A block of mass M is executing simple harmonic motion on a smooth horizontal surface under the restoring force of an ideal spring. Its total mechanical energy is E . At the exact instant the block crosses its mean position, a piece of putty of mass M 3 is dropped vertically onto the block and sticks to it. The new total mechanical energy of the oscillator is:

Step-by-step solution

At the mean position, the potential energy of the spring-mass system is zero, and its total mechanical energy is entirely kinetic. Let v be the velocity of the block at the mean position. E = 1 2 M v^2 When the putty of mass M 3 is dropped onto the block, there is no external horizontal force acting on the system. Thus, the horizontal linear momentum is conserved during this perfectly inelastic collision. Let v' be the new velocity of the combined mass. Mv = (M + M 3 )v' Mv = 4M 3 v' v' = 3v 4 The new total mechani