| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Composite solid with hemisphere and cylinder/cone |
| Difficulty | Standard +0.3 This is a standard centre of mass question requiring application of the formula for composite bodies made of uniform laminas. While it involves multiple shapes (rectangle and triangle) and requires careful coordinate setup, the method is routine and well-practiced in M2. The multi-part structure guides students through the solution systematically, making it slightly easier than average. |
| Answer | Marks |
|---|---|
| 7 (i) | T = 42m(0.5 – 0.4) / 0.4 |
| Answer | Marks |
|---|---|
| Y = 3.7m AG | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | 3 | T = 10.5m |
| (ii) | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| ω =4.58rads−1 | M1 | |
| A1 | 2 | ft cv of T |
| Page 6 | Mark Scheme | Syllabus |
| GCE A LEVEL – May/June 2014 | 9709 | 52 |
| ( iii) | mv 2 /0.4=T × (0.4/0.5)±2m |
| Answer | Marks |
|---|---|
| or v = 1.6 ms | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | 3 | Either case |
| (iv) | Y = 10m |
| Answer | Marks |
|---|---|
| v = 2 ms | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | 3 | Fresh value of Y not Y(i) |
Question 7:
--- 7 (i) ---
7 (i) | T = 42m(0.5 – 0.4) / 0.4
10m = Y + T × (0.3 / 0.5)
Y = 3.7m AG | B1
M1
A1 | 3 | T = 10.5m
(ii) | 2
mω ×0.4=T×(0.4/0.5)
ω =4.58rads−1 | M1
A1 | 2 | ft cv of T
Page 6 | Mark Scheme | Syllabus | Paper
GCE A LEVEL – May/June 2014 | 9709 | 52
( iii) | mv 2 /0.4=T × (0.4/0.5)±2m
–1
v = 2.04 ms
–1
or v = 1.6 ms | M1
A1
A1 | 3 | Either case
(iv) | Y = 10m
2
10m=mv /0.4
–1
v = 2 ms | B1
M1
A1 | 3 | Fresh value of Y not Y(i)
Reject or ignore –2\includegraphics{figure_7}
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\hfill \mbox{\textit{CAIE M2 2014 Q7}}