2 problems found
A particle moves on a smooth triangular horizontal surface \(AOB\) with angle \(AOB = 30^\circ\). The surface is bounded by two vertical walls \(OA\) and \(OB\) and the coefficient of restitution between the particle and the walls is \(e\), where \(e < 1\). The particle, which is initially at point \(P\) on the surface and moving with velocity \(u_1\), strikes the wall \(OA\) at \(M_1\), with angle \(PM_1A = \theta\), and rebounds, with velocity \(v_1\), to strike the wall \(OB\) at \(N_1\), with angle \(M_1N_1B = \theta\). Find \(e\) and \(\displaystyle {v_1 \over u_1}\) in terms of \(\theta\). The motion continues, with the particle striking side \(OA\) at \(M_2\), \(M_3\), \( \ldots \) and striking side \(OB\) at \(N_2\), \(N_3\), \(\ldots \). Show that, if \(\theta < 60^\circ\,\), the particle reaches \(O\) in a finite time.
A particle rests at a point \(A\) on a horizontal table and is joined to a point \(O\) on the table by a taut inextensible string of length \(c\). The particle is projected vertically upwards at a speed \(64\surd(6gc)\). It next strikes the table at a point \(B\) and rebounds. The coefficient of restitution for any impact between the particle and the table is \({1\over 2}\). After rebounding at \(B\), the particle will rebound alternately at \(A\) and \(B\) until the string becomes slack. Show that when the string becomes slack the particle is at height \(c/2\) above the table. Determine whether the first rebound between \(A\) and \(B\) is nearer to \(A\) or to \(B\).
Solution: \begin{align*} \text{N2}(radially): && T + mg \cos \theta &= m\frac{v^2}{r} \\ \Rightarrow && v^2-gc \cos \theta &\geq 0 \\ \text{COE}: && \frac12 m u^2 &= \frac12mv^2 + mgc\cos \theta \\ \Rightarrow && u^2 &= v^2 + 2gc\cos \theta \\ && u^2 &\geq gc \cos \theta+2gc\cos \theta \\ \Rightarrow && u^2 &\geq 3gc\cos \theta \end{align*} Therefore it will complete bounces with the string taught if it leaves the table with \(u^2 \geq 3gc\). After \(6\) bounces it will leave the table with speed \(\sqrt{6gc} > \sqrt{3gc}\) and after \(7\) bounces it will leave the table with speed \(\sqrt{\frac{3}{2} gc} < \sqrt{3 gc}\). When it leaves the table with speed \(\sqrt{\tfrac32 gc}\), the string will go slack when \begin{align*} && \tfrac32 gc &= 3gc \cos \theta \\ \Rightarrow && \cos \theta &= \frac{1}{2} \end{align*} ie at a height \(c\cos \theta = \frac12c\) above the table. Once the string goes slack, the particle travels under circular motion, \begin{align*} && u^2 &= \frac12 gc \\ \Rightarrow && u_\rightarrow &= \sqrt{\tfrac12 gc} \cos \theta \\ && u_{\uparrow} &= \sqrt{\tfrac12 gc} \sin \theta \\ \Rightarrow && s &= ut - \tfrac12 gt^2 \\ \Rightarrow && -\frac{c}{2} &= \sqrt{\tfrac12 gc} \frac{\sqrt{3}}{2} t - \tfrac12 g t^2 \\ \Rightarrow && t &= \sqrt{\frac{c}{g}} \left ( \frac{\sqrt{6}+\sqrt{22}}{4} \right) \\ \Rightarrow && s_{\rightarrow} &= \tfrac12 \sqrt{\tfrac12 gc} \sqrt{\frac{c}{g}} \left ( \frac{\sqrt{6}+\sqrt{22}}{4} \right) - \frac{\sqrt{3}}{2}c \\ &&&= \left ( \frac{\sqrt{11}-3\sqrt{3}}{8} \right)c \\ \end{align*} We need to establish whether this position is positive or negative (ie if we cross the centre line. Clearly \(\sqrt{11} < 3\sqrt{3}\) so we haven't crossed the centre line and we land closer to where we took off. Since it's the 7th take off, this is closer to \(B\).