| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Minor (Further Mechanics Minor) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Vertical circle – surface contact (sphere/track, leaving surface) |
| Difficulty | Standard +0.3 This is a connected particles problem on a curved surface requiring equilibrium analysis (resolving forces tangentially), energy conservation with two particles, and stating modeling assumptions. While it involves multiple steps and careful geometry, the techniques are standard for Further Mechanics: resolving in appropriate directions, applying conservation of energy to a system, and recognizing standard assumptions. The question is methodical rather than requiring novel insight, making it slightly easier than average for Further Maths mechanics. |
| Spec | 3.03k Connected particles: pulleys and equilibrium3.03l Newton's third law: extend to situations requiring force resolution6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (a) | Let the tension in the string be T N. |
| T = 5 g | B1 | 1.1 |
| T 1 0 g s i n = | M1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| s i n 12 3 0 = = | A1 | 2.2a |
| Answer | Marks |
|---|---|
| (b) | 110v2 +15v2 |
| Answer | Marks |
|---|---|
| 360 | M1 |
| Answer | Marks |
|---|---|
| B1 | 3.4 |
| Answer | Marks |
|---|---|
| 1.1 | Attempt at conservation of energy; |
| Answer | Marks | Guidance |
|---|---|---|
| v = 0 .8 0 5 (m s-1) | A1 | 2.2b |
| Answer | Marks |
|---|---|
| (c) | e.g. we have assumed that in the subsequent motion that Q |
| Answer | Marks | Guidance |
|---|---|---|
| AP as P is small and or the pulley size is negligible | B1 | 2.4 |
Question 4:
4 | (a) | Let the tension in the string be T N.
T = 5 g | B1 | 1.1
T 1 0 g s i n = | M1 | 1.1 | Attempt to resolve tangentially.
Condone cos for sin
s i n 12 3 0 = = | A1 | 2.2a
[3]
(b) | 110v2 +15v2
2 2
10
=10g(2cos35−2cos45)−5g 22
360 | M1
B1
B1 | 3.4
3.1a
1.1 | Attempt at conservation of energy;
two PE terms present and at least
one KE term.
Correct PE term for Q
Correct PE term for P
v = 0 .8 0 5 (m s-1) | A1 | 2.2b | 0.8047065…
[4]
(c) | e.g. we have assumed that in the subsequent motion that Q
does not reach the pulley before P arrives at the point
where 45
or the length of string from AP is equal to the arc length
AP as P is small and or the pulley size is negligible | B1 | 2.4
[1]4 The diagram shows two particles P and Q , of masses 10 kg and 5 kg respectively, which are attached to the ends of a light inextensible string. The string is taut and passes over a small smooth pulley. The pulley is fixed at the highest point A on a smooth curved surface, the vertical cross-section of which is a quadrant of a circle with centre O and radius 2 m . Particle Q hangs vertically below the pulley and P is in contact with the surface, where the angle AOP is equal to $\theta ^ { \circ }$. The pulley, P and Q all lie in the same vertical plane.\\
\includegraphics[max width=\textwidth, alt={}, center]{cad8805d-59f6-4ed2-81f4-9e8c749461f5-4_499_492_559_251}
Throughout this question you may assume that there are no resistances to the motion of either P or Q and the force acting on P due to the tension in the string is tangential to the curved surface at P .
\begin{enumerate}[label=(\alph*)]
\item Given that P is in equilibrium at the point where $\theta = \alpha$, determine the value of $\alpha$.
Particle P is now released from rest at the point on the surface where $\theta = 35$, and starts to move downwards on the surface. In the subsequent motion it is given that P does not leave the surface.
\item By considering energy, determine the speed of P at the instant when $\theta = 45$.
\item State one modelling assumption you have made in determining the answer to part (b).
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2023 Q4 [8]}}