**Part (a)**

R ↑: $T_A \cos 60° = T_B \cos 60° + mg$

$T_A = T_B + 2mg$

NL2 along radius: $T_A \cos 30° + T_B \cos 30° = ma\cos 30°\omega^2$

$T_A + T_B = ma\omega^2$

$T_A = \frac{1}{2}ma\omega^2 + 2mg$

$T_B = \frac{1}{2}ma\omega^2 - 2mg$

$T_A = \frac{1}{2}ma\omega^2 + 2mg < 3mg$

$\omega^2 < \frac{4g}{a}$

$T_B = \frac{1}{2}ma\omega^2 - 2mg > 0$

$\omega^2 > \frac{2g}{a}$

$S = \frac{2\pi a}{\omega} \Rightarrow \pi\sqrt{\frac{a}{g}} < S < \pi\sqrt{\frac{2a}{g}}$

$k = 2$

| M1A1 |
| M1A1A1 |
| dM1A1 |
| A1 |
| M1 |
| M1 |
| dM1A1cso |
| (12) |

**Note:** Solutions using $\omega = \frac{2\pi}{S}$ or $\omega = \frac{2\pi}{T}$

R ↑: $T_A \cos 60° = T_B \cos 60° + mg$ | M1A1 |
$T_A = T_B + 2mg$

$T_A + T_B = ma\left(\frac{2\pi}{S}\right)^2$ | M1A1A1 |

$T_A = \frac{1}{2}ma\left(\frac{2\pi}{S}\right)^2 + 2mg$ | dM1A1 |

$T_B = \frac{1}{2}ma\left(\frac{2\pi}{S}\right)^2 - 2mg$ | A1 |

$T_A = \frac{1}{2}ma\left(\frac{2\pi}{S}\right)^2 + 2mg < 3mg$ | M1 |

$S^2 > \frac{\pi^2 a}{g}$

$T_B = \frac{1}{2}ma\left(\frac{2\pi}{S}\right)^2 - 2mg > 0$ | M1 |

$S^2 < \frac{2\pi^2 a}{g}$

$\Rightarrow \pi\sqrt{\frac{a}{g}} < S < \pi\sqrt{\frac{2a}{g}}$

$k = 2$ | dM1A1cso |
| (12) |

**Guidance:** The final M mark is for using $S = \frac{2\pi}{\omega}$ and must only be awarded when both inequalities have been used to obtain the final result.

**Solutions using $T_A = 3mg$ and $T_B = 0$:**

If 2 cases are considered, (i) with $T_A = 3mg$ and (ii) with $T_B = 0$, first 8 marks are available but no more.

If equations are formed including $T_A = 3mg$ and $T_B = 0$ in the same equation, there may be marks gained before the substitution is made but once the substitution is made there are no further marks available.

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