| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Session | Specimen |
| Marks | 5 |
| Topic | Simple Harmonic Motion |
| Type | Small oscillations: simple pendulum (particle on string) |
| Difficulty | Challenging +1.2 This is a standard pendulum energy/SHM question requiring multiple techniques (energy conservation, circular motion, small angle approximation for SHM) across four parts. While it involves several steps and the SHM approximation requires some care, all techniques are routine for Further Maths students and the question provides significant scaffolding through its structure. It's moderately harder than average due to the multi-part nature and FM content, but doesn't require novel insight. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration |
**Question 6**
**(i)** Gain in Kinetic Energy = Loss in Potential Energy:
$$0.5m(1.5\dot{\theta})^2 = m \times 10 \times 1.5(\cos\theta - \cos 0.1)$$
$$\left(\frac{\mathrm{d}\theta}{\mathrm{d}t}\right)^2 = \frac{40}{3}\{\cos\theta - \cos(0.1)\}$$ (AG) M1A1, A1 **[3 marks]**
**(ii)(a)** $0.5mu^2 = m \times 10 \times 1.5(1 - \cos 0.1)$ M1
$\Rightarrow u^2 = 30(1-\cos 0.1) \Rightarrow u = 0.39$ A1 **[2 marks]**
**(b)** Taking tension in string as $T$ newtons and applying Newton II:
$T = 0.2 \times 10\cos 0.05 + 0.2 \times 1.5 \times \dfrac{40}{3}(\cos 0.05 - \cos 0.1)$ M1A1
$T = 2.0$ A1 **[3 marks]**
**(iii)** Differentiating result in (i) with respect to $t$:
$2\dot{\theta}\ddot{\theta} = -\dfrac{40}{3}\sin\theta\dot{\theta}$ M1A1
$\Rightarrow \ddot{\theta} \approx -\dfrac{20}{3}\theta$ when $\theta$ is small $\Rightarrow$ SHM A1 **[3 marks]**
**(iv)** Time is $\frac{1}{4}$ period $= \frac{1}{4} \times 2\pi \times \sqrt{\dfrac{3}{20}} = \frac{1}{2}\pi\sqrt{\dfrac{3}{20}}$ ($= 0.608$) seconds M1A1 **[2 marks]**6 A simple pendulum consists of a light inextensible string of length 1.5 m with a small bob of mass 0.2 kg at one end. When suspended from a fixed point and hanging at rest under gravity, the bob is given a horizontal speed of $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and it comes instantaneously to rest when the string makes an angle of 0.1 rad with the vertical. At time $t$ seconds after projection the string makes an angle $\theta$ with the vertical.\\
(i) Show that, neglecting air resistance,
$$\left( \frac { \mathrm { d } \theta } { \mathrm {~d} t } \right) ^ { 2 } = \frac { 40 } { 3 } \{ \cos \theta - \cos ( 0.1 ) \}$$
(ii) Find, correct to 2 significant figures,
\begin{enumerate}[label=(\alph*)]
\item the value of $u$,
\item the tension in the string when $\theta = 0.05 \mathrm { rad }$.\\
(iii) By differentiating the above equation for $\left( \frac { \mathrm { d } \theta } { \mathrm { d } t } \right) ^ { 2 }$, or otherwise, show that the motion of the bob can be modelled approximately by simple harmonic motion.\\
(iv) Hence find the value of $t$ at which the bob first comes instantaneously to rest.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 Q6 [5]}}