| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | L-shaped or composite rectangular lamina |
| Difficulty | Standard +0.3 This is a standard centre of mass question from M2 requiring application of the formula for composite bodies. While it involves multiple parts and careful calculation with moments, it follows a routine procedure taught in all M2 courses with no novel problem-solving or geometric insight required, making it slightly easier than average. |
| Answer | Marks |
|---|---|
| 2 (i) | 10cos30 × 1.2sinθ – 10sin30 × 1.2cosθ |
| Answer | Marks |
|---|---|
| θ = 47(.0) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | 4 | Creating a 3 term solvable equation in |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) | µ = (10cos30 – 6)/(10sin30) | |
| µ = 0.532 | M1 | |
| A1 | 2 | For using F = µR with a reasonable |
Question 2:
--- 2 (i) ---
2 (i) | 10cos30 × 1.2sinθ – 10sin30 × 1.2cosθ
= 6 × 0.8sinθ
5.5923..sinθ = 6cosθ
θ = 47(.0)
OR
10 × 1.2sin(θ – 30) = 6 × 0.8sinθ or
10 × 1.2cos(120 – θ) = 6 × 0.8sinθ
5.5923..sinθ = 6cosθ
θ = 47(.0) | M1
A1
M1
A1
M1
A 1
M1
A1 | 4 | Creating a 3 term solvable equation in
sinθ and cosθ
Creating a 3 term solvable equation in
sinθ and cosθ
(ii) | µ = (10cos30 – 6)/(10sin30)
µ = 0.532 | M1
A1 | 2 | For using F = µR with a reasonable
attempt to find F and R\includegraphics{figure_2}
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\hfill \mbox{\textit{CAIE M2 2014 Q2}}