CAIE M2 2010 November — Question 1 3 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2010
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeRotating disc with friction
DifficultyModerate -0.8 This is a straightforward application of circular motion formulas requiring only direct substitution into F = mω²r with all values given. It's a single-step calculation with no conceptual challenges or problem-solving required, making it easier than average but not trivial since students must recognize which formula applies.
Spec6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r

A horizontal circular disc rotates with constant angular speed \(9 \text{ rad s}^{-1}\) about its centre \(O\). A particle of mass \(0.05 \text{ kg}\) is placed on the disc at a distance \(0.4 \text{ m}\) from \(O\). The particle moves with the disc and no sliding takes place. Calculate the magnitude of the resultant force exerted on the particle by the disc. [3]

Question 1:
AnswerMarks
12
0.05 x 9 x 0.4 = HM1
H = 1.62 NA1
R = 1.70 NA1
[3]1.622 +0.52
Question 1:
1 | 2
0.05 x 9 x 0.4 = H | M1
H = 1.62 N | A1
R = 1.70 N | A1
[3] | 1.622 +0.52
A horizontal circular disc rotates with constant angular speed $9 \text{ rad s}^{-1}$ about its centre $O$. A particle of mass $0.05 \text{ kg}$ is placed on the disc at a distance $0.4 \text{ m}$ from $O$. The particle moves with the disc and no sliding takes place. Calculate the magnitude of the resultant force exerted on the particle by the disc. [3]

\hfill \mbox{\textit{CAIE M2 2010 Q1 [3]}}