| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2015 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of lamina by integration |
| Difficulty | Standard +0.3 This is a standard M3 centre of mass question requiring routine application of formulas for a lamina bounded by a curve. The integration is straightforward (powers of x only), and students are explicitly told to use the standard method. While it requires multiple integrations and careful bookkeeping, it involves no conceptual difficulty beyond applying memorized formulas. |
| Spec | 6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Area \(= \int_1^3 y\,dx = \int_1^3 \frac{3}{x^2}\,dx = \left[-3x^{-1}\right]_1^3 = -1-(-3) = 2\) | B1 | Correct area of \(R\) (may be embedded in working) |
| (i) \(\int_1^3 xy\,dx = \int_1^3 x \times \frac{3}{x^2}\,dx = \int_1^3 \frac{3}{x}\,dx\) | M1 | Attempting the integral \(\int_1^3 xy\,dx\) (integration to be seen) |
| \(\left[3\ln x\right]_1^3 = 3\ln 3\) | A1 | Correct integration and limits (substitution not needed) |
| \(\bar{x} = \frac{3\ln 3}{2} \quad (= 1.647...)\) | M1A1 | M1 divide by their area - denominator must be an area; A1 correct value for \(\bar{x}\) - exact or decimal 1.6 or better |
| (ii) \(\int_1^3 \frac{1}{2}y^2\,dx = \int_1^3 \frac{1}{2} \times \frac{9}{x^4}\,dx\) | M1 | Attempting the integral \(\int_1^3 \frac{1}{2}y^2\,dx\) or \(\int_1^3 y^2\,dx\) (integration to be seen) |
| \(\frac{9}{2}\left[-\frac{x^{-3}}{3}\right]_1^3 = \frac{9}{2}\left[-\frac{1}{81}+\frac{1}{3}\right] = 1\frac{4}{9}\) | M1A1 | Correct integration and limits shown |
| \(\bar{y} = \frac{1\frac{4}{9}}{2} = \frac{13}{18} \quad (= 0.722...)\) | DM1A1 (9) | DM1 divide by their area - must have used \(\int \frac{1}{2}y^2\,dx\); A1 correct value for \(\bar{y}\) - exact or decimal 0.72 or better |
# Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Area $= \int_1^3 y\,dx = \int_1^3 \frac{3}{x^2}\,dx = \left[-3x^{-1}\right]_1^3 = -1-(-3) = 2$ | B1 | Correct area of $R$ (may be embedded in working) |
| **(i)** $\int_1^3 xy\,dx = \int_1^3 x \times \frac{3}{x^2}\,dx = \int_1^3 \frac{3}{x}\,dx$ | M1 | Attempting the integral $\int_1^3 xy\,dx$ (integration to be seen) |
| $\left[3\ln x\right]_1^3 = 3\ln 3$ | A1 | Correct integration and limits (substitution not needed) |
| $\bar{x} = \frac{3\ln 3}{2} \quad (= 1.647...)$ | M1A1 | M1 divide by their area - denominator must be an area; A1 correct value for $\bar{x}$ - exact or decimal 1.6 or better |
| **(ii)** $\int_1^3 \frac{1}{2}y^2\,dx = \int_1^3 \frac{1}{2} \times \frac{9}{x^4}\,dx$ | M1 | Attempting the integral $\int_1^3 \frac{1}{2}y^2\,dx$ or $\int_1^3 y^2\,dx$ (integration to be seen) |
| $\frac{9}{2}\left[-\frac{x^{-3}}{3}\right]_1^3 = \frac{9}{2}\left[-\frac{1}{81}+\frac{1}{3}\right] = 1\frac{4}{9}$ | M1A1 | Correct integration and limits shown |
| $\bar{y} = \frac{1\frac{4}{9}}{2} = \frac{13}{18} \quad (= 0.722...)$ | DM1A1 (9) | DM1 divide by their area - must have used $\int \frac{1}{2}y^2\,dx$; A1 correct value for $\bar{y}$ - exact or decimal 0.72 or better |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3706a02d-95c6-4e7a-bf38-88b338d77892-03_547_671_260_648}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A uniform lamina is in the shape of the region $R$ which is bounded by the curve with equation $y = \frac { 3 } { x ^ { 2 } }$, the lines $x = 1$ and $x = 3$, and the $x$-axis, as shown in Figure 1.
The centre of mass of the lamina has coordinates $( \bar { x } , \bar { y } )$.\\
Use algebraic integration to find\\
(i) the value of $\bar { x }$,\\
(ii) the value of $\bar { y }$.\\
\hfill \mbox{\textit{Edexcel M3 2015 Q2 [9]}}