| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion with applied force on slope |
| Difficulty | Moderate -0.3 This is a standard M1 mechanics problem involving forces on an inclined plane with friction. It follows a routine template: resolve perpendicular to find normal reaction, calculate friction using μR, then resolve parallel to find acceleration. All steps are straightforward applications of Newton's laws with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03e Resolve forces: two dimensions3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Diagram with forces \(R\), \(T\), \(F\), \(0.7g\) correctly labelled | B1 | Accept \(W\) or \(mg\) (or 6.86) for weight; arrows and labels needed (can replace \(W\) with 2 correct components) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(R = 0.7g\cos 22°\) | M1 | Component of weight attempted |
| A1 | All correct, including signs | |
| \(R = 6.36\) N | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(F = 0.25 \times 6.36\) | M1 | |
| \(F = 1.59\) N | A1 | CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(5.6 - 0.7g\sin 22° - 1.59 = 0.7a\) | M1, A2 | 4 terms with weight component attempted; A marks \(-1\) each error, accept \(\pm 0.7a\) |
| \(a = 2.06 \text{ ms}^{-2}\) | A1F | FT one error, accept \(\pm\) |
## Question 4:
### Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| Diagram with forces $R$, $T$, $F$, $0.7g$ correctly labelled | B1 | Accept $W$ or $mg$ (or 6.86) for weight; arrows and labels needed (can replace $W$ with 2 correct components) | **Total: 1** |
### Part (b)
| Working | Marks | Guidance |
|---------|-------|----------|
| $R = 0.7g\cos 22°$ | M1 | Component of weight attempted |
| | A1 | All correct, including signs |
| $R = 6.36$ N | A1 | CAO | **Total: 3** |
### Part (c)
| Working | Marks | Guidance |
|---------|-------|----------|
| $F = 0.25 \times 6.36$ | M1 | |
| $F = 1.59$ N | A1 | CAO | **Total: 2** |
### Part (d)
| Working | Marks | Guidance |
|---------|-------|----------|
| $5.6 - 0.7g\sin 22° - 1.59 = 0.7a$ | M1, A2 | 4 terms with weight component attempted; A marks $-1$ each error, accept $\pm 0.7a$ |
| $a = 2.06 \text{ ms}^{-2}$ | A1F | FT one error, accept $\pm$ | **Total: 4** |
---4 A block is being pulled up a rough plane inclined at an angle of $22 ^ { \circ }$ to the horizontal by a rope parallel to the plane, as shown in the diagram.
The mass of the block is 0.7 kg , and the tension in the rope is $T$ newtons.\\
\includegraphics[max width=\textwidth, alt={}, center]{cfe0bdbc-35e3-485f-a922-b652a72f4c95-3_264_460_1649_779}
\begin{enumerate}[label=(\alph*)]
\item Draw a diagram to show the forces acting on the block.
\item Show that the normal reaction force between the block and the plane has magnitude 6.36 newtons, correct to three significant figures.
\item The coefficient of friction between the block and the plane is 0.25 . Find the magnitude of the frictional force acting on the block during its motion.
\item The tension in the rope is 5.6 newtons. Find the acceleration of the block.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2006 Q4 [10]}}