| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2019 |
| Session | Specimen |
| Marks | 4 |
| Topic | Projectiles |
| Type | Projectile on inclined plane |
| Difficulty | Challenging +1.2 Part (a) is a standard projectile trajectory derivation requiring routine application of kinematic equations. Part (b) requires treating the trajectory as a quadratic in tan α and applying the discriminant condition, which is a moderately clever algebraic manipulation but follows a clear path once recognized. Part (c) combines the geometric insight of finding where the incline meets the envelope of trajectories, requiring students to connect multiple concepts, but the question guides them explicitly to the method. Overall, this is above-average difficulty due to the multi-step reasoning and the non-standard approach in parts (b) and (c), but it's well within reach for competent Further Maths students and doesn't require exceptional insight. |
| Spec | 3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
**Question 12(a)**
B1 for $x = 20\cos\alpha\, t$; B1 for $y = 20\sin\alpha t - 5t^2$ [B1B1]
$y = 20\sin\alpha \cdot \dfrac{x}{20\cos\alpha} - 5\left(\dfrac{x}{20\cos\alpha}\right)^2 = x\tan\alpha - \dfrac{x^2}{80}(1 + \tan^2\alpha)$ **AG** [M1A1]
**Total: 4 marks**
**Question 12(b)**
$x^2\tan^2\alpha - 80x\tan\alpha + x^2 + 80y = 0$ (Can be implied by what follows.) [B1]
Real roots $\Rightarrow 6400x^2 - 4x^2(x^2 + 80y) \geqslant 0$ [M1A1]
$\Rightarrow 1600 - x^2 - 80y \geqslant 0 \Rightarrow y \leqslant 20 - \dfrac{x^2}{80}$, $(x \neq 0)$ [A1]
**Total: 4 marks**
**Question 12(c)**
$x^2 + 80x\tan 30 - 1600 = 0$ [B1]
$x^2 + \dfrac{80}{\sqrt{3}}x - 1600 = 0$ [B1]
$x = -\dfrac{40}{\sqrt{3}} \pm \sqrt{\dfrac{1600}{3} + 1600}$ [B1]
$= -\dfrac{40}{\sqrt{3}} + 2 \times \dfrac{40}{\sqrt{3}} = \dfrac{40}{\sqrt{3}}$ (ignore − at this stage) [B1 implied]
$\Rightarrow R = x \div \cos 30 = \dfrac{80}{3}$ (A0 if both solutions retained) [A1]
**OR Alternative solution:** [M1A1]
$x = R\cos 30°$ and $y = R\sin 30°$:
$y = 20 - \dfrac{x^2}{80} \Rightarrow R\sin 30 = 20 - \dfrac{R^2(1-\sin^2 30)}{80}$
$\therefore 0.75R^2 + 40R - 1600 = 0 \Rightarrow (0.5R + 40)(1.5R - 40) = 0$ [M1]
$\Rightarrow R = \dfrac{80}{3}$ (A0 if both solutions retained) [A1]
**Total: 4 marks**12 A particle is projected from the origin with speed $20 \mathrm {~ms} ^ { - 1 }$ at an angle $\alpha$ above the horizontal.
\begin{enumerate}[label=(\alph*)]
\item Prove that the equation of its trajectory is
$$y = x \tan \alpha - \frac { x ^ { 2 } } { 80 } \left( l + \tan ^ { 2 } \alpha \right) .$$
\item Regarding the equation of the trajectory as a quadratic equation in $\tan \alpha$, show that $\tan \alpha$ has real values provided that
$$y \leqslant 20 - \frac { x ^ { 2 } } { 80 } .$$
\item A plane is inclined at an angle of $30 ^ { \circ }$ to the horizontal. The line $l$, with equation $y = x \tan 30 ^ { \circ }$, is a line of greatest slope in the plane. The particle is projected from the origin with speed $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point on the plane, in the vertical plane containing $l$. By considering the intersection of $l$ with the curve $y = 20 - \frac { x ^ { 2 } } { 80 }$, find the maximum range up this inclined plane.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2019 Q12 [4]}}