| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Year | 2015 |
| Session | June |
| Marks | 14 |
| 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, vector resolution in two reference frames, and the non-trivial constraint that the particle strikes perpendicular to the plane. The derivation of the relationship 2tan(α)tan(θ)=1 requires insight into velocity components, and parts (ii)(a) and (ii)(b) demand sustained algebraic manipulation with trigonometric identities across multiple steps. While the techniques are A-level standard, the problem requires significantly more geometric insight and algebraic stamina than typical projectile questions. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
**Question 12(i) and 12(ii)**
**(i)**
Taking axes along and perpendicular to plane:
$\dot{x} = u\cos\theta - gt\sin\alpha$; $\quad \dot{x} = 0 \Rightarrow t = \dfrac{u\cos\theta}{g\sin\alpha}$ — M1, A1 (Equation for $\dot{x}$, equated to 0; Find (eliminate) $t$)
$y = ut\sin\theta - \dfrac{1}{2}gt^2\cos\alpha$; $\quad y = 0 \Rightarrow t = \dfrac{2u\sin\theta}{g\cos\alpha}$ — M1, A1 (Equation for $y$, equated to 0; Find (eliminate) $t$)
$\dfrac{u\cos\theta}{g\sin\alpha} = \dfrac{2u\sin\theta}{g\cos\alpha} \Rightarrow 2\tan\alpha\tan\theta = 1$ (AG) — M1A1 (Equate and simplify; get AG)
**[6]**
**(ii)(a)**
$x = u\cos\theta \cdot \dfrac{u\cos\theta}{g\sin\alpha} - \dfrac{1}{2}g\sin\alpha \cdot \dfrac{u^2\cos^2\theta}{g^2\sin^2\alpha}$ — M1
$\Rightarrow x\sin\alpha = \dfrac{u^2}{2g}\cos^2\theta$ — A1 (Correct expression for $x\sin\alpha$)
But $\cos^2\theta = \dfrac{1}{1+\tan^2\theta} = \dfrac{1}{1 + \frac{1}{4\tan^2\alpha}} = \dfrac{4\tan^2\alpha}{1 + 4\tan^2\alpha}$ — M1A1 ($\cos\theta$ in terms of $\tan\alpha$)
$\Rightarrow x\sin\alpha = \dfrac{2u^2\tan^2\alpha}{g(1 + 4\tan^2\alpha)}$ (AG) — A1 (Obtain given answer)
**[5]**
**(ii)(b)**
Time of flight $= \dfrac{u\cos\theta}{g\sin\alpha} = \dfrac{u}{g\sin\alpha}\sqrt{\dfrac{4\tan^2\alpha}{1+4\tan^2\alpha}}$ — M1 (Find $t$ in terms of $\tan\alpha$)
$= \dfrac{2u}{g\cos\alpha\sqrt{1+4\tan^2\alpha}} = \dfrac{2u\sec\alpha}{g\sqrt{1+4\tan^2\alpha}}$ (AG) — M1A1 (Simplify to given answer)
(N.B. Other orders of doing this may be seen.)
**[3]**12 Points $A$ and $B$ lie on a line of greatest slope of a plane inclined at an angle $\alpha$ to the horizontal, with $B$ above $A$. A particle is projected from $A$ with speed $u$ at an angle $\theta$ to the plane and subsequently strikes the plane at right angles at $B$.\\
(i) Show that $2 \tan \alpha \tan \theta = 1$.\\
(ii) In either order, show that
\begin{enumerate}[label=(\alph*)]
\item the vertical height of $B$ above $A$ is $\frac { 2 u ^ { 2 } \tan ^ { 2 } \alpha } { g \left( 1 + 4 \tan ^ { 2 } \alpha \right) }$,
\item the time of flight from $A$ to $B$ is $\frac { 2 u \sec \alpha } { g \sqrt { 1 + 4 \tan ^ { 2 } \alpha } }$.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2015 Q12 [14]}}