| Distance travelled by Ali: $1.5(t + 10)$ | M1A1 | |
| Distance travelled by Beth: $2t$ and correct triangle seen or implied | B1 | |
| Cosine rule: $(2t)^2 = 75^2 + 1.5^2(t + 10)^2 - 2 \times 75 \times 1.5(t + 10)\cos 45$ | M1A1 | |
| $1.75t^2 + 114.1t - 4259 = 0$ | M1 | |
| $t = \frac{-114.1 + \sqrt{114.1^2 + 4 \times 1.75 \times 4259}}{3.5} = 26.5$ | A1 | |
| Sine rule: $\frac{\sin\alpha}{1.5(t + 10)} = \frac{\sin 45}{2t}$ | M1A1 | |
| $\frac{\sin\alpha}{1.5 \times 36.5} = \frac{\sin 45}{2 \times 26.5} \Rightarrow \alpha = 46.9°$ to side $PQ$ or equivalent | M1A1 | |

### Alt 5 Alt
| Position vector of A: $\begin{pmatrix}\frac{1.5}{2}(t + 10)\\ \frac{1.5}{2}(t + 10)\end{pmatrix}$ or with $t$ | B1 | |
| Position vector of B: $\begin{pmatrix}2t\sin\alpha\\ 75 - 2t\cos\alpha\end{pmatrix}$ value for time consistent | M1A1 | |
| Equate components: | M1A1 | |
| Form equation in $t$: $4t^2 = \frac{9}{4} \times \frac{1}{2}(t + 10)^2 + (75 - \frac{3}{2\sqrt{2}}(t - 10))^2$ | M1A1 | |
| Simplify and solve: $14t^2 + 912.8t - 34072 = 0$ | M1 | |
| $t = 26.5$ | A1 | |
| Substitute $t$ and solve for $\alpha$ | M1 | |
| $\Rightarrow \alpha = 46.9°$ to side $PQ$ | A1 | |

### Alt 5 Alt (using triangle of velocities)
| Using distances: $\tan\theta = \frac{15/\sqrt{2}}{75 - 15/\sqrt{2}}$ $\theta = 9.35°$ | M1A1 | |
| $\alpha = 45° + \theta = 54.35°$ | | |
| Distance to travel at relative velocity: $\sqrt{(15/\sqrt{2})^2 + (75 - 15/\sqrt{2})^2} = \sqrt{10.61^2 + 64.39^2} = 65.3$ (m) | B1 | |
| Using relative velocities: $\frac{\sin\alpha}{\sin\beta} = \frac{2}{1.5}$ their $\alpha, \beta$ | M1A1 | |
| $\beta = 37.5°$ | | |
| $\Rightarrow$ Beth should travel at $\theta + \beta = 46.9°$ to side $PQ$ or equivalent | M1A1 | |
| Relative velocity: $\frac{v}{\sin(180° - \alpha - \beta)} = \frac{2}{\sin\alpha}$ | M1A1 | |
| $v = 2.46$ (ms$^{-1}$) | | |
| Time to intercept $= \frac{65.3}{2.46} = 26.5$ (s) | M1A1 | |

## Question 6a
| $v^2 = kg(5e^{x/2k} - 4) \Rightarrow \frac{v^2}{2} = \frac{kg}{2}(5e^{x/2k} - 4)$ | | |
| $\Rightarrow \frac{d}{dv}(\frac{v^2}{2}) = -\frac{1}{2k}\frac{kg}{2}(5e^{x/2k}) = -\frac{5g}{4}e^{-x/2k}$ | M1 | |
| From $v^2: 5ge^{x/2k} = \frac{v^2}{k} + 4g \Rightarrow \frac{d}{dv}(\frac{v^2}{2}) = -(\frac{v^2}{4k} + g)$ | M1 | |
| $\Rightarrow ma = -(\frac{mv^2}{4k} + mg)$ | A1 | |
| So resistance is $\frac{mv^2}{4k}$ | A1 | Given answer |

## Question 6b
| At max height $v = 0$ | M1 | |
| $\Rightarrow (5e^{x/2k} - 4) = 0$, $e^{x/2k} = \frac{4}{5}$, $x = 2k\ln(\frac{5}{4})$ | M1A1 | |

