| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Suspended lamina equilibrium angle |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question involving composite shapes (triangle with triangle removed) and suspended lamina equilibrium. Part (a) requires routine application of the composite body formula with symmetry simplifying calculations. Part (b) uses basic trigonometry to find the angle when the centre of mass hangs vertically below the suspension point. Both parts follow well-established procedures taught in M2 with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.04b Find centre of mass: using symmetry6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Ratio of areas \(54 : 24 : 30\) (or equivalent) | B1 | |
| Distance of centre of mass from \(AC\): \(3a\), \(2a\) | B1 | The two B marks can be implied by a correct moments equation |
| Moments about \(AC\): \(30d = 3a \times 54 - 2a \times 24\ (= 114a)\) | M1 | All terms need to be there. Must be subtracting. Allow with \(g\) as common factor. Allow use of axis parallel to \(AC\). Allow in vector form. Correct unsimplified equation |
| \(d = \dfrac{114a}{30} = 3.8a\) | A1, A1 | Accept any equivalent form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Correct triangle and use of \(\tan^{-1}\) or equivalent | M1 | Find a relevant angle using their \(d\). Condone ratio the wrong way up. Allow \(\tan^{-1}\!\left(\dfrac{\text{their } d}{\text{their } \bar{x}}\right)\) |
| \(\tan^{-1}\!\left(\dfrac{3.8a}{6a}\right)\), \(\tan^{-1}\!\left(\dfrac{6a}{3.8a}\right)\), \(32.3°\) or \(57.65°\) | A1 | |
| Required angle \(= \tan^{-1}\!\left(\dfrac{9a}{6a}\right) - \tan^{-1}\!\left(\dfrac{3.8a}{6a}\right) = 23.96...= 24°\) | DM1 | Correct method for required angle. Dependent on previous M mark |
| \(\theta = 24°\) | A1 | Only. The question asks for answers to nearest degree |
# Question 3:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Ratio of areas $54 : 24 : 30$ (or equivalent) | B1 | |
| Distance of centre of mass from $AC$: $3a$, $2a$ | B1 | The two B marks can be implied by a correct moments equation |
| Moments about $AC$: $30d = 3a \times 54 - 2a \times 24\ (= 114a)$ | M1 | All terms need to be there. Must be subtracting. Allow with $g$ as common factor. Allow use of axis parallel to $AC$. Allow in vector form. Correct unsimplified equation |
| $d = \dfrac{114a}{30} = 3.8a$ | A1, A1 | Accept any equivalent form |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct triangle and use of $\tan^{-1}$ or equivalent | M1 | Find a relevant angle using their $d$. Condone ratio the wrong way up. Allow $\tan^{-1}\!\left(\dfrac{\text{their } d}{\text{their } \bar{x}}\right)$ |
| $\tan^{-1}\!\left(\dfrac{3.8a}{6a}\right)$, $\tan^{-1}\!\left(\dfrac{6a}{3.8a}\right)$, $32.3°$ or $57.65°$ | A1 | |
| Required angle $= \tan^{-1}\!\left(\dfrac{9a}{6a}\right) - \tan^{-1}\!\left(\dfrac{3.8a}{6a}\right) = 23.96...= 24°$ | DM1 | Correct method for required angle. Dependent on previous M mark |
| $\theta = 24°$ | A1 | Only. The question asks for answers to nearest degree |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{47420c50-c232-41e9-8c4d-a890d86ea933-04_814_1127_219_411}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The uniform lamina $A B C D E F$, shown shaded in Figure 1, is symmetrical about the line through $B$ and $E$. It is formed by removing the isosceles triangle $F E D$, of height $6 a$ and base $8 a$, from the isosceles triangle $A B C$ of height $9 a$ and base $12 a$.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $a$, the distance of the centre of mass of the lamina from $A C$.
The lamina is freely suspended from $A$ and hangs in equilibrium.
\item Find, to the nearest degree, the size of the angle between $A B$ and the downward vertical.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2014 Q3 [9]}}