# Question 2:

## Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $V = mga\sin\theta$ with datum level through A | B1 | |
| $+ \frac{1}{2}\frac{mg}{a}(2a\cos\theta)^2$ | M1* | Genuine attempt at extension + substitution into $\frac{1}{2}kx^2$ |
| $V = mga(\sin\theta + 1 + \cos 2\theta) = mga(\sin\theta + 2\cos^2\theta)$ | A1 | |
| $\frac{dV}{d\theta} = mga\cos\theta - 4mga\cos\theta\sin\theta$ | M1 dep* | Differentiates their $V$ of the correct form |
| $= mga\cos\theta(1 - 4\sin\theta)$ | E1 | |
| **[5]** | | |

## Part (ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $mga\cos\theta(1 - 4\sin\theta) = 0$ | M1 | |
| $\cos\theta = 0 \Rightarrow \theta = \frac{1}{2}\pi, -\frac{1}{2}\pi$ | A1 | |
| $\sin\theta = \frac{1}{4} \Rightarrow \theta = \sin^{-1}\frac{1}{4}$ | A1 | For this root and rejection of $\pi - \arcsin(0.25)$; accept either no reason given for rejection or $\sin\theta = 1/4$ only and the interval for $\theta$ stated |
| $\frac{d^2V}{d\theta^2} = -mga\sin\theta - 4mga\cos 2\theta$ | M1 | May use $\frac{dV}{d\theta} = mga(\cos\theta - 2\sin 2\theta)$ giving $-mga(\sin\theta + 4\cos 2\theta)$ or $-mga(\sin\theta + 4 - 8\sin^2\theta)$ |
| When $\theta = \frac{1}{2}\pi$, $\frac{d^2V}{d\theta^2} = -mga + 4mga > 0$ so stable | A1 | For all A marks accept (as a minimum) correct $V''$ with corresponding sign of $V''$ + correct conclusion |
| When $\theta = -\frac{1}{2}\pi$, $\frac{d^2V}{d\theta^2} = mga + 4mga > 0$ so stable | A1 | |
| When $\theta = \sin^{-1}\frac{1}{4}$, $\frac{d^2V}{d\theta^2} = -\frac{1}{4}mga - \frac{7}{2}mga < 0$ so unstable | A1 | |
| **[7]** | | |

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