Edexcel M3 2007 January — Question 3 9 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2007
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeHorizontal elastic string on rough surface
DifficultyStandard +0.8 This M3 question requires applying conservation of energy with elastic potential energy, friction work, and careful algebraic manipulation. While the setup is standard (horizontal elastic string with friction), students must correctly identify that extension = 4a/3 - a = a/3, calculate elastic PE using λx²/2a, and use work-energy principle to find friction coefficient. The multi-step energy accounting and algebraic manipulation elevate this above routine M3 questions but it follows a recognizable framework.
Spec3.03t Coefficient of friction: F <= mu*R model6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle

3. A particle \(P\) of mass \(m\) is attached to one end of a light elastic string, of natural length \(a\) and modulus of elasticity 3.6 mg . The other end of the string is fixed at a point \(O\) on a rough horizontal table. The particle is projected along the surface of the table from \(O\) with speed \(\sqrt { } ( 2 a g )\). At its furthest point from \(O\), the particle is at the point \(A\), where \(O A = \frac { 4 } { 3 } a\).
  1. Find, in terms of \(m , g\) and \(a\), the elastic energy stored in the string when \(P\) is at \(A\).
  2. Using the work-energy principle, or otherwise, find the coefficient of friction between \(P\) and the table.

AnswerMarks Guidance
(a) E.P.E. = \(\frac{1}{2} \cdot \frac{3.6mg}{a} \cdot x^2 = \frac{1}{2} \cdot \frac{3.6mg}{a} \left(\frac{a}{3}\right)^2 = 0.2 mga\)M1 A1 ↓ A1 (3)
(b) Friction = \(\mu mg \Rightarrow\) work done by friction = \(\mu mg\left(\frac{4a}{3}\right)\)M1 A1
Work-energy: \(\frac{1}{2}m \cdot 2ga = \mu mgd + 0.2mga\) (3 relevant terms)M1 A1√ ↓ M1 A1 (6)
Solving to find \(\mu\): \(\mu = 0.6\)
(b) 1st M1: allow for attempt to find work done by frictional force (i.e. not just finding friction). 2nd M1: "relevant" terms, i.e. energy or work terms! A1 f.t. on their work done by friction Guidance note
(a) E.P.E. = $\frac{1}{2} \cdot \frac{3.6mg}{a} \cdot x^2 = \frac{1}{2} \cdot \frac{3.6mg}{a} \left(\frac{a}{3}\right)^2 = 0.2 mga$ | M1 A1 ↓ A1 (3) |

(b) Friction = $\mu mg \Rightarrow$ work done by friction = $\mu mg\left(\frac{4a}{3}\right)$ | M1 A1 |

Work-energy: $\frac{1}{2}m \cdot 2ga = \mu mgd + 0.2mga$ (3 relevant terms) | M1 A1√ ↓ M1 A1 (6) |

Solving to find $\mu$: $\mu = 0.6$ | |

(b) 1st M1: allow for attempt to find work done by frictional force (i.e. not just finding friction). 2nd M1: "relevant" terms, i.e. energy or work terms! A1 f.t. on their work done by friction | | Guidance note

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3. A particle $P$ of mass $m$ is attached to one end of a light elastic string, of natural length $a$ and modulus of elasticity 3.6 mg . The other end of the string is fixed at a point $O$ on a rough horizontal table. The particle is projected along the surface of the table from $O$ with speed $\sqrt { } ( 2 a g )$. At its furthest point from $O$, the particle is at the point $A$, where $O A = \frac { 4 } { 3 } a$.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $m , g$ and $a$, the elastic energy stored in the string when $P$ is at $A$.
\item Using the work-energy principle, or otherwise, find the coefficient of friction between $P$ and the table.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3 2007 Q3 [9]}}