1
\begin{enumerate}[label=(\alph*)]
\item 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.
\begin{enumerate}[label=(\roman*)]
\item 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 $8 \mathrm {~ms} ^ { - 1 }$.
\item Find the mass of the second sphere.
\end{enumerate}\item 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.
\begin{enumerate}[label=(\roman*)]
\item 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 .
\item Find the modulus of elasticity of this second rope.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{68cbb8bb-2898-4812-a221-6ea5363b0812-3_559_705_262_680}
\captionsetup{labelformat=empty}
\caption{Fig. 2.1}
\end{center}
\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 .
\begin{enumerate}[label=(\roman*)]
\item Show that $k = 2$.

A uniform solid S is formed by rotating the region R through $2 \pi$ radians about the $x$-axis.
\item 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]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{68cbb8bb-2898-4812-a221-6ea5363b0812-3_455_794_1521_639}
\captionsetup{labelformat=empty}
\caption{Fig. 2.2}
\end{center}
\end{figure}
\item 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.]
\item 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]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{68cbb8bb-2898-4812-a221-6ea5363b0812-4_668_262_255_904}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\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 $x \mathrm {~m}$.\\
(i) Show that $\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } = - 10 ( x - 4 )$.\\
(ii) Write down the value of $x$ when P is at the centre of its motion.
\item Find the amplitude and the period of the oscillations.
\item Find the velocity of P when $t = 1.2$.
\end{enumerate}
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{OCR MEI M3 2016 Q1 [19]}}