Answer:
0.075 m
Explanation:
The picture of the problem is missing: find it in attachment.
At first, block A is released at a distance of 
h = 0.75 m
above block B. According to the law of conservation of energy, its initial potential energy is converted into kinetic energy, so we can write:

where
 is the acceleration due to gravity
 is the acceleration due to gravity
 is the mass of the block
 is the mass of the block
 is the speed of the block A just before touching block B
 is the speed of the block A just before touching block B
Solving for the speed,

Then, block A collides with block B. The coefficient of restitution in the collision is given by:

where:
e = 0.7 is the coefficient of restitution in this case
 is the final velocity of block B
 is the final velocity of block B
 is the final velocity of block A
 is the final velocity of block A

 is the initial velocity of block B
 is the initial velocity of block B
Solving,
 
 
Re-arranging it,
 (1)
 (1)
Also, the total momentum must be conserved, so we can write:

where

And substituting (1) and all the other values,

This is the velocity of block B after the collision. Then, its kinetic energy is converted into elastic potential energy of the spring when it comes to rest, according to

where
k = 600 N/m is the spring constant
x is the compression of the spring
And solving for x,
