| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Modelling assumptions and refinements |
| Difficulty | Standard +0.3 This is a standard M1 mechanics question on motion on a slope with two models (smooth then rough surface). It requires routine application of F=ma, resolving forces parallel/perpendicular to the slope, and understanding of friction. The 'show that' parts guide students through calculations, and the conceptual questions about what happens after rest are straightforward applications of equilibrium conditions. Slightly above average difficulty due to the two-part modelling comparison, but all techniques are standard M1 content with no novel problem-solving required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.2a = -0.2 \times 9.8 \sin 20°\) → \(a = -9.8 \sin 20° = -3.35 \text{ ms}^{-2}\) | M1, A1, A1 (Total: 3) | Two term equation of motion with weight resolved. Correct equation. Correct acceleration from correct working. SC No negative sign but otherwise correct award M1A1A0. Allow \(a = -g \sin 20°\). |
| Answer | Marks | Guidance |
|---|---|---|
| \(0 = 4^2 + 2 \times (-3.35)s\) → \(s = \frac{16}{6.7} = 2.39 \text{ m}\) | M1, A1, A1 (Total: 3) | Use of constant acceleration equation with \(v = 0\) and \(u = 4\). Correct equation. Correct distance. |
| Answer | Marks | Guidance |
|---|---|---|
| The puck slides back down the slope as the puck is at rest and the resultant force is now acting down the slope / no friction / smooth slope. | B1, E1 (Total: 2) | Slides back down. Acceptable explanation. |
| Answer | Marks | Guidance |
|---|---|---|
| \(R = 0.2 \times 9.8 \cos 20°\) → \(F = 0.5 \times 0.2 \times 9.8 \cos 20° = 0.921 \text{ N}\) | M1, M1, A1 (Total: 3) | Finding normal reaction by resolving. Must see a trig term. Use of \(F = \mu R\). Correct friction from correct working. |
| Answer | Marks | Guidance |
|---|---|---|
| \(0.2a = -0.921 - 0.2 \times 9.8 \sin 20°\) → \(a = -7.96 \text{ ms}^{-2}\) | M1, A1, A1 (Total: 3) | Three term equation of motion with the weight resolved. Correct equation. Correct acceleration (with or without the minus sign, applied to both A1 marks). |
| Answer | Marks | Guidance |
|---|---|---|
| The puck stays at rest because the friction has a maximum of 0.921 and the component of the weight down the slope is less (0.670). | B1, dE1 (Total: 2) | Stays at rest. Acceptable explanation. |
### Part (a)(i)
$0.2a = -0.2 \times 9.8 \sin 20°$ → $a = -9.8 \sin 20° = -3.35 \text{ ms}^{-2}$ | M1, A1, A1 (Total: 3) | Two term equation of motion with weight resolved. Correct equation. Correct acceleration from correct working. SC No negative sign but otherwise correct award M1A1A0. Allow $a = -g \sin 20°$.
### Part (a)(ii)
$0 = 4^2 + 2 \times (-3.35)s$ → $s = \frac{16}{6.7} = 2.39 \text{ m}$ | M1, A1, A1 (Total: 3) | Use of constant acceleration equation with $v = 0$ and $u = 4$. Correct equation. Correct distance.
### Part (a)(iii)
The puck slides back down the slope as the puck is at rest and the resultant force is now acting down the slope / no friction / smooth slope. | B1, E1 (Total: 2) | Slides back down. Acceptable explanation.
### Part (b)(i)
$R = 0.2 \times 9.8 \cos 20°$ → $F = 0.5 \times 0.2 \times 9.8 \cos 20° = 0.921 \text{ N}$ | M1, M1, A1 (Total: 3) | Finding normal reaction by resolving. Must see a trig term. Use of $F = \mu R$. Correct friction from correct working.
### Part (b)(ii)
$0.2a = -0.921 - 0.2 \times 9.8 \sin 20°$ → $a = -7.96 \text{ ms}^{-2}$ | M1, A1, A1 (Total: 3) | Three term equation of motion with the weight resolved. Correct equation. Correct acceleration (with or without the minus sign, applied to both A1 marks).
### Part (b)(iii)
The puck stays at rest because the friction has a maximum of 0.921 and the component of the weight down the slope is less (0.670). | B1, dE1 (Total: 2) | Stays at rest. Acceptable explanation.
---5 A puck, of mass 0.2 kg , is placed on a slope inclined at $20 ^ { \circ }$ above the horizontal, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{217f0e3e-9d1b-41f1-8299-f56d073fbbeb-3_280_773_1249_623}
The puck is hit so that initially it moves with a velocity of $4 \mathrm {~ms} ^ { - 1 }$ directly up the slope.
\begin{enumerate}[label=(\alph*)]
\item A simple model assumes that the surface of the slope is smooth.
\begin{enumerate}[label=(\roman*)]
\item Show that the acceleration of the puck up the slope is $- 3.35 \mathrm {~m} \mathrm {~s} ^ { - 2 }$, correct to three significant figures.
\item Find the distance that the puck will travel before it comes to rest.
\item What will happen to the puck after it comes to rest?
Explain why.
\end{enumerate}\item A revised model assumes that the surface is rough and that the coefficient of friction between the puck and the surface is 0.5 .
\begin{enumerate}[label=(\roman*)]
\item Show that the magnitude of the friction force acting on the puck during this motion is 0.921 N , correct to three significant figures.
\item Find the acceleration of the puck up the slope.
\item What will happen to the puck after it comes to rest in this case?
Explain why.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M1 2008 Q5 [16]}}