| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Frame with straight rod/wire components only |
| Difficulty | Standard +0.3 This is a straightforward centre of mass problem requiring standard techniques: finding the COM of three uniform rods using symmetry and the midpoint formula, then applying v = rω. The geometry is simple (square frame with one side removed), and both parts are direct applications of learned methods with no novel problem-solving required. Slightly easier than average due to the symmetric setup and routine calculations. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(3x_G = 2x0.3 + 1x0.6\) OR \(3x_G = 2x0.3 + 0\) OR \(3x_G = 4x0.3\) OR \(3x_G = 1x0.3 + 1x0.6 + 0\) OR \(3x_G = 4x0.3 - 1x0.3\) | M1 | Table of moments idea. M0 for reducing to 1D problem. Masses/weights may be included. |
| \(x_G = 0.4\) (from AD) OR \(x_G = 0.2\) (from BC) | A1 | |
| \(y_a = 0.3m\) from AB or CD | A1 | |
| \(AG^2 = 0.4^2 + 0.3^2\) | M1 | Pythagoras with 2 appropriate distances. This may only be seen in (ii), allow M1A1 in this case. |
| \(AG = 0.5\) m | A1 | |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = 0.5x3\) | M1 | Allow use of candidate's 0.2, 0.4, 0.3, 0.5 |
| \(v = 1.5\) m s\(^{-1}\) | A1 | |
| [2] |
**Part (i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $3x_G = 2x0.3 + 1x0.6$ OR $3x_G = 2x0.3 + 0$ OR $3x_G = 4x0.3$ OR $3x_G = 1x0.3 + 1x0.6 + 0$ OR $3x_G = 4x0.3 - 1x0.3$ | M1 | Table of moments idea. M0 for reducing to 1D problem. Masses/weights may be included. |
| $x_G = 0.4$ (from AD) OR $x_G = 0.2$ (from BC) | A1 | |
| $y_a = 0.3m$ from AB or CD | A1 | |
| $AG^2 = 0.4^2 + 0.3^2$ | M1 | Pythagoras with 2 appropriate distances. This may only be seen in (ii), allow M1A1 in this case. |
| $AG = 0.5$ m | A1 | |
| | **[5]** | |
**Part (ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = 0.5x3$ | M1 | Allow use of candidate's 0.2, 0.4, 0.3, 0.5 |
| $v = 1.5$ m s$^{-1}$ | A1 | |
| | **[2]** | |1\\
\includegraphics[max width=\textwidth, alt={}, center]{941c0c81-a74f-49c0-acb7-1c23266fc2c8-02_378_471_260_836}
A uniform square frame $A B C D$ has sides of length 0.6 m . The side $A D$ is removed from the frame, and the open frame $A B C D$ is attached at $A$ to a fixed point (see diagram).\\
(i) Calculate the distance of the centre of mass of the open frame from $A$.
The open frame rotates about $A$ in the plane $A B C D$ with angular speed $3 \mathrm { rad } \mathrm { s } ^ { - 1 }$.\\
(ii) Calculate the speed of the centre of mass of the open frame.
\hfill \mbox{\textit{OCR M2 2011 Q1 [7]}}