By Law of conservation of momentum
#vecp_"initial"=vecp_"final"#
Let floating speed of the astronaut be #vecu m/s#. There will always a velocity associated with the astronaut due to gravitation pull of the earth or of the planet/satellite it is circling about.
Since the astronaut is holding the object while floating. Therefore,
#vecp_"initial"=("mass of astronaut "M_A+"mass of object "m_@)xx vecu#
Let us assume that he is floating along #x#-axis.
Let him throw the object at an angle #theta# with the #x#-axis. Due to Newton's third law of motion, there will be reaction and he will be thrown back in the opposite direction.
Let #v_@# be the final velocity of the object, and #v_A# be the final velocity of the astronaut.
#vecp_"final"=M_A*v_A+m_@*v_@#
As initially #y# component of momentum was zero, hence
#vecp_("final"y)=M_A*v_(Ay)+m_@*v_(@y)=0# ......(1)
Equating the orthogonal #x# component of momentum we obtain
#vecp_("final"x)=M_A*v_(Ax)+m_@*v_(@x)= (M_A+m_@)xx vecu# ....(2)
Assuming initially the frame of reference is the astronaut himself and also that he throws the object in #x# direction of this frame of reference.
There is zero #y# component of the momentum before and after throwing of the object. Equation (1) gives us no result, as it becomes 0=0. From (2) we obtain
#M_A*v_(Ax)+m_@*v_(@x)=0#
Inserting given values
#75*v_A=-3*2#
or #|v_A|=(3*2)/75#
Change in speed of the astronaut #=0.08m/s#
