| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| 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 with two straightforward parts: (a) uses symmetry and the given centre of mass distance to find cos θ through basic coordinate geometry, and (b) applies the equilibrium condition for a suspended lamina with an attached particle. Both parts follow routine M2 procedures with no novel insight required, making it slightly easier than average. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use total height \(A\) to \(C\); \(a + a\cos\theta\) seen | B1 | |
| The centres of mass of the rhombuses lie on a straight line passing through the centre of mass | M1 | Using symmetry of figure; condone sin/cos confusion |
| \(0.9a = \frac{1}{2}(a + a\cos\theta)\) | A2 | \(-1\) each error |
| \(\cos\theta = 0.8\) | A1 | Given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distance from \(A\) to centre of rhombus \(= a\cos\frac{\theta}{2}\) | B1 | |
| \(\lambda\cdot a\cos\frac{\theta}{2}\cdot\cos\frac{\theta}{2} + \lambda\cdot a\cos\frac{\theta}{2}\cos\frac{\theta}{2} = 2\lambda\times0.9a\) | M1, A2 | Taking moments about axis through \(A\) parallel to \(FB\); condone sin/cos confusion; \(-1\) each error |
| \(\cos^2\frac{\theta}{2} = 0.9\) | ||
| \(\cos\theta = 2\cos^2\frac{\theta}{2} - 1 = 0.8\) | A1 | Given answer — from exact working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distance from \(A\) to centre of rhombus \(= a\cos\frac{\theta}{2}\) | B1 | |
| The centres of mass of the rhombuses lie on a straight line through the centre of mass | M1 | Using symmetry; condone sin/cos confusion |
| \(a\cos\frac{\theta}{2}\times\cos\frac{\theta}{2} = 0.9a\) | A2 | \(-1\) each error |
| \(\cos^2\frac{\theta}{2} = 0.9\) | ||
| \(\cos\theta = 2\cos^2\frac{\theta}{2} - 1 = 0.8\) | A1 | Given answer — from exact working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Correct division of lamina with correct mass ratios (table with EFBC, EDC, FAB, lamina rows showing \(2a^2\sin\theta\), \(a^2\sin\theta\cos\theta\), \(a^2\sin\theta\cos\theta\), \(2a^2\sin\theta\) and positions \(a\cos\theta+\frac{a}{2}\), \(a+\frac{2}{3}a\cos\theta\), \(\frac{2}{3}a\cos\theta\), \(0.9a\)) | B1 | Correct division of lamina with correct mass ratios |
| \(2a^2\sin\theta\left(a\cos\theta+\frac{a}{2}\right)-a^2\sin\theta\cos\theta\left(\frac{2}{3}a\cos\theta+a-\frac{2}{3}a\cos\theta\right)=2a^2\sin\theta\times0.9a\) | M1 A2 | Moments equation – addition and subtraction of terms consistent with their division. -1 each error |
| \(\cos\theta=0.8\) | A1 | *Given answer* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Correct division with EDF & BCD, AFD & ADB, lamina: masses \(2\times\frac{1}{2}a^2\sin\theta\), \(2\times\frac{1}{2}a^2\sin\theta\), \(2a^2\sin\theta\); positions \(\frac{2}{3}a(1+\cos\theta)\), \(\frac{1}{3}a(1+\cos\theta)\), \(0.9a\) | B1 | Correct division of lamina with correct mass ratios |
| \(2\times\frac{1}{3}a(1+\cos\theta)\times\frac{1}{2}a^2\sin\theta+2\times\frac{2}{3}a(1+\cos\theta)\times\frac{1}{2}a^2\sin\theta=2a^2\sin\theta\times0.9a\) | M1 A2 | Moments equation addition and subtraction of terms consistent with their division. -1 each error |
| \(\cos\theta=0.8\) | A1 | *Given answer* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(AM=a\cos\theta\) | B1 | Let M be the midpoint of FB. Centre of mass lies at the midpoint of DM |
| \(DM=a-a\cos\theta\) | ||
| \(0.9a=a\cos\theta+\frac{1}{2}(a-a\cos\theta)\) | M1 A2 | |
| \(0.4a=\frac{1}{2}a\cos\theta\), \(\cos\theta=0.8\) | A1 | *Given answer* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.4a\) seen or implied (from diagram with measurements \(0.1a\), \(0.4a\), \(0.5a\)) | B1 | \(0.4a\) seen or implied |
