| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| Topic | Projectiles |
| Type | Projectile on inclined plane |
| Difficulty | Challenging +1.8 This is a challenging projectile motion problem requiring coordinate transformation to an inclined plane and vector decomposition. Part (i) demands deriving a non-trivial trigonometric relationship for impact angle, requiring careful resolution of velocity components and geometric insight. Part (ii) extends this to find a specific launch angle. While the techniques are A-level standard (projectile equations, trigonometry), the inclined plane context and the algebraic manipulation needed to prove the given formula elevate this significantly above routine projectile questions. |
| Spec | 1.05b Sine and cosine rules: including ambiguous case1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=13.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \( | v_y | = V\sin(\theta - 45°) = \frac{V}{\sqrt{2}}(\sin\theta - \cos\theta)\) — B1 |
**(i)** Take the origin as the point of projection and axes parallel and perpendicular to the plane. Let speed of projection be $V$.
On landing: $0 = V\sin(\theta - 45°)t - g\cos 45°\, t^2$ — M1A1
Use of compound angle formulae (at any stage). — M1
$\Rightarrow t = \frac{2}{5\sqrt{2}}\left\{\frac{V}{\sqrt{2}}(\sin\theta - \cos\theta)\right\} = \frac{V}{5}(\sin\theta - \cos\theta)$ — A1
$v_x = V\cos(\theta - 45°) - g\sin 45° \cdot \frac{V}{5}(\sin\theta - \cos\theta)$ — M1
$= \frac{V}{\sqrt{2}}(\cos\theta + \sin\theta) - \frac{2V}{\sqrt{2}}(\sin\theta - \cos\theta)$
$= \frac{V}{\sqrt{2}}(3\cos\theta - \sin\theta)$ — A1
$|v_y| = V\sin(\theta - 45°) = \frac{V}{\sqrt{2}}(\sin\theta - \cos\theta)$ — B1
$\tan\phi = \frac{\sin\theta - \cos\theta}{3\cos\theta - \sin\theta} = \frac{\tan\theta - 1}{3 - \tan\theta}$ (AG) — M1A1 [9]
**(ii)** Landing horizontally $\Rightarrow \tan\phi = 1$ — M1
$\Rightarrow 1 = \frac{\tan\theta - 1}{3 - \tan\theta} \Rightarrow \tan\theta = 2 \Rightarrow \theta = 63.4°$ — A1 [2]
**[Total: 11]**3 A particle is projected at an angle $\theta$ above the horizontal from the foot of a plane which is inclined at $45 ^ { \circ }$ to the horizontal. Subsequently the particle impacts on the plane at a higher point.\\
(i) Prove that the angle at which the particle strikes the plane is $\phi$, where
$$\tan \phi = \frac { \tan \theta - 1 } { 3 - \tan \theta }$$
(ii) Find the angle to the horizontal at which the particle would have to be projected if it were to strike the plane horizontally.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2010 Q3 [11]}}