A rigid sphere of density \(\rho\) and radius \(a\) is released from rest when its centre is at a height \(h\) above a large horizontal rubber sheet. Assuming that the part of the sheet outside the circle of contact with the sphere remains at the same horizontal level throughout the ensuing motion and that the tension \(T\) in the sheet is constant, show that if the vertical penetration \(x\) into the sheet is not greater than \(a\), then \(x\) satisfies the equation \begin{equation*} \frac{d^2x}{dt^2} = \frac{3T}{2\rho a^4}(x^2-2ax) + g, \end{equation*} where \(g\) is the acceleration due to gravity. Hence show that the sphere will penetrate to a depth \(ka\) (\(k < 1\)) if the tension is given by \begin{equation*} T = \frac{2g\rho a(h+ka-a)}{3k^2-k^3}. \end{equation*} [It may be assumed that the force per unit area exerted by the sheet at any point of contact is normal to it and of magnitude \(2T/a\).]
A uniform rod \(AB\) of mass \(m\) is at rest and is set in motion by parallel impulses \(J\) and \(K\) applied at \(A\) and \(B\) in a direction perpendicular to the rod. Prove that \(A\) starts to move with velocity \((4J-2K)/m\) and \(B\) with velocity \((4K-2J)/m\). Three equal uniform rods, \(AB\), \(BC\), \(CD\), each of mass \(m\), are freely jointed together at \(B\) and \(C\): they are at rest on a smooth table with the rods lying along three sides of a square \(ABCD\). If the framework is set in motion by an impulse \(J\) in the direction \(AB\) applied at \(A\) to the rod \(AB\), show that the velocity with which \(D\) starts to move is \(J/6m\).
A particle of mass \(m\) at \(\mathbf{r}\) is rotating about the origin \(O\) with angular velocity \(\boldsymbol{\omega}\). It is constrained to lie on a smooth sphere with centre at \(O\) and is acted upon by an external force \(\mathbf{F}\). Starting from the vector equation of linear motion, prove that \begin{equation*} \frac{d\mathbf{h}}{dt} = \mathbf{G}, \end{equation*} where \(\mathbf{h} = m[\mathbf{r} \times (\boldsymbol{\omega} \times \mathbf{r})]\) and \(\mathbf{G}\) is the moment about \(O\) of the force \(\mathbf{F}\). Hence show that if \(\mathbf{G}\) is perpendicular to \(\mathbf{h}\) then \(|\mathbf{h}|\) is constant. Show also that the kinetic energy \(T\) of the particle may be expressed in the form \(T = \frac{1}{2}\boldsymbol{\omega} \cdot \mathbf{h}\).
A particle of mass \(m\) is attached to the midpoint of a light elastic string of modulus \(\lambda\) and of unstretched length \(2l\), the ends of which are fixed at the same horizontal level a distance \(2b\) apart. Show that the particle can hang in equilibrium at a depth \(a\) below this level, where \(a\) satisfies the equation \begin{equation*} a\left(1-\frac{l}{\sqrt{(a^2+b^2)}}\right) = \frac{mgl}{2\lambda}. \end{equation*} The particle is pulled a small distance vertically downwards from this equilibrium position and is then released. Show that the period of the resulting small oscillations is \(2\pi/\omega\) where \begin{equation*} \omega^2 = \frac{g}{a}\left(\frac{1-(b^2/l^2)}{1-l/c}\right) \end{equation*} and \(c = \sqrt{(a^2+b^2)}\).
A function \(f(x)\) has all its derivatives non-zero in some interval. It can be calculated with a maximum error \(\epsilon\), \(\epsilon\) being independent of \(x\). It is desired to evaluate its derivative \(f'(x)\) at some point \(x = x_0\) in that interval, using the result \(f'(x_0) \approx f_1(x_0)\), where \begin{equation*} f_1(x_0) \equiv \frac{f(x_0+h)-f(x_0-h)}{2h}. \end{equation*} By expanding \(f(x_0 \pm h)\) in Taylor series, and assuming that \(h\) is sufficiently small for each term of the Taylor series to be considerably smaller than the one before, show that the value of \(h\) which minimises the possible error in \(f'(x_0)\) is proportional to \(\epsilon^{\frac{1}{3}}\). Obtain the constant of proportionality in terms of an appropriate derivative of \(f\).
A uniform cube of mass \(m\) lies at rest on a smooth horizontal table. A small, smooth sphere of mass \(km\), \(k < 1\), is projected with speed \(u\), and at an angle of elevation \(\alpha\), from a point \(P\) on the table. It hits a vertical face of the cube at a point \(Q\), rebounds, and lands back at the point of projection. The vertical plane through \(P\) and \(Q\) bisects those edges of the cube which it intersects. If \(e\) is the coefficient of restitution between the sphere and the cube, show that \(e > k\) and that \(P\) must be at a distance \begin{equation*} \frac{u^2(e-k)\sin 2\alpha}{g(1+e)} \end{equation*} from the initial position of that face of the cube which is struck, where \(g\) is the acceleration due to gravity.
\(A\) and \(B\) play a series of games the results of which are independent. In each game, \(A\) has probability \(p\) of winning, \(B\) probability \(q\), where \(q = 1-p\), and each pays the winner one unit. Supposing that \(A\) starts with \(n\) units and \(B\) with \(N\), let \(c_n\) denote the probability that \(A\) loses all his money. By writing down a relation between \(c_{n+1}\), \(c_n\) and \(c_{n-1}\), show that \begin{equation*} c_n = \frac{(q/p)^N-(q/p)^n}{(q/p)^N-1} \quad (p \neq q) \end{equation*} and \begin{equation*} c_n = 1-(n/N) \quad (p = q). \end{equation*} A casino can be considered as an opponent with infinite reserves of capital. Find the probability that a compulsive gambler \(A\), with only a finite amount of capital, will eventually lose all his money to the casino, and say whether British establishments \((p = q)\) differ in this respect from those on the Continent \((p < q)\).