| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Major (Further Mechanics Major) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Conical pendulum – horizontal circle in free space (no surface) |
| Difficulty | Standard +0.3 This is a standard conical pendulum problem requiring resolution of forces and application of circular motion principles. Part (a) involves simple vertical equilibrium (T cos θ = mg), while part (b) uses horizontal circular motion (T sin θ = mω²r). Both are routine applications of well-practiced techniques with straightforward algebra, making this slightly easier than average for Further Maths students who have covered circular motion. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | T c o s 0 .2 g = |
| Answer | Marks | Guidance |
|---|---|---|
| 375 | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | r 0 .7 5 s i n = | B1 |
| T s i n 0 .2 ( 0 .7 5 s i n ) 2 = | M1 | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| 1 0 = | A1 | 1.1 |
Question 3:
3 | (a) | T c o s 0 .2 g = | M1 | 1.1 | Resolving vertically – correct number of terms and T
resolved. M0 if 0.2 used as the weight. Allow with
either T or 15
49
cos= =82
375 | A1 | 1.1 | For reference: 82.491881… (no indication of why this
is the greatest value is required) allow awrt 82 and
awrt 1.44 (radians) - condone 8 2 but not < 82
Ignore incorrect use of inequality symbols if 82 given
as answer
[2]
(b) | r 0 .7 5 s i n = | B1 | 1.1 | Allow a value of 0.74 (or better) to imply this mark
T s i n 0 .2 ( 0 .7 5 s i n ) 2 = | M1 | 3.3 | Applying N2L horizontally – correct number of terms
and T resolved (allow with either T or 15). Allow
2 v
with r for radius and a r 2 = or a = but not just a
r
1 0 = | A1 | 1.1 | Condone awrt 10 from correct working (might see
10.018… from using r = 0.74 and 8 2 = ) condone
1 0 but not 10
Ignore incorrect use of inequality symbols if 10 given
as answer
[3]\includegraphics{figure_3}
The diagram shows a particle P, of mass 0.2 kg, which is attached by a light inextensible string of length 0.75 m to a fixed point O.
Particle P moves with constant angular speed $\omega \text{ rad s}^{-1}$ in a horizontal circle with centre vertically below O. The string is inclined at an angle $\theta$ to the vertical.
The greatest tension that the string can withstand without breaking is 15 N.
\begin{enumerate}[label=(\alph*)]
\item Find the greatest possible value of $\theta$, giving your answer to the nearest degree. [2]
\item Determine the greatest possible value of $\omega$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2023 Q3 [5]}}