| Exam Board | Edexcel |
|---|---|
| Module | FM2 AS (Further Mechanics 2 AS) |
| Session | Specimen |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Centre of mass with variable parameter |
| Difficulty | Challenging +1.2 This is a multi-part centre of mass question requiring decomposition into standard shapes, application of the composite body formula, and working backwards to find parameters. While it has many parts (7 marks typical), each step uses standard FM2 techniques: calculating centroids of rectangles/triangles, removing mass (negative contribution), and using equilibrium geometry. The symmetry argument in (c)(i) requires some insight, but overall this is a straightforward application of learned methods rather than requiring novel problem-solving. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Rel. Mass: \(2 \quad 5 \quad 1 \quad 8\) | B1 | For correct relative masses |
| \(y\): \(2 \quad 0.5 \quad 1.5 \quad \bar{y}\) | B1 | For correct \(y\) values |
| \(x\): \(0.5 \quad 2.5 \quad 4.5 \quad \bar{x}\) | B1 | For correct \(x\) values |
| \((2\times2)+(5\times0.5)+(1\times1.5) = 8\bar{y}\) | M1 | For a moments equation, correct no. of terms, condone sign errors |
| \(\bar{y} = 1\) | A1* | For a correct given answer |
| \((2\times0.5)+(5\times2.5)+(1\times4.5) = 8\bar{x}\) | M1 | For a moments equation, correct no. of terms |
| \(\bar{x} = 2.25\) | A1 | For 2.25 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of correct strategy to solve the problem by use of 'moments equation' | M1 | For a moments equation, correct no. of terms, condone sign errors |
| \((8\times2.25) - (2\pi r^2\times0.5) = (8-2\pi r^2)2.5\) | A1ft | For a correct equation, follow through on their \(\bar{x}\) |
| Solving for \(r\) | M1 | For solving for \(r\) |
| \(r = \frac{1}{\sqrt{2\pi}} = 0.399\) | A1 | For 0.399 or 0.40 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Since \(\bar{y}\) for original plate is 1, holes must be symmetrically placed about the line \(y=1\) | B1 | For consideration of symmetry about \(y=1\) |
| \(a = 1.5\) | B1 | For \(a = 1.5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of tan from an appropriate triangle | M1 | For use of tan from an appropriate triangle |
| \(\tan\alpha = \frac{2}{1.5} = \frac{4}{3}\) | A1ft | For a correct equation, follow through on their \(a\) |
| \(\alpha = 53.1°\) | A1 | For a correct angle |
## Question 3:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Rel. Mass: $2 \quad 5 \quad 1 \quad 8$ | B1 | For correct relative masses |
| $y$: $2 \quad 0.5 \quad 1.5 \quad \bar{y}$ | B1 | For correct $y$ values |
| $x$: $0.5 \quad 2.5 \quad 4.5 \quad \bar{x}$ | B1 | For correct $x$ values |
| $(2\times2)+(5\times0.5)+(1\times1.5) = 8\bar{y}$ | M1 | For a moments equation, correct no. of terms, condone sign errors |
| $\bar{y} = 1$ | A1* | For a correct given answer |
| $(2\times0.5)+(5\times2.5)+(1\times4.5) = 8\bar{x}$ | M1 | For a moments equation, correct no. of terms |
| $\bar{x} = 2.25$ | A1 | For 2.25 |
**Total: 7 marks**
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of correct strategy to solve the problem by use of 'moments equation' | M1 | For a moments equation, correct no. of terms, condone sign errors |
| $(8\times2.25) - (2\pi r^2\times0.5) = (8-2\pi r^2)2.5$ | A1ft | For a correct equation, follow through on their $\bar{x}$ |
| Solving for $r$ | M1 | For solving for $r$ |
| $r = \frac{1}{\sqrt{2\pi}} = 0.399$ | A1 | For 0.399 or 0.40 |
**Total: 4 marks**
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Since $\bar{y}$ for original plate is 1, holes must be symmetrically placed about the line $y=1$ | B1 | For consideration of symmetry about $y=1$ |
| $a = 1.5$ | B1 | For $a = 1.5$ |
**Total: 2 marks**
### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of tan from an appropriate triangle | M1 | For use of tan from an appropriate triangle |
| $\tan\alpha = \frac{2}{1.5} = \frac{4}{3}$ | A1ft | For a correct equation, follow through on their $a$ |
| $\alpha = 53.1°$ | A1 | For a correct angle |
**Total: 3 marks**3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5bfd2018-ea46-4ea5-9cf7-4210d125a91c-07_611_1146_280_456}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows the shape and dimensions of a template $O P Q R S T U V$ made from thin uniform metal.\\
$O P = 5 \mathrm {~m} , P Q = 2 \mathrm {~m} , Q R = 1 \mathrm {~m} , R S = 1 \mathrm {~m} , T U = 2 \mathrm {~m} , U V = 1 \mathrm {~m} , V O = 3 \mathrm {~m}$.\\
Figure 1 also shows a coordinate system with $O$ as origin and the $x$-axis and $y$-axis along $O P$ and $O V$ respectively. The unit of length on both axes is the metre.
The centre of mass of the template has coordinates $( \bar { x } , \bar { y } )$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that $\bar { y } = 1$
\item Find the value of $\bar { x }$.
A new design requires the template to have its centre of mass at the point (2.5,1). In order to achieve this, two circular discs, each of radius $r$ metres, are removed from the template which is shown in Figure 1, to form a new template $L$. The centre of the first disc is ( $0.5,0.5$ ) and the centre of the second disc is ( $0.5 , a$ ) where $a$ is a constant.
\end{enumerate}\item Find the value of $r$.
\item \begin{enumerate}[label=(\roman*)]
\item Explain how symmetry can be used to find the value of $a$.
\item Find the value of $a$.
The template $L$ is now freely suspended from the point $U$ and hangs in equilibrium.
\end{enumerate}\item Find the size of the angle between the line $T U$ and the horizontal.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 AS Q3 [16]}}