| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile on inclined plane |
| Difficulty | Standard +0.8 This is a challenging M3 projectile motion problem requiring coordinate geometry with an inclined plane. Part (a) is standard trajectory derivation (5 marks), but part (b) requires finding the intersection of the parabolic path with a 45° inclined plane 6m away, involving simultaneous equations and careful geometric reasoning. The multi-step nature, geometric complexity, and need to work with the inclined plane coordinate system make this significantly harder than average A-level questions, though it remains within the M3 syllabus scope. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships3.02h Motion under gravity: vector form3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 14\cos 30° \cdot t\) and \(y = 14\sin 30° \cdot t - \frac{1}{2}gt^2\) | B1 B1 | B1: Correct horizontal eqn. B1: Correct vertical eqn. |
| \(t = \frac{x}{14\cos 30°}\) | M1 | M1: Making \(t\) the subject of \(x\) |
| \(y = 14\sin 30° \times \frac{x}{14\cos 30°} - \frac{1}{2}(0.8)\left(\frac{x}{14\cos 30°}\right)^2\) | dM1 | dM1: Elimination of \(t\) from their \(y\) |
| \(y = x\tan 30° - \frac{x^2}{40\cos^2 30°}\) and \(y = \frac{x\sqrt{3}}{3} - \frac{x^2}{30}\) | A1 | A1: CSO, AG must see the line above final answer, OE |
| \(y = x - 6\) | M1 | M1: For seeing \(y = x \pm 6\) |
| \(x - 6 = \frac{x\sqrt{3}}{3} - \frac{x^2}{30}\) | dM1 | dM1: Substituting \(x \pm 6\) into the given eqn. |
| \(x^2 + (30 - 10\sqrt{3})x - 180 = 0\) | A1 | A1: Correct simplified quadratic |
| \(x = \frac{-(30-10\sqrt{3}) \pm \sqrt{(30-10\sqrt{3})^2 - 4 \times 1(-180)}}{2 \times 1}\) | dM1 | dM1: Solving quadratic |
| \(x = 8.50\) (Accept \(8.5\) or AWRT \(8.50\); \(x = -21.2\) or exact equivalent not needed) | A1 | A1: CAO, accept 8.5 or AWRT 8.50 |
| \(PQ = \frac{8.499 - 6}{\cos 45°}\) | dM1 | dM1: Correct exp. for \(PQ\), FT their \(+ve\) x-value |
| \(PQ = 3.53\) or \(\frac{5\sqrt{2}}{2} \text{ (m)}\) | A1 | A1: CAO, AWRT 3.53 or exact value, allow 3.54 |
| Answer | Marks | Guidance |
|---|---|---|
| Let \(PQ = d\). Then \(x = 6 + d\cos 45°\) and \(y = d\sin 45°\) | M1 M1 A1 | M1: Expression for \(x\) in terms of \(d\). M1: Expression for \(y\) in terms of \(d\). A1: Both correct |
| \(d\sin 45° = (6 + d\cos 45°)\tan 30° - \frac{(6 + d\cos 45°)^2}{30}\) | dM1 | dM1: Substituting \(x \pm 6\) into given expression |
| \(d^2\cos^2 45° + (30\sin 45° - 30\cos 45° \tan 30° + 12\cos 45°)d - (180\tan 30° - 36) = 0\) | A1 | A1: Correct simplified quadratic |
| \(d = \frac{-(30\sin 45° - 30\cos 45° \tan 30° + 12\cos 45°) \pm \sqrt{(30\sin 45° - 30\cos 45° \tan 30° + 12\cos 45°)^2 - 4(\cos^2 45°)(36 - 180\tan 30°)}}{2\cos^2 45°}\) | dM1 | dM1: Solution of their quadratic eqn. |
| \(d = 3.53 \text{ m}\) (Allow \(3.54\) m) | A1 | A1: CAO, AWRT 3.53 or exact value, allow 3.54 |
