## Question 3(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| $V = mga\sin\theta\ (+c)$ | B1 | GPE (allow any level for GPE $= 0$) |
| $+ \frac{\lambda}{2(5a)}\left(2a - 2a\sin\theta\right)^2$ | M1 | EPE using $\frac{\lambda x^2}{2a}$ with $5a$ for the natural length and genuine attempt at the extension |
| | A1 | Correct (AEF) e.g. $mga\sin\theta + \frac{2\lambda a}{5}(1 - 2\sin\theta + \sin^2\theta)$ |
| Hence $\frac{dV}{d\theta} = mga\cos\theta$ | M1 | Differentiate their $V$ (at least one of their terms correctly differentiated — $V$ must be GPE + EPE) |
| $+ \frac{\lambda}{10a}\times 2\left(2a - 2a\sin\theta\right)\times -2a\cos\theta$ | A1 | AEF e.g. $mga\cos\theta + \frac{2\lambda a}{5}(-4\cos\theta + 2\sin\theta\cos\theta)$ |
| $= a\cos\theta\left(mg - \frac{4\lambda}{5}(1 - \sin\theta)\right)$ | E1 [6] | AG |

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## Question 3(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\theta = \frac{1}{2}\pi \Rightarrow \frac{dV}{d\theta} = 0$ | M1 | Or $V' = 0 \therefore a\cos\theta = 0 \Rightarrow \theta = \pi/2$ |
| Hence equilibrium | E1 | |
| $\frac{d^2V}{d\theta^2} = -a\sin\theta\left(mg - \frac{4\lambda}{5}(1-\sin\theta)\right)$ | M1 | Differentiate |
| $+ a\cos\theta\left(\frac{4\lambda}{5}\cos\theta\right)$ | A1 | Correct |
| When $\theta = \frac{1}{2}\pi$: $\frac{d^2V}{d\theta^2} = -a(mg) + 0 < 0$ | M1 | Substitute $\theta = \frac{1}{2}\pi$ into their $\frac{d^2V}{d\theta^2}$ |
| Therefore unstable | E1 [6] | Must establish $< 0$ |

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## Question 3(iii):

| Answer | Mark | Guidance |
|--------|------|----------|
| Taking moments about A | M1 | Or use $T = \frac{\lambda x}{l_0}$ with their $x$ and $5mg = 4\lambda(1-\sin\theta)$ |
| $mga\cos\theta = 2aT\cos\theta \Rightarrow T = \frac{1}{2}mg$ | A1 | |
| Resolving vertically $R = \frac{1}{2}mg$ | A1ft [3] | For ft must be in terms of $m$ and $g$ only |

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## Question 3(iv):

| Answer | Mark | Guidance |
|--------|------|----------|
| $mg - \frac{4\lambda}{5}(1-\sin\theta) = 0$ | B1 | |
| $\Rightarrow \sin\theta = \frac{4\lambda - 5mg}{4\lambda}$ | B1 | AEF — making $\sin\theta$ the subject or a form that allows consideration of values of $\sin\theta$ |
| Since $0 < \theta < \pi$, $\sin\theta > 0$ | M1 | Consideration of $\sin\theta$ in the interval $0 < \theta < \pi$ (maybe implicit) |
| $\Rightarrow \lambda > \frac{5mg}{4}$ | E1 [4] | Must have considered why $\sin\theta > 0$ or relative position of B with respect to A |

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## Question 3(v):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{d^2V}{d\theta^2} = a\left(\frac{4\lambda}{5}(1 + \sin\theta - 2\sin^2\theta) - mg\sin\theta\right)$ | M1* | Consider second differential of $V$ and substitute $\sin\theta = 1 - \frac{5mg}{4\lambda}$ |
| $= a\left(\frac{4\lambda}{5}\left(1 + \frac{4\lambda-5mg}{4\lambda} - 2\left(\frac{4\lambda-5mg}{4\lambda}\right)^2\right) - mg\left(\frac{4\lambda-5mg}{4\lambda}\right)\right)$ | A1 | …in terms of $\lambda$, $m$ only; CAO in terms of $\lambda$, $m$ only (AEF) |
| $= \frac{mga}{4\lambda}(8\lambda - 5mg) > 0$ | M1 dep* | Consider sign using result from (iv) and their $\frac{d^2V}{d\theta^2}$ in terms of $\lambda$, $m$ only |
| because $\lambda > \frac{5mg}{4}$, therefore stable | E1 [4] | |

**OR:**

| Answer | Mark | Guidance |
|--------|------|----------|
| Substituting $mg - \frac{4\lambda}{5}(1-\sin\theta)$ in $V''$ and obtaining $V''$ in terms of $\cos^2\theta$ only | M1* | |
| $\frac{d^2V}{d\theta^2} = \frac{4\lambda}{5}a\cos^2\theta$ | A1 | CAO |
| $\frac{d^2V}{d\theta^2} > 0$ in the interval $0 < \theta < \pi\ldots$ | M1 dep* | Considering sign of their $\frac{d^2V}{d\theta^2} = \frac{4\lambda}{5}a\cos^2\theta$ for $0 < \theta < \pi$ |
| …except at $\theta = \frac{\pi}{2}$ but this is the other equilibrium position. | E1 [4] | Stable — with reference to $\theta \neq \frac{\pi}{2}$ |