| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | June |
| Marks | 15 |
| 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 testing standard inclined plane concepts with smooth/rough surfaces. Part (a) requires basic force resolution and Newton's second law (routine calculation). Part (b) involves standard SUVAT, friction force calculation, and coefficient of friction—all textbook exercises. The modelling discussion in (b)(iv) requires minimal insight. No novel problem-solving or complex multi-step reasoning needed. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03a Force: vector nature and diagrams3.03b Newton's first law: equilibrium3.03d Newton's second law: 2D vectors3.03i Normal reaction force3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Diagram showing \(R\) or \(N\) pointing upward from the slope, with \(mg\) or \(W\) or \(3g\) pointing vertically downward, with arrows and labels. | B1 | Correct diagram with arrows and labels |
| Total: 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(3a = 3g \sin 30°\) | M1 | Two term equation of motion |
| \(a = g \sin 30° = 4.9 \text{ ms}^{-2}\) | A1 | AG Correct acceleration from correct working. (Allow \(a = g \sin 30°\)) |
| Total: — marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(s = \frac{1}{2} a \times 2^2\) | M1 | Constant acceleration equation with \(u = 0\) |
| \(a = 2.5 \text{ ms}^{-2}\) | A1 | AG Correct answer from correct working. (Use of \(v = 5\) must be justified) |
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(3 \times 2.5 = 3g \sin 30° - F\) | M1 | Three term equation of motion |
| \(F = 3g \sin 30° - 7.5 = 7.20 \text{ N (to 3 sf)}\) | A1 | Correct equation |
| A1 | Correct \(F\). Accept 7.2 N | |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(R = 3g \cos 30° = (25.46)\) | M1 | Resolving perpendicular to the slope to find \(R\) |
| \(7.2 = \mu \times 3g \cos 30°\) | A1 | Correct \(R\) |
| M1 | Use of \(F = \mu R\) | |
| A1F | Correct expression | |
| \(\mu = \frac{7.2}{3g \cos 30°} = 0.283\) | A1F | Correct \(\mu\). Accept 0.282. (Follow through from incorrect \(F\) from above, but not an incorrect \(R\)) |
| Total: 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Reduce \(a\), as the air resistance would reduce the magnitude of the resultant force or because the air resistance increases as the velocity increases towards its terminal value | B1 | Reduces |
| B1 | Explanation. Second B1 dependent on the first B1 mark | |
| Total: 15 marks |
**6(a)(i)**
| Diagram showing $R$ or $N$ pointing upward from the slope, with $mg$ or $W$ or $3g$ pointing vertically downward, with arrows and labels. | B1 | Correct diagram with arrows and labels |
| **Total: 5 marks** | | |
**6(a)(ii)**
| $3a = 3g \sin 30°$ | M1 | Two term equation of motion |
| $a = g \sin 30° = 4.9 \text{ ms}^{-2}$ | A1 | AG Correct acceleration from correct working. (Allow $a = g \sin 30°$) |
| **Total: — marks** | | |
**6(b)(i)**
| $s = \frac{1}{2} a \times 2^2$ | M1 | Constant acceleration equation with $u = 0$ |
| $a = 2.5 \text{ ms}^{-2}$ | A1 | AG Correct answer from correct working. (Use of $v = 5$ must be justified) |
| **Total: 2 marks** | | |
**6(b)(ii)**
| $3 \times 2.5 = 3g \sin 30° - F$ | M1 | Three term equation of motion |
| $F = 3g \sin 30° - 7.5 = 7.20 \text{ N (to 3 sf)}$ | A1 | Correct equation |
| | A1 | Correct $F$. Accept 7.2 N |
| **Total: 3 marks** | | |
**6(b)(iii)**
| $R = 3g \cos 30° = (25.46)$ | M1 | Resolving perpendicular to the slope to find $R$ |
| $7.2 = \mu \times 3g \cos 30°$ | A1 | Correct $R$ |
| | M1 | Use of $F = \mu R$ |
| | A1F | Correct expression |
| $\mu = \frac{7.2}{3g \cos 30°} = 0.283$ | A1F | Correct $\mu$. Accept 0.282. (Follow through from incorrect $F$ from above, but not an incorrect $R$) |
| **Total: 5 marks** | | |
**6(b)(iv)**
| Reduce $a$, as the air resistance would reduce the magnitude of the resultant force or because the air resistance increases as the velocity increases towards its terminal value | B1 | Reduces |
| | B1 | Explanation. Second B1 dependent on the first B1 mark |
| **Total: 15 marks** | | |6 A box, of mass 3 kg , is placed on a slope inclined at an angle of $30 ^ { \circ }$ to the horizontal. The box slides down the slope. Assume that air resistance can be ignored.
\begin{enumerate}[label=(\alph*)]
\item A simple model assumes that the slope is smooth.
\begin{enumerate}[label=(\roman*)]
\item Draw a diagram to show the forces acting on the box.
\item Show that the acceleration of the box is $4.9 \mathrm {~ms} ^ { - 2 }$.
\end{enumerate}\item A revised model assumes that the slope is rough. The box slides down the slope from rest, travelling 5 metres in 2 seconds.
\begin{enumerate}[label=(\roman*)]
\item Show that the acceleration of the box is $2.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\item Find the magnitude of the friction force acting on the box.
\item Find the coefficient of friction between the box and the slope.
\item In reality, air resistance affects the motion of the box. Explain how its acceleration would change if you took this into account.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M1 2007 Q6 [15]}}