Edexcel M3 — Question 3 9 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeElastic string on smooth inclined plane
DifficultyStandard +0.8 Part (a) is a standard equilibrium problem with elastic strings requiring Hooke's law and resolving forces (routine for M3). Part (b) requires energy conservation with elastic potential energy over multiple phases (slack string, then stretched), finding maximum extension beyond equilibrium—this involves setting up and solving a quadratic equation with careful bookkeeping of energy terms, which is moderately challenging but follows standard M3 methodology.
Spec6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle

A particle of mass \(m\) kg is attached to the end \(B\) of a light elastic string \(AB\). The string has natural length \(l\) m and modulus of elasticity \(\lambda\) N. \includegraphics{figure_3} The end \(A\) is attached to a fixed point on a smooth plane inclined at an angle \(\alpha\) to the horizontal, as shown, and the particle rests in equilibrium with the length \(AB = \frac{5l}{4}\) m.
  1. Show that \(\lambda = 4 mg \sin \alpha\). [3 marks]
The particle is now moved and held at rest at \(A\) with the string slack. It is then gently released so that it moves down the plane along a line of greatest slope.
  1. Find the greatest distance from \(A\) that the particle reaches down the plane. [6 marks]

AnswerMarks
(a) \(T = mg\sin\alpha\)M1 A1 A1
\(\frac{\lambda}{4} = mg\sin\alpha\)
\(\lambda = 4mg\sin\alpha\)
(b) E.P.E. gained = grav. P.E. lost: \(\frac{4mg\sin\alpha}{2l}(d-l)^2 = mgd\sin\alpha\)M1 A1 A1
\(2d^2 - 5ld + 2l^2 = 0\)
\((2d-l)(d-2l) = 0\)
\(d = 2l\) m9 marks total
**(a)** $T = mg\sin\alpha$ | M1 A1 A1
$\frac{\lambda}{4} = mg\sin\alpha$ | 
$\lambda = 4mg\sin\alpha$ | 

**(b)** E.P.E. gained = grav. P.E. lost: $\frac{4mg\sin\alpha}{2l}(d-l)^2 = mgd\sin\alpha$ | M1 A1 A1
$2d^2 - 5ld + 2l^2 = 0$ | 
$(2d-l)(d-2l) = 0$ | 
$d = 2l$ m | 9 marks total
A particle of mass $m$ kg is attached to the end $B$ of a light elastic string $AB$. The string has natural length $l$ m and modulus of elasticity $\lambda$ N.

\includegraphics{figure_3}

The end $A$ is attached to a fixed point on a smooth plane inclined at an angle $\alpha$ to the horizontal, as shown, and the particle rests in equilibrium with the length $AB = \frac{5l}{4}$ m.

\begin{enumerate}[label=(\alph*)]
\item Show that $\lambda = 4 mg \sin \alpha$. [3 marks]
\end{enumerate}

The particle is now moved and held at rest at $A$ with the string slack. It is then gently released so that it moves down the plane along a line of greatest slope.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the greatest distance from $A$ that the particle reaches down the plane. [6 marks]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3  Q3 [9]}}