CAIE M1 2015 November — Question 4 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2015
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPulley systems
TypeLimiting equilibrium on incline
DifficultyStandard +0.3 This is a standard M1 pulley problem with limiting equilibrium in both directions. Students must resolve forces parallel to the plane, apply friction limits (F ≤ μR), and solve two inequalities to find the range of m. While it requires careful consideration of two limiting cases (about to slip up vs down), the techniques are routine for M1 and the algebra is straightforward.
Spec3.03l Newton's third law: extend to situations requiring force resolution3.03m Equilibrium: sum of resolved forces = 03.03t Coefficient of friction: F <= mu*R model

4 \includegraphics[max width=\textwidth, alt={}, center]{2a91fb7a-0eaf-4c50-8a2c-4755c0b44c17-2_499_784_1617_685} Blocks \(P\) and \(Q\), of mass \(m \mathrm {~kg}\) and 5 kg respectively, are attached to the ends of a light inextensible string. The string passes over a small smooth pulley which is fixed at the top of a rough plane inclined at \(35 ^ { \circ }\) to the horizontal. Block \(P\) is at rest on the plane and block \(Q\) hangs vertically below the pulley (see diagram). The coefficient of friction between block \(P\) and the plane is 0.2 . Find the set of values of \(m\) for which the two blocks remain at rest.

Question 4:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(F = 0.2 \times mg\cos 35\)B1 Maximum value of \(F\)
M1For resolving forces along the plane in either case
\(5g - mg\sin 35 - 0.2\,mg\cos 35 = 0\)A1 Equilibrium, on the point of moving up the plane
\(5g - Mg\sin 35 + 0.2\,Mg\cos 35 = 0\)A1 Equilibrium, on the point of moving down the plane
\(m = 6.78\) or \(M = 12.2\)M1 For solving either
\(6.78 \leqslant \text{mass} \leqslant 12.2\)A1 (6)
## Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = 0.2 \times mg\cos 35$ | B1 | Maximum value of $F$ |
| | M1 | For resolving forces along the plane in either case |
| $5g - mg\sin 35 - 0.2\,mg\cos 35 = 0$ | A1 | Equilibrium, on the point of moving up the plane |
| $5g - Mg\sin 35 + 0.2\,Mg\cos 35 = 0$ | A1 | Equilibrium, on the point of moving down the plane |
| $m = 6.78$ or $M = 12.2$ | M1 | For solving either |
| $6.78 \leqslant \text{mass} \leqslant 12.2$ | A1 (6) | |

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\includegraphics[max width=\textwidth, alt={}, center]{2a91fb7a-0eaf-4c50-8a2c-4755c0b44c17-2_499_784_1617_685}

Blocks $P$ and $Q$, of mass $m \mathrm {~kg}$ and 5 kg respectively, are attached to the ends of a light inextensible string. The string passes over a small smooth pulley which is fixed at the top of a rough plane inclined at $35 ^ { \circ }$ to the horizontal. Block $P$ is at rest on the plane and block $Q$ hangs vertically below the pulley (see diagram). The coefficient of friction between block $P$ and the plane is 0.2 . Find the set of values of $m$ for which the two blocks remain at rest.

\hfill \mbox{\textit{CAIE M1 2015 Q4 [6]}}