| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Particle just remains at rest |
| Difficulty | Standard +0.3 This is a standard two-part friction problem on an inclined plane requiring resolution of forces parallel and perpendicular to the slope, application of F=μR at limiting equilibrium, and checking equilibrium conditions. The calculations are straightforward with given sin α = 0.1, though part (ii) requires understanding that if μ found in (i) exceeds tan α, the box remains at rest. Slightly above average due to the two-part reasoning structure, but still a textbook exercise. |
| 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 | Marks | Guidance |
| \([W\sin\alpha + F = 40]\) | M1 | For resolving forces parallel to the plane |
| \(F = 40 - 300 \times 0.1 \quad (= 10)\) | A1 | |
| \(R = 300\sqrt{(1 - 0.1^2)} \quad (= 298.496...)\) | B1 | |
| M1 | For using \(\mu = F/R\) | |
| Coefficient is 0.0335 | A1 | 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| [The component of weight (30 N) is greater than the frictional force (10 N)] | M1 | For comparing the weight component parallel to the plane and the frictional force or for using Newton's 2nd Law and finding the acceleration |
| Box does not remain in equilibrium | A1 | 2 marks |
## Question 4:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[W\sin\alpha + F = 40]$ | M1 | For resolving forces parallel to the plane |
| $F = 40 - 300 \times 0.1 \quad (= 10)$ | A1 | |
| $R = 300\sqrt{(1 - 0.1^2)} \quad (= 298.496...)$ | B1 | |
| | M1 | For using $\mu = F/R$ |
| Coefficient is 0.0335 | A1 | 5 marks |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| [The component of weight (30 N) is greater than the frictional force (10 N)] | M1 | For comparing the weight component parallel to the plane and the frictional force or for using Newton's 2nd Law and finding the acceleration |
| Box does not remain in equilibrium | A1 | 2 marks |
---4 A box of mass 30 kg is at rest on a rough plane inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = 0.1$, acted on by a force of magnitude 40 N . The force acts upwards and parallel to a line of greatest slope of the plane. The box is on the point of slipping up the plane.\\
(i) Find the coefficient of friction between the box and the plane.
The force of magnitude 40 N is removed.\\
(ii) Determine, giving a reason, whether or not the box remains in equilibrium.
\hfill \mbox{\textit{CAIE M1 2013 Q4 [7]}}