CAIE M2 2014 June — Question 5

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2014
SessionJune
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeL-shaped or composite rectangular lamina
DifficultyStandard +0.3 This is a standard centre of mass question requiring students to apply the formula for the centre of mass of a composite body by treating the removed section as negative mass. While it involves multiple steps (finding individual centres of mass and combining them), the method is routine and well-practiced in M2, making it slightly easier than average for A-level.
\includegraphics{figure_5}
Question 5:
AnswerMarks
5 (i)OG(arc) = 1.8sin(π/2)/(π/2)
OX(1.8×2+π×1.8)=1.1459×π×1.8
AnswerMarks
OX = 0.7(00) mB1
M1
AnswerMarks Guidance
A13 1.1459.. or 3.6/π
0.70017..
AnswerMarks
(ii)OY = 1.8tan22
OG(lamina) = 2×1.8sin(π/2)/(3π/2)
1.8tan22 × (W + 27.5) =
0.7W + 0.763943 × 27.5
AnswerMarks
W = 37(.3)NB1
B1
M1
A1
AnswerMarks Guidance
A15 C of M solid = 0.727247..m from O
C of M lamina = 0.763943.. or 2.4/π
27.5 4×1.8/ 3π −0.727247 =
       
W(0.727247 – 0.70017)
Accept to 2sf as sensitive to rounding
error
Question 5:
--- 5 (i) ---
5 (i) | OG(arc) = 1.8sin(π/2)/(π/2)
OX(1.8×2+π×1.8)=1.1459×π×1.8
OX = 0.7(00) m | B1
M1
A1 | 3 | 1.1459.. or 3.6/π
0.70017..
(ii) | OY = 1.8tan22
OG(lamina) = 2×1.8sin(π/2)/(3π/2)
1.8tan22 × (W + 27.5) =
0.7W + 0.763943 × 27.5
W = 37(.3)N | B1
B1
M1
A1
A1 | 5 | C of M solid = 0.727247..m from O
C of M lamina = 0.763943.. or 2.4/π
27.5 4×1.8/ 3π −0.727247 =
       
W(0.727247 – 0.70017)
Accept to 2sf as sensitive to rounding
error
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