| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Particles at coordinate positions |
| Difficulty | Standard +0.3 This is a standard M3 centre of mass question requiring integration to find area and centroid coordinates. Part (a) is routine integration with given limits (roots at x=0,2). Part (b) applies standard formulas for x̄ and ȳ using moments. While it requires careful integration and formula recall, it follows a well-practiced procedure with no novel problem-solving or geometric insight needed. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(A = \int_0^2 (2x - x^2) dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}\) | M1 A1; A1 (cso) | 4 marks |
| (b) \(\bar{x} = 1\) (by symmetry) | B1 | |
| \(\frac{4}{3}\bar{y} = \frac{1}{2}\int_0^2 y^2 dx = \frac{1}{2}\int_0^2 (2x - x^2)^2 dx = \frac{1}{2}\int_0^2 (4x^2 - 4x^3 + x^4) dx\) | M1; A1 | |
| \(= \frac{1}{2}\left[\frac{4x^3}{3} - x^4 + \frac{x^5}{5}\right]\) | A1 | |
| \(\frac{4}{3}\bar{y} = \frac{1}{2}\left[\frac{4x^3}{3} - x^4 + \frac{x^5}{5}\right]_0^2 = \frac{8}{15}\) | ||
| \(\bar{y} = \frac{8}{15} \times \frac{3}{4} = \frac{2}{5}\) accept exact equivalents | A1 | 5 marks [9] |
**(a)** $A = \int_0^2 (2x - x^2) dx = \left[x^2 - \frac{x^3}{3}\right]_0^2 = 4 - \frac{8}{3} = \frac{4}{3}$ | M1 A1; A1 (cso) | 4 marks
**(b)** $\bar{x} = 1$ (by symmetry) | B1 |
$\frac{4}{3}\bar{y} = \frac{1}{2}\int_0^2 y^2 dx = \frac{1}{2}\int_0^2 (2x - x^2)^2 dx = \frac{1}{2}\int_0^2 (4x^2 - 4x^3 + x^4) dx$ | M1; A1 |
$= \frac{1}{2}\left[\frac{4x^3}{3} - x^4 + \frac{x^5}{5}\right]$ | A1 |
$\frac{4}{3}\bar{y} = \frac{1}{2}\left[\frac{4x^3}{3} - x^4 + \frac{x^5}{5}\right]_0^2 = \frac{8}{15}$ | |
$\bar{y} = \frac{8}{15} \times \frac{3}{4} = \frac{2}{5}$ accept exact equivalents | A1 | 5 marks [9]
---The rudder on a ship is modelled as a uniform plane lamina having the same shape as the region $R$ which is enclosed between the curve with equation $y = 2x - x^2$ and the $x$-axis.
\begin{enumerate}[label=(\alph*)]
\item Show that the area of $R$ is $\frac{4}{3}$. [4]
\item Find the coordinates of the centre of mass of the lamina. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2007 Q1 [9]}}