Figure shows a block P of mass m resting on a horizontal smooth floor at a distance l from a rigid wall. Block is pushed toward right by a distance 3l/2 and released, when block passes from its mean position another block of mass m1 is placed on it which sticks to it due to friction. Find the ratio of m1m so that the combined block just collides with the left wall.

When block P is released from rest from a distance 3l/2 toward right from mean position, this will be the amplitude of oscillation, so velocity of block when passing from its mean position is given asv=Aω=3l2km  As ω=kmIf mass m1 is added to it, and just after that if velocity of combined block becomes v1, from momentum conservation we have          mv=m+m1v1 or  v1=mm+m13l2kmIf this is the velocity of combined block at mean position, it must be given asv1=A1ω1  Now ω1=km+m1where  A1 and ω1 are the new amplitude and angular frequency of SHM of the block. It is given that combined block just reaches the left wall thus the new amplitude of oscillation must be l so we havemm+m1⋅3l2km=l km+m1or  3m2m+m1=1or  9m=4m+4m1or  m1=54m⇒m1m=1.25