CAIE M1 2012 November — Question 6 9 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2012
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMotion on a slope
TypeRange of forces for equilibrium
DifficultyStandard +0.3 This is a standard mechanics problem requiring resolution of forces on an inclined plane and application of friction inequalities (F ≤ μR). While it involves multiple steps (resolving perpendicular and parallel to plane, considering two friction cases), the method is routine and well-practiced in M1. The calculation is straightforward with no conceptual surprises, making it slightly easier than average.
Spec3.03m Equilibrium: sum of resolved forces = 03.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes

6 \includegraphics[max width=\textwidth, alt={}, center]{631ddcd9-17c0-4a15-8671-40788c3a84d3-3_255_511_794_817} The diagram shows a particle of mass 0.6 kg on a plane inclined at \(25 ^ { \circ }\) to the horizontal. The particle is acted on by a force of magnitude \(P \mathrm {~N}\) directed up the plane parallel to a line of greatest slope. The coefficient of friction between the particle and the plane is 0.36 . Given that the particle is in equilibrium, find the set of possible values of \(P\).

AnswerMarks Guidance
\([P = \pm F + 0.6g\sin 25°]\)M1 For resolving forces in the direction of \(P\)
\(P_{\max} = F + 0.6g\sin 25°\) or '\(P = F + 0.6g\sin 25°\) when the particle is about to slide upwards'A1
\(P_{\min} = -F + 0.6g\sin 25°\) or '\(P = -F + 0.6g\sin 25°\) when the particle is about to slide downwards'A1
\(R = 0.6g\cos 25°\)B1
\([F = 0.36 \times 0.6g\cos 25°]\)M1 For using \(F = \mu R\)
\([P_{\max} = 0.36 \times 0.6g\cos 25° + 0.6g\sin 25°, P_{\min} = -0.36 \times 0.6g\cos 25° + 0.6g\sin 25°]\)DM1 For substituting for \(F\) to obtain values of \(P_{\max}\) and \(P_{\min}\)
\(P_{\max} = 4.49, P_{\min} = 0.578\) (accept 0.58)A1 Dependent on first M mark
\(\quad\)M1 For identifying range of value for equilibrium; AEF; Accept 0.58 instead of 0.578 and accept \(<\) instead of \(\leq\)
Set of values is \(\{P; 0.578 \leq P \leq 4.49\}\)A1 9 marks total
$[P = \pm F + 0.6g\sin 25°]$ | M1 | For resolving forces in the direction of $P$

$P_{\max} = F + 0.6g\sin 25°$ or '$P = F + 0.6g\sin 25°$ when the particle is about to slide upwards' | A1 |

$P_{\min} = -F + 0.6g\sin 25°$ or '$P = -F + 0.6g\sin 25°$ when the particle is about to slide downwards' | A1 |

$R = 0.6g\cos 25°$ | B1 |

$[F = 0.36 \times 0.6g\cos 25°]$ | M1 | For using $F = \mu R$

$[P_{\max} = 0.36 \times 0.6g\cos 25° + 0.6g\sin 25°, P_{\min} = -0.36 \times 0.6g\cos 25° + 0.6g\sin 25°]$ | DM1 | For substituting for $F$ to obtain values of $P_{\max}$ and $P_{\min}$
$P_{\max} = 4.49, P_{\min} = 0.578$ (accept 0.58) | A1 | Dependent on first M mark

$\quad$ | M1 | For identifying range of value for equilibrium; AEF; Accept 0.58 instead of 0.578 and accept $<$ instead of $\leq$
Set of values is $\{P; 0.578 \leq P \leq 4.49\}$ | A1 | 9 marks total

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\includegraphics[max width=\textwidth, alt={}, center]{631ddcd9-17c0-4a15-8671-40788c3a84d3-3_255_511_794_817}

The diagram shows a particle of mass 0.6 kg on a plane inclined at $25 ^ { \circ }$ to the horizontal. The particle is acted on by a force of magnitude $P \mathrm {~N}$ directed up the plane parallel to a line of greatest slope. The coefficient of friction between the particle and the plane is 0.36 . Given that the particle is in equilibrium, find the set of possible values of $P$.

\hfill \mbox{\textit{CAIE M1 2012 Q6 [9]}}