CAIE M1 2021 November — Question 3 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2021
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeEquilibrium of particle under coplanar forces
DifficultyModerate -0.3 This is a standard equilibrium problem requiring resolution of forces in two perpendicular directions. Students must apply sin α = 3/5 to find cos α = 4/5, then set up and solve two simultaneous equations. While it involves multiple steps and trigonometry, it follows a routine textbook method with no novel insight required, making it slightly easier than average.
Spec1.05g Exact trigonometric values: for standard angles3.03m Equilibrium: sum of resolved forces = 0
3 \includegraphics[max width=\textwidth, alt={}, center]{e1b91e54-a3ae-436c-a4f7-7095891f7034-04_519_616_260_762} Coplanar forces of magnitudes \(24 \mathrm {~N} , P \mathrm {~N} , 20 \mathrm {~N}\) and 36 N act at a point in the directions shown in the diagram. The system is in equilibrium. Given that \(\sin \alpha = \frac { 3 } { 5 }\), find the values of \(P\) and \(\theta\).
Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
Attempt at resolving in any directionM1 Correct number of terms. No substitution for \(\alpha\) required.
\(P\cos\theta = (36-24)\cos 36.9\) or \(P\cos\theta = (36-24) \times 0.8\)A1
\(P\sin\theta + 20 = (24+36)\sin 36.9 = 14.4 + 21.6\) or \(P\sin\theta + 20 = 60 \times 0.6 = 36\)A1
\(P\cos\theta = 9.6,\ P\sin\theta = 16 \Rightarrow P = \sqrt{16^2 + 9.6^2}\)M1 Correct method for solving equations for \(P\). OE
\(\theta = \tan^{-1}\!\left(\dfrac{16}{9.6}\right)\)M1 Correct method for solving equations for \(\theta\). OE e.g. \(\theta = \tan^{-1}\!\left(\dfrac{5}{3}\right)\)
\(P = 18.7\), \(\theta = 59[.0]\)A1 Allow \(P = \dfrac{16\sqrt{34}}{5}\). Allow \(P = 18.6\). 
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Attempt at resolving in any direction | M1 | Correct number of terms. No substitution for $\alpha$ required. |
| $P\cos\theta = (36-24)\cos 36.9$ or $P\cos\theta = (36-24) \times 0.8$ | A1 | |
| $P\sin\theta + 20 = (24+36)\sin 36.9 = 14.4 + 21.6$ or $P\sin\theta + 20 = 60 \times 0.6 = 36$ | A1 | |
| $P\cos\theta = 9.6,\ P\sin\theta = 16 \Rightarrow P = \sqrt{16^2 + 9.6^2}$ | M1 | Correct method for solving equations for $P$. OE |
| $\theta = \tan^{-1}\!\left(\dfrac{16}{9.6}\right)$ | M1 | Correct method for solving equations for $\theta$. OE e.g. $\theta = \tan^{-1}\!\left(\dfrac{5}{3}\right)$ |
| $P = 18.7$, $\theta = 59[.0]$ | A1 | Allow $P = \dfrac{16\sqrt{34}}{5}$. Allow $P = 18.6$. |
3\\
\includegraphics[max width=\textwidth, alt={}, center]{e1b91e54-a3ae-436c-a4f7-7095891f7034-04_519_616_260_762}

Coplanar forces of magnitudes $24 \mathrm {~N} , P \mathrm {~N} , 20 \mathrm {~N}$ and 36 N act at a point in the directions shown in the diagram. The system is in equilibrium.

Given that $\sin \alpha = \frac { 3 } { 5 }$, find the values of $P$ and $\theta$.\\

\hfill \mbox{\textit{CAIE M1 2021 Q3 [6]}}
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