| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile on inclined plane |
| Difficulty | Challenging +1.8 This is a challenging M3 projectile problem requiring coordinate transformation to an inclined plane, multi-step derivation involving trigonometric identities, and optimization. While the question provides helpful identity hints and is structured with clear parts, it demands sophisticated manipulation of projectile equations in a non-standard coordinate system and extended algebraic reasoning across multiple steps—significantly harder than routine mechanics questions but within reach for well-prepared M3 students. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(v_y^2 = u^2\sin^2\theta - 2g\cos\alpha \cdot y\) | M1 A1 | |
| \(0 = u^2\sin^2\theta - 2g\cos\alpha \cdot y_{\max}\) | m1 | |
| \(y_{\max} = \frac{u^2\sin^2\theta}{2g\cos\alpha}\) | A1F | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(0 = u\sin\theta - g\cos\alpha \cdot t\) | M1 | |
| \(t = \frac{u\sin\theta}{g\cos\alpha}\) | A1 | |
| \(y_{\max} = u\sin\theta\left(\frac{u\sin\theta}{g\cos\alpha}\right) - \frac{1}{2}g\cos\alpha\left(\frac{u\sin\theta}{g\cos\alpha}\right)^2\) | m1 | |
| \(y_{\max} = \frac{u^2\sin^2\theta}{2g\cos\alpha}\) | A1F | 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(u\sin\theta \cdot t - \frac{1}{2}g\cos(\alpha)t^2 = 0\) | M1 | |
| \(t = \frac{2u\sin\theta}{g\cos\alpha}\) | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(x = u\cos\theta \cdot t - \frac{1}{2}g\sin(-\alpha)t^2\) | M1 A1 | |
| \(R = u\cos\theta\left(\frac{2u\sin\theta}{g\cos\alpha}\right) + \frac{1}{2}g\sin\alpha\left(\frac{2u\sin\theta}{g\cos\alpha}\right)^2\) | M1 | |
| \(= \frac{2u^2\cos\theta\sin\theta\cos\alpha + 2u^2\sin\alpha\sin^2\theta}{g\cos^2\alpha}\) | m1 | Dependent on both M1s |
| \(= \frac{2u^2\sin\theta(\cos\theta\cos\alpha + \sin\theta\sin\alpha)}{g\cos^2\alpha}\) | A1F | |
| \(= \frac{2u^2\sin\theta\cos(\theta-\alpha)}{g\cos^2\alpha}\) | A1 | 6 marks Answer given |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\overline{OP} = \frac{2u^2\sin\theta\cos(\theta-\alpha)}{g\cos^2\alpha}\) | ||
| \(= \frac{2u^2 \cdot \frac{1}{2}[\sin(2\theta-\alpha)+\sin\alpha]}{g\cos^2\alpha}\) | M1 A1 | |
| \(\overline{OP}\) is max when \(\sin(2\theta - \alpha) = 1\) | M1 | |
| \(\overline{OP}_{\max} = \frac{u^2(1+\sin\alpha)}{g\cos^2\alpha}\) | A1F | |
| \(\overline{OP}_{\max} = \frac{u^2(1+\sin\alpha)}{g(1-\sin^2\alpha)}\) | ||
| \(\overline{OP}_{\max} = \frac{u^2}{g(1-\sin\alpha)}\) | A1 | 5 marks Answer given |
## Question 7(a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $v_y^2 = u^2\sin^2\theta - 2g\cos\alpha \cdot y$ | M1 A1 | |
| $0 = u^2\sin^2\theta - 2g\cos\alpha \cdot y_{\max}$ | m1 | |
| $y_{\max} = \frac{u^2\sin^2\theta}{2g\cos\alpha}$ | A1F | **4 marks** |
**Alternative method:**
| Working/Answer | Marks | Guidance |
|---|---|---|
| $0 = u\sin\theta - g\cos\alpha \cdot t$ | M1 | |
