| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2012 |
| Session | June |
| Marks | 12 |
| Topic | Projectiles |
| Type | Projectile on inclined plane |
| Difficulty | Challenging +1.8 This is a sophisticated projectile motion problem requiring derivation of trajectory equation, discriminant analysis for the envelope of trajectories, and geometric reasoning about maximum range on an inclined plane. While the individual techniques (trajectory derivation, quadratic discriminant) are standard A-level, the multi-part structure, the novel approach of treating tan α as the variable, and the geometric insight needed in part (iii) to connect the bounding parabola to maximum range elevate this significantly above typical mechanics questions. However, it's still within reach for well-prepared Further Maths students following clear scaffolding. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=13.02i Projectile motion: constant acceleration model |
**Question 12(i)**
$x = 20\cos\alpha\, t \quad y = 20\sin\alpha\, t - 5t^2$ (Allow $g$ or 9.8 here for B mark.) **B1B1**
$y = 20\sin\alpha \cdot \frac{x}{20\cos\alpha} - 5\left(\frac{x}{20\cos\alpha}\right)^2 = x\tan\alpha - \frac{x^2}{80}(1 + \tan^2\alpha)$ (AG) **M1A1** [4]
**Question 12(ii)**
$x^2\tan^2\alpha - 80x\tan\alpha + x^2 + 80y = 0$ (Can be implied by what follows.) **B1**
Real roots $\Rightarrow 6400x^2 - 4x^2(x^2 + 80y) \geq 0$ **M1A1**
$\Rightarrow 1600 - x^2 - 80y \geq 0 \Rightarrow y \leq 20 - \frac{x^2}{80}$, $(x \neq 0)$. **A1** [4]
**Question 12(iii)**
$x = R\cos\beta$ and $y = R\sin\beta \Rightarrow y = x\tan\beta$
In $y = 20 - \frac{x^2}{80} \Rightarrow R\sin\beta = 20 - \frac{R^2(1-\sin^2\beta)}{80}$ **M1A1**
$\therefore R^2(1 - \sin^2\beta) + 80R\sin\beta - 1600 = 0 \Rightarrow (R[1-\sin\beta]+40)(R[1+\sin\beta]-40)$ **M1**
$\Rightarrow R = \frac{40}{1+\sin\beta}$ (up) or $\frac{-40}{1-\sin\beta}$ (down) **A1** [4] **[12]**12 A projectile is launched from the origin with speed $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ 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 2012 Q12 [12]}}