# Question 6(i):

e.g. $R$ has no weight/mass; There is no friction; There is no vertical force at $R$; String has minimum length/elastic energy | B1 | For any one contributory reason **Total: 1**

## Question 6(ii):

$V = \frac{1}{2}\left(\frac{mg}{a}\right)(2a\sin\theta)^2 + mga\cos\theta$ | B2 | Give B1 for correct EE or PE
$= mga(2\sin^2\theta + \cos\theta)$ |  |
$\frac{dV}{d\theta} = mga(4\sin\theta\cos\theta - \sin\theta)$ | B1 ft, B1 ft | Diffn of $\sin^2\theta$ (or $\cos^2\theta$); Diffn of $\cos\theta$ (or $\sin\theta$)
For equilibrium, $mga\sin\theta(4\cos\theta - 1) = 0$ | M1 |
$\theta = 1.32$ (or $75.5°$) | A1 | Allow $\cos^{-1}\frac{1}{4}$, $76°$; Ignore $\theta=0$ if stated **Total: 6**

SR If done by taking moments, M1A1 for $\frac{mg(2a\sin\theta)}{a}(2a\cos\theta) = mg(a\sin\theta)$; A1 for $\theta = 1.32$ (Max 3 out of 6)

## Question 6(iii):

$\frac{d^2V}{d\theta^2} = mga\cos\theta(4\cos\theta - 1) - 4mga\sin^2\theta$ | B2 ft | Give B1 if just one error; Only B1 ft if work is simpler
When $\cos\theta = \frac{1}{4}$:
$\frac{d^2V}{d\theta^2} = -4mga\sin^2\theta \quad (= -\frac{15}{4}mga) < 0$ | M1 |
Equilibrium is unstable | A1 | Fully correct working only **Total: 4**

OR When $\theta < 1.32$, $\frac{dV}{d\theta} > 0$ | B1 ft |
When $\theta > 1.32$, $\frac{dV}{d\theta} < 0$ | B1 ft |
$V$ has a maximum; Equilibrium is unstable | M1, A1 | Fully correct working, and convincing demonstration of signs above

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