| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 15 |
| 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 mechanics problem involving forces on an inclined plane with friction. While it requires resolving forces in two directions, applying F=ma, and understanding friction conditions, these are routine techniques for M1 students. The multi-part structure guides students through the problem systematically. The most challenging aspect is part (b) requiring comparison of weight component with maximum friction, but this is a standard 'show that' question testing understanding of static vs kinetic friction. Overall, slightly easier than average due to its structured nature and use of well-practiced M1 techniques. |
| Spec | 3.03e Resolve forces: two dimensions3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(R(\downarrow)\) \(N + 24\cos 60° = 2g\cos 30°\) \(\Rightarrow N = 16.97 - 12 = 4.97 \text{ N}\) \(\Rightarrow F = 0.4 \cdot 4.97 = 1.99 \text{ N}\) \(R(\rightarrow)\) \(2a = 24\cos 30° - 2g\cos 60° - 1.99\) \(\Rightarrow a \approx 4.5 \text{ m s}^{-2}\) | M1 A1 A1, M1 A1, M1 A1 | 8 marks |
| (b) \(R(\downarrow)\) \(N' = 2g\cos 30° = 16.97\) \(\Rightarrow F'_{\max} = 0.4 \cdot 16.97 = 6.79 \text{ N}\) Component of weight down plane \(= 2g\sin 30° = 9.8 \text{ N}\) | M1 A1, M1 | Marks up to (c) below |
| (c) \(9.8 > F'_{\max} \Rightarrow\) net force down plane \(\Rightarrow\) parcel moves \(2f = 9.8 - 6.79, \Rightarrow f \approx 1.5 \text{ m s}^{-2}\) | A1 | 4 marks (total for b&c) |
**(a)** $R(\downarrow)$ $N + 24\cos 60° = 2g\cos 30°$ $\Rightarrow N = 16.97 - 12 = 4.97 \text{ N}$ $\Rightarrow F = 0.4 \cdot 4.97 = 1.99 \text{ N}$ $R(\rightarrow)$ $2a = 24\cos 30° - 2g\cos 60° - 1.99$ $\Rightarrow a \approx 4.5 \text{ m s}^{-2}$ | M1 A1 A1, M1 A1, M1 A1 | 8 marks
**(b)** $R(\downarrow)$ $N' = 2g\cos 30° = 16.97$ $\Rightarrow F'_{\max} = 0.4 \cdot 16.97 = 6.79 \text{ N}$ Component of weight down plane $= 2g\sin 30° = 9.8 \text{ N}$ | M1 A1, M1 | Marks up to (c) below
**(c)** $9.8 > F'_{\max} \Rightarrow$ net force down plane $\Rightarrow$ parcel moves $2f = 9.8 - 6.79, \Rightarrow f \approx 1.5 \text{ m s}^{-2}$ | A1 | 4 marks (total for b&c)\includegraphics{figure_3}
A small parcel of mass $2$ kg moves on a rough plane inclined at an angle of $30°$ to the horizontal. The parcel is pulled up a line of greatest slope of the plane by means of a light rope which it attached to it. The rope makes an angle of $30°$ with the plane, as shown in Fig. 3. The coefficient of friction between the parcel and the plane is $0.4$.
Given that the tension in the rope is $24$ N,
\begin{enumerate}[label=(\alph*)]
\item find, to 2 significant figures, the acceleration of the parcel. [8]
\end{enumerate}
The rope now breaks. The parcel slows down and comes to rest.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that, when the parcel comes to this position of rest, it immediately starts to move down the plane again. [4]
\item Find, to 2 significant figures, the acceleration of the parcel as it moves down the plane after it has come to this position of instantaneous rest. [3]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q7 [15]}}