CAIE M1 2018 November — Question 1 4 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2018
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeSmooth ring on string
DifficultyStandard +0.3 This is a straightforward equilibrium problem requiring resolution of forces in two perpendicular directions. Students need to identify the three forces (weight, tension, applied force), use the given angles (45° and 90°), and solve two simultaneous equations. While it requires careful diagram interpretation and systematic force resolution, it's a standard M1 equilibrium question with no conceptual surprises—slightly easier than average due to the convenient angles and clear setup.
Spec3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces

A smooth ring \(R\) of mass \(m\) kg is threaded on a light inextensible string \(ARB\). The ends of the string are attached to fixed points \(A\) and \(B\) with \(A\) vertically above \(B\). The string is taut and angle \(ARB = 90°\). The angle between the part \(AR\) of the string and the vertical is \(45°\). The ring is held in equilibrium in this position by a force of magnitude \(2.5\) N, acting on the ring in the direction \(BR\) (see diagram). Calculate the tension in the string and the mass of the ring. [4] \includegraphics{figure_1}

Question 1:
AnswerMarks Guidance
1[T cos 45 + T cos 45 = 2.5 cos 45] M1
T = 1.25 NA1
[2.5 sin 45 = mg]M1 For resolving vertically
Mass of ring = 0.177 kgA1 Allow m = √2/8
First alternative method for Q1
AnswerMarks Guidance
[2.5 = T + mg cos 45]M1 Resolve forces along BR
[T = mg cos 45]M1 Resolve forces perpendicular to BR and eliminate T or m
T = 1.25 NA1
Mass of ring = 0.177 kgA1 Allow m = √2/8
Second alternative method for Q1
2Tcos45 2.5 mg
= =
sin135 sin90 sin135
or
2.5−T T mg
= =
AnswerMarks Guidance
sin135 sin135 sin90M1 Attempt to apply Lami’s theorem,
M1All three terms of Lami attempted
T = 1.25 NA1
Mass of ring = 0.177 kgA1 Allow m = √2/8
4
AnswerMarks Guidance
QuestionAnswer Marks
Question 1:
1 | [T cos 45 + T cos 45 = 2.5 cos 45] | M1 | For resolving horizontally
T = 1.25 N | A1
[2.5 sin 45 = mg] | M1 | For resolving vertically
Mass of ring = 0.177 kg | A1 | Allow m = √2/8
First alternative method for Q1
[2.5 = T + mg cos 45] | M1 | Resolve forces along BR
[T = mg cos 45] | M1 | Resolve forces perpendicular to BR and eliminate T or m
T = 1.25 N | A1
Mass of ring = 0.177 kg | A1 | Allow m = √2/8
Second alternative method for Q1
2Tcos45 2.5 mg
= =
sin135 sin90 sin135
or
2.5−T T mg
= =
sin135 sin135 sin90 | M1 | Attempt to apply Lami’s theorem,
M1 | All three terms of Lami attempted
T = 1.25 N | A1
Mass of ring = 0.177 kg | A1 | Allow m = √2/8
4
Question | Answer | Marks | Guidance
A smooth ring $R$ of mass $m$ kg is threaded on a light inextensible string $ARB$. The ends of the string are attached to fixed points $A$ and $B$ with $A$ vertically above $B$. The string is taut and angle $ARB = 90°$. The angle between the part $AR$ of the string and the vertical is $45°$. The ring is held in equilibrium in this position by a force of magnitude $2.5$ N, acting on the ring in the direction $BR$ (see diagram). Calculate the tension in the string and the mass of the ring. [4]

\includegraphics{figure_1}

\hfill \mbox{\textit{CAIE M1 2018 Q1 [4]}}