| Exam Board | Edexcel |
|---|---|
| Module | FM2 AS (Further Mechanics 2 AS) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Folded lamina |
| Difficulty | Challenging +1.2 This is a multi-step Further Maths mechanics problem requiring understanding of centre of mass for composite shapes and equilibrium of suspended bodies. Part (a) requires geometric insight about symmetry/invariance under folding, while part (b) involves standard equilibrium calculations with trigonometry. The setup is moderately complex but the techniques are standard for FM2, placing it above average difficulty but not requiring exceptional insight. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| In the folding process, each point of the lamina remains the same distance from \(CD\) | B1 | Any equivalent explanation e.g. folding doesn't change the mass distribution relative to \(CD\). A calculation to verify is not the same as an explanation. Allow use of 'vertical' for \(CD\). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For the folded lamina: \(\bar{x} = 2a \; (= d_2)\) oe | B1 | Seen anywhere |
| Area ratios correct (e.g. Large triangle \(ACE\): \(27a^2\), Removed triangle \(BCD\): \(12a^2\), Added triangle \(BCD\): \(12a^2\), Folded lamina: \(27a^2\)) | B1 | Any equivalent form for the mass (area) ratios |
| Distances from \(EA\) correct (e.g. \(3a\), \(5a\), \(a\), \(\bar{y}\)) | B1 | Or correct distances from an alternative axis parallel to \(AE\) e.g. \(BD\) |
| Moments about \(EA\): \(27 \times 3a - 12 \times 5a + 12 \times a = 27\bar{y}\) | M1 | Moments about \(AE\) or a parallel axis. Need all terms. Must be dimensionally correct. Condone sign errors. |
| Correct unsimplified moments equation ft on their 'table' | A1ft | Correct unsimplified moments equation ft on their 'table' |
| \(\bar{y} = \dfrac{11a}{9}\) | A1 | Correct (for their axis) only |
| \(\theta = \tan^{-1}\dfrac{4a - \bar{x}}{3a - \bar{y}} \left(= \tan^{-1}\dfrac{9}{8}\right)\) or \((90° - \theta) = \tan^{-1}(\text{reciprocal})\) | M1 | Correct use of trigonometry to find a relevant angle |
| \(\alpha = \tan^{-1}\dfrac{4a - \bar{x}}{3a - \bar{y}} + \tan^{-1}\dfrac{2}{3}\) or oe | M1 | Correct strategy for the required angle |
| \(= 82°\) (nearest degree) | A1 | Correct answer only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{BA} \cdot \overrightarrow{BG} = \dfrac{2}{9}\begin{pmatrix}-9\\-8\end{pmatrix} \cdot \begin{pmatrix}2\\-3\end{pmatrix} \left(= \dfrac{4}{3}\right)\) | M1 | Correct use of trigonometry to find a relevant angle |
| \(\cos\alpha = \dfrac{\frac{4}{3}}{\frac{2}{9}\sqrt{145}\sqrt{13}} (= 0.138...)\) | M1 | Correct strategy for the required angle |
| \(\theta = 82°\) | A1 | Correct answer only |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| In the folding process, each point of the lamina remains the same distance from $CD$ | B1 | Any equivalent explanation e.g. folding doesn't change the mass distribution relative to $CD$. A calculation to verify is not the same as an explanation. Allow use of 'vertical' for $CD$. |
**(1 mark)**
---
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For the folded lamina: $\bar{x} = 2a \; (= d_2)$ oe | B1 | Seen anywhere |
| Area ratios correct (e.g. Large triangle $ACE$: $27a^2$, Removed triangle $BCD$: $12a^2$, Added triangle $BCD$: $12a^2$, Folded lamina: $27a^2$) | B1 | Any equivalent form for the mass (area) ratios |
| Distances from $EA$ correct (e.g. $3a$, $5a$, $a$, $\bar{y}$) | B1 | Or correct distances from an alternative axis parallel to $AE$ e.g. $BD$ |
| Moments about $EA$: $27 \times 3a - 12 \times 5a + 12 \times a = 27\bar{y}$ | M1 | Moments about $AE$ or a parallel axis. Need all terms. Must be dimensionally correct. Condone sign errors. |
| Correct unsimplified moments equation ft on their 'table' | A1ft | Correct unsimplified moments equation **ft** on their 'table' |
| $\bar{y} = \dfrac{11a}{9}$ | A1 | Correct (for their axis) only |
| $\theta = \tan^{-1}\dfrac{4a - \bar{x}}{3a - \bar{y}} \left(= \tan^{-1}\dfrac{9}{8}\right)$ or $(90° - \theta) = \tan^{-1}(\text{reciprocal})$ | M1 | Correct use of trigonometry to find a relevant angle |
| $\alpha = \tan^{-1}\dfrac{4a - \bar{x}}{3a - \bar{y}} + \tan^{-1}\dfrac{2}{3}$ or oe | M1 | Correct strategy for the required angle |
| $= 82°$ (nearest degree) | A1 | Correct answer only |
**Alternative for final 3 marks:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{BA} \cdot \overrightarrow{BG} = \dfrac{2}{9}\begin{pmatrix}-9\\-8\end{pmatrix} \cdot \begin{pmatrix}2\\-3\end{pmatrix} \left(= \dfrac{4}{3}\right)$ | M1 | Correct use of trigonometry to find a relevant angle |
| $\cos\alpha = \dfrac{\frac{4}{3}}{\frac{2}{9}\sqrt{145}\sqrt{13}} (= 0.138...)$ | M1 | Correct strategy for the required angle |
| $\theta = 82°$ | A1 | Correct answer only |
**(9 marks) — Total: (10 marks)**4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{269e7aef-d7b7-4c3b-8d55-5a00696c97cc-14_888_1322_294_374}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The uniform triangular lamina $A B C D E$ is such that angle $C E A = 90 ^ { \circ } , C E = 9 a$ and $E A = 6 a$. The point $D$ lies on $C E$, with $D E = 3 a$. The point $B$ on $C A$ is such that $D B$ is parallel to $E A$ and $D B = 4 a$. The triangular lamina is folded along the line $D B$ to form the folded lamina $A B D E C F$, as shown in Figure 2.
The distance of the centre of mass of the triangular lamina from $D C$ is $d _ { 1 }$\\
The distance of the centre of mass of the folded lamina from $D C$ is $d _ { 2 }$
\begin{enumerate}[label=(\alph*)]
\item Explain why $d _ { 1 } = d _ { 2 }$
The folded lamina is freely suspended from $B$ and hangs in equilibrium with $B A$ inclined at an angle $\alpha$ to the downward vertical through $B$.
\item Find, to the nearest degree, the size of angle $\alpha$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 AS 2019 Q4 [10]}}