| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2009 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Composite solid with standard shapes - calculation only |
| Difficulty | Standard +0.8 This M3 centre of mass question requires finding the COM of a solid of revolution using integration (part a), then applying equilibrium conditions for a suspended composite body (part b). While the integration is standard, combining the two solids and using the suspension condition requires careful geometric reasoning and multi-step calculation, placing it moderately above average difficulty. |
| Spec | 4.08d Volumes of revolution: about x and y axes6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\int y^2\,dx = \int(4-x^2)^2\,dx = \int(16 - 8x^2 + x^4)\,dx = 16x - \frac{8x^3}{3} + \frac{x^5}{5}\) | M1 A1 | |
| \(\left[16x - \frac{8x^3}{3} + \frac{x^5}{5}\right]_0^2 = \frac{256}{15}\) | M1 A1 | |
| \(\int xy^2\,dx = \int x(4-x^2)^2\,dx = \int(16x - 8x^3 + x^5)\,dx = 8x^2 - 2x^4 + \frac{x^6}{6}\) | M1 A1 | |
| \(\left[8x^2 - 2x^4 + \frac{x^6}{6}\right]_0^2 = \frac{32}{3}\) | M1 A1 | |
| \(\bar{x} = \frac{\int xy^2\,dx}{\int y^2\,dx} = \frac{32}{3} \times \frac{15}{256} = \frac{5}{8}\) | M1 A1 | (10) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(A \times \bar{x} = (\pi r^2 l) \times \frac{l}{2}\) | M1 | |
| \(\frac{256}{15}\pi \times \frac{5}{8} = \pi \times 16l \times \frac{l}{2}\) | A1 ft | |
| \(l = \frac{2\sqrt{3}}{3}\) | M1 A1 | accept exact equivalents or awrt 1.15 (4) [14] |
## Question 6:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\int y^2\,dx = \int(4-x^2)^2\,dx = \int(16 - 8x^2 + x^4)\,dx = 16x - \frac{8x^3}{3} + \frac{x^5}{5}$ | M1 A1 | |
| $\left[16x - \frac{8x^3}{3} + \frac{x^5}{5}\right]_0^2 = \frac{256}{15}$ | M1 A1 | |
| $\int xy^2\,dx = \int x(4-x^2)^2\,dx = \int(16x - 8x^3 + x^5)\,dx = 8x^2 - 2x^4 + \frac{x^6}{6}$ | M1 A1 | |
| $\left[8x^2 - 2x^4 + \frac{x^6}{6}\right]_0^2 = \frac{32}{3}$ | M1 A1 | |
| $\bar{x} = \frac{\int xy^2\,dx}{\int y^2\,dx} = \frac{32}{3} \times \frac{15}{256} = \frac{5}{8}$ | M1 A1 | **(10)** |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $A \times \bar{x} = (\pi r^2 l) \times \frac{l}{2}$ | M1 | |
| $\frac{256}{15}\pi \times \frac{5}{8} = \pi \times 16l \times \frac{l}{2}$ | A1 ft | |
| $l = \frac{2\sqrt{3}}{3}$ | M1 A1 | accept exact equivalents or awrt 1.15 **(4) [14]** |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8374fa0f-cb28-497f-8696-877d7d0762f1-09_433_376_242_781}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
The region $R$ is bounded by part of the curve with equation $y = 4 - x ^ { 2 }$, the positive $x$-axis and the positive $y$-axis, as shown in Figure 3. The unit of length on both axes is one metre. A uniform solid $S$ is formed by rotating $R$ through $360 ^ { \circ }$ about the $x$-axis.
\begin{enumerate}[label=(\alph*)]
\item Show that the centre of mass of $S$ is $\frac { 5 } { 8 } \mathrm {~m}$ from $O$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8374fa0f-cb28-497f-8696-877d7d0762f1-09_702_584_1138_676}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a cross section of a uniform solid $P$ consisting of two components, a solid cylinder $C$ and the solid $S$. The cylinder $C$ has radius 4 m and length $l$ metres. One end of $C$ coincides with the plane circular face of $S$. The point $A$ is on the circumference of the circular face common to $C$ and $S$. When the solid $P$ is freely suspended from $A$, the solid $P$ hangs with its axis of symmetry horizontal.
\item Find the value of $l$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2009 Q6 [14]}}