| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion down rough slope |
| Difficulty | Moderate -0.8 This is a standard M1 mechanics question testing routine application of Newton's laws on an inclined plane. All parts follow textbook procedures: resolving forces perpendicular and parallel to the slope, applying F=μR, and using F=ma. No problem-solving insight required, just methodical application of learned techniques with straightforward numerical values. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03d Newton's second law: 2D vectors3.03e Resolve forces: two dimensions3.03t Coefficient of friction: F <= mu*R model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Diagram showing: Weight \(mg\) (downward), Normal reaction \(N\) (perpendicular to slope), Friction \(F\) (up the slope) | B1 | All three forces correctly labelled and directed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(N = mg\cos 40°\) | M1 | Resolving perpendicular to slope |
| \(N = 3 \times 9.8 \times \cos 40° \approx 22.5\) N | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F = \mu N = 0.2 \times N\) | M1 | Use of \(F=\mu N\) |
| \(F = 0.2 \times 22.5 \approx 4.50\) N | A1 | ft their (b) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(ma = mg\sin 40° - F\) | M1 | Newton's second law along slope |
| \(3a = 3(9.8)\sin 40° - 4.50\) | M1 | Correct substitution |
| \(a \approx 4.80 \text{ ms}^{-2}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| The box is modelled as a particle (no air resistance/box treated as point mass) | B1 | Accept any valid assumption e.g. no air resistance |
# Question 3:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Diagram showing: Weight $mg$ (downward), Normal reaction $N$ (perpendicular to slope), Friction $F$ (up the slope) | B1 | All three forces correctly labelled and directed |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $N = mg\cos 40°$ | M1 | Resolving perpendicular to slope |
| $N = 3 \times 9.8 \times \cos 40° \approx 22.5$ N | A1 | cao |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F = \mu N = 0.2 \times N$ | M1 | Use of $F=\mu N$ |
| $F = 0.2 \times 22.5 \approx 4.50$ N | A1 | ft their (b) |
## Part (d)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $ma = mg\sin 40° - F$ | M1 | Newton's second law along slope |
| $3a = 3(9.8)\sin 40° - 4.50$ | M1 | Correct substitution |
| $a \approx 4.80 \text{ ms}^{-2}$ | A1 | cao |
## Part (e)
| Answer/Working | Marks | Guidance |
|---|---|---|
| The box is modelled as a particle (no air resistance/box treated as point mass) | B1 | Accept any valid assumption e.g. no air resistance |
I can see these are answer space pages from an exam paper (P58718/Jan13/MM1B), but the actual mark scheme is not shown in these images - they only show blank answer spaces for Questions 3, 4, and 5.
However, I can work out the solutions from the question content visible:
---3 A box, of mass 3 kg , is placed on a rough slope inclined at an angle of $40 ^ { \circ }$ to the horizontal. It is released from rest and slides down the slope.
\begin{enumerate}[label=(\alph*)]
\item Draw a diagram to show the forces acting on the box.
\item Find the magnitude of the normal reaction force acting on the box.
\item The coefficient of friction between the box and the slope is 0.2 . Find the magnitude of the friction force acting on the box.
\item Find the acceleration of the box.
\item State an assumption that you have made about the forces acting on the box.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2013 Q3 [9]}}