A block of mass M is oscillating on a frictionless horizontal surface, attached to a spring of force constant k . The amplitude of oscillation is A . When the block passes exactly through its mean position, a piece of putty of mass m is dropped vertically onto it and sticks to it. If the new amplitude of oscillation becomes A 2 , what is the ratio m M ?

Step-by-step solution

At the mean position, the velocity of the block is maximum, given by v = A = A k M . When the putty of mass m is dropped vertically, there is no external horizontal force. By conservation of linear momentum in the horizontal direction: Mv = (M+m)v' v' = M M+m v The new total mechanical energy of the system is entirely kinetic at the mean position: E' = 1 2 (M+m)v'^2 = 1 2 (M+m) ( M M+m v )^2 = M M+m ( 1 2 Mv^2 ) = M M+m E Since total energy is proportional to the square of the amplitude ( E = 1 2 kA^2 ), we have: 1