# Question 7(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $V\sin\beta = e_1 U\sin\alpha$ | B1 | 3.4 – $e_1 U\sin\alpha$ seen from a relevant equation or diagram |
| $V\cos\beta = U\cos\alpha$ | B1 | 3.4 – $U\cos\alpha$ seen in a relevant equation or diagram |
| Eliminate $U$ and $V$ from two equations | M1 | 1.1b – A clear method using **two equations** to eliminate $U$ and $V$ |
| $\tan\beta = e_1\tan\alpha$ * | A1* | 2.2a – Given answer correctly obtained. Must include both equations showing how to reach both $\tan\beta$ and $e_1\tan\alpha$. Accept $\tan\beta = e_1\tan\alpha$ or $e_1\tan\alpha = \tan\beta$ |
| **Total** | **(4)** | |

# Question 7(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan\gamma = e_2\tan(90°-\beta)$, so $\tan\gamma = e_2\cot\beta$, $\cot\gamma = \dfrac{\tan\beta}{e_2}$ | B1 | 1.1b – Form correct expression for $\tan\gamma$ or $\cot\gamma$ in terms of $e_2$ and $\beta$ or $(90-\beta)$. May quote result from (a) or obtain again |
| $\tan\gamma = e_2 \times \dfrac{1}{\tan\beta} = e_2 \times \dfrac{1}{e_1\tan\alpha}$ | M1 | 3.1b – **Use result from (a)** to eliminate $\tan\beta$ and form equation in $\alpha$, $\gamma$, $e_1$, $e_2$ |
| $e_1\tan\alpha = e_2\cot\gamma$ * | A1* | 2.2a – Given answer correctly obtained. Must include replacement of $\tan\beta$ and rearrangement. Accept $e_1\tan\alpha = e_2\cot\gamma$ or $e_2\cot\gamma = e_1\tan\alpha$ |
| **Total** | **(3)** | |

# Question 7(c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $(\text{angle } APQ + \text{angle } AQP) = (180°-\alpha-\beta)+\{180°-(90°-\beta)-\gamma\} = 270°-\alpha-\gamma$. Otherwise: angle $PAQ = \alpha+\gamma-90°$ | M1 | 1.1b – Clear attempt to find angle sum (condone slips) or another relevant starting point e.g. expression for angle $PAQ$ |
| To return to $A$: angle $APQ$ + angle $AQP < 180°$ since $APQ$ is a triangle. Otherwise: angle $PAQ > 0$ | M1 | 3.1b – Clear statement to form inequality e.g. correct angle sum $< 180°$ is acceptable, or angle $PAQ > 0$ |
| $270°-\alpha-\gamma < 180° \Rightarrow \alpha > 90°-\gamma$ | A1 | 1.1b – Correct simplified inequality in correct form |
| $\tan\alpha > \tan(90°-\gamma)$ – correct method using addition formulae | M1 | 1.1b – Correct method to form inequality in tan or cot |
| $\dfrac{e_2\cot\gamma}{e_1} > \cot\gamma$ | M1 | 1.1b – Using part (b) to eliminate the angles |
| $e_2 > e_1$ * | A1* | 2.2a – Given answer correctly obtained |
| **Total** | **(6)** | |

# Question 7(d):

| Answer/Working | Mark | Guidance |
|---|---|---|
| From (b), $\alpha = 90°-\gamma$, so it moves parallel to $AP$, e.g. parallel to the initial velocity | B1 | 2.4 |
| **Total** | **(1)** | |

## Question 7:

**Part (c) alt — Use of trig identity**

| Answer/Working | Mark | Guidance |
|---|---|---|
| (angle $APQ$ + angle $AQP$) $= (180° - \alpha - \beta) + \{180° - (90° - \beta) - \gamma\} = 270 - \alpha - \gamma$ | M1 | |
| To return to $A$, (angle $APQ$ + angle $AQP$) $< 180°$, since $APQ$ is a triangle | M1 | |
| $\tan(\alpha + \gamma) = \dfrac{\tan\alpha + \tan\gamma}{1 - \tan\alpha\tan\gamma}$ and $\tan\alpha = \dfrac{e_2\cot\gamma}{e_1}$ or $\tan\alpha = \dfrac{e_2}{e_1\tan\gamma}$, leads to $\tan(\alpha+\gamma) = \dfrac{e_2 + e_1\tan^2\gamma}{e_1\tan\gamma - e_2\tan\gamma}$ oe | A1 | |
| $180 > (\alpha+\gamma) > 90 \Rightarrow \tan(\alpha+\gamma) < 0 \Rightarrow \dfrac{e_2 + e_1\tan^2\gamma}{e_1\tan\gamma - e_2\tan\gamma} < 0$ | M1 | Condone if '$180 >$' is not stated again |
| Since numerator $> 0$, therefore $e_1\tan\gamma - e_2\tan\gamma < 0$ | M1 | |
| $e_2 > e_1$ * | A1 | |

**Part (d)**

| Answer/Working | Mark | Guidance |
|---|---|---|
| $\alpha = 90° - \gamma$, so it moves parallel to $AP$; or $\alpha = 90° - \gamma$, so it moves parallel to the initial velocity | B1 | Use given information from (b) to make any equivalent statement with a correct reason and no incorrect statements. Do not accept 'it moves parallel to the initial speed' |