CAIE M1 2023 November — Question 2 5 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2023
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeSmooth ring on string
DifficultyStandard +0.3 This is a straightforward three-force equilibrium problem requiring resolution of forces in two perpendicular directions and solving simultaneous equations. The setup is clearly defined with given angles, and the method is standard M1 content with no conceptual subtleties—slightly easier than average since it's a direct application of equilibrium conditions without complications like friction or multiple bodies.
Spec3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces

\includegraphics{figure_2} The diagram shows a smooth ring \(R\), of mass \(m\) kg, threaded on a light inextensible string. A horizontal force of magnitude 2 N acts on \(R\). The ends of the string are attached to fixed points \(A\) and \(B\) on a vertical wall. The part \(AR\) of the string makes an angle of 30° with the vertical, the part \(BR\) makes an angle of 40° with the vertical and the string is taut. The ring is in equilibrium. Find the tension in the string and find the value of \(m\). [5]

Question 2:
AnswerMarks Guidance
2Attempt to resolve in at least one direction to form an equation. *M1
sign errors; allow sin/cos mix;
allow with different T ’s.
AnswerMarks Guidance
Tsin30+Tsin40−2=0A1 If different T’s then allow
M1A1A0 max.
AnswerMarks Guidance
Tcos30−Tcos40−mg=0A1 Allow with their T.
Attempt to solve for T or mDM1 From equation(s) with correct
number of relevant terms.
AnswerMarks Guidance
Tension T = 1.75, m = 0.0175A1 T =1.7501 m=0.017497
awrt 1.75 for T www, and awrt
0.0175 for m www.
5
AnswerMarks Guidance
QuestionAnswer Marks
Question 2:
2 | Attempt to resolve in at least one direction to form an equation. | *M1 | Correct number of terms; allow
sign errors; allow sin/cos mix;
allow with different T ’s.
Tsin30+Tsin40−2=0 | A1 | If different T’s then allow
M1A1A0 max.
Tcos30−Tcos40−mg=0 | A1 | Allow with their T.
Attempt to solve for T or m | DM1 | From equation(s) with correct
number of relevant terms.
Tension T = 1.75, m = 0.0175 | A1 | T =1.7501 m=0.017497
awrt 1.75 for T www, and awrt
0.0175 for m www.
5
Question | Answer | Marks | Guidance
\includegraphics{figure_2}

The diagram shows a smooth ring $R$, of mass $m$ kg, threaded on a light inextensible string. A horizontal force of magnitude 2 N acts on $R$. The ends of the string are attached to fixed points $A$ and $B$ on a vertical wall. The part $AR$ of the string makes an angle of 30° with the vertical, the part $BR$ makes an angle of 40° with the vertical and the string is taut. The ring is in equilibrium.

Find the tension in the string and find the value of $m$. [5]

\hfill \mbox{\textit{CAIE M1 2023 Q2 [5]}}