| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2022 |
| Session | January |
| Marks | 7 |
| 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 is a standard M3 centre of mass problem requiring integration to find ȳ for a curved lamina. While the setup is routine (identifying bounds, applying the standard formula ȳ = ∫y·dA/∫dA), it requires careful algebraic manipulation of the polynomial y = x(x+a), correct integration of terms up to x⁴, and simplification to reach a clean answer in terms of a. The multi-step calculation and algebraic complexity place it moderately above average difficulty. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Area \(= \int_0^a (x^2 + ax)\,dx = \left[\frac{1}{3}x^3 + \frac{1}{2}ax^2\right]_0^a = \frac{5a^3}{6}\) | M1A1 | Attempt area by integration. Powers of both terms increase by 1. A1: Correct area. |
| \(\int \frac{1}{2}y^2\,dx = \int_0^a \frac{1}{2}(x^4 + 2ax^3 + a^2x^2)\,dx\) | M1 | Use \(\int \frac{1}{2}y^2\,dx\) to give \(\int_0^2 \frac{1}{2}(x^4 + 2ax^3 + a^2x^2)\,dx\). Limits not needed. Squaring must be correct. Condone missing \(\frac{1}{2}\) or any multiple. |
| \(= \frac{1}{2}\left[\frac{1}{5}x^5 + \frac{a}{2}x^4 + \frac{a^2}{3}x^3\right]_0^a \left(= \frac{31a^5}{60}\right)\) | DM1A1 | Attempt integration (powers of at least 2 terms increase by 1). Depends on second M mark. Correct integration and correct limits shown. |
| \(\bar{y} = \dfrac{\int \frac{1}{2}y^2\,dx}{\int y\,dx} = \dfrac{31a^5}{60} \div \dfrac{5a^3}{6} = \dfrac{31a^2}{50}\) | M1A1 (7) | M1: Use correct formula. A1: Correct answer. |
# Question 1:
| Working/Answer | Marks | Guidance |
|---|---|---|
| Area $= \int_0^a (x^2 + ax)\,dx = \left[\frac{1}{3}x^3 + \frac{1}{2}ax^2\right]_0^a = \frac{5a^3}{6}$ | M1A1 | Attempt area by integration. Powers of both terms increase by 1. A1: Correct area. |
| $\int \frac{1}{2}y^2\,dx = \int_0^a \frac{1}{2}(x^4 + 2ax^3 + a^2x^2)\,dx$ | M1 | Use $\int \frac{1}{2}y^2\,dx$ to give $\int_0^2 \frac{1}{2}(x^4 + 2ax^3 + a^2x^2)\,dx$. Limits not needed. Squaring must be correct. Condone missing $\frac{1}{2}$ or any multiple. |
| $= \frac{1}{2}\left[\frac{1}{5}x^5 + \frac{a}{2}x^4 + \frac{a^2}{3}x^3\right]_0^a \left(= \frac{31a^5}{60}\right)$ | DM1A1 | Attempt integration (powers of at least 2 terms increase by 1). Depends on second M mark. Correct integration and correct limits shown. |
| $\bar{y} = \dfrac{\int \frac{1}{2}y^2\,dx}{\int y\,dx} = \dfrac{31a^5}{60} \div \dfrac{5a^3}{6} = \dfrac{31a^2}{50}$ | M1A1 (7) | M1: Use correct formula. A1: Correct answer. |
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bd1e2b07-4a87-49d6-addd-c9f67467ef2f-02_472_750_255_660}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A uniform lamina is in the shape of the region $R$.\\
Region $R$ is bounded by the curve with equation $y = x ( x + a )$ where $a$ is a positive constant, the positive $x$-axis and the line with equation $x = a$, as shown shaded in Figure 1.
Find the $\boldsymbol { y }$ coordinate of the centre of mass of the lamina.
\hfill \mbox{\textit{Edexcel M3 2022 Q1 [7]}}