Edexcel M1 2005 January — Question 5 13 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2005
SessionJanuary
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPulley systems
TypeParticle on rough horizontal surface, particle hanging
DifficultyStandard +0.3 This is a standard M1 pulley problem requiring SUVAT equations, Newton's second law for two connected particles, and friction calculations. While it involves multiple steps (kinematics → dynamics → friction coefficient), each step follows routine procedures taught in M1 with no novel problem-solving required. The multi-part structure and 13 total marks indicate moderate length, but the techniques are all textbook applications, making it slightly easier than the average A-level question.
Spec3.02d Constant acceleration: SUVAT formulae3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model

\includegraphics{figure_4} A block of wood \(A\) of mass 0.5 kg rests on a rough horizontal table and is attached to one end of a light inextensible string. The string passes over a small smooth pulley \(P\) fixed at the edge of the table. The other end of the string is attached to a ball \(B\) of mass 0.8 kg which hangs freely below the pulley, as shown in Figure 4. The coefficient of friction between \(A\) and the table is \(\mu\). The system is released from rest with the string taut. After release, \(B\) descends a distance of 0.4 m in 0.5 s. Modelling \(A\) and \(B\) as particles, calculate
  1. the acceleration of \(B\), [3]
  2. the tension in the string, [4]
  3. the value of \(\mu\). [5]
  4. State how in your calculations you have used the information that the string is inextensible. [1]

Question 5:
AnswerMarks
5(a) ‘s = ut + ½at2’ for B: 0.4 = ½ a(0.5)2 M1 A1
a = 3.2 m s–2 A1
(3)
(b) N2L for B: 0.8g – T = 0.8 x 3.2 M1 A1√
T = 5.28 or 5.3 N M1 A1
(4)
(c) A: F = μ x 0.5g B1
N2L for A: T – F = 0.5a M1 A1
Sub and solve μ = 0.75 or 0.751 M1 A1
(5)
(d) Same acceleration for A and B. B1
(1)
Question 5:
5 | (a) ‘s = ut + ½at2’ for B: 0.4 = ½ a(0.5)2 M1 A1
a = 3.2 m s–2 A1
(3)
(b) N2L for B: 0.8g – T = 0.8 x 3.2 M1 A1√
↓
T = 5.28 or 5.3 N M1 A1
(4)
(c) A: F = μ x 0.5g B1
N2L for A: T – F = 0.5a M1 A1
↓
Sub and solve μ = 0.75 or 0.751 M1 A1
(5)
(d) Same acceleration for A and B. B1
(1)
\includegraphics{figure_4}

A block of wood $A$ of mass 0.5 kg rests on a rough horizontal table and is attached to one end of a light inextensible string. The string passes over a small smooth pulley $P$ fixed at the edge of the table. The other end of the string is attached to a ball $B$ of mass 0.8 kg which hangs freely below the pulley, as shown in Figure 4. The coefficient of friction between $A$ and the table is $\mu$. The system is released from rest with the string taut. After release, $B$ descends a distance of 0.4 m in 0.5 s. Modelling $A$ and $B$ as particles, calculate

\begin{enumerate}[label=(\alph*)]
\item the acceleration of $B$, [3]
\item the tension in the string, [4]
\item the value of $\mu$. [5]
\item State how in your calculations you have used the information that the string is inextensible. [1]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2005 Q5 [13]}}