**(a)**
| $x = 14\cos 30° \cdot t$ and $y = 14\sin 30° \cdot t - \frac{1}{2}gt^2$ | B1 B1 | B1: Correct horizontal eqn. B1: Correct vertical eqn. |
| $t = \frac{x}{14\cos 30°}$ | M1 | M1: Making $t$ the subject of $x$ |
| $y = 14\sin 30° \times \frac{x}{14\cos 30°} - \frac{1}{2}(0.8)\left(\frac{x}{14\cos 30°}\right)^2$ | dM1 | dM1: Elimination of $t$ from their $y$ |
| $y = x\tan 30° - \frac{x^2}{40\cos^2 30°}$ and $y = \frac{x\sqrt{3}}{3} - \frac{x^2}{30}$ | A1 | A1: CSO, AG must see the line above final answer, OE |
| $y = x - 6$ | M1 | M1: For seeing $y = x \pm 6$ |
| $x - 6 = \frac{x\sqrt{3}}{3} - \frac{x^2}{30}$ | dM1 | dM1: Substituting $x \pm 6$ into the given eqn. |
| $x^2 + (30 - 10\sqrt{3})x - 180 = 0$ | A1 | A1: Correct simplified quadratic |
| $x = \frac{-(30-10\sqrt{3}) \pm \sqrt{(30-10\sqrt{3})^2 - 4 \times 1(-180)}}{2 \times 1}$ | dM1 | dM1: Solving quadratic |
| $x = 8.50$ (Accept $8.5$ or AWRT $8.50$; $x = -21.2$ or exact equivalent not needed) | A1 | A1: CAO, accept 8.5 or AWRT 8.50 |
| $PQ = \frac{8.499 - 6}{\cos 45°}$ | dM1 | dM1: Correct exp. for $PQ$, FT their $+ve$ x-value |
| $PQ = 3.53$ or $\frac{5\sqrt{2}}{2} \text{ (m)}$ | A1 | A1: CAO, AWRT 3.53 or exact value, allow 3.54 |
**Total: 12 marks**
---
## Alternative Solution for Question 3(b)
| Let $PQ = d$. Then $x = 6 + d\cos 45°$ and $y = d\sin 45°$ | M1 M1 A1 | M1: Expression for $x$ in terms of $d$. M1: Expression for $y$ in terms of $d$. A1: Both correct |
| $d\sin 45° = (6 + d\cos 45°)\tan 30° - \frac{(6 + d\cos 45°)^2}{30}$ | dM1 | dM1: Substituting $x \pm 6$ into given expression |
| $d^2\cos^2 45° + (30\sin 45° - 30\cos 45° \tan 30° + 12\cos 45°)d - (180\tan 30° - 36) = 0$ | A1 | A1: Correct simplified quadratic |
| $d = \frac{-(30\sin 45° - 30\cos 45° \tan 30° + 12\cos 45°) \pm \sqrt{(30\sin 45° - 30\cos 45° \tan 30° + 12\cos 45°)^2 - 4(\cos^2 45°)(36 - 180\tan 30°)}}{2\cos^2 45°}$ | dM1 | dM1: Solution of their quadratic eqn. |
| $d = 3.53 \text{ m}$ (Allow $3.54$ m) | A1 | A1: CAO, AWRT 3.53 or exact value, allow 3.54 |
---A ball is projected from a point $O$ on horizontal ground with speed $14 \text{ m s}^{-1}$ at an angle of elevation $30°$ above the horizontal. The ball travels in a vertical plane through the point $O$ and hits a point $Q$ on a plane which is inclined at $45°$ to the horizontal. The point $O$ is $6$ metres from $P$, the foot of the inclined plane, as shown in the diagram. The points $O$, $P$ and $Q$ lie in the same vertical plane. The line $PQ$ is a line of greatest slope of the inclined plane.
\includegraphics{figure_3}
\begin{enumerate}[label=(\alph*)]
\item During its flight, the horizontal and upward vertical distances of the ball from $O$ are $x$ metres and $y$ metres respectively.
Show that $x$ and $y$ satisfy the equation
$$y = x\frac{\sqrt{3}}{3} - \frac{x^2}{30}$$
Use $\cos 30° = \frac{\sqrt{3}}{2}$ and $\tan 30° = \frac{\sqrt{3}}{3}$. [5 marks]
\item Find the distance $PQ$. [7 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2016 Q3 [12]}}