| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: string becomes slack |
| Difficulty | Standard +0.8 This is a standard Further Maths vertical circle problem requiring energy conservation and circular motion dynamics. Students must find initial velocity from impulse, apply energy equations to find speed at angle θ, then determine tension using centripetal force. The slack condition (T=0) requires solving a transcendental equation. While multi-step, it follows a well-established method taught in FM1/Further Mechanics courses. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.03f Impulse-momentum: relation6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | 20 = 4u ⇒ u = 5 |
| Answer | Marks |
|---|---|
| sf) | B1 |
| Answer | Marks |
|---|---|
| [7] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | = 50 |
| Answer | Marks |
|---|---|
| necessary) | Assuming zero PE level at initial |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (b) | 4v2 |
| Answer | Marks |
|---|---|
| ∴θ=113.3° or 1.98 rads | M1 |
| Answer | Marks |
|---|---|
| [3] | 2.1 |
| Answer | Marks |
|---|---|
| 3.2a | NII for radial direction. T could |
| Answer | Marks |
|---|---|
| conservation of energy | v2 =9.32+15.68cosθ |
Question 6:
6 | (a) | 20 = 4u ⇒ u = 5
Initial energy = 1×4×52
2
Energy at θ= 1×4×v2+4g×0.8(1−cosθ)
2
2v2+15.68=50⇒v2 =17.16
v2 17.16
Radial: a = =
r
0.8 0.8
π
Tangential: ma =−mgsin
t
3
2 429)2
3g
a= − + =23.067... so the
2 20
magnitude of the acceleration is 23.1ms–2 (3
sf) | B1
B1
M1
A1
M1
M1
A1
[7] | 1.1
1.1
1.1
1.1
3.1b
3.1b
1.1 | = 50
Attempt to derive total ME at
general or specific angle
Equating energies to derive a
value for v2
Correct form for centripetal
acceleration and use of v2
NII for tangential direction with
weight resolved (– not
necessary) | Assuming zero PE level at initial
level of P
v = 4.142...
a =21.45
r
3g
a =− =−8.4870...
t
2
6 | (b) | 4v2
Radial: T −4gcosθ=
0.8
v2 =52−2g×0.8(1−cosθ)
−7.84cosθ=9.32+15.68cosθ
9.32
∴cosθ=−
23.52
∴θ=113.3° or 1.98 rads | M1
M1
A1
[3] | 2.1
2.1
3.2a | NII for radial direction. T could
be set to 0. Correct form of a .
r
v2 in terms of cosθ from
conservation of energy | v2 =9.32+15.68cosθ6 A particle $P$ of mass 4 kg is attached to one end of a light inextensible string of length 0.8 m . The other end of the string is attached to a fixed point $O . P$ is at rest vertically below $O$ when it experiences a horizontal impulse of magnitude 20 Ns . In the subsequent motion the angle the string makes with the downwards vertical through $O$ is denoted by $\theta$ (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{c6445493-9802-46ca-b7eb-7738a831d9ee-4_387_502_1434_255}
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the acceleration of $P$ at the first instant when $\theta = \frac { 1 } { 3 } \pi$ radians.
\item Determine the value of $\theta$ at which the string first becomes slack.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2021 Q6 [10]}}