| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of solid of revolution |
| Difficulty | Standard +0.8 This is a standard M3/Further Mechanics centre of mass problem requiring integration of x·πy² and πy² with y=9-x², then finding their ratio. While it involves multiple steps (finding volume, moment, then dividing), the technique is routine for M3 students and follows a standard formula. The algebra is moderately involved but straightforward, placing it somewhat above average difficulty. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(V = \pi\int_0^3(9-x^2)^2\,dx = \pi\int_0^3(81-18x^2+x^4)\,dx\) | M1 | |
| \(= \pi\left[81x - 6x^3 + \frac{x^5}{5}\right]_0^3 = \frac{648}{5}\pi\) | M1 A1 | |
| \(\int_0^3\pi(9-x^2)^2 x\,dx = \frac{\pi}{6}\left[-(9-x^2)^3\right]_0^3\) | M1 A1 | OR equivalent method shown |
| \(= \frac{\pi}{6}\left[0+(9)^3\right] = \frac{243}{2}\pi\) | M1 | |
| \(= \frac{243}{2}\pi\) | A1 | |
| \(\bar{x} = \dfrac{\frac{243}{2}}{\frac{648}{5}} = \frac{15}{16}\) (accept \(0.94\)) | M1 A1 | (9) Total: 9 |
# Question 2:
| Working/Answer | Marks | Notes |
|---|---|---|
| $V = \pi\int_0^3(9-x^2)^2\,dx = \pi\int_0^3(81-18x^2+x^4)\,dx$ | M1 | |
| $= \pi\left[81x - 6x^3 + \frac{x^5}{5}\right]_0^3 = \frac{648}{5}\pi$ | M1 A1 | |
| $\int_0^3\pi(9-x^2)^2 x\,dx = \frac{\pi}{6}\left[-(9-x^2)^3\right]_0^3$ | M1 A1 | OR equivalent method shown |
| $= \frac{\pi}{6}\left[0+(9)^3\right] = \frac{243}{2}\pi$ | M1 | |
| $= \frac{243}{2}\pi$ | A1 | |
| $\bar{x} = \dfrac{\frac{243}{2}}{\frac{648}{5}} = \frac{15}{16}$ (accept $0.94$) | M1 A1 | **(9) Total: 9** |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{826ad8ff-6e5c-4224-88ba-e78b79d1bc21-03_438_661_223_644}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The shaded region $R$ is bounded by the curve with equation $y = 9 - x ^ { 2 }$, the positive $x$-axis and the positive $y$-axis, as shown in Figure 1. A uniform solid $S$ is formed by rotating $R$ through $360 ^ { \circ }$ about the $x$-axis.
Find the $x$-coordinate of the centre of mass of $S$.\\
\hfill \mbox{\textit{Edexcel M3 2011 Q2 [9]}}