| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2013 |
| Session | June |
| Marks | 6 |
| Topic | Projectiles |
| Type | Two moving objects interception (non-projectile) |
| Difficulty | Standard +0.8 This is a relative velocity problem requiring vector decomposition with bearings, optimization to find the closest approach condition (perpendicular relative velocity), and time calculation. It demands solid understanding of 2D kinematics, coordinate geometry with bearings, and the geometric insight that closest approach occurs when relative position is perpendicular to relative velocity—more sophisticated than standard projectile questions but accessible to strong Further Maths students. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case3.02e Two-dimensional constant acceleration: with vectors |
(i) Uses $A$'s velocity perpendicular to $_AV_B$ for closest approach. **M1**
Calculates relevant angle from (12, 16, 20) ~ (3, 4, 5) triangle. **M1A1**
e.g. $\cos^{-1}\left(\frac{4}{5}\right) = 36.87°$
Finds bearing: $300° + 36.87° = 337°$ (nearest degree). **A1** [4]
(ii) Magnitude of $_AV_B = \sqrt{20^2 - 16^2} = 12$ **M1A1**
Uses distance triangle to find length of travel along relative path:
$15\cos(120° - 53.13°)° = 5.892$ km **M1A1**
Time $= \frac{5.892}{12} \times 60 = 29.5$ minutes **M1A1** [6]10 Ship $A$ is 15 km due south of ship $B$. Ship $B$ is travelling at $20 \mathrm {~km} \mathrm {~h} ^ { - 1 }$ on a bearing of $300 ^ { \circ }$. Ship $A$ is travelling at $16 \mathrm {~km} \mathrm {~h} ^ { - 1 }$. Find\\
(i) the bearing, to the nearest degree, that $A$ must take in order to get as close as possible to $B$, [4]\\
(ii) the time, in minutes, that it takes for the ships to be as close as possible.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2013 Q10 [6]}}