Edexcel M1 2014 January — Question 4 11 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2014
SessionJanuary
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMotion on a slope
TypeString at angle to slope
DifficultyStandard +0.3 This is a standard M1 mechanics problem requiring resolution of forces in two directions (parallel and perpendicular to slope) and application of friction law F=μR. While it involves multiple components and algebraic manipulation with the friction coefficient, it follows a completely routine method taught in all M1 courses with no novel insight required. The tan α = 3/4 setup is standard bookwork.
Spec3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes

4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{fade35da-8dca-4d98-a07c-ed3a173fccda-08_396_483_214_735} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} A fixed rough plane is inclined to the horizontal at an angle \(\alpha\), where \(\tan \alpha = \frac { 3 } { 4 }\) A small box of mass \(m\) is at rest on the plane. A force of magnitude \(k m g\), where \(k\) is a constant, is applied to the box. The line of action of the force is at angle \(\alpha\) to the line of greatest slope of the plane through the box, as shown in Figure 1, and lies in the same vertical plane as this line of greatest slope. The coefficient of friction between the box and the plane is \(\mu\). The box is on the point of slipping up the plane. By modelling the box as a particle, find \(k\) in terms of \(\mu\).

Question 4:
AnswerMarks Guidance
\(\cos\alpha = \frac{4}{5}\) or \(\sin\alpha = \frac{3}{5}\)B1; B1 B1 for use of \(F = \mu R\); B1 for \(\cos\alpha = 0.8\) or \(\sin\alpha = 0.6\) seen
\(kmg\cos\alpha - mg\sin\alpha = F\)M1 A1 A1 M1 resolving parallel to plane; A1 correct equation (\(-1\) each error, omission of both \(g\)'s is 1 error); N.B. if \(\cos 4/5\) or \(\sin 3/5\) written, treat as 1 A error but allow recovery
\(mg\cos\alpha - kmg\sin\alpha = R\)M1 A1 A1 M1 resolving perpendicular to plane; A1 correct equation (same rules apply)
equation in \(k\) and \(\mu\) onlyDM1 DM1 dependent on first two M marks
\(k = \dfrac{3+4\mu}{4+3\mu}\)DM1 A1 (11) Fourth DM1 dependent on third M1 for solving for \(k\); A1 for \(\dfrac{3+4\mu}{4+3\mu}\) oe
## Question 4:

$\cos\alpha = \frac{4}{5}$ or $\sin\alpha = \frac{3}{5}$ | B1; B1 | B1 for use of $F = \mu R$; B1 for $\cos\alpha = 0.8$ or $\sin\alpha = 0.6$ seen

$kmg\cos\alpha - mg\sin\alpha = F$ | M1 A1 A1 | M1 resolving parallel to plane; A1 correct equation ($-1$ each error, omission of both $g$'s is 1 error); N.B. if $\cos 4/5$ or $\sin 3/5$ written, treat as 1 A error but allow recovery

$mg\cos\alpha - kmg\sin\alpha = R$ | M1 A1 A1 | M1 resolving perpendicular to plane; A1 correct equation (same rules apply)

equation in $k$ and $\mu$ only | DM1 | DM1 dependent on first two M marks

$k = \dfrac{3+4\mu}{4+3\mu}$ | DM1 A1 (11) | Fourth DM1 dependent on third M1 for solving for $k$; A1 for $\dfrac{3+4\mu}{4+3\mu}$ oe

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4.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{fade35da-8dca-4d98-a07c-ed3a173fccda-08_396_483_214_735}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

A fixed rough plane is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac { 3 } { 4 }$ A small box of mass $m$ is at rest on the plane. A force of magnitude $k m g$, where $k$ is a constant, is applied to the box. The line of action of the force is at angle $\alpha$ to the line of greatest slope of the plane through the box, as shown in Figure 1, and lies in the same vertical plane as this line of greatest slope. The coefficient of friction between the box and the plane is $\mu$. The box is on the point of slipping up the plane. By modelling the box as a particle, find $k$ in terms of $\mu$.\\

\hfill \mbox{\textit{Edexcel M1 2014 Q4 [11]}}