Problems

Two small spheres \(A\) and \(B\) of equal mass \(m\) are suspended in contact by two light inextensible strings of equal length so that the strings are vertical and the line of centres is horizontal. The coefficient of restitution between the spheres is \(e\). The sphere \(A\) is drawn aside through a very small distance in the plane of the strings and allowed to fall back and collide with the other sphere \(B\), its speed on impact being \(u\). Explain briefly why the succeeding collisions will all occur at the lowest point. (Hint: Consider the periods of the two pendulums involved.) Show that the speed of sphere \(A\) immediately after the second impact is \(\frac{1}{2}u(1+e^{2})\) and find the speed, then, of sphere \(B\).

Consider a simple pendulum of length \(l\) and angular displacement \(\theta\), which is {\bf not} assumed to be small. Show that $$ {1\over 2}l \left({\d\theta\over \d t}\right)^2 = g(\cos\theta -\cos\gamma)\,, $$ where \(\gamma\) is the maximum value of \(\theta\). Show also that the period \(P\) is given by $$ P= 2 \sqrt{l\over g} \int_0^\gamma \left( \sin^2(\gamma/2)-\sin^2(\theta/2) \right)^{-{1\over 2}} \,\d\theta \,. $$ By using the substitution \(\sin(\theta/2)=\sin(\gamma/2) \sin\phi\), and then finding an approximate expression for the integrand using the binomial expansion, show that for small values of \(\gamma\) the period is approximately $$ 2\pi \sqrt{l\over g} \left(1+{\gamma^2\over 16}\right) \,. $$