| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Marks | 16 |
| 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 lamina with attached particles. Part (a) is straightforward calculation, parts (b) and (c) are 'show that' questions requiring systematic moment calculations, and part (d) applies equilibrium conditions with a given angle. While multi-step, it follows a predictable template with no novel geometric insight required, making it slightly easier than average. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| (a) \(\frac{1}{2}a\), since masses on \(AD\) are equal to mass at \(B\) | A1 | |
| (b) \(y\) coords. taken vert. from \(AB\) | M2 A1 | |
| \(\bar{y} = \dfrac{16ma}{20m} = \dfrac{4}{5}a\) | M1 A1 | |
| Table: lamina \(8m\), \(y=a\), \(my=8ma\); particle at \(A\): \(2m\), \(0\), \(0\); particle at \(B\): \(6m\), \(0\), \(0\); particle at \(D\): \(4m\), \(2a\), \(8ma\); total \(20m\), \(\bar{y}\), \(16ma\) | ||
| (c) \(x\) coords. taken horiz. from \(AD\) | M1 A1 | |
| \(\bar{x} = \dfrac{(10+k)ma}{(20+k)m} = \dfrac{(10+k)a}{(20+k)}\) | M1 A1 | |
| Table: lamina \(8m\), \(x=\frac{a}{2}\), \(mx=4ma\); particle at \(A\): \(2m\), \(0\), \(0\); particles at \(B\): \((6+k)m\), \(a\), \((6+k)ma\); particle at \(D\): \(4m\), \(0\), \(0\); total \((20+k)m\), \(\bar{x}\), \((10+k)ma\) | ||
| (d) new \(\bar{y} = \dfrac{16ma}{(20+k)m} = \dfrac{16a}{(20+k)}\) | M2 A1 | |
| \(\tan45° = \dfrac{16a}{(10+k)a}\) \(\therefore\) \(1 = \dfrac{16}{10+k}\) giving \(k = 6\) | M2 A1 | (16) |
## Question 6:
| Answer/Working | Marks | Notes |
|---|---|---|
| **(a)** $\frac{1}{2}a$, since masses on $AD$ are equal to mass at $B$ | A1 | |
| **(b)** $y$ coords. taken vert. from $AB$ | M2 A1 | |
| $\bar{y} = \dfrac{16ma}{20m} = \dfrac{4}{5}a$ | M1 A1 | |
| Table: lamina $8m$, $y=a$, $my=8ma$; particle at $A$: $2m$, $0$, $0$; particle at $B$: $6m$, $0$, $0$; particle at $D$: $4m$, $2a$, $8ma$; total $20m$, $\bar{y}$, $16ma$ | | |
| **(c)** $x$ coords. taken horiz. from $AD$ | M1 A1 | |
| $\bar{x} = \dfrac{(10+k)ma}{(20+k)m} = \dfrac{(10+k)a}{(20+k)}$ | M1 A1 | |
| Table: lamina $8m$, $x=\frac{a}{2}$, $mx=4ma$; particle at $A$: $2m$, $0$, $0$; particles at $B$: $(6+k)m$, $a$, $(6+k)ma$; particle at $D$: $4m$, $0$, $0$; total $(20+k)m$, $\bar{x}$, $(10+k)ma$ | | |
| **(d)** new $\bar{y} = \dfrac{16ma}{(20+k)m} = \dfrac{16a}{(20+k)}$ | M2 A1 | |
| $\tan45° = \dfrac{16a}{(10+k)a}$ $\therefore$ $1 = \dfrac{16}{10+k}$ giving $k = 6$ | M2 A1 | **(16)** |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0ea2267e-6c46-4a4f-9a38-c242de57901d-4_433_282_196_726}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
Figure 3 shows a uniform rectangular lamina $A B C D$ of mass $8 m$ in which the sides $A B$ and $B C$ are of length $a$ and $2 a$ respectively. Particles of mass $2 m , 6 m$ and $4 m$ are fixed to the lamina at the points $A , B$ and $D$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Write down the distance of the centre of mass from $A D$.
\item Show that the distance of the centre of mass from $A B$ is $\frac { 4 } { 5 } a$.
Another particle of mass $k m$ is attached to the lamina at the point $B$.
\item Show that the distance of the centre of mass from $A D$ is now given by $\frac { ( 10 + k ) a } { 20 + k }$.\\
(4 marks)\\
Given that when the lamina is suspended freely from the point $A$ the side $A B$ makes an angle of $45 ^ { \circ }$ with the vertical,
\item find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 Q6 [16]}}