| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Deriving trajectory equation |
| Difficulty | Moderate -0.3 This is a standard M3 projectiles question requiring derivation of the trajectory equation from parametric equations (a routine procedure taught explicitly) followed by straightforward substitution to find range. The 'show that' format and the two-part structure make it slightly easier than average, though it requires multiple steps and careful algebraic manipulation. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = -\frac{1}{2}gt^2 + 20\sin 30 \cdot t\) | M1A1 | |
| \(x = 20\cos 30 \cdot t\) | M1 | |
| \(t = \frac{x}{20\cos 30}\) | A1 | |
| \(y = -\frac{1}{2}g\frac{x^2}{400\cos^2 30} + 20\sin 30 \cdot \frac{x}{20\cos 30}\) | M1 | |
| \(y = x\tan 30 - \frac{gx^2}{800\cos^2 30°}\) | A1 | AG |
| 6 |
| Answer | Marks | Guidance |
|---|---|---|
| \(2.5 = x\tan 30 - \frac{9.8x^2}{800\cos^2 30}\) | M1A1 | Substituting and tidying up |
| \(9.8x^2 - 346x + 1500 = 0\) | M1 | |
| \(x = \frac{346 \pm \sqrt{119716 - 58800}}{19.6}\) | A1F | |
| \(= 30.3\) (or \(30.2\)) \(\&\) \(5.06\) (or \(5.05\)) answer: \(30.3\text{m}\) (or \(30.2\text{m}\)) | A1F | At least 3 s.f. required |
| 4 |
| Answer | Marks |
|---|---|
| No air resistance, the ball is a particle etc. | B1B1 |
| 2 | |
| 12 | Total |
### Part (a)
$y = -\frac{1}{2}gt^2 + 20\sin 30 \cdot t$ | M1A1 |
$x = 20\cos 30 \cdot t$ | M1 |
$t = \frac{x}{20\cos 30}$ | A1 |
$y = -\frac{1}{2}g\frac{x^2}{400\cos^2 30} + 20\sin 30 \cdot \frac{x}{20\cos 30}$ | M1 |
$y = x\tan 30 - \frac{gx^2}{800\cos^2 30°}$ | A1 | AG
| | **6** |
### Part (b)
$2.5 = x\tan 30 - \frac{9.8x^2}{800\cos^2 30}$ | M1A1 | Substituting and tidying up
$9.8x^2 - 346x + 1500 = 0$ | M1 |
$x = \frac{346 \pm \sqrt{119716 - 58800}}{19.6}$ | A1F |
$= 30.3$ (or $30.2$) $\&$ $5.06$ (or $5.05$) answer: $30.3\text{m}$ (or $30.2\text{m}$) | A1F | At least 3 s.f. required
| | **4** |
### Part (c)
No air resistance, the ball is a particle etc. | B1B1 |
| | **2** |
| | **12** | **Total** |5 A football is kicked from a point $O$ on a horizontal football ground with a velocity of $20 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle of elevation of $30 ^ { \circ }$. During the motion, the horizontal and upward vertical displacements of the football from $O$ are $x$ metres and $y$ metres respectively.
\begin{enumerate}[label=(\alph*)]
\item Show that $x$ and $y$ satisfy the equation
$$y = x \tan 30 ^ { \circ } - \frac { g x ^ { 2 } } { 800 \cos ^ { 2 } 30 ^ { \circ } }$$
\item On its downward flight the ball hits the horizontal crossbar of the goal at a point which is 2.5 m above the ground. Using the equation given in part (a), find the horizontal distance from $O$ to the goal.\\
(4 marks)\\
\includegraphics[max width=\textwidth, alt={}, center]{f8c04360-f54b-4d08-aee9-fe28612918d0-3_330_1411_1902_303}
\item State two modelling assumptions that you have made.
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2006 Q5 [13]}}