| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Composite solid with standard shapes - calculation only |
| Difficulty | Standard +0.3 This is a standard M3 centre of mass question requiring volume of revolution integration and composite body calculations using standard formulas. Part (a) involves routine integration with a 'show that' structure, and part (b) applies the standard composite centre of mass formula with a hemisphere (whose centre of mass position is a standard result). While it requires multiple steps, all techniques are textbook exercises with no novel insight needed. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Moments: \(\pi \int_1^2 xy^2 dx = V\bar{x}\) or \(\int_1^2 xy^2 dx = \bar{x}\int_1^2 y^2 dx\) | M1 | |
| \(\int_1^2 y^2 dx = \int_1^2 \frac{1}{4x^4} dx = \left[-\frac{1}{12x^3}\right]_1^2\) (= \(\frac{7}{96}\)) (either) | M1 A1 | |
| \(\int_1^2 xy^2 dx = \int_1^2 \frac{1}{4x^3} dx = \left[-\frac{1}{8x^2}\right]_1^2\) (= \(\frac{3}{32}\)) (both) | A1 | |
| Solving to find \(\bar{x}\) (= \(\frac{9}{7}\)) \(\Rightarrow\) required dist = \(\frac{9}{7} - 1 = \frac{2}{7}\) m (*) | ↓ M1 A1 cso (6) | |
| (b) | H | S |
| Mass | \((\rho)\frac{2}{3}\pi\left(\frac{1}{2}\right)^3\) | \((\rho)\frac{7\pi}{96}\) |
| \(=\left[\frac{1}{12}(\rho)\pi\right]\) | \(=\left[\frac{5}{32}(\rho)\pi\right]\) | |
| Dist of CM from base | \(\frac{19}{16}\) m | \(\frac{5}{7}\) m |
| Moments: | \(=\left[\frac{1}{12}(\rho)\pi\right]\left(\frac{19}{16}\right) + (\rho)\frac{7\pi}{96}\left(\frac{5}{7}\right) = \left[\frac{5}{32}(\rho)\pi\right]\bar{x}\) | |
| \(\bar{x} = \frac{29}{30}\) m or 0.967 m (awrt) | ||
| B1, M1 | ||
| B1 B1 | ||
| M1 A1 | ||
| A1 (7) | ||
| Allow distances to be found from different base line if necessary | Guidance note |
(a) Moments: $\pi \int_1^2 xy^2 dx = V\bar{x}$ or $\int_1^2 xy^2 dx = \bar{x}\int_1^2 y^2 dx$ | M1 |
$\int_1^2 y^2 dx = \int_1^2 \frac{1}{4x^4} dx = \left[-\frac{1}{12x^3}\right]_1^2$ (= $\frac{7}{96}$) (either) | M1 A1 |
$\int_1^2 xy^2 dx = \int_1^2 \frac{1}{4x^3} dx = \left[-\frac{1}{8x^2}\right]_1^2$ (= $\frac{3}{32}$) (both) | A1 |
Solving to find $\bar{x}$ (= $\frac{9}{7}$) $\Rightarrow$ required dist = $\frac{9}{7} - 1 = \frac{2}{7}$ m (*) | ↓ M1 A1 cso (6) |
(b) | H | S | T |
| --- | --- | --- |
| Mass | $(\rho)\frac{2}{3}\pi\left(\frac{1}{2}\right)^3$ | $(\rho)\frac{7\pi}{96}$ | H + S |
| | $=\left[\frac{1}{12}(\rho)\pi\right]$ | | $=\left[\frac{5}{32}(\rho)\pi\right]$ |
| Dist of CM from base | $\frac{19}{16}$ m | $\frac{5}{7}$ m | $\bar{x}$ |
| Moments: | $=\left[\frac{1}{12}(\rho)\pi\right]\left(\frac{19}{16}\right) + (\rho)\frac{7\pi}{96}\left(\frac{5}{7}\right) = \left[\frac{5}{32}(\rho)\pi\right]\bar{x}$ | | |
| | $\bar{x} = \frac{29}{30}$ m or 0.967 m (awrt) | | |
| B1, M1 |
| B1 B1 |
| M1 A1 |
| A1 (7) |
Allow distances to be found from different base line if necessary | | Guidance note
---6.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\includegraphics[alt={},max width=\textwidth]{25b3ece7-69ed-4ec4-a6c7-4cd83ec2cc5e-09_515_1015_319_477}
\end{center}
\end{figure}
The shaded region $R$ is bounded by the curve with equation $y = \frac { 1 } { 2 x ^ { 2 } }$, the $x$-axis and the lines $x = 1$ and $x = 2$, as shown in Figure 4. The unit of length on each axis is 1 m . A uniform solid $S$ has the shape made by rotating $R$ through $360 ^ { \circ }$ about the $x$-axis.
\begin{enumerate}[label=(\alph*)]
\item Show that the centre of mass of $S$ is $\frac { 2 } { 7 } \mathrm {~m}$ from its larger plane face.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\includegraphics[alt={},max width=\textwidth]{25b3ece7-69ed-4ec4-a6c7-4cd83ec2cc5e-09_616_431_1420_778}
\end{center}
\end{figure}
A sporting trophy $T$ is a uniform solid hemisphere $H$ joined to the solid $S$. The hemisphere has radius $\frac { 1 } { 2 } \mathrm {~m}$ and its plane face coincides with the larger plane face of $S$, as shown in Figure 5. Both $H$ and $S$ are made of the same material.
\item Find the distance of the centre of mass of $T$ from its plane face.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2007 Q6 [13]}}