CAIE M1 2017 June — Question 3 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2017
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeParticle suspended by strings
DifficultyModerate -0.3 This is a standard equilibrium problem requiring resolution of forces in two perpendicular directions and solving simultaneous equations. The setup is straightforward with clearly defined angles, and the method (resolve horizontally and vertically) is a routine textbook technique for M1. Slightly easier than average due to the direct application of standard methods with no geometric complications or proof elements.
Spec3.03m Equilibrium: sum of resolved forces = 0

3 \includegraphics[max width=\textwidth, alt={}, center]{4941e074-2f93-4a0c-80ba-0ca96a48389e-04_442_584_255_778} Two light inextensible strings are attached to a particle of weight 25 N . The strings pass over two smooth fixed pulleys and have particles of weights \(A \mathrm {~N}\) and \(B \mathrm {~N}\) hanging vertically at their ends. The sloping parts of the strings make angles of \(30 ^ { \circ }\) and \(40 ^ { \circ }\) respectively with the vertical (see diagram). The system is in equilibrium. Find the values of \(A\) and \(B\).

Question 3:
AnswerMarks Guidance
AnswerMark Guidance
*EITHER:*(M1) Resolve horizontally and/or vertically at the 25 N weight
\(A\cos 30 + B\cos 40 = 25\)A1
\(A\sin 30 = B\sin 40\)A1
M1Solve for \(A\) and/or \(B\)
\(A = 17.1\)A1
\(B = 13.3\)A1
*OR:*(M1) Attempt Lami's theorem
\(\dfrac{25}{\sin 70} = \dfrac{A}{\sin 140} = \dfrac{B}{\sin 150}\)A1 One correct equation
A1A second correct equation
M1Solve for \(A\) and/or \(B\)
\(A = 17.1\)A1
\(B = 13.3\)A1
Total: 6
## Question 3:

| Answer | Mark | Guidance |
|--------|------|----------|
| *EITHER:* | (M1) | Resolve horizontally and/or vertically at the 25 N weight |
| $A\cos 30 + B\cos 40 = 25$ | A1 | |
| $A\sin 30 = B\sin 40$ | A1 | |
| | M1 | Solve for $A$ and/or $B$ |
| $A = 17.1$ | A1 | |
| $B = 13.3$ | A1 | |
| *OR:* | (M1) | Attempt Lami's theorem |
| $\dfrac{25}{\sin 70} = \dfrac{A}{\sin 140} = \dfrac{B}{\sin 150}$ | A1 | One correct equation |
| | A1 | A second correct equation |
| | M1 | Solve for $A$ and/or $B$ |
| $A = 17.1$ | A1 | |
| $B = 13.3$ | A1 | |
| **Total: 6** | | |

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\includegraphics[max width=\textwidth, alt={}, center]{4941e074-2f93-4a0c-80ba-0ca96a48389e-04_442_584_255_778}

Two light inextensible strings are attached to a particle of weight 25 N . The strings pass over two smooth fixed pulleys and have particles of weights $A \mathrm {~N}$ and $B \mathrm {~N}$ hanging vertically at their ends. The sloping parts of the strings make angles of $30 ^ { \circ }$ and $40 ^ { \circ }$ respectively with the vertical (see diagram). The system is in equilibrium. Find the values of $A$ and $B$.\\

\hfill \mbox{\textit{CAIE M1 2017 Q3 [6]}}