| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Vertical circle – string/rod (tension and energy) |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring energy conservation, circular motion dynamics, and careful geometric analysis across multiple stages of motion. It involves non-trivial setup (string over peg), deriving a given result, then using tension ratios to find initial speed and subsequent motion. The multi-part nature with connected results and the need to track geometry changes makes this substantially harder than typical A-level mechanics questions. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
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One end of a light inextensible string of length $\frac { 3 } { 2 } a$ is attached to a fixed point $O$ on a horizontal surface. The other end of the string is attached to a particle $P$ of mass $m$. The string passes over a small fixed smooth peg $A$ which is at a distance $a$ vertically above $O$. The system is in equilibrium with $P$ hanging vertically below $A$ and the string taut. The particle is projected horizontally with speed $u$ (see diagram). When $P$ is at the same horizontal level as $A$, the tension in the string is $T$. Show that $T = \frac { 2 m } { a } \left( u ^ { 2 } - a g \right)$.
The ratio of the tensions in the string immediately before, and immediately after, the string loses contact with the peg is $5 : 1$.\\
(i) Show that $u ^ { 2 } = 5 a g$.\\
(ii) Find, in terms of $m$ and $g$, the tension in the string when $P$ is next at the same horizontal level as $A$.
\hfill \mbox{\textit{CAIE FP2 2015 Q10 EITHER}}