| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Limiting equilibrium both directions |
| Difficulty | Standard +0.3 This is a standard two-part limiting equilibrium problem on an inclined plane requiring resolution of forces parallel and perpendicular to the slope, and application of F ≤ μR in both directions. The calculations are straightforward with given tan α = 7/24 allowing easy determination of sin α and cos α. While it requires careful sign management and understanding of friction direction reversal, it follows a routine mechanics template with no novel insight required. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (Resolving forces parallel to slope with three terms) | M1 | For resolving forces parallel to slope with three terms |
| \(F + W\sin\alpha = 7.2\) | A1 | |
| \([\mu \times 7.5\cos\alpha \geqslant 7.2 - 7.5\sin\alpha]\) | M1 | For using \(F \leqslant \mu R\) |
| \(\mu \geqslant 17/24\) | A1 | AG [Total: 4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([7.2 + 7.5 \times (7/25) - \mu(7.5 \times 24/25) > 0]\) | M1 | For using resultant force down the plane \(> 0\) and \(F = \mu R\) |
| \(\mu < 31/24\) | A1 | AG [Total: 2] |
## Question 3 (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| (Resolving forces parallel to slope with three terms) | M1 | For resolving forces parallel to slope with three terms |
| $F + W\sin\alpha = 7.2$ | A1 | |
| $[\mu \times 7.5\cos\alpha \geqslant 7.2 - 7.5\sin\alpha]$ | M1 | For using $F \leqslant \mu R$ |
| $\mu \geqslant 17/24$ | A1 | **AG** [Total: 4] |
## Question 3 (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[7.2 + 7.5 \times (7/25) - \mu(7.5 \times 24/25) > 0]$ | M1 | For using resultant force down the plane $> 0$ and $F = \mu R$ |
| $\mu < 31/24$ | A1 | **AG** [Total: 2] |
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ffefbc81-402f-4048-8741-23c8bae30d5a-2_231_485_1238_486}
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\caption{Fig. 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ffefbc81-402f-4048-8741-23c8bae30d5a-2_206_485_1263_1174}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
A block of weight 7.5 N is at rest on a plane which is inclined to the horizontal at angle $\alpha$, where $\tan \alpha = \frac { 7 } { 24 }$. The coefficient of friction between the block and the plane is $\mu$. A force of magnitude 7.2 N acting parallel to a line of greatest slope is applied to the block. When the force acts up the plane (see Fig. 1) the block remains at rest.\\
(i) Show that $\mu \geqslant \frac { 17 } { 24 }$.
When the force acts down the plane (see Fig. 2) the block slides downwards.\\
(ii) Show that $\mu < \frac { 31 } { 24 }$.
\hfill \mbox{\textit{CAIE M1 2014 Q3 [6]}}