## Question 6c
| $x = 0$, $v = \sqrt{kg}$ | B1 | |
| Differential equation in $v$ and $t$: $\frac{dv}{dt} = -(g + \frac{v^2}{4k})$ | B1 | |
| Separate variables: $-\int\frac{1}{4k}dt = \int\frac{1}{4kg + v^2}dv$ | M1 | |
| Integrate: $\frac{T}{4k} = \frac{1}{\sqrt{4kg}}\tan^{-1}(\frac{v}{\sqrt{4kg}})|_{\sqrt{kg}}^0$ | M1A1 | |
| Use limits: $T = \frac{4k}{\sqrt{4kg}}(\tan^{-1}\frac{1}{2} - \tan^{-1}0) = \frac{4k}{\sqrt{g}}\arctan(\frac{1}{2})$ | M1A1 | Given answer |

## Question 7a
| Impulse on A: $I = 2(i + 3j - 3i - j)$ | M1A1 | |
| $= -4i + 4j = 4(-i + j)$ | A1 | |
| Impulse parallel to l.o.c., hence l.o.c. parallel to $-i + j$ | A1 | Given answer |

## Question 7b
| Impulse equal and opposite: $4i - 4j = 3(v + i - 2j)$ | M1A1 | |
| $3v = i + 2j$, $v = \frac{1}{3}(i + 2j)$ | A1 | |

### Alt
| Alt using CLM: $2(3i + j) + 3(-i + 2j) = 2(i + 3j) + 3v$ | M1A1 | |
| $3v = i + 2j$, $v = \frac{1}{3}(i + 2j)$ | A1 | |

## Question 7c
| Components of velocities parallel to $-i + j$: | | |
| A before: $(3i + j) \cdot \frac{1}{\sqrt{2}}(-i + j) = \frac{-2}{\sqrt{2}}$ | | |
| A after: $(i + 3j) \cdot \frac{1}{\sqrt{2}}(-i + j) = \frac{2}{\sqrt{2}}$ | | |
| B before: $(-i + 2j) \cdot \frac{1}{\sqrt{2}}(-i + j) = \frac{3}{\sqrt{2}}$ | | |
| B after: $\frac{1}{3}(i + 2j) \cdot \frac{1}{\sqrt{2}}(-i + j) = \frac{1}{3\sqrt{2}}$ | M1A3 | follow through from 7(b) |
| NB: the marks are all available if the unit vector $(\sqrt{2})$ is not used. | | |
| Coefficient of restitution: $\frac{1}{\sqrt{2}}(2 - \frac{1}{3}) = \frac{e}{\sqrt{2}}(3 + 2)$ | M1 | |
| $e = \frac{1}{3}$ | A1 | |

### Alternative (non-vector form)
| Components parallel to the line of centres: | | |
| A before: $-\sqrt{10}\cos(135° - \beta) = -\sqrt{10}(-\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{10}} + \frac{1}{\sqrt{2}} \cdot \frac{3}{\sqrt{10}}) = -\frac{2}{\sqrt{2}}$ | | |
| A after: $\sqrt{10}\cos(135° - \beta) = \sqrt{10}(-\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{10}} + \frac{1}{\sqrt{2}} \cdot \frac{3}{\sqrt{10}}) = \frac{2}{\sqrt{2}}$ | | |
| B before: $\sqrt{5}\cos(45° - \alpha) = \sqrt{5}(\frac{1}{\sqrt{2}} \cdot \frac{2}{\sqrt{5}} + \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{5}}) = \frac{3}{\sqrt{2}}$ | | |
| B after: $\frac{\sqrt{5}}{3}\cos(45° + \alpha) = \frac{\sqrt{5}}{3}(\frac{1}{\sqrt{2}} \cdot \frac{2}{\sqrt{5}} - \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{5}}) = \frac{1}{3\sqrt{2}}$ | | |
| follow through from 7(b) | | |
| Coefficient of restitution: $\frac{1}{\sqrt{2}}(2 - \frac{1}{3}) = \frac{e}{\sqrt{2}}(3 + 2)$ | M1 | |
| $e = \frac{1}{3}$ | A1 | |