| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2020 |
| Session | Specimen |
| Marks | 2 |
| 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 kinematics. Part (a) involves resolving forces and using F=mrω², part (b) requires eliminating tension between vertical and horizontal equations, and part (c) is a straightforward 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 |
**Question 8(a):**
$T\sin\theta = ml\sin\omega^2 \Rightarrow T = ml\omega^2$ [M1A1]
**Total: 2 marks**
**Question 8(b):**
$T\dfrac{h}{l} = mg \Rightarrow T = \dfrac{mgl}{h}$ [B1]
$\Rightarrow \omega^2 h = g$ [B1]
**Total: 2 marks**
**Question 8(c):**
Time is $\dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{h}{g}}$ [M1A1]
**Total: 2 marks**8\\
\includegraphics[max width=\textwidth, alt={}, center]{f4acd242-eb78-4124-bfa2-fdecaa188690-4_614_741_1548_662}
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).
\begin{enumerate}[label=(\alph*)]
\item Find the tension in the string in terms of $m , l$ and $\omega$.
\item Show that $\omega ^ { 2 } h = g$.
\item Deduce an expression in terms of $g$ and $h$ for the time taken for $P$ to complete one full circle during its motion.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2020 Q8 [2]}}