Edexcel M3 2022 January — Question 1 6 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2022
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeElastic string equilibrium and statics
DifficultyStandard +0.3 This is a standard M3 elastic strings problem requiring application of Hooke's law and equilibrium conditions to two connected particles. While it involves multiple steps and careful bookkeeping of extensions in different sections, the method is routine: resolve forces at each particle, apply Hooke's law to each string section, and solve simultaneous equations. No novel insight required, just systematic application of standard techniques.
Spec6.02h Elastic PE: 1/2 k x^2

  1. A light elastic string \(A B\) has natural length \(11 a\) and modulus of elasticity \(6 m g\)
A particle of mass \(4 m\) is attached to the point \(C\) on the string where \(A C = 8 a\) and a particle of mass \(2 m\) is attached to the end \(B\) \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{a1365c54-4910-449b-b270-c56c1bc5a751-02_581_202_429_957} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} The end \(A\) of the string is attached to a fixed point and the string hangs vertically below \(A\) with the particle of mass \(4 m\) in equilibrium at the point \(P\) and the particle of mass \(2 m\) in equilibrium at the point \(Q\), as shown in Figure 1.
  1. Find the length \(A P\)
  2. Find the length \(P Q\)

Question 1:
Part (a):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(T_{AP} = 6mg\)M1 For resolving vertically for the system
\(\frac{6mgx}{8a} = 6mg\)M1 Use of Hooke's Law to set up equation using tension; M0 if \(11a\) used for natural length
\(AP = 16a\)A1 cao
Part (b):
AnswerMarks Guidance
Working/AnswerMark Guidance
\(2mg = \frac{6mgy}{3a}\)M1 M0 if \(11a\) used for natural length
\(y = a\)A1 cao
\(PQ = 4a\)A1 cao
# Question 1:

## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $T_{AP} = 6mg$ | M1 | For resolving vertically for the system |
| $\frac{6mgx}{8a} = 6mg$ | M1 | Use of Hooke's Law to set up equation using tension; M0 if $11a$ used for natural length |
| $AP = 16a$ | A1 | cao |

## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $2mg = \frac{6mgy}{3a}$ | M1 | M0 if $11a$ used for natural length |
| $y = a$ | A1 | cao |
| $PQ = 4a$ | A1 | cao |

---
\begin{enumerate}
  \item A light elastic string $A B$ has natural length $11 a$ and modulus of elasticity $6 m g$
\end{enumerate}

A particle of mass $4 m$ is attached to the point $C$ on the string where $A C = 8 a$ and a particle of mass $2 m$ is attached to the end $B$

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{a1365c54-4910-449b-b270-c56c1bc5a751-02_581_202_429_957}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

The end $A$ of the string is attached to a fixed point and the string hangs vertically below $A$ with the particle of mass $4 m$ in equilibrium at the point $P$ and the particle of mass $2 m$ in equilibrium at the point $Q$, as shown in Figure 1.\\
(a) Find the length $A P$\\
(b) Find the length $P Q$

\hfill \mbox{\textit{Edexcel M3 2022 Q1 [6]}}