| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Range of forces for equilibrium |
| Difficulty | Standard +0.3 This is a standard mechanics problem requiring resolution of forces on an inclined plane and application of friction laws. While it involves multiple components (weight resolution, normal reaction, limiting friction) and two cases, the method is routine and well-practiced in M1 courses. The calculations are straightforward with no conceptual surprises, making it slightly easier than average. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | For resolving forces parallel to the plane (either case) | |
| \([R = 197, F = 63.0]\) | M1 | For using \(F = 0.32R\) and \(R = 20g\cos10°\) (or \(20g\sin10°\) if part of consistent sin/cos interchange) |
| \(P = F + 20g\sin10°\) | A1 | |
| Least magnitude is \(97.8\) N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P = F - 20g\sin10°\) | A1ft | [6] ft with \(P\cos10°\) instead of P or sign error or cos instead of sin in component of weight |
| Least magnitude is \(28.3\) N | A1 |
| Answer | Marks |
|---|---|
| For \(P = F + 20\sin10°\) in (i) and \(P = F - 20\sin10°\) in (ii) | B1 |
| M1 | For using \(F = 0.32R\) and \(R = 20\cos10°\) |
| Least magnitude is \(9.78\) N in (i) and \(2.83\) in (ii) | A1 |
# Question 2:
## Part (i)
| M1 | For resolving forces parallel to the plane (either case)
$[R = 197, F = 63.0]$ | M1 | For using $F = 0.32R$ and $R = 20g\cos10°$ (or $20g\sin10°$ if part of consistent sin/cos interchange)
$P = F + 20g\sin10°$ | A1 |
Least magnitude is $97.8$ N | A1 |
## Part (ii)
$P = F - 20g\sin10°$ | A1ft | [6] ft with $P\cos10°$ instead of P or sign error or cos instead of sin in component of weight
Least magnitude is $28.3$ N | A1 |
**SR (for candidates who omit g) (max 3/6):**
For $P = F + 20\sin10°$ in **(i)** and $P = F - 20\sin10°$ in **(ii)** | B1 |
| M1 | For using $F = 0.32R$ and $R = 20\cos10°$
Least magnitude is $9.78$ N in **(i)** and $2.83$ in **(ii)** | A1 |
---2 A block of mass 20 kg is at rest on a plane inclined at $10 ^ { \circ }$ to the horizontal. A force acts on the block parallel to a line of greatest slope of the plane. The coefficient of friction between the block and the plane is 0.32 . Find the least magnitude of the force necessary to move the block,\\
(i) given that the force acts up the plane,\\
(ii) given instead that the force acts down the plane.
\hfill \mbox{\textit{CAIE M1 2008 Q2 [6]}}