CAIE M1 2015 November — Question 2 5 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2015
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFriction
TypeRing on horizontal rod equilibrium
DifficultyModerate -0.3 This is a straightforward limiting equilibrium problem requiring resolution of forces in two directions and application of F=μR. The setup is standard with given coefficient of friction and angle, requiring only systematic application of equilibrium conditions with no geometric insight or multi-step problem-solving—slightly easier than average A-level mechanics.
Spec3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03t Coefficient of friction: F <= mu*R model

2 \includegraphics[max width=\textwidth, alt={}, center]{f23ea8e7-9b81-4192-8c20-8c46aabfecca-2_195_719_1114_712} A ring of mass 0.2 kg is threaded on a fixed rough horizontal rod and a light inextensible string is attached to the ring at an angle \(\alpha\) above the horizontal, where \(\cos \alpha = 0.96\). The ring is in limiting equilibrium with the tension in the string \(T \mathrm {~N}\) (see diagram). Given that the coefficient of friction between the ring and the rod is 0.25 , find the value of \(T\).
[0pt] [5]

Question 2:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(F = T\cos\alpha = 0.96T\)B1
\(R = 0.2g - T\sin\alpha = 2 - 0.28T\)B1
\(0.96T = 0.25(2 - 0.28T)\)M1 For using \(F = \mu R\)
\((0.96 + 0.07)T = 0.5 \rightarrow T = \ldots\)M1 For solving resultant equation for \(T\)
\(T = 0.485\)A1 Total: 5 marks
## Question 2:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = T\cos\alpha = 0.96T$ | B1 | |
| $R = 0.2g - T\sin\alpha = 2 - 0.28T$ | B1 | |
| $0.96T = 0.25(2 - 0.28T)$ | M1 | For using $F = \mu R$ |
| $(0.96 + 0.07)T = 0.5 \rightarrow T = \ldots$ | M1 | For solving resultant equation for $T$ |
| $T = 0.485$ | A1 | Total: 5 marks |

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\includegraphics[max width=\textwidth, alt={}, center]{f23ea8e7-9b81-4192-8c20-8c46aabfecca-2_195_719_1114_712}

A ring of mass 0.2 kg is threaded on a fixed rough horizontal rod and a light inextensible string is attached to the ring at an angle $\alpha$ above the horizontal, where $\cos \alpha = 0.96$. The ring is in limiting equilibrium with the tension in the string $T \mathrm {~N}$ (see diagram). Given that the coefficient of friction between the ring and the rod is 0.25 , find the value of $T$.\\[0pt]
[5]

\hfill \mbox{\textit{CAIE M1 2015 Q2 [5]}}