| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Equilibrium on slope with force at angle to slope |
| Difficulty | Standard +0.3 This is a standard M1 friction problem on an inclined plane with straightforward resolution of forces. While it requires careful component resolution in two directions and algebraic manipulation to reach the given expression, it follows a completely standard method with no novel insight required. The multi-part structure guides students through the solution systematically, and part (d) is a simple modelling critique. Slightly easier than average due to its routine nature. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Resolve \(\parallel\) to plane: \(\mu R + mg \sin 30° = kmg \cos 30°\) | M1 A1 | |
| Resolve \(\perp\) to plane: \(R + kmg \sin 30° = mg \cos 30°\) | M1 A1 | |
| \(\mu mg\left(\frac{\sqrt{3}}{2} - \frac{k}{2}\right) = mg\left(k\frac{\sqrt{3}}{2} - \frac{1}{2}\right)\) giving \(\mu = \frac{k\sqrt{3}-1}{\sqrt{3}-k}\) | M1 A1 A1 | 7 marks |
| (b) With \(k = \frac{3\sqrt{3}}{2}\), \(\mu = \left(\frac{9}{2}-1\right) + \frac{4\sqrt{3}}{2} - \frac{2}{4\sqrt{3}} = \frac{2\sqrt{3}}{12} \cdot \frac{\sqrt{3}}{6}\) | M1 A1 A1 A1 | 4 marks |
| (c) Force down plane = \(\frac{1}{2}mg\). Max. friction = \(\frac{\sqrt{3}}{6} \times mg\frac{\sqrt{3}}{2} = \frac{1}{4}mg\) | B1 M1 A1 | |
| so moves down with acceleration \(\frac{1}{4}g = 2.45 \text{ ms}^{-2}\) | M1 A1 | 5 marks |
| (d) P is shown as a ball, in which case it would roll | B1 | total: 16 |
(a) Resolve $\parallel$ to plane: $\mu R + mg \sin 30° = kmg \cos 30°$ | M1 A1 |
Resolve $\perp$ to plane: $R + kmg \sin 30° = mg \cos 30°$ | M1 A1 |
$\mu mg\left(\frac{\sqrt{3}}{2} - \frac{k}{2}\right) = mg\left(k\frac{\sqrt{3}}{2} - \frac{1}{2}\right)$ giving $\mu = \frac{k\sqrt{3}-1}{\sqrt{3}-k}$ | M1 A1 A1 | 7 marks
(b) With $k = \frac{3\sqrt{3}}{2}$, $\mu = \left(\frac{9}{2}-1\right) + \frac{4\sqrt{3}}{2} - \frac{2}{4\sqrt{3}} = \frac{2\sqrt{3}}{12} \cdot \frac{\sqrt{3}}{6}$ | M1 A1 A1 A1 | 4 marks
(c) Force down plane = $\frac{1}{2}mg$. Max. friction = $\frac{\sqrt{3}}{6} \times mg\frac{\sqrt{3}}{2} = \frac{1}{4}mg$ | B1 M1 A1 |
so moves down with acceleration $\frac{1}{4}g = 2.45 \text{ ms}^{-2}$ | M1 A1 | 5 marks
(d) P is shown as a ball, in which case it would roll | B1 | total: 16A particle $P$, of mass $m$, is in contact with a rough plane inclined at 30° to the horizontal as shown. A light string is attached to $P$ and makes an angle of 30° with the plane. When the tension in this string has magnitude $kmg$, $P$ is just on the point of moving up the plane.
\includegraphics{figure_7}
\begin{enumerate}[label=(\alph*)]
\item Show that $\mu$, the coefficient of friction between $P$ and the plane, is $\frac{k\sqrt{3} - 1}{\sqrt{3} - k}$. [7 marks]
\item Given further that $k = \frac{3\sqrt{3}}{7}$, deduce that $\mu = \frac{\sqrt{3}}{6}$. [3 marks]
\end{enumerate}
The string is now removed.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Determine whether $P$ will move down the plane and, if it does, find its acceleration. [5 marks]
\item Give a reason why the way in which $P$ is shown in the diagram might not be consistent with the modelling assumptions that have been made. [1 mark]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q7 [16]}}