| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2016 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of lamina by integration |
| Difficulty | Standard +0.8 This M3 question requires setting up and evaluating a centre of mass integral for a triangular lamina, finding the equation of the hypotenuse, then computing the y-coordinate of the centroid using ∫y dA / ∫dA. While the integration itself is straightforward (polynomial), the setup requires careful geometric reasoning and the question explicitly demands algebraic integration rather than using the standard centroid formula, making it moderately challenging for this level. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int xy\,dx = \int\left(-\frac{2x^2}{3}+6x\right)dx\) | M1 | Attempting correct form for \(\int xy\,dx\) using equation of a line. Limits not needed |
| \(= \left[-\frac{2}{9}x^3+3x^2\right]_0^9\) | DM1, A1 | Attempting integration (limits not needed); correct integration and correct limits shown. Equation of line must be correct |
| \(= -162+243-0 = 81\) | A1 | Substitute correct limits to obtain 81 |
| \(\bar{x} = \frac{81}{27} = 3\) | M1, A1cso (6) | Divide value from integration by area of triangle; \(\bar{x}=3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \frac{2}{3}x\) | — | Equation of line |
| \(\int xy\,dx = \int\frac{2x^2}{3}\,dx = \left[\frac{2}{9}x^3\right]_0^9 = 162\) | M1DM1A1, A1 | |
| \(\bar{x} = \frac{162}{27}\ (=6)\); dist from \(AB = 9-6=3\) | M1, A1 |
## Question 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int xy\,dx = \int\left(-\frac{2x^2}{3}+6x\right)dx$ | M1 | Attempting correct form for $\int xy\,dx$ using equation of a line. Limits not needed |
| $= \left[-\frac{2}{9}x^3+3x^2\right]_0^9$ | DM1, A1 | Attempting integration (limits not needed); correct integration and correct limits shown. Equation of line must be correct |
| $= -162+243-0 = 81$ | A1 | Substitute correct limits to obtain 81 |
| $\bar{x} = \frac{81}{27} = 3$ | M1, A1cso (6) | Divide value from integration by area of triangle; $\bar{x}=3$ |
**ALT 1** (using $C$ as origin, $AB$ parallel to $y$-axis):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \frac{2}{3}x$ | — | Equation of line |
| $\int xy\,dx = \int\frac{2x^2}{3}\,dx = \left[\frac{2}{9}x^3\right]_0^9 = 162$ | M1DM1A1, A1 | |
| $\bar{x} = \frac{162}{27}\ (=6)$; dist from $AB = 9-6=3$ | M1, A1 | |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4c1c51ff-6ae8-402d-b303-b656d26e4230-03_430_739_324_607}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a uniform triangular lamina $A B C$ in which $A B = 6 \mathrm {~cm} , B C = 9 \mathrm {~cm}$ and angle $A B C = 90 ^ { \circ }$. The centre of mass of the lamina is $G$. Use algebraic integration to find the distance of $G$ from $A B$.\\
(6)\\
\hfill \mbox{\textit{Edexcel M3 2016 Q2 [6]}}