| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Range of forces for equilibrium |
| Difficulty | Moderate -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 force components and the limiting friction condition, it follows a well-established method with straightforward calculations—slightly easier than average due to its routine nature and clear structure. |
| Spec | 3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Correct 4 force diagram | B1 | Angles shown. \(F\) either up/down slope. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Attempt to resolve forces parallel to the plane | M1 | Three terms, allow sign errors. |
| \(P + F = 12g\sin 25\ [= 50.7]\) | A1 | Must have correct direction of \(F\) here. |
| \(R = 12g\cos 25\ [= 108.8]\) | B1 | |
| \(F = 120\cos 25 \times 0.35\ [= 38.1]\); \(P + 38.1 = 120\sin 25\) | M1 | Attempt to solve for \(P\) using equations with correct number of terms. \(R\) must be a single-term component of \(12g\). |
| \(P = 12.6\) | A1 | \(P = 12.64926\ldots\) Allow \(P = 12.7\) |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct 4 force diagram | B1 | Angles shown. $F$ either up/down slope. |
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt to resolve forces parallel to the plane | M1 | Three terms, allow sign errors. |
| $P + F = 12g\sin 25\ [= 50.7]$ | A1 | Must have correct direction of $F$ here. |
| $R = 12g\cos 25\ [= 108.8]$ | B1 | |
| $F = 120\cos 25 \times 0.35\ [= 38.1]$; $P + 38.1 = 120\sin 25$ | M1 | Attempt to solve for $P$ using equations with correct number of terms. $R$ must be a single-term component of $12g$. |
| $P = 12.6$ | A1 | $P = 12.64926\ldots$ Allow $P = 12.7$ |4 A particle of mass 12 kg is stationary on a rough plane inclined at an angle of $25 ^ { \circ }$ to the horizontal. A force of magnitude $P \mathrm {~N}$ acting parallel to a line of greatest slope of the plane is used to prevent the particle sliding down the plane. The coefficient of friction between the particle and the plane is 0.35 .
\begin{enumerate}[label=(\alph*)]
\item Draw a sketch showing the forces acting on the particle.
\item Find the least possible value of $P$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q4 [6]}}