| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2016 |
| Session | Specimen |
| Topic | Circular Motion 1 |
| Type | Period or time for one revolution |
| Difficulty | Standard +0.3 This is a standard conical pendulum problem requiring application of Newton's second law in circular motion and basic trigonometry. Part (i) involves resolving forces and using F=mrω², part (ii) requires eliminating tension between vertical and horizontal equations (straightforward algebra), and part (iii) is a direct application of T=2π/ω. All steps are routine for Further Maths students with no novel insight required, making it slightly easier than average. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
NOT FOUND
(Question 8 in the mark scheme concerns a rod resting against a wall, not a particle on a string.)8\\
\includegraphics[max width=\textwidth, alt={}, center]{c4bbba86-2968-4247-b300-357217cf213b-4_670_819_548_621}
A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $l$. The other end of the string is attached to a fixed point $A$. The particle moves with constant angular speed $\omega$ in a horizontal circle whose centre is at a distance $h$ vertically below $A$ (see diagram).\\
(i) Find the tension in the string in terms of $m , l$ and $\omega$.\\
(ii) Show that $\omega ^ { 2 } h = g$.\\
(iii) Deduce an expression in terms of $g$ and $h$ for the time taken for $P$ to complete one full circle during its motion.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2016 Q8}}