| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| 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 M1 limiting equilibrium problem requiring resolution of forces parallel and perpendicular to a slope in two scenarios, followed by applying F=ma. While it involves multiple parts and careful sign management, the techniques are routine for mechanics students with no novel problem-solving required—slightly easier than average due to its methodical, textbook nature. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resolving forces parallel to the plane | M1 | For resolving forces parallel to the plane |
| \(F = 5.9 - 6.1\sin\alpha\) | A1 | |
| \(R = 6.1\cos\alpha\) | B1 | |
| \([5.9 - 6.1\sin\alpha \leq \mu(6.1\cos\alpha)]\) | M1 | For using \(F \leq \mu R\) |
| \(\mu > \frac{4}{5}\) | A1 [5] | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([6.1 \times (11/61) + 5.9 - \mu \times 6.1 \times (60/61) > 0]\) | M1 | For using \(F = \mu R\) and 'net downward force \(> 0\)' |
| \(\mu < \frac{7}{6}\) | A1 [2] | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([6.1 \times (11/61) + 5.9 - \mu \times 6.1 \times (60/61) = 0.61 \times 1.7]\) | M1 | For using Newton's 2nd law and \(F = \mu R\) |
| \(\mu = 0.994\) | A1 [2] |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolving forces parallel to the plane | M1 | For resolving forces parallel to the plane |
| $F = 5.9 - 6.1\sin\alpha$ | A1 | |
| $R = 6.1\cos\alpha$ | B1 | |
| $[5.9 - 6.1\sin\alpha \leq \mu(6.1\cos\alpha)]$ | M1 | For using $F \leq \mu R$ |
| $\mu > \frac{4}{5}$ | A1 [5] | AG |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[6.1 \times (11/61) + 5.9 - \mu \times 6.1 \times (60/61) > 0]$ | M1 | For using $F = \mu R$ and 'net downward force $> 0$' |
| $\mu < \frac{7}{6}$ | A1 [2] | AG |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[6.1 \times (11/61) + 5.9 - \mu \times 6.1 \times (60/61) = 0.61 \times 1.7]$ | M1 | For using Newton's 2nd law and $F = \mu R$ |
| $\mu = 0.994$ | A1 [2] | |
---6
\begin{tikzpicture}[scale=1.2, thick]
% Inclined plane
\coordinate (O) at (0,0);
\coordinate (B) at (6,0);
\coordinate (T) at (6,1.1);
\draw[thick] (O) -- (B) -- (T) -- cycle;
% Angle alpha
\draw (1.2,0) arc[start angle=0, end angle={atan(1.1/6)}, radius=1.2];
\node at (1.6,0.12) {$\alpha$};
% Block on the slope
\pgfmathsetmacro{\slopeangle}{atan(1.1/6)}
\coordinate (block) at ($(O)!0.5!(T)$);
\begin{scope}[shift={(block)}, rotate=\slopeangle]
\draw[fill=gray!30] (-0.4,-0.02) rectangle (0.4,0.45);
\end{scope}
% Weight arrow
\draw[->, >=stealth] (block) -- ++(0,-1.0) node[right] {$6.1$ N};
% Force arrow up the slope
\draw[->, >=stealth] ($(block)+(\slopeangle:0.45)$) -- ++(\slopeangle:1.2) node[above] {$5.9$ N};
% Label
\node[below] at (3,-0.5) {\textbf{Fig.\ 1}};
\end{tikzpicture}
\begin{tikzpicture}[scale=1.2, thick]
% Inclined plane
\coordinate (O) at (0,0);
\coordinate (B) at (6,0);
\coordinate (T) at (6,1.1);
\draw[thick] (O) -- (B) -- (T) -- cycle;
% Angle alpha
\draw (1.2,0) arc[start angle=0, end angle={atan(1.1/6)}, radius=1.2];
\node at (1.6,0.12) {$\alpha$};
% Block on the slope
\pgfmathsetmacro{\slopeangle}{atan(1.1/6)}
\coordinate (block) at ($(O)!0.65!(T)$);
\begin{scope}[shift={(block)}, rotate=\slopeangle]
\draw[fill=gray!30] (-0.4,-0.02) rectangle (0.4,0.45);
\end{scope}
% Weight arrow
\draw[->, >=stealth] (block) -- ++(0,-1.0) node[right] {$6.1$ N};
% Force arrow down the slope
\draw[->, >=stealth] ($(block)+(\slopeangle+180:0.45)$) -- ++(\slopeangle+180:1.2) node[above] {$5.9$ N};
% Label
\node[below] at (3,-0.5) {\textbf{Fig.\ 2}};
\end{tikzpicture}
A block of weight 6.1 N is at rest on a plane inclined at angle $\alpha$ to the horizontal, where $\tan \alpha = \frac { 11 } { 60 }$. The coefficient of friction between the block and the plane is $\mu$. A force of magnitude 5.9 N acting parallel to a line of greatest slope is applied to the block.\\
(i) When the force acts up the plane (see Fig. 1) the block remains at rest. Show that $\mu \geqslant \frac { 4 } { 5 }$.\\
(ii) When the force acts down the plane (see Fig. 2) the block slides downwards. Show that $\mu < \frac { 7 } { 6 }$.\\
(iii) Given that the acceleration of the block is $1.7 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ when the force acts down the plane, find the value of $\mu$.
\hfill \mbox{\textit{CAIE M1 2012 Q6 [9]}}