| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Topic | Projectiles |
| Type | Projectile clearing obstacle |
| Difficulty | Standard +0.3 This is a standard A-level mechanics projectile motion question requiring derivation of the range formula (routine calculus/kinematics), then solving simultaneous equations using symmetry. The symmetry insight (range = 100m) is straightforward given the wall positions, and the algebra is mechanical. Slightly above average due to the multi-step nature and need to apply the trajectory equation twice, but all techniques are standard textbook material. |
| Spec | 1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Use of \(x = Vt\cos\theta\) and \(y = Vt\sin\theta - \frac{1}{2}gt^2\) | M1 A1 | |
| Solving \(y = 0\) for \(t\) and substitute in \(x\) formula | M1 | |
| \(R = \frac{2V^2\sin\theta\cos\theta}{g} = \frac{V^2\sin 2\theta}{g}\) | A1 AG | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| (symmetry of the trajectory) implies \(R = 100\text{m}\) | B1 | [1] |
| Answer | Marks |
|---|---|
| \(V = \sqrt{1000}\left(= 10\sqrt{10}\right) \text{ ms}^{-1}\) | BV ∨ |
| Answer | Marks | Guidance |
|---|---|---|
| Solving \(30 = 10\sqrt{10}r\sin\frac{\pi}{4}\) | M1 | |
| Obtain \(t = \frac{3}{\sqrt{5}}\) or \(t = \frac{x}{V\cos\theta}\) and substitute later | A1 M1 | |
| Obtain \(h = 21\text{m}\) | A1 | [5] |
**Part (i)**
Use of $x = Vt\cos\theta$ and $y = Vt\sin\theta - \frac{1}{2}gt^2$ | M1 A1
Solving $y = 0$ for $t$ and substitute in $x$ formula | M1
$R = \frac{2V^2\sin\theta\cos\theta}{g} = \frac{V^2\sin 2\theta}{g}$ | A1 AG | [4]
**Part (ii)**
(symmetry of the trajectory) implies $R = 100\text{m}$ | B1 | [1]
**Part (iii)**
$V = \sqrt{1000}\left(= 10\sqrt{10}\right) \text{ ms}^{-1}$ | BV ∨
$\left(= \sqrt{g \times \text{their}R}\right)$
Solving $30 = 10\sqrt{10}r\sin\frac{\pi}{4}$ | M1
Obtain $t = \frac{3}{\sqrt{5}}$ or $t = \frac{x}{V\cos\theta}$ and substitute later | A1 M1
Obtain $h = 21\text{m}$ | A1 | [5]\includegraphics{figure_11}
A projectile is fired from a point $O$ in a horizontal plane, with initial speed $V$, at an angle $\theta$ to the horizontal (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Show that the range of the projectile on the horizontal plane is
$$\frac{2V^2 \sin \theta \cos \theta}{g}.$$ [4]
\end{enumerate}
There are two vertical walls, each of height $h$, at distances 30 m and 70 m, respectively, from $O$ with bases on the horizontal plane. The value of $\theta$ is $45°$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumii}{1}
\item If the projectile just clears both walls, state the range of the projectile. [1]
\item Hence find the value of $V$ and of $h$. [5]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2011 Q11 [10]}}