CAIE M1 2017 June — Question 5 8 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2017
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFriction
TypeParticle on inclined plane - force at angle to slope
DifficultyChallenging +1.2 This is a multi-step friction problem requiring resolution of forces in two directions, consideration of limiting friction inequality (F ≤ μR), and solving simultaneous inequalities to find a range for P. While it involves several components (inclined plane, angled force, friction limits), the techniques are standard M1 material with no novel insight required—just careful systematic application of equilibrium conditions and friction laws.
Spec3.03e Resolve forces: two dimensions3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes

\includegraphics{figure_5} A particle of mass \(0.12\) kg is placed on a plane which is inclined at an angle of \(40°\) to the horizontal. The particle is kept in equilibrium by a force of magnitude \(P\) N acting up the plane at an angle of \(30°\) above a line of greatest slope, as shown in the diagram. The coefficient of friction between the particle and the plane is \(0.32\). Find the set of possible values of \(P\). [8]

Question 5:
AnswerMarks Guidance
5M1 Resolve perpendicular to the plane, three terms
R + P sin 30 = 0.12g cos 40A1 R does not need to be the subject
F = 0.32RM1 Use F = µR
[P cos 30 + F = 0.12g sin 40]
AnswerMarks Guidance
minM1 About to slip down, 3 terms
[P cos 30 – F = 0.12g sin 40]
AnswerMarks Guidance
maxM1 About to slip up, 3 terms
[P cos 30 = 0.12g sin 40
±0.32 (0.12g cos 40 – P sin 30)]
OR
[Pcos 30 ± 0.32R = 0.12g sin 40
R + P sin 30 = 0.12g cos 40]
AnswerMarks Guidance
Must reach P =… in either methodM1 Substitute for F and solve for P in either case, 4 terms
OR solve a pair of simultaneous equations (each with 3
terms) in R and P for P in one of the cases
P = 1.04 P = 0.676
AnswerMarks Guidance
max minA1 For either correct
0.676 ⩽ P ⩽ 1.04A1
Total:8
QuestionAnswer Marks
Question 5:
5 | M1 | Resolve perpendicular to the plane, three terms
R + P sin 30 = 0.12g cos 40 | A1 | R does not need to be the subject
F = 0.32R | M1 | Use F = µR
[P cos 30 + F = 0.12g sin 40]
min | M1 | About to slip down, 3 terms
[P cos 30 – F = 0.12g sin 40]
max | M1 | About to slip up, 3 terms
[P cos 30 = 0.12g sin 40
±0.32 (0.12g cos 40 – P sin 30)]
OR
[Pcos 30 ± 0.32R = 0.12g sin 40
R + P sin 30 = 0.12g cos 40]
Must reach P =… in either method | M1 | Substitute for F and solve for P in either case, 4 terms
OR solve a pair of simultaneous equations (each with 3
terms) in R and P for P in one of the cases
P = 1.04 P = 0.676
max min | A1 | For either correct
0.676 ⩽ P ⩽ 1.04 | A1
Total: | 8
Question | Answer | Marks | Guidance
\includegraphics{figure_5}

A particle of mass $0.12$ kg is placed on a plane which is inclined at an angle of $40°$ to the horizontal. The particle is kept in equilibrium by a force of magnitude $P$ N acting up the plane at an angle of $30°$ above a line of greatest slope, as shown in the diagram. The coefficient of friction between the particle and the plane is $0.32$. Find the set of possible values of $P$.
[8]

\hfill \mbox{\textit{CAIE M1 2017 Q5 [8]}}