Edexcel M1 — Question 4 10 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPulley systems
TypeThree or more connected particles
DifficultyStandard +0.8 This is a multi-part pulley problem requiring careful consideration of three connected particles with one (C) resting inside another (B). Students must recognize that C and B move together, apply Newton's second law to multiple bodies, resolve forces in different directions, and handle the non-trivial reaction force calculation. The conceptual leap that C experiences the same acceleration as B while having a reduced normal reaction is above-average difficulty for M1.
Spec3.03d Newton's second law: 2D vectors3.03f Weight: W=mg3.03o Advanced connected particles: and pulleys

4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{4fe54579-ac77-46f9-85e1-2e95963d6b3e-3_467_348_201_708} \captionsetup{labelformat=empty} \caption{Fig. 1}
\end{figure} Figure 1 shows a weight \(A\) of mass 6 kg connected by a light, inextensible string which passes over a smooth, fixed pulley to a box \(B\) of mass 5 kg . There is an object \(C\) of mass 3 kg resting on the horizontal floor of box \(B\). The system is released from rest. Find, giving your answers in terms of \(g\),
  1. the acceleration of the system,
  2. the force on the pulley.
  3. Show that the reaction between \(C\) and the floor of \(B\) is \(\frac { 18 } { 7 } \mathrm {~g}\) newtons.

Part (a)
AnswerMarks
Eqn. of motion for A: \(T - 6g = 6a\) (1)M1
Eqn. of motion for B & C: \(8g - T = 8a\) (2)M1
\((1) + (2)\) gives \(2g = 14a\) i.e. \(a = \frac{g}{7}\) m\(s^{-2}\)M1 A1
Part (b)
AnswerMarks
Sub. \(a\) into (1) to get \(T = 6a + 6g = \frac{6g}{7} + 6g\)M1
Force on pulley = \(2T = \frac{96g}{7}\)M1 A1
Part (c)
AnswerMarks Guidance
Resolve \(\downarrow\) for C: \(3g - R = 3 \times \frac{g}{7}\)M1
\(R = 3g - \frac{3g}{7} = \frac{18g}{7}\)M1 A1 (10)
**Part (a)**
Eqn. of motion for A: $T - 6g = 6a$ (1) | M1 |
Eqn. of motion for B & C: $8g - T = 8a$ (2) | M1 |
$(1) + (2)$ gives $2g = 14a$ i.e. $a = \frac{g}{7}$ m$s^{-2}$ | M1 A1 |

**Part (b)**
Sub. $a$ into (1) to get $T = 6a + 6g = \frac{6g}{7} + 6g$ | M1 |
Force on pulley = $2T = \frac{96g}{7}$ | M1 A1 |

**Part (c)**
Resolve $\downarrow$ for C: $3g - R = 3 \times \frac{g}{7}$ | M1 |
$R = 3g - \frac{3g}{7} = \frac{18g}{7}$ | M1 A1 | (10)
4.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{4fe54579-ac77-46f9-85e1-2e95963d6b3e-3_467_348_201_708}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}

Figure 1 shows a weight $A$ of mass 6 kg connected by a light, inextensible string which passes over a smooth, fixed pulley to a box $B$ of mass 5 kg . There is an object $C$ of mass 3 kg resting on the horizontal floor of box $B$.

The system is released from rest. Find, giving your answers in terms of $g$,
\begin{enumerate}[label=(\alph*)]
\item the acceleration of the system,
\item the force on the pulley.
\item Show that the reaction between $C$ and the floor of $B$ is $\frac { 18 } { 7 } \mathrm {~g}$ newtons.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1  Q4 [10]}}