| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | June |
| Marks | 16 |
| 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 the trajectory equation, solving a quadratic in tan θ, and finding final velocity direction. While it involves multiple steps and algebraic manipulation, all techniques are routine for this level with no novel problem-solving required. The 'show that' in part (b)(i) provides the answer, making it slightly easier than average. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=13.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(x = u\cos\theta \cdot t \Rightarrow t = \frac{x}{u\cos\theta}\) | M1 | Horizontal equation of motion |
| \(y = u\sin\theta \cdot t - \frac{1}{2}gt^2\) | M1 | Vertical equation of motion |
| Substitute \(t\): \(y = u\sin\theta \cdot \frac{x}{u\cos\theta} - \frac{1}{2}g\left(\frac{x}{u\cos\theta}\right)^2\) | M1 A1 | Substitution |
| \(y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}\) | A1 | |
| \(y = x\tan\theta - \frac{gx^2(1+\tan^2\theta)}{2u^2}\) | A1 | Using \(\sec^2\theta = 1+\tan^2\theta\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Substitute \(x=5\), \(y=0.5\), \(u=8\), \(g=9.8\) | M1 | |
| \(0.5 = 5\tan\theta - \frac{9.8 \times 25(1+\tan^2\theta)}{128}\) | A1 | |
| \(1.909\tan^2\theta - 5\tan\theta + 2.409 = 0\) | M1 | Form quadratic in \(\tan\theta\) |
| \(\tan\theta = \frac{5 \pm \sqrt{25 - 4(1.909)(2.409)}}{2(1.909)}\) | M1 | Apply quadratic formula |
| \(\tan\theta = 2.045...\) or \(0.643...\) giving \(\theta \approx 63.1°\) or \(32.6°\) | A1 | Both values confirmed to 3 s.f. |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(v_x = 8\cos 63.1° = 3.613...\) | M1 | Horizontal component of velocity at basket |
| \(v_y = 8\sin 63.1° - 9.8t\) where \(t = \frac{5}{v_x}\) | M1 | Vertical component |
| \(t = \frac{5}{3.613} = 1.384\) s | A1 | |
| \(v_y = 8\sin 63.1° - 9.8(1.384) = -6.42\) | A1 | |
| \(\alpha = \arctan\left(\frac{6.42}{3.613}\right) \approx 60°\) below horizontal | A1 | Direction stated clearly |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| The basketball is modelled as a particle (air resistance ignored / no spin) | B1 | Accept reasonable modelling assumption |
# Question 3:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $x = u\cos\theta \cdot t \Rightarrow t = \frac{x}{u\cos\theta}$ | M1 | Horizontal equation of motion |
| $y = u\sin\theta \cdot t - \frac{1}{2}gt^2$ | M1 | Vertical equation of motion |
| Substitute $t$: $y = u\sin\theta \cdot \frac{x}{u\cos\theta} - \frac{1}{2}g\left(\frac{x}{u\cos\theta}\right)^2$ | M1 A1 | Substitution |
| $y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$ | A1 | |
| $y = x\tan\theta - \frac{gx^2(1+\tan^2\theta)}{2u^2}$ | A1 | Using $\sec^2\theta = 1+\tan^2\theta$ |
## Part (b)(i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Substitute $x=5$, $y=0.5$, $u=8$, $g=9.8$ | M1 | |
| $0.5 = 5\tan\theta - \frac{9.8 \times 25(1+\tan^2\theta)}{128}$ | A1 | |
| $1.909\tan^2\theta - 5\tan\theta + 2.409 = 0$ | M1 | Form quadratic in $\tan\theta$ |
| $\tan\theta = \frac{5 \pm \sqrt{25 - 4(1.909)(2.409)}}{2(1.909)}$ | M1 | Apply quadratic formula |
| $\tan\theta = 2.045...$ or $0.643...$ giving $\theta \approx 63.1°$ or $32.6°$ | A1 | Both values confirmed to 3 s.f. |
## Part (b)(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $v_x = 8\cos 63.1° = 3.613...$ | M1 | Horizontal component of velocity at basket |
| $v_y = 8\sin 63.1° - 9.8t$ where $t = \frac{5}{v_x}$ | M1 | Vertical component |
| $t = \frac{5}{3.613} = 1.384$ s | A1 | |
| $v_y = 8\sin 63.1° - 9.8(1.384) = -6.42$ | A1 | |
| $\alpha = \arctan\left(\frac{6.42}{3.613}\right) \approx 60°$ below horizontal | A1 | Direction stated clearly |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| The basketball is modelled as a particle (air resistance ignored / no spin) | B1 | Accept reasonable modelling assumption |3 A player projects a basketball with speed $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at an angle $\theta$ above the horizontal. The basketball travels in a vertical plane through the point of projection and goes into the basket. During the motion, the horizontal and upward vertical displacements of the basketball from the point of projection are $x$ metres and $y$ metres respectively.\\
\includegraphics[max width=\textwidth, alt={}, center]{3a1726d9-1b0c-41de-8b43-56019e18aac1-06_737_937_513_550}
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $y$ in terms of $x , u , g$ and $\tan \theta$.
\item The player projects the basketball with speed $8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point 0.5 metres vertically below and 5 metres horizontally from the basket.
\begin{enumerate}[label=(\roman*)]
\item Show that the two possible values of $\theta$ are approximately $63.1 ^ { \circ }$ and $32.6 ^ { \circ }$, correct to three significant figures.
\item Given that the player projects the basketball at $63.1 ^ { \circ }$ to the horizontal, find the direction of the motion of the basketball as it enters the basket. Give your answer to the nearest degree.
\end{enumerate}\item State a modelling assumption needed for answering parts (a) and (b) of this question.\\
(1 mark)
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2013 Q3 [16]}}