| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2020 |
| Session | Specimen |
| Marks | 4 |
| Topic | Projectiles |
| Type | Projectile on inclined plane |
| Difficulty | Challenging +1.2 Part (a) is a standard derivation of projectile trajectory equation using 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 prompt. Part (c) combines the envelope of trajectories with inclined plane geometry—this requires synthesis of parts (a)-(b) and geometric insight, but the question explicitly guides students to consider the intersection. Overall, this is above-average difficulty due to the multi-step reasoning and the envelope concept, but the structured parts and explicit hints keep it accessible for Further Maths students. |
| Spec | 3.02i 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) [blank]
$\Rightarrow R = x \div \cos 30 = \dfrac{80}{3}$ (A0 if both solutions retained) [A1]
**OR Alternative solution:** [M1A1]
$x = R\cos 30°$ and $y = R\sin 30°$:
$y = 20 - \dfrac{x^2}{80} \Rightarrow R\sin 30 = 20 - \dfrac{R^2(1-\sin^2 30)}{80}$ [M1]
$\therefore 0.75R^2 + 40R - 1600 = 0 \Rightarrow (0.5R + 40)(1.5R - 40) = 0$
$\Rightarrow R = \dfrac{80}{3}$ (A0 if both solutions retained) [A1]
**Available marks: 4**12 A particle is projected from the origin with speed $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ 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 $20 \mathrm {~ms} ^ { - 1 }$ 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 2020 Q12 [4]}}