| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion down rough slope |
| Difficulty | Moderate -0.8 This is a straightforward mechanics problem requiring standard application of Newton's second law on an inclined plane with friction. Students resolve forces parallel and perpendicular to the slope, calculate friction force (μR), then find deceleration and use SUVAT equations. All steps are routine textbook exercises with no problem-solving insight required, making it easier than average but not trivial due to the multi-step calculation. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| \(F = 0.2 \times 6g \cos 8\) | B1 | |
| \([6g \sin 8 - F = 6a]\) | M1 | For use of Newton's second law |
| Deceleration is \(0.589 \text{ m s}^{-2}\) | A1 | 3; Accept \(a = -0.589\) |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | For use of \(0 = u^2 + 2as\) | |
| Distance is \(7.64 \text{ m}\) | A1 | 2 |
**(i)**
$F = 0.2 \times 6g \cos 8$ | B1 |
$[6g \sin 8 - F = 6a]$ | M1 | For use of Newton's second law
Deceleration is $0.589 \text{ m s}^{-2}$ | A1 | 3; Accept $a = -0.589$
**(ii)**
| M1 | For use of $0 = u^2 + 2as$
Distance is $7.64 \text{ m}$ | A1 | 2
---2 A block of mass 6 kg is sliding down a line of greatest slope of a plane inclined at $8 ^ { \circ }$ to the horizontal. The coefficient of friction between the block and the plane is 0.2 .\\
(i) Find the deceleration of the block.\\
(ii) Given that the initial speed of the block is $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, find how far the block travels.
\hfill \mbox{\textit{CAIE M1 2011 Q2 [5]}}