**(a)**

$\frac{1}{2}mU^2 - \frac{1}{2}mv^2 = 2mga$ | M1A1 |

$T + mg = m\frac{v^2}{a}$ | M1A1 |

$T = \frac{mU^2 - 4mga}{a} - mg$ | DM1 |

$T = \frac{mU^2 - 5mga}{a}$ | A1 |

$T \geq 0 \Rightarrow U^2 \geq 5ga$ | DM1 |

$U \geq \sqrt{5ag}$ * | A1 | (8)

**(b)**

At top: $T = \frac{9mga - 5mga}{a} = 4mg$ | M1(either tension)A1 |

At bottom: $T' - mg = \frac{mU^2}{a}$ | A1 |

$kT = mg + \frac{9mag}{a} = 10mg$ | DM1 |

$k = \frac{10mg}{4mg} = \frac{5}{2}$ | A1 | (5) [13] |

**Notes for Question 6(a):**

M1 | for an energy equation, from the bottom to the top. A difference of KE terms and a PE term needed. From bottom to a general point gets M0 until a value for $\theta$ at the top is used. $v^2 = u^2 + 2as$ scores M0 |

A1 | for all terms correct (inc signs) |

M1 | for NL2 along the radius at the top. Two forces and mass $\times$ acceleration needed. Accel can be in either form here. But see NB at end of (a) |

A1 | for a fully correct equation. Acceleration should be $\frac{v^2}{a}$ now. |

M1 dep | for eliminating $v$ (vel at top) between the two equations. Dependent on both previous M marks. If $v$ is set $= 0$, award M0 |

A1 | for a correct expression for $T$ |

M1 dep | for using $T \geq 0$ to obtain an inequality for $U^2$ or $U$. Allow with $>$. Dependent on all previous M marks. |

A1 cso | for $U \geq \sqrt{5ag}$ * Watch square root! Give A0 if $>$ seen on previous line. |

**NB:** The second and fourth M marks (and their As if earned) can be given together if $mg \leq m\frac{v^2}{a}$ is seen |

**(b)**

M1 | for obtaining an expression for the tension at the top or at the bottom, no need to substitute for $U$ yet. |

A1 | Substitute for $U$ and obtain one correct tension (4mg at top or 10mg at bottom) |

A1 | for the other tension correct |

M1 dep | for using tension at bottom $= k \times$ tension at top and solving for $k$ |

A1 cso | for $k = \frac{5}{2}$ oe |

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