| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Equilibrium on slope with force parallel to slope |
| Difficulty | Moderate -0.3 This is a straightforward mechanics problem requiring resolution of forces on an inclined plane in two scenarios. Part (i) involves standard force resolution with weight components, tension, and friction (4 marks of routine calculation). Part (ii) applies the limiting friction condition F=μR, which is a direct application of a standard formula. The problem requires no novel insight—just systematic application of well-practiced techniques for inclined plane problems. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| \([F + T = 8 \times 10\sin20°]\) | M1 | For resolving forces parallel to the plane |
| Frictional component is 14.4 N | A1 | |
| \([R = 80\cos20°]\) | M1 | For resolving forces normal to the plane |
| Normal component is 75.2 N | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \([\Tcos20° + F\cos20° = R\sin20° \text{ and}\) | (M1) | For resolving forces horizontally and vertically |
| \(\Tsin20° + F\sin20° + R\cos20° = 8g]\) | ||
| \([\tan20° = (13\cos20° + F\cos20°) \div\) | (M1) | For attempting to solve for F or R |
| \((80 - 13\sin20° - F\sin20°) \to\) | ||
| \(F = 80\sin20° - 13 \text{ or}\) | ||
| \(\tan20° = (80 - R\cos20° - 13\sin20°) \div\) | ||
| \((R\sin20° - 13\cos20°) \to R = 80\cos20°]\) | ||
| Frictional component is 14.4 N | (A1) | |
| Normal component is 75.2 N | (A1) |
| Answer | Marks | Guidance |
|---|---|---|
| \(F = 8 \times 10\sin20° \text{ or } \mu = \tan20°\) | B1ft | ft following consistent sin/cos mix in (i) for F = \(8 \times 10\cos20°\) or \(\mu = \tan70°\) |
| Coefficient is 0.364 (accept 0.36) | B1 | [2] |
**(i)**
$[F + T = 8 \times 10\sin20°]$ | M1 | For resolving forces parallel to the plane
Frictional component is 14.4 N | A1 |
$[R = 80\cos20°]$ | M1 | For resolving forces normal to the plane
Normal component is 75.2 N | A1 | [4]
**SR (max 3 out of 4) for consistent sin/cos exchange** – method marks as above and **A1 (only) for F = 62.2 and R = 27.4**
**Alternative scheme for part (i):**
$[\Tcos20° + F\cos20° = R\sin20° \text{ and}$ | (M1) | For resolving forces horizontally and vertically
$\Tsin20° + F\sin20° + R\cos20° = 8g]$ | |
$[\tan20° = (13\cos20° + F\cos20°) \div$ | (M1) | For attempting to solve for F or R
$(80 - 13\sin20° - F\sin20°) \to$ | |
$F = 80\sin20° - 13 \text{ or}$ | |
$\tan20° = (80 - R\cos20° - 13\sin20°) \div$ | |
$(R\sin20° - 13\cos20°) \to R = 80\cos20°]$ | |
Frictional component is 14.4 N | (A1) |
Normal component is 75.2 N | (A1) |
**(ii)**
$F = 8 \times 10\sin20° \text{ or } \mu = \tan20°$ | B1ft | ft following consistent sin/cos mix in (i) for F = $8 \times 10\cos20°$ or $\mu = \tan70°$
Coefficient is 0.364 (accept 0.36) | B1 | [2]\includegraphics{figure_4}
A block of mass 8 kg is at rest on a plane inclined at 20° to the horizontal. The block is connected to a vertical wall at the top of the plane by a string. The string is taut and parallel to a line of greatest slope of the plane (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Given that the tension in the string is 13 N, find the frictional and normal components of the force exerted on the block by the plane. [4]
\end{enumerate}
The string is cut; the block remains at rest, but is on the point of slipping down the plane.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item Find the coefficient of friction between the block and the plane. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2009 Q4 [6]}}