CAIE M1 2018 June — Question 5 7 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2018
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMotion on a slope
TypeRange of forces for equilibrium
DifficultyStandard +0.8 This question requires understanding that friction can act in either direction depending on whether P acts up or down the slope, setting up two equilibrium equations for maximum and minimum P, then using the constraint that P_max = 2P_min to solve for μ. It involves multiple conceptual steps beyond routine equilibrium problems and requires insight into how friction direction changes with applied force direction.
Spec3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes
5 A particle of mass 20 kg is on a rough plane inclined at an angle of \(60 ^ { \circ }\) to the horizontal. Equilibrium is maintained by a force of magnitude \(P \mathrm {~N}\) acting on the particle, in a direction parallel to a line of greatest slope of the plane. The greatest possible value of \(P\) is twice the least possible value of \(P\). Find the value of the coefficient of friction between the particle and the plane.
Question 5:
AnswerMarks Guidance
AnswerMarks Guidance
\(R = 20g\cos 60\ [= 100]\)B1
\(F = \mu\times 20g\cos 60\ [= 100\mu]\)M1 Use \(F = \mu R\)
M1Resolve along plane in either case
\((P_{\max} =)\ 20g\sin 60 + F\)A1 One correct equation
\((P_{\min} =)\ 20g\sin 60 - F\)A1 Second correct equation
\(20g\sin 60 + F = 2(20g\sin 60 - F)\)M1 Use of \(P_{\max} = 2P_{\min}\) to give four term equation in \(F\) or \(\mu\) or \(P\)
\(\mu = \frac{\sqrt{3}}{3} = 0.577\)A1
Alternative if \(P_{\min}\) acts down the plane:
\(P_{\min} = F - 20g\sin 60\)A1
\(20g\sin 60 + F = 2(F - 20g\sin 60)\)M1
\(\mu = 3\sqrt{3} = 5.196\)A1 
## Question 5:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $R = 20g\cos 60\ [= 100]$ | B1 | |
| $F = \mu\times 20g\cos 60\ [= 100\mu]$ | M1 | Use $F = \mu R$ |
| | M1 | Resolve along plane in either case |
| $(P_{\max} =)\ 20g\sin 60 + F$ | A1 | One correct equation |
| $(P_{\min} =)\ 20g\sin 60 - F$ | A1 | Second correct equation |
| $20g\sin 60 + F = 2(20g\sin 60 - F)$ | M1 | Use of $P_{\max} = 2P_{\min}$ to give four term equation in $F$ or $\mu$ or $P$ |
| $\mu = \frac{\sqrt{3}}{3} = 0.577$ | A1 | |
| **Alternative if $P_{\min}$ acts down the plane:** | | |
| $P_{\min} = F - 20g\sin 60$ | A1 | |
| $20g\sin 60 + F = 2(F - 20g\sin 60)$ | M1 | |
| $\mu = 3\sqrt{3} = 5.196$ | A1 | |

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5 A particle of mass 20 kg is on a rough plane inclined at an angle of $60 ^ { \circ }$ to the horizontal. Equilibrium is maintained by a force of magnitude $P \mathrm {~N}$ acting on the particle, in a direction parallel to a line of greatest slope of the plane. The greatest possible value of $P$ is twice the least possible value of $P$. Find the value of the coefficient of friction between the particle and the plane.\\

\hfill \mbox{\textit{CAIE M1 2018 Q5 [7]}}
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