Where did 2mv come from in the solution to the question below? 

I know that this is an inelastic collission, which is why the two objects have the same final velocity. However, I don't understand why they equated the V1(M+m) to 2mv. Where did they get 2mv from?
 A: When a man in frictionless surface throws the ball in forward direction, by conservation of linear momentum he gets pushed back (exactly the case in space where astronaut throws something back to move foreward).Here,when man throws the ball, the momentum of ball and man are exactly equal and their velocities are in opposite direction. But you need to note one thing that when the ball strikes the wall and rebounds, we consider it as an elastic collision. So, the velocity of ball after rebound is same as that of velocity with which it strikes the wall but the direction is just opposite. Here the opposite direction is the direction in which the man had already been moving.
As the man catches the ball (being inelastic collision) the momentum of ball gets transferred to combined system i.e.(man+ball).
A: Straight after the man throws the ball, his velocity is determined by conservation of linear momentum, that is, the momentum of the man recoiling is equal to the momentum of the ball leaving his hand:
$Mv_1=mv$
After the ball bounces elastically off the wall, it returns towards the man at opposite velocity (equal in magnitude, opposite direction), so applying conservation of momentum again, at the instant when he recatches the ball, we get:
$(M+m)V_1=Mv_1+mv$
ie: momentum of (man holding ball) = momentum of man + momentum of ball 
The $2mv$ comes from substituting $Mv_1=mv$ (momentum of man = momentum of ball) from the first conservation of momentum equation. That is,
$(M+m)V_1=Mv_1+mv = 2mv$
which gives:
$V_1=\frac{2mv}{(M+m)}$
A: When the man throws the ball, both the ball and the man get equal momentum in the opposite directions. Since the collusion is elastic, i.e: no loss in energy, the ball rebounds with momentum of the same magnitude but in the opposite direction. At this point, both the ball and the man have momentum in the same direction with equal mangitude. 
When the man catches the ball, the total momentum of the man-ball system is equal to the sum of the momentums of the ball and the man taken individually. Since the momentum of the ball and the man are of same magnitiude and same direction, they add up to give you twice the momentum of the ball.

