CAIE M1 2022 June — Question 2 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2022
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeResultant of coplanar forces
DifficultyModerate -0.3 This is a straightforward vector resolution problem requiring students to resolve four forces into components, sum them, and find the magnitude and direction of the resultant. While it involves multiple forces and some calculation, it's a standard textbook exercise testing routine application of component methods with no conceptual challenges or novel problem-solving required.
Spec3.03p Resultant forces: using vectors
\includegraphics{figure_2} Coplanar forces of magnitudes \(60\text{N}\), \(20\text{N}\), \(16\text{N}\) and \(14\text{N}\) act at a point in the directions shown in the diagram. Find the magnitude and direction of the resultant force. [6]
Question 2:
AnswerMarks Guidance
2Resolving either direction M1
X 20cos601416cos50 14.2846A1
Y 6020sin6016sin50 30.42278A1
R 14.28462 30.422782M1 Attempt to solve for R; one missing term in total
30.42278
 tan1    tan12.1297 
 14.2846 
 14.2846 
OR  tan1    tan10.4596 
AnswerMarks Guidance
30.42278M1 Attempt to solve for 𝜃 or 𝛼; one missing term in total
R = 33.6 N
Direction is 64.8° above the 14 N force
or 25.2° above the negative 𝑥-axis
or 25.2° left of the 60 N force
or bearing 335°
AnswerMarks Guidance
or 115° anticlockwise from the positive x-axisA1 Both correct.
OE; allow 64.9, 25.1
Giving an angle only is insufficient.
Direction may be seen on a diagram, with minimum of arrow
on resultant.
Arrows on both components only is A0 as it doesn’t show the
direction of the resultant.
However the direction is stated, it must be able to be drawn
uniquely.
6
AnswerMarks Guidance
QuestionAnswer Marks 
Question 2:
2 | Resolving either direction | M1 | 3 terms; allow sign errors and allow sin/cos mix
X 20cos601416cos50 14.2846 | A1
Y 6020sin6016sin50 30.42278 | A1
R 14.28462 30.422782 | M1 | Attempt to solve for R; one missing term in total
30.42278
 tan1    tan12.1297 
 14.2846 
 14.2846 
OR  tan1    tan10.4596 
30.42278 | M1 | Attempt to solve for 𝜃 or 𝛼; one missing term in total
R = 33.6 N
Direction is 64.8° above the 14 N force
or 25.2° above the negative 𝑥-axis
or 25.2° left of the 60 N force
or bearing 335°
or 115° anticlockwise from the positive x-axis | A1 | Both correct.
OE; allow 64.9, 25.1
Giving an angle only is insufficient.
Direction may be seen on a diagram, with minimum of arrow
on resultant.
Arrows on both components only is A0 as it doesn’t show the
direction of the resultant.
However the direction is stated, it must be able to be drawn
uniquely.
6
Question | Answer | Marks | Guidance
\includegraphics{figure_2}

Coplanar forces of magnitudes $60\text{N}$, $20\text{N}$, $16\text{N}$ and $14\text{N}$ act at a point in the directions shown in the diagram.

Find the magnitude and direction of the resultant force. [6]

\hfill \mbox{\textit{CAIE M1 2022 Q2 [6]}}
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