| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion down smooth slope |
| Difficulty | Moderate -0.8 This is a straightforward two-stage kinematics problem with standard mechanics setup. Part (i) requires resolving forces on a smooth slope (mg sin 30°) and applying v = u + at. Part (ii) uses the velocity from the slope as initial velocity on rough ground with constant friction, then applies v² = u² + 2as. All steps are routine applications of SUVAT equations with no conceptual challenges or novel problem-solving required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| 2(i) | a = gsin30 = 5 | 1 |
| 2.5 = 0 + 5t | 1 | M1 |
| t = 0.5 Time = 0.5 s | 1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2(ii) | v2 = 0 + 2 × 5 × 3 = 30 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 0 = 30 + 2 × (–2) × s | 1 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Distance = 7.5 m | 1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| v2 = 0 + 2 × 5 × 3 = 30 | 1 | (B1) |
| 0.5 × 0.5 × 30 = 1 × distance | 1 | (M1) |
| Distance = 7.5 m | 1 | (A1) |
| Answer | Marks | Guidance |
|---|---|---|
| PE lost = 0.5 × 10 × 3 sin30 = 7.5 | 1 | (B1) |
| 7.5 = 1 × distance | 1 | (M1) |
| Distance = 7.5 m | 1 | (A1) |
Question 2:
--- 2(i) ---
2(i) | a = gsin30 = 5 | 1 | B1
2.5 = 0 + 5t | 1 | M1 | Using v = u + at
t = 0.5 Time = 0.5 s | 1 | A1
3
--- 2(ii) ---
2(ii) | v2 = 0 + 2 × 5 × 3 = 30 | 1 | B1
–1 = 0.5a → a = –2
0 = 30 + 2 × (–2) × s | 1 | M1 | For applying Newton’s second law to the particle and
using v2 = u2 + 2as
Distance = 7.5 m | 1 | A1
First alternative method for 2(ii)
v2 = 0 + 2 × 5 × 3 = 30 | 1 | (B1)
0.5 × 0.5 × 30 = 1 × distance | 1 | (M1) | KE lost = WD against Friction
Distance = 7.5 m | 1 | (A1)
Second alternative method for 2(ii)
PE lost = 0.5 × 10 × 3 sin30 = 7.5 | 1 | (B1) | Using PE lost = mgh
7.5 = 1 × distance | 1 | (M1) | PE lost = WD against Friction
Distance = 7.5 m | 1 | (A1)
3
MarksA particle of mass 0.5 kg starts from rest and slides down a line of greatest slope of a smooth plane. The plane is inclined at an angle of 30° to the horizontal.
\begin{enumerate}[label=(\roman*)]
\item Find the time taken for the particle to reach a speed of 2.5 m s$^{-1}$. [3]
\item Find the distance that the particle travels along the ground before it comes to rest. [3]
\end{enumerate}
When the particle has travelled 3 m down the slope from its starting point, it reaches rough horizontal ground at the bottom of the slope. The frictional force acting on the particle is 1 N.
\hfill \mbox{\textit{CAIE M1 Q2 [6]}}