Edexcel M3 2007 January — Question 7 16 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2007
SessionJanuary
Marks16
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimple Harmonic Motion
TypeComplete motion cycle with slack phase
DifficultyChallenging +1.2 This is a standard M3/Further Mechanics SHM question with elastic strings requiring multiple techniques (equilibrium conditions, SHM equations, energy conservation, and motion under gravity). While it involves several parts and the slack phase adds complexity, the solution path follows well-established methods taught in M3. The calculations are straightforward once the approach is identified, making it moderately above average difficulty but not requiring novel insight.
Spec6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle

  1. A particle \(P\) of mass 0.25 kg is attached to one end of a light elastic string. The string has natural length 0.8 m and modulus of elasticity \(\lambda \mathrm { N }\). The other end of the string is attached to a fixed point \(A\). In its equilibrium position, \(P\) is 0.85 m vertically below \(A\).
    1. Show that \(\lambda = 39.2\).
    The particle is now displaced to a point \(B , 0.95 \mathrm {~m}\) vertically below \(A\), and released from rest.
  2. Prove that, while the string remains stretched, \(P\) moves with simple harmonic motion of period \(\frac { \pi } { 7 } \mathrm {~s}\).
  3. Calculate the speed of \(P\) at the instant when the string first becomes slack. The particle first comes to instantaneous rest at the point \(C\).
  4. Find, to 3 significant figures, the time taken for \(P\) to move from \(B\) to \(C\).

AnswerMarks Guidance
(a) \(T = \frac{\lambda}{0.8}(0.05) = 0.25g\)M1
\(\lambda = \frac{(0.8)(0.25g)}{0.05} = 39.2\) (*)A1 (2)
(b) \(T = \frac{39.2}{0.8}(x + 0.05)\)M1
\(mg - T = ma\) (3 term equn)M1
\(0.25g - \frac{39.2}{0.8}(x + 0.05) = 0.25\ddot{x}\) (or equivalent)A1
\(\ddot{x} = -196x\)A1 ↓ M1 A1 cso (6)
SHM with period \(\frac{2\pi}{\omega} = \frac{2\pi}{14} = \frac{\pi}{7}\) s (*)
(c) \(v = 14\sqrt{(0.1)^2 - (0.05)^2}\)M1 A1√
\(= 1.21(24\ldots) \approx 1.21\) m s\(^{-1}\) (3 s.f.) Accept \(7\sqrt{3}/10\)A1 (3)
(d) Time \(T\) under gravity = \(\frac{1.21 \ldots}{g}\) (= 0.1237 s)B1√
Complete method for time \(T'\) from B to slack.M1 A1
[\(\uparrow\) e.g. \(\frac{\pi}{28} + t\), where \(0.05 = 0.1\sin 14t\) OR \(T'\), where \(-0.05 = 0.1\cos 14T'\)]
\(T'' = 0.1496\)sA1
Total time = \(T + T' = 0.273\) sA1 (5)
(b) 1st M1 must have extn as \(x + k\) with \(k \neq 0\) (but allow M1 if e.g. \(x + 0.15\)), or must justify later Guidance note
For last four marks, must be using \(\ddot{x}\) (not a) Guidance note
(c) Using \(x = 0\) is M0 Guidance note
(d) M1 – must be using distance for when string goes slack. Using \(x = -0.1\) (i.e. assumed end of the oscillation) is M0 Guidance note
(a) $T = \frac{\lambda}{0.8}(0.05) = 0.25g$ | M1 |

$\lambda = \frac{(0.8)(0.25g)}{0.05} = 39.2$ (*) | A1 (2) |

(b) $T = \frac{39.2}{0.8}(x + 0.05)$ | M1 |

$mg - T = ma$ (3 term equn) | M1 |

$0.25g - \frac{39.2}{0.8}(x + 0.05) = 0.25\ddot{x}$ (or equivalent) | A1 |

$\ddot{x} = -196x$ | A1 ↓ M1 A1 cso (6) |

SHM with period $\frac{2\pi}{\omega} = \frac{2\pi}{14} = \frac{\pi}{7}$ s (*) | |

(c) $v = 14\sqrt{(0.1)^2 - (0.05)^2}$ | M1 A1√ |

$= 1.21(24\ldots) \approx 1.21$ m s$^{-1}$ (3 s.f.) Accept $7\sqrt{3}/10$ | A1 (3) |

(d) Time $T$ under gravity = $\frac{1.21 \ldots}{g}$ (= 0.1237 s) | B1√ |

Complete method for time $T'$ from B to slack. | M1 A1 |

[$\uparrow$ e.g. $\frac{\pi}{28} + t$, where $0.05 = 0.1\sin 14t$ OR $T'$, where $-0.05 = 0.1\cos 14T'$] | |

$T'' = 0.1496$s | A1 |

Total time = $T + T' = 0.273$ s | A1 (5) |

(b) 1st M1 must have extn as $x + k$ with $k \neq 0$ (but allow M1 if e.g. $x + 0.15$), or must justify later | | Guidance note

For last four marks, **must** be using $\ddot{x}$ (not a) | | Guidance note

(c) Using $x = 0$ is M0 | | Guidance note

(d) M1 – must be using distance for when string goes slack. Using $x = -0.1$ (i.e. assumed end of the oscillation) is M0 | | Guidance note
\begin{enumerate}
  \item A particle $P$ of mass 0.25 kg is attached to one end of a light elastic string. The string has natural length 0.8 m and modulus of elasticity $\lambda \mathrm { N }$. The other end of the string is attached to a fixed point $A$. In its equilibrium position, $P$ is 0.85 m vertically below $A$.\\
(a) Show that $\lambda = 39.2$.
\end{enumerate}

The particle is now displaced to a point $B , 0.95 \mathrm {~m}$ vertically below $A$, and released from rest.\\
(b) Prove that, while the string remains stretched, $P$ moves with simple harmonic motion of period $\frac { \pi } { 7 } \mathrm {~s}$.\\
(c) Calculate the speed of $P$ at the instant when the string first becomes slack.

The particle first comes to instantaneous rest at the point $C$.\\
(d) Find, to 3 significant figures, the time taken for $P$ to move from $B$ to $C$.\\

\hfill \mbox{\textit{Edexcel M3 2007 Q7 [16]}}