Outside the International Sapce Station, a 60 kg astronaut holding a 4.0 kg object (both initially at rest) throws the object at 10 m/s relative to the space station. A 50 kg astronaut, initially at rest, catches the object. What is the speed of the separation of the two astronauts?

Answer 1

because of the need to conserve momentum, the 60 kg astronaut acquires a velocity V1 such that
60 V2 = 4.0 * 10 m/s as soon as he throws the object. Thus V1 = 2/3 m/s

When the object is caught by the 50 kg astronaut, he or she acquires a velocity such that
4.0 x 2/3 = (50 + 4) V2
V2 = (8/3)/54 = 0.15 m/s

Check my numbers and reasoning