| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2015 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of lamina by integration |
| Difficulty | Challenging +1.2 This is a Further Maths M3 question on centre of mass by integration. Part (a) requires standard integration of a polynomial to find centroid coordinates. Parts (b)(i) and (b)(ii) involve solids of revolution with a clever composite body approach, but the integrations are routine (involving √(x²+16) which suggests substitution or standard forms). While requiring multiple techniques and careful setup, these are well-practiced M3 methods without requiring novel geometric insight. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area \(= \int_0^a \frac{x^2(a-x)}{a^2}\,dx\) | M1 | |
| \(= \left[\frac{x^3}{3a} - \frac{x^4}{4a^2}\right]_0^a \left(= \frac{a^2}{12}\right)\) | A1 | |
| \(\int xy\,dx\) | M1 | |
| \(= \int_0^a \frac{x^3(a-x)}{a^2}\,dx = \left[\frac{x^4}{4a} - \frac{x^5}{5a^2}\right]_0^a \left(= \frac{a^3}{20}\right)\) | A1 | |
| \(\bar{x} = \frac{\frac{1}{20}a^3}{\frac{1}{12}a^2} = \frac{3a}{5}\) | A1 | |
| \(\int \frac{1}{2}y^2\,dx = \int_0^a \frac{x^4(a-x)^2}{2a^4}\,dx\) | M1 | For \(\int \ldots y^2\,dx\) |
| \(= \left[\frac{x^5}{10a^2} - \frac{x^6}{6a^3} + \frac{x^7}{14a^4}\right]_0^a \left(= \frac{a^3}{210}\right)\) | A2 | Give A1 if just one error (e.g. omission of factor \(\frac{1}{2}\)) |
| \(\bar{y} = \frac{\frac{1}{210}a^3}{\frac{1}{12}a^2} = \frac{2a}{35}\) \((\approx 0.0571a)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Volume is \(\int_0^3 \pi(x^2+16)\,dx\) | M1 | \(\pi\) may be omitted throughout |
| \(= \pi\left[\frac{x^3}{3}+16x\right]_0^3 \quad (=57\pi)\) | A1 | |
| \(\int \pi x y^2\,dx\) | M1 | |
| \(=\int_0^3 \pi x(x^2+16)\,dx = \pi\left[\frac{x^4}{4}+8x^2\right]_0^3 \quad \left(=\frac{369}{4}\pi\right)\) | A1 | |
| \(\bar{x} = \dfrac{\frac{369}{4}\pi}{57\pi} = \dfrac{123}{76} \quad (\approx 1.62)\) | A1 | |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Volume of \(A\) and \(B\) combined is \(\pi \times 5^2 \times 3 = 75\pi\) | M1 | CM of composite body |
| \((18\pi)\bar{x}_B + (57\pi)\!\left(\dfrac{123}{76}\right) = (75\pi)(1.5)\) | A2 | Give A1 if just one error |
| OR | ||
| \(\int_0^3 \pi x\!\left(25-(x^2+16)\right)dx\) | M1 | |
| \(= \dfrac{81}{4}\pi\) | A1 | |
| \((18\pi)\bar{x}_B = \dfrac{81}{4}\pi\) | A1 FT | |
| \(\bar{x}_B = \dfrac{9}{8} \quad (=1.125)\) | A1 | CAO |
| [4] |
# Question 4:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area $= \int_0^a \frac{x^2(a-x)}{a^2}\,dx$ | M1 | |
| $= \left[\frac{x^3}{3a} - \frac{x^4}{4a^2}\right]_0^a \left(= \frac{a^2}{12}\right)$ | A1 | |
| $\int xy\,dx$ | M1 | |
| $= \int_0^a \frac{x^3(a-x)}{a^2}\,dx = \left[\frac{x^4}{4a} - \frac{x^5}{5a^2}\right]_0^a \left(= \frac{a^3}{20}\right)$ | A1 | |
| $\bar{x} = \frac{\frac{1}{20}a^3}{\frac{1}{12}a^2} = \frac{3a}{5}$ | A1 | |
| $\int \frac{1}{2}y^2\,dx = \int_0^a \frac{x^4(a-x)^2}{2a^4}\,dx$ | M1 | For $\int \ldots y^2\,dx$ |
| $= \left[\frac{x^5}{10a^2} - \frac{x^6}{6a^3} + \frac{x^7}{14a^4}\right]_0^a \left(= \frac{a^3}{210}\right)$ | A2 | Give A1 if just one error (e.g. omission of factor $\frac{1}{2}$) |
| $\bar{y} = \frac{\frac{1}{210}a^3}{\frac{1}{12}a^2} = \frac{2a}{35}$ $(\approx 0.0571a)$ | A1 | |
## Question 4(b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Volume is $\int_0^3 \pi(x^2+16)\,dx$ | M1 | $\pi$ may be omitted throughout | Condone consistent use of $2\pi y^2$ etc |
| $= \pi\left[\frac{x^3}{3}+16x\right]_0^3 \quad (=57\pi)$ | A1 | |
| $\int \pi x y^2\,dx$ | M1 | |
| $=\int_0^3 \pi x(x^2+16)\,dx = \pi\left[\frac{x^4}{4}+8x^2\right]_0^3 \quad \left(=\frac{369}{4}\pi\right)$ | A1 | |
| $\bar{x} = \dfrac{\frac{369}{4}\pi}{57\pi} = \dfrac{123}{76} \quad (\approx 1.62)$ | A1 | |
| **[5]** | | |
---
## Question 4(b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Volume of $A$ and $B$ combined is $\pi \times 5^2 \times 3 = 75\pi$ | M1 | CM of composite body |
| $(18\pi)\bar{x}_B + (57\pi)\!\left(\dfrac{123}{76}\right) = (75\pi)(1.5)$ | A2 | Give A1 if just one error | FT values from (i) |
| **OR** | | |
| $\int_0^3 \pi x\!\left(25-(x^2+16)\right)dx$ | M1 | |
| $= \dfrac{81}{4}\pi$ | A1 | |
| $(18\pi)\bar{x}_B = \dfrac{81}{4}\pi$ | A1 FT | |
| $\bar{x}_B = \dfrac{9}{8} \quad (=1.125)$ | A1 | CAO |
| **[4]** | | |4
\begin{enumerate}[label=(\alph*)]
\item A uniform lamina occupies the region bounded by the $x$-axis and the curve $y = \frac { x ^ { 2 } ( a - x ) } { a ^ { 2 } }$ for $0 \leqslant x \leqslant a$. Find the coordinates of the centre of mass of this lamina.
\item The region $A$ is bounded by the $x$-axis, the $y$-axis, the curve $y = \sqrt { x ^ { 2 } + 16 }$ and the line $x = 3$. The region $B$ is bounded by the $y$-axis, the curve $y = \sqrt { x ^ { 2 } + 16 }$ and the line $y = 5$. These regions are shown in Fig. 4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{70a2c3ce-7bdb-4ddd-92fc-f7dcbdfdcfaf-5_604_460_605_792}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Find the $x$-coordinate of the centre of mass of the uniform solid of revolution formed when the region $A$ is rotated through $2 \pi$ radians about the $x$-axis.
\item Using your answer to part (i), or otherwise, find the $x$-coordinate of the centre of mass of the uniform solid of revolution formed when the region $B$ is rotated through $2 \pi$ radians about the $x$-axis.
\section*{END OF QUESTION PAPER}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2015 Q4 [18]}}