| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion down rough slope |
| Difficulty | Moderate -0.3 This is a standard M1 mechanics problem requiring resolution of forces on a slope, application of F=ma with friction, and use of constant acceleration equations. While it involves multiple steps, the techniques are routine and well-practiced at this level with no novel problem-solving required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R = mg\cos 40\) | B1 | |
| Use of \(F = \mu R\) | B1 | Seen or implied (can be on diagram) |
| \(mg\sin 40 - F = \pm ma\) | M1A1 | M1 for resolving parallel to plane, correct no. of terms, \(mg\) resolved; \(F\) does not need to be substituted |
| \(acc = 2.55\ (\text{m s}^{-2})\) or \(2.5\ (\text{m s}^{-2})\) | A1 | Must be positive; if \(m\) is given specific numerical value, can score max B1B1M1A0A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(v^2 = u^2 + 2as = 2 \times a \times 3\); Speed at \(B\) is \(3.9\ (\text{m s}^{-1})\) or \(3.91\ (\text{m s}^{-1})\) | M1A1 | M1 for complete method for finding speed (usually \(v^2 = u^2 + 2as\)) |
# Question 2:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R = mg\cos 40$ | B1 | |
| Use of $F = \mu R$ | B1 | Seen or implied (can be on diagram) |
| $mg\sin 40 - F = \pm ma$ | M1A1 | M1 for resolving parallel to plane, correct no. of terms, $mg$ resolved; $F$ does not need to be substituted |
| $acc = 2.55\ (\text{m s}^{-2})$ or $2.5\ (\text{m s}^{-2})$ | A1 | Must be **positive**; if $m$ is given specific numerical value, can score max B1B1M1A0A0 |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $v^2 = u^2 + 2as = 2 \times a \times 3$; Speed at $B$ is $3.9\ (\text{m s}^{-1})$ or $3.91\ (\text{m s}^{-1})$ | M1A1 | M1 for complete method for finding speed (usually $v^2 = u^2 + 2as$) |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ed659098-c1cf-4ee1-a12a-bf8b6c42db95-03_435_840_269_561}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A rough plane is inclined at $40 ^ { \circ }$ to the horizontal. Two points $A$ and $B$ are 3 metres apart and lie on a line of greatest slope of the inclined plane, with $A$ above $B$, as shown in Figure 2. A particle $P$ of mass $m \mathrm {~kg}$ is held at rest on the plane at $A$. The coefficient of friction between $P$ and the plane is $\frac { 1 } { 2 }$. The particle is released.
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of $P$ down the plane.
\item Find the speed of $P$ at $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2014 Q2 [7]}}