| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Lamina suspended in equilibrium |
| Difficulty | Challenging +1.2 This is a standard Further Maths mechanics question requiring integration to find centre of mass coordinates and then applying equilibrium conditions. Part (a) involves computing ȳ using the standard formula with a cubic function (straightforward integration), while part (b) requires basic geometry/trigonometry with the suspended lamina. The techniques are well-practiced in FM2, though the multi-step nature and Further Maths context place it slightly above average difficulty. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct strategy | M1 | Complete strategy for \(\bar{y}\): moments equation, use of limits and division by area |
| \(8\bar{y} = \frac{1}{2}\int y^2\,dx = \frac{1}{2}\int\left\{\frac{(x-2)^6}{16}+(x-2)^3+4\right\}dx\) | M1 | Moments equation to obtain terms of the correct form (with or without limits). Integral must be in terms of \(x\) only or \(y\) only. Allow if area (8) not seen |
| \(= \frac{1}{2}\left[\frac{(x-2)^7}{7\times16}+\frac{(x-2)^4}{4}+4x\right]_0^4\) | A1 | Correct unsimplified answer (with or without limits). Allow if area (8) not seen |
| \(8\bar{y} = \frac{1}{2}\left[\frac{8}{7}+4+16+\frac{8}{7}-4-0\right] = \frac{64}{7}\), \(\quad \bar{y} = \frac{8}{7}\) * | A1* | Use moments equation and given area to deduce given answer from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(8\bar{x} = \int\left(\frac{x(x-2)^3}{4}+2x\right)dx\) | M1 | Relevant integral in terms of \(x\) only or \(y\) only (with or without limits). Allow if area (8) not seen |
| \(= \left[\frac{x(x-2)^4}{16}-\frac{(x-2)^5}{80}+x^2\right]_0^4\) | A1 | Correct unsimplified form after integration (with or without limits). Allow if area (8) not seen |
| \(= \frac{64}{16}-\frac{32}{80}+16-\frac{32}{80} = 19.2\) | M1 | Complete process to find \(\bar{x}\) following relevant integral |
| \(\bar{x} = 2.4\) | A1 | Correct answer |
| Complete strategy to find \(\theta\) | M1 | Complete strategy to find \(\theta\) e.g. find \(\bar{x}\) and then use trig to find appropriate angle |
| \(\tan\theta = \frac{4-\bar{x}}{4-\frac{8}{7}} = \left(\frac{14}{25}\right)\) | A1ft | Use the model to find a relevant angle. Follow their \(\bar{x}\) |
| \(\theta = 29.2°\) (Accept 29) | A1 | 2 s.f. or better: 29.2488... |
## Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct strategy | M1 | Complete strategy for $\bar{y}$: moments equation, use of limits and division by area |
| $8\bar{y} = \frac{1}{2}\int y^2\,dx = \frac{1}{2}\int\left\{\frac{(x-2)^6}{16}+(x-2)^3+4\right\}dx$ | M1 | Moments equation to obtain terms of the correct form (with or without limits). Integral must be in terms of $x$ only or $y$ only. Allow if area (8) not seen |
| $= \frac{1}{2}\left[\frac{(x-2)^7}{7\times16}+\frac{(x-2)^4}{4}+4x\right]_0^4$ | A1 | Correct unsimplified answer (with or without limits). Allow if area (8) not seen |
| $8\bar{y} = \frac{1}{2}\left[\frac{8}{7}+4+16+\frac{8}{7}-4-0\right] = \frac{64}{7}$, $\quad \bar{y} = \frac{8}{7}$ * | A1* | Use moments equation and given area to deduce **given answer** from correct working |
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## Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $8\bar{x} = \int\left(\frac{x(x-2)^3}{4}+2x\right)dx$ | M1 | Relevant integral in terms of $x$ only or $y$ only (with or without limits). Allow if area (8) not seen |
| $= \left[\frac{x(x-2)^4}{16}-\frac{(x-2)^5}{80}+x^2\right]_0^4$ | A1 | Correct unsimplified form after integration (with or without limits). Allow if area (8) not seen |
| $= \frac{64}{16}-\frac{32}{80}+16-\frac{32}{80} = 19.2$ | M1 | Complete process to find $\bar{x}$ following relevant integral |
| $\bar{x} = 2.4$ | A1 | Correct answer |
| Complete strategy to find $\theta$ | M1 | Complete strategy to find $\theta$ e.g. find $\bar{x}$ and then use trig to find appropriate angle |
| $\tan\theta = \frac{4-\bar{x}}{4-\frac{8}{7}} = \left(\frac{14}{25}\right)$ | A1ft | Use the model to find a relevant angle. Follow their $\bar{x}$ |
| $\theta = 29.2°$ (Accept 29) | A1 | 2 s.f. or better: 29.2488... |
---\begin{enumerate}
\item Numerical (calculator) integration is not acceptable in this question.
\end{enumerate}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9b995178-a4be-4d5a-95f8-6c2978ff01b3-08_547_550_303_753}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The shaded region $O A B$ in Figure 2 is bounded by the $x$-axis, the line with equation $x = 4$ and the curve with equation $y = \frac { 1 } { 4 } ( x - 2 ) ^ { 3 } + 2$. The point $A$ has coordinates (4, 4) and the point $B$ has coordinates $( 4,0 )$.
A uniform lamina $L$ has the shape of $O A B$. The unit of length on both axes is one centimetre. The centre of mass of $L$ is at the point with coordinates $( \bar { x } , \bar { y } )$.
Given that the area of $L$ is $8 \mathrm {~cm} ^ { 2 }$,\\
(a) show that $\bar { y } = \frac { 8 } { 7 }$
The lamina is freely suspended from $A$ and hangs in equilibrium with $A B$ at an angle $\theta ^ { \circ }$ to the downward vertical.\\
(b) Find the value of $\theta$.
\hfill \mbox{\textit{Edexcel FM2 2019 Q3 [11]}}