| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2020 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: string becomes slack |
| Difficulty | Challenging +1.2 This is a standard vertical circle problem requiring energy conservation and circular motion equations. Part (a) is routine application of energy methods, and part (b) requires setting tension equal to the breaking value using the standard circular motion formula. The multi-step nature and need to combine two key principles elevates it slightly above average, but it follows a well-established template for Further Mechanics questions. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (a) | 1 1 |
| Answer | Marks |
|---|---|
| v2 =50.7−14.7cosθ | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.3 |
| Answer | Marks |
|---|---|
| 2.1 | Attempt at GPE |
| Answer | Marks |
|---|---|
| must be shown | In this solution θ is the angle |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | (b) | v2 |
| Answer | Marks |
|---|---|
| θ = 97.5° or 1.70 rad | M1 |
| Answer | Marks |
|---|---|
| [4] | 3.3 |
| Answer | Marks |
|---|---|
| 1.1 | Using F = ma in radial |
| Answer | Marks |
|---|---|
| 1029 | Allow sin instead of cos |
Question 3:
3 | (a) | 1 1
×2.8×62+2.8×9.8(0.75−0.75cosθ)= ×2.8v2
2 2
50.4+20.58−20.58cosθ=1.4v2
v2 =50.7−14.7cosθ | M1
M1
A1
[3] | 3.3
1.1
2.1 | Attempt at GPE
0.75cos2.8gwith correct theta
and length
Conservation of energy, allow
sign errors
AG An intermediate step
must be shown | In this solution θ is the angle
between OA and the upwards
vertical through O
3 | (b) | v2
T +2.8×9.8cosθ=2.8×
0.75
50.7−14.7cosθ
200+2.8×9.8cosθ=2.8×
0.75
θ = 97.5° or 1.70 rad | M1
A1
M1
A1
[4] | 3.3
1.1
2.2a
1.1 | Using F = ma in radial
direction with tension,
resolved weight and an
expression for centripetal
acceleration
Using their expression for v2
and breaking tension of string
134
cosθ=− =−0.1802...
1029 | Allow sin instead of cos
0.75cos2.8gwith correct theta
and length3 One end of a light inextensible string of length 0.75 m is attached to a particle $A$ of mass 2.8 kg . The other end of the string is attached to a fixed point $O$. $A$ is projected horizontally with speed $6 \mathrm {~ms} ^ { - 1 }$ from a point 0.75 m vertically above $O$ (see Fig. 3). When $O A$ makes an angle $\theta$ with the upward vertical the speed of $A$ is $v \mathrm {~ms} ^ { - 1 }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{831ba5da-df19-43bb-b163-02bbddb4e2b8-2_388_220_1790_244}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Show that $v ^ { 2 } = 50.7 - 14.7 \cos \theta$.
\item Given that the string breaks when the tension in it reaches 200 N , find the angle that $O A$ turns through between the instant that $A$ is projected and the instant that the string breaks.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2020 Q3 [7]}}