OCR MEI M3 2016 June — Question 1

Exam BoardOCR MEI
ModuleM3 (Mechanics 3)
Year2016
SessionJune
TopicDimensional Analysis
1
  1. In an investigation, small spheres are dropped into a long column of a viscous liquid and their terminal speeds measured. It is thought that the terminal speed \(V\) of a sphere depends on a product of powers of its radius \(r\), its weight \(m g\) and the viscosity \(\eta\) of the liquid, and is given by $$V = k r ^ { \alpha } ( m g ) ^ { \beta } \eta ^ { \gamma } ,$$ where \(k\) is a dimensionless constant.
    1. Given that the dimensions of viscosity are \(\mathrm { ML } ^ { - 1 } \mathrm {~T} ^ { - 1 }\) find \(\alpha , \beta\) and \(\gamma\). A sphere of mass 0.03 grams and radius 0.2 cm has a terminal speed of \(6 \mathrm {~ms} ^ { - 1 }\) when falling through a liquid with viscosity \(\eta\). A second sphere of radius 0.25 cm falling through the same liquid has a terminal speed of \(8 \mathrm {~ms} ^ { - 1 }\).
    2. Find the mass of the second sphere.
  2. A manufacturer is testing different types of light elastic ropes to be used in bungee jumping. You may assume that air resistance is negligible. A bungee jumper of mass 80 kg is connected to a fixed point A by one of these elastic ropes. The natural length of this rope is 25 m and its modulus of elasticity is 1600 N . At one instant, the jumper is 30 m directly below A and he is moving vertically upwards at \(15 \mathrm {~ms} ^ { - 1 }\). He comes to instantaneous rest at a point B , with the rope slack.
    1. Find the distance AB . The same bungee jumper now tests a second rope, also of natural length 25 m . He falls from rest at A . It is found that he first comes instantaneously to rest at a distance 54 m directly below A .
    2. Find the modulus of elasticity of this second rope. \begin{figure}[h]
      \includegraphics[alt={},max width=\textwidth]{68cbb8bb-2898-4812-a221-6ea5363b0812-3_559_705_262_680} \captionsetup{labelformat=empty} \caption{Fig. 2.1}
      \end{figure} The region R shown in Fig. 2.1 is bounded by the curve \(y = k ^ { 2 } - x ^ { 2 }\), for \(0 \leqslant x \leqslant k\), and the coordinate axes. The \(x\)-coordinate of the centre of mass of a uniform lamina occupying the region R is 0.75 .
    3. Show that \(k = 2\). A uniform solid S is formed by rotating the region R through \(2 \pi\) radians about the \(x\)-axis.
    4. Show that the centre of mass of S is at \(( 0.625,0 )\). Fig. 2.2 shows a solid T made by attaching the solid S to the base of a uniform solid circular cone C . The cone \(C\) is made of the same material as \(S\) and has height 8 cm and base radius 4 cm . \begin{figure}[h]
      \includegraphics[alt={},max width=\textwidth]{68cbb8bb-2898-4812-a221-6ea5363b0812-3_455_794_1521_639} \captionsetup{labelformat=empty} \caption{Fig. 2.2}
      \end{figure}
    5. Show that the centre of mass of T is at a distance of 6.75 cm from the vertex of the cone. [You may quote the standard results that the volume of a cone is \(\frac { 1 } { 3 } \pi r ^ { 2 } h\) and its centre of mass is \(\frac { 3 } { 4 } h\) from its vertex.]
    6. The solid T is suspended from a point P on the circumference of the base of C . Find the acute angle between the axis of symmetry of T and the vertical. \begin{figure}[h]
      \includegraphics[alt={},max width=\textwidth]{68cbb8bb-2898-4812-a221-6ea5363b0812-4_668_262_255_904} \captionsetup{labelformat=empty} \caption{Fig. 3}
      \end{figure} One end of a light elastic string, of natural length 2.7 m and modulus of elasticity 54 N , is attached to a fixed point L . The other end of the string is attached to a particle P of mass 2.5 kg . One end of a second light elastic string, of natural length 1.7 m and modulus of elasticity 8.5 N , is attached to P . The other end of this second string is attached to a fixed point M , which is 6 m vertically below L . This situation is shown in Fig. 3. The particle P is released from rest when it is 4.2 m below L . Both strings remain taut throughout the subsequent motion. At time \(t \mathrm {~s}\) after P is released from rest, its displacement below L is \(x \mathrm {~m}\).
    7. Show that \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = - 10 ( x - 4 )\).
    8. Write down the value of \(x\) when P is at the centre of its motion.
    9. Find the amplitude and the period of the oscillations.
    10. Find the velocity of P when \(t = 1.2\).
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