| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Two-part friction scenarios |
| Difficulty | Moderate -0.3 This is a standard two-part friction problem requiring resolution of forces and application of F=μR. Part (i) involves showing a given coefficient (straightforward equilibrium with limiting friction), while part (ii) applies the same coefficient to find acceleration using F=ma. The calculations are routine for M1 level with no conceptual surprises, making it slightly easier than average. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| Part (i) | ||
| \(12 + 15\sin30° = R\) | M1 | For resolving forces vertically |
| \(F = 15\cos30°\) | A1 | |
| \(B1\) | ||
| \([\mu = \frac{15\cos30°}{(12 + 15\sin30°)}]\) | M1 | For using \(\mu = \frac{F}{R}\) |
| Coefficient is 0.666 | A1 | 5 |
| Part (ii) | ||
| \([F = 0.666(12 - 15\sin30°)]\) | B1 | |
| \(15\cos30° - F = 1.2a\) | M1 | For using Newton's second law |
| Acceleration is 8.33 m s\(^{-2}\) | A1 | 4 |
| **Part (i)** | | |
|---|---|---|
| $12 + 15\sin30° = R$ | M1 | For resolving forces vertically |
| $F = 15\cos30°$ | A1 | |
| $B1$ | |
| $[\mu = \frac{15\cos30°}{(12 + 15\sin30°)}]$ | M1 | For using $\mu = \frac{F}{R}$ |
| Coefficient is 0.666 | A1 | 5 | AG |
| **Part (ii)** | | |
|---|---|---|
| $[F = 0.666(12 - 15\sin30°)]$ | B1 | |
| $15\cos30° - F = 1.2a$ | M1 | For using Newton's second law |
| Acceleration is 8.33 m s$^{-2}$ | A1 | 4 |5
\begin{figure}[h]
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\caption{Fig. 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{efa7175f-832b-4cd3-82ab-52e402115081-3_317_522_922_1155}
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\caption{Fig. 2}
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\end{figure}
A small ring of weight 12 N is threaded on a fixed rough horizontal rod. A light string is attached to the ring and the string is pulled with a force of 15 N at an angle of $30 ^ { \circ }$ to the horizontal.\\
(i) When the angle of $30 ^ { \circ }$ is below the horizontal (see Fig. 1), the ring is in limiting equilibrium. Show that the coefficient of friction between the ring and the rod is 0.666 , correct to 3 significant figures.\\
(ii) When the angle of $30 ^ { \circ }$ is above the horizontal (see Fig. 2), the ring is moving with acceleration $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$. Find the value of $a$.
\hfill \mbox{\textit{CAIE M1 2009 Q5 [9]}}