**Question 10(i)(a)**

- $-mg\sin\theta = mr\ddot{\theta}$ cwo **M1** NII tangentially. Ignore radial

- $\ddot{\theta} = -\frac{g}{r}\sin\theta \approx -\frac{g}{r}\theta$ so approx. SHM **A2** **A1** for $\ddot{\theta}$ correct including signs; **A1** for $\sin\theta \approx \theta$ and stating SHM

- Period $2\pi \div \omega = 2.35(095)$ (seconds) **B1** awrt 2.35 or $\frac{1}{5}\pi\sqrt{14}$

**Alternative 1 (Hor & Vert):**

- $-T\sin\theta = ma\cos\theta$ and **M1** NII ($\uparrow$) and ($\leftrightarrow$)

- $T\cos\theta - mg\sin\theta = ma\sin\theta$ cwo **A1** both fully correct

- $\ddot{\theta} = -\frac{g}{r}\sin\theta \approx -\frac{g}{r}\theta$ so approx. SHM **A1** eliminating $T$ and using $\sin\theta \approx \theta$ and $a = r\ddot{\theta}$ and stating SHM

- Period $2\pi \div \omega = 2.35(095)$ (seconds) **B1** awrt 2.35 or $\frac{1}{5}\pi\sqrt{14}$

**Alternative 2 ($x$):**

- $-mg\sin\theta = m\ddot{x}$ cwo **M1** NII tangentially. Ignore radial

- $\ddot{x} = -g\sin\theta \approx -g\theta = -\frac{g}{r}x$ so approx. SHM **A2** **A1** for $\ddot{x}$ correct including signs; **A1** for using $\sin\theta \approx \theta$ and $x = r\theta$ (or just $x \approx r\sin\theta$) and stating SHM

- Period $2\pi \div \omega = 2.35(095)$ (seconds) **B1** awrt 2.35 or $\frac{1}{5}\pi\sqrt{14}$

**Alternative 3 (energy):**

- $\frac{1}{2}mv^2 + mgr - mgr\cos\theta = \text{const}$ **M1** adding KE and PE (condone sign errors)

- $v = r\dot{\theta}$ (or this and $\dot{v} = r\ddot{\theta}$) used to derive DE **M1** eliminating $v$ (and/or $\dot{v}$)

- $mr^2\dot{\theta}\ddot{\theta} + mgr\sin\theta\dot{\theta} = 0 \Rightarrow \ddot{\theta} + \frac{g}{r}\theta \approx 0$ so approx. SHM **A1** differentiating implicitly, using $\sin\theta \approx \theta$ and rearranging and stating SHM

- Period $2\pi \div \omega = 2.35(095)$ (seconds) **B1** awrt 2.35 or $\frac{1}{5}\pi\sqrt{14}$

**Question 10(i)(b)**

- $\dot{\theta}^2 = \omega^2(\theta_0^2 - \theta^2)$ **M1** SHM energy equation in terms of $\theta$ and $\dot{\theta}$ used

- $= \frac{10}{1.4}(0.3^2 - 0.2^2)$ **A1** FT for correct equation with their $\omega$

- $v = r\dot{\theta} = 1.4\times0.5976...$ **M1** using $v = r\dot{\theta}$

- $v = 0.836(66...)$ or $\sqrt{70}/10$ or awrt 0.837 **A1** converting to $v$

**Alternative 1 ($x$):**

- $v^2 = \omega^2(a^2 - x^2)$ **M1** SHM energy equation in terms of $v$ and $x$ used

- $x = 1.4\theta$ soi **M1** either $x = 0.28$ or $a = 0.42$

- $v^2 = \frac{10}{1.4}(0.42^2 - 0.28^2)$ **A1** FT for correct equation with their $\omega$

- $v = 0.836(66...)$ or $\sqrt{70}/10$ or awrt 0.837 **A1**

**Alternative 2 ($\theta$ solution):**

- Solution to SHM equation is $\theta = A\cos\omega t$ ($+B\sin\omega t$) so initial conditions $\Rightarrow \theta = 0.3\cos\omega t$ **M1** particular solution, their $\omega$

- Substituting $\omega t = \cos^{-1}\frac{2}{3}$ into $\dot{\theta} = \pm0.3\omega\sin\omega t$ **M1** Award if $\sin^{-1}$ into $\pm0.3\omega\cos\omega t$

- $v = r\dot{\theta} = 1.4\times0.5976...$ **M1** using $v = r\dot{\theta}$

- $v = 0.836(66...)$ or $\sqrt{70}/10$ or awrt 0.837 **A1** converting to $v$ (must be +ve)

**Alternative 3 ($x$ solution):**

- Gen sol to SHM eqn is $x = A\cos\omega t$ ($+B\sin\omega t$) **M1**

- so initial conditions $\Rightarrow x = 0.42\cos\omega t$ **A1** using $t=0 \Rightarrow$ both $v=0$ and $x = 1.4\times0.3$

- Substituting $\omega t = \cos^{-1}\frac{2}{3}$ into $\dot{x} = \pm0.42\omega\sin\omega t$ **M1** Award if $\sin^{-1}$ in $\pm0.42\omega\cos\omega t$

- $v = 0.836(66...)$ or $\sqrt{70}/10$ or awrt 0.837 **A1** must be +ve

**Question 10(ii)**

- KE gained = PE lost **M1**

- $\frac{1}{2}mv^2 = mg(1.4\cos0.2 - 1.4\cos0.3)$ **A1** correct equation

- $[v^2 = 0.692]$ $v = 0.832(13)$ **A1** awrt 0.832

- % difference $0.544\%$ **A1** awrt 0.544 (could be from correct calculations of $\dot{\theta}$)