CAIE M1 2024 June — Question 3 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2024
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 A-level mechanics students practice extensively. The 6 marks reflect multiple steps rather than conceptual difficulty.
Spec3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces
\includegraphics{figure_3} Four coplanar forces of magnitude \(P\) N, 10 N, 16 N and 2 N act at a point in the directions shown in the diagram. It is given that the forces are in equilibrium. Find the values of \(\theta\) and \(P\). [6]
Question 3:
AnswerMarks Guidance
3Resolving either direction to get an equation *M1
sin/cos mix.
10cos252cos4016sin
9.063077871.53208888616sin
 
7.53098898416sin
 
AnswerMarks
  sin0.4706868115  A1
P10sin2516cos2sin40
P4.22618261716cos1.285575219
 
AnswerMarks Guidance
 P5.51175783716cos A1 This may be with their.
10cos252cos40
Attempt to solve for sin1  
AnswerMarks Guidance
 16 DM1 From equation(s) with correct number of relevant terms.
Must be a numerical expression for .
AnswerMarks Guidance
Attempt to solve for P10sin2516costheir 2sin40DM1 From equation(s) with correct number of relevant terms.
Using their.
Must be a numerical expression for P.
AnswerMarks Guidance
28.1AND P19.6A1 28.07888819 and 19.6285636.
AWRT 28.1 and AWRT 19.6 from correct work.
6
AnswerMarks Guidance
QuestionAnswer Marks 
Question 3:
3 | Resolving either direction to get an equation | *M1 | Correct number of relevant terms; allow sign errors; allow
sin/cos mix.
10cos252cos4016sin
9.063077871.53208888616sin
 
7.53098898416sin
 
  sin0.4706868115   | A1
P10sin2516cos2sin40
P4.22618261716cos1.285575219
 
 P5.51175783716cos  | A1 | This may be with their.
10cos252cos40
Attempt to solve for sin1  
 16  | DM1 | From equation(s) with correct number of relevant terms.
Must be a numerical expression for .
Attempt to solve for P10sin2516costheir 2sin40 | DM1 | From equation(s) with correct number of relevant terms.
Using their.
Must be a numerical expression for P.
28.1AND P19.6 | A1 | 28.07888819 and 19.6285636.
AWRT 28.1 and AWRT 19.6 from correct work.
6
Question | Answer | Marks | Guidance
\includegraphics{figure_3}

Four coplanar forces of magnitude $P$ N, 10 N, 16 N and 2 N act at a point in the directions shown in the diagram. It is given that the forces are in equilibrium.

Find the values of $\theta$ and $P$. [6]

\hfill \mbox{\textit{CAIE M1 2024 Q3 [6]}}
This paper (7 questions)
View full paper
Q1 3 Q2 5 Q3 6 Q4 7 Q5 8 Q6 11 Q7 10