| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2023 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of solid of revolution |
| Difficulty | Standard +0.3 This is a standard M3 centre of mass question requiring routine application of volume and centre of mass formulas for solids of revolution. Students must integrate y² for volume and xy² for centre of mass using y = 1 + √x, which involves straightforward polynomial integration after expansion. The 'show that' format and algebraic simplification add minor complexity, but this is a textbook exercise testing formula recall and integration technique rather than problem-solving insight. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\pi\int_0^1(1+\sqrt{x})^2\,dx\) | M1 | Use of \(\pi\int_0^1(1+\sqrt{x})^2\,dx\). Limits not needed. \(\pi\) is required. |
| \(=\pi\left[x+\frac{4}{3}x^{\frac{3}{2}}+\frac{1}{2}x^2\right]_0^1\) | A1 | Correct integration – limits not needed |
| \(=\frac{17\pi}{6}\) m³ | A1* | Correct answer with units. Limits must be seen. Accept \(\frac{17}{6}\pi\) m³ |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\pi\int_0^1 x(1+\sqrt{x})^2\,dx\) | M1 | Use of \(\pi\int_0^1 x(1+\sqrt{x})^2\,dx\). Limits not needed (\(\pi\)'s will cancel so may not be seen) |
| \(=\pi\left[\frac{1}{2}x^2+\frac{4}{5}x^{\frac{5}{2}}+\frac{1}{3}x^3\right]_0^1\) | A1 | Correct integration – limits not needed |
| \(=\frac{49\pi}{30}\) | A1 | Correct unsimplified with or without \(\pi\) (may see \(\frac{1}{2}+\frac{4}{5}+\frac{1}{3}-0\)) |
| \(\bar{x}=\frac{\frac{49\pi}{30}}{\frac{17\pi}{6}}\) | dM1 | Correct expression with their numerator (consistent \(\pi\) - seen in neither or both) |
| \(=\frac{49}{85}\) m | A1* | Correct answer with units |
# Question 1:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\pi\int_0^1(1+\sqrt{x})^2\,dx$ | M1 | Use of $\pi\int_0^1(1+\sqrt{x})^2\,dx$. Limits not needed. $\pi$ is required. |
| $=\pi\left[x+\frac{4}{3}x^{\frac{3}{2}}+\frac{1}{2}x^2\right]_0^1$ | A1 | Correct integration – limits not needed |
| $=\frac{17\pi}{6}$ m³ | A1* | Correct answer with units. Limits must be seen. Accept $\frac{17}{6}\pi$ m³ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\pi\int_0^1 x(1+\sqrt{x})^2\,dx$ | M1 | Use of $\pi\int_0^1 x(1+\sqrt{x})^2\,dx$. Limits not needed ($\pi$'s will cancel so may not be seen) |
| $=\pi\left[\frac{1}{2}x^2+\frac{4}{5}x^{\frac{5}{2}}+\frac{1}{3}x^3\right]_0^1$ | A1 | Correct integration – limits not needed |
| $=\frac{49\pi}{30}$ | A1 | Correct unsimplified with or without $\pi$ (may see $\frac{1}{2}+\frac{4}{5}+\frac{1}{3}-0$) |
| $\bar{x}=\frac{\frac{49\pi}{30}}{\frac{17\pi}{6}}$ | dM1 | Correct expression with their numerator (consistent $\pi$ - seen in neither or both) |
| $=\frac{49}{85}$ m | A1* | Correct answer with units |
---1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{631b78c4-2763-4a1e-9d30-2f301fe3af2e-02_703_561_280_753}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The shaded region R is bounded by the x -axis, the line with equation $\mathrm { x } = 1$, the curve with equation $y = 1 + \sqrt { x }$ and the y-axis, as shown in Figure 1. The unit of length on both of the axes is 1 m .
The region R is rotated through $2 \pi$ radians about the x-axis to form a solid of revolution which is used to model a uniform solid $S$.
Show, using the model and algebraic integration, that
\begin{enumerate}[label=(\alph*)]
\item the volume of $S$ is $\frac { 17 \pi } { 6 } \mathrm {~m} ^ { 3 }$
\item the centre of mass of $S$ is $\frac { 49 } { 85 } \mathrm {~m}$ from 0 .\\
\includegraphics[max width=\textwidth, alt={}, center]{631b78c4-2763-4a1e-9d30-2f301fe3af2e-02_2264_41_314_1987}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2023 Q1 [8]}}