| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Minor (Further Mechanics Minor) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Vertical circle – string/rod (tension and energy) |
| Difficulty | Challenging +1.2 This is a standard vertical circular motion problem with friction requiring energy methods across two scenarios to find an unknown initial speed. While it involves multiple steps (energy equation for first scenario to find friction, then apply to second scenario), the approach is methodical and the physics concepts (conservation of energy with work done against friction) are standard A-level Further Maths mechanics. The geometry is straightforward and the 'comment on validity' part is routine. More challenging than basic projectile motion but less demanding than problems requiring novel geometric insight or complex force analysis. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | Let the constant friction force have magnitude F (N). |
| Gain in GPE = 0.3 𝑔 sin 25 | B1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| Distance travelled from A to B = 23 06 50 2 π = 43 16 π | B1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 12 0 .3 4 2 − 43 16 π F = 0 .3 g s i n 2 5 ( F = 0 .3 2 3 5 1 ) | M1 | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| 12 0 .3 v 2 − F π = 0 | M1 | 3.4 |
| v = 2 .6 0 3 0 | A1 | 2.2a |
| Answer | Marks |
|---|---|
| (b) | As the bead moves around the hoop, the normal reaction |
| Answer | Marks | Guidance |
|---|---|---|
| remain constant. | B1 | 3.5a |
Question 3:
3 | (a) | Let the constant friction force have magnitude F (N).
Gain in GPE = 0.3 𝑔 sin 25 | B1 | 1.1 | May be embedded in WEP
equation.
( )
Distance travelled from A to B = 23 06 50 2 π = 43 16 π | B1 | 1.1 | May be embedded in WEP
equation.
12 0 .3 4 2 − 43 16 π F = 0 .3 g s i n 2 5 ( F = 0 .3 2 3 5 1 ) | M1 | 3.3 | WEP all correct using their gain in
GPE and their arc length.
12 0 .3 v 2 − F π = 0 | M1 | 3.4
v = 2 .6 0 3 0 | A1 | 2.2a
[5]
(b) | As the bead moves around the hoop, the normal reaction
force on the bead is likely to change, thus friction unlikely to
remain constant. | B1 | 3.5a
[1]3 A rough circular hoop, with centre O and radius 1 m , is fixed in a vertical plane. A , B and C are points on the hoop such that A and C are at the same horizontal level as O , and OB makes an angle of $25 ^ { \circ }$ above the horizontal, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{9b624694-edb6-4000-838f-3557e078952d-4_650_729_404_251}
A bead P of mass 0.3 kg is threaded onto the hoop. P is projected vertically downwards from A on two separate occasions.
\begin{itemize}
\item The first time, when P is projected with a speed of $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, it first comes to rest at B .
\item The second time, when P is projected with a speed of $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, it first comes to rest at C .
\end{itemize}
The situation is modelled by assuming that during the motion of P the magnitude of the frictional force exerted by the hoop on P is constant.
\begin{enumerate}[label=(\alph*)]
\item Determine the value of $v$.
\item Comment on the validity of the modelling assumption used in this question.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2022 Q3 [6]}}