Edexcel M1 — Question 6 14 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPulley systems
TypeForce on pulley from string
DifficultyChallenging +1.2 This is a multi-part connected particles problem requiring Newton's second law, energy methods, and projectile motion. While it involves several steps and careful tracking of the system through different phases, the techniques are standard M1 fare (finding tension and acceleration, then using kinematics/energy). The pulley force calculation requires resolving forces at an angle, which adds modest complexity. Overall, this is above-average difficulty for M1 but uses only routine methods without requiring novel insight.
Spec3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys

\includegraphics{figure_2} Figure 2 shows a particle \(A\) of mass 5 kg, lying on a smooth horizontal table which is 0.9 m above the floor. A light inextensible string of length 0.7 m connects \(A\) to a particle \(B\) of mass 2 kg. The string passes over a smooth pulley which is fixed to the edge of the table and \(B\) hangs vertically 0.4 m below the pulley. When the system is released from rest,
  1. show that the magnitude of the force exerted on the pulley is \(\frac{10\sqrt{5}}{7}\) g N. [7 marks]
  2. find the speed with which \(A\) hits the pulley. [3 marks]
When \(A\) hits the pulley, the string breaks and \(B\) subsequently falls freely under gravity.
  1. Find the speed with which \(B\) hits the ground. [4 marks]

(a) eqn. of motion for A: \(T = 5a\) (1)
eqn. of motion for Q: \(2g - T = 2a\) (2)
(1) + (2) gives \(2g = 7a\) i.e. \(a = \frac{2g}{7}\)
from (1), \(T = 5a = \frac{10}{7}g\) N
so force on pulley = \(\sqrt{T^2 + T^2} = T\sqrt{2}\)
AnswerMarks Guidance
\(= \frac{10\sqrt{2}}{7}g\) NM1 M1
(b) \(s = 0.3, u = 0, a = \frac{2}{7}g\) use \(v^2 = u^2 + 2as\)
AnswerMarks Guidance
\(v^2 = \frac{6}{35}g\) i.e. \(v = \sqrt{\frac{6}{35}g} = \sqrt{1.68} = 1.30\) ms\(^{-1}\) (3sf)M1 M1 A1
(c) B has 0.2 m left to fall
for B: \(u^2 = \frac{6}{35}g, s = 0.2, a = g\) use \(v^2 = u^2 + 2as\)
AnswerMarks Guidance
\(v^2 = \frac{6}{35}g + 2g(0.2)\) ∴ \(v^2 = 5.6; v = 2.4\) ms\(^{-1}\) (1dp)B1 M1
**(a)** eqn. of motion for A: $T = 5a$ (1)
eqn. of motion for Q: $2g - T = 2a$ (2)
(1) + (2) gives $2g = 7a$ i.e. $a = \frac{2g}{7}$
from (1), $T = 5a = \frac{10}{7}g$ N
so force on pulley = $\sqrt{T^2 + T^2} = T\sqrt{2}$
$= \frac{10\sqrt{2}}{7}g$ N | M1 | M1 | A1 | A1 | M1 | M1 A1 |

**(b)** $s = 0.3, u = 0, a = \frac{2}{7}g$ use $v^2 = u^2 + 2as$
$v^2 = \frac{6}{35}g$ i.e. $v = \sqrt{\frac{6}{35}g} = \sqrt{1.68} = 1.30$ ms$^{-1}$ (3sf) | M1 | M1 A1 |

**(c)** B has 0.2 m left to fall
for B: $u^2 = \frac{6}{35}g, s = 0.2, a = g$ use $v^2 = u^2 + 2as$
$v^2 = \frac{6}{35}g + 2g(0.2)$ ∴ $v^2 = 5.6; v = 2.4$ ms$^{-1}$ (1dp) | B1 | M1 | M1 A1 | (14) |

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\includegraphics{figure_2}

Figure 2 shows a particle $A$ of mass 5 kg, lying on a smooth horizontal table which is 0.9 m above the floor. A light inextensible string of length 0.7 m connects $A$ to a particle $B$ of mass 2 kg. The string passes over a smooth pulley which is fixed to the edge of the table and $B$ hangs vertically 0.4 m below the pulley.

When the system is released from rest,

\begin{enumerate}[label=(\alph*)]
\item show that the magnitude of the force exerted on the pulley is $\frac{10\sqrt{5}}{7}$ g N. [7 marks]
\item find the speed with which $A$ hits the pulley. [3 marks]
\end{enumerate}

When $A$ hits the pulley, the string breaks and $B$ subsequently falls freely under gravity.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the speed with which $B$ hits the ground. [4 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1  Q6 [14]}}