Another way to solve the problem is the use of 4-vectors formalism, if you know it. Given a particle of mass M, having velocity ##\vec v ## in an inertial frame ( the “laboratory”) , its 4-impulse is defined as :
$$P = (\gamma M c , \gamma M \vec v ) = (\frac{E}{c} , \vec p) $$
Note that mass is invariant , doesn’t change with velocity. This is an old argument of discussion, if mass is to be considered increasing with velocity or not . The modern point of view is that mass is invariant with velocity. But this is a good subject for another discussion, now doesn’t matter.
In your exercise, the given mass M is at rest in the reference frame of the laboratory, so its 4-impulse is:
##P=( Mc,0)##
because its velocity is zero and ##\gamma =1## . So its total energy is only rest energy , that is : ##E = Mc^2##, and the time component of P is just ## E/c = Mc## . The space component is zero : ## \vec p = 0 ## .
Now M splits into ##m_{1}##, which goes left with speed ##v_{1}## and relevant factor ##\gamma_{1}##; and ##m_{2}##, which goes right with speed ##v_{2}## and relevant factor ##\gamma_{2}##.
This implies that they have 4-impulse :
##P_{1} = (\gamma_{1} m_{1} c , \gamma_{1} m_{1} \vec v_{1} )##
##P_{2} = (\gamma_{2} m_{2} c , \gamma_{2} m_{2} \vec v_{2} )##
Now , it must be : ## P = P_{1} + P_{2} ##
from which , first of all, we see that the vector sum of spatial terms , that is momenta, is zero ; and it is almost obvious because the system is isolated, as in non relativistic mechanics.
But more important is that :
$$ Mc = \gamma_{1} m_{1} c + \gamma_{2} m_{2} c = (\gamma_{1} m_{1} + \gamma_{2} m_{2} ) c $$
Now we can do two considerations .
First : multiplying for c both members , obtain that the rest energy ## Mc^2## of the original particle equals the sum of total energies of the two masses generated by the splitting. And this is logical : where does the energy of each particle come from ? Obvious , from the rest energy of the original particle. But each particle must have rest energy ## m_{i}c^2## ( i = 1,2) and kinetic energy too.
If splitting were relativistic, that is the two generated particles had a relativistic speed, the solution would be rather difficult. But we have small velocities, so the energy of each particle must be the sum of rest energy ##mc^2## and traditional kinetic energy ##1/2 *m *v^2## .
second: consider that, for small speed, we can assume:
##\gamma \approx (1+ \frac {v^2}{2c^2} ## , as already suggested in reply #2. Do this for each of the two masses; in the end, you will have that:
$$M = m_{1} + m_{2} + \frac {1}{c^2} * (K_{1} + K_{2} )$$
where (...) are the classical kinetic energies.
And now you can answer your questions.