| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Lamina with attached triangle |
| Difficulty | Standard +0.3 This is a standard two-part centre of mass question requiring decomposition into simple shapes (square and triangle), calculation of individual centroids, and then applying the composite formula. Part (ii) adds a suspension problem requiring basic trigonometry. All techniques are routine for M2 level with no novel insights required, making it slightly easier than average. |
| Spec | 6.04c Composite bodies: centre of mass6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\bar{x} = 9\) | B1 | ignore any working |
| c of m of \(\Delta\) 4 cm above BD | B1 | 8 cm below C/see their diagram |
| \((324 + 108)(m)\bar{y} =\) | M1 | \(432\bar{y} = 108x8 + 18^2(12+9)\) |
| \(324(m)x9 + 108(m)x(18+4)\) | from C | |
| \(432\bar{y}\) | A1 | left hand side |
| \(324 \times 9 \quad (18^2 \times 9)\) | A1 | 1st term on right hand side \(2916\) |
| \(108 \times (18+4)\) | A1 | 2nd term on right hand side \(2376\) |
| \(\bar{y} = 12.25\) | A1 | 7 marks total; \(5292 \div 432\) or \(49/4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\tan\theta = 5.75/9\) | M1 | must be \(\ldots/9\) |
| \(\theta = 32.6°\) or \(147.4°\) | A1\(\checkmark\) | 2 marks total; \(\checkmark\tan^{-1}((18\text{-their }\bar{y})/9)\) or \(180°\); 9 marks total |
# Question 5:
## Part (i):
| Working | Mark | Guidance |
|---------|------|----------|
| $\bar{x} = 9$ | B1 | ignore any working |
| c of m of $\Delta$ 4 cm above BD | B1 | 8 cm below C/see their diagram |
| $(324 + 108)(m)\bar{y} =$ | M1 | $432\bar{y} = 108x8 + 18^2(12+9)$ |
| $324(m)x9 + 108(m)x(18+4)$ | | from C |
| $432\bar{y}$ | A1 | left hand side |
| $324 \times 9 \quad (18^2 \times 9)$ | A1 | 1st term on right hand side $2916$ |
| $108 \times (18+4)$ | A1 | 2nd term on right hand side $2376$ |
| $\bar{y} = 12.25$ | A1 | 7 marks total; $5292 \div 432$ or $49/4$ |
## Part (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $\tan\theta = 5.75/9$ | M1 | must be $\ldots/9$ |
| $\theta = 32.6°$ or $147.4°$ | A1$\checkmark$ | 2 marks total; $\checkmark\tan^{-1}((18\text{-their }\bar{y})/9)$ or $180°$; 9 marks total |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{d6d87705-be4b-407d-b699-69fb441d88a7-3_657_549_1219_799}
A uniform lamina $A B C D E$ consists of a square and an isosceles triangle. The square has sides of 18 cm and $B C = C D = 15 \mathrm {~cm}$ (see diagram).\\
(i) Taking $x$ - and $y$-axes along $A E$ and $A B$ respectively, find the coordinates of the centre of mass of the lamina.\\
(ii) The lamina is freely suspended from $B$. Calculate the angle that $B D$ makes with the vertical.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d6d87705-be4b-407d-b699-69fb441d88a7-4_441_1355_265_394}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
A light inextensible string of length 1 m passes through a small smooth hole $A$ in a fixed smooth horizontal plane. One end of the string is attached to a particle $P$, of mass 0.5 kg , which hangs in equilibrium below the plane. The other end of the string is attached to a particle $Q$, of mass 0.3 kg , which rotates with constant angular speed in a circle of radius 0.2 m on the surface of the plane (see Fig. 1).\\
\hfill \mbox{\textit{OCR M2 2006 Q5 [9]}}