| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Session | Specimen |
| Marks | 12 |
| Topic | Projectiles |
| Type | Projectile on inclined plane |
| Difficulty | Challenging +1.2 This is a standard Further Maths mechanics problem on projectile motion on an inclined plane. Parts (i) and (ii) are bookwork results requiring coordinate system setup and algebraic manipulation but following a well-established method. Part (iii) adds modest challenge by requiring the perpendicularity condition and trigonometric manipulation, but the approach is systematic. The 12 total marks and multi-step nature elevate it above average A-level, though it remains a textbook-style question without requiring novel insight. |
| Spec | 1.05q Trig in context: vectors, kinematics, forces3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \((t \neq 0) \therefore t = \frac{2V\sin(\alpha-\beta)}{g\cos\beta}\) | M1, A1, A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(= \frac{2V^2\sin(\alpha-\beta)}{g\cos\beta}\left(\cos(\alpha-\beta) - \frac{\sin\beta\sin(\alpha-\beta)}{\cos\beta}\right) = \frac{2V^2\sin(\alpha-\beta)\cos\alpha}{g\cos^2\beta}\) | M1, A1, A1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow \tan\alpha = \cot\beta + 2\tan\beta\) | M1, A1, A1, A1 | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow x = \frac{2V^2\cos^2\alpha}{g}\left(\frac{\sin\alpha}{\cos\alpha} - \frac{\sin\beta}{\cos\beta}\right) = \frac{2V^2\cos\alpha\sin(\alpha-\beta)}{g\cos\beta}\) | M1, A1 | |
| \(T = \frac{x}{V\cos\alpha} = \frac{2V\sin(\alpha-\beta)}{g\cos\beta}\) | M1, A1, A1 | (AG) |
| \(R = \frac{x}{\cos\beta} = \frac{2V^2\cos\alpha\sin(\alpha-\beta)}{g\cos^2\beta}\) | M1, A1 | 7 |
| Answer | Marks |
|---|---|
| \(V_x = V\cos\alpha\) and \(V_y = V\sin\alpha - 2V\cdot\frac{\sin(\alpha-\beta)}{\cos\beta} = -V\left(\frac{\sin\alpha\cos\beta - 2\sin(\alpha-\beta)}{\cos\beta}\right)\) | B1, B1 |
| \(\frac{V_x}{-V_y} = \tan\beta \Rightarrow \frac{\cos\alpha\cos\beta}{2\sin(\alpha-\beta) - \sin\alpha\cos\beta} = \frac{\sin\beta}{\cos\beta}\) | M1 |
| \(\Rightarrow \cos\alpha\cos^2\beta = 2\sin\beta(\sin\alpha\cos\beta - \cos\alpha\sin\beta) - \sin\alpha\sin\beta\cos\beta\) | A1 |
| \(\Rightarrow \cos\alpha\cos^2\beta = \sin\alpha\sin\beta\cos\beta - 2\cos\alpha\sin^2\beta\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow \tan\alpha = \frac{1+2\tan^2\beta}{\tan\beta} = \cot\beta + 2\tan\beta\) | A1 | 5 |
### (i)
Perpendicular to the plane: $0 = V\sin(\alpha-\beta)t - \frac{1}{2}g\cos\beta t^2$
$(t \neq 0) \therefore t = \frac{2V\sin(\alpha-\beta)}{g\cos\beta}$ | M1, A1, A1 | **3**
### (ii)
Parallel to the plane:
$x = V\cos(\alpha-\beta) \cdot \frac{2V\sin(\alpha-\beta)}{g\cos\beta} - \frac{1}{2}g\sin\beta \cdot \frac{4V^2\sin^2(\alpha-\beta)}{g^2\cos^2\beta}$
$= \frac{2V^2\sin(\alpha-\beta)}{g\cos\beta}\left(\cos(\alpha-\beta) - \frac{\sin\beta\sin(\alpha-\beta)}{\cos\beta}\right) = \frac{2V^2\sin(\alpha-\beta)\cos\alpha}{g\cos^2\beta}$ | M1, A1, A1 | **4**
### (iii)
Velocity component parallel to plane is zero
$0 = V\cos(\alpha-\beta) - g\sin\beta \cdot \frac{2V\sin(\alpha-\beta)}{g\cos\beta}$
$\Rightarrow 2\tan(\alpha-\beta) = \cot\beta$
$\Rightarrow 2\tan\alpha - 2\tan\beta = \cot\beta(1 + \tan\alpha\tan\beta)$
$\Rightarrow \tan\alpha = \cot\beta + 2\tan\beta$ | M1, A1, A1, A1 | **5**
## Question 4 - Alternative Solution
### (i) & (ii)
$y = x\tan\alpha - \frac{gx^2}{2V^2\cos^2\alpha} = x\tan\beta$
$\Rightarrow x = \frac{2V^2\cos^2\alpha}{g}\left(\frac{\sin\alpha}{\cos\alpha} - \frac{\sin\beta}{\cos\beta}\right) = \frac{2V^2\cos\alpha\sin(\alpha-\beta)}{g\cos\beta}$ | M1, A1 |
$T = \frac{x}{V\cos\alpha} = \frac{2V\sin(\alpha-\beta)}{g\cos\beta}$ | M1, A1, A1 | (AG) |
$R = \frac{x}{\cos\beta} = \frac{2V^2\cos\alpha\sin(\alpha-\beta)}{g\cos^2\beta}$ | M1, A1 | **7**
### (iii)
$V_x = V\cos\alpha$ and $V_y = V\sin\alpha - 2V\cdot\frac{\sin(\alpha-\beta)}{\cos\beta} = -V\left(\frac{\sin\alpha\cos\beta - 2\sin(\alpha-\beta)}{\cos\beta}\right)$ | B1, B1 |
$\frac{V_x}{-V_y} = \tan\beta \Rightarrow \frac{\cos\alpha\cos\beta}{2\sin(\alpha-\beta) - \sin\alpha\cos\beta} = \frac{\sin\beta}{\cos\beta}$ | M1 |
$\Rightarrow \cos\alpha\cos^2\beta = 2\sin\beta(\sin\alpha\cos\beta - \cos\alpha\sin\beta) - \sin\alpha\sin\beta\cos\beta$ | A1 |
$\Rightarrow \cos\alpha\cos^2\beta = \sin\alpha\sin\beta\cos\beta - 2\cos\alpha\sin^2\beta$ | A1 |
$\Rightarrow 1 = \tan\alpha\tan\beta - 2\tan^2\beta$
$\Rightarrow \tan\alpha = \frac{1+2\tan^2\beta}{\tan\beta} = \cot\beta + 2\tan\beta$ | A1 | **5**A particle is projected with velocity $V$ at an angle $\alpha$ to the horizontal up a plane inclined at $\beta$ to the horizontal, where $\alpha > \beta$.
\begin{enumerate}[label=(\roman*)]
\item Show that the time of flight is $\frac{2V \sin(\alpha - \beta)}{g \cos \beta}$. [3]
\item Show that the range on the inclined plane is $\frac{2V^2 \sin(\alpha - \beta) \cos \alpha}{g \cos^2 \beta}$. [4]
\item If the particle strikes the plane at right angles, prove that $\tan \alpha = \cot \beta + 2 \tan \beta$. [5]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 Q4 [12]}}