# Question 10(a):

**Finding KE after falling ka from PE:**

$mg \times ka - \frac{1}{2}(3mg/2)(ka-a)^2/a$ | M1=[A1 | Energy equation with EPE term

$= mga(k - 3k^2/4 + 3k/2 - \frac{3}{4})$

$= \frac{1}{4}mga(10k - 3 - 3k^2)$ **A.G.** | A1 | 3 marks total

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**Finding equation for k (or ka) at B using KE = 0:**

$3k^2 - 10k + 3 = 0$ (A.E.F.) | M1 |

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**Finding both roots and selecting root > 1:**

Roots 3 and $\frac{1}{3}$, so $k = 3$ **A.G.** | M1 A1 | 3 marks total

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**Finding time $t_1$ to fall $a$ from $A$ (under no tension):**

$a = \frac{1}{2}gt_1^2$, $t_1 = \sqrt{(2a/g)}$ | B1 |

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**Finding ext. $e$ or distance fallen at equilibrium point O:**

$mg = 3mge/2a$, $e = 2a/3$ or $AO = 5a/3$ | B1 |

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**State or find SHM equation at x below (or above) O:**

$m\frac{d^2x}{dt^2} = mg - 3mg(x+e)/2a$ | M1 |

$\frac{d^2x}{dt^2} = -3gx/2a$ | A1 |

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**State or use correct amplitude $x_0$ and $\omega^2$:**

$x_0 = 4a/3$ and $\omega^2 = 3g/2a$ | B1 |

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**Finding time $t_2$ to fall from $a$ to $3a$ below A:**

$[-]\ e = x_0\cos\omega t_2$ | M1 |

$t_2 = \sqrt{(2a/3g)}\cos^{-1}(-\frac{1}{2})$ (A.E.F.) | A1 |

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**Finding total time $t_1 + t_2$:**

$\sqrt{(2a/g)} + (2\pi/3)\sqrt{(2a/3g)}$ **A.G.** | A1 | 8 marks, 14 total

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# Question 10(b):

**Finding 4 summations required:**

$\Sigma x = 12 + p$, $\Sigma x^2 = 38 + p^2$

$\Sigma y = 23$, $[\Sigma y^2 = 125]$

$\Sigma xy = 63 + 2p$ | M1 |

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**EITHER: Substitute $b_1 = 1$ in formula for gradient:**

$63 + 2p - (12+p)\times 23/5$

$= 38 + p^2 - (12+p)^2/5$

or $39 - 13p = 4p^2 - 24p + 46$ | M1 A1 A1 | A1 for each side

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**OR: Substitute in normal equations:**

$5k + (12+p) = 23$ and $(12+p)k + (38+p^2) = 63 + 2p$ | (M1 A1 A1) | A1 for each equation

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**Obtain and solve quadratic for p (A.E.F):**

$0.8p^2 - 2.2p + 1.4 = 0$

or $4p^2 - 11p + 7 = 0$ | M1 A1 |

$p = 1,\ 1.75$ (or $7/4$) | A1, A1 | 8 marks total

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# Question 10(b)(i):

**Finding equation for correlation coefficient with $p = 1$:**

**EITHER:**

$r = (65 - 13\times 23/5)\ /\ \sqrt{\{(39 - 13^2/5)(125 - 23^2/5)\}}$ | M1 A1 |

$= 5.2\ /\ \sqrt{(5.2\times 19.2)}$ or $1.04\ /\ \sqrt{(1.04\times 3.84)}$

**OR:**

$r^2 = 1\times(65 - 13\times 23/5)/(125 - 23^2/5)$

or $1\times(39 - 13^2/5)/(125 - 23^2/5)$

$= 5.2/19.2$ or $1.04/3.84$ | M1 A1 |

**Evaluate r:**

$r = \sqrt{39/12}$ or $0.52$ | A1 | 3 marks total

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# Question 10(b)(ii):

**EITHER: Recall or find gradient $b_2$ of line:**

$b_2 = 5.2/19.2$ or $r^2\ [= 0.2708]$ | M1 |

**Find regression line of x on y:**

$x - 13/5 = b_2(y - 23/5)$

$x = (13/48)y + 65/48$

or $0.271y + 1.35$ | M1, A1 |

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**OR: Use normal equations for $x = a_2 + b_2 y$:**

$5a_2 + 23b_2 = 13$ and $23a_2 + 125b_2 = 65$ | (M1), (M1) |

**Solve for $a_2$, $b_2$:**

$b_2 = 13/48$ or $0.271$

And $a_2 = 65/48$ or $1.35$ | (A1) | 3 marks, 14 total