| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Projectiles |
| Type | Projectile on inclined plane |
| Difficulty | Challenging +1.2 This is a structured M3 projectile problem with guided steps using given identities. Part (a) requires resolving forces and deriving range (standard but multi-step), part (b) uses a provided identity to find maximum (calculus routine once set up), and part (c) requires recognizing perpendicular impact conditions. The guidance through identities and 'show that' format makes this more accessible than it initially appears, though it requires solid mechanics foundations and careful algebra. |
| 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 | Marks | Guidance |
| \(y = ut\sin\theta - \frac{1}{2}gt^2\cos\theta\) | M1A1 | |
| \(y = 0 \Rightarrow t = \frac{2u\sin\theta}{g\cos\alpha}\) | A1F | |
| \(x = ut\cos\theta - \frac{1}{2}gt^2\sin\alpha\) | M1A1 | |
| \(R = u\dfrac{2u\sin\theta}{g\cos\alpha}\cos\theta - \frac{1}{2}g\left(\dfrac{2u\sin\theta}{g\cos\alpha}\right)^2\sin\alpha\) | M1 | |
| \(R = \dfrac{2u^2\sin\theta\cos(\theta+\alpha)}{g\cos^2\alpha}\) | m1 A1 | 8 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(R = \dfrac{2u^2 \times \frac{1}{2}[\sin(2\theta+\alpha)+\sin(-\alpha)]}{g\cos^2\alpha}\) | B1 | |
| \(R\) is maximum when \(\sin(2\theta+\alpha) = 1\) | M1 | |
| or \(2\theta + \alpha = \dfrac{\pi}{2}\) | ||
| \(\therefore\quad \theta = \dfrac{\pi}{4} - \dfrac{\alpha}{2}\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(y = 0 \Rightarrow t = \dfrac{2u\sin\theta}{g\cos\alpha}\) | M1 A2,1 | For using \(y=0\) and \(\dot{x}=0\); A2 for both correct |
| \(\dot{x} = 0 \Rightarrow t = \dfrac{u\cos\theta}{g\sin\alpha}\) | ||
| \(\dfrac{2u\sin\theta}{g\cos\alpha} = \dfrac{u\cos\theta}{g\sin\alpha}\) | ||
| \(2\tan\theta = \cot\alpha\) | A1 | 4 marks |
# Question 7:
## Part 7(a):
| Working | Marks | Guidance |
|---------|-------|---------|
| $y = ut\sin\theta - \frac{1}{2}gt^2\cos\theta$ | M1A1 | |
| $y = 0 \Rightarrow t = \frac{2u\sin\theta}{g\cos\alpha}$ | A1F | |
| $x = ut\cos\theta - \frac{1}{2}gt^2\sin\alpha$ | M1A1 | |
| $R = u\dfrac{2u\sin\theta}{g\cos\alpha}\cos\theta - \frac{1}{2}g\left(\dfrac{2u\sin\theta}{g\cos\alpha}\right)^2\sin\alpha$ | M1 | |
| $R = \dfrac{2u^2\sin\theta\cos(\theta+\alpha)}{g\cos^2\alpha}$ | m1 A1 | 8 marks | Dependent on M1s; Answer given |
## Part 7(b):
| Working | Marks | Guidance |
|---------|-------|---------|
| $R = \dfrac{2u^2 \times \frac{1}{2}[\sin(2\theta+\alpha)+\sin(-\alpha)]}{g\cos^2\alpha}$ | B1 | |
| $R$ is maximum when $\sin(2\theta+\alpha) = 1$ | M1 | |
| or $2\theta + \alpha = \dfrac{\pi}{2}$ | | |
| $\therefore\quad \theta = \dfrac{\pi}{4} - \dfrac{\alpha}{2}$ | A1 | 3 marks | Answer given |
## Part 7(c):
| Working | Marks | Guidance |
|---------|-------|---------|
| $y = 0 \Rightarrow t = \dfrac{2u\sin\theta}{g\cos\alpha}$ | M1 A2,1 | For using $y=0$ and $\dot{x}=0$; A2 for both correct |
| $\dot{x} = 0 \Rightarrow t = \dfrac{u\cos\theta}{g\sin\alpha}$ | | |
| $\dfrac{2u\sin\theta}{g\cos\alpha} = \dfrac{u\cos\theta}{g\sin\alpha}$ | | |
| $2\tan\theta = \cot\alpha$ | A1 | 4 marks | Answer given; N.B. problem arose affecting marking — see Examination Report |7 A particle is projected from a point on a plane which is inclined at an angle $\alpha$ to the horizontal. The particle is projected up the plane with velocity $u$ at an angle $\theta$ above the plane. The motion of the particle is in a vertical plane containing a line of greatest slope of the inclined plane.\\
\includegraphics[max width=\textwidth, alt={}, center]{daea0765-041a-4569-a535-f90fe4708313-5_401_748_516_644}
\begin{enumerate}[label=(\alph*)]
\item Using the identity $\cos ( A + B ) = \cos A \cos B - \sin A \sin B$, show that the range up the plane is
$$\frac { 2 u ^ { 2 } \sin \theta \cos ( \theta + \alpha ) } { g \cos ^ { 2 } \alpha }$$
\item Hence, using the identity $2 \sin A \cos B = \sin ( A + B ) + \sin ( A - B )$, show that, as $\theta$ varies, the range up the plane is a maximum when $\theta = \frac { \pi } { 4 } - \frac { \alpha } { 2 }$.
\item Given that the particle strikes the plane at right angles, show that
$$2 \tan \theta = \cot \alpha$$
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2007 Q7 [15]}}