| 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 +0.8 This is a centre of mass problem requiring integration to find the centroid of a region bounded by a curve and coordinate axes. While the integration itself is straightforward (polynomial functions), students must correctly set up the moment integrals and apply the centroid formulas, which involves multiple steps and careful algebraic manipulation. This is moderately above average difficulty for A-level, as it combines calculus with mechanics concepts and requires systematic problem-solving rather than routine application. |
| Answer | Marks |
|---|---|
| 6 (i) | 0.6dv / dt =0.6g – 3v |
| Answer | Marks |
|---|---|
| t = 0.738 s | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | 5 | Newton's Second Law |
| Answer | Marks |
|---|---|
| (ii) | 0.6vdv / dx = –3v |
| Answer | Marks |
|---|---|
| x = 0.39 m | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | 3 | Newton's Second Law |
Question 6:
--- 6 (i) ---
6 (i) | 0.6dv / dt =0.6g – 3v
∫1/(10−5v)dv=∫dt
−1ln(10−5v)=t(+c)
5
Finds c or uses limits twice
t = 0.738 s | B1
M1
A1
M1
A1 | 5 | Newton's Second Law
0.6∫1/ ( 0.6g−3v) dv=∫dt
0.6ln(0.6g−3v )=t (+c )
−3
(ii) | 0.6vdv / dx = –3v
∫0.2dv=−∫dx
x = 0.39 m | B1
M1
A1 | 3 | Newton's Second Law
Integration with use of limits or finding c\includegraphics{figure_6}
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\hfill \mbox{\textit{CAIE M2 2014 Q6}}