1 problem found
A particle, \(A\), is dropped from a point \(P\) which is at a height \(h\) above a horizontal plane. A~second particle, \(B\), is dropped from \(P\) and first collides with \(A\) after \(A\) has bounced on the plane and before \(A\) reaches \(P\) again. The bounce and the collision are both perfectly elastic. Explain why the speeds of \(A\) and \(B\) immediately before the first collision are the same. The masses of \(A\) and \(B\) are \(M\) and \(m\), respectively, where \(M>3m\), and the speed of the particles immediately before the first collision is \(u\). Show that both particles move upwards after their first collision and that the maximum height of \(B\) above the plane after the first collision and before the second collision is \[ h+ \frac{4M(M-m)u^2}{(M+m)^2g}\,. \]