## 6a.
Energy: $\frac{1}{2}m \times 144 = \frac{1}{2}m \times 64 + mgh$ | M1 A2 |
$h = 4.1$ or $4.08$ (m) | A1 |
(4) | |
**N.B.** If they find an h, using a non-energy method, they score nothing in (a) BUT it can be used in (b), (c) and (d) without penalty. | |

## 6b.
Vertical distance: $h = 12\sin\alpha \times 1.5 - \frac{g}{2} \times 1.5^2$ (their $h$) | M1 A1 ft |
$(\sin\alpha = 0.839)$ | $\alpha = 57.1°$ ($57°$) (0.996 rads) | A1 |
(3) | |

**Alt:** Use $12\cos\alpha = 8\cos\beta$ and $-8\sin\beta = 12\sin\alpha - 1.5g$ and eliminate $\beta$ to give equation in ONE trig ratio for $\alpha$ | M1 A1 ft |
$(\sin\alpha = 0.839)$ | $\alpha = 57.1°$ ($57°$) (0.996 rads) | A1 |
(3) | |

## 6c.
Horizontal cpts: $12\cos\alpha = 8\cos\beta$ (with their $\alpha$ for the A mark) | M1 A1 ft |
$\beta = 35.4°$ ($35°$) (0.617 or 0.62 rads) | A1 |
(3) | |

**Alt:** $\tan\beta = \frac{12\sin\alpha - 1.5g}{12\cos\alpha}(= -0.709....)$ | M1 A1 ft |
$\beta = 35.4°$ ($35°$) (0.617 rads or 0.62 rads) | A1 |
(3) | |

**Alt:** $h = \frac{(12\sin\alpha - 8\sin\beta) \times 1.5}{2}$ | M1 A1 ft |
$\beta = 35.4°$ ($35°$) (0.617 rads or 0.62 rads) | A1 |
(3) | |

**Alt:** $-8\sin\beta = 12\sin\alpha - (9.8 \times 1.5)$ | M1 A1 ft |
$\beta = 35.4°$ ($35°$) (0.617 rads or 0.62 rads) | A1 |
(3) | |

## 6d.
Correct strategy e.g. $2(1.5 - T)$ where $T =$ time to max height | M1 |
Correct equations e.g. Time to max height: $T = \frac{12\sin\alpha}{g} = 1.028$ | DM1 A1 |
$t = 2\left(1.5 - \frac{12\sin\alpha}{g}\right)$ | |
Correct answer: $t = 0.94$ or $0.945$ (s) | A1 |

**Alt 1: Time between the two points when $v = 8$** | M1 |
$0 = 8\sin\beta t - \frac{1}{2}gt^2 = \left(t\left(8\sin\beta - \frac{1}{2}gt\right)\right)$ | DM1 A1 |
$t = 0.94$, $0.945$ | A1 |

**Alt 2: Find the difference between the two times when $h = 4.08$** | M1 |
$h = 12\sin\alpha t - \frac{1}{2}gt^2$ for their h, $\alpha$ | DM1 A1 |

**Alt 3: Use $v = u + at$ (vert) between the two points** | M1 |
$8\sin\beta = -8\sin\beta + gt$ | DM1 A1 |
$t = \frac{16\sin\beta}{g} = 0.94, 0.945$ | A1 |
(4) | |
[14] | |

**Notes for Q6**

6(a)
- M1 for an energy equation with correct no. of terms etc. (M0 if no energy equation used)
- First A1 and Second A1 for a correct equation.
- Third A1 for 4.1 or 4.08 (m)

6(b)
- M1 for a complete method to find an equation in $\sin\alpha$ only, using their h. (N.B. This equation could be obtained by getting two equations in $\alpha$ and $\beta$, which don't require h, and eliminating $\beta$, see above)
- First A1 ft for a correct equation, ft on their h (if used)
- Second A1 for $57°$, $57.1°$, $1.0$, $1.00$ rad

6(c)
- M1 for a complete method to find equation in $\alpha$ and $\beta$ only
- **Either by equating horizontal velocity components:** $12\cos\alpha = 8\cos\beta$
- **Or by finding the vertical velocity component at B, $v_y$, and then using** $\tan\beta = \frac{|v_y|}{12\cos\alpha}$ or $\sin\beta = \frac{|v_y|}{8}$.
- First A1 ft for a correct equation.
- N.B. $v_y$ can be found in a number of ways:
  - $v = u + at: v_y = 12\sin\alpha - 1.5g$
  - $v^2 = u^2 + 2as: v_y = \sqrt{(12\sin\alpha)^2 - 2g \times \text{their}h}$
  - $s = vt - \frac{1}{2}at^2: h = 1.5v_y + \frac{1}{2}g \times 1.5^2 \Rightarrow v_y = \frac{2h - 3g}{3}$
  - $s = \left(\frac{u + v}{2}\right)t: h = \left(\frac{12\sin\alpha + v_y}{2}\right) \times 1.5 \Rightarrow v_y = \frac{4h}{3} - 12\sin\alpha$

- Second A1 for $\beta = 35.4°$ ($35°$) (0.617 rads or 0.62 rads)

6(d)
- First M1 for a strategy to solve the problem
- Second DM1, dependent on first M, for attempt to set up an equation based on the strategy
- First A1 for a correct equation in t.
- Second A1 for 0.94 or 0.945

---