| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of solid of revolution |
| Difficulty | Standard +0.3 This is a standard M3/Further Mechanics question on centre of mass of a solid of revolution. Part (a) is routine volume calculation using the standard formula V = π∫y²dx. Part (b) requires the standard centre of mass formula x̄ = (∫x·πy²dx)/V. Both parts involve straightforward integration of exponential functions with no conceptual surprises—slightly easier than average due to the clean exponential function and standard bookwork application. |
| Spec | 4.08d Volumes of revolution: about x and y axes6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\pi\int_0^1 e^{-2x}\,dx = \frac{\pi}{-2}\left[e^{-2x}\right]_0^1\) | M1 A1 | M1: using \(V = \pi\int y^2\,dx = \pi\int e^{-2x}\,dx\) and attempting integration. Limits not needed for M1. A1: correct integration, correct limits must be shown |
| \(= \frac{\pi}{2}(1-e^{-2})\) PRINTED ANSWER | A1cso | Must be seen in this form |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\pi\int_0^1 xe^{-2x}\,dx = \pi\left[\frac{-1}{2}xe^{-2x}\right]_0^1 - \pi\int_0^1\frac{-1}{2}e^{-2x}\,dx\) | M1 A1 | M1: attempting integration by parts, limits not needed yet. A1: correct result with or without limits |
| \(= \pi\left(-\frac{1}{2}e^{-2} + \frac{1}{2}\left[\frac{-1}{2}e^{-2x}\right]_0^1\right)\) | M1dep A1ft | M1dep: attempting the next integral. A1ft: substituting correct limits |
| \(= \pi\left(-\frac{1}{2}e^{-2} - \frac{1}{4}(e^{-2}-1)\right)\) | ||
| \(= \pi\left(\frac{1}{4} - \frac{3}{4}e^{-2}\right)\) | A1cao | |
| \(\bar{x} = \dfrac{\pi\left(\frac{1}{4}-\frac{3}{4}e^{-2}\right)}{\frac{\pi}{2}(1-e^{-2})} = \dfrac{1}{2}\cdot\dfrac{(e^2-3)}{(e^2-1)}\) | M1 A1 | M1: using \(\bar{x} = \frac{(\pi)\int xy^2\,dx}{(\pi)\int y^2\,dx}\), must be correct way up. A1: must be in terms of \(e\), only 2 terms in each of numerator and denominator, no fractions in either |
## Question 4(a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\pi\int_0^1 e^{-2x}\,dx = \frac{\pi}{-2}\left[e^{-2x}\right]_0^1$ | M1 A1 | M1: using $V = \pi\int y^2\,dx = \pi\int e^{-2x}\,dx$ and attempting integration. Limits not needed for M1. A1: correct integration, correct limits must be shown |
| $= \frac{\pi}{2}(1-e^{-2})$ **PRINTED ANSWER** | A1cso | Must be seen in this form |
---
## Question 4(b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\pi\int_0^1 xe^{-2x}\,dx = \pi\left[\frac{-1}{2}xe^{-2x}\right]_0^1 - \pi\int_0^1\frac{-1}{2}e^{-2x}\,dx$ | M1 A1 | M1: attempting integration by parts, limits not needed yet. A1: correct result with or without limits |
| $= \pi\left(-\frac{1}{2}e^{-2} + \frac{1}{2}\left[\frac{-1}{2}e^{-2x}\right]_0^1\right)$ | M1dep A1ft | M1dep: attempting the next integral. A1ft: substituting correct limits |
| $= \pi\left(-\frac{1}{2}e^{-2} - \frac{1}{4}(e^{-2}-1)\right)$ | | |
| $= \pi\left(\frac{1}{4} - \frac{3}{4}e^{-2}\right)$ | A1cao | |
| $\bar{x} = \dfrac{\pi\left(\frac{1}{4}-\frac{3}{4}e^{-2}\right)}{\frac{\pi}{2}(1-e^{-2})} = \dfrac{1}{2}\cdot\dfrac{(e^2-3)}{(e^2-1)}$ | M1 A1 | M1: using $\bar{x} = \frac{(\pi)\int xy^2\,dx}{(\pi)\int y^2\,dx}$, must be correct way up. A1: must be in terms of $e$, only 2 terms in each of numerator and denominator, no fractions in either |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2c0bb9ea-31a6-42f1-9e2e-d792eee8fd10-05_568_620_269_653}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of the region $R$ bounded by the curve with equation $y = \mathrm { e } ^ { - x }$, the line $x = 1$, the $x$-axis and the $y$-axis. A uniform solid $S$ is formed by rotating $R$ through $2 \pi$ radians about the $x$-axis.
\begin{enumerate}[label=(\alph*)]
\item Show that the volume of $S$ is $\frac { \pi } { 2 } \left( 1 - \mathrm { e } ^ { - 2 } \right)$.
\item Find, in terms of e, the distance of the centre of mass of $S$ from $O$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2014 Q4 [10]}}