| \(\cos\theta=\frac{0.4a}{0.5a}\) | M1 A2 | Trig ratio for \(\theta\). Condone sin/cos confusion. Correct unsimplified expression |
| \(=0.8\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(kW(a\cos\theta)=W(0.9a-a\cos\theta)\) | M1 A2 | Taking moments about \(B\). Must have both terms. Condone trig & sign errors. -1 each error |
| \(0.8kW=0.1W\), \(k=\frac{1}{8}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Centre of mass is on \(BF\) | M1 | Taking moments about A |
| \(0.9aW=(k+1)W\times a\cos\theta=(k+1)\times0.8aW\) | A2 | -1 each error |
| \(k=\frac{1}{8}\) | A1 |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use total height $A$ to $C$; $a + a\cos\theta$ seen | B1 | |
| The centres of mass of the rhombuses lie on a straight line passing through the centre of mass | M1 | Using symmetry of figure; condone sin/cos confusion |
| $0.9a = \frac{1}{2}(a + a\cos\theta)$ | A2 | $-1$ each error |
| $\cos\theta = 0.8$ | A1 | **Given answer** |
**Alternative 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance from $A$ to centre of rhombus $= a\cos\frac{\theta}{2}$ | B1 | |
| $\lambda\cdot a\cos\frac{\theta}{2}\cdot\cos\frac{\theta}{2} + \lambda\cdot a\cos\frac{\theta}{2}\cos\frac{\theta}{2} = 2\lambda\times0.9a$ | M1, A2 | Taking moments about axis through $A$ parallel to $FB$; condone sin/cos confusion; $-1$ each error |
| $\cos^2\frac{\theta}{2} = 0.9$ | | |
| $\cos\theta = 2\cos^2\frac{\theta}{2} - 1 = 0.8$ | A1 | **Given answer** — from exact working |
**Alternative 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance from $A$ to centre of rhombus $= a\cos\frac{\theta}{2}$ | B1 | |
| The centres of mass of the rhombuses lie on a straight line through the centre of mass | M1 | Using symmetry; condone sin/cos confusion |
| $a\cos\frac{\theta}{2}\times\cos\frac{\theta}{2} = 0.9a$ | A2 | $-1$ each error |
| $\cos^2\frac{\theta}{2} = 0.9$ | | |
| $\cos\theta = 2\cos^2\frac{\theta}{2} - 1 = 0.8$ | A1 | **Given answer** — from exact working |
> Working backwards from $\cos\theta = 0.8$ to deduce that the distance is $0.9a$ is acceptable for 5/5.
## Question Alt 3:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct division of lamina with correct mass ratios (table with EFBC, EDC, FAB, lamina rows showing $2a^2\sin\theta$, $a^2\sin\theta\cos\theta$, $a^2\sin\theta\cos\theta$, $2a^2\sin\theta$ and positions $a\cos\theta+\frac{a}{2}$, $a+\frac{2}{3}a\cos\theta$, $\frac{2}{3}a\cos\theta$, $0.9a$) | B1 | Correct division of lamina with correct mass ratios |
| $2a^2\sin\theta\left(a\cos\theta+\frac{a}{2}\right)-a^2\sin\theta\cos\theta\left(\frac{2}{3}a\cos\theta+a-\frac{2}{3}a\cos\theta\right)=2a^2\sin\theta\times0.9a$ | M1 A2 | Moments equation – addition and subtraction of terms consistent with their division. -1 each error |
| $\cos\theta=0.8$ | A1 | *Given answer* |
---
## Question Alt 4:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Correct division with EDF & BCD, AFD & ADB, lamina: masses $2\times\frac{1}{2}a^2\sin\theta$, $2\times\frac{1}{2}a^2\sin\theta$, $2a^2\sin\theta$; positions $\frac{2}{3}a(1+\cos\theta)$, $\frac{1}{3}a(1+\cos\theta)$, $0.9a$ | B1 | Correct division of lamina with correct mass ratios |
| $2\times\frac{1}{3}a(1+\cos\theta)\times\frac{1}{2}a^2\sin\theta+2\times\frac{2}{3}a(1+\cos\theta)\times\frac{1}{2}a^2\sin\theta=2a^2\sin\theta\times0.9a$ | M1 A2 | Moments equation addition and subtraction of terms consistent with their division. -1 each error |
| $\cos\theta=0.8$ | A1 | *Given answer* |
---
