CAIE M1 2022 June — Question 4 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2022
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeEquilibrium of particle under coplanar forces
DifficultyStandard +0.3 This is a standard equilibrium problem requiring resolution of forces in two perpendicular directions to form two simultaneous equations. While it involves trigonometry and solving for two unknowns, it's a routine textbook exercise with a clear method that students practice extensively. The 6 marks reflect straightforward algebraic manipulation rather than conceptual difficulty, making it slightly easier than average.
Spec3.03b Newton's first law: equilibrium3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 0
\includegraphics{figure_4} Three coplanar forces of magnitudes 20 N, 100 N and \(F\) N act at a point. The directions of these forces are shown in the diagram. Given that the three forces are in equilibrium, find \(F\) and \(\alpha\). [6]
Question 4:
AnswerMarks Guidance
4Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
4Attempt to resolve in any direction M1
Correct number of terms.
F cos α – 20 cos 40 – 100 sin 20 = 0
AnswerMarks
[F cos α = 15.320…+ 34.202… = 49.5229…]A1
F sin α + 20 sin 40 – 100 cos 20 = 0
AnswerMarks
[F sin α = 93.969… − 12.855… = 81.1135…]A1
49.52292 81.11352
AnswerMarks Guidance
F M1 OE; Attempt to solve for F; one term missing in total
81.1135
α tan1
 
AnswerMarks Guidance
49.5229M1 OE; Attempt to solve for α; one term missing in total
F = 95(.0), α = 58.6A1 F = 95.0364… and α = 58.5943…
Alternative mark scheme for question 4: For candidates who use cosine and/or sine rule
AnswerMarks Guidance
Attempt at cosine rule from triangle of forcesM1 Must use lengths 100 and 20 with a suitable angle
F2 1002 202 210020cos70A1 Correct
F = 95[.0]A1
95.0364 20 95.0364 100
 OR 
AnswerMarks Guidance
sin70 sin sin70 sinM1 Attempt at sin rule
A1where 70 where 40
α = 58.6A1 α = 58.5943…
QuestionAnswer Marks
4Alternative mark scheme for question 4: For candidates who resolve in other directions
Attempt to resolve (e.g. parallel or perpendicular to 100 N)M1 For resolving. Allow sin/cos mix. Allow sign error.
Correct number of terms.
Fsin2020sin201000 Fsin2093.159
AnswerMarks
 A1
Fcos2020cos200 Fcos2018.793
AnswerMarks Guidance
 A1
F  93.1592 18.7932M1 OE; Attempt to solve for F; one term missing in total
93.159
tan1 20
 
AnswerMarks Guidance
18.793M1 OE; Attempt to solve for α; one term missing in total
F = 95[.0], α = 58.6A1 F = 95.0364… and α = 58.5943…
6
AnswerMarks Guidance
QuestionAnswer Marks 
Question 4:
4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
4 | Attempt to resolve in any direction | M1 | For resolving. Allow sin/cos mix. Allow sign error.
Correct number of terms.
F cos α – 20 cos 40 – 100 sin 20 = 0
[F cos α = 15.320…+ 34.202… = 49.5229…] | A1
F sin α + 20 sin 40 – 100 cos 20 = 0
[F sin α = 93.969… − 12.855… = 81.1135…] | A1
49.52292 81.11352
F  | M1 | OE; Attempt to solve for F; one term missing in total
81.1135
α tan1
 
49.5229 | M1 | OE; Attempt to solve for α; one term missing in total
F = 95(.0), α = 58.6 | A1 | F = 95.0364… and α = 58.5943…
Alternative mark scheme for question 4: For candidates who use cosine and/or sine rule
Attempt at cosine rule from triangle of forces | M1 | Must use lengths 100 and 20 with a suitable angle
F2 1002 202 210020cos70 | A1 | Correct
F = 95[.0] | A1
95.0364 20 95.0364 100
 OR 
sin70 sin sin70 sin | M1 | Attempt at sin rule
A1 | where 70 where 40
α = 58.6 | A1 | α = 58.5943…
Question | Answer | Marks | Guidance
4 | Alternative mark scheme for question 4: For candidates who resolve in other directions
Attempt to resolve (e.g. parallel or perpendicular to 100 N) | M1 | For resolving. Allow sin/cos mix. Allow sign error.
Correct number of terms.
Fsin2020sin201000 Fsin2093.159
  | A1
Fcos2020cos200 Fcos2018.793
  | A1
F  93.1592 18.7932 | M1 | OE; Attempt to solve for F; one term missing in total
93.159
tan1 20
 
18.793 | M1 | OE; Attempt to solve for α; one term missing in total
F = 95[.0], α = 58.6 | A1 | F = 95.0364… and α = 58.5943…
6
Question | Answer | Marks | Guidance
\includegraphics{figure_4}

Three coplanar forces of magnitudes 20 N, 100 N and $F$ N act at a point. The directions of these forces are shown in the diagram.

Given that the three forces are in equilibrium, find $F$ and $\alpha$. [6]

\hfill \mbox{\textit{CAIE M1 2022 Q4 [6]}}
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