| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Major (Further Mechanics Major) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Smooth ring on rotating string |
| Difficulty | Standard +0.8 This is a 3D circular motion problem requiring resolution of forces in both horizontal and vertical directions, geometric reasoning to find angles, and application of Newton's second law for circular motion. While the setup is more complex than typical A-level mechanics (involving a string through a ring with different tensions on each side), the solution follows standard methods once the geometry is understood. The 8 marks and multi-step nature place it above average difficulty, but it's still a structured problem testing core Further Mechanics concepts without requiring exceptional insight. |
| Spec | 3.03d Newton's second law: 2D vectors3.03n Equilibrium in 2D: particle under forces6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (i) | Let (cid:84)= angle between AR and vertical and |
| Answer | Marks |
|---|---|
| Substitute to give T (cid:32)6.37 AG | B1 |
| Answer | Marks |
|---|---|
| [4] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | N |
| Answer | Marks |
|---|---|
| (ii) | 0.27v2 |
| Answer | Marks |
|---|---|
| The speed is 6.57 ms-1 | M1 |
| Answer | Marks |
|---|---|
| [4] | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | E |
Question 5:
5 | (i) | Let (cid:84)= angle between AR and vertical and
(cid:68)= angle between BR and vertical
4 5
cos(cid:84)(cid:32) cos(cid:68)(cid:32) or equivalent
5 13
Tcos(cid:84)(cid:32)Tcos(cid:68)(cid:14)0.27g
Substitute to give T (cid:32)6.37 AG | B1
M1
A1
A1
[4] | 1.1
3.3
2.1
1.1 | N
Resolve vertically
(ii) | 0.27v2
Tsin(cid:84)(cid:14)Tsin(cid:68)(cid:32)
1.2
Solve: v(cid:32)6.57
The speed is 6.57 ms-1 | M1
A1
M1
A1
[4] | 3.3
1.1
3.4
1.1 | E
N2L in radial direction
M
EliminateFig. 5 shows a light inextensible string of length 3.3 m passing through a small smooth ring R. The ends of the string are attached to fixed points A and B, where A is vertically above B. The ring R has mass 0.27 kg and is moving with constant speed in a horizontal circle of radius 1.2 m. The distances AR and BR are 2 m and 1.3 m respectively.
\includegraphics{figure_5}
\begin{enumerate}[label=(\roman*)]
\item Show that the tension in the string is 6.37 N. [4]
\item Find the speed of R. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Major Q5 [8]}}