## Question Alt 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $AM=a\cos\theta$ | B1 | Let M be the midpoint of FB. Centre of mass lies at the midpoint of DM |
| $DM=a-a\cos\theta$ | | |
| $0.9a=a\cos\theta+\frac{1}{2}(a-a\cos\theta)$ | M1 A2 | |
| $0.4a=\frac{1}{2}a\cos\theta$, $\cos\theta=0.8$ | A1 | *Given answer* |
---
## Question Alt 6:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.4a$ seen or implied (from diagram with measurements $0.1a$, $0.4a$, $0.5a$) | B1 | $0.4a$ seen or implied |
| $\cos\theta=\frac{0.4a}{0.5a}$ | M1 A2 | Trig ratio for $\theta$. Condone sin/cos confusion. Correct unsimplified expression |
| $=0.8$ | A1 | |
---
## Question 4(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $kW(a\cos\theta)=W(0.9a-a\cos\theta)$ | M1 A2 | Taking moments about $B$. Must have both terms. Condone trig & sign errors. -1 each error |
| $0.8kW=0.1W$, $k=\frac{1}{8}$ | A1 | |
**Alt 4(b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Centre of mass is on $BF$ | M1 | Taking moments about A |
| $0.9aW=(k+1)W\times a\cos\theta=(k+1)\times0.8aW$ | A2 | -1 each error |
| $k=\frac{1}{8}$ | A1 | |
---4.
\begin{figure}[h]
\begin{center}
\begin{tikzpicture}[line cap=round, line join=round]
% Define parameters for the shape
\def\a{4} % Length of the side 'a'
\def\thetaAngle{40} % Angle \theta from the vertical axis
% Define coordinates based on the geometric properties
% A is the origin
\coordinate (A) at (0,0);
% D is straight up from A, at distance 'a'
\coordinate (D) at (0,\a);
% B is rotated from the vertical by -\theta, at distance 'a' from A
\coordinate (B) at (90-\thetaAngle:\a);
% F is rotated from the vertical by \theta, at distance 'a' from A
\coordinate (F) at (90+\thetaAngle:\a);
% C is vertically above B by distance 'a'
\coordinate (C) at ($(B)+(0,\a)$);
% E is vertically above F by distance 'a'
\coordinate (E) at ($(F)+(0,\a)$);
% Draw the main solid boundary lines
\draw[thick] (A) -- (B) -- (C) -- (D) -- (E) -- (F) -- cycle;
% Draw the dashed line of symmetry
\draw[thick, dashed] (A) -- (D);
% Add vertex labels
\node[below, yshift=-2pt] at (A) {$A$};
\node[right, xshift=2pt] at (B) {$B$};
\node[above right] at (C) {$C$};
\node[above, yshift=2pt] at (D) {$D$};
\node[above left] at (E) {$E$};
\node[left, xshift=-2pt] at (F) {$F$};
% Add side length labels 'a'
% The positions are chosen to match the reference image visually
\path (A) -- (B) node[midway, below right=1pt] {$a$};
\path (B) -- (C) node[midway, right=2pt] {$a$};
\path (D) -- (C) node[midway, above left=1pt] {$a$};
\path (D) -- (E) node[midway, above right=1pt] {$a$};
\path (E) -- (F) node[midway, left=2pt] {$a$};
\path (F) -- (A) node[midway, below left=1pt] {$a$};
% Add angle arcs and labels for \theta
\def\arcRad{0.6} % Radius for the angle arcs
% Right angle \theta
\draw (90-\thetaAngle:\arcRad) arc (90-\thetaAngle:90:\arcRad);
\node at (90-\thetaAngle/2:\arcRad+0.25) {$\theta$};
% Left angle \theta
\draw (90:\arcRad) arc (90:90+\thetaAngle:\arcRad);
\node at (90+\thetaAngle/2:\arcRad+0.25) {$\theta$};
\end{tikzpicture}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
The uniform plane lamina $A B C D E F$ shown in Figure 1 is made from two identical rhombuses. Each rhombus has sides of length $a$ and angle $B A D =$ angle $D A F = \theta$. The centre of mass of the lamina is $0.9 a$ from $A$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\cos \theta = 0.8$
The weight of the lamina is $W$. A particle of weight $k W$ is fixed to the lamina at the point $A$. The lamina is freely suspended from $B$ and hangs in equilibrium with $D A$ horizontal.
\item Find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2015 Q4 [9]}}