| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2022 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Centre of mass with variable parameter |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question with two parts: (a) showing a given result using composite bodies (straightforward calculation with parameter k), and (b) using equilibrium conditions with a given angle to find k. The techniques are routine for M2 - dividing into rectangles, finding individual centres of mass, combining them, then applying the suspension principle that the centre of mass lies vertically below the pivot. The algebra is manageable and the question follows a familiar template, making it slightly easier than average overall. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M(\(PV\)) | M1 | Allow use of parallel axis. Terms dimensionally consistent. Could be seen as part of a vector equation. Condone error(s) in distance(s) |
| \(a \times 2ka^2 + \left(2a + \frac{1}{2}ka\right)2ka^2 = \bar{x} \times (2ka^2 + 2ka^2)\) | A1 | Correct unsimplified equation |
| \(2\bar{x} = a + 2a + \frac{1}{2}ka \Rightarrow \bar{x} = \frac{6+k}{4}a\) * | A1* | Obtain given answer from correct working |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M(\(PR\)) | M1 | Allow use of parallel axis. Terms dimensionally consistent. Could be seen as part of vector equation in (a) but needs to be used here. Condone error(s) in distance(s). If working from \(VU\) they might assume c of m of \(QRST\) lies on their axis |
| \(\frac{1}{2}ka \times 2ka^2 + a \times 2ka^2 = \bar{y} \times 4ka^2\) | A1 | Correct unsimplified equation |
| \(\bar{y} = \frac{k+2}{4}a\) | A1 | Correct answer (\(\pm\)) seen or implied. Accept distance from \(VU = \pm\frac{3k-2}{4}a\) or distance from \(TS = \pm\frac{6-k}{4}a\) |
| Use angle to form equation in \(k\) | M1 | Correct use of given ratio. Allow reciprocal |
| \(\frac{7}{15} = \frac{\bar{y}}{\bar{x}} = \frac{(k+2)a}{4} \times \frac{4}{(6+k)a}\) | A1 | Correct unsimplified equation using given \(\bar{x}\) and \(\bar{y}\), e.g. \(\frac{ka - \bar{y}_{VU}}{\bar{x}}\) or \(\frac{2a - \bar{y}_{TS}}{\bar{x}}\) |
| \(\Rightarrow k = \frac{3}{2} (=1.5)\) | A1 | Correct only |
| Total | 6 |
# Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| M($PV$) | M1 | Allow use of parallel axis. Terms dimensionally consistent. Could be seen as part of a vector equation. Condone error(s) in distance(s) |
| $a \times 2ka^2 + \left(2a + \frac{1}{2}ka\right)2ka^2 = \bar{x} \times (2ka^2 + 2ka^2)$ | A1 | Correct unsimplified equation |
| $2\bar{x} = a + 2a + \frac{1}{2}ka \Rightarrow \bar{x} = \frac{6+k}{4}a$ * | A1* | Obtain **given answer** from correct working |
| **Total** | **3** | |
---
# Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| M($PR$) | M1 | Allow use of parallel axis. Terms dimensionally consistent. Could be seen as part of vector equation in (a) but needs to be used here. Condone error(s) in distance(s). If working from $VU$ they might assume c of m of $QRST$ lies on their axis |
| $\frac{1}{2}ka \times 2ka^2 + a \times 2ka^2 = \bar{y} \times 4ka^2$ | A1 | Correct unsimplified equation |
| $\bar{y} = \frac{k+2}{4}a$ | A1 | Correct answer ($\pm$) seen or implied. Accept distance from $VU = \pm\frac{3k-2}{4}a$ or distance from $TS = \pm\frac{6-k}{4}a$ |
| Use angle to form equation in $k$ | M1 | Correct use of given ratio. Allow reciprocal |
| $\frac{7}{15} = \frac{\bar{y}}{\bar{x}} = \frac{(k+2)a}{4} \times \frac{4}{(6+k)a}$ | A1 | Correct unsimplified equation using given $\bar{x}$ and $\bar{y}$, e.g. $\frac{ka - \bar{y}_{VU}}{\bar{x}}$ or $\frac{2a - \bar{y}_{TS}}{\bar{x}}$ |
| $\Rightarrow k = \frac{3}{2} (=1.5)$ | A1 | Correct only |
| **Total** | **6** | |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1732eb73-8c16-4a45-8d3b-a88e659e47ea-16_588_871_219_539}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The uniform lamina $P Q R S T U V$ shown in Figure 2 is formed from two identical rectangles, $P Q U V$ and $Q R S T U$.\\
The rectangles have sides $P Q = R S = 2 a$ and $P V = Q R = k a$.
\begin{enumerate}[label=(\alph*)]
\item Show that the centre of mass of the lamina is $\left( \frac { 6 + k } { 4 } \right) a$ from $P V$
The lamina is freely suspended from $P$ and hangs in equilibrium with $P R$ at an angle of $\alpha$ to the downward vertical.
Given that $\tan \alpha = \frac { 7 } { 15 }$
\item find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2022 Q6 [9]}}