Three particles are simultaneously projected under gravity \(g\) in different directions from the same point. Show that after a time \(t\) they are at the vertices of a triangle of area proportional to \(t^2\). If the initial velocities of two of the particles are in the same vertical plane and are of magnitudes \(u, v\) and elevations \(\alpha, \beta\) respectively, show that the plane of this triangle will pass through the point of projection after a time $$\frac{2uv\sin(\beta - \alpha)}{g(u\cos\alpha - v\cos\beta)},$$ assuming this to be positive.
Solution: Suppose the positions of the particles are \(\mathbf{x}_1, \mathbf{x}_2,\mathbf{x}_3\), and \(\mathbf{g} = \begin{pmatrix} 0 \\ 0 \\ -g \end{pmatrix}\) relative to the initial point as the origin. \begin{align*} && \mathbf{x}_1 &= \mathbf{u}_1t - \frac12 \mathbf{g} t^2 \\ && \mathbf{x}_2 &= \mathbf{u}_2t - \frac12 \mathbf{g} t^2 \\ && \mathbf{x}_3 &= \mathbf{u}_3t - \frac12 \mathbf{g} t^2 \\ \end{align*} Then, \begin{align*} A &= |\frac12 (\mathbf{u}_2t - \frac12 \mathbf{g} t^2 - (\mathbf{u}_1t - \frac12 \mathbf{g} t^2)) \times (\mathbf{u}_3t - \frac12 \mathbf{g} t^2 - (\mathbf{u}_1t - \frac12 \mathbf{g} t^2))| \\ &= \frac12 |(\mathbf{u}_2t - \mathbf{u}_1t) \times (\mathbf{u}_3t -\mathbf{u}_1t )| \\ &= \frac{t^2}{2} |(\mathbf{u}_2 - \mathbf{u}_1) \times (\mathbf{u}_3-\mathbf{u}_1 )| \end{align*}
A smooth wire is in the form of one bay of a cycloid (with intrinsic equation \(s = 4a\sin\psi\)) vertical and concavity upwards. A bead of mass \(m\) slides down the wire from rest at the highest point under the action of gravity \(g\). Find the reaction of the wire when the particle reaches the point where the tangent is inclined at angle \(\psi\) to the horizontal. Show also that the magnitude of the resultant acceleration of the particle remains constant throughout the motion.
A uniform chain of total mass \(m\) and length \(l\) is released from rest when held vertically with its lower end just touching the bottom of the interior of a bucket of mass \(M\). When half of the chain has fallen into the bucket and lies coiled at the base, the bucket is the distance the bucket has fallen and \(y\) is the length of chain remaining above the base of the bucket. Show that the linear momentum of the system is given by \[ (M+m)\dot{x} - my\dot{y}/l \], and determine the velocity of the vertical part of the chain relative to the bucket at this stage in terms of \(y\).
(i) Obtain the expression $$\frac{1}{2}(m_1 + m_2)V^2 + \frac{1}{2}\frac{m_1m_2}{m_1 + m_2}v^2$$ for the kinetic energy of the translatory motion of two bodies of masses \(m_1\) and \(m_2\), where \(V\) is the velocity of the centre of mass and \(v\) the relative velocity of the centres of mass of the two bodies separately. (ii) A thin uniform tube of mass \(M\) rests on a smooth horizontal table, and within it is a smooth particle of mass \(m\) resting close to the middle point. If the tube is suddenly set rotating about its centre, which is initially at rest, show that when the particle leaves the tube the angular velocity will be reduced to \((M + m)/(M + 4m)\) of the initial value.
In the finite motion of a simple pendulum of length \(l\) under gravity \(g\), the inclination to the vertical oscillates between \(-\alpha\) and \(+\alpha\). Show that the total period of oscillation is given by $$4(l/g)^{1/2}\int_0^{\pi/2}(1 - \sin^2\frac{1}{2}\alpha\sin^2\psi)^{-1/2}d\psi.$$ If \(\alpha\) is sufficiently small for the integrand to be expanded in powers of \(\sin\frac{1}{2}\alpha\), show that to order \(\alpha^2\) the period is $$2\pi(l/g)^{1/2}(1 + \frac{1}{16}\alpha^2),$$ and find also the next term, in \(\alpha^4\).
A straight river of unit width is flowing with speed \(w\), and a swan starts and swims across, always endeavouring to reach the point \(O\) on the opposite bank directly opposite its starting point. The speed of the swan relative to the water is constant and equal to \(u\) (where \(u > w\)). If \(Or\) is taken along the river bank to correspond perpendicular to it across the river, show that the path of the swan has equation $$2x = y^{1-c} - y^{1+c},$$ where \(c = w/u\).
A groove of semicircular cross-section and radius \(b\) is cut round a right circular cylinder of radius \(a\) (where \(a > b\)). (The centres of the semicircular cross-sections lie in a circle perpendicular to the axis of the cylinder.) Show that the surface of the groove is \(2\pi^2ab - 4\pi b^2\). Find also the volume of material removed. Explain the relation between the two results.
A bag contains a large number of red, white and blue dice in equal numbers. If \(n\) are drawn at random, show that the probability \(P(n,r)\) of drawing exactly \(r\) red dice is equal to the term containing \((\frac{1}{3})^r(\frac{2}{3})^{n-r}\) in the binomial expansion of \((\frac{1}{3} + \frac{2}{3})^n\). If \(r\) dice are thrown, find the probability \(Q(r,s)\) of throwing exactly \(s\) sixes. If \(n\) dice are drawn from the bag and the red dice drawn are thrown, show that the probability of throwing exactly \(s\) sixes is $$\sum_{t=0}^{n-s} P(n, s+t)Q(s+t, s)$$ and prove that this is equal to a term in a binomial expansion. Explain why a binomial distribution is obtained.
Objects \(\ldots, \langle -2 \rangle, \langle -1 \rangle, \langle 0 \rangle, \langle 1 \rangle, \langle 2 \rangle, \ldots\) are given. Two objects \(\langle p \rangle\) and \(\langle q \rangle\) are equal if \(p-q\) is a multiple of 4. Prove that
The equations of motion of a particle in a plane, referred to rectangular axes \(Ox, Oy\) in the plane, are $$\ddot{x} = ky, \quad \ddot{y} = -kx.$$ Show that the equations of motion become $$\ddot{x'} = ky', \quad \ddot{y'} = -kx',$$ and are thus unaltered in form, when referred to axes \(Ox', Oy'\) derived from the first by rotation about \(O\) through an arbitrary angle \(\alpha\). Are the equations of motion unaltered in form when referred to axes \(Ox, Oy\) derived from \(Ox, Oy\) by reflection in an arbitrary line \(x/\cos\beta = y/\sin\beta\) through the origin? Make a similar investigation for the equations $$\ddot{x} = k(x\ddot{y} - y\ddot{x})y, \quad \ddot{y} = -k(x\ddot{y} - y\ddot{x})x.$$