CAIE M1 2022 June — Question 3 5 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2022
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFriction
TypeSingle angled force - find limiting friction or coefficient
DifficultyStandard +0.3 This is a standard M1 friction problem requiring resolution of forces in two directions, application of F ≤ μR, and solving a linear equation. The angle is given in convenient form (sin α = 0.28), making calculations straightforward. It's slightly above average difficulty due to requiring careful setup of the limiting friction condition, but follows a well-practiced method with no conceptual surprises.
Spec3.03e Resolve forces: two dimensions3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces

A crate of mass 300 kg is at rest on rough horizontal ground. The coefficient of friction between the crate and the ground is 0.5. A force of magnitude \(X\) N, acting at an angle \(\alpha\) above the horizontal, is applied to the crate, where \(\sin \alpha = 0.28\). Find the greatest value of \(X\) for which the crate remains at rest. [5]

Question 3:
AnswerMarks Guidance
3For attempt at resolving horizontally or vertically M1
number of terms.
AnswerMarks Guidance
R = 300g – 0.28X or R300gXsin16.3A1 α = 16.26…
0.96X – F = 0 or 0.96X 0.5(300gXsin)0
AnswerMarks Guidance
Or X cos 16.3 – F = 0 or X cos 16.30.5(300gXsin)0A1 Or using their F
Use of F = 0.5RM1 Use to get an equation in X only. Allow sin/cos mix. Allow sign
error. Allow g missing.
Must be from 2 term R, which is a linear combination of 300(g)
and a component of X
AnswerMarks
X = 1360 [1363.63…]A1
5
AnswerMarks Guidance
QuestionAnswer Marks
Question 3:
3 | For attempt at resolving horizontally or vertically | M1 | Allow sin/cos mix. Allow sign error. Allow g missing. Correct
number of terms.
R = 300g – 0.28X or R300gXsin16.3 | A1 | α = 16.26…
0.96X – F = 0 or 0.96X 0.5(300gXsin)0
Or X cos 16.3 – F = 0 or X cos 16.30.5(300gXsin)0 | A1 | Or using their F
Use of F = 0.5R | M1 | Use to get an equation in X only. Allow sin/cos mix. Allow sign
error. Allow g missing.
Must be from 2 term R, which is a linear combination of 300(g)
and a component of X
X = 1360 [1363.63…] | A1
5
Question | Answer | Marks | Guidance
A crate of mass 300 kg is at rest on rough horizontal ground. The coefficient of friction between the crate and the ground is 0.5. A force of magnitude $X$ N, acting at an angle $\alpha$ above the horizontal, is applied to the crate, where $\sin \alpha = 0.28$.

Find the greatest value of $X$ for which the crate remains at rest. [5]

\hfill \mbox{\textit{CAIE M1 2022 Q3 [5]}}