| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Particle on inclined plane |
| Difficulty | Moderate -0.8 This is a straightforward SUVAT application with standard kinematics equations (s=ut+½at², v²=u²+2as) followed by basic resolution of forces (a=g sin θ). All steps are routine M1 procedures with no problem-solving insight required, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| \(1.08 = \frac{1}{2}a \times 1.2^2\), \(a = \frac{1.08 \times 2}{1.44} = 1.5 \text{ m s}^{-2}\) | M1, M1, A1 | Using \(s = ut + \frac{1}{2}at^2\) with \(u = 0\), \(s = 1.08\) and \(t = 1.2\). Solving for \(a\). Correct acceleration. Accept \(\frac{3}{2}\) or 1.50. |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = 0 + 1.5 \times 1.2 = 1.8 \text{ m s}^{-1}\) OR \(1.08 = \frac{1}{2}(0 + v) \times 1.2\), \(v = \frac{2 \times 1.08}{1.2} = 1.8 \text{ m s}^{-1}\) OR \(v = \sqrt{2 \times 1.5 \times 1.08} = 1.8 \text{ m s}^{-1}\) | M1, A1 | Use of a constant acceleration equation to find \(v\), with \(t = 1.2\) or \(s = 1.08\). Correct speed. Accept 1.80. |
| Answer | Marks | Guidance |
|---|---|---|
| Resolving parallel to slope: \(mg \sin \alpha = ma\), \(g \sin \alpha = 1.5\), \(\alpha = \sin^{-1}\left(\frac{1.5}{9.8}\right) = 8.80° \approx 9°\) | M1, A1; dM1, A1 | M1: Resolving parallel to the slope to obtain a two term equation, with their acceleration from part (a). Allow cos instead of sin. Must include \(\pm g\). A1: Correct equation. May include \(m\). dM1: Solving for \(\alpha\) to obtain an angle. A1: Correct angle to nearest degree. Allow methods based on \(\sin \alpha = \frac{1.5k}{9.8k}\) where \(k\) can have any value. Using \(g = 9.81\) gives the same angle. |
**3. (a)**
$1.08 = \frac{1}{2}a \times 1.2^2$, $a = \frac{1.08 \times 2}{1.44} = 1.5 \text{ m s}^{-2}$ | M1, M1, A1 | Using $s = ut + \frac{1}{2}at^2$ with $u = 0$, $s = 1.08$ and $t = 1.2$. Solving for $a$. Correct acceleration. Accept $\frac{3}{2}$ or 1.50. | 3 marks
**3. (b)**
$v = 0 + 1.5 \times 1.2 = 1.8 \text{ m s}^{-1}$ OR $1.08 = \frac{1}{2}(0 + v) \times 1.2$, $v = \frac{2 \times 1.08}{1.2} = 1.8 \text{ m s}^{-1}$ OR $v = \sqrt{2 \times 1.5 \times 1.08} = 1.8 \text{ m s}^{-1}$ | M1, A1 | Use of a constant acceleration equation to find $v$, with $t = 1.2$ or $s = 1.08$. Correct speed. Accept 1.80. | 2 marks
**3. (c)**
Resolving parallel to slope: $mg \sin \alpha = ma$, $g \sin \alpha = 1.5$, $\alpha = \sin^{-1}\left(\frac{1.5}{9.8}\right) = 8.80° \approx 9°$ | M1, A1; dM1, A1 | M1: Resolving parallel to the slope to obtain a two term equation, with their acceleration from part (a). Allow cos instead of sin. Must include $\pm g$. A1: Correct equation. May include $m$. dM1: Solving for $\alpha$ to obtain an angle. A1: Correct angle to nearest degree. Allow methods based on $\sin \alpha = \frac{1.5k}{9.8k}$ where $k$ can have any value. Using $g = 9.81$ gives the same angle. | 4 marks
**Total for Question 3: 9 marks**
---3 A toy car is placed at the top of a ramp. After the car has been released from rest, it travels a distance of 1.08 metres down the ramp, in a time of 1.2 seconds.
Assume that there is no resistance to the motion of the car.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the acceleration of the car while it is moving down the ramp.
\item Find the speed of the car, when it has travelled 1.08 metres down the ramp.
\item Find the angle between the ramp and the horizontal, giving your answer to the nearest degree.\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2016 Q3 [4]}}