| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile passing through given point |
| Difficulty | Standard +0.3 This is a standard M3 projectile question requiring derivation of trajectory equation (routine manipulation of parametric equations) and solving a quadratic in tan θ. The 'show that' part guides students through the algebra, and part (b) is straightforward substitution followed by basic reasoning about time of flight. Slightly above average due to the algebraic manipulation required, but follows a well-practiced template. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02f Solve quadratic equations: including in a function of unknown3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Horizontal: \(x = 40\cos\theta \cdot t\) so \(t = \dfrac{x}{40\cos\theta}\) | B1 | Correct horizontal equation |
| Vertical: \(y = 40\sin\theta \cdot t - \dfrac{1}{2}(10)t^2\) | M1 | Using correct vertical equation of motion |
| Substituting \(t = \dfrac{x}{40\cos\theta}\) into vertical equation | M1 | Substitution |
| \(y = x\tan\theta - \dfrac{10x^2}{2 \times 1600\cos^2\theta}\) | A1 | Correct simplified form |
| \(y = x\tan\theta - \dfrac{x^2}{320\cos^2\theta}\) | ||
| Using \(\sec^2\theta = 1 + \tan^2\theta\): \(y = x\tan\theta - \dfrac{x^2(1+\tan^2\theta)}{320}\) | M1 | Using \(\sec^2\theta\) identity |
| Rearranging: \(x^2\tan^2\theta - 320x\tan\theta + (x^2 + 320y) = 0\) | A1 | Correct completion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Substituting \(x = 150\), \(y = 8\): \(22500\tan^2\theta - 48000\tan\theta + 24560 = 0\) | M1 | Substituting correct values |
| Dividing: \(\tan^2\theta - \dfrac{48000}{22500}\tan\theta + \dfrac{24560}{22500} = 0\) | A1 | Correct equation in \(\tan\theta\) |
| Solving quadratic: \(\tan\theta = \dfrac{48000 \pm \sqrt{48000^2 - 4(22500)(24560)}}{2(22500)}\) | M1 | Using quadratic formula |
| \(\tan\theta \approx 1.678\) or \(\tan\theta \approx 0.655\) | A1 | Both correct values |
| \(\theta \approx 59.2°\) or \(\theta \approx 33.2°\) | A1 | Both angles correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Smaller angle \(\theta \approx 33.2°\) gives shortest time | B1 | Correct value chosen |
| Smaller \(\theta\) gives larger horizontal component of velocity, so ball reaches \(x = 150\) in less time | B1 | Valid reason |
# Question 3:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Horizontal: $x = 40\cos\theta \cdot t$ so $t = \dfrac{x}{40\cos\theta}$ | B1 | Correct horizontal equation |
| Vertical: $y = 40\sin\theta \cdot t - \dfrac{1}{2}(10)t^2$ | M1 | Using correct vertical equation of motion |
| Substituting $t = \dfrac{x}{40\cos\theta}$ into vertical equation | M1 | Substitution |
| $y = x\tan\theta - \dfrac{10x^2}{2 \times 1600\cos^2\theta}$ | A1 | Correct simplified form |
| $y = x\tan\theta - \dfrac{x^2}{320\cos^2\theta}$ | | |
| Using $\sec^2\theta = 1 + \tan^2\theta$: $y = x\tan\theta - \dfrac{x^2(1+\tan^2\theta)}{320}$ | M1 | Using $\sec^2\theta$ identity |
| Rearranging: $x^2\tan^2\theta - 320x\tan\theta + (x^2 + 320y) = 0$ | A1 | Correct completion |
## Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Substituting $x = 150$, $y = 8$: $22500\tan^2\theta - 48000\tan\theta + 24560 = 0$ | M1 | Substituting correct values |
| Dividing: $\tan^2\theta - \dfrac{48000}{22500}\tan\theta + \dfrac{24560}{22500} = 0$ | A1 | Correct equation in $\tan\theta$ |
| Solving quadratic: $\tan\theta = \dfrac{48000 \pm \sqrt{48000^2 - 4(22500)(24560)}}{2(22500)}$ | M1 | Using quadratic formula |
| $\tan\theta \approx 1.678$ or $\tan\theta \approx 0.655$ | A1 | Both correct values |
| $\theta \approx 59.2°$ or $\theta \approx 33.2°$ | A1 | Both angles correct |
## Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Smaller angle $\theta \approx 33.2°$ gives shortest time | B1 | Correct value chosen |
| Smaller $\theta$ gives larger horizontal component of velocity, so ball reaches $x = 150$ in less time | B1 | Valid reason |3 (In this question, use $g = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.)\\
A golf ball is hit from a point $O$ on a horizontal golf course with a velocity of $40 \mathrm {~ms} ^ { - 1 }$ at an angle of elevation $\theta$. The golf ball travels in a vertical plane through $O$. During its flight, the horizontal and upward vertical distances of the golf ball from $O$ are $x$ and $y$ metres respectively.
\begin{enumerate}[label=(\alph*)]
\item Show that the equation of the trajectory of the golf ball during its flight is given by
$$x ^ { 2 } \tan ^ { 2 } \theta - 320 x \tan \theta + \left( x ^ { 2 } + 320 y \right) = 0$$
\item \begin{enumerate}[label=(\roman*)]
\item The golf ball hits the top of a tree, which has a vertical height of 8 m and is at a horizontal distance of 150 m from $O$.
Find the two possible values of $\theta$.
\item Which value of $\theta$ gives the shortest possible time for the golf ball to travel from $O$ to the top of the tree? Give a reason for your choice of $\theta$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M3 2011 Q3 [13]}}