4}e^{2x}\right]_0^{\ln 3} = 2$ | A1 | For $\frac{1}{4}e^{2x}$ |
| $\bar{y} = \frac{2}{2} = 1$ | A1 | If area wrong, SC1 for $\bar{x} = \frac{3\ln 3 - 2}{\text{area}}$ and $\bar{y} = \frac{2}{\text{area}}$ |
| | **Total: 9** | |
## Part (b)(i):
| Working | Marks | Guidance |
|---------|-------|----------|
| Volume is $\int \pi y^2\,dx = \int_2^a \pi\frac{36}{x^4}\,dx$ | M1 | $\pi$ may be omitted throughout |
| $= \pi\left[-\frac{12}{x^3}\right]_2^a = \pi\left(\frac{3}{2} - \frac{12}{a^3}\right)$ | A1 | |
| $\int \pi xy^2\,dx = \int_2^a \pi\frac{36}{x^3}\,dx$ | M1 | |
| $= \pi\left[-\frac{18}{x^2}\right]_2^a = \pi\left(\frac{9}{2} - \frac{18}{a^2}\right)$ | A1 | |
| $\bar{x} = \frac{\int \pi xy^2\,dx}{\int \pi y^2\,dx}$ | M1 | |
| $= \frac{\pi\left(\frac{9}{2} - \frac{18}{a^2}\right)}{\pi\left(\frac{3}{2} - \frac{12}{a^3}\right)} = \frac{3(a^3 - 4a)}{a^3 - 8}$ | E1 | |
| | **Total: 6** | |
## Part (b)(ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| Since $a > 2$, $4a > 8$ | M1 | Condone $\geq$ instead of $>$ throughout |
| so $a^3 - 4a < a^3 - 8$ | A1 | |
| Hence $\bar{x} = \frac{3(a^3-4a)}{a^3-8} < 3$ i.e. CM is less than 3 units from O | E1 | Fully acceptable explanation. Dependent on M1A1 |
| OR As $a\to\infty$, $\bar{x} = \frac{3(1-4a^{-2})}{1-8a^{-3}} \to 3$ | M1A1 | Accept $\bar{x} \approx \frac{3a^3}{a^3} \to 3$ etc (M1 for $\bar{x}\to 3$ stated, but A1 requires correct justification) |
| Since $\bar{x}$ increases as $a$ increases, $\bar{x}$ is less than 3 | E1 | |
| | **Total: 3** | |4
\begin{enumerate}[label=(\alph*)]
\item A uniform lamina occupies the region bounded by the $x$-axis, the $y$-axis, the curve $y = \mathrm { e } ^ { x }$ for $0 \leqslant x \leqslant \ln 3$, and the line $x = \ln 3$. Find, in an exact form, the coordinates of the centre of mass of this lamina.
\item A region is bounded by the $x$-axis, the curve $y = \frac { 6 } { x ^ { 2 } }$ for $2 \leqslant x \leqslant a$ (where $a > 2$ ), the line $x = 2$ and the line $x = a$. This region is rotated through $2 \pi$ radians about the $x$-axis to form a uniform solid of revolution.
\begin{enumerate}[label=(\roman*)]
\item Show that the $x$-coordinate of the centre of mass of this solid is $\frac { 3 \left( a ^ { 3 } - 4 a \right) } { a ^ { 3 } - 8 }$.
\item Show that, however large the value of $a$, the centre of mass of this solid is less than 3 units from the origin.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2009 Q4 [18]}}