CAIE M1 2013 November — Question 1 4 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2013
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeParticle suspended by strings
DifficultyModerate -0.8 This is a straightforward equilibrium problem requiring resolution of forces in two perpendicular directions with a given angle. The solution involves basic trigonometry (finding sin α and cos α from tan α), then applying T cos α = mg and T sin α = F. It's simpler than average A-level mechanics as it's a standard two-force equilibrium with no complications, though it does require careful handling of the 8-15-17 triangle.
Spec3.03b Newton's first law: equilibrium3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 0

1 \includegraphics[max width=\textwidth, alt={}, center]{3e58aa5a-3789-4aaf-8656-b5b98cd7f693-2_291_591_255_776} A particle \(P\) of mass 0.3 kg is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point \(X\). A horizontal force of magnitude \(F \mathrm {~N}\) is applied to the particle, which is in equilibrium when the string is at an angle \(\alpha\) to the vertical, where \(\tan \alpha = \frac { 8 } { 15 }\) (see diagram). Find the tension in the string and the value of \(F\).

Question 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(T\cos\alpha = mg\)M1 For resolving forces vertically
Tension is \(3.4\) NA1
\(F = T\sin\alpha\)M1 For resolving forces horizontally
\(F = 1.6\)A1 [4]
## Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $T\cos\alpha = mg$ | M1 | For resolving forces vertically |
| Tension is $3.4$ N | A1 | |
| $F = T\sin\alpha$ | M1 | For resolving forces horizontally |
| $F = 1.6$ | A1 [4] | |

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\includegraphics[max width=\textwidth, alt={}, center]{3e58aa5a-3789-4aaf-8656-b5b98cd7f693-2_291_591_255_776}

A particle $P$ of mass 0.3 kg is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point $X$. A horizontal force of magnitude $F \mathrm {~N}$ is applied to the particle, which is in equilibrium when the string is at an angle $\alpha$ to the vertical, where $\tan \alpha = \frac { 8 } { 15 }$ (see diagram). Find the tension in the string and the value of $F$.

\hfill \mbox{\textit{CAIE M1 2013 Q1 [4]}}