| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Session | Specimen |
| Marks | 2 |
| Topic | Projectiles |
| Type | Two moving objects interception (non-projectile) |
| Difficulty | Standard +0.8 This is a relative velocity problem requiring vector decomposition, optimization (quickest interception), and geometric reasoning across multiple parts. While the mechanics are A-level standard, the 'as quickly as possible' constraint requires insight into velocity triangles and the two-part structure demands careful spatial reasoning, placing it moderately above average difficulty. |
| Spec | 3.02e Two-dimensional constant acceleration: with vectors |
**Question 1**
**(i)** Regarding $A$ being at rest:
$$\frac{\sin(45° - \theta)}{20} = \frac{\sin 135°}{35}$$
$$\Rightarrow \sin(45° - \theta) = \frac{4}{7\sqrt{2}}$$
$$\Rightarrow 45° - \theta = 23.83...$$
$$\Rightarrow \theta = 21.16... \Rightarrow \text{bearing is } 021°$$ (AG)
M1A1, M1, A1 **[4 marks]**
**(ii)** $t = \dfrac{8}{35 \sin 21.16...}$
$A$ moves $20t = \dfrac{160}{35 \sin 21.16...} = 12.7 \text{ km} = 13 \text{ km (nearest km)}$
M1A1, M1A1 **[4 marks]**
**Alternative for (i) and (ii):**
Let $B$ steer on a bearing $\theta$; at time $t$:
$35\cos\theta \times t = 20t + 8$ M1A1
$35\sin\theta \times t = 8$ A1
$1225\cos^2\theta \times t^2 = 400t^2 + 320t + 64$, $\ 1225\sin^2\theta \times t^2 = 64$
Adding: $1225t^2 = 400t^2 + 320t + 64 \Rightarrow 825t^2 - 320t - 128 = 0$ M1
Whence $t = 0.6329...$ A1
$\sin\theta = \dfrac{8}{35 \times 0.6329...} \Rightarrow \theta = 21.16... \Rightarrow \text{bearing is } 021°$ (AG) M1A1
$A$ moves $20t = 12.7 \text{ km} = 13 \text{ km (nearest km)}$ A1 **(8)**1 A ship $A$ is steaming north at $20 \mathrm {~km} \mathrm {~h} ^ { - 1 }$. Initially a ship $B$ is at a distance 8 km due west of $A$, and is steaming on a course such that it will take up a position 8 km directly ahead of $A$ as quickly as possible.\\
(i) Given that the maximum speed of B is $35 \mathrm {~km} \mathrm {~h} ^ { - 1 }$, show that the bearing of this course is $021 ^ { \circ }$, correct to the nearest degree.\\
(ii) Find the distance that $A$ moves between the instants when $B$ is due west of $A$ and when $B$ is due north of $A$, giving your answer to the nearest kilometre.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 Q1 [2]}}