Figure shown a block P of mass m resting on a smooth floor at a distance l from a regid wall. Block is pushed towards right by a distance `3//2` and released. When block passes from its mean position another block of mass `m_(1)` so that the combined block just collides with the left wall.

Explanation

When block p is released from rest from a distance `3l//2` towards right from mean position, this will be the amplitude of oscllation. So velocity of block when passing from mean position is given as
`v = A omega = (3l)/(2) sqrt((k)/(m))` `[as omega = sqrt((k)/(m))]`
If mean `m_(1)` is added to it and just after if velocity of combined block become `v_(1)` from momentum conservation we have
`mv = (m + m_(1)) v_(1)`
or `v_(1) = (m)/((m + m_(1))) ((3l)/(2) sqrt((k)/(m)))`
If this the velocity of combined block at mean position , it nust be given as
`v_(1) = A omega_(1) [now omega_(1) = sqrt((k)/(m + m_(1)))]`
Where `A_(1) and omega_(1)` are the new amplitude and angular frequency , repectvely, 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 have
`(m)/((m + m_(1)) . (3l)/(2) sqrt((k)/(m)) = l_(1) sqrt((k)/(m + m_(1)))`
or `(3 sqrtm)/(2sqrt (m + m_(1))) = 1`
or `9m = 4m + 4m_(1) or m_(1) = (5)/(4) m`