You hold glider AA of mass 0.125 kgkg and glider BB of mass 0.375 kgkg at rest on an air track with a compressed spring of negligible mass between them. When you release the gliders, the spring pushes them apart. Once the gliders are no longer in contact with the spring, glider AA is moving to the right at 0.600 m/sm/s. What is the velocity (magnitude and direction) of glider BB at this time

The magnitude of the velocity of glider B is 0.2m/s and the direction is the negative direction  

Explanation:

Inelastic Collision

Given data

mass of glider A m1= 0.125kg

initial velocity u1=0

final velocity v1= 0.600 m/s

mass of glider B m2= 0.375kg

initial velocity u2=0

final velocity v2=?

We know that the expression for the conservation of momentum is given as

m1u1+m2u2=m1v1+m2v2

since u1=u2=u=0m/s

u(m1+m2)=m1v1+m2v2

substituting we have

0(0.125+0.0375)=0.125*0.6+0.375*v2

0=0.075+0.375v2

0.375v2=-0.075

v2=-0.075/0.375

v2=-0.2m/s

 The magnitude of the velocity of glider B is 0.2m/s and the direction is the negative direction      

your answer is, orbit around the nucleus