| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Two-part friction scenarios |
| Difficulty | Standard +0.3 This is a standard two-part friction problem requiring resolution of forces and application of F=μR in limiting equilibrium, followed by Newton's second law. The calculations are straightforward with given sin θ = 5/13, requiring only routine mechanics techniques with no novel problem-solving insight. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([0.195\cos\theta = F]\) | M1 | For resolving forces horizontally |
| \(F = 0.195\cos22.6 = 0.195 \times \frac{12}{13} = 0.18 = \frac{9}{50}\) | A1 | |
| \([R = 0.24 + 0.195\sin\theta]\) | M1 | For resolving forces vertically |
| \(R = 0.24 + 0.195\sin22.6 = 0.24 + 0.195 \times \frac{5}{13} = 0.315 = \frac{63}{200}\) | A1 | |
| M1 | For using \(\mu = F/R\) | |
| Coefficient \(\mu = 4/7\) or \(0.571\) | A1 | |
| Total: 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R = 0.24 - 0.195\sin 22.6\) | ||
| \(= 0.24 - 0.195 \times \frac{5}{13}\) | ||
| \(= 0.165 = \frac{33}{200}\) | B1 | |
| M1 | For using Newton's second law for motion along the rod | |
| \(0.195 \times \frac{12}{13} - \left(\frac{4}{7}\right) \times 0.165 = 0.024a\) | A1 | |
| Acceleration is \(3.57 \text{ ms}^{-2}\) | A1 | Allow acceleration \(= 25/7\) |
# Question 6:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[0.195\cos\theta = F]$ | M1 | For resolving forces horizontally |
| $F = 0.195\cos22.6 = 0.195 \times \frac{12}{13} = 0.18 = \frac{9}{50}$ | A1 | |
| $[R = 0.24 + 0.195\sin\theta]$ | M1 | For resolving forces vertically |
| $R = 0.24 + 0.195\sin22.6 = 0.24 + 0.195 \times \frac{5}{13} = 0.315 = \frac{63}{200}$ | A1 | |
| | M1 | For using $\mu = F/R$ |
| Coefficient $\mu = 4/7$ or $0.571$ | A1 | |
| | **Total: 6** | |
# Question 6 (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = 0.24 - 0.195\sin 22.6$ | | |
| $= 0.24 - 0.195 \times \frac{5}{13}$ | | |
| $= 0.165 = \frac{33}{200}$ | B1 | |
| | M1 | For using Newton's second law for motion along the rod |
| $0.195 \times \frac{12}{13} - \left(\frac{4}{7}\right) \times 0.165 = 0.024a$ | A1 | |
| Acceleration is $3.57 \text{ ms}^{-2}$ | A1 | Allow acceleration $= 25/7$ |
---6
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{48f66bd5-33c1-4ce9-85f9-69faf10e871c-4_149_410_306_518}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{48f66bd5-33c1-4ce9-85f9-69faf10e871c-4_133_406_260_1210}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
A small ring of mass 0.024 kg is threaded on a fixed rough horizontal rod. A light inextensible string is attached to the ring and the string is pulled with a force of magnitude 0.195 N at an angle of $\theta$ with the horizontal, where $\sin \theta = \frac { 5 } { 13 }$. When the angle $\theta$ is below the horizontal (see Fig. 1) the ring is in limiting equilibrium.\\
(i) Find the coefficient of friction between the ring and the rod.
When the angle $\theta$ is above the horizontal (see Fig. 2) the ring moves.\\
(ii) Find the acceleration of the ring.
\hfill \mbox{\textit{CAIE M1 2015 Q6 [10]}}