| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Topic | Circular Motion 2 |
| Type | Elastic string in circular motion |
| Difficulty | Challenging +1.2 This is a standard Further Maths mechanics problem requiring systematic application of circular motion principles, Hooke's law, and resolution of forces. While it involves multiple steps and algebraic manipulation across three parts, each step follows predictable methods without requiring novel insight. The guided structure ('show that') reduces problem-solving demand, making it moderately above average difficulty but well within typical Further Maths scope. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.05c Horizontal circles: conical pendulum, banked tracks |
**(i)** Horizontally: $T\sin\theta = 5(0.2 + x)\sin\theta\,\omega^2 \Rightarrow T = (1 + 5x)\omega^2$ (AG) — M1A1 [2]
**(ii)(a)** Vertically: $T\cos\theta = 5g$ — B1
Hooke's Law: $T = \frac{100x}{0.2} = 500x$ — M1A1
$\Rightarrow 500x\cos\theta = 5g \Rightarrow \cos\theta = \frac{1}{10x}$ (AG) — A1 [4]
**(b)** $x \to 0.1 \Rightarrow \cos\theta \to 1 \Rightarrow \theta \to 0$ — B1
$5(0.2 + 0.1)\omega^2 > 50 \times 10 \times 0.1 \Rightarrow \omega^2 > \frac{100}{3} \Rightarrow \omega > \frac{10\sqrt{3}}{3}$ (AG) — B1 [2]
**(iii)** $\omega^2 = 50 \Rightarrow 0.2 + x = 2x \Rightarrow x = 0.2$, $l = 0.4$ — B1
$\cos\theta = \frac{1}{2}$, $\sin\theta = \frac{\sqrt{3}}{2}$ — B1
$r = \frac{2}{5} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{5}$ (0.346) — B1 [3]
**[Total: 11]**4 One end of a light elastic string of natural length 0.2 m and modulus of elasticity 100 N is attached to a fixed point $A$. The other end is attached to a particle of mass 5 kg . The particle moves with angular speed $\omega$ radians per second in a horizontal circle with the centre vertically below $A$. The string makes an angle $\theta$ with the vertical.
\begin{enumerate}[label=(\roman*)]
\item By considering the horizontal component of the tension in the string, show that the tension in the string is $( 1 + 5 x ) \omega ^ { 2 } \mathrm {~N}$, where $x$ is the extension, in metres, of the string.
\item (a) By considering vertical forces and also Hooke's law, deduce that $\cos \theta = \frac { 1 } { 10 x }$.\\
(b) Show that $\omega > \frac { 10 \sqrt { 3 } } { 3 }$.
\item When the value of $\omega$ is $5 \sqrt { 2 }$, find the radius of the circular motion.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2010 Q4 [11]}}