Blocks A and B have masses 40 kg and 60 kg respectively. They are placed on a smooth surface and the spring connected between them is stretched by 1.5m. If they are released from rest, determine the speeds of both blocks at the instant the spring becomes unstretched.

Explanation

Let, both block start moving with velocity `V_(1)` and `V_(2)` as shown in figure
Since no horizontal force on system so, applying momentum
conservation
`0=40V_(1)-60V_(2) " "2V_(1)=3V_(2)" "`....(1)
Now applying energy conservation, Loss in potential energy = gain in kinetic energy
`(1)/(2)kx^(2)=(1)/(2)m_(1)V_(1)^(2)+(1)/(2)m_(2)V_(2)^(2)`
`(1)/(2)xx 600 xx (1.5)^(2)`
`=(1)/(2) xx 40xx V_(1)^(2)+(1)/(2) xx 60 xx V_(2)^(2)" "` .........(2)
Solving euation (1) and (2) we get,
`V_(1)=4.5 m//s, V_(2)=3m//s`.