CAIE M1 2006 June — Question 5 8 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2006
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPulley systems
TypeForce on pulley from string
DifficultyStandard +0.3 This is a standard M1 pulley problem requiring resolution of forces at the pulley, equilibrium equations, and friction at limiting equilibrium. The three parts involve straightforward application of T = 4N from pulley geometry, friction F = μR, and then F = ma for the accelerating system. Slightly above average due to the pulley force resolution in part (i), but otherwise routine mechanics.
Spec3.03k Connected particles: pulleys and equilibrium3.03m Equilibrium: sum of resolved forces = 03.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model

5 \includegraphics[max width=\textwidth, alt={}, center]{b5873699-d207-4cad-9518-1321dc429c15-3_305_599_1717_774} Particles \(P\) and \(Q\) are attached to opposite ends of a light inextensible string. \(P\) is at rest on a rough horizontal table. The string passes over a small smooth pulley which is fixed at the edge of the table. \(Q\) hangs vertically below the pulley (see diagram). The force exerted on the string by the pulley has magnitude \(4 \sqrt { } 2 \mathrm {~N}\). The coefficient of friction between \(P\) and the table is 0.8 .
  1. Show that the tension in the string is 4 N and state the mass of \(Q\).
  2. Given that \(P\) is on the point of slipping, find its mass. A particle of mass 0.1 kg is now attached to \(Q\) and the system starts to move.
  3. Find the tension in the string while the particles are in motion.

AnswerMarks Guidance
(i) \(T = 4\sqrt{2}\cos 45° = 4\) NB1 2
Mass of \(Q\) is \(0.4\) kgB1
(ii) \(4 = 0.8 \times m_P \times 10\)M1 For using \(F = T\), \(F = \mu R\) (or \(m_Q g = \mu R\)) and \(R = mg\)
Mass of \(P\) is \(0.5\) kgA1 2
(iii)M1 For applying Newton's second law to \(P\) or to \(Q\)
\(T - 0.8 \times 0.5g = 0.5a\)A1 ft
\(0.5g - T = 0.5a\)
Alternative to either of the two A1 marks above:
Tension is \(4.5\) NA1 4
**(i)** $T = 4\sqrt{2}\cos 45° = 4$ N | B1 | 2 |
Mass of $Q$ is $0.4$ kg | B1 | |

**(ii)** $4 = 0.8 \times m_P \times 10$ | M1 | For using $F = T$, $F = \mu R$ (or $m_Q g = \mu R$) and $R = mg$ |
| | |
Mass of $P$ is $0.5$ kg | A1 | 2 |

**(iii)** | M1 | For applying Newton's second law to $P$ or to $Q$ |
$T - 0.8 \times 0.5g = 0.5a$ | A1 ft | |
$0.5g - T = 0.5a$ | | |
| | Alternative to either of the two A1 marks above: |
Tension is $4.5$ N | A1 | 4 | $5 - 4 = (0.5 + 0.5)a$ ... B1 |

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\includegraphics[max width=\textwidth, alt={}, center]{b5873699-d207-4cad-9518-1321dc429c15-3_305_599_1717_774}

Particles $P$ and $Q$ are attached to opposite ends of a light inextensible string. $P$ is at rest on a rough horizontal table. The string passes over a small smooth pulley which is fixed at the edge of the table. $Q$ hangs vertically below the pulley (see diagram). The force exerted on the string by the pulley has magnitude $4 \sqrt { } 2 \mathrm {~N}$. The coefficient of friction between $P$ and the table is 0.8 .\\
(i) Show that the tension in the string is 4 N and state the mass of $Q$.\\
(ii) Given that $P$ is on the point of slipping, find its mass.

A particle of mass 0.1 kg is now attached to $Q$ and the system starts to move.\\
(iii) Find the tension in the string while the particles are in motion.

\hfill \mbox{\textit{CAIE M1 2006 Q5 [8]}}