| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion with applied force on slope |
| Difficulty | Standard +0.3 This is a standard M1 mechanics question on forces on a slope with three routine parts: (a) resolving forces in equilibrium to verify a given result, (b) finding normal reaction using standard resolution, and (c) applying F=ma with a changed force direction. All parts use straightforward resolution of forces parallel and perpendicular to the plane with no novel problem-solving required. Slightly easier than average due to the guided structure and given answer in part (a). |
| Spec | 3.03d Newton's second law: 2D vectors3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| R (\(\parallel\) plane): \(49\cos\theta = 6g\sin 30\) | M1 A1 | |
| \(\Rightarrow \cos\theta = \frac{3}{5}\) * | A1 (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| R (\(\perp\) plane): \(R = 6g\cos 30 + 49\sin\theta\) | M1 A1 | |
| \(R \approx 90.1 \text{ or } 90\text{ N}\) | DM1 A1 (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| R (\(\parallel\) to plane): \(49\cos 30 - 6g\sin 30 = 6a\) | M1 A2,1,0 | |
| \(\Rightarrow a \approx 2.17 \text{ or } 2.2 \text{ ms}^{-2}\) | A1 (4) | Total: 11 |
## Question 4:
### Part (a)
| Working | Marks | Notes |
|---------|-------|-------|
| R ($\parallel$ plane): $49\cos\theta = 6g\sin 30$ | M1 A1 | |
| $\Rightarrow \cos\theta = \frac{3}{5}$ * | A1 (3) | |
### Part (b)
| Working | Marks | Notes |
|---------|-------|-------|
| R ($\perp$ plane): $R = 6g\cos 30 + 49\sin\theta$ | M1 A1 | |
| $R \approx 90.1 \text{ or } 90\text{ N}$ | DM1 A1 (4) | |
### Part (c)
| Working | Marks | Notes |
|---------|-------|-------|
| R ($\parallel$ to plane): $49\cos 30 - 6g\sin 30 = 6a$ | M1 A2,1,0 | |
| $\Rightarrow a \approx 2.17 \text{ or } 2.2 \text{ ms}^{-2}$ | A1 (4) | **Total: 11** |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7ba14d10-1b57-4930-8d65-f21088c5d513-06_305_607_246_701}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A particle $P$ of mass 6 kg lies on the surface of a smooth plane. The plane is inclined at an angle of $30 ^ { \circ }$ to the horizontal. The particle is held in equilibrium by a force of magnitude 49 N , acting at an angle $\theta$ to the plane, as shown in Figure 1. The force acts in a vertical plane through a line of greatest slope of the plane.
\begin{enumerate}[label=(\alph*)]
\item Show that $\cos \theta = \frac { 3 } { 5 }$.
\item Find the normal reaction between $P$ and the plane.
The direction of the force of magnitude 49 N is now changed. It is now applied horizontally to $P$ so that $P$ moves up the plane. The force again acts in a vertical plane through a line of greatest slope of the plane.
\item Find the initial acceleration of $P$.
$\_\_\_\_$}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2008 Q4 [11]}}