Martians come in two colours, blue and green, the proportion of blue Martians in the (effectively infinite) population being \(p\). A psychologist, interested in assessing the mean IQ among Martians, decides to take a random sample of size 8 from the population, and to use the average \(I\) of their IQ's as an estimate of the overall mean in the population. His eager young assistant, however, suggests that it may be better to sample \(S_1\) and \(S_2\) of blue and green Martians, \(S_1 + S_2 = 8\), and to use \begin{align*} J = pI_1 + (1-p)I_2 \end{align*} as the estimate of the overall mean IQ, where \(I_1\) and \(I_2\) are the averages of the IQ's in the blue and the green samples, respectively. Show that if the IQ of blue Martians has mean \(\mu_1\) and variance \(\sigma_1^2\), and that of green Martians has mean \(\mu_2\) and variance \(\sigma_2^2\), then both \(I\) and \(J\) are unbiased estimators of the mean IQ in the population; and that, for the best choice of the ratio \(S_1: S_2\) (to be determined), the variance of \(J\) is less than that of \(I\), unless \(\mu_1 = \mu_2\).
Three particles of unit mass lie always on a straight line; they can however pass through each other without hindrance. Each attracts each other according to an inverse cube law of force, e.g. the force on the first due to the second is \((x_2 - x_1)^{-3}\) in some units, being the distance along the line. Show that the quantity \begin{align*} 2E = \dot{x}_1^2 + \dot{x}_2^2 + \dot{x}_3^2 - (x_1-x_2)^{-2} - (x_2-x_3)^{-2} - (x_3-x_1)^{-2} \end{align*} is constant in the motion. Show also that \begin{align*} \frac{d^2}{dt^2}(x_1^2 + x_2^2 + x_3^2) = 4E. \end{align*} Hence show that, if the particles start from rest at finite distances apart, they will reach their common centre of gravity simultaneously after a finite time.
A ring of weight \(mg\) is free to move on a fixed smooth horizontal rod. A light inextensible string of length \(2l\) is attached to the ring at one end. Its other end is attached to a particle of weight \(mg\). The system is held with the particle just below the rod and with the string just taut and lying along the underside of the rod. The system is released from rest in this position. Express the velocities of the ring and the particle in terms of the angle \(\theta\) made by the string with the horizontal during motion of the system in which the string remains taut. For such motion, show that \begin{align*} l\dot{\theta}^2 = \frac{2g \sin \theta}{1 + \cos^2 \theta} \end{align*} and evaluate the tension in the string as a function of \(\theta\). Hence show that for \(0 < \theta < \pi\) the string never becomes slack. Show also that the particle follows an elliptic path.
A heavy uniform chain of weight \(w\) per unit length rests in a vertical plane on a fixed rough circular cylinder of radius \(a\), the axis of which is horizontal and at right angles to the plane in which the chain rests. The coefficient of friction is \(\mu = 1\) and the chain is on the point of slipping. By consideration of an element of the chain whose ends \(P\) and \(Q\) are at the extremities of radii of the cylinder which make angles \(\theta\) and \(\theta + \delta\theta\) with the upward vertical, derive a differential equation for \(T(\theta)\), the tension in the chain at \(P\), and verify that \begin{align*} T(\theta) = wa \sin \theta + ce^{\theta} \end{align*} satisfies it for \(c\) constant. If one end of the chain is at \(\theta = 0\) and a length \(s\) hangs vertically beyond the lowest point of contact between the chain and the cylinder, show that \(s = a\).
Two planets circle around their common centre of gravity \(C\) under the influence of Newtonian gravity; the effect of their parent sun can be neglected. Obtain the total energy of their mutual motion, and the total angular momentum about \(C\), as functions of the planets' masses and separation. Miners from the initially less massive planet take ore from the other planet back to their home planet, at a slow, fairly steady, rate. Assuming the two orbits always remain circular, and that the planets are small compared with their separation, show that as the relative masses of the planets change so will their separation, and it will reach a minimum when the two masses are equal. As the planets start to separate at later times, an ecologist suggests that the journey could be kept short, and hence fuel saved, if equivalent masses of unwanted material were shipped back at the same rate, thus keeping the separation at its minimum. Do you agree that this would save fuel in the long run?
Relative to an observer \(O\), a point \(A\) of a rigid body has velocity \(\mathbf{u}\). Another point \(P\) of the body, at position \(\mathbf{r}\) relative to \(A\), will have relative to \(O\) a velocity \begin{align*} \mathbf{u} + \boldsymbol{\omega} \times \mathbf{r}, \end{align*} where \(\boldsymbol{\omega}\) is the angular velocity of the body, which is independent of the position of \(A\) or \(P\). (i) If \(\boldsymbol{\omega}\) is non-zero, a line \(L\) may be defined to be the set of points whose position vectors \(\mathbf{x}\) relative to \(A\) satisfy \begin{align*} \mathbf{x} = \frac{\boldsymbol{\omega} \times \mathbf{u}}{|\boldsymbol{\omega}|^2} + \lambda\boldsymbol{\omega}, \end{align*} where \(\lambda\) is an arbitrary real parameter. If \(B\) lies on \(L\) and \(\mathbf{s}\) is the position vector of \(P\) relative to \(B\), show that the velocity of \(P\) relative to \(O\) may be written \begin{align*} \mathbf{V} + \boldsymbol{\omega} \times \mathbf{s}, \end{align*} where \(\mathbf{V}\) is parallel to \(\boldsymbol{\omega}\), and show that the magnitude of \(\mathbf{V}\) is \((\boldsymbol{\omega} \cdot \mathbf{u})/|\boldsymbol{\omega}|\). (Such a motion is called a screw motion with axis \(L\).) (ii) Another screw motion, defined by \(L'\), \(\mathbf{V}'\), and \(\boldsymbol{\omega}'\), is superimposed, where \(\boldsymbol{\omega} + \boldsymbol{\omega}' \neq \mathbf{0}\). Verify that the resulting motion is also a screw motion, defined by \(L''\), \(\mathbf{V}''\) and \(\boldsymbol{\omega}''\), where \begin{align*} \boldsymbol{\omega}'' &= \boldsymbol{\omega} + \boldsymbol{\omega}',\\ \mathbf{V}'' &= \frac{\boldsymbol{\omega}'}{|\boldsymbol{\omega}''|^2}\{\boldsymbol{\omega}'' \cdot (\mathbf{V} + \mathbf{V}') + (\mathbf{x} - \mathbf{x}') \cdot (\boldsymbol{\omega} \times \boldsymbol{\omega}')\}, \end{align*} and \(\mathbf{x}\), \(\mathbf{x}'\) are any two points on \(L\), \(L'\) respectively.