| $t = \frac{u\sin\theta}{g\cos\alpha}$ | A1 | |
| $y_{\max} = u\sin\theta\left(\frac{u\sin\theta}{g\cos\alpha}\right) - \frac{1}{2}g\cos\alpha\left(\frac{u\sin\theta}{g\cos\alpha}\right)^2$ | m1 | |
| $y_{\max} = \frac{u^2\sin^2\theta}{2g\cos\alpha}$ | A1F | **4 marks** |
## Question 7(b)(i):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $u\sin\theta \cdot t - \frac{1}{2}g\cos(\alpha)t^2 = 0$ | M1 | |
| $t = \frac{2u\sin\theta}{g\cos\alpha}$ | A1 | **2 marks** |
## Question 7(b)(ii):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $x = u\cos\theta \cdot t - \frac{1}{2}g\sin(-\alpha)t^2$ | M1 A1 | |
| $R = u\cos\theta\left(\frac{2u\sin\theta}{g\cos\alpha}\right) + \frac{1}{2}g\sin\alpha\left(\frac{2u\sin\theta}{g\cos\alpha}\right)^2$ | M1 | |
| $= \frac{2u^2\cos\theta\sin\theta\cos\alpha + 2u^2\sin\alpha\sin^2\theta}{g\cos^2\alpha}$ | m1 | Dependent on both M1s |
| $= \frac{2u^2\sin\theta(\cos\theta\cos\alpha + \sin\theta\sin\alpha)}{g\cos^2\alpha}$ | A1F | |
| $= \frac{2u^2\sin\theta\cos(\theta-\alpha)}{g\cos^2\alpha}$ | A1 | **6 marks** Answer given |
## Question 7(b)(iii):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\overline{OP} = \frac{2u^2\sin\theta\cos(\theta-\alpha)}{g\cos^2\alpha}$ | | |
| $= \frac{2u^2 \cdot \frac{1}{2}[\sin(2\theta-\alpha)+\sin\alpha]}{g\cos^2\alpha}$ | M1 A1 | |
| $\overline{OP}$ is max when $\sin(2\theta - \alpha) = 1$ | M1 | |
| $\overline{OP}_{\max} = \frac{u^2(1+\sin\alpha)}{g\cos^2\alpha}$ | A1F | |
| $\overline{OP}_{\max} = \frac{u^2(1+\sin\alpha)}{g(1-\sin^2\alpha)}$ | | |
| $\overline{OP}_{\max} = \frac{u^2}{g(1-\sin\alpha)}$ | A1 | **5 marks** Answer given |7 A projectile is fired with speed $u$ from a point $O$ on a plane which is inclined at an angle $\alpha$ to the horizontal. The projectile is fired at an angle $\theta$ to the inclined plane and moves in a vertical plane through a line of greatest slope of the inclined plane. The projectile lands at a point $P$, lower down the inclined plane, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{eed9842d-cd89-4eb7-b5ba-9380971be196-5_415_1098_495_463}
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $u , g , \theta$ and $\alpha$, the greatest perpendicular distance of the projectile from the plane.
\item \begin{enumerate}[label=(\roman*)]
\item Find, in terms of $u , g , \theta$ and $\alpha$, the time of flight from $O$ to $P$.
\item By using the identity $\cos A \cos B + \sin A \sin B = \cos ( A - B )$, show that the distance $O P$ is given by $\frac { 2 u ^ { 2 } \sin \theta \cos ( \theta - \alpha ) } { g \cos ^ { 2 } \alpha }$.
\item Hence, by using the identity $2 \sin A \cos B = \sin ( A + B ) + \sin ( A - B )$ or otherwise, show that, as $\theta$ varies, the maximum possible distance $O P$ is $\frac { u ^ { 2 } } { g ( 1 - \sin \alpha ) }$.\\
(5 marks)
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M3 2008 Q7 [17]}}