| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Topic | Circular Motion 2 |
| Type | Vertical circle: tension at specific point |
| Difficulty | Standard +0.8 This is a Further Mechanics vertical circle problem requiring energy conservation to derive a velocity equation, then applying Newton's second law with circular motion dynamics to find when tension reaches a critical value. It involves multiple steps (energy equation, tension equation with centripetal force, solving for angle) and requires careful handling of the geometry and sign conventions, making it moderately challenging but still a standard FM question type. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods |
2 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 $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point 0.75 m vertically above $O$ (see Fig. 2). When $O A$ makes an angle $\theta$ with the upward vertical the speed of $A$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
$\xrightarrow [ A \text { a } ] { 6 \mathrm {~m} \mathrm {~s} ^ { - 1 } }$
Fig. 2
\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 2021 Q2 [7]}}