| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion down smooth slope |
| Difficulty | Standard +0.8 This M3 projectile motion question requires resolving forces on an inclined plane, deriving time of flight from perpendicular motion, then using trigonometric identities to simplify the range expression. While systematic, it demands careful coordinate resolution, algebraic manipulation, and non-trivial use of the given identity—more challenging than standard projectile questions but follows established M3 techniques. |
| Spec | 1.05l Double angle formulae: and compound angle formulae3.02i Projectile motion: constant acceleration model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Perpendicular to plane: \(0 = u\sin\alpha \cdot T - \frac{1}{2}g\cos\theta \cdot T^2\) | M1 | Resolving perpendicular to inclined plane, equation of motion |
| \(0 = u\sin\alpha - \frac{1}{2}g\cos\theta \cdot T\) | A1 | Correct equation (dividing by \(T\)) |
| \(u = \frac{gT\cos\theta}{2\sin\alpha}\) | A1 | Correct expression for \(u\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Along the plane: \(OP = u\cos\alpha \cdot T + \frac{1}{2}g\sin\theta \cdot T^2\) | M1 | Resolving along plane, displacement equation |
| Substituting \(T = \frac{2u\sin\alpha}{g\cos\theta}\) | M1 | Substituting their \(T\) from part (a) |
| \(OP = u\cos\alpha \cdot \frac{2u\sin\alpha}{g\cos\theta} + \frac{1}{2}g\sin\theta \cdot \frac{4u^2\sin^2\alpha}{g^2\cos^2\theta}\) | A1 | Correct substitution |
| \(OP = \frac{2u^2\sin\alpha\cos\alpha}{g\cos\theta} + \frac{2u^2\sin^2\alpha\sin\theta}{g\cos^2\theta}\) | A1 | Correct expansion |
| \(OP = \frac{2u^2\sin\alpha}{g\cos^2\theta}(\cos\alpha\cos\theta + \sin\alpha\sin\theta)\) | M1 | Factoring, using identity \(\cos(X-Y) = \cos X\cos Y + \sin X\sin Y\) |
| \(OP = \frac{2u^2\sin\alpha\cos(\alpha-\theta)}{g\cos^2\theta}\) | A1 | Correct completion of proof |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Perpendicular to plane: $0 = u\sin\alpha \cdot T - \frac{1}{2}g\cos\theta \cdot T^2$ | M1 | Resolving perpendicular to inclined plane, equation of motion |
| $0 = u\sin\alpha - \frac{1}{2}g\cos\theta \cdot T$ | A1 | Correct equation (dividing by $T$) |
| $u = \frac{gT\cos\theta}{2\sin\alpha}$ | A1 | Correct expression for $u$ |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Along the plane: $OP = u\cos\alpha \cdot T + \frac{1}{2}g\sin\theta \cdot T^2$ | M1 | Resolving along plane, displacement equation |
| Substituting $T = \frac{2u\sin\alpha}{g\cos\theta}$ | M1 | Substituting their $T$ from part (a) |
| $OP = u\cos\alpha \cdot \frac{2u\sin\alpha}{g\cos\theta} + \frac{1}{2}g\sin\theta \cdot \frac{4u^2\sin^2\alpha}{g^2\cos^2\theta}$ | A1 | Correct substitution |
| $OP = \frac{2u^2\sin\alpha\cos\alpha}{g\cos\theta} + \frac{2u^2\sin^2\alpha\sin\theta}{g\cos^2\theta}$ | A1 | Correct expansion |
| $OP = \frac{2u^2\sin\alpha}{g\cos^2\theta}(\cos\alpha\cos\theta + \sin\alpha\sin\theta)$ | M1 | Factoring, using identity $\cos(X-Y) = \cos X\cos Y + \sin X\sin Y$ |
| $OP = \frac{2u^2\sin\alpha\cos(\alpha-\theta)}{g\cos^2\theta}$ | A1 | Correct completion of proof |
---5 A particle is projected from a point $O$ on a plane which is inclined at an angle $\theta$ to the horizontal. The particle is projected down the plane with velocity $u$ at an angle $\alpha$ above the plane. The particle first strikes the plane at a point $P$, as shown in the diagram. 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]{3a1726d9-1b0c-41de-8b43-56019e18aac1-12_389_789_557_639}
\begin{enumerate}[label=(\alph*)]
\item Given that the time of flight from $O$ to $P$ is $T$, find an expression for $u$ in terms of $\theta , \alpha , T$ and $g$.
\item Using the identity $\cos ( X - Y ) = \cos X \cos Y + \sin X \sin Y$, show that the distance $O P$ is given by $\frac { 2 u ^ { 2 } \sin \alpha \cos ( \alpha - \theta ) } { g \cos ^ { 2 } \theta }$.\\
(6 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2013 Q5 [10]}}