| One correct triangle | B1 | |
| Two triangles combined using their common w. (seen or implied) | B1 | |
| $12\cos 30° = x\cos 45°$ | M1 | Horizontal components equal |
| $x = 12\sqrt{\frac{3}{2}} (= 6\sqrt{6} = 14.6......)$ | A1 | |
| $w^2 = 12^2 + x^2 - 2 \times 12 \times \cos 45°$ | M1 | Cosine rule |
| $w = 10.5 \text{ (km h}^{-1})$ | A1 | |
| EITHER: Sine Rule: $\frac{\sin\theta}{x} = \frac{\sin 45°}{w}$ or $\frac{\sin\theta}{\sin 60°} = \frac{12}{w}$ | M1 | |
| $\theta = 81.2°$ | A1 | |
| Direction 261° | A1 | |
| OR: $\uparrow w\cos\theta = 12 - 12\cos 30°$ $\leftrightarrow w\sin\theta = 12\cos 30°$ | | |
| $\Rightarrow \tan\theta = \frac{\cos 30°}{1 - \cos 30°}$ | (M1) | |
| $\theta = 81.2°$ | (A1) | |
| Direction 261° | (A1) | |
| | | (9) |

## Question 5alt

| $\mathbf{w} = \begin{pmatrix} x\cos 45° \\ 12 - x\cos 45° \end{pmatrix}$ | B1 | w expressed as a vector |
| $\mathbf{w} = \begin{pmatrix} 12\cos 30° \\ -12\sin 30° + y \end{pmatrix}$ | B1 | Second expression of w as a vector |
| $12\cos 30° = x\cos 45°$ | M1 | Horizontal components equal |
| $x\cos 45° = 6\sqrt{3}$ | A1 | Or $x = 6\sqrt{6}$ |
| $\|\mathbf{w}\|^2 = 3 \times 36 + (12 - 6\sqrt{3})^2$ | M1 | Use of Pythagoras |
| $\|\mathbf{w}\| = 10.5 \text{ (km h}^{-1})$ | A1 | |
| $\tan\theta = \frac{6\sqrt{3}}{12 - 6\sqrt{3}}$ | M1 | Correct method for direction of w |
| $\theta = 81.2°$ | A1 | |
| Direction 261° | A1 | |
| | | (9) [9] |

## Question 6a

| $\frac{7e}{0.8} = 0.2g$ | B1 | $T = mg$ at equilibrium ($e = 0.224$) |
| $0.2\ddot{x} = 0.2g - \frac{7(x+e)}{0.8} - 2v$ | M1 | Equation of motion. No missing/additional terms. Condone sign error(s). Allow with m and l |
| | A1 | At most one error. m and l substituted |
| $\left(0.2\ddot{x} = -\frac{7x}{0.8} - 2\dot{x}\right)$ | A1 | Correct unsimplified equation. m and l substituted |
| $\ddot{x} + 10\dot{x} + 43.75x = 0$ | A1 | *given answer* |
| | | (5) |

## Question 6b

| AE: $m^2 + 10m + 43.75 = 0$, $m = -5 \pm \frac{5\sqrt{3}}{2}$ | M1 | Form and solve AE |
| $x = e^{-5t}\left(A\cos\frac{5\sqrt{3}}{2}t + B\sin\frac{5\sqrt{3}}{2}t\right)$ | A1 | |
| $t = 0, x = 0.2 \Rightarrow A = 0.2$ | B1 | |
| $\dot{x} = -5e^{-5t}(A\cos\omega t + B\sin\omega t) + e^{-5t}(-A\omega\sin\omega t + B\omega\cos\omega t)$ | M1 | |
| $0 = -5A + B\omega = -1 + B\omega$, $B = \frac{2}{5\sqrt{3}}$ | M1 | Use of $t = 0, \dot{x} = 0$ to find B |
| $x = \frac{e^{-5t}}{5}\left(\cos\frac{5\sqrt{3}}{2}t + \frac{2}{\sqrt{3}}\sin\frac{5\sqrt{3}}{2}t\right)$ | A1 | |
| | | (6) |

## Question 6c

| instantaneous rest when $t = \frac{\pi}{\omega}$ (s) | M1 | Use of periodic time $= \frac{2\pi}{\omega}$ |
| $= \frac{2\pi}{5\sqrt{3}} = 0.73$ | A1 | |
| | | (2) |

## Question 6c alt

| $\dot{x} = -5e^{-5t}\left(\frac{1}{5}\cos\omega t + \frac{2}{5\sqrt{3}}\sin\omega t\right) + e^{-5t}\left(-\frac{1}{5}\omega\sin\omega t + \frac{2\omega}{5\sqrt{3}}\cos\omega t\right)$ | | |
| $= 3e^{-5t}\sin\frac{5\sqrt{3}}{2}t = 0$ | M1 | Solve $\dot{x} = 0$ |
| $\Rightarrow \frac{5\sqrt{3}}{2}t = \pi$, $t = \frac{2\pi}{5\sqrt{3}} = 0.73\text{s}$ | A1 | |
| | | (2) [13] |

## Question 7a

| PE of mass at C $= -3mg \times 4a\cos\theta$ | B1 | |
| PE of rods $= -2mg \times a\cos\theta - 2mg \times 3a\cos\theta$ | B1 | |
| Extension in the spring $= 4a\cos\theta - 2a$ | B1 | |
| $\frac{7mg(4a\cos\theta - 2a)^2}{4a} - 20mga\cos\theta$ | M1 | Total PE |
| $= 7mga(4\cos^2\theta - 4\cos\theta + 1) - 20mga\cos\theta$ | | |
| $= 28mga\cos^2\theta - 48mga\cos\theta + \text{constant}$ | A1 | *given answer* |
| | | (5) |

## Question 7b

| $\frac{dV}{d\theta} = -56mga\cos\theta\sin\theta + 48mga\sin\theta$ | M1 | Differentiate |
| $8\sin\theta(-7\cos\theta + 6) = 0$ | M1 | $\frac{dV}{d\theta} = 0$ and solve for θ |
| $\Rightarrow \theta = \cos^{-1}\frac{6}{7}(31°, 0.54r)$ | A1 | |
| | | (4) |

## Question 7c

| $\frac{d^2V}{d\theta^2} = -56mga(\cos^2\theta - \sin^2\theta) + 48mga\cos\theta$ | M1 | Differentiate to obtain second derivative |
| $= -56mga\left(2 \times \frac{36}{49} - 1\right) + 48mga \times \frac{6}{7}$ | DM1 | Find value of second derivative when $\theta = \cos^{-1}\frac{6}{7}$. Dependent on preceding M1 |
| $= \frac{728}{49}mga > 0$, stable | A1 | (14.8...mga) |
| | | (4) [13] |