| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2016 |
| Session | Specimen |
| Marks | 12 |
| Topic | Projectiles |
| Type | Projectile on inclined plane |
| Difficulty | Challenging +1.8 This is a sophisticated multi-part projectile question requiring: (i) standard trajectory derivation, (ii) algebraic manipulation treating the trajectory as a quadratic in tan α with discriminant analysis to find the envelope/bounding parabola, and (iii) geometric insight to find maximum range on an inclined plane by intersecting with the envelope. Part (iii) requires recognizing that maximum range occurs at the boundary of possible trajectories and involves non-trivial algebraic manipulation. While the individual techniques are A-level standard, the conceptual sophistication (envelope of trajectories, geometric optimization) and multi-step reasoning place this well above average difficulty. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=13.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow R_{\text{down}} = \frac{ | -40\sin\beta - 40 | }{\cos^2\beta} = \frac{40(1+\sin\beta)}{1-\sin^2\beta} = \frac{40}{1-\sin\beta}\) B1 |
**(i)** $x = 20\cos\alpha t$, $y = 20\sin\alpha t - 5t^2$ **B1 B1**
$y = 20\sin\alpha \cdot \frac{x}{20\cos\alpha} - 5\left(\frac{x}{20\cos\alpha}\right)^2 = x\tan\alpha - \frac{x^2}{80}(1 + \tan^2\alpha)$ AG **M1 A1**
**(ii)** $x^2\tan^2\alpha - 80x\tan\alpha + x^2 + 80y = 0$ (Can be implied by what follows.) **B1**
Real roots $\Rightarrow 6400x^2 - 4x^2(x^2 + 80y) \geqslant 0$ **M1 A1**
$\Rightarrow 1600 - x^2 - 80y \geqslant 0 \Rightarrow y \leqslant 20 - \frac{x^2}{80}$, $(x \neq 0)$. **A1**
**(iii)** $x = R\cos\beta$ and $y = R\sin\beta \Rightarrow y = x\tan\beta$.
In $y = 20 - \frac{x^2}{80} \Rightarrow R\sin\beta = 20 - \frac{R^2(1-\sin^2\beta)}{80}$ **M1 A1**
$\therefore R^2(1-\sin^2\beta) + 80R\sin\beta - 1600 = 0 \Rightarrow (R[1-\sin\beta] + 40)(R[1+\sin\beta] - 40) = 0$ **M1**
$\Rightarrow R = \frac{40}{1+\sin\beta}$ (up) or $\frac{-40}{1-\sin\beta}$ (down) **A1**
**Alternative solution:**
$x^2 + 80\tan\beta x - 1600 = 0$ **B1**
$\Rightarrow x = -40\tan\beta \pm 40\sec\beta$ **B1**
$\Rightarrow R_{\text{up}} = \frac{-40\sin\beta + 40}{\cos^2\beta} = \frac{40(1-\sin\beta)}{1-\sin^2\beta} = \frac{40}{1+\sin\beta}$ **B1**
$\Rightarrow R_{\text{down}} = \frac{|-40\sin\beta - 40|}{\cos^2\beta} = \frac{40(1+\sin\beta)}{1-\sin^2\beta} = \frac{40}{1-\sin\beta}$ **B1**12 A projectile is launched from the origin with speed $20 \mathrm {~ms} ^ { - 1 }$ at an angle $\alpha$ above the horizontal.\\
(i) Prove that the equation of its trajectory is
$$y = x \tan \alpha - \frac { x ^ { 2 } } { 80 } \left( 1 + \tan ^ { 2 } \alpha \right) .$$
(ii) Regarding the equation of the trajectory as a quadratic equation in $\tan \alpha$, show that $\tan \alpha$ has real values provided that
$$y \leqslant 20 - \frac { x ^ { 2 } } { 80 } .$$
(iii) A plane is inclined at an angle $\beta$ to the horizontal. The line $l$, with equation $y = x \tan \beta$, is a line of greatest slope in the plane. A particle is projected from a point on the plane, in the vertical plane containing $l$. By considering the intersection of $l$ with the bounding parabola $y = 20 - \frac { x ^ { 2 } } { 80 }$, deduce that the maximum range up, or down, this inclined plane is $\frac { 40 } { 1 + \sin \beta }$, or $\frac { 40 } { 1 - \sin \beta }$, respectively.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2016 Q12 [12]}}