Edexcel M1 2002 June — Question 7 15 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2002
SessionJune
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPulley systems
TypeParticle on rough incline connected to particle on horizontal surface or other incline
DifficultyStandard +0.3 This is a standard M1 pulley system question requiring Newton's second law for connected particles, followed by kinematics with friction. Part (a) is routine application of F=ma to both particles with given answer to verify. Part (b) uses standard equations of motion. The question involves multiple steps but uses well-practiced techniques with no novel insight required, making it slightly easier than average for M1.
Spec3.03e Resolve forces: two dimensions3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys3.03v Motion on rough surface: including inclined planes6.02i Conservation of energy: mechanical energy principle

\includegraphics{figure_3} Particles \(A\) and \(B\), of mass \(2m\) and \(m\) respectively, are attached to the ends of a light inextensible string. The string passes over a small smooth pulley fixed at the edge of a rough horizontal table. Particle \(A\) is held on the table, while \(B\) rests on a smooth plane inclined at \(30°\) to the horizontal, as shown in Fig. 3. The string is in the same vertical plane as a line of greatest slope of the inclined plane. The coefficient of friction between \(A\) and the table is \(\mu\). The particle \(A\) is released from rest and begins to move. By writing down an equation of motion for each particle,
  1. show that, while both particles move with the string taut. Each particle has an acceleration of magnitude \(\frac{1}{5}(1 - 4\mu)g\). [7]
When each particle has moved a distance \(h\), the string breaks. The particle \(A\) comes to rest before reaching the pulley. Given that \(\mu = 0.2\),
  1. find, in terms of \(h\), the total distance moved by \(A\). [6]
For the model described above,
  1. state two physical factors, apart from air resistance, which could be taken into account to make the model more realistic. [2]

(a)
AnswerMarks Guidance
\(R = 2mg \Rightarrow F = 2\mu mg\)B1
\(A: T - 2\mu mg = 2ma\)M1 A1
\(B: mg \times \frac{1}{2} - T = ma\)M1 A1
Eliminating \(T\): \(3ma = \frac{1}{2}mg - 2\mu mg\)M1
\(a = \frac{1}{6}(1 - 4\mu)g\) \((\ast)\)A1 (7)
(b)
AnswerMarks Guidance
\(\mu = 0.2 \Rightarrow a = \frac{1}{30}g\)B1
when string breaks: \(v^2 = 2 \times \frac{1}{30}g \times h = \frac{1}{15}gh\)M1 A1
\(A\) decelerating with deceleration \(f \Rightarrow 2mf = 2\mu mg\)
\(f = \mu g = \frac{1}{5}g\)B1
Hence distance travelled during deceleration is given by \(\frac{1}{15}gh = 2 \times \frac{1}{2}gd\)M1
\(\Rightarrow d = \frac{1}{15}h\)
\(\therefore\) Total distance \(= \frac{2}{5}h\)A1 cso (6)
(c)
AnswerMarks Guidance
Any two from: weight of pulley; friction at pulley; friction on slope; weight of string; string extensible; 'spin' of particleB1 B1 (2)
Total: (15 marks)
Note: \((\ast)\) indicates final line is given on the paper; cso = correct solution only
## (a)
$R = 2mg \Rightarrow F = 2\mu mg$ | B1 |

$A: T - 2\mu mg = 2ma$ | M1 A1 |

$B: mg \times \frac{1}{2} - T = ma$ | M1 A1 |

Eliminating $T$: $3ma = \frac{1}{2}mg - 2\mu mg$ | M1 |

$a = \frac{1}{6}(1 - 4\mu)g$ $(\ast)$ | A1 | (7) |

## (b)
$\mu = 0.2 \Rightarrow a = \frac{1}{30}g$ | B1 |

when string breaks: $v^2 = 2 \times \frac{1}{30}g \times h = \frac{1}{15}gh$ | M1 A1 |

$A$ decelerating with deceleration $f \Rightarrow 2mf = 2\mu mg$ | |

$f = \mu g = \frac{1}{5}g$ | B1 |

Hence distance travelled during deceleration is given by $\frac{1}{15}gh = 2 \times \frac{1}{2}gd$ | M1 |

$\Rightarrow d = \frac{1}{15}h$ | |

$\therefore$ Total distance $= \frac{2}{5}h$ | A1 cso | (6) |

## (c)
Any two from: weight of pulley; friction at pulley; friction on slope; weight of string; string extensible; 'spin' of particle | B1 B1 | (2) |

**Total: (15 marks)**

---

**Note:** $(\ast)$ indicates final line is given on the paper; cso = correct solution only
\includegraphics{figure_3}

Particles $A$ and $B$, of mass $2m$ and $m$ respectively, are attached to the ends of a light inextensible string. The string passes over a small smooth pulley fixed at the edge of a rough horizontal table. Particle $A$ is held on the table, while $B$ rests on a smooth plane inclined at $30°$ to the horizontal, as shown in Fig. 3. The string is in the same vertical plane as a line of greatest slope of the inclined plane. The coefficient of friction between $A$ and the table is $\mu$. The particle $A$ is released from rest and begins to move.

By writing down an equation of motion for each particle,

\begin{enumerate}[label=(\alph*)]
\item show that, while both particles move with the string taut. Each particle has an acceleration of magnitude $\frac{1}{5}(1 - 4\mu)g$. [7]
\end{enumerate}

When each particle has moved a distance $h$, the string breaks. The particle $A$ comes to rest before reaching the pulley. Given that $\mu = 0.2$,

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find, in terms of $h$, the total distance moved by $A$. [6]
\end{enumerate}

For the model described above,

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item state two physical factors, apart from air resistance, which could be taken into account to make the model more realistic. [2]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2002 Q7 [15]}}