| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Block on rough horizontal surface – equilibrium (finding friction, normal reaction, or coefficient of friction) |
| Difficulty | Standard +0.3 This is a standard M1 equilibrium problem requiring resolution of forces in two perpendicular directions, followed by application of F=μR and F=ma. The angle is given in component form (sin/cos provided), eliminating trigonometric complexity. All steps are routine textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors3.03m 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 | Marks | Guidance |
| M1 | For resolving forces in the x direction or the y direction | |
| \(F_x - 6.1 - 5 \times 0.28 = 0\) and \(F_y + 4.8 - 5 \times 0.96 = 0\) | A1 | |
| Frictional force acts parallel to x axis and to the right | A1 | |
| \(F_y = 0 \rightarrow F = F_x \rightarrow\) Frictional force has magnitude \(7.5\ \text{N}\) | A1 [4] | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([\mu = 7.5/(1.25 \times 10)]\) | M1 | For using \(F = \mu R\) and \(R = mg\) |
| Coefficient is \(0.6\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([7.5 - 8.6 - 1.4 = 1.25a \rightarrow a = -2]\) | M1 | For applying Newton's second law |
| Magnitude of acceleration is \(2\ \text{ms}^{-2}\) | A1 | |
| Direction of acceleration is parallel to x axis and to the left | B1 [3] |
## Question 5:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| | M1 | For resolving forces in the x direction or the y direction |
| $F_x - 6.1 - 5 \times 0.28 = 0$ and $F_y + 4.8 - 5 \times 0.96 = 0$ | A1 | |
| Frictional force acts parallel to x axis and to the right | A1 | |
| $F_y = 0 \rightarrow F = F_x \rightarrow$ Frictional force has magnitude $7.5\ \text{N}$ | A1 **[4]** | AG |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[\mu = 7.5/(1.25 \times 10)]$ | M1 | For using $F = \mu R$ and $R = mg$ |
| Coefficient is $0.6$ | A1 **[2]** | |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[7.5 - 8.6 - 1.4 = 1.25a \rightarrow a = -2]$ | M1 | For applying Newton's second law |
| Magnitude of acceleration is $2\ \text{ms}^{-2}$ | A1 | |
| Direction of acceleration is parallel to x axis and to the left | B1 **[3]** | |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{8d64372d-0b9a-4b93-8c41-7096c813f714-4_620_623_255_760}
A small block of mass 1.25 kg is on a horizontal surface. Three horizontal forces, with magnitudes and directions as shown in the diagram, are applied to the block. The angle $\theta$ is such that $\cos \theta = 0.28$ and $\sin \theta = 0.96$. A horizontal frictional force also acts on the block, and the block is in equilibrium.\\
(i) Show that the magnitude of the frictional force is 7.5 N and state the direction of this force.\\
(ii) Given that the block is in limiting equilibrium, find the coefficient of friction between the block and the surface.
The force of magnitude 6.1 N is now replaced by a force of magnitude 8.6 N acting in the same direction, and the block begins to move.\\
(iii) Find the magnitude and direction of the acceleration of the block.
\hfill \mbox{\textit{CAIE M1 2011 Q5 [9]}}