Edexcel M3 — Question 3 8 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Marks8
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TopicHooke's law and elastic energy
TypeElastic string equilibrium and statics
DifficultyStandard +0.3 This is a standard M3 elastic strings question testing Hooke's law, EPE formula, and energy considerations. Part (a) requires routine equilibrium (T=mg), part (b) applies the EPE formula directly, part (c) calculates work done by weight, and part (d) asks for conceptual understanding of energy dissipation when lowering gently. All techniques are textbook standard with no novel problem-solving required, making it slightly easier than average.
Spec6.02a Work done: concept and definition6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle

A particle \(P\) of mass \(m\) kg is attached to one end of a light elastic string of natural length 0·5 m and modulus of elasticity \(\frac{mg}{2}\) N. The other end of the string is attached to a fixed point \(O\) and \(P\) hangs vertically below \(O\).
  1. Find the stretched length of the string when \(P\) rests in equilibrium. [3 marks]
  2. Find the elastic potential energy stored in the string in the equilibrium position. [2 marks]
\(P\), which is still attached to the string, is now held at rest at \(O\) and then lowered gently into its equilibrium position.
  1. Find the work done by the weight of the particle as it moves from \(O\) to the equilibrium position. [2 marks]
  2. Explain the discrepancy between your answers to parts (b) and (c). [1 mark]

AnswerMarks
(a) \(mg = \frac{mge}{2(0.5)}\)M1 A1 M1 A1
\(e = 0.5 \times 2 = 1\) m
\(OP = 1.5\) m
(b) P.E. \(= \frac{mg(1)^2}{2(2)(0.5)} = \frac{mg}{2}\) JM1 A1 M1 A1
(c) Work done \(= mg \times 1.5 = \frac{3mg}{2}\) J
(d) Grav. P.E. lost > elastic P.E. gained, because the weight does work in moving the supportive force in (c)B1
Total: 8 marks
**(a)** $mg = \frac{mge}{2(0.5)}$ | M1 A1 M1 A1 | 
$e = 0.5 \times 2 = 1$ m | | 
$OP = 1.5$ m | | 

**(b)** P.E. $= \frac{mg(1)^2}{2(2)(0.5)} = \frac{mg}{2}$ J | M1 A1 M1 A1 | 

**(c)** Work done $= mg \times 1.5 = \frac{3mg}{2}$ J | | 

**(d)** Grav. P.E. lost > elastic P.E. gained, because the weight does work in moving the supportive force in (c) | B1 | 

**Total: 8 marks**
A particle $P$ of mass $m$ kg is attached to one end of a light elastic string of natural length 0·5 m and modulus of elasticity $\frac{mg}{2}$ N. The other end of the string is attached to a fixed point $O$ and $P$ hangs vertically below $O$.

\begin{enumerate}[label=(\alph*)]
\item Find the stretched length of the string when $P$ rests in equilibrium. [3 marks]
\item Find the elastic potential energy stored in the string in the equilibrium position. [2 marks]
\end{enumerate}

$P$, which is still attached to the string, is now held at rest at $O$ and then lowered gently into its equilibrium position.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the work done by the weight of the particle as it moves from $O$ to the equilibrium position. [2 marks]
\item Explain the discrepancy between your answers to parts (b) and (c). [1 mark]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3  Q3 [8]}}