A block explodes into three pieces of equal mass. Piece A has speed v after the explosion, and pieces B and C have speed 2v . What is the angle between the direction of the piece A and the piece B?

Here we need the conservation of momentum law. I suppose that before the explosion the block was in rest thus its momentum was zero. The law states that after the explosion the total momentum of the pieces is also zero (if we ignore the mass and speed of explosives).
Momentum is a vector quantity and the total momentum is a vector sum. Denote the mass of each piece as m. Then
m vec V_1+m vec V_2+m vec V_3=0,  or  vec V_1+vec V_2+vec V_3=0
(look at the picture attached).
 
Projections on the x and y axes give us
-v+2v cos alpha + 2v cos beta =0,  or  2(cos alpha + cos beta)=1,  and
2v sin alpha = 2v sin beta,  so  alpha = beta.
 
Thus from the first equation we have
4cos alpha=1, or  alpha=arccos(1/4) approx 75.5^@.
The angle in question is  180^@-alpha approx 104.5^@.
 
http://www.ux1.eiu.edu/~cfadd/1350/09Mom/ConsMom.html