| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Minor (Further Mechanics Minor) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Vertical circle – string/rod (tension and energy) |
| Difficulty | Standard +0.3 This is a straightforward vertical circular motion question requiring standard energy methods. Part (i) is simple GPE calculation using trigonometry, part (ii) is basic conservation of energy, part (iii) is conceptual understanding, and part (iv) adds work done against resistance—all routine applications with clear structure and no novel problem-solving required. Slightly easier than average due to explicit guidance and standard techniques. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (i) | 60(cid:117)9.8(cid:117)9(cid:11)1(cid:16)cos45(cid:12) |
| Answer | Marks |
|---|---|
| so change in GPE is 1550 J (3 s.f.) | M1 |
| Answer | Marks |
|---|---|
| [3] | 1.1a |
| Answer | Marks |
|---|---|
| 1.1 | GPE change is mgh |
| Answer | Marks |
|---|---|
| N | GPE is gravitational |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (ii) | 1 (cid:167) 2(cid:183) |
| Answer | Marks |
|---|---|
| so v = 7.1879… so speed is 7.19ms -1 (3 s.f.) | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | E |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (iii) | Smaller, because work is done against the resistance, so the |
| final kinetic energy is lower. | I |
| Answer | Marks | Guidance |
|---|---|---|
| [1] | 3.5b | |
| 5 | (iv) | C |
| Answer | Marks |
|---|---|
| v = 7.0980… so speed is 7.10ms -1 (3 s.f.) | M1 |
| Answer | Marks |
|---|---|
| [5] | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | W-E equation with KE, GPE and WD |
| Answer | Marks |
|---|---|
| Correct, accept only sign errors | W-E is work-energy |
Question 5:
5 | (i) | 60(cid:117)9.8(cid:117)9(cid:11)1(cid:16)cos45(cid:12)
J
(cid:167) 2(cid:183)
= 5292(cid:117)(cid:168)1(cid:16) (cid:184) = 1549.99 …
(cid:168) (cid:184)
2
(cid:169) (cid:185)
so change in GPE is 1550 J (3 s.f.) | M1
A1
A1
[3] | 1.1a
1.1
1.1 | GPE change is mgh
Finding h
N | GPE is gravitational
potential energy
5 | (ii) | 1 (cid:167) 2(cid:183)
(cid:117)60(cid:117)v2 (cid:32) 5292(cid:117)(cid:168)1(cid:16) (cid:184)
2 (cid:168) 2 (cid:184)
(cid:169) (cid:185)
so v = 7.1879… so speed is 7.19ms -1 (3 s.f.) | M1
A1
[2] | 3.3
M
1.1 | E
Conservation of energy
5 | (iii) | Smaller, because work is done against the resistance, so the
final kinetic energy is lower. | I
E1
[1] | 3.5b
5 | (iv) | C
1 1
(cid:117)60(cid:117)v2 (cid:16) (cid:117)60(cid:117)1.52 E
2 2
P
(cid:167) 2(cid:183) 1
= 5292(cid:117)(cid:168)1(cid:16) (cid:184)(cid:16)15(cid:117) (cid:117)2(cid:117)(cid:83)(cid:117)9
(cid:168) (cid:184)
2 S8
(cid:169) (cid:185)
v = 7.0980… so speed is 7.10ms -1 (3 s.f.) | M1
A1
B1
A1
A1
[5] | 3.3
1.1
3.4
1.1
1.1 | W-E equation with KE, GPE and WD
KE terms correct
WD against resistance
Correct, accept only sign errors | W-E is work-energy
KE is Kinetic Energy
WD is work done5 A young man of mass 60 kg swings on a trapeze. A simple model of this situation is as follows.
The trapeze is a light seat suspended from a fixed point by a light inextensible rope. The man's centre of mass, G , moves on an arc of a circle of radius 9 m with centre O , as shown in Fig. 5. The point C is 9 m vertically below O . B is a point on the arc where angle COB is $45 ^ { \circ }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{54711a46-83ce-4fb9-b6d3-53b264725c74-4_383_371_552_852}
\captionsetup{labelformat=empty}
\caption{Fig. 5}
\end{center}
\end{figure}
(i) Calculate the gravitational potential energy lost by the man if he swings from B to C .
In this model it is also assumed that there is no resistance to the man's motion and he starts at rest from B.\\
(ii) Using an energy method, find the man's speed at C .
A new model is proposed which also takes into account resistance to the man's motion.\\
(iii) State whether you would expect any such model to give a larger, smaller or the same value for the man's speed at C . Give a reason for your answer.
A particular model takes account of the resistance by assuming that there is a force of constant magnitude 15 N always acting in the direction opposing the man's motion. This new model also takes account of the man 'pushing off' along the arc from B to C with a speed of $1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(iv) Using an energy method, find the man's speed at C .
\hfill \mbox{\textit{OCR MEI Further Mechanics Minor Q5 [11]}}