| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Modelling assumptions and refinements |
| Difficulty | Moderate -0.8 This is a straightforward M1 mechanics question requiring standard resolution of forces on a slope (part a), use of constant acceleration formula s=ut+½at² (part b), and recognition that friction reduces acceleration (part c). All techniques are routine with no problem-solving insight needed, making it easier than average A-level questions. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03f Weight: W=mg3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(4a = 4g\sin 40°\) | M1 | Resolving and application of Newton's second law. Allow \(\cos 40°\) |
| A1 | Correct expression | |
| \(a = g\sin 40° = 6.30 \text{ ms}^{-2}\) AG | A1 | Correct result from correct working. Must see 6.30 not 6.3. Just seeing \(g\sin 40° = 6.30 \text{ ms}^{-2}\) scores full marks. Use of \(g = 9.81\) gives 6.31, M1A1A0, but don't penalise again on the same script |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.9 = 0 + \frac{1}{2} \times a \times 0.6^2\) | M1 | Use of a constant acceleration equation to find \(a\), with \(s = 0.9\), \(u = 0\) and \(t = 0.6\) |
| A1 | Correct equation | |
| \(a = \frac{0.9 \times 2}{0.6^2} = 5 \text{ ms}^{-2}\) | A1 | Correct acceleration |
| ALT Method: | ||
| \(0.9 = \frac{1}{2}(0 + v) \times 0.6\), \(v = 3\) | No marks at this stage | |
| \(3 = 0 + 0.6a\) | (M1A1) | M1: Constant acceleration equation with \(u = 0\) and \(t = 0.6\). A1: Correct equation |
| \(a = 5 \text{ ms}^{-2}\) | (A1) | Correct acceleration |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| The acceleration is reduced because of air resistance or the fact that there is friction | B1 | Must mention air resistance/resistive forces or friction. Do not allow "air friction" |
| Total: 1 mark |
## Question 3:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4a = 4g\sin 40°$ | M1 | Resolving and application of Newton's second law. Allow $\cos 40°$ |
| | A1 | Correct expression |
| $a = g\sin 40° = 6.30 \text{ ms}^{-2}$ **AG** | A1 | Correct result from correct working. Must see 6.30 not 6.3. Just seeing $g\sin 40° = 6.30 \text{ ms}^{-2}$ scores full marks. Use of $g = 9.81$ gives 6.31, M1A1A0, but don't penalise again on the same script |
| **Total: 3 marks** | | |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.9 = 0 + \frac{1}{2} \times a \times 0.6^2$ | M1 | Use of a constant acceleration equation to find $a$, with $s = 0.9$, $u = 0$ and $t = 0.6$ |
| | A1 | Correct equation |
| $a = \frac{0.9 \times 2}{0.6^2} = 5 \text{ ms}^{-2}$ | A1 | Correct acceleration |
| **ALT Method:** | | |
| $0.9 = \frac{1}{2}(0 + v) \times 0.6$, $v = 3$ | | No marks at this stage |
| $3 = 0 + 0.6a$ | (M1A1) | M1: Constant acceleration equation with $u = 0$ and $t = 0.6$. A1: Correct equation |
| $a = 5 \text{ ms}^{-2}$ | (A1) | Correct acceleration |
| **Total: 3 marks** | | |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| The acceleration is reduced because of air resistance or the fact that there is friction | B1 | Must mention air resistance/resistive forces or friction. Do not allow "air friction" |
| **Total: 1 mark** | | |
---3 A box of mass 4 kg is held at rest on a plane inclined at an angle of $40 ^ { \circ }$ to the horizontal. The box is then released and slides down the plane.
\begin{enumerate}[label=(\alph*)]
\item A simple model assumes that the only forces acting on the box are its weight and the normal reaction from the plane. Show that, according to this simple model, the acceleration of the box would be $6.30 \mathrm {~m} \mathrm {~s} ^ { - 2 }$, correct to three significant figures.
\item In fact, the box moves down the plane with constant acceleration and travels 0.9 metres in 0.6 seconds. By using this information, find the acceleration of the box.
\item Explain why the answer to part (b) is less than the answer to part (a).
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2009 Q3 [7]}}