| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile on inclined plane |
| Difficulty | Challenging +1.2 This is a standard M3 projectile-on-inclined-plane problem requiring coordinate transformation and range formula application. While it involves multiple steps (resolving in tilted coordinates, applying kinematic equations, finding maximum perpendicular distance), these are well-practiced techniques for M3 students with no novel insight required. The problem is slightly above average difficulty due to the inclined plane context and two-part structure, but follows a predictable template. |
| Spec | 3.02i Projectile motion: constant acceleration model |
| Answer | Marks |
|---|---|
| Axes taken along and perpendicular to plane | |
| Along plane: \(x = u\cos30° \cdot t - \frac{1}{2}g\sin20° \cdot t^2\) | M1 A1 |
| Perp to plane: \(y = u\sin30° \cdot t - \frac{1}{2}g\cos20° \cdot t^2\) | M1 A1 |
| At \(T\): \(y = 0 \Rightarrow t = \frac{2u\sin30°}{g\cos20°} = \frac{u}{g\cos20°}\) | M1 |
| \(OT = 200\): \(200 = u\cos30° \cdot \frac{u}{g\cos20°} - \frac{1}{2}g\sin20° \cdot \left(\frac{u}{g\cos20°}\right)^2\) | M1 |
| \(u \approx 47.0 \text{ m s}^{-1}\) | A1 |
| Answer | Marks |
|---|---|
| Max perpendicular distance when \(\frac{dy}{dt} = 0\) | M1 |
| \(u\sin30° = g\cos20° \cdot t \Rightarrow t = \frac{u\sin30°}{g\cos20°}\) | A1 |
| \(y_{max} = \frac{u^2\sin^230°}{2g\cos20°}\) | M1 |
| \(y_{max} \approx 30.1 \text{ m}\) | A1 |
## Question 6:
### Part (a):
| Axes taken along and perpendicular to plane | | |
| Along plane: $x = u\cos30° \cdot t - \frac{1}{2}g\sin20° \cdot t^2$ | M1 A1 | |
| Perp to plane: $y = u\sin30° \cdot t - \frac{1}{2}g\cos20° \cdot t^2$ | M1 A1 | |
| At $T$: $y = 0 \Rightarrow t = \frac{2u\sin30°}{g\cos20°} = \frac{u}{g\cos20°}$ | M1 | |
| $OT = 200$: $200 = u\cos30° \cdot \frac{u}{g\cos20°} - \frac{1}{2}g\sin20° \cdot \left(\frac{u}{g\cos20°}\right)^2$ | M1 | |
| $u \approx 47.0 \text{ m s}^{-1}$ | A1 | |
### Part (b):
| Max perpendicular distance when $\frac{dy}{dt} = 0$ | M1 | |
| $u\sin30° = g\cos20° \cdot t \Rightarrow t = \frac{u\sin30°}{g\cos20°}$ | A1 | |
| $y_{max} = \frac{u^2\sin^230°}{2g\cos20°}$ | M1 | |
| $y_{max} \approx 30.1 \text{ m}$ | A1 | |6 A projectile is fired from a point $O$ on a plane which is inclined at an angle of $20 ^ { \circ }$ to the horizontal. The projectile is fired up the plane with velocity $u \mathrm {~ms} ^ { - 1 }$ at an angle of $30 ^ { \circ }$ to the inclined plane. The projectile travels in a vertical plane containing a line of greatest slope of the inclined plane.
The projectile hits a target $T$ on the inclined plane.\\
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\begin{enumerate}[label=(\alph*)]
\item Given that $O T = 200 \mathrm {~m}$, determine the value of $u$.
\item Find the greatest perpendicular distance of the projectile from the inclined plane.\\
(4 marks)\\
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{0590950d-145c-4ae2-bc3c-f61a9433d158-18_2486_1714_221_153}
\end{center}
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{0590950d-145c-4ae2-bc3c-f61a9433d158-19_2486_1714_221_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2011 Q6 [11]}}