| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Particle attached to lamina - find mass/position |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question involving a trapezium lamina with straightforward geometry and a particle attachment problem. Part (a) requires routine application of centre of mass formulas by splitting the trapezium into rectangles/triangles, while part (b) involves a simple moment equilibrium condition about the suspension point. The geometry is clean (right angles, simple multiples of 'a'), and the methods are textbook-standard with no novel insight required. Slightly easier than average due to the guided 'show that' in part (a)(i) and the straightforward setup. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Rectangle + Triangle table: Mass ratio 2, 1, 3; From \(AB\): \(a\), \(\frac{4}{3}a\), \(\bar{y}\) | B1 | Also accept Square − Triangle approach with correct values |
| \(2a + 1\cdot\frac{4}{3}a = 3\bar{y}\) | M1 | Moments about \(AB\) or parallel axis. Need all three terms and correct signs |
| \(\Rightarrow \bar{y} = \frac{10}{9}a\) | A1 | Answer Given — need some evidence e.g. \(\frac{10}{3}a = 3\bar{y}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Rectangle + Triangle table: From \(AD\): \(\frac{1}{2}a\), \(\frac{4}{3}a\), \(\bar{x}\) | B1 | 2nd B1 on epen. Award if seen anywhere in response to part (a) |
| \(2\cdot\frac{1}{2}a + 1\cdot\frac{4}{3}a = 3\bar{x}\) | M1 | Moments about \(AD\) or parallel axis. Need all 3 terms and correct signs |
| \(\Rightarrow \bar{x} = \frac{7}{9}a\) | A1 | Accept \(0.778a\) or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(M(\text{pivot}):\ kMga = 3Mg\frac{a}{9}\) or \(M(A): 3M\times\frac{10a}{9} = (3+k)Ma\) or \(M(C): 3M\times\frac{8a}{9} + kM\times 2a = (3+k)Ma\) | M1 | Form equation in \(k\). Condone \(g\) omitted on both sides. Must be using \(\frac{10a}{9}\) (distances along \(AD\) from \(AB\)) |
| \(kMga = 3Mg\frac{a}{9}\) | A1 | Correct unsimplified equation (with or without \(g\)) |
| \(k = \frac{1}{3}\) | A1 | Accept equivalent fractions or 0.333 or better |
# Question 2:
## Part (a)(i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Rectangle + Triangle table: Mass ratio 2, 1, 3; From $AB$: $a$, $\frac{4}{3}a$, $\bar{y}$ | B1 | Also accept Square − Triangle approach with correct values |
| $2a + 1\cdot\frac{4}{3}a = 3\bar{y}$ | M1 | Moments about $AB$ or parallel axis. Need all three terms and correct signs |
| $\Rightarrow \bar{y} = \frac{10}{9}a$ | A1 | **Answer Given** — need some evidence e.g. $\frac{10}{3}a = 3\bar{y}$ |
## Part (a)(ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Rectangle + Triangle table: From $AD$: $\frac{1}{2}a$, $\frac{4}{3}a$, $\bar{x}$ | B1 | 2nd B1 on epen. Award if seen anywhere in response to part (a) |
| $2\cdot\frac{1}{2}a + 1\cdot\frac{4}{3}a = 3\bar{x}$ | M1 | Moments about $AD$ or parallel axis. Need all 3 terms and correct signs |
| $\Rightarrow \bar{x} = \frac{7}{9}a$ | A1 | Accept $0.778a$ or better |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $M(\text{pivot}):\ kMga = 3Mg\frac{a}{9}$ or $M(A): 3M\times\frac{10a}{9} = (3+k)Ma$ or $M(C): 3M\times\frac{8a}{9} + kM\times 2a = (3+k)Ma$ | M1 | Form equation in $k$. Condone $g$ omitted on both sides. Must be using $\frac{10a}{9}$ (distances along $AD$ from $AB$) |
| $kMga = 3Mg\frac{a}{9}$ | A1 | Correct unsimplified equation (with or without $g$) |
| $k = \frac{1}{3}$ | A1 | Accept equivalent fractions or 0.333 or better |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{36cced0d-f982-4534-a3fe-13c32fb37f5b-04_538_625_251_657}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A uniform lamina is in the shape of a trapezium $A B C D$ with $A B = a , D A = D C = 2 a$ and angle $B A D =$ angle $A D C = 90 ^ { \circ }$, as shown in Figure 1.
The centre of mass of the lamina is at the point $G$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that the distance of $G$ from $A B$ is $\frac { 10 a } { 9 }$.
\item Find the distance of $G$ from $A D$.
The mass of the lamina is $3 M$. A particle of mass $k M$ is now attached to the lamina at $B$. The lamina is freely suspended from the midpoint of $A D$ and hangs in equilibrium with $A D$ horizontal.
\end{enumerate}\item Find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2017 Q2 [9]}}