| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2002 |
| Session | June |
| 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 a slope with friction. Students need to resolve forces, calculate friction (F = μR), and compare component forces—all routine M1 techniques with no problem-solving insight required. The small angle and simple coefficient make calculations easy. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For using \(N = mg\cos\alpha\) [\(5g\cos 12° (= 48.9)\)] and \(F = \mu N\) [\(0.2 \times 48.9\)] | M1 | Accept absence of \(g\) and/or sin/cos mix |
| Frictional force is \(9.78\) N (\(9.59\) from \(g = 9.8\) and \(9.60\) from \(g = 9.81\)) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Component of weight \(= 5g\sin 12° (= 10.4)\) (ft absence of \(g\) and/or sin/cos mix only) | B1 ft | B1 can be earned in (i) |
| For comparing component of weight with frictional force or for finding the acceleration (\(0.123\)) using both the component of weight and the frictional force | M1 | |
| Alternative: For comparing \(\mu\) with \(\tan 12°\) or for comparing the 'angle' of friction with angle of inclination | M1 | |
| \(0.2 < \tan 12°\) or \(\tan^{-1} 0.2 < 12°\) | A1 | Illustration: \(5a = 1.04 - 9.78 \rightarrow a < 0\) speed decreasing scores A1 ft; \(5a = 10.4 + 9.78 \rightarrow a > 0\) speed increasing scores A0 |
| Speed increasing (ft for arithmetic errors only) | A1 ft | Radians: Can score both M marks, allow one A mark for both \(8.44\) and \(-26.8\) (or \(-27\) or \(-30\)) (max 3 out of 5) |
# Question 2:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For using $N = mg\cos\alpha$ [$5g\cos 12° (= 48.9)$] and $F = \mu N$ [$0.2 \times 48.9$] | M1 | Accept absence of $g$ and/or sin/cos mix |
| Frictional force is $9.78$ N ($9.59$ from $g = 9.8$ and $9.60$ from $g = 9.81$) | A1 | |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Component of weight $= 5g\sin 12° (= 10.4)$ (ft absence of $g$ and/or sin/cos mix only) | B1 ft | B1 can be earned in (i) |
| For comparing component of weight with frictional force or for finding the acceleration ($0.123$) using both the component of weight and the frictional force | M1 | |
| **Alternative:** For comparing $\mu$ with $\tan 12°$ or for comparing the 'angle' of friction with angle of inclination | M1 | |
| $0.2 < \tan 12°$ or $\tan^{-1} 0.2 < 12°$ | A1 | Illustration: $5a = 1.04 - 9.78 \rightarrow a < 0$ speed decreasing scores A1 ft; $5a = 10.4 + 9.78 \rightarrow a > 0$ speed increasing scores A0 |
| Speed increasing (ft for arithmetic errors only) | A1 ft | Radians: Can score both M marks, allow one A mark for both $8.44$ and $-26.8$ (or $-27$ or $-30$) (max 3 out of 5) |
---2 A basket of mass 5 kg slides down a slope inclined at $12 ^ { \circ }$ to the horizontal. The coefficient of friction between the basket and the slope is 0.2 .\\
(i) Find the frictional force acting on the basket.\\
(ii) Determine whether the speed of the basket is increasing or decreasing.
\hfill \mbox{\textit{CAIE M1 2002 Q2 [5]}}