| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Horizontal force on slope |
| Difficulty | Standard +0.3 This is a standard M1 equilibrium problem on a slope with a horizontal force. While it requires resolving forces in two directions and using friction at limiting equilibrium, the setup is straightforward with tan α given explicitly. The multi-part structure and calculation steps place it slightly above average, but it follows a well-practiced method with no novel insight required. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Diagram showing \(R\), \(F\), \(P\) (horizontal), \(1.1g\) (downward) correctly on inclined plane | B2 | \(-1\) e.e.o.o., labels not needed (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(F = \frac{1}{2}R\) | B1 | |
| \((\uparrow)\): \(R\cos\alpha + F\sin\alpha = mg\) | M1 A2 | |
| \(R = \frac{1.1g}{\cos\alpha + \frac{1}{2}\sin\alpha} = 9.8\ \text{N}\) | M1 A1 | (6) |
| \((\rightarrow)\): \(P + \frac{1}{2}R\cos\alpha = R\sin\alpha\) | M1 A2 | |
| \(P = R\!\left(\sin\alpha - \frac{1}{2}\cos\alpha\right)\) | M1 | |
| \(= 1.96\) | A1 | (5) [13] |
## Question 5:
### Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Diagram showing $R$, $F$, $P$ (horizontal), $1.1g$ (downward) correctly on inclined plane | B2 | $-1$ e.e.o.o., labels not needed **(2)** |
### Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $F = \frac{1}{2}R$ | B1 | |
| $(\uparrow)$: $R\cos\alpha + F\sin\alpha = mg$ | M1 A2 | |
| $R = \frac{1.1g}{\cos\alpha + \frac{1}{2}\sin\alpha} = 9.8\ \text{N}$ | M1 A1 | **(6)** |
| $(\rightarrow)$: $P + \frac{1}{2}R\cos\alpha = R\sin\alpha$ | M1 A2 | |
| $P = R\!\left(\sin\alpha - \frac{1}{2}\cos\alpha\right)$ | M1 | |
| $= 1.96$ | A1 | **(5) [13]** |
---5.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{86bb11a4-b409-49b1-bffb-d0e3727d345c-07_352_834_300_551}
\end{center}
\section*{Figure 2}
A small package of mass 1.1 kg is held in equilibrium on a rough plane by a horizontal force. The plane is inclined at an angle $\alpha$ to the horizontal, where $\tan \alpha = \frac { 3 } { 4 }$. The force acts in a vertical plane containing a line of greatest slope of the plane and has magnitude $P$ newtons, as shown in Figure 2.
The coefficient of friction between the package and the plane is 0.5 and the package is modelled as a particle. The package is in equilibrium and on the point of slipping down the plane.
\begin{enumerate}[label=(\alph*)]
\item Draw, on Figure 2, all the forces acting on the package, showing their directions clearly.
\item \begin{enumerate}[label=(\roman*)]
\item Find the magnitude of the normal reaction between the package and the plane.
\item Find the value of $P$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2009 Q5 [13]}}