| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | String at angle to slope |
| Difficulty | Standard +0.3 This is a standard M1 statics problem on an inclined plane with a force at an angle. It requires resolving forces in two directions (parallel and perpendicular to the plane) and applying equilibrium conditions, but follows a routine procedure with no novel insight needed. The given sin α simplifies calculations, and all three parts follow directly from standard force resolution—slightly easier than average due to its straightforward structure. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03t Coefficient of friction: F <= mu*R model |
| Answer | Marks | Guidance |
|---|---|---|
| \([R + 0.6\sin \alpha = 0.5g \cos \alpha]\) | M1 | For resolving forces perpendicular to the plane |
| Normal component is 4.63(2) N | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(F + 0.6\cos \alpha = 0.5g \sin \alpha\) | M1 | For resolving forces parallel to a line of greatest slope |
| A1 | ||
| Frictional component is 0.824 N | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mu = F/R\) | M1 | For using \(\mu = F/R\) |
| Coefficient is 0.178 | A1 ft | 2 |
**(i)**
$[R + 0.6\sin \alpha = 0.5g \cos \alpha]$ | M1 | For resolving forces perpendicular to the plane
Normal component is 4.63(2) N | A1 | 2
**(ii)**
$F + 0.6\cos \alpha = 0.5g \sin \alpha$ | M1 | For resolving forces parallel to a line of greatest slope
| A1 |
Frictional component is 0.824 N | A1 | 3
**(iii)**
$\mu = F/R$ | M1 | For using $\mu = F/R$
Coefficient is 0.178 | A1 ft | 23\\
\includegraphics[max width=\textwidth, alt={}, center]{2bb3c9bb-60f0-440d-a148-b4db3478ca31-2_241_535_1247_806}
A particle $P$ of mass 0.5 kg rests on a rough plane inclined at angle $\alpha$ to the horizontal, where $\sin \alpha = 0.28$. A force of magnitude 0.6 N , acting upwards on $P$ at angle $\alpha$ from a line of greatest slope of the plane, is just sufficient to prevent $P$ sliding down the plane (see diagram). Find\\
(i) the normal component of the contact force on $P$,\\
(ii) the frictional component of the contact force on $P$,\\
(iii) the coefficient of friction between $P$ and the plane.
\hfill \mbox{\textit{CAIE M1 2012 Q3 [7]}}