OCR MEI M3 2006 January — Question 3 18 marks

Exam BoardOCR MEI
ModuleM3 (Mechanics 3)
Year2006
SessionJanuary
Marks18
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TopicHooke's law and elastic energy
TypeSimple harmonic motion with elastic string
DifficultyStandard +0.3 This is a standard M3 elastic strings and SHM question with straightforward application of Hooke's law, energy conservation, and deriving the SHM equation. Part (i) is routine equilibrium, (ii) uses F=ma with extension, (iii) is standard energy method, (iv) derives the standard SHM equation from forces, and (v) applies standard SHM solution. All parts follow textbook methods with no novel insight required, making it slightly easier than average.
Spec3.02d Constant acceleration: SUVAT formulae4.10f Simple harmonic motion: x'' = -omega^2 x6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle
3 A light elastic rope has natural length 15 m . One end of the rope is attached to a fixed point O and the other end is attached to a small rock of mass 12 kg . When the rock is hanging in equilibrium vertically below O , the length of the rope is 15.8 m .
  1. Show that the modulus of elasticity of the rope is 2205 N . The rock is pulled down to the point 20 m vertically below O , and is released from rest in this position. It moves upwards, and comes to rest instantaneously, with the rope slack, at the point A .
  2. Find the acceleration of the rock immediately after it is released.
  3. Use an energy method to find the distance OA. At time \(t\) seconds after release, the rope is still taut and the displacement of the rock below the equilibrium position is \(x\) metres.
  4. Show that \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = - 12.25 x\).
  5. Write down an expression for \(x\) in terms of \(t\), and hence find the time between releasing the rock and the rope becoming slack.
3 A light elastic rope has natural length 15 m . One end of the rope is attached to a fixed point O and the other end is attached to a small rock of mass 12 kg .

When the rock is hanging in equilibrium vertically below O , the length of the rope is 15.8 m .\\
(i) Show that the modulus of elasticity of the rope is 2205 N .

The rock is pulled down to the point 20 m vertically below O , and is released from rest in this position. It moves upwards, and comes to rest instantaneously, with the rope slack, at the point A .\\
(ii) Find the acceleration of the rock immediately after it is released.\\
(iii) Use an energy method to find the distance OA.

At time $t$ seconds after release, the rope is still taut and the displacement of the rock below the equilibrium position is $x$ metres.\\
(iv) Show that $\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = - 12.25 x$.\\
(v) Write down an expression for $x$ in terms of $t$, and hence find the time between releasing the rock and the rope becoming slack.

\hfill \mbox{\textit{OCR MEI M3 2006 Q3 [18]}}
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Q1 18 Q2 18 Q3 18 Q4 18